- #1
dgreenheck
- 23
- 0
Homework Statement
Suppose that u(x,y) is a solution of Laplace's equation. If [itex]\theta[/itex] is a fixed real number, define the function v(x,y) = u(xcos[itex]\theta[/itex] - ysin[itex]\theta[/itex], xsin[itex]\theta[/itex] + ycos[itex]\theta[/itex]). Show that v(x,y) is a solution of Laplace's equation.
Homework Equations
Laplace's equation: uxx + uyy = 0.
Separated solutions
X''(x) - [itex]\lambda[/itex]X(x)=0.
Y''(y) - [itex]\lambda[/itex]Y(y)=0.Solutions for [itex]\lambda[/itex] > 0
X(x) = A1ekx + A2e-kx
Y(y) = A3cosky + A4sinky.Solutions for [itex]\lambda[/itex] < 0
Y(y) = A1ekx + A2e-kx
X(x) = A3cosky + A4sinky.The Attempt at a Solution
I began by trying to analyze each of the cases ([itex]\lambda[/itex]>0, [itex]\lambda[/itex]<0, [itex]\lambda[/itex]=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[itex]\theta[/itex]cos2[itex]\theta[/itex] for uxx and -k2sin2[itex]\theta[/itex]cos2[itex]\theta[/itex] 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.