Find harmonic function

Question

Find harmonic function in $U=[0,\pi]^2\subset R^2$ such that \begin{cases} u(x,0)=3\sin(x)+7\sin(17x)\\ u(x,\pi)=0\\ u(0,y)=\sin(3y)+17\sin(7y)\\ u(\pi,y)=0 \end{cases}

I tried to use separation of variables and look for solution in form $$u(x,y)=X(x)Y(y)$$ But I don't know how to satisfy 1. and 3. condition.

Answer 1

Harmonic functions satisfy $\Delta u = 0$. Let $u(x,y) = X(x) Y(y)$ then $X'' Y + X Y'' = 0$. Rearranging, $X''/X = - Y''/Y = C$ where $C$ is a constant.

Consider first $C > 0$, and define $\mu^2 = C$, then $X'' - \mu^2 X = 0$. Solving yields $X(x) = A \sinh [\mu (\pi - x)]$. Similarly $Y(y) = B\sin \mu y$. Plugging in the boundary condition $Y(\pi) = B \sin \mu \pi = 0$ implies that $\mu_n = n$ for $n \in \mathbb Z$. Hence a solution is given by $$ u(x,y) = \sum_n A_n \sinh[n(\pi - x)] \sin n y. $$ Plugging in the condition $u(0,y)$ implies that $A_n = 0$ unless $n = 3,7$, hence $$ u(x,y) = \frac{\sinh[3(\pi-x)]}{\sinh 3\pi} \sin 3y + \frac{17 \sinh[7(\pi-x)]}{\sinh 7\pi} \sin 7y. $$ Can you finish the rest?

Answer 2

You can't perform separation of variables just yet. It's not really possible to match both of the inhomogeneous boundaries to the same product, as you have probably tried. What we usually do, is to solve two separate problems, such that

$$ \begin{matrix} \begin{cases} \nabla^2u_1=0 \\ u_1(x,0) = 3\sin x + 7\sin (17x) \\ u_1(x,\pi) = 0 \\ u_1(0,y) = 0 \\ u_1(\pi,y) = 0 \end{cases} && \begin{cases} \nabla^2u_2=0 \\ u_2(x,0) = 0 \\ u_2(x,\pi) = 0 \\ u_2(0,y) = \sin(3y) + 17\sin(7y) \\ u_2(\pi,y) = 0 \end{cases} \end{matrix} $$

You can check that $u = u_1 + u_2$ solves the original problem.

For some more intuition, let's solve the first problem in $u_1$. Separating the variables:

\begin{align} X'' + \lambda X = 0, & \quad X(0) = X(\pi) = 0 \\ Y'' - \lambda Y = 0, & \quad Y(\pi) = 0 \end{align}

which gives

\begin{align} X(x) &= \sin(nx) \\ Y(y) &= \sinh\big(n(\pi-y)\big) \end{align}

where $\lambda_n=n^2$. Notice, the two homogeneous B.C.s on $X(x)$ allows us to "lock" the solution to a specific $\sin$ function. This wouldn't have been possible in the original problem.

So the general solution for the first problem is

$$ u_1(x,y) = \sum_{n=1}^\infty a_n \sin(nx) \sinh\big(n(\pi-y)\big) $$

The final inhomogeneous boundary gives

$$ u_1(x,0) = \sum_{n=1}^\infty a_n \sinh(n\pi) \sin(nx) = 3\sin x + 7\sin(17x) $$

You can make your own conclusion about what happens here.

The second problem is nearly identical to the first, except the $\sin$ series is on $y$ instead of $x$.