I have the following partial differential equation:
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
Best Answer
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 $\partial \Omega$.