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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.

UfmdFkiF
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2 Answers2

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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?

Gregory
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  • Try plugging it into the ODE $X'' - \mu^2 X = 0$. Does it also satisfy $X(\pi) = 0$? – Gregory Jun 23 '18 at 19:15
  • I've done a lot of these before. I know its a linear combo of $e^{\mu x}$ and $e^{-\mu x}$ which can also be sinh and cosh. – Gregory Jun 23 '18 at 19:17
  • The 0 boundary condition helps me choose which one I want. – Gregory Jun 23 '18 at 19:19
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    How do we know $X(x)=A\sinh[\mu(\pi-x)]$ is only possible result? – UfmdFkiF Jun 23 '18 at 20:05
  • There are theorems which describe existence and uniqueness of harmonic functions with boundary conditions. – Gregory Jun 23 '18 at 20:10
  • This may be useful in explaining... https://math.stackexchange.com/questions/156505/uniqueness-theorem-for-harmonic-function – Gregory Jun 23 '18 at 20:11
  • I disagree. The process can also be done by swapping $x$ and $y$ are plugging in the b.c. to complete the solution. – Gregory Jun 23 '18 at 20:34
  • You're within your right to downvote, I did mention that the OP can finish the rest. – Gregory Jun 23 '18 at 20:36
  • I don't see where the OP said that, but ok. I see questions on satisfying BC 1 and 3 of which I did one of them. – Gregory Jun 23 '18 at 20:38
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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 $x$.

Dylan
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