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.

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    $\begingroup$ I think $u(x)=\delta(x-1)$ is a solution. $\endgroup$ – Mario Carneiro Dec 16 '12 at 8:03

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

$$ \begin{eqnarray*} u \left( x \right) & = & \int_0^{\infty} u \left( t \right) u \left( \frac{x}{t} \right) dt \end{eqnarray*}, $$


$$ \int_{0}^{\infty}x^{s-1}u(x)dx = \int_{0}^{\infty}x^{s-1} \int_0^{\infty} u \left( t \right) u \left(\frac{x}{t} \right) dt\, dx $$

$$ \implies U(s)= \int_{0}^{\infty}u(t) \int_0^{\infty} x^{s-1}u \left(\frac{x}{t} \right) {d} x\, dt $$

Using the change of variables $x=ty$ for the inner integral, we have

$$ \implies U(s)= \int_{0}^{\infty}u(t)t^s dt\int_0^{\infty} x^{s-1}u \left(y\right) dy = U(s+1)U(s).$$

  • $\begingroup$ @Fabian: Thanks for the comment. $\endgroup$ – Mhenni Benghorbal Dec 17 '12 at 1:39
  • $\begingroup$ So this proves that there are no other solutions? And could you spell out the integrals in the forward Mellin transform? I'm curious, but I've not seen Mellin transforms used in diff eqs. $\endgroup$ – Mario Carneiro Dec 18 '12 at 9:48
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    $\begingroup$ @MhenniBenghorbal Thanks a lot for your answer and help! $\endgroup$ – Learner Dec 18 '12 at 14:39
  • $\begingroup$ @MhenniBenghorbal Do you have a recommendation on a reference for someone beginning to learn about Mellin transforms and their applications? $\endgroup$ – Learner Dec 18 '12 at 14:40
  • $\begingroup$ @Learner: Here are reference1 reference2. $\endgroup$ – Mhenni Benghorbal Dec 18 '12 at 15:21

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.)

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    $\begingroup$ I am not sure how much it helps to look at blobs at particular positions given the fact that the integral equation is nonlinear. So in the end, we cannot simply sum the individual contributions. $\endgroup$ – Fabian Dec 16 '12 at 11:29
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    $\begingroup$ I haven't assumed anything of the sort. For $x$ large, one of the terms will be small and nearly constant (assuming the solution is smooth) while the other ranges over the blob. The net result will be an integral over the blob (which is a constant) times the small value of the function at the other end. The sum over two terms is taking advantage of the fact that I have two regions (in $t$ space) which contribute to the integral, and the rest is second-order small. Imagine integrating over two disjoint rectangle functions. I don't have to worry about any nonlinearity in the integral itself. $\endgroup$ – Mario Carneiro Dec 16 '12 at 11:41
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    $\begingroup$ I understand your reasoning. If I take $u(x)$ to have a blob everything is fine. I just do not understand why you believe that the solution of the equation (which is what we are after) will have such a blob? $\endgroup$ – Fabian Dec 16 '12 at 11:44
  • $\begingroup$ Well, it can't blow up at $0$ or $\infty$, and it's reasonable to assume it's positive and has some sort of (inexact) symmetry $t\mapsto1/t$, so I expect it to get big, and then die off again. Hence "blob". (Picture the function $e^{-\log^2x}$.) (BTW, exact symmetry is ruled out by the weight function $1$, which makes the $(1,\infty)$ interval contribute more than the $(0,1)$ interval.) $\endgroup$ – Mario Carneiro Dec 16 '12 at 11:47
  • $\begingroup$ More importantly, it is consistent to assume these things about the function, in the sense that my asymptotic form suggests a blob-like structure. If other presuppositions about the form of the solution causes the equation itself to reinforce those presuppositions, then there may be another avenue for a solution. $\endgroup$ – Mario Carneiro Dec 16 '12 at 11:55

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