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Energy conservation

We consider the 1D \(p\)-system:

\[\begin{align*} \label{nel_pde} \epsilon_t(x,t)-u_x(x,t) & = 0 \\ \rho(x)u(x,t)_t - \sigma(\epsilon(x,t),x)_x & = 0.\end{align*}\]

Let \(\eta\) denote the energy density:

\[\begin{align*} \eta(u,\epsilon,x) = \frac{1}{2}\rho(x)u^2 + \int_0^\epsilon \sigma(s,x)ds.\end{align*}\]

It is straightforward to see that \(\eta\) is conserved for smooth solutions:

\[\begin{align*} \frac{d}{dt} \int_{-\infty}^\infty \eta dx & = \int_{-\infty}^\infty \eta_t dx \\ & = \int_{-\infty}^\infty \left(\rho(x) u u_t + \frac{d}{dt}\int_0^{\epsilon(x,t)} \sigma(s,x) ds \right) dx \\ & = \int_{-\infty}^\infty \left(\rho u u_t + \sigma(\epsilon,x)\epsilon_t\right) dx \\ & = \int_{-\infty}^\infty \left(u\sigma_x + \sigma u_x\right) dx = \int_{-\infty}^\infty (\sigma u)_x dx = 0.\end{align*}\]

Here I’ve assumed differentiability of \(u\) and \(\sigma\), which will hold if we consider \(\rho\) and \(\sigma\) to be smooth functions with respect to \(x\). I think the result will still hold if they are discontinuous, but proving it is a bit more tedious. Thus, the solution of ([nel_pde]) satisfies energy conservation, no matter how the coefficients vary with respect to \(x\). In particular, energy will be conserved even for media that lead to effective dissipation.

Dissipation

Now suppose we add a dissipative term to ([nel_pde]):

\[\begin{align*} \epsilon_t(x,t)-u_x(x,t) & = 0 \\ \rho(x)u(x,t)_t - \sigma(\epsilon(x,t),x)_x & = u_{xx}.\end{align*}\]

Now, following the same approach, we obtain

\[\begin{align*} \frac{d}{dt} \int_{-\infty}^\infty \eta dx & = \int_{-\infty}^\infty \eta_t dx \\ & = \int_{-\infty}^\infty \left(\rho(x) u u_t + \frac{d}{dt}\int_0^{\epsilon(x,t)} \sigma(s,x) ds \right) dx \\ & = \int_{-\infty}^\infty \left(\rho u u_t + \sigma(\epsilon,x)\epsilon_t\right) dx \\ & = \int_{-\infty}^\infty \left(uu_{xx} + u\sigma_x + \sigma u_x\right) dx \\ & = \int_{-\infty}^\infty \left(u u_{xx} + (\sigma u)_x\right) dx \\ & = \int_{\infty}^\infty u u_{xx} dx \\ & = \int_{\infty}^\infty ( (u u_x)_x - (u_x)^2 ) dx \\ & = \int_{\infty}^\infty - (u_x)^2 dx.\end{align*}\]

This integral is strictly negative for any non-constant solution.

There seems to be a contradiction here. If the homogenized equations are dissipative, but the first-order hyperbolic system conserves energy, then the two must exhibit very different behavior. How can this be?

Second-derivative terms in the homogenized system

Let’s look at the second-derivative terms that appear in the homogenized equations. For simplicity, we consider a linear stress relation: \(\sigma(\epsilon, x)=K(x)\epsilon\). From LeVeque & Yong eqn. (5.17) we have

\[\begin{align*} \overline{K^{-1}}\sigma_t-u_x & = -C u_{xx} \\ \overline{\rho}u_t - \sigma_x & = C \sigma_{xx}.\end{align*}\]

The energy density (for the linear, homogenized medium) is

\[\begin{align*} \eta(u,\epsilon,x) = \frac{1}{2}\overline{\rho}u^2 + \frac{1}{2\overline{K^{-1}}}\epsilon^2\end{align*}\]

Now, following the same approach, and using the spatially averaged stress relation \(\overline{K^{-1}}\sigma(\epsilon) = \epsilon\), we obtain

\[\begin{align*} \frac{d}{dt} \int_{-\infty}^\infty \eta dx & = \int_{-\infty}^\infty \eta_t dx \\ & = \int_{-\infty}^\infty \left(\overline{\rho}u u_t + \frac{1}{\overline{K^{-1}}} \epsilon \epsilon_t\right) dx \\ & = \int_{-\infty}^\infty \left(\overline{\rho}u u_t + \overline{K^{-1}}\sigma \sigma_t \right) dx \\ & = \int_{-\infty}^\infty \left(u(\sigma_x + C \sigma_{xx}) + \sigma (u_x - C u_{xx})\right) dx \\ & = \int_{-\infty}^\infty \left((\sigma u)_x + C ( u \sigma_{xx} - \sigma u_{xx} \right) dx \\\end{align*}\]

Using the relations

\[\begin{align*} u\sigma_{xx} & = (u\sigma_x)_x - u_x \sigma_x \\ \sigma u_{xx} & = (\sigma u_x)_x - u_x \sigma_x,\end{align*}\]

we obtain

\[\begin{align*} \frac{d}{dt} \int_{-\infty}^\infty \eta dx & = \int_{-\infty}^\infty ((u\sigma_x)_x + (\sigma u_x)_x -u_x\sigma_x + u_x\sigma_x) dx = 0.\end{align*}\]

Thus the terms that look like dissipation in the homogenized equations are not dissipative at all! Indeed, if we consider just those terms, we have the system

\[\begin{align*} \label{simpdiss} u_t & = \sigma_{xx} \\ \sigma_t & = -u_{xx}.\end{align*}\]

These can be combined to yield \[u_{tt} = - u_{xxxx}\] which has solutions of the form \(u(x,t) = \exp(i(x\pm t))\), so we could say that it is "hyperbolic". Looking at it a different way, system ([simpdiss]) can be written as \(q_t + A q_{xx}\), where \(A\) is diagonalizable with pure imaginary eigenvalues. This is a natural generalization of the definition of hyperbolicity to second-order systems.

One lesson from this is that our intuition from scalar PDEs does not necessarily apply to systems.