# Finding decaying solution for a Non linear Ode

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

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

## 2 Answers

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.

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 $$y=\frac{e^u-e^{-u}}{\sqrt2}=\frac{2\sqrt2Ce^{\pm ax}}{1-C^2e^{\pm 2ax}}=\pm\frac{\sqrt2}{\sinh(a(x-x_0))}$$

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 $$v(x)=\pm a(x-x_0)$$.

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.

• Actually the discontinuous solution you derived has precisely the behavior I wanted as this non linear poisson equation describes a potential. What is strange though is the potential can have a singularity at any x_0 and this boundary condition does not fix x_0 to be zero. Anyway, thanks this was very helpful. – Human Aug 14 at 10:31
• @Human : The equation is autonomous, any shifted solution is again a solution. You would have to impose some symmetry condition, as example, if you do not want to directly fix the position of the singularity. – LutzL Aug 14 at 10:44

$$\displaystyle 2\frac{d^2y} {dx^2}\frac{dy} {dx}=2a^2(y+y^3)\frac{dy} {dx}$$ $$\left(\frac{dy} {dx}\right)^2=a^2(y^2+\frac12 y^4)+c$$ 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.

• You are right. I should have worded it better. This equation is the non linear poisson equation and I wanted a potential that vanished at infinity. What is strange is that the boundary condition does not fix the singularity to be at zero and it has to be argued on physical grounds. Anyway, this answer was very helpful. – Human Aug 14 at 10:38