Laplace's equation of a disk of radius 1

Question

I am asked to show that $$u(r,\theta)=\frac{2}{\pi}\arctan\left(\frac{2r \sin\theta}{1-r^2}\right)$$ satisfies the Laplace equation in a disk of radius 1.

I have tried to take the partial derivatives with respect to $r$ and $\theta$ and plug into the equation of a disk with polar coordinates. I have tried for hours and cannot figure this problem out.

Please help!

Robert Z · Accepted Answer · 2016-12-03T13:24:24.210

2

Hint. You should verify that $$r\frac{\partial}{\partial r}\left(r \frac{\partial f}{\partial r}\right)+\frac{\partial^2f}{\partial \theta^2}=0.$$ If you prefer to use cartesian coordinates let $$g(x,y)=\arctan\left(\frac{2y}{1-x^2-y^2}\right)$$ (the factor $2/\pi$ can be left out) and verify $$\frac{\partial^2 g}{\partial x^2}+\frac{\partial^2 g}{\partial y^2}=0.$$ In order to simplify the long computation you could consider $$g(x,y)=\arctan\left(A(x,y)\right)\quad \mbox{ with }\quad A(x,y)=\frac{2y}{1-x^2-y^2}. $$ Its laplacian is zero iff $$-2A(A_x^2+A_y^2)+(1+A^2)\cdot(A_{xx}+A_{yy})=0.$$

edited Dec 03 '16 at 13:24

answered Dec 03 '16 at 13:03

Robert Z

145,942

I tried the polar and could not get it. The partials were too crazy for me to understand. – David House Dec 03 '16 at 13:09
That will probably help. How did you derive the cartesian coordinates of the two variable function? – David House Dec 03 '16 at 13:18
@David House To avoid long computations you could break it in two parts.See my edited answer. It works. – Robert Z Dec 03 '16 at 13:28
I will try that. Thank you. – David House Dec 03 '16 at 13:31
@David House Note that $x=r\cos(\theta)$ and $y=r\sin(\theta)$. – Robert Z Dec 03 '16 at 13:31
Part 2. asks to show that as r →1, u(r,θ) = 1 on the upper semicircle and u(r,θ) = -1 on the lower semicircle. Any hints for this one? – David House Dec 03 '16 at 13:33
@David House Evaluate $\lim_{t\to\pm\infty}\arctan(t)$. – Robert Z Dec 03 '16 at 13:39

score 0 · Answer 2 · answered Dec 03 '16 at 13:26

0

When I saw "tried for hours" got me worried. So without wasting time doing manually, I opted to do it with Maple.

PDE:=1/r*Diff(r*diff(u(r,theta),r),r)+1/r^2*diff(u(r,theta),theta$2);

ans:=u(r, theta)=2/Pi*arctan(2*r*sin(theta)/(1-r^2));

pdetest(ans, PDE)

0

The output of the pdetest command is 0 shows that your proposed function $u(t,\theta)$ is a solution to the Laplace equation in polar coordinates.

answered Dec 03 '16 at 13:26

zhk

585
7
26

unfortunately, I have to do it by hand. – David House Dec 03 '16 at 13:27

Laplace's equation of a disk of radius 1

2 Answers2