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The momentum equation of the continuum model: local form rho D_t u = rho b + div sigma

Physical scope

Conditional on the adopted mass-balance and momentum-balance postulates and on the Axiom of Countable Choice used by the transport and divergence machinery: the force-form balance on every admissible region, together with the continuity equation and the transport identity, is equivalent to the local momentum equation rho D_t u = rho b + div sigma on the current regions, where sigma is the stress matrix produced by the balance principles. The result is a conditional model deduction; it asserts no existence of solutions, no constitutive relation and no measured value.

Statement

Assume the Axiom of Countable Choice ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let the model of Adopted balance principles of the continuum model: force and couple-free torque balance on every admissible region be given with the mass-balance postulate of Adopted mass balance: the mass of every material subregion is constant in time: a motion $χ$ with spatial velocity $u$ and material acceleration $a = D_{t} u$ (Continuum motions, spatial velocity, material acceleration and swept regions, Euclidean spatial calculus: gradient, divergence and material derivative), a $C^{1}$ positive density field $ρ$ (The density field of the continuum model and the mass of a material subregion), continuous body forces and $C^{1}$ tractions (Adopted force model of the continuum branch: continuous body forces and Cauchy contact tractions), and the balance principles (F) and (T). Let $J_{0} \subseteq I$ have compact closure in the time window. For localization, use the material subregions whose current images are bounded $C^{1}$ domains with closure in the open field domain (Bounded C1 domains and their outward normals), with boundary surface integral as in Surface integration on compact C1 hypersurfaces; this class includes the inverse images of all sufficiently small current balls.

Let $σ$ be the stress matrix of Existence, uniqueness and symmetry of the stress matrix from the adopted balance principles on the current regions (whose existence and symmetry follow from the same balance principles), and define the row divergence

$g_{i} : = \sum_{j = 1}^{3} \partial_{j} σ_{i j} (i = 1, 2, 3) .$

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

$\begin{matrix} ρ (t, x) D_{t} u (t, x) = ρ (t, x) b (t, x) + g (t, x), & (M) \end{matrix}$

and for a symmetric $σ$ the row divergence $g$ coincides with the column divergence, so (M) may be written $ρ D_{t} u = ρ b + div σ$ . Conversely, with the same kinematics, positive density satisfying the continuity equation, and a $C^{1}$ symmetric stress representing the traction, (M) implies force balance (F) on every admissible region and integral momentum balance (L) on every admissible smooth material image. Thus these force and integral momentum formulations are equivalent to (M) under the stated regularity and continuity hypotheses. The statement is a conditional deduction from the adopted balances; it asserts no solution existence, no boundary condition, no constitutive relation and no empirical value.

Facts & Assumptions

Given: ${A C}_{ω}$ ; the model fields; the mass-balance postulate; the balance principles (F) and (T); the admissibility of smooth current-image test regions, including inverse images of sufficiently small current balls.

[F1]

Stress matrix and symmetry (Existence, uniqueness and symmetry of the stress matrix from the adopted balance principles): on every bounded region inside the current configuration there is a unique continuous (indeed $C^{1}$ ) stress matrix with $t (t, y, n) = σ (t, y) n$ , $t (t, y, - n) = - t (t, y, n)$ and $σ (t, y)^{T} = σ (t, y)$ .

[F2]

Continuity equation and transport (The continuity equation of the continuum model: local form of mass balance, the material form, and the incompressibility reading, The transport identity and the material derivative along label worldlines for the adopted continuum model): under the mass-balance postulate, $\partial_{t} ρ + div (ρ u) = 0$ , and for every material subregion the componentwise transport identity holds for the field $ρ u_{i}$ ; together with the product rule this gives $\frac{d}{d t} \int_{χ_{t} (B^{'})} ρ u_{i} d x = \int_{χ_{t} (B^{'})} ρ (D_{t} u)_{i} d x$ .

[F3]

Local momentum balance (Local momentum balance: $ρ D_{t} u = ρ b + div σ$ from the integral balance on admissible material subregions): if $ρ$ satisfies (K) $\partial_{t} ρ + div (ρ u) = 0$ on the current regions and if (L), for each $t \in J_{0}$ and material subregion whose current image is a bounded $C^{1}$ domain with closure in the field domain, $\frac{d}{d t} \int_{χ_{t} (B^{'})} ρ u d x = \int_{χ_{t} (B^{'})} ρ b d x + \int_{\partial χ_{t} (B^{'})} σ ν d S$ , then $ρ D_{t} u = ρ b + g$ on $J_{0} \times χ_{t} (B)$ with $g_{i} = \sum_{j} \partial_{j} σ_{i j}$ .

[F4]

Balance in force form (Adopted balance principles of the continuum model: force and couple-free torque balance on every admissible region clause (iii)): for every admissible region $D$ at time $t$ , $\int_{\partial D} t (t, y, ν_{D}) d S + \int_{D} (ρ b - ρ a) (t, y) d y = 0$ , with $a = D_{t} u$ .

[F5]

Algebra (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 $g (c) \neq 0$ ): the product rule for the divergence and the componentwise computation of the row divergence; for a symmetric matrix the row and column divergences agree.

[F6]

Piecewise- $C^{1}$ divergence theorem (Specified finite piecewise C1 boundary presentations, Divergence for finite piecewise C1 presentations): for a bounded domain with a specified finite piecewise- $C^{1}$ presentation and a $C^{1}$ vector field on a neighborhood of its closure, the integral of its divergence equals the outward flux over the presented faces. Applied to each row of a $C^{1}$ stress field, this gives $\int_{\partial D} σ ν_{D} d S = \int_{D} g d x$ with $g_{i} = \sum_{j} \partial_{j} σ_{i j}$ .

Proof

technique · direct; convert the force-form balance on material subregions into the integral momentum balance with the transport identity and the continuity equation, then instantiate the mathematical localization theorem

1.1givenF1F4algebra

Fix a material subregion $B^{'}$ and $t \in J_{0}$ . By hypothesis $χ_{t} (B^{'})$ is an admissible region at time $t$ ; the force balance [F4] with $D : = χ_{t} (B^{'})$ reads $\int_{\partial χ_{t} (B^{'})} t (t, y, ν) d S + \int_{χ_{t} (B^{'})} (ρ b - ρ D_{t} u) d y = 0$ . Replacing $t (t, y, ν) = σ (t, y) ν$ by [F1] and rearranging gives $\int_{χ_{t} (B^{'})} ρ D_{t} u d y = \int_{χ_{t} (B^{'})} ρ b d y + \int_{\partial χ_{t} (B^{'})} σ ν d S$ .

1.2givenF1F2F5F6algebra

Conversely suppose (M) holds with the same continuous fields and a $C^{1}$ stress representing $t$ . At fixed time extend the stress to the field neighborhood by its columns $σ e_{j} = t (e_{j})$ , which are $C^{1}$ ; on the current region this agrees with [F1], and continuity carries the traction identity to the boundary of every admissible $D \subseteq Ω_{t}$ . By [F6], applied componentwise to these rows, $\int_{\partial D} σ ν_{D} d S = \int_{D} g d x$ . Integrating (M) therefore gives $\int_{\partial D} t d S + \int_{D} (ρ b - ρ a) d x = 0$ , precisely (F). On a smooth material image, [F2] identifies $\int_{D} ρ a$ with the derivative of its momentum integral, giving (L) too. This proves the converse without assuming either balance in that direction.

2.1givenstep 1.1F2F5algebra

The transport identity of [F2] applied componentwise to the field $ρ u_{i}$ and the continuity equation give $\frac{d}{d t} \int_{χ_{t} (B^{'})} ρ u_{i} d x = \int_{χ_{t} (B^{'})} (\partial_{t} (ρ u_{i}) + div (ρ u_{i} u)) d x = \int_{χ_{t} (B^{'})} ρ (D_{t} u)_{i} d x$ ; comparing with step 1.1 shows that the integral momentum balance (L) of [F3] holds for every material subregion and every $t \in J_{0}$ .

3.1givenstep 2.1F2F3

The continuity equation (K) of [F3] holds by [F2] under the mass-balance postulate, and the admissible test class includes inverse images of all sufficiently small current balls, whose images are bounded $C^{1}$ domains with closure in the field domain; step 2.1 supplies (L) for every such admissible pair. Applying [F3] therefore gives $ρ (t, x) D_{t} u (t, x) = ρ (t, x) b (t, x) + g (t, x)$ for every $(t, x) \in J_{0} \times χ_{t} (B)$ , with $g$ the row divergence.

4.1step 3.1step 1.2F1F5algebra∎

Reading: by [F1] the matrix $σ$ is symmetric, so $\sum_{j} \partial_{j} σ_{i j} = \sum_{j} \partial_{j} σ_{j i}$ and the row divergence equals the column divergence; with the dimensions of Continuum body data, the fixed inertial frame and the dimensions of the fields, both sides of (M) have dimension $μ ℓ^{- 2} τ^{- 2}$ (force per volume), so the equation is dimensionally consistent. No further conclusion is drawn.

Remarks

Two equivalent forms. Steps 1.1-4.1 show that the force-form balance on material subregions, the continuity equation and the transport identity together are equivalent to the integral momentum balance (L); the local equation (M) is the localization of either form. In a model in which the mass balance carries source terms, the continuity equation changes and the equivalence with (M) is lost.
What is not claimed. Equation (M) is an identity satisfied by any motion of the model; it neither asserts that motions exist for given data nor determines them without a constitutive law relating $σ$ to the motion. The elastic and fluid items below supply such laws, and the local well-posedness theorem of this page uses them.

The momentum equation of the continuum model: local form rho D_t u = rho b + div sigma

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources