# Math Help - PDE seperation of variables (advanced problem)

1. ## PDE seperation of variables (advanced problem)

I have the following PDE:

$\frac{\partial C}{\partial x}=a\frac{\partial^2 C}{\partial y^2}$
where a is constant

with boundary conditions:

$C(0,y)=1,\ \ \ \ \frac{\partial C}{\partial y}(x,0)=bC,\ \ \ \ \frac{\partial C}{\partial y}(x,1)=0$
b is also constant.

I know that separation of variables is required and have a general solution for the X problem:

$X=Ae^{-\lambda^{2}ax}$

Where I am struggling is a general solution for Y, I have tried a sum of sin and cos but am having no luck with getting the constants after superposition as the boundary values just aren't helpful at removing any of the constants.
Any help would be greatly appreciated.

2. Originally Posted by pfarnall
I have the following PDE:

$\frac{\partial C}{\partial x}=a\frac{\partial^2 C}{\partial y^2}$
where a is constant

with boundary conditions:

$C(0,y)=1,\ \ \ \ {\color{red}\frac{\partial C}{\partial y}(x,0)=bC},\ \ \ \ \frac{\partial C}{\partial y}(x,1)=0$
b is also constant.
That boundary condition doesn't make sense to me. The term on the right should be a constant, not a function. Is bC supposed to be bC(x,0)?

If so, then the solution for the y problem will be of the form $Y = A\cos(\lambda y)+B\sin(\lambda y)$, and $Y' = B\lambda\cos(\lambda y) - A\lambda\sin(\lambda y)$. The boundary condition Y'(1) = 0 tells you that $B\cos\lambda = A\sin\lambda$. So take $A = \cos\lambda,\ B = \sin\lambda$. Then $Y' = \lambda(\sin\lambda\cos(\lambda y) - \cos\lambda\sin(\lambda y))$.

Now the boundary condition Y'(0) = bY(0) (if that is what it should be), tells you that $\lambda\sin\lambda = b\cos\lambda$. So the eigenvalue equation for $\lambda$ is $\tan\lambda = b/\lambda$. That has infinitely many solutions, but unfortunately you can't solve them explicitly.

3. Originally Posted by Opalg
That boundary condition doesn't make sense to me. The term on the right should be a constant, not a function. Is bC supposed to be bC(x,0)?

If so, then the solution for the y problem will be of the form $Y = A\cos(\lambda y)+B\sin(\lambda y)$, and $Y' = B\lambda\cos(\lambda y) - A\lambda\sin(\lambda y)$. The boundary condition Y'(1) = 0 tells you that $B\cos\lambda = A\sin\lambda$. So take $A = \cos\lambda,\ B = \sin\lambda$. Then $Y' = \lambda(\sin\lambda\cos(\lambda y) - \cos\lambda\sin(\lambda y))$.

Now the boundary condition Y'(0) = bY(0) (if that is what it should be), tells you that $\lambda\sin\lambda = b\cos\lambda$. So the eigenvalue equation for $\lambda$ is $\tan\lambda = b/\lambda$. That has infinitely many solutions, but unfortunately you can't solve them explicitly.
Apologies, yes the boundary condition is as you've assumed, I should have made that clearer. I got down to the equations you have but didn't think to try those values of A and B. Seems a little silly of me in hindsight.
Thanks very much for the help, very much appreciated.