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?
Best Answer
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 $$ \int_0^L u(x,t)dx = \int_0^{2\pi} \exp(-t)\sin(x) dx = 0 \quad (t \ge 0). $$