# Does an analytic solution exist for this equation?

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!

• Would it help to plot lambda as a function of x ? Mar 28, 2020 at 12:07
• That does show that for negative $\lambda$ $x<a$ whilst for positive $\lambda$ $x>a$. This is actually consistent with the result that I am after, so that is comforting. Mar 28, 2020 at 12:25
• WolframAlpha says that lambda approaches 0.254189 as x approaches 1, and lambda approaches -0.593109 as x approaches 0. It has some more information as well Mar 28, 2020 at 12:53
• I would guess that lambda is some sort of logarithmic function, or equivalenty that x is some exponent of lambda. I can't get an exact form though, that first term on the lhs is really bugging me. Mar 28, 2020 at 13:02
• Those limits are 0.3 ln 7/3 and -0.7 ln 7/3 Mar 28, 2020 at 13:07

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

• The value for 0.7 is out. I found Wolfie got it wrong there too Mar 31, 2020 at 0:43

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 $$y=1-x$$.

• Wouldn't you need to have$$\lambda = -a \ln{\frac{a}{1-a}} - \frac{\ln{\left(\frac{1-a}{a}\right)^2}}{\ln{\frac{(1-a)x}{a}}},$$ to match the boundary conditions? Mar 28, 2020 at 19:15