1
$\begingroup$

$u_{t} - u_{xx} + cu = f $ on $U_{T}$

$u = g$ on $\Gamma_{T}$

Problem is Show uniqueness of solution. I tried to find it with energy method, but the method doesn't working since $\frac{\partial }{\partial t}e(t) = 2\int_{U}ww_{t} + w^{2} dx = -\int_{U}|Dw|^{2}dx+e(t)$ Could you give me a hint?

Edited; Proof : Just let $v(x,t) = e^{ct}u(x,t)$ then I can use energy method since $v(x,t)$ solves non homogeneous heat equation, which I can deal with!

$\endgroup$
  • $\begingroup$ What are $U_T$ and $\Gamma_T$? $\endgroup$ – Paul Feb 7 '14 at 2:19
  • $\begingroup$ @Paul Just $U_{T} = \mathbb{R}^{n}\times (0,T)$, $\Gamma_{T} = \mathbb{R}^{n}\times (t=0)$. I think I prove it! $\endgroup$ – user124697 Feb 7 '14 at 2:27
1
$\begingroup$

Your method works fine. Here's another idea:

Suppose that $u_1$ and $u_2$ are two solutions to the variational heat equation as given. It follows that $v := u_1-u_2$ satisfies

$v_{t} - v_{xx} + cv = 0 $ on $U_{T}$

$v = 0$ on $\Gamma_{T}$

(Why is this the case?). It follows that $v(x,t) = 0$ (why is this the case)? That is, $u_1(x,t) - u_2(x,t) = 0$, which is to say that $u_1$ and $u_2$ are identical solutions. We conclude that the solution to this IVP is unique.

$\endgroup$
  • $\begingroup$ Actually I know answer of first question, but not second question. Why v(x,t) =0? $\endgroup$ – user124697 Feb 7 '14 at 2:36
  • 1
    $\begingroup$ Does the energy method not work for $v$? If not, I guess you could just note that $e^{ct}v(x,t)$ solves the heat equation to come to the same conclusion. $\endgroup$ – Omnomnomnom Feb 7 '14 at 2:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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