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Problem: Consider a heat equation $$u_t - u_{xx} = 0,$$ with $x \in [0,L]$ and $t > 0$. In addition, be also the full $$E(t) = \int_0^L u(x,t)dx.$$ If $u$ is a function that satisfies a Dirichlet condition $u(0,t) = u(L,t) = 0$, then explain why $E(t)$ is not constant.

Idea: The idea is to show that the only solution to the heat problem with Dirichlet condition presented such that $E(t)$ is constant is $u \equiv 0$. So, I tried to take the following approach: $$0 = E_t(t) = \int_0^L u_t(x,t)dx = \int_0^L u_{xx}(x,t)dx = u_x(L,t) - u_x(0,t).$$ Then we would have $$\dfrac{d}{dt}[u_x(L,t) - u_x(0,t)] = 0 \ \ \Rightarrow \ \ \dfrac{d}{dx}[u_t(L,t) - u_t(0,t)] = 0 \ \ \Rightarrow \ \ u_t(L,t) - u_t(0,t) = C,$$ where $C$ is a constant. If $C \neq 0$, then the result is immediate. The problem is that if $C = 0$, then I couldn't finish. Do you have any ideas to help?

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  • $\begingroup$ The implications you have are wrong - the expression $u(L,T)-u(0,t)$ doesn't depend on $x$, so it doesn't make much sense to take the $x$-derivative. And also $u_t(L,T)-u_t(0,t) = 0$ just because $u(0,t) = 0 = u(L,t)$ for any $t$. $\endgroup$ Commented Nov 21, 2021 at 15:08
  • $\begingroup$ I thought about it too, but I couldn't get around the problem. Do you have any suggestions on how I can do this? $\endgroup$
    – Santos
    Commented Nov 21, 2021 at 15:13
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    $\begingroup$ I expect you're misinterpreting the question - as Gerd showed, it's certainly possible for $E(t)$ to be constant even if $u$ isn't identically zero. I'd guess that either there are some extra assumptions you're meant to use (e.g. $u\ge 0$), or else you're just supposed to explain why $E(t)$ isn't necessarily constant (in contrast to the Neumann b.c. case). But really your best bet is to ask your instructor, since we can't possibly know what's the intent behind the question. $\endgroup$
    – Carmeister
    Commented Nov 22, 2021 at 0:23

1 Answer 1

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Am I missing something? Set $L=2\pi$ and $u(x,t)=\exp(-t)\sin(x)$. Then $u(0,t)=u(L,t)=0$, $$ u_t(x,t) -u_{xx}(x,t)= - \exp(-t)\sin(x) + \exp(-t)\sin(x)=0 $$ and $$ \int_0^L u(x,t)dx = \int_0^{2\pi} \exp(-t)\sin(x) dx = 0 \quad (t \ge 0). $$

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  • $\begingroup$ But in the case that $L$ is arbitrary? Does this solution you pointed out have unity (Unless the identically null solution)? $\endgroup$
    – Santos
    Commented Nov 21, 2021 at 18:12
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    $\begingroup$ This solution is the unique solution with initial-boundary condition $u(x,0)=\sin(x)$ ($x \in [0,2\pi]$) and $u(0,t)=u(2\pi,t)=0$ $(t \ge 0)$. You can transform this example to any $L>0$: $u(x,t)=\exp(-4\pi^2t/L^2)\sin(2\pi x/L)$. $\endgroup$
    – Gerd
    Commented Nov 21, 2021 at 18:55

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