Take the 2-minute tour ×
Mathematica Stack Exchange is a question and answer site for users of Mathematica. It's 100% free, no registration required.

I have to numerically solve a nonlinear partial integro-differential equation using Mathematica. This is my equation,

$$\frac{\partial y(x,t)}{\partial t}=\int_{-\infty}^\infty K_0(|x-u|) \frac{\partial^2 y(u,t)}{\partial u^2}\mathrm du+\sin\,y(x,t)$$

$K$ is the modified Bessel function of the 2nd kind and answer $y$ is a function of $x,t$.

what I want to do is to first try to solve it with some built-in functionality with MATLAB or MATHEMATICA if possible. If not, what is your recommendation to go around this IDE. any help would be really appreciated.

share|improve this question
1  
I tried to $\TeX$ up your equation, but it isn't clear which partial derivatives you're taking. Where did the equation come from? –  J. M. Aug 17 '12 at 4:05
4  
What I would try to do is to first remove the second derivative under the integral by integrating by parts twice, and then integrate your equation over time numerically iteratively, discretizing the time variable and computing the value of y(t+dt) from y(t) using your equation (its r.h.s) with y(t,x) obtained at the previous step. I actually used similar methods with success in the past, also in Mathematica. –  Leonid Shifrin Aug 17 '12 at 13:03
    
Ahmad says that y is a function of x and t. Why does the integral show it to be a function of u? So u is a function of x and perhaps t? –  drN Aug 19 '12 at 23:44
1  
Guys,first of all I should say that K(x,u) is first modified Bessel function and u is just a variable under the Integral.secondly I haven't tried your answers so far but I think Leonid's is understandable.@ Leonid:Does any solver exist in mathematica for this? –  Ahmad Sheikhzada Aug 20 '12 at 11:16
1  
Maybe this Handbook of integral equations book could provide some clue. –  Silvia Aug 23 '12 at 4:34
show 5 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.