Re: How could I solve this PDE

From: Paul Abbott (paul_at_physics.uwa.edu.au)
Date: 09/20/04 

Date: Mon, 20 Sep 2004 12:39:24 +0800

In article <cikl5n$8t7$1@nntp.itservices.ubc.ca>,
 israel@math.ubc.ca (Robert Israel) wrote:

> In article <685cf493.0409170554.63134442@posting.google.com>,
> Nas <nasimusus@yahoo.com> wrote:
> |>Hi every one. I would like to find function A(x,t) by solving this PDE:
>
> |>dA/dt=cosh(ax).dA/dx + b.cosh(ax).t
>
> |>"a" and "b" are constants. Does anyone has any idea of how I should deal
> |>with it?
>
> Maple's pdsolve gives a rather complicated solution:
>
> A(x, t) = -(b*x*a^2*t+2*b*x*a*arctan(exp(a*x))-2*b*arctan(exp(a*x))*
> ln(-I*(1+exp(2*a*x))^(1/2)-I+exp(a*x))+I*b*polylog(2,(1+I*exp(a*x))^2/
> (1+exp(2*a*x)))-2*b*arctan(exp(a*x))*ln(I*(1+exp(2*a*x))^(1/2)-I+exp(a*x))+
> 2*b*arctan(exp(a*x))*ln(2*I)+2*b*arctan(exp(a*x))*ln(-I+exp(a*x))-
> I*b*polylog(2,-(1+I*exp(a*x))^2/(1+exp(2*a*x)))-
> _F1((t*a+2*arctan(exp(a*x)))/a)*a^2)/a^2

Mathematica's DSolve gives an equivalent and similarly complicated
solution:

 A[x,t] -> (a^2 C[1][t + (2 ArcTan[Tanh[(a x)/2]])/a] -
  b (t x a^2 + 2 ArcTan[Tanh[(a x)/2]] (a x - Log[1/(1 + I/E^(a x))] +
  Log[1/(1 - I E^(a x))]) -
  I PolyLog[2, -I E^(2 I ArcTan[Tanh[(a x)/2]])] +
  I PolyLog[2, I E^(2 I ArcTan[Tanh[(a x)/2]])]))/a^2

Cheers,
Paul 

-- 
Paul Abbott                                   Phone: +61 8 6488 2734
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