We study the existence of positive solutions to the boundary-value problem ... Key Words: ordinary differential equation; existence of solutions; multi-point.
Journal of Mathematical Analysis and Applications 256, 556–567 (2001) doi:10.1006/jmaa.2000.7320, available online at http://www.idealibrary.com on
Existence of Solutions of Nonlinear m-Point Boundary-Value Problems Ruyun Ma1 Department of Mathematics, Northwest Normal University, Lanzhou 730070, Gansu, People’s Republic of China
and Nelson Castaneda Department of Mathematical Sciences, Central Connecticut State University, New Britain, Connecticut 06050 Submitted by William F. Ames Received April 10, 2000
We study the existence of positive solutions to the boundary-value problem u + atf u = 0 x 0 =
m−2 i=1
bi x ξi
t ∈ 0 1 x1 =
m−2 i=1
ai xξi
where ξ ∈ 0 1 with 0 < ξ1 < ξ2 < · · · < ξm−2 < 1 ai bi ∈ 0 ∞ with 0 < m−2 i m−2 i=1 ai < 1, and i=1 bi < 1. We show the existence of at least one positive solution if f is either superlinear or sublinear by applying the fixed point theorem in cones. © 2001 Academic Press Key Words: ordinary differential equation; existence of solutions; multi-point boundary-value problems; Leray–Schauder continuation theorem.
1. INTRODUCTION The study of multi-point boundary-value problems for linear second order ordinary differential equations was initiated by Il’in and Moiseev [5]. 1
Supported by the Natural Science Foundation of China. 556
0022-247X/01 $35.00 Copyright © 2001 by Academic Press All rights of reproduction in any form reserved.
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Since then, nonlinear multi-point boundary-value problems have been studied by several authors using the Leray–Schauder Continuation Theorem, nonlinear alternatives of Leray–Schauder, coincidence degree theory, and fixed point theorem in cones. For example, Gupta [4] studied the existence of solutions for the generalized multi-point boundary-value problem x t = gt xt x t + et x0 =
n−2 i=1
hi xτi
x 1 =
ae t ∈ 0 1 m−2 i=1
ki x ξi
(1.1) (1.2)
Ruyun Ma [7] showed the existence of positive solutions for a three-point boundary-value problem of Dirichlet type ˆ u = 0 u + atf x0 = 0
t ∈ 0 1
x1 = αxη
(1.3) (1.4)
under the conditions that 0 < ηα < 1. We refer the reader to [1–3, 5–6] for other recent results on nonlinear multi-point boundary-value problems. In this paper, we are concerned with the existence of positive solutions to the generalized boundary-value problem u + atf u = 0 x 0 =
m−2 i=1
bi x ξi
t ∈ 0 1 x1 =
m−2 i=1
ai xξi
(1.5) (1.6)
where ξi ∈ 0 1 with 0 < ξi < ξ2 < · · · < ξm−2 < 1, and ai bi ∈ 0 ∞. Our purpose here is to give some existence results for positive solutions to (1.5)–(1.6), assuming that m−2 (A1) ai bi ∈ 0 ∞ satisfy 0 < m−2 i=1 ai < 1 and i=1 bi < 1, and f is either superlinear or sublinear. Our proof here is based upon the fixed point theorem in cones. From now on, we assume the following: (A2) f ∈ C 0 ∞ 0 ∞ (A3) a ∈ C 0 1 0 ∞ and there exists t0 ∈ 0 1 such that at0 > 0. Set
f u f u f∞ = lim u→∞ u u Then f0 = 0 and f∞ = ∞ correspond to the superlinear case, and f0 = ∞ and f∞ = 0 correspond to the sublinear case. By the positive solution of (1.5)–(1.6) we understand a function ut which is positive on 0 < t < 1 and satisfies the differential equation (1.5) and the boundary conditions (1.6). f0 = lim+ u→0
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The main result of this paper is the following Theorem 1. Assume (A1), (A2), and (A3) hold. Then the problem (1.5)–(1.6) has at least one positive solution in the case (i) f0 = 0 and f∞ = ∞ (superlinear) or (ii) f0 = ∞ and f∞ = 0 (sublinear). The proof of the above theorem is based upon an application of the following fixed point theorem in cones [10]. Theorem 2. Let E be a Banach space, and let K ⊂ E be a cone. Assume 1 ⊂ 2 , and let 1 2 are open bounded subsets of E with 0 ∈ 1 2 \1 −→ K A K ∩ be a completely continuous operator such that (i)
u ∈ K ∩ ∂1 , and Au ≥ u
Au ≤ u
u ∈ K ∩ ∂1 , and Au ≤ u 2 \1 . Then A has a fixed point in K ∩ (ii) Au ≥ u
u ∈ K ∩ ∂2 ; or u ∈ K ∩ ∂2 .
2. THE PRELIMINARY LEMMAS m−2 Lemma 1. Let 1 − m−2 1 − b i i=1 i=1 ai = 0. Then for y ∈ C 0 1 , the problem u + yt = 0 x 0 =
m−2 i=1
bi x ξi
has a unique solution ut = − where A=
m−2 i=1
bi
ξi
m−2 i=1
B=
1− −
0
i=1
m−2 i=1
t
0
x1 =
(2.1)
m−2
ai xξi
i=1
t − sys ds + At + B
ys ds
bi − 1 1
1 m−2
t ∈ 0 1
ai bi
0
ξi
m−2 i=1
0
bi − 1
(2.3)
(2.4)
1 − sys ds −
ys ds
(2.2)
m−2 i=1
1−
m−2 i=1
ai ξi
ai
ξi
0
ξi − sys ds
(2.5)
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Lemma 2. Let (A1) be satisfied. If y ∈ C 0 1 and y ≥ 0, then the unique solution u of the problem (2.1)–(2.2) satisfies u ≥ 0 Proof.
From (A1) and (2.3), we know that
u0 = B =
1−
1 m−2 i=1
ai
1
0
1 − sys ds − m−2 i=1
− ≥
1−
1 m−2 i=1
1−
1 m−2 i=1
ξi
i=1
m−2
ai
i=1
m−2 ai
bi
m−2
ai
i=1
1
0
i=1
bi
i=1
−
1
ξi
i=1
ys ds
ai
0
ys ds
ξi
0
ξi − sys ds
1−
bi − 1
ξi
m−2
ai
0
m−2
m−2
ai ξi
i=1
1 − sys ds −
m−2
− ≥
t ∈ 0 1
m−2 i=1
1−
bi − 1
ai
m−2 i=1
ξi
0
ξi − sys ds
ai ξi
1 − sys ds
m−2 i=1
bi
ξi
m−2 i=1
0
ys ds
bi − 1
1−
m−2 i=1
ai ξi
≥ 0
(2.6)
From the fact that u t = −yt ≤ 0, we know that the graph of ut is concave down on 0 1. So, if u1 ≥ 0, then the concavity of u and (2.6) imply that u ≥ 0 for t ∈ 0 1 . If u1 < 0, then we claim that there exists i0 ∈ 1 m − 2 such that uξi0 < u1 < 0
(2.7)
(In fact, if uξi ≥ u1, for all i ∈ 1 m − 2, then by the assumption 0 < m−2 i=1 ai < 1 we have that u1 =
m−2 i=1
ai uξi ≥
m−2 i=1
ai u1 > u1
a contradiction!) Clearly (2.7) contradicts the concavity of u.
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m−2 Lemma 3. Let ai bi ∈ 0 ∞ satisfy m−2 i=1 ai > 1 and i=1 bi < 1. If y ∈ C 0 1 and yt ≥ 0 for t ∈ 0 1, then (2.1)–(2.2) has no nontrivial solutions satisfying u ≥ 0
t ∈ 0 1
Proof. From the fact that u t = −yt ≤ 0, we know that the graph of ut is concave down on (0,1) and u t is nonincreasing on [0,1]. This m−2 together with the assumption that i=1 bi < 1 and the boundary condition u 0 = m−2 i=1 bi u ξi implies that u 0 ≤ 0
(2.8)
(In fact, if u 0 > 0, then (A1) implies that there exists ξ˜ ∈ ξi ξm−2 such that u ξ˜ > u 1 > 0. This contradicts the fact that u t is nonincreasing on [0 1].) Moreover u t ≤ 0
for t ∈ 0 1
(2.9)
This means that u0 ≥ ut ≥ u1
for t ∈ 0 1
(2.10)
Now assume to the contrary that ut ≥ 0 on 0 1 and there exists τ > 0 such that uτ > 0. If u1 > 0, then uξi > 0 for i = 1 m − 2 and u1 =
m−2 i=1
ai uξi ≥
m−2 i=1
ai minuξi i = 1 m − 2
> minuξi i = 1 m − 2 This contradicts (2.10). If u1 = 0, then we know from the assumption m−2 i=1 ai > 1 that there exists i∗ ∈ 1 m − 2 such that ai∗ > 0
uξi∗ = 0
(2.11)
and uτ > 0
for some τ ∈ 0 ξi∗ ∪ ξi∗ 1
If τ ∈ 0 ξi∗ , then uτ > uξi∗ = u1 = 0, which contradicts the concavity of u. If τ ∈ ξi∗ 1, then 0 = u1 = uξi∗ < uτ, which contradicts the concavity of u again. m−2 In the rest of the paper, we assume that 0 < m−2 i=1 ai < 1 and i=1 bi < 1. Moreover, we will work in the Banach space C 0 1 , and only the sup norm is used.
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Lemma 4. Let (A1) hold. If y ∈ C 0 1 and y ≥ 0, then the unique solution u of the problem (2.1)–(2.2) satisfies inf ut ≥ γu
t∈ 01
where γ=
m−2
ai 1 − ξi m−2 1 − i=1 ai ξi i=1
(2.12)
Proof. From the fact that u t = −yt ≤ 0, we know that the graph of ut is concave down on (0 1) and u t is nonincreasing on 0 1 . This m−2 together with the assumption that i=1 bi < 1 and the boundary condition u 0 = m−2 i=1 bi u ξi implies that u 0 ≤ 0
(2.13)
and moreover u t ≤ 0
for t ∈ 0 1
(2.14)
This implies that u = u0
min ut = u1
t∈ 01
(2.15)
For all i ∈ 1 m − 2, we have from the concavity of u that u0 ≤ u1 +
u1 − uξi 0 − 1 1 − ξi
(2.16)
or u01 − ξi ≤ u11 − ξi + uξi − u1 (2.17) m−2 This together with the boundary condition x1 = i=1 ai xξi implies that 1 − m−2 i=1 ai u0 ≤ m−2 + 1 u1 i=1 ai 1 − ξi 1 − m−2 i=1 ai ξi ≤ m−2 + 1 u1 (2.18) i=1 ai 1 − ξi Thus min ut ≥
t∈ 01
This completes the proof.
m−2
ai 1 − ξi u1 m−2 1 − i=1 ai ξi i=1
(2.19)
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ma and castaneda 3. PROOF OF THE MAIN THEOREM
Proof of Theorem 1. Superlinear Case. Suppose then that f0 = 0 and f∞ = ∞. We wish to show the existence of a positive solution of (1.5)–(1.6). Now (1.5)–(1.6) has a solution y = yt if and only if y solves the operator equation m−2 ξi t bi 0 asf ys ds yt = − t − sasf ys ds + t i=1 m−2 0 i=1 bi − 1 1 m−2 ξi 1 + 1 − sasf ys ds − ai ξi − s m−2 0 1 − i=1 ai 0 i=1 m−2 ξi bi 0 asf ys ds × asf ys ds − i=1 m−2 i=1 bi − 1 m−2 def × 1− ai ξi = Ayt (3.1) i=1
Denote K = y y ∈ C 0 1 y ≥ 0 min yt ≥ γy 0≤t≤1
(3.2)
where γ is defined in (2.12). It is obvious that K is a cone in C 0 1 . Moreover, by Lemma 4, AK ⊂ K. It is also easy to check that A K → K is completely continuous. Now since f0 = 0, we may choose H1 > 0 so that f y ≤ "y, for 0 < y < H1 , where " > 0 satisfies m−2 ξi 1 " i=1 bi 0 as ds ≤ 1 (3.3) 1 − sas ds + 0 1 − m−2 1 − m−2 i=1 ai i=1 bi Thus, if y ∈ K and y = H1 , then from (3.1) and (3.3), we get 1 1 1 − sasf ys ds Ayt ≤ 0 1 − m−2 i=1 ai m−2 ξi m−2 i=1 bi 0 asf ys ds 1− a i ξi + 1 − m−2 i=1 i=1 bi 1 1 ≤ 1 − sas"ys ds 0 1 − m−2 i=1 ai m−2 ξi m−2 bi 0 as"ys ds 1 − a ξ + i=1 i i 1 − m−2 i=1 i=1 bi
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1 1 ≤" 1 − sas ds m−2 1 − i=1 ai 0 m−2 ξi i=1 bi 0 as ds y + 1 − m−2 i=1 bi
563
(3.4)
Now if we let 1 = y ∈ C 0 1 y < H1
(3.5)
then (3.4) shows that Ay ≤ y, for y ∈ K ∩ ∂1 . 2 > 0 such that f u ≥ ρu, for Further, since f∞ = ∞, there exists H u ≥ H2 , where ρ > 0 is chosen so that ργ
m−2
ai 1 − ξi
i=1
1
0
asds ≥ 1
(3.6)
2 /γ and 2 = y ∈ C 0 1 y < H2 . Then Let H2 = max2H1 H y ∈ K and y = H2 implies 2 min yt ≥ γy ≥ H
0≤t≤1
and so Ay0 =
1−
1 m−2 i=1
ai
1
0
−
1 − sasf ys ds m−2 i=1
− ≥
m−2 i=1
=
m−2 i=1
≥
m−2 i=1
≥ ργ
ai
1
0
ai 1 − ξi
i=1
m−2 i=1
ξi
0
ξi − sasf ys ds
ξ m−2 bi 0 i asf ys ds ai ξi 1− m−2 i=1 i=1 bi − 1
1 − sasf ys ds −
ai 1 − ξi
m−2
ai
m−2 i=1
1
0
1
0
ai 1 − ξi
ai
1
0
ξi − sasf ys ds
asf ys ds asρys ds
1
0
as dsy
(3.7)
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Hence, for y ∈ K ∩ ∂2 , Ay ≥ ργ
m−2 i=1
ai 1 − ξi
1
0
as dsy ≥ y
Therefore, by the first part of the Fixed Point Theorem, it follows that A 2 \1 , such that H1 ≤ u ≤ H2 . This completes has a fixed point in K ∩ the superlinear part of the theorem. Sublinear Case. Suppose next that f0 = ∞ and f∞ = 0. We first choose H3 > 0 such that f y ≥ My for 0 < y < H3 , where Mγ
m−2 i=1
ai 1 − ξi
1
0
as ds ≥ 1
(3.8)
By using the method to get (3.7), we can get that 1 1 Ay0 = 1 − sasf ys ds 0 1 − m−2 i=1 ai m−2 ξi − ai ξi − sasf ys ds 0
i=1
− ≥
m−2 i=1
≥ Mγ
ai 1 − ξi m−2 i=1
m−2
1
0
i=1
ξ m−2 bi 0 i asf ys ds 1 − a ξ m−2 i i i=1 i=1 bi − 1
asf ys ds
ai 1 − ξi
1
0
as dsy
≥ H3
(3.9)
Thus, we may let 3 = y ∈ C 0 1 y < H3 so that Ay ≥ y
y ∈ K ∩ ∂3
4 > 0 so that f y ≤ λy for y ≥ H 4 , Now, since f∞ = 0, there exists H where λ > 0 satisfies m−2 1
1 1 i=1 bi 0 as ds λ ≤ 1 (3.10) 1 − sas ds + 0 1 − m−2 1 − m−2 i=1 ai i=1 bi
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We consider two cases: Case (i). Suppose f is bounded, say f y ≤ N for all y ∈ 0 ∞. In this case, choose 1
1 H4 = max 2H3 N 1 − sas ds 0 1 − m−2 i=1 ai m−2 1 i=1 bi 0 asds + 1 − m−2 i=1 bi so that for y ∈ K with y = H4 we have from (3.1) that 1 1 Ayt ≤ 1 − sasf ys ds m−2 1 − i=1 ai 0 m−2 ξi m−2 i=1 bi 0 asf ys ds 1− ai ξ i + 1 − m−2 i=1 i=1 bi
1 1 ≤ 1 − sasf ys ds 0 1 − m−2 i=1 ai m−2 1 i=1 bi 0 asf ys ds + 1 − m−2 i=1 bi m−2 1
1 1 i=1 bi 0 as ds ≤N 1 − sas ds + 0 1 − m−2 1 − m−2 i=1 ai i=1 bi ≤ H4 and therefore Ay ≤ y. Case (ii). If f is unbounded, then we know from (A2) that there is 4 such that H4 H4 > max2H3 γ1 H f y ≤ f H4
for
0 < y ≤ H4
(We are able to do this since f is unbounded.) Then for y ∈ K and y = H4 we have 1 1 Ayt ≤ 1 − sasf ys ds 0 1 − m−2 i=1 ai m−2 ξi m−2 i=1 bi 0 asf ys ds + 1− ai ξi 1 − m−2 i=1 i=1 bi
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≤
1−
1 m−2 i=1
ai
1
0
+
≤
1−
1 m−2 i=1
ai
1
0
+
1 − sasf H4 ds m−2 i=1
bi
1
asf H4 ds m−2 1 − i=1 bi 0
1 − sasλH4 ds m−2 i=1
bi
1
asλH4 ds m−2 1 − i=1 bi 0
≤ H4 Therefore, in either case we may put 4 = y ∈ C 0 1 y < H4 and for y ∈ K ∩ ∂4 we may have Ay ≤ y. By the second part of the Fixed Point Theorem, it follows that BVP (1.1)–(1.2) has a positive solution. Therefore, we have completed the proof of Theorem 1. ACKNOWLEDGMENT The author is very grateful to the referee for his or her valuable suggestions.
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