Unsuccessful attempts to solve a linear second order PDE system

  • Thread starter jmk9
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Main Question or Discussion Point

Dear All,

I am trying to solve the following system of PDEs
[tex]
\frac{\partial{A}}{\partial{t}}= a_{2}\frac{\partial{{^{2}}A}}{\partial{x^{2}}}-a_{1}\frac{\partial{A}}{\partial{x}}-a_{0}A+b_{0}B
[/tex]

[tex]
\frac{\partial{B}}{\partial{t}}= b_{2}\frac{\partial{{^{2}}B}}{\partial{x^{2}}}-b_{1}\frac{\partial{B}}{\partial{x}}-b_{0}B+a_{0}A
[/tex]

where

[tex]
A(0,0)=A_0\delta(x)\delta(t) \\ B(0,0)=B_0\delta(x)\delta(t)
[/tex]

The system models two packets of ionised particles which interconvert with with reaction rates a0 and b0 whilst diffusing and drifting with constant velocity in the presence of a weak electric field.

I have tried a direct method using integral transforms. Taking the Fourier transform over x and the Laplace transform over t, solving for B and subsequently for A and subsequently calculating the inverse Laplace transform (after ignoring one a0b0 term) and then taking the inverse FT I get

[tex]
A(x,t)=\frac{A_{0}e^{-\frac{(x-a_{1}t)^{2}}{4a_{2}t}-a_{0}t}}{(2\pi)^{\frac{3}{2}}}\int_{-\infty}^{\infty}{e^{-a_{2}t(k-\frac{i(x-a_{1}t)}{2a_{2}t})^{2}}dk}\\ +\frac{b_{0}B_{0}}{(2\pi)^{\frac{3}{2}}}\int_{-\infty}^{\infty}{\frac{e^{-a_{2}t(k-\frac{i(x-a_{1}t)}{2a_{2}t})^{2}-\frac{(x-a_{1}t)^{2}}{4a_{2}t}-a_{0}t}-e^{-b_{2}t(k-\frac{i(x-b_{1}t)}{2b_{2}t})^{2}-\frac{(x-b_{1}t)^{2}}{4b_{2}t}-b_{0}t}}{(b_{2}-a_{2})k^{2}+i(b_{1}-a_{1})k+b_{0}-a_{0}}dk}
[/tex]

The first integral can be performed. The second though causes me some grief; the best I can do is do some algerba and reduce it to a sum of four integrals of the form

[tex]
\int{\frac{e^{-cz^{2}}}{z-\alpha}dz}
[/tex]

which then again I cannot solve... If anyone can help here, that would be great.

Looking in the literature, Schummers et al. {Phys. Rev. A 7:689 (1973)} report a solution which is appropriate for modelling the situations that I am studying using slightly more complicated initial value conditions. Unfortunately the proof of the solution is not in the public domain so I cannot read it for myself. But they do state "The coupled differential equations were solved by successive applications of the Green's function representing depletion of a species n into species n+1."
My understanding of the Green's function method is very superficial, so I am very uncertain whether the following is correct. In any case, for species (A) above, the Green's function is given by

[tex] \mathfrak{L}G_A=\frac{\partial{G_A}}{\partial{t}}- a_{2}\frac{\partial{{^{2}}G_A}}{\partial{x^{2}}}+a_{1}\frac{\partial{G_A}}{\partial{x}}+a_{0}G_A = \delta(x-\chi)\delta(t-\tau)
[/tex]


By definition, the RHS of the above equation means it is zero everywhere unless x=chi and t=tau in which case it equals one. Solving for these two cases I obtain

[tex]
\mathfrak{L}G_A(x-\chi,t-\tau)=0\Rightarrow G_{A0}(x-\chi,t-\tau)=\frac{e^{-\frac{((x-\chi)-a_1(t-\tau))^2}{4a_2(t-\tau)}-a_0(t-\tau)}}{4\sqrt{\pi a_2(t-\tau)}}
[/tex]

[tex]
\mathfrak{L}G_A(x-\chi,t-\tau)=1
[/tex]
[tex]\Rightarrow G_{A1}(x-\chi,t-\tau)=\frac{e^{-\frac{((x-\chi)-a_1(t-\tau))^2}{4a_2(t-\tau)}-a_0(t-\tau)}}{4\sqrt{\pi a_2(t-\tau)}}+\frac{1-e^{-a_0(t-\tau)}}{a_0}
[/tex]


With analogous expressions for G_B.
Now I am not so sure whether this result is even correct or what exactly to do next. According to the method the solution is given by

[tex]
A(x,t)=\iint{G_A(x-\chi, t-\tau)[A_0\delta(\chi)\delta(\tau)+b_0B(\chi,\tau)]d\chi d\tau}
[/tex]
[tex]
=A_0G_A(x,t)+b_0\iint{G_A(x-\chi, t-\tau)B(\chi,\tau)d\chi d\tau}
[/tex]

[tex]
B(x,t)=\iint{G_B(x-\chi, t-\tau)[B_0\delta(\chi)\delta(\tau)+a_0A(\chi,\tau)]d\chi d\tau}
[/tex]

[tex]
=B_0G_B(x,t)+b_0\iint{G_B(x-\chi, t-\tau)A(\chi,\tau)d\chi d\tau}
[/tex]

So now I have another system which I do not know how to treat... and so I am crying now desperately for help!!!
If there are other methods for solving this particular problem please do suggest them. And also I would be eternally grateful if you could let me know if I am on the right path for the Green's function strategy and how I can proceed to get an expression...

Thanks in advance!

[PS when I first asked one of my lecturers at uni to show me how to solve this a couple of years ago he derived a nice series solution which unfortunately does not converge very fast and the terms of the expansion take a very long time to compute, so that the final formula is not very useful, unfortunately...]

BTW, some formulae do not display correctly, probably a bug somewhere?
 

Answers and Replies

  • #2
11
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oh, the eqns display fine now that I posted them
 
  • #3
11
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So far I have had no replies whatsoever... shall I assume that nobody is ever going to look at it?
 
  • #4
Entropee
Gold Member
133
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I actually wrote this down and gave it to someone in the math lab at my school. We will see what he comes up with later tonight.
 
  • #5
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Dear Entropee,

Thank you very much, if you friend/colleague finds something that would be great!!! This system has been doing my head in for a long time!

J
 
  • #6
Entropee
Gold Member
133
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Ugh haha one of them couldn't figure it out, and one of them knows how but he says it takes too long and won't do it. He said to set it up with matrices. Not that that helps LOL. Good luck man.
 
  • #7
11
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Oh, I see. Well thanks for taking this on anyway, much appreciated. Do you think you can ask him what steps I should take, roughly?

I have managed to see a solution to something similar in modelling flow-reactors... apparently they somehow set it up with difference equations... But I am not familiar with that method.

There exists a solution in the literature, but the authors refer to work not in the public domain. So much for that. Anyway, they use the Green's function method above, so assuming that I have stated the problem correctly all that remains now is some algebra... by the way they do not solve the resulting integral analytically.

This system is insanity trans-substantiated in an equation.....
 

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