BIFURCATIONS OF PLANAR SLIDING HOMOCLINICS

We suppose the following conditions: (i) f−(0)= 0, and D f−(0) has no eigenvalues on the imaginary axis, (ii) there are two solutions γ−(s), γ+(s) of ż = f−(z), y ≤ 1 defined on R− = (−∞,0], R+ = [0,+∞), respectively, such that lims→±∞ γ±(s)= 0 and γ±(s)= (x±(s), y±(s)) with y±(0)= 1, x−(0) < x+(0). Moreover, f±(z)= ( f±1(z), f±2(z)) with f±1(x,1) > 0, f+2(x,1) < 0 for x−(0)≤ x ≤ x+(0). Furthermore, f−2(x,1) > 0 for x−(0)≤ x < x+(0), f−2(x+(0),1)= 0, and ∂x f−2(x+(0),1) < 0. Assumptions (i) and (ii) mean that (1.1) for ε = 0 has a sliding homoclinic solution γ, created by γ±, to a hyperbolic equilibrium 0. We are interested in the bifurcation of γ to bounded solutions on R of (1.1) under the perturbation εg(z, t,ε).

The plan of the paper is as follows.In Section 2, we study (1.1) by using functional methods based on [4] along with the implicit function theorem [5].In Section 3, we generalize results of Section 2 to systems with multiple discontinuous levels.Final Section 4 is devoted to a concrete system of piece-vice linear systems with periodic perturbations.
Sliding periodic solutions of discontinuous differential equations are investigated in [1-3] with both analytical and numerical methods.Qualitative properties of discontinuous systems are studied in [6].Bifurcations for planar discontinuous ordinary differential systems with small periodic perturbations from homoclinic solutions transversally intersecting levels of discontinuity are studied in [7] to generalize the well-known Melnikov method for a smooth case [4] to a discontinuous one.We note that bifurcations from sliding homoclinic solutions, studied in this paper, are different to [4,7].

Bifurcation result
In this section, we find conditions under which γ persists in (1.1) for ε = 0 small.For this purpose, we consider (1.1) as a system in R 3 defined by while on y = 1 (cf.[1,6]), we consider the system where q ± = (q ±1 , q ±2 ).We first study the system Proof.We consider the Banach space with the usual sup-norm • .We put where v = (v 1 ,v 2 ).Next, the system has an exponential dichotomy on R − (cf.[4]), that is, there are positive constants K, a and a projection P : R 2 → R 2 such that where V − (s), V − (0) = I is the fundamental matrix solution of (2.6).Moreover, since γ− (s) solves (2.6), and it is bounded on R − , and γ− (0) is transversal to the x-axis, we can suppose (cf.[4]) that Im(I − P) = R γ− (0) and ImP is the x-axis.Then, (2.5) can be rewritten as a fixed point problem on the Banach space X, where for ε small, we can solve (2.8) by using the implicit function theorem to obtain a unique small solution v(s,α,ε) of (2.8), and so 3).The proof is finished.

.14)
If h(x,s,ε) is the right-hand side of (2.2), then conditions (i) and (ii) imply that h(x,s,ε) > 0 for any x − (0) ≤ x ≤ x + (0) and ε small.Then assumption (ii) gives the solvability of the equation for the function s + (α,ε) > 0, where x(s) solves (2.2) and (2.14).So, s + (α,ε) is the time when the sliding motion of (2.2) is ending.We put Finally, we consider the initial value problem (2.17) That is the initial value problem We note that γ + (0) = (ϕ + (α,0),1) and we look for a solution z of (2.18) near to γ where By shifting the time Now we study the problem has an exponential dichotomy on R + (cf.[4]), that is, there are positive constants M, b and a projection Q : R 2 → R 2 such that where V + (s), V + (0) = I is the fundamental matrix solution of (2.25).Moreover, since γ+ (s) solves (2.25) and it is bounded on R + , we can suppose (cf.[4]) that ImQ = R γ+ (0) and Im(I − Q) is orthogonal to the line R γ+ (0).On the other hand, condition (ii) implies that Consequently, Q is the orthogonal projection onto the x-axis.Let Γ = γ+ (0) ⊥ be a nonzero orthogonal vector onto γ+ (0).Now, for simplicity, we can take Γ = (0,1).So, Im( We note that is a basis of a space of bounded solutions on R + of the adjoint system (cf.[4]) Proof.A general form of a bounded solution of equation Then using the initial condition w(0) = u, we get the equation which implies ( Consequently, the unique bounded solution of (2.24) on R + is given by (2.38) Then, (2.32) follows directly from (2.38).The proof is finished.
On the other hand, from the definition of function s + (α,ε) in (2.15), we see that ∂ α s + (α,0) = 0. So, simple roots of M(α) are in one-to-one correspondence with simple roots of the function (2.45) Summarizing we arrive at the following result.

Example
We present in this section an illustrative example.Let a + be the unique (positive) solution of the equation   Function (4.9) has two different simple roots over the period 2π.By applying Theorem 3.2, we get the existence of two bounded solutions of (3.1) with (4.3) near to γ, which is homoclinic to a small hyperbolic 2π-periodic solution of (3.1) with (4.3).