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Cm Math Continuum

1 · Prerequisites

2 · Summary

The continuum layer of the non-relativistic classical mechanics mathematics collection, all of it mathematics: the Euclidean spatial calculus and continuum motions; the transport theorem for the stated $C^{2}$ family of diffeomorphisms; the localization of mass and momentum balance and the control-volume identities; the Cauchy-tetrahedron representation of a balanced traction field by a stress matrix and the symmetry of that matrix under couple-free angular balance; explicit linear elastic and fluid constitutive models with exact plane-wave, channel-flow, Bernoulli and Kelvin-circulation results; the two characteristic scalar conservation laws of Burgers' equation with weak and entropy solutions; the classical $C^{2}$ fixed-end string for $C^{5}$ / $C^{4}$ data solved by its sine series; and the proved finite-time continuum limit of a cyclic spring chain. Every item carries domain: mathematics and depends only on mathematical suppliers; nothing here asserts that any continuum model describes a physical body.

The transport layer begins with Euclidean spatial calculus: gradient, divergence and material derivative and Continuum motions, spatial velocity, material acceleration and swept regions, which fix the objects: gradient, divergence and material derivative for time-dependent velocities using the spatial derivative at fixed time, in the standard Euclidean coordinates, and a motion as a $C^{2}$ family of diffeomorphisms with positive Jacobian on the material region. The derivative of the determinant along an invertible C¹ matrix curve: $(\det A)^{'} = \det A tr (A^{- 1} A^{'})$ proves the Jacobi formula $(\det A)^{'} = \det A tr (A^{- 1} A^{'})$ , and The transport theorem for a $C^{2}$ family of diffeomorphisms: $\frac{d}{d t} \int_{ψ_{t} (B)} f = \int_{ψ_{t} (B)} (\partial_{t} f + div (f w))$ combines it with the change-of-variables theorem to transport integrals over moving regions for the stated $C^{2}$ family of diffeomorphisms. Localization: a continuous function whose integrals vanish over a neighbourhood-basis family is identically zero records the localization step used by every balance statement.

The balance layer is Local mass balance: conservation on every material subregion implies $\partial_{t} ρ + div (ρ u) = 0$ , Local momentum balance: $ρ D_{t} u = ρ b + div σ$ from the integral balance on admissible material subregions, A continuous traction field balanced on every piecewise $C^{1}$ region is given by a continuous stress matrix, and $t (- n) = - t (n)$ , Angular-momentum balance without couples forces the stress matrix to be symmetric and Control-volume balance: mass and momentum rates for a moving non-material region: they are conditional localization theorems, each stating its own integral balance or contact hypothesis explicitly; none derives a balance law from particle mechanics. On them rest the adopted constitutive branches The small-strain isotropic linear elastic constitutive law and its strain energy and Incompressible Newtonian and barotropic inviscid fluid constitutive models, the strain-energy decomposition The isotropic elastic strain energy is $μ ∣ e_{0} ∣_{F}^{2} + (λ / 2 + μ / 3) (tr e)^{2}$ , hence positive definite under $μ > 0$ , $λ + 2 μ / 3 > 0$ , the exact plane-wave solutions Exact plane-wave solutions of the isotropic linear elastic model, with longitudinal and transverse speeds, the barotropic gradient form and Bernoulli law Barotropic inviscid momentum in gradient form, Bernoulli along streamlines and the irrotational constant with the circulation theorem Kelvin circulation: the circulation around a $C^{1}$ material loop is constant in the barotropic inviscid model, and the exact channel flows Exact plane Poiseuille and Couette flows of the incompressible Newtonian model, with flux and ansatz uniqueness of the explicitly adopted steady differential equations. Scalar conservation laws are treated in Weak and entropy solutions of the scalar conservation law $u_{t} + (u^{2} / 2)_{x} = 0$ , which fixes the weak and entropy solution classes under the declared countable-choice measure foundation, with test functions smooth up to the initial boundary and allowed nonzero initial trace, and in The method of characteristics for $u_{t} + u u_{x} = 0$ and the finite blow-up time of the gradient and Riemann shock and rarefaction waves for $u_{t} + (u^{2} / 2)_{x} = 0$ are weak solutions, and the shock is an entropy solution, which prove the characteristic gradient blow-up criterion and the shock and rarefaction Riemann solutions, including all convex $C^{1}$ entropy pairs as well as the Kružkov family.

The one-dimensional wave layer is self-contained: d'Alembert's formula solves the wave equation, is the unique $C^{2}$ solution, and the fixed-end sine modes solve it exactly solves the free wave equation and exhibits the fixed-end modes; the sine coefficients and their orthogonality and completeness are developed in Sine coefficients, sine series and the odd periodic extension, Sine orthogonality on a half period and the full-period integrals of odd $2 L$ -periodic functions, Fejer completeness: a continuous one-periodic function with all integrals against the characters vanishing is zero and Repeated integration by parts gives $∣ b_{n} (F) ∣ \leq 2 L^{5} M_{5} / (π^{5} n^{5})$ for odd $2 L$ -periodic $C^{5}$ functions and $∣ b_{n} (G) ∣ \leq 2 L^{4} M_{4} / (π^{4} n^{4})$ for $C^{4}$ , with Uniform convergence of a $C^{2}$ sequence with its first and second derivatives on a rectangle passes $C^{2}$ regularity and termwise differentiation to the limit supplying the two-variable differentiation pass; The classical $C^{2}$ fixed-end string: the sine series solves the wave equation with the given data, uniquely, with an explicit tail bound assembles the unique $C^{2}$ solution with an explicit $O (N^{- 4})$ tail bound. Finally The cyclic spring chain, its discrete energy, and the finite-regularity periodic comparison class and Exact dispersion relation $ω^{2} = (4 c^{2} / a^{2}) \sin^{2} (κ a / 2)$ for the cyclic chain and its continuum error bound describe the discrete chain, The centered second difference differs from the second derivative by at most $M a^{2} / 12$ when $∣ u^{''''} ∣ \leq M$ and Discrete periodic summation by parts, the scaled Cauchy-Schwarz inequality, and the discrete energy identity supply the two estimates, and Finite-time continuum limit of the cyclic spring chain: uniform $O (a^{3 / 2})$ nodal and uniform interpolation convergence proves the finite-time uniform limit.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

DefinitionDefinition: AI-adaptedProof: Not applicableOpen item page →

Euclidean spatial calculus: gradient, divergence and material derivative

Definition

We use one-based component labels: for $x \in R^{3}$ , $x_{i} : = x (i - 1)$ for $1 \leq i \leq 3$ , and $e_{i}$ denotes the imported standard unit vector at index $i - 1$ . Thus $\partial_{i}$ here is the imported partial derivative $\partial_{i - 1}$ , and an entry labelled $(i, j)$ of a $3 \times 3$ matrix is its imported entry $(i - 1, j - 1)$ . All component formulas below use this relabelling; the underlying maps, matrices, inner product and trace are unchanged.

Throughout, $R^{3}$ carries its standard ordered orthonormal basis $e_{1}, e_{2}, e_{3}$ , its Euclidean inner product $⟨ \cdot, \cdot ⟩_{2}$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

Let $O \subseteq R^{3}$ be open, let $a \in O$ and let $v : O \to R^{3}$ be of class $C^{1}$ ( $C^{k}$ Euclidean maps and diffeomorphisms), so that the total derivative $D v (a) : R^{3} \to R^{3}$ is a linear map (The total (Fréchet) derivative $D f (a)$ as the linear first-order approximation with $o (∥ h ∥_{2})$ remainder, A linear map $L : R^{m} \to R^{n}$ in Euclidean coordinates) whose matrix in the standard bases is $(\partial_{j} v_{i} (a))_{1 \leq i, j \leq 3}$ (The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case, A total derivative computes every directional derivative, and its matrix is the Jacobian). The divergence of $v$ at $a$ is

$div v (a) : = tr (D v (a)) = \sum_{i = 1}^{3} \partial_{i} v_{i} (a),$

the trace of the total derivative (The trace $tr (A)$ as the sum of the diagonal entries), read in the standard basis of $M_{3} (R)$ (The vector space $M_{m \times n} (F) : = F^{m \times n}$ of $m$ by $n$ matrices over a field, with entrywise operations).

For a $C^{1}$ scalar field $f : O \to R$ and $a \in O$ , the gradient $\nabla f (a) \in R^{3}$ is the vector of partial derivatives of The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case; it is the unique vector with $D f (a) h = ⟨ \nabla f (a), h ⟩_{2}$ for every $h \in R^{3}$ (Directional derivatives and partial derivatives of a map $U \subseteq R^{m} \to R^{n}$ , For a differentiable scalar field, $D_{v} f (a) = ⟨ \nabla f (a), v ⟩$ and the unit direction of steepest ascent is the normalized gradient). We write $⟨ \nabla f (a), v (a) ⟩_{2} = : \nabla f (a) \cdot v (a)$ .

Let $I \subseteq R$ be a nonempty open interval and let $u : I \times O \to R^{3}$ and $f : I \times O \to R$ be $C^{1}$ . For $(t, x) \in I \times O$ write $D_{x} f (t, x) : R^{3} \to R$ for the total derivative of $y \mapsto f (t, y)$ at fixed $t$ , and $\nabla_{x} f (t, x)$ for its gradient. The derivative $D f (t, x)$ of the full map on $I \times O$ acts on $R \times R^{3}$ ; it is distinct from $D_{x} f$ . Define the partial time derivative $\partial_{t} f (t, x) : = \frac{d}{d s} ∣_{s = t} f (s, x)$ and the material derivative of $f$ along $u$

$D_{t} f (t, x) : = \partial_{t} f (t, x) + D_{x} f (t, x) (u (t, x)) = \partial_{t} f (t, x) + \nabla_{x} f (t, x) \cdot u (t, x) .$

Spatial derivatives $D u$ , $D w$ , $\nabla f$ and $div u$ of time-dependent fields always mean $D_{x} u$ , $D_{x} w$ , $\nabla_{x} f$ and spatial divergence at fixed time; a stationary field $u (x)$ is included by setting $u (t, x) = u (x)$ . The same definitions apply on any open spacetime domain, by restriction to a product neighborhood of each point.

For a $C^{1}$ vector field $w : I \times O \to R^{3}$ the material derivative $D_{t} w$ is defined componentwise, and $div w (t, x)$ means the divergence of $x \mapsto w (t, x)$ at $x$ at each fixed $t$ .

Remarks

All objects above are defined in the single fixed ordered orthonormal basis $(e_{1}, e_{2}, e_{3})$ ; gradients and divergences are not attached to any physical frame, unit system or observer. Coordinates are pure numbers here. A physical reading (length units, time units, a chosen inertial chart) is additional structure supplied by a consuming model, never by this definition.
The divergence is the trace of the total derivative, not a sum of partial derivatives read in an arbitrary coordinate system; the two agree in the standard orthonormal coordinates used here, and both expressions are recorded so that no later item may silently use a different convention. The gradient likewise depends on the chosen Euclidean inner product.
The chain rule (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ ) gives, for a $C^{1}$ curve $γ : J \to O$ with $γ^{'} (t) = u (t, γ (t))$ , the identity $\frac{d}{d t} f (t, γ (t)) = D_{t} f (t, γ (t))$ whenever $f$ is $C^{1}$ ; this is the sense in which $D_{t}$ differentiates along the flow of $u$ and is used by the transport theorem of this collection.
Integrals and derivatives of fields are taken where they are defined: the material derivative requires $f$ to be $C^{1}$ on a neighbourhood of the point in question, and no extension of $u$ or $f$ is implicit. Boundary behaviour is never assumed from this definition alone.

DefinitionDefinition: AI-adaptedProof: Not applicableOpen item page →

Continuum motions, spatial velocity, material acceleration and swept regions

Definition

Let $B \subseteq R^{3}$ be a material region (nonempty, bounded, open, Material regions, densities and their mass) and let $I \subseteq R$ be a nonempty open interval (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

A motion of $B$ on $I$ consists of an open set $O \subseteq R^{3}$ with $\overline{B} \subseteq O$ (Open ball, closed ball and sphere in a metric space) together with a map

$χ : I \times O ⟶ R^{3}$

such that:

$χ$ is of class $C^{2}$ ( $C^{k}$ Euclidean maps and diffeomorphisms);
for every $t \in I$ the map $χ_{t} : = χ (t, \cdot) : O \to R^{3}$ is of class $C^{1}$ , is injective, and has $D_{X} χ_{t} (X)$ invertible for every $X \in O$ ;
writing

$J (t, X) : = \det (D_{X} χ_{t} (X))$

with the determinant of For $n \geq 1$ , the determinant over a commutative ring by the Leibniz formula, and $∣ \det A ∣$ for a real matrix, one has $J (t, X) > 0$ for every $(t, X) \in I \times B$ .

We call $X \in B$ the material label (reference position), $x = χ (t, X)$ the position at time $t$ of the label $X$ , and

$Ω_{t} : = χ_{t} (B)$

the current (material) region at time $t$ . Condition 2 makes each $χ_{t} : O \to χ_{t} (O)$ a $C^{1}$ diffeomorphism onto the open set $χ_{t} (O)$ (An injective regular $C^{k}$ map is a $C^{k}$ diffeomorphism onto its image).

The spatial velocity field of the motion is

$u : {(t, x) : t \in I, x \in χ_{t} (O)} ⟶ R^{3}, u (t, χ (t, X)) : = \partial_{t} χ (t, X),$

and the spatial acceleration field is the material derivative of $u$ in the sense of Euclidean spatial calculus: gradient, divergence and material derivative,

$a (t, x) : = D_{t} u (t, x) = \partial_{t} u (t, x) + D u (t, x) u (t, x) .$

For a compact subinterval $K \subseteq I$ the swept region of $B$ over $K$ is

$Σ_{K} : = {(t, x) : t \in K, x \in χ_{t} (\overline{B})},$

and for a scalar field $f$ on a neighbourhood of $Σ_{K}$ the spatial field $f (t, x)$ of a material quantity $F (t, X)$ is defined by $f (t, χ (t, X)) = F (t, X)$ .

Well-definedness

Put $Φ (t, X) : = (t, χ (t, X))$ on $I \times O$ . By condition 1, $Φ$ is $C^{1}$ and its total derivative at $(t, X)$ is the block map $(s, Y) \mapsto (s, \partial_{t} χ (t, X) s + D_{X} χ_{t} (X) Y)$ , invertible because $D_{X} χ_{t} (X)$ is invertible. If $Φ (t, X) = Φ (t^{'}, X^{'})$ , then $t = t^{'}$ and injectivity of $χ_{t}$ (condition 2) gives $X = X^{'}$ ; so $Φ$ is injective and, by An injective regular $C^{k}$ map is a $C^{k}$ diffeomorphism onto its image, $Φ$ corestricts to a $C^{1}$ diffeomorphism from $I \times O$ onto the open set $Φ (I \times O)$ . On that set the inverse map

$Λ : = Φ^{- 1}, Λ (t, x) = (t, χ_{t}^{- 1} (x)),$

is $C^{1}$ by the same corollary. Hence $u (t, x) = \partial_{t} χ (t, Λ (t, x))$ is a composition of $C^{1}$ maps, so $u$ is $C^{1}$ on its open domain by the chain rule (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ ), and $a = D_{t} u$ is continuous there. Finally $Σ_{K} = Φ (K \times \overline{B})$ is compact: it is the continuous image of a compact set.

Remarks

The positive Jacobian in condition 3 is required only on $I \times B$ ; the definition itself is purely mathematical and asserts neither conservation of mass nor any physical identification of $B$ with a body of matter. Physical readings (a constitutive law, a mass balance, a traction model) are separate adopted hypotheses in consuming items.
Boundary regularity of $B$ is not part of this definition. Statements that integrate over $χ_{t} (B)$ or $\partial χ_{t} (B)$ carry their own hypotheses; the transport theorem of this collection states the boundary regularity it uses, and the local balance statements say when $χ_{t} (B)$ is required to be a bounded $C^{1}$ domain.
The notation $χ_{t} (O)$ keeps the domain of $u$ honest: $u (t, \cdot)$ is defined on the image of the whole reference chart $O$ , not merely on $Ω_{t}$ ; this is what allows differentiation of $u$ at points of the current region with room to spare. Values of $u$ are spatial quantities, whereas $\partial_{t} χ$ is a material quantity; the identity $u (t, χ (t, X)) = \partial_{t} χ (t, X)$ is the definition that relates them, and no equality of the two off the image is claimed.

LemmaStatement: AI-adaptedProof: AI-adaptedOpen item page →

The derivative of the determinant along an invertible C¹ matrix curve: $(\det A)^{'} = \det A tr (A^{- 1} A^{'})$

Statement

Let $n \geq 1$ , let $I \subseteq R$ be a nonempty open interval, and let

$A : I ⟶ M_{n} (R), A (t) = (a_{i j} (t)),$

be a matrix-valued curve whose entries $a_{i j} : I \to R$ are differentiable at $t_{0} \in I$ (The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set, The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral); write $A^{'} (t_{0}) : = (a_{i j}^{'} (t_{0}))$ (Finite rectangular matrices over a commutative ring, their entries, rows and columns). Suppose $A (t)$ is invertible for every $t \in I$ (A finite square real matrix is invertible if and only if its determinant is nonzero, If $\det (A)$ is a unit, then $A^{- 1} = \det (A)^{- 1} adj (A)$ ). Then $t \mapsto \det A (t)$ is differentiable at $t_{0}$ (For $n \geq 1$ , the determinant over a commutative ring by the Leibniz formula, and $∣ \det A ∣$ for a real matrix) and

$(\det A)^{'} (t_{0}) = \det A (t_{0}) tr ⁣ (A (t_{0})^{- 1} A^{'} (t_{0})),$

with the trace of The trace $tr (A)$ as the sum of the diagonal entries. In particular, a $C^{1}$ matrix curve with invertible values has $(\det A)^{'} = \det A tr (A^{- 1} A^{'})$ on all of $I$ ( $C^{k}$ Euclidean maps and diffeomorphisms).

Facts & Assumptions

Given: Integers $n \geq 1$ ; an open interval $I$ ; a matrix curve $A$ with entries differentiable at $t_{0} \in I$ ; invertibility of $A (t)$ for all $t \in I$ .

[F1]

The Leibniz determinant $\det : M_{n} (R) \to R$ is column-multilinear, alternating and normalized, and $\det (X Y) = \det X \det Y$ for all $X, Y \in M_{n} (R)$ (The Leibniz determinant is column-multilinear, alternating and normalized over every commutative ring, For same-sized finite square matrices over a commutative ring, $\det (A B) = \det (A) \det (B)$ , For $n \geq 1$ , the determinant over a commutative ring by the Leibniz formula, and $∣ \det A ∣$ for a real matrix).

[F2]

If $A$ is invertible, then $\det A \neq 0$ , $A^{- 1}$ exists, and $A^{- 1} = \det (A)^{- 1} adj (A)$ (A finite square real matrix is invertible if and only if its determinant is nonzero, If $\det (A)$ is a unit, then $A^{- 1} = \det (A)^{- 1} adj (A)$ ).

[F3]

For $0 \leq i < n$ , replacing the $i$ -th column of the identity by $w \in R^{n}$ produces a matrix whose determinant is the $i$ -th entry $w_{i}$ : the column expansion over the standard basis has one nonzero term, by alternation and normalization [F1]. Consequently, for invertible $B$ and $b \in R^{n}$ , the matrix $B^{(i)}$ obtained from $B$ by replacing its $i$ -th column by $b$ satisfies

$\det B^{(i)} = \det B (B^{- 1} b)_{i} .$

Indeed $B^{(i)} = B E$ where $E$ is the identity with $i$ -th column $B^{- 1} b$ , so [F1] gives $\det B^{(i)} = \det B \det E = \det B (B^{- 1} b)_{i}$ .

[F4]

The trace is linear, and $tr (X Y) = tr (Y X)$ for square matrices of matching size (Trace is a linear functional on $M_{n} (F)$ , For $A \in M_{m \times n} (F)$ and $B \in M_{n \times m} (F)$ , $tr (A B) = tr (B A)$ ).

[F5]

Differentiability of the entries means that for $h \to 0$ , $a_{i j} (t_{0} + h) = a_{i j} (t_{0}) + h a_{i j}^{'} (t_{0}) + o (h)$ ; equivalently $A (t_{0} + h) = B + h H + r (h)$ with $B = A (t_{0})$ , $H = A^{'} (t_{0})$ and $\max_{i, j} ∣ r_{i j} (h) ∣ / ∣ h ∣ \to 0$ (The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set, Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ ).

Proof

technique · direct; expand the determinant by column-multilinearity and keep the terms up to first order

1.1givenF2F5choose

Put $B : = A (t_{0})$ , $H : = A^{'} (t_{0})$ and, for small $h \neq 0$ with $t_{0} + h \in I$ , write $A (t_{0} + h) = B + h H + r (h)$ with the remainder $r (h)$ of [F5]. By [F2], $B$ is invertible. Since $r_{i j} (h) = o (h)$ entrywise, choose $0 < ε_{0} \leq 1$ such that $\max_{i, j} ∣ r_{i j} (h) ∣ \leq ∣ h ∣$ for $0 < ∣ h ∣ < ε_{0}$ , and fix such an $h$ .

1.2F1givenalgebra

For fixed $h$ , expand $\det A (t_{0} + h)$ by column-multilinearity [F1]: the $i$ -th column is $b_{i} + h h_{i} + r_{i} (h)$ , so $\det A (t_{0} + h) = \sum_{S \subseteq {0, \dots, n - 1}} \det (c_{0}^{S}, \dots, c_{n - 1}^{S}), c_{i}^{S} = {\begin{cases} h h_{i} + r_{i} (h), & i \in S, \\ b_{i}, & i \notin S, \end{cases}$ the sum being over the $2^{n}$ subsets $S$ . The term $S = \emptyset$ is $\det B$ .

1.3F3F4algebra

For $S = {i}$ a single index, [F3] applied to $b : = h h_{i} + r_{i} (h)$ gives $\det B (B^{- 1} (h h_{i} + r_{i} (h)))_{i}$ . Summing over $0 \leq i < n$ and using linearity of $B^{- 1}$ and of the trace [F4], $\sum_{i} \det B (B^{- 1} (h h_{i} + r_{i} (h)))_{i} = \det B (h tr (B^{- 1} H) + tr (B^{- 1} r (h))) .$

2.1F1step 1.1algebra

For $∣ S ∣ \geq 2$ , the Leibniz formula [F1] expresses $\det (c^{S})$ as $n!$ signed products. Each product has $∣ S ∣$ entries from increment columns and $n - ∣ S ∣$ entries from $B$ . Put $M : = \max {1, \max_{i, j} ∣ H_{i j} ∣ + 1}$ and $P : = \max {1, \max_{i, j} ∣ B_{i j} ∣}$ . Step 1.1 bounds each increment entry by $∣ h ∣ M$ , so for $∣ h ∣ \leq 1$ each product has magnitude at most $∣ h ∣^{∣ S ∣} M^{∣ S ∣} P^{n - ∣ S ∣} \leq h^{2} (M P)^{n}$ . Summing over all $S$ and permutations gives $∣ \sum_{∣ S ∣ \geq 2} \det (c^{S}) ∣ \leq 2^{n} n! (M P)^{n} h^{2}$ . When $n = 1$ the sum is empty and equals zero.

3.1step 1.3step 2.1F4F5algebra∎

Combining the preceding steps, $\det A (t_{0} + h) - \det B - h \det B tr (B^{- 1} H) = \det B tr (B^{- 1} r (h)) + O (h^{2}) .$ The right-hand side is $o (h)$ : $∥ tr (B^{- 1} r (h)) ∥ \leq n \max_{i, j} ∣ (B^{- 1} r (h))_{i j} ∣ \leq C^{'} ∣ r (h) ∣_{\max} = o (∣ h ∣)$ by [F5], and $O (h^{2}) = o (h)$ . Hence $(\det A)^{'} (t_{0}) = \det B tr (B^{- 1} H)$ , which is the stated formula.

Remarks

The formula is exactly the Jacobi formula for the derivative of the determinant. Its use in this collection is for $n = 3$ and $A (t) = D_{X} χ_{t} (X)$ , where the chain rule identifies $A^{'} (t) = (D u (t, χ_{t} (X))) A (t)$ , so that $(\det A)^{'} = \det A div u$ ; that composition is carried out in the transport theorem, not here.
Only differentiability of the entries at the point $t_{0}$ is used; continuity of $t \mapsto tr (A (t)^{- 1} A^{'} (t))$ is not asserted by this lemma and is not needed for the pointwise derivative.

LemmaStatement: AI-adaptedProof: AI-adaptedOpen item page →

Localization: a continuous function whose integrals vanish over a neighbourhood-basis family is identically zero

Statement

Assume the Axiom of Countable Choice ${A C}_{ω}$ , which supplies the measure properties of Lebesgue measure used below (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Let $W \subseteq R^{3}$ be open and let $H : W \to R$ be continuous in the Euclidean metric: for every $y \in W$ and $ε > 0$ there is $δ > 0$ such that $z \in W$ , $∥ z - y ∥_{2} < δ$ imply $∣ H (z) - H (y) ∣ < ε$ . Let $F$ be a family of nonempty open subsets of $W$ such that

$\begin{matrix} for every y_{0} \in W and every neighbourhood V of y_{0} there is A \in F with y_{0} \in A \subseteq V . & (L) \end{matrix}$

Suppose the signed Lebesgue integral is defined and $\int_{A} H d x = 0$ for every $A \in F$ (Integral over a measurable subset, Integrable real and complex functions, and their integrals, Lebesgue measurable sets, the family $L (R^{n})$ , and the restricted set function $λ_{n}$ ). Then $H = 0$ on $W$ .

Facts & Assumptions

Given: The Axiom of Countable Choice, the open set $W$ , the continuous $H$ , the family $F$ satisfying (L), and the defined signed integrals $\int_{A} H = 0$ .

[F1]

If $y_{0} \in W$ and $A \subset W$ is open with $y_{0} \in A$ , then some $σ > 0$ has $B_{σ} (y_{0}) \subseteq A$ ; if $σ$ is chosen also so that $\overline{B} (y_{0}, ρ) \subseteq W$ for every $ρ < σ$ , then for every $0 < r < σ$ the half-open box

$Q_{r} : = \prod_{j = 0}^{2} (y_{0, j} - \frac{r}{2}, y_{0, j} + \frac{r}{2}]$

has $\overline{Q_{r}} \subseteq B_{σ} (y_{0}) \subseteq A$ because every point of $\overline{Q_{r}}$ is within distance $\frac{\sqrt{3}}{2} r < σ$ of $y_{0}$ ; its Lebesgue measure is $λ_{3} (Q_{r}) = r^{3} > 0$ (Half-open boxes in $R^{n}$ and their volume, A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included).

[F2]

Order and linearity of the Lebesgue integral: if $0 \leq g \leq h$ pointwise and both are integrable on a measurable set, then $\int g \leq \int h$ ; and $\int - H = - \int H$ whenever $H$ is integrable (Monotonicity and nonnegative homogeneity of the nonnegative integral, The Lebesgue integral is linear on $L^{1} (μ)$ ).

[F3]

If $H$ is continuous and positive throughout a nonempty compact set $K \subseteq W$ , then it attains its minimum on $K$ and that minimum is positive (A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value, Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point). Every closed box $\overline{Q_{r}}$ is compact by Heine--Borel (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line).

Proof

technique · direct; a positive value would force a positive integral over a small box

1.1givenF1F3choose

Suppose for contradiction that $H (y_{0}) \neq 0$ for some $y_{0} \in W$ . If $H (y_{0}) > 0$ , continuity and openness give $σ > 0$ with $\overline{B} (y_{0}, σ) \subseteq W$ and $H > 0$ on this closed ball; if $H (y_{0}) < 0$ , apply the same argument to $- H$ . Put $V : = B_{σ} (y_{0})$ .

2.1step 1.1givenF1F3

Choose $A \in F$ with $y_{0} \in A \subseteq V$ by (L). Since $A$ is open, [F1] gives a box $Q_{r}$ with $\overline{Q_{r}} \subseteq A$ . In the positive case, $H > 0$ on $V$ , so [F3] applied to the compact set $\overline{Q_{r}} \subset V$ gives $ε : = \min_{\overline{Q_{r}}} H > 0$ ; in the negative case apply [F3] to $- H$ .

3.1step 2.1F1F2algebra∎

In the case $H > 0$ on $V$ : since $0 \leq ε 1_{Q_{r}} \leq H$ on $A$ , [F2] and [F1] give $\int_{A} H d x \geq \int_{Q_{r}} H d x \geq ε λ_{3} (Q_{r}) > 0,$ contradicting $\int_{A} H = 0$ . In the case $H < 0$ on $V$ , applying the same argument to $- H$ , whose integral over $A$ also vanishes by the linearity clause of [F2], gives $\int_{A} (- H) > 0$ , again a contradiction. Hence $H (y_{0}) = 0$ for every $y_{0} \in W$ .

Remarks

The family $F$ is exactly the family for which vanishing region integrals are available in the applications: images of small reference boxes under a diffeomorphism, small Euclidean balls, or bounded $C^{1}$ domains shrinking to a point. The hypothesis (L) is what makes "choose $A$ inside any neighbourhood" legitimate; no countable choice is used, since $A$ is chosen for a single specified neighbourhood.
For an $R^{n}$ -valued continuous $H$ the statement and proof apply to each component separately; the applications in this collection use the scalar or the componentwise form.

TheoremStatement: AI-adaptedProof: AI-adaptedOpen item page →

The transport theorem for a $C^{2}$ family of diffeomorphisms: $\frac{d}{d t} \int_{ψ_{t} (B)} f = \int_{ψ_{t} (B)} (\partial_{t} f + div (f w))$

Statement

Assume the Axiom of Countable Choice ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )), as used by the change-of-variables theorem below. Let $I \subseteq R$ be a nonempty open interval (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length), let $B \subseteq R^{3}$ be a material region (Material regions, densities and their mass) and let $ψ : I \times O \to R^{3}$ be a motion of $B$ on $I$ in the sense of Continuum motions, spatial velocity, material acceleration and swept regions, with associated spatial velocity

$w (t, ψ (t, X)) = \partial_{t} ψ (t, X)$

and Jacobian $J (t, X) = \det D_{X} ψ_{t} (X)$ (Continuum motions, spatial velocity, material acceleration and swept regions, Euclidean spatial calculus: gradient, divergence and material derivative).

Let $J_{0} \subseteq I$ be an open interval with $\overline{J_{0}} \subseteq I$ , put

$Σ : = {(t, x) : t \in \overline{J_{0}}, x \in ψ_{t} (\overline{B})},$

and let $f : U \to R$ be of class $C^{1}$ ( $C^{k}$ Euclidean maps and diffeomorphisms) on an open set $U \subseteq R \times R^{3}$ containing $Σ$ . Then

$F (t) : = \int_{ψ_{t} (B)} f (t, x) d x$

is defined and differentiable on $J_{0}$ (Integral over a measurable subset, Integrable real and complex functions, and their integrals), and for every $t \in J_{0}$

$F^{'} (t) = \int_{ψ_{t} (B)} (\partial_{t} f + div (f w)) (t, x) d x,$

with the divergence and spatial derivative of Euclidean spatial calculus: gradient, divergence and material derivative and the Lebesgue integral of Lebesgue measurable sets, the family $L (R^{n})$ , and the restricted set function $λ_{n}$ . The same formula holds componentwise for $f$ replaced by a $C^{1}$ vector field.

Facts & Assumptions

Given: ${A C}_{ω}$ ; an open interval $I$ ; a motion $ψ$ of $B$ on $I$ with spatial velocity $w$ ; an open interval $J_{0}$ with $\overline{J_{0}} \subseteq I$ ; a $C^{1}$ function $f$ on an open neighbourhood of the swept region $Σ$ .

[F1]

For each $t \in I$ the map $ψ_{t} : O \to ψ_{t} (O)$ is a $C^{1}$ diffeomorphism onto an open set, and $u$ and $w$ are $C^{1}$ on their open domains (Continuum motions, spatial velocity, material acceleration and swept regions).

[F2]

Change of variables: if $T : U_{0} \to V_{0}$ is a $C^{1}$ diffeomorphism between open subsets of $R^{3}$ and $h \in L^{1} (V_{0})$ , then $\int_{V_{0}} h d y = \int_{U_{0}} h (T (x)) ∣ \det D T (x) ∣ d x$ ; under the present hypotheses $\det D_{X} ψ_{t} = J > 0$ on $I \times B$ (A C^1 diffeomorphism satisfies the change-of-variables formula for L^1 functions, The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F3]

The determinant-derivative lemma: for the $C^{1}$ matrix curve $t \mapsto A (t) : = D_{X} ψ_{t} (X)$ with invertible values, $(\det A)^{'} = \det A tr (A^{- 1} A^{'})$ (The derivative of the determinant along an invertible C¹ matrix curve: $(\det A)^{'} = \det A tr (A^{- 1} A^{'})$ ).

[F4]

Differentiation under the integral sign: if $g : B \times J_{0} \to R$ has $x \mapsto g (x, t)$ integrable for every $t$ , $t \mapsto g (x, t)$ differentiable for almost every $x$ , $x \mapsto \partial_{t} g (x, t)$ measurable, and $∣ \partial_{t} g ∣ \leq G$ for an integrable $G$ , then $t \mapsto \int_{B} g$ is differentiable with derivative $\int_{B} \partial_{t} g$ (Differentiation under the integral sign).

[F5]

The chain rule and the product rule: $C^{1}$ compositions are $C^{1}$ with the usual total-derivative rule; finite sums and products of differentiable functions satisfy the algebra rules (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ , Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , $C^{k}$ Euclidean maps and diffeomorphisms).

[F6]

On each compact cylinder $\overline{J_{1}} \times \overline{B}$ used in the local proof, continuous functions are bounded and attain their maxima (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line, A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value, Open ball, closed ball and sphere in a metric space).

[F7]

For a $C^{2}$ map $ψ$ on an open spacetime domain, each coordinate mixed partial $\partial_{X_{j}} \partial_{t} ψ_{i}$ equals $\partial_{t} \partial_{X_{j}} ψ_{i}$ . At a point, the rectangular increment over a small positive $t$ - and $X_{j}$ -interval, divided by the rectangle area, is the average of either ordered mixed partial by two applications of the one-variable fundamental theorem; continuity makes both averages converge to their point values (The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ , A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion).

[F8]

A bounded measurable subset of $R^{3}$ has finite Lebesgue measure: it is contained in a bounded box, whose measure is its finite volume, and Lebesgue measure is monotone as a measure (A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included, Assuming countable choice, $L (R^{n})$ is a sigma-algebra containing every elementary set and $λ_{n}$ is a complete measure extending elementary volume).

Proof

technique · direct; pull the moving integral back to the fixed region $B$, differentiate under the integral sign, and push forward again

1.1givenF1F5F6construct

Fix $τ \in J_{0}$ . Choose a bounded open interval $J_{1}$ with $τ \in J_{1}$ and $\overline{J_{1}} \subseteq J_{0}$ . Its swept set $Σ_{1} : = {(t, x) : t \in \overline{J_{1}}, x \in ψ_{t} (\overline{B})}$ is compact and lies in $U$ , since it is contained in $Σ$ . The continuous map $(t, X) \mapsto (t, ψ (t, X))$ therefore maps $\overline{J_{1}} \times \overline{B}$ into $U$ . Compactness gives an open neighbourhood $O_{1}$ of $\overline{B}$ in $O$ such that $(t, ψ (t, X)) \in U$ for $t \in \overline{J_{1}}$ , $X \in O_{1}$ : finitely many product neighbourhoods inside the inverse image of $U$ cover the compact cylinder, or equivalently its positive distance from the closed complement supplies a uniform spatial margin. Define $g : J_{1} \times O_{1} ⟶ R, g (t, X) : = f (t, ψ (t, X)) J (t, X) .$ By [F1] and [F5], $g$ is $C^{1}$ , and $\partial_{t} g$ is bounded on the compact cylinder $\overline{J_{1}} \times \overline{B}$ by [F6]. The argument below is applied on this local interval; compactness of the whole $\overline{J_{0}}$ is not required.

2.1F1F2F8step 1.1algebra

For $t \in J_{1}$ , [F2] applied to the $C^{1}$ diffeomorphism $ψ_{t} : B \to ψ_{t} (B)$ and the integrable function $f (t, \cdot)$ restricted to $ψ_{t} (B)$ gives $F (t) = \int_{ψ_{t} (B)} f (t, x) d x = \int_{B} f (t, ψ (t, X)) J (t, X) d X = \int_{B} g (t, X) d X .$ These integrals are finite because $B$ and $ψ_{t} (B)$ are bounded measurable sets of finite measure by [F8], while the integrands are continuous and bounded on their compact closures.

2.2F1F3F5F7step 1.1algebra

Differentiate $g$ in time at $(t, X)$ . By [F5], $\partial_{t} [f (t, ψ (t, X))] = (\partial_{t} f) (t, ψ (t, X)) + D_{x} f (t, ψ (t, X)) (\partial_{t} ψ (t, X)) = (\partial_{t} f) (t, ψ (t, X)) + ⟨ \nabla f, w ⟩ (t, ψ (t, X))$ , where the second form uses $w (t, ψ (t, X)) = \partial_{t} ψ (t, X)$ . For $A (t) = D_{X} ψ_{t} (X)$ , [F7] gives $A^{'} (t) = D_{X} \partial_{t} ψ (t, X)$ . Differentiating $w (t, ψ (t, X)) = \partial_{t} ψ (t, X)$ with respect to $X$ and applying the chain rule yields $D_{x} w (t, ψ (t, X)) A (t) = D_{X} \partial_{t} ψ (t, X) = A^{'} (t)$ . By [F3], $(\partial_{t} J) (t, X) = J (t, X) tr (D_{x} w (t, ψ (t, X))) = J (t, X) (div w) (t, ψ (t, X))$ . Hence $\begin{matrix} \partial_{t} g (t, X) = [(\partial_{t} f + ⟨ \nabla f, w ⟩ + f div w) \circ (t, ψ (t, X))] J (t, X) . & (*) \end{matrix}$

3.1givenF5step 2.2algebra

The product rule for a scalar field $f$ and a vector field $w$ , read off componentwise from the definitions of gradient and divergence, gives $div (f w) = \sum_{i} \partial_{i} (f w_{i}) = \sum_{i} (\partial_{i} f) w_{i} + f \sum_{i} \partial_{i} w_{i} = ⟨ \nabla f, w ⟩ + f div w .$ Therefore ( $*$ ) reads $\partial_{t} g (t, X) = (\partial_{t} f + div (f w)) (t, ψ (t, X)) J (t, X)$ .

4.1F4step 1.1step 2.1step 3.1F6

Apply [F4] to $g$ on $X = B$ over the interval $J_{1}$ with the integrable dominating constant $\sup_{\overline{J_{1}} \times \overline{B}} ∣ \partial_{t} g ∣ < + \infty$ (legitimate because $B$ has finite Lebesgue measure and the supremum is finite by step 1.1 and [F6]): the function $t \mapsto \int_{B} g (t, X) d X = F (t)$ is differentiable on $J_{1}$ with $F^{'} (t) = \int_{B} \partial_{t} g (t, X) d X = \int_{B} (\partial_{t} f + div (f w)) (t, ψ (t, X)) J (t, X) d X .$

5.1F2step 4.1algebra∎

Change variables back with [F2], applied to $h : = (\partial_{t} f + div (f w)) (t, \cdot)$ restricted to $ψ_{t} (B)$ , which is integrable by continuity on the compact closure and the diffeomorphism $ψ_{t}$ : $\int_{B} (\partial_{t} f + div (f w)) (t, ψ (t, X)) J (t, X) d X = \int_{ψ_{t} (B)} (\partial_{t} f + div (f w)) (t, x) d x .$ Combining with step 4.1 gives the stated formula at $τ$ ; since $τ \in J_{0}$ was arbitrary, it holds on all of $J_{0}$ .

Remarks

The material motion of Continuum motions, spatial velocity, material acceleration and swept regions is the special case $ψ = χ$ , $w = u$ ; the theorem is stated for a general $C^{2}$ family so that the control-volume identities can reuse it with an auxiliary boundary velocity. Positivity of $J$ on $B$ is used to rewrite $∣ \det D ψ_{t} ∣$ as $J$ on the reference region; without that positivity the pullback would instead use $∣ J ∣$ .
The proof is local at each time in $J_{0}$ : a compact inner time cylinder supplies the integrable majorant. The entire interval $J_{0}$ need not be bounded. No claim is made about a time interval whose closure meets the boundary of $I$ or about times at which the diffeomorphism family degenerates.
The proof uses the change-of-variables theorem, which assumes ${A C}_{ω}$ ; the axiom is declared here and in the consuming items. No other choice principle is used.

TheoremStatement: AI-adaptedProof: AI-adaptedOpen item page →

Local mass balance: conservation on every material subregion implies $\partial_{t} ρ + div (ρ u) = 0$

Statement

Assume the Axiom of Countable Choice ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )), as used by the transport theorem. Let $I \subseteq R$ be a nonempty open interval, let $B \subseteq R^{3}$ be a material region, let $χ : I \times O \to R^{3}$ be a motion of $B$ on $I$ with spatial velocity $u$ (Continuum motions, spatial velocity, material acceleration and swept regions), and let $J_{0} \subseteq I$ be an open interval with $\overline{J_{0}} \subseteq I$ . Define the swept region

$Σ : = {(t, x) : t \in \overline{J_{0}}, x \in χ_{t} (\overline{B})} .$

Write $D_{χ} : = {(t, x) : t \in I, x \in χ_{t} (O)}$ for the open domain of the spatial velocity $u$ (Continuum motions, spatial velocity, material acceleration and swept regions). Let $U \subseteq D_{χ}$ be open and contain the swept region $Σ$ . Let $ρ : U \to (0, \infty)$ be of class $C^{1}$ ( $C^{k}$ Euclidean maps and diffeomorphisms); call it a density field. A material subregion is a nonempty bounded open $B^{'} \subseteq B$ with $\overline{B^{'}} \subseteq B$ (Material regions, densities and their mass); for such $B^{'}$ the same $χ$ is a motion of $B^{'}$ .

(i) Localization. Assume for every material subregion $B^{'}$ and every $t \in J_{0}$

$\begin{matrix} \frac{d}{d t} \int_{χ_{t} (B^{'})} ρ (t, x) d x = 0, & (CM) \end{matrix}$

It is sufficient to assume this identity only for those pairs $(t, B^{'})$ for which $χ_{t} (B^{'})$ is an open Euclidean ball with closure inside $χ_{t} (B)$ . Under either version of the conservation hypothesis,

$\partial_{t} ρ (t, x) + div (ρ u) (t, x) = 0$

for every $(t, x) \in J_{0} \times χ_{t} (B)$ , with the material derivative and divergence of Euclidean spatial calculus: gradient, divergence and material derivative.

(ii) Converse. Conversely, if $\partial_{t} ρ + div (ρ u) = 0$ on an open neighbourhood of $Σ$ in $U$ , then (CM) holds for every material subregion $B^{'}$ and every $t \in J_{0}$ .

Facts & Assumptions

Given: ${A C}_{ω}$ ; the motion $χ$ ; the open interval $J_{0}$ ; the $C^{1}$ density field $ρ$ ; the conservation hypothesis (CM) in (i), or the local equation in (ii).

[F1]

Transport theorem: for every material subregion $B^{'}$ and every function $f$ of class $C^{1}$ on a neighbourhood of its swept region, $\frac{d}{d t} \int_{χ_{t} (B^{'})} f = \int_{χ_{t} (B^{'})} (\partial_{t} f + div (f u))$ at each $t$ in the interior of the relevant interval (The transport theorem for a $C^{2}$ family of diffeomorphisms: $\frac{d}{d t} \int_{ψ_{t} (B)} f = \int_{ψ_{t} (B)} (\partial_{t} f + div (f w))$ , Continuum motions, spatial velocity, material acceleration and swept regions).

[F2]

Localization lemma: if $H$ is continuous on an open $W \subseteq R^{3}$ and $\int_{A} H d x = 0$ for every member $A$ of a family $F$ of nonempty open subsets of $W$ such that every point of $W$ has members of $F$ contained in each of its neighbourhoods, then $H = 0$ on $W$ (Localization: a continuous function whose integrals vanish over a neighbourhood-basis family is identically zero). The supporting order rule for the nonnegative Lebesgue integral and the positivity of box volumes are those cited there (Monotonicity and nonnegative homogeneity of the nonnegative integral, The Lebesgue integral is linear on $L^{1} (μ)$ , A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included, Half-open boxes in $R^{n}$ and their volume, Integral over a measurable subset).

[F3]

For each fixed $t$ the map $χ_{t}$ is a homeomorphism of $O$ onto the open set $χ_{t} (O)$ and is continuous at each point $X_{0}$ ; every neighbourhood of $y_{0} = χ_{t} (X_{0})$ contains the image of a sufficiently small box around $X_{0}$ (Continuum motions, spatial velocity, material acceleration and swept regions, Open ball, closed ball and sphere in a metric space, Half-open boxes in $R^{n}$ and their volume).

[F4]

On every compact inner time cylinder $K \times \overline{B}$ , with $K \subseteq J_{0}$ compact, and on compact spatial neighbourhoods, continuous functions are bounded (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line, A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value).

Proof

technique · direct; transport the vanishing integral to the reference region, then localize with an elementary open-set argument

1.1givenF1algebra

Assume (CM), either on all material subregions or on the stated current-ball test pairs. Fix a test pair $(t, B^{'})$ to which the chosen hypothesis applies. The domain of $f : = ρ$ is the open set $U$ , which contains $Σ$ , and $ρ$ is $C^{1}$ , so [F1] applies to the motion of $B^{'}$ and gives $0 = \frac{d}{d t} \int_{χ_{t} (B^{'})} ρ (t, x) d x = \int_{χ_{t} (B^{'})} (\partial_{t} ρ + div (ρ u)) (t, x) d x .$

2.1step 1.1givenF3construct

Put $H : = \partial_{t} ρ + div (ρ u)$ , a continuous function on the common open field domain. Fix $t \in J_{0}$ and an open ball $A$ with $\overline{A} \subseteq W_{t} : = χ_{t} (B)$ . Its preimage $B^{'} : = χ_{t}^{- 1} (A)$ is nonempty, open and bounded; indeed $χ_{t}^{- 1} (\overline{A})$ is compact and contained in $B$ , so $\overline{B^{'}} \subseteq B$ . Thus $B^{'}$ is a material subregion and $(t, B^{'})$ is a permitted current-ball test pair. Step 1.1 gives $\int_{A} H (t, \cdot) d x = 0$ .

3.1step 2.1F2F3construct

The family $F_{t}$ of all open balls with closures in the open $W_{t}$ contains, around each point, a member inside any prescribed neighbourhood: choose a sufficiently small positive radius. Step 2.1 gives vanishing integrals on every member. Applying [F2] to the continuous function $H (t, \cdot)$ and this family yields $H (t, y) = 0$ for every $y \in W_{t}$ .

4.1step 1.1step 3.1F1F4algebra∎

Steps 1.1 and 3.1 prove (i): for every $t \in J_{0}$ and every $y_{0} \in χ_{t} (B)$ , $\partial_{t} ρ + div (ρ u) = 0$ at $(t, y_{0})$ . Conversely, if $\partial_{t} ρ + div (ρ u) = 0$ on a neighbourhood of $Σ$ , then for every material subregion $B^{'}$ , [F1] applies to $f = ρ$ on the swept region of $B^{'}$ (contained in $Σ$ ), and its integrand vanishes identically, so the derivative in (CM) is zero for every $t \in J_{0}$ . This proves (ii).

Remarks

Positivity of $ρ$ is part of the density reading and is not used by the implication (i); only $C^{1}$ regularity on $U$ and the conservation hypothesis enter. The converse also uses only the stated regularity.
The two implications are conditional on the adopted conservation law (CM) or on the local equation; neither direction asserts that a physical density is conserved. Localization needs the full stated neighbourhood family: all small current balls suffice. Conservation on only one chosen region is insufficient.
The statement is local in the current region: it concludes the equation at every point of $χ_{t} (B)$ for $t \in J_{0}$ , and says nothing on $\partial χ_{t} (B)$ or at the endpoints of $I$ , where the transport theorem is not asserted.

TheoremStatement: AI-adaptedProof: AI-adaptedOpen item page →

Local momentum balance: $ρ D_{t} u = ρ b + div σ$ from the integral balance on admissible material subregions

Statement

Assume the Axiom of Countable Choice ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )), as used by the transport and divergence theorems. Let $I \subseteq R$ be a nonempty open interval, let $B \subseteq R^{3}$ be a material region, let $χ : I \times O \to R^{3}$ be a motion of $B$ on $I$ with spatial velocity $u$ and material derivative $D_{t}$ (Continuum motions, spatial velocity, material acceleration and swept regions, Euclidean spatial calculus: gradient, divergence and material derivative), and let $J_{0} \subseteq I$ be an open interval with $\overline{J_{0}} \subseteq I$ . Define the swept region $Σ = {(t, x) : t \in \overline{J_{0}}, x \in χ_{t} (\overline{B})}$ .

Write $D_{χ} : = {(t, x) : t \in I, x \in χ_{t} (O)}$ for the open domain of the spatial velocity $u$ (Continuum motions, spatial velocity, material acceleration and swept regions), and take $U \subseteq D_{χ}$ open and containing $Σ$ . Let $ρ : U \to (0, \infty)$ be a $C^{1}$ density field, let $σ : U \to M_{3} (R)$ be a $C^{1}$ field (the stress field of Traction fields and stress matrices representing them), and let $b : U \to R^{3}$ be continuous. Suppose:

(K) $ρ$ satisfies $\partial_{t} ρ + div (ρ u) = 0$ at every $(t, x)$ with $t \in J_{0}$ and $x \in χ_{t} (B)$ ;

(L) for every material subregion $B^{'}$ (nonempty bounded open with $\overline{B^{'}} \subseteq B$ ) and every $t \in J_{0}$ for which $χ_{t} (B^{'})$ is a bounded $C^{1}$ domain (Bounded C1 domains and their outward normals) , the balance is

$\frac{d}{d t} \int_{χ_{t} (B^{'})} ρ (t, x) u (t, x) d x = \int_{χ_{t} (B^{'})} ρ (t, x) b (t, x) d x + \int_{\partial χ_{t} (B^{'})} σ (t, y) ν (y) d S (y),$

where $ν$ is the outward unit normal and the surface integral is that of Surface integration on compact C1 hypersurfaces.

Define the row divergence of $σ$ componentwise by

$g (t, x) : = (\sum_{j} \partial_{j} σ_{1 j}, \sum_{j} \partial_{j} σ_{2 j}, \sum_{j} \partial_{j} σ_{3 j}) (t, x) .$

Then for every $(t, x) \in J_{0} \times χ_{t} (B)$ ,

$ρ (t, x) D_{t} u (t, x) = ρ (t, x) b (t, x) + g (t, x) .$

Facts & Assumptions

Given: ${A C}_{ω}$ ; the motion $χ$ ; $ρ, σ, b$ with hypotheses (K) and (L).

[F1]

Transport theorem, componentwise: for each component $i$ , $\frac{d}{d t} \int_{χ_{t} (B^{'})} ρ u_{i} = \int_{χ_{t} (B^{'})} (\partial_{t} (ρ u_{i}) + div (ρ u_{i} u))$ , for every material subregion $B^{'}$ and $t \in J_{0}$ (The transport theorem for a $C^{2}$ family of diffeomorphisms: $\frac{d}{d t} \int_{ψ_{t} (B)} f = \int_{ψ_{t} (B)} (\partial_{t} f + div (f w))$ , Continuum motions, spatial velocity, material acceleration and swept regions). Here $ρ u$ is a $C^{1}$ vector field on $U$ by $C^{k}$ Euclidean maps and diffeomorphisms and Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ .

[F2]

Divergence theorem on a bounded $C^{1}$ domain: for $F \in C^{1} (\overline{Ω}; R^{3})$ , $\int_{Ω} div F d x = \int_{\partial Ω} F \cdot ν d S$ , with interior-up-to-boundary regularity (Divergence on a bounded C1 Euclidean domain, Bounded C1 domains and their outward normals). Applied to the rows $F_{i} : = (σ_{i 1}, σ_{i 2}, σ_{i 3})$ on $Ω = χ_{t} (B^{'})$ , whose closure lies in $U$ , it gives $\int_{\partial χ_{t} (B^{'})} σ ν d S = \int_{χ_{t} (B^{'})} g d x$ componentwise; the hypothesis (L) provides the surface integral with the outward normal convention of Traction fields and stress matrices representing them.

[F3]

For a $C^{1}$ vector field $u$ and a $C^{1}$ scalar $ρ$ ,

$\partial_{t} (ρ u_{i}) + div (ρ u_{i} u) - u_{i} (\partial_{t} ρ + div (ρ u)) = ρ \partial_{t} u_{i} + ρ \sum_{j} u_{j} \partial_{j} u_{i},$

as a direct computation from the product rule: $div (ρ u_{i} u) = u_{i} div (ρ u) + ρ \sum_{j} u_{j} \partial_{j} u_{i}$ (Euclidean spatial calculus: gradient, divergence and material derivative, Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , A total derivative computes every directional derivative, and its matrix is the Jacobian).

[F4]

Localization lemma: if $H$ is continuous on an open $W \subseteq R^{3}$ and $\int_{A} H d x = 0$ for every member $A$ of a family of nonempty open subsets of $W$ containing, below every neighbourhood of every point, a member around that point, then $H = 0$ on $W$ (Localization: a continuous function whose integrals vanish over a neighbourhood-basis family is identically zero); the supporting positivity of box volumes and order rule are those cited there (A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included, Half-open boxes in $R^{n}$ and their volume, Monotonicity and nonnegative homogeneity of the nonnegative integral, The Lebesgue integral is linear on $L^{1} (μ)$ , Open ball, closed ball and sphere in a metric space).

Proof

technique · direct; insert the continuity equation into the transported momentum integral and use the divergence theorem, then localize

1.1givenF1F2algebra

Fix a material subregion $B^{'}$ and $t \in J_{0}$ whose current image $χ_{t} (B^{'})$ is a bounded $C^{1}$ domain. By [F1] and [F2], hypothesis (L) reads componentwise $0 = \frac{d}{d t} \int_{χ_{t} (B^{'})} ρ u_{i} d x - \int_{χ_{t} (B^{'})} ρ b_{i} d x - \int_{χ_{t} (B^{'})} g_{i} d x = \int_{χ_{t} (B^{'})} (\partial_{t} (ρ u_{i}) + div (ρ u_{i} u) - ρ b_{i} - g_{i}) d x .$

2.1givenF3step 1.1algebra

Define the continuous field $H_{i} : = \partial_{t} (ρ u_{i}) + div (ρ u_{i} u) - ρ b_{i} - g_{i}$ on $U$ (continuity by Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ and componentwise differentiation of $C^{1}$ fields). At points with $t \in J_{0}$ and $x \in χ_{t} (B)$ , (K) and [F3] give $H_{i} = ρ \partial_{t} u_{i} + ρ \sum_{j} u_{j} \partial_{j} u_{i} - ρ b_{i} - g_{i} = ρ (D_{t} u)_{i} - ρ b_{i} - g_{i},$ the last equality by the definition of $D_{t} u$ in Euclidean spatial calculus: gradient, divergence and material derivative.

3.1step 1.1step 2.1F4givenconstruct

Fix $t \in J_{0}$ and $i \in {1, 2, 3}$ . Let $A$ be any open Euclidean ball with $\overline{A} \subseteq W_{t} : = χ_{t} (B)$ . Its preimage $B^{'} : = χ_{t}^{- 1} (A)$ is nonempty, open and bounded with $\overline{B^{'}} \subseteq B$ , since the inverse image of $\overline{A}$ is compact inside $B$ . Its current image $A$ is a bounded $C^{1}$ domain: the spherical boundary has the usual local graph charts and outward radial normal. Thus (L) applies to this pair, and step 1.1 gives $\int_{A} H_{i} (t, \cdot) d x = 0$ . These balls form a neighbourhood family at every point of $W_{t}$ . By [F4] the continuous $H_{i} (t, \cdot)$ therefore vanishes on $W_{t}$ .

4.1step 2.1step 3.1algebra∎

By step 2.1, $H_{i} = ρ (D_{t} u)_{i} - ρ b_{i} - g_{i}$ ; step 3.1 therefore gives $ρ (t, y_{0}) (D_{t} u)_{i} (t, y_{0}) = ρ (t, y_{0}) b_{i} (t, y_{0}) + g_{i} (t, y_{0})$ for $i = 1, 2, 3$ at every $(t, y_{0}) \in J_{0} \times χ_{t} (B)$ , which is the stated vector identity.

Remarks

The conclusion is conditional on the integral balance (L) and the continuity equation (K); it is a localization theorem, not a derivation of the balance law. Hypothesis (L) also states the boundary regularity that makes the surface integral and the divergence theorem meaningful; without it the surface term is not defined.
The row divergence convention $g_{i} = \sum_{j} \partial_{j} σ_{i j}$ is exactly the vector whose component $i$ appears in the divergence theorem applied to the $i$ -th row of $σ$ ; it is the matrix divergence consistent with the traction convention $t (n) = σ n$ of Traction fields and stress matrices representing them. For a symmetric stress matrix the row and column divergences coincide; their agreement is not assumed here.
No symmetry of $σ$ , no constitutive relation and no boundary condition are used or asserted. The equation is local on the current region only.

TheoremStatement: AI-adaptedProof: AI-adaptedOpen item page →

A continuous traction field balanced on every piecewise $C^{1}$ region is given by a continuous stress matrix, and $t (- n) = - t (n)$

Statement

Assume ${A C}_{ω}$ where the surface and divergence machinery of the cited items assumes it. Let $O \subseteq R^{3}$ be open, let $t : O \times S^{2} \to R^{3}$ be continuous, where $S^{2} = {n \in R^{3} : ∥ n ∥_{2} = 1}$ is the unit sphere, and let $g : O \to R^{3}$ be continuous and bounded, with $C_{g} : = \sup_{O} ∥ g ∥_{2} < + \infty$ . Suppose that for every bounded piecewise $C^{1}$ Euclidean domain $D$ with a specified finite presentation (Specified finite piecewise C1 boundary presentations) whose closure lies in $O$ and whose associated pieces lie in $O$ ,

$\begin{matrix} \int_{\partial D} t (y, ν_{D} (y)) d S (y) + \int_{D} g (y) d y = 0, & (B) \end{matrix}$

where $ν_{D}$ is the actual outward unit normal off the edge set of the presentation, the boundary term is the sum of the surface integrals over the specified faces (Surface integration on compact C1 hypersurfaces), and values of the normal on the edge set are irrelevant.

Then there is a unique continuous map

$σ : O ⟶ M_{3} (R)$

with

$t (y, n) = σ (y) n for every (y, n) \in O \times S^{2},$

the product being matrix-vector multiplication (The vector space $M_{m \times n} (F) : = F^{m \times n}$ of $m$ by $n$ matrices over a field, with entrywise operations, Rectangular matrix multiplication and the identity matrix $I_{n}$ , including zero-sized shapes). Explicitly, the columns of $σ (y)$ are the vectors $t (y, e_{1}), t (y, e_{2}), t (y, e_{3})$ ; and necessarily

$t (y, - n) = - t (y, n) for every (y, n) \in O \times S^{2} .$

Facts & Assumptions

Given: The open set $O$ , the continuous traction field $t$ , the bounded continuous field $g$ , and the balance hypothesis (B).

[F1]

A bounded piecewise $C^{1}$ domain is a nonempty bounded open set with compact faces covering its boundary, a compact edge set of surface measure zero on each face, and, off the edges, a local $C^{1}$ graph description with the domain on one side; the boundary integral is the sum of the face surface integrals with the actual outward normals (Specified finite piecewise C1 boundary presentations, Surface integration on compact C1 hypersurfaces). Boxes and tetrahedra with closures in $O$ and vertices in general position are examples: their faces are compact planar triangles or rectangles, their edges form a compact set meeting each face in a segment of zero surface measure, and off the edges the boundary is locally a plane. The cube and tetrahedron constructions below therefore fall under (B).

[F2]

For a continuous $h$ on a compact set $K$ and a compact surface piece $F$ contained in $K$ with finite area $S (F) > 0$ , the face average satisfies

$∣ \frac{1}{S (F)} \int_{F} h d S - h (y_{0}) ∣ \leq \sup_{y \in F} ∥ h (y) - h (y_{0}) ∥_{2} .$

Indeed the definition of the surface integral and the order rules for the nonnegative Lebesgue integral bound $\int_{F} ∥ h - h (y_{0}) ∥_{2} d S$ by the supremum times $S (F)$ , and the signed integral obeys the triangle inequality (Surface integration on compact C1 hypersurfaces, Monotonicity and nonnegative homogeneity of the nonnegative integral, The Lebesgue integral is linear on $L^{1} (μ)$ ). If moreover all points of $F$ are within distance $ρ$ of $y_{0}$ , then by uniform continuity of $h$ on a compact neighbourhood of $y_{0}$ the right-hand side tends to $0$ as $ρ \to 0$ (Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous, Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line).

[F3]

(Geometry of the two test bodies.) Fix an orthonormal basis $(n, p, q)$ of $R^{3}$ and $h > 0$ .

The thin box $D_{h} : = y_{0} + {a n + b p + c q : ∣ a ∣ < h^{2}, ∣ b ∣ < h, ∣ c ∣ < h}$ has two faces of area $4 h^{2}$ with outward normals $\pm n$ and four lateral faces of area $4 h^{3}$ with outward normals $\pm p, \pm q$ ; its volume is $8 h^{4}$ . These areas are the products of the corresponding side lengths $2 h^{2}, 2 h, 2 h$ .
Let $n \in S^{2}$ have no zero component, define $s_{i} : = 1$ when $n_{i} > 0$ and $s_{i} : = - 1$ when $n_{i} < 0$ , and put $P_{i} : = y_{0} + (h / n_{i}) e_{i}$ , and let $T_{h}$ be the open tetrahedron with vertices $y_{0}, P_{1}, P_{2}, P_{3}$ . Equivalently, in coordinates $v = y - y_{0}$ it is given by $s_{i} v_{i} > 0$ and $n \cdot v < h$ . Its coordinate face $F_{i}$ has outward normal $- s_{i} e_{i}$ and area $A ∣ n_{i} ∣$ , where $A : = h^{2} / (2 ∣ n_{1} n_{2} n_{3} ∣)$ ; its oblique face $F_{0}$ has outward normal $n$ and area $A$ . Indeed the coordinate-face normal points out of $s_{i} v_{i} > 0$ , and the oblique-face normal points out of $n \cdot v < h$ . The volume is $A h / 3$ . The reflection $v_{i} \mapsto s_{i} v_{i}$ is an isometry taking this tetrahedron to the positive-coordinate tetrahedron with intercepts $h / ∣ n_{i} ∣$ , so its face areas are unchanged. Alternatively the cross-product formula for triangle area gives $\frac{1}{2} ∥ (P_{2} - P_{1}) \times (P_{3} - P_{1}) ∥_{2} = h^{2} / (2 ∣ n_{1} n_{2} n_{3} ∣)$ , since $\sum_{i} n_{i}^{2} = 1$ (The cross product is bilinear, alternating, and orthogonal to both factors, The squared cross-product norm is the Gram determinant of two vectors). The face opposite $P_{i}$ is a right triangle with legs $h / ∣ n_{j} ∣, h / ∣ n_{k} ∣$ , and the volume is one third of the oblique-face area times its distance $h$ from $y_{0}$ .

[F4]

The columns of a matrix $σ \in M_{3} (R)$ determine it, and $(σ n)_{k} = \sum_{i} σ_{k i} n_{i}$ . A map into $M_{3} (R)$ is continuous if and only if its three columns are continuous.

Proof

technique · direct; shrink a box and a right tetrahedron about an arbitrary point and pass to the limit in the balance (B)

1.1givenF1choose

Fix $y_{0} \in O$ and choose $r > 0$ with $\overline{B} (y_{0}, r) \subseteq O$ (Open ball, closed ball and sphere in a metric space). The set $\overline{B} (y_{0}, r) \times S^{2}$ is compact (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line, A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value), so $t$ is uniformly continuous there, and $∥ t ∥_{2}$ is bounded there by some $C_{t} < + \infty$ (Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous, A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value).

2.1givenF1F2F3step 1.1algebra

(Reversed normals.) Fix $n \in S^{2}$ and extend it to an orthonormal basis $(n, p, q)$ . For $0 < h < \min {1, r / 3}$ use the thin box of F3, whose closure lies in $B (y_{0}, r)$ . Its two large-face averages ${\overline{t}}_{n}^{\pm}$ satisfy ${\overline{t}}_{n}^{\pm} \to t (y_{0}, \pm n)$ by [F2], since each face lies within distance $\sqrt{h^{4} + 2 h^{2}} \leq \sqrt{3} h$ of $y_{0}$ . The four lateral-face contributions have total norm at most $16 C_{t} h^{3}$ , and $∥ \int_{D_{h}} g d y ∥_{2} \leq 8 C_{g} h^{4}$ . Thus (B), divided by $4 h^{2}$ , gives $∥ {\overline{t}}_{n}^{+} + {\overline{t}}_{n}^{-} ∥_{2} \leq 4 C_{t} h + 2 C_{g} h^{2}$ . Letting $h ↓ 0$ yields $\begin{matrix} t (y_{0}, n) + t (y_{0}, - n) = 0. & (R) \end{matrix}$

3.1givenF1F2F3step 1.1step 2.1algebra

(Signed tetrahedron identity.) Fix $n \in S^{2}$ with $n_{i} \neq 0$ for all $i$ and put $s_{i} = 1$ if $n_{i} > 0$ and $s_{i} = - 1$ if $n_{i} < 0$ . For $0 < h < r \min_{i} ∣ n_{i} ∣ / 2$ the signed tetrahedron of F3 has closure in $O$ . Divide its balance (B) by the oblique-face area $A$ . The coordinate-face area ratios are $∣ n_{i} ∣$ , their actual outward normals are $- s_{i} e_{i}$ , and the volume ratio is $h / 3$ . All face points tend to $y_{0}$ as $h ↓ 0$ , so [F2] and the bound $∥ \int_{T_{h}} g ∥_{2} / A \leq C_{g} h / 3$ give $t (y_{0}, n) + \sum_{i = 1}^{3} ∣ n_{i} ∣ t (y_{0}, - s_{i} e_{i}) = 0.$ By (R), $t (y_{0}, - s_{i} e_{i}) = - s_{i} t (y_{0}, e_{i})$ for either sign of $s_{i}$ . Since $∣ n_{i} ∣ s_{i} = n_{i}$ , this is $\begin{matrix} t (y_{0}, n) = \sum_{i = 1}^{3} n_{i} t (y_{0}, e_{i}) . & (T) \end{matrix}$

4.1step 3.1givenalgebra

(Zero components.) Let $n \in S^{2}$ be arbitrary. For $ε > 0$ put $v_{ε, i} : = n_{i}$ when $n_{i} \neq 0$ and $v_{ε, i} : = ε$ when $n_{i} = 0$ , and set $n_{ε} = v_{ε} / ∥ v_{ε} ∥_{2}$ . Every component of $n_{ε}$ is nonzero, its nonzero original components retain their signs, and $n_{ε} \to n$ . Apply (T) from step 3.1 to $n_{ε}$ . Continuity of $t (y_{0}, \cdot)$ and of the finite linear combination $\sum_{i} n_{i} t (y_{0}, e_{i})$ then gives (T) for $n$ , including all normals with zero components.

5.1F4step 4.1constructalgebra

(Matrix representation and uniqueness.) Define $σ (y_{0}) \in M_{3} (R)$ by $σ (y_{0}) e_{i} : = t (y_{0}, e_{i})$ . Then for every $n \in S^{2}$ , (T) reads $t (y_{0}, n) = \sum_{i} n_{i} t (y_{0}, e_{i}) = σ (y_{0}) n$ . Since $y_{0} \in O$ was arbitrary, this defines $σ : O \to M_{3} (R)$ . If $σ_{1}, σ_{2}$ both represent $t$ , then $(σ_{1} - σ_{2}) e_{i} = 0$ for $i = 1, 2, 3$ and the columns determine the matrix, so $σ_{1} = σ_{2}$ : the representation is unique.

6.1step 5.1F4given∎

(Continuity.) Each column map $y \mapsto σ (y) e_{i} = t (y, e_{i})$ is continuous on $O$ because $t$ is; hence each entry $y \mapsto σ_{k i} (y)$ is continuous, and $σ$ is continuous as a map into $M_{3} (R)$ (finite-dimensional, entrywise continuity). This proves the existence, uniqueness and continuity clauses.

Remarks

The mathematical content is conditional: the conclusion is that the assumed balance (B) for all piecewise $C^{1}$ test regions forces the contact interaction to be linear in the normal and to satisfy the reversed-normal rule. It is not a derivation of (B) itself, which is an adopted contact-balance hypothesis in any continuum model.
The hypothesis that (B) holds for boxes and tetrahedra is used only through the two test families; simultaneously the surface integral over the four lateral faces is controlled by the boundedness of $t$ on the compact neighbourhood. No symmetry of $σ$ is asserted here: linearity in $n$ alone does not make $σ$ symmetric, and the separate angular-momentum balance is needed for that.
A body couple density or surface couple field contributes to angular-momentum balance; its presence alone does not invalidate the pure-force balance (B). Whenever (B) retains the stated test-region hypotheses, the thin-box argument still forces the reversed-normal traction rule. This theorem makes no claim about additional couple tractions or their representation.

TheoremStatement: AI-adaptedProof: AI-adaptedOpen item page →

Angular-momentum balance without couples forces the stress matrix to be symmetric

Statement

Assume the Axiom of Countable Choice ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )), as used by the divergence and surface-integral machinery. Let $O \subseteq R^{3}$ be open, let $σ : O \to M_{3} (R)$ be of class $C^{1}$ , and let $b : O \to R^{3}$ be continuous. Suppose that for every bounded $C^{1}$ domain $D$ with $\overline{D} \subseteq O$ (Bounded C1 domains and their outward normals), with outward unit normal $ν$ and the surface integral of Surface integration on compact C1 hypersurfaces:

$\begin{matrix} \int_{\partial D} σ ν d S + \int_{D} b d x = 0 & (LB) \end{matrix}$

and

$\begin{matrix} \int_{\partial D} y \times (σ ν) d S + \int_{D} y \times b d x = 0. & (AB) \end{matrix}$

Then

$σ (y)^{T} = σ (y) for every y \in O .$

Facts & Assumptions

Given: ${A C}_{ω}$ ; the open set $O$ ; the $C^{1}$ field $σ$ ; the continuous field $b$ ; the balances (LB) and (AB).

[F1]

Divergence theorem: for $F \in C^{1} (\overline{D}; R^{3})$ on a bounded $C^{1}$ domain $D$ , $\int_{D} div F d x = \int_{\partial D} F \cdot ν d S$ (Divergence on a bounded C1 Euclidean domain). Applying it to the rows $F_{i} = (σ_{i 1}, σ_{i 2}, σ_{i 3})$ gives $\int_{\partial D} σ ν d S = \int_{D} g d x$ with $g_{i} : = \sum_{j} \partial_{j} σ_{i j}$ .

[F2]

Cross product as a bilinear alternating map: $(a \times b)_{i} = \sum_{j, k} ε_{i j k} a_{j} b_{k}$ , where $ε_{i j k} = 0$ if indices repeat, $+ 1$ for an even permutation of $(1, 2, 3)$ , and $- 1$ for an odd permutation; and the product rule differentiates $y_{j} σ_{k l}$ as $\partial_{l} (y_{j} σ_{k l}) = δ_{j l} σ_{k l} + y_{j} \partial_{l} σ_{k l}$ (The cross product in $R^{3}$ , The cross product is bilinear, alternating, and orthogonal to both factors, Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , A total derivative computes every directional derivative, and its matrix is the Jacobian, The vector space $M_{m \times n} (F) : = F^{m \times n}$ of $m$ by $n$ matrices over a field, with entrywise operations).

[F3]

Localization lemma: a continuous function on an open set whose integral vanishes over every member of a neighbourhood-basis family is identically zero; the family of all small balls qualifies (Localization: a continuous function whose integrals vanish over a neighbourhood-basis family is identically zero, Open ball, closed ball and sphere in a metric space).

[F4]

The integrals in (LB) and (AB) are defined componentwise and are finite for $C^{1}$ data on a bounded $C^{1}$ domain (Integral over a measurable subset, The Lebesgue integral is linear on $L^{1} (μ)$ ).

Proof

technique · direct; differentiate the two integral balances by the divergence theorem and localize

1.1givenF1F3F4algebra

Fix a bounded $C^{1}$ domain $D$ with $\overline{D} \subseteq O$ . By [F1] applied to the three rows of $σ$ , (LB) becomes $\int_{D} (g + b) d x = 0, g_{i} : = \sum_{j} \partial_{j} σ_{i j} .$ Since this holds for every bounded $C^{1}$ domain with closure in $O$ , it holds in particular for every open ball $B (y_{0}, r)$ with $\overline{B} (y_{0}, r) \subseteq O$ ; the function $y \mapsto g (y) + b (y)$ is continuous, so [F3] gives $\begin{matrix} g + b = 0 on O . & (M) \end{matrix}$

2.1givenF1F2step 1.1algebra

Fix a bounded $C^{1}$ domain $D$ with $\overline{D} \subseteq O$ and a component index $i \in {1, 2, 3}$ . By [F2], and applying the divergence theorem [F1] to the components $F_{l}^{(i)} : = (ε_{i j k} y_{j} σ_{k l})_{l}$ of the vector field whose $l$ -th entry is $ε_{i j k} y_{j} σ_{k l}$ (summed over $j, k$ ), $(\int_{\partial D} y \times (σ ν) d S)_{i} = \int_{D} \sum_{j, k, l} ε_{i j k} (δ_{j l} σ_{k l} + y_{j} \partial_{l} σ_{k l}) d x = \int_{D} (\sum_{j, k} ε_{i j k} σ_{k j} + (y \times g)_{i}) d x .$ Hence (AB) reads $\int_{D} (\sum_{j, k} ε_{i j k} σ_{k j} + (y \times (g + b))_{i}) d x = 0.$ By (M), the second term vanishes identically, leaving $\int_{D} (\sum_{j, k} ε_{i j k} σ_{k j}) d x = 0$ for every bounded $C^{1}$ domain with closure in $O$ .

3.1step 2.1F3algebra∎

Apply step 2.1 to open balls with closure in $O$ and localize with [F3]: the continuous function $y \mapsto \sum_{j, k} ε_{i j k} σ_{k j} (y)$ vanishes on $O$ for each $i$ . For $i = 1$ this reads $σ_{32} - σ_{23} = 0$ , for $i = 2$ it reads $σ_{13} - σ_{31} = 0$ , and for $i = 3$ it reads $σ_{21} - σ_{12} = 0$ ; hence $σ^{T} = σ$ on $O$ .

Remarks

The content is conditional: (LB) and (AB) are adopted balance hypotheses for bodies without body couples or surface couples. If a couple density $c$ is present, (AB) must add $\int_{D} c d x$ and the conclusion weakens to the local relation $ε_{i j k} σ_{k j} + c_{i} = 0$ ; no symmetry follows in that case, and this item asserts symmetry only in the couple-free balance.
The orbital terms cancel exactly because both balances use the same $b$ : this is what the subtraction of (M) inside the angular integral expresses. Using a different body-force convention in the two balances would invalidate the cancellation and is not part of this statement.
The center of the angular balance is taken to be the origin of the fixed coordinates; translating the center adds to both integrals a term proportional to the total balance (LB) and therefore does not change the conclusion. No rotational frame or physical interpretation is used.

TheoremStatement: AI-adaptedProof: AI-adaptedOpen item page →

Control-volume balance: mass and momentum rates for a moving non-material region

Statement

$D_{χ} : = {(t, x) : t \in I, x \in χ_{t} (O)}$

for the open domain on which $u$ is defined (Continuum motions, spatial velocity, material acceleration and swept regions). Let $J_{0} \subseteq I$ be an open interval with compact closure in $I$ , and let $U \subseteq D_{χ}$ be open and contain the swept region of $B$ over $\overline{J_{0}}$ .

Let $ρ : U \to (0, \infty)$ and $σ : U \to M_{3} (R)$ be $C^{1}$ and let $b : U \to R^{3}$ be continuous. Define $g_{i} : = \sum_{j} \partial_{j} σ_{i j}$ and suppose, on $U$ :

$\begin{matrix} \partial_{t} ρ + div (ρ u) = 0 and ρ D_{t} u = ρ b + g . & (KM) \end{matrix}$

Let $ψ : I \times O^{'} \to R^{3}$ be a second motion of a material region $B^{'} \subseteq B$ (with the same reference chart domain $O^{'} = O$ used for $B$ ; write $w$ for its spatial velocity), and suppose that for every $t \in J_{0}$ the control region $C_{t} : = ψ_{t} (B^{'})$ is a bounded $C^{1}$ domain (Bounded C1 domains and their outward normals) with outward unit normal $ν$ and boundary surface integral as in Surface integration on compact C1 hypersurfaces.

Assume also that the control sweep $Σ_{C} : = {(t, x) : t \in \overline{J_{0}}, x \in ψ_{t} (\overline{B^{'}})}$ is contained in $U$ . Thus the fields and the local equations (KM) are defined on a neighbourhood of every closed control region. No condition that the control motion coincide with the material motion is imposed.

Then for every $t \in J_{0}$ :

$\begin{matrix} \frac{d}{d t} \int_{C_{t}} ρ d x = - \int_{\partial C_{t}} ρ (u - w) \cdot ν d S, & (CV1) \end{matrix}$

and componentwise

$\begin{matrix} \frac{d}{d t} \int_{C_{t}} ρ u_{i} d x = \int_{C_{t}} ρ b_{i} d x + (\int_{\partial C_{t}} σ ν d S)_{i} - \int_{\partial C_{t}} ρ u_{i} (u - w) \cdot ν d S (i = 1, 2, 3) . & (CV2) \end{matrix}$

Facts & Assumptions

Given: ${A C}_{ω}$ ; the two motions $χ, ψ$ ; the fields $ρ, σ, b$ satisfying (KM); the boundary regularity of the control region.

[F1]

Transport theorem with the family $ψ$ : for every $C^{1}$ function $f$ on a neighbourhood of the swept region of $B^{'}$ over $\overline{J_{0}}$ , $\frac{d}{d t} \int_{C_{t}} f = \int_{C_{t}} (\partial_{t} f + div (f w))$ (The transport theorem for a $C^{2}$ family of diffeomorphisms: $\frac{d}{d t} \int_{ψ_{t} (B)} f = \int_{ψ_{t} (B)} (\partial_{t} f + div (f w))$ ), applied to $f = ρ$ and componentwise to $f = ρ u_{i}$ .

[F2]

Divergence theorem: for a bounded $C^{1}$ domain $C$ and $F \in C^{1} (\overline{C}; R^{3})$ , $\int_{C} div F = \int_{\partial C} F \cdot ν d S$ (Divergence on a bounded C1 Euclidean domain). Applied to $F = ρ (w - u)$ and to the rows $F_{i} = (σ_{i 1}, σ_{i 2}, σ_{i 3})$ , it gives $\int_{C} div (ρ (w - u)) = \int_{\partial C} ρ (w - u) \cdot ν d S$ and $\int_{C} g d x = \int_{\partial C} σ ν d S$ (Traction fields and stress matrices representing them).

[F3]

Product rule identity: for $C^{1}$ data satisfying (K),

$\partial_{t} (ρ u_{i}) + div (ρ u_{i} u) = u_{i} (\partial_{t} ρ + div (ρ u)) + ρ (D_{t} u)_{i} = ρ (D_{t} u)_{i},$

and $div (ρ u_{i} w) = u_{i} div (ρ w) + ρ w \cdot \nabla u_{i}$ componentwise (Euclidean spatial calculus: gradient, divergence and material derivative, Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , A total derivative computes every directional derivative, and its matrix is the Jacobian).

Proof

technique · direct; write each rate as a transport integral and use (KM) plus the divergence theorem

1.1givenF1F2F3algebra

Fix $t \in J_{0}$ . The control-sweep hypothesis puts ${t} \times ψ_{t} (\overline{B^{'}})$ compactly inside $U$ . Continuity of $ψ$ gives a small open time interval around $t$ whose closed sweep is still in $U$ , so the local transport argument [F1] applies with $f = ρ$ and the control motion $ψ$ . Thus $\frac{d}{d t} \int_{C_{t}} ρ d x = \int_{C_{t}} (\partial_{t} ρ + div (ρ w)) d x = \int_{C_{t}} div (ρ (w - u)) d x,$ since $div (ρ w) - div (ρ u) = div (ρ (w - u))$ and $div (ρ u) = - \partial_{t} ρ$ by (K). By [F2], the last integral is $- \int_{\partial C_{t}} ρ (u - w) \cdot ν d S$ , which is (CV1).

2.1givenF1F3step 1.1algebra

Fix $i \in {1, 2, 3}$ . By [F1] with $f = ρ u_{i}$ , $\frac{d}{d t} \int_{C_{t}} ρ u_{i} d x = \int_{C_{t}} (\partial_{t} (ρ u_{i}) + div (ρ u_{i} w)) d x .$ By [F3], $div (ρ u_{i} w) = div (ρ u_{i} u) + div (ρ u_{i} (w - u))$ , and by [F3] with (K) and (KM) the first term equals $ρ (D_{t} u)_{i} = ρ b_{i} + g_{i}$ on $C_{t}$ . Hence $\frac{d}{d t} \int_{C_{t}} ρ u_{i} d x = \int_{C_{t}} (ρ b_{i} + g_{i}) d x + \int_{C_{t}} div (ρ u_{i} (w - u)) d x .$

3.1step 2.1F2algebra∎

Apply [F2] to the last two terms of step 2.1: $\int_{C_{t}} g_{i} d x = (\int_{\partial C_{t}} σ ν d S)_{i}$ and $\int_{C_{t}} div (ρ u_{i} (w - u)) d x = - \int_{\partial C_{t}} ρ u_{i} (u - w) \cdot ν d S$ . Substituting gives (CV2) for each $i$ .

Remarks

The control region need not be material: $ψ$ is an arbitrary motion of the reference region $B^{'}$ , and $w$ is the velocity of its points. The relative velocity $u - w$ in the flux terms records the transport of mass and momentum across the moving boundary. Taking $ψ = χ$ recovers the material statements, in which the relative velocity vanishes and both rates reduce to the material integrals of (K) and (KM).
The identities fix the inertial conventions: the momentum flux uses the spatial momentum density $ρ u$ times the normal component of the relative velocity, and the traction term is the surface integral of $σ ν$ . No stress symmetry, constitutive relation or boundary condition is used or asserted; the only regularities required are those stated.

DefinitionDefinition: AI-adaptedProof: Not applicableOpen item page →

The small-strain isotropic linear elastic constitutive law and its strain energy

Definition

Let $B \subseteq R^{3}$ be a material region (Material regions, densities and their mass) and let $O \subseteq R^{3}$ be open with $\overline{B} \subseteq O$ . A displacement field on $B$ is a map $ξ : O \to R^{3}$ of class $C^{2}$ ( $C^{k}$ Euclidean maps and diffeomorphisms); its total derivative $D ξ (X) : R^{3} \to R^{3}$ at $X \in O$ is linear (The total (Fréchet) derivative $D f (a)$ as the linear first-order approximation with $o (∥ h ∥_{2})$ remainder) and has matrix $(\partial_{j} ξ_{i} (X))$ (The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case).

Write $Sym (3) : = {A \in M_{3} (R) : A^{T} = A}$ for the real symmetric $3 \times 3$ matrices (The vector space $M_{m \times n} (F) : = F^{m \times n}$ of $m$ by $n$ matrices over a field, with entrywise operations, The transpose $A^{T}$ of a matrix) and $I_{3}$ for the identity (Rectangular matrix multiplication and the identity matrix $I_{n}$ , including zero-sized shapes). The linear strain of the displacement field is the symmetric-matrix field

$e (X) : = \frac{D ξ (X) + D ξ (X)^{T}}{2} \in Sym (3), e_{i j} (X) = \frac{\partial_{j} ξ_{i} (X) + \partial_{i} ξ_{j} (X)}{2},$

the symmetric part of the Jacobian matrix, and its trace is $tr e = \sum_{i} \partial_{i} ξ_{i}$ (The trace $tr (A)$ as the sum of the diagonal entries). The Frobenius inner product of $A, C \in M_{3} (R)$ is $⟨ A, C ⟩_{F} : = tr (A^{T} C) = \sum_{i, j} A_{i j} C_{i j}$ , with squared norm $∣ A ∣_{F}^{2} = ⟨ A, A ⟩_{F}$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ); the deviatoric part of $e$ is

$e_{0} : = e - \frac{tr e}{3} I_{3} .$

An isotropic linear elastic constitutive law is a pair $(λ, μ) \in R \times R$ of Lamé parameters with

$μ > 0, λ + \frac{2 μ}{3} > 0.$

Its stress matrix field is

$σ (X) : = λ (tr e (X)) I_{3} + 2 μ e (X) \in Sym (3),$

and its strain-energy density is the real function of $e$

$W (e) : = \frac{λ}{2} (tr e)^{2} + μ tr (e^{2}) .$

Remarks

These data are an adopted constitutive model, not a consequence of the balance laws of this collection. The pair $(λ, μ)$ and the linear relation between $e$ and $σ$ are model parameters and model structure; no physical material is asserted to obey them, and no measurement of $λ$ or $μ$ is asserted by this definition.
The matrix $σ$ is symmetric for every displacement field because $e$ is symmetric; symmetry of the stress is thus automatic in this model branch and does not require the separate angular-momentum balance. This is a property of the model, not a general theorem about stress.
The hypotheses $μ > 0$ and $λ + 2 μ / 3 > 0$ are exactly the positivity hypotheses proved sufficient for $W \geq 0$ with equality only at $e = 0$ in the companion item; the constitutive definition itself assumes only these inequalities and makes no further definiteness claim.
Units are not part of the definition. A physical reading assigns length units to $ξ$ and hence dimensionless strain and a stress unit to $σ$ ; the mathematical statements below use the matrices alone, and any unit convention belongs to a consuming model.

LemmaStatement: AI-adaptedProof: AI-adaptedOpen item page →

The isotropic elastic strain energy is $μ ∣ e_{0} ∣_{F}^{2} + (λ / 2 + μ / 3) (tr e)^{2}$ , hence positive definite under $μ > 0$ , $λ + 2 μ / 3 > 0$

Statement

Let $e \in Sym (3)$ with deviatoric part $e_{0} = e - \frac{tr e}{3} I_{3}$ and Lamé parameters $λ, μ \in R$ with $μ > 0$ and $λ + \frac{2 μ}{3} > 0$ (The small-strain isotropic linear elastic constitutive law and its strain energy). Then

$W (e) = μ ∣ e_{0} ∣_{F}^{2} + (\frac{λ}{2} + \frac{μ}{3}) (tr e)^{2} \geq 0,$

and $W (e) = 0$ if and only if $e = 0$ . Here $∣ A ∣_{F}^{2} = tr (A^{T} A)$ is the squared Frobenius norm (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ , The trace $tr (A)$ as the sum of the diagonal entries).

Facts & Assumptions

Given: A symmetric $3 \times 3$ matrix $e$ and Lamé parameters $λ, μ$ with the stated inequalities.

[F1]

Definitions: $W (e) = \frac{λ}{2} (tr e)^{2} + μ tr (e^{2})$ , $e_{0} = e - \frac{tr e}{3} I_{3}$ , and $∣ A ∣_{F}^{2} = tr (A^{T} A)$ is the sum over the entries squared (The small-strain isotropic linear elastic constitutive law and its strain energy, The trace $tr (A)$ as the sum of the diagonal entries, The vector space $M_{m \times n} (F) : = F^{m \times n}$ of $m$ by $n$ matrices over a field, with entrywise operations).

[F2]

Trace identities: $tr$ is linear, $tr (X Y) = tr (Y X)$ for square matrices of equal size, and $tr (I_{3}) = 3$ ; transposition satisfies $(A^{T})^{T} = A$ and $(A + B)^{T} = A^{T} + B^{T}$ (Trace is a linear functional on $M_{n} (F)$ , For $A \in M_{m \times n} (F)$ and $B \in M_{n \times m} (F)$ , $tr (A B) = tr (B A)$ , Rectangular matrix multiplication and the identity matrix $I_{n}$ , including zero-sized shapes, The transpose $A^{T}$ of a matrix).

Proof

technique · direct; expand the Frobenius norm of the deviatoric part and compare with $W$

1.1F1F2algebra

Since $e$ is symmetric, $e^{T} = e$ and $tr (e^{2}) = tr (e^{T} e) = ∣ e ∣_{F}^{2}$ . Also $tr (e_{0}) = 0$ because $tr$ is linear and $tr I_{3} = 3$ .

2.1F1F2step 1.1algebra

Expand $∣ e_{0} ∣_{F}^{2} = tr ((e - \frac{tr e}{3} I_{3})^{2}) = tr (e^{2}) - 2 \frac{tr e}{3} tr (e) + \frac{(tr e)^{2}}{9} tr (I_{3}) = ∣ e ∣_{F}^{2} - \frac{2}{3} (tr e)^{2} + \frac{1}{3} (tr e)^{2} = ∣ e ∣_{F}^{2} - \frac{1}{3} (tr e)^{2}$ , using linearity of the trace and step 1.1.

3.1F1step 1.1step 2.1algebra

Substituting $tr (e^{2}) = ∣ e ∣_{F}^{2} = ∣ e_{0} ∣_{F}^{2} + \frac{1}{3} (tr e)^{2}$ from the preceding steps into [F1] gives $W (e) = \frac{λ}{2} (tr e)^{2} + μ (∣ e_{0} ∣_{F}^{2} + \frac{1}{3} (tr e)^{2}) = μ ∣ e_{0} ∣_{F}^{2} + (\frac{λ}{2} + \frac{μ}{3}) (tr e)^{2} .$

4.1step 3.1givenalgebra∎

Both coefficients are positive: $μ > 0$ , and $\frac{λ}{2} + \frac{μ}{3} = \frac{3 λ + 2 μ}{6} = \frac{3}{6} (λ + \frac{2 μ}{3}) > 0$ by hypothesis. Since both summands are nonnegative, $W (e) \geq 0$ . If $W (e) = 0$ , then $μ ∣ e_{0} ∣_{F}^{2} = 0$ and $(tr e)^{2} = 0$ ; as $∣ e_{0} ∣_{F}^{2}$ is a sum of squares, $e_{0} = 0$ , and $tr e = 0$ , so $e = e_{0} + \frac{tr e}{3} I_{3} = 0$ . Conversely $W (0) = 0$ .

Remarks

The decomposition is the standard deviatoric/volumetric split: $W$ is the sum of a shear term controlled by $μ$ and a volumetric term controlled by the bulk modulus $K = λ + \frac{2 μ}{3}$ . The hypotheses are exactly the inequalities needed for both coefficients to be positive; they are strictly weaker than requiring $λ > 0$ .
No stress-strain inversion, no bounds on the stress matrix, and no constitutive interpretation beyond the stated model are asserted. The lemma is a statement about symmetric matrices.

TheoremStatement: AI-adaptedProof: AI-adaptedOpen item page →

Exact plane-wave solutions of the isotropic linear elastic model, with longitudinal and transverse speeds

Statement

Let $ρ_{0} > 0$ and Lam'e parameters $λ, μ \in R$ with $μ > 0$ and $λ + \frac{2 μ}{3} > 0$ , let $a, k \in R^{3}$ and $ω \in R ∖ {0}$ , and define the plane wave

$ξ : R \times R^{3} \to R^{3}, ξ (t, x) : = a \cos (⟨ k, x ⟩_{2} - ω t),$

with the Euclidean inner product $⟨ \cdot, \cdot ⟩_{2}$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ). For each fixed $t$ , the spatial displacement map is $ξ_{t} (x) : = ξ (t, x)$ . Define the strain and stress fields using the spatial derivative at fixed time:

$e (t, x) : = \frac{1}{2} (D_{x} ξ (t, x) + (D_{x} ξ (t, x))^{T}), σ (t, x) : = λ (tr e (t, x)) I_{3} + 2 μ e (t, x) .$

These are the pointwise constitutive formulas of the isotropic linear elastic model (The small-strain isotropic linear elastic constitutive law and its strain energy); in particular $D_{x} ξ$ is the total derivative of $x \mapsto ξ_{t} (x)$ , not the derivative of the full spacetime map.

Then the componentwise identity

$ρ_{0} \partial_{t}^{2} ξ_{i} = \sum_{j = 1}^{3} \partial_{j} σ_{i j} (i = 1, 2, 3)$

holds on all of $R \times R^{3}$ if and only if

$\begin{matrix} ρ_{0} ω^{2} a = μ ∣ k ∣_{2}^{2} a + (λ + μ) ⟨ k, a ⟩_{2} k . & (D) \end{matrix}$

When (D) holds and $a, k$ are nonzero, define the phase speed by $c_{p h} : = ∣ ω ∣ / ∣ k ∣_{2}$ . If $a = α k$ for some $α \neq 0$ (longitudinal polarization), then $ρ_{0} ω^{2} = (λ + 2 μ) ∣ k ∣_{2}^{2}$ and $c_{p h}^{2} = (λ + 2 μ) / ρ_{0}$ . If $⟨ k, a ⟩_{2} = 0$ (transverse polarization), then $ρ_{0} ω^{2} = μ ∣ k ∣_{2}^{2}$ and $c_{p h}^{2} = μ / ρ_{0}$ . The constitutive positivity hypotheses imply $λ + 2 μ = (λ + 2 μ / 3) + 4 μ / 3 > 0$ and $μ > 0$ , so both branch phase speeds are real and positive. (When $k = 0$ , (D) has no solution with $a \neq 0$ and $ω \neq 0$ .)

Facts & Assumptions

Given: The parameters $ρ_{0}, λ, μ$ , the vectors $a, k$ , the frequency $ω \neq 0$ , and the field $ξ$ as constructed.

[F1]

Sine and cosine have the usual derivatives, $\sin^{'} = \cos$ , $\cos^{'} = - \sin$ , and $\sin (0) = 0$ , $\cos (0) = 1$ ; compositions with affine maps are differentiable by the chain rule (The derivatives of sine and cosine are cosine and minus sine, Sine and cosine defined by their real power series, The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ , Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , A total derivative computes every directional derivative, and its matrix is the Jacobian, $C^{k}$ Euclidean maps and diffeomorphisms).

[F2]

The strain and stress are computed from the partial derivatives of $ξ$ as in the constitutive definition; $tr$ and the identity matrix are those of The trace $tr (A)$ as the sum of the diagonal entries and Rectangular matrix multiplication and the identity matrix $I_{n}$ , including zero-sized shapes (The small-strain isotropic linear elastic constitutive law and its strain energy, The vector space $M_{m \times n} (F) : = F^{m \times n}$ of $m$ by $n$ matrices over a field, with entrywise operations).

Proof

technique · direct computation; compare the coefficients of the single scalar factor $\cos(\langle k,x\rangle_2-\omega t)$

1.1givenF1algebra

Put $θ (t, x) : = ⟨ k, x ⟩_{2} - ω t$ . Then $\partial_{t} θ = - ω$ and $\partial_{j} θ = k_{j}$ , so by [F1] and the chain rule $\partial_{t} ξ_{i} = a_{i} ω \sin θ, \partial_{j} ξ_{i} = - a_{i} k_{j} \sin θ, \partial_{t}^{2} ξ_{i} = - a_{i} ω^{2} \cos θ .$

2.1step 1.1F2algebra

From step 1.1, $e_{i j} = \frac{1}{2} (\partial_{j} ξ_{i} + \partial_{i} ξ_{j}) = - \frac{1}{2} (a_{i} k_{j} + a_{j} k_{i}) \sin θ$ and $tr e = - ⟨ k, a ⟩_{2} \sin θ$ . Hence $σ_{i j} = - \sin θ (λ ⟨ k, a ⟩_{2} δ_{i j} + μ (a_{i} k_{j} + a_{j} k_{i})) .$

3.1step 2.1F1algebra

Differentiate step 2.1 and use [F1]: $\sum_{j} \partial_{j} σ_{i j} = - \cos θ \sum_{j} k_{j} (λ ⟨ k, a ⟩_{2} δ_{i j} + μ (a_{i} k_{j} + a_{j} k_{i})) = - \cos θ (μ ∣ k ∣_{2}^{2} a_{i} + (λ + μ) ⟨ k, a ⟩_{2} k_{i}),$ where $\sum_{j} k_{j} δ_{i j} = k_{i}$ , $\sum_{j} k_{j}^{2} = ∣ k ∣_{2}^{2}$ and $\sum_{j} k_{j} a_{j} = ⟨ k, a ⟩_{2}$ .

4.1step 1.1step 3.1algebra

The identity $ρ_{0} \partial_{t}^{2} ξ_{i} = \sum_{j} \partial_{j} σ_{i j}$ for all $(t, x)$ reads, by steps 1.1 and 3.1, $\cos θ (ρ_{0} ω^{2} a_{i} - μ ∣ k ∣_{2}^{2} a_{i} - (λ + μ) ⟨ k, a ⟩_{2} k_{i}) = 0$ for all $(t, x)$ . Choosing $t = 0$ , $x = 0$ gives $\cos θ = 1$ , so the bracket vanishes; conversely vanishing of the bracket makes the identity hold everywhere. This is exactly (D).

5.1step 4.1givenalgebra∎

Assume (D) and $a, k \neq 0$ . If $a = α k$ with $α \neq 0$ , then (D) reads $ρ_{0} ω^{2} α k = μ ∣ k ∣_{2}^{2} α k + (λ + μ) α ∣ k ∣_{2}^{2} k = α ∣ k ∣_{2}^{2} (λ + 2 μ) k$ , so $ρ_{0} ω^{2} = (λ + 2 μ) ∣ k ∣_{2}^{2}$ . If $⟨ k, a ⟩_{2} = 0$ , then (D) reads $ρ_{0} ω^{2} a = μ ∣ k ∣_{2}^{2} a$ , so $ρ_{0} ω^{2} = μ ∣ k ∣_{2}^{2}$ . Dividing by $ρ_{0} ∣ k ∣_{2}^{2}$ gives the phase-speed formulas. Positivity follows from $λ + 2 μ = (λ + 2 μ / 3) + 4 μ / 3 > 0$ and $μ > 0$ .

Remarks

The theorem verifies exact solutions of the adopted constant-coefficient PDE; it does not assert that every solution of the elastic model is a plane wave, that a wave packet propagates unchanged, or that any material realises the model. General initial-value problems for the elastic system are not solved here.
The identity (D) is the dispersion relation of the plane-wave ansatz. The two eigenvalue branches computed in the last step are the longitudinal and transverse speeds; they require $λ + 2 μ > 0$ and $μ > 0$ respectively, and no claim is made about $λ$ alone.

DefinitionDefinition: AI-adaptedProof: Not applicableOpen item page →

Incompressible Newtonian and barotropic inviscid fluid constitutive models

Definition

Let $I \subseteq R$ be a nonempty open interval, let $O \subseteq R^{3}$ be open, and let $u : I \times O \to R^{3}$ be a $C^{1}$ spatial velocity field with the conventions of Euclidean spatial calculus: gradient, divergence and material derivative. Write $D u (t, x)$ for the total derivative in the spatial variable and $div u = tr (D u)$ (The trace $tr (A)$ as the sum of the diagonal entries). For $A \in M_{3} (R)$ define its transpose by $(A^{T})_{i j} : = A_{j i}$ for $1 \leq i, j \leq 3$ , using the same one-based component labels; call $A$ symmetric when $A_{i j} = A_{j i}$ for every such pair.

(a) Incompressible Newtonian fluid. Model data are a $C^{1}$ scalar field $p : I \times O \to R$ (pressure) and a real number $η > 0$ (dynamic viscosity; a dimensionless mathematical parameter until a unit system is fixed). The model consists of the incompressibility constraint

$div u (t, x) = 0 ((t, x) \in I \times O)$

and the Newtonian stress field

$σ (t, x) : = - p (t, x) I_{3} + η (D u (t, x) + D u (t, x)^{T}) \in M_{3} (R),$

with $(D u (t, x)^{T})_{i j} = \partial_{i} u_{j} (t, x)$ and $I_{3}$ the identity (Rectangular matrix multiplication and the identity matrix $I_{n}$ , including zero-sized shapes, The vector space $M_{m \times n} (F) : = F^{m \times n}$ of $m$ by $n$ matrices over a field, with entrywise operations). The entries of the second term are $η (\partial_{j} u_{i} + \partial_{i} u_{j})$ , so $σ$ is symmetric for every $(t, x)$ .

(b) Barotropic inviscid fluid. Model data are a $C^{1}$ function $p : (0, \infty) \to R$ (pressure law) and a $C^{1}$ scalar map $V : I \times O \to R$ . Calling $V$ a body-force potential means that its associated vector field is defined by $b (t, x) : = - \nabla_{x} V (t, x)$ ; this field is continuous, and in a consuming momentum equation the body-force density is $ρ b = - ρ \nabla_{x} V$ . This datum is separate from the stress law. The enthalpy of the pressure law is the $C^{1}$ function

$h (ρ) : = \int_{1}^{ρ} \frac{p^{'} (s)}{s} d s (ρ > 0),$

which is characterised by $h (1) = 0$ and $h^{'} (ρ) = p^{'} (ρ) / ρ$ for every $ρ > 0$ (The integral function $F (x) : = \int_{a}^{x} f$ of an integrable $f$ , The first fundamental theorem: if $f$ is integrable on $[a, b]$ and continuous at $c$ , then $F^{'} (c) = f (c)$ ; in particular a continuous $f$ has $F$ as a primitive). For a $C^{1}$ positive density field $ρ : I \times O \to (0, \infty)$ the branch's stress field is

$σ (t, x) : = - p (ρ (t, x)) I_{3},$

and the branch is inviscid (no velocity-gradient term) and barotropic (the stress depends on position and time only through the density).

Remarks

Both branches are adopted constitutive models. The presence or absence of a viscosity term, the value of $η$ (The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set conventions do not enter), the pressure law and the incompressibility constraint are model structure, not consequences of the momentum balance or of the transport theorem. In particular, incompressibility is an additional scalar equation imposed by the model, and no material is asserted to obey it.
The sign convention is that of Traction fields and stress matrices representing them: the traction across a surface with unit normal $n$ is $σ n$ , so the isotropic term $- p I_{3}$ produces the compressive traction $- p n$ ; positive $p$ is a pressure in this convention. A viscous stress proportional to $D u + D u^{T}$ is the classical Newtonian form, asserted here for the stated $C^{1}$ field class only.
In (b) the pressure law is defined on the positive reals only; any statement using this branch must supply a positive density field, and no continuation of $p$ to $ρ \leq 0$ or through a vacuum is part of this definition. The enthalpy is an integral of a continuous integrand on the compact interval $[1, ρ]$ (The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral); no path or choice issue arises.
Units are interpretive: after a unit system is fixed the values of $p$ and the parameter $η$ acquire units, and $h$ acquires energy-per-mass units. The mathematical items of this collection use the fields themselves and attach no unit names; a physical reading belongs to a consuming model.

TheoremStatement: AI-adaptedProof: AI-adaptedOpen item page →

Barotropic inviscid momentum in gradient form, Bernoulli along streamlines and the irrotational constant

Statement

Let $I \subseteq R$ be a nonempty open interval and $O \subseteq R^{3}$ open. Let $u : I \times O \to R^{3}$ and $ρ : I \times O \to (0, \infty)$ be $C^{1}$ , let $p : (0, \infty) \to R$ be $C^{1}$ with enthalpy $h (ρ_{0}) = \int_{1}^{ρ_{0}} p^{'} (s) / s d s$ , and let $V : I \times O \to R$ be $C^{1}$ (Incompressible Newtonian and barotropic inviscid fluid constitutive models). Write $σ : = - p (ρ) I_{3}$ and suppose that, on $I \times O$ ,

$\begin{matrix} \partial_{t} ρ + div (ρ u) = 0 and ρ D_{t} u = ρ (- \nabla V) + g, g_{i} : = \sum_{j} \partial_{j} σ_{i j} . & (M) \end{matrix}$

(The local equation (M) is the conclusion of Local momentum balance: $ρ D_{t} u = ρ b + div σ$ from the integral balance on admissible material subregions under its integral balance and continuity hypotheses; here it is the adopted local form of the barotropic inviscid model.)

(i) On $I \times O$ ,

$D_{t} u = - \nabla (h (ρ) + V) .$

(ii) Suppose in addition that the flow is steady: $\partial_{t} u = 0$ , $\partial_{t} ρ = 0$ and $\partial_{t} V = 0$ on $I \times O$ . A streamline on an open interval $J \subseteq I$ is a $C^{1}$ curve $γ : J \to O$ satisfying $γ^{'} (t) = u (t, γ (t))$ . For every streamline on a nonempty $J$ ,

$\frac{d}{d t} (h (ρ) + V + \frac{1}{2} ∣ u ∣_{2}^{2}) (t, γ (t)) = 0,$

equivalently $u \cdot \nabla (h (ρ) + V + \frac{1}{2} ∣ u ∣_{2}^{2}) = 0$ on $I \times O$ : the quantity is constant along every streamline.

(iii) If the steady hypotheses of (ii) hold and in addition the spatial velocity gradient $D u (t, x)$ is symmetric at every $(t, x)$ (the irrotational condition used in this collection), then

$\nabla (h (ρ) + V + \frac{1}{2} ∣ u ∣_{2}^{2}) = 0 on I \times O,$

so the full derivative of this quantity vanishes and it is constant on every Euclidean ball contained in $I \times O$ , hence on each connected component of $I \times O$ .

Facts & Assumptions

Given: The fields $u, ρ, p, V$ ; equation (M); in (ii) steadiness; in (iii) symmetry of $D u$ .

[F1]

The divergence of $σ = - p (ρ) I_{3}$ is $g_{i} = - \partial_{i} p (ρ)$ : since $σ_{i j} = - p (ρ) δ_{i j}$ , $\sum_{j} \partial_{j} σ_{i j} = - \partial_{i} p (ρ)$ (Incompressible Newtonian and barotropic inviscid fluid constitutive models, Rectangular matrix multiplication and the identity matrix $I_{n}$ , including zero-sized shapes).

[F2]

Chain rule for the composition $h \circ ρ$ and product rule identities: $\nabla (h (ρ)) = h^{'} (ρ) \nabla ρ = (p^{'} (ρ) / ρ) \nabla ρ$ , and $u \cdot (D u u) = \frac{1}{2} u \cdot \nabla (∣ u ∣_{2}^{2})$ (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ , Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , A total derivative computes every directional derivative, and its matrix is the Jacobian, The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

[F3]

If $D u$ is symmetric, then $D u u = \nabla (\frac{1}{2} ∣ u ∣_{2}^{2})$ : $(D u u)_{i} = \sum_{j} \partial_{j} u_{i} u_{j} = \sum_{j} \partial_{i} u_{j} u_{j} = \partial_{i} \frac{1}{2} \sum_{j} u_{j}^{2}$ (Euclidean spatial calculus: gradient, divergence and material derivative, The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ , A total derivative computes every directional derivative, and its matrix is the Jacobian).

[F4]

A $C^{1}$ curve with $γ^{'} (t) = u (t, γ (t))$ differentiates a $C^{1}$ scalar field $φ$ by $\frac{d}{d t} φ (t, γ (t)) = \partial_{t} φ (t, γ (t)) + \nabla φ (t, γ (t)) \cdot γ^{'} (t)$ (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ , A total derivative computes every directional derivative, and its matrix is the Jacobian, The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set).

[F5]

A function locally constant on an open set is constant on each connected component: on a connected component, every nonempty level set and its complement are relatively open. A connected set is one that is not a union of two nonempty disjoint relatively open sets, and a connected component is a maximal connected subset.

Proof

technique · direct; divide the momentum equation by the positive density, recognize the enthalpy gradient, and project along streamlines

1.1givenF1F2algebra

By [F1], (M) reads componentwise $ρ (D_{t} u)_{i} = - ρ \partial_{i} V - \partial_{i} p (ρ)$ . Dividing by $ρ > 0$ and using [F2], $(D_{t} u)_{i} = - \partial_{i} V - \frac{p^{'} (ρ)}{ρ} \partial_{i} ρ = - \partial_{i} V - \partial_{i} (h (ρ)),$ because $h^{'} (ρ) = p^{'} (ρ) / ρ$ . This is (i).

2.1step 1.1F2givenalgebra

By (i) and [F2], $u \cdot \nabla (h (ρ) + V) = - u \cdot D_{t} u = - u \cdot \partial_{t} u - \frac{1}{2} u \cdot \nabla (∣ u ∣_{2}^{2})$ . Under steadiness $\partial_{t} u = 0$ , so $u \cdot \nabla (h (ρ) + V + \frac{1}{2} ∣ u ∣_{2}^{2}) = 0.$

3.1step 2.1F2F4givenalgebra

For the streamline $γ$ , [F4] gives $\frac{d}{d t} Q (t, γ (t)) = \partial_{t} Q + u \cdot \nabla Q$ with $Q : = h (ρ) + V + \frac{1}{2} ∣ u ∣_{2}^{2}$ ; by [F2] and steadiness, $\partial_{t} Q = h^{'} (ρ) \partial_{t} ρ + \partial_{t} V + u \cdot \partial_{t} u = 0$ . Step 2.1 makes the spatial term vanish, so the derivative is $0$ . This is (ii).

4.1step 1.1F3F5givenalgebra∎

Under the symmetry hypothesis, [F3] gives $D_{t} u = \partial_{t} u + \nabla (\frac{1}{2} ∣ u ∣_{2}^{2})$ ; with the steady condition $\partial_{t} u = 0$ and (i), $\nabla Q = \nabla (h (ρ) + V) + \nabla (\frac{1}{2} ∣ u ∣_{2}^{2}) = - D_{t} u + D_{t} u = 0,$ where $Q = h (ρ) + V + \frac{1}{2} ∣ u ∣_{2}^{2}$ . Steadiness also gives $\partial_{t} Q = 0$ , so the full derivative $D Q$ vanishes. A function with zero total derivative on a convex open set is constant (A totally differentiable map with zero derivative on a convex open set is constant); every Euclidean ball in $R \times R^{3}$ is convex (Open ball, closed ball and sphere in a metric space, The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ), so $Q$ is constant on every ball in $I \times O$ . It is therefore locally constant, and [F5] gives constancy on each connected component. This is (iii).

Remarks

The steady hypotheses of (ii) and (iii) are part of the statement. A time-dependent or sheared barotropic flow need not satisfy Bernoulli along streamlines. The term “irrotational” in (iii) is defined here by the explicit condition $D u = (D u)^{T}$ ; no curl convention is needed. No global constant is asserted in (ii): different streamlines may carry different constants.
Only the stated smoothness and the local equation (M) are used. The physical reading — an inviscid barotropic fluid with conservative body force — is the model branch of Incompressible Newtonian and barotropic inviscid fluid constitutive models; the theorem itself is a conditional consequence of (M).

TheoremStatement: AI-adaptedProof: AI-adaptedOpen item page →

Kelvin circulation: the circulation around a $C^{1}$ material loop is constant in the barotropic inviscid model

Statement

Let $O \subseteq R^{3}$ be open, let $I \subseteq R$ be a nonempty open interval, let $u : I \times O \to R^{3}$ and $ρ : I \times O \to (0, \infty)$ be $C^{1}$ , let $p : (0, \infty) \to R$ be $C^{1}$ on the positive density interval $(0, \infty)$ , with enthalpy $h (r) : = \int_{1}^{r} p^{'} (s) / s d s$ for $r > 0$ as in Incompressible Newtonian and barotropic inviscid fluid constitutive models, and let $V : I \times O \to R$ be $C^{1}$ , with the barotropic inviscid stress $σ = - p (ρ) I_{3}$ ; assume the local equations (M) of Barotropic inviscid momentum in gradient form, Bernoulli along streamlines and the irrotational constant on $I \times O$ :

$\partial_{t} ρ + div (ρ u) = 0, ρ D_{t} u = ρ (- \nabla V) + g, g_{i} = \sum_{j} \partial_{j} σ_{i j},$

so that, by part (i) of that theorem, $D_{t} u = - \nabla (h (ρ) + V)$ on $I \times O$ .

Let $t_{0} < t_{1}$ with $[t_{0}, t_{1}] \subseteq I$ and let $γ : [t_{0}, t_{1}] \times [0, 1] \to O$ be $C^{1}$ with

$γ (t, 0) = γ (t, 1) and \partial_{t} γ (t, s) = u (t, γ (t, s)) ((t, s) \in [t_{0}, t_{1}] \times [0, 1]) .$

Here $C^{1}$ on the closed parameter rectangle means that $γ$ is continuous and its first partial derivatives exist and are continuous there, with one-sided derivatives on the edges.

Then the circulation

$Γ (t) : = \int_{0}^{1} \sum_{i = 1}^{3} u_{i} (t, γ (t, s)) \partial_{s} γ_{i} (t, s) d s$

is constant on $[t_{0}, t_{1}]$ . The integral is an ordinary Riemann integral of a continuous real function; a singleton time interval has a trivially constant circulation and requires no time differentiation.

Facts & Assumptions

Given: The fields $u, ρ, p, V$ and the local equation; the material loop $γ$ .

[F1]

By Barotropic inviscid momentum in gradient form, Bernoulli along streamlines and the irrotational constant(i), $D_{t} u = - \nabla (h (ρ) + V)$ on $I \times O$ ; the enthalpy satisfies $h \in C^{1}$ .

[F2]

Chain rule for the composite $s \mapsto φ (t, γ (t, s))$ and for $t \mapsto u (t, γ (t, s))$ : $\partial_{s} (u (t, γ)) = (D u) (t, γ) \partial_{s} γ$ , and $\partial_{t} (u (t, γ)) = \partial_{t} u (t, γ) + (D u) (t, γ) \partial_{t} γ$ (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ , Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , A total derivative computes every directional derivative, and its matrix is the Jacobian, Euclidean spatial calculus: gradient, divergence and material derivative).

[F3]

The identity $u \cdot (D u v) = \frac{1}{2} \nabla (∣ u ∣_{2}^{2}) \cdot v$ for $C^{1}$ fields: componentwise, $\sum_{i} u_{i} \partial_{j} u_{i} v_{j} = \frac{1}{2} \partial_{j} (\sum_{i} u_{i}^{2}) v_{j}$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ , A total derivative computes every directional derivative, and its matrix is the Jacobian, Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ ).

[F4]

If $g$ and $\partial_{t} g$ are continuous on a compact rectangle, FTC gives $h^{- 1} (g (t + h, s) - g (t, s)) = h^{- 1} \int_{t}^{t + h} \partial_{t} g (τ, s) d τ$ . Uniform continuity makes these quotients converge uniformly in $s$ to $\partial_{t} g (t, s)$ ; the uniform integral-error bound therefore gives $\frac{d}{d t} \int g (t, s) d s = \int \partial_{t} g (t, s) d s$ . The same argument with the coordinate roles exchanged permits differentiation in $s$ of $\int_{t_{0}}^{t} U (τ, s) d τ$ when $U$ and $\partial_{s} U$ are continuous; endpoints use one-sided derivatives (The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ , A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion, Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous, Uniformly close integrable functions have integrals differing by at most the interval length times their uniform error).

[F5]

Fundamental theorem of calculus: $\int_{0}^{1} \partial_{s} F (s) d s = F (1) - F (0)$ for $C^{1}$ $F$ ; and $\int_{0}^{1} \partial_{s} F = 0$ when $F (0) = F (1)$ (The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ , The first fundamental theorem: if $f$ is integrable on $[a, b]$ and continuous at $c$ , then $F^{'} (c) = f (c)$ ; in particular a continuous $f$ has $F$ as a primitive, The integral with oriented limits: $\int_{a}^{a} f : = 0$ and $\int_{b}^{a} f : = - \int_{a}^{b} f$ , For $a < c < b$ : $f$ is integrable on $[a, b]$ if and only if it is integrable on $[a, c]$ and on $[c, b]$ , and then $\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{b} f$ ; with the oriented form for arbitrary $a, b, c$ ).

Proof

technique · direct; differentiate the parameterized circulation and recognize the integrand as a total $s$-derivative

1.1givenF2F4algebra

Put $U (t, s) : = u (t, γ (t, s))$ . The chain rule makes $U$ a $C^{1}$ map, with $\partial_{t} U = \partial_{t} u + (D u) \partial_{t} γ$ and $\partial_{s} U = (D u) \partial_{s} γ$ . Although $γ$ is only assumed $C^{1}$ , its material equation supplies the needed mixed derivative: from $γ (t, s) = γ (t_{0}, s) + \int_{t_{0}}^{t} U (τ, s) d τ$ , differentiating in $s$ gives $\partial_{s} γ (t, s) = \partial_{s} γ (t_{0}, s) + \int_{t_{0}}^{t} \partial_{s} U (τ, s) d τ$ by [F4]. Therefore $\partial_{t} \partial_{s} γ = \partial_{s} U = (D u) \partial_{s} γ$ , continuously. Thus the circulation integrand $U \cdot \partial_{s} γ$ has a continuous time derivative, and [F4] gives $Γ^{'} (t) = \int_{0}^{1} [(\partial_{t} u + (D u) \partial_{t} γ) \cdot \partial_{s} γ + u \cdot (D u) \partial_{s} γ] d s$ for $t$ in the time interior.

2.1step 1.1F2F3algebra

Since $\partial_{t} γ = u (t, γ)$ , the first term in step 1.1 is $(\partial_{t} u + (D u) u) \cdot \partial_{s} γ = (D_{t} u) \cdot \partial_{s} γ$ . The second term is exactly $u \cdot (D u) \partial_{s} γ = \partial_{s} (\frac{1}{2} ∣ u (t, γ) ∣_{2}^{2})$ by the chain rule and [F3]. Hence $Γ^{'} (t) = \int_{0}^{1} [(D_{t} u) \cdot \partial_{s} γ + \partial_{s} (\frac{1}{2} ∣ u ∣_{2}^{2})] d s .$ No equality between $(D u) u$ and $(D u)^{T} u$ , and no symmetry of $D u$ , is used.

3.1step 2.1F1F2F5algebra∎

Substituting [F1], $Γ^{'} (t) = \int_{0}^{1} [- \nabla (h (ρ) + V) \cdot \partial_{s} γ + \partial_{s} \frac{1}{2} ∣ u ∣_{2}^{2}] d s = - \int_{0}^{1} \partial_{s} [(h (ρ) + V) (t, γ (t, s))] d s + [\frac{1}{2} ∣ u ∣_{2}^{2}]_{s = 0}^{s = 1}$ by the chain rule [F2] and the fundamental theorem [F5]. Since $γ (t, 1) = γ (t, 0)$ , both boundary terms vanish, so $Γ^{'} (t) = 0$ for every $t \in (t_{0}, t_{1})$ ; [F4] also gives continuous one-sided endpoint derivatives, so The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ yields $Γ (t) - Γ (t_{0}) = \int_{t_{0}}^{t} Γ^{'} (τ) d τ = 0$ on $[t_{0}, t_{1}]$ .

Remarks

The loop is material: its points move with the fluid velocity, which is the hypothesis $\partial_{t} γ = u (γ)$ . The theorem is conditional on the barotropic inviscid local equation and uses the gradient form $D_{t} u = - \nabla (h + V)$ ; viscosity, non-barotropic pressure or a non-conservative body force invalidate the argument.
The circulation is defined as an oriented line integral with the parameterization $(t, s) \mapsto γ (t, s)$ ; a different parameterization of the same oriented loop gives the same value by Substitution: if $φ$ is differentiable on $[c, d]$ with $φ^{'}$ integrable and $f$ is continuous on an interval containing $φ ([c, d])$ , then $\int_{φ (c)}^{φ (d)} f = \int_{c}^{d} (f \circ φ) φ^{'}$ for a C1 orientation-preserving change of parameter with endpoints fixed. No claim is made about the existence of loops with prescribed circulation, nor about vortex dynamics.

TheoremStatement: AI-adaptedProof: AI-adaptedOpen item page →

d'Alembert's formula solves the wave equation, is the unique $C^{2}$ solution, and the fixed-end sine modes solve it exactly

Statement

Let $c > 0$ .

(a) The formula. For every $f \in C^{2} (R)$ and $g \in C^{1} (R)$ , the function

$u (t, x) : = \frac{f (x - c t) + f (x + c t)}{2} + \frac{1}{2 c} \int_{x - c t}^{x + c t} g (s) d s ((t, x) \in R^{2})$

is of class $C^{2}$ ( $C^{k}$ Euclidean maps and diffeomorphisms) and satisfies

$u_{t t} = c^{2} u_{x x}, u (0, x) = f (x), u_{t} (0, x) = g (x)$

for every $x \in R$ (The integral with oriented limits: $\int_{a}^{a} f : = 0$ and $\int_{b}^{a} f : = - \int_{a}^{b} f$ , The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set).

(b) Uniqueness. If $u \in C^{2} (R^{2})$ satisfies $u_{t t} = c^{2} u_{x x}$ with $u (0, \cdot) = f$ and $u_{t} (0, \cdot) = g$ for $f \in C^{2} (R)$ , $g \in C^{1} (R)$ , then $u$ equals the formula in (a).

(c) Fixed ends and modes. Let $L > 0$ and, for each integer $n \geq 1$ ,

$u_{n} (t, x) : = \sin ⁣ (\frac{n π x}{L}) \cos ⁣ (\frac{n π c t}{L}) .$

Then each $u_{n}$ is $C^{\infty}$ on $R^{2}$ , satisfies $u_{t t} = c^{2} u_{x x}$ , vanishes at $x = 0$ and $x = L$ for every $t$ , and has $u_{n} (0, x) = \sin (n π x / L)$ , $\partial_{t} u_{n} (0, x) = 0$ . Every finite sum $\sum_{n \leq N} α_{n} u_{n} + β_{n} v_{n}$ with the companion modes $v_{n} (t, x) : = \sin (n π x / L) \sin (n π c t / L) / (n π c / L)$ also solves the equation, vanishes at both ends and has the corresponding finite initial data.

Facts & Assumptions

Given: The wave speed $c > 0$ ; data $f \in C^{2} (R)$ , $g \in C^{1} (R)$ in (a), (b); $L > 0$ and the modes in (c).

[F1]

The integral function $F (x) : = \int_{0}^{x} g$ is $C^{2}$ with $F^{'} = g$ , and for $a < b$ the oriented integral satisfies $\int_{a}^{b} g = F (b) - F (a)$ , $\int_{a}^{b} = - \int_{b}^{a}$ , and additivity over subintervals; continuous functions are integrable on compact intervals (The integral function $F (x) : = \int_{a}^{x} f$ of an integrable $f$ , The first fundamental theorem: if $f$ is integrable on $[a, b]$ and continuous at $c$ , then $F^{'} (c) = f (c)$ ; in particular a continuous $f$ has $F$ as a primitive, The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ , The integral with oriented limits: $\int_{a}^{a} f : = 0$ and $\int_{b}^{a} f : = - \int_{a}^{b} f$ , For $a < c < b$ : $f$ is integrable on $[a, b]$ if and only if it is integrable on $[a, c]$ and on $[c, b]$ , and then $\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{b} f$ ; with the oriented form for arbitrary $a, b, c$ , A function integrable on $[a, b]$ is integrable on every closed subinterval, A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion, Integrable functions on $[a, b]$ form a set closed under sums and scalar multiples, and $\int_{a}^{b} (λ f + μ g) = λ \int_{a}^{b} f + μ \int_{a}^{b} g$ ).

[F2]

Chain rule and algebra of derivatives for compositions with the affine maps $(t, x) \mapsto x \pm c t$ (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ , Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , $C^{k}$ Euclidean maps and diffeomorphisms).

[F3]

Trigonometric values and derivatives: $\sin^{'} = \cos$ , $\cos^{'} = - \sin$ , $\sin (n π) = 0$ and $\cos (0) = 1$ (The derivatives of sine and cosine are cosine and minus sine, The zero sets of sine and cosine and the least positive common period 2 pi, Parity and the Pythagorean identity for sine and cosine, Sine and cosine defined by their real power series).

[F4]

If $q$ is $C^{2}$ on an open neighbourhood of a point $(t, x)$ , then $q_{t x} = q_{x t}$ there. For small $h, k > 0$ , the divided rectangular increment of $q$ over $[t, t + h] \times [x, x + k]$ equals the average of each mixed partial in its corresponding iterated order, by the one-variable fundamental theorem; continuity of the second partials makes the two averages tend to the respective mixed partial values, which must therefore agree (The first fundamental theorem: if $f$ is integrable on $[a, b]$ and continuous at $c$ , then $F^{'} (c) = f (c)$ ; in particular a continuous $f$ has $F$ as a primitive, A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion).

Proof

technique · direct; differentiate the explicit formula, then split the solution into the two characteristic combinations for uniqueness, and verify the modes by substitution

1.1givenF1F2algebra

By [F1] write $F (x) = \int_{0}^{x} g$ ; then $u (t, x) = \frac{f (x - c t) + f (x + c t)}{2} + \frac{F (x + c t) - F (x - c t)}{2 c}$ . By [F2] (affine chain rule) and [F1], this is $C^{2}$ in $(t, x)$ and $u_{t} = \frac{- c f^{'} (x - c t) + c f^{'} (x + c t)}{2} + \frac{g (x + c t) + g (x - c t)}{2}, u_{x} = \frac{f^{'} (x - c t) + f^{'} (x + c t)}{2} + \frac{g (x + c t) - g (x - c t)}{2 c} .$

1.2givenF2F4algebra

(Uniqueness.) For a $C^{2}$ solution $u$ set $v : = u_{t} + c u_{x}$ and $w : = u_{t} - c u_{x}$ . By [F2] and the mixed-partial identity [F4], $v_{t} - c v_{x} = u_{t t} + c u_{x t} - c u_{t x} - c^{2} u_{x x} = 0$ and $w_{t} + c w_{x} = u_{t t} - c u_{x t} + c u_{t x} - c^{2} u_{x x} = 0$ . Hence $s \mapsto v (s, x_{0} - c s)$ is constant for every fixed $x_{0}$ : its derivative is $v_{t} - c v_{x} = 0$ by [F2], so $v (t, x) = v (0, x + c t) = g (x + c t) + c f^{'} (x + c t)$ for every $(t, x)$ . Likewise $w (t, x) = w (0, x - c t) = g (x - c t) - c f^{'} (x - c t)$ .

2.1step 1.1F1F2algebra

Differentiating again by [F2] and [F1], $u_{t t} = \frac{c^{2} f^{''} (x - c t) + c^{2} f^{''} (x + c t)}{2} + \frac{c g^{'} (x + c t) - c g^{'} (x - c t)}{2}, u_{x x} = \frac{f^{''} (x - c t) + f^{''} (x + c t)}{2} + \frac{g^{'} (x + c t) - g^{'} (x - c t)}{2 c} .$ Hence $u_{t t} = c^{2} u_{x x}$ . Evaluating step 1.1 at $t = 0$ gives $u (0, x) = f (x)$ and $u_{t} (0, x) = \frac{- c f^{'} (x) + c f^{'} (x)}{2} + \frac{g (x) + g (x)}{2} = g (x)$ . This proves (a).

2.2step 1.2F1F2algebra

From step 1.2, $u_{t} = \frac{v + w}{2} = \frac{g (x + c t) + g (x - c t)}{2} + \frac{c f^{'} (x + c t) - c f^{'} (x - c t)}{2}$ . Since $s \mapsto u (s, x)$ is differentiable with derivative $u_{t} (s, x)$ , the fundamental theorem [F1] gives $u (t, x) - f (x) = \int_{0}^{t} u_{t} (s, x) d s$ . Substituting step 1.2 and integrating with [F1] (the substitutions $y = x + c s$ and $y = x - c s$ are affine) yields $\int_{0}^{t} \frac{g (x + c s) + g (x - c s)}{2} d s = \frac{F (x + c t) - F (x - c t)}{2 c}, \int_{0}^{t} \frac{c f^{'} (x + c s) - c f^{'} (x - c s)}{2} d s = \frac{f (x + c t) + f (x - c t)}{2} - f (x) .$ Adding and cancelling $f (x)$ identifies $u$ with the formula of (a). This proves (b).

3.1givenF2F3algebra∎

(Modes.) Fix $n \geq 1$ and put $θ : = n π x / L$ , $φ : = n π c t / L$ . By [F2], $u_{n}$ is $C^{\infty}$ with $u_{n, t t} = - (n π c / L)^{2} u_{n}$ and $u_{n, x x} = - (n π / L)^{2} u_{n}$ , so $u_{n, t t} = c^{2} u_{n, x x}$ . By [F3], $u_{n} (t, 0) = \sin 0 = 0$ and $u_{n} (t, L) = \sin (n π) \cos φ = 0$ ; $u_{n} (0, x) = \sin θ \cos 0 = \sin θ$ and $\partial_{t} u_{n} (0, x) = - (n π c / L) \sin θ \sin 0 = 0$ . The companion mode $v_{n} = (L / (n π c)) \sin θ \sin φ$ satisfies $v_{n, t t} = - (n π c / L)^{2} v_{n}$ and $v_{n, x x} = - (n π / L)^{2} v_{n}$ by the same computation, has $v_{n} (0, x) = 0$ and $\partial_{t} v_{n} (0, x) = \sin θ$ , and vanishes at both ends. Finite sums solve the linear equation and the boundary conditions because derivatives and finite sums commute; they have the corresponding finite initial data. This proves (c).

Remarks

Parts (a) and (b) are stated on all of $R^{2}$ for $C^{2}$ data; no weaker regularity, no distributional solutions, and no boundary conditions enter. The formula requires $f \in C^{2}$ , $g \in C^{1}$ exactly so that the displayed second derivatives exist and are continuous.
The modes in (c) are exact solutions of the equation and of the two endpoint conditions; they are not asserted to exhaust the solutions with fixed ends, and no infinite-series representation is claimed here. The companion modes $v_{n}$ are normalized so that $\partial_{t} v_{n} (0, \cdot) = \sin (n π x / L)$ ; their amplitude factor has time units divided by the corresponding velocity when units are declared in a consuming model.

DefinitionDefinition: AI-adaptedProof: Not applicableOpen item page →

The cyclic spring chain, its discrete energy, and the finite-regularity periodic comparison class

Definition

Let $N \geq 2$ be an integer, let $a > 0$ and $c > 0$ be real parameters, and let $L : = N a$ . Indices are read modulo $N$ : $z_{j + N} : = z_{j}$ .

(a) The chain. A state of the chain is a pair $(q, p) \in R^{N} \times R^{N}$ , where $q = (q_{1}, \dots, q_{N})$ are the positions and $p = (q_{1}^{'}, \dots, q_{N}^{'})$ the velocities of the $N$ nodes. The cyclic spring-chain system is the first-order system

$q_{j}^{'} = p_{j}, p_{j}^{'} = \frac{c^{2}}{a^{2}} (q_{j + 1} - 2 q_{j} + q_{j - 1}) (j \in Z / N),$

equivalently the second-order system $q_{j}^{''} = \frac{c^{2}}{a^{2}} (q_{j + 1} - 2 q_{j} + q_{j - 1})$ for a $C^{2}$ curve $q : I \to R^{N}$ . Its discrete energy is

$E (q, p) : = \frac{1}{2} \sum_{j = 1}^{N} (p_{j}^{2} + \frac{c^{2}}{a^{2}} (q_{j + 1} - q_{j})^{2}) .$

(b) Discrete norms and differences. For $z, w \in R^{N}$ define the scaled inner product and norm

$⟨ z, w ⟩_{a} : = a \sum_{j = 1}^{N} z_{j} w_{j}, ∥ z ∥_{a} : = ⟨ z, z ⟩_{a}^{1 / 2},$

and for the discrete difference $(D_{+} z)_{j} : = (z_{j + 1} - z_{j}) / a$ . The discrete Laplacian is $(Δ_{a} z)_{j} : = (z_{j + 1} - 2 z_{j} + z_{j - 1}) / a^{2}$ .

(c) The comparison class. Let $I \subseteq R$ be a nonempty open interval and let $u : I \times R \to R$ be a function that is $C^{2}$ in the first variable and $C^{4}$ in the second, is $L$ -periodic in $x$ ( $u (t, x + L) = u (t, x)$ for all $(t, x)$ ), and satisfies

$u_{t t} (t, x) = c^{2} u_{x x} (t, x) ((t, x) \in I \times R) .$

For a compact subinterval $[0, T] \subseteq I$ put $M_{T} : = \sup {∣ \partial_{x}^{4} u (t, x) ∣ : (t, x) \in [0, T] \times R}$ (finite when continuous derivatives are bounded on the compact time interval, as assumed in the items that use it), and for $a = L / N$ define the sampled initial data

$q_{j} (0) : = u (0, j a), q_{j}^{'} (0) : = u_{t} (0, j a) (j = 1, \dots, N) .$

Remarks

This is a purely mathematical definition. The mechanical reading — $N$ equal masses $m$ , nearest-neighbour springs of stiffness $k$ with $c^{2} = k a^{2} / m$ , a closed chain — is a model interpretation attached by a consuming physical item; the mathematical results of this collection use only the ODE system and the comparison class.
The cyclic indexing (index $N + 1$ is index $1$ ) makes the chain closed and makes summation by parts in the discrete variable exact; it is part of the model, not a boundary condition. Other boundary conditions are separate models and are not covered by this definition.
The comparison class requires a globally defined periodic solution with the regularity in clause (c); the items using it establish existence of the chain solutions and compare the two evolutions on a compact time interval. No uniqueness or continuation statement for $u$ is part of this definition.

TheoremStatement: AI-adaptedProof: AI-adaptedOpen item page →

Exact dispersion relation $ω^{2} = (4 c^{2} / a^{2}) \sin^{2} (κ a / 2)$ for the cyclic chain and its continuum error bound

Statement

Let $N \geq 2$ , $a > 0$ , $c > 0$ , let $κ, ω \in R$ satisfy $κ N a \in 2 π Z$ , and let $A \in R ∖ {0}$ . Consider the travelling cosine ansatz for the cyclic chain of The cyclic spring chain, its discrete energy, and the finite-regularity periodic comparison class:

$q_{j} (t) : = A \cos (κ j a - ω t) (j \in Z / N) .$

(a) The curve $q : R \to R^{N}$ solves $q_{j}^{''} = \frac{c^{2}}{a^{2}} (q_{j + 1} - 2 q_{j} + q_{j - 1})$ for all $t \in R$ and all $j$ if and only if

$\begin{matrix} ω^{2} = \frac{4 c^{2}}{a^{2}} \sin^{2} ⁣ (\frac{κ a}{2}) . & (DR) \end{matrix}$

(b) If (DR) holds and $∣ κ a ∣ \leq 1$ , then

$∣ ω^{2} - c^{2} κ^{2} ∣ \leq \frac{c^{2} κ^{4} a^{2}}{12},$

so the continuum dispersion $ω^{2} = c^{2} κ^{2}$ is recovered with an explicit error bound of second order in the spacing $a$ and fourth order in the wave number $κ$ .

Facts & Assumptions

Given: $N, a, c, κ, ω, A$ with $A \neq 0$ and the ansatz $q_{j} (t) = A \cos (κ j a - ω t)$ .

[F1]

Trigonometric addition formulas, $1 - \cos u = 2 \sin^{2} (u / 2)$ and the bound $∣ \cos r ∣ \leq 1$ (The addition formulas for sine and cosine, Parity and the Pythagorean identity for sine and cosine, Sine and cosine defined by their real power series). The additional bound $∣ \sin r ∣ \leq ∣ r ∣$ follows locally: by the derivative identity [F2] and integral evaluation [F3], $\sin r = \int_{0}^{r} \cos s d s$ , so $∣ \sin r ∣ \leq \int_{\min (0, r)}^{\max (0, r)} ∣ \cos s ∣ d s \leq ∣ r ∣$ .

[F2]

Derivatives: $(\cos)^{'} = - \sin$ , $(\sin)^{'} = \cos$ ; affine compositions differentiate by the chain rule (The derivatives of sine and cosine are cosine and minus sine, The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ , Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set, $C^{k}$ Euclidean maps and diffeomorphisms).

[F3]

Fundamental theorem of calculus: for $C^{1}$ functions on intervals, $\int_{r}^{s} φ^{'} = φ (s) - φ (r)$ and $∣ \int_{r}^{s} ψ ∣ \leq \int_{\min (r, s)}^{\max (r, s)} ∣ ψ ∣$ for oriented integrals (The first fundamental theorem: if $f$ is integrable on $[a, b]$ and continuous at $c$ , then $F^{'} (c) = f (c)$ ; in particular a continuous $f$ has $F$ as a primitive, The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ , The integral with oriented limits: $\int_{a}^{a} f : = 0$ and $\int_{b}^{a} f : = - \int_{a}^{b} f$ ).

Proof

technique · direct; evaluate both sides of the ODE on the ansatz and bound the trigonometric remainder

1.1givenF1F2algebra

Since $κ N a \in 2 π Z$ , $q_{j + N} (t) = q_{j} (t)$ for every integer $j$ and every $t$ by cosine periodicity, so the ansatz is well-defined on the cyclic sites and its shifted values agree across the seam. Put $θ_{j} : = κ j a - ω t$ . By [F2], $q_{j}^{''} (t) = - ω^{2} A \cos θ_{j}$ . By [F1], $\cos (θ_{j} + κ a) + \cos (θ_{j} - κ a) = 2 \cos θ_{j} \cos (κ a)$ , so $q_{j + 1} - 2 q_{j} + q_{j - 1} = A \cos θ_{j} (2 \cos (κ a) - 2) = - 2 A \cos θ_{j} (1 - \cos (κ a)) = - 4 A \cos θ_{j} \sin^{2} ⁣ (\frac{κ a}{2}) .$

1.2givenF1F3

For the error bound, first prove $∣ \sin z - z ∣ \leq ∣ z ∣^{3} / 6$ for all $z$ . Write $\sin z - z = \int_{0}^{z} (\cos s - 1) d s$ and $\cos s - 1 = - \int_{0}^{s} \sin r d r$ ; by [F3] and $∣ \sin r ∣ \leq ∣ r ∣$ , $∣ \cos s - 1 ∣ \leq s^{2} / 2$ ; hence $∣ \sin z - z ∣ \leq \int_{0}^{∣ z ∣} s^{2} / 2 d s = ∣ z ∣^{3} / 6$ .

2.1step 1.1givenalgebra

The ODE holds for all $t, j$ if and only if $- ω^{2} A \cos θ_{j} = - \frac{4 c^{2}}{a^{2}} A \cos θ_{j} \sin^{2} ⁣ (\frac{κ a}{2})$ for all $t, j$ . Since $A \neq 0$ and $\cos θ_{j} = 1$ for $θ_{j} = 0$ (a value attained for appropriate $t$ , for instance $t = κ j a / ω$ when $ω \neq 0$ , and the identity is also immediate at $t = 0$ , $j \equiv 0$ ), the coefficients must agree; conversely (DR) makes the two sides equal identically. If $ω = 0$ the same comparison at $t = 0$ and $j = 0$ gives the same conclusion. This proves (a).

3.1step 1.2step 2.1F1algebra∎

Put $z : = κ a / 2$ and assume $∣ z ∣ \leq 1 / 2$ (equivalently $∣ κ a ∣ \leq 1$ ). Then $∣ 4 \sin^{2} z - (2 z)^{2} ∣ = 4 ∣ \sin z - z ∣ ∣ \sin z + z ∣ \leq 4 \cdot \frac{∣ z ∣^{3}}{6} \cdot 2 ∣ z ∣ = \frac{4 ∣ z ∣^{4}}{3},$ using $∣ \sin z ∣ \leq ∣ z ∣$ in $∣ \sin z + z ∣ \leq 2 ∣ z ∣$ . Substituting $∣ z ∣ = ∣ κ a ∣ / 2$ gives $∣ 4 \sin^{2} (κ a / 2) - (κ a)^{2} ∣ \leq (κ a)^{4} / 12$ . Multiplying by $c^{2} / a^{2}$ and using (DR) yields the claimed bound $∣ ω^{2} - c^{2} κ^{2} ∣ \leq c^{2} κ^{4} a^{2} / 12$ .

Remarks

The exact relation (DR) is bounded between $0$ and $4 c^{2} / a^{2}$ ; the long-wavelength branch has $ω^{2} \to c^{2} κ^{2}$ as $κ a \to 0$ , and the bound of (b) makes that limit quantitative. The bound is stated for $∣ κ a ∣ \leq 1$ only; no claim is made for short waves.
The ansatz is a real travelling wave of constant amplitude; it is not asserted that every chain solution is a superposition of such waves, and no stability or energy estimate for the ansatz is claimed here. The continuum comparison of this item is strictly local in wave number.

Remark

The scalar remainder calculation in steps 1.2 and 3.1 proves $∣ \sin z - z ∣ \leq ∣ z ∣^{3} / 6$ and $∣ 4 \sin^{2} z - 4 z^{2} ∣ \leq 4 ∣ z ∣^{4} / 3$ for every real $z$ : those inequalities use only $∣ \sin z ∣ \leq ∣ z ∣$ , not the additional long-wave restriction of clause (b). They can therefore be used directly to check a specified compatible mode outside that restricted clause.

LemmaStatement: AI-adaptedProof: AI-adaptedOpen item page →

The centered second difference differs from the second derivative by at most $M a^{2} / 12$ when $∣ u^{''''} ∣ \leq M$

Statement

Let $J \subseteq R$ be a nonempty open interval (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length), let $u : J \to R$ be of class $C^{4}$ ( $C^{k}$ Euclidean maps and diffeomorphisms), let $x \in J$ and $a > 0$ satisfy $[x - a, x + a] \subseteq J$ , and put

$M : = \sup_{s \in [x - a, x + a]} ∣ u^{''''} (s) ∣ < + \infty .$

Then

$∣ \frac{u (x + a) - 2 u (x) + u (x - a)}{a^{2}} - u^{''} (x) ∣ \leq \frac{M a^{2}}{12} .$

Facts & Assumptions

Given: The open interval $J$ , the $C^{4}$ function $u$ , the point $x \in J$ and $a > 0$ with $[x - a, x + a] \subseteq J$ , and the bound $M$ .

[F1]

Fundamental theorem of calculus: for a $C^{1}$ function $φ$ on an interval and points $r < s$ in it, $\int_{r}^{s} φ^{'} = φ (s) - φ (r)$ ; moreover $\int_{s}^{r} = - \int_{r}^{s}$ , and the integral is additive over subintervals (The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ , The integral with oriented limits: $\int_{a}^{a} f : = 0$ and $\int_{b}^{a} f : = - \int_{a}^{b} f$ , For $a < c < b$ : $f$ is integrable on $[a, b]$ if and only if it is integrable on $[a, c]$ and on $[c, b]$ , and then $\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{b} f$ ; with the oriented form for arbitrary $a, b, c$ ).

[F2]

The bound $M$ is the maximum of the continuous function $∣ u^{''''} ∣$ on the compact set $[x - a, x + a]$ , so $∣ u^{''''} ∣ \leq M$ there (A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value, Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line, Open ball, closed ball and sphere in a metric space); derivatives and sums compose by the chain and algebra rules (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ , Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set).

[F3]

(Second-order integral remainder.) If $φ \in C^{2}$ on an interval containing $[0, a]$ , then $φ (a) = φ (0) + φ^{'} (0) a + \int_{0}^{a} (a - s) φ^{''} (s) d s$ . By the product rule, $\frac{d}{d s} ((a - s) φ^{'} (s)) = - φ^{'} (s) + (a - s) φ^{''} (s)$ ; integrating this identity with [F1] and rearranging gives the displayed formula.

Proof

technique · direct; apply the second-order integral remainder to two affine scans of $u$ and bound the double remainder

1.1givenF1F2F3algebra

Fix $s \in [0, a]$ and apply [F3] to $φ_{+} (r) : = u (x + r)$ on $[0, s]$ : since $φ_{+}^{''} (r) = u^{''} (x + r)$ and $φ_{+}^{'} (0) = u^{'} (x)$ , $u (x + s) = u (x) + u^{'} (x) s + \int_{0}^{s} (s - r) u^{''} (x + r) d r .$ Applying [F3] to $φ_{-} (r) : = u (x - r)$ on $[0, s]$ , with $φ_{-}^{''} (r) = u^{''} (x - r)$ , $u (x - s) = u (x) - u^{'} (x) s + \int_{0}^{s} (s - r) u^{''} (x - r) d r .$

2.1step 1.1F1F3algebra

Add the two identities of step 1.1 with $s = a$ and subtract $2 u (x)$ : $u (x + a) - 2 u (x) + u (x - a) = \int_{0}^{a} (a - s) (u^{''} (x + s) + u^{''} (x - s)) d s .$ Apply [F3] once more to the $C^{2}$ function $r \mapsto ψ (r) : = u^{''} (x + r) + u^{''} (x - r)$ on $[0, s]$ , whose value and first derivative at $0$ are $ψ (0) = 2 u^{''} (x)$ and $ψ^{'} (0) = 0$ : $u^{''} (x + s) + u^{''} (x - s) = 2 u^{''} (x) + \int_{0}^{s} (s - r) (ψ^{''} (r)) d r = 2 u^{''} (x) + \int_{0}^{s} (s - r) (u^{''''} (x + r) + u^{''''} (x - r)) d r .$

3.1step 2.1F1F2algebra∎

Substitute step 2.1 into step 2.1's first identity to obtain $u (x + a) - 2 u (x) + u (x - a) - a^{2} u^{''} (x) = \int_{0}^{a} (a - s) \int_{0}^{s} (s - r) (u^{''''} (x + r) + u^{''''} (x - r)) d r d s .$ Bounding the inner integral by $2 M s^{2} / 2 = M s^{2}$ via [F2] and [F1] (monotone bounds for the Riemann integral), and then $\int_{0}^{a} (a - s) M s^{2} d s = M a^{4} / 12$ , gives $∣ u (x + a) - 2 u (x) + u (x - a) - a^{2} u^{''} (x) ∣ \leq \frac{M a^{4}}{12} .$ Dividing by $a^{2} > 0$ proves the statement.

Remarks

The constant $1 / 12$ is sharp for the class $C^{4}$ with a uniform fourth-derivative bound: the estimate is an equality for a quartic profile $u (x) = x^{4} / 24$ with $u^{''''} = 1$ after comparing the two required bounds, and the factor arises solely from $\int_{0}^{a} (a - s) s^{2} d s$ . The lemma states only the upper bound.
Only derivatives up to order four on the closed interval $[x - a, x + a]$ are used; the function need not be periodic and no boundary conditions enter.

LemmaStatement: AI-adaptedProof: AI-adaptedOpen item page →

Discrete periodic summation by parts, the scaled Cauchy-Schwarz inequality, and the discrete energy identity

Statement

Use the notation $z_{j + N} = z_{j}$ , $(D_{+} z)_{j} = (z_{j + 1} - z_{j}) / a$ , $(Δ_{a} z)_{j} = (z_{j + 1} - 2 z_{j} + z_{j - 1}) / a^{2}$ , $⟨ z, w ⟩_{a} = a \sum_{j} z_{j} w_{j}$ , $∥ z ∥_{a}^{2} = ⟨ z, z ⟩_{a}$ and $E_{a} (q, p) = \frac{1}{2} (∥ p ∥_{a}^{2} + c^{2} ∥ D_{+} q ∥_{a}^{2})$ of The cyclic spring chain, its discrete energy, and the finite-regularity periodic comparison class.

(a) Summation by parts. For all $z, w \in R^{N}$ ,

$⟨ D_{+} z, D_{+} w ⟩_{a} = - ⟨ Δ_{a} z, w ⟩_{a} .$

(b) Scaled Cauchy-Schwarz. For all $z, w \in R^{N}$ , $∣ ⟨ z, w ⟩_{a} ∣ \leq ∥ z ∥_{a} ∥ w ∥_{a}$ (On $R^{n}$ , abstract Cauchy–Schwarz is exactly the published finite-sum Cauchy–Schwarz inequality, The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

(c) Energy identity. Let $I \subseteq R$ be an open interval, let $q : I \to R^{N}$ be $C^{2}$ and let $R : I \to R^{N}$ be continuous with

$q_{j}^{''} = \frac{c^{2}}{a^{2}} (q_{j + 1} - 2 q_{j} + q_{j - 1}) + R_{j} = c^{2} (Δ_{a} q)_{j} + R_{j} (j \in Z / N) .$

Then $t \mapsto E_{a} (q (t), q^{'} (t))$ is $C^{1}$ on $I$ with

$\frac{d}{d t} E_{a} (q (t), q^{'} (t)) = a \sum_{j = 1}^{N} R_{j} (t) q_{j}^{'} (t),$

and consequently $∣ \frac{d}{d t} E_{a} ∣ \leq ∥ R (t) ∥_{a} \sqrt{2 E_{a} (t)}$ whenever $E_{a} \geq 0$ (which always holds).

Facts & Assumptions

Given: $N \geq 2$ , $a > 0$ , $c > 0$ , vectors $z, w$ ; in (c), the $C^{2}$ curve $q$ and the continuous field $R$ satisfying the stated equation.

[F1]

Index shifts in a finite cyclic sum change nothing: $\sum_{j = 1}^{N} z_{j + 1} = \sum_{j = 1}^{N} z_{j}$ , and sums split as $\sum_{j} (α_{j} - β_{j}) = \sum_{j} α_{j} - \sum_{j} β_{j}$ (The cyclic spring chain, its discrete energy, and the finite-regularity periodic comparison class).

[F2]

Cauchy-Schwarz in coordinates (On $R^{n}$ , abstract Cauchy–Schwarz is exactly the published finite-sum Cauchy–Schwarz inequality); the scaled form (b) follows by applying it to $\sqrt{a} z$ and $\sqrt{a} w$ .

[F3]

$E_{a}$ is a finite sum of squares of $C^{1}$ functions, hence $C^{1}$ , and its derivative is computed by the product rule and linearity of the finite sum (Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set, Integrable functions on $[a, b]$ form a set closed under sums and scalar multiples, and $\int_{a}^{b} (λ f + μ g) = λ \int_{a}^{b} f + μ \int_{a}^{b} g$ ).

Proof

technique · direct; shift indices in the finite sum and differentiate the quadratic energy

1.1givenF1algebra

Compute $a \sum_{j} (D_{+} z)_{j} (D_{+} w)_{j} = \frac{1}{a} \sum_{j} (z_{j + 1} - z_{j}) (w_{j + 1} - w_{j})$ . Splitting the sum and shifting the index in the first product term, $\frac{1}{a} \sum_{j} (z_{j + 1} - z_{j}) (w_{j + 1} - w_{j}) = \frac{1}{a} \sum_{j} (z_{j} - z_{j - 1}) w_{j} - \frac{1}{a} \sum_{j} (z_{j + 1} - z_{j}) w_{j} = - \frac{1}{a} \sum_{j} (z_{j + 1} - 2 z_{j} + z_{j - 1}) w_{j},$ which is $- ⟨ Δ_{a} z, w ⟩_{a}$ . This proves (a).

1.2F2algebra

For (b), apply [F2] to the vectors $(\sqrt{a} z_{j})$ and $(\sqrt{a} w_{j})$ .

2.1F3step 1.1algebra

Assume (c). By [F3], $\frac{d}{d t} E_{a} (q (t), q^{'} (t)) = a \sum_{j} (q_{j}^{'} (t) q_{j}^{''} (t) + c^{2} (D_{+} q)_{j} (t) (D_{+} q^{'})_{j} (t)) .$ Applying (a) with $z = q (t)$ , $w = q^{'} (t)$ turns the second term into $- c^{2} a \sum_{j} (Δ_{a} q)_{j} q_{j}^{'}$ . Hence $\frac{d}{d t} E_{a} = a \sum_{j} q_{j}^{'} (q_{j}^{''} - c^{2} (Δ_{a} q)_{j}) = a \sum_{j} R_{j} q_{j}^{'} .$

3.1step 2.1step 1.2algebra∎

By (b), $∣ a \sum_{j} R_{j} q_{j}^{'} ∣ = ∣ ⟨ R, q^{'} ⟩_{a} ∣ \leq ∥ R ∥_{a} ∥ q^{'} ∥_{a} \leq ∥ R ∥_{a} \sqrt{2 E_{a}}$ , because $∥ q^{'} ∥_{a}^{2} \leq 2 E_{a}$ .

Remarks

The identity (a) is the discrete analogue of integration by parts on the circle; cyclicity is exactly what makes the boundary term vanish. On a chain with free or fixed ends the boundary terms would appear explicitly, and such models are not covered by this statement.
The energy identity (c) reduces the discrete energy rate to the pairing of the perturbation $R$ with the velocity; it holds for any continuous $R$ , including the Taylor remainder of the continuum limit.

TheoremStatement: AI-adaptedProof: AI-adaptedOpen item page →

Finite-time continuum limit of the cyclic spring chain: uniform $O (a^{3 / 2})$ nodal and uniform interpolation convergence

Statement

Let $L > 0$ , $c > 0$ , let $N \geq 2$ be an integer and $a : = L / N$ , and let $[0, T]$ be a compact time interval, with $T \geq 0$ . Let

$u : [0, T] \times R \to R$

be $C^{2}$ in $t$ and $C^{4}$ in $x$ ( $C^{k}$ Euclidean maps and diffeomorphisms), $L$ -periodic in $x$ , and satisfy $u_{t t} = c^{2} u_{x x}$ on $[0, T] \times R$ ; put

$M : = \sup_{[0, T] \times R} ∣ \partial_{x}^{4} u ∣ < + \infty, M_{2} : = \sup_{[0, T] \times R} ∣ \partial_{x}^{2} u ∣ < + \infty .$

Let $q : [0, T] \to R^{N}$ be the (by (i), unique) solution of the cyclic spring-chain system of The cyclic spring chain, its discrete energy, and the finite-regularity periodic comparison class with the sampled initial data $q_{j} (0) = u (0, j a)$ , $q_{j}^{'} (0) = u_{t} (0, j a)$ , and put

$e_{j} (t) : = q_{j} (t) - u (t, j a), C : = \frac{c^{2} \sqrt{L} M}{12} .$

Then:

Here smoothness on the closed time interval means restriction of a map defined on an open neighbourhood, with the indicated derivatives extending continuously to the interval. A chain solution on $[0, T]$ is the restriction of a solution on an open time interval containing $[0, T]$ ; the sampled derivative $q_{j}' (0)$ is the derivative of that solution. Thus derivatives remain defined when $T = 0$ . Uniqueness concerns the restricted state $(q, q')$ .

(i) the solution $q$ exists and is unique on $[0, T]$ ;

(ii) for every $t \in [0, T]$ ,

$∥ e (t) ∥_{a} \leq \frac{C a^{2} t^{2}}{2}, ∥ e^{'} (t) ∥_{a} \leq C a^{2} t;$

(iii) consequently $\max_{j} ∣ e_{j} (t) ∣ \leq \frac{1}{2} C a^{3 / 2} t^{2}$ for $t \in [0, T]$ , so the nodal error tends to $0$ uniformly as $a \to 0$ ;

(iv) let $I_{a} q (t, \cdot)$ be the piecewise-linear function on $[0, L]$ that equals $q_{j} (t)$ at $x = j a$ and is extended $L$ -periodically; then for the periodic extension of $u$ ,

$\sup_{x \in [0, L]} ∣ I_{a} q (t, x) - u (t, x) ∣ \leq \max_{j} ∣ e_{j} (t) ∣ + \frac{M_{2} a^{2}}{8} (0 \leq t \leq T),$

so the interpolated chain displacement converges to $u$ uniformly on $[0, T] \times [0, L]$ as $a \to 0$ .

Facts & Assumptions

Given: $L, c, N, a, T$ ; the smooth periodic solution $u$ with the bounds $M, M_{2}$ ; the sampled initial data.

[F1]

Linear matrix ODEs with continuous coefficients have unique solutions on a compact interval (Linear matrix ODEs have unique global solutions on a fixed interval); a first-order system $z^{'} = A z$ with constant $A$ is of this form, and its solutions are $z (t) = Y (t) z (0)$ for the matrix solution $Y$ with $Y (0) = I$ (First-order systems, initial value problems, and solutions on intervals, Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ ).

[F2]

Apply The centered second difference differs from the second derivative by at most $M a^{2} / 12$ when $∣ u^{''''} ∣ \leq M$ to the one-variable function $x \mapsto u (t, x)$ at each $x = j a$ . This gives $∣ Δ_{a} u (t, j a) - u_{x x} (t, j a) ∣ \leq M a^{2} / 12$ for every $j, t$ .

[F3]

Discrete summation by parts and the energy identity of Discrete periodic summation by parts, the scaled Cauchy-Schwarz inequality, and the discrete energy identity: for the chain perturbation $R$ defined below, $\frac{d}{d t} E_{a} = a \sum_{j} R_{j} e_{j}^{'}$ and $∣ \frac{d}{d t} E_{a} ∣ \leq ∥ R ∥_{a} \sqrt{2 E_{a}}$ ; the scaled Cauchy–Schwarz inequality is that item's part (b) (On $R^{n}$ , abstract Cauchy–Schwarz is exactly the published finite-sum Cauchy–Schwarz inequality).

[F4]

Linear interpolation error: if $φ \in C^{2}$ on $[x_{j}, x_{j + 1}]$ and $∣ φ^{''} ∣ \leq M_{2}$ there, then the chord $ℓ$ through the endpoint values satisfies $∣ φ (x) - ℓ (x) ∣ \leq M_{2} a^{2} / 8$ . Put $g = φ - ℓ$ , so $g (x_{j}) = g (x_{j + 1}) = 0$ and $∣ g^{''} ∣ \leq M_{2}$ . For an interior $x$ , set $c : = g (x) / ((x - x_{j}) (x - x_{j + 1}))$ and $H (s) : = g (s) - c (s - x_{j}) (s - x_{j + 1})$ . Then $H$ vanishes at $x_{j}, x, x_{j + 1}$ ; Rolle's theorem twice gives $H^{''} (ξ) = 0$ for some $ξ \in (x_{j}, x_{j + 1})$ . Hence $g (x) = \frac{1}{2} g^{''} (ξ) (x - x_{j}) (x - x_{j + 1})$ , and $∣ (x - x_{j}) (x - x_{j + 1}) ∣ \leq a^{2} / 4$ gives the stated bound. Endpoints are immediate (Rolle's theorem: if $a < b$ , $f$ is continuous on $[a, b]$ , differentiable at every point of $(a, b)$ , and $f (a) = f (b)$ , then $f^{'} (c) = 0$ for some $c \in (a, b)$ , The first fundamental theorem: if $f$ is integrable on $[a, b]$ and continuous at $c$ , then $F^{'} (c) = f (c)$ ; in particular a continuous $f$ has $F$ as a primitive, The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set).

[F5]

Norms: $∥ z ∥_{a}^{2} = a \sum_{j} z_{j}^{2}$ , so $\max_{j} ∣ z_{j} ∣ \leq a^{- 1 / 2} ∥ z ∥_{a}$ . For continuous $r : [0, t] \to R^{N}$ , $∥ \int_{0}^{t} r (s) d s ∥_{a} \leq \int_{0}^{t} ∥ r (s) ∥_{a} d s$ : the scaled Cauchy--Schwarz inequality gives the triangle inequality for $∥ \cdot ∥_{a}$ , so this bound holds for every Riemann sum and passes to the integral limit (The cyclic spring chain, its discrete energy, and the finite-regularity periodic comparison class, On $R^{n}$ , abstract Cauchy–Schwarz is exactly the published finite-sum Cauchy–Schwarz inequality, The first fundamental theorem: if $f$ is integrable on $[a, b]$ and continuous at $c$ , then $F^{'} (c) = f (c)$ ; in particular a continuous $f$ has $F$ as a primitive, For $a < c < b$ : $f$ is integrable on $[a, b]$ if and only if it is integrable on $[a, c]$ and on $[c, b]$ , and then $\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{b} f$ ; with the oriented form for arbitrary $a, b, c$ ).

Proof

technique · direct; compare the chain to the sampled solution, bound the Taylor remainder by the fourth derivative, and integrate the energy identity

1.1givenF1construct

Write the chain as the first-order system $z^{'} = B z$ on $R^{2 N}$ with $z = (q, p)$ and the constant matrix $B (q, p) = (p, \frac{c^{2}}{a^{2}} (q_{j + 1} - 2 q_{j} + q_{j - 1}))_{j}$ . By [F1] applied to the matrix ODE with $C (t) = B$ on $[- 1, T + 1]$ and initial time $0$ , the initial value problem $z^{'} = B z$ , $z (0) = (q (0), q^{'} (0))$ has a unique solution on $[0, T]$ , and its first component is the unique solution $q$ of the chain equations with the sampled data. The solution is smooth on the larger interval by repeated differentiation of its constant-coefficient equation. Restricting to $[0, T]$ proves (i), including $T = 0$ ; in that case all errors and energy at $0$ vanish by the sampled state and every bound is immediate.

2.1givenF2step 1.1algebra

Put $u_{j} (t) : = u (t, j a)$ and define $R_{j} (t) : = c^{2} (Δ_{a} u_{j} (t) - u_{x x} (t, j a)) = c^{2} Δ_{a} u_{j} (t) - u_{t t} (t, j a),$ the last equality by $u_{t t} = c^{2} u_{x x}$ . By [F2], $∣ R_{j} (t) ∣ \leq c^{2} M a^{2} / 12$ for every $j, t$ , hence $∥ R (t) ∥_{a} = (a \sum_{j} R_{j} (t)^{2})^{1 / 2} \leq \frac{c^{2} M a^{2}}{12} \sqrt{a N} = C a^{2} .$ Since $e_{j} = q_{j} - u_{j}$ , the identities $u_{j}^{''} = c^{2} (Δ_{a} u)_{j} - R_{j}$ and $q_{j}^{''} = c^{2} (Δ_{a} q)_{j}$ give $e_{j}^{''} = c^{2} (Δ_{a} e)_{j} + R_{j}$ ; moreover $e_{j} (0) = 0$ and $e_{j}^{'} (0) = 0$ by the sampled initial data.

3.1givenF3F5step 2.1algebra

Put $E_{a} (t) : = \frac{1}{2} (∥ e^{'} (t) ∥_{a}^{2} + c^{2} ∥ D_{+} e (t) ∥_{a}^{2})$ , which is $\geq 0$ with $E_{a} (0) = 0$ . By [F3] (the energy identity with the perturbation $R$ of step 2.1), $\frac{d}{d t} \sqrt{2 E_{a} (t) + δ} = \frac{E_{a}^{'} (t)}{\sqrt{2 E_{a} (t) + δ}} \leq \frac{∥ R (t) ∥_{a} \sqrt{2 E_{a} (t)}}{\sqrt{2 E_{a} (t) + δ}} \leq ∥ R (t) ∥_{a} \leq C a^{2}$ for every $δ > 0$ . Integrating from $0$ and letting $δ ↓ 0$ gives $\sqrt{2 E_{a} (t)} \leq C a^{2} t$ ; in particular $∥ e^{'} (t) ∥_{a} \leq C a^{2} t$ . Integrating $e (t) = e (0) + \int_{0}^{t} e^{'} (s) d s$ and using [F5] gives $∥ e (t) ∥_{a} \leq \int_{0}^{t} C s a^{2} d s = \frac{1}{2} C a^{2} t^{2}$ . This is (ii).

4.1step 3.1F4F5algebra∎

By [F5], $\max_{j} ∣ e_{j} (t) ∣ \leq a^{- 1 / 2} ∥ e (t) ∥_{a} \leq \frac{1}{2} C a^{3 / 2} t^{2}$ , which is (iii). For (iv), fix $t \in [0, T]$ and $x = j a + s$ with $0 \leq s \leq a$ . The interpolant of the samples of $u$ satisfies the chord bound [F4] on $[j a, (j + 1) a]$ with $φ = u (t, \cdot)$ , and $∣ I_{a} q (t, x) - I_{a} u (t, x) ∣ \leq \max_{j} ∣ e_{j} (t) ∣$ because both interpolations are convex combinations of their nodal values; hence $\sup_{x} ∣ I_{a} q (t, x) - u (t, x) ∣ \leq \max_{j} ∣ e_{j} (t) ∣ + \frac{M_{2} a^{2}}{8} .$ By (iii) the right-hand side tends to $0$ as $a \to 0$ uniformly on $[0, T]$ .

Remarks

The convergence is finite-time and uniform only for the stated smooth $L$ -periodic regime; the rate is the nodal $O (a^{3 / 2})$ together with the interpolation error $O (a^{2})$ , and no uniform bound on $∥ q ∥_{\infty}$ is claimed beyond what the estimates give. Rough data, non-periodic boundary conditions and nonlinear chains are separate problems.
The mechanical interpretation $m = ρ a$ , $k = T_{t e n s i o n} / a$ , $c^{2} = k a^{2} / m$ is a model reading of the parameters; the theorem compares the two mathematical evolutions and asserts no physical identification of the chain with any material string.
The estimate is derived from the energy identity, so it requires only the perturbation $R$ to be continuous and the chain to solve exactly the linear system; no numerical integrator error is analyzed, and floating-point roundoff is not part of the statement.

TheoremStatement: AI-adaptedProof: AI-adaptedOpen item page →

The method of characteristics for $u_{t} + u u_{x} = 0$ and the finite blow-up time of the gradient

Statement

Let $f \in C^{2} (R)$ ( $C^{k}$ Euclidean maps and diffeomorphisms) be bounded (so $∥ f ∥_{\infty} < + \infty$ , A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value, Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line) and suppose that

$f^{'} (a) \geq - K for every a \in R, and f^{'} (a_{*}) = - K for some a_{*} \in R,$

with $K > 0$ . Put $T_{0} : = 1 / K$ , and for $0 \leq t < T_{0}$ define $φ_{t} (a) : = a + t f (a)$ .

(i) For every $t \in [0, T_{0})$ the map $φ_{t} : R \to R$ is a $C^{1}$ diffeomorphism onto $R$ (An injective regular $C^{k}$ map is a $C^{k}$ diffeomorphism onto its image), and with $a (t, x) : = φ_{t}^{- 1} (x)$ the characteristic label,

$u (t, x) : = f (a (t, x)), (t, x) \in [0, T_{0}) \times R,$

is $C^{1}$ jointly in $(t, x)$ , satisfies $u (0, x) = f (x)$ and solves Burgers' equation

$\begin{matrix} u_{t} (t, x) + u (t, x) u_{x} (t, x) = 0 on (0, T_{0}) \times R . & (B) \end{matrix}$

(ii) For every $a_{0} \in R$ the function $t \mapsto u (t, φ_{t} (a_{0})) = f (a_{0})$ is constant, and the gradient along this characteristic, $q (t) : = u_{x} (t, φ_{t} (a_{0}))$ , satisfies $q^{'} = - q^{2}$ and $q (t) = f^{'} (a_{0}) / (1 + t f^{'} (a_{0}))$ for $0 \leq t < T_{0}$ .

(iii) Put $x_{*} : = φ_{T_{0}} (a_{*}) = a_{*} + T_{0} f (a_{*})$ and let $a (t) : = a (t, x_{*})$ for $0 \leq t < T_{0}$ . The level set $S : = {a : φ_{T_{0}} (a) = x_{*}}$ is a nonempty compact interval $[α, β]$ . Put $a^{†} : = \min {\max {x_{*}, α}, β}$ . Then $a (t) \to a^{†}$ (in particular $a (t) \to a_{*}$ when $S = {a_{*}}$ ), $f^{'} (a (t)) \to - K$ , and

$\sup_{x \in R} ∣ u_{x} (t, x) ∣ \geq ∣ u_{x} (t, x_{*}) ∣ = \frac{∣ f^{'} (a (t)) ∣}{1 + t f^{'} (a (t))} ⟶ + \infty (t ↑ T_{0}),$

while $∣ u (t, x_{*}) ∣ = ∣ f (a (t)) ∣ \leq ∥ f ∥_{\infty}$ stays bounded. Thus the classical $C^{1}$ solution obtained by characteristics has a gradient blow-up at the finite time $T_{0}$ although the solution values and the characteristic locations remain finite; $C^{1}$ solvability up to time $T_{0}$ is not asserted.

Facts & Assumptions

Given: The bounded $f \in C^{2} (R)$ , the constant $K > 0$ with $f^{'} \geq - K$ and $f^{'} (a_{*}) = - K$ , and $T_{0} = 1 / K$ .

[F1]

For $a < b$ , $f (b) - f (a) = \int_{a}^{b} f^{'} (s) d s \geq - K (b - a)$ (The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ , The integral with oriented limits: $\int_{a}^{a} f : = 0$ and $\int_{b}^{a} f : = - \int_{a}^{b} f$ ).

[F2]

A continuous function on $R$ with limits $+ \infty$ and $- \infty$ at the two ends is surjective onto $R$ (Intermediate value theorem, by bisection with a canonical left-half rule: a continuous function on $[a, b]$ takes every value between $f (a)$ and $f (b)$ ); a $C^{1}$ injective map with nowhere-vanishing derivative is a $C^{1}$ diffeomorphism onto its image (An injective regular $C^{k}$ map is a $C^{k}$ diffeomorphism onto its image, $C^{k}$ Euclidean maps and diffeomorphisms).

[F3]

Chain rule in two variables and the algebra of derivatives (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ , A total derivative computes every directional derivative, and its matrix is the Jacobian, Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set).

[F4]

Mean value theorem: a differentiable real function with strictly positive derivative on an interval is strictly increasing there (The mean value theorem, as the case $g (x) = x$ of Cauchy's: for $f$ continuous on $[a, b]$ with $a < b$ and differentiable on $(a, b)$ there is $c \in (a, b)$ with $f (b) - f (a) = f^{'} (c) (b - a)$ ).

[F5]

Every bounded sequence of real numbers has a convergent subsequence (Bolzano-Weierstrass: every bounded real sequence has a convergent subsequence).

Proof

technique · direct; invert the characteristic map for $t<T_0$ and differentiate the implicit relation

1.1givenF1F2F3algebra

Fix $t \in [0, T_{0})$ . If $a < b$ and $φ_{t} (a) = φ_{t} (b)$ , then $b - a = - t (f (b) - f (a)) \leq t K (b - a) < b - a$ by [F1] and $t K < 1$ , a contradiction; hence $φ_{t}$ is injective. Moreover $φ_{t} (a) = a + t f (a) \to \pm \infty$ as $a \to \pm \infty$ because $f$ is bounded and $a \to \pm \infty$ ; by [F2] it is surjective. Finally $φ_{t}^{'} (a) = 1 + t f^{'} (a) \geq 1 - t K > 0$ , so [F2] makes $φ_{t}$ a $C^{1}$ diffeomorphism onto $R$ .

1.2givenF1F2algebra

Endpoint level set. Let $F_{0} (a) : = a + T_{0} f (a)$ . Its derivative is $1 + T_{0} f^{'} (a) \geq 0$ , so [F1] makes it nondecreasing. Since $f$ is bounded, $F_{0} (a) \to \pm \infty$ as $a \to \pm \infty$ . Its level set $S = F_{0}^{- 1} (x_{*})$ is therefore a nonempty compact interval $[α, β]$ , containing $a_{*}$ . If $α < β$ , then $F_{0}$ is constant on this interval, so $f^{'} = - K$ in its interior and, by continuity, at its endpoints. If $α = β = a_{*}$ the same derivative value follows from the hypothesis. Thus $f^{'} = - K$ on all of $S$ .

2.1step 1.1F3F4algebra

Fix $(t_{0}, x_{0})$ with $0 \leq t_{0} < T_{0}$ and put $a_{0} : = a (t_{0}, x_{0})$ . By step 1.1, $1 + t_{0} f^{'} (a_{0}) > 0$ . Continuity gives an open rectangle $Q$ around $(t_{0}, a_{0})$ on which $1 + t f^{'} (a) > 0$ . For each fixed $t$ in this rectangle, [F4] shows that $a \mapsto φ_{t} (a)$ is strictly increasing on its $a$ -interval. Thus $Φ (t, a) : = (t, φ_{t} (a))$ is injective on $Q$ , and its derivative has determinant $1 + t f^{'} (a) > 0$ . The injective-regular corollary therefore makes $Φ ∣_{Q}$ a $C^{1}$ diffeomorphism onto its open image. This also applies at $t_{0} = 0$ , because $Q$ may extend to small negative times and the determinant remains positive locally. On the portion with $t \geq 0$ , the local inverse agrees with the globally defined $a (t, x) = φ_{t}^{- 1} (x)$ by uniqueness of the inverse. Hence $a (t, x)$ is $C^{1}$ in a relative neighbourhood of every point of $[0, T_{0}) \times R$ , and differentiating the identity $φ_{t} (a (t, x)) = x$ is now legitimate: $\partial_{x} a (t, x) = \frac{1}{1 + t f^{'} (a (t, x))}, \partial_{t} a (t, x) = - \frac{f (a (t, x))}{1 + t f^{'} (a (t, x))} .$ By [F3], $u = f \circ a$ is $C^{1}$ jointly, $u (0, x) = f (x)$ , and $u_{t} = - f^{'} (a) \frac{f (a)}{1 + t f^{'} (a)}, u_{x} = \frac{f^{'} (a)}{1 + t f^{'} (a)}, u_{t} + u u_{x} = \frac{f^{'} (a)}{1 + t f^{'} (a)} (- f (a) + f (a)) = 0.$ This proves (i).

2.2step 1.1step 1.2givenF5algebra

Label limit. The identity $a (t) + t f (a (t)) = x_{*}$ and boundedness of $f$ keep $a (t)$ in a fixed compact interval. Every subsequential limit $a_{0}$ as $t \to T_{0}$ from below satisfies $F_{0} (a_{0}) = x_{*}$ , hence belongs to $S$ . For $a \in S$ , $f (a) = K (x_{*} - a)$ , so $φ_{t} (a) = (1 - t K) a + t K x_{*}$ . If $x_{*} < α$ , this gives $φ_{t} (α) > x_{*}$ and strict monotonicity of $φ_{t}$ from step 1.1 forces $a (t) < α$ ; its only possible limit in $S$ is $α$ . Similarly, if $x_{*} > β$ , $a (t) > β$ and its only possible limit in $S$ is $β$ . If $x_{*} \in S$ , then $f (x_{*}) = 0$ and $φ_{t} (x_{*}) = x_{*}$ , so uniqueness of the inverse gives $a (t) = x_{*}$ . Thus every subsequential limit as $t \to T_{0}$ from below equals $a^{†}$ . If $a (t)$ did not converge to $a^{†}$ , there would be $ε > 0$ and a sequence $t_{m} \to T_{0}$ from below with $∣ a (t_{m}) - a^{†} ∣ \geq ε$ . Boundedness and [F5] give a convergent subsequence of labels, contradicting the identified limit. Therefore $a (t) \to a^{†}$ , and by step 1.2 and continuity, $f^{'} (a (t)) \to - K$ .

3.1step 2.1F3algebra

Fix $a_{0} \in R$ and put $x (t) : = φ_{t} (a_{0}) = a_{0} + t f (a_{0})$ ; then $a (t, x (t)) = a_{0}$ , so $u (t, x (t)) = f (a_{0})$ is constant. By step 2.1, $q (t) = u_{x} (t, x (t)) = f^{'} (a_{0}) / (1 + t f^{'} (a_{0}))$ , and differentiating this quotient with [F3], $q^{'} (t) = - \frac{f^{'} (a_{0})^{2}}{(1 + t f^{'} (a_{0}))^{2}} = - q (t)^{2} .$ This proves (ii).

4.1step 1.1step 2.1step 3.1step 2.2givenalgebra∎

Fixed-spatial gradient blow-up. Step 2.1 gives $u_{x} (t, x_{*}) = f^{'} (a (t)) / (1 + t f^{'} (a (t)))$ . Its denominator is positive for $t < T_{0}$ by step 1.1 and tends to $1 - T_{0} K = 0$ , while its numerator tends to $- K \neq 0$ by step 2.2. Hence $∣ u_{x} (t, x_{*}) ∣ \to + \infty$ , and the supremum over $x$ diverges as well. Also $∣ u (t, x_{*}) ∣ = ∣ f (a (t)) ∣ \leq ∥ f ∥_{\infty}$ . Independently, along the specified characteristic $φ_{t} (a_{*})$ , step 3.1 gives $u_{x} (t, φ_{t} (a_{*})) = - K / (1 - t K)$ . Thus the fixed-spatial and material-characteristic gradient blow-up claims both hold even when the endpoint map has a flat level interval; no injectivity at $T_{0}$ was assumed.

Remarks

The hypothesis that the minimum of $f^{'}$ is attained at $a_{*}$ and equals $- K$ is used only to locate the blow-up; for merely $f^{'} \geq - K$ the solution still exists for $t < 1 / K$ , and the bound is sharp in general.
At $T_{0}$ the characteristic map may have a flat level interval; even when it is injective its inverse is no longer regular at the critical label. For $t > T_{0}$ its derivative is negative at $a_{*}$ , while boundedness of $f$ preserves its limits at infinity, so it is not injective. The classical $C^{1}$ construction stops at $T_{0}$ . The theorem constructs a classical $C^{1}$ solution on $[0, T_{0})$ with unbounded gradient; it does not claim that no weak continuation exists, and the weak solutions are treated separately.
The statement is purely mathematical; no physical identification of $u$ with a fluid velocity or of the blow-up with an observed phenomenon is made.

DefinitionDefinition: AI-adaptedProof: Not applicableOpen item page →

Weak and entropy solutions of the scalar conservation law $u_{t} + (u^{2} / 2)_{x} = 0$

Definition

Assume ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )) for the actual Lebesgue measure foundation used by the bounded Borel weak/entropy class.

Write $F (u) : = \frac{1}{2} u^{2}$ and let $λ_{2}$ be two-dimensional Lebesgue measure on $R^{2}$ (Lebesgue measurable sets, the family $L (R^{n})$ , and the restricted set function $λ_{n}$ ).

Every space-time expression written as $\int_{0}^{\infty} ⁣ ⁣ \int_{R} (\dots) d x d t$ below means the single signed Lebesgue integral over $[0, \infty) \times R$ with respect to $λ_{2}$ ; it is not an iterated-integral convention. The initial-data integral is with respect to one-dimensional Lebesgue measure $λ_{1}$ . In (W) and (E), each integrand is Borel measurable, bounded, and supported in the compact support of the test function (or its initial trace). A compact support is contained in a bounded rectangle of finite Lebesgue measure, by A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included; monotonicity and homogeneity of the nonnegative integral give $\int ∣ q ∣ \leq ∥ q ∥_{\infty} λ_{d} (K) < \infty$ for such an integrand $q$ supported in $K$ (Monotonicity and nonnegative homogeneity of the nonnegative integral). Thus all displayed signed integrals are defined as Lebesgue integrals of integrable functions (Integrable real and complex functions, and their integrals).

A test function is a function $φ : [0, \infty) \times R \to R$ of class $C^{1}$ up to the boundary: it is the restriction of a $C^{1}$ function on an open neighborhood of this closed half-space, and its support within the half-space is compact. No vanishing or continuation by zero for $t < 0$ is required. Thus the trace $φ (0, x)$ may be nonzero. Derivatives at $t = 0$ are the continuous one-sided derivatives. Write $C_{c}^{1} ([0, \infty) \times R)$ for this class; admissible means belonging to this class and nonnegative means $φ \geq 0$ on the half-space.

(a) Weak solutions. Let $f : R \to R$ be bounded and Borel measurable (Extended-real-valued measurable functions) and let $u : (0, \infty) \times R \to R$ be bounded and Borel measurable. Then $u$ is a weak solution of

$\begin{matrix} u_{t} + (F (u))_{x} = 0 with u (0, \cdot) = f & (CL) \end{matrix}$

when

$\begin{matrix} \int_{0}^{\infty} ⁣ ⁣ \int_{R} [u (t, x) φ_{t} (t, x) + F (u (t, x)) φ_{x} (t, x)] d x d t + \int_{R} f (x) φ (0, x) d x = 0 & (W) \end{matrix}$

for every test function $φ$ (Integral over a measurable subset, Integrable real and complex functions, and their integrals, The Lebesgue integral is linear on $L^{1} (μ)$ ). The integrals exist because $u, f$ are bounded and the test functions have compact support; the identity depends on $u$ and $f$ only through their almost-everywhere classes.

(b) Entropy pairs. An entropy pair is a pair $(η, q)$ of $C^{1}$ functions $η, q : R \to R$ (The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set) such that

$q^{'} (u) = η^{'} (u) F^{'} (u) = u η^{'} (u) for every u \in R;$

the function $η$ is then called an entropy and $q$ its entropy flux. The entropy is convex when $η ((1 - a) r + a s) \leq (1 - a) η (r) + a η (s)$ for all real $r, s$ and $a \in [0, 1]$ .

(c) Entropy solutions. A weak solution $u$ of (CL) is an entropy solution when, for every convex entropy pair $(η, q)$ with $q^{'} = u η^{'}$ , the distributional inequality

$(η (u))_{t} + (q (u))_{x} \leq 0$

holds, by definition meaning that

$\begin{matrix} \int_{0}^{\infty} ⁣ ⁣ \int_{R} [η (u) φ_{t} + q (u) φ_{x}] d x d t + \int_{R} η (f (x)) φ (0, x) d x \geq 0 & (E) \end{matrix}$

for every nonnegative test function $φ$ .

(d) Riemann data. For reals $u_{L}, u_{R}$ the Riemann initial data are

$f (x) : = {\begin{cases} u_{L}, & x < 0, \\ u_{R}, & x \geq 0. \end{cases}$

Remarks

The sign convention is fixed by (E): with $φ \geq 0$ the identity says that the distribution $η (u)_{t} + q (u)_{x}$ is a nonpositive measure. Under the opposite convention the inequality and the displayed integral would both flip sign; the two conventions are not mixed anywhere in this collection.
The distinguished family of entropies used in the examples is $η_{k} (u) : = ∣ u - k ∣$ , which is convex and only Lipschitz at $u = k$ , together with $q_{k} (u) = sgn (u - k) (u^{2} - k^{2}) / 2$ off $u = k$ and $q_{k} (k) = 0$ . Here $sgn (r) : = 1$ for $r > 0$ , $0$ for $r = 0$ , and $- 1$ for $r < 0$ . The identity $q_{k}^{'} = u η_{k}^{'}$ holds off $u = k$ , and the pair is applied to piecewise $C^{1}$ solutions for which the distributional computation is carried out piecewise. The definition of a general entropy solution in (c) is stated for $C^{1}$ entropies; the piecewise application to $η_{k}$ is part of the corresponding theorem and is checked there.
The definition asserts neither existence nor uniqueness of weak or entropy solutions, and no selection principle. Integrals are Lebesgue integrals over the stated domains and the test-function class is exactly as declared; evaluating a weak solution at $t = 0$ pointwise is not part of the definition, only the initial term in (W).

TheoremStatement: AI-adaptedProof: AI-adaptedOpen item page →

Riemann shock and rarefaction waves for $u_{t} + (u^{2} / 2)_{x} = 0$ are weak solutions, and the shock is an entropy solution

Statement

Assume ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )) for the Lebesgue weak formulation and the imported divergence theorem.

Let $F (u) = \frac{1}{2} u^{2}$ and use the weak and entropy formulations of Weak and entropy solutions of the scalar conservation law $u_{t} + (u^{2} / 2)_{x} = 0$ .

(a) Shock. Let $u_{L} > u_{R}$ , put $s : = (u_{L} + u_{R}) / 2$ and

$u (t, x) : = {\begin{cases} u_{L}, & x < s t, \\ u_{R}, & x \geq s t, \end{cases} (t > 0),$

with Riemann initial data $f$ equal to $u_{L}$ for $x < 0$ and $u_{R}$ for $x \geq 0$ . Values on the shock line or at the single initial jump point do not change the integral identities. Then $u$ is a weak solution of $u_{t} + (F (u))_{x} = 0$ with initial data $f$ .

(b) Entropy of the shock. For every $k \in R$ , set $η_{k} (u) : = ∣ u - k ∣$ and $q_{k} (u) : = sgn (u - k) (u^{2} - k^{2}) / 2$ for $u \neq k$ , with $q_{k} (k) : = 0$ . Then the entropy inequality (E) of Weak and entropy solutions of the scalar conservation law $u_{t} + (u^{2} / 2)_{x} = 0$ holds for the pair $(η_{k}, q_{k})$ , with interface coefficient

$s [η_{k}] - [q_{k}] = {\begin{cases} - (k - u_{L}) (k - u_{R}), & u_{R} \leq k \leq u_{L}, \\ 0, & k \notin [u_{R}, u_{L}], \end{cases} \geq 0,$

where $[h] : = h (u_{R}) - h (u_{L})$ . The shock also satisfies (E) for every convex $C^{1}$ entropy pair of the Definition, and is an entropy solution in its full declared sense.

(c) Rarefaction. Let $u_{L} < u_{R}$ and define

$u (t, x) : = {\begin{cases} u_{L}, & x \leq u_{L} t, \\ x / t, & u_{L} t < x < u_{R} t, \\ u_{R}, & x \geq u_{R} t, \end{cases} (t > 0) .$

Then $u$ is continuous on $(0, \infty) \times R$ , is a weak solution of $u_{t} + (F (u))_{x} = 0$ with the Riemann initial data, and satisfies the entropy inequality (E) for every convex $C^{1}$ entropy pair and for every pair $(η_{k}, q_{k})$ , $k \in R$ (indeed with equality), and is an entropy solution.

Facts & Assumptions

Given: Countable choice, the reals $u_{L} \neq u_{R}$ ; the constructions above.

[F1]

Interface computation. Let $s \in R$ and let $u$ be bounded, continuous (Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point) on the two open half-planes ${x - s t < 0}$ and ${x - s t > 0}$ with $C^{1}$ extensions to each closed half-plane, and suppose $u_{t} + (F (u))_{x} = 0$ classically on each open half-plane. Let $f$ be the initial trace. Then for every test function $φ$ ,

$\int_{0}^{\infty} ⁣ ⁣ \int_{R} [u φ_{t} + F (u) φ_{x}] d x d t + \int_{R} f (x) φ (0, x) d x = \int_{0}^{\infty} φ (t, s t) (s [u] - [F (u)]) d t,$

with $[h] : = h_{+} - h_{-}$ the jump from the left trace to the right trace across the line $x = s t$ . Proof. Put $J : = (u φ, F (u) φ)$ on $(0, \infty) \times R$ , and let $D_{\pm} : = {(t, x) : 0 < t < T, \pm (x - s t) > 0, ∣ x ∣ < R}$ for $T, R$ large enough that the support of $φ$ avoids the top and lateral boundaries; it may meet the initial boundary. These are bounded piecewise $C^{1}$ domains, and $div J = u φ_{t} + F (u) φ_{x}$ on each (the classical equation kills the derivative of $u$ ). The divergence theorem on a plane polygon (Divergence for finite piecewise C1 presentations, Specified finite piecewise C1 boundary presentations, Surface integration on compact C1 hypersurfaces) gives $\int \int_{D_{\pm}} div J = \int_{\partial D_{\pm}} J \cdot ν d S$ . The outward normals are $(- 1, 0)$ on the bottom segments ${t = 0}$ , $(0, - 1)$ or $(0, 1)$ on the far vertical sides (where $φ = 0$ ), $(s, - 1) / \sqrt{1 + s^{2}}$ on the interface from $D_{+}$ and $(- s, 1) / \sqrt{1 + s^{2}}$ from $D_{-}$ . Summing the two identities, the vertical sides contribute $0$ , the bottom segments contribute $- \int f φ (0, x) d x$ because $J \cdot (- 1, 0) = - u φ$ , and the interface contributes $\int_{0}^{\infty} φ (t, s t) (s [u] - [F (u)]) d t$ : indeed, with $d S = \sqrt{1 + s^{2}} d t$ , $J \cdot ν_{+} = [s u_{+} - F (u_{+})] φ / \sqrt{1 + s^{2}}$ and $J \cdot ν_{-} = [- s u_{-} + F (u_{-})] φ / \sqrt{1 + s^{2}}$ . Rearranging gives the displayed identity. ( $C^{k}$ Euclidean maps and diffeomorphisms, The Lebesgue integral is linear on $L^{1} (μ)$ , Integral over a measurable subset.)

[F2]

If a continuous function $h$ on $[0, \infty)$ satisfies $\int_{0}^{\infty} h (t) φ (t, s t) d t = 0$ for every test function $φ$ , then $h = 0$ : the restrictions $t \mapsto φ (t, s t)$ of test functions contain every compactly supported $C^{1}$ function of $t$ . [Given directly: for any compactly supported $C^{1}$ $ψ (t)$ take $φ (t, x) = ψ (t) χ (x - s t)$ with a bump $χ \in C_{c}^{1} (R)$ , $χ (0) = 1$ .]

[F3]

For $u (t, x) = x / t$ on the fan $u_{L} t < x < u_{R} t$ : $u_{t} = - x / t^{2}$ , $u_{x} = 1 / t$ , hence $u_{t} + u u_{x} = 0$ ; the constant branches satisfy the equation trivially. The pair $(η_{k}, q_{k})$ has $q_{k}^{'} = u η_{k}^{'}$ off $u = k$ ; on a strip $t \geq ε > 0$ , $u \neq k$ except on one line, and $η_{k} \circ u$ , $q_{k} \circ u$ are piecewise $C^{1}$ and Lipschitz, so the classical chain rule holds off that line and the distribution has no singular part there (Weak and entropy solutions of the scalar conservation law $u_{t} + (u^{2} / 2)_{x} = 0$ , $C^{k}$ Euclidean maps and diffeomorphisms, The Lebesgue integral is linear on $L^{1} (μ)$ ).

Proof

technique · direct; the two profiles are piecewise $C^1$ with a linear interface, so [F1] computes their weak residual, and the entropy residual is evaluated in closed form

1.1givenF1algebra

On each of the two open regions $x < s t$ and $x > s t$ the function $u$ of (a) is constant, so the equation holds classically there. The traces across $x = s t$ are $u_{-} = u_{L}$ (left) and $u_{+} = u_{R}$ (right), and the initial trace is the Riemann data $f$ . By [F1], the weak residual of (a) integrates $φ (t, s t)$ against $s [u] - [F (u)] = s (u_{R} - u_{L}) - \frac{1}{2} (u_{R}^{2} - u_{L}^{2}) = (u_{R} - u_{L}) (s - \frac{u_{L} + u_{R}}{2}) = 0.$ Hence the weak identity (W) holds for every test function, and $u$ is a weak solution.

1.2givenF1F3algebra

For (c), first integrate on $ε < t < T$ , splitting the truncated domain into the left constant region, the fan, and the right constant region, with $T$ and the spatial cutoff outside the support of the test function. There $u = x / t$ is $C^{1}$ on the closed fan and satisfies $u_{t} + u u_{x} = 0$ by [F3]; the constant pieces do too. Apply the divergence theorem to each piece. On a shared fan edge the outward normals are opposite and the traces of $u$ , $F (u)$ and $φ$ agree, so the two boundary integrals cancel. The bottom term is $- \int u (ε, x) φ (ε, x) d x$ . As $ε ↓ 0$ , $u (ε, x)$ differs from $f (x)$ only between the rays $x = u_{L} ε$ and $x = u_{R} ε$ , an interval of length $(u_{R} - u_{L}) ε$ , and it is uniformly bounded by $\max (∣ u_{L} ∣, ∣ u_{R} ∣)$ . Thus the difference in the bottom integral tends to zero; continuity of $φ$ on its compact support replaces $φ (ε, x)$ by $φ (0, x)$ . The integrands over $0 < t < ε$ are bounded on a fixed compact spatial support, so their integrals also tend to zero. Taking the limit gives (W) with the actual initial term.

2.1givenstep 1.1F1algebra

For (b), apply the same computation to the pair $(η_{k}, q_{k})$ in place of $(u, F (u))$ : on each open region the function $u$ is constant, so $(η_{k} (u))_{t} + (q_{k} (u))_{x} = 0$ classically there, and [F1] gives $\int_{0}^{\infty} ⁣ ⁣ \int_{R} [η_{k} φ_{t} + q_{k} φ_{x}] + \int_{R} η_{k} (f) φ (0) = \int_{0}^{\infty} φ (t, s t) (s [η_{k}] - [q_{k}]) d t .$ For $u_{R} < k < u_{L}$ a direct evaluation gives $[η_{k}] = 2 k - u_{L} - u_{R}$ and $[q_{k}] = (2 k^{2} - u_{L}^{2} - u_{R}^{2}) / 2$ , whence $s [η_{k}] - [q_{k}] = - (k - u_{L}) (k - u_{R}) \geq 0$ . For $k \geq u_{L}$ or $k \leq u_{R}$ the same expressions give $s [η_{k}] - [q_{k}] = 0$ . In every case the right-hand side is $\geq 0$ for nonnegative test functions, which is (E).

3.1step 1.1step 2.1F1algebra

Let $(η, q)$ be any convex $C^{1}$ entropy pair. On either side of the shock the entropy residual vanishes classically, and [F1] applied to $(η (u), q (u))$ gives the residual coefficient $s [η] - [q] = \int_{u_{R}}^{u_{L}} (z - s) η^{'} (z) d z = \int_{u_{R}}^{u_{L}} (z - s) (η^{'} (z) - η^{'} (s)) d z \geq 0.$ The middle equality uses $s = (u_{L} + u_{R}) / 2$ , hence $\int_{u_{R}}^{u_{L}} (z - s) d z = 0$ . For any $a < b$ , convexity gives $η (a + h) \leq (1 - h / (b - a)) η (a) + (h / (b - a)) η (b)$ for $0 < h < b - a$ ; dividing by $h$ after subtracting $η (a)$ and taking $h ↓ 0$ yields $η^{'} (a) \leq (η (b) - η (a)) / (b - a)$ . Applying convexity to $b - h$ and taking the same limit gives $(η (b) - η (a)) / (b - a) \leq η^{'} (b)$ . Thus $η^{'}$ is nondecreasing, so $(z - s) (η^{'} (z) - η^{'} (s)) \geq 0$ for every $z$ . This proves (E) for every declared convex $C^{1}$ pair, in addition to the Kružkov computation of step 2.1.

4.1step 1.2step 3.1F1F3algebra∎

On the rarefaction fan every $C^{1}$ entropy pair has $\partial_{t} η (u) + \partial_{x} q (u) = η^{'} (u) (u_{t} + u u_{x}) = 0$ ; its traces are continuous across the fan edges. For a Kružkov pair split additionally along $x = k t$ when that line lies inside the fan; the two entropy and flux traces agree there too. Apply the truncated-domain argument of step 1.2 to these entropy fields. Since $u$ ranges in the compact interval $[u_{L}, u_{R}]$ , $η (u)$ and $q (u)$ are bounded and the initial difference is supported on an interval of length $(u_{R} - u_{L}) ε$ ; the same limit establishes (E) with equality and its nonzero initial trace for every stated pair. Both profiles are therefore entropy solutions for the full class of the Definition, with the asserted additional Kružkov identities.

Remarks

Parts (a) and (c) construct two explicit weak solutions of the same conservation law with different jump behaviour: the shock is discontinuous and satisfies the entropy inequalities of (b), while the rarefaction is continuous and satisfies them with equality. The theorem does not claim uniqueness or that all weak solutions are entropy solutions; a general existence or contraction theory is not part of this statement.
The entropy family $η_{k}$ is only Lipschitz at $u = k$ ; the computation is carried out at the level of continuous piecewise $C^{1}$ functions, for which the interface formula [F1] applies verbatim. The Riemann initial data are bounded and Borel, so the weak formulation is in force.

TheoremStatement: AI-adaptedProof: AI-adaptedOpen item page →

Exact plane Poiseuille and Couette flows of the incompressible Newtonian model, with flux and ansatz uniqueness

Statement

The flux integrals below are ordinary Riemann integrals of continuous profiles on $[0, h]$ .

Let $I \subseteq R$ be a nonempty open time interval, let $ρ > 0$ , $η > 0$ and $h > 0$ be constants, and let $O : = R \times (0, h) \times R$ with spatial coordinates $(x, y, z)$ . Let $v \in C ([0, h]) \cap C^{2} ((0, h))$ , with endpoint values given by its continuous extension, and $p_{s p} \in C^{1} (O)$ . Define stationary spacetime fields $u : I \times O \to R^{3}$ and $p : I \times O \to R$ by $u (t, x, y, z) : = (v (y), 0, 0)$ and $p (t, x, y, z) : = p_{s p} (x, y, z)$ . Here $D_{x} u$ denotes the spatial total derivative at fixed $t$ , $(u \cdot \nabla) u : = D_{x} u u$ , and $Δ u : = (\sum_{j = 1}^{3} \partial_{j}^{2} u_{i})_{i = 1}^{3}$ acts componentwise (Euclidean spatial calculus: gradient, divergence and material derivative). The incompressible Newtonian equations of Incompressible Newtonian and barotropic inviscid fluid constitutive models with no body force and steady data read

$\begin{matrix} div u = 0, ρ (u \cdot \nabla) u = - \nabla p + η Δ u, & (NS) \end{matrix}$

where the second equation is the adopted stationary momentum equation with $σ = - p I_{3} + η (D_{x} u + (D_{x} u)^{T})$ .

For this item, “irrotational” means that the spatial velocity gradient $D_{x} u$ is symmetric.

(a) Plane Poiseuille flow. Let $G \in R$ , $p_{0} \in R$ and

$v (y) : = \frac{G y (h - y)}{2 η}, p_{s p} (x, y, z) : = p_{0} - G x .$

Then $u$ is $C^{\infty}$ on $I \times O$ , satisfies (NS) there, and $v (0) = v (h) = 0$ . Its flux per unit width is

$\int_{0}^{h} v (y) d y = \frac{G h^{3}}{12 η} .$

Moreover, within the ansatz $u = (v (y), 0, 0)$ , $p = p_{0} - G x$ with the same constant $G$ and the no-slip conditions $v (0) = v (h) = 0$ , this is the only solution: any two such profiles differ by a linear function vanishing at both endpoints, hence coincide.

(b) Plane Couette flow. Let $V \in R$ , let $v (y) : = V y / h$ , and let $p_{s p}$ be constant. Then the stationary $u, p$ defined above satisfy (NS) on $I \times O$ and $v (0) = 0$ , $v (h) = V$ . If $V \neq 0$ , the velocity gradient $D_{x} u$ is not symmetric, so the flow is not irrotational by the definition above.

Facts & Assumptions

Given: The explicitly adopted steady system (NS), constants $ρ, η, h > 0$ , $G, p_{0}, V \in R$ , and the profiles above.

[F1]

Adopted equation input: (NS) is the differential system being solved here, not a conclusion drawn from an unspecified integral balance. To obtain (NS) from Local momentum balance: $ρ D_{t} u = ρ b + div σ$ from the integral balance on admissible material subregions, one must additionally impose its continuity hypothesis (K), integral momentum hypothesis (L), admissible domains and regularity. For the present exact-solution computation (NS) itself is adopted. Constitutive substitution: for the branch (a) of Incompressible Newtonian and barotropic inviscid fluid constitutive models, $σ = - p I_{3} + η (D_{x} u + (D_{x} u)^{T})$ and $div u = 0$ ; the adopted momentum equation is $ρ D_{t} u = div σ$ in the absence of body force, and for stationary data $D_{t} u = D_{x} u u$ . For $u = (v (y), 0, 0)$ , $div u = \partial_{x} v = 0$ , and $div (D_{x} u + (D_{x} u)^{T}) = Δ u$ because $div u = 0$ ; hence $div σ = - \nabla p + η Δ u$ (Euclidean spatial calculus: gradient, divergence and material derivative, The trace $tr (A)$ as the sum of the diagonal entries, Rectangular matrix multiplication and the identity matrix $I_{n}$ , including zero-sized shapes).

[F2]

For $u = (v (y), 0, 0)$ with $v$ independent of $x$ , $(u \cdot \nabla) u = D_{x} u u = 0$ ; the only nonzero entry of $D_{x} u$ is $\partial_{y} u_{1} = v^{'} (y)$ , and $Δ u = (v^{''} (y), 0, 0)$ (A total derivative computes every directional derivative, and its matrix is the Jacobian, Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ ).

[F3]

If $w \in C ([0, h]) \cap C^{2} ((0, h))$ and $w^{''} = 0$ on $(0, h)$ , then $w$ is affine on $(0, h)$ : the fundamental theorem applied on each compact subinterval shows $w^{'}$ is constant there, and overlapping subintervals give one constant throughout $(0, h)$ . Continuity extends that affine formula to $[0, h]$ , so $w (0) = w (h) = 0$ implies $w = 0$ (The first fundamental theorem: if $f$ is integrable on $[a, b]$ and continuous at $c$ , then $F^{'} (c) = f (c)$ ; in particular a continuous $f$ has $F$ as a primitive, The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ , The integral with oriented limits: $\int_{a}^{a} f : = 0$ and $\int_{b}^{a} f : = - \int_{a}^{b} f$ , For $a < c < b$ : $f$ is integrable on $[a, b]$ if and only if it is integrable on $[a, c]$ and on $[c, b]$ , and then $\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{b} f$ ; with the oriented form for arbitrary $a, b, c$ , Integrable functions on $[a, b]$ form a set closed under sums and scalar multiples, and $\int_{a}^{b} (λ f + μ g) = λ \int_{a}^{b} f + μ \int_{a}^{b} g$ , A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion, The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set).

Proof

technique · direct; solve the reduced ordinary differential equation and evaluate the flux

1.1givenF1F2algebra

With $u = (v (y), 0, 0)$ , [F2] gives $(u \cdot \nabla) u = 0$ and $Δ u = (v^{''}, 0, 0)$ ; also $- \nabla p = (G, 0, 0)$ for $p = p_{0} - G x$ . Hence (NS) reduces to $0 = G + η v^{''} (y)$ and $\partial_{y} p = 0$ , i.e. $v^{''} (y) = - \frac{G}{η}, p = p_{0} - G x .$ For part (b) with $p$ constant the first equation becomes $v^{''} = 0$ .

2.1step 1.1F3algebra

(Poiseuille.) Apply [F3] to $w (y) : = v (y) + \frac{G}{2 η} y^{2}$ , which satisfies $w^{''} = 0$ on $(0, h)$ and is continuous on $[0, h]$ . Thus $w (y) = A y + B$ ; the conditions $v (0) = v (h) = 0$ give $B = 0$ and $A = \frac{G h}{2 η}$ , so $v (y) = \frac{G y (h - y)}{2 η}$ , the stated profile; it is $C^{\infty}$ and satisfies both equations of step 1.1. Computing the flux, $\int_{0}^{h} v = \frac{G}{2 η} (h \int_{0}^{h} y d y - \int_{0}^{h} y^{2} d y) = \frac{G}{2 η} (\frac{h^{3}}{2} - \frac{h^{3}}{3}) = \frac{G h^{3}}{12 η} .$ If two profiles $v_{1}, v_{2}$ both solve $v^{''} = - G / η$ with vanishing endpoints, their difference $w$ has $w^{''} = 0$ and $w (0) = w (h) = 0$ ; by [F3], $w = 0$ .

3.1step 1.1F2algebra∎

(Couette.) For $v (y) = V y / h$ we have $v^{''} = 0$ , so step 1.1 is satisfied with constant $p$ ; the boundary values are $v (0) = 0$ and $v (h) = V$ . Finally $D_{x} u$ has the single nonzero entry $\partial_{y} u_{1} = V / h$ in row $1$ , column $2$ , while $\partial_{x} u_{2} = 0$ , so $D_{x} u \neq (D_{x} u)^{T}$ when $V \neq 0$ : the antisymmetric part is nonzero and the flow is not irrotational by the definition above.

Remarks

The no-slip boundary conditions are adopted model conditions, not consequences of the equations; the uniqueness statement is only within the stated steady parallel ansatz with the same driving constant $G$ . General nonlinear initial-value problems, finite-Reynolds stability, entrance effects and three-dimensional perturbations are not treated.
The Couette computation exhibits explicitly that a laminar shear flow has nonzero antisymmetric velocity gradient; this is why the irrotational Bernoulli branch of the barotropic item does not apply to it. The two flows here are exact solutions of the adopted constant-coefficient model and carry no empirical claim about any real fluid.

DefinitionDefinition: AI-adaptedProof: Not applicableOpen item page →

Sine coefficients, sine series and the odd periodic extension

Definition

Let $L > 0$ . A function $F : R \to R$ is $L$ -periodic when $F (x + L) = F (x)$ for all $x \in R$ , and odd when $F (- x) = - F (x)$ for all $x$ .

(a) Sine coefficients. Let $F : [0, L] \to R$ be continuous (Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point); a function on a larger domain is used through its restriction to $[0, L]$ , with no periodicity required for the coefficient definition. In particular both $L$ -periodic and odd $2 L$ -periodic functions are admissible. Then $\int_{0}^{L} ∣ F ∣ < \infty$ (A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion). For every integer $n \geq 1$ the $n$ -th sine coefficient of $F$ is

$b_{n} (F) : = \frac{2}{L} \int_{0}^{L} F (x) \sin ⁣ (\frac{n π x}{L}) d x \in R,$

an ordinary Riemann integral of a continuous function on a compact interval (The integral with oriented limits: $\int_{a}^{a} f : = 0$ and $\int_{b}^{a} f : = - \int_{a}^{b} f$ , Sine and cosine defined by their real power series). The sine series of $F$ is the series of functions

$\sum_{n \geq 1} b_{n} (F) \sin ⁣ (\frac{n π x}{L}),$

whose $N$ -th partial sum is $S_{N}^{s} F (x) : = \sum_{n \leq N} b_{n} (F) \sin (n π x / L)$ .

(b) Odd periodic extension. Let $F : [0, L] \to R$ be continuous with $F (0) = F (L) = 0$ . The odd $2 L$ -periodic extension of $F$ is the function on $R$ defined by

$F^{o d d} (x) : = {\begin{cases} F (x), & 0 \leq x \leq L, \\ - F (- x), & - L \leq x \leq 0, \end{cases} F^{o d d} (x + 2 L) : = F^{o d d} (x) .$

Equivalently, $F^{o d d}$ is the unique $2 L$ -periodic odd function whose restriction to $[0, L]$ is $F$ (the two definitions agree at $0$ and $L$ because both values vanish; the extension is well defined and continuous).

(c) Given a specified wave speed $c > 0$ , the angular frequency of mode $n$ is $ω_{n} : = c n π / L$ ; its fundamental ( $n = 1$ ) value is $c π / L$ .

Remarks

All coefficients are ordinary integrals of continuous functions on compact intervals; no choice principle is needed, and no convergence of the sine series is asserted by the definition. Sine coefficients are real because $F$ and the sine functions are real-valued.
The normalization in $b_{n}$ is the coefficient normalization for the orthogonal family ${\sin (n π x / L)}_{n \geq 1}$ with squared norm $L / 2$ : $\int_{0}^{L} \sin (n π x / L) \sin (k π x / L) d x = (L / 2) δ_{n k}$ , the factor $(2 / L)$ being the inverse of the squared norm $L / 2$ ; the orthogonality is proved in the companion lemma, not assumed here.
A function that is odd and $2 L$ -periodic automatically satisfies $F (0) = F (L) = 0$ by oddness and periodicity; the extension in (b) is a construction, not a claim that $F$ already has these symmetries.

LemmaStatement: AI-adaptedProof: AI-adaptedOpen item page →

Sine orthogonality on a half period and the full-period integrals of odd $2 L$ -periodic functions

Statement

All integrals in this statement and proof are ordinary Riemann integrals of continuous functions on compact intervals, as in Sine coefficients, sine series and the odd periodic extension.

Let $L > 0$ . Then:

(a) For all integers $n, k \geq 1$ ,

$\int_{0}^{L} \sin (\frac{n π x}{L}) \sin (\frac{k π x}{L}) d x = \frac{L}{2} δ_{n k} .$

(b) Let $W : R \to R$ be continuous (Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point), odd and $2 L$ -periodic. Then

$\int_{0}^{2 L} W (x) d x = 0, \int_{0}^{2 L} W (x) \cos (\frac{k π x}{L}) d x = 0 for every integer k,$

and, for every integer $k \geq 1$ ,

$\frac{1}{2 L} \int_{0}^{2 L} W (x) \sin (\frac{k π x}{L}) d x = \frac{1}{L} \int_{0}^{L} W (x) \sin (\frac{k π x}{L}) d x .$

Facts & Assumptions

Given: The period $L > 0$ ; the integers $n, k$ in (a); the continuous odd $2 L$ -periodic $W$ in (b).

[F1]

Product-to-sum identities and values: $2 \sin a \sin b = \cos (a - b) - \cos (a + b)$ , $\sin (m π) = 0$ and $\cos (m π) = (- 1)^{m}$ for integers $m$ , and $\sin 0 = 0$ , $\cos 0 = 1$ (The addition formulas for sine and cosine, The zero sets of sine and cosine and the least positive common period 2 pi, Parity and the Pythagorean identity for sine and cosine, Sine and cosine defined by their real power series).

[F2]

The sine and cosine integrands in this proof are continuous on each compact interval, hence Riemann integrable (A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion). For $m \neq 0$ , $\int_{0}^{L} \cos (m π x / L) d x = (L / (m π)) [\sin (m π) - 0] = 0$ , and for $m = 0$ the integral is $L$ ; the values follow from the fundamental theorem (The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ , The integral with oriented limits: $\int_{a}^{a} f : = 0$ and $\int_{b}^{a} f : = - \int_{a}^{b} f$ , Integrable functions on $[a, b]$ form a set closed under sums and scalar multiples, and $\int_{a}^{b} (λ f + μ g) = λ \int_{a}^{b} f + μ \int_{a}^{b} g$ ).

[F3]

Additivity and change of variable on a period: for a continuous $2 L$ -periodic $g$ , $\int_{0}^{2 L} g = \int_{0}^{L} g + \int_{L}^{2 L} g$ , and substituting $x = 2 L - u$ in the second integral gives $\int_{L}^{2 L} g (x) d x = \int_{0}^{L} g (2 L - u) d u = \int_{0}^{L} g (- u) d u$ by periodicity (For $a < c < b$ : $f$ is integrable on $[a, b]$ if and only if it is integrable on $[a, c]$ and on $[c, b]$ , and then $\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{b} f$ ; with the oriented form for arbitrary $a, b, c$ , The integral with oriented limits: $\int_{a}^{a} f : = 0$ and $\int_{b}^{a} f : = - \int_{a}^{b} f$ , Substitution: if $φ$ is differentiable on $[c, d]$ with $φ^{'}$ integrable and $f$ is continuous on an interval containing $φ ([c, d])$ , then $\int_{φ (c)}^{φ (d)} f = \int_{c}^{d} (f \circ φ) φ^{'}$ , The first fundamental theorem: if $f$ is integrable on $[a, b]$ and continuous at $c$ , then $F^{'} (c) = f (c)$ ; in particular a continuous $f$ has $F$ as a primitive).

Proof

technique · direct; product-to-sum and elementary substitutions on a period

1.1givenF1F2algebra

For $n, k \geq 1$ , [F1] gives $2 \sin (n π x / L) \sin (k π x / L) = \cos ((n - k) π x / L) - \cos ((n + k) π x / L)$ . Integrating over $[0, L]$ and using [F2]: if $n \neq k$ both frequencies are nonzero, so the integral is $0$ ; if $n = k$ the first term contributes $L$ and the second (frequency $2 n \neq 0$ ) contributes $0$ , so the integral is $L / 2$ . This proves (a).

1.2givenF3algebra

Since $W$ is odd and $2 L$ -periodic, [F3] gives $\int_{0}^{2 L} W = \int_{0}^{L} W + \int_{0}^{L} W (- u) d u = \int_{0}^{L} W - \int_{0}^{L} W = 0$ .

1.3givenF1F3algebra

Likewise, $\int_{0}^{2 L} W (x) \cos (k π x / L) d x = \int_{0}^{L} W (x) \cos (k π x / L) d x + \int_{0}^{L} W (- u) \cos (k π (2 L - u) / L) d u$ . Now $W (- u) = - W (u)$ and $\cos (k π (2 L - u) / L) = \cos (2 k π - k π u / L) = \cos (k π u / L)$ ; the two integrals are equal and opposite, so the total is $0$ .

2.1givenF1F3algebra∎

For the sine integral, split at $L$ : $\int_{0}^{2 L} W (x) \sin (k π x / L) d x = \int_{0}^{L} (\dots) d x + \int_{0}^{L} W (- u) \sin (k π (2 L - u) / L) d u$ . Here $\sin (k π (2 L - u) / L) = \sin (2 k π - k π u / L) = - \sin (k π u / L)$ , and $W (- u) = - W (u)$ ; the second integral equals $\int_{0}^{L} W (u) \sin (k π u / L) d u$ . Hence the full-period integral is twice the half-period integral, which is the displayed identity.

Remarks

The sine functions $\sin (n π x / L)$ , $n \geq 1$ , are the eigenfunctions of the fixed-end problem in the elementary sense of satisfying the two endpoint conditions and the second-derivative eigenequation; part (a) records only their mutual orthogonality, not completeness, which is proved separately with Fejer means.
The identities of part (b) are the reason an odd $2 L$ -periodic function has no cosine component in its complex Fourier expansion; the statement is computational and uses no expansion theory.

LemmaStatement: AI-adaptedProof: AI-adaptedOpen item page →

Fejer completeness: a continuous one-periodic function with all integrals against the characters vanishing is zero

Statement

Let $h : R \to C$ be one-periodic, meaning $h (t + 1) = h (t)$ for all real $t$ , and continuous, meaning that both real functions $Re h, Im h : R \to R$ satisfy Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point and suppose that

$\int_{0}^{1} h (t) e_{k} (- t) d t = 0 for every k \in Z,$

All integrals here are componentwise Riemann integrals: for continuous $p = p_{1} + i p_{2}$ , $\int_{a}^{b} p : = \int_{a}^{b} p_{1} + i \int_{a}^{b} p_{2}$ , with the real oriented convention of The integral with oriented limits: $\int_{a}^{a} f : = 0$ and $\int_{b}^{a} f : = - \int_{a}^{b} f$ ; both integrals exist by A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion. Here $e_{k} (t) : = e^{2 π i k t}$ (The complex exponential by its power series). Then $h = 0$ on $R$ .

Facts & Assumptions

Given: The continuous one-periodic $h$ and the vanishing of all the displayed integrals.

[F1]

The Fejer kernel is $F_{N} (t) = \frac{1}{N + 1} ∣ \sum_{j = 0}^{N} e_{j} (t) ∣^{2}$ and equals $\sum_{∣ k ∣ \leq N} (1 - \frac{∣ k ∣}{N + 1}) e_{k} (t)$ ; it is nonnegative with $\int_{0}^{1} F_{N} = 1$ (The Fejer kernel is a positive approximate identity, Cesaro and Abel means of a Fourier series).

[F2]

The Fejer means $σ_{N} h = h * F_{N}$ converge to $h$ uniformly: $\sup_{x} ∣ σ_{N} h (x) - h (x) ∣ \to 0$ (Cesaro and Abel means of a Fourier series, Fejer means converge uniformly for continuous periodic functions).

[F3]

For continuous one-periodic complex $φ$ , affine substitution applied to each real component gives $\int_{0}^{1} φ (x - t) d t = \int_{x - 1}^{x} φ (u) d u = \int_{0}^{1} φ (u) d u$ . For the second equality, write $x = m + r$ with $m \in Z$ , $0 \leq r < 1$ ; translate by $m$ , split $[r - 1, r]$ at $0$ , and translate $[r - 1, 0]$ by $1$ to $[r, 1]$ , using periodicity and additivity (Substitution: if $φ$ is differentiable on $[c, d]$ with $φ^{'}$ integrable and $f$ is continuous on an interval containing $φ ([c, d])$ , then $\int_{φ (c)}^{φ (d)} f = \int_{c}^{d} (f \circ φ) φ^{'}$ , The integral with oriented limits: $\int_{a}^{a} f : = 0$ and $\int_{b}^{a} f : = - \int_{a}^{b} f$ , For $a < c < b$ : $f$ is integrable on $[a, b]$ if and only if it is integrable on $[a, c]$ and on $[c, b]$ , and then $\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{b} f$ ; with the oriented form for arbitrary $a, b, c$ ).

[F4]

Character periodicity: $\exp (i π) = - 1$ by $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ , hence $\exp (2 π i) = 1$ by exponential addition. Iteration gives $\exp (2 π i k) = 1$ for $k \geq 0$ ; for $k < 0$ , $\exp (2 π i k) \exp (- 2 π i k) = \exp (0) = 1$ gives the same conclusion. Thus $e_{k} (t + 1) = e_{k} (t)$ . Exponential addition: $e_{k} (x - t) = e_{k} (x) e_{k} (- t) = e_{k} (x) e_{- k} (t)$ ( $\exp (z + w) = \exp z \exp w$ , and the complex exponential extends the real exponential, The complex exponential by its power series).

[F5]

Continuous real components are Riemann integrable on compact intervals. The componentwise convention makes the integral complex-linear: expand complex multiplication into its two real components and apply real linearity and additivity (A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion, Integrable functions on $[a, b]$ form a set closed under sums and scalar multiples, and $\int_{a}^{b} (λ f + μ g) = λ \int_{a}^{b} f + μ \int_{a}^{b} g$ , For $a < c < b$ : $f$ is integrable on $[a, b]$ if and only if it is integrable on $[a, c]$ and on $[c, b]$ , and then $\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{b} f$ ; with the oriented form for arbitrary $a, b, c$ ).

Proof

technique · direct; expand the Fejer convolution with the finite trigonometric polynomial kernel and use the vanishing integrals

1.1givenF1F5algebra

Fix $N \geq 0$ . Since $σ_{N} h = h * F_{N} = \int_{0}^{1} h (x - t) F_{N} (t) d t$ and $F_{N}$ is the finite sum of [F1], [F5] gives $σ_{N} h (x) = \sum_{∣ k ∣ \leq N} (1 - \frac{∣ k ∣}{N + 1}) \int_{0}^{1} h (x - t) e_{k} (t) d t .$

1.2givenF3F4algebra

For each $k$ , substitute $t = x - u$ (the integrand $h (x - t) e_{k} (t)$ is one-periodic in $t$ ): by [F3] and [F4], $\int_{0}^{1} h (x - t) e_{k} (t) d t = \int_{0}^{1} h (u) e_{k} (x - u) d u = e_{k} (x) \int_{0}^{1} h (u) e_{- k} (u) d u = 0,$ the last equality by the hypothesis of the lemma.

2.1step 1.1step 1.2F2algebra∎

Steps 1.1 and 1.2 give $σ_{N} h (x) = 0$ for every $x \in R$ and every $N \geq 0$ . By [F2], $h (x) = \lim_{N} σ_{N} h (x) = 0$ for every $x$ . This proves the lemma.

Remarks

All functions here are continuous on the compact torus, so every integral displayed is an ordinary integral of a continuous periodic function, and the instances of the Fourier-mean machinery used are the choice-free ones noted in Cesaro and Abel means of a Fourier series; no selection principle is invoked in the argument as applied to continuous $h$ .

LemmaStatement: AI-adaptedProof: AI-adaptedOpen item page →

Uniform convergence of a $C^{2}$ sequence with its first and second derivatives on a rectangle passes $C^{2}$ regularity and termwise differentiation to the limit

Statement

Let $T, L > 0$ and $R : = [0, T] \times [0, L]$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length). For each $k \in N$ let $u_{k}$ be of class $C^{2}$ on an open neighbourhood of $R$ ( $C^{k}$ Euclidean maps and diffeomorphisms, Directional derivatives and partial derivatives of a map $U \subseteq R^{m} \to R^{n}$ ). Suppose that on $R$ :

$u_{k} \to u, \partial_{t} u_{k} \to v, \partial_{x} u_{k} \to w, \partial_{t}^{2} u_{k} \to v_{2}, \partial_{t} \partial_{x} u_{k} \to z, \partial_{x}^{2} u_{k} \to w_{2}$

pointwise for $u$ and uniformly for the five derivative sequences (Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point). Then $u$ is $C^{2}$ on the interior of $R$ , all listed limits are continuous on $R$ , and on the interior

$\partial_{t} u = v, \partial_{x} u = w, \partial_{t}^{2} u = v_{2}, \partial_{t} \partial_{x} u = \partial_{x} \partial_{t} u = z, \partial_{x}^{2} u = w_{2},$

the derivatives extending continuously to $R$ .

Facts & Assumptions

Given: The rectangle $R$ , the $C^{2}$ sequence $u_{k}$ and the stated convergences.

[F1]

One-variable theorem: if $f_{k} : [a, b] \to R$ are continuously differentiable, $f_{k} (c) \to A$ at one point and $f_{k}^{'} \to g$ uniformly on $[a, b]$ , then $f_{k} \to f$ uniformly for a differentiable $f$ with $f^{'} = g$ (If continuously differentiable functions converge at one point and their derivatives converge uniformly on a closed interval, then the functions converge uniformly to a differentiable function whose derivative is the derivative limit).

[F2]

A uniform limit of continuous functions is continuous (The uniform limit of continuous real-valued functions on a metric space is continuous, Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point).

[F3]

$C^{2}$ on an open set means: the first and second partial derivatives exist and are continuous; if the first partial derivatives are continuous, the map is totally differentiable ( $C^{k}$ Euclidean maps and diffeomorphisms, If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative, A total derivative computes every directional derivative, and its matrix is the Jacobian).

[F4]

If $g$ is $C^{2}$ on an open neighbourhood of a point $(t, x)$ , then its two mixed partials agree there. For small $h, k > 0$ , the rectangular increment $g (t + h, x + k) - g (t + h, x) - g (t, x + k) + g (t, x)$ , divided by $h k$ , is both the average of $\partial_{x} \partial_{t} g$ and the average of $\partial_{t} \partial_{x} g$ over the rectangle, by applying the one-variable fundamental theorem in the two orders; continuity of both second partials makes the respective averages tend to their values at $(t, x)$ as $h, k ↓ 0$ (The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ , A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion).

Proof

technique · direct; apply the one-variable uniform-differentiation theorem in each variable separately

1.1givenF1

Fix $x \in [0, L]$ and apply [F1] to the sequence $t \mapsto u_{k} (t, x)$ on $[0, T]$ with $c = 0$ : $u_{k} (0, x) \to u (0, x)$ and $\partial_{t} u_{k} (\cdot, x) \to v (\cdot, x)$ uniformly on $[0, T]$ (the convergence of $\partial_{t} u_{k}$ is uniform on all of $R$ , hence on the slice). We obtain: $t \mapsto u (t, x)$ is differentiable on $[0, T]$ with $\partial_{t} u (t, x) = v (t, x)$ , and $u_{k} (\cdot, x) \to u (\cdot, x)$ uniformly on $[0, T]$ . Applying the same statement to $\partial_{t} u_{k} (\cdot, x)$ with its $t$ -derivative $\partial_{t}^{2} u_{k} (\cdot, x) \to v_{2} (\cdot, x)$ , $t \mapsto v (t, x)$ is differentiable with $\partial_{t} v (t, x) = v_{2} (t, x)$ .

1.2givenF1F4

Fix $t \in [0, T]$ and apply [F1] in the variable $x$ to $x \mapsto u_{k} (t, x)$ and to $x \mapsto \partial_{x} u_{k} (t, x)$ : $x \mapsto u (t, x)$ has $\partial_{x} u (t, x) = w (t, x)$ , and $x \mapsto w (t, x)$ has $\partial_{x} w (t, x) = w_{2} (t, x)$ . Applying [F1] in $x$ to $x \mapsto \partial_{t} u_{k} (t, x)$ is legitimate because its $x$ -derivative equals $\partial_{t} \partial_{x} u_{k}$ by [F4]; the limits are $v (t, x)$ and $z (t, x)$ , so $\partial_{x} v (t, x) = z (t, x)$ . Applying [F1] in $t$ to the sequence $t \mapsto \partial_{x} u_{k} (t, x)$ , whose values converge to $w (t, x)$ and whose derivatives converge uniformly to $z (t, x)$ , gives $\partial_{t} w (t, x) = z (t, x)$ .

2.1step 1.1step 1.2F2F3F4∎

The functions $v, w, v_{2}, z, w_{2}$ are continuous on $R$ as uniform limits of continuous functions [F2]. Fix $(t_{0}, x_{0}) \in R$ . By continuity, $v$ and $w$ are bounded on a relative neighbourhood of $(t_{0}, x_{0})$ . The integral-evaluation theorem The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ , with continuous slice derivatives integrable by A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion and reversed endpoints handled by The integral with oriented limits: $\int_{a}^{a} f : = 0$ and $\int_{b}^{a} f : = - \int_{a}^{b} f$ , and steps 1.1-1.2 give, for $(t, x)$ in that neighbourhood, $u (t, x) - u (t_{0}, x_{0}) = \int_{x_{0}}^{x} w (t_{0}, s) d s + \int_{t_{0}}^{t} v (s, x) d s .$ Thus $∣ u (t, x) - u (t_{0}, x_{0}) ∣ \leq M_{w} ∣ x - x_{0} ∣ + M_{v} ∣ t - t_{0} ∣$ for local bounds $M_{w}, M_{v}$ , proving continuity of $u$ on $R$ , including its boundary. Steps 1.1-1.2 identify the first and second partial derivatives of $u$ with these continuous functions; in particular $\partial_{t} \partial_{x} u = \partial_{x} \partial_{t} u = z$ . By [F3], $u$ is $C^{2}$ on the interior of $R$ with the stated derivatives, all extending continuously to $R$ .

Remarks

Only partial derivatives up to order two are needed, and all hypotheses are uniform on the compact rectangle. The lemma is used for the sine-series solution of the string, whose partial sums carry the differentiated convergences estimated uniformly.
The conclusion does not require the mixed partial derivatives to be assumed continuous in advance beyond being limits of continuous functions; their agreement follows from the two identifications with the same limit $z$ (the one-variable theorem applied in the two variable orders).

LemmaStatement: AI-adaptedProof: AI-adaptedOpen item page →

Repeated integration by parts gives $∣ b_{n} (F) ∣ \leq 2 L^{5} M_{5} / (π^{5} n^{5})$ for odd $2 L$ -periodic $C^{5}$ functions and $∣ b_{n} (G) ∣ \leq 2 L^{4} M_{4} / (π^{4} n^{4})$ for $C^{4}$

Statement

Let $L > 0$ . Let $F : R \to R$ be odd and $2 L$ -periodic of class $C^{5}$ and let $G : R \to R$ be odd and $2 L$ -periodic of class $C^{4}$ (Sine coefficients, sine series and the odd periodic extension, $C^{k}$ Euclidean maps and diffeomorphisms); put

$M_{5} : = \sup_{R} ∣ F^{(5)} ∣ < + \infty, M_{4} : = \sup_{R} ∣ G^{(4)} ∣ < + \infty .$

Then for every integer $n \geq 1$ , with the sine coefficients $b_{n}$ of Sine coefficients, sine series and the odd periodic extension,

$∣ b_{n} (F) ∣ \leq \frac{2 L^{5} M_{5}}{π^{5} n^{5}}, ∣ b_{n} (G) ∣ \leq \frac{2 L^{4} M_{4}}{π^{4} n^{4}} .$

Facts & Assumptions

Given: The period $L > 0$ , the functions $F, G$ and the bounds $M_{5}, M_{4}$ ; an integer $n \geq 1$ .

[F1]

Integration by parts: for $C^{1}$ functions $p, q$ on $[0, L]$ , their derivatives and products are continuous and hence integrable there; therefore $\int_{0}^{L} p q^{'} = [p q]_{0}^{L} - \int_{0}^{L} p^{'} q$ (A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion, If $u, v$ are differentiable on $[a, b]$ with $u^{'}, v^{'}$ integrable, then $\int_{a}^{b} u v^{'} = u (b) v (b) - u (a) v (a) - \int_{a}^{b} u^{'} v$ , The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ , The integral with oriented limits: $\int_{a}^{a} f : = 0$ and $\int_{b}^{a} f : = - \int_{a}^{b} f$ , For $a < c < b$ : $f$ is integrable on $[a, b]$ if and only if it is integrable on $[a, c]$ and on $[c, b]$ , and then $\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{b} f$ ; with the oriented form for arbitrary $a, b, c$ , Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ ).

[F2]

Boundary values: an odd $2 L$ -periodic function $W$ satisfies $W (0) = 0$ and $W (L) = W (- L) = - W (L)$ , hence $W (L) = 0$ ; its even-order derivatives are again odd $2 L$ -periodic, so $W^{''} (0) = W^{''} (L) = W^{''''} (0) = W^{''''} (L) = 0$ whenever these derivatives exist. Also $\sin (n π \cdot 0 / L) = \sin 0 = 0$ and $\sin (n π L / L) = \sin (n π) = 0$ , while $\cos (0) = 1$ and $\cos (n π) = (- 1)^{n}$ (Sine and cosine defined by their real power series, The derivatives of sine and cosine are cosine and minus sine, The zero sets of sine and cosine and the least positive common period 2 pi, Sine coefficients, sine series and the odd periodic extension).

[F3]

The bounds $M_{5}, M_{4}$ dominate the corresponding derivatives everywhere, so in particular on $[0, L]$ ; and $∣ b_{n} ∣ = ∣ \frac{2}{L} \int_{0}^{L} F \sin (n π x / L) d x ∣ \leq \frac{2}{L} \int_{0}^{L} ∣ F ∣$ when used with the sup bound (Absolute value in an ordered field, A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value, Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line, Open ball, closed ball and sphere in a metric space, Integrable functions on $[a, b]$ form a set closed under sums and scalar multiples, and $\int_{a}^{b} (λ f + μ g) = λ \int_{a}^{b} f + μ \int_{a}^{b} g$ ).

Proof

technique · direct; integrate by parts five times (four times for $G$) and use the vanishing boundary values of the odd derivatives

1.1givenF1F2algebra

Put $I_{0} : = \int_{0}^{L} F (x) \sin (n π x / L) d x$ and $I_{j} : = \int_{0}^{L} F^{(j)} (x) \cdot (\sin if j even, \cos if j odd) (n π x / L) d x$ for $j = 1, \dots, 5$ . Applying [F1] successively, each integration by parts produces the boundary term $[F^{(j)} (x) \cdot (\pm \frac{L}{n π}) (the other trigonometric factor)]_{0}^{L},$ which vanishes by [F2]: at $j$ even the boundary factor is $\cos$ , nonzero at the endpoints, but the multiplying derivative $F^{(j)}$ is odd, hence vanishes at $0$ and $L$ ; at $j$ odd the boundary factor is $\sin$ , which vanishes at $0$ and $L$ . Therefore $I_{0} = \frac{L}{n π} I_{1}$ , $I_{1} = - \frac{L}{n π} I_{2}$ , $I_{2} = \frac{L}{n π} I_{3}$ , $I_{3} = - \frac{L}{n π} I_{4}$ , $I_{4} = \frac{L}{n π} I_{5}$ , and chaining, $I_{0} = (\frac{L}{n π})^{5} I_{5} .$

2.1step 1.1F3algebra∎

Since $∣ I_{5} ∣ \leq \int_{0}^{L} ∣ F^{(5)} ∣ d x \leq L M_{5}$ , step 1.1 gives $∣ b_{n} (F) ∣ = \frac{2}{L} ∣ I_{0} ∣ \leq \frac{2}{L} (\frac{L}{n π})^{5} L M_{5} = \frac{2 L^{5} M_{5}}{π^{5} n^{5}}$ . The same computation with four integrations by parts and $G^{(4)}$ in place of $F^{(5)}$ gives $∣ b_{n} (G) ∣ \leq \frac{2 L^{4} M_{4}}{π^{4} n^{4}}$ .

Remarks

The powers $n^{- 5}$ and $n^{- 4}$ are exactly the number of integrations by parts available; from $C^{k}$ odd periodic functions one obtains $O (n^{- k})$ (with the parity pattern of the boundary terms as used above). No sharpness is claimed.
The estimates are uniform in $n$ , and their summands are summable; this is what permits the termwise differentiation of the string series in the companion theorem. The constant depends on $L$ , on the sup norms and on the fixed exponent, not on $n$ .

TheoremStatement: AI-adaptedProof: AI-adaptedOpen item page →

The classical $C^{2}$ fixed-end string: the sine series solves the wave equation with the given data, uniquely, with an explicit tail bound

Statement

Let $L > 0$ , $c > 0$ . Let $F : R \to R$ be odd and $2 L$ -periodic of class $C^{5}$ and let $G : R \to R$ be odd and $2 L$ -periodic of class $C^{4}$ (Sine coefficients, sine series and the odd periodic extension, $C^{k}$ Euclidean maps and diffeomorphisms), with $M_{5} : = \sup ∣ F^{(5)} ∣ < \infty$ , $M_{4} : = \sup ∣ G^{(4)} ∣ < \infty$ . Put

$a_{n} : = b_{n} (F), d_{n} : = \frac{b_{n} (G)}{ω_{n}}, ω_{n} : = \frac{c n π}{L} (n \geq 1),$

and define, for $(t, x) \in R^{2}$ ,

$\begin{matrix} u (t, x) : = \sum_{n \geq 1} [a_{n} \cos (ω_{n} t) + d_{n} \sin (ω_{n} t)] \sin (\frac{n π x}{L}) . & (S) \end{matrix}$

Then:

(i) the series (S) and the termwise series of $\partial_{t} u, \partial_{x} u, \partial_{t}^{2} u, \partial_{t} \partial_{x} u, \partial_{x}^{2} u$ converge absolutely and uniformly on every strip $[- T, T] \times [0, L]$ ; the sum $u$ is $C^{2}$ , is odd and $2 L$ -periodic in $x$ , satisfies $u_{t t} = c^{2} u_{x x}$ , obeys the fixed-end conditions $u (t, 0) = u (t, L) = 0$ , and has

$u (0, x) = F (x), u_{t} (0, x) = G (x) (x \in R) .$

(ii) (Uniqueness.) If $v$ is $C^{2}$ on $[0, T] \times [0, L]$ for some $T > 0$ , satisfies $v_{t t} = c^{2} v_{x x}$ there, $v (t, 0) = v (t, L) = 0$ for all $t$ , and $v (0, \cdot) = v_{t} (0, \cdot) = 0$ , then $v = 0$ on $[0, T] \times [0, L]$ .

(iii) (Tail bound.) For every $N \geq 1$ and every $(t, x) \in [- T, T] \times [0, L]$ ,

$∣ u (t, x) - \sum_{n \leq N} [a_{n} \cos (ω_{n} t) + d_{n} \sin (ω_{n} t)] \sin (\frac{n π x}{L}) ∣ \leq C (N + 1)^{- 4},$

where $C : = 2 (\frac{2 L^{5} M_{5}}{π^{5}} + \frac{2 L^{5} M_{4}}{c π^{5}})$ .

Facts & Assumptions

Given: $L, c > 0$ ; the odd periodic data $F, G$ with the stated regularities and bounds.

[F1]

Coefficient decay: $∣ a_{n} ∣ \leq \frac{2 L^{5} M_{5}}{π^{5} n^{5}}$ and $∣ b_{n} (G) ∣ \leq \frac{2 L^{4} M_{4}}{π^{4} n^{4}}$ , hence $∣ d_{n} ∣ \leq \frac{2 L^{5} M_{4}}{c π^{5} n^{5}}$ (Repeated integration by parts gives $∣ b_{n} (F) ∣ \leq 2 L^{5} M_{5} / (π^{5} n^{5})$ for odd $2 L$ -periodic $C^{5}$ functions and $∣ b_{n} (G) ∣ \leq 2 L^{4} M_{4} / (π^{4} n^{4})$ for $C^{4}$ ).

[F2]

The modes $\sin (n π x / L) \cos (ω_{n} t)$ and $\sin (n π x / L) \sin (ω_{n} t) / (ω_{n})$ solve $u_{t t} = c^{2} u_{x x}$ , vanish at $x = 0, L$ , and have initial data $(\sin (n π x / L), 0)$ and $(0, \sin (n π x / L))$ respectively (d'Alembert's formula solves the wave equation, is the unique $C^{2}$ solution, and the fixed-end sine modes solve it exactly(c), The derivatives of sine and cosine are cosine and minus sine, The zero sets of sine and cosine and the least positive common period 2 pi).

[F3]

Weierstrass majorant test: a series of functions majorised by a summable sequence converges absolutely and uniformly (The Weierstrass M-test gives absolute pointwise convergence and uniform convergence of a function series, A series of real-valued functions and its pointwise and uniform convergence through its partial sums).

[F4]

The rectangle differentiation lemma: uniform convergence of a $C^{2}$ sequence together with its first and second partial derivatives implies the limit is $C^{2}$ with those derivatives (Uniform convergence of a $C^{2}$ sequence with its first and second derivatives on a rectangle passes $C^{2}$ regularity and termwise differentiation to the limit).

[F5]

Orthogonality and completeness: $\int_{0}^{L} \sin (n π x / L) \sin (k π x / L) d x = (L / 2) δ_{n k}$ and the full-period identities for odd $2 L$ -periodic functions (Sine orthogonality on a half period and the full-period integrals of odd $2 L$ -periodic functions); a continuous one-periodic function with all character integrals zero vanishes, hence a continuous odd $2 L$ -periodic real function with all sine coefficients zero vanishes: rescale $x = 2 L y$ by Substitution: if $φ$ is differentiable on $[c, d]$ with $φ^{'}$ integrable and $f$ is continuous on an interval containing $φ ([c, d])$ , then $\int_{φ (c)}^{φ (d)} f = \int_{c}^{d} (f \circ φ) φ^{'}$ and use $e^{- i r} = \cos r - i \sin r$ ( $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ ) (Fejer completeness: a continuous one-periodic function with all integrals against the characters vanishing is zero).

[F6]

For $N \geq 1$ and integers $m > N + 1$ , decreasing $x^{- 5}$ gives $\sum_{n = N + 2}^{m} n^{- 5} \leq \int_{N + 1}^{m} x^{- 5} d x \leq \frac{1}{4} (N + 1)^{- 4}$ . Passing to the limit of the nonnegative partial sums and adding $(N + 1)^{- 5}$ gives $\sum_{n > N} n^{- 5} \leq 2 (N + 1)^{- 4}$ , using only proper integrals (The integral test: for $f \geq 0$ nonincreasing on $[0, \infty)$ , $\sum_{k} f (k)$ converges if and only if the sequence $(\int_{0}^{N} f)_{N}$ is bounded, with $\int_{0}^{N} f \leq \sum_{k < N} f (k) \leq f (0) + \int_{0}^{N} f$ , For $a < c < b$ : $f$ is integrable on $[a, b]$ if and only if it is integrable on $[a, c]$ and on $[c, b]$ , and then $\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{b} f$ ; with the oriented form for arbitrary $a, b, c$ , The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ ).

[F7]

For a $C^{1}$ function $H$ on a compact time-space rectangle, the time difference quotient $Q_{h} (x) : = (H (t + h, x) - H (t, x)) / h$ equals $h^{- 1} \int_{t}^{t + h} \partial_{t} H (s, x) d s$ by the fundamental theorem. Uniform continuity of $\partial_{t} H$ on the rectangle makes $Q_{h} \to \partial_{t} H (t, \cdot)$ uniformly in $x$ . These functions are continuous and integrable, so Uniformly close integrable functions have integrals differing by at most the interval length times their uniform error gives convergence of their integrals and $\frac{d}{d t} \int_{0}^{L} H (t, x) d x = \int_{0}^{L} \partial_{t} H (t, x) d x$ (The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ , A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion, Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous, Uniformly close integrable functions have integrals differing by at most the interval length times their uniform error).

[F8]

For a $C^{2}$ function on an open neighbourhood of a point $(t, x)$ , the mixed partials $v_{t x}$ and $v_{x t}$ agree. The divided rectangular increment over $[t, t + h] \times [x, x + k]$ equals the averages of the two ordered mixed partials by applying the one-variable fundamental theorem in each order; continuity makes both averages tend to their point values (The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ , A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion).

[F9]

For $C^{2}$ data, integration by parts in $x$ on $[0, L]$ gives $\int_{0}^{L} v_{x} v_{x t} d x = [v_{x} v_{t}]_{0}^{L} - \int_{0}^{L} v_{x x} v_{t} d x = - \int_{0}^{L} v_{x x} v_{t} d x$ when $v_{t} (t, 0) = v_{t} (t, L) = 0$ , using [F8] to identify $v_{x t} = \partial_{x} v_{t}$ (If $u, v$ are differentiable on $[a, b]$ with $u^{'}, v^{'}$ integrable, then $\int_{a}^{b} u v^{'} = u (b) v (b) - u (a) v (a) - \int_{a}^{b} u^{'} v$ , The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ , The integral with oriented limits: $\int_{a}^{a} f : = 0$ and $\int_{b}^{a} f : = - \int_{a}^{b} f$ , For $a < c < b$ : $f$ is integrable on $[a, b]$ if and only if it is integrable on $[a, c]$ and on $[c, b]$ , and then $\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{b} f$ ; with the oriented form for arbitrary $a, b, c$ , Integrable functions on $[a, b]$ form a set closed under sums and scalar multiples, and $\int_{a}^{b} (λ f + μ g) = λ \int_{a}^{b} f + μ \int_{a}^{b} g$ , The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set).

Proof

technique · direct; majorise the differentiated series, identify the initial data by completeness, and prove uniqueness by the energy identity

1.1givenF1F2F3algebra

By [F1] the coefficient sequences satisfy $∣ a_{n} ∣ + ∣ d_{n} ∣ \leq C_{1} n^{- 5}$ with $C_{1} : = \frac{2 L^{5} M_{5}}{π^{5}} + \frac{2 L^{5} M_{4}}{c π^{5}}$ . Since $∣ \sin ∣ \leq 1$ , $∣ \cos ∣ \leq 1$ (Parity and the Pythagorean identity for sine and cosine) and $∣ \partial_{t}^{2} ∣ \leq ω_{n}^{2}$ , $∣ \partial_{t} \partial_{x} ∣ \leq ω_{n} (n π / L)$ , $∣ \partial_{x}^{2} ∣ \leq (n π / L)^{2}$ termwise (The derivatives of sine and cosine are cosine and minus sine), each of the five differentiated series is majorised termwise by a constant multiple of $n^{- 3}$ : $ω_{n}^{2} (∣ a_{n} ∣ + ∣ d_{n} ∣) \leq (\frac{c π}{L})^{2} C_{1} n^{- 3}, ω_{n} \frac{n π}{L} (∣ a_{n} ∣ + ∣ d_{n} ∣) \leq \frac{c π^{2}}{L^{2}} C_{1} n^{- 3}, (\frac{n π}{L})^{2} (∣ a_{n} ∣ + ∣ d_{n} ∣) \leq \frac{π^{2}}{L^{2}} C_{1} n^{- 3} .$ These majorants are independent of $t$ and $x$ , so [F3] gives absolute uniform convergence of the series and all five differentiated series on $R^{2}$ , in particular on every stated strip.

1.2givenF7F8F9algebra

(Uniqueness.) Let $v$ be as in (ii) and put $E (t) : = \frac{1}{2} \int_{0}^{L} (v_{t} (t, x)^{2} + c^{2} v_{x} (t, x)^{2}) d x$ , which is finite for each $t$ because the integrand is continuous on the compact interval (A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion). By [F7] applied to this integrand, $E$ is differentiable and its derivative is the integral of its time derivative. Using [F9], $E^{'} (t) = \int_{0}^{L} (v_{t} v_{t t} + c^{2} v_{x} v_{x t}) d x = \int_{0}^{L} v_{t} (v_{t t} - c^{2} v_{x x}) d x = 0,$ where the boundary term vanishes because $\partial_{t} v (t, 0) = \partial_{t} v (t, L) = 0$ : the endpoint conditions $v (\cdot, 0) = v (\cdot, L) = 0$ hold identically in $t$ , so their time derivatives vanish. The integral-evaluation theorem The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ gives $E (t) - E (0) = \int_{0}^{t} E^{'} (s) d s = 0$ , so $E$ is constant; $E (0) = 0$ because $v (0, \cdot) = v_{t} (0, \cdot) = 0$ ; so $v_{t} (t, x) = v_{x} (t, x) = 0$ for all $(t, x)$ by continuity and nonnegativity of the integrand. Thus $v$ is constant on the connected rectangle and, vanishing at $t = 0$ , is identically $0$ . This is (ii).

2.1step 1.1F2F4algebra

Let $u_{N}$ be the $N$ -th partial sum of (S). Fix any compact rectangle in $R^{2}$ and translate its coordinates to the form required by [F4]. Step 1.1 gives uniform convergence there of the partial sums and all first and second partial-derivative series, so [F4] makes the limit $u$ $C^{2}$ on the rectangle interior with those termwise derivatives. Such rectangles cover $R^{2}$ . Since every summand satisfies $u_{n, t t} = c^{2} u_{n, x x}$ by [F2] and the differentiated series converge uniformly, $u_{t t} = c^{2} u_{x x}$ . Also every summand vanishes at $x = 0$ and $x = L$ , so $u (t, 0) = u (t, L) = 0$ , and each term is odd and $2 L$ -periodic in $x$ , so $u$ is too.

2.2step 1.1F1F3F5algebra

(Initial data.) Let $R : = F - \sum_{n} a_{n} \sin (n π x / L)$ ; uniform convergence from step 1.1 makes $R$ continuous, odd and $2 L$ -periodic. For each $k \geq 1$ , the uniform integral-error bound and [F5] give $\int_{0}^{L} R (x) \sin (k π x / L) d x = \frac{L}{2} b_{k} (F) - \frac{L}{2} a_{k} = 0$ . Rescaling $x = 2 L y$ and using the exponential identity in [F5], the character integral of $R (2 L \cdot)$ at positive $k$ is $- \frac{i}{L} \int_{0}^{L} R (x) \sin (k π x / L) d x = 0$ ; its cosine part vanishes by oddness. The $k = 0$ integral vanishes by the mean identity of [F5], and negative $k$ gives the opposite sine sign with the same zero integral. Thus every character integral vanishes and Fejer completeness [F5] yields $R = 0$ . For $G$ , its coefficient series converges uniformly by the $n^{- 4}$ bound [F1], so the same argument gives $G = \sum_{n} b_{n} (G) \sin (n π x / L)$ . Evaluating (S) and its uniformly convergent time-derivative series at $t = 0$ gives $u (0, x) = F (x)$ and $u_{t} (0, x) = G (x)$ for all real $x$ .

3.1step 1.1F1F6algebra∎

(Tail bound.) For fixed $(t, x)$ the tail of (S) is bounded in absolute value by $\sum_{n > N} (∣ a_{n} ∣ + ∣ d_{n} ∣) \leq C_{1} \sum_{n > N} n^{- 5} \leq 2 C_{1} (N + 1)^{- 4} = C (N + 1)^{- 4}$ by [F1], the triangle inequality and [F6]. This is (iii).

Remarks

The theorem solves the fixed-end wave problem for $C^{5}$ / $C^{4}$ odd periodic data with an explicit uniform tail bound; it does not treat merely continuous or $L^{2}$ data, non-periodic compatibility conditions beyond the odd extension, or the $L^{2}$ convergence of the full trigonometric system. The uniqueness clause is stated for $C^{2}$ solutions on a strip and uses only the energy identity.
The exact constants in (iii) inherit the constants of the coefficient decay; they are not asserted to be optimal, and no numerical error is claimed. The initial-data convergence is uniform on $R$ because the same convergence mode produced the series.

5 · Examples, experiments and counterexamples

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Cm Math Continuum

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

Euclidean spatial calculus: gradient, divergence and material derivative

Definition

Remarks

Continuum motions, spatial velocity, material acceleration and swept regions

Definition

Well-definedness

Remarks

The derivative of the determinant along an invertible C¹ matrix curve: (det⁡A)′=det⁡A tr⁡(A−1A′)(detA)′=detAtr(A−1A′)

Statement

Facts & Assumptions

Proof

Remarks

Localization: a continuous function whose integrals vanish over a neighbourhood-basis family is identically zero

Statement

Facts & Assumptions

Proof

Remarks

The transport theorem for a C2C2 family of diffeomorphisms: ddt∫ψt(B)f=∫ψt(B)(∂tf+div⁡(fw))dtd​∫ψt​(B)​f=∫ψt​(B)​(∂t​f+div(fw))

Statement

Facts & Assumptions

Proof

Remarks

Local mass balance: conservation on every material subregion implies ∂tρ+div⁡(ρu)=0∂t​ρ+div(ρu)=0

Statement

Facts & Assumptions

Proof

Remarks

Local momentum balance: ρDtu=ρb+div⁡σρDt​u=ρb+divσ from the integral balance on admissible material subregions

Statement

Facts & Assumptions

Proof

Remarks

A continuous traction field balanced on every piecewise C1C1 region is given by a continuous stress matrix, and t(−n)=−t(n)t(−n)=−t(n)

Statement

Facts & Assumptions

Proof

Remarks

Angular-momentum balance without couples forces the stress matrix to be symmetric

Statement

Facts & Assumptions

Proof

Remarks

Control-volume balance: mass and momentum rates for a moving non-material region

Statement

Facts & Assumptions

Proof

Remarks

The small-strain isotropic linear elastic constitutive law and its strain energy

Definition

Remarks

The isotropic elastic strain energy is μ∣e0∣F2+(λ/2+μ/3)(tr⁡e)2μ∣e0​∣F2​+(λ/2+μ/3)(tre)2, hence positive definite under μ>0μ>0, λ+2μ/3>0λ+2μ/3>0

Statement

Facts & Assumptions

Proof

Remarks

Exact plane-wave solutions of the isotropic linear elastic model, with longitudinal and transverse speeds

Statement

Facts & Assumptions

Proof

Remarks

Incompressible Newtonian and barotropic inviscid fluid constitutive models

Definition

Remarks

Barotropic inviscid momentum in gradient form, Bernoulli along streamlines and the irrotational constant

Statement

Facts & Assumptions

Proof

Remarks

Kelvin circulation: the circulation around a C1C1 material loop is constant in the barotropic inviscid model

Statement

Facts & Assumptions

Proof

Remarks

d'Alembert's formula solves the wave equation, is the unique C2C2 solution, and the fixed-end sine modes solve it exactly

Statement

The derivative of the determinant along an invertible C¹ matrix curve: $(\det A)^{'} = \det A tr (A^{- 1} A^{'})$

The transport theorem for a $C^{2}$ family of diffeomorphisms: $\frac{d}{d t} \int_{ψ_{t} (B)} f = \int_{ψ_{t} (B)} (\partial_{t} f + div (f w))$

Local mass balance: conservation on every material subregion implies $\partial_{t} ρ + div (ρ u) = 0$

Local momentum balance: $ρ D_{t} u = ρ b + div σ$ from the integral balance on admissible material subregions

A continuous traction field balanced on every piecewise $C^{1}$ region is given by a continuous stress matrix, and $t (- n) = - t (n)$

The isotropic elastic strain energy is $μ ∣ e_{0} ∣_{F}^{2} + (λ / 2 + μ / 3) (tr e)^{2}$ , hence positive definite under $μ > 0$ , $λ + 2 μ / 3 > 0$

Kelvin circulation: the circulation around a $C^{1}$ material loop is constant in the barotropic inviscid model

d'Alembert's formula solves the wave equation, is the unique $C^{2}$ solution, and the fixed-end sine modes solve it exactly

Exact dispersion relation $ω^{2} = (4 c^{2} / a^{2}) \sin^{2} (κ a / 2)$ for the cyclic chain and its continuum error bound

The centered second difference differs from the second derivative by at most $M a^{2} / 12$ when $∣ u^{''''} ∣ \leq M$

Finite-time continuum limit of the cyclic spring chain: uniform $O (a^{3 / 2})$ nodal and uniform interpolation convergence

The method of characteristics for $u_{t} + u u_{x} = 0$ and the finite blow-up time of the gradient

Weak and entropy solutions of the scalar conservation law $u_{t} + (u^{2} / 2)_{x} = 0$

Riemann shock and rarefaction waves for $u_{t} + (u^{2} / 2)_{x} = 0$ are weak solutions, and the shock is an entropy solution

Sine orthogonality on a half period and the full-period integrals of odd $2 L$ -periodic functions

Uniform convergence of a $C^{2}$ sequence with its first and second derivatives on a rectangle passes $C^{2}$ regularity and termwise differentiation to the limit

Repeated integration by parts gives $∣ b_{n} (F) ∣ \leq 2 L^{5} M_{5} / (π^{5} n^{5})$ for odd $2 L$ -periodic $C^{5}$ functions and $∣ b_{n} (G) ∣ \leq 2 L^{4} M_{4} / (π^{4} n^{4})$ for $C^{4}$

The classical $C^{2}$ fixed-end string: the sine series solves the wave equation with the given data, uniquely, with an explicit tail bound