# Determine whether ODE admits a unique solution locally

Consider the initial value problem, $$5y'' + \bigg(\frac{y'}{x}\bigg)^2 + 4y^2 = 0, \;\; y(0) = 0, \; y'(0) = 0.$$

Determine whether or not this ODE has a unique solution in the neighborhood of the origin.

It seems like we can not use Picard-Lindelof here because of the nonlinear term, namely that it is not Lipschitz at the origin. Beyond this, I have no intuition for how one might prove or disprove this. Any ideas? Thanks!

EDIT: Looking into it some, I've seen some people suggest using asymptotic methods near the origin as a way to show that one (or more) solutions exist. Any thoughts?

• Any progress on this? – nbubis Aug 5 '17 at 12:09
• I haven't had too much time to look into the asymptotic methods for showing uniqueness, hopefully this week I'll get around to looking further into it. It seems like it has good potential though. – Indefinite Aug 6 '17 at 18:18
• Besides $y=0$, there is a formal solution of the form $y(x)=\sum_{n=2}^\infty a_n x^n$ with $a_2=-5/2$, $a_3$ arbitrary and the remaining $a_n$ uniquely determined. It seems to me that the series are convergent, but this would have to be proved... – Helmut Mar 22 '18 at 22:55
• @nbubis I like the problem very much and have wriiten a solution. The method can be applied to similar problems. – Helmut Mar 24 '18 at 9:31
• I am wondering whether there is a reason why my solution is not accepted? – Helmut Oct 10 '18 at 10:26

Claim: There is a one parameter family of convergent series solutions of the form $$y(x)=-\tfrac52x^2+b\,x^3+z(x)\mbox{ where }z(x)\in x^4\CC[[x]]\mbox{ and }b\mbox{ is arbitrary.}$$ To show this, we insert $y(x)=ax^2+b\,x^3+z(x)$, $z(x)\in M:=x^4\CC[[x]]$, in the equation, compare coefficients and find in a first step the conditions $10a+4a^2=0$ and $30b+12ab=0$. The first one is satisfied for $a=0$ which probably leads to $y=0$ and by $a=-5/2$ which we use from now on. The second equation is then satisfied by any $b$.
Inserting now $y(x)=-\frac52x^2+b\,x^3+z$ and multiplying by $x^2$, we obtain the equation $$5x^2z''-10xz'+9b^2x^4+6bx^2z'+z'^2+x^2(-\tfrac52x^2+bx^3+z)^2=0$$ for a series $z\in M$.
This leads to consider the operator $L:M\to M$, $u\to 5x^2u''-10xu'$ which maps $u(x)=\sum_{k=4}^\infty u_kx^k$ to $\sum_{k=4}^\infty (5k-15)ku_kx^k$. Therefore it has an inverse $K=L^{-1}:M\to M$ which maps $u(x)$ to $\sum_{k=4}^\infty \frac{u_k}{(5k-15)k}x^k$. The above equation can now rewritten in ''fixed point form'' for $r=Lz$. $$r=-9b^2x^4-6bx^2(Kr)'-(Kr)'^2-x^2(-\tfrac52x^2+bx^3+Kr)^2=:F(x,r).$$ The existence of a unique formal solution $r\in M$ now follows by comparing the coefficients of $x^k$, $k=4,5,...$ on both sides. On the left hand side, we have $r_k$. The coefficient $[F(x,r)]_k$ of $x^k$ on the right hand side only depends upon $r_4,...,r_{k-1}$. This means that $r_k$ are determined recursively.
In order to prove convergence of $r(x)$ we introduce a notation of majorant series. For $f(x)=\sum_{n=0}^\infty f_nx^n\in\CC[[x]]$ and $g(x))=\sum_{n=0}^\infty g_nx^n\in\RR_+[[x]]$ we write $f\ll g$ if $|f_n|\leq g_n$ for all nonnegative integers $n$. This relation is compatible with addition, multiplication and differentiation. For the operator $K$ we show easily $$f\ll g\mbox{ implies }Kf\ll g, (Kf)'\ll \frac1x g$$ for $f,g$ divisible by $x^4.$ This leads us to the majorant equation $$g=9|b|^2x^4+6|b|x g+\left(\frac gx\right)^2+x^2(\tfrac52x^2+|b|x^3+g)^2=:H(x,g)$$ It has a formal solution in $x^4\RR_+[[x]]$ determined by $g_k=[H(x,g)]_k$, $k=4,5,...$.
The main point is that, by construction, $r\ll g$. A formal proof proceeds by induction but I omit it here. It remains to show the convergence of $g$ (which implies that of $r$ and then that of $z$ and $y$). We put $g=x^4 h$ and obtain the equation $$h=9|b|^2+6|b|x h + x^2h^2+(\tfrac52x+|b|x^2+x^3h)^2$$ By the implicit function theorem, this equation has a holomorphic solution $h$ with $h(0)=9|b|^2$. Its power series is the a convergent formal solution of the $h$-equation. Thus also $g=x^4h$ is convergent and the proof is complete.