Diffusion eq. with periodic BC using method of images

Breuno
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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
 
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Breuno said:
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! :smile:

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. :smile:
 
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
 
Breuno said:

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.
 
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
 
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?
 
To solve this, I first used the units to work out that a= m* a/m, i.e. t=z/λ. This would allow you to determine the time duration within an interval section by section and then add this to the previous ones to obtain the age of the respective layer. However, this would require a constant thickness per year for each interval. However, since this is most likely not the case, my next consideration was that the age must be the integral of a 1/λ(z) function, which I cannot model.
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