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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


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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})\ $. – Andrew Jul 6 '11 at 14:16

it's very late and probably you know the answer already, but in case others are wondering about the same question, I thought I'd write a few lines.

If you want to solve $-\Delta u=f$ on $\mathbb{R}^n$, then $u(x)=\int_{\mathbb{R}^n}\Phi(x-y)f(y)dy$, where

$$\Phi(x)= \left\{ \begin{array}{ll} -\frac{1}{2\pi}\mathrm{log}(|x|) & \mbox{if } n = 2 \\ \frac{1}{n(n-2)\alpha(n)}\frac{1}{|x|^{n-2}} & \mbox{if } n \geq 3 \end{array} \right.$$

A possible reference is Chapter 2.2 of Evans: Partial Differential Equations (in particular pp. 20-25). There you find a detailed proof of this statement in the case that $f \in C_{c}^2(\mathbb{R}^n)$, that is, $f$ is twice continuosly differentiable with compact support. In particular, Evans takes care about your (justified) worries about "differentiating through the integral sign". On the other hand, I don't know if there is any argument for the existence of this integral kernel a priori.

However, as Andrew pointed out already, the integral kernel you have written down is the Poisson kernel for $\mathbb{R}_{+}^{n}$, which is designed to solve Laplace's equation ($\Delta u =0$) on $\mathbb{R}_{+}^{n}$, and NOT, as its name might suggest, Poisson's equation ($-\Delta u =f$), which you want to solve (supposedly). If you want to solve Laplace's equation on $\mathbb{R}_{+}^{n}$ with boundary condition $u=g$ on $\partial \mathbb{R}_{+}^{n}$ instead, then the solution is

$$u(x)=\int_{\partial \mathbb{R}_{+}^{n}}K(x,y)g(y)dy$$

where $K(x,y)$ is the Poisson kernel


You can read about all of that and more, again, in Evans, Chapter 2.2.4.b (in particular on page 37).

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