Finding a Solution to a Equation that Ends up as a Weird Repeating Series

Question

I need to find the solution to this equation that ends up in a weird repeating series. The equation in question is: $$ \ln(y)=\frac{K}{\alpha}+\frac {x^{2}}{2\alpha\sigma}+\frac{\ln(\ln(y))}{2\alpha} $$

Another variation of the same problem is: $$ \frac{y^{2\alpha}}{\ln(y)}=Ae^{x^2/\sigma} $$

Note that those equal signs are supposed to be asymptotic expansions where $x\to\infty$. So if I end up getting a bunch of lower order terms that go slow enough (mainly slower than a constant), they can be thrown out.

Edit: I'm looking for the solution to be an asymptotic expansion for y~y(x) up to an order that's less than a constant. I also need y to be increasing very quickly. The answer has been found, but requires use of the -1 branch for the Product Log function.

Answer 1

Let's look at the fixed point of $u\to a+b\log(u)$, so $$u=a+b\log u\\e^u=e^a u^{b}\\u^{-b}e^u=e^a\\ue^{-u/b}=e^{-a/b}\\-\frac{u}be^{-u/b}=-\frac1be^{-a/b}\\-\frac{u}b=W\left(-\frac{e^{-a/b}}b\right)\\u=-bW\left(-\frac{e^{-a/b}}b\right)$$ where $W$ is the product logarithm, which is only defined for $-e^{-a/b}/b\ge-1/e$ (and requires a branch cut on $(-1/e,0)$).

In our case, we have $\log(y)$ the fixed point of $u\to a+b\log(u)$ where $a=\frac{K}\alpha+\frac{x^2}{2\alpha\sigma}$ and $b=\frac1{2\alpha}$ so the above gives us $$\log(y)=-\frac1{2\alpha}W_0\left(-2\alpha e^{-2K-x^2/\sigma}\right)\\y(x)=\exp\left(-\frac1{2\alpha}W_0\left(-2\alpha e^{-2K-x^2/\sigma}\right)\right)$$

There's probably a way to clean this up, but eh.

Note that for large $x$ we have that $W_0(x)\sim \log(x)-\log(\log(x))$.

Answer 2

For the first equation $$\ln(y)=\frac{K}{\alpha}+\frac {x^{2}}{2\alpha\sigma}+\frac{\ln(\ln(y))}{2\alpha}$$ let us define $y=e^z$ and $A=\frac{K}{\alpha}+\frac {x^{2}}{2\alpha\sigma}$. This transforms the equation to $$z=A+\frac{\log(z)}{2\alpha}$$ the solution of which being given in terms of Lambert function $$z=-\frac{W\left(-2 \alpha e^{-2 \alpha A}\right)}{2 \alpha }$$ $$\log(y)=-\frac{W\left(-2 \alpha e^{-(2 K+\frac{x^2}{\sigma })}\right)}{2 \alpha }$$ and $x$ is supposed to be very large. What you could use if that, for small values of $t$ $$W(t)=t-t^2+\frac{3 t^3}{2}+O\left(t^4\right)$$ So, limited to first order, $$\log(y)\approx e^{-(2 K+\frac{x^2}{\sigma })}$$