# The solution of the heat equation is unique

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

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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.
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