Question about analytic solution to a Poisson Boltzmann problem

Question

So I'm reading this paper and this paper, both of which solve the Poisson Boltzmann equation for a sphere, the problem: $$\frac{d^2y}{dx^2}+\frac{2}{x}\frac{dy}{dx}=\sinh(y)$$ subject to the boundary conditions $$y(x\to\infty)=y'(x\to\infty)=y''(x\to\infty)=\cdots=0$$ $$\frac{dy}{dx}\bigg|_{x=\kappa R}=-s$$ for some constant value $s$. Some variable changes have been made here but they aren't really important for my question. Both papers argue that the planar case reduces to:

$$\frac{d^2y}{dx^2}=\sinh(y)$$

with the intention of using this to obtain an approximate analytic solution. This next part is where I get confused. The papers suggest that for the purpose of free energy calcluation, we only need to consider the region near the surface. They claim (without showing intermediate steps) that integrating the above expression gives: $$s=-\frac{dy}{dx}\bigg|_{x=\kappa R}=2\sinh\bigg(\frac{y_0}{2}\bigg)$$ where $y_0$ is the value of $y$ at $x=\kappa R$.

My question is: how did they make that leap? I understand that you could write:

$$\frac{d^2y}{dx^2}=\sinh(y)$$ $$\frac{d^2y}{dx^2}\frac{dy}{dx}=\sinh(y)\frac{dy}{dx}$$ $$\int_{\kappa R}^\infty\frac{d^2y}{dx^2}\frac{dy}{dx}dx=\int_{\kappa R}^\infty\sinh(y)\frac{dy}{dx}dx$$ $$\frac{1}{2}\bigg(-\frac{dy}{dx}\bigg|_{x=\kappa R}\bigg)^2=\int_{\kappa R}^\infty\sinh(y)dy$$ But the integral on the right hand side confuses me. (I would expect it to have the form of $\cosh$ and even so, the integral should be divergent unless I'm picking incorrect bounds or something anyways.)

Could anyone help me figure out how they got their answer? I'm sure they're right, but they both seemed to get there independently. In addition, the first paper uses an expansion along the lines of: $$y=y^{(0)}+\frac{1}{\kappa R}y^{(1)}+\frac{1}{\kappa^2R^2}y^{(2)}$$ (they use different variable names but it's the same thing) and collects terms of $\frac{1}{(\kappa R)^n}$ to find a useful solution. Does anyone know what $y^{(n)}$ might denote in this context? It almost looks like a Taylor expansion, but it doesn't seem like they use it this way. (Eq. 5 of the paper)

All help would be greatly appreciated.

Answer 1

I guess I beat everyone to an answer, so I'll add it here. Basically, I realized that there was a hyperbolic trig identity that comes into play. Furthermore, as commented, the bounds of integration are important. The line from my question written:

$$\frac{1}{2}\bigg(-\frac{dy}{dx}\bigg|_{x=\kappa R}\bigg)^2=\int_{\kappa R}^\infty\sinh(y)dy$$

Should be rewritten:

$$\frac{1}{2}\bigg(\frac{dy}{dx}\bigg|_{x=\kappa R}\bigg)^2=\int_{y(\kappa R)}^{y(\infty)}\sinh(y)dy$$

$$\frac{1}{2}\bigg(\frac{dy}{dx}\bigg|_{x=\kappa R}\bigg)^2=\cosh\big(y(\infty)\big)-\cosh(y_0)$$

$$\frac{1}{2}\bigg(\frac{dy}{dx}\bigg|_{x=\kappa R}\bigg)^2=1-\cosh(y_0)$$

Then we use the identity that

$$\sinh\bigg(\frac{x}{2}\bigg)=sgn(x)\sqrt{\frac{\cosh(x)-1}{2}}$$

to show the desired result.