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I really need help solving this :

$$y_{xx}-\left(y^{3}-y\right)-\varepsilon\frac{1}{2}\left(1-y^{2}\right)=0 $$ With boundary conditions : $$ y(\pm \infty )=-1 $$

I need to find a solution that is accurate up to $O(\epsilon)$ i all so know that $$ y^{(k)}(\pm \infty) = 0\;\;\;\;\; for\; \; k>0 $$ Thanks alot!

Please note that holding boundary condition is not easy and very important, the answer given below is a great effort but unfortunatly it is not a good answer. any help would be great!

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    $\begingroup$ Have you tried the regular perturbation expansion $v = v_0 + \epsilon v_1 + \epsilon^2 v_2 + \ldots$, where $v_i$ are functions of $x$, exclusively, and $v_0(\pm \infty) = -1$, $v_1(\pm \infty) = 1/4$, $v_i(\pm \infty ) = 0$ for $i \geq 2$? You only need to solve up to first order so $v \sim v_0 + \epsilon v_1$. I would love to elaborate some more on my comment, but I'm at an airport now! $\endgroup$ – Dmoreno Jul 7 '15 at 3:32
  • $\begingroup$ @Dmoreno Thanks for you answer! But regular pertubations end up with the same answer of eather a constant function or $\tanh[frac{x}{\sqrt{2}} $ Which does not keep boundary conditions. Thus "making" me chosse the constant, but then next orders are constants too! and the numerical result of this is not a constant graph! it look a aliitle bit like a gaussian $\endgroup$ – Simba Jul 7 '15 at 7:30
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$$\frac{d^2v}{dx^2}-\left(v^{3}-v\right)-\varepsilon\frac{1}{2}\left(1-v^{2}\right)=0$$ This is an autonomous ODE. So, the change of function is : $\frac{dv}{dx}=f(v)$

$\frac{d^2v}{dx^2}=\frac{df}{dv}\frac{dv}{dx}=f'f$

$$f'f-\left(v^{3}-v\right)-\varepsilon\frac{1}{2}\left(1-v^{2}\right)=0$$

$$\frac{1}{2}\frac{d(f^2)}{dv}=\left(v^{3}-v\right)+\varepsilon\frac{1}{2}\left(1-v^{2}\right)=0$$

$$f^2=\frac{1}{2}v^4-v^2+\varepsilon\left(v-\frac{1}{3}v^{3}\right)+c_1$$

$$\frac{dv}{dx}=\sqrt{\frac{1}{2}v^4-v^2+\varepsilon\left(v-\frac{1}{3}v^{3}\right)+c_1}$$

Especially in case of $\varepsilon=0$ with conditions $v(\pm\infty)=-1$ and $v'(\pm\infty)=0$ :

$\frac{1}{2}(-1)^4-(-1)^2+c_1=0$ hense $c_1=\frac{1}{2}$

$$x=\int \frac{dv}{\sqrt{\frac{1}{2}v^4-v^2+\varepsilon\left(v-\frac{1}{3}v^{3}\right)+\frac{1}{2}}}+c_2$$

The exact analytical result is a very complicated formula involving an elliptic integral of the first kind. Further approximate calculus can be done in considering a series expansion relatively to $\varepsilon$, at least up to $O(\varepsilon^2)$.

We search an approximate on the form : $$x=F_0(v)+\varepsilon F_1(v)+O(\varepsilon^2)$$

$$F_0(v)=\int \frac{dv}{\sqrt{\frac{1}{2}v^4-v^2+\frac{1}{2}}}+c_2=-\sqrt{2}\tanh^{-1}(v)+c_2$$

$$F_1(v)=\int\frac{\sqrt{2}v(v^2-3)}{3(v^2-1)^3}dv=-\frac{v^2-2}{6(v^2-1)^2}+c_3$$

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  • $\begingroup$ @JJacuelin Thanks for you answer! Can you explane how would you do a series expansion reltively to $\epsilon$ ? in which part would you recommend trying this? i dont need an analitic solution, just up to $ O(\epsilon)$ $\endgroup$ – Simba Jul 6 '15 at 6:55
  • $\begingroup$ and by the way, i think the expression $ \varepsilon\left(v-\frac{2}{3}v^{3}\right) $ should be $ \varepsilon\left(v-\frac{1}{3}v^{3}\right) $ $\endgroup$ – Simba Jul 6 '15 at 7:06
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    $\begingroup$ You are right, it is $\frac{1}{3}$. I corrected my first answer. $\endgroup$ – JJacquelin Jul 6 '15 at 14:25
  • $\begingroup$ Your answer is wonderfull! But, im trying to get the inverse connection, I cant seem to find a way of manipulating the $ \frac{dv}{dx} $ to get a $ O(\epsilon) $, can you think of something ? $\endgroup$ – Simba Jul 6 '15 at 15:12
  • $\begingroup$ @ JJacquelin In addition, please notice that the arcTan(v) connection does not hold the boundary conditions... $\endgroup$ – Simba Jul 6 '15 at 15:14

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