Continuation of bounded polynomial ODE solutions

Write each component as a finite sum $F_i(x)={\sum }_{\alpha \in E_i}{c}_{i\alpha }{x}^{\alpha }$. On any bounded coordinate box, for each exponent $q\ge 1$ the identity $a^q-b^q=(a-b){\sum }_{j=0}^{q-1}{a}^{q-1-j}{b}^j$ bounds the difference of each monomial at two points by a finite constant times their Euclidean distance; telescoping over the $N$ factors and summing the finitely many monomials proves that $F$ is locally Lipschitz there. It is also continuous, so [F1] applies at every state.

Put $R_{+}=\max (1,R)$ and $M_R=({\sum }_{i=1}^N({\sum }_{\alpha \in E_i}\mathrm{\mid }{c}_{i\alpha }\mathrm{\mid }{R}_{+}^{\mathrm{\mid }\alpha \mathrm{\mid }}{)}^2{)}^{1\mathrm{/}2}$. Since $\mathrm{\mid }{x}_j(t)\mathrm{\mid }\le R$ for each coordinate, the finite polynomial sums give $\mathrm{\parallel }F(x(t)){\mathrm{\parallel }}_2\le M_R$ for all $t<T_{\ast }$.

For $0\le s<t<T_{\ast }$, apply [F3] to $x{\mathrm{\mid }}_{[s,t]}$ to obtain $\mathrm{\parallel }x(t)-x(s){\mathrm{\parallel }}_2\le M_R(t-s)$. Thus, for $t_n=T_{\ast }(1-1\mathrm{/}n)$, $n\ge 2$, each coordinate sequence $(x_i(t_n){)}_{n\ge 2}$ is Cauchy because $\mathrm{\mid }{x}_i(t_n)-x_i(t_m)\mathrm{\mid }\le \mathrm{\parallel }x(t_n)-x(t_m){\mathrm{\parallel }}_2\le M_R\mathrm{\mid }{t}_n-t_m\mathrm{\mid }$. By [F2], each coordinate has a real limit ${\xi }_i$; set $x_{\ast }=({\xi }_1,\dots ,{\xi }_N)\in {\mathbb{R}}^N$. For any $\epsilon >0$, choose for each of the finitely many coordinates an index after which $\mathrm{\mid }{x}_i(t_n)-{\xi }_i\mathrm{\mid }<\epsilon \mathrm{/}\sqrt{N}$, and take the maximum of those indices. Then $\mathrm{\parallel }x(t_n)-x_{\ast }{\mathrm{\parallel }}_2^2={\sum }_{i=1}^N\mathrm{\mid }{x}_i(t_n)-{\xi }_i{\mathrm{\mid }}^2<{\epsilon }^2$, so $x(t_n)\to x_{\ast }$. For any $t<T_{\ast }$ and all sufficiently large $n$ with $t_n\ge t$, the same Lipschitz bound gives $\mathrm{\parallel }x(t)-x_{\ast }{\mathrm{\parallel }}_2\le M_R(t_n-t)+\mathrm{\parallel }x(t_n)-x_{\ast }{\mathrm{\parallel }}_2$; letting $n\to \mathrm{\infty }$ yields $\mathrm{\parallel }x(t)-x_{\ast }{\mathrm{\parallel }}_2\le M_R(T_{\ast }-t)$, so $x(t)\to x_{\ast }$. The closed-ball bound passes to the limit, hence $\mathrm{\parallel }{x}_{\ast }{\mathrm{\parallel }}_2\le R$.

By step 1.1 the polynomial field is locally Lipschitz near $x_{\ast }$. Apply [F1] to the IVP at $(T_{\ast },x_{\ast })$ to obtain a local solution $y$ on an open interval about $T_{\ast }$.

By [F4], $x(t)=x_0+{\int }_0^{t}F(x(s))ds$ for $t<T_{\ast }$ and $y(t)=x_{\ast }+{\int }_{T_{\ast }}^{t}F(y(s))ds$ near $T_{\ast }$. The limit in step 3.1 and continuity of $F$ extend the first identity to $t=T_{\ast }$. Define $\tilde{x}=x$ before $T_{\ast }$ and $\tilde{x}=y$ from $T_{\ast }$ onward, assigning the common value $x_{\ast }$ at $T_{\ast }$. The joined curve is continuous and the two integral identities add to its Volterra equation on $[0,T_{\ast }+ϵ)$ for some $ϵ>0$; [F4] therefore makes it a $C^1$ solution of the same ODE extending $x$.

The extension in step 5.1 agrees with the original curve at every $t<T_{\ast }$ and is defined beyond $T_{\ast }$. This proves the statement.