Solving PDEs using Fouries Series ?

[Makveger]

Solving PDEs using Fouries Series ?

Hello
I am trying to solve 2D Laplace's equation (\nabla²u) using Fourier series using these boundary conditions for a square domain of length L:
u(x, 0) = 0
u(0, y) = 0
u(L, y) = 0
u(x,L) = U_o

After solving the 2 ODEs(separating variables method) the solution is in this form (using the boundary conditions except the last one):
u(x,y) = E*Sin(nπx/L)*Sinh(nπy/L)

And I'm stuck here,after using the 4th boundary conditions to convert the above function into a series
How can I convert this formula into a series??
How can the solution be the sum of all the values of u for all the values of n(Where n =1,2,3,...) ??

 

[JJacquelin]

u(0, y) = 0 and u(x,L) = Uo
Then, can you clearly state u(0,L)= ?

u(L, y) = 0 and u(x,L) = Uo
Then, can you clearly state u(L,L)= ?

 

[kuan]

I think the sinuous functions in x direction are the basis you choose. However, for the y direction, it must be both sinuous and co-sinuous functions. When you get the matrix representation in y direction, I think we need to impost the boundary condition on both sides. The simple way would be change the first row and the last row by the boundary condition in the so called physical space.

 

[JJacquelin]

Roughly I agree with the proposal of Kuan. But I am afraid that it will be not so simple.
In fact, around the points of (0 , L) and (L , L), a special beaviour occurs. That is why I asked Makvenger to clarify the bounding conditions on these particular points. Up to now, Makvenger gave no answer. So I think that it would be premature to go further in the mathematical development.
Without more relevant information, the solution have to include not only sinusoidal fuctions, but also Heaviside functions.

 

[kuan]

JJacquelin, I see what you mean! For me, I simply ignored these 2 points at (0 , L) and (L , L). No matter what conditions they are, the Fourier series is going to diverge pointwisely at at (0 , L) and (L , L).

 

[JJacquelin]

OK. Kuan, that's right.
But, is it allowed to ignore what appened close to whese points ?
If this is of no importance regarding to the physical problem, OK, the solution expessed only in terms of sinusoidal functions will be sufficient.
If not, the solution has to be more complicated, in order to completely fulfill the boundary conditions.
So, the answer of Makveger is a key point.

 

[kuan]

Yes, JJacquelin. You are right, I totally agree with you!

 

[yus310]

lapace or separation of variables works here.. either way is good.

 

[naman chauhan]

Hey can somebody tell me the Fourier transform of e^-(pi*t^2)?

 

[JJacquelin]

Hey can somebody tell me the Fourier transform of e^-(pi*t^2)?

WolframAlpha can :
http://www.wolframalpha.com/input/?i=Fourier+transform+exp(-pi*t^2)

 

[naman chauhan]

Thanks for your help. But can you please help me out with the mathematics. I have tried it but got struck in the integral.

 

[LCKurtz]

Hello
I am trying to solve 2D Laplace's equation (\nabla²u) using Fourier series using these boundary conditions for a square domain of length L:
u(x, 0) = 0
u(0, y) = 0
u(L, y) = 0
u(x,L) = U_o

After solving the 2 ODEs(separating variables method) the solution is in this form (using the boundary conditions except the last one):
u(x,y) = E*Sin(nπx/L)*Sinh(nπy/L)

And I'm stuck here,after using the 4th boundary conditions to convert the above function into a series
How can I convert this formula into a series??
How can the solution be the sum of all the values of u for all the values of n(Where n =1,2,3,...) ??

You haven't written your $u(x,y)$ correctly. You have $X_n(x)=\sin\frac{n\pi x}{L}$ and $Y_n(y) =\sinh\frac {n\pi y}{L}$. So your prospective solution is$$
u(x,y) = \sum_{n=1}^\infty c_nX_n(x)Y_n(y) = \sum_{n=1}^\infty c_n\sin\frac{n\pi x}{L}
\sinh\frac {n\pi y}{L}$$Now apply your fourth BC $u(x,L)=U_0$:$$
U_0 =\sum_{n=1}^\infty c_n\sin\frac{n\pi x}{L}
\sinh\frac {n\pi L}{L}$$Now use your half range sine expansion to get the constants. Notice that what you usually call $a_n$ in a sine expansion is $a_n = c_n\sinh(n\pi)$ in this problem.

 

[Mute]

WolframAlpha can :
http://www.wolframalpha.com/input/?i=Fourier+transform+exp(-pi*t^2)

For anyone using wolframalpha or mathematica to compute Fourier transforms, be aware that the default convention is to use the unitary-transformation convention: factors of $1/\sqrt{2\pi}$ outside both the transform and the inverse. Typically in physics the factors are 1 in front of the transform and $1/2\pi$ in front of the inverse.

(Overall the difference doesn't matter as long as you use a consistent convention, or unless you want to compare to a particular solution which uses a particular convention).