Analytic ODE systems from majorants

Put $w=v-v_0$, adjoin $z_0=t$, and write $z=(z_0,w)\in {\mathbb{R}}^{N+1}$. The autonomous system is $z^{\mathrm{\prime }}=A(z)=(1,H(z_0,v_0+w))$, $z(0)=0$. A formal series $z={\sum }_{q\ge 1}{a}_q{t}^q$ has the unique recursion $(q+1)a_{q+1}=[t^q]A(z)$. This coefficient depends only on $a_1,\dots ,a_q$: each substituted series has zero constant term, so no positive-degree factor can use $a_{q+1}$. Its zeroth coordinate is exactly $t$.

With $L=N+1$, F1 supplies $M\ge 1,r>0$ majorising every component of $A$ by $M\mathrm{/}(1-{\sum }_{i=0}^N{Z}_i\mathrm{/}r)$. The scalar equation $W-\frac{L}{2r}{W}^2=Mt$, $W(0)=0$, has an analytic branch by F3: the derivative in $W$ at zero is one. Its coefficient recursion $W=Mt+\frac{L}{2r}{W}^2$ gives $[t]W=M$ and every subsequent coefficient nonnegative. Differentiating the identity yields $W^{\mathrm{\prime }}=M\mathrm{/}(1-LW\mathrm{/}r)$ near zero, since the denominator there is one. Solving this quadratic on the branch through zero gives $W=(r\mathrm{/}L)(1-\sqrt{1-2LMt\mathrm{/}r})$. The square root is the analytic branch with value one at zero, so a positive convergence neighborhood is explicit; its positive real endpoint $t_{\ast }=r\mathrm{/}(2LM)$ is singular, since differentiating $W-L{W}^2\mathrm{/}(2r)=Mt$ there would give $0=M$ if W extended analytically through it. No entire majorant is claimed.

Compare the recursion in step 1.1 with the vector whose L components are W. At degree zero the initial coefficients agree. If $\mathrm{\mid }[t^j]z_i\mathrm{\mid }\le [t^j]W$ for $j\le q$, F2 bounds $[t^q]A_i(z)$ by $[t^q]M\mathrm{/}(1-LW\mathrm{/}r)=(q+1)[t^{q+1}]W$. Dividing by $q+1>0$ gives the next coefficient bound. This induction proves $z_i\ll W$ at every degree.

Choose $s>0$ strictly inside the convergence radius of W. Its coefficients obey $[t^q]W\le C{s}^{-q}$, where $C={\sum }_q[t^q]W{s}^q<\mathrm{\infty }$. Step 2.1 transfers this bound to z. F4 and F2 show the series sums and its compositions differentiate as in the formal recursion on a smaller interval, so $z^{\mathrm{\prime }}-A(z)=0$. The recursion also forces the Taylor series of every analytic solution; hence two such germs coincide. Finally, for the original nonnegative H with $v_0=0$, the recursion $(q+1)[t^{q+1}]v=[t^q]H(t,v)$ consists solely of nonnegative sums and products from its zero initial coefficient, so every coefficient of v is nonnegative.