Fourth-order equation

http://jipam.vu.edu.au/

Volume 5, Issue 1, Article 13, 2004

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

ZHANBING BAI, WEIGAO GE, AND YIFU WANG DEPARTMENT OFAPPLIEDMATHEMATICS,

BEIJINGINSTITUTE OFTECHNOLOGY, BEIJING100081, PEOPLE’SREPUBLIC OFCHINA.

[email protected]

Received 17 September, 2003; accepted 23 January, 2004 Communicated by A.M. Fink

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 prob- lem. 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.

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

2000 Mathematics Subject Classification. 34B15, 34B10.

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

ISSN (electronic): 1443-5756 c

2004 Victoria University. All rights reserved.

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.

124-03

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 prob- lem

u⁽⁴⁾+βu⁰⁰−αu=f(t, u), 0< t <1, u(0) =u(1) =u⁰⁰(0) =u⁰⁰(1) = 0,

wheref : [0,1]×R⁺ →R⁺is continuous,α, β ∈Randβ <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 pos- itive solutions of nonlinear fourth-order equation and the first eigenline of linear fourth-order equation.

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] Letf(x)be continuous for a ≤ x≤ b and letc < λ1 =π²/(b−a)². Letu satisfies

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

Assume thatu(x₁) = u(x₂) = 0 wherea ≤ x₁ < x₂ ≤ b andu(x) 6= 0forx₁ ≤ x ≤ x₂. If eitherf(x)≥ 0for allx ∈[x₁, x₂]orf(x)≤ 0for allx ∈[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].

Proof. It follows by Lemma 2.1.

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.

Therefore, by the use of Lemma 2.2, 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 thatu(0) = 0, u(1) = 0implies thatu(x) ≥ 0in

[0,1].

Remark 2.4. 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].

Lemma 2.5. [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.

3. THEMONOTONEMETHOD

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

For given a, b ∈ R satisfying a/π² + b/π⁴ + 1 > 0, a² − 4b ≥ 0, a > −2π² and f : [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 sayβ is a lower solution for the problem (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.

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 ifα, 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 3.2. If there existαandβ, 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, andx∈[0,1], where a, b ∈ Rsatisfya/π² +b/π⁴+ 1 > 0, a²−4b ≥ 0, a > −2π² andr = (a−√

a²−4b)/2,

then there exist two monotone sequences{α_n}and{β_n}, non-increasing and non-decreasing, respectively, withα₀ = αand β₀ = β, which converge uniformly to the extremal solutions in [β, α]of the problem (1.1) – (1.2).

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].

Sincea/π² +b/π⁴+ 1 > 0, with the use of Lemma 2.5 and Fredholm Alternative [9], the problem (3.6) – (3.7) has a unique solutionu. DefineT :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, for ζ ∈ C, set ω = T ζ. By the definition of α, β and C, 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)

(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 thatα≥ω. Analogously, there holdsω≥β.

By the proof of Lemma 2.3, 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. Letu₁ =T η₁, u₂ = T η₂, where η₁, η₂ ∈C satisfyη₁ ≤ η₂ andη₁⁰⁰+r(α−β)≥ η⁰⁰₂. We show

(3.13) u₁ ≤u₂, u⁰⁰₁ +r(α−β)≥u⁰⁰₂. In fact, by (3.4), (3.5), and the definition ofu₁, u₂,

L(u₂−u₁)(x) = f₁(x, η₂(x), η₂⁰⁰(x))−f₁(x, η₁(x), η₁⁰⁰(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 thatu₁ ≤u₂. Similar to Step 1, we can easily prove u⁰⁰₁ +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), α⁰⁰_n−1(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)

(3.17) αn(0) =αn(1) =α_n⁰⁰(0) =α⁰⁰_n(1) = 0.

Analogously,

β_n⁽⁴⁾(x) =f₁(x, β_n−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 >0 depending 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 eachn ∈N, there existsξn∈(0,1) such that

(3.21) α⁰⁰⁰_n(ξ_n) = 0.

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

Z x

ξn

α⁽⁴⁾_n (s)ds| ≤M_α,β.

By combining (3.15) and (3.17), we can similarly get that there isC_α,β >0depending 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{αn}is bounded inC³[0,1].

Similarly,{βn}is bounded inC³[0,1].

Now, by using the fact that {α_n} and {β_n} are bounded in C³[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, {αn},{βn} 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 = 0 are upper and lower solutions of (3.25) – (3.26), respectively. Lettinga=−5, b = 4, then all assumptions of Theorem 3.2 are fulfilled. Hence the problem (3.25) – (3.26) has at least one solutionu, which satisfies0≤u≤sinπx.

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