Convergence of the Goursat majorant

Admissible choices exist: $\rho =\min (1,r\mathrm{/}(2dNM))$ gives $a\ge 1\mathrm{/}2$. Fix any rho allowed by the statement, so $a>0$. The analytic function $P(y,q)=aq-b{q}^2-M\mathrm{/}(1-y\mathrm{/}r)+M$ vanishes at $(0,0)$ and has $P_q(0,0)=a>0$. F1 gives g. Writing $g={\sum }_{k\ge 1}{c}_k{y}^k$ yields $c_k=(b\mathrm{/}a){\sum }_{i=1}^{k-1}{c}_i{c}_{k-i}+(M\mathrm{/}a)r^{-k}$. At k=1 the sum is empty and $c_1=M\mathrm{/}(ar)>0$; successive coefficients are nonnegative.

The function $g(\sigma \mathrm{/}\rho +Nv)$ is analytic near zero, and its power-series coefficients in sigma,v are nonnegative by the finite binomial expansion and step 1.1. F2 therefore gives an analytic v with nonnegative coefficients. Moreover $v^{\mathrm{\prime }}(0)=g(0)=0$. Choose a positive sigma-radius small enough that $\mathrm{\mid }\sigma \mathrm{/}\rho +Nv(\sigma )\mathrm{\mid }$ is inside the actual convergence radius of g and less than r, and that $\mathrm{\mid }b{v}^{\mathrm{\prime }}(\sigma )\mathrm{\mid }<1$. These inequalities hold by continuity and their zero origin values.

Put $q=v^{\mathrm{\prime }}(\sigma )$ and $y=\sigma \mathrm{/}\rho +Nv(\sigma )$. The algebraic identity of step 1.1 is equivalent to $(q+M)(1-bq)=M\mathrm{/}(1-y\mathrm{/}r)$, hence $q=M\mathrm{/}((1-y\mathrm{/}r)(1-bq))-M$ on the neighborhood of step 2.1. With $\sigma =t+\rho {\sum }_i{x}_i$ we have ${\mathrm{\partial }}_t{U}_j=q$, ${\mathrm{\partial }}_{x_i}{U}_j=\rho q$, ${\sum }_j{U}_j=Nv$, and ${\sum }_{i,j}{\mathrm{\partial }}_{x_i}{U}_j=dN\rho q$. Substitution gives exactly the system G of F3.

On a sufficiently small polydisc in (t,x), $\mathrm{\mid }t\mathrm{\mid }+\rho {\sum }_i\mathrm{\mid }{x}_i\mathrm{\mid }$ is smaller than the radius selected in step 2.1, so this substitution converges absolutely. The coefficients of U and of its trace are nonnegative because those of v and the linear form are nonnegative. At the origin, $U_j=v(0)=0$ and ${\mathrm{\partial }}_{x_i}{U}_j=\rho v^{\mathrm{\prime }}(0)=0$. Thus every hypothesis on U in F3 holds and U dominates the formal zero-data solution.