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I am really stuck in this proof. Any hint or suggestion from you would be appreciated.

Let $U \subset \Bbb{R^{n}}$ be open, bounded, with smooth boundary $\partial U$, and $ T \gt 0$. Use energy method to prove that the initial boundary-value problem

$u_{t}-\Delta u =f$ in $U_{T}$

$\frac{\partial u}{\partial \nu}=0$ on $\partial U \times [0,T]$

$u=g$ on $U \times \{t=0\}$

,where $f=f(x,t),g=g(x) $ are given smooth functions, has at most one solution $u \in C^{2}_{1}$ $(\overline U_{T})$

dmtri
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Yang
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1 Answers1

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HINT: Start by assume you have two solutions $u_{1}$ and $u_{2}$ for the given problem, and show that $u=u_{1}-u_{2}$ is identically 0, because $u$ satisfies $u_{t}=\Delta u$, with $u$ having 0 boundary data. The trick to showing $u$ is the zero function is to integrate $u_{t}-\Delta u$ against $u$, for example, and use something like $ u \Delta u = \nabla\cdot (u\nabla u)-\nabla u\cdot\nabla u$. Then follow your nose using Calculus identities and use the fact that you have 0 boundary data.

So suppose $u_{1}$ and $u_{2}$ are solutions of the problem. Then $u=u_{1}-u_{2}$ satisfies $$ u_{t}-\Delta u = 0,\;\;\; \frac{\partial u}{\partial\nu}=0 \mbox{ on } \partial U \times [0,T], \;\;\; u = 0 \mbox{ on } U \times \{ t= 0\}. $$ Therefore, $$ 0=\int_{U}uu_{t}-u\Delta u\,dx = \int_{U}uu_{t}-\nabla\cdot(u\nabla u)+|\nabla u|^{2}\,dx $$ $$ = \int_{U}uu_{t}+|\nabla u|^{2}\,dx-\int_{\partial U}u\frac{\partial u}{\partial\nu}\,dS(x), $$ where $dS$ is the surface area element on $\partial U$. (This is Gauss' Divergence Theorem.) Because $\frac{\partial u}{\partial \nu}=0$ on $\partial U$ and $2uu_{t}=(u^{2})_{t}$, the above becomes $$ 0 = \int_{U}\frac{1}{2}\frac{\partial}{\partial t}u^{2}+|\nabla u|^{2}dx $$ Now integrate the above on $[0,t']$ for any $0 \le t' \le T$, and use the fact that $u(x,t)|_{t=0}=0$: $$ 0 = \int_{U}\frac{1}{2}|u(x,t')|^{2}\,dx +\int_{0}^{t'}\int_{U}|\nabla u(x,t)|^{2}dxdt. $$ Both terms on the right are non-negative and, therefore, both must be separately 0. Just considering the first term leads to the conclusion that $u=0$ for all $x \in U$ and $0 \le t' \le T$ because $u$ is continuous. Therefore $u_{1}=u_{2}$ as desired.

Disintegrating By Parts
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  • Thanks. However, I have a question. I know how to show $u=u_1-u_2 =0$ in $U_{T}$ and $U \times {t=0}$. Can you give me some ideas on how to show u is zero on another boundary. Thank [email protected]. – Yang Dec 06 '13 at 16:18
  • Once you know that the integral of a continuous function is 0 over the open region, then you know that the function must be 0. For example, if you can show that the square of the length of the gradient is 0 over the open region, then the gradient must be 0 because it is assumed continuous. So you conclude that the function must be a constant because all of its first order derivatives are 0. You can use line integrals to conclude this. Etc. – Disintegrating By Parts Dec 06 '13 at 19:51
  • I noticed that I wrote the integral of a continuous function when I meant integral of the square of a continuous function. Big difference, and the latter applies to your case. – Disintegrating By Parts Dec 07 '13 at 22:24
  • Can I ask you how I can apply the energy method to do this proof? [email protected]. – Yang Dec 08 '13 at 04:14
  • I added more detail for you. – Disintegrating By Parts Dec 08 '13 at 07:01