- #1

- 11

- 0

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?