8
$\begingroup$

In our lecture we have shown that $\forall f \in L^2(\mathbb{R}^n_+) $ there is a unique $ u $ in the Sobolev space $ H^2(\mathbb{R}^n_+) $ satisfying $ -\Delta u = f. $

Now in our exercise sheet we are asked to show that there is an integral kernel $ \Phi $ such that $ u(x) = \int_{\mathbb{R}^n} \ \Phi (x-y) \ f(y) \ dy $.

Wikipedia tells me that there is an integral kernel and that it is of the form \begin{equation*} \Phi(x) \ = \ const. \ \cdot \ \frac{x_n}{({\sum_{i = 1}^{n} x_i^2})^{n/2}} \end{equation*}

So now to my question:

How can you show that this is indeed an integral kernel for poisson's equation? In particular, how can you differentiate under the integral sign and "take the Laplacian" of $ \Phi $ at $ x - y = 0 $ ? Moreover, do you know a priori that there has to be such an integral kernel?

Thanks a lot in advance, I would really appreciate your help!

Best regards

Phil

  • 3
    Does all conditions for $u$ are written down here? Solution of the problem formulated is not unique since no boundary condition is specified. The Poisson kernel $\Phi$ is for the Cauchy problem $\Delta u=0$ in $\mathbb R^n_+$, $u(x_1,\ldots,x_{n-1},0)=\psi(x_1,\ldots,x_{n-1})\ $.2011-07-06

1 Answers 1