Closed-form solution of the following equation:

Question

I'm looking for a closed-form solution for $x$ in the following equation:

$(v^2 - 3x^2)\cdot \ln(\frac{1}{x}) - 2x^2 = 0$,

where $v$ is a constant $ \in [0, 1]$ and $ 0 \leq x \leq v $.

I thought about using the Lambert $W$ function, but couldn't transform the equation into the correct form.

Solving the problem numerically with a computer is no problem; see the graph below. On the horizontal axis are the different values of $v$; on the vertical axis, the solutions $x^{*}$.

Graph

In case it helps, here's a table of some $(v, x^{*})$ pairs:

$v$	$x^{*}$
$0$	$0$
$0.25$	$0.12556794085822331$
$0.5$	$0.23849380900899855$
$0.75$	$0.34035742432223576$
$1$	$0.43122112794123896$

Answer 1

1.)

$$(v^2-3x^2)\ln(\frac{1}{x})-2x^2=0$$ $x\to\frac{1}{x}$: $$(v^2-\frac{3}{x^2})\ln(x)-\frac{2}{x^2}=0$$ $$v^2x^2\ln(x)-3\ln(x)-2=0$$ We see, your equation is related to an equation whose left-hand side is a function term that is an algebraic expression of both $x$ and $\ln(x)$. Liouville and Ritt proved that such kind of functions doesn't have elementary inverses: How can we show that $A(z,e^z)$ and $A(\ln (z),z)$ have no elementary inverse?
That means we cannot solve the equation for $x$ by rearranging it by only applying finite numbers of only elementary functions/operations which we can read from the equation and apply them from the left to the left-hand side and the right-hand-side of the equation.
But if the equation has solutions that are elementary numbers, is another mathematical problem.

2.)

$$(v^2-3x^2)\ln\left(\frac{1}{x}\right)-2x^2=0$$ $$v^2\ln\left(\frac{1}{x}\right)-3x^2\ln\left(\frac{1}{x}\right)-2x^2=0$$ $x\to e^\frac{t}{2}$, $\forall t,x\in\mathbb{R}$: $$\left(\frac{3}{2}t-2\right)e^t-\frac{1}{2}v^2t=0$$ $$\left(\frac{3}{2}t-2\right)e^t=\frac{1}{2}v^2t$$ $$\frac{3t-4}{v^2t}e^t=1$$ $$\frac{t-\frac{4}{3}}{t}e^t=\frac{1}{3}v^2$$

We see, we have the variable $t$ in a non-linear rational expression. We cannot apply Lambert W therefore. But we can apply a Generalized Lambert W:

$$t=W\left(^\frac{4}{3}_{\ 0};\frac{1}{3}v^2\right)$$
$t\to 2\ln(x)$: $$2\ln(x)=W\left(^\frac{4}{3}_{\ 0};\frac{1}{3}v^2\right)$$ $$x=e^{W\left(^\frac{4}{3}_{\ 0};\frac{1}{3}v^2\right)}$$

We have a closed form for $x$ and the representations of Generalized Lambert W give some hints for calculating $x$.

See the references below.

[Mezö 2017] Mezö, I.: On the structure of the solution set of a generalized Euler-Lambert equation. J. Math. Anal. Appl. 455 (2017) (1) 538-553

[Mezö/Baricz 2017] Mezö, I.; Baricz, A.: On the generalization of the Lambert W function. Transact. Amer. Math. Soc. 369 (2017) (11) 7917–7934 (On the generalization of the Lambert W function with applications in theoretical physics. 2015)

[Castle 2018] Castle, P.: Taylor series for generalized Lambert W functions. 2018

Answer 2

Your equation can be rearranged as $$\begin{align} \ln(x) &= \frac{2x^2}{3x^2-v^2} \\ x &= \exp \left(\frac{2x^2}{3x^2-v^2}\right) \end{align}$$

This doesn't look to me like something that would have a nice closed form. Mathematica's Solve and Reduce functions both fail to make any progress even on the slightly simplified version $$\begin{align} \ln(x) &= \exp \left(\frac{x^2}{x^2-1}\right). \end{align}$$ This makes me expect there is no satisfying closed form. I'd be glad to be proven wrong though!