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$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!

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

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.

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Actually I know answer of first question, but not second question. Why v(x,t) =0? – user124697 Feb 7 '14 at 2:36
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. – Omnomnomnom Feb 7 '14 at 2:40

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