Note: Use the prime notation for derivatives, so the derivative of $X$ is written as $X^\prime$. Do NOT use $X^\prime(x)$. The variable $\lambda$ is typed lambda.

Solve Laplace's equation with homogeneous vertical boundary conditions:

$\displaystyle \frac{\partial^2 u}{\partial x^2} +\frac{\partial^2 u}{\partial y^2}=0, 0< x < a, 0 < y < b$

$\displaystyle u(0,y)=0, \hskip 10pt u(a,y)=0, 0 < y < b$

$\displaystyle u(x,0) = f(x), \hskip 10pt u(x,b) = g(x),0

We try for a separable solution $u(x,y) = X(x)Y(y)$, plugging $XY$ into the PDE for $u$ we get

$=0$

The PDE can be separated into an ODE in X and an ODE in Y (put all multiplicative constants in the DE for Y):

= = $-\lambda$

where $\lambda$ is a constant. The differential equations can be separated because they are independent of each other.

The boundary conditions in the PDE translate into initial conditions for the differential equation in X:

DE in X: $= 0$

Initial Conditions (IC's) in X:

$X($$) =$
$X($$) =$

We have solved this S-L problem before. For each $n=1,2,3,\ldots$ there is, up to a multiplicative constant, one basic solution, $X_n$, which is a trigonometric function with period $2a/n$:

$X_n(x) =$ .

The $\lambda$ corresponding to this solution is $\lambda_n =$. With $\lambda=\lambda_n$, the corresponding differential equation for $Y$ is:

DE in Y: $= 0$

Using A and B as the constants, the solution for Y with $\lambda=\lambda_n$ is a linear combination of hyperbolic functions $A\cosh(\cdots)+B\sinh(\cdots)$

$Y_n(y) =$ .

The PDE is linear so it admits a series solution that is a linear combination of the solutions $X_n(x)Y_n(y)$

$u(x,y) = \displaystyle\sum\limits_{n=1}^\infty\left( A_n \cosh\left(\frac{n\pi}{a}y\right) +B_n \sinh\left(\frac{n\pi}{a}y\right)\right) \sin\left( \frac{n\pi}{a}x\right)$.
Now we can apply the lateral boundary conditions:
$u(x,0) = \displaystyle\sum\limits_{n=1}^\infty A_n\sin\left( \frac{n\pi}{a}x\right) =f(x)$
$u(x,b) = \displaystyle\sum\limits_{n=1}^\infty\left( A_n \cosh\left(\frac{n\pi}{a}b\right) +B_n \sinh\left(\frac{n\pi}{a}b\right)\right) \sin\left( \frac{n\pi}{a}x\right)=g(x)$
The first series we recognize as a Fourier series on [0,a], so

 $A_n =$ ---------- $\displaystyle\int$ $f(x)$ dx

The second series is also a Fourier series with coefficient

 $\displaystyle A_n \cosh\left(\frac{n\pi}{a}b\right) +B_n \sinh\left(\frac{n\pi}{a}b\right)=$ ---------- $\displaystyle\int$ $g(x)$ dx

Which we solve for $B_n$
 $B_n=$ $\hskip 40pt 1$----------------------------------- $\Biggl($ ---------- $\displaystyle\int$ $g(x)$ $dx - A_n$ $\Biggr)$

You can earn partial credit on this problem.