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Analytical solution to integro-differential equation

Hi all,

I'm trying to find an analytical solution to the following integro-differential equation:

[itex]
a f'(x)\int_0^x f(x)dx + b f'(x) + a [f(x)]^2 - a f(x) = 0
[/itex]

with initial condition:
[itex]
f(0)=1
[/itex]

This is a simplified problem for which I know the solution: $f(x)=1$.
I'm trying to find a general method to solve this equation that I can use for more complex problems. The main difficulty is the product of the differential and the integral.
Can anyone point me in the right direction? Integral transforms (e.g. Laplace) seem to be the general way to tackle integro-differential equations but I'm not sure how to apply those here.

Many thanks in advance,
Maarten

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I'm not sure that this can be solved analytically. One thing you can try is dividing with f'(x) (assuming it's not zero) and then differentiating (the other case, f'(x)=0, makes your job very easy). That will get rid of the integral, but although the expression you get is not really easy to work with, it's reduced to a second order differential equation.

Hi ! There are two obvious solutions  f(x)=0  and f(x)=1. In order to show that there are no other solution, let : $F(x)=\int_0^x f(x)dx$ Then, bring back F(x), f(x)=F'(x) and f'(x)=F''(x) into the ODE This leads to a second order ODE, rather easy to solve, thanks to the conditions F'(0)=f(0)=1 and F(0)=0.

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Analytical solution to integro-differential equation

Quote by mbraakhekke 					 				 				Integral transforms (e.g. Laplace) seem to be the general way to tackle integro-differential equations but I'm not sure how to apply those here.
Integral transforms are typically only useful for linear equations. This equation is nonlinear, so the typical integral transforms are not likely to work at all. (The laplace transform won't).

Quote by JJacquelin 					 				 				Hi ! There are two obvious solutions  f(x)=0  and f(x)=1. In order to show that there are no other solution, let : $F(x)=\int_0^x f(x)dx$ Then, bring back F(x), f(x)=F'(x) and f'(x)=F''(x) into the ODE This leads to a second order ODE, rather easy to solve, thanks to the conditions F'(0)=f(0)=1 and F(0)=0.
This is what I would do.

So, I suppose that you did it. 		 		  		 		 			  			 				 					Attached Thumbnails

Thanks to all for your suggestions.  @JJacquelin: your derivation is very helpful. The remaining equation without the boundary conditions: $(F'(x)-1)(F(x)+k)=C$ should be also solvable for more complex situations. Mathematica delivers a general solution in terms of the Lambert W (product log) function. Anyway, this is something I can work with.  Thanks again, Maarten

Hello !  I agree. Without the boundary conditions, the general solution f(x) involves the Lambert-W function : 		 		  		 		 			  			 				 					Attached Thumbnails

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