Separation of variables on upper half-plane issue

[loesung]

Given the Cauchy problem
[tex]
\begin{cases}
u_{xx}+u_{yy}=0, & 0<y<\infty, x\in\mathbb{R} \\
u(x,0)=0 & \mbox{on }\{ x_2=0\}\\
\frac{\partial u}{\partial y}=\frac{1}{n}\sin (n y) & \mbox{on }\{ x_2=0\}\\
\end{cases}[/tex]

I am given that, by using using separation of variables, the solution ought to be
[tex]u(x,y)=\frac{1}{n^2}\sinh ny \sin nx. \hspace{0.6 in } (*)[/tex]
The upper half plane is really freaking me out . Do we need to make a (Moebius) transformation before solving? I was told to just use separation of variables, but I am not sure how to deal with an unbounded domain.


I know that the solution in general has the form

[tex]\sum_{j=0}^\infty c_j \sin(j n x )\sinh (j n x),[/tex]
In particular I am not sure how handle the coefficients here when the domain is the upper half plane. Based on the given solution above, I would expect that

[tex]c_n=\int_0^{something}\frac{1}{n}\cos{nx}\Rightarrow
\cdots \Rightarrow \frac{1}{n^2},[/tex]

(I am not sure about the limits of the integral, or if this is even correct)

So then the solution would be

[tex]u(x,y)=\frac{1}{n^2}\sin{nx}\sinh{ny}. [/tex]

The lack of any detail here just highlights that I am not sure what I am doing! I would appreciate any advice!

Thanks in advance,



Los

(by the way, this is Hadamard's example of an ill-posed problem, but that's not my issue).

 

[gtoguy97]

Yes, you are correct in your approach. Using separation of variables, we can write the solution as u(x,y) = \sum_{j=0}^\infty c_j \sin(j n x )\sinh (j n y). To find the coefficients, we use the given boundary conditions. For the second boundary condition, we have\frac{\partial u}{\partial y} = \sum_{j=0}^\infty j n c_j \sin(jnx)\cosh(jny)|_{y=0} = \frac{1}{n}\sin(nx). This gives us the coefficient $c_n=\frac{1}{n^2}$. Then, using this value for $c_n$, the solution is given by u(x,y) = \frac{1}{n^2}\sin(nx)\sinh(ny). This is the same result as given in (*).