The rescaled pulse endpoint map converges to the kick $K(q,p)=(q,p+q^2-a)$ in $C^1$, and the period map is smooth, symplectic and $C^1$-close to $F$

At $\delta =0$ the forward solution through $(q_0,p_0)$ is $z_0(s)=(q_0,p_0+(q_0^2-a){\int }_0^{s}w)$. For initial data in $S$, $\mathrm{\mid }{q}_0\mathrm{\mid }\le R+1$ and $\mathrm{\mid }{q}_0^2-a\mathrm{\mid }\le (R+1{)}^2$; hence $\mathrm{\mid }{p}_0+(q_0^2-a){\int }_0^{s}w\mathrm{\mid }\le M-1$ on $[0,1]$. Thus this reference solution lies at least distance $1$ from the boundary of the stated tube $T$. Its endpoint is $K(q_0,p_0)$.

On $T$ and for $\mathrm{\mid }\delta \mathrm{\mid }\le 1$, there are constants $B_0,L_0$ depending only on $a,w$ such that $\mathrm{\mid }{f}_{\delta }-f_0\mathrm{\mid }\le B_0\mathrm{\mid }\delta \mathrm{\mid }$ and $\mathrm{\parallel }{D}_z{f}_{\delta }\mathrm{\parallel }\le L_0$: one may take $B_0=M+R+2$ and $L_0=1+2\mathrm{\parallel }w{\mathrm{\parallel }}_{\mathrm{\infty }}(R+2)$. Up to the first exit from $T$, subtracting the integral equations gives $\mathrm{\mid }{z}_{\delta }(s)-z_0(s)\mathrm{\mid }\le B_0\mathrm{\mid }\delta \mathrm{\mid }+L_0{\int }_0^s\mathrm{\mid }{z}_{\delta }-z_0\mathrm{\mid }$, so Gronwall gives $\sup \mathrm{\mid }{z}_{\delta }-z_0\mathrm{\mid }\le B_0{e}^{L_0}\mathrm{\mid }\delta \mathrm{\mid }$. Choose $d_0>0$ with $B_0{e}^{L_0}{d}_0<1\mathrm{/}2$. An exit on $[0,1]$ is impossible by the distance margin in step 1.1. A maximal solution terminating before the end without exiting is extendible by [F1]. Therefore every datum in $S$ has a solution on $[0,1]$ staying in $T$.

Larger tubes establish the domains needed for inversion. Set $b:=R+2$, $Q:=[-b,b{]}^2$, $B:=b+b^2+a+2$ and $E:=[-B,B{]}^2$. For every datum in $E$, the forward reference solution has $\mathrm{\mid }q\mathrm{\mid }\le B$, $\mathrm{\mid }p\mathrm{\mid }\le B+B^2+a$. For a terminal datum $y=(y_q,y_p)\in E$ at $s=1$, the reference backward solution is $(y_q,y_p-(y_q^2-a){\int }_s^{1}w)$ with exactly the same bounds. Both reference families therefore have distance at least $1$ from the boundary of $T_E:=[-B-1,B+1]\times [-B-B^2-a-1,B+B^2+a+1]$. Apply the integral comparison of step 2.1 on this tube, in the backward case after $\tau =1-s$ so that the field is $-f_{\delta }(1-\tau ,z)$. This field has the same boundedness and Lipschitz estimates. With a smaller threshold $d_E>0$, both evolutions exist over the whole unit interval for all their data in $E$, and their state differences from the reference evolutions are uniformly $O(\mathrm{\mid }\delta \mathrm{\mid })$. All constants depend only on $a,w$. In particular $K^{-1}(Q)\subset \mathrm{int}E$, since $K^{-1}(y)=(y_q,y_p-y_q^2+a)$ and its coordinates are bounded by $b$ and $b+b^2+a=B-2$. This uses the actual inverse preimage, without assuming it lies in $Q$.

These families are smooth in initial or terminal data and $\delta$ on neighbourhoods of the compact data sets after decreasing $d_E$. To justify the whole interval from the local assertion [F3], cover each reference graph by finitely many local time-state-parameter neighbourhoods where [F3] gives smooth local solution operators, and subdivide $[0,1]$ into finitely many sufficiently short successive intervals. The graph stays inside the tube with a positive margin by step 3.1; the field and its state derivative are bounded on a slightly enlarged compact tube, so one common short interval is admissible at every graph point. At each subdivision endpoint local existence and smooth dependence apply to the endpoint data from the preceding interval. Their finite composition is smooth and equals the unique solution. Shrinking the initial-data and parameter neighbourhoods finitely many times keeps each intermediate value in its admissible neighbourhood. Finitely many such initial-data neighbourhoods cover each compact data set, yielding one positive parameter threshold. The same argument for the time-reflected field gives the backward family. Thus $K_{\delta }$ and the backward endpoint map $L_{\delta }(y):=z(0)$ for $z(1)=y$ are smooth near $E$ and $Q$, respectively, for all sufficiently small $\mathrm{\mid }\delta \mathrm{\mid }$, including $0$.

The state comparisons in steps 2.1 and 3.1 also hold in $C^1$. Indeed, on either compact tube, $D_z{f}_{\delta }-D_z{f}_0$ is bounded by $\mathrm{\mid }\delta \mathrm{\mid }$, and $D_z{f}_0$ is uniformly Lipschitz in $z$, so along compared trajectories $\mathrm{\parallel }{D}_z{f}_{\delta }(s,z_{\delta })-D_z{f}_0(s,z_0)\mathrm{\parallel }\le C\mathrm{\mid }\delta \mathrm{\mid }$. By [F3] their derivative matrices solve the variational equations with the same initial matrix $I$. Gronwall first bounds both matrices by $e^L$, then, after subtracting their integral equations, bounds the difference by $C{e}^{2L}\mathrm{\mid }\delta \mathrm{\mid }$, with fixed constants for the tube. The identical argument for the reflected field controls derivatives in terminal data. Consequently there are finite positive $C_f,C_b$ with ${\sup }_Q(\mathrm{\mid }{K}_{\delta }-K\mathrm{\mid }+\mathrm{\parallel }D{K}_{\delta }-DK\mathrm{\parallel })\le C_f\mathrm{\mid }\delta \mathrm{\mid }$ and ${\sup }_Q(\mathrm{\mid }{L}_{\delta }-K^{-1}\mathrm{\mid }+\mathrm{\parallel }D{L}_{\delta }-D{K}^{-1}\mathrm{\parallel })\le C_b\mathrm{\mid }\delta \mathrm{\mid }$. Increasing $C_f$ includes the forward estimates on $S$ in the original tube; hence clause 1 follows.

Choose ${\delta }_1>0$ below all previous thresholds and with $C_b{\delta }_1<1$. Then $L_{\delta }(Q)\subset \mathrm{int}E$, because $K^{-1}(Q)$ has coordinate margin at least $2$ in $E$. Put $V:=(-b,b{)}^2$ and $O:=\mathrm{int}E$. Uniqueness of the original and reflected equations shows that $K_{\delta }$ is injective on $O$: solutions sharing their terminal value coincide when evolved backwards. Its differential is invertible by the following direct calculation: the rescaled field is $J\mathrm{\nabla }h$ for $h(s,q,p)=\delta (p^2+q^2)\mathrm{/}2+w(s)(aq-q^3\mathrm{/}3)$; hence $A=D_z{f}_{\delta }=J\mathrm{Hess}h$ with symmetric Hessian and $A^{\mathsf{T}}J+JA=0$. The variational matrix satisfies $(Y^{\mathsf{T}}JY{)}^{\mathrm{\prime }}=0$ and $Y(0)=I$, so $Y^{\mathsf{T}}JY=J$. A vector in the kernel of $Y$ must then lie in the kernel of the invertible matrix $J$, so $Y$ is invertible. By The Euclidean inverse function theorem, the smooth map $K_{\delta }$ on the open set $O$ has C1 local inverses at every point; A local inverse of a $C^k$ regular map is $C^k$ upgrades these to smooth local inverses, which agree by injectivity; therefore $K_{\delta }:O\to K_{\delta }(O)$ is a smooth diffeomorphism onto its open image. For every $y\in V$, its backward trajectory starts at $L_{\delta }(y)\in O$ and ends at $y$; hence $V\subset K_{\delta }(O)$ and the inverse on $V$ is exactly $L_{\delta }$. Define $N:=R_0^{-1}(O)$, an open neighbourhood of $S$. Then $G_{\delta }=K_{\delta }\circ R_0:N\to K_{\delta }(O)$ is a smooth diffeomorphism, and clause 2 follows from step 5.1 and $R_0(Q)=Q$, since $R_0$ is a linear isometry.

The matrix computation of step 6.1 also gives symplecticity. The rescaled field is $J\mathrm{\nabla }h$ with $h(s,q,p)=\delta (p^2+q^2)\mathrm{/}2+w(s)(aq-q^3\mathrm{/}3)$. Its derivative matrix is $A=J\mathrm{Hess}h$ with symmetric Hessian. Since $J^{\mathsf{T}}=-J$ and $J^2=-I$, direct multiplication gives $A^{\mathsf{T}}J+JA=0$. Therefore the variational matrix satisfies $(Y^{\mathsf{T}}JY{)}^{\mathrm{\prime }}=Y^{\mathsf{T}}(A^{\mathsf{T}}J+JA)Y=0$ and $Y(0)=I$, giving $Y^{\mathsf{T}}JY=J$. This identity forces $Y$ invertible (a vector in its kernel must lie in the kernel of $J$). Also $D{R}_0=J$ satisfies $J^{\mathsf{T}}JJ=J$. Applying the chain rule to the composition yields $D{G}_{\delta }^{\mathsf{T}}JD{G}_{\delta }=J$ on $N$, proving clause 3.

Since $S_w(V)=V$, the conjugated inverse is defined on the open neighbourhood $V$ of $S$ and has the explicit expression $H_{\delta }=R_0\circ S_w\circ L_{\delta }\circ S_w$, using $S_w{R}_0^{-1}=R_0{S}_w$. At $\delta =0$ this is $S_w{F}^{-1}{S}_w=F$. Fixed linear isometries preserve both state and derivative comparison norms, so step 5.1 gives ${\sup }_S(\mathrm{\mid }{H}_{\delta }-F\mathrm{\mid }+\mathrm{\parallel }D{H}_{\delta }-DF\mathrm{\parallel })\le C_b\delta$. The composition is a smooth diffeomorphism onto its image, and clause 4 holds with $C_2:=C_b$ and clause 2 with $C_1:=C_f$.

Increase $C_1,C_2$ to positive constants if needed and decrease ${\delta }_1$ to include every threshold above. For $0<\delta \le {\delta }_1$, each coordinate difference is bounded by the Euclidean difference norm and each derivative entry difference by the operator norm. Thus both maps satisfy the six componentwise closeness inequalities with the common bound $\bar{C}\delta$, where $\bar{C}:=\max (C_1,C_2)$. The forward tube conclusion was established independently in steps 1.1-2.1 and was preserved throughout the enlarged-domain argument.