Does an analytic solution exist for this equation?

Question

I am working on a problem in chaos theory where I am after a certain scaling exponent. I have devised a method for finding this exponent analytically, however it involves solving the following equation for $x$:

$$\ln{\frac{1-a}{1-x}}+ \ln{\frac{(1-a)x}{(1-x)a}}\left[\frac{\lambda}{\ln{\frac{1-a}{a}}}-a \right] = 0.$$

This equation is only defined for $x \in [0,1]$. The constant $a$ is fixed to 0.7 and $\lambda$ is an independent parameter that can be varied in the range $[-0.5, 0.5]$.

Obviously the equation is solved for $x=a$; but some numerical inspection shows there is another root which depends on $\lambda$. This is the root that I am looking for.

Sadly, both me and Mathematica are stumped by this equation. It could very well be that no analytical solution exists and I just need to take a different approach, but I figured it couldn't hurt to ask around here.

If anyone has any ideas that would be awesome!

Answer 1

I do not think that you could get a closed form for $x$ and numerical methods will be required.

To isolate the important terms, let $$k=\frac{\lambda}{\ln{\frac{1-a}{a}}}-a \qquad \text{and} \qquad b=(k+1)\log(1-a)-k\log(a) $$ which make that we look for the non-trivial zero $(x=a)$ of the function $$f(x)=b+k\log(x)-(k+1)\log(1-x)$$

When $\lambda$ is "close" to its lower bound, $x$ is "small". Expanding $f(x)$ around $x=0$ gives $$f(x)=b+k \log (x)+(k+1) x+O\left(x^2\right)$$ Ignoring the higher order terms, the "approximate" solution is $$x=\frac k {k+1}\,W\left(\frac{k+1}{k}e^{-\frac{b}{k}}\right)$$ where appears Lambert function.

Similarly, when $\lambda$ is "close" to its upper bound, $x$ is "close" to $1$. Expanding $f(x)$ around $x=1$ gives $$f(x)=b-k (1-x)-(k+1) \log (1-x)+O\left((1-x)^2\right)$$Ignoring the higher order terms, the "approximate" solution is $$x=1-\frac {k+1}k\,W\left(\frac k{k+1}e^{-\frac{b}{k+1}}\right)$$ When $\lambda$ is "close" to $0$, $x \sim \lambda$.

For example, with $a=0.7$ and $\lambda=-0.3$, the approximation gives $x=0.0843$ while the exact solution, obtained using Newton method, is $0.0850$.

Edit

From a practical point of view, for a given $a$, I should build a table of $\lambda$ as a function of $x$ (this is a direct calculation) and, later, for a given $\lambda$, interpolate in the table to obtain a starting guess $x_0$ for starting Newton iteration.

Using again $a=0.7$, the table would be $$\left( \begin{array}{cc} x & \lambda \\ 0.01 & -0.407234 \\ 0.02 & -0.381465 \\ 0.03 & -0.363124 \\ 0.04 & -0.348276 \\ 0.05 & -0.335532 \\ 0.10 & -0.287362 \\ 0.15 & -0.251336 \\ 0.20 & -0.221038 \\ 0.25 & -0.194133 \\ 0.30 & -0.169460 \\ 0.35 & -0.146334 \\ 0.40 & -0.124303 \\ 0.45 & -0.103039 \\ 0.50 & -0.082283 \\ 0.55 & -0.061814 \\ 0.60 & -0.041424 \\ 0.65 & -0.020900 \\ 0.70 & +0.254189 \\ 0.75 & +0.021582 \\ 0.80 & +0.044279 \\ 0.85 & +0.068787 \\ 0.90 & +0.096449 \\ 0.95 & +0.130807 \\ 0.96 & +0.139368 \\ 0.97 & +0.149046 \\ 0.98 & +0.160548 \\ 0.99 & +0.175825 \end{array} \right)$$

Answer 2

Near $x=0$, the dominant term is $\ln x$, so the expression inside square brackets is zero and $\lambda = a\ln\left({1-a\over a}\right)$. Near $x=1$, the dominant term is $\ln 1-x$, and $ \lambda =(a-1)\ln\left({1-a\over a}\right)$.

I think the next order term for small $x$ gives $$\lambda = (1-a)\ln\left({a\over 1-a}\right)-\frac{\ln (1-a)\ln((1-a)/a)}{\ln((1-a)x/a)}$$ So basically $O(1/ln x)$

And similar order for small $y=1-x$.