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Consider the heat equation

\begin{equation} u_t - u_{xx} = 0, \ \ \ x \in [0, L],\ t > 0. \end{equation} and \begin{equation} E(t) = \int^L_0 (u(x,t))^2 dx. \end{equation} If $u$ satisfies the neumann condition $u_x(0, t) = u_x(L, t) = 0$, show that $E(t)$ is constant.

Attempt: Integrating by parts, we have \begin{equation} \frac{1}{2}E'(t) = \int^L_0 uu_t dx = \int^L_0 uu_{xx} dx = uu_x|^L_0 - \int^L_0 u_x^2 dx. \end{equation}

Integrating the term $\int^L_0 u_x^2 dx$ by parts, i arrive at the same place. Help!

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    $\begingroup$ Hint: Multiply the equation by $u$ and then integrate. $\endgroup$
    – xpaul
    Commented Dec 6, 2021 at 23:54

1 Answer 1

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$E(t)$ is indeed non-constant: e.g.

$$ u(x, t) = \exp \left(-\frac{\pi^2t}{L^2} \right) \cos \left( \frac{\pi x}{L}\right)$$

satisfies the heat equation $u_t = u_{xx}$ and $$ u_x ( 0, t) = u_x (L, t) = 0,$$ but

$$E(t) = \int_0^L u^2(x, t)dx = \exp \left(-\frac{2\pi^2t}{L^2} \right)\int_0^L \cos ^2\left( \frac{\pi x}{L}\right) dx $$

is not constant in $t$. What you can show is only that $E(t)$ is non-increasing (as you did).

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