# Thread: A nonlinear integro-differential equation in 2D

1. ## A nonlinear integro-differential equation in 2D

I hope to find a closed-form solution for the following equation:

(I believe it's properly classified as 'homogenous nonlinear Fredholm integro-differential equation in two dimentions')

$\displaystyle \frac{\partial{T(x,t)}}{\partial{t}} = \int_{0}^{1}{K[x-T(u,t)]f_0(u)du}$

where $\displaystyle f_0(x)$ is a known piecewise polinomial function (spline) of degree 3, and $\displaystyle K(x)$ is a known even function. I know it might sound an ill-posed problem, but it would suffice if there was some sufficiently broad class of functions for $\displaystyle K(x)$ for which the thing could be solvable. Cubic splines, again, would be ideal! I tried to substitute in a parametrized polinomial $\displaystyle f_0(x)$ and $\displaystyle K(x)$ into the equation and got a sum of sums of integrals of powers of $\displaystyle T(u,t)$,x and $\displaystyle u$. But I don't know what to do next...

Some books on Integral equations treat this sort of problems, but I've got the nasty partial derivative on the left side. I could find some papers on integro-differential equations of this kind, but they seem to deal with some special cases and in 1 dimension, and frankly too obsucre for my level of expertise (recent graduate in applied maths, rudimentary knowledge of mathematical physics)

I have reasons to prefer analytical solution, but, of course, numeric methods would be my last resort.

I would appreciate ANY form of help.

Thank you for your attention!

_________________

In case you care where does this problem originate from:
I came up with it doing research in image processing (to keep up with the field's notorious habbit of applying mathematical apparatus to it's problems in precarious ways). Basically, it is sort of a continuous analog of the n-body problem of physics, except that I unceremoniously replaced acceleration with velocity in the left part. It can be interpreted as equation describing translocation ($\displaystyle T(x,t)$ - is position of 'point' that was at postion $\displaystyle x$ at the moment $\displaystyle t_0$) of continuous matter on a line, governed by gravitation-like mutual repulsion of it's particles ($\displaystyle K(x)$ is the 'law', would be $\displaystyle \frac{1}{x^2}$ in Newton's gravitation), where $\displaystyle f_0(x)=f(x,t_0)$ is the initial density distribution. The equation was transformed from it's original formulation using the interrelation between $\displaystyle T(x,t)$ and $\displaystyle f(x,t)$, to eliminate the last one.

That all might sound lunatic, but I actually constructed a programm to solve it through naive simulation, and it prooved to yield the result I expected.

2. You mentioned numeric methods, so I'd like to offer a suggestion: First I'd listen to what it's tellin' me: The rate at which the function of two variables is changing in the t direction is the "sum" of some "attenuation" of the function at $\displaystyle t$ in the range $\displaystyle 0\leq x\leq 1$. The attenuation is of course the composite function $\displaystyle K[x-T(u,t)]f_0(u)$. So right off the bat, I'd put this one on the back burner and look at the simpler initial interval problem:

$\displaystyle \frac{\partial{T(x,t)}}{\partial{t}} = \int_{0}^{1}{T(u,t)du},\quad T(x,0)=g(x),\quad 0\leq x \leq 1$

To me, that's conceptually the same and note how I need to know what the function $\displaystyle T(x,t)$ is not at a point as in an ordinary IVP, but along the interval of integration at t=0. Now, if I start at t=0 and keep a running tally of what the function looks like at the previous time period, I can numerically integrate the integral and treat this like an ordinary differential equation and solve it numerically. So, I'd then break up the interval $\displaystyle 0\leq x \leq 1$ into say for starters, just ten points. Let me pick just one of the points $\displaystyle x_0=0.5$. Now for that point, I solve the ordinary IVP numerically:

$\displaystyle \frac{dT(x_0,t)}{dt}= \int_{0}^{1}T(u,t)du$

via Runge-Kutta. Now, do that ten times at each time period, and always before going to the next time period, numerically integrate a least-square fit of the data points you calculated for the previous time period. This numerical value, then becomes the approximation for the derivative for the next time step.

I'd work on this simple one a while to see what happens and if the results look encouraging, I would then gradually build up the integrand to the form of the original problem. That is, I'd first add the $\displaystyle f_0$ part maybe, say $\displaystyle f(x)=x^3$. Then next try a simple form for K such as $\displaystyle K(x)=\cos(x)$. Keep playing with it that way until I build up to the problem I need and always be tolerant of the fact that this may not work or the approach may break down at some point.

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