# Help solving heat equation with Neumann Boundary Conditions with different domain

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
Homework Helper
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.