A really difficult heat-diffusion problem 
A very long beam with a quadratic cross-section
$0\le x\le L$,
$0\le y\le L$ lies on a plane with a temperature of 0 degrees. The other
surfaces of the beam are in contact with air which holds 10 degrees
temperature. Determine the stationary temperature distribution in the
beam, assuming it is of infinite length $-\infty\le z\le \infty$.

The solution:
I am sure that there are two ways to solve this, one using the heat equation, and one using the Laplace equation.
The heat equation requires a time-dimension, which is not given, however they term "stationary" says it all. $\frac{\partial}{\partial t}=0$. So the heat equation for this should be:
$$\nabla^2 u=0$$
which in fact becomes a Laplace equation, where $u=u(x,y,z)$.
There is a problem with ansatz, it is not easy to make an ansatz here, but I shall try. Consider that the x and y dimensions at the origin are both in contact with a zero-temperature surface and a 10-degree temperature air, then we could consider both to be non-zero at the origin, if we use Kelvin degrees. So we assume both are cosinoid, and therefore have the ansatz:
$$u(x,y,z)=\cos\frac{n\pi}{L}x\cos\frac{m\pi}{L}y\cdot u(z)$$
The operator is then:
$$\nabla^2\bigg(\cos\frac{n\pi}{L}x\cos\frac{m\pi}{L}y\cdot u(z)\bigg):=-\frac{\cos\frac{m\pi x}{L}\cos\frac{n\pi x}{L}\big(\pi^2(m^2+n^2)\big)}{L^2}u(z) -u_{zz} $$
This gives
Insert in the PDE:
$$u_{zz} +\frac{\cos\frac{m\pi x}{L}\cos\frac{n\pi x}{L}(\pi^2(m^2+n^2)\big)}{L^2}u(z) =0$$
Let $$f(x,y)=\frac{\cos\frac{m\pi x}{L}\cos\frac{n\pi x}{L}\big(\pi^2(m^2+n^2)\big)}{L^2}$$, then we have the "solution":
$$u(z)=C_1\cos (f(x,y)\cdot z)+C_2\sin(f(x,y)\cdot z)$$
Insert in the original ansatz and expand:
$$u(x,y,z)=\cos\frac{n\pi}{L}x\cos\frac{m\pi}{L}y\cdot \big(C_1\cos \bigg(\frac{\cos\frac{m\pi x}{L}\cos\frac{n\pi x}{L}\big(\pi^2(m^2+n^2)\big)}{L^2}\cdot z\bigg)+C_2\sin\bigg(\frac{\cos\frac{m\pi x}{L}\cos\frac{n\pi x}{L}\big(\pi^2(m^2+n^2)\big)}{L^2}\cdot z\bigg)$$
Looking only at a level curve of $u(x,y,z)$, with $z=0$, it looks like this

But the function $u(x,y,z)$ is weird, because each time a level curve is made with x or y dimension equal to zero, then the z-term becomes 1.
In fact, considering a similar ODE problem:
$$y''+\cos xy'=0$$ gives a rather exotic solution, so solving the given PDE must be impossible, analytically.
Does anyone have an idea on how to solve this problem?
Thanks
 A: By Sals suggestion, I add the proposed solution.
Let the boundaries of the heat-diffusion problem be defined as

We ignore the z-dimension.
The PDE is given therefore by:
$$\nabla^2 u=0 \ \ \ \ \ \ 0\le x\le L, \ \ \ \ 0\le y\le L\\
u(0,y)=10,\ \ \ \ u(L,y)=10  \ \ \ \  \ (1)\\
u(x,0)=0, \ \ \ \ u(x,L)=10 \ \ \ \ \ \ (2)$$
The IC in (1) are Dirichlet homogeneous on x, thus a suitable ansatz is $u(x,y)=\sin\frac{n\pi x}{L}u(y)$
We insert in the PDE and get:
$$-\bigg(\frac{n\pi}{L}\bigg)^2\sin\frac{n\pi x}{L}u(y)+\sin\frac{n\pi x}{L}u_{yy}=0$$
which gives the ODE:
$$u(y)_{yy}-\bigg(\frac{n\pi}{L}\bigg)^2u(y)=0$$
This gives the solution $$u(y)=C_1\cosh\frac{n\pi}{L}y+C_2\sinh\frac{n\pi}{L}y$$
Use I.C. from (2) and obtain:
$$u(y)=\frac{10}{\sinh n\pi}\sinh\frac{n\pi}{L}y$$
Then, the final solution is:
$$u(x,y)=\frac{10}{\sinh n\pi}\sinh\frac{n\pi}{L}y\sin\frac{n\pi}{L}x$$
Which has the plots


Any comment or eventual improvements are welcome!
A: The BVP is well posed and the solution may be represented as an infinite sum. The mistake is in the ansatz $u(x,y)=u(y)\sin(n\pi x/L)$, because the BCs at $x=0$ and $x=L$ are non-homogeneous.
Let $u(x,y)=v(x,y)+C$, with $C=10$. Now $v$ also satisfies Laplace's equation in the square but with the simpler BCs
$$\tag{1}
v(x,0)=-C\qquad,\qquad v(x,L)=v(0,y)=v(L,y)=0
$$
Solving (1) by separation of variables $v(x,y)=X(x)Y(y)$ is straightforward see eg here
. The result is
$$\tag{2}
v(x,y)=\sum\limits_{n=1}^\infty c_n\sin(n\pi x/L)\sinh(n\pi-n\pi y/L)\\
c_n=\frac{2}{L\sinh(n\pi)}\int\limits_0^L dx \ \sin(n\pi x/L)v(x,0)
$$
With $v(x,0)=-C$ we find
$$\tag{3}
c_n=-\frac{2C}{n\pi \sinh(n\pi)}\left(1-(-1)^n\right)
$$
Inserting (3) into (2) we have
$$\tag{4}
v(x,y)=-4C\sum\limits_{n=1,3,5\cdots}\frac{\sinh(n\pi-n\pi y/L)}{n\pi\sinh(n\pi)}\sin(n\pi x/L)
$$
Before plotting, it's useful to switch to dimensionless variables. Define $\chi=x/L$, $\eta=y/L$, $\phi=u/C$. The BVP is now posed on $0<\chi<1$, $0<\eta<1$, with BCs $\phi=1$ and $\phi=0$. The solution is
$$\tag{5}
\phi(\chi,\eta)=1-\sum\limits_{n=1,3,5,\cdots} \frac{4 \sinh(n\pi-n\pi \eta)}{n\pi \sinh(n\pi)}\sin(n\pi \chi)
$$
Here is a plot of the $n=1$ term (left) and the partial sum to $11$ terms (right)

