# Unusual Differential Equation for CDF

Consider the following differential equation

$$F(cx) = F(x) + x F'(x)$$

for $c>1$.

1. Does this differential equation belong to a some well known class?

2. Is there a way to find all the solutions $F(\cdot)$ of this equation that are also cumulative distribution functions?

3. $F(x) = x^a$ for a properly chosen $a$ is a solution. Is it unique in the class of cumulative distribution functions?

P.S. It is a repost from https://math.stackexchange.com/questions/565758/differential-equation-for-cdf

You do not tell the range of $x$. Distribution functions are usually defined on the real line, while $x^a$ on the real line is not a distribution function for any $a$.
Anyway, here is a way to solve your equation. Set $F(x)=\phi(\log x)$ and then $\log x=t$. You obtain $$\phi(a+t)=\phi(t)+\phi'(t),$$ where $a=\log c$. This is a linear differential-difference equations, and there is a large literature on equations of this type. The usual method is to apply Fourier transform. It gives $$e^{iaz}\Phi(z)=\Phi(z)+iz\Phi(z),$$ So $\Phi(z)(e^{iaz}-iz-1)=0.$ This means that $\Phi$ is a linear combination of delta-functions sitting at the roots of the equation $e^{iaz}-iz-1$. So $\phi$ is an exponential sum, and $F$ is a linear combination of (complex) powers.
• My bad, $F(x) = x^a$ is a cdf of a power distribution with support in [0,1]. – pzryumov Nov 26 '13 at 3:41
• Then perhaps you explain the definition of cdf ? The standard definition that I know is an increasing function on the real line with $F(x)\to 1$ as $x\to+\infty$ and $F(x)\to 0$ as $x\to-\infty$. And what do you mean by cdf? – Alexandre Eremenko Nov 26 '13 at 4:41
If you substitute $x=e^t$, $c=e^a$, then your equation transforms to $$F(t+a)=F(t)+F'(t).$$ This kind of delay equation is well studied.