# Finding solutions of the differential equation $x\frac{d}{d x}\left(x\frac{d y}{d x}\right)=4(y-x^{2}y+y^{3})$

In my research I have come across the following non-linear differential equation: $$x\frac{d}{d x}\left(x\frac{d y}{d x}\right)=4(y-x^{2}y+y^{3})$$ I want to find the general solution of this equation for $$x\geq0$$, but didn't manage to do it. I have found some facts about the solutions by investigating the asymptotics:

1. For $$x\ll1$$, the solutions go like $$y=Cx^2$$ for some constant $$C$$, or diverge.
2. For $$x\gg1$$, the solutions go like $$y=x$$, or go to $$0$$ (probably after oscillating, found by numerics), or diverge.

In particular, I am interested in a one of the solutions. By looking upon the numerics, I believe that there is only one solution that goes like $$y=x$$ for $$x\gg1$$, and it does not diverge at $$x=0$$. I didn't manage to prove this by now. I am interested in finding this solution analytically, or if impossible, just find appropriate $$C$$ analytically.

Thanks!

• Did you try to change the independent variable $x=e^t$? Commented Jul 30, 2020 at 21:25
• Yes, It was helpful in order to investigate the asymptotics for $x\ll1$, but for arbitrary $x$ it gives $$\frac{d^{2}y}{dt^{2}}=4\left(y-e^{2t}y+y^{3}\right)$$ which I did not find easier to solve.
– Roy
Commented Jul 30, 2020 at 23:35
• All I can currently do is to show that at least one solution with the properties you required exists. But since it is just a minimizer of some quite fancy energy functional (actually, even worse than that: it is actually the limit of a sequence of minimizers on finite intervals; its energy on the whole line diverges to $-\infty$), I doubt very much that anything explicit can be done here. Let me know if you are interested in that existence proof. Commented Jul 31, 2020 at 1:25
• Looking at the plot of $z=f(t,y)=4y(1-e^{2t}+y^2)$, $y$ can't be asymptotic to $e^t$ for ever. That would either make $y=y(t)$ infinitely steep or make it cross the line $\sqrt{e^{2t}-1}≈e^{t}$ and make $z<0$, which would make $y=y(t)$ oscillatory. It would oscillate between the two positive and negative regions. Commented Jul 31, 2020 at 3:58
• @fedja I am interested in your existence proof. Nice if you give the details. Thanks Commented Aug 1, 2020 at 12:33

Mathematica's command DSolve cannot solve your ODE in closed form; so, apparently, a closed-form solution does not exist.

However, it will shown here that, in line with your conjecture, for each real $$c$$ your ODE has solutions $$y$$ such that $$y(x)=(c+o(1))x^2$$ as $$x\to0+$$.

Indeed, your ODE is $$\begin{equation*} x^2 y''+x y'=4(y-x^2y+y^3) \tag{1} \end{equation*}$$ for real $$x>0$$. Assuming the initial conditions $$y(0+)=y'(0+)=0$$ and using the Taylor formulas $$y'(x)=x\int_0^1 v(sx)\,ds$$ and $$y(x)=x^2\int_0^1 (1-s)v(sx)\,ds$$ with $$v:=y''$$, we rewrite (1) as $$\begin{equation*} v(x)=\int_0^1 K(x,s)v(sx)\,ds+4x^4\Big(\int_0^1 (1-s)v(sx)\,ds\Big)^3, \tag{2} \end{equation*}$$ where $$K(x,s):=[4(1-x^2)(1-s)-1]s.$$ For any real $$c$$, setting now $$\begin{equation*} v(x)=c+x w(x), \end{equation*}$$ we rewrite (2) as $$\begin{equation*} w(x)=F(w)(x):=-2cx+\int_0^1 K(x,s)w(sx)\,ds+4x^3\Big(\int_0^1 (1-s)(c+xw(sx))\,ds\Big)^3. \tag{3} \end{equation*}$$ Take now any real $$m>0$$ and $$h>0$$, and let $$\begin{equation*} W_{m,h}:=\{w\in C[0,h]\colon\|w\|\le m\}, \end{equation*}$$ where $$\|w\|:=\max_{x\in[0,h]}|w(x)|$$.

Since $$|K(x,s)|$$ is convex in $$x^2$$, for $$x\in[0,1]$$ we have $$\begin{equation*} \int_0^1 |K(x,s)|\,ds\le\max\Big(\int_0^1 |K(0,s)|\,ds,\int_0^1 |K(1,s)|\,ds\Big) =\max(19/48,1/2)=1/2. \end{equation*}$$ So, for any $$w\in W_{m,h}$$ $$\begin{equation*} \|F(w)\|\le2|c|h+m/2+h^3(|c|+hm)^3/2\le m \tag{4} \end{equation*}$$ if $$h>0$$ is small enough -- which will be assumed in what follows. So, $$F$$ maps $$W_{m,h}$$ into $$W_{m,h}$$. Moreover, if for some $$w$$ and $$u$$ in $$W_{m,h}$$ we have $$\|w-u\|\le t$$ for some real $$t\ge0$$, then similarly to (4) we get $$\begin{equation*} \|F(w)-F(u)\|\le t/2+3h^3(|c|+hm)^2ht/2\le\tfrac23\,t \end{equation*}$$ if $$h>0$$ is small enough. So, $$F$$ is a contractive map of $$W_{m,h}$$ into $$W_{m,h}$$, and hence $$F$$ has a fixed point. That is, equation (3) has a solution $$w\in W_{m,h}$$.

Thus, ODE (1) has a solution $$y$$ with $$y''(x)=c+x w(x)$$ for $$x\in[0,h]$$ and $$w\in W_{m,h}$$, and with $$y(0+)=y'(0+)=0$$. So, $$y(x)=cx^2/2+O(x^3)$$ as $$x\to0+$$. $$\Box$$

• wouldn't the small-$x$ behavior already follow upon linearization of the ODE? the linearized ODE has solution $c J_2(2x)=cx^2/2+{\cal O}(x^3)$. Commented Jul 31, 2020 at 11:15
• Thank you losif, nice to see the proof of the conjecture! Nevertheless, it does not give new details about the eqution, details that I still hope we can find.
– Roy
Commented Jul 31, 2020 at 12:02
• @CarloBeenakker : The linearization needs to be justified. The answer here may be viewed, in part, as such a justification. (However, the method -- which is rather standard -- should work, not just for this ODE, but in a general setting, without even an implicit association with linearization. In view of the Schauder fixed point theorem, one does not even need the right-hand-side function $F$ to be contractive.) Commented Jul 31, 2020 at 13:08

Is this too naive? To obtain the large-$$x$$ behavior, I take a series expansion of $$y(x)$$ in powers of $$1/x$$, $$y(x)=c x + a_0 + a_1/x + a_2/x^2+ a_3/x^3 +\cdots,$$ substitute into $$R(x)=x\frac{d}{d x}\left(x\frac{d y}{d x}\right)-4(y-x^{2}y+y^{3}),$$ expand $$R(x)$$ in powers of $$1/x$$ and demand that the leading order terms vanish. It follows that $$c=1$$, $$a_{n}=0$$ for any even $$n$$, while for odd $$n$$ I find $$a_1=-\frac{3}{8},\;\;a_3=-\frac{9}{128},a_5=-\frac{99}{1024},\;\;a_7=-\frac{11637}{32768},\;\;a_9=-\frac{627669}{262144},$$ so coefficients of the form $$a_{2n+1}=-b_n/2^{2n+1}$$ with integer $$b_n$$.

• I believe that this is indeed true for solutions that does not go to $0$ or diverge at infinity. This is the way I found the asymptotics for $x\gg1$ I wrote about - but this is not the only solution, there are vanishing and diverging solutions too. Nevertheless, I do not see how to use it to prove that the solution that does not vanish or diverge at infinity is unique, and how to find the constant $C$ of the $x\ll1%$ asymptotics of this solution, or show it exists.
– Roy
Commented Jul 31, 2020 at 12:22
• Good question. It could happen that the convergence radius of the series in $1/x$ is zero ... so one would need a statement about the asymptotics of the $a_n$ to settle that. Commented Jul 31, 2020 at 14:26
• @MichaelEngelhardt : I'd guess the radius of convergence of the series expansion is $0$ in this case. Of course, we could use just an asymptotic expansion for $y(x)$, provided that we have the matching asymptotic expansions for $y'(x)$ and $y''(x)$. But of course, we'd need to show that a solution $y$ admitting such asymptotic expansions exists. Commented Jul 31, 2020 at 15:00