Diffusion eq. with periodic BC using method of images

[Breuno]

Homework Statement

Considering the periodic boundary conditions (given below) I am supposed to find the solution T(x,t) with the initial condition T(x,0)=$$\delta$$(x) Also I am limited to use method of images so I can't use separation of variables unfortunately.

Homework Equations

The boundary conditions are give by:
$$T(x=-L/2,t)=T(x=L/2,t)$$

$$\frac{\partial T}{\partial x}(x=-L/2)=\frac{\partial T}{\partial x}(x=L/2)$$

The Attempt at a Solution

I've only started and for the initial condition using method of images I get:

$$T(x,t)=\sum{(-1)^{n}\ T_{g}(x+n*L,t)}$$

where the sum goes from -infinity to infinity.

My problem is how to implement the periodic boundary conditions into the problem.
In my textbook it says that using theese kind of boundary conditions in 1-D is equivalent to transforming the coordinates from a line to a circle. What does that mean?

I'd much appreciate it if you gave me a hint on how to solve this

Thanks
/Simon

 

[tiny-tim]

Welcome to PF!

The boundary conditions are give by:
$$T(x=-L/2,t)=T(x=L/2,t)$$

$$\frac{\partial T}{\partial x}(x=-L/2)=\frac{\partial T}{\partial x}(x=L/2)$$
…
In my textbook it says that using theese kind of boundary conditions in 1-D is equivalent to transforming the coordinates from a line to a circle. What does that mean?

Hi Simon! Welcome to PF!

It just means that under those boundary conditions, the function repeats itself whenever x increases by L.

So it's the same as a function on a circle with perimeter L.

 

[Breuno]

Thanks for the welcome =)

Ok so the function repeats itself when x increases by L. How do I use this when "mirroring"?

Since the delta-function has alternating signs (regarding the initial condition) for every other mirror image. Does this goes for the BC as well?

A lot of confusion here since I don't know the exact properties of the method of images. If anyone has a link where it is explained I'd appreciate it :P

 

[gabbagabbahey]

I've only started and for the initial condition using method of images I get:

$$T(x,t)=\sum{(-1)^{n}\ T_{g}(x+n*L,t)}$$

where the sum goes from -infinity to infinity.

What exactly is $T_{g}(x+n*L,t)$? I assume you are summing over n?

What x interval are you trying to find the solution on? [-L/2,L/2] perhaps? The method of images entails adding additional "image sources" outside of the region that you are looking for a solution on. These extra sources are placed such that the solution T(x,t) due to all of the sources will satisfy the boundary conditions.

 

[Breuno]

Yea sorry I forgot to write that I sum over n. Tg is just the gaussian solution to the diffusion eq. And $$\int^{-\infty}_{\infty} Tg(x,t)dx=1$$

 

[gabbagabbahey]

Okay, so

$$T_g(x,t)=(4 \pi kt )^{-\frac{1}{2}}e^{\frac{-x^2}{4kt}}$$

where $k$ is the diffusion constant?

You know that using the method of images is going to involve adding image sources, so say you place one at $x=x_0$ such that $T(x,0;x_0)=\delta (x-x_0)$ what then would $T(x,t;x_0)$ due to just that source be?