Seeking an explicit solution (using Lambert W?) for $\lambda$ in $2\lambda = \sinh(\lambda)+(1+2\beta)(\cosh(\lambda)-1)$

Question

Let $\beta\in \mathbb{R}$, I'm looking for an explicit solution in terms of $\lambda$ to the following equation:

$$2\lambda = \sinh(\lambda)+(1+2\beta)(\cosh(\lambda)-1)$$

It seems, numerically, that for $\beta\in [-1,0]$ the equation has three real solutions $(0,\lambda^*_1,\lambda^*_2)$ and for $\beta>0$ or $\beta<-1$ there are two solutions $(0,\lambda^*_3).$

Could there exist expressions of these solutions in terms of Lambert functions by any chance? See my answer below for a conjecture about the general form.

Answer 1

$$(\beta+1)e^\lambda+\beta e^{-\lambda}-2\lambda-2\beta-1=0$$

The equation is not in a standard form: $e^{-\lambda}$ disturbs. We multiply the equation by $e^\lambda$ therefore.

$$(\beta+1)\left(e^\lambda\right)^2-2\lambda e^{\lambda}-2\beta e^{\lambda}-e^{\lambda}+\beta=0$$

The standard form $c_1\left(e^{\lambda}\right)^2+c_2e^{\lambda}+c_3\lambda e^{\lambda}+c_4\lambda+c_5=0$ has explicit solutions in terms of elementary functions and/or LambertW e.g. for $c_1=c_3=0$, $c_1=c_4=0$, $c_2=c_3=0$, $c_3=c_4=0$, $c_4=c_5=0$ and $c_1=c_2=c_5=0$.

For $\beta=-1$, we get the solutions $\lambda=0$ and $λ=W\left(−1,−\frac{1}{2}e^{-\frac{1}{2}}\right)+\frac{1}{2}$.
For $\beta=-0.5$, we get i.a. the solution $\lambda=0$.
For $\beta=0$, we get the solutions $\lambda=0$ and $λ=−W\left(−1,−\frac{1}{2}e^{-\frac{1}{2}}\right)−\frac{1}{2}$.
For $\beta=0.5$, we get i.a. the solution $\lambda=0$.

Answer 2

Too long for a comment.

I really wonder how you did arrive to this conjecture (I guess that crazy is appropriate) but it seems to be numerically correct only for the range $-1.5 \leq \beta \leq 0.5$ if using the $W_0(.)$ branch.

From a numerical point of view, excluding the trivial $\lambda=0$, I should rewrite the equation as $$\beta =\frac{2 \lambda -\sinh (\lambda )-\cosh (\lambda )+1}{2 (\cosh (\lambda )-1)}$$ and the rhs looks like an hyperbola with horizontal asymptotes $0^-$ when $\lambda \to - \infty$ and $-1^+$ when $\lambda \to + \infty$. This explains one root for $\beta >0$ and two roots when $-1< \beta <0$.

Concerning approximations, using Taylor expansions around $\lambda=0$, we have (this seems to be rather good ) $$\beta=\frac{1}{\lambda }-\frac{1}{2}-\frac{\lambda }{4}+\frac{7 \lambda ^3}{720}+O\left(\lambda ^4\right) $$

Answer 3

Too long for a comment

Hi Claude,

Here is how I found this conjecture. Let me tell you first that the following calculations do not seem to make sense, so do not consider it at a rigorous level.

We recall that want to solve the following equation for any $\alpha \in \mathbb{R}$ (this is an equivalent form as above):

\begin{equation} (1+\beta) (e^\lambda -1)-\lambda=\beta (1-e^{-\lambda})+\lambda \end{equation}

We write it as \begin{equation} \frac{1+\beta}{1+2\beta} e^\lambda +\frac{\beta}{1+2\beta}e^{-\lambda}-(\frac{\lambda}{\beta+\frac{1}{2}}+1)=0 \end{equation} We introduce a Bernoulli random variable $p$ which takes value $+1$ with probability $ \frac{1+\beta}{1+2\beta}$ and $-1$ with probability $ \frac{\beta}{1+2\beta}$.

Note that this might imply some conditions on $\beta$ for the random variable to be properly defined.

We rewrite the equation to solve as \begin{equation} \mathbb{E}_p\left[e^{p \lambda}-(1+\frac{\lambda}{\beta+\frac{1}{2}}) \right]=0 \end{equation}

We now solve the reduced equation

\begin{equation} e^{p \lambda}-(1+\frac{\lambda}{\beta+\frac{1}{2}}) =0 \end{equation} The solution is \begin{equation} \lambda_p=-(\beta+\frac{1}{2}) -\frac{W\left(- p( \beta +\frac{1}{2}) e^{-p\left(\beta +\frac{1}{2}\right)}\right)}{p} \end{equation}

There might be multiple solutions due to the two real branches of the Lambert $W$ function.

My conjecture is then that the solution to my first problem is the expected value of $\lambda_p$ \begin{equation} \lambda=\mathbb{E}_p[\lambda_p]=-(\beta+\frac{1}{2}) +\frac{\beta W\left( e^{\beta +\frac{1}{2}} ( \beta +\frac{1}{2})\right)- (\beta +1) W\left(- e^{-\beta -\frac{1}{2}} ( \beta +\frac{1}{2})\right)}{2 \beta +1} \end{equation}

There is no mathematical ground for this conjecture but somehow it seems to do the job. The question is why ?

Answer 4

The equation can easily written in the form $$ (\beta+1)e^{2\lambda}-2\lambda e^{\lambda}-2\beta e^{\lambda}-e^{\lambda}=-\beta.\tag 1 $$ Assume the function $\lambda(x)$ is such $$ \lambda e^{\lambda}=P(\beta), $$ where $P(x)$ will be set later. Then equation (1) becomes $$ (\beta+1)e^{2\lambda}-(2\beta+1)e^{\lambda}=2P(\beta)-\beta.\tag 2 $$ Hence solving (2) with repect to $e^{\lambda}$ we get $$ e^{\lambda}=Q(\beta):=\frac{2\beta+1\pm\sqrt{1+8(\beta+1)P(\beta)}}{2(\beta+1)}.\tag 3 $$ Multyplying both sides of (3) with $\lambda$, we get $$ P(\beta)=\frac{2\beta+1\pm\sqrt{1+8(\beta+1)P(\beta)}}{2(\beta+1)}\lambda. $$ Hence $$ \lambda=W\left(P(\beta)\right)=\frac{2(\beta+1)P(\beta)}{2(\beta+1)\pm\sqrt{1+8(\beta+1)P(\beta)}-1} $$ Inverting $P(\beta)$ and solving with repect to $P^{(-1)}(\beta)$, we get : $$ P^{(-1)}(\beta)=-\frac{\beta(\beta-W(\beta)-2W(\beta)^2)}{(\beta-W(\beta))^2},\tag 4 $$ where $W(z)$ is Lambert function i.e. $W(z)e^{W(z)}=z$. Hence if $L(\beta)$ is ''the known function'' $$ L(\beta):=-\frac{\beta(\beta-W(\beta)-2W(\beta)^2)}{(\beta-W(\beta))^2},\tag 5 $$ its inverse will be $P(\beta)$. Hence $$ P(\beta)=L^{(-1)}(\beta) $$ and the solution of (1) is $$ \lambda=W\left(P(\beta)\right)=W\left(L^{(-1)}(\beta)\right).\tag 6 $$

NOTES. The solution you have found $$ \lambda=-\left(\beta+\frac{1}{2}\right)+\frac{\beta}{2\beta+1}W\left(e^{\beta+\frac{1}{2}}\left(\beta+\frac{1}{2}\right)\right)-\frac{\beta+1}{2\beta+1}W\left(-e^{-\beta-\frac{1}{2}}\left(\beta+\frac{1}{2}\right)\right)\tag 7 $$ by using the identity $$ W(xe^{x})=x\tag 8 $$ becomes $$ \lambda=-\left(\beta+\frac{1}{2}\right)+\frac{\beta}{2\beta+1}\left(\beta+\frac{1}{2}\right)-\frac{\beta+1}{2\beta+1}\left(-\beta-\frac{1}{2}\right)=0 $$ and is valid for all $\beta$.

CONTINUING

I have ploted the solution you conjectured. (In Mathematica Program), $x\in\textbf{C}$, $k\in\textbf{Z}$.
$$ W[k,x]:=ProductLog[k,x] $$ $$ S[k,x]:=-(x+1/2)+x/(2x+1)W[k,(x+1/2)Exp[x+1/2]]-(x+1)/(2x+1)W[k,(-x-1/2)Exp[-x-1/2]] $$ Ecxept for the value $k=0$, where it becomes $\lambda=S[0,x]=0$ (trivialy set $\lambda=0$ in (1)), for all $x\in [-3/2,1/2]$; the formula $\lambda=S[k,x]$ is not a solution. Hence there is no conjecture and all this by numerical verification.