# Laplace equation

by Logarythmic
Tags: equation, laplace
 P: 286 Does $$\nabla ^2 u(r,\theta) = 0$$ with the boundary conditions $$u(1,\theta) = u(2,\theta) = \sin^2 \theta$$ have any solutions? This was a problem on my exam but someone must have written the conditions wrong, or am I stupid?
 P: 1,069 Why not? It's a well-defined, well-behaved boundary condition over an enclosed space (the region between two concentric spheres). The general solution in spherical coordinates is sum (Ai r^i + Bi / r^(i+1))Pi(cos theta). Plug in r=1 and r=2, and rewrite the sin^2 theta as 1 - cos^2 theta... then you only need the first few Legendre polynomials and you can solve for the coefficients.
 P: 286 But the problem is to be done in plane polar coordinates...?
P: 1,069

## Laplace equation

Ah, right, sorry, I'm used to looking at these things in spherical coordinates... oh well, same wine, different bottle.

The general solutions look like A0 ln r + B0 + sum (Ai r^i + Bi / r^i)(Ci cos (i theta) + Di sin (i theta)). Use the double angle formula to get your cos^2 term.
 P: 286 How am I ever gonna get $$u(2,\theta)$$, wich contains an $$ln 2$$, to equal $$\sin^2 \theta$$? And how can you possibly write $$\sin^2 \theta$$ as a linear combination of sin and cos?
 PF Patron HW Helper Sci Advisor Thanks P: 11,935 $$\sin^{2}(\theta)=\frac{1}{2}-\frac{\cos(2\theta)}{2}$$
P: 1,069
 Quote by Logarythmic How am I ever gonna get $$u(2,\theta)$$, wich contains an $$ln 2$$, to equal $$\sin^2 \theta$$?