# Heat equation problem with Dirichlet condition

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?

• 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 $t$. Commented Nov 21, 2021 at 15:08
• I thought about it too, but I couldn't get around the problem. Do you have any suggestions on how I can do this? Commented Nov 21, 2021 at 15:13
• 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. Commented Nov 22, 2021 at 0:23

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 $$\int_0^L u(x,t)dx = \int_0^{2\pi} \exp(-t)\sin(x) dx = 0 \quad (t \ge 0).$$
• But in the case that $L$ is arbitrary? Does this solution you pointed out have unity (Unless the identically null solution)? Commented Nov 21, 2021 at 18:12
• 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 $L>0$: $u(x,t)=\exp(-4\pi^2t/L^2)\sin(2\pi x/L)$.