Positive Periodic Solution for Second-Order Singular Semipositone Differential Equations

Definition 1. We say that (4) admits the antimaximum principle if (6) has a unique T-periodic solution for any l ∈ ℂ(ℝ/Tℤ) and the unique T-periodic solution x_l(t) > 0 for all t if l≻0.

Lemma 2 (see [14], Corollary 2.6.)Assume that a≢0 and the following two inequalities are satisfied:

where [a(s)] ₊ = max {a(s), 0}. Then the Green’s function G(t, s), associated with (5), is positive for all (t, s)∈[0, T]×[0, T].

Lemma 3. Assume Ω is an open subset of a convex set K in a normed linear space X and p ∈ Ω. Let $$ be a compact and continuous map. Then one of the following two conclusions holds:

• (I)

  T has at least one fixed point in $$.

• (II)

  There exists x ∈ ∂Ω and 0 < λ < 1 such that x = λTx + (1 − λ)p.

Theorem 4. Suppose that (4) satisfies (A) and

Furthermore, assume that there exist three constants M, R₀, r > Mw^*/ι such that:

• (H₁)

  F(t, x, y) = f(t, x, y) + M ≥ 0 for all (t, x, y)∈[0, T]×(0, r] × R.

• (H₂)

  f(t, x, y) ≥ g₀(x) for (t, x, y)∈[0, T]×(0, R₀] × R, where the nonincreasing continuous function g₀(x) > 0 satisfies $$ and $$.

• (H₃)

  0 ≤ F(t, x, y)≤(g(x) + h(x))ϱ(|y|),  for  all  (t, x, y)∈[0, T]×(0, r] × R, where g(·) > 0 is nonincreasing in (0, r] and h(·)/g(·) ≥ 0,  ϱ(·) ≥ 0 are nondecreasing in (0, r].

• (H₄)
  $$ ()

•  

  where

  $$ ()

•  

  Then (1) has at least one positive periodic solution u(t) with 0 < ∥u + Mw∥≤r.

Proof. For convinence, let us write Z(t) = x(t) − Mw(t), Z_n(t) = x_n(t) − Mw(t), where w(t) = (ℒ1)(t). Let

First we show that has a solution x satisfying (5), 0 < ∥x∥≤r and Z(t) > 0 for t ∈ [0, T]. If this is true, it is easy to see that Z(t) will be a positive solution of (1)–(5) with 0 < ∥Z + Mw∥≤r.

Choose n₀ ∈ {1,2, …} such that 1/n₀ < r, and then let N₀ = {n₀, n₀ + 1 + ⋯}.

Consider the family of equations

where λ ∈ [0,1],  n ∈ N₀,  x ∈ B_r = {x : ∥x∥<r} and F_n(t, x, y) = F(t, max {1/n, x}, y).

A T-periodic solution of (17) is just a fixed of the operator equation

where p = 1/n and T_n is a completely continuous operator defined by where we have used the fact

We claim that for any T-periodic solution x_n(t) of (17) satisfies

Note that the solution x_n(t) of (17) is also satisfies the following equivalent equation Integrating (22) from 0 to T, we obtain

By the periodic boundary conditions, we have x^′(t₀) = 0 for some t₀ ∈ [0, T]. Therefore,

where we have used the assumption (11) and ∥x_n∥<r. Therefore, which implies that (21) holds. In particular, let λF_n(t, Z(t), Z^′(t)) + a(t)/n = 1 in (17), we have

Choose n₁ ∈ N₀ such that 1/n₁ ≤ R₁, and then let N₁ = {n₁, n₁ + 1, …}. The following lemma holds.

Lemma 5. There exists an integer n₂ > n₁ large enough such that, for all n ∈ N₂ = {n₂, n₂ + 1, …},

Proof. The lower bound in (27) is established by using the strong force condition of f(t, x, y). By condition (H₂), there exists R₁ ∈ (0, R₀) and a continuous function $$ such that

for all (t, x, y)∈[0, T]×(0, R₁] × R, where $$ satisfies also the strong force condition like in (H₂).

For n ∈ N₁, let α_n = min _0≤t≤TZ_n(t),  β_n = max _0≤t≤TZ_n(t).

If α_n ≥ R₁, due to n ∈ N₁, (27) holds.

If α_n < R₁, we claim that, for all n ∈ N₁,

Otherwise, suppose that β_n ≤ R₁ for some n ∈ N₁. Then it is easy to verify In fact, if 1/n ≤ Z_n(t) ≤ R₁, we obtain from (28) and, if Z_n(t) ≤ 1/n, we have Integrating (22) (with λ = 1) from 0 to T, we deduce that where estimation (30) and the fact ∥x_n∥<r are used. This is a contradiction. Hence (29) holds.

Due to α_n < R₁, that is, α_n = min _0≤t≤T[x_n(t) − Mw(t)] = x_n(a_n) − Mw(a_n) < R₁ for some a_n ∈ [0, T]. By (29), there exists c_n ∈ [0, T] (without loss of generality, we assume a_n < c_n.) such that x_n(c_n) = Mw(c_n) + R₁ and x_n(t) ≤ Mw(t) + R₁ for a_n ≤ t ≤ c_n.

It can be checked that

where $$ is defined by (15).

In fact, if t ∈ [a_n, c_n] is such that 1/n ≤ Z_n(t) ≤ R₁, we have

and, if t ∈ [a_n, c_n] is such that Z_n(t) ≤ 1/n, we have So (34) holds.

Using (17) (with λ = 1) for x_n(t) and the estimation (34), we have, for t ∈ [a_n, c_n]

As $$, $$ for all t ∈ [a_n, c_n], so Z_n(t) is strictly increasing on [a_n, c_n]. We use ξ_n to denote the inverse function of Z_n restricted to [a_n, c_n].

Suppose that (27) does not hold, that is, for some n ∈ N₁, Z_n(t) < 1/n < R₁. Then there would exist b_n ∈ (a_n, c_n) such that Z_n(b_n) = 1/n and

Multiplying (17) (with λ = 1) by $$ and integrating from b_n to c_n, we obtain

By the facts ∥x_n∥ < r,  $$,  ∥w^′∥ ≤ r and the definition of Z_n(t), we can obtain $$, together with ∥x_n∥<r, implies that the second term and the third term are bounded. The first term is

which is also bounded. As a consequence, there exists a B₁ > 0 such that On the other hand, by (H₂), we can choose n₂ ∈ N₁ large enough such that for all n ∈ N₂ = {n₂, n₂ + 1, …}. So (27) holds.

Lemma 6. There exist a constant δ > 0 such that, for all n ∈ N₂,

Proof. Multiplying (17) (with λ = 1) by $$ and integrating from a_n to c_n, we obtain

In the same way as in the proof of (41), one way readily prove that the right-hand side of the above equality is bounded. On the other hand, if n ∈ N₂, by (H₂),

if α_n → 0₊. Thus we know that there exists a constant δ > 0 such that α_n ≥ δ. Hence (43) holds.

Next, we will prove (17) has periodic solution x_n(t).

For ιr > 0, we can choose n₃ ∈ N₂ such that 1/n₃ < ιr, which together with (H₄) imply

Let N₃ = {n₃, n₃ + 1, …}. For n ∈ N₃, consider (17).

Next we claim that any fixed point x_n of (18) for any λ ∈ [0,1] must satisfy ∥x_n∥≠r. So, by using the Leray-Schauder alternative principle, (17) (with λ = 1) has a periodic solution x_n(t). Otherwise, assume that x_n is a fixed point x_n of (18) for some λ ∈ [0,1] such that ∥x_n∥ = r. Note that

For n ∈ N₃, we have By (27) and assumption (H₃), for all t ∈ [0, T] and n ∈ N₃, we have Therefore, This is a contradiction to the choice of n₃ and the claim is proved.

The fact ∥x_n∥<r and $$ show that $$ is a bounded and equicontinuous family on [0, T]. Now Arzela-Ascoli Theorem guarantees that $$ has a subsequence $$, converging uniformly on [0, T] to a function x ∈ C [0, T]. From the fact ∥x_n∥<r and x_n(t) > δ, x satisfies δ ≤ x(t) ≤ r for all t. Moreover, $$ satisfies the integral equation

Letting k → ∞, we arrive at where the uniform continuity of F(t, x, y) on [0, T]×[δ, r]×[−(r + M)L, (r + M)L] is used. Therefore, x is a positive periodic solution of (16) and Z(t) = x(t) − Mw(t) ≥ δ. Thus we complete the prove of Theorem 4.

Corollary 7. Let the nonlinearity in (1) be

where α > 1,  β > 0,  1 > γ ≥ 0,  μ > 0 is a positive parameter, e(t) is a T-periodic function.

• (i)

  If β + γ < 1, then (1) has at least one positive periodic solution for each μ > 0.

• (ii)

  If β + γ ≥ 1, then (1) has at least one positive periodic solution for each 0 < μ < μ₁, where μ₁ is some positive constant.

Proof. We will apply Theorem 4 with M = max _0≤t≤T | e(t)| and g(x) = x^−α,  h(x) = μx^β + 2M,  ϱ(y) = 1+|y|^γ. Then condition (H₁)–(H₃) are satisfied and existence condition (H₄) becomes

So (1) has at least one positive periodic solution for Note that μ₁ = ∞ if β + γ < 1 and μ₁ < ∞ if β + γ ≥ 1. We have the desired results.