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Let $U$ be a region in $\mathbb{R}^n$, $T>0$ be a constant and $U_T=U×(0,T)$.

Pick any smooth function $g:U\to\mathbb{R}$ and define $A=\{v:U\to\mathbb{R}\text{ smooth }|\,v(x)=g(x)\text{ for all }x\in\partial U$}.

Now pick any $h\in A$ and suppose that $u: U_T\to\mathbb{R}$ is a smooth solution of the following initial boundary value problem of the heat equation, i.e., $$\begin{cases}u_t (x,t)=\Delta u(x,t)&\text{ for all }(x,t)\in U_T\\ u(x,t)=g(x)&\text{ for all }(x,t)\in\partial U \times (0,T)\\ u(x,0)=h(x)&\text{ for all }x\in U, \end{cases}$$

with initial data $h$ and boundary data $g.$

Define an energy $E[u](t)=\frac{1}{2} \int_U \vert Du(x,t)\vert^2dx$, which gives a well-defined finite number for every $t∈(0,T)$.

Problem $\mathbf1$: Prove that $\frac{d}{dt}E[u](t)\leq0$ for all $t∈(0,T)$ and that $\frac{d}{dt}E[u](t)=0$ for some $t∈(0,T)$ if and only if $\Delta u(x,t)=0$ for all $x\in U$.

Here I've got \begin{align*} \frac{d}{dt}E[u](t)&=\int_U Du_t Du\,dx\\ &=-\int_U u_t \Delta u\,dx + \int_ {\partial U} \frac{\partial U}{\partial \nu} u_t\,dS(x)\\ &= -\int_U (u_t)^2\,dx + \int_ {\partial U} \frac{\partial U}{\partial \nu} u_t \,dS(x). \end{align*}

Then how can I say that it's less or equal to zero?

Problem $\mathbf2$: Then use $\frac{d}{dt}E[u](t)\leq0$ for all $t\in(0,T)$ to establish that the initial boundary value problem above has at most one smooth solution $u: U_T\to\mathbb{R}$.

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  • $\begingroup$ Which part are you having problems with? You will need to integrate by parts to show that $\frac{d}{dt} E[u](t)=-\int_U |u_t|^2 dx\le 0$. For the uniqueness, take two solutions $u_1$ and $u_2$ and show $E[(u_1-u_2)](t)\equiv 0$, as that energy is nonnegative, initally zero and (weakly) decreasing. $\endgroup$
    – Kusma
    Sep 20, 2018 at 10:02
  • $\begingroup$ Thanks for your help. :) I have edited the question and added the part I'm stucking at. I see a bit different in the integrate results.. ah $\endgroup$ Sep 20, 2018 at 10:23
  • $\begingroup$ Note that $u_t=0$ on the boundary because $u$ satisfies $u(x,t)=g(x)$ on the boundary. $\endgroup$
    – Kusma
    Sep 20, 2018 at 11:14
  • $\begingroup$ oh right thanks heaps :) $\endgroup$ Sep 20, 2018 at 11:31

1 Answer 1

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Let's see that $\frac{d}{dt}E[u](t)≤0$, with equality if and only if $\Delta u(x,t)=0.$ $$\frac{d}{dt}E[u](t)=\frac{1}{2}\frac{d}{dt}\int_{U}\sum_{i=1}^{n}(\partial_iu(x,t))^2dx= \int_{U}\sum_{i=1}^{n}\partial_iu(x,t)\partial_t\partial_iu(x,t)dx=\\ \sum_{i=1}^{n}\int_{U}\partial_iu(x,t)\partial_i\partial_tu(x,t)dx\stackrel{(1)}{=}\\ -\sum_{i=1}^{n}\int_{U}\partial_i\partial_iu(x,t)\partial_tu(x,t)dx+ \sum_{i=1}^{n}\int_{\partial U}\partial_iu(x,t)\partial_tu(x,t)\nu_i(x) dS(x)\stackrel{(2)}{=}\\ -\int_{U}(\Delta u(x,t))^2dx+\sum_{i=1}^{n}\int_{\partial U}\partial_iu(x,t)\partial_tg(x)\nu_i(x) dS(x)=-\int_{U}(\Delta u(x,t))^2dx\leq0.$$ Where the equality $(1)$ is just integration by parts and in $(2)$ we are using the fact that $u$ satisfies the heat equation and the boundary condition.
For the second part let $u$, $v$ be solutions of the initial boundary value problem. Using the first part we know that the energy is nonincreasing. If we show that $E[u-v](0)=0$ then $E[u-v](t)=0$ for every $t$, being $E$ nonnegative. $$E[u-v](0)=\frac{1}{2}\int_{U}\vert Du(x,0)-Dv(x,0)\vert^2 dx= \frac{1}{2}\int_{U}\vert Dh(x)-Dh(x)\vert^2 dx=0.$$ This means that $Du(x,t)=Dv(x,t)$ for every $x$ and $t$ and also $\partial_t u(x,t)=\partial_t v(x,t)$, thanks to the first part. Thus $u(x,t)=v(x,t)+c$ for some constant $c$. Using the initial value you get $c=0$

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  • $\begingroup$ Thanks so much for your help. I've got it. :) $\endgroup$ Sep 20, 2018 at 12:12

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