# The initial boundary value problem of the heat equation

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 $$u: U_T\to\mathbb{R}$$.

• 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. Sep 20, 2018 at 10:02
• 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 Sep 20, 2018 at 10:23
• Note that $u_t=0$ on the boundary because $u$ satisfies $u(x,t)=g(x)$ on the boundary. Sep 20, 2018 at 11:14
• oh right thanks heaps :) Sep 20, 2018 at 11:31

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 $c=0$