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A multiplicity result of positive radial solutions for a multiparameter elliptic system on an exterior domain
Nonlinear Analysis 45 (2001) 597 – 611
www.elsevier.nl/locate/na
A multiplicity result of positive radial solutions for a multiparameter elliptic system on an exterior domain Yong-Hoon Lee Department of Mathematics, Pusan National University, Pusan 609-735, South Korea Received 2 March 1999; accepted 10 September 1999
Keywords: Semilinear elliptic system; Positive radial solution; Exterior domain; Upper solution; Lower solution; Fixed point index
1. Introduction The existence and multiplicity of positive radial solutions for elliptic problems of the form 3u + k(|x|)f(u(x)) = 0 with suitable boundary conditions has been studied by several authors recently. One may refer, with more references therein, to Lan and Webb [4] for scalar equations de
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Y.H. Lee / Nonlinear Analysis 45 (2001) 597 – 611
2 inequality in R+ with superlinear growth at ∞; they proved that there exists ∗ ¿ 0 such that the problem has at least two, at least one or no positive radial solutions according to 0 ¡ ¡ ∗ ; =∗ or ¿ ∗ : Thus, it is interesting to consider the system which is de
3u + k1 (|x|)f(u; v) = 0; (P)
3v + k2 (|x|)g(u; v) = 0 in ?; u(x) = 0 = v(x) on |x| = r0 ; lim u(x) = 0 = lim v(x); |x|→∞
|x|→∞
2 \{(0; 0)} without any where n ≥ 3: Throughout this paper, we consider (; ) ∈ R+ further mention and assume ki ∈ C([r0 ; ∞); R+ ); not vanishing identically on any sub2 ; R0+ ); where we denote R+ = [0; ∞); R0+ = (0; ∞) interval of [r0 ; ∞) and f; g ∈ C(R+ 2 and R+ = [0; ∞) × [0; ∞). Moreover, under assumptions; ∞ (H1 ) r0 rki (r) dr ¡ ∞; i = 1; 2: 2 , (H2 ) f and g are nondecreasing on R+ i.e. f(u1 ; v1 ) ≤ f(u2 ; v2 ) and g(u1 ; v1 ) ≤ g(u2 ; v2 ) whenever (u1 ; v1 ) ≤ (u2 ; v2 ); where 2 the inequality on R+ can be understood componentwise:
(H3 )
f∞ ,
f(u; v) = ∞; (u;v)→∞ u + v lim
g∞ ,
g(u; v) = ∞; (u;v)→∞ u + v lim
we prove that there exists (∗ ; ∗ ) ¿ (0; 0) such that problem (P) has at least two, at least one or no positive radial solutions according to (; ) ¡ (∗ ; ∗ ); (; ) ∈ G 2 \{(0; 0)}: = ∗ ; ≤ ∗ } ∪ {(; ) ∈ or (; ) ¿ (∗ ; ∗ ); where G = {(; ) ∈ R+ 2 ∗ ∗ R+\{(0; 0)}: ≤ ; = }: Proofs of our theorems are based on upper and lower solutions argument and the
v (t) + q2 (t)g(u(t); v(t)) = 0;
0 ¡ t ¡ 1;
u(0) = u(1) = v(0) = v(1) = 0: where qi can be given as qi (t) =
r02 (1 − t)−2(n−1)=n−2 ki (r0 (1 − t)−1=(n−2) ); (n − 2)2
i = 1; 2:
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We know that the existence of positive solutions for (S) guarantees the existence of positive radial solutions for (P). We notice that by (H1 ); qi ∈ C([0; 1); R+ ) satisfy 1 (1 − s)qi (s) ds ¡ ∞: It is interesting to note that the function qi are either regular 0 or singular at t = 1 depending on the asymptotic behaviour of the coeJcient function ki in (P), see Section 4 for further details. We will focus on proving for the singular case, since then the result for regular case is followed with less restriction. For this reason, we extend our argument for the case that qi may be singular at both end points t = 0 and 1 in the remaining sections. The paper is organized as follows; in Section 2, we introduce and prove a theorem of upper and lower solutions method for singular systems. In Section 3, we prove the existence and nonexistence parts of the result and in Section 4, we
2. Upper and lower solutions method In this section, we prove a theorem of upper and lower solutions method for a singular second-order ordinary diHerential system. Let us consider u (t) + F(t; u(t); v(t)) = 0; (S1 )
v (t) + G(t; u(t); v(t)) = 0;
0 ¡ t ¡ 1;
u(0) = u(1) = v(0) = v(1) = 0; where F; G : D → R are continuous functions and D ⊂(0; 1) × R2 . A solution (u; v) of (S1 ) means functions u; v ∈ C([0; 1]; R) ∩ C 2 ((0; 1); R) such that (t; u(t); v(t)) ∈ D for all t ∈ (0; 1) and u and v satisfy (S1 ): Lemma 1. Assume that there exist hF ; hG ∈ C((0; 1); R+ ) such that (C1 )
|F(t; u; v)| ≤ hF (t);
|G(t; u; v)| ≤ hG (t);
for all (t; u; v) ∈ (0; 1) × R2 and hF and hG satisfy 1 1 s(1 − s)hF (s) ds ¡ ∞; CG , s(1 − s)hG (s) ds ¡ ∞: (C2 ) CF , 0
0
Then (S1 ) has a solution. Proof. Let us de
b(u; v)(t) =
0
1
K(t; s)G(s; u(s); v(s)) ds;
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Y.H. Lee / Nonlinear Analysis 45 (2001) 597 – 611
where K(t; s) is the Green’s function explicitly written as s(1 − t) for 0 ≤ s ≤ t; K(t; s) = t(1 − s) for t ≤ s ≤ 1: Also let T (u; v)(t) = (a(u; v)(t); b(u; v)(t)): Then T : X , C([0; 1]; R) × C([0; 1]; R) → X is well-de
t s
+ ,
s
!
xhF (x) d x +
0
t s t
0
!
1
!
(1 − x)hF (x) d x
xhG (x) d x +
’F (!) d! +
t
s
1
!
d!
(1 − x)hG (x) d x
d!
’G (!) d!:
We note that ’F and ’G do not depend on (u; v) in B and thus the equicontinuity of TB follows the fact ’F ; ’G ∈ L1 (0; 1). Integrating by parts, we obtain 1 t 1 |’F (s)| ds = lim− (1 − t) shF (s) ds + lim+ t (1 − s)hF (s) ds 0
t→1
0
+2
1
0
≤ lim
t→1−
0
t→0
t
s(1 − s)hF (s) ds t
s(1 − s)hF (s) ds + lim+ t→0
t
1
s(1 − s)hF (s) ds
Y.H. Lee / Nonlinear Analysis 45 (2001) 597 – 611
+2 ≤4 Similarly, we get
1 0
0
1
0 1
601
s(1 − s)hF (s) ds
s(1 − s)hF (s) ds = 4CF :
|’G (s)| ds ≤ 4CG ; and the proof is complete.
Denition 1. For $u ; $v ∈ C([0; 1]; R)∩C 2 ((0; 1); R); we say ($u ; $v ) is a lower solution of (S1 ) if (t; $u (t); $v (t)) ∈ D for all t ∈ (0; 1) and $u (t) + F(t; $u (t); $v (t)) ≥ 0; $v (t) + G(t; $u (t); $v (t)) ≥ 0;
t ∈ (0; 1);
$u (0); $u (1); $v (0); $v (1) ≤ 0: We also de
for all t ∈ [0; 1]:
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Y.H. Lee / Nonlinear Analysis 45 (2001) 597 – 611
where pi are given by p1 (t; u; v) = max{$u (t); min{u; %u (t)}}; p2 (t; u; v) = max{$v (t); min{v; %v (t)}}: Then F ∗ ; G ∗ : (0; 1) × R2 → R are continuous and satisfy condition (C1 ) and (C2 ) in Lemma 1. Let us consider the problem u (t) + F ∗ (t; u(t); v(t)) = 0; (M1 )
v (t) + G ∗ (t; u(t); v(t)) = 0;
0 ¡ t ¡ 1;
u(0) = u(1) = v(0) = v(1) = 0: If all solutions (u; v) of (M1 ) satisfy ($u (t); $v (t)) ≤ (u(t); v(t)) ≤ (%u (t); %v (t))
for all t ∈ [0; 1];
then by Lemma 1, the proof is complete. Suppose that the argument is not true, so let us assume that ($u ; $v ) Ä (u; v) on [0; 1]: Then for either u or v; let us say u; we may choose an interval [t1 ; t2 ] ⊂(0; 1); by boundary conditions of u and $u ; such that u(t1 ) = $u (t1 );
u(t2 ) = $u (t2 );
u(t) ¡ $u (t)
for all t ∈ (t1 ; t2 ):
Thus $u − u has a positive maximum on (t1 ; t2 ). On the other hand, by (a5 ) we obtain ($u − u) ≥ F ∗ (t; u(t); v(t)) − F(t; $u (t); $v (t)) = F(t; $u (t); p2 (t; u(t); v(t))) − F(t; $u (t); $v (t)) ≥ 0 for all t ∈ (t1 ; t2 ): Thus by maximum principle, this contradicts that $u −u has a positive maximum on (t1 ; t2 ): The validity that (u; v) ≤ (%u ; %v ) on (0; 1) can be shown by a similar way. 1 Remark 1. (i) If we assume, instead of (a4 ); the condition 0 shF (s) ds ¡ ∞; 1 shG (s) ds ¡ ∞; then the pair of solutions u; v ∈ C 1 ((0; 1]; R): On the other hand, if 01 1 (1 − s)hF (s) ds ¡ ∞; 0 (1 − s)hG (s) ds ¡ ∞; then u; v ∈ C 1 ([0; 1); R): 0 (ii) Theorem 1 is valid for arbitrary
v (t) + q2 (t)g(u(t); v(t)) = 0; u(0) = u(1) = v(0) = v(1) = 0:
0 ¡ t ¡ 1;
Y.H. Lee / Nonlinear Analysis 45 (2001) 597 – 611
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In what follows, we assume that qi ∈ C((0; 1); R+ ); i = 1; 2 do not vanish identically on any subinterval of (0; 1): The following is the main theorem of this section. Theorem 2. Assume 1 (H) qi are singular at t = 0 and 1 and 0 s(1 − s)qi (s) ds ¡ ∞; i = 1; 2. (H2 ) f and g are quasi-monotone nondecreasing w.r.t. v and u; respectively. (H3 ) f∞ = ∞ = g∞ . Then there exists (∗ ; ∗ ) ¿ (0; 0) such that (S) has at least one positive solution for (; ) ≤ (∗ ; ∗ ) and no solution for (; ) ¿ (∗ ; ∗ ): We set up an operator equation for
B (u; v)(t) ,
0
1
K(s; t)q2 (s)g(u(s); v(s)) ds:
Also de
min
1=4≤t≤3=4
(u(t) + v(t)) ≥
1 4 (u; v)
:
Then C and K are cones in X . We need some lemmas to prove Theorem 2. 2 Lemma 2. Assume (H). Then for all (; ) ∈ R+ \{(0; 0)}; T; is completely continuous on X and T; (C) ⊂ K:
Proof. The complete continuities of A ; B and T; can be shown by the same way as in the proof of Lemma 1 and we prove T; (C) ⊂ K: Let (u; v) ∈ C; then for t ∈ [ 14 ; 34 ]; using the following fact for the Green’s function; K(t; s) ≥ 14 K(!; s); for all !; s ∈ [0; 1]; we obtain 1 1 A (u; v)(t) ≥ K(!; s)q1 (s)f(u(s); v(s)) ds = A (u; v)(!) 4 0 4 for all ! ∈ [0; 1]: Thus we get min
1=4≤t≤3=4
A (u; v)(t) ≥ 14 A (u; v)∞ :
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Y.H. Lee / Nonlinear Analysis 45 (2001) 597 – 611
Similarly, min
1=4≤t≤3=4
B (u; v)(t) ≥ 14 B (u; v)∞ :
The following lemma can be proved by simple calculations. One may refer to Dalmasso [1] for the proof in detail. Lemma 3. Assume (H). Then the problem; for i = 1; 2 (Li )
u (t) + t(1 − t)qi (t) = 0;
0 ¡ t ¡ 1;
u(0) = 0 = u(1)
has a unique solution ’i ∈ C 1 [0; 1] ∩ C 2 (0; 1) satisfying ci ’i (t) ≤ t(1 − t) ≤ di ’i (t) for some ci ; di ¿ 0; i = 1; 2: Since each component of solutions of (S) are not guaranteed of C 1 [0; 1]; integration by part cannot be applied directly and we need to prove the following lemma. Lemma 4. Assume (H). Let (u; v) be a solution of (S) and ’i be a solution of (Li ). Then 1 1 u(s)’1 (s) ds; u (s)’1 (s) ds = 0
0
0
1
v (s)’2 (s) ds =
0
1
v(s)’2 (s) ds:
Proof. We
t→0+
By Lemma 3, we get the following inequalities: t0 |’1 (+)u (+)| ≤ ’1 (+) u (s) ds + u (t0 ) +
t0 1 1 +(1 − +) |u (s)| ds + +(1 − +)|u (t0 )| c1 c1 + t0 1 1 ≤ (1 − +) sq1 (s)f(u(s); v(s)) ds + +(1 − +)|u (t0 )|: c1 c 1 +
≤
Thus by (H), for given , ¿ 0 and
+ ˜ t→0
Y.H. Lee / Nonlinear Analysis 45 (2001) 597 – 611
Finally, 0
1
u (s)’1 (s) ds = lim
+ ˜ t; t→0
1−t˜
t
605
u (s)’1 (s) ds
= − lim+ ’1 (t)u (t) + lim ’1 (1 − t˜)u (1 − t˜) + ˜ t→0
t→0
+ lim
+ ˜ t; t→0
= lim
+ ˜ t; t→0
t
1−t˜
t 1−t˜
u(s)’1 (s) ds
u(s)’1 (s) ds =
0
1
u(s)’1 (s) ds:
We can show the other equality by the same way and the proof is complete. The following lemma gives the a priori bound of solutions for problem (S). 2 Lemma 5. Assume (H) and (H3 ). Let R be a compact subset of R+ \{(0; 0)}: Then there exists a constant bR ¿ 0 such that for all (; ) ∈ R and for all possible positive solutions (u; v) of (S) at (; ); one has
(u; v) ¡ bR : Proof. Suppose by contradiction that there is a sequence (un ; vn ) of positive solutions of (S) with corresponding (n ; n ) belonging to R such that (un ; vn ) → ∞: Then by Lemma 2, (un ; vn ) ∈ K and thus min
1=4 ≤ t ≤ 3=4
(un (t) + vn (t)) ≥ 14 (un ∞ + vn ∞ ):
(1)
Considering a subsequence if necessary, we
for all u + v ≥ R1 ;
where - satis
(2)
for all n
and m = min{K(t; s): (t; s) ∈ [ 14 ; 34 ] × [ 14 ; 34 ]}: Thus for t ∈ [ 14 ; 34 ]; using (1) and (2), we have 1 K(t; s)q1 (s)f(un (s); vn (s)) ds un ∞ ≥ un (t) = n 0
≥ n
3=4
1=4
K(t; s)q1 (s)f(un (s); vn (s)) ds
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Y.H. Lee / Nonlinear Analysis 45 (2001) 597 – 611
≥ n m
3=4
1=4
≥ n m≥
n m4
q1 (s)f(un (s); vn (s)) ds
3=4
1=4
q1 (s)(un (s) + vn (s)) ds
3=4
1=4
q1 (s)(un ∞ + vn ∞ ) ds
¿ un ∞ + vn ∞ ≥ un ∞ for suJciently large n. This is a contradiction. Proof of the case n = 0; n ¿ 0 for suJciently large n can be done by using g∞ = ∞ and the proof is complete. Before proving Theorem 2, we note that by (H) and (H2 ); we can use Theorem 1 only checking condition (a1 ) in the theorem. Proof of Theorem 2. The proof consists of three claims. Claim 1. Problem (S) has a positive solution for certain (; ): Let %i (t) =
1 0
K(t; s)qi (s) ds be the unique solution of
u (t) + qi (t) = 0; u(0) = 0 = u(1): Let Mf = maxt∈[0; 1] f(%1 (t); %2 (t)) and Mg = maxt∈[0; 1] g(%1 (t); %2 (t)): Then Mf and Mg ¿ 0; since f; g ¿ 0: %1 (t) + q1 (t)f(%1 (t); %2 (t)) = q1 (t)[f(%1 (t); %2 (t)) − 1] ≤ 0; %2 (t) + q2 (t)g(%1 (t); %2 (t)) = q2 (t)[g(%1 (t); %2 (t)) − 1] ≤ 0; whenever (; ) ≤ (1=Mf ; 1=Mg ): This shows that (%1 ; %2 ) is an upper solution of (S). On the other hand, (0,0) is obviously a lower solution of (S) and (0; 0) ≤ (%1 ; %2 ): Thus by Theorem 1, (S) has a positive solution for (0; 0) ¡ (; ) ≤ (1=Mf ; 1=Mg ): 2 De
for all u; v ≥ 0:
(3)
Let (; ) ∈ S and (u; v) be a positive solution of (S) at (; ); then by Lemmas 3, 4
Y.H. Lee / Nonlinear Analysis 45 (2001) 597 – 611
and (3), we get the following inequalities: 1 s(1 − s)q1 (s)u(s) ds ≤ mf d1 mf 0
0
≤ d1
1
0
= −d1 = d1
1
0
’1 (s)q1 (s)(u(s) + v(s)) ds
’1 (s)q1 (s)f(u(s); v(s)) ds
1
0
1
607
’1 (s)u (s) ds = −d1
0
1
’1 (s)u(s) ds
s(1 − s)q1 (s)u(s) ds:
1 Since 0 ¡ 0 s(1 − s)q1 (s)u(s) ds ¡ ∞; ≤ d1 =mf : Similarly, we can get ≤ d2 =mg : Therefore S is bounded above by (d1 =mf ; d2 =mg ): Claim 3. (S; ≤) has a maximal element. Let P be a chain in S: By Zorn’s Lemma, it is enough to show that P has an upper bound in S: Without loss of generality, we may choose a distinct sequence {(n ; n )} ⊂ P such that (n ; n ) ≤ (n+1 ; n+1 ); n=1; 2; : : : : By Claim 2, two sequences {(n )} and {(n )} converge to, say, P and P ; respectively. If (P ; P ) ∈ S; then the proof is done. Since the sequence {(n ; n )} is bounded above, it belongs to a compact 2 \{(0; 0)} and Lemma 5 implies that the corresponding solutions {(un ; vn )} subset of R+ are uniformly bounded in X: By the compactness of the integral operators A and B ; the sequence {(un ; vn )} has a subsequence converging to, say, (uP ; vP ) ∈ X: We can easily show, by the Lebesgue convergence theorem, that (uP ; vP ) is a solution of (S) at (P ; P ): Thus (P ; P ) ∈ S. Let (∗ ; ∗ ) be a maximal element of S and (u∗ ; v∗ ); the corresponding positive solution of (S). Then for all (; ) ≤ (∗ ; ∗ ); problem (S) has a positive solution at (; ); since (u∗ ; v∗ ) and (0; 0) are an upper solution and a lower solution, respectively. This concludes the proof. 4. Multiplicity In this section, we show the existence of the second positive solution for (; ) ¡ (∗ ; ∗ ), where (∗ ; ∗ ) is a maximal element of S. Let (u∗ ; v∗ ) be a positive solution of (S) at (∗ ; ∗ ). Lemma 6. For (; ) ¡ (∗ ; ∗ ); there exists ,0 ¿ 0 such that (u∗ + ,; v∗ + ,) is an upper solution of (S) at (; ); for all 0 ¡ , ≤ ,0 . Proof. Let M be a constant such that f(u∗ (t); v∗ (t)) ≥ M ¿ 0;
g(u∗ (t); v∗ (t)) ≥ M ¿ 0:
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Then by Lemma 4 and the uniform continuity of f and g, there exists ,0 ¿ 0 such that M (∗ − ) |f(u∗ (t) + ,; v∗ (t) + ,) − f(u∗ (t); v∗ (t))| ¡ ; |g(u∗ (t) + ,; v∗ (t) + ,) − g(u∗ (t); v∗ (t))| ¡
M (∗ − )
for all t ∈ [0; 1] and 0 ¡ , ≤ ,0 . Let u,∗ (t) = u∗ (t) + , and v,∗ (t) = v∗ (t) + ,, then u,∗ (0); u,∗ (1); v,∗ (0); v,∗ (1) ¿ 0 and u,∗ (t) + q1 (t)f(u,∗ (t); v,∗ (t)) = u∗ (t) + q1 (t)f(u∗ (t) + ,; v∗ (t) + ,) = − ∗ q1 (t)f(u∗ (t); v∗ (t)) + q1 (t)f(u∗ (t) + ,; v∗ (t) + ,) = q1 (t)[f(u∗ (t) + ,; v∗ (t) + ,) − f(u∗ (t); v∗ (t))] −(∗ − )q1 (t)f(u∗ (t); v∗ (t)) ¡ M (∗ − )q1 (t) − (∗ − )q1 (t)f(u∗ (t); v∗ (t)) = (∗ − )q1 (t)[M − f(u∗ (t); v∗ (t))] ≤ 0 for all t ∈ [0; 1]. The inequalities for v,∗ can be shown similarly and the proof is complete. We note that for 0 ¡ , ≤ ,0 ; (u,∗ ; v,∗ ) satis
v,∗ (t) ¿
0
1
K(t; s)q2 (s)g(u,∗ (s); v,∗ (s)) ds:
We introduce the following two lemmas for
Y.H. Lee / Nonlinear Analysis 45 (2001) 597 – 611
609
We now state and prove our main result. Theorem 3. Assume (H); (H2 ) and (H3 ). Then there exists (∗ ; ∗ ) ¿ (0; 0) such that problem (S) has at least two positive solution for (; ) ¡ (∗ ; ∗ ); at least one positive solution for (; ) ∈ G and no positive solution for (; ) ¿ (∗ ; ∗ ); where 2 2 \{(0; 0)}: = ∗ ; ≤ ∗ } ∪ {(; ) ∈ R+ \{(0; 0)}: ≤ ∗ ; = ∗ }: G = {(; ) ∈ R+ Proof. By Theorem 2, it is enough to show the existence of the second positive solution of (S) for (; ) ¡ (∗ ; ∗ ). Let u,∗ (t) = u∗ (t) + ,; v,∗ (t) = v∗ (t) + ,, where , is given in Lemma 6. Let ? = {(u; v) ∈ X : − , ¡ u(t) ¡ u,∗ (t); −, ¡ v(t) ¡ v,∗ (t); t ∈ [0; 1]}, then ? is bounded open in X , 0 ∈ ? and T; : K ∩ ? → K is condensing, since it is completely continuous. Let (u; v) ∈ K ∩ @?, then there exists t0 ∈ [0; 1] such that either u(t0 ) = u,∗ (t0 ) or v(t0 ) = v,∗ (t0 ). Suppose that u(t0 ) = u,∗ (t0 ): Then by (H2 ) and Lemma 6, 1 A (u; v)(t0 ) = K(t0 ; s)q1 (s)f(u(s); v(s)) ds 0
≤
1
K(t0 ; s)q1 (s)f(u∗ (s) + ,; v∗ (s) + ,) ds
0
¡ u∗ (t0 ) + , = u(t0 ) ≤ 1u(t0 ) for all 1 ≥ 1. Similarly, for the case v(t0 ) = v,∗ (t0 ), we can get B (u; v)(t0 ) ¡ 1v(t0 ); for all 1 ≥ 1. Thus T; (u; v) = 1(u; v), for all (u; v) ∈ K ∩ @? and all 1 ≥ 1 and by Lemma 7, i(T; ; K ∩ ?; K) = 1: Let R = max{bR ; 4R1 ; (u,∗ ; v,∗ )}, where bR is given in Lemma 5 with R a compact 2 subset of R+ \{(0; 0)} containing (; ) and R1 is given in (2) for n = in the proof of Lemma 5. Let KR = {(u; v) ∈ K: (u; v) ¡ R}. Then by Lemma 5, T; (u; v) = (u; v) for (u; v) ∈ @KR . Furthermore, if (u; v) ∈ @KR ; then min (u(t) + v(t)) ≥ 14 (u; v) ≥ R1 :
t∈[1=4; 3=4]
Thus by (2), f(u(t); v(t)) ≥ -(u(t) + v(t)), for all t ∈ [ 14 ; 34 ] and 3=4 q1 (s)f(u(s); v(s)) ds A (u; v)(t) ≥ m 1=4
≥ m≥
m4
3=4
1=4
q1 (s)(u(s) + v(s)) ds
3=4
1=4
¿ (u; v):
q1 (s)(u; v) ds
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Y.H. Lee / Nonlinear Analysis 45 (2001) 597 – 611
Therefore T; (u; v) ≥ A (u; v)∞ ≥ (u; v) and by Lemma 8, i(T; ; KR ; K) = 0: Consequently, by the additivity of the
3v + k2 (|x|)g(u; v) = 0 in ?; u(x) = 0 = v(x) on |x| = r0 ; lim u(x) = 0 = lim v(x);
|x|→∞
|x|→∞
where ; ¿ 0; n ≥ 3. As we mentioned in Section 1, qi are given by qi (t) =
r02 (1 − t)(−2(n−1))=(n−2) ki (r0 (1 − t)−1=(n−2) ); (n − 2)2
i = 1; 2
and (H1 ) corresponds to condition (H) via the transformations. We need to consider the following two cases for ki : Case I: limr→∞ r 2(n−1) ki (r) ¡ ∞. In this case, limt→1− qi (t) ¡ ∞ so that qi can be extended continuously on [0; 1] and the problem becomes regular. Thus applying Theorem 5, we obtain desired result described below. Case II: limr→∞ r 2(n−1) ki (r) = ∞. In this case, limt→1− qi (t) = ∞ and the problem becomes singular. To apply Theorem 4, we need to add condition (H1 ). Since functions ki in Case I also satis
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611
Corollary 1. Assume (H1 ) – (H3 ). Then there exists (∗ ; ∗ ) ¿ (0; 0) such that problem (P) has at least two positive radial solutions for (; ) ¡ (∗ ; ∗ ); at least one positive radial solution for (; ) ∈ G and no positive radial solution for (; ) ¿ (∗ ; ∗ ); 2 2 \ {(0; 0)}: ≤ where G = {(; ) ∈ R+ \ {(0; 0)}: = ∗ ; ≤ ∗ } ∪ {(; ) ∈ R+ ∗ ∗ ; = }. Remark 2. According to our result in this paper, we do not give any information on 2 2 the areas like {(; ) ∈ R+ \{(0; 0)}: ≤ \{(0; 0)}: ¿ ∗ ; ≤ ∗ } or {(; ) ∈ R+ ∗ ∗ ; ¿ }. Reminding that the multiparameter set S of solutions, in the proof of 2 Theorem 2, is bounded above in R+ , we leave some questions; ∗ ∗ 1. Is the maximal element ( ; ) of S unique? 2. Is the following statement true? 2 There exists a curve P which separates R+ into two (relatively) open sets O1 and O2 such that problem (P) has at least two positive radial solutions for (; ) ∈ O1 ; at least one positive radial solution for (; ) ∈ P and no positive radial solution for (; ) ∈ O2 . Acknowledgements The author wishes to acknowledge the
www.elsevier.nl/locate/na
A multiplicity result of positive radial solutions for a multiparameter elliptic system on an exterior domain Yong-Hoon Lee Department of Mathematics, Pusan National University, Pusan 609-735, South Korea Received 2 March 1999; accepted 10 September 1999
Keywords: Semilinear elliptic system; Positive radial solution; Exterior domain; Upper solution; Lower solution; Fixed point index
1. Introduction The existence and multiplicity of positive radial solutions for elliptic problems of the form 3u + k(|x|)f(u(x)) = 0 with suitable boundary conditions has been studied by several authors recently. One may refer, with more references therein, to Lan and Webb [4] for scalar equations de
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Y.H. Lee / Nonlinear Analysis 45 (2001) 597 – 611
2 inequality in R+ with superlinear growth at ∞; they proved that there exists ∗ ¿ 0 such that the problem has at least two, at least one or no positive radial solutions according to 0 ¡ ¡ ∗ ; =∗ or ¿ ∗ : Thus, it is interesting to consider the system which is de
3u + k1 (|x|)f(u; v) = 0; (P)
3v + k2 (|x|)g(u; v) = 0 in ?; u(x) = 0 = v(x) on |x| = r0 ; lim u(x) = 0 = lim v(x); |x|→∞
|x|→∞
2 \{(0; 0)} without any where n ≥ 3: Throughout this paper, we consider (; ) ∈ R+ further mention and assume ki ∈ C([r0 ; ∞); R+ ); not vanishing identically on any sub2 ; R0+ ); where we denote R+ = [0; ∞); R0+ = (0; ∞) interval of [r0 ; ∞) and f; g ∈ C(R+ 2 and R+ = [0; ∞) × [0; ∞). Moreover, under assumptions; ∞ (H1 ) r0 rki (r) dr ¡ ∞; i = 1; 2: 2 , (H2 ) f and g are nondecreasing on R+ i.e. f(u1 ; v1 ) ≤ f(u2 ; v2 ) and g(u1 ; v1 ) ≤ g(u2 ; v2 ) whenever (u1 ; v1 ) ≤ (u2 ; v2 ); where 2 the inequality on R+ can be understood componentwise:
(H3 )
f∞ ,
f(u; v) = ∞; (u;v)→∞ u + v lim
g∞ ,
g(u; v) = ∞; (u;v)→∞ u + v lim
we prove that there exists (∗ ; ∗ ) ¿ (0; 0) such that problem (P) has at least two, at least one or no positive radial solutions according to (; ) ¡ (∗ ; ∗ ); (; ) ∈ G 2 \{(0; 0)}: = ∗ ; ≤ ∗ } ∪ {(; ) ∈ or (; ) ¿ (∗ ; ∗ ); where G = {(; ) ∈ R+ 2 ∗ ∗ R+\{(0; 0)}: ≤ ; = }: Proofs of our theorems are based on upper and lower solutions argument and the
v (t) + q2 (t)g(u(t); v(t)) = 0;
0 ¡ t ¡ 1;
u(0) = u(1) = v(0) = v(1) = 0: where qi can be given as qi (t) =
r02 (1 − t)−2(n−1)=n−2 ki (r0 (1 − t)−1=(n−2) ); (n − 2)2
i = 1; 2:
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599
We know that the existence of positive solutions for (S) guarantees the existence of positive radial solutions for (P). We notice that by (H1 ); qi ∈ C([0; 1); R+ ) satisfy 1 (1 − s)qi (s) ds ¡ ∞: It is interesting to note that the function qi are either regular 0 or singular at t = 1 depending on the asymptotic behaviour of the coeJcient function ki in (P), see Section 4 for further details. We will focus on proving for the singular case, since then the result for regular case is followed with less restriction. For this reason, we extend our argument for the case that qi may be singular at both end points t = 0 and 1 in the remaining sections. The paper is organized as follows; in Section 2, we introduce and prove a theorem of upper and lower solutions method for singular systems. In Section 3, we prove the existence and nonexistence parts of the result and in Section 4, we
2. Upper and lower solutions method In this section, we prove a theorem of upper and lower solutions method for a singular second-order ordinary diHerential system. Let us consider u (t) + F(t; u(t); v(t)) = 0; (S1 )
v (t) + G(t; u(t); v(t)) = 0;
0 ¡ t ¡ 1;
u(0) = u(1) = v(0) = v(1) = 0; where F; G : D → R are continuous functions and D ⊂(0; 1) × R2 . A solution (u; v) of (S1 ) means functions u; v ∈ C([0; 1]; R) ∩ C 2 ((0; 1); R) such that (t; u(t); v(t)) ∈ D for all t ∈ (0; 1) and u and v satisfy (S1 ): Lemma 1. Assume that there exist hF ; hG ∈ C((0; 1); R+ ) such that (C1 )
|F(t; u; v)| ≤ hF (t);
|G(t; u; v)| ≤ hG (t);
for all (t; u; v) ∈ (0; 1) × R2 and hF and hG satisfy 1 1 s(1 − s)hF (s) ds ¡ ∞; CG , s(1 − s)hG (s) ds ¡ ∞: (C2 ) CF , 0
0
Then (S1 ) has a solution. Proof. Let us de
b(u; v)(t) =
0
1
K(t; s)G(s; u(s); v(s)) ds;
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Y.H. Lee / Nonlinear Analysis 45 (2001) 597 – 611
where K(t; s) is the Green’s function explicitly written as s(1 − t) for 0 ≤ s ≤ t; K(t; s) = t(1 − s) for t ≤ s ≤ 1: Also let T (u; v)(t) = (a(u; v)(t); b(u; v)(t)): Then T : X , C([0; 1]; R) × C([0; 1]; R) → X is well-de
t s
+ ,
s
!
xhF (x) d x +
0
t s t
0
!
1
!
(1 − x)hF (x) d x
xhG (x) d x +
’F (!) d! +
t
s
1
!
d!
(1 − x)hG (x) d x
d!
’G (!) d!:
We note that ’F and ’G do not depend on (u; v) in B and thus the equicontinuity of TB follows the fact ’F ; ’G ∈ L1 (0; 1). Integrating by parts, we obtain 1 t 1 |’F (s)| ds = lim− (1 − t) shF (s) ds + lim+ t (1 − s)hF (s) ds 0
t→1
0
+2
1
0
≤ lim
t→1−
0
t→0
t
s(1 − s)hF (s) ds t
s(1 − s)hF (s) ds + lim+ t→0
t
1
s(1 − s)hF (s) ds
Y.H. Lee / Nonlinear Analysis 45 (2001) 597 – 611
+2 ≤4 Similarly, we get
1 0
0
1
0 1
601
s(1 − s)hF (s) ds
s(1 − s)hF (s) ds = 4CF :
|’G (s)| ds ≤ 4CG ; and the proof is complete.
Denition 1. For $u ; $v ∈ C([0; 1]; R)∩C 2 ((0; 1); R); we say ($u ; $v ) is a lower solution of (S1 ) if (t; $u (t); $v (t)) ∈ D for all t ∈ (0; 1) and $u (t) + F(t; $u (t); $v (t)) ≥ 0; $v (t) + G(t; $u (t); $v (t)) ≥ 0;
t ∈ (0; 1);
$u (0); $u (1); $v (0); $v (1) ≤ 0: We also de
for all t ∈ [0; 1]:
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Y.H. Lee / Nonlinear Analysis 45 (2001) 597 – 611
where pi are given by p1 (t; u; v) = max{$u (t); min{u; %u (t)}}; p2 (t; u; v) = max{$v (t); min{v; %v (t)}}: Then F ∗ ; G ∗ : (0; 1) × R2 → R are continuous and satisfy condition (C1 ) and (C2 ) in Lemma 1. Let us consider the problem u (t) + F ∗ (t; u(t); v(t)) = 0; (M1 )
v (t) + G ∗ (t; u(t); v(t)) = 0;
0 ¡ t ¡ 1;
u(0) = u(1) = v(0) = v(1) = 0: If all solutions (u; v) of (M1 ) satisfy ($u (t); $v (t)) ≤ (u(t); v(t)) ≤ (%u (t); %v (t))
for all t ∈ [0; 1];
then by Lemma 1, the proof is complete. Suppose that the argument is not true, so let us assume that ($u ; $v ) Ä (u; v) on [0; 1]: Then for either u or v; let us say u; we may choose an interval [t1 ; t2 ] ⊂(0; 1); by boundary conditions of u and $u ; such that u(t1 ) = $u (t1 );
u(t2 ) = $u (t2 );
u(t) ¡ $u (t)
for all t ∈ (t1 ; t2 ):
Thus $u − u has a positive maximum on (t1 ; t2 ). On the other hand, by (a5 ) we obtain ($u − u) ≥ F ∗ (t; u(t); v(t)) − F(t; $u (t); $v (t)) = F(t; $u (t); p2 (t; u(t); v(t))) − F(t; $u (t); $v (t)) ≥ 0 for all t ∈ (t1 ; t2 ): Thus by maximum principle, this contradicts that $u −u has a positive maximum on (t1 ; t2 ): The validity that (u; v) ≤ (%u ; %v ) on (0; 1) can be shown by a similar way. 1 Remark 1. (i) If we assume, instead of (a4 ); the condition 0 shF (s) ds ¡ ∞; 1 shG (s) ds ¡ ∞; then the pair of solutions u; v ∈ C 1 ((0; 1]; R): On the other hand, if 01 1 (1 − s)hF (s) ds ¡ ∞; 0 (1 − s)hG (s) ds ¡ ∞; then u; v ∈ C 1 ([0; 1); R): 0 (ii) Theorem 1 is valid for arbitrary
v (t) + q2 (t)g(u(t); v(t)) = 0; u(0) = u(1) = v(0) = v(1) = 0:
0 ¡ t ¡ 1;
Y.H. Lee / Nonlinear Analysis 45 (2001) 597 – 611
603
In what follows, we assume that qi ∈ C((0; 1); R+ ); i = 1; 2 do not vanish identically on any subinterval of (0; 1): The following is the main theorem of this section. Theorem 2. Assume 1 (H) qi are singular at t = 0 and 1 and 0 s(1 − s)qi (s) ds ¡ ∞; i = 1; 2. (H2 ) f and g are quasi-monotone nondecreasing w.r.t. v and u; respectively. (H3 ) f∞ = ∞ = g∞ . Then there exists (∗ ; ∗ ) ¿ (0; 0) such that (S) has at least one positive solution for (; ) ≤ (∗ ; ∗ ) and no solution for (; ) ¿ (∗ ; ∗ ): We set up an operator equation for
B (u; v)(t) ,
0
1
K(s; t)q2 (s)g(u(s); v(s)) ds:
Also de
min
1=4≤t≤3=4
(u(t) + v(t)) ≥
1 4 (u; v)
:
Then C and K are cones in X . We need some lemmas to prove Theorem 2. 2 Lemma 2. Assume (H). Then for all (; ) ∈ R+ \{(0; 0)}; T; is completely continuous on X and T; (C) ⊂ K:
Proof. The complete continuities of A ; B and T; can be shown by the same way as in the proof of Lemma 1 and we prove T; (C) ⊂ K: Let (u; v) ∈ C; then for t ∈ [ 14 ; 34 ]; using the following fact for the Green’s function; K(t; s) ≥ 14 K(!; s); for all !; s ∈ [0; 1]; we obtain 1 1 A (u; v)(t) ≥ K(!; s)q1 (s)f(u(s); v(s)) ds = A (u; v)(!) 4 0 4 for all ! ∈ [0; 1]: Thus we get min
1=4≤t≤3=4
A (u; v)(t) ≥ 14 A (u; v)∞ :
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Y.H. Lee / Nonlinear Analysis 45 (2001) 597 – 611
Similarly, min
1=4≤t≤3=4
B (u; v)(t) ≥ 14 B (u; v)∞ :
The following lemma can be proved by simple calculations. One may refer to Dalmasso [1] for the proof in detail. Lemma 3. Assume (H). Then the problem; for i = 1; 2 (Li )
u (t) + t(1 − t)qi (t) = 0;
0 ¡ t ¡ 1;
u(0) = 0 = u(1)
has a unique solution ’i ∈ C 1 [0; 1] ∩ C 2 (0; 1) satisfying ci ’i (t) ≤ t(1 − t) ≤ di ’i (t) for some ci ; di ¿ 0; i = 1; 2: Since each component of solutions of (S) are not guaranteed of C 1 [0; 1]; integration by part cannot be applied directly and we need to prove the following lemma. Lemma 4. Assume (H). Let (u; v) be a solution of (S) and ’i be a solution of (Li ). Then 1 1 u(s)’1 (s) ds; u (s)’1 (s) ds = 0
0
0
1
v (s)’2 (s) ds =
0
1
v(s)’2 (s) ds:
Proof. We
t→0+
By Lemma 3, we get the following inequalities: t0 |’1 (+)u (+)| ≤ ’1 (+) u (s) ds + u (t0 ) +
t0 1 1 +(1 − +) |u (s)| ds + +(1 − +)|u (t0 )| c1 c1 + t0 1 1 ≤ (1 − +) sq1 (s)f(u(s); v(s)) ds + +(1 − +)|u (t0 )|: c1 c 1 +
≤
Thus by (H), for given , ¿ 0 and
+ ˜ t→0
Y.H. Lee / Nonlinear Analysis 45 (2001) 597 – 611
Finally, 0
1
u (s)’1 (s) ds = lim
+ ˜ t; t→0
1−t˜
t
605
u (s)’1 (s) ds
= − lim+ ’1 (t)u (t) + lim ’1 (1 − t˜)u (1 − t˜) + ˜ t→0
t→0
+ lim
+ ˜ t; t→0
= lim
+ ˜ t; t→0
t
1−t˜
t 1−t˜
u(s)’1 (s) ds
u(s)’1 (s) ds =
0
1
u(s)’1 (s) ds:
We can show the other equality by the same way and the proof is complete. The following lemma gives the a priori bound of solutions for problem (S). 2 Lemma 5. Assume (H) and (H3 ). Let R be a compact subset of R+ \{(0; 0)}: Then there exists a constant bR ¿ 0 such that for all (; ) ∈ R and for all possible positive solutions (u; v) of (S) at (; ); one has
(u; v) ¡ bR : Proof. Suppose by contradiction that there is a sequence (un ; vn ) of positive solutions of (S) with corresponding (n ; n ) belonging to R such that (un ; vn ) → ∞: Then by Lemma 2, (un ; vn ) ∈ K and thus min
1=4 ≤ t ≤ 3=4
(un (t) + vn (t)) ≥ 14 (un ∞ + vn ∞ ):
(1)
Considering a subsequence if necessary, we
for all u + v ≥ R1 ;
where - satis
(2)
for all n
and m = min{K(t; s): (t; s) ∈ [ 14 ; 34 ] × [ 14 ; 34 ]}: Thus for t ∈ [ 14 ; 34 ]; using (1) and (2), we have 1 K(t; s)q1 (s)f(un (s); vn (s)) ds un ∞ ≥ un (t) = n 0
≥ n
3=4
1=4
K(t; s)q1 (s)f(un (s); vn (s)) ds
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Y.H. Lee / Nonlinear Analysis 45 (2001) 597 – 611
≥ n m
3=4
1=4
≥ n m≥
n m4
q1 (s)f(un (s); vn (s)) ds
3=4
1=4
q1 (s)(un (s) + vn (s)) ds
3=4
1=4
q1 (s)(un ∞ + vn ∞ ) ds
¿ un ∞ + vn ∞ ≥ un ∞ for suJciently large n. This is a contradiction. Proof of the case n = 0; n ¿ 0 for suJciently large n can be done by using g∞ = ∞ and the proof is complete. Before proving Theorem 2, we note that by (H) and (H2 ); we can use Theorem 1 only checking condition (a1 ) in the theorem. Proof of Theorem 2. The proof consists of three claims. Claim 1. Problem (S) has a positive solution for certain (; ): Let %i (t) =
1 0
K(t; s)qi (s) ds be the unique solution of
u (t) + qi (t) = 0; u(0) = 0 = u(1): Let Mf = maxt∈[0; 1] f(%1 (t); %2 (t)) and Mg = maxt∈[0; 1] g(%1 (t); %2 (t)): Then Mf and Mg ¿ 0; since f; g ¿ 0: %1 (t) + q1 (t)f(%1 (t); %2 (t)) = q1 (t)[f(%1 (t); %2 (t)) − 1] ≤ 0; %2 (t) + q2 (t)g(%1 (t); %2 (t)) = q2 (t)[g(%1 (t); %2 (t)) − 1] ≤ 0; whenever (; ) ≤ (1=Mf ; 1=Mg ): This shows that (%1 ; %2 ) is an upper solution of (S). On the other hand, (0,0) is obviously a lower solution of (S) and (0; 0) ≤ (%1 ; %2 ): Thus by Theorem 1, (S) has a positive solution for (0; 0) ¡ (; ) ≤ (1=Mf ; 1=Mg ): 2 De
for all u; v ≥ 0:
(3)
Let (; ) ∈ S and (u; v) be a positive solution of (S) at (; ); then by Lemmas 3, 4
Y.H. Lee / Nonlinear Analysis 45 (2001) 597 – 611
and (3), we get the following inequalities: 1 s(1 − s)q1 (s)u(s) ds ≤ mf d1 mf 0
0
≤ d1
1
0
= −d1 = d1
1
0
’1 (s)q1 (s)(u(s) + v(s)) ds
’1 (s)q1 (s)f(u(s); v(s)) ds
1
0
1
607
’1 (s)u (s) ds = −d1
0
1
’1 (s)u(s) ds
s(1 − s)q1 (s)u(s) ds:
1 Since 0 ¡ 0 s(1 − s)q1 (s)u(s) ds ¡ ∞; ≤ d1 =mf : Similarly, we can get ≤ d2 =mg : Therefore S is bounded above by (d1 =mf ; d2 =mg ): Claim 3. (S; ≤) has a maximal element. Let P be a chain in S: By Zorn’s Lemma, it is enough to show that P has an upper bound in S: Without loss of generality, we may choose a distinct sequence {(n ; n )} ⊂ P such that (n ; n ) ≤ (n+1 ; n+1 ); n=1; 2; : : : : By Claim 2, two sequences {(n )} and {(n )} converge to, say, P and P ; respectively. If (P ; P ) ∈ S; then the proof is done. Since the sequence {(n ; n )} is bounded above, it belongs to a compact 2 \{(0; 0)} and Lemma 5 implies that the corresponding solutions {(un ; vn )} subset of R+ are uniformly bounded in X: By the compactness of the integral operators A and B ; the sequence {(un ; vn )} has a subsequence converging to, say, (uP ; vP ) ∈ X: We can easily show, by the Lebesgue convergence theorem, that (uP ; vP ) is a solution of (S) at (P ; P ): Thus (P ; P ) ∈ S. Let (∗ ; ∗ ) be a maximal element of S and (u∗ ; v∗ ); the corresponding positive solution of (S). Then for all (; ) ≤ (∗ ; ∗ ); problem (S) has a positive solution at (; ); since (u∗ ; v∗ ) and (0; 0) are an upper solution and a lower solution, respectively. This concludes the proof. 4. Multiplicity In this section, we show the existence of the second positive solution for (; ) ¡ (∗ ; ∗ ), where (∗ ; ∗ ) is a maximal element of S. Let (u∗ ; v∗ ) be a positive solution of (S) at (∗ ; ∗ ). Lemma 6. For (; ) ¡ (∗ ; ∗ ); there exists ,0 ¿ 0 such that (u∗ + ,; v∗ + ,) is an upper solution of (S) at (; ); for all 0 ¡ , ≤ ,0 . Proof. Let M be a constant such that f(u∗ (t); v∗ (t)) ≥ M ¿ 0;
g(u∗ (t); v∗ (t)) ≥ M ¿ 0:
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Y.H. Lee / Nonlinear Analysis 45 (2001) 597 – 611
Then by Lemma 4 and the uniform continuity of f and g, there exists ,0 ¿ 0 such that M (∗ − ) |f(u∗ (t) + ,; v∗ (t) + ,) − f(u∗ (t); v∗ (t))| ¡ ; |g(u∗ (t) + ,; v∗ (t) + ,) − g(u∗ (t); v∗ (t))| ¡
M (∗ − )
for all t ∈ [0; 1] and 0 ¡ , ≤ ,0 . Let u,∗ (t) = u∗ (t) + , and v,∗ (t) = v∗ (t) + ,, then u,∗ (0); u,∗ (1); v,∗ (0); v,∗ (1) ¿ 0 and u,∗ (t) + q1 (t)f(u,∗ (t); v,∗ (t)) = u∗ (t) + q1 (t)f(u∗ (t) + ,; v∗ (t) + ,) = − ∗ q1 (t)f(u∗ (t); v∗ (t)) + q1 (t)f(u∗ (t) + ,; v∗ (t) + ,) = q1 (t)[f(u∗ (t) + ,; v∗ (t) + ,) − f(u∗ (t); v∗ (t))] −(∗ − )q1 (t)f(u∗ (t); v∗ (t)) ¡ M (∗ − )q1 (t) − (∗ − )q1 (t)f(u∗ (t); v∗ (t)) = (∗ − )q1 (t)[M − f(u∗ (t); v∗ (t))] ≤ 0 for all t ∈ [0; 1]. The inequalities for v,∗ can be shown similarly and the proof is complete. We note that for 0 ¡ , ≤ ,0 ; (u,∗ ; v,∗ ) satis
v,∗ (t) ¿
0
1
K(t; s)q2 (s)g(u,∗ (s); v,∗ (s)) ds:
We introduce the following two lemmas for
Y.H. Lee / Nonlinear Analysis 45 (2001) 597 – 611
609
We now state and prove our main result. Theorem 3. Assume (H); (H2 ) and (H3 ). Then there exists (∗ ; ∗ ) ¿ (0; 0) such that problem (S) has at least two positive solution for (; ) ¡ (∗ ; ∗ ); at least one positive solution for (; ) ∈ G and no positive solution for (; ) ¿ (∗ ; ∗ ); where 2 2 \{(0; 0)}: = ∗ ; ≤ ∗ } ∪ {(; ) ∈ R+ \{(0; 0)}: ≤ ∗ ; = ∗ }: G = {(; ) ∈ R+ Proof. By Theorem 2, it is enough to show the existence of the second positive solution of (S) for (; ) ¡ (∗ ; ∗ ). Let u,∗ (t) = u∗ (t) + ,; v,∗ (t) = v∗ (t) + ,, where , is given in Lemma 6. Let ? = {(u; v) ∈ X : − , ¡ u(t) ¡ u,∗ (t); −, ¡ v(t) ¡ v,∗ (t); t ∈ [0; 1]}, then ? is bounded open in X , 0 ∈ ? and T; : K ∩ ? → K is condensing, since it is completely continuous. Let (u; v) ∈ K ∩ @?, then there exists t0 ∈ [0; 1] such that either u(t0 ) = u,∗ (t0 ) or v(t0 ) = v,∗ (t0 ). Suppose that u(t0 ) = u,∗ (t0 ): Then by (H2 ) and Lemma 6, 1 A (u; v)(t0 ) = K(t0 ; s)q1 (s)f(u(s); v(s)) ds 0
≤
1
K(t0 ; s)q1 (s)f(u∗ (s) + ,; v∗ (s) + ,) ds
0
¡ u∗ (t0 ) + , = u(t0 ) ≤ 1u(t0 ) for all 1 ≥ 1. Similarly, for the case v(t0 ) = v,∗ (t0 ), we can get B (u; v)(t0 ) ¡ 1v(t0 ); for all 1 ≥ 1. Thus T; (u; v) = 1(u; v), for all (u; v) ∈ K ∩ @? and all 1 ≥ 1 and by Lemma 7, i(T; ; K ∩ ?; K) = 1: Let R = max{bR ; 4R1 ; (u,∗ ; v,∗ )}, where bR is given in Lemma 5 with R a compact 2 subset of R+ \{(0; 0)} containing (; ) and R1 is given in (2) for n = in the proof of Lemma 5. Let KR = {(u; v) ∈ K: (u; v) ¡ R}. Then by Lemma 5, T; (u; v) = (u; v) for (u; v) ∈ @KR . Furthermore, if (u; v) ∈ @KR ; then min (u(t) + v(t)) ≥ 14 (u; v) ≥ R1 :
t∈[1=4; 3=4]
Thus by (2), f(u(t); v(t)) ≥ -(u(t) + v(t)), for all t ∈ [ 14 ; 34 ] and 3=4 q1 (s)f(u(s); v(s)) ds A (u; v)(t) ≥ m 1=4
≥ m≥
m4
3=4
1=4
q1 (s)(u(s) + v(s)) ds
3=4
1=4
¿ (u; v):
q1 (s)(u; v) ds
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Y.H. Lee / Nonlinear Analysis 45 (2001) 597 – 611
Therefore T; (u; v) ≥ A (u; v)∞ ≥ (u; v) and by Lemma 8, i(T; ; KR ; K) = 0: Consequently, by the additivity of the
3v + k2 (|x|)g(u; v) = 0 in ?; u(x) = 0 = v(x) on |x| = r0 ; lim u(x) = 0 = lim v(x);
|x|→∞
|x|→∞
where ; ¿ 0; n ≥ 3. As we mentioned in Section 1, qi are given by qi (t) =
r02 (1 − t)(−2(n−1))=(n−2) ki (r0 (1 − t)−1=(n−2) ); (n − 2)2
i = 1; 2
and (H1 ) corresponds to condition (H) via the transformations. We need to consider the following two cases for ki : Case I: limr→∞ r 2(n−1) ki (r) ¡ ∞. In this case, limt→1− qi (t) ¡ ∞ so that qi can be extended continuously on [0; 1] and the problem becomes regular. Thus applying Theorem 5, we obtain desired result described below. Case II: limr→∞ r 2(n−1) ki (r) = ∞. In this case, limt→1− qi (t) = ∞ and the problem becomes singular. To apply Theorem 4, we need to add condition (H1 ). Since functions ki in Case I also satis
Y.H. Lee / Nonlinear Analysis 45 (2001) 597 – 611
611
Corollary 1. Assume (H1 ) – (H3 ). Then there exists (∗ ; ∗ ) ¿ (0; 0) such that problem (P) has at least two positive radial solutions for (; ) ¡ (∗ ; ∗ ); at least one positive radial solution for (; ) ∈ G and no positive radial solution for (; ) ¿ (∗ ; ∗ ); 2 2 \ {(0; 0)}: ≤ where G = {(; ) ∈ R+ \ {(0; 0)}: = ∗ ; ≤ ∗ } ∪ {(; ) ∈ R+ ∗ ∗ ; = }. Remark 2. According to our result in this paper, we do not give any information on 2 2 the areas like {(; ) ∈ R+ \{(0; 0)}: ≤ \{(0; 0)}: ¿ ∗ ; ≤ ∗ } or {(; ) ∈ R+ ∗ ∗ ; ¿ }. Reminding that the multiparameter set S of solutions, in the proof of 2 Theorem 2, is bounded above in R+ , we leave some questions; ∗ ∗ 1. Is the maximal element ( ; ) of S unique? 2. Is the following statement true? 2 There exists a curve P which separates R+ into two (relatively) open sets O1 and O2 such that problem (P) has at least two positive radial solutions for (; ) ∈ O1 ; at least one positive radial solution for (; ) ∈ P and no positive radial solution for (; ) ∈ O2 . Acknowledgements The author wishes to acknowledge the