Existence of Solutions of Nonlinear Mixed Two-Point Boundary Value Problems for Third-Order Nonlinear Differential Equation

Definition 2.1. Suppose the functions α(x) and β(x) ∈ C⁽³⁾[a, b] satisfy

Then α(x) and β(x) are, respectively, called the lower and upper solutions of the BVP (1.3).

Definition 2.2. Let C[D × ℝ, ℝ] denote the class of continuous functions from D × ℝ into ℝ, and let f(x, y, y′, y^′′) ∈ C[D × ℝ, ℝ] and α(x), β(x) ∈ C⁽³⁾[a, b] be lower and upper solutions of BVP (1.3). Suppose that there is a function W(s) ∈ C[ℝ₊, (0, +∞)] such that

for every (x, y, y′, y^′′) ∈ D × ℝ, where with Then we say that f satisfies Nagumo’s condition on the set D relative to α(x),   β(x).

Lemma 2.3. The boundary value problem

has a Green function with

Lemma 2.4. Assume that (H₁),   (H₂) hold. Then for any solution y of y^′′′ = f(x, y, y′, y^′′) with α(x) ≤ y(x) ≤ β(x),   α′(x) ≤ y′(x) ≤ β′(x) on [a, b], there exists a constant N > 0 depending only on α,   β,   W, such that

and one calls N is Nagumo’s constant.

Lemma 2.5. Assume that (H₁)–(H₃) hold. Then for any constant A ∈ [α(a), β(a)],   B ∈ [α′(a), β′(a)],   C ∈ [α′(b), β′(b)], the boundary value problem

at least has a solution y ∈ C⁽³⁾[a, b], with on [a, b].

Proof. By Lemma 2.3, it is clear that BVP (2.12) is equivalent to integral equation

where W(x) is a polynomial satisfying y^′′′ = 0,    y(a) = A,   y^′(a) = B,   y′(b) = C.

Denote

where N is Nagumo’s constant.

Assume

Then F(x, y, y′, y^′′) is bounded and continuous on [a, b] × ℝ³. Suppose |F(x, y, y′, y^′′)| ≤ M; |W⁽ⁱ⁾(x)| ≤ K  (i = 0,1, 2), x ∈ [a, b].

Now, define an operator T on the set E = C⁽²⁾[[a, b], ℝ] by

If y ∈ E, the norm is defined by It is clear that

This shows that T maps the closed, bounded, and convex set

into itself. Also, T is continuous and (Ty)′ is bounded. All of these considerations imply that T is completely continuous by Ascoli’s theorem. Schauder’ fixed point theorem then yields the fixed point y of T on B. In other words, the following boundary value problem has a solution y ∈ C⁽³⁾[a, b], and satisfying α′(x) ≤ y′(x) ≤ β′(x) and α(a) ≤ y(a) ≤ β(a), we have α(x) ≤ y(x) ≤ β(x), x ∈ [a, b].

In the following we prove that

In fact, if it is invalid, there is no harm in setting the right inequality to be not true (the case that the left inequality is not true can be proved in the same way). By the assumption, if y′(x) > β′(x), for some x ∈ [a, b], then there is a x₀ ∈ (a, b) such that

Now, let

They imply that α(x₀) ≤ ω(x₀) ≤ β(x₀) and We have which contradicts (2.24); hence, (2.22) is true.

Further, by the definition of F,   y is a solution of the boundary value problem

Because there is a ξ ∈ [a, b], such that

so [a, b] has a maximal subinterval [c, d] with interior point ξ, for any x ∈ [c, d], |y^′′(x)| ≤ m. Hence, for x ∈ [c, d], y(x) is the solution of BVP (2.12). And by Lemma 2.4, we have |y^′′(x)| ≤ N < m; this contracts that [c, d] is the maximal subinterval, so we know [c, d] = [a, b]. Consequently, y is a solution of BVP (2.12).

Theorem 3.1. Assume (H₁)–(H₄),   (H₆) hold, then BVP

has a solution y ∈ C⁽³⁾[a, b], satisfying on [a, b].

Proof. By Lemma 2.5, we know that the boundary value problem

has a solution y ∈ C⁽³⁾[a, b], with on [a, b]. For any A ∈ [α(a), β(a)],   B ∈ [α′(a), β′(a)].

For fixed A, if B = α′(a), then y^′′(a) ≥ α^′′(a),   y^′′(b) ≤ α^′′(b). By (H₆), we know

On the other hand, if B = β′(a), then y^′′(a) ≤ β^′′(a),   y^′′(b) ≥ β^′′(b). By (H₆), we have

Define the following sets:

Obviously, Π(y) is nonempty. If the theorem is not true, we know that M₁ and M₁ are all nonempty, and M₁ ⋃ M₂ = [α′(a), β′(a)]. we claim that M₁ is closed. To see this, let B_n ∈ M₁, with B_n → B₀  (n → ∞). Consider the following boundary value problem: By Lemma 2.5, it is known that, for every n ∈ N, BVP (3.8) has a solution y_n(x) ∈ C⁽³⁾[a, b], satisfying and, by Lemma 2.4, we know $$.

Clearly, sequences {y_n(x)}, $$, $$ are uniformly bounded and equicontinuous on [a, b]. Consequently, there exists a subsequence of y_n(x) which converges uniformly on [a, b], to a solution y₀(x) of the BVP:

with By assumption, equality cannot occur, so that and thus B₀ ∈ M₁. Consequently, M₁ is closed. Likewise, we may show M₂ is closed. This is a contradiction and proves the theorem.

Theorem 3.2. Assume (H₁)–(H₆) hold, then BVP

has a solution y ∈ C⁽³⁾[a, b], satisfying on [a, b].

Example 4.1. Consider the following linear boundary value problem

It is easy to know that α(x) = −xe^x − 1,   β(x) = xe^x + 1 are lower and upper solutions of the linear boundary value problem, respectively, where

and all assumptions of Theorem 3.2 hold. So the linear boundary value problem has a solution y(x) satisfying Obviously, the trivial solution of the linear boundary value problem is one.

Example 4.2. Consider nonlinear boundary value problem

It is easy to verify that α(x) = −sin (x),   β(x) = 0 are lower and upper solutions of the nonlinear boundary value problem, respectively, where

and all assumptions of Theorem 3.2 hold, so the BVP has a solution y(x) satisfying