Classical continua with explicit constitutive assumptions — Examples

Assume ${\mathrm{A}\mathrm{C}}_{\omega }$ (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)) for the cited Lebesgue transport and continuity suppliers.

Let $L_0>0$ have dimension length and let $B=(0,L_0{)}^3\subseteq {\mathbb{R}}^3$ be the reference cube (the unit-cube case is $L_0=1\mathrm{m}$). Fix an expansion rate $\alpha >0$ with $[\alpha ]={\tau }^{-1}$ and put $I=(-1\mathrm{/}\alpha ,\mathrm{\infty })$ and define the uniform expansion

$\chi (t,X):=((1+\alpha t)X_1,(1+\alpha t)X_2,(1+\alpha t)X_3)((t,X)\in I\times {\mathbb{R}}^3)\mathrm{.}$

Then $\chi$ is a motion of $B$ on $I$ in the sense of Continuum motions, spatial velocity, material acceleration and swept regions (with $O={\mathbb{R}}^3$): it is $C^{\mathrm{\infty }}$, each ${\chi }_t$ is a diffeomorphism with $D_X{\chi }_t=(1+\alpha t)I_3$ and $J(t,X)=(1+\alpha t{)}^3>0$ on $I\times B$. Adopt the model of Adopted continuum kinematic model: body data, smooth motions as families of label worldlines, the continuum idealization and frame covariance with these data and the constant-reference-density field

$\rho (t,x):=\frac{{\rho }_0}{(1+\alpha t{)}^3}({\rho }_0>0)$

on all of $I\times {\mathbb{R}}^3$. This is a positive $C^{\mathrm{\infty }}$ scalar field with $[{\rho }_0]=[\rho ]=\mu {\mathrm{\ell }}^{-3}$, so it is defined on an open neighborhood of every swept closure. Its values outside the current body are an explicit mathematical extension. Then:

1. the spatial velocity is $u(t,x)=\alpha x\mathrm{/}(1+\alpha t)$ and $\mathrm{div}u(t,x)=3\alpha \mathrm{/}(1+\alpha t)$ (Euclidean spatial calculus: gradient, divergence and material derivative);
2. the model mass of $B$ at time $t$ is ${\int }_{{\chi }_t(B)}\rho dx={\rho }_0{L}_0^3$ for every $t\in I$: the mass-balance assumption of Adopted mass balance: the mass of every material subregion is constant in time holds for every material subregion, since more generally $m(t,B^{\mathrm{\prime }})={\rho }_0\mathrm{vol}(B^{\mathrm{\prime }})$, and the continuity equation ${\mathrm{\partial }}_t\rho +\mathrm{div}(\rho u)=0$ of The continuity equation of the continuum model: local form of mass balance, the material form, and the incompressibility reading holds identically on $I\times {\mathbb{R}}^3$;
3. for the field $f(t,x):=x_1^2$ the transport identity of The transport identity and the material derivative along label worldlines for the adopted continuum model is verified exactly: with $F(t)={\int }_{{\chi }_t(B)}f(t,x)dx$ one has $F(t)=L_0^5(1+\alpha t{)}^5\mathrm{/}3$ and $F^{\mathrm{\prime }}(t)=5\alpha L_0^5(1+\alpha t{)}^4\mathrm{/}3={\int }_{{\chi }_t(B)}({\mathrm{\partial }}_{t}f+\mathrm{div}(fu))dx$ for every $t\in I$;
4. along each label the density satisfies $\rho (t,\chi (t,X))(1+\alpha t{)}^3={\rho }_0$, the constant-product form of density times volume of part (ii) of the continuity theorem, and the material derivative is $D_t\rho (t,\chi (t,X))=-3\alpha {\rho }_0\mathrm{/}(1+\alpha t{)}^4=-\rho \mathrm{div}u$, as the material form of that theorem requires.

• The example is a closed-form check of the model's kinematic and mass-balance machinery on expanding flow data; the divergence $3\alpha \mathrm{/}(1+\alpha t)$ is exactly the relative volume rate ${\mathrm{\partial }}_{t}J\mathrm{/}J$ of part (iii) of The continuity equation of the continuum model: local form of mass balance, the material form, and the incompressibility reading, which the computation exhibits explicitly.

• Nothing here is a measurement or a material law: the density field was chosen to make the mass balance hold, and the example verifies the model's identities, not the adequacy of the model for any material.

Adopt the isotropic linear elastic branch of Adopted isotropic linear elastic constitutive model with its small-strain regime and strain energy with reference density ${\rho }_0>0$ of dimension $\mu {\mathrm{\ell }}^{-3}$ and moduli $\lambda ,{\mu }_s$ of dimension $\mu {\mathrm{\ell }}^{-1}{\tau }^{-2}$, with ${\mu }_s>0$ and $\lambda +\frac{2{\mu }_s}{3}>0$ (Continuum body data, the fixed inertial frame and the dimensions of the fields). Use $I=\mathbb{R}$, $O={\mathbb{R}}^3$ and let $x_1$ be the reference position in metres and $t$ time in seconds. The amplitude $A\mathrm{\ne }0$ has dimension $\mathrm{\ell }$, and the wave number $\kappa \mathrm{\ne }0$ has dimension ${\mathrm{\ell }}^{-1}$. Define the positive longitudinal speed and angular frequency by $c_L:=\sqrt{(\lambda +2{\mu }_s)\mathrm{/}{\rho }_0}$ and $\omega :=\kappa c_L$; then $[c_L]=\mathrm{\ell }{\tau }^{-1}$, $[\omega ]={\tau }^{-1}$ and the phase is dimensionless. Let

$\xi (t,x):=A\cos (\kappa x_1-\omega t)e_1,$

with $A\mathrm{\ne }0$, $\kappa \mathrm{\ne }0$ and $e_1$ the first standard basis vector: a longitudinal plane wave travelling along $e_1$. This is a solution of the linear reference-field model for every declared $A,\kappa$; physical small-strain use requires small $\mathrm{\mid }A\kappa \mathrm{\mid }$. No finite-motion diffeomorphism is inferred: its candidate placement has determinant $1-A\kappa \sin (\kappa x_1-\omega t)$, which need not be positive for these unrestricted algebraic data. Then:

1. the linear strain is $e_{11}(t,x)=-A\kappa \sin (\kappa x_1-\omega t)$ with all other entries zero, so $\mathrm{tr}e=e_{11}$ and $e_0=e-\frac{1}{3}(\mathrm{tr}e)I_3$ has Frobenius squared norm $\frac{2}{3}{e}_{11}^2$;
2. the stress matrix is $\sigma (t,x)=(\lambda +2{\mu }_s)e_{11}(t,x)e_1{e}_1^{\mathsf{T}}+\lambda e_{11}(t,x)(e_2{e}_2^{\mathsf{T}}+e_3{e}_3^{\mathsf{T}});$
3. the displacement equation (E) of The displacement equation of the linear elastic branch: rho_0 partial_t^2 xi = rho_0 b + mu_s Laplacian xi + (lambda + mu_s) grad div xi with $b\equiv 0$ holds exactly: ${\rho }_0{\mathrm{\partial }}_t^2{\xi }_1=(\lambda +2{\mu }_s){\mathrm{\partial }}_1^2{\xi }_1$;
4. the strain-energy density, of dimension $\mu {\mathrm{\ell }}^{-1}{\tau }^{-2}$, is $W(e)=(\frac{\lambda }{2}+{\mu }_s)e_{11}^2={\mu }_s\frac{2}{3}{e}_{11}^2+(\frac{\lambda }{2}+\frac{{\mu }_s}{3})e_{11}^2\ge 0$, with equality only where $e_{11}=0$; both coefficients in the decomposition are positive under the branch's hypotheses, and the squared speed $c_L^2=(\lambda +2{\mu }_s)\mathrm{/}{\rho }_0$ is positive with $[c_L]=\mathrm{\ell }{\tau }^{-1}$.

• The computation exhibits the branch's structure concretely: the wave is longitudinal (the displacement is parallel to the propagation direction) and its speed is the longitudinal speed of Plane elastic waves in the adopted model: the dispersion condition and the longitudinal and transverse speeds; the transverse branch, with speed $\sqrt{{\mu }_s\mathrm{/}{\rho }_0}$, is not recomputed here.

• The energy density computed is the model's strain energy of the branch (The small-strain isotropic linear elastic constitutive law and its strain energy); it is a nonnegative function of the strain and is zero exactly when the linear strain vanishes, under the branch's positivity hypotheses. Rigid translations and infinitesimal rigid rotations both have zero linear strain; for this particular longitudinal wave, zero energy is exactly $e_{11}=0$. No measured modulus or speed appears.

Let $L>0$ be a length, $c>0$ a speed, $A_{\ast }>0$ a displacement scale with $[A_{\ast }]=\mathrm{\ell }$, and $\beta \in \mathbb{R}\setminus \{0\}$ dimensionless. Let $[x]=\mathrm{\ell }$ and $[t]=\tau$, and consider the fixed-end string model of The string model: d'Alembert solutions, fixed-end modes and uniqueness for the adopted one-dimensional continuum with initial displacement and velocity

$F(x):=A_{\ast }[\sin (\frac{\pi x}{L})+\beta \sin (\frac{3\pi x}{L}\left)\right],G(x):=0.$

Then $F$ is odd and $2L$-periodic of class $C^{\mathrm{\infty }}$ and $G$ is odd and $2L$-periodic of class $C^4$, with $M_4=\sup \mathrm{\mid }{G}^{(4)}\mathrm{\mid }=0$ and $M_5=\sup \mathrm{\mid }{F}^{(5)}\mathrm{\mid }\le A_{\ast }(\pi \mathrm{/}L{)}^5(1+243\mathrm{\mid }\beta \mathrm{\mid })=:{\bar{M}}_5$ (Sine coefficients, sine series and the odd periodic extension). Put ${\omega }_n:=n\pi c\mathrm{/}L$ and

$u(t,x):=A_{\ast }[\cos ({\omega }_{1}t)\sin (\frac{\pi x}{L})+\beta \cos ({\omega }_{3}t)\sin (\frac{3\pi x}{L}\left)\right]\mathrm{.}$

Then:

1. the sine coefficients of $F$ are $b_1(F)=A_{\ast }$, $b_3(F)=\beta A_{\ast }$ and $b_n(F)=0$ otherwise, and the coefficients of $G$ all vanish; hence the series solution of The classical $C^2$ fixed-end string: the sine series solves the wave equation with the given data, uniquely, with an explicit tail bound is exactly the displayed finite sum $u$;
2. $u$ is $C^{\mathrm{\infty }}$ on ${\mathbb{R}}^2$, satisfies $u_{tt}=c^2{u}_{xx}$, vanishes at $x=0$ and $x=L$ for every $t$, and has $u(0,\cdot )=F$, $u_t(0,\cdot )=0$; by the uniqueness clause of the theorem it is the unique $C^2$ fixed-end solution with these data;
3. the d'Alembert formula of d'Alembert's formula solves the wave equation, is the unique $C^2$ solution, and the fixed-end sine modes solve it exactly with data $(F,0)$ gives the same field: $\frac{1}{2}[F(x-ct)+F(x+ct)]=u(t,x)$ by the product-to-sum identity (The addition formulas for sine and cosine, Parity and the Pythagorean identity for sine and cosine);
4. the tail bound at $N=1$ is verified: with $S_{1}u=A_{\ast }\cos ({\omega }_{1}t)\sin (\pi x\mathrm{/}L)$ one has $\mathrm{\mid }u-S_{1}u\mathrm{\mid }=A_{\ast }\mathrm{\mid }\beta \mathrm{\mid }\mathrm{\mid }\cos ({\omega }_{3}t)\mathrm{\mid }\mathrm{\mid }\sin (3\pi x\mathrm{/}L)\mathrm{\mid }\le A_{\ast }\mathrm{\mid }\beta \mathrm{\mid }$, while the theorem gives $\mathrm{\mid }u-S_{N}u\mathrm{\mid }\le C(N+1{)}^{-4}$ with $C\le \bar{C}:=4A_{\ast }(1+243\mathrm{\mid }\beta \mathrm{\mid })$, so the explicit upper estimate is $A_{\ast }(1+243\mathrm{\mid }\beta \mathrm{\mid })\mathrm{/}4$ at $N=1$, which also dominates the direct bound $A_{\ast }\mathrm{\mid }\beta \mathrm{\mid }$ for every $\beta$;
5. the mode frequencies satisfy $[{\omega }_n]={\tau }^{-1}$ under $[c]=\mathrm{\ell }{\tau }^{-1}$ and $[L]=\mathrm{\ell }$ (Continuum body data, the fixed inertial frame and the dimensions of the fields).

• The two-mode data make every series finite, so the check isolates the theorem's content — the mode frequencies, the boundary and initial conditions, uniqueness, the d'Alembert correspondence and the tail constant — without requiring convergence arguments beyond the supplier theorem.

• Modes with larger $n$ oscillate faster and the displayed bound is deliberately conservative at small $N$; the $O(N^{-4})$ decay becomes informative for large mode numbers and smooth data, as the theorem states. No measured frequency, tension or density appears.

Take the periodic string model of The continuum string model as the proved limit of the mass-spring chain, with explicit convergence and dispersion error with physical circumference $L_{\ast }=2\pi \mathrm{m}$, linear mass density ${\rho }_{\mathrm{\ell },\ast }=1\mathrm{k}\mathrm{g}{\mathrm{m}}^{-1}$ and tension ${\tau }_{s,\ast }=1\mathrm{N}=1\mathrm{k}\mathrm{g}\mathrm{m}{\mathrm{s}}^{-2}$. Then $c_{\ast }:=\sqrt{{\tau }_{s,\ast }\mathrm{/}{\rho }_{\mathrm{\ell },\ast }}=1\mathrm{m}{\mathrm{s}}^{-1}$. Use reference units ${\mathrm{\ell }}_{\ast }=1\mathrm{m}$, ${\tau }_{\ast }=1\mathrm{s}$ and displacement scale $A_{\ast }=1\mathrm{m}$. In the equations below, $x,t,L,c$ denote numerical values in these units, so $L=2\pi$, $c=1$, and the dimensionless displacement $u=u_{\mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}}\mathrm{/}{A}_{\ast }$ has the exact travelling-standing solution with physical wave number $1{\mathrm{m}}^{-1}$ and angular frequency $1{\mathrm{s}}^{-1}$

$u(t,x):=A\sin x\cos t(A>0),$

where $A$ is dimensionless. Physical displacements and their error bounds are obtained by multiplying the displayed values by $A_{\ast }$. The normalized density and tension values are ${\rho }_{\mathrm{\ell }}={\rho }_{\mathrm{\ell },\ast }\mathrm{/}(1\mathrm{k}\mathrm{g}{\mathrm{m}}^{-1})=1$ and ${\tau }_s={\tau }_{s,\ast }\mathrm{/}(1\mathrm{N})=1$; these are numerical values of different dimensioned quantities. The normalized field is $2\pi$-periodic and solves $u_{tt}=u_{xx}$. For $N\ge 2$ put $a=2\pi \mathrm{/}N$ and consider the cyclic spring chain of The cyclic spring chain, its discrete energy, and the finite-regularity periodic comparison class with parameter $c=1$ and spacing $a$, i.e. the normalized node mass is $m={\rho }_{\mathrm{\ell }}a$ in units of $1\mathrm{k}\mathrm{g}$ and the normalized spring constant is $k_s={\tau }_s\mathrm{/}a$ in units of $1\mathrm{N}{\mathrm{m}}^{-1}$. Then:

1. $M=\sup \mathrm{\mid }{\mathrm{\partial }}_x^{4}u\mathrm{\mid }=A$, $M_2=\sup \mathrm{\mid }{\mathrm{\partial }}_x^{2}u\mathrm{\mid }=A$ and the theorem's constant is $C=\sqrt{2\pi }A\mathrm{/}12\approx 0.20889A$;
2. the exact chain solution with the sampled data is $q_j(t)=A\sin (ja)\cos ({\omega }_{a}t)$ with ${\omega }_a=2\sin (a\mathrm{/}2)\mathrm{/}a$ (positive for $0<a<\pi$), the exact nodal error is $e_j(t)=A\sin (ja)[\cos ({\omega }_{a}t)-\cos t]$, and the theorem's bounds read $\mathrm{\parallel }e(t){\mathrm{\parallel }}_a\le C{a}^2{t}^2\mathrm{/}2$, ${\max }_j\mathrm{\mid }{e}_j(t)\mathrm{\mid }\le \frac{1}{2}C{a}^{3\mathrm{/}2}{t}^2$ and ${\sup }_x\mathrm{\mid }{I}_{a}q-u\mathrm{\mid }\le {\max }_j\mathrm{\mid }{e}_j\mathrm{\mid }+A{a}^2\mathrm{/}8$;
3. at $N=100$ (so $a=2\pi \mathrm{/}100\approx 0.062832$) and $t=1$ the evaluated bounds are $\mathrm{\parallel }e(1){\mathrm{\parallel }}_a\le 0.0004124A$, ${\max }_j\mathrm{\mid }{e}_j(1)\mathrm{\mid }\le 0.001645A$ and $\sup \mathrm{\mid }{I}_{a}q-u\mathrm{\mid }\le 0.002139A$, while the exact error satisfies $\mathrm{\mid }{e}_j(1)\mathrm{\mid }\le A\mathrm{\mid }{\omega }_a-1\mathrm{\mid }\approx 0.00016449A$, so the theoretical bound is verified and visibly conservative;
4. the exact dispersion relation is ${\omega }_a^2=(4\mathrm{/}{a}^2){\sin }^2(a\mathrm{/}2)$, and the theorem's bound $\mathrm{\mid }{\omega }_a^2-1\mathrm{\mid }\le a^2\mathrm{/}12$ gives $\le 0.00032899$ at $N=100$, while the exact value satisfies $\mathrm{\mid }{\omega }_a^2-1\mathrm{\mid }\approx 0.00032894$.