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

 

[blue_raver22]

Hi Max,

Thank you for sharing your problem with us. I understand that you are trying to solve the heat equation with Neumann boundary conditions in a different domain than what you are used to. It looks like you have already made some progress with separating the variables and finding the general solution for the spatial part. However, you are stuck with the boundary conditions and are unsure of how to proceed.

Firstly, I want to assure you that this is not a stupid question and it is completely normal to encounter difficulties when solving mathematical problems. I would suggest taking a step back and reviewing the Neumann boundary conditions. These conditions specify that the derivative of the solution with respect to the spatial variable must be zero at the boundaries. In your case, the boundaries are at x = -1 and x = 1.

Now, let's consider the case where A = 0. In this case, the solution for the spatial part would only involve the cosine term, which would result in a constant. However, this would not satisfy the boundary conditions since the derivative of a constant is always zero. Therefore, we can conclude that A cannot be equal to zero.

Next, let's consider the case where B = 0. This would result in the solution for the spatial part only involving the sine term, which would also result in a constant. Again, this would not satisfy the boundary conditions. Therefore, B cannot be equal to zero either.

So, how do we proceed? We can use the fact that both A and B cannot be equal to zero to our advantage. Since we know that both sine and cosine cannot be zero at the same time, we can set up a system of equations to solve for the value of lambda. Using the equations you have derived, we can write:

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

Since we know that A and B cannot be zero, we can divide both equations by A and B respectively to get:

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

Now, we can use the trigonometric identities \sin(\theta) = 0 when \theta = n\pi and \cos(\theta) = 0 when \theta = \frac{(2n+1)\pi}{2}. Substituting these values into our equations, we get:

\sqrt{\lambda} = n\pi