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I would like to know how to solve the first exercise of chapter 7 of Evans' Partial Differential Equations, second edition. The problem goes like this:

Let $U\subset\mathbb{R}^n$ be an open and bounded set, with smooth boundary, and let $T>0$. Prove that there is at most one smooth solution of this initial/boundary value problem for the heat equation with Neumann boundary conditions $$\begin{cases}u_t-\Delta u=f&\text{in }U_T\\ \frac{\partial u}{\partial\nu}=0&\text{in }\partial U\times[0,T]\\ u=g&\text{in }U\times\{t=0\} \end{cases}$$

This is what I've got so far:

Suppose $u$ and $v$ are two regular solutions of the given problem. Then $u-v$ is a regular solution to the problem $$\begin{cases}u_t-\Delta u=0&\text{in }U_T\\ \frac{\partial u}{\partial\nu}=0&\text{in }\partial U\times[0,T]\\ u=0&\text{in }U\times\{t=0\} \end{cases}$$ If $u$ is a solution to the last problem, on the one hand, we have that $$\begin{aligned} \int_{U\times(0,T)}\Delta udx&=\int_0^T\left(\int_U\Delta u(x,\tau)dx\right)d\tau&\text{ (Fubini's theorem)}\\ &=\int_0^T\left(\int_{\partial U}\frac{\partial u}{\partial\nu}(x,\tau)dS\right)d\tau&\text{ (Green's formula)}\\ &=\int_{\partial U\times(0,T)}\frac{\partial u}{\partial\nu}dS&\text{ (Fubini's theorem)}\\ &=0 \end{aligned}$$ Because, by hypothesis, we know that $\partial u/\partial\nu$ is identically equal to zero in $\partial U\times[0,T]$.

While on the other hand $$\int_{U\times(0,T)}u_tdx=\int_{U\times(0,T)}\Delta udx=0\Rightarrow u_t=0\text{ in }U\times(0,T)\Rightarrow u\text{ is constant in }t\in (0,T)$$

Since $u=0$ in $U\times\{t=0\}$ and we are supposing this is a regular solution, by continuity we can conclude that $u$ is identically zero in $U\times [0,T]$. But recall that the difference of two regular solutions of the first problem is a solution to the second, so if there is a regular solution to the original problem, it must be unique.

What do you think? Are there flaws in my proof? Is there another method to prove this result? Thanks in advance for your help.

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  • $\begingroup$ You may find this post interesting $\endgroup$
    – EditPiAf
    Dec 2, 2020 at 23:10

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