2
$\begingroup$

Consider the following heat equation in $\Bbb R^n$: $$ \begin{align} u_t - \Delta u = 0,&\quad (x,t)\in B_1(0)\times (0,\infty)\\ u = u_0(x),&\quad (x,t)\in B_1(0)\times \{t=0\}\\ \frac12u+\partial_r u = 0,&\quad (x,t)\in\partial B_1(0)\times (0,\infty) \end{align} $$ where $u_0\in C^2(B_1(0))$, and suppose $u$ is a $C^2$ solution to this problem.

1). Show that $$\int_0^\infty\left(\int_{\partial B_1(0)} u^2(s,x) dS(x)\right)ds\le \int_{B_1(0)}u_0^2(x)dx<\infty.$$ 2). Show that $$\int_0^\infty \left(\int_{B_1(0)}u^2(s,x)dx\right)ds<\infty.$$

Since the integration is over the time domain $(0,\infty)$, a natural thought would be to relate the integrand $\int_{\partial B_1(0)} u^2(s,x) dS(x)$ or $\int_{B_1(0)}u^2(s,x)dx$ to the time-derivative of something.

Alternatively we can use Fubini's theorem to exchange the integration order. But that would lead to integrate $u^2$ over time, which possibly needs the invocation of the concrete form of $u$ (to estimate its growth). But that's something I want to avoid because I want to do everything a priori.

How can I proceed now?

$\endgroup$

1 Answer 1

1
$\begingroup$

As $u$ is $C^2$ you can multiply the equation by $u$. Then integrating with respect to $x$ $$\int_{B_1} \partial_t u u dx- \int_{B_1} \Delta u u dx =0 $$ i.e $$\frac{1}{2} \partial_t \int_{B_1} |u|^2 dx + \int_{B_1} \left| \nabla u \right|^2 dx - \int_{\partial B_1} \partial_r u \cdot u ~dS=0.$$ using the boundary conditions $$\int_{\partial B_1} \partial_r u \cdot u ~dS = - \frac{1}{2} \int_{\partial B_1} u \cdot u ~dS.$$ So finally integrating for $t$ from $0$ to $T$ you obtain $$\frac{1}{2} \int_{B_1} |u|^2(T,x) dx-\frac{1}{2} \int_{B_1} |u|^2(0,x) dx + \int_0^T\int_{B_1} \left| \nabla u \right|^2 dxdt+\frac{1}{2} \int_0^T \int_{\partial B_1} u^2~dSdt=0$$ i.e \begin{align}\frac{1}{2} \int_0^T \int_{\partial B_1} u^2~dSdt&=\frac{1}{2} \int_{B_1} |u|^2(0,x) dx-\underbrace{\frac{1}{2} \int_{B_1} |u|^2(T,x) dx}_{\geq 0}-\underbrace{\int_0^T\int_{B_1} \left| \nabla u \right|^2 dxdt}_{\geq 0} \\ &\leq\frac{1}{2} \int_{B_1} |u|^2(0,x) dx\end{align} thus the result of question 1) taking $T \to \infty$.

For the second question you now have $$ \int_0^\infty \int_{\partial B_1} u^2~dSdt < + \infty$$ $$ \int_0^\infty \int_{B_1} \left| \nabla u \right|^2 dx dt < + \infty$$ and by Poincaré inequality, there exist a constant $C>0$ such that for all $t$, $$\int_{B_1} u^2 dx \leq C \left( \int_{\partial B_1} u^2~dS+ \int_{B_1} \left| \nabla u \right|^2 dx\right).$$

It remains only to integrate over $t$.

$\endgroup$

You must log in to answer this question.