Solving a non-linear integro-differential equation

I am trying to solve the following equation

$$f^2(x) - g^2(x) = \alpha\int_0^x f(u) (x-u)du$$

For $\alpha=0$ we get $f=g$. I would like to see how the solution moves away from $g$ when I increase the value of $\alpha$.

In order to derive a closed-form solution I first tried Laplace transforms since $$\mathcal{L} \left\{ \int_0^\cdot f(u) (\cdot-u)du \right\}(s) = \frac{1}{s^2} \mathcal{L}\left\{ f\right\}(s) =: \frac{F(s)}{s^2}$$ The problem is that I did not found any general formula for the Laplace transform of $f^2$.

Is there any ways to relate the Laplace transform of $f^2$ to $F$ (or a polynomial/series in $F$) ? If not is there any other method to solve this problem analytically ?

• You can not use Laplace technique for this problem. – Mhenni Benghorbal Apr 20 '14 at 23:47
• Yes it seems rather hopeless. Is there any other way to solve it besides numerical methods ? – vanna Apr 21 '14 at 16:26
• 1) For $\alpha$=0, you did not get all solutions. Squareroots are multivalued and all... 2) Series methods. – ex0du5 Jun 2 '14 at 5:44
• Is uniqueness obvious? – username Jun 7 '14 at 14:14
• Since you say you want to see how the solution moves away from $g$ when $\alpha$ goes away from zero it seems to me that the method of choice here should be perturbation theory ( see @JJacquelin answer below ). This is a very simple and fast approach and gives you an answer to your problem. Usually a simple first order expansion is good enough for these purposes. Finding an anlytical solution seems like a waste of time (and is probably not possible) – Winther Jun 8 '14 at 17:01

With the function $g(x)$ supposed to be given, on can express the unknown function $f(x)$ as a series of the parameter $\alpha$. This allows to see how $f(x)$ moves away from $g(x)$ when the value of $\alpha$ incresses :

$f(x) = g(x)+\alpha g_1(x) + \alpha^2 g_2(x)+...$

The analytical form of $g_1(x)$ and $g_2(x)$ are shown below. One could continue with the same method to compute the next terms, but this will be more and more arduous.

On a purely formal way, one can derive the recurrence formula giving $g_k(x)$ from the preceeding terms $g(x), g_1(x), g_2(x), ..., g_{k-1}(x)$ But this involves multiple integrals which will be arduous to compute, depending on the given $g(x)$. This is also without considering the question of the convergence of so complicated series.

I will just expand a little on what doraemonpaul posted, as I do think that post is useful for most real-world reasons to ask this type of question.

Write

$$f(x) = \sum_{j=0}^{\infty} f_j x^j$$

(and similarly $g(x) = \sum_{j=0}^{\infty} g_j x^j$).

Then we know that

$$f'(x) = \sum_{j=0}^{\infty} (j+1) f_{j+1} x^j$$

$$(f'(x))^2 = \sum_{j=0}^{\infty} (j+1) f_{j+1} x^j \sum_{k=0}^{\infty} (k+1) f_{k+1} x^k$$

$$= \sum_{j=0}^{\infty} \sum_{k=0}^{\infty} (j+1) (k+1) f_{j+1} f_{k+1} x^{j+k}$$

$$= \sum_{n=0}^{\infty} x^n \sum_{j=0}^n (j+1) (n-j+1) f_{j+1} f_{n-j+1}$$

and

$$f''(x) = \sum_{j=0}^{\infty} (j+2) (j+1) f_{j+2} x^j$$

$$f(x) f''(x) = \sum_{j=0}^{\infty} f_j x^j \sum_{k=0}^{\infty} (k+2) (k+1) f_{k+2} x^k$$

$$= \sum_{j=0}^{\infty} \sum_{k=0}^{\infty} (k+2) (k+1) f_j f_{k+2} x^{j+k}$$

$$= \sum_{n=0}^{\infty} x^n \sum_{j=0}^n (n-j+2) (n-j+1) f_j f_{n-j+2}$$

(and, of course, the same forms for g). Now taking this and plugging into the equation from doraemonpaul (after rescaling $\alpha = 2 \beta$) that:

$$f(x) f''(x) + (f'(x))^2 - \beta f(x) = g(x) g''(x) + (g'(x))^2$$

and equating like powers of $x^n$ you get the following sequence of equations $\forall_{n \in \mathbb{N}}$:

$$-\beta f_n + \sum_{j=0}^n [(n-j+2) (n-j+1) f_j f_{n-j+2} + (j+1) (n-j+1) f_{j+1} f_{n-j+1}] = \sum_{j=0}^n [(n-j+2) (n-j+1) g_j g_{n-j+2} + (j+1) (n-j+1) g_{j+1} g_{n-j+1}]$$

So, as you can see, you can calculate each Taylor expansion coefficient of g in terms of the coefficients of f by starting with the equation for n=0 and working up (with $g_0$, $g_1$ to choose). So the characterization of the solution is pretty complete with the diffy que. This is sufficient for most real-world requirements here. Do you need something else?

Hint:

$f^2(x)-g^2(x)=\alpha\int_0^xf(u)(x-u)~du$

$f^2(x)-g^2(x)=\alpha\left(x\int_0^xf(u)~du-\int_0^xuf(u)~du\right)$

$2f(x)f'(x)-2g(x)g'(x)=\alpha\left(\int_0^xf(u)~du+xf(x)-xf(x)\right)$

$2f(x)f'(x)-2g(x)g'(x)=\alpha\int_0^xf(u)~du$

$2f(x)f''(x)+2(f'(x))^2-2g(x)g''(x)-2(g'(x))^2=\alpha f(x)$

• Not really sure I see your point here. This leads to a non-linear ODE that is, to my knowledge, not analytically solvable. – vanna May 1 '14 at 11:01
• I had that first part in this form: $f^2 -g^2 = \alpha f \star \mbox{id} \Rightarrow D(f^2) - D(g^2) = \alpha f \star D \, \mbox{id} = \alpha f \star 1 = \alpha \int\limits_0^xf(u)\,du$, but was not sure if it could be trusted. – mvw Jun 9 '14 at 0:41