Worked check of the transport identity and the continuity equation on a uniform expansion

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.