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Existence results for a fourth order multipoint boundary value problem at resonance
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Existence results for a fourth order multipoint boundary value problem at resonance S.A. Iyase Department of Mathematics, Covenant University, Canaanland P.M.B. 1023, Ota, Ogun State, Nigeria Received 1 November 2013; received in revised form 22 June 2015; accepted 15 August 2015
Abstract In this paper we present some existence results for a fourth order multipoint boundary value problem at resonance. Our main tools are based on the coincidence degree theory of Mawhin. c 2015 The Author. Production and Hosting by Elsevier B.V. on behalf of Nigerian Mathematical Society. This is an open access ⃝ article under the CC BY-NC-ND license (http://creativecommons.org/licenses/by-nc-nd/4.0/). Keywords: Fourth order; Multipoint boundary value problem; Resonance; Coincidence degree
1. Introduction In this paper, we shall discuss the solvability of the multipoint boundary value problem x (iv) (t) = f (t, x(t), x ′ (t), x ′′ (t), x ′′′ (t)) x(0) =
m−2
αi x(ξi )
x ′ (0) = x ′′ (0) = 0,
(1.1) x(1) = x(η)
(1.2)
i=1
where f : [0, 1] × R4 → R is a continuous function αi (1 ≤ i ≤ m − 2) ∈ R, 0 < ξ1 ≤ ξ2 ≤ · · · < ξm−2 < 1 and η ∈ (0, 1). Multipoint boundary value problems of ordinary differential equations arise in a variety of different areas of Applied Mathematics, Physics and Engineering. For example Bridges of small sizes are often designed with two supported points, which leads to a standard two-point boundary condition and bridges of Large sizes are sometimes contrived with multipoint supports which corresponds to a multipoint boundary condition. Boundary value problem (1.1)–(1.2) is called a problem at resonance if L x = x (iv) (t) = 0 has non-trivial solutions under the boundary conditions (1.2) that is, when dim ker L ≥ 1. On the interval [0, 1] second order and third order boundary value problems at resonance have been studied by many authors (see [1–4]) and references therein.
Peer review under responsibility of Nigerian Mathematical Society. E-mail address: [email protected]. http://dx.doi.org/10.1016/j.jnnms.2015.08.003 c 2015 The Author. Production and Hosting by Elsevier B.V. on behalf of Nigerian Mathematical Society. This is an open access 0189-8965/⃝ article under the CC BY-NC-ND license (http://creativecommons.org/licenses/by-nc-nd/4.0/).
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Although the existing literature on solutions of multipoint boundary value problems is quite large, to the best of our knowledge there are few papers that have investigated the existence of solutions of fourth order multipoint boundary value problems at resonance. Our motivation for this paper is derived from these previous results. In what follows, we shall use the classical spaces C k [0, 1], k = 1, 2, 3. For x ∈ C 3 [0, 1] we use the norm |x|∞ = maxt∈[0,1] |x(t)|. We denote the norm in L 1 [0, 1] by | |1 and on L 2 [0, 1] by | |2 . We will use the Sobolev spaces W 4,1 (0, 1) which may be defined by W 4,1 (0, 1) = {x : [0, 1] −→ R : x, x ′ , x ′′ , x ′′′ } are absolutely continuous on [0, 1] with x (iv) ∈ L 1 [0, 1]. 2. Preliminaries Consider the linear equation L x = x (iv) (t) = 0 x(0) =
m−2
(2.1)
αi x(ξi ),
x ′ (0) = x ′′ (0) = 0,
x(1) = x(η).
(2.2)
i=1
If we consider a solution of the form x(t) =
3
ai t i ,
(2.3)
ai ∈ R.
i=0
Then this solution exists if and only if a3 (1 − η3 ) = 0,
η ∈ (0, 1).
(2.4)
In this case (2.1)–(2.2) has non-trivial solutions. Hence if L x = y then L is not invertible. Therefore, the problem is said to be at resonance. We shall prove existence results for the boundary value problem (1.1)–(1.2) under the condition (2.4). We shall apply the continuation Theorem of Mawhin [5] to get our results. We present some preliminaries needed to understand this continuation Theorem. Let X and Z be real Banach spaces and L : dom L ⊂ X −→ Z be a linear operator which is Fredholm of index zero and P : X −→ X, Q : Z −→ Z be continuous projections such that I m P = ker L , ker Q = I m L
and
X = ker L ⊕ ker P
Z = I m L ⊕ I m Q. It follows that L|dom L∩ker P −→ I m L is invertible and we write the inverse of this map by K p . Let Ω be an open bounded subset of X such that dom L ∩ Ω ̸= Φ and let N : Ω¯ −→ Z be an L-compact mapping, that is, the maps Q N (Ω¯ ) is bounded and K p (I − Q)N : Ω¯ −→ X is compact. In order to obtain our existence results we shall use the following fixed point Theorem of Mawhin. Theorem 2.1 (See [5]). Let L be a Fredholm operator of index zero and let N be L-compact on Ω¯ . Assume that the following conditions are satisfied (i) L x ̸= λN x for every (x, λ) ∈ [(dom L\ker L) ∩ ∂Ω × (0, 1)] (ii) N x ̸∈ I m L for every x ∈ ker L ∩ ∂Ω (iii) deg(J Q N |ker L∩∂ Ω ; Ω ∩ ker L , 0) ̸= 0 where Q : Z → Z is a continuous projection as above and J : I m Q → ker L is an isomorphism. Then the equation L x = N x has at least one solution in dom L ∩ Ω¯ . We shall prove existence results for the boundary value problem (1.1)-(1.2) when m−2
αi ξi3 = 0
i=1
and
m−2
αi = 1.
i=1
Let X = C 3 [0, 1], Z = L 1 [0, 1]. Let L : dom L ⊂ X −→ Z be defined by L x = x (iv)
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where dom L = x ∈ W
(0, 1), x(0) =
4,1
m−2
αi x(ξi ), x (0) = x (0) = 0, x(1) = x(η) . ′
′′
i=1
We define N : X −→ Z by setting N = f (t, x(t), x ′ (t), x ′′ (t), x ′′′ (t)). Then the boundary value problem (1.1)–(1.2) can be put in the form (2.5)
L x = N x. In what follows we shall use the following lemmas. m−2 m−2 Lemma 2.1. If i=1 αi = 1 then there exists l ∈ {0, 1, 2, . . . , m − 4} such that i=1 αi ξil+4 ̸= 0. Proof. Follows the same procedure as in [1]. m−2 m−2 Lemma 2.2. If i=1 αi = 1, i=1 αi ξi3 = 0 then m−2 ξi s τ2 τ1 (A) I m L = y ∈ Z : i=1 αi 0 0 0 0 y(v)dvdτ1 dτ2 ds = 0 (B) L : dom L ⊂ X −→ Z is a Fredholm operator of index zero. Proof. We will show that the problem x (iv) (t) = y
for y ∈ Z
(2.6)
has a solution x(t) satisfying x(0) =
m−2
αi x(ξi ),
x ′ (0) = x ′′ (0) = 0,
(2.7)
x(1) = x(η)
i=1
if and only if m−2
αi
ξi
s
τ2
τ1
y(v)dvdτ1 dτ2 ds = 0. 0
i=1
0
0
(2.8)
0
Suppose (2.6) has a solution x(t) satisfying (2.7) then from (2.6) we have t s τ2 τ1 t2 t3 x(t) = x(0) + x ′ (0)t + x ′′ (0) + x ′′′ (0) + y(v)dvdτ1 dτ2 ds. 2 6 0 0 0 0 m−2 m−2 Using i=1 αi = 1, i=1 αi ξi3 = 0 we obtain m−2 ξi s τ2 τ1 αi y(v)dvdτ1 dτ2 ds = 0, for y ∈ Z . i=1
0
Now suppose m−2 αi i=1
0
ξi
0
0
s
0
τ2
τ1
y(v)dvdτ1 dτ2 ds = 0. 0
0
0
Let x(t) = c −
t3 1 − η3
1 s
η
τ2
τ1
t
0
0
0
0
0
0
0
τ2
τ1
y(v)dvdτ1 dτ2 ds 0
0
where c is an arbitrary constant. Then x(t) is a solution of (2.6) with m−2 ξi s τ2 τ1 αi y(v)dvdτ1 dτ2 ds = 0. i=1
s
y(v)dvdτ1 dτ2 ds + 0
0
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For y ∈ Z , we define the projection Q : Z −→ Z by m−2 ξi s τ2 τ1 A l t αi (Qy)(t) = y(v)dvdτ1 dτ2 ds m−2 0 0 0 0 l+4 i=1 αi ξi i=1
where A = (l + 1)(l + 2)(l + 3)(l + 4).
(2.9)
Let y1 = y − Qy, that is y1 ∈ ker Q. Then by direct calculations we have m−2 ξi s τ2 τ1 αi y1 (v)dvdτ1 dτ2 ds i=1
0
0
0
0
=
m−2 i=1
αi
ξi
s
0
τ2
0
0
τ1
0
ξi s τ2 τ1 A vl dvdτ1 dτ2 ds = 0. y(v)dvdτ1 dτ2 ds 1 − m−2 0 0 0 0 αi ξil+4 i=1
So, y1 ∈ I m L. Hence Z = I m L + I m Q. Since I m L ∩ I m Q = {0} we obtain Z = I m L ⊕ I m Q. Now ker L = {x ∈ dom L : x = c, c ∈ R}. Hence, dim ker L = dim I m Q = 1.
Hence L is a Fredholm operator of index zero. Let P : X → X be defined by t ∈ [0, 1].
P x(t) = x(0), Lemma 2.3. If written as
m−2 i=1
αi = 1,
−t 3 K p y(t) = 1 − η3
m−2
1 s
αi ξi3 = 0. Then the generalized inverse K p : I m L −→ dom L ∩ ker P can be
i=1
τ1
τ2
t
s
τ1
τ2
y(v)dvdτ1 dτ2 ds +
η
0
0
y(v)dvdτ1 dτ2 ds. 0
0
0
0
0
Proof. For any y ∈ I m L, we have (L K p )y(t) = (K p y(t))(iv) = y(t) and for x ∈ dom L ∩ ker P, one has −t 3 1 − η3 t s τ2 τ1
(K p L)x(t) = K p (x iv ) = + =
0 −t 3
1 − η3
0
0
η
1 s 0
τ2
0
τ1
x (iv) (v)dvdτ1 dτ2 ds
0
x (iv) (v)dvdτ1 dτ2 ds
0
(1 − η2 ) ′′ (1 − η3 ) ′′′ x(1) − x(η) − (1 − η )x (0) − x (0) − x (0) 2 6
+ x(t) − x(0) − t x ′ (0) −
3
′
t 2 ′′ x (0) − t 3 x ′′′ (0). 2
Since x ∈ dom L ∩ ker P, P x(t) = x(0) = 0. Also x ′ (0) = x ′′ (0) = 0.
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Thus, (K p L)x(t) = x(t). Hence, K p = (L|dom L∩ker P )−1 . 3. Main results m−2 m−2 Theorem 3.1. Let i=1 αi = 1, i=1 αi ξi3 = 0 and let f : [0, 1] × R4 → R be a continuous function and suppose that f has the decomposition f (t, x, y, w, z) = g(t, x, y, w, z) + h(t, x, y, w, z) (H1 ) Assume there exists M1 > 0 such that for all x ∈ dom L\ker L if x(t) > M1 , t ∈ [0, 1] then m−2 ξi s τ2 τ1 αi [ f (v, x(v), x ′ (v), x ′′ (v), x ′′′ (v))]dvdτ2 dτ2 ds ̸= 0 0
i=1
0
0
(3.1)
0
(H2 ) zg(t, x, y, w, z) ≤ 0
for all (t, x, y, w, z) ∈ [0, 1] × R4
(3.2)
(a) |h(t, x, y, w, z)| ≤ M{|x|r + |y| + |w| + |z|θ } for 0 < r, θ < 1
(3.3)
(b) z[ f (t, x, y, w, z)] ≤ (|z|2 + 1)[D(t, x, y, w) + m(t)] where D(t, x, y, w) is bounded on bounded sets and m(t) ∈ L 1 [0, 1]. (H3 ) There exists N ∗ > 0 such that for all c ∈ R, |c| > N ∗ then either m−2 ξi s τ2 τ1 cA l ·t αi [ f (v, c, 0, 0, 0)]dvdτ1 dτ2 ds < 0 m−2 0 0 0 0 i=1 αi ξil+4
(3.4)
(3.5)
i=1
or cA m−2 i=1
· tl
αi ξil+4
m−2 i=1
αi
0
ξi
0
s
0
τ2
τ1
[ f (v, c, 0, 0, 0)]dvdτ1 dτ2 ds > 0
0
Then (1.1)–(1.2) has at least one solution in C 3 [0, 1] provided M< where B =
4(B 2
Bπ 2 + π 2 + 4)
√ 4 + π 2.
Proof. Set Ω1 = {x ∈ dom K \ker L : L x = λN x, λ ∈ [0, 1]} for x ∈ Ω1 . Since L x = λN x, then λ ̸= 0, N x ∈ I m L = ker Q hence m−2 ξi s τ2 τ1 αi { f (v, x(v), x ′ (v), x ′′ (v), x ′′′ (v))}dvdτ1 dτ2 ds = 0. i=1
0
0
0
0
(3.6)
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Thus by (H1 ) there exists t0 ∈ [0, 1] such that |x(t0 )| ≤ M1 . Therefore, t |x(t)| ≤ |x(t0 )| + |x ′ (s)|ds, t ∈ [0, 1] t0
|x|∞ ≤ M1 + |x |∞ . ′
(3.7)
We note that for x(1) = x(η) there exists t1 ∈ (η, 1) such that x ′ (t1 ) = 0 and from x ′ (t1 ) = x ′ (0) = 0 there exists t2 ∈ (0, t1 ) such that x ′′ (t2 ) = 0 and x ′′ (0) = x ′′ (t2 ) there exists t3 ∈ (0, t2 ) such that x ′′′ (t3 ) = 0. Hence for x ∈ Ω1 we have t t t x ′′′ (s)x (iv) (s)ds = λ x ′′′ (s)g(s, x, x ′ , x ′′ , x ′′′ )ds + λ x ′′′ (s)h(s, x, x ′ , x ′′ , x ′′′ )ds t3
t3
1 ′′′ 2 |x |2 ≤ 2
1
t3
|x ′′′ | |h(t, x, x ′ , x ′′ , x ′′′ )|dt.
0
Using the Cauchy inequality |ab| ≤ we have 1 0
εa 2 b2 + 2 2ε
for ε > 0
ε |x | |h(t, x, x , x , x )|dt ≤ 2 ′′′
′
′′
′′′
0
1
1 |x | dt + 2ε ′′′ 2
1
|h(t, x, x ′ , x ′′ , x ′′′ )|dt.
0
From condition (H2a ) we obtain the estimate |h(t, x, y, w, z)|2 ≤ 4M 2 {|x|2r + |y|2 + |w|2 + |z|2θ }. Therefore, 1 ′′′ 2 ε ′′′ 2 2M 2 2r 2M 2 ′′′ 2θ 2M 2 ′ 2 2M 2 ′′ 2 |x |2 − |x |2 ≤ |x|2 + |x |2 + |x |2 + |x |2 . 2 2 ε ε ε ε From Holder’s inequality we get 8M 2 2M 2 2r 1 ε 32M 2 ′′′ 2 ′′′ 2θ − |x | ≤ − − |x| + |x | . 2 2 2 1 2 2 ε επ 4 επ 2 Since 0 ≤ θ, r < 1 we infer the existence of a constant M2 such that |x ′′′ |22 < M2
(3.8)
provided ε 32M 2 8M 2 1 > + + . 2 2 επ 4 επ 2 √ 2 The choice ε = 4M π4+π minimizes the right hand side of (3.9) with a minimum value √ B = 4 + π 2. Hence (3.8) holds provided M<
(3.9) 2M(B+π 2 +4) Bπ 2
where
Bπ 2 . 4(B 2 + π 2 + 4)
From (3.8) and x ′ (0) = x ′′ (0) = 0 we get |x ′ |∞ < M3 |x |∞ < M4 , ′′
for some M3 > 0 M4 > 0
(3.10) (3.11)
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and from (3.7) we derive |x|∞ ≤ M1 + |x ′ |∞ < M1 + M3 = M5 .
(3.12)
Now using condition (H2b ) of Theorem 3.1 we get x ′′′ x iv ≤ D(t, x, x ′ , x ′′ ) + m(t) |x ′′′ |2 + 1 and hence
′′′
loge |x | ≤
t
t3
t 1 x ′′′ (s)x (iv) (s) ′′′ 2 ds = loge (|x (s)| + 1) ≤ D + |m|1 2 |x ′′′ |2 + 1 t3
(3.13)
where the constant D depends on M3 , M4 and M5 . Since x ′′′ (t3 ) = 0 we infer from (3.13) that |x ′′′ |∞ < e N0 = M6
(3.14)
where N0 = D + |m|1 . Hence ∥x∥ = max{|x|∞ , |x ′ |∞ , |x ′′ |∞ , |x ′′′ |∞ ≤ max{M4 , M3 , M5 , M6 }}. Therefore, Ω1 is bounded. Let Ω2 = {x ∈ ker L : N x ∈ I m L}. For x ∈ Ω2 , we have x = c ∈ R, thus m−2 ξi s τ2 τ1 αi [ f (v, x(v), 0, 0, 0)]dvdτ1 dτ2 ds = 0. i=1
0
0
0
(3.15)
0
Then we have by H3 and (3.15) that ∥x∥ = c ≤ N ∗ which shows that Ω2 is bounded. We define the isomorphism J : I m Q −→ ker L by J (c) = c,
c ∈ R.
If (3.5) holds we set Ω3 = {x ∈ ker L : −λx + (1 − λ)J Q N x = 0}, λ ∈ [0, 1] For c0 ∈ Ω3 , we obtain (1 − λ)A λc0 = · tl m−2 αi ξil+4
m−2 i=1
αi
0
ξi
0
s
τ2
0
τ1
[ f (v, c0 , 0, 0, 0)]dvdτ1 dτ2 ds
0
i=0
if λ = 1 then c0 = 0 and if |c0 | > N ∗ then from (3.5) we have ξi s τ2 τ1 (1 − λ)c0 A l m−2 2 λc0 = ·t αi [ f (v, c0 , 0, 0, 0)]dvdτ1 dτ2 ds < 0 m−2 0 0 0 0 l+4 i=1 αi ξi i=0
which contradicts λc02 ≥ 0. Thus Ω3 is bounded. If (3.6) holds, then let Ω3 = {x ∈ ker L : λx + (1 − λ)J Q N x = 0, λ ∈ [0, 1]}. Following the above argument we can show that Ω3 is bounded.
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3 Ω ⊂ Ω . By the Arzela–Ascoli Theorem we can show that Let Ω be a bounded open subset of X such that ∪i=1 i K p (I − Q)N : Ω −→ X is compact [5]. So N is L-compact. Thus we have shown that
(i) L x ̸= λN x for every (x, λ) ∈ [dom L \ker L ∩∂Ω × (0, 1)] (ii) N x ̸∈ I m L for every x ∈ ker L ∩ ∂Ω . Finally we shall prove that (iii) of Theorem 2.1 is satisfied. Define H (x, λ) = ±λx + (1 − λ)Q N x, we have H (x, 1) = ±x,
H (x, 0) = Q N x.
Thus H (x, λ) is a homotopy from the identity ±I to Q N and is such that H (x, λ) ̸= 0 for every x ∈ ∂Ω ∩ ker L. Therefore deg(J Q N |ker L∩∂ Ω , Ω ∩ ker L , 0) = deg(H (·, 0), Ω ∩ ker L , 0) = deg(H (·, 1), Ω ∩ ker L , 0) = deg(±I, Ω ∩ ker L , 0) ̸= 0. Then by Theorem 2.1 L x = N x has at least one solution in dom L ∩ Ω¯ . In other words (1.1)–(1.2) has at least one solution in C 3 [0, 1]. References [1] [2] [3] [4] [5]
Du Z, Lin Xiaojie, Ge Weigao. On a third order multipoint boundary value problem at resonance. J. Math. Anal. Appl. 2005;302:217–29. Feng W, Webb JRL. Solvability of three-point boundary value problems at resonance. Nonlinear Anal. TMA 2001;47:2407–18. Gupta CP. A second order m-point boundary value problem at resonance. Nonlinear Anal. TMA 1995;24:1483–9. Nagle RK, Pothoven KL. On a third-order nonlinear bvp at resonance. J. Math. Anal. Appl. 1995;195:148–59. Mawhin J. Topological degree methods in non-linear bvp. In: NSF CBMS regional conference series in math. 40. Providence: American Math. Soc.; 1979.
Further reading [1] [2] [3] [4] [5] [6]
Agarwal R. On fourth order bvp arising on beam analysis. Differential Integral Equations 1989;2:91–110. Aftabizadeh AR. Existence and uniqueness theorems for fourth order boundary value problems. J. Math. Anal. Appl. 1986;116:415–26. Aftabizadeh A, Gupta CP, Xu J. Existence and uniqueness theorems for three points bvp. SIAM J. Math. Anal. Appl. 1989;20(3):716–26. Feng W, Webb JR. Solvability of m-point bvp with nonlinear growth. J. Math. Anal. Appl. 1997;212:467–80. Gupta CP. Existence and uniqueness theorems for the bending of an elastic beam equation. Appl. Anal. 1988;926:289–304. Iyase SA. Existence and Uniqueness theorems for a class of three point fourth-order boundary value problems. J. Nigerian Math. Soc. 2010; 29:239–50. [7] Iyase SA. Existence and uniqueness of solutions of a fourth order four point bvp. J. Nigerian Math. Soc. 1999;18:31–40. [8] Liu B. Positive solutions of fourth order two point bvp. Appl. Math. Comput. 2004;148:107–20.
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Existence results for a fourth order multipoint boundary value problem at resonance S.A. Iyase Department of Mathematics, Covenant University, Canaanland P.M.B. 1023, Ota, Ogun State, Nigeria Received 1 November 2013; received in revised form 22 June 2015; accepted 15 August 2015
Abstract In this paper we present some existence results for a fourth order multipoint boundary value problem at resonance. Our main tools are based on the coincidence degree theory of Mawhin. c 2015 The Author. Production and Hosting by Elsevier B.V. on behalf of Nigerian Mathematical Society. This is an open access ⃝ article under the CC BY-NC-ND license (http://creativecommons.org/licenses/by-nc-nd/4.0/). Keywords: Fourth order; Multipoint boundary value problem; Resonance; Coincidence degree
1. Introduction In this paper, we shall discuss the solvability of the multipoint boundary value problem x (iv) (t) = f (t, x(t), x ′ (t), x ′′ (t), x ′′′ (t)) x(0) =
m−2
αi x(ξi )
x ′ (0) = x ′′ (0) = 0,
(1.1) x(1) = x(η)
(1.2)
i=1
where f : [0, 1] × R4 → R is a continuous function αi (1 ≤ i ≤ m − 2) ∈ R, 0 < ξ1 ≤ ξ2 ≤ · · · < ξm−2 < 1 and η ∈ (0, 1). Multipoint boundary value problems of ordinary differential equations arise in a variety of different areas of Applied Mathematics, Physics and Engineering. For example Bridges of small sizes are often designed with two supported points, which leads to a standard two-point boundary condition and bridges of Large sizes are sometimes contrived with multipoint supports which corresponds to a multipoint boundary condition. Boundary value problem (1.1)–(1.2) is called a problem at resonance if L x = x (iv) (t) = 0 has non-trivial solutions under the boundary conditions (1.2) that is, when dim ker L ≥ 1. On the interval [0, 1] second order and third order boundary value problems at resonance have been studied by many authors (see [1–4]) and references therein.
Peer review under responsibility of Nigerian Mathematical Society. E-mail address: [email protected]. http://dx.doi.org/10.1016/j.jnnms.2015.08.003 c 2015 The Author. Production and Hosting by Elsevier B.V. on behalf of Nigerian Mathematical Society. This is an open access 0189-8965/⃝ article under the CC BY-NC-ND license (http://creativecommons.org/licenses/by-nc-nd/4.0/).
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Although the existing literature on solutions of multipoint boundary value problems is quite large, to the best of our knowledge there are few papers that have investigated the existence of solutions of fourth order multipoint boundary value problems at resonance. Our motivation for this paper is derived from these previous results. In what follows, we shall use the classical spaces C k [0, 1], k = 1, 2, 3. For x ∈ C 3 [0, 1] we use the norm |x|∞ = maxt∈[0,1] |x(t)|. We denote the norm in L 1 [0, 1] by | |1 and on L 2 [0, 1] by | |2 . We will use the Sobolev spaces W 4,1 (0, 1) which may be defined by W 4,1 (0, 1) = {x : [0, 1] −→ R : x, x ′ , x ′′ , x ′′′ } are absolutely continuous on [0, 1] with x (iv) ∈ L 1 [0, 1]. 2. Preliminaries Consider the linear equation L x = x (iv) (t) = 0 x(0) =
m−2
(2.1)
αi x(ξi ),
x ′ (0) = x ′′ (0) = 0,
x(1) = x(η).
(2.2)
i=1
If we consider a solution of the form x(t) =
3
ai t i ,
(2.3)
ai ∈ R.
i=0
Then this solution exists if and only if a3 (1 − η3 ) = 0,
η ∈ (0, 1).
(2.4)
In this case (2.1)–(2.2) has non-trivial solutions. Hence if L x = y then L is not invertible. Therefore, the problem is said to be at resonance. We shall prove existence results for the boundary value problem (1.1)–(1.2) under the condition (2.4). We shall apply the continuation Theorem of Mawhin [5] to get our results. We present some preliminaries needed to understand this continuation Theorem. Let X and Z be real Banach spaces and L : dom L ⊂ X −→ Z be a linear operator which is Fredholm of index zero and P : X −→ X, Q : Z −→ Z be continuous projections such that I m P = ker L , ker Q = I m L
and
X = ker L ⊕ ker P
Z = I m L ⊕ I m Q. It follows that L|dom L∩ker P −→ I m L is invertible and we write the inverse of this map by K p . Let Ω be an open bounded subset of X such that dom L ∩ Ω ̸= Φ and let N : Ω¯ −→ Z be an L-compact mapping, that is, the maps Q N (Ω¯ ) is bounded and K p (I − Q)N : Ω¯ −→ X is compact. In order to obtain our existence results we shall use the following fixed point Theorem of Mawhin. Theorem 2.1 (See [5]). Let L be a Fredholm operator of index zero and let N be L-compact on Ω¯ . Assume that the following conditions are satisfied (i) L x ̸= λN x for every (x, λ) ∈ [(dom L\ker L) ∩ ∂Ω × (0, 1)] (ii) N x ̸∈ I m L for every x ∈ ker L ∩ ∂Ω (iii) deg(J Q N |ker L∩∂ Ω ; Ω ∩ ker L , 0) ̸= 0 where Q : Z → Z is a continuous projection as above and J : I m Q → ker L is an isomorphism. Then the equation L x = N x has at least one solution in dom L ∩ Ω¯ . We shall prove existence results for the boundary value problem (1.1)-(1.2) when m−2
αi ξi3 = 0
i=1
and
m−2
αi = 1.
i=1
Let X = C 3 [0, 1], Z = L 1 [0, 1]. Let L : dom L ⊂ X −→ Z be defined by L x = x (iv)
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where dom L = x ∈ W
(0, 1), x(0) =
4,1
m−2
αi x(ξi ), x (0) = x (0) = 0, x(1) = x(η) . ′
′′
i=1
We define N : X −→ Z by setting N = f (t, x(t), x ′ (t), x ′′ (t), x ′′′ (t)). Then the boundary value problem (1.1)–(1.2) can be put in the form (2.5)
L x = N x. In what follows we shall use the following lemmas. m−2 m−2 Lemma 2.1. If i=1 αi = 1 then there exists l ∈ {0, 1, 2, . . . , m − 4} such that i=1 αi ξil+4 ̸= 0. Proof. Follows the same procedure as in [1]. m−2 m−2 Lemma 2.2. If i=1 αi = 1, i=1 αi ξi3 = 0 then m−2 ξi s τ2 τ1 (A) I m L = y ∈ Z : i=1 αi 0 0 0 0 y(v)dvdτ1 dτ2 ds = 0 (B) L : dom L ⊂ X −→ Z is a Fredholm operator of index zero. Proof. We will show that the problem x (iv) (t) = y
for y ∈ Z
(2.6)
has a solution x(t) satisfying x(0) =
m−2
αi x(ξi ),
x ′ (0) = x ′′ (0) = 0,
(2.7)
x(1) = x(η)
i=1
if and only if m−2
αi
ξi
s
τ2
τ1
y(v)dvdτ1 dτ2 ds = 0. 0
i=1
0
0
(2.8)
0
Suppose (2.6) has a solution x(t) satisfying (2.7) then from (2.6) we have t s τ2 τ1 t2 t3 x(t) = x(0) + x ′ (0)t + x ′′ (0) + x ′′′ (0) + y(v)dvdτ1 dτ2 ds. 2 6 0 0 0 0 m−2 m−2 Using i=1 αi = 1, i=1 αi ξi3 = 0 we obtain m−2 ξi s τ2 τ1 αi y(v)dvdτ1 dτ2 ds = 0, for y ∈ Z . i=1
0
Now suppose m−2 αi i=1
0
ξi
0
0
s
0
τ2
τ1
y(v)dvdτ1 dτ2 ds = 0. 0
0
0
Let x(t) = c −
t3 1 − η3
1 s
η
τ2
τ1
t
0
0
0
0
0
0
0
τ2
τ1
y(v)dvdτ1 dτ2 ds 0
0
where c is an arbitrary constant. Then x(t) is a solution of (2.6) with m−2 ξi s τ2 τ1 αi y(v)dvdτ1 dτ2 ds = 0. i=1
s
y(v)dvdτ1 dτ2 ds + 0
0
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For y ∈ Z , we define the projection Q : Z −→ Z by m−2 ξi s τ2 τ1 A l t αi (Qy)(t) = y(v)dvdτ1 dτ2 ds m−2 0 0 0 0 l+4 i=1 αi ξi i=1
where A = (l + 1)(l + 2)(l + 3)(l + 4).
(2.9)
Let y1 = y − Qy, that is y1 ∈ ker Q. Then by direct calculations we have m−2 ξi s τ2 τ1 αi y1 (v)dvdτ1 dτ2 ds i=1
0
0
0
0
=
m−2 i=1
αi
ξi
s
0
τ2
0
0
τ1
0
ξi s τ2 τ1 A vl dvdτ1 dτ2 ds = 0. y(v)dvdτ1 dτ2 ds 1 − m−2 0 0 0 0 αi ξil+4 i=1
So, y1 ∈ I m L. Hence Z = I m L + I m Q. Since I m L ∩ I m Q = {0} we obtain Z = I m L ⊕ I m Q. Now ker L = {x ∈ dom L : x = c, c ∈ R}. Hence, dim ker L = dim I m Q = 1.
Hence L is a Fredholm operator of index zero. Let P : X → X be defined by t ∈ [0, 1].
P x(t) = x(0), Lemma 2.3. If written as
m−2 i=1
αi = 1,
−t 3 K p y(t) = 1 − η3
m−2
1 s
αi ξi3 = 0. Then the generalized inverse K p : I m L −→ dom L ∩ ker P can be
i=1
τ1
τ2
t
s
τ1
τ2
y(v)dvdτ1 dτ2 ds +
η
0
0
y(v)dvdτ1 dτ2 ds. 0
0
0
0
0
Proof. For any y ∈ I m L, we have (L K p )y(t) = (K p y(t))(iv) = y(t) and for x ∈ dom L ∩ ker P, one has −t 3 1 − η3 t s τ2 τ1
(K p L)x(t) = K p (x iv ) = + =
0 −t 3
1 − η3
0
0
η
1 s 0
τ2
0
τ1
x (iv) (v)dvdτ1 dτ2 ds
0
x (iv) (v)dvdτ1 dτ2 ds
0
(1 − η2 ) ′′ (1 − η3 ) ′′′ x(1) − x(η) − (1 − η )x (0) − x (0) − x (0) 2 6
+ x(t) − x(0) − t x ′ (0) −
3
′
t 2 ′′ x (0) − t 3 x ′′′ (0). 2
Since x ∈ dom L ∩ ker P, P x(t) = x(0) = 0. Also x ′ (0) = x ′′ (0) = 0.
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Thus, (K p L)x(t) = x(t). Hence, K p = (L|dom L∩ker P )−1 . 3. Main results m−2 m−2 Theorem 3.1. Let i=1 αi = 1, i=1 αi ξi3 = 0 and let f : [0, 1] × R4 → R be a continuous function and suppose that f has the decomposition f (t, x, y, w, z) = g(t, x, y, w, z) + h(t, x, y, w, z) (H1 ) Assume there exists M1 > 0 such that for all x ∈ dom L\ker L if x(t) > M1 , t ∈ [0, 1] then m−2 ξi s τ2 τ1 αi [ f (v, x(v), x ′ (v), x ′′ (v), x ′′′ (v))]dvdτ2 dτ2 ds ̸= 0 0
i=1
0
0
(3.1)
0
(H2 ) zg(t, x, y, w, z) ≤ 0
for all (t, x, y, w, z) ∈ [0, 1] × R4
(3.2)
(a) |h(t, x, y, w, z)| ≤ M{|x|r + |y| + |w| + |z|θ } for 0 < r, θ < 1
(3.3)
(b) z[ f (t, x, y, w, z)] ≤ (|z|2 + 1)[D(t, x, y, w) + m(t)] where D(t, x, y, w) is bounded on bounded sets and m(t) ∈ L 1 [0, 1]. (H3 ) There exists N ∗ > 0 such that for all c ∈ R, |c| > N ∗ then either m−2 ξi s τ2 τ1 cA l ·t αi [ f (v, c, 0, 0, 0)]dvdτ1 dτ2 ds < 0 m−2 0 0 0 0 i=1 αi ξil+4
(3.4)
(3.5)
i=1
or cA m−2 i=1
· tl
αi ξil+4
m−2 i=1
αi
0
ξi
0
s
0
τ2
τ1
[ f (v, c, 0, 0, 0)]dvdτ1 dτ2 ds > 0
0
Then (1.1)–(1.2) has at least one solution in C 3 [0, 1] provided M< where B =
4(B 2
Bπ 2 + π 2 + 4)
√ 4 + π 2.
Proof. Set Ω1 = {x ∈ dom K \ker L : L x = λN x, λ ∈ [0, 1]} for x ∈ Ω1 . Since L x = λN x, then λ ̸= 0, N x ∈ I m L = ker Q hence m−2 ξi s τ2 τ1 αi { f (v, x(v), x ′ (v), x ′′ (v), x ′′′ (v))}dvdτ1 dτ2 ds = 0. i=1
0
0
0
0
(3.6)
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Thus by (H1 ) there exists t0 ∈ [0, 1] such that |x(t0 )| ≤ M1 . Therefore, t |x(t)| ≤ |x(t0 )| + |x ′ (s)|ds, t ∈ [0, 1] t0
|x|∞ ≤ M1 + |x |∞ . ′
(3.7)
We note that for x(1) = x(η) there exists t1 ∈ (η, 1) such that x ′ (t1 ) = 0 and from x ′ (t1 ) = x ′ (0) = 0 there exists t2 ∈ (0, t1 ) such that x ′′ (t2 ) = 0 and x ′′ (0) = x ′′ (t2 ) there exists t3 ∈ (0, t2 ) such that x ′′′ (t3 ) = 0. Hence for x ∈ Ω1 we have t t t x ′′′ (s)x (iv) (s)ds = λ x ′′′ (s)g(s, x, x ′ , x ′′ , x ′′′ )ds + λ x ′′′ (s)h(s, x, x ′ , x ′′ , x ′′′ )ds t3
t3
1 ′′′ 2 |x |2 ≤ 2
1
t3
|x ′′′ | |h(t, x, x ′ , x ′′ , x ′′′ )|dt.
0
Using the Cauchy inequality |ab| ≤ we have 1 0
εa 2 b2 + 2 2ε
for ε > 0
ε |x | |h(t, x, x , x , x )|dt ≤ 2 ′′′
′
′′
′′′
0
1
1 |x | dt + 2ε ′′′ 2
1
|h(t, x, x ′ , x ′′ , x ′′′ )|dt.
0
From condition (H2a ) we obtain the estimate |h(t, x, y, w, z)|2 ≤ 4M 2 {|x|2r + |y|2 + |w|2 + |z|2θ }. Therefore, 1 ′′′ 2 ε ′′′ 2 2M 2 2r 2M 2 ′′′ 2θ 2M 2 ′ 2 2M 2 ′′ 2 |x |2 − |x |2 ≤ |x|2 + |x |2 + |x |2 + |x |2 . 2 2 ε ε ε ε From Holder’s inequality we get 8M 2 2M 2 2r 1 ε 32M 2 ′′′ 2 ′′′ 2θ − |x | ≤ − − |x| + |x | . 2 2 2 1 2 2 ε επ 4 επ 2 Since 0 ≤ θ, r < 1 we infer the existence of a constant M2 such that |x ′′′ |22 < M2
(3.8)
provided ε 32M 2 8M 2 1 > + + . 2 2 επ 4 επ 2 √ 2 The choice ε = 4M π4+π minimizes the right hand side of (3.9) with a minimum value √ B = 4 + π 2. Hence (3.8) holds provided M<
(3.9) 2M(B+π 2 +4) Bπ 2
where
Bπ 2 . 4(B 2 + π 2 + 4)
From (3.8) and x ′ (0) = x ′′ (0) = 0 we get |x ′ |∞ < M3 |x |∞ < M4 , ′′
for some M3 > 0 M4 > 0
(3.10) (3.11)
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and from (3.7) we derive |x|∞ ≤ M1 + |x ′ |∞ < M1 + M3 = M5 .
(3.12)
Now using condition (H2b ) of Theorem 3.1 we get x ′′′ x iv ≤ D(t, x, x ′ , x ′′ ) + m(t) |x ′′′ |2 + 1 and hence
′′′
loge |x | ≤
t
t3
t 1 x ′′′ (s)x (iv) (s) ′′′ 2 ds = loge (|x (s)| + 1) ≤ D + |m|1 2 |x ′′′ |2 + 1 t3
(3.13)
where the constant D depends on M3 , M4 and M5 . Since x ′′′ (t3 ) = 0 we infer from (3.13) that |x ′′′ |∞ < e N0 = M6
(3.14)
where N0 = D + |m|1 . Hence ∥x∥ = max{|x|∞ , |x ′ |∞ , |x ′′ |∞ , |x ′′′ |∞ ≤ max{M4 , M3 , M5 , M6 }}. Therefore, Ω1 is bounded. Let Ω2 = {x ∈ ker L : N x ∈ I m L}. For x ∈ Ω2 , we have x = c ∈ R, thus m−2 ξi s τ2 τ1 αi [ f (v, x(v), 0, 0, 0)]dvdτ1 dτ2 ds = 0. i=1
0
0
0
(3.15)
0
Then we have by H3 and (3.15) that ∥x∥ = c ≤ N ∗ which shows that Ω2 is bounded. We define the isomorphism J : I m Q −→ ker L by J (c) = c,
c ∈ R.
If (3.5) holds we set Ω3 = {x ∈ ker L : −λx + (1 − λ)J Q N x = 0}, λ ∈ [0, 1] For c0 ∈ Ω3 , we obtain (1 − λ)A λc0 = · tl m−2 αi ξil+4
m−2 i=1
αi
0
ξi
0
s
τ2
0
τ1
[ f (v, c0 , 0, 0, 0)]dvdτ1 dτ2 ds
0
i=0
if λ = 1 then c0 = 0 and if |c0 | > N ∗ then from (3.5) we have ξi s τ2 τ1 (1 − λ)c0 A l m−2 2 λc0 = ·t αi [ f (v, c0 , 0, 0, 0)]dvdτ1 dτ2 ds < 0 m−2 0 0 0 0 l+4 i=1 αi ξi i=0
which contradicts λc02 ≥ 0. Thus Ω3 is bounded. If (3.6) holds, then let Ω3 = {x ∈ ker L : λx + (1 − λ)J Q N x = 0, λ ∈ [0, 1]}. Following the above argument we can show that Ω3 is bounded.
8
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3 Ω ⊂ Ω . By the Arzela–Ascoli Theorem we can show that Let Ω be a bounded open subset of X such that ∪i=1 i K p (I − Q)N : Ω −→ X is compact [5]. So N is L-compact. Thus we have shown that
(i) L x ̸= λN x for every (x, λ) ∈ [dom L \ker L ∩∂Ω × (0, 1)] (ii) N x ̸∈ I m L for every x ∈ ker L ∩ ∂Ω . Finally we shall prove that (iii) of Theorem 2.1 is satisfied. Define H (x, λ) = ±λx + (1 − λ)Q N x, we have H (x, 1) = ±x,
H (x, 0) = Q N x.
Thus H (x, λ) is a homotopy from the identity ±I to Q N and is such that H (x, λ) ̸= 0 for every x ∈ ∂Ω ∩ ker L. Therefore deg(J Q N |ker L∩∂ Ω , Ω ∩ ker L , 0) = deg(H (·, 0), Ω ∩ ker L , 0) = deg(H (·, 1), Ω ∩ ker L , 0) = deg(±I, Ω ∩ ker L , 0) ̸= 0. Then by Theorem 2.1 L x = N x has at least one solution in dom L ∩ Ω¯ . In other words (1.1)–(1.2) has at least one solution in C 3 [0, 1]. References [1] [2] [3] [4] [5]
Du Z, Lin Xiaojie, Ge Weigao. On a third order multipoint boundary value problem at resonance. J. Math. Anal. Appl. 2005;302:217–29. Feng W, Webb JRL. Solvability of three-point boundary value problems at resonance. Nonlinear Anal. TMA 2001;47:2407–18. Gupta CP. A second order m-point boundary value problem at resonance. Nonlinear Anal. TMA 1995;24:1483–9. Nagle RK, Pothoven KL. On a third-order nonlinear bvp at resonance. J. Math. Anal. Appl. 1995;195:148–59. Mawhin J. Topological degree methods in non-linear bvp. In: NSF CBMS regional conference series in math. 40. Providence: American Math. Soc.; 1979.
Further reading [1] [2] [3] [4] [5] [6]
Agarwal R. On fourth order bvp arising on beam analysis. Differential Integral Equations 1989;2:91–110. Aftabizadeh AR. Existence and uniqueness theorems for fourth order boundary value problems. J. Math. Anal. Appl. 1986;116:415–26. Aftabizadeh A, Gupta CP, Xu J. Existence and uniqueness theorems for three points bvp. SIAM J. Math. Anal. Appl. 1989;20(3):716–26. Feng W, Webb JR. Solvability of m-point bvp with nonlinear growth. J. Math. Anal. Appl. 1997;212:467–80. Gupta CP. Existence and uniqueness theorems for the bending of an elastic beam equation. Appl. Anal. 1988;926:289–304. Iyase SA. Existence and Uniqueness theorems for a class of three point fourth-order boundary value problems. J. Nigerian Math. Soc. 2010; 29:239–50. [7] Iyase SA. Existence and uniqueness of solutions of a fourth order four point bvp. J. Nigerian Math. Soc. 1999;18:31–40. [8] Liu B. Positive solutions of fourth order two point bvp. Appl. Math. Comput. 2004;148:107–20.