PDE: Laplace's Equation solutions

[dgreenheck]

Homework Statement

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.

Homework Equations

Laplace's equation: u_xx + u_yy = 0.

Separated solutions

​

X''(x) - \lambdaX(x)=0.

​

Y''(y) - \lambdaY(y)=0.

Solutions for \lambda > 0

​

X(x) = A₁e^kx + A₂e^-kx

​

Y(y) = A₃cosky + A₄sinky.

Solutions for \lambda < 0

​

Y(y) = A₁e^kx + A₂e^-kx

​

X(x) = A₃cosky + A₄sinky.

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\thetacos²\theta for u_xx and -k²sin²\thetacos²\theta for u_yy. Would this be a valid way of proving the statement to avoid doing all the work? Or is there a better way? Thanks.

 

[Hootenanny]

My thinking is that I can somehow just differentiate the arguments for v(x,y) so I would get a factor of k²sin²\thetacos²\theta for u_xx and -k²sin²\thetacos²\theta for u_yy. Would this be a valid way of proving the statement to avoid doing all the work? Or is there a better way? Thanks.

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.

 

[bcyalcin]

Separated solutions
X''(x) - λX(x)=0.
Y''(y) + λY(y)=0.there must be "+" , not "-"

 

[bcyalcin]

by the way

therefore other equation will be changed

 

[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.