A nonlinear "Fredholm" integral equation Consider the integral equation
\begin{eqnarray*}
  u \left( x \right) & = & \int_0^{\infty} u \left( t \right) u \left(
  \frac{x}{t} \right) \mathrm{d} t
\end{eqnarray*}
where the objective is to solve for $u \left( x \right)$ for $x > 0$. I know
about Fredholm integral equations, and this equation looks like some kind of
non-linear Fredholm equation.
How is it possible to solve such an equation? Any hints or references are
welcome.
 A: You can solve the given integral equation using the Mellin transform techniques. If you take the Mellin transform to both sides of the equation w.r.t $x$, you get
$$ U(s)=U(s)U(s+1)\implies U(s)(1-U(s+1))=0$$
$$\implies U(s)=0\quad \mathrm{or}\quad U(s+1)=1. $$
Taking the inverse mellin transform of the above yields the two solutions

$$ u(x) = 0 \quad \mathrm{or} \quad u(x) = \frac{\delta(x-1)}{x}. $$

Deriving the Mellin Transform of the equation: the Mellin transform of afunction $(x)$ is given by

$$ F(s) = \int_{0}^{\infty}x^{s-1}f(x)dx. $$

Taking the Mellin transform of the equation
$$ \begin{eqnarray*}
u \left( x \right) & = & \int_0^{\infty} u \left( t \right) u \left(
\frac{x}{t} \right) dt
\end{eqnarray*}, $$
yields
$$ \int_{0}^{\infty}x^{s-1}u(x)dx = \int_{0}^{\infty}x^{s-1} \int_0^{\infty} u \left( t \right) u \left(\frac{x}{t} \right) dt\, dx  $$ 
$$ \implies U(s)= \int_{0}^{\infty}u(t) \int_0^{\infty} x^{s-1}u \left(\frac{x}{t} \right) {d} x\, dt $$
Using the change of variables $x=ty$ for the inner integral, we have
$$ \implies U(s)= \int_{0}^{\infty}u(t)t^s dt\int_0^{\infty} x^{s-1}u \left(y\right) dy = U(s+1)U(s).$$
A: Since $u(x)$ is integrable, let's suppose there is a blob at $a$ of width $\Delta a\ll a$ (modeled as a rectangle of area $b$) and it goes to $0$ as $x\to\infty$ and $x\to0$. Then for large $x\not\approx a^2$,
$$\begin{align}
u(x)=\int_0^\infty u(t)u(x/t)\,dt & \approx\int_a^{a+\Delta a} u(t)u(x/t)\,dt+\int_{\frac x{a+\Delta a}}^{x/a} u(t)u(x/t)\,dt \\
& \approx b\,u(x/a)+b\frac x{a^2}u(x/a)
\end{align}$$
For large $x$, we can ignore the first term, and let $v(s)=\log u(a^{s+2})$ so that
$$v(s)=\log(b(1+a^s))+v(s-1)\approx \log b+s\log a+v(s-1).$$
Thus $v$ is quadratic, and in particular $v(s)=\frac s2\log(a^{s+1}b^2)+c$ so that
$$\begin{align}
u(x)\approx A(a^{s+1}b^2)^{s/2} & =A\exp\bigg(\Big(\frac{\log x}{2\log a}-1\Big)\log\!\frac{xb^2}a\!\bigg) \\
& =Bx^{\frac32+\log_ab+\frac12\log_ax},\qquad\mbox{as }x\to\infty.
\end{align}$$
For small $x$, the first term dominates, and a similar process yields
$$u(x)\approx Cx^{\log_ab},\qquad\mbox{as }x\to0.$$
Of course, this is all assuming $a\neq1$, and it's only an asymptotic form. It does guide the guesswork (assuming an analytic solution exists), though, especially considering the $x^{\log x}$ falloff.
Note also that the most dominant term in the "falloff" is $x^{1/2\log_ax}$, which actually goes to $\infty$ for $a>1$. This would cause the integral to diverge, so we are forced to conclude $a\leq1$. Similarly, for small $x$, we need $\log_ab>-1$ for the integral to converge, so $\log b<-\log a\Rightarrow b<a^{-1}\geq1$.
What happens if $a=1$? The same derivation as before applies, but now the equation is
$u(x)\approx b(1+x)u(x),$ with no solution. Thus there is no solution except for $u(x)=0$ in this range. My earlier suggestion of $u(x)=\delta(x-1)$ falls under this category. Let's check it: If $x\neq1$, then
$$\begin{align}
0=u(x) & =\int_0^\infty\delta(t-1)\delta(x/t-1)\,dt \\
& =\int_{1-\epsilon}^{1+\epsilon}\delta(t-1)\delta(x/t-1)\,dt+\int_{x-\epsilon}^{x+\epsilon}\delta(t-1)\delta(x/t-1)\,dt \\
& =\int_{1-\epsilon}^{1+\epsilon}\delta(t-1)\cdot0\,dt+\int_{x-\epsilon}^{x+\epsilon}0\cdot\delta(x/t-1)\,dt=0
\end{align}$$
And if $x\approx1$, we can't evaluate it, but we can check that it integrates to $1$:
$$\begin{align}
1=\int_0^\infty u(x)\,dx & =\int_0^\infty\int_0^\infty\delta(t-1)\delta(x/t-1)\,dt\,dx \\
& =\int_0^\infty\delta(t-1)\int_0^\infty t\delta(x-t)\,dx\,dt \\
& =\int_0^\infty\delta(t-1)t\,dt=1
\end{align}$$
So this is a true solution to the equation, if not a very nice one. (Also 0 is a solution.)
(I may edit this later with more special cases.)
