Stuck on Laplace's Equation 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!
 A: Using the maximum principle one can conclude that a solution to $\Delta u = 0$ in a set $\Omega$ satisfying:
$u = 0$ on some portion of the boundary and $u_{\nu} = 0$ (the normal derivative) on the other portion; is identically $0$.
More specifically, assume $u$ is not constant. The strong maximum principle rules out the existence of an interior maximum or minimum. If a maximum or minimum occurs on the portion where $u_{\nu}$ vanishes, this is a contradiction of the Hopf lemma. Thus, the maximum and minimum points occur on the portion of the boundary where $u = 0$.
Thus we have uniqueness of solutions in your setting; all you must do is find one, and you're done!
A: In fact your Professor's first step is using a technique of variable transformation so that some of the conditions convert from inhomogeneous to homogeneous.
In fact for a PDE of $\theta(x,y)$ with some of the conditions $\theta(0,y)=f(y)$ and $\theta(a,y)=g(y)$ , the variable transformation $\theta(x,y)=v(x,y)+w(x,y)$ when choosing $v(x,y)=f(y)+\dfrac{x}{a}(g(y)-f(y))$ can convert to a PDE of $w(x,y)$ with some of the conditions $w(0,y)=0$ and $w(a,y)=0$ .
This technique is also mentioned in http://maths.swan.ac.uk/staff/vl/pdes-course.pdf#page=32 and http://maths.swan.ac.uk/staff/vl/pdes-course.pdf#page=38.
In this question, since $a=1$ , $f(y)=-1$ and $g(y)=1$ , so $2x-1$ comes from this way.
The corresponding PDE of $w(x,y)$ in this question is $-w_{yy}-w_{xx}=0$ .
Since in this question, $\theta_y(x,y)=w_y(x,y)$ , the other corresponding conditions $w_y(x,0)$ and $w_y(x,1)$ are unchanged their results.
A: I can't currently respond to your comment with a comment (I need to register officially), so I'll do so with an answer.
To determine uniqueness of solutions for $\Delta u = 0$, the most common technique is to apply a maximum principle; this works for the Dirichlet problem (prescribed boundary values) and for some mixed Dirichlet/Neumann conditions (like in your problem); some very general conditions for uniqueness can be found in, for instance, the exercises in Ch. 3 of Gilbarg-Trudinger.
We definitely don't have uniqueness for arbitrary mixed Dirichlet/Neumann boundary conditions or for all problems with leading order term $\Delta u$; indeed, if we have a solution to a problem with Neumann conditions we can get another by adding a constant. Or the problem $\Delta u + u = 0$ with $0$ boundary condition is generally not unique, having both $0$ and some eigenfunction of $\Delta$ as a solution. I hope this helps!
