Green function region of convergence nonlinear differential equation Given a second order nonlinear differential equation with the boundary values:
$y'' + \frac{1}{x}y' + y^2 = 0$, with $y'(0)= 0, y(1)=1$
How would one estimate the region of convergence given by:
$$\max\limits_{x,t \in<0,1>}|G(x,t)|\max\limits_{y \in (-\infty,\infty)}|f'(y)| < 1$$
Any reference would be welcome because I'm new to Green functions.
 A: The standard method first homogenizes the problem, for instance by setting $y=x^3+u$ so that $u(0)=0=u(1)$. The equation changes to
$$
0=(6x+u'')+(3x+\tfrac1xu) +(x^3+u)^2=u''+\tfrac1xu+2x^3u+x^6+9x+u^2
$$
In the comments you identified the linear operator $L[u]=u''+\frac1xu'$ as the one to invert per Green's function. 
The next step is to determine basis solutions of $L[u]=0\iff (xu')'=0\iff u=C\ln(x)+D$ that satisfy one of the homogeneous boundary conditions, for instance $u_1(0)=1$, $u_1'(0)=0$ and $u_2(1)=0$, $u_2'(1)=1$. This gives $u_1(x)=1$ and $u_2(x)=\ln(x)$. 
Then $\tilde G(s,x)=u_1(\min(s,x))u_2(\max(s,x))$ is continuous and $\tilde G_x(s,x)=\chi_{[s,1]}(x)\frac1x$ has a jump from $0$ to $\frac1s$ at $x=s$, so that the Green function is obtained by reducing this to the unit jump in 
$$G(s,x)=s\ln(\max(s,x)).$$ 
The associated operator is 
$$G[u](x)=\int_0^1G(x,t)u(t)\,dt=x\ln(x)\int_0^xu(s)\,ds+\int_x^1s\ln(s)\,u(s)\,ds$$ 
so that $L[G[u]])=u$. Applying this to the full differential equation gives the fixed-point integral equation
$$
u=-G[2x^3u+x^6+9x+u^2]
$$
One would have to check if the singularity at $x=0$ prevents the finiteness for continuous functions $u$ with $u(0)=u(1)=0$.
