Solving Integral Equation for u(x)

[foxjwill]

Homework Statement

Solve for u(x):

$$0 = e^{2\int u(x) dx} + u(x) e^{\int u(x) dx} - a(x)$$

Homework Equations

The Attempt at a Solution

I tried using the quadratic formula,

$$e^{\int u(x) dx} = \frac{-u(x) \pm \sqrt{u^2(x) + 4a(x)}}{2}$$

, converting to log notation and differentiating, but from there I didn't know how to solve for u(x). I thought maybe I could use something on the lines of the log-definitions of the inverse trig functions. Any ideas?

 

[olgranpappy]

what are the limits on the integration? if they are fixed then $\int dx u(x)$ is just a number (call it C) and
$$u=(a-e^{2 C})/e^{C}$$

 

[coomast]

You could convert it to a differential equation and try to solve this, however it turns out that this new differential equation is severely determined by the unknown function a(x). In order to do this, set:

$$e^{\int u(x)dx}=f(x)$$

Thus:

$$u(x)=\frac{1}{f(x)}\frac{df(x)}{dx}$$

And putting this into the equation gives:

$$\frac{df(x)}{dx}+[f(x)]^2=a(x)$$

This is a Riccati equation, which can be transformed into a linear one by transforming:

$$f(x)=\frac{1}{u(x)}\frac{du(x)}{dx}$$

So:

$$\frac{df(x)}{dx}=-\frac{1}{[u(x)]^2}\left(\frac{du(x)}{dx}\right)^2 +\frac{1}{u(x)} \frac{d^2u(x)}{dx^2}$$

The equation becomes:

$$\frac{d^2u(x)}{dx^2}-u(x)\cdot a(x)=0$$

And this one can be solved if a(x) is known. P.e. a(x)=-1 gives sin and cos functions, a(x)=x gives airy functions, a(x)=1 gives hyperbolic ones, etc.

This might be the complete wrong idea, but I don't see it in another way.

[Edit] Someone was faster...

 

[foxjwill]

You could convert it to a differential equation and try to solve this, however it turns out that this new differential equation is severely determined by the unknown function a(x). In order to do this, set:

$$e^{\int u(x)dx}=f(x)$$

Thus:

$$u(x)=\frac{1}{f(x)}\frac{df(x)}{dx}$$

And putting this into the equation gives:

$$\frac{df(x)}{dx}+[f(x)]^2=a(x)$$

This is a Riccati equation, which can be transformed into a linear one by transforming:

$$f(x)=\frac{1}{u(x)}\frac{du(x)}{dx}$$

So:

$$\frac{df(x)}{dx}=-\frac{1}{[u(x)]^2}\left(\frac{du(x)}{dx}\right)^2 +\frac{1}{u(x)} \frac{d^2u(x)}{dx^2}$$

The equation becomes:

$$\frac{d^2u(x)}{dx^2}-u(x)\cdot a(x)=0$$

And this one can be solved if a(x) is known. P.e. a(x)=-1 gives sin and cos functions, a(x)=x gives airy functions, a(x)=1 gives hyperbolic ones, etc.

This might be the complete wrong idea, but I don't see it in another way.

[Edit] Someone was faster...

Actually, I got the integral equation by trying to solve

$$\frac{d^2u(x)}{dx^2}-u(x)\cdot a(x)=0$$

 

[coomast]

foxjwill, to my knowledge there is no solution in terms of a(x). As I pointed out the solution depends so heavily on this function a(x) that you can't solve it without knowing it explicitly. The few examples I gave did show this, no? A sin or cos function compared to a hyperbolic one or even an Airy function (which is closely related to the functions of Bessel) are so different, even for the simple assumed functions of a(x) equal to -1, 1 and x. Maybe there is an explicit integral representation of the solution, but I think it will be closely related to the one you originally posted.

 

[olgranpappy]

also, you could try using Green's functions and getting an approximation... the utility of this approach probably depends on the form of a(x). E.g., find the "homogeneous" solutions
$$\frac{d^2 f}{dx^2}=0$$
and the "free" green's function
$$\frac{d^2}{dx^2}G(x,x')=\delta(x-x')$$
and then consider the term
$$u(x)a(x)$$
as an inhomogeneous term so that the "solution" is given by
$$u(x)=f(x)+\int dx' G(x,x')a(x')u(x')$$

Then, supposing u(x) is only a little different from f(x) once can develop succesive approximations for u(x) as
$$u(x)\approx f(x) + \int dx' G(x,x')a(x')f(x')+\int dx' G(x,x')a(x')\int dx'' G(x',x'')a(x'')f(x'')+\ldots$$