# PDE: Laplace's Equation solutions

1. Sep 16, 2011

### dgreenheck

1. The problem statement, all variables and given/known data

Suppose that u(x,y) is a solution of Laplace's equation. If $\theta$ is a fixed real number, define the function v(x,y) = u(xcos$\theta$ - ysin$\theta$, xsin$\theta$ + ycos$\theta$). Show that v(x,y) is a solution of Laplace's equation.

2. Relevant equations

Laplace's equation: uxx + uyy = 0.

Separated solutions

X''(x) - $\lambda$X(x)=0.

Y''(y) - $\lambda$Y(y)=0.

Solutions for $\lambda$ > 0

X(x) = A1ekx + A2e-kx

Y(y) = A3cosky + A4sinky.

Solutions for $\lambda$ < 0

Y(y) = A1ekx + A2e-kx

X(x) = A3cosky + A4sinky.

3. The attempt at a solution

I began by trying to analyze each of the cases ($\lambda$>0, $\lambda$<0, $\lambda$=0) for the solution. But working these out would take forever and I know it isn't the most elegant way of doing it. My thinking is that I can somehow just differentiate the arguments for v(x,y) so I would get a factor of k2sin2$\theta$cos2$\theta$ for uxx and -k2sin2$\theta$cos2$\theta$ for uyy. Would this be a valid way of proving the statement to avoid doing all the work? Or is there a better way? Thanks.

2. Sep 16, 2011

### Hootenanny

Staff Emeritus
That is perfectly fine and the way I would do it. It is also likely to be the way intended by whomever wrote the problem.

3. Jan 7, 2012

### bcyalcin

Separated solutions
X''(x) - λX(x)=0.
Y''(y) + λY(y)=0.

there must be "+" , not "-"

4. Jan 7, 2012

### bcyalcin

by the way

therefore other equation will be changed

5. Jan 7, 2012

### TachyonRunner

Your second strategy would perhaps work best. Since you are given that u satisfies Laplace's equation, it is perhaps a good idea to plug v into Laplace's equation and try to get it in terms of u. Your main tool for that will be the chain rule for scalar fields.