Possible analytical solution to $g''(x)=\alpha\left[g(x)^3-g(x)\right]+\beta g(x) e^{-\kappa x}$

Question

kind of related to a previous question of mine. I am describing a physical phenomena related to charged molecules and am interested in the following quantity:

$$\xi=\int_0^{\infty}\left[1-g(x)^2\right]\text{d}x$$

where $g(x)$ is the solution to the following differential equation: $$g''(x)=\alpha\left[g(x)^3-g(x)\right]+\beta g(x) e^{-\kappa x}$$ with boundary conditions $g(0)=0$ and $g(\infty)=1$. For $\beta=0$ this has a nice analytical solution $g(x)=\tanh{\frac{\sqrt{\alpha}x}{\sqrt{2}}}$ with $\xi=\frac{\sqrt{2}}{\sqrt{\alpha}}$.

What I have tried so far: getting an approximate solution by rewriting equation 2 to $1-g(x)^2$ and integrating, splitting the integration between $0\leq x \leq l$ and $x>l$, where $l$ is the region where the exponential term of equation 2 is dominating. I haven't managed to prove the value of $l$ to make this work.

What I did find through some "educated" and accidental guesses is $\xi \approx \frac{\sqrt{2}}{\sqrt{\alpha}}+\frac{\ln{\left[\frac{\beta}{\kappa \sqrt{2\alpha}}+1\right]}}{\kappa}$, which is surprisingly accurate. However, I am in no form able to derive something close to this.

This has me thinking that there might be a solution in the form of $g(x)=g(x)_{\beta=0}+c_1 y(x)$.

What I'm wondering is if there is an analytical solution to the actual differential equation, and if so, could someone point me in the correct direction such that I can (learn to) solve it. If not, what might be other methods to obtain an approximate expression for $\xi$. Thank you.

Edit: I have some software which solves a lattice-based model for this physical problem and found that indeed for a certain region $0\leq x \leq l$ $g(x)$ is very small and almost stationary, after this point the $\tanh$ solution seems to describe $g(x)$ with high accuracy.

Edit: Analysis of the parameters $\alpha$ and $\kappa$ shows that $\frac{2}{\alpha}>\kappa^2$. In general $\kappa>0$,and $\alpha<\kappa<\beta$.

Edit: For the physicists with us, the problem here is seemingly equivalent to describing the spin order parameter in a correlated magnetic field $\beta g(x) e^{-\kappa x}$ see for instance Lubensky, T. C., and Morton H. Rubin. "Critical phenomena in semi-infinite systems. II. Mean-field theory." Physical Review B 12.9 (1975): 3885.

Answer 1

This is by no means a full answer however it is too long for a comment and I think it is of value. We give general solutions in two special cases.

1. $\beta= 0 $. Here this is an autonomous equation and using entry 2.9.2-1 in 1 we get:

\begin{equation} x + C_2 = \int \frac{d g}{\sqrt{\alpha g^2 (g^2/2-1) + C_1}} \end{equation}

The solution above reduces to the solution given in the body of the question if $C_2=0$ and $C_1 = \alpha/2 $. the required boundary conditions are satisfied.

2. $\alpha = 0 $. This is a translation-dilatation equation. Using entry 2.9.1-2.27 in 1 we get one reduces it firstly to a Riccati equation and then to a linear second order ODE which is solved in terms of Bessel functions. The result reads:

\begin{equation} g(x) = C_1 I_0(2 \frac{\sqrt{\beta}}{\kappa} \sqrt{e^{-\kappa x}}) + C_2 K_0(2 \frac{\sqrt{\beta}}{\kappa} \sqrt{e^{-\kappa x}}) \end{equation}

If $C_2=0$ and $C_1=1$ then $g(\infty) = 1$ as required however the value at zero is in general not zero unless some quantization conditions are imposed on the parameters.

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Update: I was thinking that one might find a general solution to this ODE by using a perturbation approach, i.e. by postulating that $g(x) := \sum\limits_{n=0}^\infty f_p(x) \cdot \beta^p $ with $f_0(x) := \tanh(\sqrt{\alpha/2} x) $. In here I found the first order correction $f_1(x) $. This quantity satisfies the following ODE:

\begin{equation} f^{''}_1(x)+ \alpha \left(1- 3 \tanh(\sqrt{\alpha/2} x)^2\right) f_1(x) - e^{-\kappa x} \tanh(\sqrt{\alpha/2} x) = 0 \quad (i) \end{equation}

Now by substituting for $u = \exp(\sqrt{\alpha/2} x)$ we found that $f_1(u) = 1/\sqrt{u} {\tilde f}_1(u) $ where

\begin{equation} {\tilde f}^{''}_1(u)+ \alpha \left(-\frac{15}{4 u^2} + \frac{24}{(1+u^2)^2}\right) {\tilde f}_1(u) + \frac{2}{\alpha} u^{-\frac{3}{2} - \frac{\sqrt{2} \kappa}{\sqrt{\alpha}}} \frac{1-u^2}{1+u^2} = 0 \quad (ii) \end{equation}

The solution to $(ii)$ is given as :

\begin{equation} {\tilde f}_1(u) = C_1 {\tilde q}_1(u) + C_2 {\tilde q}_2(u) + \int\limits_0^u \frac{1}{{\mathfrak W}(u)} \left| \begin{array}{lll} {\tilde q}_1(\xi) & {\tilde q}_2(\xi) \\ {\tilde q}_1(u) & {\tilde q}_2(u) \end{array} \right| rhs(\xi) d\xi \quad (iii) \end{equation}

where

\begin{eqnarray} {\tilde q}_1(u) &:=& \frac{u^{5/2}}{(1+u^2)^2} \\ {\tilde q}_2(u) &:=& \frac{-1-8 u^2+8 u^6 + u^8 + 24 u^4 \log(u)}{4 u^{3/2} (1+u^2)^2} \\ {\mathfrak W}(u) &:=& 1 \\ rhs(u) &:= & -\frac{2}{\alpha} u^{-\frac{3}{2} - \frac{\sqrt{2} \kappa}{\sqrt{\alpha}}} \frac{1-u^2}{1+u^2} \end{eqnarray}

The integrals in $(ii)$ can be expressed in terms of the Gaussian hypergeometric function and its first derivatives by parameters.

alpha =.; b =.; Clear[f]; Clear[f1]; Clear[f2];
g[x_] := Tanh[Sqrt[alpha/2] x] + b f1[x];
rem = Collect[
  Expand[((D[#, {x, 2}] - alpha # (#^2 - 1) - b # Exp[-k x]) & /@ {g[
        x]})] /. b^n_ :> 0 /; n > 1, b, Simplify]
rem = rem/b;
rem = (rem /. Derivative[2][f1][x_] :> alpha/2 u D[u D[f1[u], u], u] /. 
      x :> Sqrt[2/alpha] Log[u] /. 
     f1[Sqrt[2] Sqrt[1/alpha] Log[u]] :> f1[u]) // FunctionExpand;
rem = Simplify[rem, Assumptions -> alpha > 0];
rem /= Coefficient[rem, Derivative[2][f1][u]];
rem1 = Collect[rem, Derivative[n_][f1][u], Simplify]

f1[u_] := 1/Sqrt[u] f2[u];
rem1 /= Coefficient[rem1, Derivative[2][f2][u]];
rem2 = Collect[rem1, {f2[u], Derivative[n_][f2][u]}, Apart];
rem2 = First[rem2]
msol = (f2[u] /. 
   First@DSolve[(f2^\[Prime]\[Prime])[u] + 
       Coefficient[rem2, f2[u]] f2[u] == 0, f2[u], u])
rhs[u_] = (rem2 /. f2[u] :> 0 /. Derivative[2][f2][u] :> 0) // 
   Simplify;
{ff1[u_], ff2[u_]} = {Coefficient[msol, C[1]], 
   Coefficient[msol, C[2]]};

(*The Wronskian*)
W = Det[{{ff1[u], ff2[u]}, {D[ff1[u], u], D[ff2[u], u]}}] // Simplify

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Update 1: As a matter of fact one can write down a recurrence relation for the series expansion solution in powers of $\beta$. So we have $g(x):= \sum\limits_{p=0}^\infty f_p(x) \cdot \beta^p$ with $f_0(x):=\tanh(\sqrt{\alpha/2} x)$. Then we have:

\begin{equation} f_p(x) :=C_1 {\tilde q}_1(x) + C_2 {\tilde q}_2(x) + \int\limits_0^x \frac{1}{{\mathfrak W}(\xi)}\left| \begin{array}{lll} {\tilde q}_1(\xi) & {\tilde q}_2(\xi) \\ {\tilde q}_1(x) & {\tilde q}_2(x) \end{array}\right| rhs(\xi) d\xi \end{equation}

where

\begin{eqnarray} {\tilde q}_1(x) &:=& \frac{e^{\sqrt{2} \sqrt{\alpha } x}}{\left(e^{\sqrt{2} \sqrt{\alpha } x}+1\right)^2} \\ {\tilde q}_2(x) &:=& \frac{e^{-\sqrt{2} \sqrt{\alpha } x} \left(12 \sqrt{2} \sqrt{\alpha } x e^{2 \sqrt{2} \sqrt{\alpha } x}-8 e^{\sqrt{2} \sqrt{\alpha } x}+8 e^{3 \sqrt{2} \sqrt{\alpha } x}+e^{4 \sqrt{2} \sqrt{\alpha } x}-1\right)}{\left(e^{\sqrt{2} \sqrt{\alpha } x}+1\right)^2}\\ {\mathfrak W}(x)&:=& 2 \sqrt{2\alpha} \\ rhs(x) &:=& \alpha \sum\limits_{\begin{array}{l} p_1+p_2+p_3=p \\ 0 \le p_1\le p-1 \\ 0 \le p_2 \le p-1 \\ 0 \le p_3 \le p-1 \end{array}} f_{p_1}(x) f_{p_2}(x) f_{p_3}(x) \quad + \quad f_{p-1}(x) e^{-\kappa x} \end{eqnarray} for $p\ge 1$.

I wouldn't guarantee that it is possible to evaluate all those integrals in closed form but at least one could do that for $p=1,2$ (we almost did it already for $p=1$ above) and then evaluate the integrals numerically for higher values of $p$.

q1[x_] := (E^(
  Sqrt[2] Sqrt[alpha] x)) /(1 + E^(Sqrt[2] Sqrt[alpha] x))^2;
q2[x_] := (
  E^(-Sqrt[2] Sqrt[alpha]
     x) (-1 - 8 E^(Sqrt[2] Sqrt[alpha] x) + 
     8 E^(3 Sqrt[2] Sqrt[alpha] x) + E^(4 Sqrt[2] Sqrt[alpha] x) + 
     24 (Sqrt[alpha] x)/Sqrt[2] E^(2 Sqrt[2] Sqrt[alpha] x) ))/ ((1 + 
    E^(Sqrt[2] Sqrt[alpha] x))^2);
g[x_] := C[1] q1[x] + C[2] q2[x];
((# (alpha - 3 alpha Tanh[(Sqrt[alpha] x)/Sqrt[2]]^2) + 
      D[#, {x, 2}]) & /@ {g[x]}) // Simplify

Det[{{q1[x], q2[x]}, {D[q1[x], x], D[q2[x], x]}}] // Simplify

1: Andrey D. Polyanin, Valentin F. Zaitsev, Handbook of exact solutions for ordinary differential equations, Chapman & Hall/CRC 2003

Answer 2

It is implicit in the comment by Sangchul Lee that you might have a lower bound on $\xi$. If you have $g$ positive and concave then the equation gives $$\alpha(g^2-1)+\beta e^{-\kappa x} \le0.$$ Integration then gives $$\xi \ge \frac{\beta}{\alpha\kappa}.$$

Answer 3

HINT.

If $\;\beta=0,\;$ then, looking for OP solution, $$4g'g''=4\alpha(g^3-g)g',$$ $$2g'^2=\alpha(g^4-2g^2+1),$$ $$g'=\pm a(g^2-1),\quad a=\sqrt{\dfrac{\alpha}2},$$ $$\pm ax+\ln D=\dfrac12\ln\left|\dfrac{g-1}{g+1}\right|,$$ $$\dfrac{g-1}{g+1}=D^2e^{\pm2ax},$$ $$g=\dfrac{De^{\pm2ax}-1}{De^{\pm2ax}+1}.\tag1$$ I.e. $\,g=\coth ax\,$ is the solution too.

Then we can try $$g=\dfrac{f(x)e^{2ax}-1}{f(x)e^{\mp ax}+1},$$ with the equation

$$(f(x)+e^{-2ax})f''(x)-2f'^2(x)-4a(f(x)-e^{-2ax})f'(x)$$ $$=\dfrac b2e^{2ax-kx}(f(x)+e^{-2ax})^2(f(x)-e^{-2ax}),$$

$$\left(\dfrac{f(x)}{f(x)+e^{-2ax}}\right)''-4a\dfrac{f(x)f'(x)}{(f(x)+e^{-2ax})^2}=\dfrac b2e^{2ax-kx}(f(x)-e^{-2ax}).\tag2$$