Mel’nikov Methods and Homoclinic Orbits in Discontinuous Systems

Abstract

We consider a discontinuous system exhibiting a, possibly non-smooth, homoclinic trajectory. We assume that the critical point lies on the discontinuity level. We study the persistence of such a trajectory when the system is subject to a smooth non-autonomous perturbation. We use a Mel’nikov type approach and we introduce conditions which enable us to reformulate the problem in the setting of smooth systems so that we can follow the outline of the classical theory.

Appendix

In this appendix we give the explicit proof of a roughness result in exponential dichotomies, which is probably known by the experts but for which we are not able to give a precise reference.

Let $t \mapsto \varvec{A}(t)$ be a piecewise continuous $n\times n$ matrix valued function defined on $\mathbb{R }$. We recall that the linear differential equation

$$\begin{aligned} \dot{\vec {x}}= \varvec{A}(t) {\vec {x}}\end{aligned}$$

(5.1)

is said to have an exponential dichotomy on an interval $J$ if there are projections $\varvec{P}$ and positive constants $k, \alpha _1,\alpha _2$ such that

$$\begin{aligned} \begin{array}{lll} \Vert \mathbf{X }(t)\varvec{P}\mathbf{X }^{-1}(s)\Vert \le k {\, \mathrm e}^{-\alpha _1(t-s)} &{} \text{ for }&{} s,t \in J \text{ with } s\le t \\ \Vert \mathbf{X }(t)(\mathbb{I }- \mathbf{P })\mathbf{X }^{-1}(s)\Vert \le k {\, \mathrm e}^{-\alpha _2(s-t)} &{} \text{ for }&{} s,t \in J \text{ with } t\le s \end{array} \end{aligned}$$

(5.2)

where $\mathbf{X }(t)$ is the fundamental matrix of equation (5.1).

Assume that system (5.1) has an exponential dichotomy on $J$ with constant $k$ and exponents $\alpha _1,\alpha _2$, and let $\beta _1,\beta _2$ be positive constants such that $\beta _1<\alpha _1$ and $\beta _2<\alpha _2$. The well known roughness property of exponential dichotomies (see e.g. [19, p. 133]) implies that if $t \mapsto \varvec{B}(t)$ is a piecewise continuous $n\times n$ matrix valued function such that $\Vert \varvec{B}(\cdot )\Vert _\infty <\delta$, then for $\delta$ sufficiently small the perturbed system

$$\begin{aligned} \dot{\vec {x}}= [\varvec{A}(t)+\varvec{B}(t)] {\vec {x}}\end{aligned}$$

(5.3)

has an exponential dichotomy on $J$ with projection $\varvec{Q}$, constant $k^{\prime }$ and exponents $\beta _1,\beta _2$. We want to show that if we further assume that the function $\varvec{B}$ is $L^1$ then Eq. (5.3) has an exponential dichotomy on $J$ with the same exponents $\alpha _1,\alpha _2$.

More precisely we will prove the following result.

Proposition 5.1

Assume that system (5.1) has an exponential dichotomy on $J=[0,+\infty )$ with projection $\varvec{P}$, constant $k$ and exponents $\alpha _1,\alpha _2$. Let $\varvec{B} \in L^1[0,+\infty )$ be a piecewise continuous $n\times n$ matrix valued function, and fix $\bar{T}>0$ such that $\int _{\bar{T}}^{\infty } \Vert \varvec{B}(\tau )\Vert \, d\tau < \frac{1}{k}$. Then there is $\delta _0$ such that if $\Vert \varvec{B}(\cdot )\Vert _\infty <\delta <\delta _0$ the perturbed system (5.3) has an exponential dichotomy on $[\bar{T}, +\infty )$ (and, consequently, on $J$) with projection $\varvec{Q}$, constant $k^{\prime }$ and the same exponents $\alpha _1,\alpha _2$.

First of all we prove the following Gronwall-like lemma.

Lemma 5.2

Let $\phi : [\bar{T}, +\infty ) \rightarrow \mathbb{R }$ be a bounded, continuous function such that, for $t \ge \bar{T}$,

$$\begin{aligned} \phi (t) \le k_1 {\, \mathrm e}^{-\alpha _1 t} + k_2 \int \limits _{\bar{T}}^t {\, \mathrm e}^{-\alpha _1 (t-\tau )} b(\tau ) \phi (\tau ) d\tau + k_2 \int \limits _t^\infty {\, \mathrm e}^{-\alpha _2 (\tau -t)} b(\tau ) \phi (\tau ) d\tau , \end{aligned}$$

where $k_1, k_2, \alpha _1, \alpha _2$ are positive constants, $b \in L^1[0,+\infty )$ is a nonnegative piecewise continuous function, and $\bar{T}>0$ is such that $\int _{\bar{T}}^{\infty } b(\tau ) \, d\tau < \frac{1}{4k_2}$. Then, there is $c>0$ such that

$$\begin{aligned} \phi (t) \le c k_1 {\, \mathrm e}^{-\alpha _1 t}, \qquad t \ge \bar{T}. \end{aligned}$$

Proof

Consider the corresponding integral equation

$$\begin{aligned} \psi (t)\!=\! k_1 {\, \mathrm e}^{-\alpha _1 t} \!+\! k_2 \int \limits _{\bar{T}}^t {\, \mathrm e}^{-\alpha _1 (t-\tau )} b(\tau ) \psi (\tau ) \, d\tau \!+\! k_2 \int \limits _t^\infty {\, \mathrm e}^{-\alpha _2 (\tau -t)} b(\tau ) \psi (\tau ) \, d\tau , \end{aligned}$$

(5.4)

for $t \ge \bar{T}$. Then, any bounded continuous solution $\psi$ is twice differentiable and is in fact a solution of the differential equation

$$\begin{aligned} \psi ^{\prime \prime } +(\alpha _1-\alpha _2)\psi ^{\prime }- [\alpha _1\alpha _2 -k_2 (\alpha _1+\alpha _2)b(t)] \psi (t)=0. \end{aligned}$$

We have

$$\begin{aligned} \psi (t) = c_1 {\, \mathrm e}^{-\alpha _1 t} + c_2 {\, \mathrm e}^{\alpha _2 t} +k_2\int \limits _{\bar{T}}^t \left( {\, \mathrm e}^{-\alpha _1 (t-\tau )}-{\, \mathrm e}^{\alpha _2 (t-\tau )}\right) b(\tau ) \psi (\tau ) d\tau , \qquad t \ge \bar{T}, \end{aligned}$$

Since we are assuming that $\psi$ is bounded we have

$$\begin{aligned} c_2 = k_2 \int \limits _{\bar{T}}^\infty {\, \mathrm e}^{-\alpha _2 \tau } b(\tau ) \psi (\tau ) \, d\tau , \end{aligned}$$

thus

$$\begin{aligned} \psi (t)&= \left( k_2\int \limits _t^\infty {\, \mathrm e}^{-\alpha _2 \tau } b(\tau ) \psi (\tau ) \, d\tau \right) {\, \mathrm e}^{\alpha _2 t} \\&+ \left( c_1 + k_2 \int \limits _{\bar{T}}^t {\, \mathrm e}^{\alpha _1 \tau } b(\tau ) \psi (\tau ) \, d\tau \right) {\, \mathrm e}^{-\alpha _1 t}, \qquad t \ge \bar{T}, \end{aligned}$$

Let us show that $\psi (t) = O({\, \mathrm e}^{-\alpha _1 t})$ as $t \rightarrow +\infty$. For this purpose we define the Banach space

$$\begin{aligned} X_{1}=\left\{ \psi \in C([\bar{T},+\infty ),\mathbb{R }) | \sup _{t \ge \bar{T}}[ |\psi (t)| e^{\alpha _1 t}]<\infty \right\} \end{aligned}$$

endowed with the norm $\displaystyle |\psi |_{\alpha _1} = \sup _{t \ge \bar{T}}[ |\psi (t)| e^{\alpha _1 t}]$.

We define a linear operator $\mathcal T$ acting on $X_{1}$ as follows

$$\begin{aligned} \mathcal T (\psi ) (t)&= \left( k_2\int \limits _t^\infty {\, \mathrm e}^{-\alpha _2 \tau } b(\tau ) \psi (\tau ) \, d\tau \right) {\, \mathrm e}^{\alpha _2 t} \\&+ \left( c_1 + k_2\int \limits _{\bar{T}}^t {\, \mathrm e}^{\alpha _1 \tau } b(\tau ) \psi (\tau ) d\tau \right) {\, \mathrm e}^{-\alpha _1 t}, \qquad t \ge \bar{T}, \end{aligned}$$

Let us show that $\mathcal T$ is a contraction. We have

$$\begin{aligned} \Vert \mathcal T (\psi )(t) \Vert \le \left( c_1 + 2 k_2 \Vert b\Vert _{L^1([\bar{T},+\infty ))} \Vert \psi \Vert _{\alpha _1} \right) {\, \mathrm e}^{-\alpha _1 t}, \qquad t \ge \bar{T}, \end{aligned}$$

therefore the operator $\mathcal{T }$ is well defined. Moreover if $\psi _1, \psi _2 \in X_{1}$ we have

$$\begin{aligned} \Vert \mathcal{T } (\psi _2)(t) - \mathcal{T } (\psi _1)(t) \Vert \le 2k_2 \Vert b\Vert _{L^1([\bar{T},+\infty ))} \Vert \psi _2-\psi _1\Vert _{\alpha _1} {\, \mathrm e}^{-\alpha _1 t}, \qquad t \ge \bar{T}. \end{aligned}$$

Since $2k_2\int _{\bar{T}}^{\infty } b(\tau ) \, d\tau <\frac{1}{2}$ by assumption, $\mathcal{T }$ is a contraction, hence has a unique fixed point $\bar{\psi }\in X_{1}$. It follows that

$$\begin{aligned} \bar{\psi }(t) \le c k_1 {\, \mathrm e}^{-\alpha _1 t}, \qquad t \ge \bar{T} \end{aligned}$$

for some constant $c$.

Moreover for any constant $L>\dfrac{k_1}{k_2(1-\Vert b\Vert _{L^1([\bar{T},+\infty ))})}$ we have

$$\begin{aligned} L \ge k_1 {\, \mathrm e}^{-\alpha _1 t} + k_2 \int _{\bar{T}}^t {\, \mathrm e}^{-\alpha _1 (t-\tau )} b(\tau ) L \, d\tau + k_2 \int _t^\infty {\, \mathrm e}^{-\alpha _2 (\tau -t)} b(\tau ) L \, d\tau , \end{aligned}$$

for any $t \ge \bar{T}$. So if we choose $L \ge \sup _{t \ge \bar{T}} \phi (t)$ by the upper lower solution method we find a solution $\tilde{\psi }(t)$ of (5.4) such that $\phi (t) \le \tilde{\psi }(t) \le L$ for any $t \ge \bar{T}$. Since $\bar{\psi }(t)$ is uniquely determined we find $\tilde{\psi }(t)=\bar{\psi }(t)$ and the result follows.$\square$

With a similar argument we prove the following lemma.

Lemma 5.3

Let $k_1, k_2, \alpha _1, \alpha _2, b$ and $\bar{T}$ be as in Lemma 5.2.

1) Let $\phi : [\bar{T}, +\infty ) \rightarrow \mathbb{R }$ be a bounded, continuous function such that, for all $t \ge s \ge \bar{T}$,

$$\begin{aligned} \phi (t) \le k_1 {\, \mathrm e}^{-\alpha _1 (t-s)} + k_2 \int \limits _s^t {\, \mathrm e}^{-\alpha _1 (t-\tau )} b(\tau ) \phi (\tau ) \, d\tau + k_2 \int \limits _t^\infty {\, \mathrm e}^{-\alpha _2 (\tau -t)} b(\tau ) \phi (\tau ) \, d\tau , \end{aligned}$$

then, there is $c>0$ such that

$$\begin{aligned} \phi (t) \le c k_1 {\, \mathrm e}^{-\alpha _1 (t-s)}, \qquad t \ge s \ge \bar{T}. \end{aligned}$$

2) Let $\phi : [\bar{T}, +\infty ) \rightarrow \mathbb{R }$ be a bounded, continuous function such that, for all $s \ge t \ge \bar{T}$,

$$\begin{aligned} \phi (t) \le k_1 {\, \mathrm e}^{-\alpha _2 (s-t)} + k_2 \int \limits _{\bar{T}}^t {\, \mathrm e}^{-\alpha _1 (t-\tau )} b(\tau ) \phi (\tau ) \, d\tau + k_2 \int _t^s {\, \mathrm e}^{-\alpha _2 (\tau -t)} b(\tau ) \phi (\tau ) \, d\tau , \end{aligned}$$

then, there is $c^{\prime }>0$ such that

$$\begin{aligned} \phi (t) \le c^{\prime } k_1 {\, \mathrm e}^{-\alpha _2 (s-t)}, \qquad s \ge t \ge \bar{T}. \end{aligned}$$

We now give the

Proof of Proposition 5.1

Assume that $\delta _0 \le \dfrac{\alpha }{4k}$. We define a linear operator $\mathcal T$, acting on the Banach space of bounded continuous functions $C_b^0([\bar{T},+\infty ),\mathbb{R }^{n})$ with the standard supremum norm $\Vert \cdot \Vert _\infty$. Let $\vec {\xi } \in \mathbb{R }^n$ be fixed and let

$$\begin{aligned} \mathcal T (\varvec{u}) (t)&= \mathbf{X }(t) \varvec{P} \vec {\xi } + \int _{\bar{T}}^{t} \mathbf{X }(t) \varvec{P} \mathbf{X }^{-1}(\tau ) \varvec{B}(\tau ) \vec {u}(\tau ) d\tau \\&-\int _{t}^{\infty } \mathbf{X }(t) [\mathbb{I }- \mathbf{P }] \mathbf X ^{-1}(\tau ) \varvec{B}(\tau ) \vec {u}(\tau ) \, d\tau \, , \quad t \ge \bar{T}. \end{aligned}$$

Then,

$$\begin{aligned} \Vert \mathcal T (\vec {u})(t)\Vert&\le k {\, \mathrm e}^{-\alpha _1 t} \Vert \vec {\xi }\Vert + \left( \int \limits _{\bar{T}}^{t} k {\, \mathrm e}^{-\alpha _1 (t-\tau )} \delta \, d\tau + \int \limits _{t}^{\infty } k {\, \mathrm e}^{-\alpha _2 (\tau -t)} \delta \, d\tau \right) \Vert \vec {u}\Vert _\infty \\&\le k_1 \Vert \vec {\xi }\Vert + \dfrac{2k}{\alpha } \delta \Vert \vec {u}\Vert _\infty \, , \quad t \ge \bar{T}. \end{aligned}$$

Moreover,

$$\begin{aligned} \Vert \mathcal T (\vec {u}_2)(t)-\mathcal T (\vec {u}_1)(t)\Vert \le \dfrac{2k}{\alpha } \delta \Vert \vec {u}_2 - \vec {u}_1\Vert _\infty , \quad t \ge \bar{T}. \end{aligned}$$

Since $\dfrac{2k}{\alpha } \delta <1$ it follows that $\mathcal T$ is a contraction and hence has a unique fixed point.

Such a fixed point $\vec {\hat{u}}$ verifies, for $t \ge s \ge \bar{T}$,

$$\begin{aligned} \vec {\hat{u}}(t)&= \mathbf{X }(t) \varvec{P} \mathbf{X }^{-1}(s) \vec {\hat{u}}(s) + \int \limits _{\bar{T}}^{t} \mathbf{X }(t) \varvec{P} \mathbf{X }^{-1}(\tau ) \varvec{B}(\tau ) \vec {\hat{u}}(\tau ) \, d\tau \nonumber \\&-\int \limits _{t}^{\infty } \mathbf{X }(t) [\mathbb{I }-\mathbf{P }] \mathbf{X }^{-1}(\tau ) \varvec{B}(\tau ) \vec {\hat{u}}(\tau ) \, d\tau . \end{aligned}$$

(5.5)

Then we have, for any $t \ge s \ge \bar{T}$,

$$\begin{aligned} \Vert \vec {\hat{u}}(t)\Vert&\le k {\, \mathrm e}^{-\alpha _1 t} \Vert \vec {\hat{u}} (s)\Vert + k \int \limits _{\bar{T}}^{t} {\, \mathrm e}^{-\alpha _1 (t-\tau )} \Vert \varvec{B}(\tau )\Vert \Vert \vec {\hat{u}}(\tau )\Vert \, d\tau \\&+\,k \int \limits _{t}^{\infty } {\, \mathrm e}^{-\alpha _2 (\tau -t)} \Vert \varvec{B}(\tau )\Vert \Vert \vec {\hat{u}}(\tau )\Vert \, d\tau \end{aligned}$$

We apply Lemma 5.3 with $\Vert \vec {u}(t)\Vert$ in place of $\phi (t)$ and $\Vert \varvec{B}(t)\Vert$ in place of $b(t)$. Hence there is $c>0$ such that

$$\begin{aligned} \Vert \vec {u}(t)\Vert \le ck {\, \mathrm e}^{-\alpha _1 (t-s)}\,\Vert \vec {u}(s)\Vert , \qquad t \ge s \ge \bar{T}. \end{aligned}$$

Analogously, one can prove that there is $c^{\prime }>0$ such that

$$\begin{aligned} \Vert \vec {u}(t)\Vert \le c^{\prime }k {\, \mathrm e}^{-\alpha _2 (s-t)}\,\Vert \vec {u}(s)\Vert , \qquad s \ge t \ge \bar{T}. \end{aligned}$$

Following an argument similar to [19, Lemma 7.4] we obtain the estimates of the exponential dichotomy on $[\bar{T}, + \infty )$ and, consequently, on $[0, + \infty )$.

We recall that when $\varvec{A}(t)\equiv \varvec{A}$ is a constant function, we have that (5.1) admits exponential dichotomy in the whole of $\mathbb{R }$ if and only if $\varvec{A}$ has no eigenvalues with real part equal to $0$. Let us denote by $\lambda _u$ and $\lambda _s$ the eigenvalues of $\varvec{A}$ respectively with smallest positive real part and with largest negative real part. If $\lambda _u$ and $\lambda _s$ are real and simple then the exponents of the dichotomy are exactly $\lambda _u$ and $\lambda _s$ and the constant is $1$. Similarly if $\lambda _u$ and $\lambda _s$ are semisimple (as we assume in this paper), that is, $\lambda _u=a+ib$ and $\lambda _s=-c+id$ with $a,c>0$, then the exponents of the dichotomy are again exactly $\text{ Re }(\lambda _u)=a$ and $\text{ Re }(\lambda _s)=-c$. On the other hand, if $\lambda _u$ and $\lambda _s$ have algebraic multiplicity larger than geometric multiplicity, then (5.1) admits exponential dichotomy in the whole of $\mathbb{R }$ but with exponents $\lambda ^+$ and $\lambda ^-$ where $0>\lambda ^->-\text{ Re }(\lambda _s)$ and $0<\lambda ^+<\text{ Re }(\lambda _u)$.