Solving Heat Equation w/ Neumann BCs Different Domain

[maxtor101]

Hi guys!

I'm to find the solution to

\frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2}

Subject to an initial condition

u(x,0) = u_0(x) = a \exp(- \frac{x^2}{2c^2})

And Neumann boundary conditions

\frac{\partial u}{\partial x} (-1,t) = \frac{\partial u}{\partial x} (1,t) = 0

I can usually do this no problem assuming the domain is for instance [0,L], but I get stuck with this one :

Using separation of variables :

u(x,t) = f(x)g(t)

This yields:

\frac{1}{g} \frac{dg}{dt} = \frac{1}{f} \frac{d^2f}{dx^2} = -\lambda


Spatial Part:

\frac{1}{f} \frac{d^2f}{dx^2} = -\lambda

\frac{d^2f}{dx^2} + \lambda f = 0

Therefore :

f(x) = A \cos(\sqrt{\lambda} x) + B \sin(\sqrt{\lambda} x)

And since I'm considering Neumann Boundary conditions I get the derivative of this

f \prime (x) = -A \sqrt{\lambda}\sin(\sqrt{\lambda} x) + B \sqrt{\lambda} \cos(\sqrt{\lambda} x)


So, f \prime (-1) = 0

This gives:

A \sin(\sqrt{\lambda}) + B \cos(\sqrt{\lambda}) = 0

And for f \prime (1) = 0

I get :

-A \sin(\sqrt{\lambda}) + B \cos(\sqrt{\lambda}) = 0

So from these two equations I can conclude that:

Firstly by just adding the two equations

B \cos(\sqrt{\lambda}) = 0

So either B = 0 or \cos(\sqrt{\lambda}) = 0

Now substituting B \cos(\sqrt{\lambda}) = 0 back into A \sin(\sqrt{\lambda}) + B \cos(\sqrt{\lambda}) = 0

I also get
A \sin(\sqrt{\lambda}) = 0

So either A = 0 or \sin(\sqrt{\lambda}) = 0

Obviously \sin(\sqrt{\lambda}) and \cos(\sqrt{\lambda}) can't both equal zero, so how do I approach this...

Apologies if this is a stupid question..
Any help would be greatly appreciated
Max

 

[hunt_mat]

Have you tried a transform method? I might be tempted to try a Laplace transform on the t variable. I think that this problem requires a transform solution rather than a separation of variable solution.

 

[HallsofIvy]

sin(\sqrt{\lambda}) and cos(\sqrt{\lambda}) don't both have to be 0: one of A and B can be 0 and still give you a non-trivial solution.

 

[maxtor101]

Ah ok I see! Thanks for your help.

So I can choose for example in case 1: A=0 , B=B . case2: B=0 , A=A