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On the existence of solutions of p -Laplacian m -point boundary value problem at resonance
Nonlinear Analysis 70 (2009) 1557–1564 www.elsevier.com/locate/na
On the existence of solutions of p-Laplacian m-point boundary value problem at resonanceI Yanling Zhu ∗ , Kai Wang Department of Applied Mathematics, Anhui University of Finance & Economics, Bengbu 233000, Anhui Province, PR China Received 24 June 2007; accepted 14 February 2008
Abstract By using the theory of coincidence degree, we study a kind of solutions of p-Laplacian m-point boundary value problem at resonance in the following form 0 0 0 (φ p (u )) (t) = f (t, u, u ), 0 < t < 1, m−2 X 0 ai u(ξi ), u(1) = u (0) = 0, i=1
Pm−2 where m ≥ 3, ai > 0 (i = 1, 2, . . . , m − 2), 0 < ξ1 < ξ2 < · · · < ξm−2 < 1 such that i=1 ai = 1. A result on the existence of solutions is obtained. The degrees of two variables x1 , x2 in the function f (t, x1 , x2 ) are allowable to be bigger than 1. c 2008 Elsevier Ltd. All rights reserved.
MSC: 34B15 Keywords: m-point boundary value problem; Theory of coincidence degree; p-Laplacian
1. Introduction The second-order boundary value problem has been extensively studied in papers [1–8]. In 1960, Hartman in [7] proved the two-dimensional system as follows 00 u = f (t, u), 0 < t < T, (1.1) u(0) = u(T ) = 0. The condition (H) : ∃M1 > 0, such that h f (t, u), ui ≥ 0,
∀t ∈ [0, T ], |u| = M1
I The work is supported by the Natural Science Foundation of Educational Bureau of Anhui Province (No. KJ2008B235) and Youth Foundation of Anhui University of Finance and Economics (No. ACKTQ0748ZC). ∗ Corresponding author. Tel.: +86 552 4128667. E-mail addresses: [email protected], [email protected] (Y. Zhu).
c 2008 Elsevier Ltd. All rights reserved. 0362-546X/$ - see front matter doi:10.1016/j.na.2008.02.035
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Y. Zhu, K. Wang / Nonlinear Analysis 70 (2009) 1557–1564
is imposed on f (t, u). In 2000, Jean Mawhin in [8] extended the result of Hartman. The author studied the boundary value problem as follows (φ p (u 0 ))0 (t) = f (t, u), 0 < t < T, (1.2) u(0) = u(T ) = 0. But for p-Laplacian m-point boundary value problem, as far as we know, the problem has been studied far less often. The reason for this is that we cannot apply the theory of coincidence degree directly since the p-Laplacian operator (φ p (u 0 ))0 = (|u 0 | p−2 u 0 )0 is not linear with respect to u except p = 2. In this paper, we study the existence of solutions of p-Laplacian m-point boundary value problem 0 0 0 (φ p (u )) (t) = f (t, u, u ), 0 < t < 1, m−2 X (1.3) 0 u(1) = ai u(ξi ), u (0) = 0, i=1
Pm−2 where m ≥ 3, ai > 0, i = 1, 2, . . . , m − 2, 0 < ξ1 < ξ2 < · · · < ξm−2 < 1 such that i=1 ai = 1. We translate BVP (1.3) into a right form and overcome some difficulties in using the theory of coincidence degree directly such as seeking D(L) and Im L. The significance is that the degrees of two variables x1 , x2 in the function f (t, x1 , x2 ) are allowable to be bigger than 1. By using the theory of coincidence degree, we obtain a new result to guarantee the existence of solutions. Furthermore, an example is given to demonstrate our result. 2. Some preliminaries In order to use Mawhin’s continuation theorem, We first rewrite BVP (1.3) in the following form 0 x (t) = φq (x2 (t)), 10 x2 (t) = f (t, x1 (t), φq (x2 (t))), 0 = φq (x2 (0)), m−2 X ai x1 (ξi ), 0 = x1 (1) −
(2.1)
i=1
where 1/ p + 1/q = 1. We can easily see if x(t) = (x1 (t), x2 (t))T is a solution of (2.1), then x1 (t) is a solution of BVP(1.3). We set the following notations: X = {x = (x1 , x2 )> ∈ C[0, 1] × C[0, 1]} with the norm kxk = max{|x1 |0 , |x2 |0 }, where|x|0 = maxt∈[0,1] |x(t)|. Y = {y = (y1 , y2 , c1 , c2 )> ∈ C[0, 1] × C[0, 1] × R 2 } with the norm kyk = max{|y1 |0 , |y2 |0 , |c1 |, |c2 |}. Clearly, X and Y are Banach spaces. We define operator L and N in the following form respectively: L : D(L) → Y : L x = L(x1 , x2 )> = (x10 , x20 , 0, 0)> , where ( ) m−2 X D(L) = x = (x1 , x2 )> ∈ C 1 [0, 1] × C 1 [0, 1]|x10 (t) = φq (x2 (t)), x1 (1) = ai x1 (ξi ), x2 (0) = 0 , i=1
( Im L =
) Z 1 m−2 m−2 X Z 1 Z s X > y = (y1 , y2 , 0, 0) ∈ Y ai y1 (s)ds = 0, ai φq y2 (u)du ds = 0 , i=1 0 ξi ξi i=1
and N : X → Y , φq (x2 (t)) f (t, x1 (t), φq (x2 (t))) x1 (t) φq (x2 (0)) . (N x)(t) = N = m−2 x2 (t) X x1 (1) − ai x1 (ξi )
i=1
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Then (2.1) is equivalent to operator equation L x = N x. We can also prove that ker L = R 2 , Y /Im L = R 2 , so L is a Fredholm operator with index zero. Let project operators P, Q as follows, respectively, P : X → ker L , P(x) = P(x1 , x2 )> = (x1 (1), x2 (1))> ; Q : Y → ImQ ⊂ R 4 , Qy = Q(y1 , y2 , c1 , c2 )> =
1−
1 m−2 P
m−2 X
ai ξi
Z ai
1 ξi
i=1
>
y1 (s)ds, 1−
i=1
1 m−2 P
m−2 X
ai ξi
Z ai
i=1
1
ξi
φq
Z 0
s
y2 (u)du ds, 0, 0 .
i=1
Set L p = L | D(L)∩ker P and K = L −1 p : Im L → D(L) denotes the inverse of L p , then K y = K (y1 , y2 , 0, 0) = − >
Z
1
y1 (s)ds, −
t
Z
1
!> y2 (s)ds
.
(2.2)
t
From (2.2) it is easy to see that N is L-compact on Ω , where Ω is an open bounded set in X . Lemma ([9]). Suppose that X and Y are two Banach spaces, and L : D(L) ⊂ X → Y is a Fredholm operator with index zero. Furthermore, Ω ⊂ X is an open bounded set and N : Ω → Y is L-compact on Ω . If (1) L x 6= λN x, ∀x ∈ ∂Ω ∩ D(L), λ ∈ (0, 1); (2) N x 6∈ Im L , ∀x ∈ ∂Ω ∩ ker L; (3) deg{J Q N , Ω ∩ ker L , 0} 6= 0, where J : ImQ → ker L is an isomorphism, then the equation L x = N x has a solution in Ω ∩ D(L). 3. Main result Theorem 3.1. Suppose p > 2 and the following conditions are satisfied (H1 ) There is a constant D > 0 such that f (t, x1 , x2 ) > 0,
∀t ∈ [0, 1], x1 > D, x2 ≥ 0;
f (t, x1 , x2 ) < 0,
∀t ∈ [0, 1], x1 < −D, x2 ≤ 0.
(H2 ) The function f has the decomposition f (t, x1 , x2 ) = g(t, x1 , x2 ) + h 1 (t, x1 ) + h 2 (t, x2 ) such that x2 g(t, x1 , x2 ) ≤ −β|x2 |n+1 , x2 h 2 (t, x2 ) < 0,
∀(t, x1 , x2 ) ∈ [0, 1] × R 2 ,
∀(t, x2 ) ∈ [0, 1] × R,
and lim
sup
|x|→+∞ t∈[0,1]
|h 1 (t, x)| ≤ r1 , |x|n
where n ≥ 1, r1 ≥ 0, β > r1 , are all constants, g(t, x1 , x2 ) is continuous on [0, 1] × R × R, and h i (t, xi ), i ∈ {1, 2}, is continuous on [0, 1] × R. Then (2.1) has at least one solution. Pm−2 Proof. We can easily obtain i=1 ai ξi < 1. Consider the operator equation: L x = λN x, λ ∈ (0, 1). Set Ω1 = {x ∈ X : L x = λN x, λ ∈ (0, 1)}, ∀x(t) = (x1 , x2 )> ∈ Ω1 , then 0 x (t) = λφq (x2 (t)), 10 x2 (t) = λ f (t, x1 (t), φq (x2 (t))), 0 = λφ"q (x2 (0)), # m−2 X 0 = λ x1 (1) − ai x1 (ξi ) . i=1
(3.1)
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Y. Zhu, K. Wang / Nonlinear Analysis 70 (2009) 1557–1564
From the first equation of (3.1) we can know that x2 (t) = ϕ p ( λ1 x10 (t)), which together with the second equation of (3.1) yields (φ p (x 0 (t)))0 = λ p f t, x1 (t), 1 x 0 (t) , ∀t ∈ (0, 1), 1 λ 1 (3.2) m−2 X 0 (0) = 0, x (1) = x a x (ξ ). i i 1 1 1 i=1
Now we show that |x1 (0)| ≤ D.
(3.3)
Suppose the contrary, then |x1 (0)| ≥ D. Without loss of generality, we assume that x1 (0) > D. From the second equation of (3.1) and condition (H1 ), we can get 1 x20 (t) = λ f 0, x1 (0), x10 (0) = λ f (0, x1 (0), 0) > 0. λ By the continuity of x2 (t) on [0, 1], we obtain ∃δ > 0,
such that x20 (t) > 0,
∀t ∈ (0, δ).
(3.4)
In view of the first equation of (3.1), we see that x10 (0) = λφq (x2 (0)) = 0, i.e., x2 (0) = 0. By (3.4) we have ∀t ∈ (0, δ),
x2 (t) > 0,
i.e., x10 (t) > 0.
Now we prove x10 (t) > 0,
∀t ∈ (0, 1).
(3.5)
Otherwise, ∃ t0 ∈ (0, 1) such that x10 (t0 ) = 0, so x2 (t0 ) = 0, i.e., x20 (t0 ) ≤ 0.
(3.6)
We see t
Z
x1 (t) = x1 (0) +
0
x10 (s)ds > x1 (0) > D,
∀t ∈ [0, t0 ],
so x20 (t)
= λf
1 0 t, x1 (t), x1 (t) > 0, λ
∀t ∈ [0, t0 ],
which is a contradiction to (3.6). So (3.5) holds. However x1 (1) =
m−2 X
ai x1 (ξi ) < x1 (ξm − 2) < x1 (1),
i=1
which is a contradiction. This implies that (3.3) holds. Thus Z t 0 |x1 (t)| ≤ |x1 (0)| + x1 (s)ds 0
Z ≤ D+ 0
1
|x10 (t)|dt
∀t ∈ [0, 1].
(3.7)
On the other hand, multiplying the two sides of the first equation of (3.2) by φ p (x10 (t)) and integrating them over [0, 1], we get Z 1 2 1 1 0 0 p 0 φ p (x1 (t)) = λ φ p (x1 (t)) f t, x1 (t), x1 (t) dt 2 λ 0
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Y. Zhu, K. Wang / Nonlinear Analysis 70 (2009) 1557–1564 1
1 t, x1 (t), x10 (t) dt λ 0 Z 1 1 |x10 (t)| p−2 x10 (t)g t, x1 (t), x10 (t) dt = λp λ 0 Z 1 Z + λp |x10 (t)| p−2 x10 (t)h 1 (t, x1 (t))dt + λ p = λp
Z
|x10 (t)| p−2 x10 (t) f
0
1
0
1 |x10 (t)| p−2 x10 (t)h 2 t, x10 (t) dt. λ
(3.8)
By condition (H2 ), we know n+1 Z 1 Z 1 2 1 p+1 0 p−2 1 0 0 p dt ≤ − φ p (x1 (1)) + λ λ β |x1 (t)| |x10 (t)| p−2 x10 (t)h 1 (t, x1 (t))dt λ x1 (t) 2 0 0 Z 1 1 0 0 p−2 0 p |x1 (t)| x1 (t)h 2 t, x1 (t) dt, +λ λ 0 i.e., λ
p−n
β
Z 0
1
|x10 (t)| p+n−1 dt
≤λ
p
1
Z 0
≤ λp
1
Z 0
|x10 (t)| p−2 x10 (t)h 1 (t, x1 (t))dt
Z 0
1
|x10 (t)| p−2 x10 (t)h 2
1 0 t, x1 (t) dt λ
|x10 (t)| p−1 |h 1 (t, x1 (t))|dt. β−r1 3 ,
It follows from condition (H2 ) that for ε = |h 1 (t, x1 )| ≤ (r1 + ε)|x1 | ,
+λ
p
(3.9)
there exists a constant ρ > D such that
∀t ∈ [0, 1], |x1 | > ρ.
n
So β
Z
1
0
|x10 (t)| p+n−1 dt
≤λ
n
1
Z 0
|x10 (t)| p−1 (r1
≤ (r1 + ε)|x1 |n0
1
Z 0
≤ (r1 + ε) D +
+ ε)|x1 (t)| dt + λ n
n
0
|x10 (t)| p−1 dt + h 1, ρ
Z 0
1
1
Z
|x10 (t)|dt
!n Z
1 0
1
Z 0
|x10 (t)| p−1 h 1, ρ dt
|x10 (t)| p−1 dt
|x10 (t)| p−1 dt + h 1, ρ
Z 0
1
|x10 (t)| p−1 dt,
(3.10)
where h 1, ρ = max t∈[0,1] |h 1 (t, x)|. |x|≤ρ
From the knowledge of mathematical analysis, we know that there exists a constant δ ∈ (0, 1) such that (1 + x)n < 1 + (n + 1)x, R1 Case 1: If 0 |x10 (t)|dt = 0 or we get β
Z 0
1
∀x ∈ (0, δ), n > 0. R1 0
D |x10 (t)|dt
(3.11)
≥ δ, from (3.7) we see that |x10 (t)| ≤ D + D/δ. Then by (3.10) and (3.11) !n Z Z 1 1 |x10 (t)|dt 0 p−1 |x1 (t)| dt + h 1, ρ |x10 (t)| p−1 dt 1+ D 0 0 !Z Z Z 1 1 n+1 1 0 1+ |x1 (t)|dt |x10 (t)| p−1 dt + h 1, ρ |x10 (t)| p−1 dt D 0 0 0 ! p−1 ! p−1 Z 1 Z 1 p+n−1 p+n−1 |x10 (t)| p+n−1 dt + h 1, ρ |x10 (t)| p+n−1 dt R1
|x10 (t)| p+n−1 dt
≤ (r1 + ε)D
n
≤ (r1 + ε)D n
≤ (r1 + ε)D n
0
0
+ (r1 + ε)D n−1 (n + 1)
0
Z 0
1
! |x10 (t)| p+n−1 dt
p p+n−1
.
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Y. Zhu, K. Wang / Nonlinear Analysis 70 (2009) 1557–1564
So there is a constant M0 > 0 such that Case 2: If β
Z 0
1
R1 0
|x10 (t)| p+n−1 dt ≤ M0 .
≤ δ, it follows from (3.10) and (3.11) that
D R1 0 |x 0 1 (t)|dt
|x10 (t)| p+n−1 dt 1
Z
≤ (r1 + ε)
0
≤ (r1 + ε)
0 1
Z
≤ (r1 + ε)
0
!n
|x10 (t)|dt Z
1
0
≤ (r1 + ε)
1
+ (r1 + ε)(n + 1)D
Z 0
1 In view of ε = β−r 3 , From (3.13) we have
|x1 (t)| ≤ D +
!n Z
1
p+n−2 p+n−1
1
Z 0
1
|x10 (t)|dt
0
|x10 (t)|dt
!n−1 Z
1
0
Z
1
|x10 (t)| p−1 dt
0
!Z
1 0
|x10 (t)| p−1 dt + h 1,ρ
|x10 (t)| p+n−1 dt + h 1,ρ
0
!n Z
D
(n + 1)D 1+ R1 0 0 |x 1 (t)|dt
|x10 (t)|dt
+ (r1 + ε)(n + 1)D Z
1+ R1 0
1
Z
!n |x10 (t)|dt
|x10 (t)| p−1 dt Z
1
0
+ h 1, ρ
0
Z + h 1, ρ
0
1
1
|x10 (t)| p−1 dt
|x10 (t)| p−1 dt
|x10 (t)| p−1 dt
|x10 (t)| p−1 dt !
1
Z
p−1 p+n−1
|x10 (t)| p+n−1 dt
0
! p+n−2 p+n−1
.
|x10 (t)| p+n−1 dt
< 1, we can see that there is a constant M1 > 0 such that !
1 p+n−1
R1 0
|x10 (t)| p+n−1 dt < M1 .
1
≤ D + M1p+n−1 .
|x10 (t)| p+n−1 dt
So either Case 1 or Case 2, we obtain 1
|x1 (t)| ≤ max{D + D/δ, D + M1p+n−1 } := M2 Z 1 |x10 (t)| p+n−1 dt ≤ max{M0 , M1 } := M3 .
(3.12)
0
Multiplying the two sides of the first equation of (3.2) by φ p ( λ1 x10 (t)) and integrating them from 0 to t, we get 2 Z t 1 0 1 0 1 0 1 p φp x (t) =λ φp x (t) f s, x1 (s), x1 (s) ds 2 λ 1 λ 1 λ 0 Z t 1 = λ2− p |x10 (s)| p−2 x10 (s)g s, x1 (s), x10 (s) ds λ 0 Z t Z t 1 + λ2− p |x10 (s)| p−2 x10 (s)h 1 (s, x1 (s))ds + λ2− p |x10 (s)| p−2 x10 (s)h 2 s, x10 (s) ds λ 0 0 Z t Z 1 ≤ λ2− p |x10 (s)| p−1 |h 1 (s, x1 (s))|ds ≤ λ2− p h 1,M2 |x10 (t)| p−1 dt 0
≤ λ2− p h 1,M2
0
Z 0
1
! |x10 (t)| p+n−1 dt
p−1 p+n−1
,
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Y. Zhu, K. Wang / Nonlinear Analysis 70 (2009) 1557–1564
where h 1,M2 = max t∈[0,1] |h 1 (t, x)|. In view of p < 2, ε = (β − r1 )/3, |x|≤M2
φp
2 p−1 1 0 x1 (t) ≤ 2h 1,M2 M3p+n−1 := M4 , λ
R1 0
|x10 (s)| p+n−1 ds ≤ M3 , we have
∀t ∈ [0, 1],
i.e., 1 1 0 x (t) ≤ M 2 p−2 := M5 . 4 λ 1 0 From the second equation of (3.1) we easily have x2 (0) = 0 and Z 1 Z 1 1 0 0 |x2 (t)|dt ≤ f t, x1 (t), λ x1 (t) dt ≤ f M2 ,M5 , 0 0 where f M2 ,M5 = max t∈[0,1],|x1 |≤M2 | f (t, x1 , x2 )|. Then |x2 |≤M5
|x2 (t)| ≤ |x2 (0)| +
1
Z 0
|x20 (t)|dt ≤ f M2 ,M5 := M6 .
Let M = max{M2 , M6 } + 1, Ω = {x ∈ X : |x1 |0 < M, |x2 |0 < M}, so condition (1) of Lemma is satisfied. Now we show that N x 6∈ Im L , ∀x ∈ ∂Ω ∩ ker L. If not, there exists x0 = (x1 , x2 )> ∈ ∂Ω ∩ ker L such that N x0 = (φq (x2 ), f (t, x1 , φq (x2 )), φq (x2 ), 0)> ∈ Im L . So Q N x0 = 0. But from the definition of operator Q we get Q N x0 = φq (x2 ),
1−
1 m−2 P
m−2 X
ai ξi
Z ai
i=1
1
ξi
φq
Z 0
s
>
f (u, x1 , φq (x2 )du)ds, 0, 0 = 0.
i=1
So x2 = 0. If x1 > D, from condition (H1 ) we have f (t, x1 , x2 ) > 0 and then m−2 X Z 1 Z s ai φq f (u, x1 , φq (x2 )du)ds > 0, i=1
ξi
0
which is a contradiction. Similarly, we also have a contradiction when x1 < −D. Thus |x1 | ≤ D < M, x2 = 0, which is a contradiction to x0 = (x1 , x2 )T ∈ ∂Ω . So condition (2) of Lemma is satisfied. Define the isomorphism J : ImQ → ker L , J (c1 , c2 , 0, 0)> = (c2 , c1 )> . Let H (x, µ) = µx + (1 − µ)J Q N x,
∀(x, µ) ∈ Ω¯ × [0, 1].
Then we have
µx1 + H (x, µ) =
m−2 P R1 1−µ ai ξi m−2 P 1− ai ξi i=1
f (s, x1 , φq (x2 ))ds 6= 0, i=1 q−2 µx2 + (1 − µ)|x2 | x2
∀(x, µ) ∈ (∂Ω ∩ ker L) × [0, 1]. Hence deg{J Q N , Ω ∩ ker L , 0} = deg{H (x, 0), Ω ∩ ker L , 0} = deg{H (x, 1), Ω ∩ ker L , 0} = deg{I, Ω ∩ ker L , 0} 6= 0. So condition (3) of Lemma is satisfied. By applying Lemma, we conclude that equation L x = N x has a solution x(t) = (x1 (t), x2 (t))T on Ω ∩ D(L), i.e., Eq. (1.3) has a solution x1 (t).
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Y. Zhu, K. Wang / Nonlinear Analysis 70 (2009) 1557–1564
As an application, we consider the following equation: 0 0 4 05 5 03 (ϕ 3 (x (t))) = −(6 + x (t))x (t) + x (t) − x (t), 2 1 1 3 2 0 x(1) = x + x . x (0) = 0, 4 2 4 3
t ∈ (0, 1), (3.13)
Corresponding to BVP(1.3), we have p = 3/2, g(t, x1 , x2 ) = −(6 + x14 )x25 , h 1 (t, x1 ) = x15 , h 2 (t, x2 ) = −x23 , a1 = 1/4, a2 = 3/4, β = 6, r1 = 5 and we can easily verify the conditions of Theorem 3.1 hold. So we know that BVP(1.3) has at least one solution. References [1] V.A. Il’in, E.I. Moiseev, Nonlocal boundary value problems of the second kind for a Sturm–Liouville operator, J. Differential Equations 23 (1987) 979–987. [2] C.P. Gupta, A second order m-point boundary value problem at resonance, Nonlinear Anal. 24 (1995) 1483–1489. [3] C.P. Gupta, Solvability of a multiple boundary value problem at resonance, Results Math. 28 (1995) 270–276. [4] R.Y. Ma, Existence theorems for second order m-point boundary value problems, J. Math. Anal. Appl. 211 (1997) 545–555. [5] W. Feng, J.R.L. Webb, Solvability of m-point boundary value problems with nonlinear growth, J. Math. Anal. Appl. 212 (1997) 467–480. [6] C.P. Gupta, S.K. Ntouyas, P.Ch. Tsamatos, Solvability of an m-point boundary value problem for second order ordinary differential equations, J. Math. Anal. Appl. 189 (1995) 575–584. [7] Ph. Hartman, On boundary value problems for systems of ordinary nonlinear second order differential equations, Trans. Amer. Math. Soc. 96 (1960) 493–509. [8] J. Mawhin, Some boundary value problems for Hartman-type perturbations of the ordinary vector p-Laplacian, Nonlinear Anal. 40 (2000) 497–503. [9] R. Gaines, J. Mawhin, Coincidence Degree and Nonlinear Differential Equations, Springer, Berlin, 1977.
On the existence of solutions of p-Laplacian m-point boundary value problem at resonanceI Yanling Zhu ∗ , Kai Wang Department of Applied Mathematics, Anhui University of Finance & Economics, Bengbu 233000, Anhui Province, PR China Received 24 June 2007; accepted 14 February 2008
Abstract By using the theory of coincidence degree, we study a kind of solutions of p-Laplacian m-point boundary value problem at resonance in the following form 0 0 0 (φ p (u )) (t) = f (t, u, u ), 0 < t < 1, m−2 X 0 ai u(ξi ), u(1) = u (0) = 0, i=1
Pm−2 where m ≥ 3, ai > 0 (i = 1, 2, . . . , m − 2), 0 < ξ1 < ξ2 < · · · < ξm−2 < 1 such that i=1 ai = 1. A result on the existence of solutions is obtained. The degrees of two variables x1 , x2 in the function f (t, x1 , x2 ) are allowable to be bigger than 1. c 2008 Elsevier Ltd. All rights reserved.
MSC: 34B15 Keywords: m-point boundary value problem; Theory of coincidence degree; p-Laplacian
1. Introduction The second-order boundary value problem has been extensively studied in papers [1–8]. In 1960, Hartman in [7] proved the two-dimensional system as follows 00 u = f (t, u), 0 < t < T, (1.1) u(0) = u(T ) = 0. The condition (H) : ∃M1 > 0, such that h f (t, u), ui ≥ 0,
∀t ∈ [0, T ], |u| = M1
I The work is supported by the Natural Science Foundation of Educational Bureau of Anhui Province (No. KJ2008B235) and Youth Foundation of Anhui University of Finance and Economics (No. ACKTQ0748ZC). ∗ Corresponding author. Tel.: +86 552 4128667. E-mail addresses: [email protected], [email protected] (Y. Zhu).
c 2008 Elsevier Ltd. All rights reserved. 0362-546X/$ - see front matter doi:10.1016/j.na.2008.02.035
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Y. Zhu, K. Wang / Nonlinear Analysis 70 (2009) 1557–1564
is imposed on f (t, u). In 2000, Jean Mawhin in [8] extended the result of Hartman. The author studied the boundary value problem as follows (φ p (u 0 ))0 (t) = f (t, u), 0 < t < T, (1.2) u(0) = u(T ) = 0. But for p-Laplacian m-point boundary value problem, as far as we know, the problem has been studied far less often. The reason for this is that we cannot apply the theory of coincidence degree directly since the p-Laplacian operator (φ p (u 0 ))0 = (|u 0 | p−2 u 0 )0 is not linear with respect to u except p = 2. In this paper, we study the existence of solutions of p-Laplacian m-point boundary value problem 0 0 0 (φ p (u )) (t) = f (t, u, u ), 0 < t < 1, m−2 X (1.3) 0 u(1) = ai u(ξi ), u (0) = 0, i=1
Pm−2 where m ≥ 3, ai > 0, i = 1, 2, . . . , m − 2, 0 < ξ1 < ξ2 < · · · < ξm−2 < 1 such that i=1 ai = 1. We translate BVP (1.3) into a right form and overcome some difficulties in using the theory of coincidence degree directly such as seeking D(L) and Im L. The significance is that the degrees of two variables x1 , x2 in the function f (t, x1 , x2 ) are allowable to be bigger than 1. By using the theory of coincidence degree, we obtain a new result to guarantee the existence of solutions. Furthermore, an example is given to demonstrate our result. 2. Some preliminaries In order to use Mawhin’s continuation theorem, We first rewrite BVP (1.3) in the following form 0 x (t) = φq (x2 (t)), 10 x2 (t) = f (t, x1 (t), φq (x2 (t))), 0 = φq (x2 (0)), m−2 X ai x1 (ξi ), 0 = x1 (1) −
(2.1)
i=1
where 1/ p + 1/q = 1. We can easily see if x(t) = (x1 (t), x2 (t))T is a solution of (2.1), then x1 (t) is a solution of BVP(1.3). We set the following notations: X = {x = (x1 , x2 )> ∈ C[0, 1] × C[0, 1]} with the norm kxk = max{|x1 |0 , |x2 |0 }, where|x|0 = maxt∈[0,1] |x(t)|. Y = {y = (y1 , y2 , c1 , c2 )> ∈ C[0, 1] × C[0, 1] × R 2 } with the norm kyk = max{|y1 |0 , |y2 |0 , |c1 |, |c2 |}. Clearly, X and Y are Banach spaces. We define operator L and N in the following form respectively: L : D(L) → Y : L x = L(x1 , x2 )> = (x10 , x20 , 0, 0)> , where ( ) m−2 X D(L) = x = (x1 , x2 )> ∈ C 1 [0, 1] × C 1 [0, 1]|x10 (t) = φq (x2 (t)), x1 (1) = ai x1 (ξi ), x2 (0) = 0 , i=1
( Im L =
) Z 1 m−2 m−2 X Z 1 Z s X > y = (y1 , y2 , 0, 0) ∈ Y ai y1 (s)ds = 0, ai φq y2 (u)du ds = 0 , i=1 0 ξi ξi i=1
and N : X → Y , φq (x2 (t)) f (t, x1 (t), φq (x2 (t))) x1 (t) φq (x2 (0)) . (N x)(t) = N = m−2 x2 (t) X x1 (1) − ai x1 (ξi )
i=1
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Y. Zhu, K. Wang / Nonlinear Analysis 70 (2009) 1557–1564
Then (2.1) is equivalent to operator equation L x = N x. We can also prove that ker L = R 2 , Y /Im L = R 2 , so L is a Fredholm operator with index zero. Let project operators P, Q as follows, respectively, P : X → ker L , P(x) = P(x1 , x2 )> = (x1 (1), x2 (1))> ; Q : Y → ImQ ⊂ R 4 , Qy = Q(y1 , y2 , c1 , c2 )> =
1−
1 m−2 P
m−2 X
ai ξi
Z ai
1 ξi
i=1
>
y1 (s)ds, 1−
i=1
1 m−2 P
m−2 X
ai ξi
Z ai
i=1
1
ξi
φq
Z 0
s
y2 (u)du ds, 0, 0 .
i=1
Set L p = L | D(L)∩ker P and K = L −1 p : Im L → D(L) denotes the inverse of L p , then K y = K (y1 , y2 , 0, 0) = − >
Z
1
y1 (s)ds, −
t
Z
1
!> y2 (s)ds
.
(2.2)
t
From (2.2) it is easy to see that N is L-compact on Ω , where Ω is an open bounded set in X . Lemma ([9]). Suppose that X and Y are two Banach spaces, and L : D(L) ⊂ X → Y is a Fredholm operator with index zero. Furthermore, Ω ⊂ X is an open bounded set and N : Ω → Y is L-compact on Ω . If (1) L x 6= λN x, ∀x ∈ ∂Ω ∩ D(L), λ ∈ (0, 1); (2) N x 6∈ Im L , ∀x ∈ ∂Ω ∩ ker L; (3) deg{J Q N , Ω ∩ ker L , 0} 6= 0, where J : ImQ → ker L is an isomorphism, then the equation L x = N x has a solution in Ω ∩ D(L). 3. Main result Theorem 3.1. Suppose p > 2 and the following conditions are satisfied (H1 ) There is a constant D > 0 such that f (t, x1 , x2 ) > 0,
∀t ∈ [0, 1], x1 > D, x2 ≥ 0;
f (t, x1 , x2 ) < 0,
∀t ∈ [0, 1], x1 < −D, x2 ≤ 0.
(H2 ) The function f has the decomposition f (t, x1 , x2 ) = g(t, x1 , x2 ) + h 1 (t, x1 ) + h 2 (t, x2 ) such that x2 g(t, x1 , x2 ) ≤ −β|x2 |n+1 , x2 h 2 (t, x2 ) < 0,
∀(t, x1 , x2 ) ∈ [0, 1] × R 2 ,
∀(t, x2 ) ∈ [0, 1] × R,
and lim
sup
|x|→+∞ t∈[0,1]
|h 1 (t, x)| ≤ r1 , |x|n
where n ≥ 1, r1 ≥ 0, β > r1 , are all constants, g(t, x1 , x2 ) is continuous on [0, 1] × R × R, and h i (t, xi ), i ∈ {1, 2}, is continuous on [0, 1] × R. Then (2.1) has at least one solution. Pm−2 Proof. We can easily obtain i=1 ai ξi < 1. Consider the operator equation: L x = λN x, λ ∈ (0, 1). Set Ω1 = {x ∈ X : L x = λN x, λ ∈ (0, 1)}, ∀x(t) = (x1 , x2 )> ∈ Ω1 , then 0 x (t) = λφq (x2 (t)), 10 x2 (t) = λ f (t, x1 (t), φq (x2 (t))), 0 = λφ"q (x2 (0)), # m−2 X 0 = λ x1 (1) − ai x1 (ξi ) . i=1
(3.1)
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Y. Zhu, K. Wang / Nonlinear Analysis 70 (2009) 1557–1564
From the first equation of (3.1) we can know that x2 (t) = ϕ p ( λ1 x10 (t)), which together with the second equation of (3.1) yields (φ p (x 0 (t)))0 = λ p f t, x1 (t), 1 x 0 (t) , ∀t ∈ (0, 1), 1 λ 1 (3.2) m−2 X 0 (0) = 0, x (1) = x a x (ξ ). i i 1 1 1 i=1
Now we show that |x1 (0)| ≤ D.
(3.3)
Suppose the contrary, then |x1 (0)| ≥ D. Without loss of generality, we assume that x1 (0) > D. From the second equation of (3.1) and condition (H1 ), we can get 1 x20 (t) = λ f 0, x1 (0), x10 (0) = λ f (0, x1 (0), 0) > 0. λ By the continuity of x2 (t) on [0, 1], we obtain ∃δ > 0,
such that x20 (t) > 0,
∀t ∈ (0, δ).
(3.4)
In view of the first equation of (3.1), we see that x10 (0) = λφq (x2 (0)) = 0, i.e., x2 (0) = 0. By (3.4) we have ∀t ∈ (0, δ),
x2 (t) > 0,
i.e., x10 (t) > 0.
Now we prove x10 (t) > 0,
∀t ∈ (0, 1).
(3.5)
Otherwise, ∃ t0 ∈ (0, 1) such that x10 (t0 ) = 0, so x2 (t0 ) = 0, i.e., x20 (t0 ) ≤ 0.
(3.6)
We see t
Z
x1 (t) = x1 (0) +
0
x10 (s)ds > x1 (0) > D,
∀t ∈ [0, t0 ],
so x20 (t)
= λf
1 0 t, x1 (t), x1 (t) > 0, λ
∀t ∈ [0, t0 ],
which is a contradiction to (3.6). So (3.5) holds. However x1 (1) =
m−2 X
ai x1 (ξi ) < x1 (ξm − 2) < x1 (1),
i=1
which is a contradiction. This implies that (3.3) holds. Thus Z t 0 |x1 (t)| ≤ |x1 (0)| + x1 (s)ds 0
Z ≤ D+ 0
1
|x10 (t)|dt
∀t ∈ [0, 1].
(3.7)
On the other hand, multiplying the two sides of the first equation of (3.2) by φ p (x10 (t)) and integrating them over [0, 1], we get Z 1 2 1 1 0 0 p 0 φ p (x1 (t)) = λ φ p (x1 (t)) f t, x1 (t), x1 (t) dt 2 λ 0
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Y. Zhu, K. Wang / Nonlinear Analysis 70 (2009) 1557–1564 1
1 t, x1 (t), x10 (t) dt λ 0 Z 1 1 |x10 (t)| p−2 x10 (t)g t, x1 (t), x10 (t) dt = λp λ 0 Z 1 Z + λp |x10 (t)| p−2 x10 (t)h 1 (t, x1 (t))dt + λ p = λp
Z
|x10 (t)| p−2 x10 (t) f
0
1
0
1 |x10 (t)| p−2 x10 (t)h 2 t, x10 (t) dt. λ
(3.8)
By condition (H2 ), we know n+1 Z 1 Z 1 2 1 p+1 0 p−2 1 0 0 p dt ≤ − φ p (x1 (1)) + λ λ β |x1 (t)| |x10 (t)| p−2 x10 (t)h 1 (t, x1 (t))dt λ x1 (t) 2 0 0 Z 1 1 0 0 p−2 0 p |x1 (t)| x1 (t)h 2 t, x1 (t) dt, +λ λ 0 i.e., λ
p−n
β
Z 0
1
|x10 (t)| p+n−1 dt
≤λ
p
1
Z 0
≤ λp
1
Z 0
|x10 (t)| p−2 x10 (t)h 1 (t, x1 (t))dt
Z 0
1
|x10 (t)| p−2 x10 (t)h 2
1 0 t, x1 (t) dt λ
|x10 (t)| p−1 |h 1 (t, x1 (t))|dt. β−r1 3 ,
It follows from condition (H2 ) that for ε = |h 1 (t, x1 )| ≤ (r1 + ε)|x1 | ,
+λ
p
(3.9)
there exists a constant ρ > D such that
∀t ∈ [0, 1], |x1 | > ρ.
n
So β
Z
1
0
|x10 (t)| p+n−1 dt
≤λ
n
1
Z 0
|x10 (t)| p−1 (r1
≤ (r1 + ε)|x1 |n0
1
Z 0
≤ (r1 + ε) D +
+ ε)|x1 (t)| dt + λ n
n
0
|x10 (t)| p−1 dt + h 1, ρ
Z 0
1
1
Z
|x10 (t)|dt
!n Z
1 0
1
Z 0
|x10 (t)| p−1 h 1, ρ dt
|x10 (t)| p−1 dt
|x10 (t)| p−1 dt + h 1, ρ
Z 0
1
|x10 (t)| p−1 dt,
(3.10)
where h 1, ρ = max t∈[0,1] |h 1 (t, x)|. |x|≤ρ
From the knowledge of mathematical analysis, we know that there exists a constant δ ∈ (0, 1) such that (1 + x)n < 1 + (n + 1)x, R1 Case 1: If 0 |x10 (t)|dt = 0 or we get β
Z 0
1
∀x ∈ (0, δ), n > 0. R1 0
D |x10 (t)|dt
(3.11)
≥ δ, from (3.7) we see that |x10 (t)| ≤ D + D/δ. Then by (3.10) and (3.11) !n Z Z 1 1 |x10 (t)|dt 0 p−1 |x1 (t)| dt + h 1, ρ |x10 (t)| p−1 dt 1+ D 0 0 !Z Z Z 1 1 n+1 1 0 1+ |x1 (t)|dt |x10 (t)| p−1 dt + h 1, ρ |x10 (t)| p−1 dt D 0 0 0 ! p−1 ! p−1 Z 1 Z 1 p+n−1 p+n−1 |x10 (t)| p+n−1 dt + h 1, ρ |x10 (t)| p+n−1 dt R1
|x10 (t)| p+n−1 dt
≤ (r1 + ε)D
n
≤ (r1 + ε)D n
≤ (r1 + ε)D n
0
0
+ (r1 + ε)D n−1 (n + 1)
0
Z 0
1
! |x10 (t)| p+n−1 dt
p p+n−1
.
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Y. Zhu, K. Wang / Nonlinear Analysis 70 (2009) 1557–1564
So there is a constant M0 > 0 such that Case 2: If β
Z 0
1
R1 0
|x10 (t)| p+n−1 dt ≤ M0 .
≤ δ, it follows from (3.10) and (3.11) that
D R1 0 |x 0 1 (t)|dt
|x10 (t)| p+n−1 dt 1
Z
≤ (r1 + ε)
0
≤ (r1 + ε)
0 1
Z
≤ (r1 + ε)
0
!n
|x10 (t)|dt Z
1
0
≤ (r1 + ε)
1
+ (r1 + ε)(n + 1)D
Z 0
1 In view of ε = β−r 3 , From (3.13) we have
|x1 (t)| ≤ D +
!n Z
1
p+n−2 p+n−1
1
Z 0
1
|x10 (t)|dt
0
|x10 (t)|dt
!n−1 Z
1
0
Z
1
|x10 (t)| p−1 dt
0
!Z
1 0
|x10 (t)| p−1 dt + h 1,ρ
|x10 (t)| p+n−1 dt + h 1,ρ
0
!n Z
D
(n + 1)D 1+ R1 0 0 |x 1 (t)|dt
|x10 (t)|dt
+ (r1 + ε)(n + 1)D Z
1+ R1 0
1
Z
!n |x10 (t)|dt
|x10 (t)| p−1 dt Z
1
0
+ h 1, ρ
0
Z + h 1, ρ
0
1
1
|x10 (t)| p−1 dt
|x10 (t)| p−1 dt
|x10 (t)| p−1 dt
|x10 (t)| p−1 dt !
1
Z
p−1 p+n−1
|x10 (t)| p+n−1 dt
0
! p+n−2 p+n−1
.
|x10 (t)| p+n−1 dt
< 1, we can see that there is a constant M1 > 0 such that !
1 p+n−1
R1 0
|x10 (t)| p+n−1 dt < M1 .
1
≤ D + M1p+n−1 .
|x10 (t)| p+n−1 dt
So either Case 1 or Case 2, we obtain 1
|x1 (t)| ≤ max{D + D/δ, D + M1p+n−1 } := M2 Z 1 |x10 (t)| p+n−1 dt ≤ max{M0 , M1 } := M3 .
(3.12)
0
Multiplying the two sides of the first equation of (3.2) by φ p ( λ1 x10 (t)) and integrating them from 0 to t, we get 2 Z t 1 0 1 0 1 0 1 p φp x (t) =λ φp x (t) f s, x1 (s), x1 (s) ds 2 λ 1 λ 1 λ 0 Z t 1 = λ2− p |x10 (s)| p−2 x10 (s)g s, x1 (s), x10 (s) ds λ 0 Z t Z t 1 + λ2− p |x10 (s)| p−2 x10 (s)h 1 (s, x1 (s))ds + λ2− p |x10 (s)| p−2 x10 (s)h 2 s, x10 (s) ds λ 0 0 Z t Z 1 ≤ λ2− p |x10 (s)| p−1 |h 1 (s, x1 (s))|ds ≤ λ2− p h 1,M2 |x10 (t)| p−1 dt 0
≤ λ2− p h 1,M2
0
Z 0
1
! |x10 (t)| p+n−1 dt
p−1 p+n−1
,
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Y. Zhu, K. Wang / Nonlinear Analysis 70 (2009) 1557–1564
where h 1,M2 = max t∈[0,1] |h 1 (t, x)|. In view of p < 2, ε = (β − r1 )/3, |x|≤M2
φp
2 p−1 1 0 x1 (t) ≤ 2h 1,M2 M3p+n−1 := M4 , λ
R1 0
|x10 (s)| p+n−1 ds ≤ M3 , we have
∀t ∈ [0, 1],
i.e., 1 1 0 x (t) ≤ M 2 p−2 := M5 . 4 λ 1 0 From the second equation of (3.1) we easily have x2 (0) = 0 and Z 1 Z 1 1 0 0 |x2 (t)|dt ≤ f t, x1 (t), λ x1 (t) dt ≤ f M2 ,M5 , 0 0 where f M2 ,M5 = max t∈[0,1],|x1 |≤M2 | f (t, x1 , x2 )|. Then |x2 |≤M5
|x2 (t)| ≤ |x2 (0)| +
1
Z 0
|x20 (t)|dt ≤ f M2 ,M5 := M6 .
Let M = max{M2 , M6 } + 1, Ω = {x ∈ X : |x1 |0 < M, |x2 |0 < M}, so condition (1) of Lemma is satisfied. Now we show that N x 6∈ Im L , ∀x ∈ ∂Ω ∩ ker L. If not, there exists x0 = (x1 , x2 )> ∈ ∂Ω ∩ ker L such that N x0 = (φq (x2 ), f (t, x1 , φq (x2 )), φq (x2 ), 0)> ∈ Im L . So Q N x0 = 0. But from the definition of operator Q we get Q N x0 = φq (x2 ),
1−
1 m−2 P
m−2 X
ai ξi
Z ai
i=1
1
ξi
φq
Z 0
s
>
f (u, x1 , φq (x2 )du)ds, 0, 0 = 0.
i=1
So x2 = 0. If x1 > D, from condition (H1 ) we have f (t, x1 , x2 ) > 0 and then m−2 X Z 1 Z s ai φq f (u, x1 , φq (x2 )du)ds > 0, i=1
ξi
0
which is a contradiction. Similarly, we also have a contradiction when x1 < −D. Thus |x1 | ≤ D < M, x2 = 0, which is a contradiction to x0 = (x1 , x2 )T ∈ ∂Ω . So condition (2) of Lemma is satisfied. Define the isomorphism J : ImQ → ker L , J (c1 , c2 , 0, 0)> = (c2 , c1 )> . Let H (x, µ) = µx + (1 − µ)J Q N x,
∀(x, µ) ∈ Ω¯ × [0, 1].
Then we have
µx1 + H (x, µ) =
m−2 P R1 1−µ ai ξi m−2 P 1− ai ξi i=1
f (s, x1 , φq (x2 ))ds 6= 0, i=1 q−2 µx2 + (1 − µ)|x2 | x2
∀(x, µ) ∈ (∂Ω ∩ ker L) × [0, 1]. Hence deg{J Q N , Ω ∩ ker L , 0} = deg{H (x, 0), Ω ∩ ker L , 0} = deg{H (x, 1), Ω ∩ ker L , 0} = deg{I, Ω ∩ ker L , 0} 6= 0. So condition (3) of Lemma is satisfied. By applying Lemma, we conclude that equation L x = N x has a solution x(t) = (x1 (t), x2 (t))T on Ω ∩ D(L), i.e., Eq. (1.3) has a solution x1 (t).
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Y. Zhu, K. Wang / Nonlinear Analysis 70 (2009) 1557–1564
As an application, we consider the following equation: 0 0 4 05 5 03 (ϕ 3 (x (t))) = −(6 + x (t))x (t) + x (t) − x (t), 2 1 1 3 2 0 x(1) = x + x . x (0) = 0, 4 2 4 3
t ∈ (0, 1), (3.13)
Corresponding to BVP(1.3), we have p = 3/2, g(t, x1 , x2 ) = −(6 + x14 )x25 , h 1 (t, x1 ) = x15 , h 2 (t, x2 ) = −x23 , a1 = 1/4, a2 = 3/4, β = 6, r1 = 5 and we can easily verify the conditions of Theorem 3.1 hold. So we know that BVP(1.3) has at least one solution. References [1] V.A. Il’in, E.I. Moiseev, Nonlocal boundary value problems of the second kind for a Sturm–Liouville operator, J. Differential Equations 23 (1987) 979–987. [2] C.P. Gupta, A second order m-point boundary value problem at resonance, Nonlinear Anal. 24 (1995) 1483–1489. [3] C.P. Gupta, Solvability of a multiple boundary value problem at resonance, Results Math. 28 (1995) 270–276. [4] R.Y. Ma, Existence theorems for second order m-point boundary value problems, J. Math. Anal. Appl. 211 (1997) 545–555. [5] W. Feng, J.R.L. Webb, Solvability of m-point boundary value problems with nonlinear growth, J. Math. Anal. Appl. 212 (1997) 467–480. [6] C.P. Gupta, S.K. Ntouyas, P.Ch. Tsamatos, Solvability of an m-point boundary value problem for second order ordinary differential equations, J. Math. Anal. Appl. 189 (1995) 575–584. [7] Ph. Hartman, On boundary value problems for systems of ordinary nonlinear second order differential equations, Trans. Amer. Math. Soc. 96 (1960) 493–509. [8] J. Mawhin, Some boundary value problems for Hartman-type perturbations of the ordinary vector p-Laplacian, Nonlinear Anal. 40 (2000) 497–503. [9] R. Gaines, J. Mawhin, Coincidence Degree and Nonlinear Differential Equations, Springer, Berlin, 1977.