Local well-posedness of a supplied truncated coupled RG system

Since $\mathrm{\Omega }$ is open and $(t_0,x_0)\in \mathrm{\Omega }$, choose $r,h_0>0$ so the compact cylinder $K=[t_0-h_0,t_0+h_0]\times \bar{B}(x_0,2r)$ lies in $\mathrm{\Omega }$. Continuity of $\beta$ on $K$ gives a finite bound $M$, and [F3] gives a single state-Lipschitz constant $L$ there. Choose $0<h\le h_0$ with $h(M+1)\le r$ and $h(L+1)<1$; then $hM\le r$ and $Lh<1$.

For every $y\in B(x_0,r)$, the cylinder $[t_0-h,t_0+h]\times \bar{B}(y,r)$ lies in $K$. Its vector-field and Lipschitz bounds are at most $M$ and $L$. Applying [F4] with these bounds and the inequalities from step 1.1 gives a unique solution $x_y$ on this same interval, with graph in that cylinder. The common-interval uniqueness clause of [F4] gives agreement on every larger common interval containing $t_0$. The same reasoning applies to every $z\in B(x_0,r)$.

Take $y,z\in B(x_0,r)$ and put $a=\mathrm{\parallel }y-z{\mathrm{\parallel }}_2$ and $u(t)=\mathrm{\parallel }{x}_y(t)-x_z(t){\mathrm{\parallel }}_2$. By [F2] and [F5], $x_y(t)-x_z(t)=y-z+{\int }_{t_0}^{t}g(s)ds$, where $g(s):=\beta (s,x_y(s))-\beta (s,x_z(s))$ is continuous on the closed interval between $t_0$ and $t$. For continuous $g:[a_0,b_0]\to {\mathbb{R}}^{2N}$ with $a_0\le b_0$, define its vector integral componentwise by ${\int }_{a_0}^{b_0}g=({\int }_{a_0}^{b_0}{g}_1,\dots ,{\int }_{a_0}^{b_0}{g}_{2N})$, with each coordinate integral the Riemann integral. If $v={\int }_{a_0}^{b_0}g\mathrm{\ne }0$, set $e=v\mathrm{/}\mathrm{\parallel }v{\mathrm{\parallel }}_2$; using the Euclidean inner product $⟨p,q⟩={\sum }_{j=1}^{2N}{p}_j{q}_j$ and linearity of the scalar Riemann integral gives $\mathrm{\parallel }v{\mathrm{\parallel }}_2={\int }_{a_0}^{b_0}⟨e,g(s)⟩ds$. Since $\mathrm{\parallel }e{\mathrm{\parallel }}_2=1$ and $0\le \mathrm{\parallel }g(s)-⟨e,g(s)⟩e{\mathrm{\parallel }}_2^2=\mathrm{\parallel }g(s){\mathrm{\parallel }}_2^2-⟨e,g(s){⟩}^2$, we have $⟨e,g(s)⟩\le \mathrm{\parallel }g(s){\mathrm{\parallel }}_2$, so $\mathrm{\parallel }v{\mathrm{\parallel }}_2\le {\int }_{a_0}^{b_0}\mathrm{\parallel }g(s){\mathrm{\parallel }}_{2}ds$; the same inequality is immediate if $v=0$. Reversing orientation changes the vector integral only by a sign. Since both graphs lie in $K$, [F3] gives $\mathrm{\parallel }g(s){\mathrm{\parallel }}_2\le Lu(s)$ for each $s$ between $t$ and $t_0$; the triangle inequality and the preceding estimate, applied on $[\min (t,t_0),\max (t,t_0)]$, yield $u(t)\le a+L{\int }_{\min (t,t_0)}^{\max (t,t_0)}u(s)ds\mathrm{.}$ Moreover, each curve remains within radius $r$ of its own initial value; because $y,z\in B(x_0,r)$, this gives $u(t)\le 4r$.

If $t\mathrm{\ne }{t}_0$, set $d=\mathrm{\mid }t-t_0\mathrm{\mid }$, $\epsilon =\mathrm{sgn}(t-t_0)$, and $v(q)=u(t_0+\epsilon q)$ for $0\le q\le d$. Step 3.1 gives $v(q)\le a+L{\int }_0^{q}v(s)ds$ and $v\le U:=4r$. Repeated substitution, by induction on $m\ge 1$, yields $v(d)\le a{\sum }_{j=0}^{m-1}\frac{(Ld{)}^j}{j!}+U\frac{(Ld{)}^m}{m!}\mathrm{.}$ For $m=1$, use $v\le U$ in the integral remainder. The induction step integrates each power $s^j$ and uses ${\int }_0^d{s}^{j}ds=d^{j+1}\mathrm{/}(j+1)$.

For fixed $Ld$, the remainder $(Ld{)}^m\mathrm{/}m!$ tends to zero since the ratio of successive terms is $Ld\mathrm{/}(m+1)\to 0$, and the partial sums converge to $e^{Ld}$. Letting $m\to \mathrm{\infty }$ in step 4.1 gives $u(t)\le a{e}^{L\mathrm{\mid }t-t_0\mathrm{\mid }}$. At $t=t_0$ the same estimate is equality. This is the required continuous-dependence bound.