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Suppose $u_t=u_{xx}$ on $(0,1)\times(0,\infty)$ and $\int_0^1u(x,0)dx=0$ with Neumann boundary condition $u_x(0,t)=u_x(1,t)=0$. Show that $\int_0^1u^2(x,t)dx\to0$ as $t\to\infty$.

The $t$-derivative of $\int_0^1u(x,t)dx=0$ is 0, so it is constantly 0. Also, $\int_0^1 u^2(x,t)dx$ is decreasing, but why should it go to 0?

abc
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  • Something's wrong with the problem as you've stated it. Set $u(x,t)=1/2-x.$ This is a solution, and $\int_0^1 u(x,0)\ dx=0$. But the integral of $u^2$ does not go to zero. – Potato Mar 30 '14 at 07:44
  • forgot to write the boundary condition – abc Mar 30 '14 at 07:53

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