Finding decaying solution for a nonlinear ODE

Question

I am trying to solve the non linear ODE

$$\displaystyle \frac{d^2y} {dx^2}=a^2(y+y^3)$$

With the boundary conditions that it vanish at $\pm\infty$. I am thinking that it might be better to deal with it an integral equation

$$\displaystyle \left(\frac{d^2} {dx^2}-a^2 \right) y =a^2y^3$$

The Green's function will be for that of the modified Hemholtz equation in 1D and the after stacking the solution to the homogenous equation(following approach of Lippmann Schwinger equation) I can write it in integral form as

$$\displaystyle y(x)=e^{-a|x|}-\frac{1}{2a}\int_{-\infty}^{\infty}e^{-a|x-x_{1}|}y^{3}(x_{1})dx_{1}$$

Now I have no idea how to solve this integral equation. For the Lippmann Schwinger equation Neumann series can be used should I do the same with this one or is there some better method for these kind of integral equations?

Answer 1

You can integrate once, after multiplying with $2y'$ $$ y'^2=a^2y^2+\frac{a^2}2y^4+c $$ Vanishing at infinity implies coming to rest there, so that the solution that you seek has $c=0$. Then $$ y'=\pm ay\sqrt{1+\frac12y^2} $$ This has the trivial solution $y=0$. No other solution in both sign variants can change its sign. All further considerations assume that the solution is either entirely positive or entirely negative.

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Substitute $y=\sqrt2\sinh(u)$ and assume $y\ne 0$, $u\ne 0$ $$ \sqrt2\cosh(u)u'=\pm a\sqrt2\sinh(u)\cosh(u)\implies \frac{2e^uu'}{e^{2u}-1}=\pm a \\~\\ \frac{e^u-1}{e^u+1}=Ce^{\pm ax}\implies e^u=\frac{1+Ce^{\pm ax}}{1-Ce^{\pm ax}} $$ so that finally $$ y=\frac{e^u-e^{-u}}{\sqrt2}=\frac{2\sqrt2Ce^{\pm ax}}{1-C^2e^{\pm 2ax}}=\pm\frac{\sqrt2}{\sinh(a(x-x_0))} $$

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One could restart now with $y(x)=\frac{\sqrt{2}}{\sinh(v(x))}$ to get $$ \frac{2\cosh^2(v)v'^2}{\sinh^4(v)}=\frac{2a^2}{\sinh^2(v)}\left(1+\frac1{\sinh^2(v)}\right)\implies v'^2=a^2, $$ which indeed directly gives $v(x)=\pm a(x-x_0)$.

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As any of these solutions has a pole at $x=x_0$, they do not solve the given problem, the stationary solution $y=0$ is the only one satisfying all conditions.

Answer 2

$$\displaystyle 2\frac{d^2y} {dx^2}\frac{dy} {dx}=2a^2(y+y^3)\frac{dy} {dx}$$ $$\left(\frac{dy} {dx}\right)^2=a^2(y^2+\frac12 y^4)+c$$ In the wording of the problem the sentence : "With the boundary conditions that it vanish at ±∞ " is somehow ambiguous for me.

If one understand it as " $y(x)$ is continuous from $(x=-\infty\:,\:y=0)$ to $(x=+\infty\:,\:y=0)$ " then it is implied that $y'(\pm\infty)=0$ then $c=0$. $$\left(\frac{dy} {dx}\right)^2=a^2(y^2+\frac12 y^4)$$ With change of variable $X=e^{ax}$ it is easy to solve this ODE which is separable. The result is : $$y(x)=\frac{4C}{e^{-ax}-2C^2e^{ax}}$$ In case of $C\neq 0$ the function $y(x)$ is discontinuous at $x=x_m=-\frac{1}{2a}\ln(2C^2)$ to which $|y|\to\infty$. This contradicts the supposed specification of continuity. So the unique solution of the problem with the specified boundary conditions is $y(x)=0$.

If the above interpretation of the boundary conditions is not what you mean, please define it more clearly.