Hypothesis (H₀). The right-hand side $f:{\mathbb R}\times\Omega\times\tilde\Lambda\to{\mathbb R}^d$ of (C_λ) is a Carathéodory function with the following properties: For almost every $t\in{\mathbb R}$ and each $\lambda\in\tilde\Lambda$ the function $f(t,\cdot,\lambda):\Omega\to{\mathbb R}^d$ is differentiable with continuous partial derivative $D_2f(t,\cdot):\Omega\times\tilde\Lambda\to{\mathbb R}^{d\times d}$ such that for all bounded $B\subseteq\Omega$ one has

(2.1)\begin{equation} {\mathop{\textrm{ess sup}}\limits_{t\in{\mathbb R}}} \sup_{x\in B}\left|D_2^jf(t,x,\lambda)\right| \lt \infty \;\text{for all }\lambda\in\tilde\Lambda \end{equation}

for all $j\in\left\{0,1\right\}$. Moreover, for each $\lambda_0\in\tilde\Lambda$ and ɛ > 0 there exists a δ > 0 with

\begin{equation*} \left|x-y\right| \lt \delta \quad\Rightarrow\quad {\mathop{\textrm{ess sup}}\limits_{t\in{\mathbb R}}}\left|D_2^jf(t,x,\lambda)-D_2^jf(t,y,\lambda_0)\right| \lt \varepsilon \end{equation*}

for all $x,y\in\Omega$ and $\lambda\in B_\delta(\lambda_0)$.

Hypothesis (H₁). The Carathéodory equation (C_λ) has a family $(\phi_\lambda)_{\lambda\in\tilde\Lambda}$ of bounded permanent solutions $\phi_\lambda:{\mathbb R}\to\Omega$ such that for every ɛ > 0, $\lambda_0\in\tilde\Lambda$ there is a δ > 0 with

\begin{equation*} d(\lambda,\lambda_0) \lt \delta \quad\Rightarrow\quad \sup_{t\in{\mathbb R}}\left|\phi_\lambda(t)-\phi_{\lambda_0}(t)\right| \lt \varepsilon\;\text{for all }\lambda\in\tilde\Lambda, \end{equation*}

and there exists a $\bar\rho \gt 0$ with $\inf_{t\in{\mathbb R}}\operatorname{dist}_{\partial\Omega}(\phi_\lambda(t)) \gt \bar\rho$ for all $\lambda\in\tilde\Lambda$.

Proof. The argument essentially follows [Reference Pötzsche36, Corollary 2.1].

Proof. Let $\lambda\in\tilde\Lambda$ be fixed. The assumption $(H_1)$ directly yields $\phi_\lambda\in L^\infty({\mathbb R},\Omega)$. As a solution to (C_λ), ϕ_λ is absolutely continuous, hence the strong derivative $\dot\phi_\lambda$ exists a.e. in ${\mathbb R}$ with $\dot\phi_\lambda(t)\equiv f(t,\phi_\lambda(t),\lambda)$. Thus, since (2.1) yields that $f(\cdot,\phi_\lambda(\cdot),\lambda)$ is essentially bounded on ${\mathbb R}$ and we deduce $\dot\phi_\lambda\in L^\infty({\mathbb R})$, i.e. $\phi_\lambda\in W^{1,\infty}({\mathbb R},\Omega)$.

(a) Assume $\phi\in L^\infty({\mathbb R},\Omega)$ is an entire solution of (C_λ) in ${\mathcal B}_{\bar\rho}(\phi_\lambda)$. Then ϕ and ϕ_λ both satisfy (2.2) and the Fundamental Theorem of Calculus [Reference Leoni27, p. 85, Theorem 3.30] yields that $\phi,\phi_\lambda$ are absolutely continuous (on any bounded subinterval of ${\mathbb R}$). This yields that the strong derivatives $\dot\phi$, $\dot\phi_\lambda$ exist a.e. in ${\mathbb R}$. Thus, $\delta:=\phi-\phi_\lambda\in L^\infty({\mathbb R})$ fulfills the identity $\dot\delta(t)+\dot\phi_\lambda(t)\equiv f(t,\delta(t)+\phi_\lambda(t),\lambda)$ a.e. on ${\mathbb R}$. Hence, $\left\|\phi-\phi_\lambda\right\|_\infty \lt \bar\rho$ and the fact $ \dot\delta(t) \equiv f(t,\delta(t)+\phi_\lambda(t),\lambda)-f(t,\phi_\lambda(t),\lambda), $ a.e. on ${\mathbb R}$ has two consequences: First, there exists a bounded set $B\subseteq\Omega$ such that the inclusion $\phi_\lambda(t)+\theta\delta(t)\in B$ holds for all $t\in{\mathbb R}$, $\theta\in[0,1]$ due to the convexity of Ω. Whence the Mean Value Theorem [Reference Zeidler46, p. 243, Theorem 4.C for n = 1] implies

\begin{align*} \left|\dot\delta(t)\right| &= \left|\int_0^1D_2f(t,\theta\delta(t)+\phi_\lambda(t),\lambda)\,{\mathrm d}\theta\delta(t)\right|\\ &\leq \int_0^1\left|D_2f(t,\theta\delta(t)+\phi_\lambda(t),\lambda)\,{\mathrm d}\theta\right|\left\|\delta\right\|_\infty \end{align*}

and thanks to $(H_0)$ the right-hand side of this inequality is essentially bounded in $t\in{\mathbb R}$, i.e. $\delta\in W^{1,\infty}({\mathbb R})$ holds. Second, δ defines an entire solution of the equation of perturbed motion $\dot x=f(t,x+\phi_\lambda(t),\lambda)-f(t,\phi_\lambda(t),\lambda)$, which implies $G(\delta,\lambda)=0$.

Conversely, let $\psi\in L^\infty({\mathbb R})$ be strongly differentiable a.e. in ${\mathbb R}$ with $\left\|\psi\right\|_\infty \lt \bar\rho$ and $G(\psi,\lambda)=0$, i.e. $\dot\psi(t)=f(t,\psi(t)+\phi_\lambda(t),\lambda)-f(t,\phi_\lambda(t),\lambda)$ holds for a.a. $t\in{\mathbb R}$. First, $\dot\psi(t)+\dot\phi_\lambda(t)=f(t,\psi(t)+\phi_\lambda(t),\lambda)$ a.e. in ${\mathbb R}$ implies that $\psi+\phi_\lambda$ is a bounded entire solution of (C_λ) in ${\mathcal B}_{\bar\rho}(\phi_\lambda)$. Second, as above one establishes $\dot\psi\in L^\infty({\mathbb R})$ and therefore the inclusion $\psi\in W^{1,\infty}({\mathbb R})$ results.

(b) can be shown analogously.

Proof. Throughout, let $\lambda\in\tilde\Lambda$ be fixed.

$(c)\Rightarrow(a)$ Because ϕ_λ is hyperbolic, (V_λ) has an exponential dichotomy on ${\mathbb R}$ with projector P_λ and growth rate α > 0. Due to the explicit form (2.7) from Theorem 2.2 the invertibility of the Fréchet derivative $D_1G(0,\lambda)$ means that for each $g\in L^\infty({\mathbb R})$ there exists a unique solution $\psi\in W^{1,\infty}({\mathbb R})$ of the perturbed variation equation

(V_λ,g)\begin{align} \dot x=A(t,\lambda)x+g(t). \end{align}

In order to verify this, with Green’s function (2.6) we define

\begin{align*} \psi:{\mathbb R}&\to{\mathbb R}^d,& \psi(t)&:=\int_{\mathbb R}\Gamma_{P_\lambda}(t,s)g(s)\,{\mathrm d} s. \end{align*}

As in [Reference Aulbach, Wanner, Aulbach and Colonius4, proof of Lemma 3.2] one shows that ψ is actually a solution of (V_λ,g). Moreover, the dichotomy estimates (2.4) yield $\psi\in L^\infty({\mathbb R})$. Hence, since the solution identity $\dot\psi(t)\equiv A(t,\lambda)\psi(t)+g(t)$ holds a.e. in ${\mathbb R}$ we obtain from Lemma 2.1 and the inclusion $g\in L^\infty({\mathbb R})$ that also $\dot\psi$ is essentially bounded, i.e. $\psi\in W^{1,\infty}({\mathbb R})$. It remains to show that ψ is uniquely determined by the inhomogeneity $g\in L^\infty({\mathbb R})$. If $\bar\psi\in L^\infty({\mathbb R})$ is another bounded entire solution to (V_λ,g), then the difference $\psi-\bar\psi\in L^\infty({\mathbb R})$ solves the variation equation (V_λ). Due to our hyperbolicity assumption, (V_λ) has exponential dichotomies on ${\mathbb R}_+$ and on ${\mathbb R}_-$ with projector P_λ satisfying ${\mathbb R}^d=R(P_\lambda(0))\oplus N(P_\lambda(0))$. This yields $R(P_\lambda(0))\cap N(P_\lambda(0))=\left\{0\right\}$ and the dynamical characterization (2.5) implies that the trivial solution is the unique bounded entire solution to the variation equation (V_λ); thus $\psi=\bar\psi$. This shows that $D_1G(0,\lambda):W^{1,\infty}({\mathbb R})\to L^\infty({\mathbb R})$ is invertible and Banach’s Isomorphism Theorem [Reference Zeidler46, pp. 179–180, Proposition 1] yields the claim.

$(a)\Rightarrow(b)$ Repeating the arguments yielding (a) it remains to show that inhomogeneities $g\in L_0^\infty({\mathbb R})$ imply $\psi\in W_0^{1,\infty}({\mathbb R})$. Thereto, note $\psi=\psi_1+\psi_2$ with

\begin{align*} \psi_1(t)&:=\int_{-\infty}^t\Phi_\lambda(t,s)P_\lambda(s)g(s)\,{\mathrm d} s,& \psi_2(t)&:=-\int_t^{\infty}\Phi_\lambda(t,s)[I_d-P_\lambda(s)]g(s)\,{\mathrm d} s, \end{align*}

and choose ɛ > 0. Then, there exists a real T > 0 such that $\left|g(s)\right| \lt \tfrac{\alpha}{K}\tfrac{\varepsilon}{2}$ holds for a.a. $s\in{\mathbb R}\setminus(-T,T)$. On the one hand, provided we choose $T_1\geq T$ sufficiently large that $\tfrac{K}{\alpha}\left\|g\right\|_\infty e^{\alpha(T-t)} \lt \tfrac{\varepsilon}{2}$ for all $t\geq T_1$, then this leads to the estimates

\begin{align*} \left|\psi_1(t)\right| &\leq \int_{-\infty}^T\left|\Phi_\lambda(t,s)P_\lambda(s)g(s)\,{\mathrm d} s\right|+\int_T^\infty\left|\Phi_\lambda(t,s)P_\lambda(s)g(s)\,{\mathrm d} s\right|\\ &\stackrel{(2.4)}{\leq} K\int_{-\infty}^Te^{-\alpha(t-s)}\left|g(s)\right|\,{\mathrm d} s+K\int_T^\infty e^{-\alpha(t-s)}\left|g(s)\right|\,{\mathrm d} s\\ &\leq K\int_{-\infty}^Te^{-\alpha(t-s)}\,{\mathrm d} s\left\|g\right\|_\infty+\alpha\int_T^\infty e^{-\alpha(t-s)}\,{\mathrm d} s\frac{\varepsilon}{2}\\ &\leq \frac{K}{\alpha}e^{\alpha(T-t)}\left\|g\right\|_\infty+\frac{\varepsilon}{2} \lt \varepsilon\;\text{for all } t\geq T_1,\\ \left|\psi_1(t)\right| &\leq \int_{-\infty}^t\left|\Phi_\lambda(t,s)P_\lambda(s)g(s)\right|\,{\mathrm d} s \stackrel{(2.4)}{\leq} \int_{-\infty}^te^{-\alpha(t-s)}\,{\mathrm d} s\varepsilon = \varepsilon\;\text{for all } t\leq-T \end{align*}

and in conclusion $\lim_{t\to\pm\infty}\left|\psi_1(t)\right|=0$. On the other hand, if we furthermore choose $T_1\geq T$ so large that $\frac{K}{\alpha}\left\|g\right\|_\infty e^{\alpha(t+T)} \lt \frac{\varepsilon}{2}$ for all $t\leq-T_1$, then

\begin{align*} \left|\psi_2(t)\right| &\leq \int_t^{\infty}\left|\Phi_\lambda(t,s)[I_d-P_\lambda(s)]g(s)\right|\,{\mathrm d} s \stackrel{(2.4)}{\leq} \alpha\int_t^\infty e^{\alpha(t-s)}\varepsilon = \varepsilon \;\text{for all } t\geq T,\\ \left|\psi_2(t)\right| &\leq \int_t^{-T}\left|\Phi_\lambda(t,s)[I_d-P_\lambda(s)]g(s)\right|\,{\mathrm d} s+ \int_{-T}^\infty\left|\Phi_\lambda(t,s)[I_d-P_\lambda(s)]\right|\,{\mathrm d} s\\ &\stackrel{(2.4)}{\leq} K\int_{-T}^\infty e^{\alpha(t-s)}\left|g(s)\right|\,{\mathrm d} s+K\int_{-T}^\infty e^{\alpha(t-s)}\left|g(s)\right|\,{\mathrm d} s\\ &\leq \alpha\int_{-T}^\infty e^{\alpha(t-s)}\,{\mathrm d} s\frac{\varepsilon}{2}+K\int_{-T}^\infty e^{\alpha(t-s)}\,{\mathrm d} s\left\|g\right\|_\infty\\ &\leq \frac{\varepsilon}{2}+\frac{K}{\alpha}e^{\alpha(t+T)}\left\|g\right\|_\infty \lt \varepsilon \;\text{for all } t\leq-T \end{align*}

also guarantee $\lim_{t\to\pm\infty}\left|\psi_2(t)\right|=0$. We conclude that $\psi\in L_0^\infty({\mathbb R})$ holds. Moreover, from the identity $\dot\psi(t)\equiv A(t,\lambda)\psi(t)+g(t)$ a.e. on ${\mathbb R}$, Lemma 2.1(a) and $g\in L_0^\infty({\mathbb R})$ results $\lim_{t\to\pm\infty}|\dot\psi(t)|=0$ and consequently $\psi\in W_0^{1,\infty}({\mathbb R})$.

$(b)\Rightarrow(c)$ We first note that the transition matrix $\Phi_\lambda:{\mathbb R}\times{\mathbb R}\to{\mathbb R}^{d\times d}$ of (V_λ) is an evolution family on the Banach space ${\mathbb R}^d$ in the language of e.g. [Reference Sasu39, Reference Sasu40] (note that the essential boundedness guaranteed by Lemma 2.1(a) and [Reference Aulbach, Wanner, Aulbach and Colonius4, Lemma 2.9] yield that there exists a $\omega\geq 0$ so that $\left|\Phi_\lambda(t,s)\right|\leq e^{\omega(t-s)}$ for all $s\leq t$). In particular, the proof of [Reference Sasu39, Theorem 4.8] establishes that for each $g\in L_0^\infty({\mathbb R})$ there exists a unique solution $\phi\in L_0^\infty({\mathbb R})$ of the integral equation

(2.8)\begin{equation} \phi(t)=\Phi_\lambda(t,\tau)\phi(\tau)+\int_\tau^t\Phi_\lambda(t,s)g(s)\,{\mathrm d} s\;\text{for all }\tau\leq t. \end{equation}

Indeed, the abstract setting of [Reference Sasu40, Theorem 1.2] is met, because the function space $L_0^\infty({\mathbb R})$ possesses the following properties:

• • For each $x\in L_0^\infty({\mathbb R})$ and $\tau\in{\mathbb R}$ the shifted function $x_\tau:=x(\tau+\cdot)$ satisfies that $x_\tau\in L_0^\infty({\mathbb R})$ and $\left\|x\right\|_\infty=\left\|x_\tau\right\|_\infty$,

• • $L_0^\infty({\mathbb R})$ contains the continuous functions $x:{\mathbb R}\to{\mathbb R}^d$ having compact support,

• • $\int_\tau^t\left|x(s)\right|\,{\mathrm d} s\leq\int_\tau^t\left\|x\right\|_\infty\,{\mathrm d} s=(t-\tau)\left\|x\right\|_\infty$ for all $\tau\leq t$ and $x\in L_0^\infty({\mathbb R})$,

• • e.g. $x_0:{\mathbb R}\to{\mathbb R}$, $x_0(t):=\tfrac{1}{1+\left|t\right|}$ is continuous with $x_0\in L_0^\infty({\mathbb R})\setminus L^1({\mathbb R})$,

• • if $x,y:{\mathbb R}\to{\mathbb R}$ are measurable with $\left|x(t)\right|\leq\left|y(t)\right|$ for a.a. $t\in{\mathbb R}$ and $y\in L_0^\infty({\mathbb R})$, then $x\in L_0^\infty({\mathbb R})$.

It remains to show $\phi\in W_0^{1,\infty}({\mathbb R})$. Thereto, by the Variation of Constants [Reference Aulbach, Wanner, Aulbach and Colonius4, Theorem 2.10] the unique solution $\phi\in L_0^\infty({\mathbb R})$ of (2.8) is also the unique solution to the perturbed variation equation (V_λ,g) (satisfying $x(\tau)=\phi(\tau)$) and hence absolutely continuous on each bounded subinterval of ${\mathbb R}$. Moreover, the solution identity for (V_λ,g) implies $\dot\phi\in L_0^\infty({\mathbb R})$ due to Lemma 2.1(a). In conclusion, it results that $\phi\in W_0^{1,\infty}({\mathbb R})$.

Proof. Fix $\lambda\in\tilde\Lambda$. Above all, one can easily show that the dual variation equation (V_λ*) has the transition matrix $\Phi_\lambda^\ast(t,s):=\phi_\lambda(s,t)^T$ for all $t,s\in I$. Consequently,

\begin{align*} \Phi_\lambda^\ast(t,s)Q_\lambda(s) &\stackrel{(2.9)}{=} \phi_\lambda(s,t)^T\left[I_d-P_{\lambda}(s)^T\right] \stackrel{(2.3)}{=} (\left[I_d-P_{\lambda}(s)\right]\phi_\lambda(s,t))^T\\ &=\left(\phi_\lambda(s,t)[I_d-P_{\lambda}(t)]\right)^T = \left[I_d-P_{\lambda}(t)^T\right]\phi_\lambda(s,t)^T \stackrel{(2.9)}{=} Q_{\lambda}(t)\Phi_\lambda^\ast(t,s), \end{align*}

as well as the required dichotomy estimates (note $\left|C\right|=\left|C^T\right|$ for $C\in{\mathbb R}^{d\times d}$)

\begin{align*} \left|\Phi_\lambda^\ast(t,s)Q_\lambda(s)\right| &\stackrel{(2.9)}{=} \left|\left(\phi_\lambda(s,t)\left[I_d-P_{\lambda}(t)\right]\right)^T\right|\\ & = \left|\phi_\lambda(s,t)\left[I_d-P_{\lambda}(t)\right]\right| \stackrel{(2.4)}{\leq} Ke^{-\alpha(t-s)},\\ \left|\Phi_\lambda^\ast(s,t)\left[I_d-Q_\lambda(t)\right]\right| &\stackrel{(2.9)}{=} \left|\left(\phi_\lambda(t,s)P_{\lambda}(s)\right)^T\right|\\ & = \left|\phi_\lambda(t,s)P_{\lambda}(s)\right| \stackrel{(2.4)}{\leq} Ke^{-\alpha(t-s)} \end{align*}

for all $s\leq t$, $s,t\in I$, result, which prove that (V_λ*) has an exponential dichotomy on I with projector Q_λ. The last statement follows from [Reference Zeidler46, p. 294, Proposition 6(ii)].

Proof. Throughout, let $\lambda\in\tilde\Lambda$ be fixed.

$(c)\Rightarrow(a)$ Our assumptions imply that (V_λ) has exponential dichotomies on ${\mathbb R}_\pm$ with projectors $P_\lambda^\pm$; the corresponding growth rates may be denoted by α > 0. We establish that $D_1G(0,\lambda):W^{1,\infty}({\mathbb R})\to L^\infty({\mathbb R})$ is Fredholm.

(I) Claim: $\dim N(D_1G(0,\lambda)) \lt \infty$.

Thanks to the dynamical characterization (2.5), if we abbreviate

\begin{align*} X_+&:=R(P_\lambda^+(0)),& X_-&:=N(P_\lambda^-(0)), \end{align*}

then $X_+\cap X_-$ is precisely the subspace of all initial values $\xi\in{\mathbb R}^d$ for bounded entire solutions $\Phi_\lambda(\cdot,0)\xi$ to (V_λ). By means of the isomorphism $\xi\mapsto\Phi_\lambda(\cdot,0)\xi$ from ${\mathbb R}^d$ onto the solution space of (V_λ) one has $\dim(X_+\cap X_-)=\dim N(D_1G(\lambda,0))$.

(II) Claim: If $g\in R(D_1G(0,\lambda))$, then for all solutions $\psi\in L^\infty({\mathbb R})$ of the dual variation equation (V_λ*) one has

(2.10)\begin{equation} \int_{{\mathbb R}}\langle\psi(s),g(s)\rangle\,{\mathrm d} s=0. \end{equation}

First of all, Lemma 2.5 implies that the dual variation equation (V_λ*) has dichotomies on ${\mathbb R}_\pm$ with projectors $Q_\lambda^\pm(t)=I_d-P_\lambda^\pm(t)^T$ and growth rate α > 0. Hence the following orthogonal complements allow the dynamical characterization

(2.11)\begin{align} \begin{split} X_+^\perp&=R(Q_\lambda^+(0)) = \left\{\eta\in{\mathbb R}^d:\,\sup_{0\leq t}e^{-\gamma t}|\Phi_\lambda^\ast(t,0)\eta| \lt \infty\right\} \text{for }\gamma\in[-\alpha,\alpha),\\ X_-^\perp&=N(Q_\lambda^-(0)) = \left\{\eta\in{\mathbb R}^d:\,\sup_{t\leq 0}e^{-\gamma t}|\Phi_\lambda^\ast(t,0)\eta| \lt \infty\right\} \text{for }\gamma\in(-\alpha,\alpha]. \end{split} \end{align}

Therefore the intersection $X_+^\perp\cap X_-^\perp$ consists of initial values η giving rise to bounded entire solutions $\Phi_\lambda^\ast(\cdot,0)\eta$ to the dual variation equation (V_λ*). For each function $g\in R(D_1G(0,\lambda))$ there exists a preimage $\phi\in W^{1,\infty}({\mathbb R})$ such that the identity $g(t)\equiv\dot\phi(t)-D_2f(t,\phi_\lambda(t),\lambda)\phi(t)$ holds a.e. on ${\mathbb R}$. If $\psi\in L^\infty({\mathbb R})$ denotes a solution of (V_λ*), then the product rule implies

\begin{align*} \int_{-T}^T\langle\psi(s),g(s)\rangle\,{\mathrm d} s &= \int_{-T}^T\langle\psi(s),\dot\phi(s)-D_2f(s,\phi_\lambda(s),\lambda)\phi(s)\rangle\,{\mathrm d} s\\ &= \int_{-T}^T\langle\psi(s),\dot\phi(s)\rangle+\langle\dot\psi(s)\phi(s)\rangle\,{\mathrm d} s = \int_{-T}^T\frac{\,{\mathrm d}}{\,{\mathrm d} s}\langle\psi(s),\phi(s)\rangle\,{\mathrm d} s\\ &= \langle\psi(T),\phi(T)\rangle-\langle\psi(-T),\phi(-T)\rangle \;\text{for all } T \gt 0. \end{align*}

Since the dynamical characterization (2.11) guarantees that $\psi(t)$ decays to 0 as $t\to\pm\infty$ exponentially and $\phi\in L^\infty({\mathbb R})$ holds, one obtains (2.10) from

\begin{equation*} \int_{{\mathbb R}}\langle\psi(s),g(s)\rangle\,{\mathrm d} s=\lim_{T\to\infty}\left(\langle\psi(T),\phi(T)\rangle-\langle\psi(-T),\phi(-T)\rangle\right)=0. \end{equation*}

(III) Claim: If (2.10) holds for all solutions $\psi\in L^\infty({\mathbb R})$ of the dual variation equation (V_λ*), then $g\in R(D_1G(0,\lambda))$.

Let $g\in L^\infty({\mathbb R})$. For $\eta\in{\mathbb R}^d$ satisfying

(2.12)\begin{equation} \eta^T\left[P_\lambda^+-(I_d-P_\lambda^-)\right]=0, \end{equation}

we define

\begin{align*} \psi:{\mathbb R}&\to{\mathbb R}^{d\times d},& \psi(t)&:= \begin{cases} \Phi_\lambda^\ast(t,0)Q_\lambda^+(t)\eta,&t\geq 0,\\ \Phi_\lambda^\ast(t,0)Q_\lambda^-(t)\eta,&t\leq 0. \end{cases} \end{align*}

Then ψ is a bounded entire solution of the dual variation equation (V_λ*) and

\begin{equation*} \langle\eta,\int_{-\infty}^0P_\lambda^-(0)\Phi_\lambda(0,s)g(s)\,{\mathrm d} s+\int_0^\infty(I_d-P_\lambda^+(0))\Phi_\lambda(0,s)g(s)\,{\mathrm d} s\rangle=0 \end{equation*}

for all $\eta\in{\mathbb R}^d$ such that (2.12) holds. This, in turn, is equivalent to the fact that the linear algebraic equation

\begin{align*} &\left[P_\lambda^+(0)-(I_d-P_\lambda^-(0))\right]\xi\\ &= \int_{-\infty}^0P_\lambda^-(0)\Phi_\lambda(0,s)g(s)\,{\mathrm d} s+\int_0^\infty(I_d-P_\lambda^+(0))\Phi_\lambda(0,s)g(s)\,{\mathrm d} s \end{align*}

for $\xi\in{\mathbb R}^d$ has a solution. Consequently, with Green’s function (2.6),

\begin{align*} \phi:{\mathbb R}&\to{\mathbb R}^d,& \phi(t) &:= \begin{cases} \Phi_\lambda(t,0)P_\lambda^+(0)\xi+\int_0^\infty\Gamma_{P_\lambda^+}(t,s)g(s)\,{\mathrm d} s,&t\geq 0,\\ \Phi_\lambda(t,0)[I_d-P_\lambda^-(0)]\xi+\int_{-\infty}^0\Gamma_{P_\lambda^-}(t,s)g(s)\,{\mathrm d} s,&t\leq 0 \end{cases} \end{align*}

defines a solution of the perturbed variation equation (V_λ,g) in $W^{1,\infty}({\mathbb R})$. Due to (2.7) this means $D_1G(0,\lambda)\phi=g$, i.e. $g\in R(D_1G(0,\lambda))$.

(IV) Claim: $\operatorname{codim} R(D_1G(0,\lambda)) \lt \infty$.

For each bounded entire solution ψ of (V_λ*) we define a functional

\begin{align*} x_\psi'&\in W^{1,\infty}({\mathbb R})',& x_\psi'(x)&:=\int_{{\mathbb R}}\langle\psi(s),x(s)\rangle\,{\mathrm d} s. \end{align*}

This gives an isomorphism between $X_+^\perp\cap X_-^\perp$ and a finite-dimensional subspace of the dual space $L^\infty({\mathbb R})'$. In other words, $R(D_1G(0,\lambda))$ is the subspace of $L^\infty({\mathbb R})$ annihilated by this finite-dimensional subspace of $L^\infty({\mathbb R})'$. Thus, $R(D_1G(0,\lambda))$ is closed, $\operatorname{codim} R(D_1G(0,\lambda))=\dim(X_+^\perp\cap X_-^\perp) \lt \infty$ and $D_1G(0,\lambda)$ is Fredholm.

(V) It remains to determine the index of $D_1G(0,\lambda)$ as

\begin{align*} & \operatorname{ind} D_1G(0,\lambda)\\ &= \dim\bigl(R(P_\lambda^+(0))\cap N(P_\lambda^-(0))\bigr)-\dim\bigl(N(P_\lambda^-(0))+R(P_\lambda^+(0))\bigr)^{\perp}\\ &= \dim\left(R(P_\lambda^+(0))\cap N(P_\lambda^-(0))\right)-\left(d-\dim\bigl(N(P_\lambda^-(0))+R(P_\lambda^+(0))\bigr)\right)\\ &= \dim\left(R(P_\lambda^+(0))\cap N(P_\lambda^-(0))\right)\\ &\quad\quad\quad-\left(d-[\dim N(P_\lambda^-(0))+\dim R(P_\lambda^+(0))-\dim R(P_\lambda^+(0))\cap N(P_\lambda^-(0)) ]\right)\\ &=\dim R(P_\lambda^+(0))-\bigl(d-\dim N(P_\lambda^-(0))\bigr) = m_\lambda^--m_\lambda^+. \end{align*}

$(a)\Rightarrow(b)$ In order to show that $D_1G(0,\lambda)\in L(W_0^{1,\infty}({\mathbb R}),L_0^\infty(R))$ is Fredholm, we mimic the arguments in (a). This additionally only requires to establish the inclusions $\phi,\psi\in W_0^{1,\infty}({\mathbb R})$, provided that $g\in L_0^\infty({\mathbb R})$ holds, but the corresponding estimates result as in the proof of Theorem 2.4(b).

$(b)\Rightarrow(c)$ Referring to Lemma 2.1(a), the coefficient matrices $A(\cdot,\lambda):{\mathbb R}\to{\mathbb R}^{d\times d}$ are essentially bounded. Then the proof of [Reference Palmer31, Theorem] given for continuous $A(\cdot,\lambda)$ and the spaces $(BC^1({\mathbb R}),BC({\mathbb R}))$ literally carries over to our situation.

Hypothesis (H₂). There is a critical parameter $\lambda^\ast\in\tilde\Lambda$ so that the entire solution $\phi^\ast:=\phi_{\lambda^\ast}$ of $(C_{\lambda^\ast})$ is hyperbolic on ${\mathbb R}_+$ with projector $P_{\lambda^\ast}^+:{\mathbb R}_+\to{\mathbb R}^{d\times d}$ (Morse index $m^+$) and on ${\mathbb R}_-$ with projector $P_{\lambda^\ast}^-:{\mathbb R}_-\to{\mathbb R}^{d\times d}$ (Morse index $m^-$).

Proof. As in [Reference Sandstede38, S. 8, Lemma 1.1] the projectors $P_\lambda^\pm$ are characterized via fixed points of Lyapunov–Perron operators. Then their continuous dependence on the parameter λ is a consequence of the Uniform Contraction Principle.

Proof. Since $\lambda\mapsto P_\lambda^{\pm}(t)$ is continuous for any $t\in{\mathbb R}_\pm$ by Lemma 3.1, the assertion follows directly from [Reference Fitzpatrick and Pejsachowicz16, Proposition 6.21].

Proof. We write ${\mathbb R}^d=({\mathbb R}^{d-m^+}\times\{0\})\oplus(\{0\}\times{\mathbb R}^{m^-})$ and consider the sets $\mathrm{R}[P^{+}(0)]$ and $\mathrm{N}[P^{-}(0)]$ introduced in Lemma 3.2. Since they are are vector bundles over a contractible space, [Reference Husemoller19, p. 30, Corollary 4.8] yields the existence of morphism bundles

\begin{align*} \psi^1:\Lambda\times{\mathbb R}^{d-m^+}&\to\mathrm{R}[P^{+}(0)],& \psi^2:\Lambda\times{\mathbb R}^{m^-}&\to\mathrm{N}[P^{-}(0)] \end{align*}

such that the following diagrams are commutative:

where the vertical arrows represent the corresponding projections, and for any $\lambda\in\Lambda$ the maps $\psi^1(\lambda,\cdot):{\mathbb R}^{d-m^+}\to R(P_\lambda^+(0))$ and $\psi^2(\lambda,\cdot):{\mathbb R}^{m^-}\to N(P_\lambda^-(0))$ are linear isomorphisms. Then it is not hard to see that the functions

\begin{align*} \xi^+_1,\ldots,\xi_{d-m^+}^+:\Lambda&\to{\mathbb R}^d,& \xi^+_i(\lambda)&:=\psi^1(\lambda,e_i)\;\text{for all } 1\leq i\leq d-m^+,\\ \xi^-_1,\ldots,\xi_{m^-}^-:\Lambda&\to {\mathbb R}^d,& \xi^-_j(\lambda)&:=\psi^2(\lambda,e_{d-m^++j})\;\text{for all } 1\leq j\leq m^- \end{align*}

satisfy the claimed properties $(a$– $c)$ with the sets $\{e_1,\ldots,e_{d-m^+}\}\subset{\mathbb R}^{d-m^+}\times\{0\}$ and $\left\{e_{d-m^++1},\ldots,e_d\right\}\subset\left\{0\right\}\times{\mathbb R}^{m^-}$ derived from the standard basis $e_1,\ldots,e_d$ of ${\mathbb R}^d$. In particular, $E:\Lambda\to{\mathbb R}$ is well-defined.

Proof. The continuity of the Evans function follows immediately from Proposition 3.3 and the continuity of the determinant $\det:{\mathbb R}^{d\times d}\to{\mathbb R}$.

$(a)\Leftrightarrow(b)$ is an immediate consequence of Linear Algebra.

$(b)\Leftrightarrow(c)$ Arguing as in [Reference Coppel7, p. 19], a variation equation (V_λ) possesses an exponential dichotomy on ${\mathbb R}$ if and only if the projections satisfy (b).

Proof. (I) We first establish the claimed equivalence:

$(a)\Rightarrow(b)$ The assumption $(H_2)$ implies that $(V_{\lambda^\ast})$ has exponential dichotomies on both semiaxes with growth rate $\alpha^\ast \gt 0$. Now if $E(\lambda^\ast)=0$, then Proposition 3.4 shows that $\phi^\ast$ is nonhyperbolic, i.e. $0\in\Sigma(\lambda^\ast)$. Due to Theorem 2.4, $D_1G(0,\lambda^\ast)$ is noninvertible, but because of Theorem 2.6 Fredholm of index 0. Hence, $D_1G(0,\lambda^\ast)$ has a nontrivial kernel and from step (I) in the proof for Theorem 2.6 one obtains $\left\{0\right\}\neq R(P_{\lambda^\ast}^+(0))\cap N(P_{\lambda^\ast}^-(0))$. Because the transition matrices of (V_λ) and $\Phi_\lambda^\gamma$ of $\dot x=[A(t,\lambda)-\gamma I_d]x$ are related by $\Phi_\lambda^\gamma(t,s)=e^{\gamma(s-t)}\Phi_\lambda(t,s)$ for all $t,s\in{\mathbb R}$, the dynamical characterization (2.5) implies

\begin{equation*} \left\{0\right\} \neq \left\{\xi\in{\mathbb R}^d:\,\sup_{t\in{\mathbb R}}e^{-\gamma t}\left|\Phi_{\lambda^\ast}(t,0)\xi\right| \lt \infty\right\} = \left\{\xi\in{\mathbb R}^d:\,\sup_{t\in{\mathbb R}}\left|\Phi_{\lambda^\ast}^\gamma(t,0)\xi\right| \lt \infty\right\} \end{equation*}

for each $\gamma\in(-\alpha^\ast,\alpha^\ast)$. Consequently, the shifted equation $\dot x=[A(t,\lambda^\ast)-\gamma I_d]x$ has nontrivial bounded solutions and thus cannot possess an exponential dichotomy on ${\mathbb R}$, i.e. $\gamma\in\Sigma(\lambda^\ast)$. Since $\gamma\in(-\alpha^\ast,\alpha^\ast)$ was arbitrary, we deduce $(-\alpha^\ast,\alpha^\ast)\subseteq\Sigma(\lambda^\ast)$ and (b) results due to the compactness of spectral intervals.

$(b)\Rightarrow(a)$ Because obviously $0\in\Sigma(\lambda^\ast)$ holds, the bounded entire solution $\phi^\ast$ is nonhyperbolic and Proposition 3.4 yields (a).

(II) Let $\lambda\in\Lambda$. By means of contradiction we assume $P_{\lambda^\ast}^+(0)=I_d$ and hence the $\xi^+_1(\lambda),\ldots,\xi_d^+(\lambda)\in{\mathbb R}^d$ from Proposition 3.3 are linearly independent. Thus, $E(\lambda)\neq 0$ contradics $E(\lambda^\ast)=0$, we deduce $\dim R(P_{\lambda^\ast}^+(0)) \lt d$, hence

\begin{equation*} m^+=d-\dim R(P_{\lambda^\ast}^+(0)) \gt 0. \end{equation*}

Using a similar argument, also the assumption $P_{\lambda^\ast}^+(0)=0_d$ yields a contradiction, which yields $m^+ \lt d$. With $m^-=m^+$ it is $0 \lt m^+,m^- \lt d$.

This implies that the projectors $P_\lambda^+$ for the exponential dichotomy of (V_λ) on ${\mathbb R}_+$ are nontrivial. Then [Reference Anagnostopoulou, Pötzsche and Rasmussen2, pp. 32–33, Proposition 2.2.1] implies that ϕ_λ is unstable.

Now let $E(\lambda)\neq 0$ and hence $0\not\in\Sigma(\lambda)$ results by Proposition 3.4. First, the assumption $\Sigma(\lambda)\subset(-\infty,0)$ implies that (V_λ) has an exponential dichotomy on ${\mathbb R}_+$ with $P_\lambda(t)\equiv I_d$ resulting in the contradiction $m^+=0$. Second, assuming $\Sigma(\lambda)\subset(0,\infty)$ leads to the contradiction $m^+=d$. In conclusion, both the positive and the negative reals contains at least one spectral interval of $\Sigma(\lambda)$.

Proof. (a) The mapping $T:[a,b]\to L(W^{1,\infty}({\mathbb R}),L^{\infty}({\mathbb R}))$ is continuous because of Theorem 2.2(a). Due to Lemma 3.1 the variation equations (V_λ) have exponential dichotomies on both semiaxes ${\mathbb R}_\pm$ with respective projectors $P_\lambda^\pm$ for all $\lambda\in\Lambda$. Thus, the inclusion $T(\lambda)\in F_0(W^{1,\infty}({\mathbb R}),L^{\infty}({\mathbb R}))$ results from Theorem 2.6(a) combined with (2.2). Hence, T is a path of index 0 Fredholm operators. Furthermore, $E(a)E(b)\neq 0$ and Proposition 3.4 yield that both bounded entire solutions ϕ_a and ϕ_b are hyperbolic, i.e. the variation equations $(V_a)$ and $(V_b)$ have an exponential dichotomy on ${\mathbb R}$. Then Theorem 2.4(a) implies the inclusions

\begin{equation*} T(a),T(b)\in GL(W^{1,\infty}({\mathbb R}),L^{\infty}({\mathbb R})) \end{equation*}

and T has invertible endpoints.

We first prepare some properties of $T(\lambda)$ needed in the further steps of the proof. Throughout, we again abbreviate $A(t,\lambda):=D_2f(t,\phi_\lambda(t),\lambda)$.

(I) Claim: $N(T(\lambda))\cong R(P_\lambda^+(0))\cap N(P_\lambda^-(0))$.

This is shown in the proof of Theorem 2.6(a).

(II) Consider the formally dual operators

\begin{align*} T(\lambda)^{\ast}&:W^{1,\infty}({\mathbb R})\to L^{\infty}({\mathbb R}),& [T(\lambda)^{\ast}y](t)&:=\dot{y}(t)+A(t,\lambda)^Ty(t) \end{align*}

for a.a. $t\in{\mathbb R}$ associated to the dual variation equation (V_λ*). Thanks to Lemma 2.5, the dual variation equation (V_λ*) has exponential dichotomies on both ${\mathbb R}_\pm$ with the projectors $Q_{\lambda}^{\pm}(t):=I_d-P_{\lambda}^{\pm}(t)^T$.

(II.1) Claim: $\dim N(T(\lambda))=\dim N(T(\lambda)^\ast)$.

As in the proof of Theorem 2.6(a) (cf. (2.11)) one has

\begin{equation*} N(T(\lambda)^{\ast}) \cong R(Q^+_{\lambda}(0))\cap N(Q_\lambda^-(0)) \stackrel{(2.9)}{=} R(P_\lambda^-(0)^T)\cap N(P_\lambda^+(0)^T), \end{equation*}

while $R(P_\lambda^-(0)^T)=N(P_\lambda^-(0))^\perp$, $N(P_\lambda^+(0)^T)=R(P_\lambda^+(0))^\perp$ result from [Reference Zeidler46, p. 294, Proposition 6(ii)] and consequently the claim is established by

\begin{align*} & \dim N(T(\lambda)^\ast) = \dim\left(R(P_\lambda^-(0)^T)\cap N(P_\lambda^+(0)^T)\right)\\ &= \dim\left(N(P_\lambda^-(0))^\perp\cap R(P_\lambda^+(0))^\perp\right) = \dim\left(N(P_\lambda^-(0))+R(P_\lambda^+(0))\right)^\perp\\ &= \dim\bigl(R(P_\lambda^+(0))\cap N(P_\lambda^-(0))\bigr) \stackrel{(I)}{=} \dim N(T(\lambda)). \end{align*}

(II.2) Claim: $R(T(\lambda))\oplus N(T(\lambda)^{\ast})=L^{\infty}({\mathbb R})$.

First of all, observe that since $W^{1,\infty}({\mathbb R})\subset L^{\infty}({\mathbb R})$, it follows that $N(T(\lambda)^\ast)$ is a subset of $L^{\infty}({\mathbb R})$. Furthermore, if $g\in R(T(\lambda))$ with preimage $\phi\in W^{1,\infty}({\mathbb R})$ and functions $u\in N(T(\lambda)^{\ast})\subseteq W^{1,\infty}({\mathbb R})$, then

\begin{align*} &\int_{{\mathbb R}}\langle u(s),g(s)\rangle\,{\mathrm d} s = \int_{{\mathbb R}}\langle u(s),\dot{\phi}(s)-A(t,\lambda)\phi(s)\rangle\,{\mathrm d} s\\ &= \int_{{\mathbb R}}\langle u(s),\dot{\phi}(s)\rangle-\langle u(s),A(t,\lambda)\phi(s)\rangle\,{\mathrm d} s\\ &=\int_{{\mathbb R}}\langle u(s),\dot{\phi}(s)\rangle+\langle-u(s)A(t,\lambda)^T,\phi(s)\rangle\,{\mathrm d} s\\ &=\int_{{\mathbb R}}\langle u(s),\dot{\phi}(s)\rangle+\langle\dot{u}(s),\phi(s)\rangle\,{\mathrm d} s = \int_{{\mathbb R}}\frac{\,{\mathrm d}\langle u(s),\phi(s)\rangle}{\,{\mathrm d} s}\,{\mathrm d} s \end{align*}

due to the product rule. Now, reasoning as in the proof Theorem 2.6, one obtains

\begin{align*} & \int_{{\mathbb R}}\langle u(s),g(s)\rangle\,{\mathrm d} s = \int_{-\infty}^0\frac{\,{\mathrm d}\langle u(s),\phi(s)\rangle}{\,{\mathrm d} s}\,{\mathrm d} s+ \int_0^{\infty}\frac{\,{\mathrm d}\langle u(s),\phi(s)\rangle}{\,{\mathrm d} s}\,{\mathrm d} s\\ &= \langle u(0),\phi(0)\rangle-\lim_{t\to-\infty}\langle u(t),\phi(t)\rangle+ \lim_{t\to\infty}\langle u(t),\phi(t)\rangle-\langle u(0),\phi(0)\rangle = 0. \end{align*}

In particular, from this we obtain $R(T(\lambda))\cap N(T(\lambda)^{\ast})=\left\{0\right\}$, and combined with $\operatorname{ind} T(\lambda)=0$ and (II.1) we arrive at the desired splitting.

(III) We construct a finite-dimensional subspace $V\subseteq L^{\infty}({\mathbb R})$ complementary to each $R(T(\lambda))$, i.e., $R(T(\lambda))+V=L^{\infty}({\mathbb R})$ holds for $\lambda\in [a,b]$. Thereto, if $\lambda_0\in[a,b]$ is fixed, then because of $\dim N(T(\lambda_0)) \lt \infty$ there is a complement $W_{\lambda_0}\subset W^{1,\infty}({\mathbb R})$ with $N(T(\lambda_0))\oplus W_{\lambda_0}=W^{1,\infty}({\mathbb R})$. Now consider the bilinear operator

\begin{align*} \Sigma_{\lambda}: W_{\lambda_0}\times N(T(\lambda_0)^\ast)&\to L^{\infty}({\mathbb R}),& \Sigma_{\lambda}(w,v):=T(\lambda)w+v. \end{align*}

Because of $\Sigma_{\lambda_0}\in GL(W_{\lambda_0}\times N(T(\lambda_0)^\ast),L^{\infty}({\mathbb R}))$ it follows from the continuity of T and the openness of the set of bounded invertible operators that there exists a neighborhood $U_{\lambda_0}$ of λ ₀ in $[a,b]$ such that $\Sigma_{\lambda}\in GL(W_{\lambda_0}\times N(T(\lambda_0)^\ast),L^{\infty}({\mathbb R}))$ and

\begin{equation*} R(T(\lambda))+N(T(\lambda_0)^\ast)= L^{\infty}({\mathbb R})\;\text{for all }\lambda\in U_{\lambda_0} \end{equation*}

results. By compactness we can now cover the interval $[a,b]$ with a finite number of such neighborhoods $U_{\lambda_1},\ldots,U_{\lambda_n}\subseteq{\mathbb R}$. If

(3.1)\begin{align} V:=N(T(\lambda_1)^\ast)+N(T(\lambda_2)^\ast)+\ldots+N(T(\lambda_n)^\ast)\subset L^\infty({\mathbb R}), \end{align}

then $\dim V \lt \infty$ and $R(T(\lambda))+ V= L^{\infty}({\mathbb R})$ for all $\lambda\in [a,b]$.

(IV) This step is inspired by Step 3 in the proof of [Reference Waterstraat45, Theorem 5.3]. Keeping τ > 0 fixed, consider the family of operators

\begin{align*} S(\lambda):D(S(\lambda))&\to L^{\infty}[-\tau,\tau],& [S(\lambda)y](t)&:=\dot{y}(t)-A(t,\lambda)y(t) \end{align*}

for a.a. $t\in[-\tau,\tau]$, which due to Lemma 2.1(a) is well-defined on the domain

\begin{equation*} D(S(\lambda)) := \left\{u\in W^{1,\infty}[-\tau,\tau]\mid u(-\tau)\in N(P_\lambda^-(-\tau)),\, u(\tau)\in R(P_\lambda^+(\tau))\right\}. \end{equation*}

(IV.1) Claim: $\dim N(S(\lambda))=\dim N(T(\lambda)) \lt \infty$.

Consider the commutative diagram

(3.2)

where p abbreviates the restriction of functions in $L^{\infty}({\mathbb R})$ to $L^{\infty}[-\tau,\tau]$ given by $p(u):=u|_{[-\tau,\tau]}$ and a canonical map $i_\lambda:D(S(\lambda))\to W^{1,\infty}({\mathbb R})$ defined by extending a given function $u\in D(S(\lambda))$ to the intervals $(-\infty,-\tau)$ and $(\tau,\infty)$ as solution of (V_λ). Observe that i_λ is injective and $ i_{\lambda}\bigl(N(S(\lambda))\bigr)=N(T(\lambda)), $ holds, where the inclusion $i_{\lambda}\bigl(N(S(\lambda))\bigr)\subseteq N(T(\lambda))$ results directly due to the construction of i_λ, while $i_{\lambda}\bigl(N(S(\lambda))\bigr)\supseteq N(T(\lambda))$ as converse inclusion follows from the fact that provided $u\in N(T(\lambda))$, then $u(\tau)\in R(P_\lambda^+(\tau))$ and $u(-\tau)\in N(P_\lambda^-(-\tau))$ for any τ > 0 (recall (2.3)). Finally, this yields the assertion.

(IV.2) We decompose $[-\tau,\tau]=[-\tau,0]\cup [0,\tau]$ and define the spaces

\begin{align*} X_+&:=\{u\in W^{1,\infty}[0,\tau]\mid u(\tau)\in R(P_\lambda^+(\tau))\,\},& Y_+&:=L^{\infty}[0,\tau],\\ X_-&:=\{u\in W^{1,\infty}[-\tau,0]\mid u(-\tau)\in N(P_\lambda^-(-\tau))\,\},& Y_-&:=L^{\infty}[-\tau,0]. \end{align*}

Consider the following linear operators $S^\pm(\lambda):X_{\pm}\to Y_{\pm}$ pointwise defined as

\begin{equation*} [S^\pm(\lambda)y](t)=\dot y(t)-D_2f(t,\phi_\lambda(t),\lambda)y(t)\quad\text{for a.e.\ }t\in{\mathbb R}_\pm. \end{equation*}

Next consider the following commutative diagram

(3.3)

where the mappings $J\colon L^{\infty}[-\tau,\tau]\to Y_-\oplus Y_+$ and $J_{\lambda}\colon D(S(\lambda))\to X_-\oplus X_+$ are defined by $Ju:=(u_-,u_+)$ and $J_{\lambda}v:=(v_-,v_+)$ with $u_+$ and $v_+$ (resp. $u_-$ and $v_-$) being the corresponding restrictions to $[0,\tau]$ (resp. to $[-\tau,0])$. It is clear that J is an isomorphism, while J_λ is injective with range $R(J_{\lambda})=\left\{(v_-,v_+)\mid v_-(0)=v_+(0)\right\}$. Defining the mapping $\Sigma\colon X_-\oplus X_+\to {\mathbb R}^d$ by $\Sigma(v_-,v_+)=v_-(0)-v_+(0)$, one obtains that $R(J_{\lambda})=N(\Sigma)$. What is more, since Σ is an epimorphism, one can conclude that $ \operatorname{coker} J_{\lambda}=X_-\oplus X_+/R(J_{\lambda})=X_-\oplus X_+/N(\Sigma)\cong {\mathbb R}^d, $ and hence J_λ is Fredholm with $\operatorname{ind} J_{\lambda}=\dim N(J_{\lambda})-\dim \operatorname{coker} J_{\lambda}=0-d=-d$. Since J is an isomorphism, it follows $\operatorname{ind} J^{-1}=0$.

(IV.3) Claim: $S^+(\lambda)\colon X_+\to Y_+$ and $S^-(\lambda)\colon X_-\to Y_-$ are Fredholm with index $d-m^+$ resp. $m^-$.

For $S^+(\lambda)$ it suffices to prove that $S^+(\lambda)$ is surjective and $N(S^+(\lambda))\cong R(P_\lambda^+(\tau))$, which results from the following arguments: Consider the perturbed variation equation (V_λ,g) with inhomogeneity $g\in L^{\infty}[0,\tau]$. Due to [Reference Aulbach, Wanner, Aulbach and Colonius4, Theorem 2.10] the general solution $\bar\varphi_\lambda$ of (V_λ,g) is expressed via the Variation of Constants as

\begin{equation*} \bar\varphi_\lambda(t;\tau,\xi) = \Phi_\lambda(t,\tau)\xi+\int_{\tau}^t\Phi_\lambda(t,s)g(s)\,{\mathrm d} s\;\text{for all } t\in [0,\tau],\,\xi\in{\mathbb R}^d. \end{equation*}

Since $\Phi_\lambda$ is a bounded function on $[0,\tau]\times [0,\tau]$ and $g\in L^{\infty}[0,\tau]$ holds, standard calculations yield $\bar\varphi_\lambda(\cdot;\tau,\xi)\in W^{1,\infty}[0,\tau]$, which proves that $S^+(\lambda)$ is onto. Concerning the kernel $N(S^+(\lambda))$, note that $\Phi_\lambda(\cdot,\tau)\xi\in N(S^+(\lambda))$, we conclude

\begin{equation*} N(S^+(\lambda))\cong R(P_\lambda^+(\tau)), \end{equation*}

and finally arrive at $\operatorname{ind} S^+(\lambda)=\dim R(P_\lambda^+(\tau))=d-m^+$ because the invariance relation (2.3) implies $\dim R(P_\lambda^+(\tau))=\dim R(P_\lambda^+(0))=d-m^+$.

The argument for $S^-(\lambda)$ is dual to the above proof. Now it suffices to show that $S^-(\lambda)$ is surjective and $N(S^-(\lambda))\cong N(P_\lambda^-(-\tau))$. Thereto, for $g\in L^{\infty}[-\tau,0]$ it is

\begin{equation*} \bar\varphi_\lambda(t;-\tau,\xi)=\Phi_\lambda(t,-\tau)\xi+\int_{-\tau}^t\Phi_\lambda(t,s)g(s)\,{\mathrm d} s \;\text{for all } t\in[-\tau,0],\,\xi\in{\mathbb R}^d. \end{equation*}

From the boundedness of $\Phi_\lambda$ on the square $[-\tau,0]\times [-\tau,0]$ and $g\in L^{\infty}[-\tau,0]$ results that $\bar\varphi_\lambda(\cdot;-\tau,\xi)\in W^{1,\infty}[-\tau,0]$, which shows that $S^-(\lambda)$ is surjective. Addressing the kernel $N(S^-(\lambda))$ we have $\Phi_\lambda(\cdot,-\tau)\xi\in N(S^-(\lambda))$ and therefore $N(S^-(\lambda))\cong N(P_\lambda^-(-\tau))$, leading to $\operatorname{ind} S^-(\lambda)=\dim N(P_\lambda^-(-\tau))=m^-$ by (2.3).

(IV.4) Claim: $S(\lambda):D(S(\lambda))\to L^{\infty}[-\tau,\tau]$ is Fredholm of index 0.

Due to the composition $S(\lambda)=J^{-1}\circ (S^-(\lambda)\oplus S^+(\lambda))\circ J_{\lambda}\colon D(S(\lambda)) \to L^{\infty}[-\tau,\tau]$ the commutativity of the diagram (3.3) shows that $S(\lambda)$ is Fredholm with

\begin{align*} \operatorname{ind} S(\lambda) & = \operatorname{ind}(J^{-1}\circ (S^-(\lambda)\oplus S^+(\lambda))\circ J_{\lambda})\\ & = \operatorname{ind}(J^{-1})+\operatorname{ind} (S^-(\lambda)\oplus S^+(\lambda))+\operatorname{ind} J_{\lambda}\\ & \stackrel{(IV.2)}{=} \operatorname{ind} (S^-(\lambda)\oplus S^+(\lambda))-d\\ & = \operatorname{ind} S^-(\lambda)+\operatorname{ind} S^+(\lambda))-d \stackrel{(IV.3)}{=} n+r-d=0. \end{align*}

(V) Keeping τ > 0 fixed consider the family of operators

\begin{align*} S(\lambda)^\ast:D(S(\lambda)^\ast)&\to L^{\infty}[-\tau,\tau],& [S(\lambda)^\ast y](t)&:=\dot{y}(t)+A(t,\lambda)^Ty(t) \end{align*}

for a.a. $t\in[-\tau,\tau]$ on the domain

\begin{equation*} D(S(\lambda)^\ast) := \left\{u\in W^{1,\infty}[-\tau,\tau]\mid u(-\tau)\in N(P_\lambda^-(-\tau))^{\perp},u(\tau)\in R(P_\lambda^+(\tau))^{\perp}\right\}. \end{equation*}

Then the diagram

commutes, where $i_{\lambda}^{\ast}$ is defined according to (3.2). As above, $S(\lambda)^{\ast}$ is Fredholm of index 0 and $i_{\lambda}(N(S(\lambda)^{\ast}))=N(T(\lambda)^{\ast})$ with $\dim N(S(\lambda)^{\ast})=\dim N(T(\lambda)^{\ast})$.

(VI) Claim: $N(S(\lambda)^{\ast})\cap R(S(\lambda))=\{0\}$.

If $g\in R(S(\lambda))$ with preimage $\phi\in D(S(\lambda)^\ast)$ and $u\in N(S(\lambda)^{\ast})$, then

\begin{align*} &\int_{-\tau}^{\tau}\langle u(s),g(s)\rangle\,{\mathrm d} s=\int_{-\tau}^{\tau}\langle u(s),\dot{\phi}(s)-A(t,\lambda)\phi(s)\rangle\,{\mathrm d} s\\ &=\int_{-\tau}^{\tau}\langle u(s),\dot{\phi}(s)\rangle-\langle u(s),A(t,\lambda)\phi(s)\rangle\,{\mathrm d} s\\ &=\int_{-\tau}^{\tau}\langle u(s),\dot{\phi}(s)\rangle+\langle-u(s)A(t,\lambda)^T,\phi(s)\rangle\,{\mathrm d} s\\ &=\int_{-\tau}^{\tau}\langle u(s),\dot{\phi}(s)\rangle+\langle\dot{u}(s),\phi(s)\rangle\,{\mathrm d} s=\int_{-\tau}^{\tau}\frac{\,{\mathrm d}\langle u(s),\phi(s)\rangle}{\,{\mathrm d} s}\,{\mathrm d} s\\ &=\langle u(\tau),\phi(\tau)\rangle-\langle u(-\tau),\phi(-\tau)\rangle=0, \end{align*}

where the last equality follows from $u(\tau)\perp \phi(\tau)$ and $u(-\tau)\perp \phi(-\tau)$, which shows that $N(S(\lambda)^{\ast})\cap R(S(\lambda))=\{0\}$. Thanks to this result and $\operatorname{ind} S(\lambda)=0$ together with $\dim N(S(\lambda))=\dim N(S(\lambda)^{\ast})$, we conclude $R(S(\lambda))\oplus N(S(\lambda)^{\ast})=L^{\infty}[-\tau,\tau]$.

(VII) Repeating the arguments from Claim III, one has

\begin{equation*} R(S(\lambda))+ W= L^{\infty}[-\tau,\tau], \end{equation*}

where $W:=N(S(\lambda_1)^\ast)+\ldots+N(S(\lambda_n)^\ast)$ and the parameters $\lambda_1,\ldots,\lambda_n\in [a,b]$ are the same as in (3.1). Due to $p(N(T(\lambda)^\ast))=N(S(\lambda)^\ast)$ we conclude $p(V)=W$ with a subspace V as in (3.1). Thus the vector bundles

\begin{align*} E(T,V)&:=\left\{(\lambda,x)\in [a,b]\times W^{1,\infty}({\mathbb R})\mid T(\lambda)x\in V\right\},\\ E(S,W)&:=\left\{(\lambda,x)\in [a,b]\times D(S(\lambda))\mid S(\lambda)x\in W\right\}, \end{align*}

are well-defined and the following diagram commutes:

(3.4)

where $(i_E)_{\lambda}(x)=i_{\lambda}(x)$ and i_λ is as in (3.2).

(VIII) Claim: $\sigma(T,[a,b])=\sigma(S,[a,b])$.

We observe that $\dim W\leq\dim V$ and $i_E:E(S,W)\to E(T,V)$ is an injective bundle morphism. Then $E^0:=i_E(E(S,W))$ is a subbundle of $E(T,V)$, and since $\Lambda=[a,b]$ is compact, there is a complementary bundle E ¹ to E ⁰, i.e., $E(T,V)=E^0\oplus E^1$. We decompose $V=W_0\oplus W_1$, where $W_0:=\{\chi_{[-\tau,\tau]}u\mid\, u\in V\,\}$ and $W_1:=\{\chi_{{\mathbb R}\setminus[-\tau,\tau]}u\mid u\in V\,\}$. It follows from the construction of V and W that $p|_{W_0}:W_0\to W$ is an isomorphism, and hence $\dim W_0 =\dim W$. Now, given $\lambda\in[a,b]$ taking account the above splittings, the diagram (3.4) has the following form:

and $T(\lambda): E^0_{\lambda}\oplus E^1_{\lambda}\to W_0\oplus W_1$ can be written as operator matrix

\begin{equation*} T(\lambda) = \begin{pmatrix} T(\lambda)_{11}&T(\lambda)_{12}\\ T(\lambda)_{21}&T(\lambda)_{22} \end{pmatrix}. \end{equation*}

We prove that both mappings $T(\lambda)_{12}:E^1_{\lambda}\to W_0$ and $T(\lambda)_{21}:E^0_{\lambda}\to W_1$ are trivial. Indeed, take $u\in E^0_{\lambda}$. Since there exists a $v\in E(S,W)_{\lambda}$ with $(i_E)_{\lambda}v=u$, it follows that $T(\lambda)u$ admits the property $(T(\lambda)u)(t)=0$ for all $t\in{\mathbb R}\setminus[-\tau,\tau]$, which yields $T(\lambda)u\in W_0$. But $T(\lambda)u=(T(\lambda)_{11}u,T(\lambda)_{21}u)$, and therefore $T(\lambda)_{21}$ is trivial.

As for $T(\lambda)_{12}$, take $w\in R(T(\lambda)_{12})\subseteq W_0$. Then there is $u\in E^1_\lambda\subset W^{1,\infty}({\mathbb R})$ such that $T(\lambda)_{12} u=w$. Since $T(\lambda)_{12}u\in W_0$, it follows that $\dot{u}(t)-A(t,\lambda)u(t)=0$ for all $t\in(-\infty,-\tau)$ and for $t\in(\tau,\infty)$, which particularly shows that

\begin{align*} u(-\tau)&\in N(P_\lambda^-(-\tau)),& u(\tau)&\in R(P_\lambda^+(\tau)). \end{align*}

Hence, there exists a $v\in E(S,W)_\lambda$ such that $(i_E)_\lambda(v)=u$, which implies $u\in E^0_{\lambda}$. Thus, $u\in E^0_{\lambda}\cap E^1_{\lambda}=\{0\}$, and hence $w=T(\lambda)_{12}0=0$, establishing $T(\lambda)_{12}\equiv 0$.

Thus we have proved that $T(\lambda): E(T,V)_{\lambda}\to V$ allows the decomposition:

(3.5)\begin{equation} T(\lambda)=T(\lambda)_{11}\oplus T(\lambda)_{22}: E^0_{\lambda}\oplus E^1_{\lambda}\to W_0\oplus W_1=V. \end{equation}

Now, we have

• • $N(T(\lambda)_{22})=\{0\}$ since $N(T(\lambda))=(i_E)_{\lambda}\bigl(N(S(\lambda))\bigr)\subseteq E^0_{\lambda}$,

• • $\dim W=\dim E(S,W)_{\lambda}$ because $\operatorname{ind} S(\lambda)=0$ and $E(S,W)_{\lambda}=S(\lambda)^{-1}(W)$,

• • $\dim W_0\oplus W_1=\dim E^0_{\lambda}\oplus E^1_{\lambda}$ since $\operatorname{ind} T(\lambda)=0$,

• • $\dim E^0_{\lambda}=\dim (i_E)_{\lambda}(E(S,V))_{\lambda}=\dim E(S,V)_{\lambda}$ because $(i_E)_{\lambda}$ is injective.

Thus, $\dim E^0_{\lambda}=\dim W_0 \lt \infty$, $\dim E^1_{\lambda}=\dim W_1 \lt \infty$, and $T(\lambda)_{11}: E^0_{\lambda}\to W_0$ and $T(\lambda)_{22}:E^1_{\lambda}\to W_1$ are Fredholm of index 0. Moreover, from $N(T(\lambda)_{22})=\{0\}$ results that $T(\lambda)_{22}$ is an isomorphism and by means of (3.5) and Lemma A.1 and A.2, we obtain

(3.6)\begin{align} \begin{split} \sigma(T,[a,b])&=\sigma(T\circ\hat T,[a,b]) = \sigma((T_{11}\oplus T_{22})\circ (\hat T_1\oplus \hat T_2),[a,b]) \\ &=\sigma(T_{11}\circ\hat T_1,[a,b])\cdot\sigma(T_{22}\circ\hat T_2,[a,b])\\ &=\sigma(T_{11}\circ\hat T_1,[a,b])\cdot 1 = \sigma(T_{11}\circ\hat T_1,[a,b]), \end{split} \end{align}

where $\hat T: [a,b]\times V\to E(T,V)$,

\begin{align*} \hat T_1: [a,b]\times W_0&\to E(T_{11},W_0),& \hat T_2: [a,b]\times W_1\to E(T_{22},W_1) \end{align*}

are arbitrary bundle trivializations. If we consider

then again Lemma A.1 and A.2 imply that

(3.7)\begin{align} \sigma(T_{11}\circ\hat T_1,[a,b])=\sigma(S\circ\hat T_0,[a,b])=\sigma(S,[a,b]), \end{align}

where also $\hat T_0: [a,b]\times W\to E(S,W)$ is an arbitrary bundle trivialization. Finally, taking into account (3.6) and (3.7), we derive the claimed equality.

(IX) Claim: $\sigma(S,[a,b])=\sigma(Q,[a,b])$ for the operator

\begin{equation*} Q(\lambda):D(Q(\lambda))\subset W^{1,\infty}[-\tau,\tau]\to L^{\infty}[-\tau,\tau],\quad [Q(\lambda)u](t)=\dot{u}(t), \end{equation*}

defined on the domain

(3.8)\begin{align} D(Q(\lambda))=\left\{u\in W^{1,\infty}[-\tau,\tau]\mid u(-\tau)\in N(P_\lambda^-(0)),\, u(\tau)\in R(P_\lambda^+(0))\right\}. \end{align}

We abbreviate $\Phi_\lambda(t):=\Phi_\lambda(t,0)$, $S_\lambda:=S(\lambda)$ and similarly for further paths. With

\begin{align*} M(\lambda)&\in GL(W^{1,\infty}[-\tau,\tau]),& [M(\lambda) u](t)=\Phi_\lambda(t)u(t)\;\text{for all } t\in [-\tau,\tau], \end{align*}

we observe that $M_\lambda^{-1}S_\lambda M_\lambda$ are Fredholm of index 0 (cf. (IV.5)) and defined on

\begin{align*} &D(M_\lambda^{-1} S_\lambda M_\lambda)=\{M_\lambda^{-1}u\in W^{1,\infty}[-\tau,\tau]: u\in D(S_\lambda)\,\}\\ &=\{M_\lambda^{-1}u\in W^{1,\infty}[-\tau,\tau] : u(-\tau)\in N(P_\lambda^-(-\tau)),\, u(\tau)\in R(P_\lambda^+(\tau))\,\}\\ &=\{v\in W^{1,\infty}[-\tau,\tau] : (M_\lambda v)(-\tau)\in N(P_\lambda^-(-\tau)),\, (M_\lambda v)(\tau)\in R(P_\lambda^+(\tau))\,\}\\ &=\left\{v\in W^{1,\infty}[-\tau,\tau]:\, v(-\tau)\in N(P_\lambda^-(0)), v(\tau)\in R(P_\lambda^+(0))\right\}, \end{align*}

where we used $\Phi_\lambda(t)P_\lambda^\pm(0)\phi_\lambda(t)^{-1}=P_{\lambda}^{\pm}(t)$ for $t\in{\mathbb R}_\pm$ in the last equality (for this see (2.3)). Moreover, for $u\in D(M_\lambda^{-1} S_\lambda M_\lambda)$ one has the identity

\begin{align*} [M_\lambda^{-1} S_\lambda M_\lambda u](t) &\equiv \Phi_\lambda(t)^{-1}(\dot{\Phi}_\lambda(t)u(t)+\Phi_\lambda(t)\dot{u}(t)-A(t,\lambda)\Phi_\lambda(t)u(t))\\ &\equiv \dot{u}(t)+\Phi_\lambda(t)^{-1}(\dot{\Phi}_\lambda(t)-A(t,\lambda)\Phi_\lambda(t))u(t) \equiv \dot{u}(t) \end{align*}

a.e. on ${\mathbb R}$, i.e. $M_\lambda^{-1} S_\lambda M_\lambda=Q(\lambda)$. Hence, we obtain from Lemma A.1(c) that

\begin{align*} \sigma(S,[a,b]) &= \sigma(M^{-1},[a,b])\cdot\sigma(S,[a,b])\cdot\sigma(M,[a,b])\\ &=\sigma(M^{-1}S M,[a,b])=\sigma(Q,[a,b]). \end{align*}

(X) It is not hard to see that $L^{\infty}[-\tau,\tau]=Y_0\oplus Y_1$, where Y ₀ is the d-dimensional space of constant ${\mathbb R}^d$-valued functions and

\begin{align*} Y_1=\left\{u\in L^{\infty}[-\tau,\tau] : \int^{\tau}_{-\tau}u(s)\,{\mathrm d} s=0\,\right\}. \end{align*}

What is more, let us observe that Y ₀ is transversal to the image of Q, i.e.,

(3.9)\begin{equation} R(Q(\lambda))+Y_0=L^{\infty}[-\tau,\tau]. \end{equation}

Indeed, let $u\in L^{\infty}[-\tau,\tau]$. If we define $c,v:[-\tau,\tau]\to{\mathbb R}^d$ as

\begin{align*} c(t)&:\equiv\frac{1}{2\tau}\int_{-\tau}^{\tau}u(s)\,{\mathrm d} s,& v(t)&:=\int^{t}_{-\tau}\left(u(s)-c(s)\right)\,{\mathrm d} s, \end{align*}

then v belongs to $D(Q(\lambda))$, defined in (3.8) for all $\lambda\in [a,b]$, and

\begin{equation*} [Q(\lambda)v](t)+c(t)=u(t)-c(t)+c(t)=u(t)\;\ \text{for all } t\in[-\tau,\tau] \end{equation*}

proves (3.9). Thus $E(Q,Y_0)$ is well-defined with the fibres

\begin{align*} & E(Q,Y_0)_\lambda = Q(\lambda)^{-1}Y_0=\{u\in D(Q(\lambda)): \dot{u}(t)\equiv\text{constant}\}\\ &= \left\{u_\xi^\eta:\,u_\xi^\eta(t)=\tfrac{1}{2}\left(1+\tfrac{t}{\tau}\right)\eta+\tfrac{1}{2}\left(1-\tfrac{t}{\tau}\right)\xi\ \text{for }\xi\in N(P_\lambda^-(0)),\,\eta\in R(P_\lambda^+(0))\right\}. \end{align*}

Moreover, $Q(\lambda)$ acts on the fibers $E(Q,Y_0)_\lambda$ into Y ₀ by $Q(\lambda)u_\xi^\eta=\frac{1}{2\tau}(\eta-\xi)$. Now we are in a position to introduce the following commutative diagram:

where the mappings e_λ, $\hat L_{\lambda}$ and m are defined as follows

\begin{align*} e_\lambda:E(Q,Y_0)_{\lambda}&\to N(P_\lambda^-(0))\oplus R(P_\lambda^+(0)),& e_{\lambda}(u)&:=(u(-\tau),u(\tau))\\ m:Y_0&\to {\mathbb R}^d,& m(u)&:=2\tau u\\ \hat L_{\lambda}:N(P_\lambda^-(0))\oplus R(P_\lambda^+(0))&\to {\mathbb R}^d,& \hat L_{\lambda}(u,v)&=v-u. \end{align*}

Hence, in view of Lemma A.1 and A.2, we deduce the desired conclusion

(3.10)\begin{equation} \sigma(Q,[a,b])=\sigma(Q\circ\hat T,[a,b])=\sigma(\hat L\circ\hat T^{\hat L},[a,b]), \end{equation}

where $\hat T: [a,b]\times Y_0\to E(Q,Y_0)$ is any bundle trivialization with second bundle trivialization $\hat T^{\hat L}:[a,b]\times ({\mathbb R}^{d-m^+}\times\left\{0\right\}\oplus\left\{0\right\}\times{\mathbb R}^{m^-})\to \mathrm{N}(P^-(0))\oplus\mathrm{R}(P^+(0))$,

\begin{equation*} \hat T^{\hat L}(\lambda,x,y) := \left(\sum_{i=1}^{d-m^+} x_i \xi^+_i(\lambda), \sum_{j=1}^{m^-} y_j\xi^-_j(\lambda)\right) \end{equation*}

and the functions $\xi^+_i$, $\xi^-_j$ from Proposition 3.3 with $m^+=m^-$.

(XI) Claim: $\sigma(\hat L\circ\hat T^{\hat L},[a,b])=\operatorname{sgn} E(a)\cdot \operatorname{sgn} E(b)$.

By Lemma A.1 it results $\sigma(\hat L\circ\hat T^{\hat L},[a,b])= \operatorname{sgn}\det(\hat L_a\circ\hat T^{\hat L}_a)\cdot \operatorname{sgn} \det(\hat L_b\circ\hat T^{\hat L}_b), $ where using Proposition 3.3 one has for $\lambda\in\left\{a,b\right\}$ that

\begin{align*} & \det(\hat L_\lambda\circ\hat T^{\hat L}_\lambda) = \det(-\xi^+_1(\lambda),\ldots,-\xi_{d-m^+}^+(\lambda),\xi^-_1(\lambda),\ldots,\xi_{m^-}^-(\lambda))\\ &= (-1)^{d-m^+}\det (\xi^+_1(\lambda),\ldots,\xi_{d-m^+}^+(\lambda),\xi^-_1(\lambda),\ldots,\xi_{m^-}^-(\lambda)) = (-1)^{d-m^+} E(\lambda) \end{align*}

and thus $ \sigma(\hat L\circ\hat T^{\hat L},[a,b]) = (\operatorname{sgn} (-1)^{d-m^+})^2\operatorname{sgn} E(a)\operatorname{sgn} E(b) = \operatorname{sgn} E(a)\operatorname{sgn} E(b). $

(XII) Finally, taking into account the previous steps, one obtains

\begin{eqnarray*} \sigma(T,[a,b]) & \stackrel{\text{(VIII)}}{=} & \sigma(S,[a,b]) \stackrel{\text{(IX)}}{=} \sigma(Q,[a,b]) \stackrel{(3.10)}{=} \sigma(\hat L\circ\hat T^{\hat L},[a,b])\\ & \stackrel{\text{(XI)}}{=} & \operatorname{sgn} E(a)\cdot \operatorname{sgn} E(b), \end{eqnarray*}

which completes the proof of (a).

(b) The arguments from the above proof of part (a) carry over to the present situation with the spaces $W^{1,\infty}({\mathbb R})$ and $L^\infty({\mathbb R})$ replaced by $W_0^{1,\infty}({\mathbb R})$ resp. $L_0^\infty({\mathbb R})$, provided the respective statements (b) of Theorem 2.2, 2.4 and 2.6 are employed.

Proof. By assumption there exits a neighborhood Λ₀ of $\lambda^\ast$ such that $E(\lambda)\neq 0$ holds on $\Lambda_0\setminus\left\{\lambda^\ast\right\}$ and hence Theorem 2.4 combined with Proposition 3.4 yield that $T(\lambda)$ are nonsingular for $\lambda\neq\lambda^\ast$. Therefore, Theorem 3.6 implies

\begin{equation*} \sigma(T,[\lambda^\ast-\varepsilon,\lambda^\ast+\varepsilon]) = \operatorname{sgn}\sigma(T,[\lambda^\ast-\varepsilon,\lambda^\ast+\varepsilon]) = \operatorname{sgn} E(\lambda^\ast-\varepsilon)\cdot\operatorname{sgn} E(\lambda^\ast+\varepsilon) \end{equation*}

and passing to the limit $\varepsilon\searrow 0$ yields the claim.

Proof. This consequence of the Implicit Function Theorem [Reference Kielhöfer22, pp. 7–8, Theorem I.1.1] is established akin to [Reference Pötzsche35, Theorem 3.8] in the context of ordinary differential equations.

Proof. Above all, $\phi_\lambda\in W^{1,\infty}({\mathbb R},\Omega)$ for each $\lambda\in\tilde\Lambda$ holds due to Theorem 2.3.

We apply the abstract bifurcation Theorem A.1 to (O_λ) with the parametrized operator G from Theorem 2.2 and the Banach spaces $X=W_0^{1,\infty}({\mathbb R})$, $Y=L_0^\infty({\mathbb R})$. Indeed, because of Theorem 2.2(b) the mapping $G:U\to L_0^\infty({\mathbb R})$ is well-defined and continuous on a product $U:=\bigl\{x\in W_0^{1,\infty}({\mathbb R}):\,\left\|x\right\|_\infty \lt \rho\bigr\}^\circ\times\Lambda$. Furthermore, the partial derivative $D_1G:U\to L(W_0^{1,\infty}({\mathbb R}),L_0^\infty({\mathbb R}))$ exists as continuous function, $G(0,\lambda)\equiv 0$ holds on Λ, while Theorem 2.6(b) shows that $\lambda\mapsto D_1G(0,\lambda)$ defines a path of Fredholm operators with index 0. Since an Evans function E is assumed to change sign at $\lambda^\ast$, we readily derive from Corollary 3.7 that $\sigma(D_1G(0,\cdot),\lambda^\ast)=-1$ holds. Consequently, Theorem A.1 (with $\lambda_0=\lambda^\ast$) shows that $(0,\lambda^\ast)$ is a bifurcation point of (O_λ) and that there exists a $\delta_0 \gt 0$ and a connected set of nonzero solutions to (O_λ) in $W_0^{1,\infty}({\mathbb R})$ emanating from $(0,\lambda^\ast)$ to the surface $\left\|x\right\|_{1,\infty}=\delta$ for $\delta\in(0,\delta_0)$. Note in this context that Theorem 4.1 guarantees the equivalence

\begin{equation*} D_1G(0,\lambda)\in GL(W_0^{1,\infty}({\mathbb R}),L_0^\infty({\mathbb R})) \quad\Leftrightarrow\quad (\phi_\lambda,\lambda)\in{\mathcal T}_H. \end{equation*}

Then Theorem 2.3(b) implies that $\phi^\ast$ to $(C_{\lambda^\ast})$ bifurcates at $\lambda=\lambda^\ast$ into a set of solutions to (C_λ) homoclinic to ϕ_λ. In particular, the statements on the continuum of bifurcating bounded entire solutions to (C_λ) holds.

Proof. Because equation (4.3) is of lower triangular form the given expression for the general solution φ is due to the Variation of Constants Formula [Reference Aulbach, Wanner, Aulbach and Colonius4, Theorem 2.10]. In order to identify the bounded entire solutions of (4.3) we first note the limits

(4.4)\begin{equation} \lim_{t\to\pm\infty}\cosh t\left(\arctan\tanh\tfrac{t}{2}\mp\tfrac{\pi}{4}\right)=\mp\tfrac{1}{2}. \end{equation}

On the one hand, the representation

\begin{align*} & \cosh t\bigl[\xi_2+(\nu\xi_1^2+2\mu)\arctan\tanh\tfrac{t}{2}\bigr]\\ =& \cosh t\bigl[(\nu\xi_1^2+2\mu)\left(\arctan\tanh\tfrac{t}{2}-\tfrac{\pi}{4}\right)\bigr] + \cosh t\bigl[\xi_2+(\nu\xi_1^2+2\mu)\tfrac{\pi}{4}\bigr] \end{align*}

and (4.4) guarantee that $\sup_{t\geq 0}\left|\varphi(t;0,\xi)\right| \lt \infty$ is equivalent to

(4.5)\begin{equation} 0 = \xi_2+(\nu\xi_1^2+2\mu)\tfrac{\pi}{4}. \end{equation}

On the other hand,

\begin{align*} &\cosh t\bigl[\xi_2+(\nu\xi_1^2+2\mu)\arctan\tanh\tfrac{t}{2}\bigr]\\ = & \cosh t\bigl[(\nu\xi_1^2+2\mu)\left(\arctan\tanh\tfrac{t}{2}+\tfrac{\pi}{4}\right)\bigr] +\cosh t\bigl[\xi_2-(\nu\xi_1^2+2\mu)\tfrac{\pi}{4}\bigr] \end{align*}

combined with (4.4) ensure that $\sup_{t\leq 0}\left|\varphi(t;0,\xi)\right| \lt \infty$ holds if and only if

(4.6)\begin{equation} 0 = \xi_2-(\nu\xi_1^2+2\mu)\tfrac{\pi}{4}. \end{equation}

The relations (4.5) and (4.6) in turn are equivalent to $\nu\xi_1^2+2\mu=0$ and $\xi_2=0$.