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Consider the following problem: $$ u_{t}=u_{xx}+c^2u+u^2,\quad x\in(0,L),\, t\in(0,T)\\ u(0,t)=u(L,t)=0,\quad t\in [0,T]. $$ Define the energy as: $$ E(t)=\int_0^L u(x,t)\phi(x)dx, $$ where $\phi(x)=a\sin(bx)$ with $a,b$ chosen so that $\int_0^L \phi(x)dx=1$, $\phi$ is positive in $(0,L)$ and $\phi(L)=0$. If $cL\geq \pi$, show that $E^2\leq \frac{dE}{dt}$ and $E(0)T<1$.

By some calculations, I reach $\frac{dE}{dt}=(c^2-b^2)E(t)+\int_0^L u^2\phi dx$. But I have no idea how to process on. Can anyone give me a hint?

bdl10
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1 Answers1

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First, note that if you think of $\phi$ as a probability density (which you can do since it integrates to $1$), then $$ \int_0^L u^2 \phi dx - \left(\int_0^L u \phi dx\right)^2 = \left\langle u^2 \right\rangle - \left\langle u \right\rangle^2 = \text{var}(u) \ge 0\implies \int_0^L u^2 \phi dx\ge E^2. $$ You have shown that $\frac{dE}{dt}=(c^2-b^2)E+\int_0^Lu^2\phi dx\ge(c^2-b^2)E+E^2$.

You know that $bL = \pi$ since $\phi = a\sin(bx)$ is zero only at $0$ and $L$. If $cL\ge \pi$, then $c\ge b$ which mean $(c^2-b^2)E\ge 0$, which gives you $$\frac{dE}{dt}=(c^2-b^2)E+\int_0^Lu^2\phi dx\ge(c^2-b^2)E+E^2\ge E^2.$$

stochastic
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