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.
On
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.)
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
Deriving the Mellin Transform of the equation: the Mellin transform of afunction $(x)$ is given by
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
$$ \implies U(s)= \int_{0}^{\infty}u(t)t^s dt\int_0^{\infty} x^{s-1}u \left(y\right) dy = U(s+1)U(s).$$