# Thread:

1. ## Is this integral equation solvable? If so, how?

Hello, I'm new to the forums (actually, just found you via google) and I have 2 questions (with the other I'll post in a separate thread a little later)!

My questions is this, can the following equation be solved for :

and t sufficiently large that transient/initial conditions don't matter.

We also have charge conservation:

I tried approaching this by using a Fourier transform type solution

but when I pursue this I seem to get multiple exponential integrals and things don't seem to simplify.

Any ideas? Are there additional boundary conditions I need to impose? I have almost no experience with integral equations so all suggestions welcome!

2. No takers?

One can also try this problem with defining new variable q(x,t) such that

and choosing q(0)=q(a)=0.

Then the equation above can be written as

Then perhaps we can assume solution of the form

and so the equation becomes

Of course, I'm not sure how to evaluate

Edit:

Mathematica gives that

But I don't see how I can get from that... as the expression becomes the incredibly ugly sum:

Somehow I would suspect that the terms with x have to all go away (or my guess of solution is incorrect, but exponential should for complete set and the boundary conditions are well defined so it should be a sum). Another thing is to try expansion in terms of bessel functions instead of exponential...

Also, perhaps summation has to be not from n=1 but from n=- infinity... but q_0 = 0, for certain (no constant component).

3. It looks like an intergration by parts problem to me for some reason.

4. Originally Posted by Bwts
It looks like an intergration by parts problem to me for some reason.
Wait, what? The exponential integral cannot be simplified....

Also, by assuming simple exponential time dependence (which I'm pretty sure is correct, since our driving function is a simple oscillatory exponential) we can make this into

$i \omega \hat{q} \left(x\right) = \sigma \left( E_0 +\displaystyle {{\int_{0}}^a}\frac{d\hat{q} \left(x'\right)/dx'}{\left(x-x' \right)^{2} }dx' \right)$

Integrating this by parts (is that what you meant?) will give you

$i \omega \hat{q} \left(x\right) = \sigma \left( E_0 -2\displaystyle {{\int_{0}}^a}\frac{\hat{q} \left(x'\right)}{\left(x-x' \right)^{3} }dx' \right)$

Which I don't think is particularly better... but if someone wants to give a try from this spot.. may be.. this is the most compact formulation of the problem.. that's true..

Edit:
We can find an approximate solution by assuming that q doesn't vary rapidly with x. Then all q contribution will come from the pole at x'=x. We have

$\displaystyle {{\int_{0}}^a}\frac{\hat{q} \left(x'\right)}{\left(x-x' \right)^{3} }dx' \approx \hat{q} \left(x\right)\displaystyle {{\int_{0}}^a}\frac{1}{\left(x-x' \right)^{3} }dx' \right)= \hat{q} \left(x\right) \frac{1}{2} \left(\frac{1}{\left(x-a \right)^{2}}-\frac{1}{x^{2}}\right)$

Which means that

$i \omega \hat{q} \left(x\right) = \sigma \left( E_0 -2\hat{q} \left(x\right) \frac{1}{2} \left(\frac{1}{\left(x-a \right)^{2}}-\frac{1}{x^{2}}\right) \right)$

Simplifying we get
$i \omega \hat{q} \left(x\right) / \sigma+2\hat{q} \left(x\right) \frac{1}{2} \left(\frac{1}{\left(x-a \right)^{2}}-\frac{1}{x^{2}}\right)= E_0$

or

$\hat{q} \left(x\right) \approx \frac{ E_0}{i \omega/ \sigma+\left(\frac{1}{\left(x-a \right)^{2}}-\frac{1}{x^{2}}\right)}$
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