# Nonlinear finite elements/Entropy inequality

## Clausius-Duhem inequality

The Clausius-Duhem inequality can be expressed in integral form as

${\cfrac {d}{dt}}\left(\int _{\Omega }\rho ~\eta ~{\text{dV}}\right)\geq \int _{\partial \Omega }\rho ~\eta ~(u_{n}-\mathbf {v} \cdot \mathbf {n} )~{\text{dA}}-\int _{\partial \Omega }{\cfrac {\mathbf {q} \cdot \mathbf {n} }{T}}~{\text{dA}}+\int _{\Omega }{\cfrac {\rho ~s}{T}}~{\text{dV}}~.$ In differential form the Clusius-Duhem inequality can be written as

$\rho ~{\dot {\eta }}\geq -{\boldsymbol {\nabla }}\cdot \left({\cfrac {\mathbf {q} }{T}}\right)+{\cfrac {\rho ~s}{T}}~.$ Proof:

Assume that $\Omega$ is an arbitrary fixed control volume. Then $u_{n}=0$ and the derivative can be taken inside the integral to give

$\int _{\Omega }{\frac {\partial }{\partial t}}(\rho ~\eta )~{\text{dV}}\geq -\int _{\partial \Omega }\rho ~\eta ~(\mathbf {v} \cdot \mathbf {n} )~{\text{dA}}-\int _{\partial \Omega }{\cfrac {\mathbf {q} \cdot \mathbf {n} }{T}}~{\text{dA}}+\int _{\Omega }{\cfrac {\rho ~s}{T}}~{\text{dV}}~.$ Using the divergence theorem, we get

$\int _{\Omega }{\frac {\partial }{\partial t}}(\rho ~\eta )~{\text{dV}}\geq -\int _{\Omega }{\boldsymbol {\nabla }}\cdot (\rho ~\eta ~\mathbf {v} )~{\text{dV}}-\int _{\Omega }{\boldsymbol {\nabla }}\cdot \left({\cfrac {\mathbf {q} }{T}}\right)~{\text{dV}}+\int _{\Omega }{\cfrac {\rho ~s}{T}}~{\text{dV}}~.$ Since $\Omega$ is arbitrary, we must have

${\frac {\partial }{\partial t}}(\rho ~\eta )\geq -{\boldsymbol {\nabla }}\cdot (\rho ~\eta ~\mathbf {v} )-{\boldsymbol {\nabla }}\cdot \left({\cfrac {\mathbf {q} }{T}}\right)+{\cfrac {\rho ~s}{T}}~.$ Expanding out

${\frac {\partial \rho }{\partial t}}~\eta +\rho ~{\frac {\partial \eta }{\partial t}}\geq -{\boldsymbol {\nabla }}(\rho _{\eta })\cdot \mathbf {v} -\rho ~\eta ~({\boldsymbol {\nabla }}\cdot \mathbf {v} )-{\boldsymbol {\nabla }}\cdot \left({\cfrac {\mathbf {q} }{T}}\right)+{\cfrac {\rho ~s}{T}}$ or,

${\frac {\partial \rho }{\partial t}}~\eta +\rho ~{\frac {\partial \eta }{\partial t}}\geq -\eta ~{\boldsymbol {\nabla }}\rho \cdot \mathbf {v} -\rho ~{\boldsymbol {\nabla }}\eta \cdot \mathbf {v} -\rho ~\eta ~({\boldsymbol {\nabla }}\cdot \mathbf {v} )-{\boldsymbol {\nabla }}\cdot \left({\cfrac {\mathbf {q} }{T}}\right)+{\cfrac {\rho ~s}{T}}$ or,

$\left({\frac {\partial \rho }{\partial t}}+{\boldsymbol {\nabla }}\rho \cdot \mathbf {v} +\rho ~{\boldsymbol {\nabla }}\cdot \mathbf {v} \right)~\eta +\rho ~\left({\frac {\partial \eta }{\partial t}}+{\boldsymbol {\nabla }}\eta \cdot \mathbf {v} \right)\geq -{\boldsymbol {\nabla }}\cdot \left({\cfrac {\mathbf {q} }{T}}\right)+{\cfrac {\rho ~s}{T}}~.$ Now, the material time derivatives of $\rho$ and $\eta$ are given by

${\dot {\rho }}={\frac {\partial \rho }{\partial t}}+{\boldsymbol {\nabla }}\rho \cdot \mathbf {v} ~;~~{\dot {\eta }}={\frac {\partial \eta }{\partial t}}+{\boldsymbol {\nabla }}\eta \cdot \mathbf {v} ~.$ Therefore,

$\left({\dot {\rho }}+\rho ~{\boldsymbol {\nabla }}\cdot \mathbf {v} \right)~\eta +\rho ~{\dot {\eta }}\geq -{\boldsymbol {\nabla }}\cdot \left({\cfrac {\mathbf {q} }{T}}\right)+{\cfrac {\rho ~s}{T}}~.$ From the conservation of mass ${\dot {\rho }}+\rho ~{\boldsymbol {\nabla }}\cdot \mathbf {v} =0$ . Hence,

${\rho ~{\dot {\eta }}\geq -{\boldsymbol {\nabla }}\cdot \left({\cfrac {\mathbf {q} }{T}}\right)+{\cfrac {\rho ~s}{T}}~.}$ ### Clausius-Duhem inequality in terms of internal energy

In terms of the specific entropy, the Clausius-Duhem inequality is written as

$\rho ~{\dot {\eta }}\geq -{\boldsymbol {\nabla }}\cdot \left({\cfrac {\mathbf {q} }{T}}\right)+{\cfrac {\rho ~s}{T}}$ Show that the inequality can be expressed in terms of the internal energy as

$\rho ~({\dot {e}}-T~{\dot {\eta }})-{\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} \leq -{\cfrac {\mathbf {q} \cdot {\boldsymbol {\nabla }}T}{T}}~.$ Proof:

Using the identity ${\boldsymbol {\nabla }}\cdot (\varphi ~\mathbf {v} )=\varphi ~{\boldsymbol {\nabla }}\cdot \mathbf {v} +\mathbf {v} \cdot {\boldsymbol {\nabla }}\varphi$ in the Clausius-Duhem inequality, we get

$\rho ~{\dot {\eta }}\geq -{\boldsymbol {\nabla }}\cdot \left({\cfrac {\mathbf {q} }{T}}\right)+{\cfrac {\rho ~s}{T}}\qquad {\text{or}}\qquad \rho ~{\dot {\eta }}\geq -{\cfrac {1}{T}}~{\boldsymbol {\nabla }}\cdot \mathbf {q} -\mathbf {q} \cdot {\boldsymbol {\nabla }}\left({\cfrac {1}{T}}\right)+{\cfrac {\rho ~s}{T}}~.$ Now, using index notation with respect to a Cartesian basis $\mathbf {e} _{j}$ ,

${\boldsymbol {\nabla }}\left({\cfrac {1}{T}}\right)={\frac {\partial }{\partial x_{j}}}\left(T^{-1}\right)~\mathbf {e} _{j}=-\left(T^{-2}\right)~{\frac {\partial T}{\partial x_{j}}}~\mathbf {e} _{j}=-{\cfrac {1}{T^{2}}}~{\boldsymbol {\nabla }}T~.$ Hence,

$\rho ~{\dot {\eta }}\geq -{\cfrac {1}{T}}~{\boldsymbol {\nabla }}\cdot \mathbf {q} +{\cfrac {1}{T^{2}}}~\mathbf {q} \cdot {\boldsymbol {\nabla }}T+{\cfrac {\rho ~s}{T}}\qquad {\text{or}}\qquad \rho ~{\dot {\eta }}\geq -{\cfrac {1}{T}}\left({\boldsymbol {\nabla }}\cdot \mathbf {q} -\rho ~s\right)+{\cfrac {1}{T^{2}}}~\mathbf {q} \cdot {\boldsymbol {\nabla }}T~.$ Recall the balance of energy

$\rho ~{\dot {e}}-{\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} +{\boldsymbol {\nabla }}\cdot \mathbf {q} -\rho ~s=0\qquad \implies \qquad \rho ~{\dot {e}}-{\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} =-({\boldsymbol {\nabla }}\cdot \mathbf {q} -\rho ~s)~.$ Therefore,

$\rho ~{\dot {\eta }}\geq {\cfrac {1}{T}}\left(\rho ~{\dot {e}}-{\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} \right)+{\cfrac {1}{T^{2}}}~\mathbf {q} \cdot {\boldsymbol {\nabla }}T\qquad \implies \qquad \rho ~{\dot {\eta }}~T\geq \rho ~{\dot {e}}-{\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} +{\cfrac {\mathbf {q} \cdot {\boldsymbol {\nabla }}T}{T}}~.$ Rearranging,

${\rho ~({\dot {e}}-T~{\dot {\eta }})-{\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} \leq -{\cfrac {\mathbf {q} \cdot {\boldsymbol {\nabla }}T}{T}}~.}$ 