Applied Mathematics E-Notes, 1(2001), 77-85 \begin{matrix}c \\ ◯ \end{matrix} Available free at mirror sites of http://math2.math.nthu.edu.tw/∼amen/

Periodic Boundary Value Problems of Impulsive Differential Equations ^{∗}^{†}

Hui Fang^{‡}, Zhi-Cheng Wang^{§}

Received 1 March 2001

Abstract By means of the continuation theorem of coincidence degree theory, sufficient conditions are obtained for the existence of solutions of periodic boundary value problems involving impulsive differential equations.

In recent years, impulsive differential equations have attracted much attention since many evolution processes are subject to short term perturbations in the form of impulses. See, for instance, [1-8]. In this paper, we consider the following periodic boundary value problem of impulsive differential equation (PBVP)

\{ \begin{matrix}ẋ(t) = g(t, x(t)) + p(t), t ≠ t_{k}, & k = 1, 2, ⋯, m, \\ x(t_{k}^{+}) − x(t_{k}) = I_{k}(t_{k}, x(t_{k})), & k = 1, 2, ⋯, m, \\ x(0) = x(T), \end{matrix}  (1)

where 0 = t_{0} < t_{1} < t_{2} < ⋯ < t_{m} < t_{m+1} = T, T > 0; J = [0, T], g ∈ C(J × R, R), g(0, u) = g(T, u) for u ∈ R; I_{k} ∈ C(J × R, R) for k = 1, 2, ⋯, m; p ∈ C(J, R), and p(0) = p(T).

Next we briefly state the part of Mawhin’s coincidence degree theory that will be used in our study of PBVP (1). For further details, we refer the readers to [9, 10]. Let X and Y be real normed vector spaces, L : domL ⊂ X → Y be a linear mapping, and N : X → Y be a continuous mapping. The mapping L will be called a Fredholm mapping of index zero if dimKerL = codimImL < +∞ and ImL is closed in Y. If L is a Fredholm mapping of index zero, there then exist continuous projectors P : X → X and Q : Y → Y such that ImP = KerL and KerQ = ImL. It follows that L_{p} = L|_{domL∩KerP} : domL ∩ KerP → ImL = KerQ is one-to-one and onto ImL. We denote its inverse by K_{p}. If Ψ is an open bounded subset of X, the mapping N will be called L-compact on Ψ̄ if QN(Ψ) is bounded and K_{p}(I − Q)N : Ψ̄ → X is compact. Since ImQ is isomorphic to KerL, there exists an isomorphism Ĵ: ImQ → KerL.

Mawhin’s Continuation Theorem ([9, 10]). Let L be a Fredholm mapping of index zero and let N be L-compact on Ψ̄. Suppose (i) For each λ ∈ (0, 1), x ∈ ∂Ψ, Lx ≠ λNx; (ii) QNx ≠ 0 for each x ∈ KerL ∩ ∂Ψ; and (iii) Brouwer degree deg_{B}(ĴQN, Ψ ∩ KerL, 0) ≠ 0. Then the equation Lx = Nx has at least one solution in domL ∩ Ψ̄.

∗Mathematics Subject Classifications: 34A37

†This work is supported by the Natural Science Foundation of Yunnan Province, P. R. China

‡Department of Systems Science and Applied Mathematics, Kunming University of Science and Technology, Kunming, Yunnan 650051, P. R. China

§Department of Applied Mathematics, Hunan University, Changsha, Hunan 410082, P. R. China

Set J′ = J \ {t_{1}, ⋯, t_{m}}. Let

P C_{T} = {u : J → R| u ∈ C(J′, R), u(t_{j}^{−}), u(t_{j}^{+}) exist for
j = 1, 2, ⋯, m, and u(0) = u(T)}

and

P C_{T}^{1} = {u : J → R| u ∈ C^{1}(J′, R), u(t_{j}^{+}), u(t_{j}^{−}), u′(t_{j}^{+}), u′(t_{j}^{−}) exist,
u(t_{j}^{−}) = u(t_{j}), j = 1, 2, ⋯, m, and u(0) = u(T), u′(0) = u′(T)}

where u(t_{j}^{+}), u(t_{j}^{−}) denote the right and left limits of u(t) at t = t_{j} respectively; u′(t_{j}^{+}), u′(t_{j}^{−}) denote the right and left limits of u′(t) at t = t_{j} respectively.

For u(t) ∈ P C_{T}^{1} , it is easy to see that the left derivative u′_{−}(t_{j}) exists and is equal to u′(t_{j}^{−}). In the following, u′(t_{j}) may be understood as u′_{−}(t_{j}).

For u ∈ P C_{T}, denote its norm by ∥u∥ = sup{|u(t)| : t ∈ J}. For u ∈ P C_{T}^{1} , denote its norm by ∥u∥_{1} = ∥u∥ + ∥u′∥. One can easily prove that P C_{T} and P C_{T}^{1} are Banach spaces.

Let X = Y = P C_{T} and let L : domL = P C_{T}^{1} ⊂ X → Y be given by

Lx(t) = \{ \begin{matrix}ẋ(t), & t ≠ t_{k}, & k = 1, 2, ⋯, m \\ x(t_{k}^{+}) − x(t_{k}), & t = t_{k}, & k = 1, 2, ⋯, m \end{matrix}

It is obvious that L is a linear mapping.

THEOREM 1. Assume that ∫_{0}^{T} p(t)dt = 0 and there exist constants c_{k}, h_{k} > 0 for k = 0, 1, 2, ⋯, m, such that the following conditions hold: (i) if x ≥ c_{0}, then g(t, x) ≤ h_{0} uniformly for t ∈ J, (ii) if |x| ≥ c_{0}, then xg(t, x) > 0 uniformly for t ∈ J, (iii) if |x| ≥ c_{0}, then xI_{k}(t_{k}, x) > 0 for k = 1, 2, ⋯, m, (iv) |I_{k}(t_{k}, x)| ≤ c_{k}|x| + h_{k} for k = 1, 2, ⋯, m, and (v) ∑_{k=1}^{m} c_{k} < 1/2. Then PBVP (1) has at least one solution.

We remark that condition (i) in Theorem 1 can be replaced by the following condition: (a) if x ≤ −c_{0}, then g(t, x) ≥ −h_{0} uniformly for t ∈ J.

Before proving Theorem 1, we need the following lemmas.

LEMMA 1. H ⊂ P C_{T} is relatively compact if, and only if, every function of H is uniformly bounded on J and is equicontinuous on each J_{k} for k = 0, 1, 2, ⋯, m, where J_{0} = [0, t_{1}), J_{k} = (t_{k}, t_{k+1}) for k = 1, 2, ⋯, m − 1, and J_{m} = (t_{m}, T].

The proof is an immediate consequence of Ascoli-Arzela Theorem.

LEMMA 2. L is a Fredholm mapping of index zero.

PROOF. It is clear that

KerL = {u : u(t) ≡ c, c ∈ R}.

For y ∈ Y we define Q : Y → Y by

Qy = \frac{∑_{i=1}^{m} y(t_{i}) + ∫_{0}^{T} y(t)dt}{T + m}.

H. Fang and Z. C. Wang 79 We assert that ImL = KerQ. There are two cases to consider.

Case 1. If y ∈ ImL, then there exists x ∈ domL ⊂ X such that Lx = y i.e.

\{ \begin{matrix}ẋ(t) = y(t), & t ≠ t_{k}, k = 1, 2, ⋯, m, \\ x(t_{k}^{+}) − x(t_{k}) = y(t_{k}), & k = 1, 2, ⋯, m. \end{matrix}

Hence, we have

∫_{0}^{T} y(t)dt = ∫_{0}^{t_{1}} y(t)dt + ∑_{k=1}^{m−1} ∫_{t_{k}}^{t_{k+1}} y(t)dt + ∫_{t_{m}}^{T} y(t)dt
= x(t_{1}) − x(0) + ∑_{k=1}^{m−1}[x(t_{k+1}) − x(t_{k}^{+})] + x(T) − x(t_{m}^{+})
= x(t_{1}) + ∑_{k=1}^{m−1}[x(t_{k+1}) − x(t_{k}) − y(t_{k})] − x(t_{m}) − y(t_{m})
= − ∑_{k=1}^{m} y(t_{k}).

So, we have

∑_{k=1}^{m} y(t_{k}) + ∫_{0}^{T} y(t)dt = 0.

Therefore, y ∈ KerQ, which implies that ImL ⊂ KerQ.

Case 2. If y ∈ KerQ, then ∑_{i=1}^{m} y(t_{i}) + ∫_{0}^{T} y(t)dt = 0. Define the function x(t) by

x(t) = \{ \begin{matrix}a_{1} + ∫_{t_{1}}^{t} y(s)ds, & t ∈ [0, t_{1}]; \\ a_{i+1} + ∫_{t_{i+1}}^{t} y(s)ds, & t ∈ (t_{i}, t_{i+1}], & i = 1, 2, ⋯, m − 1; \\ a_{m+1} + ∫_{T}^{t} y(s)ds, & t ∈ (t_{m}, T]; \end{matrix}

where a_{1}, ..., a_{m+1} are given by the following system of equations

\{ \begin{matrix}∑_{i=1}^{m}(1 + t_{i} − t_{i−1})a_{i} + (T − t_{m})a_{m+1} & = & − ∑_{i=0}^{m−1} ∫_{t_{i}}^{t_{i+1}} ∫_{t_{i+1}}^{t} y(s)dsdt \\ & & − ∫_{t_{m}}^{T} ∫_{T}^{t} y(s)dsdt, \\ a_{1} − a_{2} & = & −y(t_{1}) − ∫_{t_{1}}^{t_{2}} y(s)ds, \\ · · · · · · · · & ⋯ & · · · · · · · · \\ a_{i} − a_{i+1} & = & −y(t_{i}) − ∫_{t_{i}}^{t_{i+1}} y(s)ds, \\ · · · · · · · · & ⋯ & · · · · · · · · \\ a_{m−1} − a_{m} & = & −y(t_{m−1}) − ∫_{t_{m−1}}^{t_{m}} y(s)ds, \\ a_{m} − a_{m+1} & = & −y(t_{m}) − ∫_{t_{m}}^{T} y(s)ds. \end{matrix}  (2)

Let Δ denote the determinant of coefficients of system (2). It is easy to see that Δ = (m + T)(−1)^{m} ≠ 0. Therefore, system (2) has the unique solution (a_{1}, ⋯, a_{m+1}).

It is easy to verify that x(t) ∈ P C_{T}^{1} . Next we prove that Lx = y. Clearly, we have

x′(t) = y(t), t ≠ t_{k}, k = 1, 2, ⋯, m.

By (2), x(t_{i}) = a_{i} for i = 1, 2, ⋯, m, and

x(t_{i}^{+}) = a_{i+1} + ∫_{t_{i+1}}^{t_{i}} y(s)ds, i = 1, 2, ⋯, m − 1,

x(t_{m}^{+}) = a_{m+1} + ∫_{T}^{t_{m}} y(s)ds,

we have

x(t_{i}^{+}) − x(t_{i}) = a_{i+1} + ∫_{t_{i+1}}^{t_{i}} y(s)ds − a_{i} = y(t_{i}), i = 1, 2, ⋯, m − 1,

and

x(t_{m}^{+}) − x(t_{m}) = a_{m+1} + ∫_{T}^{t_{m}} y(s)ds − a_{m} = y(t_{m}).

This proves that Lx = y. Hence, we have KerQ ⊂ ImL.

Combining Case 1 and Case 2, we have ImL = KerQ.

It is easy to verify that Q^{2}y = Qy for y ∈ Y. Therefore Q is a continuous projector. Clearly, ImL is a closed subspace of Y, and Y = ImL ⊕ ImQ. Since dimKerL = 1 and codimImL = dimImQ = 1, L is a Fredholm mapping of

index zero. The proof is complete.

Define P = Q : P C_{T} → P C_{T} by

Qy = \frac{∑_{i=1}^{m} y(t_{i}) + ∫_{0}^{T} y(t)dt}{T + m}, y ∈ P C_{T} .

From the proof of Lemma 2, we know that P = Q are continuous projectors such that X = KerL ⊕ KerP and Y = ImL ⊕ ImQ. Consequently, the restriction L_{P} of L to domL∩KerP is one-to-one and onto ImL, so that its inverse K_{P} : ImL → domL∩KerP is defined. Further, we have the following result.

LEMMA 3. K_{P} : ImL → domL ∩ KerP is a compact mapping.

PROOF. For any y ∈ ImL, set

z(t) = \{ \begin{matrix}a_{1} + ∫_{t_{1}}^{t} y(s)ds, & t ∈ [0, t_{1}]; \\ a_{i+1} + ∫_{t_{i+1}}^{t} y(s)ds, & t ∈ (t_{i}, t_{i+1}], & i = 1, 2, ⋯, m − 1; \\ a_{m+1} + ∫_{T}^{t} y(s)ds, & t ∈ (t_{m}, T]. \end{matrix}

where a_{1}, ..., a_{m+1} are given by (2). It follows from the first equation of (2) that z(t) ∈ KerP. Thus, we have K_{P} y(t) = z(t). Therefore, by Lemma 1 and the definition of K_{P}, it is easy to see that K_{P} is a compact mapping. The proof is complete.

LEMMA 4. If the conditions of Theorem 1 hold, then there exists a constant M > 0, such that every solution x(t) of the problem

Lx = λNx, λ ∈ (0, 1)

satisfies ∥x∥ ≤ M.

PROOF. Let Lx = λNx for x(t) ∈ X, i.e.

\{ \begin{matrix}ẋ(t) = λg(t, x(t)) + λp(t), t ≠ t_{k}, & k = 1, 2, ⋯, m, \\ x(t_{k}^{+}) − x(t_{k}) = λI_{k}(t_{k}, x(t_{k})), & k = 1, 2, ⋯, m. \end{matrix}  (3)

Then we have

∫_{0}^{T} ẋ(t)dt = ∫_{0}^{t_{1}} ẋ(t)dt + ∑_{k=1}^{m−1} ∫_{t_{k}}^{t_{k+1}} ẋ(t)dt + ∫_{t_{m}}^{T} ẋ(t)dt

= x(t_{1}) − x(0) + ∑_{k=1}^{m−1}[x(t_{k+1}) − x(t_{k}^{+})] + x(T) − x(t_{m}^{+})

= x(t_{1}) + ∑_{k=1}^{m−1}[x(t_{k+1}) − x(t_{k}) − λI_{k}(t_{k}, x(t_{k}))] − x(t_{m}) − λI_{m}(t_{m}, x(t_{m}))

= −λ ∑_{k=1}^{m} I_{k}(t_{k}, x(t_{k}))

= λ ∫_{0}^{T} g(t, x(t))dt,

which implies that

∫_{0}^{T} g(t, x(t))dt + ∑_{k=1}^{m} I_{k}(t_{k}, x(t_{k})) = 0  (4)

Set E_{1} = {t ∈ [0, T] : x(t) ≥ c_{0}}, E_{2} = {t ∈ [0, T] : x(t) ≤ −c_{0}}, and E_{3} = {t ∈ [0, T] : |x(t)| < c_{0}}. From (4), we have

∫_{E_{2}} g(t, x(t))dt = − ∫_{E_{1}} g(t, x(t))dt − ∫_{E_{3}} g(t, x(t))dt − ∑_{k=1}^{m} I_{k}(t_{k}, x(t_{k}))

≥ −h_{0}T − γT − ∑_{k=1}^{m} c_{k}∥x∥ − ∑_{k=1}^{m} h_{k},

where

γ = max_{t∈[0,T],|x|≤c_{0}} |g(t, x)|.

Therefore, we have

∫_{0}^{T} |g(t, x(t))|dt = ∫_{E_{1}} |g(t, x(t))|dt + ∫_{E_{2}} |g(t, x(t))|dt + ∫_{E_{3}} |g(t, x(t))|dt

= ∫_{E_{1}} |g(t, x(t))|dt − ∫_{E_{2}} g(t, x(t))dt + ∫_{E_{3}} |g(t, x(t))|dt

≤ h_{0}T + (h_{0}T + γT + ∑_{k=1}^{m} h_{k}) + ∑_{k=1}^{m} c_{k}∥x∥ + γT.

That is

∫_{0}^{T} |g(t, x(t))|dt ≤ M_{1} + ∑_{k=1}^{m} c_{k}∥x∥  (5)

where

M_{1} = 2h_{0}T + 2γT + ∑_{k=1}^{m} h_{k}.

Next we make the following claims.

Claim 1. There exists τ ∈ [0, T], such that |x(τ)| ≤ c_{0}. Indeed, assume to the contrary that |x(t)| > c_{0} for any t ∈ [0, T]. Without loss of generality, assume that x(0) > 0. Then, by x(t) ∈ C[0, t_{1}], we have x(t) > 0 for any t ∈ [0, t_{1}]. By (3) and condition (iii) of Theorem 1, we know that x(t_{1}^{+}) > 0. Therefore, we have x(t) > 0 for any t ∈ [t_{1}, t_{2}]. By using simple induction, we can prove that x(t) > 0 for any t ∈ [0, T]. So, we have x(t) > c_{0} for any t ∈ [0, T]. Hence, by conditions (ii) and (iii) of Theorem 1, we have g(t, x(t)) > 0 for any t ∈ [0, T] and I_{k}(t_{k}, x(t_{k})) > 0 for k = 1, 2, ⋯, m. This is contrary to (4). The proof of Claim 1 is complete.

Claim 2. |x(t)| ≤ |x(τ)| + ∫_{0}^{T} |ẋ(t)|dt + ∑_{k=1}^{m} |I_{k}(t_{k}, x(t_{k}))| for any t ∈ [0, T]. Indeed, without loss of generality, we assume that t ∈ (t_{k}, t_{k+1}] and τ ∈ (t_{k+q}, t_{k+q+1}], 0 ≤ q ≤ m − k. Then we have

x(t) − x(τ)
= x(t) − x(t_{k+1}) + ∑_{j=1}^{q}[x(t_{k+j}) − x(t_{k+j+1})] + x(t_{k+q+1}) − x(τ)

= ∫_{t_{k+1}}^{t} ẋ(t)dt + ∑_{j=1}^{q}[x(t_{k+j}^{+}) − x(t_{k+j+1}) − λI_{k+j}(t_{k+j}, x(t_{k+j}))] + ∫_{τ}^{t_{k+q+1}} ẋ(t)dt

= ∫_{t_{k+1}}^{t} ẋ(t)dt − ∑_{j=1}^{q} ∫_{t_{k+j}}^{t_{k+j+1}} ẋ(t)dt − λ ∑_{j=1}^{q} I_{k+j}(t_{k+j}, x(t_{k+j})) + ∫_{τ}^{t_{k+q+1}} ẋ(t)dt

= ∫_{τ}^{t} ẋ(t)dt − λ ∑_{j=1}^{q} I_{k+j}(t_{k+j}, x(t_{k+j})).

Hence, we have

|x(t)| ≤ |x(τ)| + ∫_{0}^{T} |ẋ(t)|dt + ∑_{k=1}^{m} |I_{k}(t_{k}, x(t_{k}))|.

The proof of Claim 2 is complete.

On the other hand, by (3) and (5), we obtain

∫_{0}^{T} |ẋ(t)|dt ≤ λ ∫_{0}^{T} |g(t, x(t))|dt + λ ∫_{0}^{T} |p(t)|dt ≤ M_{1} + ∑_{k=1}^{m} c_{k}∥x∥ + ∫_{0}^{T} |p(t)|dt.

Therefore, we have

|x(t)| ≤ c_{0} + M_{1} + ∑_{k=1}^{m} c_{k}∥x∥ + ∫_{0}^{T} |p(t)|dt + ∑_{k=1}^{m} c_{k}∥x∥ + ∑_{k=1}^{m} h_{k}

Hence, we have

∥x∥ ≤ \frac{1}{1 − 2 ∑_{k=1}^{m} c_{k}} [c_{0} + M_{1} + ∑_{k=1}^{m} h_{k} + ∫_{0}^{T} |p(t)|dt].  (6)

The proof is complete by taking M to be the right hand side of (6).

We now turn to the proof of Theorem 1. Let r > max{M, c_{0}} with M as given in Lemma 4 and let Ψ = {x ∈ P C_{T} : ∥x∥ < r}. Define a map N : X → Y by

Nx(t) = \{ \begin{matrix}g(t, x(t)) + p(t), & t ≠ t_{k}, & k = 1, 2, ⋯, m, \\ I_{k}(t_{k}, x(t_{k})), & t = t_{k}, & k = 1, 2, ⋯, m. \end{matrix}

We assert that N is continuous. Indeed, assume that x_{n}, x ∈ P C_{T} such that x_{n} → x. Then there exists H > 0 such that ∥x_{n}∥ ≤ H and ∥x∥ ≤ H. Given any ε > 0, it follows from the uniform continuity of g, I_{1}, ..., I_{m} on [0, T] × {x ∈ R : |x| ≤ H} that there exists 0 < δ < ε such that |g(t, x_{1}) − g(t, x_{2})| < ε and |I_{k}(t, x_{1}) − I_{k}(t, x_{2})| < ε for any x_{1}, x_{2} ∈ {x ∈ R^{n} : |x| ≤ H} with |x_{1} −x_{2}| < δ. Again, for δ > 0, there exists a positive integer K such that, for n ≥ K, ∥x_{n} − x∥ ≤ δ. Since

Nx_{n}(t) − Nx(t) = \{ \begin{matrix}g(t, x_{n}(t)) − g(t, x(t)), & t ≠ t_{k}, \\ I_{k}(t_{k}, x_{n}(t_{k})) − I_{k}(t_{k}, x(t_{k})), & t = t_{k}, \end{matrix}

for k = 1, 2, ⋯, m, we have

|Nx_{n}(t) − Nx(t)| ≤ ε, n ≥ K, t ∈ [0, T].

Hence, we have

∥Nx_{n} − Nx∥ ≤ ε, n ≥ K.

This proves that N is continuous on P C_{T}. Therefore, PBVP (1) is equivalent to the operator equation

Lx = Nx, x ∈ domL.

We now apply Mawhin’s Continuation Theorem for Ψ. In view of Lemma 3, N is L-compact on Ψ̄. By Lemma 4, it is easy to see that Lx ≠ λNx for any (x, λ) ∈ (domL ∩ ∂Ψ) × (0, 1). Note that for x ∈ ∂Ψ ∩ KerL = ∂Ψ ∩ R, we must have x = r or x = −r. Therefore, for such an x,

QNx = \frac{∑_{i=1}^{m} I_{i}(t_{i}, x) + ∫_{0}^{T} g(t, x)dt}{T + m} ≠ 0.

Let

Φ(x, μ) = μx + (1 − μ)\frac{∑_{i=1}^{m} I_{i}(t_{i}, x) + ∫_{0}^{T} g(t, x)dt}{T + m}, μ ∈ [0, 1].

Then, for any x ∈ ∂Ψ ∩ KerL, μ ∈ [0, 1], we have

xΦ(x, μ) = μx^{2} + (1 − μ)\frac{∑_{i=1}^{m} xI_{i}(t_{i}, x) + ∫_{0}^{T} xg(t, x)dt}{T + m} > 0, μ ∈ [0, 1],

which implies that Φ(x, μ) is a homotopy. Therefore, by the property of invariance under a homotopy of coincidence degree, we have

deg_{B}(ĴQN, Ψ ∩ KerL, 0) = deg_{B}(Φ(x, 0), Ψ ∩ KerL, 0)
= deg_{B}(Φ(x, 1), Ψ ∩ KerL, 0)
= deg_{B}(x, Ψ ∩ KerL, 0) = 1,

where the isomorphism Ĵ of ImQ onto KerL is the identity mapping, since ImQ = KerL. It follows from Mawhin’s Continuation Theorem that Lx = Nx has at least one solution in domL ∩ Ψ̄. Therefore, PBVP (1) has at least one solution. The proof is complete.

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