I am trying to solve the following:
$-\theta_{yy}-\theta_{xx}=0$
$\theta(0,y)=-1$
$\theta(1,y)=1$
$\theta_y(x,0)=1$
$\theta_y(x,1)=0$
I can separate this into two different problems: $\theta(x,y)=v(x,y)+w(x,y)$.
Let's call one of them A:
$-v_{yy}-v_{xx}=0$
$v(0,y)=-1$
$v(1,y)=1$
$v_y(x,0)=0$
$v_y(x,1)=0$
While the other can be B:
$-w_{yy}-w_{xx}=0$
$w(0,y)=0$
$w(1,y)=0$
$w_y(x,0)=1$
$w_y(x,1)=0$
On the other hand, Professor's first step in the solution is to say that $\theta(x,y)=2x-1+w(x,y)$. So I think he got the solution for v(x,y) substituted into the equation. I have tried to get there myself but...
Going for A, by separation of variables:
$v(x,y)=X(x)Y(y)$
so the equation turns into $-XY''-X''Y=0$
So I think I have to solve for Y first:
$Y''=-\lambda Y$
$Y'(0)=Y'(1)=0$
And solving the eigenvalue problem I get:
$\lambda_0=0; Y_0(y)=1$
$\lambda_n=(n\pi)^2; Y_n(y)=\cos(n\pi y)$
Then, I continue trying to solve for X(x):
$X''=\lambda X$
$X(0)=-1$
$X(1)=1$
I can substitute $\lambda$ so $X''=(n\pi)^2 X$. I have noticed that $\lambda_0 = 0 $ (n=0) is an eigenvalue and I get the eigenfunction $X_0(x)=2x-1$ which is OK with the solution, but I don't know how to prove that that is the only one.
I would really appreciate a piece of advice. Thank you very much!