# 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 $g:U→\mathbb{R}$ and define $A$ = {$v:U→\mathbb{R}$ smooth | $v(x)=g(x)$ for all $x$ ∈ $\partial U$}.

Now pick any $h∈A$ and suppose that $u: U_T→\mathbb{R}$ is a smooth solution of the following initial boundary value problem of the heat equation, i.e.,

$u_t (x,t)=\Delta u(x,t)$ for all $(x,t)∈U_T$,

$u(x,t)=g(x)$ for all $(x,t) ∈ \partial U × (0,T)$,

$u(x,0)=h(x)$ for all $x∈U$,

with initial data h and boundary data g.

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

1. Prove that $\frac{d}{dt}E[u](t)≤0$ 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 $x∈U$.

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

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

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

• 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. – Kusma Sep 20 '18 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 – Evelyn Venne Sep 20 '18 at 10:23
• Note that $u_t=0$ on the boundary because $u$ satisfies $u(x,t)=g(x)$ on the boundary. – Kusma Sep 20 '18 at 11:14
• oh right thanks heaps :) – Evelyn Venne Sep 20 '18 at 11:31

## 1 Answer

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 $c=0$

• Thanks so much for your help. I've got it. :) – Evelyn Venne Sep 20 '18 at 12:12