Iterative technique for a
third-order differential equation with
three-point nonlinear boundary value conditions
Xiuli Lin and Zengqin Zhao
BSchool of Mathematical Sciences, Qufu Normal University, Qufu 273165, People’s Republic of China Received 3 October 2015, appeared 18 March 2016
Communicated by Nickolai Kosmatov
Abstract. In this paper, we study the existence of extremal solutions for a nonlinear third-order differential equation with three-point nonlinear boundary value conditions.
By means of the method of upper and lower solutions and different monotone iterative techniques, the sufficient conditions which guarantee the existence of extremal solutions are given. An example illustrates the main results.
Keywords: third-order differential equation, nonlinear boundary value conditions, the method of upper and lower solutions, iterative technique.
2010 Mathematics Subject Classification: 34B15.
1 Introduction
Nonlinear boundary value conditions in differential equations can describe many phenom- ena in applied mathematics, engineering, physical or biological processes. In this paper, we consider the following third-order differential equation with three-point nonlinear boundary value conditions
−u000(t) = f(t,u(t)), t ∈[0, 1], u(0) =u00(0) =0,
p(u(1),u(ξ)) =0,
(1.1)
whereξ ∈(0, 1), f :[0, 1]×R→ Randp: R×R→ Rare continuous.
In recent years, third-order differential equations with nonlinear boundary value condi- tions have been discussed in many papers (see [1–15] and the references therein). For exam- ple, [4] considered a class of two-point nonlinear boundary value conditions by using a priori estimate, Nagumo condition, upper and lower solutions and Leray–Schauder degree. Pa- pers [7–9,13] considered some nonlinear nonlocal boundary conditions. However, according to our knowledge, for third-order differential equation, the three-point nonlinear boundary
BEmails: lin-xiuli78@163.com (Xiuli Lin), zqzhaoy@163.com (Zengqin Zhao)
value conditions in problem (1.1) are never discussed in literature. Hence the aim of this paper is to discuss this issue.
The main contributions are as follows: (a) we present problems with linear boundary value conditions, and on this basis we obtain the existence of the extremal solutions for problem (1.1) by applying the method of upper and lower solutions and monotone iterative technique;
(b) the iterative technique is not unique and an example illustrates the result.
2 Notations and preliminaries
In this section, we present some definitions and lemmas that will be used throughout the paper.
Definition 2.1. Assume ξ ∈ (0, 1), f : [0, 1]×R → R is continuous. A function u(t)is called an upper solution for problem (1.1) if it satisfies
u000(t) + f(t,u(t))≤0, t∈[0, 1], u(0) =u00(0) =0,
p(u(1),u(ξ))≥0.
Similarly, a functionu(t)is called a lower solution for problem (1.1) if it satisfies
u000(t) + f(t,u(t))≥0, t∈[0, 1], u(0) =u00(0) =0,
p(u(1),u(ξ))≤0.
Lemma 2.2. Assume thatαξ 6= 1, b∈ R and h:[0, 1]→R is continuous. Then the boundary value problem
−u000(t) =h(t), t∈ [0, 1], u(0) =u00(0) =0,
u(1) =αu(ξ) +b, has a unique solution
u(t) = b
1−αξt+ αt 1−αξ
Z 1
0 G(ξ,s)h(s)ds+
Z 1
0 G(t,s)h(s)ds, where
G(t,s) = 1 2
((1−t)(t−s2), 0≤s≤t ≤1,
t(1−s)2, 0≤t≤s ≤1. (2.1) Proof. Integrating the equation
−u000(t) =h(t), t∈[0, 1], over[0,t]for three times, we have
u(t) =−1 2
Z t
0
(t−s)2h(s)ds+ 1
2u00(0)t2+u0(0)t+u(0).
Due to the boundary conditionsu(0) =u00(0) =0,u(1) =αu(ξ) +b, it follows that
−1 2
Z 1
0
(1−s)2h(s)ds+u0(0) =αu(ξ) +b.
And then
u(t) =t(αu(ξ) +b) + 1 2t
Z 1
0
(1−s)2h(s)ds−1 2
Z t
0
(t−s)2h(s)ds
=t(αu(ξ) +b) + 1 2
Z t
0 t(1−s)2h(s)ds− 1 2
Z t
0
(t−s)2h(s)ds+1 2
Z 1
t t(1−s)2h(s)ds
=t(αu(ξ) +b) + 1 2
Z t
0
(1−t)(t−s2)h(s)ds+ 1 2
Z 1
t t(1−s)2h(s)ds
=t(αu(ξ) +b) +
Z 1
0 G(t,s)h(s)ds. (2.2)
Puttingt= ξ, we have
u(ξ) = 1 1−αξ
ξb+
Z 1
0 G(ξ,s)h(s)ds
. Substituting it into (2.2), we get
u(t) = b
1−αξt+ αt 1−αξ
Z 1
0
G(ξ,s)h(s)ds+
Z 1
0
G(t,s)h(s)ds.
Remark 2.3. It is easy to see thatG(t,s)>0 for all (t,s)∈(0, 1)×(0, 1)and u0(t) = b
1−αξ + α 1−αξ
Z 1
0 G(ξ,s)h(s)ds+
Z 1
0 G0(t,s)h(s)ds, (2.3) where
G0(t,s) = 1 2
(1−2t+s2, 0≤s≤t ≤1,
(1−s)2, 0≤t≤s ≤1. (2.4)
Remark 2.4. For (t,s)∈[0, 1]×[0, 1],
0≤ t(1−t)ϕ(s)≤G(t,s)≤ ϕ(s), (2.5) where ϕ(s) = 18(1+s)2(1−s)2.
In fact, fort ∈[s, 1],G(·,s) = 12(1−t)(t−s2)attains its maximum att= 12(1+s2)∈ [s, 1], so that
G(·,s)≤ ϕ(s) = 1
8(1+s)2(1−s)2.
For t∈[0,s], clearlyG(·,s)≤ 12s(1−s)2≤ ϕ(s) = 18(1+s)2(1−s)2. That is,G(t,s)≤ ϕ(s)for (t,s)∈[0, 1]×[0, 1].
Also, fors∈[0,t], G(t,s)
ϕ(s) =
1
2(1−t)(t−s2)
1
8(1+s)2(1−s)2 ≥ 4(1−t)(t−st)
(1+s)2(1−s)2 = 4t(1−t)
(1+s)2(1−s) ≥t(1−t).
If fors∈ [t, 1],
G(t,s) ϕ(s) =
1
2t(1−s)2
1
8(1+s)2(1−s)2 = 4t
(1+s)2 ≥t.
Since min{t,t(1−t)}= t(1−t), we have
0≤t(1−t)ϕ(s)≤ G(t,s)≤ ϕ(s).
Lemma 2.5. Assume that0<αξ <1.If u(t)∈C[0, 1], satisfying u000(t)∈C[0, 1]and
u000(t)≤0, t∈ [0, 1], u(0) =u00(0) =0, u(1)≥αu(ξ), then u(t)≥0, t∈[0, 1].
Proof. Let−u000(t) =h(t)≥0, t∈[0, 1]andu(1) =αu(ξ) +b, b≥0. Then by Lemma2.2, we get
u(t) = b
1−αξt+ αt 1−αξ
Z 1
0
G(ξ,s)h(s)ds+
Z 1
0
G(t,s)h(s)ds
≥0.
Lemma 2.6. Assume thatκ(t),µ(t)∈C[0, 1]and Z 1
0
(1+s)2(1−s)2|κ(s)|ds< 8(1−αξ)
α+1−αξ. (2.6)
Then the following linear boundary value problem
−u000(t) =κ(t)u(t) +µ(t), t∈[0, 1], u(0) =u00(0) =0,
u(1) =αu(ξ) +b,
(2.7)
has a unique solution u(t)∈ C[0, 1], where0< αξ <1,b≥0.
Proof. By Lemma2.2, problem (2.7) is equivalent to the following integral equation u(t) = b
1−αξt+ αt 1−αξ
Z 1
0 G(ξ,s) (κ(s)u(s) +µ(s))ds +
Z 1
0 G(t,s) (κ(s)u(s) +µ(s))ds=: Tu(t).
Obviously,T:C[0, 1]−→C[0, 1]. Note that by (2.5), we have for anyu,v∈C[0, 1],
|Tu(t)−Tv(t)| ≤ αt 1−αξ
Z 1
0 G(ξ,s)|κ(s)||u(s)−v(s)|ds+
Z 1
0 G(t,s)|κ(s)||u(s)−v(s)|ds
≤ 1 8
1+ αt 1−αξ
ku−vk
Z 1
0
(1+s)2(1−s)2|κ(s)|ds
≤ 1 8
α+1−αξ 1−αξ
ku−vk
Z 1
0
(1+s)2(1−s)2|κ(s)|ds=Lku−vk,
where L := 18 α+1−1−αξ
αξ
R1
0(1+s)2(1−s)2|κ(s)|ds < 1, which is easy to see from (2.6). There- fore, the operator Tis a contraction map in the spaceC[0, 1]andThas a unique fixed point in C[0, 1].
Lemma 2.7. Assume that0 < αξ < 1, κ(t)(∈ C[0, 1]) > 0 and satisfies (2.6). If u(t) ∈ C[0, 1], satisfying u000(t)∈ C[0, 1]and
u000(t) +κ(t)u(t)≤0, t ∈[0, 1], u(0) =u00(0) =0,
u(1)≥αu(ξ) then u(t)≥0, t∈ [0, 1].
Proof. Let−u000(t) =κ(t)u(t) +µ(t),t ∈[0, 1],κ(t)>0,µ(t)≥0,u(1) =αu(ξ) +b, b≥0.
Suppose that the inequality u(t) ≥ 0, t ∈ [0, 1] is not true. It means that there exists at least at∗ ∈[0, 1]such thatu(t∗)<0. Without loss of generality, we assumeu(t∗) =min{u(t): t∈[0, 1]}=ρ,ρ<0. Then by Lemma2.2and (2.5), we have
u(t) = b
1−αξt+ αt 1−αξ
Z 1
0
G(ξ,s)(κ(s)u(s) +µ(s))ds +
Z 1
0 G(t,s)(κ(s)u(s) +µ(s))ds
≥ αt 1−αξ
Z 1
0 G(ξ,s)κ(s)u(s)ds+
Z 1
0 G(t,s)κ(s)u(s)ds
≥ u(t∗) αt
1−αξ Z 1
0 ϕ(s)κ(s)ds+
Z 1
0 ϕ(s)κ(s)ds
. Lett =t∗, and note thatρ<0, 0<αξ <1, 0< ξ <1, it follows that
ρ ≥ρ αt∗
1−αξ Z 1
0 ϕ(s)κ(s)ds+
Z 1
0 ϕ(s)κ(s)ds
≥ρ α
1−αξ Z 1
0 ϕ(s)κ(s)ds+
Z 1
0 ϕ(s)κ(s)ds
. And then
1≤ α+1−αξ 8(1−αξ)
Z 1
0
(1+s)2(1−s)2κ(s)ds.
That is
Z 1
0
(1+s)2(1−s)2κ(s)ds≥ 8(1−αξ) α+1−αξ, which is in contradiction to (2.6). Henceu(t)≥0 for allt∈ [0, 1].
3 Main result
In this section, we shall apply the method of upper and lower solutions and monotone iterative technique to consider the existence of extremal solutions for problem (1.1).
Theorem 3.1. Assume that the following conditions hold.
(A1) f(t,u)is increasing with respect to u.
(A2) v0(t),w0(t) ∈C[0, 1]are lower and upper solutions of problem(1.1), respectively, and v0(t)≤ w0(t), t∈[0, 1].
(A3) There exist constantsς,τsuch that0< τξ< ςand for v0(1)≤x ≤y ≤w0(1), v0(ξ)≤ x¯≤
¯
y≤w0(ξ),
p(y, ¯y) +τ(y¯−x¯)≤ p(x, ¯x) +ς(y−x). Then problem(1.1)has extremal solutions in the sector[v0,w0], where
[v0,w0] =u∈C[0, 1]:v0(t)≤ u(t)≤w0(t), t∈ [0, 1] . Proof. Forn =0, 1, . . . , define
vn+1(t) = bn
1−αξt+ αt 1−αξ
Z 1
0 G(ξ,s)f(s,vn(s))ds+
Z 1
0 G(t,s)f(s,vn(s))ds, wn+1(t) = cn
1−αξt+ αt 1−αξ
Z 1
0 G(ξ,s)f(s,wn(s))ds+
Z 1
0 G(t,s)f(s,wn(s))ds, where
α= τ
ς, bn=vn(1)−αvn(ξ)−1
ςg(vn(1),vn(ξ)), cn=wn(1)−αwn(ξ)−1
ςg(wn(1),wn(ξ)). Then due to Lemma2.2, it is easy to show thatvn+1(t),wn+1(t)are solutions of the following boundary value problems, respectively:
−v000n+1(t) = f(t,vn(t)), t∈ [0, 1], vn+1(0) =v00n+1(0) =0,
0= p(vn(1),vn(ξ)) +ς(vn+1(1)−vn(1))−τ(vn+1(ξ)−vn(ξ)),
(3.1)
and
−w000n+1(t) = f(t,wn(t)), t ∈[0, 1], wn+1(0) =w00n+1(0) =0,
0= p(wn(1),wn(ξ)) +ς(wn+1(1)−wn(1))−τ(wn+1(ξ)−wn(ξ)).
(3.2)
Moreover, from (2.3) we have v0n+1(t) = bn
1−αξ + α 1−αξ
Z 1
0 G(ξ,s)f(s,vn(s))ds+
Z 1
0 G0(t,s)f(s,vn(s))ds, (3.3) w0n+1(t) = cn
1−αξ + α 1−αξ
Z 1
0 G(ξ,s)f(s,wn(s))ds+
Z 1
0 G0(t,s)f(s,wn(s))ds, (3.4) whereG0(t,s)is given as in (2.4).
Claim 1. The sequences vn(t),wn(t)(n ≥ 1)are lower and upper solutions of problem (1.1), respectively and the following relation holds
v0(t)≤v1(t)≤ · · · ≤vn(t)≤ · · · ≤wn(t)≤ · · · ≤w1(t)≤w0(t), t ∈[0, 1]. (3.5)
First, we prove that
v0(t)≤ v1(t)≤ w1(t)≤w0(t), t ∈[0, 1]. Letx(t) =w0(t)−w1(t). From (3.2) and(A1), we have
x000(t)≤0, t∈ [0, 1], x(0) =x00(0) =0,
x(1)≥ τ
ςx(ξ), 0<τξ<ς.
In view of Lemma 2.5, we have x(t) ≥ 0,t ∈ [0, 1], that is w0(t) ≥ w1(t). Similarly, it can be obtained thatv0(t)≤v1(t),t∈ [0, 1].
Now, letx(t) =w1(t)−v1(t). From(A1)and(A2), it follows that x000(t) = f(t,v0(t))− f(t,w0(t))≤0.
Also, x(0) =x00(0) =0 and
0= p(w0(1),w0(ξ))−p(v0(1),v0(ξ)) +ς(w1(1)−w0(1)−v1(1) +v0(1))
−τ(w1(ξ)−w0(ξ)−v1(ξ) +v0(ξ))
≤ ς(w0(1)−v0(1))−τ(w0(ξ)−v0(ξ)) +ς(w1(1)−w0(1)−v1(1) +v0(1))
−τ(w1(ξ)−w0(ξ)−v1(ξ) +v0(ξ))
= ςx(1)−τx(ξ). That is,
x(0) =x00(0) =0, x(1)≥ τ ςx(ξ).
By Lemma2.5, we havew1(t)≥ v1(t),t∈[0, 1]. And then, by induction, (3.5) holds.
In what follows, we show thatv1(t),w1(t)are lower and upper solutions of problem (1.1), respectively. From (3.1), (3.2) and(A1),(A2), it follows that
−v0001 (t) = f(t,v0(t))≤ f(t,v1(t)). Also, v1(0) =v100(0) =0 and
0= −p(v0(1),v0(ξ)) +p(v1(1),v1(ξ))−p(v1(1),v1(ξ))−ς[v1(1)−v0(1)] +τ[v1(η)−v0(ξ)]
≤ ς[v1(1)−v0(1)]−τ[v1(ξ)−v0(ξ)]−p(v1(1),v1(ξ))−ς[v1(1)−v0(1)] +τ[v1(ξ)−v0(ξ)]
= −p(v1(1),v1(ξ)),
which prove that v1(t)is a lower solution of problem (1.1). Similarly, it can be obtained that w1(t)is an upper solution of problem (1.1).
Analogously to the above arguments, using the induction method, we can show that the sequences vn(t), wn(t) (n ≥ 1)are lower and upper solutions of problem (1.1), respectively and the following relation holds
v0(t)≤v1(t)≤ · · · ≤vn(t)≤ · · · ≤wn(t)≤ · · · ≤w1(t)≤w0(t), t∈[0, 1].
Claim 2. The sequences{vn(t)},{wn(t)}uniformly converge to their limit functionsv(t),w(t), respectively.
We need to show that the sequences are bounded and equicontinuous on[0, 1]. Indeed, C1≤ v0(t)≤ · · · ≤vn(t)≤ · · · ≤ · · · ≤wn(t)· · · ≤w0(t)≤C2 (3.6) fort ∈[0, 1]andn=1, 2, . . . . That is to say that the sequences{vn(t)},{wn(t)}are uniformly bounded with respect to t. Note that {v0n(t)}, {w0n(t)} are bounded on [0, 1] by C3 > 0 because (3.3), (3.4), (3.6) and |f(t,vn)|, |p(vm,vn)| is bounded. Hence {vn(t)}, {wn(t)} are equicontinuous because for∀ε>0, t1,t2∈[0, 1]such that|t1−t2|< Mε
3, we have
|vn(t1)−vn(t2)|=|v0n(γ)||t1−t2|< ε, |wn(t1)−wn(t2)|< ε, γ∈[0, 1].
Therefore, by the Arzelà–Ascoli theorem, the sequences{vn(t)}, {wn(t)}have subsequences {vnk(t)}, {wnk(t)} which uniformly converge to their continuous limit functions v(t),w(t), respectively.
Claim 3. The limit functions v(t),w(t) are the minimal solution and maximal solution of problem (1.1), respectively.
Let u(t) ∈ [v0(t),w0(t)] be any solution of problem (1.1). We assume that the following relation holds for somek∈ N:
vk(t)≤ u(t)≤wk(t), t∈[0, 1]. Lety(t) =u(t)−vk+1(t),z(t) =wk+1(t)−u(t). Then
y000(t)≤0, t∈[0, 1], y(0) =y00(0) =0,
y(1)≥ τ
ςy(ξ), 0< τξ<ς,
and
z000(t)≤0, t∈[0, 1], z(0) =z00(0) =0,
z(1)≥ τςz(ξ), 0< τξ<ς.
This and Lemma2.5 show vk+1(t) ≤ u(t) ≤ wk+1(t). By induction, vn(t) ≤ u(t) ≤ wn(t), t∈ [0, 1]for alln∈ N. Taking the limit as n→∞, we getv(t)≤ u(t)≤w(t), t∈[0, 1]. Theorem 3.2. Assume that all assumptions of Theorem 3.1 hold. In addition, we assume that there exists q(t)>0(∈C[0, 1])such that
f(t,x)− f(t,y)≤q(t)(x−y), v0(t)≤x(t)≤ y(t)≤w0(t), t ∈[0, 1] and
Z 1
0
(1+s)2(1−s)2q(s)ds< 8(ς−τξ) ς+τ−τξ. Then the sequences{vn(t)},{wn(t)}(n≥1)satisfying
−v000n+1(t) = f(t,vn(t)) +q(t)[vn+1(t)−vn(t),], t∈ (0, 1), vn+1(0) =v00n+1(0) =0,
0= p(vn(1),vn(ξ)) +ς(vn+1(1)−vn(1))−τ(vn+1(ξ)−vn(ξ)),
and
−w000n+1(t) = f(t,wn(t)) +q(t)[wn+1(t)−wn(t),], t∈ (0, 1), wn+1(0) =wn00+1(0) =0,
0= p(wn(1),wn(ξ)) +ς(wn+1(1)−wn(1))−τ(wn+1(ξ)−wn(ξ)),
also uniformly converge to their continuous limit functions v(t),w(t), respectively. That is, v(t),w(t) are also extremal solutions for problem(1.1).
Proof. Using Lemmas 2.6 and 2.7, we can complete the proof by the same way as in Theo- rem3.1.
Example 3.3. Consider the following third-order boundary value problem
−u000(t) =u(t)sint−1138 t12, t∈ [0, 1], u(0) =u00(0) =0,
u(1)−u(12)−18u(1)u(12) =0,
(3.7)
where
f(x,y) =ysinx− 113
8 x12, p(x,y) =x−y− 1
8xy, ξ = 1 2.
It is not difficult to show that v0 = 0, w0(t) = t72 are lower and upper solutions of problem (3.7), respectively. Moreover, for t ∈ [0, 1], f(t,u) is increasing with respect to u, and for v0(1)≤x≤ y≤w0(1),v0(η)≤ x¯ ≤y¯≤ w0(η),
p(y, ¯y)−p(x, ¯x) = (y−x)−(y¯−x¯)− 1
8(yy¯−xx¯)≤(y−x)−(y¯−x¯).
Chooseς=τ=1 in Theorem3.1or ς= τ=1,q(t) =sintin Theorem3.2, problem (3.7) has extremal solutions in[v0(t),w0(t)].
Acknowledgements
The authors would like to thank the referee and Professor Kosmatov for their careful reading and kind suggestions. This work was supported by Program for Scientific Research Innovation Team in Colleges and Universities of Shandong Province, the Doctoral Program Foundation of Education Ministry of China (20133705110003), the Natural Science Foundation of Shan- dong Province of China (ZR2014AM007), the National Natural Science Foundation of China (11571197).
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