Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I haven't really understood the following proof that the solution of the heat equation is unique. Could you explain it to me?

Heat equation with Dirichlet boundary conditions:

$$\left.\begin{matrix}u_t=u_{xx}, 0<x<L, t>0\\ u(0,t)=u(L,t)=0, t>0\\ u(x,0)=f(x)=0, 0<x<L\end{matrix}\right\}(1)$$ We want to show that the solution of this problem is unique:

We suppose that the problem has two solutions, $u_1(x,t), u_2(x,t)$:

$$w(t)=\frac{1}{2}\int_0^L{|u(x,t)|^2}dx, t>0, (2)$$

$$u(x,t)=u_1(x,t)-u_2(x,t)$$ $$w(t)>0, (3)$$ $$w'(t)=\frac{1}{2} \int_0^L{(u_t u^*+u u^*_t)}dx$$ $$(1):w'(t)=\frac{1}{2} \int_0^L{(u_{xx} u^*+u u^*_{xx})}dx$$ $$\int_0^L{u_{xx}u^*}dx=u_xu^*|_0^L-\int_0^L{u_xu^*_x}dx\overset{(1)}{=} - \int_0^L{|u_x|^2}dx$$ $$w'(t)=-\int_0^L{|u_x|^2}dx \leq 0, (4)$$ We know that $u_1(x,0)=u_2(x,0)=f(x), (5)$

So $u(x,0)=u_1(x,0)-u_2(x,0)=0$

$$w(t)=w(0)+\int_0^t{w'(s)}ds \leq 0, (6)$$ $$(3),(6) \Rightarrow w(t)=0, \forall t \geq 0$$ $$u_1=u_2$$

$$$$ First of all, why do we have to take at the beginning that $$w(t)=\frac{1}{2}\int_0^L{|u(x,t)|^2}dx, t>0, $$??

Why is the derivative of $w$: $w'(t)=\frac{1}{2} \int_0^L{(u_t u^*+u u^*_t)}dx$ ??

share|cite|improve this question
up vote 3 down vote accepted

To the first question. You do not take $w>0$. You define $u=u_1-u_2$ and $w(t)=(1/2)\int_0^L|u|^2dx$. It is clear that $w\ge0$. The proof consists in proving that $w\equiv0$.

For the second question. It looks like you are taking complex values. Then (assuming $z^*$ is the complex conjugate of $z$) $|u|^2=u\,u^*$ and $|u^2|_t=u_t\,u^*+u\,u^*_t$. Finally you have to justify differentiating inside the integral.

share|cite|improve this answer
Ok!!! I got it!!! Could you explain me also the relation $(6)$? $$w(t)=w(0)+\int_0^t{w'(s)}ds$$ Why is $w$ equal to that? – Mary Star May 5 '14 at 16:14
If you are studying PDE's I am sure you must know the Fundamental Theorem of Calculus. – Julián Aguirre May 5 '14 at 17:35
Yes, you're right!!!! I had been stuck... :/ Thank you for your answer!!! – Mary Star May 5 '14 at 17:42

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.