General form of this integral functional equation? So I was doing some physics and I can recover a variant of the Maxwell distribution (the answer is a gaussian)
$$P(k)dk = \int_{-\infty}^\infty P(k-x)P(x) dx dk
$$
I was trying to extend this to the relativistic case. How would one find the general form of this probability distribution $P$? (Is it some kind of Bessel function?)
$$P(k) dk= \int_{0}^\infty P(\lambda)P(k/ \lambda) d \lambda dk
$$
Edit:
Calculations of the first claim:
$$P(k) dk = \int_{-\infty}^\infty (a^2/ \pi) e^{-a^2(k-x)^2-a^2x^2} dx dk= (a^2/ \pi) \int_{-\infty}^\infty e^{-a^2 (k^2  - 2x^2 - 2kx)}dx dk =  (a^2/ \pi) \int_{-\infty}^\infty e^{-a^2 (2(k/2 +x)^2  +k^2/2)}dx dk$$
Taking $x \to k/2 + x$
$$(a^2/ \pi) \int_{-\infty}^\infty e^{-a^2 (2x^2  +k^2/2)}dx dk= |a| \sqrt{\frac{2}{\pi}}   e^{-a^2k^2/2} dk $$
where $k \to \sqrt2 k$
$$\frac{|a|}{\sqrt{\pi}}   e^{-a^2k^2} dk = P(k) dk$$
 A: I agree with previous posters that a Gaussian is not a solution of $P*P=P$, as can be seen easily by working in Fourier space $\hat P \cdot \hat P = \hat P$ can only be solved if $\hat P =1$. Your mistake is precisely where you write $k \to \sqrt2 k$, which means you are replacing $P(k)$ with $P(\sqrt 2 k)$. I suppose you want to say something like $P(x)*P( x)=P(\sqrt 2 x)$. This shouldn't be a surprise since this is an expression for the distribution of the sum of two Gaussian random variables (as @Andrew has pointed out in the comments above).
In order to deal with equations like the second part
$$\int_{0}^\infty P(\lambda)P(k/ \lambda)d\lambda $$, one performs the equivalent of the Fourier transform, but in scale space: the Mellin transform.
$$\int_0^\infty k^\alpha \int_{0}^\infty P(\lambda)P(k/ \lambda)d\lambda dk$$ I wont go into the math, but if you look carefully, you will see that what is going on here is that ${k\over \lambda} = e^{\log(k)-\log(\lambda)}$ so $\log(k),\log(\lambda)$ behave like $k,x$ in the convolution example. Please note that you probably also have a normalization issue in the second integral as well which you should fix, before attempting to solve.
A: Notice that, in the integral
$$P(k) \; \textrm{d}k = \int_{-\infty}^{\infty} P(k-x)P(x)\; \textrm{d}x\; \textrm{d}k $$
the $\textrm{d}k$ term is really problematic. Wouldn't it be nice if we could just divide both sides by it? Well, if we do, we get the following:
$$P(k) = \int_{-\infty}^{\infty} P(k-x)P(x)\; \textrm{d}x $$
or
$$P(k) = (P*P)(k) $$
In other words, the function $P$ is precisely equal to the convolution of itself with itself!
This implies that $P(k)=0$ or $P(k)=\delta(k)$ for all $k$.
Edit 1: I've realized that $1*1$ doesn't really make any sense and just blows up to infinity, so I've replaced the latter equation with $P(k)=\delta(k)$.
Edit 2: $P(k)$ cannot be $e^{-k^2}$ because, if it was, then that would imply that $P(k)=\dfrac{\sqrt{\pi}}{2}e^{-2k}$ (check out Find the derivative of the function $f(x)=\int_{0}^\infty e^{-y^2-(x/y)^2} dy$), so you're wrong about that part. And yes, $(P*P)(k)$ returns a Gaussian-type function when $P$ itself is a Gaussian function, but it isn't the same.
Edit 3: This is the closest answer you're gonna get, OP. Suppose we have a function,
$$P(k) = \int_0^\infty P(\lambda)P\left(\dfrac{k}{\lambda}\right)\textrm{d}\lambda$$
By the substitution $u=\dfrac{k}{\lambda}$, we obtain the following:
$$P(k) = \int_0^\infty P(u)P\left(\dfrac{k}{u}\right)\cdot\dfrac{1}{u^2}\textrm{d}u$$
or
$$P(k) = \int_0^\infty P(\lambda)P\left(\dfrac{k}{\lambda}\right)\cdot\dfrac{1}{\lambda}\cdot\dfrac{1}{\lambda}\textrm{d}\lambda$$
Observe that
$$P(k) = \int_0^\infty P(e^{\ln(\lambda)})P\left(e^{\ln(k)-\ln(\lambda)}\right)\cdot\dfrac{1}{\lambda}\cdot\dfrac{1}{\lambda}\textrm{d}\lambda$$
Look at the first part of the integral and ignore the second $\dfrac{1}{\lambda}$:
$$\int_0^\infty P(e^{\ln(\lambda)})P\left(e^{\ln(k)-\ln(\lambda)}\right)\cdot\dfrac{1}{\lambda}\textrm{d}\lambda$$
Let $u = \ln(\lambda)$
$$\int_{-\infty}^\infty P(e^{u})P\left(e^{\ln(k)-u}\right)\textrm{d}u$$
Now let $G(k)=P(e^k)$
$$\int_{-\infty}^\infty G(u)G\left(\ln(k)-u\right)\textrm{d}u$$
Anyone can easily see that this is equal to
$$(G*G)(\ln(k))$$
Unfortunately, I'm not sure if I can even do better than that. Other users have given some nice answers as well, so you should check them out.
A: It is fact that if $f,g\in L^1(\textbf{R})$, then if we define the convolution of $f,g$ as
$$
(f*g)(x)=\int^{\infty}_{-\infty}f(x-t)g(t)dt,\tag 1
$$
we have
$$
\widehat{(f*g)}(\gamma)=\widehat{f}(\gamma)\widehat{g}(\gamma),\tag 2
$$
where we have used the notation $\widehat{f}(x):=\int^{\infty}_{-\infty}f(t)e^{-it x}dt$.
Hence with no loss of information we can write
$$
P(k)dk=\int^{\infty}_{-\infty}P(k-x)P(x)dxdk\Leftrightarrow
$$
$$
\int^{\infty}_{-\infty}P(k)e^{-iks}dk=\int^{\infty}_{-\infty}e^{-iks}\left(\int^{\infty}_{-\infty}P(k-x)P(x)dx\right)dk\Leftrightarrow
$$
$$
\int^{\infty}_{-\infty}P(k)e^{-iks}dk=\widehat{P}(s)\widehat{P}(s)\Leftrightarrow \widehat{P}(s)=1\Leftrightarrow P(k)=\delta(k),
$$
where $\delta(k)$ is the usual Dirac delta function.
Remark.
There are no functions $f,d$ in $L^1(\textbf{R})$, such that $f*d=f$. A more justified answer is in Charles K. Chui:"An introduction to Wavelets". Academic Press. Inc. (Hancourt Brace Jovanovich Publishers) Boston, San Diego, New York, London, Syndney, Tokyo, Toronto. 1992. pg.28
