A Computable Criterion for the Existence of Connecting Orbits in Autonomous Dynamics

Abstract

A general theorem that guarantees the existence of an orbit connecting two hyperbolic equilibria of a parametrized autonomous differential equation in ${\mathbb {R}}^n$ near a suitable approximate connecting orbit given the invertibility of a certain explicitly given matrix is proved. Numerical implementation of the theorem is described using five examples including two Sil’nikov saddle-focus homoclinic orbits and a Sil’nikov saddle-focus heteroclinic cycle.

Appendices

Appendix 1: Statement of Lemma 11.7 in Palmer [2000]

For the reader’s convenience, we present here the statement of Lemma 11.7 in Palmer [29].

Lemma

Let $E$, $F$ be Banach spaces, $O\subset E$ an open set and $\mathcal{G}:O\rightarrow F$ a $C^2$ function. Suppose $y$ is an element of $O$ for which $\Vert \mathcal{G}(y)\Vert \le \delta$ and $D\mathcal{G}(y)=L$ is invertible with $\Vert L^{-1}\Vert \le K$. Set

$$\begin{aligned} \varepsilon =2K\delta \quad \mathrm{and}\quad M=\sup \{\Vert D^2\mathcal{G}(x)\Vert :x\in O,\; \Vert x-y\Vert \le \varepsilon \}. \end{aligned}$$

Then if the closed ball of radius $\varepsilon$ around $y$ is in $O$ and

$$\begin{aligned} MK\varepsilon =2MK^2\delta <1, \end{aligned}$$

there is a unique solution $x$ of the equation $\mathcal{G}(x)=0$ satisfying $\Vert x-y\Vert \le \varepsilon$.

Appendix 2: Proofs of Lemmas

Proof of Lemma 1

Since by Eq. (5), $|x-y_k|\le \Delta _2 < R_k$, there exists $\bar{t}$ satisfying $0<\bar{t}\le h_k$ such that $|\phi ^t(x,a)-\phi ^t(y,a_0)|\le R_k$ for $0\le t\le \bar{t}$. We suppose $\bar{t}$ is maximal with this property so that either $\bar{t}=h_k$ or $\bar{t}<h_k$ and $|\phi ^{\bar{t}}(y_k,a_0)-\phi ^{\bar{t}}(y_k,a_0)|= R_k$. It follows from this that for $0\le t\le \bar{t}$

$$\begin{aligned} |\phi ^t(x,a)-y_k|\le |\phi ^t(x,a)-\phi ^t(y_k,a_0)|+|\phi ^t(y_k,a_0)-y_k|\le 2R_k, \end{aligned}$$

so that the norms of the derivatives of $f_x$ and $f_a$ at $(\phi ^t(x,a),a)$ are bounded by $M_1$ and $M_3$ respectively. Now

$$\begin{aligned} \phi ^t(x,a)-\phi ^t(y_k,a_0)=x-y+\int ^t_0 f(\phi ^u(x,a),a)-f(\phi ^u(y_k,a_0),a_0)du \end{aligned}$$

so that for $0\le t\le \bar{t}$,

$$\begin{aligned} \begin{aligned}&|\phi ^t(x,a)-\phi ^t(y_k,a_0)|\\&\quad \le |x-y_k|\\&\qquad +\int ^t_0 |f(\phi ^u(x,a),a)-f(\phi ^u(x,a),a_0)| + |f(\phi ^u(x,a_0),a_0)-f(\phi ^u(y_k,a_0),a_0)|du\\&\quad \le |x-y|+\int ^t_0 M_3|a-a_0|+M_1|\phi ^u(x,a)-\phi ^u(y_k,a_0)|du. \end{aligned} \end{aligned}$$

Then by Gronwall’s lemma

$$\begin{aligned} |\phi ^t(x,a)-\phi ^t(y_k,a_0)| \le |x-y|e^{M_1t}+M_3|a-a_0|{e^{M_1t}-1\over M_1} \le [1+M_3t]e^{M_1t}\Delta \end{aligned}$$

if $0\le t\le \bar{t}$. Then if $\bar{t}<h_k$, it would follow that $[1+M_3\bar{t}]e^{M_1\bar{t}}\Delta \ge R_k$, thus contradicting Eq. (5). Hence $\bar{t}=h_k$. It follows that $|\phi ^t(x,a)-\phi ^t(y_k,a_0)|\le R_k$ for $0\le t\le h_k$, and hence that $|\phi ^t(x,a)-y_k|\le 2R_k$ for $0\le t\le h_k$, thus proving the lemma.$\square$

Proof of Lemma 2

From Lemma 1 we know that $|\phi ^t(x,a)-y_k|\le 2R_k$ for $0\le t\le h_k$, so that the norms of the derivatives $f_x$, $f_a$, $f_{xx}$, $f_{xa}$, $f_{aa}$ at $(\phi ^t(x,a),a)$ are bounded by $M_1$, $M_2$, $M_3$, $M_4$, $M_5$ respectively. In particular, since $|f_x(\phi ^t(x,a),a)|\le M_1$ for $0\le t\le h_k$ and $Y(t)=\phi ^t_x(x,a)$ is the solution of

$$\begin{aligned} \dot{Y}=f_x(\phi ^t(x,a),a)Y,\quad Y(0)=I, \end{aligned}$$

it follows by Gronwall’s lemma that

$$\begin{aligned} | \phi ^t_x(x,a)| \le e^{M_1t},\quad 0\le t\le h_k. \end{aligned}$$

Next $y(t)=\phi ^t_{a}(x,a)$ is the solution of

$$\begin{aligned} \dot{y}=f_x(\phi ^t(x,a),a)y+f_{a}(\phi ^t(x,a),a),\quad y(0)=0 \end{aligned}$$

so that

$$\begin{aligned} y(t)=\int ^t_0\phi ^{t-s}_x(\phi ^s(x,a),a)f_{a}(\phi ^s(x,a),a)ds. \end{aligned}$$

It follows that for $0\le t\le h_k$, $|y(t)|\le \int ^t_0 e^{M_1(t-s)}M_3ds$ so that for these same $t$

$$\begin{aligned} |\phi ^{t}_{a}(x,a)|\le M_3M^{-1}_1(e^{M_1t}-1). \end{aligned}$$

Since $y(t)=\phi ^t_{xx}(x,a)(\xi ,\eta )$ is the solution of

$$\begin{aligned} \dot{y}=f_x(\phi ^t(x,a),a)y+f_{xx}(\phi ^t(x,a),a)(\phi ^t_x(x,a)\xi ,\phi ^t_x(x,a)\eta ),\quad y(0)=0 \end{aligned}$$

it follows that

$$\begin{aligned} y(t) =\int ^t_0\phi ^{t-s}_x(\phi ^s(x,a),a)f_{xx}(\phi ^s(x,a),a)(\phi ^s_x(x,a)\xi ,\phi ^s_x(x,a)\eta )ds \end{aligned}$$

so that

$$\begin{aligned} |y(t)|\le \int ^t_0e^{M_1(t-s)}M_2e^{2M_1s}|\xi |\,|\eta |ds =M_2e^{M_1t}\int ^t_0e^{M_1s}ds\,|\xi |\,|\eta | \end{aligned}$$

and hence for $0\le t\le h_k$

$$\begin{aligned} |\phi ^{t}_{xx}(x,a)|\le M_2M^{-1}_1e^{M_1t}(e^{M_1t}-1) \le M_2M^{-1}_1e^{M_1t}M_1te^{M_1t}=M_2te^{2M_1t}\le M_6te^{M_1t}.\nonumber \\ \end{aligned}$$

(36)

Since $y(t)=\phi ^t_{xa}(x,a)\xi$ is the solution of

$$\begin{aligned} \dot{y}=f_x(\phi ^t(x,a),a)y+f_{xx}(\phi ^t_x(x,a)\xi ,\phi ^t_a(x,a))+f_{xa}\phi ^t_x(x,a)\xi ,\quad y(0)=0, \end{aligned}$$

where $f_{xx}=f_{xx}(\phi ^t(x,a),a)$ etc., it follows that

$$\begin{aligned} y(t)=\int ^t_0\phi ^{t-s}_x(\phi ^s(x,a),a)[f_{xx}(\phi ^s_x(x,a)\xi ,\phi ^s_a(x,a))+f_{xa}\phi ^s_x(x,a)\xi ]ds \end{aligned}$$

so that

$$\begin{aligned} \begin{aligned}|y(t)|&\le \int ^t_0e^{M_1(t-s)}[M_2e^{M_1s}M_3M^{-1}_1(e^{M_1s}-1)+M_4e^{M_1s}]ds\,|\xi |\\&=\int ^t_0[M_2e^{M_1t}M_3M^{-1}_1(e^{M_1s}-1)+M_4e^{M_1t}]ds\,|\xi | \end{aligned} \end{aligned}$$

and hence for $0\le t\le h_k$

$$\begin{aligned} \begin{aligned} |\phi ^{t}_{xa}(x,a)|&\le M_2M_3e^{M_1t}M^{-2}_1(e^{M_1t}-M_1t-1)+tM_4e^{M_1t}\\&\le M_2M_3e^{M_1t}{1\over 2}t^2e^{M_1t}+tM_4e^{M_1t}\\&\le M_7te^{M_1t}. \end{aligned} \end{aligned}$$

(37)

Since $y(t)=\phi ^t_{aa}(x,a)$ is the solution of

$$\begin{aligned} \dot{y}=f_x(\phi ^t(x,a),a)y+f_{xx}(\phi ^t_a(x,a),\phi ^t_a(x,a))+2f_{xa}\phi ^t_a(x,a) + f_{aa},\quad y(0)=0, \end{aligned}$$

where $f_{xx}=f_{xx}(\phi ^t(x,a),a)$ etc., it follows that

$$\begin{aligned} y(t)=\int ^t_0\phi ^{t-s}_x(\phi ^s(x,a),a)[f_{xx}(\phi ^s_a(x,a),\phi ^s_a(x,a))+2f_{xa}\phi ^s_a(x,a)+ f_{aa}]ds \end{aligned}$$

so that

$$\begin{aligned} \begin{aligned}&|y(t)|\\&\quad \le \int ^t_0e^{M_1(t-s)}[M_2M^2_3M^{-2}_1(e^{M_1s}-1)^2+2M_4M_3M^{-1}_1(e^{M_1s}-1)+ M_5]ds\\&\quad =e^{M_1t}\int ^t_0[M_2M^2_3M^{-2}_1(2\cosh (M_1s)-2)+2M_4M_3M^{-1}_1(1-e^{-M_1s})+ M_5e^{-M_1s}]ds\\&\quad ={M_2M^2_3\over M^{2}_1}e^{M_1t}{2\sinh (M_1t)-2M_1t\over M_1}+2{M_4M_3\over M^{2}_1}e^{M_1t}(e^{-M_1t}+M_1t-1) +M_5{e^{M_1t}-1\over M_1} \end{aligned} \end{aligned}$$

and hence for $0\le t \le h_k$

$$\begin{aligned} \begin{aligned} |\phi ^{t}_{aa}(x,a)|&\le 2M_2M^2_3e^{M_1t}{{1\over 6}\mathrm{cosh}(M_1t)M^3_1t^3\over M^3_1}+{2M_3M_4\over M^{2}_1}e^{M_1t}{1\over 2}(M_1t)^2 +M_5{e^{M_1t}-1\over M_1}\\&= {1\over 3}M_2M^2_3e^{M_1t}\mathrm{cosh}(M_1t)t^3+M_3M_4t^2e^{M_1t}+M_5te^{M_1t}\\&\le M_{8}te^{M_1t}. \end{aligned} \end{aligned}$$

(38)

Then the lemma follows from Eqs. (36), (37) and (38).$\square$

Proof of Lemma 3

We break the solution of Eq. (16) into solving

$$\begin{aligned} \xi _{k+1}=e^{h_kA_+}\xi _k+\phi ^{h_k}_a(z_+,a_0),\quad k\ge N_1 \end{aligned}$$

(39)

and

$$\begin{aligned} \xi _{k+1}=e^{h_kA_+}\xi _k+g_{k}, \quad k\ge N_1. \end{aligned}$$

(40)

First we consider the difference equation Eq. (39). It is convenient to relate its solutions to those of the autonomous differential equation

$$\begin{aligned} \dot{x}=A_+x+f_a(z_+,a_0). \end{aligned}$$

(41)

Note first that $\phi ^t_a(z_+,a_0)$ is the solution of Eq. (41) with $x(0)=0$ so that any solution $x(t)$ of Eq. (41) satisfies

$$\begin{aligned} x(t)=e^{(t-s)A_+}x(s)+\phi ^{t-s}_a(z_+,a_0). \end{aligned}$$

(42)

Next we define the sequence $s_k$ for $k\ge N_1$ by $s_{N_1}=0$ and $s_{k+1}=s_k+h_k$ for $k\ge N_1$ and we claim that if $\xi _k$ solves Eq. (39), then

$$\begin{aligned} \xi _k=x(s_k),\quad k\ge N_1 \end{aligned}$$

where $x(t)$ is the solution of Eq. (41) with $x(0)=\xi _{N_1}$. Clearly this is true for $k=N_1$ and if it holds for some $k\ge N_1$, then by Eq. (42)

$$\begin{aligned} x(s_{k+1})=e^{h_kA_+}x(s_k)+\phi ^{h_k}_a(z_+,a_0) =e^{h_kA_+}\xi _k+\phi ^{h_k}_a(z_+,a_0)=\xi _{k+1}. \end{aligned}$$

Conversely, we can show by reversing the reasoning that if $x(t)$ is the solution of Eq. (41) with $x(0)=\xi$, then $\xi _k=x(s_k)$ is the solution of Eq. (39) with $\xi _{N_1}=\xi$. Clearly $\xi _k$ is bounded for $k\ge N_1$ if and only if $x(t)$ is bounded for $t\ge 0$.

Next we claim that Eq. (41) has a unique bounded solution $x_+(t)$ on $t\ge 0$ with $Q_+x(0)=0$ given by

$$\begin{aligned} x_+(t)=\int ^t_{0}e^{A_+(t-s)}Q_+f_a(z_+,a_0)ds -\int ^{\infty }_te^{A_+(t-s)}(I-Q_+)f_a(z_+,a_0)ds. \end{aligned}$$

Note $x_+(t)$ is well-defined and bounded since, using Eq. (3), the integrals of the norms of the integrand are bounded by

$$\begin{aligned} \left[ \int ^t_{0}K_+e^{-\alpha _+(t-s)}ds+\int ^{\infty }_tK_+e^{-\beta _+(s-t)}ds\right] |f_a(z_+,a_0)| \end{aligned}$$

so that

$$\begin{aligned} |x_+(t)|\le K_+(\alpha _+^{-1}+\beta ^{-1}_+)|f_a(z_+,a_0)|. \end{aligned}$$

(43)

That $x_+(t)$ is a solution of Eq. (41) follows by direct differentiation. It is unique because the difference between any two such solutions would be a bounded solution $x(t)$ of $\dot{x}=A_+x$ with $Q_+x(0)=0$ and so must be $0$.

Then it follows from the correspondence between the solutions of Eqs. (39) and (41) that $\xi _k=x_+(s_k)$ is the unique solution of Eq. (39) which is bounded for $k\ge N_1$ and satisfies $Q_+\xi _{N_1}=0$.

Next we consider Eq. (40). Given $\eta$ in the range of $Q_+$, we claim that the unique solution $\bar{\xi }_k$ of Eq. (40) bounded on $k\ge N_1$ such that $Q_+\bar{\xi }_{N_1}=\eta$ is given by

$$\begin{aligned} \bar{\xi }_k =e^{(s_k-s_{N_1})A_+}\eta +\sum ^k_{\ell =N_1+1}e^{(s_k-s_{\ell })A_+}Q_+g_{\ell -1} -\sum ^{\infty }_{\ell =k+1}e^{(s_k-s_{\ell })A_+}(I-Q_+)g_{\ell -1}. \end{aligned}$$

That the infinite sum is well-defined and that $\bar{\xi }_k$ is bounded is shown below; that it is a solution can be verified by direct substitution; it is unique because the difference $\xi _k$ between any two such solutions would be bounded on $k\ge N_1$ and equal $e^{(s_k-s_{N_1})A_+}(I-Q_+)\xi _{N_1}$ and therefore must be zero.

Then it follows by superposition that for each $\eta$ in the range of $Q_+$, Eq. (16) has a unique solution bounded on $k\ge N_1$ such that $Q_+\xi _{N_1}=\eta$ given by

$$\begin{aligned} \xi _k(\eta ,b)=bx_+(s_k)+\bar{\xi }_k,\quad k\ge N_1, \end{aligned}$$

(44)

where $s_k=h_{N_1}+\cdots +h_{k-1}$. Note also, using Eq. (3) again, that

$$\begin{aligned}&|\bar{\xi }_k| \le K_+|\eta | + \sum ^k_{\ell =N_1+1}K_+e^{-\alpha _+(s_k-s_{\ell })}h_{\ell -1}|\bar{g}_{\ell -1}| +\sum ^{\infty }_{\ell =k+1}K_+e^{-\beta _+(s_{\ell }-s_k)}h_{\ell -1}|\bar{g}_{\ell -1}|\nonumber \\&\quad \le K_+|\eta |+ K_+\Vert \bar{g}\Vert _{\infty }\left[ \sum ^k_{\ell =N_1+1}e^{-\alpha _+(s_k- s_{\ell })}(s_{\ell }-s_{\ell -1}) +\sum ^{\infty }_{\ell =k+1}e^{-\beta _+(s_{\ell }-s_k)}(s_{\ell }-s_{\ell -1})\right] \nonumber \\&\quad \le K_+|\eta |\nonumber \\&\qquad +\, K_+\Vert \bar{g}\Vert _{\infty }\left[ \sum ^k_{\ell =N_1+1}e^{\alpha _+h_{\ell -1}}e^{-\alpha _+(s_k- s_{\ell -1})}(s_{\ell }-s_{\ell -1}) +\sum ^{\infty }_{\ell =k+1}e^{-\beta _+(s_{\ell }-s_k)}(s_{\ell }-s_{\ell -1})\right] \nonumber \\&\quad \le K_+|\eta |+K_+\Vert \bar{g}\Vert _{\infty }\left[ e^{\alpha _+h_{\mathrm{max}}}\int ^{s_k}_{s_{N_1}}e^{-\alpha _+(s_k-t)}dt +\int ^{\infty }_{s_k}e^{-\beta _+(t-s_k)}dt\right] \nonumber \\&\qquad \le K_+|\eta |+K_+(\alpha ^{-1}_+e^{\alpha _+h_{\mathrm{max}}} + \beta ^{-1}_+)\Vert \bar{g}\Vert _{\infty }. \end{aligned}$$

(45)

Then the inequality (17) for $|\xi _k(\eta ,b)|$ follows from Eqs. (43), (44) and (45). Finally note that we have Eq. (18) where

$$\begin{aligned} \tilde{q}=-\sum ^{\infty }_{\ell =N_1+1}e^{(s_{N_1}-s_{\ell })A_+}(I-Q_+)g_{\ell -1} \end{aligned}$$

so that $Q_+\tilde{q}=0$ and

$$\begin{aligned} \begin{aligned} v_+&=\int ^{\infty }_0e^{-tA_+}(I-Q_+)f_a(z_+,a_0)dt\\&=-A_+^{-1}\int ^{\infty }_0{d\over dt}e^{-tA_+}(I-Q_+)f_a(z_+,a_0)dt\\&=-A_+^{-1}[e^{-tA_+}(I-Q_+)f_a(z_+,a_0)]^{\infty }_0\\&= A_+^{-1}(I-Q_+)f_a(z_+,a_0)\\&=H_+(I-P_r)v. \end{aligned} \end{aligned}$$

Noting that the inequality for $|\tilde{q}|$ in Eq. (19) follows as in Eq. (45), the proof of the lemma is complete.$\square$

Proof of Lemma 4

We break the solution of Eq. (20) into solving

$$\begin{aligned} \xi _{k+1}=e^{h_kA_-}\xi _k+\phi ^{h_k}_a(z_-,a_0),\quad k<-N_2 \end{aligned}$$

(46)

and

$$\begin{aligned} \xi _{k+1}=e^{h_kA_-}\xi _k+g_{k}, \quad k<-N_2. \end{aligned}$$

(47)

Note first that $\phi ^t_a(z_-,a_0)$ is the solution of the autonomous equation

$$\begin{aligned} \dot{x}=A_-x+f_a(z_-,a_0) \end{aligned}$$

(48)

with $x(0)=0$ so that any solution $x(t)$ of Eq. (48) satisfies

$$\begin{aligned} x(t)=e^{(t-s)A_-}x(s)+\phi ^{t-s}_a(z_-,a_0). \end{aligned}$$

(49)

Next we define the sequence $s_k$ for $k\le -N_2$ by $s_{-{N_2}}=0$ and $s_{k+1}=s_k+h_k$ for $k\le -N_2-1$. Then we claim that if $\xi _k$ solves Eq. (46), then

$$\begin{aligned} \xi _k=x(s_k),\quad k\le -N_2 \end{aligned}$$

where $x(t)$ is the solution of Eq. (48) with $x(0)=\xi _{-N_2}$. Clearly this is true for $k=-N_2$ and if it holds for some $k\le -N_2$, then by Eq. (49)

$$\begin{aligned} \xi _k=x(s_{k})=e^{h_{k-1}A_-}x(s_{k-1})+\phi ^{h_{k-1}}_a(z_-,a_0) \end{aligned}$$

so that

$$\begin{aligned} \begin{aligned}x(s_{k-1})&=e^{-h_{k-1}A_-}\xi _k-e^{-h_{k-1}A_-}\phi ^{h_{k-1}}_a(z_-,a_0)\\&=e^{-h_{k-1}A_-}(e^{h_{k-1}A_-}\xi _{k-1}+\phi ^{h_{k-1}}_a(z_-,a_0))-e^{-h_{k-1}A_-}\phi ^{h_{k-1}}_a(z_-,a_0))\\&=\xi _{k-1}. \end{aligned} \end{aligned}$$

Conversely, we can show by reversing the reasoning that if $x(t)$ is the solution of Eq. (48) with $x(0)=\xi$, then $\xi _k=x(s_k)$ is the solution of Eq. (46) with $\xi _{-N_2}=\xi$. Clearly $\xi _k$ is bounded if and only if $x(t)$ is bounded.

Next note that Eq. (48) has a unique bounded solution $x_-(t)$ on $t\le 0$ with $(I-Q_-)x(0)=0$ given by

$$\begin{aligned} x_-(t)=\int ^t_{-\infty }e^{A_-(t-s)}Q_-f_a(z_-,a_0)ds -\int ^{0}_te^{A_-(t-s)}(I-Q_-)f_a(z_-,a_0)ds \end{aligned}$$

and, using Eq. (3), we obtain the estimate

$$\begin{aligned} |x_-(t)|\le K_-(\alpha ^{-1}_-+\beta ^{-1}_-)|f_a(z_-,a_0)|. \end{aligned}$$

(50)

Next for each $\omega$ in the nullspace of $Q_-$, Eq. (47) has a unique solution bounded on $k\le -N_2$ such that $(I-Q_-)\xi _{-N_2}=\omega$ given by

$$\begin{aligned} \bar{\xi }_k =e^{(s_k-s_{-N_2})A_-}\omega +\sum ^k_{\ell =-\infty }e^{(s_k-s_{\ell })A_-}Q_-g_{\ell -1} -\sum ^{-N_2}_{\ell =k+1}e^{(s_k-s_{\ell })A_-}(I-Q_-)g_{\ell -1}. \end{aligned}$$

That the infinite sum is well-defined and that $\bar{\xi }_k$ is bounded is shown below; that it is a solution can be verified by direct substitution; it is unique because the difference $\xi _k$ between any two such solutions would be bounded on $k\le -N_2$ and equal $e^{(s_k-s_{-N_2})A_-}Q_-\xi _{-N_2}$ and therefore must be zero.

By the superposition principle it follows that the unique solution of Eq. (20) bounded on $k\le -N_2$ such that $(I-Q_-)\xi _{-N_2}=\omega$ is given by

$$\begin{aligned} \xi _k(\omega ,b)=bx_-(s_k)+\bar{\xi }_k,\quad k\le -N_2. \end{aligned}$$

(51)

Also using Eq. (3) again

$$\begin{aligned} \begin{aligned} |\bar{\xi }_k|&\le K_-|\omega | + \sum ^k_{\ell =-\infty }K_-e^{-\alpha _-(s_k-s_{\ell })}h_{\ell -1}|\bar{g}_{\ell -1}| +\sum ^{-N_2}_{\ell =k+1}K_-e^{-\beta _-(s_{\ell }-s_k)}h_{\ell -1}|\bar{g}_{\ell -1}|\\&\le K_-|\omega |\\&+ K_-\Vert \bar{g}\Vert _{\infty }\left[ \sum ^k_{\ell =-\infty }e^{-\alpha _-(s_k- s_{\ell })}(s_{\ell }-s_{\ell -1}) +\sum ^{-N_2}_{\ell =k+1}e^{-\beta _-(s_{\ell }-s_k)}(s_{\ell }-s_{\ell -1})\right] \\&\le K_-|\omega |\\&+ K_-\Vert \bar{g}\Vert _{\infty }\left[ \sum ^k_{\ell =-\infty }e^{\alpha _-h_{\ell -1}} e^{-\alpha _-(s_k- s_{\ell -1})}(s_{\ell }-s_{\ell -1}) +\sum ^{-N_2}_{\ell =k+1}e^{-\beta _-(s_{\ell }-s_k)}(s_{\ell }-s_{\ell -1})\right] \\&\le K_-|\omega |+K_-\Vert \bar{g}\Vert _{\infty }\left[ e^{\alpha _-h_{\mathrm{max}}}\int ^{s_k}_{-\infty }e^{-\alpha _-(s_k-t)}dt +\int ^{s_{-N_2}}_{s_k}e^{-\beta _-(t-s_k)}dt\right] \\&\le K_-|\omega |+K_-(\alpha ^{-1}_-e^{\alpha _-h_{\mathrm{max}}}+\beta ^{-1}_-)\Vert \bar{g}\Vert _{\infty }. \end{aligned} \end{aligned}$$

(52)

Then inequality Eq. (21) for $|\xi _k(\omega ,b)|$ follows from Eqs. (50), (51) and (52). Finally note that Eq. (22) holds with

$$\begin{aligned} \tilde{r}=\sum ^{-N_2}_{\ell =-\infty }e^{(s_{-N_2}-s_{\ell })A_-}Q_-g_{\ell -1} \end{aligned}$$

so that $(I-Q_-)\tilde{r}=0$ and

$$\begin{aligned} \begin{aligned} v_-&= -\int ^{\infty }_0e^{tA_-}Q_-f_a(z_-,a_0)dt\\&= -A^{-1}_-\int ^{\infty }_0{d\over dt}e^{tA_-}Q_-f_a(z_-,a_0)dt\\&= -A^{-1}_-[e^{tA_-}Q_-f_a(z_-,a_0)]^{\infty }_0\\&=A^{-1}_-Q_-f_a(z_-,a_0)\\&=H_-P_rv. \end{aligned} \end{aligned}$$

Noting that the inequality for $|\tilde{r}|$ in Eq. (23) follows as in Eq. (52), the proof of the lemma is complete.$\square$

Proof of Lemma 5

We just prove the lemma for $z_+$, as the proof for $z_-$ is similar. If we write $x=z_+ + y$, the equation $\dot{x}=f(x,a)$ becomes

$$\begin{aligned} \dot{y}=A_+y+g(y,a), \end{aligned}$$

(53)

where

$$\begin{aligned} g(y,a)=f(z_++y,a)-f(z_+,a_0)-f_x(z_+,a_0)y. \end{aligned}$$

Note that if $|y|\le \mu$, $|a-a_0|\le \Delta _1$

$$\begin{aligned} |g_y(y,a)|\!\le \! |f_x(z_++y,a)\!-\!f_x(z_+,a)|\!+\!|f_x(z_+,a)-f_x(z_+,a_0)| \le M_2|y|+M_4|a-a_0| \end{aligned}$$

and $|g(0,a)|\le M_3|a-a_0|$ so that

$$\begin{aligned} |g(y,a)|\le [M_2|y|+M_4|a-a_0|]|y|+M_3|a-a_0|. \end{aligned}$$

Note we can use the bounds $M_i$ on the norms of the derivatives of $f$ because $\mu \le \Delta _0$ and the ball of radius $\Delta _0$ with centre $z_+$ is in $U$. If $y(t)$ is a solution of Eq. (53) with $|y(t)|\le \mu$ for all $t$, then

$$\begin{aligned} y(t)=\int ^t_{-\infty }e^{(t-u)A_+}Q_+g(y(u),a)du -\int ^{\infty }_te^{-(u-t)A_+}(I-Q_+)g(y(u),a)du. \end{aligned}$$

(54)

That is, $y$ is a fixed point of the operator $S$ defined by the right side of Eq. (54). $S$ is defined on the complete metric space $X$ consisting of continuous ${\mathbb {R}}^n-$valued functions $y(t)$ with $|y(t)|\le \mu$ for all $t$, the metric being given by the supremum norm $\Vert \cdot \Vert _{\infty }$. $S$ maps $X$ into itself since, using Eq. (3),

$$\begin{aligned} \begin{aligned} |(Sy)(t)|&\le \int ^t_{-\infty }K_+e^{-\alpha _+(t-u)}[(M_2\mu +M_4|a-a_0|)\mu +M_3|a-a_0|]du\\&\quad +\int ^{\infty }_tK_+e^{-\beta _+(u-t)}[(M_2\mu +M_4|a-a_0|)\mu +M_3|a-a_0|]du \end{aligned} \end{aligned}$$

so that by Eq. (4)

$$\begin{aligned} \Vert Sy\Vert _{\infty } \le K_+(\alpha _+^{-1}+\beta _+^{-1})[(M_2\mu +M_4|a-a_0|)\mu +M_3|a-a_0|]\le \mu . \end{aligned}$$

Also by similar arguments, if $y_1$ and $y_2$ are in $X$,

$$\begin{aligned} \Vert Sy_1-Sy_2\Vert _{\infty } \le K_+(\alpha _+^{-1}+\beta _+^{-1})(M_2\mu +M_4|a-a_0|)\Vert y_1-y_2\Vert _{\infty }\le {1\over 2}\Vert y_1-y_2\Vert _{\infty }. \end{aligned}$$

Then if $y(t)$ is the unique fixed point of $S$, $x_+(t,a) =z_++y(t)$ is the unique solution of $\dot{x}=f(x,a)$ such that $|x_+(t,a)-z_+|\le \mu$ for all $t$. This completes the proof of the lemma and the paper.$\square$