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I am trying to solve the following:






I can separate this into two different problems: $\theta(x,y)=v(x,y)+w(x,y)$.

Let's call one of them A:






While the other can be B:






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:


so the equation turns into $-XY''-X''Y=0$

So I think I have to solve for Y first:

$Y''=-\lambda Y$


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):

$X''=\lambda X$



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!

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(+1): Strongly thanks to change to accept the better answer in an accepted question. You are a model user. I hope that this type of model user is appearing in SE in the future. – doraemonpaul Nov 3 '12 at 22:08
up vote 1 down vote accepted

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 and

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.

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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!

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Hi! Thank you very much for your answer. I don't have such a deep insight into PDEs, so it would really help to have a "rule" for it. From what I have understood, solutions are unique in every Laplace's Equation when its boundary conditions are $f(x,y)$? Would that be true for both Neumann and Dirichlet boundary conditions? So, can I even say that for every similar Laplace's Equation to this one, once I find a solution, it will be unique? That rule would apply to $\theta$, v and w in my problem, wouldn't it? Thank you! – Serge Sep 2 '12 at 14:44

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!

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