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!

Answer 1

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 $$-2A(A_x^2+A_y^2)+(1+A^2)\cdot(A_{xx}+A_{yy})=0.$$

Answer 2

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.