general solutions of the 2D Laplace equation

Question

I haven't studied a-level mathematics and I would like to seek some advice on how to tackle this maths problem. If you have a moment, can you show me step-by-steps for the boundary conditions I want to solve please.

Much appreciated :)

Answer 1

Actually as I am working on it, you are missing a condition on $u_y$ on the boundary to apply the Laplace transform. My previous comment was misguided, and separation of variables can be used to solve this problem.

Assume we can write $u(x,y) = X(x)Y(y)$, then we can substitute and rearrange to obtain $$\frac{X''}{X}=-\frac{Y''}{Y},$$ but the LHS is a function only of $x$ and the RHS is a function only of $y$, so for them to be equal, they must be constant, call it $-\mu^2$. We then have 2 ODEs: $$X''+\mu^2X=0, \ \ \ X'(0)=X'(1)=0$$ and $$Y''-\mu^2Y=0, \ \ \ Y(\infty)=0.$$ Notice that one of these is an eigenvalue problem and the other is just an IVP (essentially). The initial condition is left ambiguous so that it can later be applied to the true boundary condition.

I won't go through all the details of solving the ODEs, but we get $$ X(x) = b_n\cos(\mu x), \ \ \ \mu = n\pi $$

and $$ Y(y) = ce^{-\mu y}. $$ The case $\mu=0$ does not satisfy the condition on $Y(y)$, so we ignore it.

We have found a sequence of solutions $u_n(x,y) = X_n(x)Y_n(y)$ for each $n$. Since the equation is linear, the general solution will be an infinite sum of these simpler solutions, since the sum of solutions to a linear homogenous equation is also a solution. So we have $$ u(x,y) = \sum_{k=1}^\infty b_ke^{-k\pi y}\cos(k\pi x). $$ Notice that we absorbed the constant $c$ into the constants $b_n$ since both are arbitrary. This solution satisfies every condition except for the one at $y=0$, so we find that next.

$$ u(x,0) = \sum_{k=1}^\infty b_k\cos(k\pi x) = \cos(n\pi x). $$ For this specific boundary condition, we can see that the solution corresponds to $b_k = 1$ when $k=n$ and $b_k=0$ elsewhere. So our general solution is $$ u(x,y) = e^{-n\pi y}\cos(n\pi x). $$