Consider the variable coefficient, real valued wave equation

$$ u_{tt} - \nabla \cdot (c^2 \nabla u) + qu = 0, \quad u(x,0) = \phi(x), \quad u_t(x, 0) = \phi(x), $$ where $c, q \geq 0$ depend only on $x$. Then we define the total energy at time $f$ of a $C^2$ solution $u$ as

$$ E(t) = \frac{1}{2}\int_\Omega (u_t^2 + c(x)^2|\nabla u|^2 + q(x)u^2) \, dx. $$

The goal is to show that the total energy is constant given some boundary conditions. To do this, we differentiate $E$ with respect to $t$, but I have some questions about the presented derivation. The notes I'm following claim that

$$ \frac{dE}{dt}(t) = \frac{1}{2}\int_\Omega (u_tu_{tt} + c(x)^2 \nabla u \cdot \nabla u_t + q(x)uu_t)\, dx. $$

However, my calculations have that, e.g.:

$$ \frac{d}{dt} u_t^2 = 2\left(\frac{d}{dt}u_t\right)\left(\frac{d}{dt}u\right) = 2u_{tt}u_t.$$

Where does the extra factor of $2$ appear in my derivation?

With the $\nabla$ term, I'm having even more trouble recovering the solution's expression. I expand as:

$ \begin{align*} \frac{\partial}{\partial t}|\nabla u|^2 &= \frac{\partial}{\partial t}\left(\sum_{i=1}^n\left(\frac{\partial u}{\partial x_i}\right)^2 + \left(\frac{\partial u}{\partial t}\right)^2\right)\\ &= \sum_{i=1}^n 2\frac{\partial^2 u}{\partial t \partial x_i} \frac{\partial u}{\partial t} + 2 \frac{\partial^2 u}{\partial^2 t}\frac{\partial u}{\partial t} \\ \end{align*} $

which isn't the form in the notes. Any guidance in how the notes get their final expression would be much appreciated.

  • $\begingroup$ I just realized that the constant factor of $2$ is cancelled by the $\frac{1}{2}$ outside the integral. Maybe this will make the inside calculations work... $\endgroup$
    – user83387
    Feb 7, 2016 at 10:28

1 Answer 1


The notes that you are following should be corrected by erasing the initial fraction: $$ \frac{dE}{dt}(t) = \int_\Omega (u_tu_{tt} + c(x)^2 \nabla u \cdot \nabla u_t + q(x)uu_t)\, dx. $$ In particular, the derivative of $(u_t)^2$ is obtained exactly as you wrote. As for the middle term, you have instead $$ \frac{\partial}{\partial t}|\nabla u|^2 = \frac{\partial}{\partial t}\sum_{i=1}^n\left(\frac{\partial u}{\partial x_i}\right)^2= \sum_{i=1}^n 2\frac{\partial^2 u}{\partial t \partial x_i} \frac{\partial u}{\partial x_i}= \sum_{i=1}^n 2\frac{\partial u_t}{\partial x_i} \frac{\partial u}{\partial x_i}=2\nabla u \cdot \nabla u_t. $$

  • $\begingroup$ So $\nabla u$ is taking the first partials w.r.t. to only $x$? $\endgroup$
    – user83387
    Feb 7, 2016 at 10:37
  • $\begingroup$ When applying the chain rule to $\frac{\partial}{\partial t} \left(\frac{\partial u}{\partial x_i}\right)^2$ shouldn't we get $2 \frac{\partial^2 u}{\partial t \partial x_i} \frac{\partial u}{\partial t}$? $\endgroup$
    – user83387
    Feb 7, 2016 at 10:40
  • 2
    $\begingroup$ Yes, $\nabla u$ is only what I wrote and you just use the chain rule, which doesn't give $\frac{\partial u}{\partial t}$. $\endgroup$
    – John B
    Feb 7, 2016 at 10:43

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