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volume 5, issue 1, article 13, 2004.

Received 17 September, 2003;

accepted 23 January, 2004.

Communicated by:A.M. Fink

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Journal of Inequalities in Pure and Applied Mathematics

THE METHOD OF LOWER AND UPPER SOLUTIONS FOR SOME FOURTH-ORDER EQUATIONS

ZHANBING BAI, WEIGAO GE AND YIFU WANG

Department of Applied Mathematics, Beijing Institute of Technology,

Beijing 100081, People’s Republic of China.

EMail:baizhanbing@263.net

c

2000Victoria University ISSN (electronic): 1443-5756 124-03

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The Method of Lower and Upper Solutions for Some Fourth-Order Equations

Zhanbing Bai, Weigao Ge and Yifu Wang

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J. Ineq. Pure and Appl. Math. 5(1) Art. 13, 2004

Abstract

In this paper, by combining a new maximum principle of fourth-order equations with the theory of eigenline problems, we develop a monotone method in the presence of lower and upper solutions for some fourth-order ordinary differential equation boundary value problem. Our results indicate there is a relation between the existence of solutions of nonlinear fourth-order equation and the first eigenline of linear fourth-order equation.

2000 Mathematics Subject Classification:34B15, 34B10.

Key words: Maximum principle; Lower and upper solutions; Fourth-order equation.

This work is sponsored by the National Nature Science Foundation of China (10371006) and the Doctoral Program Foundation of Education Ministry of China (1999000722).

The authors thank the referees for their careful reading of the manuscript and useful suggestions.

1. Introduction

This paper consider solutions of the fourth-order boundary value problem (1.1) u⁽⁴⁾(x) = f(x, u(x), u⁰⁰(x)), 0< x <1,

(1.2) u(0) =u(1) =u⁰⁰(0) =u⁰⁰(1) = 0, wheref : [0,1]×R×R−→Ris continuous.

Many authors [1] – [8], [10], [11], [13] – [17] have studied this problem. In [1,4,6, 8,10,16], Aftabizadeh et al. showed the existence of positive solution to (1.1) – (1.2) under some growth conditions off and a non-resonance condition involving a two-parameter linear eigenvalue problem. These results are based upon the Leray–Schauder continuation method and topological degree.

In [2,5,7,11,15], Agarwal et al. considered an equation of the form u⁽⁴⁾(x) = f(x, u(x)),

with diverse kind of boundary conditions by using the lower and upper solution method.

Recently, Bai [3] and Ma et al. [14] developed the monotone method for the problem (1.1) – (1.2) under some monotone conditions off. More recently, with using Krasnosel’skii fixed point theorem, Li [13] showed the existence results of positive solutions for the following problem

u⁽⁴⁾+βu⁰⁰−αu=f(t, u), 0< t <1,

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u(0) =u(1) =u⁰⁰(0) =u⁰⁰(1) = 0,

where f : [0,1]× R⁺ → R⁺ is continuous, α, β ∈ R and β < 2π², α ≥

−β²/4, α/π⁴ +β/π² <1.

In this paper, by the use of a new maximum principle of fourth-order equation and the theory of the eigenline problem, we intend to further relax the monotone condition of f and get the iteration solution. Our results indicate there exists some relation between the existence of positive solutions of nonlinear fourth-order equation and the first eigenline of linear fourth-order equation.

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2. Maximum Principle

In this section, we prove a maximum principle for the operator L:F −→C[0,1]

defined byLu=u⁽⁴⁾−au⁰⁰+bu. Herea, b∈Rsatisfy

(2.1) a

π² + b

π⁴ + 1 >0, a²−4b≥0, a >−2π²; u∈F and

F ={u∈C⁴[0,1]|u(0) = 0, u(1) = 0, u⁰⁰(0)≤0, u⁰⁰(1) ≤0}.

Lemma 2.1. [12] Let f(x) be continuous fora ≤ x ≤ b and let c < λ₁ = π²/(b−a)². Letusatisfies

u⁰⁰(x) +cu(x) =f(x), forx∈(a, b), u(a) = u(b) = 0.

Assume that u(x₁) = u(x₂) = 0 where a ≤ x₁ < x₂ ≤ b andu(x) 6= 0for x₁ ≤ x ≤ x₂. If either f(x) ≥ 0 for all x ∈ [x₁, x₂] or f(x) ≤ 0 for all x ∈[x1, x2]andf(x)is not identically zero on[x1, x2], thenu(x)f(x)≤ 0for allx∈[x₁, x₂].

Lemma 2.2. Ifu(x)satisfies

u⁰⁰+cu(x)≥0, forx∈(a, b) u(a)≤0, u(b)≤0, wherec < λ1 =π²/(b−a)². Thenu(x)≤0,in[a, b].

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Proof. It follows by Lemma2.1.

Lemma 2.3. Ifu∈F satisfiesLu≥0, thenu≥0in[0,1].

Proof. SetAx=x⁰⁰. Asa, b∈Rsatisfy (2.1), we have that Lu=u⁽⁴⁾−au⁰⁰+bu = (A−r₂)(A−r₁)u≥0,

where r_1,2 = (a±√

a²−4b)/2 ≥ −π². In fact, r₁ = (a+√

a² −4b)/2 ≥ r2 = (a−√

a²−4b)/2. Bya/π²+b/π⁴+ 1>0, we haveaπ²+b+π⁴ >0, thusa²+ 4aπ²+ 4π⁴ > a²−4b,becausea² −4b ≥0, so

(a+ 2π²)² >(√

a²−4b)².

Combining this together witha >−2π², we can conclude a+ 2π² >√

a²−4b.

Then,r1 ≥r2 = (a−√

a²−4b)/2>−π². Lety= (A−r₁)u=u⁰⁰−r₁u, then

(A−r₂)y ≥0,

i.e.,

y⁰⁰−r2y ≥0.

On the other hand,u∈F yields that

(2.2) y(0) =u⁰⁰(0)−r1u(0) ≤0, y(1) =u⁰⁰(1)−r1u(1)≤0.

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Therefore, by the use of Lemma2.2, there exists y(x)≤0, x∈[0,1], i.e.,

u⁰⁰(x)−r₁u(x) =y(x)≤0.

This together with Lemma 2.2 and the fact that u(0) = 0, u(1) = 0 implies thatu(x)≥0in[0,1].

Remark 2.1. Observe thata, b∈Rsatisfies (2.1) if and only if

(2.3) b ≤0, a

π² + b

π⁴ + 1>0, a >−2π²; or

(2.4) b >0, a >0, a²−4b ≥0;

or

(2.5) b >0, 0> a >−2π², a π² + b

π⁴ + 1 >0, a²−4b≥0.

From (2.3) and (2.4), we can easily conclude

r₁ = a+√

a²−4b

2 ≥0.

Therefore, (2.2) can be obtained underu(0) ≥0, u(1)≥0, u⁰⁰(0)≤0, u⁰⁰(1)≤ 0, andF can be defined as

F ={u∈C⁴[0,1]|u(0)≥0, u(1) ≥0, u⁰⁰(0)≤0, u⁰⁰(1)≤0}, we refer the reader to [3,13].

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Lemma 2.4. [7] Given(a, b)∈R², the following problem u⁽⁴⁾−au⁰⁰+bu = 0,

(2.6)

u(0) =u(1) =u⁰⁰(0) =u⁰⁰(1) = 0, (2.7)

has a non-trivial solution if and only if a

(kπ)² + b

(kπ)⁴ + 1 = 0, for somek ∈N.

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3. The Monotone Method

In this section, we develop the monotone method for the fourth order two-point boundary value problem (1.1) – (1.2).

For givena, b∈Rsatisfyinga/π²+b/π⁴+ 1>0, a²−4b ≥0, a >−2π² andf : [0,1]×R×R−→R, let

(3.1) f₁(x, u, v) =f(x, u, v) +bu−av.

Then (1.1) is equal to

(3.2) Lu=f₁(x, u, u⁰⁰).

Definition 3.1. Lettingα∈ C⁴[0,1], we say thatαis an upper solution for the problem (1.1) – (1.2) ifαsatisfies

α⁽⁴⁾(x)≥f(x, α(x), α⁰⁰(x)), forx∈(0,1), α(0) = 0, α(1) = 0,

α⁰⁰(0)≤0, α⁰⁰(1)≤0.

Definition 3.2. Lettingβ ∈C⁴[0,1], we sayβis a lower solution for the prob- lem (1.1) – (1.2) ifβ satisfies

β⁽⁴⁾(x)≤f(x, β(x), β⁰⁰(x)), forx∈(0,1), β(0) = 0, β(1) = 0,

β⁰⁰(0) ≥0, β⁰⁰(1)≥0.

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Remark 3.1. Ifa, bsatisfy (2.3) or (2.4), the boundary values can be replaced by

α(0)≥0, α(1)≥0; β(0)≤0, β(1)≤0.

It is clear that if α, β are upper and lower solutions of the problem (1.1) – (1.2) respectively, α, β are upper and lower solutions of the problem (3.2) – (1.2) respectively, too.

Theorem 3.1. If there exist α andβ, upper and lower solutions, respectively, for the problem (1.1) – (1.2) which satisfy

(3.3) β ≤α and β⁰⁰+r(α−β)≥α⁰⁰, and iff : [0,1]×R×R−→Ris continuous and satisfies (3.4) f(x, u₂, v)−f(x, u₁, v)≥ −b(u₂−u₁), forβ(x)≤u₁ ≤u₂ ≤α(x), v ∈R, andx∈[0,1];

(3.5) f(x, u, v₂)−f(x, u, v₁)≤a(v₂ −v₁),

forv2+r(α−β)≥v1, α⁰⁰−r(α−β)≤v1, v2 ≤β⁰⁰+r(α−β), u∈R, and x∈[0,1], wherea, b∈Rsatisfya/π²+b/π⁴+ 1>0, a²−4b≥0, a >−2π² and r = (a −√

a²−4b)/2, then there exist two monotone sequences {α_n} and {βn}, non-increasing and non-decreasing, respectively, with α0 = α and β₀ = β, which converge uniformly to the extremal solutions in [β, α] of the problem (1.1) – (1.2).

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Proof. Consider the problem

(3.6) u⁽⁴⁾(x)−au⁰⁰(x) +bu(x) = f₁(x, η(x), η⁰⁰(x)), forx∈(0,1),

(3.7) u(0) =u(1) =u⁰⁰(0) =u⁰⁰(1) = 0, withη∈C²[0,1].

Since a/π² + b/π⁴ + 1 > 0, with the use of Lemma 2.4 and Fredholm Alternative [9], the problem (3.6) – (3.7) has a unique solution u. Define T : C²[0,1]−→C⁴[0,1]by

(3.8) T η =u.

Now, we divide the proof into three steps.

Step 1. We show

(3.9) T C ⊆C.

Here, C = {η ∈ C²[0,1] | β ≤ η ≤ α, α⁰⁰ − r(α −β) ≤ η⁰⁰ ≤ β⁰⁰+r(α−β)}is a nonempty bounded closed subset inC²[0,1].

In fact, forζ ∈C, setω =T ζ. By the definition ofα, βandC, combining (3.1), (3.4), and (3.5), we have that

(α−ω)⁽⁴⁾(x)−a(α−ω)⁰⁰(x) +b(α−ω)(x)

≥f1(x, α(x), α⁰⁰(x))−f1(x, ζ(x), ζ⁰⁰(x))

=f(x, α(x), α⁰⁰(x))−f(x, ζ(x), ζ⁰⁰(x))

−a(α−ζ)⁰⁰(x) +b(α−ζ)(x)

≥0, (3.10)

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(3.11) (α−ω)(0) = 0, (α−ω)(1) = 0,

(3.12) (α−ω)⁰⁰(0) ≤0, (α−ω)⁰⁰(1) ≤0.

With the use of Lemma 2.3, we obtain that α ≥ ω. Analogously, there holdsω ≥β.

By the proof of Lemma2.3, combining (3.10), (3.11), and (3.12), we have that

(α−ω)⁰⁰(x)−r(α−ω)(x)≤0, x∈(0,1), hence,

ω⁰⁰(x) +r(α−β)(x)≥ω⁰⁰(x) +r(α−ω)(x)≥α⁰⁰(x), forx∈(0,1), i.e.,

ω⁰⁰(x)≥α⁰⁰(x)−r(α−β)(x), forx∈(0,1).

Analogously,

ω⁰⁰(x)≤β⁰⁰(x) +r(α−β)(x), forx∈(0,1).

Thus, (3.9) holds.

Step 2. Let u₁ = T η₁, u₂ = T η₂, where η₁, η₂ ∈ C satisfy η₁ ≤ η₂ and η₁⁰⁰ + r(α−β)≥η₂⁰⁰. We show

(3.13) u1 ≤u2, u⁰⁰₁+r(α−β)≥u⁰⁰₂.

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In fact, by (3.4), (3.5), and the definition ofu₁, u₂,

L(u₂−u₁)(x) =f₁(x, η₂(x), η⁰⁰₂(x))−f₁(x, η₁(x), η⁰⁰₁(x))≥0,

(u2−u1)(0) = (u2−u1)(1) = 0, (u2−u1)⁰⁰(0) = (u2−u1)⁰⁰(1) = 0.

With the use of Lemma 2.3, we get that u1 ≤ u2. Similar to Step 1, we can easily proveu⁰⁰₁ +r(α−β)≥u⁰⁰₂. Thus, (3.13) holds.

Step 3. The sequences{α_n}and{β_n}are obtained by recurrence:

α₀ =α, β₀ =β, α_n =T αn−1, β_n=T βn−1, n= 1,2, . . . . From the results of Step 1 and Step 2, we have that

(3.14) β =β₀ ≤β₁ ≤ · · · ≤β_n≤ · · · ≤α_n ≤ · · · ≤α₁ ≤α₀ =α,

(3.15) β⁰⁰ =β₀⁰⁰, α⁰⁰ =α⁰⁰₀, α⁰⁰−r(α−β)≤α⁰⁰_n, β_n⁰⁰ ≤β⁰⁰+r(α−β).

Moreover, from the definition ofT (see (3.8)), we get

α⁽⁴⁾_n (x)−aα⁰⁰_n(x) +bα_n(x) =f₁(x, αn−1(x), α⁰⁰_n−1(x)), i.e.,

α⁽⁴⁾_n (x) =f₁(x, αn−1(x), α_n−1⁰⁰ (x)) +aα⁰⁰_n(x)−bα_n(x)

≤f₁(x, αn−1(x), α⁰⁰_n−1(x)) +a[β⁰⁰+r(α−β)](x)−bβ(x), (3.16)

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(3.17) α_n(0) =α_n(1) =α⁰⁰_n(0) =α⁰⁰_n(1) = 0.

Analogously,

β_n⁽⁴⁾(x) = f₁(x, βn−1(x), β_n−1⁰⁰ (x)) +aβ_n⁰⁰(x)−bβ_n(x)

≤f₁(x, βn−1(x), β_n−1⁰⁰ (x)) +a[β⁰⁰+r(α−β)](x)−bβ(x), (3.18)

(3.19) β_n(0) =β_n(1) =β_n⁰⁰(0) =β_n⁰⁰(1) = 0.

From (3.14), (3.15), (3.16), and the continuity off1, we have that there existsM_α,β >0depending only onαandβ(but not onnorx) such that (3.20) |α⁽⁴⁾_n (x)| ≤M_α,β, for allx∈[0,1].

Using the boundary condition (3.17), we get that for each n ∈ N, there existsξ_n ∈(0,1)such that

(3.21) α⁰⁰⁰_n(ξ_n) = 0.

This together with (3.20) yields (3.22) |α⁰⁰⁰_n(x)|=|α_n⁰⁰⁰(ξn) +

Z x

ξn

α⁽⁴⁾_n (s)ds| ≤Mα,β.

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By combining (3.15) and (3.17), we can similarly get that there isC_α,β >0 depending only onαandβ(but not onnorx) such that

(3.23) |α⁰⁰_n(x)| ≤C_α,β, for allx∈[0,1],

(3.24) |α⁰_n(x)| ≤C_α,β, for all x∈[0,1].

Thus, from (3.14), (3.22), (3.23), and (3.24), we know that{α_n}is bounded inC³[0,1]. Similarly,{β_n}is bounded inC³[0,1].

Now, by using the fact that{α_n}and{β_n}are bounded inC³[0,1], we can conclude that{α_n},{β_n}converge uniformly to the extremal solutions in[0,1]

of the problem (3.2) – (1.2). Therefore,{α_n},{β_n}converge uniformly to the extremal solutions in[0,1]of the problem (1.1) – (1.2), too.

Example 3.1. Consider the boundary value problem

(3.25) u⁽⁴⁾(x) =−5u⁰⁰(x)−(u(x) + 1)²+ sin²πx+ 1,

(3.26) u(0) =u(1) =u⁰⁰(0) =u⁰⁰(1) = 0.

It is clear that the results of [3, 7, 13, 14] can’t apply to the example. On the other hand, it is easy to check that α = sinπx, β = 0are upper and lower solutions of (3.25) – (3.26), respectively. Letting a = −5, b = 4, then all assumptions of Theorem3.1are fulfilled. Hence the problem (3.25) – (3.26) has at least one solutionu, which satisfies0≤u≤sinπx.

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