Existence of Unbounded Solutions for a Third-Order Boundary Value Problem on Infinite Intervals

Definition 2.1. A function α ∈ C²[0, +∞)∩C³(0, +∞) is called a lower solution of BVP (1.3) if

Similarly, a function β ∈ C²[0, +∞)∩C³(0, +∞) is called an upper solution of BVP (1.3) if

Definition 2.2. Given a positive function ϕ ∈ C(0, +∞) and a pair of functions α, β ∈ C¹[0, +∞) satisfying α(0) ≤ β(0) and α′(t) ≤ β′(t), t ∈ [0, +∞); a function f : [0, +∞) × R³ → R is said to satisfy the Nagumo condition with respect to the pair of functions α, β, if there exist positive functions ψ, h ∈ C[0, +∞) satisfying $$ such that

holds for all 0 ≤ t < +∞, α(t) ≤ x ≤ β(t),   α′(t) ≤ y ≤ β′(t), and z ∈ ℝ.

Let v₀(t) = 1 + t², v₁(t) = 1 + t, v₂(t) = 1 and consider the space X defined by

with the norm ∥x∥ = max {∥x∥₀, ∥x′∥₁, ∥x^′′∥₂}, where ∥x∥_i = sup _t∈[0,+∞)|x(t)/v_i(t)|. By the standard arguments, we can prove that (X, ∥·∥) is a Banach space.

Lemma 2.3. If e ∈ L¹[0, +∞), then the BVP of third-order linear differential equation

has a unique solution in X. Moreover this solution can be expressed as where

Proof. It is easy to verify that (2.6) satisfies BVP (2.5). Now we show the uniqueness. Suppose u is a solution of (2.5). Let v = u′, then we have

By a direct calculation, we obtain the general solution of the above equation: Substituting this to the boundary condition, we arrive at Therefore, (2.8) has a unique solution where Furthermore, u′ = v, u(0) = A, so The proof is complete.

Theorem 2.4 (see [1].)Let M ⊂ C_∞ = {x ∈ C[0, +∞), lim _t→+∞x(t)  exists}. Then M is relatively compact if the following conditions hold:

• (a)

  all functions from M are uniformly bounded;

• (b)

  all functions from M are equicontinuous on any compact interval of [0, +∞);

• (c)

  all functions from M are equiconvergent at infinity; that is, for any given ϵ > 0, there exists a T = T(ϵ) > 0 such that |f(t) − f(+∞)| < ϵ, for all t > T and f ∈ M.

From the above results, we can obtain the following general criteria for the relative compactness of subsets in C[0, +∞).

Theorem 2.5. Given n + 1 continuous functions ρ_i satisfying ρ_i ≥ ε > 0, i = 0,1, …, n with ε a positive constant. Let $$. Then M is relatively compact if the following conditions hold:

• (a)

  all functions from M are uniformly bounded;

• (b)

  the functions from {y_i : y_i = ρ_ix⁽ⁱ⁾, x ∈ M} are equicontinuous on any compact interval of [0, +∞), i = 0,1, 2, …, n;

• (c)

  the functions from {y_i : y_i = ρ_ix⁽ⁱ⁾, x ∈ M} are equiconvergent at infinity, i = 0,1, 2, …, n.

Proof. Set M_i = {y_i : y_i = ρ_ix⁽ⁱ⁾, x⁽ⁱ⁾ ∈ M}, then M_i ⊂ C_∞, i = 0,1, 2 … , n. From conditions (a)–(c), we have M_i is relatively compact in C_∞. Therefore, for any sequence $$, it has a convergent subsequence. Without loss of generality, we denote it this sequence. Then there exists y_i,0 ∈ M_i such that

Set x_i,0 = (1/ρ_i)  y_i,0, then $$. Noticing that all functions from M are uniformly continuous, we can obtain that $$. So M is relatively compact.

Lemma 3.1. Suppose condition (H₁) holds. And suppose further that the following condition holds:

• (H₃)

  there exists a constant γ > 1 such that sup _0≤t<+∞(1 + t) ^γϕ(t)ψ(t)<+∞.

If u is a solution of (1.3) satisfying

then there exists a constant R > 0 (without relations to u) such that ∥u^′′∥₂ ≤ R.

Proof. Let δ > 0 and R > C,

such that where C is the nonhomogeneous boundary value, M = sup _0≤t<+∞(1 + t) ^γϕ(t)ψ(t), then |u^′′(t) | ≤ R, t ∈ [0, +∞). If it is untrue, we have the following three cases.

Case 1. Consider

Without loss of generality, we suppose u^′′(t) > η, t ∈ [0, +∞). While for any T ≥ δ, which is a contraction. So there must exist t₀ ∈ [0, +∞) such that |u^′′(t₀)| ≤ η.

Case 2. Consider

Just take R = η and we can complete the proof.

Case 3. There exists [t₁, t₂]⊂[0, +∞) such that |u^′′(t₁)| = η, |u^′′(t)| > η, t ∈ (t₁, t₂] or |u^′′(t₂)| = η, |u^′′(t)| > η, t ∈ [t₁, t₂).

Suppose that u^′′(t₁) = η, u^′′(t) > η, t ∈ (t₁, t₂]. Obviously,

concludes that u^′′(t₂) ≤ R. For t₁ and t₂ are arbitrary, we have u^′′(t) ≤ max {R, η} = R, t ∈ [0, +∞).

Similarly if u^′′(t₁) = −η, u^′′(t)<−η, t ∈ (t₁, t₂], we can also obtain that u^′′(t)>−R, t ∈ [0, +∞).

Thus there exists R > 0, just related with α, β, and h, such that ∥u^′′∥₂ ≤ R.

Remark 3.2. Condition (H₃) is necessary for an a priori estimation of u^′′ in Lemma 3.1. Because the upper and lower solutions are in X, β′(t) and α′(t) are at most linearly increasing, especially at infinity. Otherwise, sup _t∈[0,+∞)α′(t) and sup _t∈[0,+∞)β′(t) may be equal to infinity.

Theorem 3.3. Suppose ϕ ∈ L¹[0, +∞) and the conditions (H₁)–(H₃) hold. Then BVP (1.3) has at least one solution u ∈ C²[0, +∞)∩C³(0, +∞) such that

Proof. Let R > 0 be the same definition in Lemma 3.1 and consider the boundary value problem

where Firstly we prove that BVP (3.10) has at least one solution u. To this end, define the operator T : X → X by By Lemma 2.3, we can see that the fixed points of T coincide with the solutions of BVP (3.10). So it is enough to prove that T has a fixed point.

We claim that T : X → X is completely continuous.

(1)  T : X → X is well defined. For any u ∈ X, ∥u∥<+∞ and it holds

where H₀ = max _{0≤s≤∥u∥}h(s). By the Lebesgue-dominated convergence theorem, we have so Tu ∈ X.

(2)  T : X → X is continuous. For any convergent sequence u_n → u in X, there exists r₁ > 0 such that sup _n∈N∥u_n∥≤r₁. Similarly, we have

so T : X → X is continuous.

(3)  T : X → X is compact. Let B be any bounded subset of X, then there exists r > 0 such that ∥u∥≤r,   for  all  u ∈ B. For any u ∈ B, one has

where H_r = max _0≤s≤rh(s), so TB is uniformly bounded. Meanwhile, for any T > 0, if t₁, t₂ ∈ [0, T], we have that is, TB is equicontinuous. From Theorem 2.5, we can see that if TB is equiconvergent at infinity, then TB is relatively compact. In fact, Then we can obtain that T : X → X is completely continuous.

By the Schäuder fixed point theorem, T has at least one fixed point u ∈ X. Next we will prove u satisfying α′(t) ≤ u′(t) ≤ β′(t), t ∈ [0, +∞). If u′(t) ≤ β′(t), t ∈ [0, +∞) does not hold, then,

Because u^′′(+∞) − β^′′(+∞) < 0, so there are two cases.

Case 1. Consider

Easily, u^′′(0⁺) − β^′′(0⁺) ≤ 0. While by the boundary condition, we have which is a contraction.

Case 2. There exists t^* ∈ (0, +∞) such that

So, u^′′(t^*) − β^′′(t^*) = 0, u^′′′(t^*) − β^′′′(t^*) ≤ 0. Unfortunately,

Consequently, u′(t) ≤ β′(t) holds for all t ∈ [0, +∞). Similarly, we can show that α′(t) ≤ u′(t) for all t ∈ [0, +∞). Noticing that α(0) ≤ A ≤ β(0), from the inequality α′(t) ≤ u′(t) ≤ β′(t), we can obtain that α(t) ≤ u(t) ≤ β(t). Lemma 3.1 guarantee that ∥u^′′∥_∞R. So,

that is, u is a solution of BVP (1.3).

Remark 3.4. For finite interval problem, it is sharp to define the lower and upper solutions satisfying α^′′(b) ≤ C and β^′′(b) ≥ C; see [15].

If f : [0, +∞) ⁴ → [0, +∞), we can establish a criteria for the existence of positive solutions.

Theorem 3.5. Let f : [0, +∞) ⁴ → [0, +∞) be continuous and ϕ ∈ L¹[0, +∞). Suppose the condition (H₂) holds and the following conditions hold.

• (P₁)

  BVP (1.3) has a pair of positive upper and lower solutions α, β ∈ X satisfying

  $$ (3.25)

• (P₂)

  For any r > 0, there exists φ_r satisfying $$ such that

  $$ (3.26)
  holds for all t ∈ [0, +∞), α(t) ≤ x ≤ β(t), α′(t) ≤ y ≤ β′(t), and 0 ≤ z ≤ r.

Then BVP (1.3) with A, B, C ≥ 0 has at least one solution such that

Proof. Choose R = (1/a)(B + β′(0)) and consider the boundary value problem (3.10) except f_R substituting by

Similarly, we can obtain that (3.10) has at least one solution u satisfying α(t) ≤ u(t) ≤ β(t) and α′(t) ≤ u′(t) ≤ β′(t), t ∈ [0, +∞). Because and u^′′(+∞) = C ≥ 0, we have Consequently, the solution u is a positive solution of (1.3).