EXISTENCE OF THE THIRD POSITIVE RADIAL SOLUTION OF A SEMILINEAR ELLIPTIC PROBLEM ON AN UNBOUNDED DOMAIN
Bong Soo Ko † and Yong-Hoon Lee‡
Abstract. We prove the multiplicity of ordered positive radial solutions for a semilinear elliptic problem defined on an exterior domain. The key argument is to prove the existence of the third solution in presence of two known solutions. For this, we obtain some partial results related to three solutions theorem for certain singular boundary value problems. Proof are mainly based on the upper and lower solutions method and degree theory.
1. Introduction
In this paper, we prove the existence of 2N − 1 distinct ordered pos- itive radial solutions for the following problem:
(P λ )
△ u + λg(|x|)f (u) = 0 in Ω = R n \ B(0, r o ),
|x|→∞ lim u(x) = 0, u = 0 on ∂Ω,
where λ is a positive real parameter, B(0, r o ) is an open ball centered at 0 with the radius r o , n ≥ 3. Assume that f ∈ C(I, R), I ⊂ R and g ∈ C 1 ([r o , ∞), R + )) satisfy the following conditions:
Received July 27, 2001.
2000 Mathematics Subject Classification: 34B15, 35J25.
Key words and phrases: existence, multiplicity, semilinear elliptic problem, singu- lar boundary value problem, upper solution, lower solution, Leray-Schauder degree, radial solution.
† This work was supported in part by the Educational Research Institute, Cheju National University 2001.
‡ This work was supported by grant R01-2000-00002 from the Basic Research
Program of the Korea Science and Engineering Foundation.
(f 0 ) there exists a positive constant M such that f (u) − f (v) ≥ −M (u − v) if u, v ∈ I, u ≥ v.
(f 1 ) There exist exactly N positive numbers 0 < a 1 < a 2 < · · · < a N
such that f (a i ) = 0 for all i = 1, · · · , N.
(f 2 ) [0, a N ] ⊂ I and there exists a positive constant K f such that f ′ (a i ) ≤ −K f for all i = 1, · · · , N.
(f 3 ) R ai
s f (u)du > 0 for all s ∈ [0, a i ) and for all i = 1, · · · , N.
(g) R ∞
r
org(r)dr < ∞.
If Ω is a bounded open domain with smooth boundary, then there have been several studies ([3], [4], [6], [7], [10], [11]) which prove generally that there exists λ o > 0 such that if λ > λ o , then problem (P λ ) has at least 2N − 1 ordered positive solutions. In those studies, the boundedness of Ω is crucial. If Ω is unbounded, we might not use the compactness on the operator or the functional which are induced from the problem (P λ ).
So the question of the multiplicity of positive solutions is nontrivial.
To obtain the radial solutions, we rewrite problem (P λ ), via transfor- mations r = |x|, s = r 2−n and t = r2−no −s
r
o2−n([8]), as (1 λ ) ( u ′′ + λq(t)f (u) = 0, 0 < t < 1
u(0) = 0 = u(1), where q is given by
q(t) = r o 2
(n − 2) 2 (1 − t)
−2(nn−2−1)g(r o (1 − t)
n−1−2).
Transforming (P λ ) to (1 λ ), we see that the coefficient function q in
(1 λ ) is of C[0, 1] if g in (P λ ) satisfies lim r→∞ r 2(n−1) g(r) < ∞ and q is
singular at t = 1 if lim r→∞ r 2(n−1) g(r) = ∞. Since the result for the
regular case can be proved by exactly the same way as the singular case
with less restrictions, our interest will be mainly focused at the singular
case and, thus, a crucial step for our goal will be to establish a theorem
of existence of the third solution for problem (1 λ ) in the case with two pairs of upper and lower solutions.
More precisely, suppose that ¯ v, ˜ v, and ¯ w, ˜ w are pairs of lower solutions and upper solutions of (1) such that ¯ v(t) ≤ ˜ w(t), ¯ v(t) ≤ ˜ v(t), ¯ w(t) ≤ ˜ w(t) for all t ∈ [0, 1] and ¯ w(t 0 ) > ˜ v(t 0 ) for some t 0 ∈ [0, 1]. If q is H¨older continuous on the interval [0, 1], then there is a solution in the ordered interval [¯ v, v] = {u ∈ C([0, 1]) : ¯ ˜ v(t) ≤ u(t) ≤ ˜ v(t), t ∈ [0, 1]} and a solution in [ ¯ w, w]. And furthermore it is known that there exists a third ˜ solution in the set [¯ v, w] \ ([¯ ˜ v, w] ∪ [ ¯ ˜ w, w]). ([1], [11]) ˜
In the case that q is singular at t = 0 and/or 1, the existence of solution given a pair of lower solution and upper solution, ¯ v and ˜ v, with
¯
v(t) ≤ ˜ v(t) for all t ∈ [0, 1], is known under additional conditions ([8], [9]). Hence, there arises a question, does there exist the third solution if there are pairs of lower solutions and upper solutions as in the preceding paragraph? As far as the authors know, the complete answer has not been made yet and some partial results for singular problems have been studied by Parter [13] and Ben-Naoum and De Coster [2], but neither are applicable in problem (P λ ). Because, in [2] we can understand necessity of strict sense of upper and lower solutions somehow. We also note that the singular function q in [13] plays the role of the coefficient of damping term. The authors are able to get another partial answer for the multiplicity for (1 λ ) using Leray-Schauder degree and Green’s function.
This paper is organized as follows: In Section 2, we introduce fun- damental theorems based on the method of upper and lower solutions.
In Section 3, we give a theorem of existence of the third solution for singular problem (1 λ ). In section 4, we study the multiplicity of positive radial solutions for problem (P λ ) on an exterior domain R n \ B(0, r).
2. Fundamental theorems for G-upper and G-lower solutions In this section, we prove a fundamental existence theorem in terms of general upper and lower solutions for singular boundary value problems of the form;
(2) ( u ′′ + f (t, u) = 0, 0 < t < 1 u(0) = A, u(1) = B,
where f : D ⊂ (0, 1) × R → R is continuous. We denote (0, ∞) by R + .
A solution u(·) means a function u ∈ C[0, 1] ∩ C 2 (0, 1) such that
(t, u(t)) ∈ D for all t ∈ (0, 1) and u ′′ (t) + f (t, u(t)) = 0 for all t ∈ (0, 1) with u(0) = A and u(1) = B.
Definition 1. α ∈ C[0, 1] ∩ C 2 (0, 1) is called a lower solution of (2) if (t, α(t)) ∈ D for all t ∈ (0, 1) and
α ′′ (t) + f (t, α(t)) ≥ 0, α(0) ≤ A, α(1) ≤ B.
Similarly, β ∈ C[0, 1] ∩ C 2 (0, 1) is called an upper solution of (2) if the above inequalities are reversed.
If α, β ∈ C[0, 1] are such that α(t) ≤ β(t) for all t ∈ [0, 1], we define the set
D β α = {(t, u) ∈ (0, 1) × R : α(t) ≤ u ≤ β(t)}.
The well-known fundamental theorem on upper and lower solutions method for problem (2) is as follows;
Theorem 1. ([9]) Let α, β be, respectively, a lower solution and an upper solution of (2) such that
(a 1 ) α(t) ≤ β(t) for all t ∈ [0, 1].
(a 2 ) D β α ⊂ D.
Assume also that there is a function h ∈ C (0, 1), R + such that (a 3 ) |f (t, u)| ≤ h(t) for all (t, u) ∈ D α β and
(a 4 ) R 1
0 s(1 − s)h(s)ds < ∞.
Then problem (2) has at least one solution u(·) such that α(t) ≤ u(t) ≤ β(t) for all t ∈ (0, 1).
We give definitions of somewhat general type of upper and lower solutions.
Definition 2. We say that a continuous function α(·) : [0, 1] → R is a G-lower solution of (2) if α ∈ C 2 (0, 1) except at finite points τ 1 , · · · , τ n
with 0 < τ 1 < · · · < τ n < 1 such that
(L 1 ) at each τ i , there existα ′ (τ i − ), α ′ (τ i + ) such that
α ′ (τ i − ) < α ′ (τ i + ) and
(L 2 ) ( α ′′ (t) + f (t, α(t)) ≥ 0 for all t ∈ (0, 1) \ {τ 1 , · · · , τ n }, α(0) ≤ A, α(1) ≤ B.
We also say that a continuous function β(·) : [0, 1] → R is a G-upper solution of (2) if β ∈ C 2 (0, 1) except at finite points σ 1 , · · · , σ m with 0 < σ 1 < · · · < σ m < 1 such that
(U 1 ) at each σ i , there exist β ′ (σ j − ), β ′ (σ j + ) and β ′ (σ j − ) > β ′ (σ j + ) and
(U 2 ) ( β ′′ (t) + f (t, β(t)) ≤ 0 for all t ∈ (0, 1) \ {σ 1 , · · · , σ m }, β(0) ≥ A, β(1) ≥ B.
The fundamental theorem on G-upper and G-lower solutions is given as follows. The proof is basically similar to that of [9] and we give it for reader’s convenience.
Theorem 2. Let α and β be, respectively, a G-lower solution and a G-upper solution of (2) satisfying (a 1 ) and (a 2 ) in Theorem 1. Also assume (a 3 ) and (a 4 ). Then (2) has at least one solution u such that
α(t) ≤ u(t) ≤ β(t) for all t ∈ (0, 1).
Proof. Define a modified function of f as follows;
F (t, u) =
f (t, β(t)) − u − β(t)
1 + u 2 if u > β(t), f (t, u) if α(t) ≤ u ≤ β(t),
f (t, α(t)) − u − α(t)
1 + u 2 if u < α(t).
Then F : (0, 1) × R → R is continuous and
(3) |F (t, u)| ≤ m(α, β) + h(t)
for all (t, u) ∈ (0, 1) × R, where m(α, β) = kαk ∞ + kβk ∞ + 1.
Consider the problem
(4) ( u ′′ + F (t, u) = 0, 0 < t < 1, u(0) = A, u(1) = B.
We claim that any solution u of (4) satisfies α(t) ≤ u(t) ≤ β(t) for all t ∈ [0, 1]. Suppose, by contradiction, that α u. So let (α − u)(t o ) = max t∈[0,1] (α − u)(t) > 0. If t o ∈ (0, 1) \ {τ 1 , · · · , τ n }. Then (α − u) ′′ (t o ) ≤ 0. Since u(t o ) < α(t o ),
0 ≥ (α − u) ′′ (t o ) = α ′′ (t o ) + F (t o , u(t o ))
= α ′′ (t o ) + f (t o , α(t o )) − u(t o ) − α(t o ) 1 + u 2 (t o )
≥ α(t o ) − u(t o ) 1 + u 2 (t o ) > 0, a contradiction.
If t o = τ i for some i = 1, · · · , n, then since α − u attains its positive maximum at τ i ,
(α − u) ′ (τ i − ) ≥ 0 and (α − u) ′ (τ i + ) ≤ 0.
Thus
0 ≤ (α − u) ′ (τ i − ) − (α − u) ′ (τ i + )
= α ′ (τ i − ) − α ′ (τ i + ).
This leads to a contradiction to the definition of G-lower solution.
Let t o = 0 or 1
0 < (α − u)(0) = α(0) − A ≤ 0, 0 < (α − u)(1) = α(0) − B ≤ 0,
a contradiction. Therefore α(t) ≤ u(t) ≤ β(t), and so we can conclude u is a solution of (2). We can prove for the case that u β by a similar fashion. We claim that (4) has at least one solution. It is well-known that problem (4) is equivalently written as
u = T u on X = C[0, 1],
where
T u(t) = A + (B − A)t + Z 1
0
G(t, s)F (s, u(s))ds and G(t, s) is the Green’s function explicitly written as
G(t, s) = ( s(1 − t), 0 ≤ s ≤ t, t(1 − s), t ≤ s ≤ 1.
By (3) and (a 4 ), T : X → X is well-defined, continuous and T X is bounded. If T is a compact operator, then the proof of the existence of a solution is done by Schauder fixed point Theorem. To show T compact, making use of Arzela-Ascoli Theorem, it suffices to show that T X is equicontinuous. Let t ∈ (0, 1), then by (3) we get
| d dt T u(t)|
≤ |B − A| + Z t
0
s|F (s, u(s))|ds + Z 1
t
(1 − s)|F (s, u(s))|ds
≤ |B − A| + m(α, β)
2 (t 2 + (1 − t) 2 ) + Z t
0
sh(s)ds + Z 1
t
(1 − s)h(s)ds , |B − A| + m(α, β)
2 (t 2 + (1 − t) 2 ) + γ(t).
If γ ∈ L 1 (0, 1), then the proof follows from that Z 1
0
|γ(s)|ds ≤ lim
t→1
−(1 − t) Z t
0
sh(s)ds + lim
t→0
+t Z 1
t
(1 − s)h(s)ds + 2
Z 1 0
s(1 − s)h(s)ds
≤ 4 Z 1
0
s(1 − s)h(s)ds < ∞.
Remark 1. It is easy to see that if we replace (a 4 ) by the condition R 1
0 sh(s)ds < ∞, then the solution u which we have found belongs to
C 1 (0, 1]. Similarly, if R 0 1 (1 − s)h(s)ds < ∞, then u ∈ C 1 [0, 1).
3. Multiplicity
In this section, we prove the existence of three solutions for the prob- lem
(5) ( u ′′ + q(t)f (u) = 0, 0 < t < 1, u(0) = 0 = u(1),
under certain assumptions for two pairs of G-lower solutions and G- upper solutions. In what follows, we assume that q ∈ C (0, 1), R + is singular at t = 0 and/or 1, and f : I ⊂ R → R is continuous. We know that problem (5) is equivalent to
u = T u on C[0, 1], where T is given by
T u(t) = Z 1
0
G(t, s)q(s)f (u(s))ds
and G(t, s) is the Green’s function given in the proof of Theorem 2.
It is well-known ([5], [8]) that T is completely continuous on C[0, 1]
if q satisfies (C)
Z 1 0
s(1 − s)q(s)ds < ∞.
Definition 3. For any u, v ∈ C[0, 1], u ≺ v means that u(t) ≤ v(t), for all t ∈ [0, 1] but u 6= v on [0, 1].
Let C 0 [0, 1] = {u ∈ C[0, 1] : u(0) = 0 = u(1)}. If q satisfies (C), then the problem
(6) ( u ′′ + q(t) = 0, 0 < t < 1, u(0) = 0 = u(1)
has the unique solution e ∈ C[0, 1] ∩ C 2 (0, 1). Let
C e [0, 1] = {u ∈ C 0 [0, 1] : γe(t) ≤ u(t) ≤ δe(t), t ∈ [0, 1] and γ, δ ∈ R}.
Then C e [0, 1] is a linear subspace of C[0, 1]. For any u ∈ C e [0, 1], we define a norm
kuk e = inf{λ > 0 : −λe(t) ≤ u ≤ λe(t), t ∈ [0, 1]}.
Then C e [0, 1] is a Banach space with respect to the norm k · k e ([1]). We claim that
T (C([0, 1])) ⊂ C e [0, 1].
Indeed, for any u ∈ C[0, 1],
(T u − δe) ′′ (t) = −q(t)f (u(t)) + δq(t)
= q(t)[−f (u(t)) + δ],
for all t ∈ (0, 1). Since T u(0) − δe(0) = 0 = T u(1) − δe(1), if we choose δ > 0 sufficiently large so that (T u − δe) ′′ (t) ≥ 0, then
T u ≤ δe.
By a similar method, we can show the existence of constant γ such that
γe ≤ T u.
Therefore,
T u ∈ C e [0, 1].
From now on, we only consider the operator T : C e [0, 1] → C e [0, 1].
Since the embedding C e [0, 1] ֒→ C[0, 1] is continuous and T : C[0, 1] → C e [0, 1] is also completely continuous, so is T on C e [0, 1]. ([1])
We now state and prove maximum principle for certain linear opera- tors.
Lemma 1. Let q ∈ C [a, b], (0, ∞), and K > 0 given. If u ∈ C 2 [a, b]
satisfies
u ′′ − Kq(t)u ≥ 0, a < t < b, u(a) ≤ 0, u(b) ≤ 0.
Then u(t) ≤ 0 for all t ∈ [a, b].
Assume, moreover, that there exists τ ∈ (a, b) such that u(τ ) = 0.
Then u(t) = 0 for all t ∈ [a, b].
Lemma 2. Let q ∈ C (a, b), (0, ∞) may be singular at t = a and/or b and K > 0 given. If u ∈ C[a, b] ∩ C 2 (a, b) satisfies
u ′′ − Kq(t)u ≥ 0, a < t < b, u(a) ≤ 0, u(b) ≤ 0.
Then u(t) ≤ 0 for all t ∈ [a, b].
Assume, moreover, that there exists τ ∈ (a, b) such that u(τ ) = 0.
Then u(t) = 0 for all t ∈ [a, b].
Here we give the proof of Lemma 2. Lemma 1 is well known. ([14]) Proof. It is not hard to show that for h ∈ C (a, b), [0, ∞)
u ∈ C[a, b] ∩ C 2 (a, b) is a solution of
u ′′ − Kq(t)u = h(t), a < t < b, u(a) = A, u(b) = B,
then u satisfies
u(t) ≤ max n
A, B, − inf t∈(a,b) h(t) K inf t∈(a,b) q(t)
o .
This fact implies the first part of Lemma 2. We also notice that if v ∈ C[t 1 , t 2 ] ∩ C 2 (t 1 , t 2 ) satisfies
v ′′ (t) − Kq(t)v(t) > 0 for t 1 < t < t 2 ,
then v cannot have a nonnegative relative maximum on (t 1 , t 2 ).
We now prove the second part by contradiction. Suppose, without loss of generality, that there exists t o ∈ (τ, b) such that u(t o ) < 0. For α > 0, let z(t) = e α(t−τ ) − 1, then z(t) > 0, = 0 or < 0 according to t > τ, t = τ or t < τ, respectively. Take ǫ with 0 < ǫ < − u(t z(to)
o
) and let v(t) = u(t) + ǫz(t). Then
v ′′ (t) − Kq(t)v(t) = u ′′ (t) − Kq(t)u(t) + ǫ[z ′′ (t) − Kq(t)z(t)], and we know that u ′′ (t) − Kq(t)u(t) ≥ 0.
If t < τ, then z ′′ (t) − Kq(t)z(t) > 0, and thus v ′′ (t) − Kq(t)v(t) > 0 for
all t ∈ (a, τ ).
If t = τ, then z ′′ (t)−Kq(t)z(t) = z ′′ (t) > 0, and thus v ′′ (t)−Kq(t)v(t) >
0 for all t ∈ (a, τ ). We notice that z ′′ (t) − Kq(t)z(t) = e α(t−τ ) [α 2 − Kq(t)] + Kq(t), and Kq(t) > 0 for all t ∈ (a, b).
If t ∈ (τ, t o ), consider an interval J = [τ − δ, t o ] for some δ > 0 with τ − δ > a. In this case, if t ∈ J and t > τ, then taking α 2 > Kq m , where q m = max J q(t), α 2 −Kq(t) ≥ α 2 −Kq m > 0. Thus v ′′ (t)−Kq(t)v(t) > 0.
Consequently,
v ′′ (t) − Kq(t)v(t) > 0 for all t ∈ J.
On the other hand, v(τ − δ) < 0 if δ is sufficiently small and v(t o ) < 0, but v(τ ) = u(τ ) = 0. This implies that the maximum of v on J occurs on the interior of J and is nonnegative. This leads to a contradiction by well known Maximum Principles and completes the proof.
The following lemma seems to be well known. However, we could not find any references for that, and so we prove it.
Lemma 3. Assume (C) and (f 0 ). Suppose that ¯ v and ˜ v are a G-lower solution and a G-upper solution satisfying ¯ v ≺ ˜ v and none of them are solutions of (5). Let w be a solution of (5) satisfying ¯ v ≺ w ≺ ˜ v. Then w satisfies
¯
v(t) < w(t) < ˜ v(t) for all t ∈ (0, 1).
Proof. Let w be a solution of problem (5) satisfying
¯
v(t) ≤ w(t) ≤ ˜ v(t) for all t ∈ (0, 1).
We show that ¯ v(t) < w(t) for all t ∈ (0, 1), and the other inequality can be shown by a similar way.
(i) At τ i , i = 1, · · · , n.
Suppose that ¯ v(τ i ) = w(τ i ) for some i, then (¯ v − w)(τ i ) = max t∈[0,1] (¯ v − w)(t), and (¯ v −w) ′ (τ i − ) ≥ 0, (¯ v −w) ′ (τ i + ) ≤ 0. Thus ¯ v ′ (τ i − )− ¯ v ′ (τ i + ) ≥ 0, and this contradiction shows that
¯
v(τ i ) < w(τ i ) for all i = 1, · · · , n.
(ii) On (τ i , τ i+1 ), i = 1, · · · , n − 1.
We get
(¯ v − w) ′′ (t) − Kq(t)(¯ v − w)(t)
≥ − q(t)[f (¯ v(t)) − f (w(t))] − Kq(t)(¯ v(t) − w(t))
≥ − M q(t)(w(t) − ¯ v(t)) − Kq(t)(¯ v(t) − w(t))
= (K − M )q(t)(w(t) − ¯ v(t)) ≥ 0 ∀t ∈ (τ i , τ i+1 ) for sufficiently large K. By (i) we get
(¯ v − w)(τ i ) < 0, (¯ v − w)(τ i+1 ) < 0.
Now suppose that there exists τ ∈ (τ i , τ i+1 ) such that ¯ v(τ ) = w(τ ), then by Lemma 1, ¯ v ≡ w on [τ i , τ i+1 ] and this contradicts to (i).
(iii) On (0, τ 1 ).
We get
(¯ v − w) ′′ (t) − Kq(t)(¯ v − w)(t) ≥ 0 on (0, τ 1 ), (¯ v − w)(0) ≤ 0, (¯ v − w)(τ 1 ) < 0.
Suppose that there exists τ ∈ (0, τ 1 ) such that ¯ v(τ ) = w(τ ), then by Lemma 2, ¯ v ≡ w on [0, τ 1 ] and this also contradicts to (i). We can show similarly that ¯ v(t) − w(t) < 0 for all t ∈ [τ n , 1). Consequently,
¯
v(t) < w(t) for all t ∈ (0, 1). We first give an existence theorem of the third solution when q is integrable. In this case, the solutions of problem (5) and the solution e of problem (6) are of C 1 ([0, 1]) and Theorem 1 and Theorem 2 are obviously make use of. We furthermore, give a restriction on the coeffi- cient function q so that lim t→0+tq(t) and lim t→1−(1 − t)q(t) exist. For example, if q is decreasing near 0 and increasing near 1, then the limits exist by the Monotone Convergence Theorem. Generally, if the limits exist, then they are 0, since q is integrable.
(1 − t)q(t) exist. For example, if q is decreasing near 0 and increasing near 1, then the limits exist by the Monotone Convergence Theorem. Generally, if the limits exist, then they are 0, since q is integrable.
Theorem 3. Assume (f 0 ) and
(H 0 ) q ∈ L 1 (0, 1) ∩ C(0, 1) and both lim t→0+tq(t) and lim t→1−(1 − t)q(t) exist.
(1 − t)q(t) exist.
Suppose also that there exist two pairs of G-lower and G-upper solutions
{¯ v, v} and { ¯ ˜ w, w} such that ¯ ˜ v ≺ ˜ v, w ¯ ≺ ˜ w, ¯ v ≺ ˜ w, w ¯ ˜v, none of them
are solutions of (5), and [¯ v(t), ˜ w(t)] ⊂ I for all t ∈ [0, 1]. Then (5) has at
least 3 distinct solutions u 1 ≺ u 2 ≺ u 3 such that u 1 ∈ [¯ v, v], u ˜ 2 ∈ [ ¯ w, w] ˜
and u 3 ∈ [¯ v, w] \ [¯ ˜ v, v] ∪ [ ¯ ˜ w, w], where [v, w] = {u ∈ C ˜ e [0, 1] : v ≺ u ≺
w}.
Proof. Let ¯ v and ˜ v be a G-lower solution and a G-upper solution of (5) and let
Ω = {u ∈ C e [0, 1] : ¯ v(t) ≤ u(t) ≤ ˜ v(t), t ∈ [0, 1]}, Ω 1 = {u ∈ C e [0, 1] : ¯ v(t) < u(t) < ˜ v(t), t ∈ (0, 1)}.
We denote Ω ◦ 1 , the interior of Ω 1 in C e ([0, 1]). We know by Theorem 2 and Lemma 3 that there exists a solution w of problem (5) satisfying w ∈ Ω 1 . Since w, e ∈ C 1 (0, 1), it is obvious that w ∈ C e ([0, 1]). We claim that w ∈ Ω ◦ 1 . For this, we first show that there exists ǫ 1 > 0 such that (7) w − ¯ v(t) > ǫ 1 e(t) for all t ∈ (0, 1).
The function w − ¯ v is of C[0, 1] and (w − ¯ v)(t) > 0 for all t ∈ (0, 1). On the other hand, the function e is of C 1 [0, 1], e > 0 and concave on (0, 1) so that
0 < e ′ (0) < ∞, −∞ < e ′ (1) < 0.
If ¯ v(t) < 0 for t = 0 and 1, then w − ¯ v(t) > 0 for all t ∈ [0, 1] and (7) is easily verified.
Let ¯ v(t) = 0 for t = 0 or 1. Assume ¯ v(0) = 0 and ¯ v(d) − w(d) < 0 for some d ∈ (0, τ 1 ). Then for given α > 0, we may choose a positive numbers ǫ so that
0 < ǫ < w(d) − ¯ v(d) exp(αd) − 1 . Let us define the function z by
z(t) = exp(αt) − 1.
Then
(¯ v − w + ǫz) ′′ (t) − Kq(t)(¯ v − w + ǫz)(t)
= (¯ v − w) ′′ (t) − Kq(t)(¯ v − w)(t) + ǫ(z ′′ (t) − Kq(t)z(t)).
As we see in the proof of Lemma 3,
(¯ v − w) ′′ (t) − Kq(t)(¯ v − w)(t) > 0 for all t ∈ (0, τ 1 )
if K > M.
On the other hand, (H 0 ) implies that tq(t) is bounded on (0, c], for arbitrarily fixed c ∈ (0, 1), thus for α > K sup{tq(t) : t ∈ (0, τ 1 ]} and t ∈ (0, d), we get the following inequalities by Mean Value Theorem;
z ′′ (t) − Kq(t)z(t) = exp(αt) α 2 − Kq(t)(1 − exp(−αt))
= exp(αt) α 2 − αK exp(−αξ)tq(t)
≥ α exp(αt)(α − Ktq(t)) ≥ 0 for some ξ ∈ (0, t). Therefore,
(¯ v − w + ǫz) ′′ (t) − Kq(t)(¯ v − w + ǫz)(t) ≥ 0
for all t ∈ (0, d) and (¯ v − w + ǫz)(0) = 0 and (¯ v − w + ǫz)(d) < 0. Lemma 2 implies that the function ¯ v − w + ǫz has the zero maximum value at t = 0 in the interval [0, d]. Thus
lim sup
t→0
+(¯ v − w + ǫz)(t)
t = lim sup
t→0
+¯
v(t) − w(t)
t + ǫα ≤ 0 and consequently
lim inf
t→0
+w(t) − ¯ v(t)
t ≥ ǫα > 0.
For the case ¯ v(1) = 0, defining z(t) = exp(α(1 − t)) − 1 and taking α > K sup{(1 − t)q(t) : t ∈ [τ n , 1)}, we obtain
lim inf
t→1
−w(t) − ¯ v(t) 1 − t > 0.
Combining this fact and properties of w − ¯ v and e described above, we get (7). Similarly, for the functions ˜ v − w and e, there exists ǫ 2 > 0 such that
(8) v ˜ − w(t) > ǫ 2 e(t) for all t ∈ (0, 1).
Therefore, by (7) and (8), we obtain w ∈ Ω ◦ 1 and, thus, Ω ◦ 1 6= ∅. If
u ∈ ∂ CeΩ ◦ 1 , the boundary of Ω ◦ 1 with respect to C e -topology and satisfies
u = T u. Then u ∈ Ω, since ∂ CeΩ ◦ 1 ⊂ Ω by the continuous embedding of
C e [0, 1] into C[0, 1]. This implies that u is a solution of problem (5) with
Ω ◦ 1 ⊂ Ω by the continuous embedding of
C e [0, 1] into C[0, 1]. This implies that u is a solution of problem (5) with
u ∈ Ω and thus, by the above argument, u ∈ Ω ◦ 1 and this contradiction shows that the Leray-Schauder degree d LS (I − T, Ω ◦ 1 , 0) is well defined.
Now we compute d LS (I −T, Ω ◦ 1 , 0). Let us consider the modified prob- lem
u ′′ + F (t, u) = 0, 0 < t < 1 u(0) = 0 = u(1),
where
F (t, u) =
q(t)f (˜ v(t)) − u − ˜ v(t)
1 + u 2 if u > ˜ v(t) q(t)f (u) if ¯ v(t) ≤ u ≤ ˜ v(t) q(t)f (¯ v(t)) − u − ¯ v(t)
1 + u 2 if u < ¯ v(t).
Then F : (0, 1) × R → R is continuous. Define the operator T u(t) = ˜
Z 1 0
G(t, s)F (s, u(s))ds.
Then ˜ T is completely continuous on C e [0, 1]. Choose a positive real con- stant δ as
δ > max n
sup
min ¯ v≤u≤max ˜ v
f (u), k¯ vk ∞ + 1
q o +kf ◦¯ vk ∞ , k˜ vk ∞ + 1
q o +kf ◦˜ vk ∞
o , where q o = inf t∈(0,1) q(t)(> 0). Then
−δe ≺ ˜ T (u) ≺ δe on [0, 1] for all u ∈ C[0, 1].
Thus Im ˜ T is contained in C e [0, 1] and bounded by δ with respect to k · k e . Choose a ball B e (0, R) in C e [0, 1] such that
Im ˜ T ∩ Ω 1 ⊂ B e (0, R) and consider the homotopy
˜ h µ u = u − µ ˜ T u.
Obviously ˜ h µ is completely continuous on C e [0, 1] for all µ ∈ [0, 1]. If u ∈ ∂B e (0, R) and ˜ h µ u = 0, then
R = kuk e = µk ˜ T uk e ≤ µδ.
Taking R big enough so that µδ < R, d LS (˜ h µ , B e (0, R), 0) is well-defined and by the homotopy invariance,
d LS (I − ˜ T , B e (0, R), 0) = d LS (I, B e (0, R), 0) = 1.
Since I − T is equivalent to I − ˜ T on Ω, so is on Ω ◦ 1 and by the excision property for I − ˜ T , we get
d LS (I − T, Ω ◦ 1 , 0) = d LS (I − ˜ T , Ω ◦ 1 , 0)
= d LS (I − ˜ T , B e (0, R), 0)
= d LS (I, B e (0, R), 0) = 1.
Let Ω ◦ 2 be the interior of
( ¯ w, w) = {u ∈ C ˜ e ([0, 1]) : ¯ w ≺ u ≺ ˜ w}
and Ω ◦ 3 be the interior of
(¯ v, w) = {u ∈ C ˜ e ([0, 1]) : ¯ v ≺ u ≺ ˜ w}
in C e ([0, 1]). Then we also have
d LS (I − T, Ω ◦ 2 , 0) = 1, d LS (I − T, Ω ◦ 3 , 0) = 1.
Thus by the excision and the additivity properties, 1 = d LS (I − T, Ω ◦ 3 , 0)
= d LS (I − T, Ω ◦ 1 , 0) + d LS (I − T, Ω ◦ 2 , 0) + d LS (I − T, Ω ◦ 3 \ (Ω ◦ 1 ∪ Ω ◦ 2 ), 0).
Therefore,
d LS (I − T, Ω ◦ 3 \ (Ω ◦ 1 ∪ Ω ◦ 2 ), 0) = −1.
This is for the proof of existence of three distinct solutions of (5).
If we choose u 1 be the minimal solution in [¯ v, v] and u ˜ 3 be the maximal solution in [ ˆ w, w], then the order of three solutions can be proved. ˜
The following theorem is another multiplicity of positive solutions
which can be applied to find 2N − 1 distinct ordered positive solutions
of the semilinear problem (P λ ) on the exterior domain.
Theorem 4. Assume that (C), (f 0 ) and f (0) > 0. Suppose also that there exist two pairs of G-lower and G-upper solutions {¯ v, ˜ v} and { ¯ w, w} ˜ satisfying the following conditions:
(i) ¯ v ≺ ˜ v, w ¯ ≺ ˜ w, v ¯ ≺ ˜ w, w ¯ ˜v, and none of them are solutions of (5);
(ii) [¯ v(t), ˜ w(t)] ⊂ I for all t ∈ [0, 1], (iii) ˜ v(i) > 0, ˜ w(i) > 0 for i = 0, 1, and
(iv) there is a small positive number δ such that ¯ v(t), ¯ w(t) ≤ 0 if 0 ≤ t ≤ δ and 0 ≤ 1 − t ≤ δ.
Then (5) has at least 3 distinct solutions u 1 ≺ u 2 ≺ u 3 such that u 1 ∈ [¯ v, v], u ˜ 2 ∈ [ ¯ w, w] and u ˜ 3 ∈ [¯ v, w] \ [¯ ˜ v, v] ∪ [ ¯ ˜ w, w]. ˜
Proof. We choose the positive function e which is the unique solution of the problem: u ′′ + q(t) = 0, 0 < t < 1 and u(0) = u(1) = 0. Then we prove this theorem in the space C e ([0, 1]). We note that the proof of this theorem can be shown using the exact same technique in Theorem 3 if we get the following: If w ∈ [¯ v, ˜ v] is a solution of (5), then w belongs to the interior of Ω 1 in the space C e [0, 1].
First, we prove that w ∈ C e ([0, 1]). To show that, we will find a positive number c so that w(t) ≤ ce(t) for all t ∈ (0, 1). It can be shown from the following calculations: By L’Hospital rule,
t→0 lim e(t) w(t) = lim
t→0
e ′ (t) w ′ (t) > 0 if w ′ (t) < ∞ as t → 0. Let w ′ (t) → ∞ as t → 0. Then
lim t→0
e ′ (t) w ′ (t) = lim
t→0
e ′′ (t) w ′′ (t) = lim
t→0
1
f (w(t)) > 0.
Similarly, we have the following limit:
t→1 lim e(t) w(t) > 0.
Thus, we get a positive constant c so that w(t) ≤ ce(t) for all t ∈ (0, 1).
Since w(t) ≥ 0 on some neighborhood of t = 0 or t = 1 in [0, 1], we can easily find a negative constant d so that de(t) ≤ w(t) for all t ∈ (0, 1).
Consequently, w ∈ C e ([0, 1]).
Secondly, we show that w lies in the interior of Ω 1 . Hence, we prove the existence of a positive number ǫ 1 so that
ǫ 1 e(t) + ¯ v(t) < w(t)
for all t ∈ (0, 1). Let
t∈(0,1) inf
w(t) − ¯ v(t) e(t) = γ.
We will show γ > 0.
Suppose that γ = 0. Then for any ǫ > 0, there is a point t ǫ ∈ (0, 1) such that
w(t ǫ ) − ¯ v(t ǫ ) e(t ǫ ) < ǫ.
Then
(∗) ǫe(t ǫ ) + ¯ v(t ǫ ) − w(t ǫ ) > 0.
We note that if either t = 0 or t = 1, then ǫe(t) + ¯ v(t) − w(t) ≤ 0.
Now, if we calculate the followings:
[ǫe(t) + ¯ v(t) − w(t)] ′′ − δ[ǫe(t) + ¯ v(t) − w(t)]
= ǫ(e ′′ (t) − δe(t)) + ¯ v ′′ (t) − w ′′ (t) − δ(¯ v(t) − w(t))
≥ ǫ(−q(t) − δe(t)) − q(t)f (¯ v(t)) + q(t)f (w(t)) − δ(¯ v(t) − w(t))
= ǫ(−q(t) − δe(t)) + q(t)[f (w(t)) − f (¯ v(t))] − δ(¯ v(t) − w(t))
≥ ǫ(−q(t) − δe(t)) + (δ − M q(t))(w(t) − ¯ v(t)).
Without loss of generality, we assume that t ǫ is a maximizer of ǫe(t) +
¯
v(t) − w(t) in [0, 1]. Let t 0 = lim inf ǫ→0 t ǫ . Suppose that t 0 6= 0, t 0 6= 1, and t 0 is not a singular interior point of ¯ v, i.e. t 0 6= τ i , i = 1, · · · , n in Definition 2. Then if we choose large δ > 0 so that δ − M q(t) > 0 on some neighborhood of t 0 in (0, 1) and choose very small ǫ > 0 so that
ǫ(−q(t) − δe(t)) + (δ − M q(t))(w(t) − ¯ v(t)) ≥ 0
on the neighborhood. It is always possible from the result of Lemma 3.
If ǫ > 0 is sufficiently small, then t ǫ belongs to the neighborhood of t 0 , and then by Maximum Principle, it leads to a contradiction.
Suppose t 0 = 0. By L’Hospital rule,
t limǫ→0
w(t ǫ ) − ¯ v(t ǫ )
e(t ǫ ) = lim
t
ǫ→0
w ′ (t ǫ )
e ′ (t ǫ ) > 0
if lim tǫ→0 e ′ (t ǫ ) < ∞, and
t limǫ→0
w ′ (t ǫ )
e ′ (t ǫ ) = lim
t
ǫ→0
w ′′ (t ǫ ) e ′′ (t ǫ )
= lim
t
ǫ→0
−q(t ǫ )f (w(t ǫ ))
−q(t ǫ )
= f (w(0)) > 0
if lim tǫ→0 e ′ (t ǫ ) = ∞. This leads to a contradiction to the fact γ = 0.
Similarly, we can get a contradiction in the case t 0 = 1.
Suppose that t 0 is a singular interior point of ¯ v. (∗) implies that
¯
v(t 0 ) = w(t 0 ). This is impossible from Lemma 3.
From condition (iii), we also can easily show the existence of a positive number ǫ 2 so that
w(t) < ˜ v(t) − ǫ 2 e(t) for all t ∈ (0, 1). Therefore,
ǫ 1 e(t) + ¯ v(t) < w(t) < ˜ v(t) − ǫ 2 e(t)
for all t ∈ (0, 1), and this implies that w ∈ Ω ◦ 1 .
4. Applications
Throughout this section, we assume that f (0) > 0. As an application of Theorem 4, we prove the existence of 2N − 1 distinct ordered positive solutions of the following problems ;
(1 λ ) ( u ′′ + λq(t)f (u) = 0, 0 < t < 1, u(0) = 0 = u(1),
where λ > 0 is a real parameter, q ∈ C((0, 1), R + ) is singular at t = 0 and/or 1.
Theorem 5. ([11]) For a, b ∈ R with a < b, consider
(9 λ ) ( u ′′ + λ˜ q(t)f (u) = 0, a < t < b,
u(a) = 0 = u(b).
Assume that ˜ q ∈ C 1 ([a, b], [0, ∞)) does not vanish identically on some subinterval of [a, b] and f satisfies (f 0 ) ∼ (f 3 ). Then there exists λ o > 0 such that for all λ > λ o , problem (9 λ ) has 2N − 1 distinct positive ordered solutions u i , u i+1
2
and u i+1 (i = 1, · · · , N − 1) such that u i (t) ≤ u i+1
2
(t) ≤ u i+1 (t)
for all t ∈ [a, b]. Moreover, for j = 1, · · · , N, u j (t) ≤ a j for all t ∈ [a, b], and u j converges to a j as λ → ∞ uniformly on every compact subset of (a, b).
Now we have a similar result for singular problem (1 λ ).
Theorem 6. Assume that q ∈ C 1 ((0, 1), (0, ∞)) satisfies (C) and f satisfies (f 0 ) ∼ (f 3 ). Then there exists λ o > 0 such that for all λ > λ o , the singular boundary value problem (1 λ ) has 2N − 1 distinct positive ordered solutions u i , u i+1
2
and u i+1 (i = 1, · · · , N − 1) such that u i (t) ≤ u i+1
2
(t) ≤ u i+1 (t) for all t ∈ [0, 1].
Proof. We prove this theorem by constructing N many G-lower solu- tions of (1 λ ). Choose a closed bounded interval [a, b] ⊂ (0, 1) with a < b.
Then by Theorem 5, there is λ 0 > 0 such that for all λ > λ o , problem (9 λ ) has N distinct positive ordered solutions v 1 , · · · , v N such that
v 1 (t) ≤ v 2 (t) ≤ · · · ≤ v N (t) for all t ∈ [a, b]
and v j (t) ≤ a j for all t ∈ [0, 1] and v j converges to a j as λ → ∞ uniformly on every compact subset of (a, b). Let
¯ u j (t) =
v j (t), a ≤ t ≤ b, 0 otherwise
for j = 1, · · · , N. Then ¯ u j is a G-lower solution and a j is a G-upper
solution of (1 λ ), respectively. Thus for each j = 1, · · · , N − 1, we have
two pairs of G-lower and G-upper solutions {¯ u j , a j } and {¯ u j+1 , a j+1 }
with ¯ u j+1 a j , none of them are solutions of (1 λ ). Since those pairs
satisfies all the conditions of Theorem 4, we obtain three solutions u j ,
u j +1
2
, u j +1 for each j such that u j ∈ [¯ u j , a j ], u j+1 ∈ [¯ u j+1 , a j +1 ], u i+1
2
∈ [¯ u j , a j+1 ] \ [¯ u j , a j ] ∪ [¯ u j+1 , a j +1 ], and
u i (t) ≤ u i+1
2