How do I find the coefficients to the solution of this PDE

Question

Before I start, I want to say that I DON’T want the solution to the problem. I just want to know how to move past this point, or if I’ve made a mistake I and if so what it was. Thank you.

For my homework I am supposed to solve the following boundary value problem.

$u_{xx}+u_{yy}=0$

$u(0,y)=0$

$u_x(1,y)=0$

$u_y(x,0)+u(x,0)=0$

$u(x,2)=100$

for: $0<x<1$

$0<y<2$ncosh(

Here is what I have done so far: I started with separation of variables.

$u(x,y)=X(x)Y(y)$

This gives 2 eigenvalue problems

$X’’(x)+\lambda X(x)=0$

$Y’’(y)-\lambda Y(y)=0$

Using the boundary values to find the eigenfunctions gives me the following:

$X_n(x)=A_nsin(\sqrt{\lambda_n}x)+ B_ncos(\sqrt{\lambda_n}x)$

$Y_n(y)= C_ncosh(\sqrt{\lambda_n}x+ D_nsinh(\sqrt{\lambda_n}x) $

$B=0, \lambda_n=(\frac{(2n+1)\pi}{2})^2$

Putting these together I get

$u(x,y)= \sum_{n=1}^{\infty} [C_ncosh(\sqrt{\lambda_n}x)+ D_nsinh(\sqrt{\lambda_n}x)] sin(\sqrt{\lambda_n}x) $ The problem is that when I then use the 3rd boundary value I get the following

$(C_n+D_n\sqrt{\lambda_n})sin(\sqrt{\lambda_n}) $ for the terms of the infinite sum.

And there doesn’t seem to be a way to get either of the coefficients B or D from that. I have no idea where to go from here

Answer 1

Okay,

So when you wrote down your system of ODE's you used the conditions $X(0)=0$ and $X'(1)=0$ correctly to find that $X_n(x) = B_n\sin{\left(\frac{(2n-1)\pi}{2}x\right)}$.

Now, using $Y'(0)+Y(0)=0$, we should see that $C\lambda + D\lambda + C + D = 0$, in other words, $C = -D$. So $Y_n(y) = C_n \sinh{\left(\frac{(2n-1)\pi}{2}y\right)}$.

Thus, we have from superposition(and condensing the coefficients into one master coefficient) and all that good stuff,

$$u(x, y) = \sum_{n=1}^{\infty}A_n \sinh{\left(\frac{(2n-1)\pi}{2}y\right)}\sin{\left(\frac{(2n-1)\pi}{2}x\right)}$$

In order to solve for the coefficients, we must play with the initial condition,

$$100 = \sum_{n=1}^{\infty}A_n \sinh{\left((2n-1)\pi\right)}\sin{\left(\frac{(2n-1)\pi}{2}x\right)}$$

By multiplying both sides by $\sin{\left(\frac{(2m-1)\pi}{2}x\right)}$, integrating both sides from $0$ to $1$, and then using the orthogonality relationship we can get a formula for $A_n$. Are you able to take it from this point?

Answer 2

Your general solution is correct

$$ u(x,y) = \sum_{n=1}^{\infty} \big(C_n \cosh (k_nx) + D_n \sinh(k_nx) \big)\sin (k_nx) $$

where $k_n = \dfrac{2n+1}{2}\pi$

Plugging the last two boundary conditions simultaneously, we get

$$ u_y(x,0) + u(x,0) = 0 = \sum_{n=1}^{\infty}(C_n + k_nD_n)\sin(k_nx) \tag{1} $$ $$ u_y(x,2) = 100 = \sum_{n=1}^{\infty} \big(C_n\cosh(2k_n) + D_n\sinh(2k_n)\big)\sin(k_nx) \tag{2} $$

$(1)$ gives $C_n + k_nD_n = 0$, so $(2)$ simplifies to

$$ \sum_{n=1}^{\infty} D_n \big(-k_n \cosh(2k_n)+\sinh(2k_n) \big)\sin(k_nx) = 100 $$

This is a Fourier sine series in $x$, which lets you solve for $D_n$ $$ D_n \big(-k_n \cosh(2k_n)+\sinh(2k_n) \big)\sin(k_nx) = \int \dots $$