Okay, I've been working at this a couple of days now, I will try to give relevant details but will omit some intermediate steps.

The problem as given says:

Consider the BVP for $u=u(x,y)$:

PDE: $u_{xx}+u_{yy}=0\ \text{ for }\ 0<x<\infty,\ 0<y<\pi$

BC: $u(x,0) = u(x,\pi) = 0,\ u(0,y) = f(y)$

The solution is to be bounded in the strip domain, $f$, $f'$ piecewise continuous. Find the separated solutions.

Use superposition and Fourier series to find the solution. Plot approximations for $f(y)=y$ and $f(y)=y(\pi-y)$.

So I used separation of variables and got $u_n=\sin(ny)\sinh(nx)$ and $u_n=\sin(ny)\cosh(nx)$.

One piece of advice I got from my instructor was that we needed to take a linear combination of them to ensure we got our rapidly decreasing exponential so I called the first one $u_{n_a}$ and the second $u_{n_b}$, then I made my linear combination $u_{n_b}-u_{n_a}=0$ which left me after some conversion to exponentials and rearranging $\sin(ny)e^{-nx}$. Then I wound up getting for real coefficients $a_0=0$ rather trivially, $a_n=0$ by orthogonality, and $b_n=\frac{e^{-nx}}{2}$.

So I'm wondering, am I on track here, would my solution for b just be the infinite sum of my $b_n$ multiplied by the $\sin(ny)e^{-nx}$ function?

I could really use some advice on where to go in this problem at this point, if anyone's able and willing. Thanks so very much to anyone who can help.

  • $\begingroup$ I'm thinking I need to not find the real series, it's not making sense to me to do that, but I'm thinking for superposition rather to find the summation $\sum^{\infty}_{n=1}a_n \sin (ny)e^{-nx} \sin (ny)$. Is this a more helpful direction to travel? $\endgroup$
    – Tristen
    Commented May 5, 2013 at 4:18

1 Answer 1


Yes, everything you did makes sense. It would be easier to work with $e^{\pm nx}\sin ny$ to begin with: these functions solve the equation and satisfy the boundary conditions. The positive exponents are rejected because we want solution to remain bounded (so that it would describe, say, the stable heat distribution in semi-infinite strip). The form of the solution ends up being
$$u(x,y)=\sum_{n=1}^\infty e^{-nx} b_n \sin ny \tag1$$ The coefficients $b_n$ are determined from the Fourier series of the initial data $f$, because plugging $x=0$ into (1) we get $u(0,y)=f(y) = \sum_{n=1}^\infty b_n \sin ny$. If you are expected to find $b_n$, the function $f$ should have been given in the problem.

  • $\begingroup$ I believe that was relegated to the final steps for finite computation and plotting, but thank you, I feel a little better about it now. $\endgroup$
    – Tristen
    Commented May 5, 2013 at 4:26

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