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I have 2 questions from Lawrence Evans' PDE book (pages 36 and 39 in the 2nd edition copy). The first is for green's function for a half space $\mathbb{R}_{+}^{n}=\{x=(x_{1},\cdots,x_{n})\in\mathbb{R}^{n}\mid x_{n}>0 \}$ where Poisson's kernel is given by $K(x,y)=\frac{2x_{n}}{n\alpha(n)}\frac{1}{\lvert x-y\rvert^{n}}$ for $x\in \mathbb{R}^{n}_{+},y\in\partial\mathbb{R}^{n}_{+}$. The second is for green's function in a unit ball $B(0,1)$ with Poisson's kernel $K(x,y)=\frac{1-\lvert x \rvert^{2}}{n\alpha(n)}\frac{1}{\lvert x-y\rvert^{n}}$ for $x\in B(0,1) ,y\in\partial B(0,1)$. How do I show that $\int_{\partial U} K(x,y) \:\mathrm{d}y =1$ where $\partial U$ is the respective boundary of the domain?

I understand that Evaluating the Poisson Kernel in the upper half space in $n$-dimensions gives a suggested solution but I am not too convinced by changing the domain to $\mathbb{R}^{n-1}$ and his evaluation of $\lvert x-y\rvert$ as $\sqrt{x_{n}^2+y_{1}^2+...+y_{n-1}^2}$.

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1 Answer 1

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The function $I(x):=\int_{\partial U} K(x,y) \:\mathrm{d}y $ is rotational symmetric, so it is constant on the boundary of any smaller ball centered at the origin. As $K(\cdot,y)$ is harmonic, so is $I$. Thus $I$ is a constant in the interior of the unit ball by maximal principal. Evaluate $I(0)$, we get $1$. Hence $I(x)=1$ for any $|x|<1$.

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