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The Green function for the Dirichlet problem for the Laplace equation in the unit disk in $\mathbb R^2$ has the following form: $$ G(x,y) = \frac{1}{2\pi}\ln \frac{|x-y|}{|x| \bigl|y - \frac{x}{|x|^2}\bigr|}. $$ The function $G$ satisfies $\Delta_x G(x,y) = \delta_y$ at fixed $y \in D$ and $G(x,y) = 0$ when $|x| = 1$ and $|y| < 1$. I'm interested in computing the value of $$ C= \sup_{|x| \leq 1} \int\limits_{|y| \leq 1} |G(x,y)| \, dy. $$ Using Matlab I found that it is approximately equal to one. Is it possible to obtain that $C = 1$ or $C \leq 1$ using analytic techniques?

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First let's find out where the logarithm changes sign. We can rewrite the logarithm to make it $$ -\frac{1}{4\pi} \log{\left( \frac{\lvert x \rvert^2 \big\lvert y-x/\lvert x \rvert^2 \big\rvert^2 }{\lvert x - y \rvert^2}\right)}. $$ Next, expand out the terms in the fraction: $$ \frac{\lvert x \rvert^2 \big\lvert y-x/\lvert x \rvert^2 \big\rvert^2 }{\lvert x - y \rvert^2} = \frac{\lvert x \rvert^2 (\lvert y \rvert^2+1/\lvert x \rvert^{2}-2x \cdot y/\lvert x \rvert^2)}{\lvert x - y \rvert^2} \\ = \frac{\lvert x \rvert^2 \lvert y \rvert^2+1-2x \cdot y}{\lvert x - y \rvert^2} \\ = 1+\frac{(1+\lvert x \rvert^2)(1+\lvert y \rvert^2)}{\lvert x - y \rvert^2}, $$ so the logarithm is always negative, and we can therefore ignore the absolute value and consider $$ \int_{\lvert y \rvert <1}\frac{1}{4\pi} \log{\left( \frac{\lvert x \rvert^2 \big\lvert y-x/\lvert x \rvert^2 \big\rvert^2 }{\lvert x - y \rvert^2}\right)} \, dy $$ We can choose coordinates so $x=(R,0)$ and $y=r(\cos{\theta},\sin{\theta})$. Then the integral becomes $$ \int_{0}^1 \int_0^{2\pi}\frac{1}{4\pi} \log{\left( \frac{ 1+r^2 R^2 -2rR\cos{\theta} }{R^2+r^2-2rR\cos{\theta}} \right)} \, r \, d\theta \, dr $$ Do the $\theta$ integral first. Splitting up the logarithm makes this easier: we have $$ \int_{0}^1 \int_0^{2\pi}\frac{1}{4\pi} \log{\left( \frac{ 1+r^2 R^2 -2rR\cos{\theta} }{R^2+r^2-2rR\cos{\theta}} \right)} \, r \, d\theta \, dr = \frac{1}{4\pi} \int_0^1 r \left( I(1,rR)-I(r,R) \right) \, dr, $$ where $$ I(a,b) = \int_0^{2\pi} \log{(a^2+b^2-2ab\cos{\theta})} \, d\theta. $$ This integral is known (a previous question about a closely related integral is here), and we have $$ I(a,b) = 4\pi \max\{ \log{\lvert a\rvert},\log{\lvert b\rvert} \}. $$ But $0 \leqslant r,R \leqslant 1$, so $I(1,rR)=0$. Therefore the integral collapses to $$ -\int_0^1 r \max\{ \log{r},\log{R} \} \, dr = -\log{R}\int_0^{R} r \, dr - \int_R^1 r\log{r} \, dr = \frac{1}{4}(1-R^2), $$

Clearly therefore the supremum is $1/4$ (and Mathematica confirms).


Alternative: now that we know we can discard the absolute value bars, the integral is the solution of $\nabla^2 u = 1$ on the unit disk with boundary condition $u=0$ on $\lvert x \rvert=1$. Since the boundary conditions have circular symmetry, we should find that $u=u(r)$, and then we can solve the differential equation $ (ru')'/r=1 $, which gives the same solution.

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  • $\begingroup$ Thank you, Chappers. By the way, there must be minus signs in the second formula (i.e. $-2xy/|x|^2$ e.t.c.). $\endgroup$
    – Appliqué
    Oct 2, 2015 at 20:17
  • $\begingroup$ Oh yes, of course. I was mixing it up with a similar calculation I was doing earlier. Thanks. $\endgroup$
    – Chappers
    Oct 2, 2015 at 20:20

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