# Thread: Help with this PDE

1. ## Help with this PDE

Hello,

I'm currently taking a course in partial differential equations. We are currently solving boundary value problems using the method of separation of variables. I'm attemping to find the seperated solutions and have gotten stuck.

Here is the problem:

Solve the BVP for the Laplace equation Uxx+Uyy=0, 0<x<pi, 0<y<pi, with boundary conditions u(0,y)=sin(y)+2sin(3y), u(pi,y)=0, u(x,0)=0, and u(x,pi)=0.

Here is my attempt at solving the problem:

Assume u(x,y)=X(x)*Y(y)
Plug into PDE:
Uxx+Uyy=0 --> X''Y + XY''=0 --> Y''/Y = -X''/X = lamda (I'll refer to lambda from now on as L)
Thus Y''=LY and X''=-LX
Plug into BC:
u(pi,y)=0=X(pi)Y(y) --> X(pi)=0
u(x,0)=0=X(x)Y(0) --> Y(0)=0
u(x,pi)=0=X(x)Y(pi) --> Y(pi)=0

Thus, we have an eigenvalue problem: Y''=LY, Y(0)=0, Y(pi)=0
The solution to the eigenvalue problem with Dirichlet B.C. in this case is:
Eigenvalues: Ln=-n^2, n=1,2,3,...
Eigenfunctions: Yn(y)=sin(ny), n=1,2,3,...

Now Xn''=-LnXn --> Xn''=n^2*Xn
The solution to this eigenvalue problem is:
Xn(x) = A*cosh(nx) + B*sinh(nx), n=1,2,3,...

Now here comes the trouble... I have to plug the boundary condition of X(pi)=0 into this
Xn(pi) = 0 = A*cosh(pi*n) + B*sinh(pi*n) , n=1,2,3,...

In all the other problems I've done, when you solve this type of equation, either A or B ends up equally zero and drops out, but there appears to be an infinite number of solutions to this linear equation, since there is only one equation with two unknowns. Any help is appreciated. Thanks!

2. Originally Posted by policebox
Hello,

I'm currently taking a course in partial differential equations. We are currently solving boundary value problems using the method of separation of variables. I'm attemping to find the seperated solutions and have gotten stuck.

Here is the problem:

Solve the BVP for the Laplace equation Uxx+Uyy=0, 0<x<pi, 0<y<pi, with boundary conditions u(0,y)=sin(y)+2sin(3y), u(pi,y)=0, u(x,0)=0, and u(x,pi)=0.

Here is my attempt at solving the problem:

Assume u(x,y)=X(x)*Y(y)
Plug into PDE:
Uxx+Uyy=0 --> X''Y + XY''=0 --> Y''/Y = -X''/X = lamda (I'll refer to lambda from now on as L)
Thus Y''=LY and X''=-LX
Plug into BC:
u(pi,y)=0=X(pi)Y(y) --> X(pi)=0
u(x,0)=0=X(x)Y(0) --> Y(0)=0
u(x,pi)=0=X(x)Y(pi) --> Y(pi)=0

Thus, we have an eigenvalue problem: Y''=LY, Y(0)=0, Y(pi)=0
The solution to the eigenvalue problem with Dirichlet B.C. in this case is:
Eigenvalues: Ln=-n^2, n=1,2,3,...
Eigenfunctions: Yn(y)=sin(ny), n=1,2,3,...

Now Xn''=-LnXn --> Xn''=n^2*Xn
The solution to this eigenvalue problem is:
Xn(x) = A*cosh(nx) + B*sinh(nx), n=1,2,3,...

Now here comes the trouble... I have to plug the boundary condition of X(pi)=0 into this
Xn(pi) = 0 = A*cosh(pi*n) + B*sinh(pi*n) , n=1,2,3,...

In all the other problems I've done, when you solve this type of equation, either A or B ends up equally zero and drops out, but there appears to be an infinite number of solutions to this linear equation, since there is only one equation with two unknowns. Any help is appreciated. Thanks!
So this gives $\displaystyle A = -\frac{B \sinh \pi n}{\cosh \pi n}$. Therefoer the solution is

$\displaystyle u(x,t) = B \sinh n x -\frac{B \sinh \pi n}{\cosh \pi n} \cosh n x = \frac{B}{\cosh \pi n} \sinh n (\pi - x)$.

3. I'm not familiar with the identity you used to simplify this expression into one term. Can you provide me with more details of it please?

Thanks.

4. Originally Posted by policebox
I'm not familiar with the identity you used to simplify this expression into one term. Can you provide me with more details of it please?

Thanks.
Try this