Laplace $2$-D Heat Conduction Consider the following steady state problem
$$\Delta T = 0,\,\,\,\, (x,y) \in \Omega, \space \space 0 \leq x \leq 4 ,\space \space \space\space 0 \leq y \leq 2 $$
$$ T(0,y) = 300, \space \space T(4,y) = 600$$
$$ \frac{\partial T}{\partial y}(x,0) = 0, \space \space \frac{\partial T}{\partial y}(x,2) = 0$$
I want to derive the analytical solution to this problem.
1) Use separation of variables.
$$\frac{X^{''}}{X}= -\frac{Y^{''}}{Y} = -\lambda $$
$$X^{''} + \lambda X = 0 \tag{1}$$
$$Y^{''} - \lambda Y = 0 \tag{2}$$
The solution to $(2$) is
$Y(y) = C_1 \cos(ay)+C_2 \sin(ay)$
We find that 
$$C_2 = 0$$ and 
$$Y^{'}(2) = C_1\alpha \sin(2\alpha) = 0 \tag{3}$$
with $(3)$ giving that $\alpha = n\frac{\pi}{2}$
so $Y$ is given by
$$Y(y) = C_1sin(\frac{n\pi}{2}y)$$
The solution to $(1)$ is
$$X(x) = Ae^{\alpha x}+Be^{-\alpha x}$$
where $\alpha$ is given by $\alpha = n\frac{\pi}{2}$ 
So the solution is:
$$u(x,y) = X(x)Y(y) = C_1sin(\alpha y)(Ae^{\alpha x}+Be^{-\alpha x}) \tag{4}$$
where $\alpha$ is given by $\alpha = n\frac{\pi}{2}$ 
Inserting the B.C. in $(4)$ gives:
$$u(0,y) \implies E_n = \frac{300}{sin(\alpha y)}$$
$$u(4,y) \implies F_n = \frac{600}{G sin(\alpha y)Ae^{4 \alpha }+Hsin(\alpha y)e^{-4 \alpha }}$$
This is how far I have come. How do I continue?

Sort of guide


*

*Transform the equation such that it'll have homogeneous boundary conditions.
I suggest to use $W(x, y) = 100 + 50x$ : this is the simplest function that has $W(0, y) = 100$, $W(2, y) = 200$ and by the way $\frac{\partial W}{\partial y} \equiv 0$. What will happen to solutions of original equation if we subtract $W(x, y)$ ? Let's see:
$$\Delta (T - W) = \Delta T - \Delta W = 0 - 0 = 0. $$
So, $T-W$ solves the equation $\Delta u = 0$, but with homogeneous boundary conditions of the same type.

*From separation of variables you have that you are trying to find solutions of form $u(x, y) = X(x) \cdot Y(y)$ such that they satisfy boundary conditions and $\Delta u = 0$. This leads to following equations:
$$ X'' = - \lambda X $$
$$ Y'' = \lambda Y $$
plus boundary conditions. Because we want to find non-trivial solutions to this equation, boundary conditions yield:
$$ u(0, y) = 0 \Leftrightarrow X(0) Y(y) = 0 \Leftrightarrow X(0) = 0 $$
$$ u(2, y) = 0 \Leftrightarrow X(2) Y(y) = 0 \Leftrightarrow X(2) = 0 $$
$$ u'_{y}(x, 0) = 0 \Leftrightarrow X(x) Y'(0) = 0 \Leftrightarrow Y'(0) = 0 $$
$$ u'_{y}(x, 4) = 0 \Leftrightarrow X(x) Y'(4) = 0 \Leftrightarrow Y'(4) = 0 $$
Then you find for what values of $\lambda$ you can satisfy these boundary conditions. You will obtain a countable (or empty) set of such 'eigenvalues' $\lambda_k$ with corresponding functions $X_k$ and $Y_k$ (each depends on two parameters -- that is because they are general solutions of second order ODE.

*Find the solution in form $U(x, y) = \sum_{k \in \mathbb{N}} X_k (x) \cdot Y_k (y) $. Any finite or infinite sum of such functions satisfy Laplace equation and boundary conditions, so you just have to determine coefficients in $X_k$ and $Y_k$. 

*Don't forget to add $W(x, y)$ to $U(x, y)$ :)
