Jtdylktuy

I'm currently looking through qualifying exam questions and one has had me stumped for a few weeks. We are asked to determine whether or not there exists a unique solution of this differential equation in a neighborhood of $x = 0$. begin{equation} y'' + frac{yy'}{x^4} + y^2 = 0 , text{ } y(0)=y'(0)=0 end{equation} I believe that the solution is unique with $y = 0$ as the only solution. I have managed to prove that a non-trivial solution must satisfy $y leq 0$, but I cannot deal with $y < 0$ case.

Edit: Thank you to Ingix for noticing a mistake in my proof, which I edited below.

Indeed, assume there exists $f(x)$ that solves the above ODE such that $f(x) > 0$ on say $(0,varepsilon_1]$, where $varepsilon_1$ is positive. Then by MVT, we see begin{equation} f'(xi) = frac{f(varepsilon_1) - f(0)}{varepsilon_1-0} > 0, xi in (0, varepsilon_1) end{equation} In particular by the continuity of $text{ }f'$, there's a neighborhood $(delta_{1}, delta_{2}] subset (0,varepsilon_1]$ such that $text{ } f' > 0$ where $delta_{1} = sup_{x in [0,varepsilon_1]}$ such that $text{ }f'(x)=0$.

Then by MVT again begin{equation} frac{f'(delta_2)-f'(delta_1)}{delta_2 - delta_1} > 0 end{equation} Therefore, there's a neighborhood $[eta_1, eta_2] subset (delta_1, delta_2)$ such that $f,f',f'' > 0$. Therefore the ODE cannot be satisfied.

I am unsure how to proceed for the case $y < 0$ because the above argument would not give a contradiction. I have tried rewriting the ODE into weak form, using $u = y^2 Rightarrow u' = 2yy'$, and I do not see any obvious transformations such as equidimensional in $x/y$, scale invariant, or autonomous to simplify the ODE. Any hints on how to proceed will be very helpful!

Obviously, $y(x)=0$ is a solution. Using the supplied informations, $y=x^au(x)$, $age 2$, $u(0)ne 0$, if a non-zero solution exists. Insert this and its derivatives to find
$$
[x^au''+2ax^{a-1}u'+a(a-1)x^{a-2}u]+frac{[x^au][x^au'+ax^{a-1}u]}{x^4}+x^{2a}u^2=0\~\
[x^2u''+2axu'+a(a-1)u]+[u][x^{a-2}u'+ax^{a-3}u]+x^{a+2}u^2=0\~\
$$
To get a non-trivial value for $u(0)$, at least two terms need to remain in this equation when setting $x=0$. This only happens for $a=3$ with the remaining equation $$6u(0)+3u(0)^2=0,$$ so that the non-trivial solution is $u(0)=-2$.

Thus for $y(x)=x^3u(x)$ the equation is
$$
x^2u''+(6+u)xu'+6u+(3+x^5)u^2=0
$$

Now one can solve for the power series coefficients of $u$ using your preferred CAS that can do power series over polynomial rings. The CAS Magma is used in the code below:

A<a>:=FunctionField(Rationals());

PS<x>:=PowerSeriesRing(A);

Pol<z>:=PolynomialRing(Rationals());

Diff:=func<w|Degree(w) lt 1 select 0 else &+[ Coefficient(w,k)*k*x^(k-1) : k in [1..Degree(w)] ]>;

prec := 40;

u := -2+0*x;

for m in [1..prec] do

    ua := u+a*x^m+O(x^(m+1));

    Dua := Diff(ua); D2ua:=Diff(Dua); 

    eqn := x^2*D2ua + (6+ua)*x*Dua + 6*ua + (3+x^5)*ua^2;

    eqn;

    u := u + Roots(Pol!Coefficient(eqn,m))[1][1]*x^m; 

    m,u;

end for;

with the result
$$
u(x)=-2 - frac{2}{17}x^5 - frac{42}{8959}x^{10}
- frac{841}{5025999}x^{15} - frac{1134494}{200418411457}x^{20}
- frac{659042819}{3546804627554529}x^{25}
- frac{60555294564493}{10115926601559333469596}x^{30}
- frac{21682330793829581}{113844637973948738866833384}x^{35}
- frac{8763668817047648604028}{1458948251016373153173581517077169}x^{40}+...
$$
This strongly suggests that the solution has the general form $y(x)=x^3v(x^5)$. Inserting again
begin{align}
y'(x)&=5x^7v'(x^5)+3x^2v(x^5)text{ and }\ y''(x)&=25x^{11}v''(x^5)+50x^6v'(x^5)+6xv(x^5)
end{align}
results in
$$
[25x^{10}v''(x^5)+50x^5v'(x^5)+6v(x^5)]+v(x^5)[5x^5v'(x^5)+3v(x^5)]+x^5v(x^5)^2=0
$$
and with $t=x^5$
$$
25t^2v''(t)+5t(10+v(t))v'(t)+6v(t)+(3+t)v(t)^2=0,
$$
which looks a little easier from a singularity point-of-view.

One could also start the power series computation directly with $y$ with the same method,

y := 0*x;

for m in [1..prec] do

    ya := y+a*x^m+O(x^(m+1));

    Dya := Diff(ya); D2ya:=Diff(Dya); 

    eqn := x^4*D2ya + ya*Dya + x^4*ya^2;

    m,eqn;

    n := Valuation(eqn);

    q := Pol!Coefficient(eqn,n);

    if Degree(q) gt 0 then

        rts := Roots(q); "a in ",[ rr[1]: rr in rts];

        r := rts[1][1];

        if r eq 0 and #rts gt 1 then r := rts[2][1]; end if;

        y := y + Roots(q)[1][1]*x^m;

    end if; 

    m,y;

end for;

which returns the same

-2*x^3 - 2/17*x^8 - 42/8959*x^13 - 841/5025999*x^18 - ...

only with more logistical effort.