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I have the following partial differential equation:

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I'm asked to prove that if $f\equiv 0$, then the total energy (kinetic energy + potential energy) of the system decreases with time.

What is the expression for the energy of this system? I know what the expression of energy is for parabolic or hyperbolic partial differential equations. But this, clearly, is neither.

UPDATE: If we define the energy to be $\frac{1}{2}(u_t)^2+\frac{1}{2}\sum\limits_{ij}a^{ij}u_{x_i}u_{x_j}$, then it seems that $\frac{dE}{dt}=-\int{d(u_t)^2}$. I don't quite understand how one gets this final expression

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    $\begingroup$ The usual way you would derive an energy equation is to multiply the wave equation by $u_t$ and integrate over all $x$. Then rewrite this as $\frac{d}{dt}\int \frac{1}{2}u_t^2 + \ldots = 0$. If there is a conserved energy this would be on the form $E'(t) = 0$. In this case you can write it as $\frac{d}{dt}\int \frac{1}{2}u_t^2 + \frac{1}{2}a_{ij}u_iu_j{\rm d}x + \int du_t^2{\rm d}x = 0$. In the absence of friction ($d=0$) you only have the first term so this would be the usual energy. $\endgroup$
    – Winther
    Apr 22, 2019 at 22:33
  • $\begingroup$ @Winther- Thank you so much!! $\endgroup$ Apr 22, 2019 at 22:56

1 Answer 1

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Let’s assume $u$ is a $C^2$ solution of \begin{align*} u_{tt}+du_t-\sum\partial_j(a^{ij}\partial_iu)=0. \end{align*} (though the $C^2$ assumption seems weird to me at this point)

We can define the energy \begin{align*} E(t)= \int_{\Omega}\frac{1}{2} \left(\frac{\partial u}{\partial t}\right)^2 +\frac{1}{2} \left(a^{ij}\frac{\partial u}{\partial x^i} \frac{\partial u}{\partial x^j}\right) dx, \end{align*} which is reasonable to be viewed as an energy since the second term of the integrand is nonnegative by the condition of $a^{ij}$.

Taking derivative with respect to $t$, \begin{align*} E’(t)&= \int_{\Omega} u_tu_{tt}+a^{ij}u_{i}u_{jt} dx\\ &= \int_{\Omega} u_t\left(-du_t+\partial_j(a^{ij}\partial_iu)\right)+a^{ij}u_{i}u_{jt} dx\\ &=-\int_{\Omega} d(u_t)^2dx + \int_{\partial \Omega} a^{ij}u_iu_t \nu_j dS\\ &=-\int_{\Omega}d (u_t)^2\leq 0, \end{align*} where we use divergence theorem and $\partial_tu=0$ on $\partial \Omega$.

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  • $\begingroup$ How did you define the energy again? Is there a way to describe energy for any PDE? $\endgroup$ Apr 22, 2019 at 22:04
  • $\begingroup$ Can you also explain the calculations? How you got $E'(t)$ as $-\int d(u_t)^2$? $\endgroup$ Apr 22, 2019 at 22:40
  • $\begingroup$ Sure! Let me edit my answer. $\endgroup$
    – Yesfun Yeh
    Apr 22, 2019 at 23:22
  • $\begingroup$ Regarding the energy for any PDE, I want to clarify that the PDE we are working with should be classified to wave equation, or hyperbolic equation. So we can, via taking $a^{ij}$ as a metric and using differential geometry language, get the above energy which is naturally generalized from the simplest wave equation $u_{tt}=u_{xx}$. $\endgroup$
    – Yesfun Yeh
    Apr 22, 2019 at 23:47

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