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Multiple solutions for a super-linear three-point boundary value problem
Nonlinear Analysis 50 (2002) 115 – 128
www.elsevier.com/locate/na
Multiple solutions for a super-linear three-point boundary value problem Bruce Calverta , Chaitan P. Guptab; ∗ a Department
b Department
of Mathematics, University of Auckland, Auckland, New Zealand of Mathematics, 084, University of Nevada, Reno, NV 89557, USA Received 8 August 1999; accepted 3 February 2001
Keywords: Multiple solutions; Super-linear three-point boundary value problem; Leray–Schauder degree
1. Introduction Let g : R → R be a continuous function such that g(x)x ¿ 0, for all x = 0; g(x)=x is increasing and lim|x|→∞ (g(x)=x) = ∞. Let p : [0; 1] × R2 → R be a function satisfying the Caratheodory conditions. Let M1 : [0; 1] × [0; ∞) → [0; ∞) be such that for each s ∈ [0; ∞); M1 (·; s) is integrable on [0; 1] and for each t ∈ [0; 1]; M1 (t; ·) is increasing 1 on [0; ∞) with s−1 0 M1 (t; s) dt → 0 as s → ∞. Suppose, further, that |p(t; x; y)| 6 M1 (t; max(|x|; |y|)) for (t; x; y) ∈ [0; 1] × R2 . Also, let ∈ (0; 1), and ∈ (0; 1) ∪ (1; ∞) be given. The object of this paper is to show the existence of multiple solutions for the super-linear three point boundary value problem x (t) + g(x(t)) = p(t; x(t); x (t)); x(0) = 0;
x( ) = x(1):
(1)
This problem, with = 1, was studied earlier by M. Henrard in [6]. Henrard uses a variant of Leray–Schauder continuation theorem, (see [4,5]), for boundary value ∗ Corresponding author. E-mail addresses: [email protected] (B. Calvert), [email protected] (C.P. Gupta).
c 2002 Elsevier Science Ltd. All rights reserved. 0362-546X/02/$ - see front matter PII: S 0 3 6 2 - 5 4 6 X ( 0 1 ) 0 0 7 3 8 - 6
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problems when the set of solutions of a homotopic family is not necessarily a priori bounded. Our paper uses some of the ideas and results of Henrard’s thesis [6]. The study of multi-point boundary value problems for second-order ordinary diEerential equations was initiated by Il’in and Moiseev in [9,10], motivated by the work of Bitsadze and SamarskiGi on non-local elliptic boundary value problems in [1–3]. A study of second-order ordinary diEerential equations involving a superlinear term was done by Capieto et al. in [4] for periodic solutions and by Capieto et al. in [5] for the Sturm–Liouville conditions. 2. Preliminary lemmas Let g : R → R be a continuous increasing function with g(0) = 0; g(x)x ¿ 0 for x = 0 and g(x)=x → ∞ as |x| → ∞. For the super-linear three-point boundary value problem (1) we may assume that g(x)=x ¿ 1 for all non-zero x by adding something to g and x to p, without any loss of generality, and we do this from now on. Let G(x) = 0 g(s) ds for x ∈ R. Under our assumptions the function G is strictly increasing on [0; ∞) and is strictly decreasing on (−∞; 0]. Also, G(x) ¿ 0 for x ∈ R; x = 0 and −1 G(0) = 0. Let G+ : [0; ∞) → [0; ∞) denote the inverse of the function G=[0; ∞). Also, −1 let G− : [0; ∞) → (−∞; 0] denote the inverse of the function G=(−∞; 0]. Note that −1 −1 (y)=y → 0 as y → ∞, and G− (y)=y → 0 as y → ∞. We shall assume in the G+ following that ∈ R is given. We consider the system of diEerential equation x = y; y = − g(x);
(2)
where denotes diEerentiation with respect to t. If, now, (x(t); y(t)) is a solution of (2) then d 2 (y + 2G(x)) = 0: dt Accordingly, it follows that if (x(t); y(t)) is a solution of (2) with x(0) = 0; y(0) = , then y2 + 2G(x(t)) = 2
for all t:
(3)
We observe from Eq. (2) that when x(t) ¿ 0; y(t) is a decreasing function of t and when x(t) 6 0; y(t) is an increasing function of t. Also, we see from (3) that if (x(t); y(t)) is a solution of (2) with x(0) = 0; y(0) = , then the graph of Eq. (3) −1 2 −1 2 passes through the points (0; ); (G− ( =2); 0); (0; −); (G+ ( =2); 0) in the x–y plane and is oval shaped. Remark 1. We may remark that if (x(t); y(t)) is a solution of (2) with x(0) = 0; y(0) = then it is unique under our assumptions on the function g. This is because of the fact that (x(t); y(t)) lies on the curve (3). Indeed, let (t) denote the negative
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of the polar angle of (x(t); y(t)) in the x–y plane. Then d=dt = (y2 + xg(x))=(x2 + y2 ) ≡ f() ¿ 0; with f seen to be continuous. This implies that (t) is an invertthat t is a strictly increasing C 1 function of ible C 1 function of t and it follows . Also, dt=d = 1=f() and t = (1=f()) d. This formula Jxes (t) uniquely and hence (x(t); y(t)) is a unique function of t. Thus, we have obtained uniqueness of the Cauchy-problem associated with (2) without requiring that the function g be locally Lipschitz. Lemma 2. Let g : R → R be a continuous increasing function with g(0) = 0; g(x)=x ¿ 1 x for x = 0; g(x)=x → ∞ as |x| → ∞; and let G(x) = 0 g(s) ds. Also; let p : [0; 1] × R × R → R be a function satisfying the Caratheodory conditions. Let M1 : [0; 1]×[0; ∞) → [0; ∞) be such that for each s ∈ [0; ∞); M1 (·; s) is integrable on [0; 1] and for each 1 t ∈ [0; 1]; M1 (t; ·) is increasing on [0; ∞) with s−1 0 M1 (t; s) dt → 0 as s → ∞. Suppose; further; that |p(t; x; y)| 6 M1 (t; max(|x|; |y|)) for (t; x; y) ∈ [0; 1] × R2 : Suppose that for ∈ [0; 1]; (x (t); y (t)) is a solution for the system of di7erential equations x (t) = y(t);
(4)
y (t) + g(x(t)) = p(t; x(t); y(t));
0 ¡ t ¡ 1;
(5)
x(0) = 0;
(6)
y(0) = :
(7)
Let ¿ 0 be given. If ¿ 0 is large enough; then for t ∈ [0; 1] we have |y (t)| 6 (1 + );
(8)
2G(x (t)) 6 2 (1 + )
(9)
and there exists a solution (x(t); y(t)) on [0; 1] for the system of di7erential equations (4)–(7) for ∈ [0; 1]: Also; d (y (t)2 + 2G(x (t)) 6 2|y (t)|M1 (t; max(|x (t)|; |y (t)|)); dt 0 ¡ t ¡ 1; ∈ [0; 1]: Proof. We note from (4) and (5) that d ((y (t))2 + 2G(x (t))) = 2g(x (t))x (t) + 2y (t)y (t) dt = 2(y (t))(p(t; x (t); y (t)) 6 2|y (t)|M1 (t; max(|x (t)|; |y (t)|))
(10)
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giving (10). Let us write h (t) = (y (t))2 + 2G(x (t)). Then, d (h (t))2 6 2h (t)M1 (t; h (t)) (11) dt since |x| 6 2G(x) for all x, in view of the assumption g(x)=x ¿ 1 for x = 0. We then get by integrating from 0 to t in (11) that t 2 2 (h (t)) 6 + 2 h (s)M1 (s; h (s)) ds: (12) 0
Let, now, k (t) = max{h (s): 0 6 s 6 t}. For x ¿ 0, let M (x):= t ¿ 0; k (t) = h (u) for some u 6 t, giving u (k (t))2 = (h (u))2 6 2 + 2 h (s)M1 (s; h (s)) ds 6 2 + 2k (t) 2
6 + 2k (t)
0
0
M1 (t; x) dt. Given
0
u
0
1
1
M1 (s; k (t)) ds M1 (s; k (t)) ds
6 2 + 2k (t)M (k (t)): Now, take suKciently large such that for a given ¿ 0, to be chosen suKciently small, 2M (x) ¡ x
for x ¿ :
Since k (t) ¿ for all t ∈ [0; 1], we get (k (t))2 6 2 + (k (t))2
and
2 ; giving 1− √ h (t) 6 1 + ; choosing small: (k (t))2 6
This gives (8), and (9). This completes the proof of the lemma. Lemma 3. Let ¿ 0 be given and M1 be as in Lemma 2. Then; there exists an A ¿ 0 such that if for ∈ [0; 1]; (x (t); y (t)) is a solution for the system of di7erential equations (4)–(7) and a t0 ∈ [0; 1] is such that x (t0 ) ¿ 0; y (t0 ) = 0: then 1 x (t0 ) 61 + 6 −1 1+ G+ (2 =2)
if || ¿ A:
Similarly; if x (t0 ) ¡ 0; y (t0 ) = 0; then x (t ) 1 0 6 −1 6 1 + if || ¿ A: 2 1 + G− ( =2)
(13)
(14)
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Proof. We Jrst claim that G(x)=x is increasing on (0; ∞). Indeed, take an x ¿ 0 and a k ∈ (0; 1). Then kx g(s) ds G(kx) = 0
=k
0
6k
x
0
x
g(kz) d z;
where s = kz
g(z) d z
= kG(x): It follows that G(kx)=kx 6 G(x)=x, and accordingly we see that G(x)=x is increasing on (0; ∞). Similarly, one can see that −(G(x)=x) is increasing on (−∞; 0). To show the right inequality of (13), we use (9) and also G(x (t0 )) x (t0 ) 6 G : 1+ 1+ To show the left inequality of (13), or equivalently 2 6 G((1 + )x (t0 )); 2 it is enough to show that 2 6 (1 + )G(x (t0 )): 2 By integrating (10), using (8), we have, with h (t) as in Lemma 2, that for t ∈ [0; 1], t 2 2 M1 (s; max(|x (s)|; |y (s)|)) ds ¿ 0: (15) (h (t)) − + 2||(1 + ) 0
Take 1 so that 21 (1 + ) 6 min(; 12 ). Take A so that for ¿ A inequalities (8), (9) hold, and when 2 1 ; ; (16) max(|x (s)|; |y (s)|) ¿ min G −1 2 8 2 2
we have, with M as in Lemma 1, M (max(|x (s)|; |y (s)|)) ¡ 1 max(|x (s)|; |y (s)|):
(17)
Now for ¿ A, using g(x)=x ¿ 1, (8) and (9), (17) gives M (max(|x (s)|; |y (s)|)) ¡ 1 (1 + )
(18)
and hence (h (t))2 ¿ 2 (1 − 2(1 + )2 1 )
(19)
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if (16) holds for all s ∈ [0; t]. This implies (y (t))2 + 2G(x (t)) ¿
2 : 2
2 2 Therefore, either (y (t))2 ¿ 4 or |x (t)| ¿ G −1 8 and 2 −1 max(|x (t)|; |y (t)|) ¿ min G ; 8 2 holds and hence (16) holds for s near t by continuity. Thus, if I is the largest interval on which (16) holds, then we must have I = [0; 1]. Hence (19) holds for all t ∈ [0; 1]. Finally, if y (t0 ) = 0 then 2G(x (t0 )) ¿ 2 (1 − 2(1 + )2 1 ) and we see that (13) holds. A similar proof works for (14). This completes the proof of the lemma. For a solution (x(t); y(t)) of the system of equations x (t) = y(t); y (t) = − g(x); x(0) = 0; let
y(0) = ;
1 ()
denote the angle given by the integral 1 x(t)y (t) − y(t)x (t) () = − dt: 1 x2 (t) + y2 (t) 0
Lemma 4. Let g : R → R be a continuous increasing function with g(x)x ¿ 0 for every x ∈ R; g(x) |x| increasing on (−∞; 0); and on (0; ∞). Then 1 (); ∈ (0; ∞); is strictly increasing on (0; ∞). Proof. Let 0 ¡ ¡ be given and let (x (t); y (t)); (x (t); y (t)) be such that x (t) = y (t); y (t) = − g(x (t)); x (0) = 0;
y (0) = ;
x (t) = y (t); y (t) = − g(x (t)); x (0) = 0;
y (0) = :
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We note that (y (t))2 + 2G(x (t)) = 2 ; (y (t))2 + 2G(x (t)) = 2 : Let denote the polar angle of the point (x ; y ) in the x–y plane and let denote the polar angle of the point (x ; y ) in the x–y plane, so that x = r cos ; y = r sin with r = x2 + y2 ; x = r cos ;
y = r sin
with r =
x 2 + y 2 :
We note that since 0 ¡ ¡ the two curves y2 + 2G(x) = 2 ; y2 + 2G(x) = 2 , cannot intersect in the x–y plane and for the same polar angle the radial coordinate on the -curve is smaller than the radial coordinate of the corresponding point on the -curve. Let t1 ; t2 and k ¿ 1 be such that (x (t2 ); y (t2 )) = k(x (t1 ); y (t1 ))
and
(t1 ) = (t2 ):
(20)
Now, we see using our assumption, g(x)=x is increasing on (0; ∞), that g(x (t2 )) g(x (t1 )) : 6 x (t1 ) x (t2 ) It then follows from (20), (21) that (y (t2 ))2 (y (t1 ))2 = (x (t2 ))2 + (y (t2 ))2 (x (t1 ))2 + (y (t1 ))2
and
x (t2 )g(x (t2 )) (x (t2 ))2 g(x (t2 )) = 2 2 (x (t2 ))2 + (y (t2 ))2 x (t2 ) (x (t2 )) + (y (t2 )) =
g(x (t2 )) (x (t1 ))2 2 2 (x (t1 )) + (y (t1 )) x (t2 )
¿
g(x (t1 )) (x (t1 ))2 (x (t1 ))2 + (y (t1 ))2 x (t1 )
=
x (t1 )g(x (t1 )) : (x (t1 ))2 + (y (t1 ))2
Accordingly, we see that −
x g(x ) + y 2 x g(x ) + y2 (t ) 6 − (t1 ); 2 x 2 + y 2 x2 + y2
which implies that (t2 ) 6 (t1 ):
(21)
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Claim. (t) ¿ (t) for all t ¿ 0. Proof. We Jrst observe that (t) ¡ 0
for all t ¿ 0;
(t) ¡ 0
for all t ¿ 0 and
! (0) = (0) = : 2 Thus, (t) and (t) are strictly decreasing functions of t. We consider the functions ! ! f (t) = − (t) and f (t) = − (t): 2 2 Now, f (t) and f (t) are strictly increasing functions of t with f (0) = 0 and f (0) = 0. Also, we see from our work above that if f (t1 ) = f (t2 ) then f (t1 ) 6 f (t2 ). We observe that the claim is equivalent to showing that f (t) 6 f (t) for all t ¿ 0. We now assert that f−1 (s) ¿ f −1 (s) for all s ¿ 0. Indeed, let t1 = f−1 (s) and t2 = f −1 (s) so that f (t1 ) = f (t2 ) = s. We see that 1 1 (f−1 ) (s) = = f (f−1 (s)) f (t1 ) ¿
1 1 = = (f −1 ) (s): f (t2 ) f (f −1 (s))
This immediately implies the assertion since f (0) = 0 and f (0) = 0. Now, to prove that f (t) 6 f (t) for all t ¿ 0, let us suppose, on the other hand, that f (t) ¿ f (t). Then, there exists a t1 ¡ t such that f (t1 ) = f (t) = s. We now get from our assertion proved above that t1 ¿ t, a contradiction. This completes the proof of the claim. Finally, we note that f (1) = 1 () and accordingly we get that 1 () ¡ 1 ( ), proving thereby that 1 is strictly increasing on (0; ∞). This completes the proof of the lemma. Remark 5. We see from Lemma 4:3 of [6] that 1 () → ∞ as → ∞, and hence there exists an even natural number k0 such that for every k ¿ k0 , k ∈ {0; 1; 2 : : :}, there exists an k ¿ 0 such that 1 (k ) = (!=2) + k!, so that if (x(t); y(t)) is a solution for the system of equations x (t) = y(t); y (t) = − g(x); x(0) = 0;
y(0) = k :
then x(t) is either maximized or minimized at t = 1.
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De nition 6. For ∈ R, we deJne a bounded open subset of R2 by 2
2 2 y : G = (x; y) ∈ R + G(x) ¡ 2 2 Writing 0 := k0 as deJned in Remark 5 and ¿ 0 we deJne another bounded open subset of R2 by 0
G+ = {(x; y) ∈ G \GL |y ¿ 0}
and a bounded open subset of C([0; 1]; R2 ) by "+ }: = {u : [0; 1] → G |u(0) ∈ G+
Let ∈ (0; 1) and ¿ 1 be given. For = k , where k is as in Remark 5, we shall and G+ have a common core with respect to the system of diEerential see that "+ equations x (t) = y(t);
(22)
y (t) = − g(x(t));
(23)
x(0) = 0;
(24)
x( ) = x(1):
(25) 2
2
We recall the meaning of common core, from [8]. We deJne U : R → R as follows. For u = (x; y) ∈ R2 , letting (x(t); y(t)) denote the solution to the Cauchy-problem x (t) = y(t); y (t) = − g(x(t)); x(0) = x;
y(0) = y;
let U (u) = u + (x; x( ) − x(1)): We say "+ and G+ have common core with respect to the three-point boundary value , there are no Jxed point of U on problem above, to mean: it has no solutions on @"+ @G+ , and for a solution u; u ∈ "+ if any only if u(0) ∈ G+ . 2 We note that u ∈ C([0; 1]; R ) belongs to the boundary of "+ ; @"+ , if and only if u(t) ∈ GL for every t ∈ [0; 1]; u(0) ∈ GL + and either there exists a t0 ∈ [0; 1] such that . u(t0 ) ∈ @G or else u(0) ∈ @G+ We suppose that u(t) = (x(t); y(t)) is a solution of
x (t) = y(t); y (t) = − g(x(t)): 2
Now, if there exists a t0 ∈ [0; 1] with u(t0 ) ∈ @G then ((y(t))2 =2) + G(x(t)) = 2 for all t ∈ [0; 1].
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Suppose that u(t) = (x(t); y(t)) is a solution of x (t) = y(t); y (t) = − g(x(t)) with x(0) = 0 and y(0) = k ; so that 1 (k ) = !2 + k! and |x(t)| is maximized at t = 1. Then we must have x( ) − x(1) = 0, since ¿ 1 by assumption. Thus, u(t) cannot be a solution of (22) – (25). Suppose, next, u(t) = (x(t); y(t)) is a solution of x (t) = y(t); y (t) = − g(x(t)) k ; u(0) ∈ @G k and u(0) ∈ @G 0 . Then we must have y(0) = 0 with x(0) = 0; u(0) ∈ @G+ and u(t) = 0 for all t ∈ [0; 1]. Thus, we see that "+ and G+ have a common core with respect to the system of diEerential Eqs. (22) – (25) when = k . We deJne a nonlinear operator P : C([0; 1]; R2 ) → C([0; 1]; R2 ) by setting for u(t) = (x(t); y(t)) ∈ C([0; 1]; R2 ), t (y(s); −g(x(s))) ds: (Pu)(t) = u(0) + (x(0); x( ) − x(1)) + 0
k+1 L k Lemma 7. The Leray–Schauder degree deg(I − P; "+ \"+ ; 0) = (−1)k ; for k su
x (t) = y(t); y (t) = − g(x(t)); x(0) = x;
y(0) = y;
U (u) = u + (x; x( ) − x(1)): Next, we deJne for u = (x; y) ∈ GL + ; s ∈ [0; 1] a function h by
h(u; s) = (−x; y − U2 (sx; y); where U2 (sx; y) denotes the second coordinate of U (sx; y). Now, for u = (x; y) ∈ GL + ; s ∈ [0; 1]; h(u; s) = (0; 0)
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if and only if y = U2 (sx; y) = U2 (0; y)
x = 0; if and only if
h(u; 1) = (0; 0): Note that h(u; 1) = (−x; y − U2 (x; y)) = (I − U )(x; y): Hence, deg(I − U; G+ ; 0): ; 0) = deg(h(·; 1); G+ ; 0) = deg(h(·; 0); G+
(26)
Let, now, for u = (u1 ; u2 ) ∈ R2 ; &(t; u) denote the solution to the Cauchy-problem x (t) = y(t); y (t) = − g(x(t)); x(0) = u1 ;
y(0) = u2
and let us deJne a mapping ': R → R; for y ∈ R by '(y) = &( ; (0; y)) − &(1; (0; y)): We see from (26) that deg(I − U; G+ ; 0) = deg(h(·; 1); G+ ; 0) = deg(h(·; 0); G+ ; 0) = deg(−I × −'; G+ ; 0)
= deg(I × '; (−; ) × (0 ; ); 0)
for some ¿ 0
= deg('; (0 ; ); 0): Now, by the deJnition of 0 and k we have that 1 (0 ) = (!=2) + k0 ! and (!=2) + k!. Let (x(t); y(t)) be a solution for the system of equations
1 (k ) =
x (t) = y(t); y (t) = − g(x); x(0) = 0;
y(0) = k :
If k is even then x(t) is maximized at t = 1 and hence '(k ) ¡ 0. Similarly, '(0 ) ¡ 0. Thus, deg('; (0 ; k ); 0) = 0. Now, if k is odd then x(t) is minimized at t = 1 and hence '(k ) ¿ 0, which implies that deg('; (0 ; k )) = 1. k+1 L k k+1 L k Thus, if k is even deg(I −P; "+ \"+ ; 0) = 1 and if k is odd deg(I −P; "+ \"+ ; 0) = − 1. This completes the proof of the lemma.
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De nition 8. For u(t) = (x(t); y(t)) ∈ C 1 ([0; 1]; R2 \{(0; 0)}) we deJne 1 x(t)y (t) − y(t)x (t) ’1 (u) = − dt; x2 (t) + y2 (t) 0 as the angle traversed from u(0) to u(1). We also deJne for ∈ [0; 1] the family of nonlinear operators P(·; ) from C([0; 1]; R2 ) × [0; 1] to C([0; 1]; R2 ) by setting for u(t) = (x(t); y(t)) ∈ C([0; 1]; R2 ) t (P(u; ))(t) = u(0)+(x(0); x( ) − x(1))+ (y(s); −g(x(s))+p(s; x(s); y(s))) ds: 0
Let us, also, denote by the open subset of C([0; 1]; R2 ) deJned by 0 )k+ = u(t) = (x(t); y(t)) ∈ C([0; 1]; R2 \GL )|y(0) ¿ 0 ; )k+
’1 (u) ∈
! 2
+ k!;
! + (k + 1)! : 2
Lemma 9. For large k; there is a bounded open subset Uk of )k+ containing the >xed point set of P(·; 1); and for all such Uk ; the Leray–Schauder degree deg(I − P(·; 1); Uk ; 0) is de>ned; that is; P(·; 1) is continuous; cl{P(u; 1) : u ∈ Uk } is compact; and deg(I − P(·; 1); Uk ; 0) = (−1)k . Proof. Let , = {u(t) = (x(t); y(t)) ∈ C([0; 1]; R2 )|P(u(t); ) = u(t)
for some ∈ [0; 1]}:
That is, u(t) ∈ , if and only if u(t) satisJes (4), (5), (24), (25) for some ∈ [0; 1]. Claim. , ∩ @)k+ is empty. Proof. We Jrst note that u(t) ∈ @)k+ implies that one of the following holds: (i) u(t) ∈ R2 \G 0 for all t and there is some t0 ∈ [0; 1]; u(t0 ) ∈ @G 0 , (ii) u(t) ∈ R2 \G 0 for all t but y(0) = 0, (iii) ’1 (u) = (!=2) + k! or ’1 (u) = (!=2) + (k + 1)!. Suppose that u(t) ∈ , ∩ @)k+ satisJes (i). By a small extension of Lemma 2, u(t) is bounded. With the lower bound on (x(t))2 + (y(t))2 for t ∈ [0; 1] from u(t) ∈ G 0 , we have an upper bound on (y(t))2 + x(t)(g(x(t)(−p(t; x(t); y(t))) (x(t))2 + (y(t))2 1 and hence we have an upper bound on ’1 (u) = 0 − (t) dt. This contradicts ’1 (u) ¿ (!=2) + k! for k large. So u(t) ∈ , ∩ @)k+ cannot satisfy (i). Suppose, next, that u(t) ∈ , ∩@)k+ satisJes (ii). Then y(0) = 0 and so x(0) = 0 gives 0 = u(0) ∈ G 0 which contradicts that u(t) ∈ @)k+ . Hence u(t) ∈ , ∩ @)k+ cannot satisfy (ii) also. − (t) =
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Finally suppose u(t) ∈ , ∩ @)k+ satisJes (iii). Now, let ¿ 0 be such that (1 + )2 ¡ and let A be as determined by Lemma 3, corresponding to this . Now, by the boundedness of ’1 (u) for u in a bounded subset of ,, as in (i), we may take k large enough such that if ’1 (u) ¿ (!=2) + k!, then = y(0) ¿ A. We then get by Lemma 3, −1 2 ( =2)=(1 + ). Let, now, x(t) supposing that x(1) ¿ 0 and y(1) = 0, that x(1) ¿ G+ be maximized at ∈ (0; 1]. Then we have, x( ) ¿ x( ) = x(1), and since y( ) = 0, −1 2 ( =2)(1 + ). It follows that 6 (1 + )2 , we get again by Lemma 3 that x( ) 6 G+ a contradiction. Similarly, we get a contradiction if x(1) ¡ 0. Thus, we must have that , ∩ @)k+ = - (empty set), and the claim is proved. Next, we note that , ∩ )k+ is bounded. Otherwise, take a sequence un in , with un → ∞, but then ’1 (un ) → ∞ by Lemma 4:3 of [6], contradicting ’1 (un ) ¡ (!=2)+ (k + 1)!. Next, we note that P : C([0; 1]; R2 )×[0; 1] → C([0; 1]; R2 ) is continuous and compact, that is, it maps bounded sets into relatively compact sets, since p(t; x; y) satisJes the Caratheodory conditions and we may apply a result of [7] for the NemytskiGi mappings. Now, take a bounded open subset Uk of )k+ containing , ∩ )k+ . By homotopy invariance we see that deg(I − P(·; 0); Uk ; 0) = deg(I − P(·; 1); Uk ; 1): However, using Lemma 4, deg(I − P(·; 0); Uk ; 0) = deg(I − P; Uk ; 0) k+1 L k = deg(I − P; "+ \"+ ; 0)
= (−1)k ; by Lemma 7. Finally, we note that this degree is independent of the choice of Uk by the excision property. This completes the proof of the lemma. 3. Main theorem Theorem 10. Let ∈ (0; 1) and ¿ 1 be given. Then; for each k su
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Proof. By Lemma 9, deg(I − P(·; 1); Uk ; 0) = (−1)k , for k suKciently large and so a 0 solution exists in )k+ . Similarly, if we deJne )k− = {u(t) ∈ C([0; 1]; R2 \GL )|y(0) ¡ 0, k ’1 (u) ∈ ((!=2) + k!; (!=2) + (k + 1)!)}, then a solution exists in )− . This completes the proof of the theorem. Remark 11. We note that the theorem above holds for 0 ¡ ¡ 1, for then we can use () and ’ (u). Remark 12. The above may be given in terms of continuation theorems of [4 – 6]. Also, one may use the degree dL (L − N (·; ); Uk ) of [6]. References [1] A.V. Bitsadze, On the theory of nonlocal boundary value problems, Soviet Math. Dokl. 30 (1) (1984) 8–10. [2] A.V. Bitsadze, On a class of conditionally solvable nonlocal boundary value problems for harmonic functions, Soviet Math. Dokl. 31 (1) (1985) 91–94. [3] A.V. Bitsadze, A.A. SamarskiGi, On some simple generalizations of linear elliptic boundary problems, Soviet. Math. Dokl. 10 (2) (1969) 398–400. [4] A. Capietto, J. Mawhin, F. Zanolin, A continuation approach to superlinear periodic boundary value problems, J. DiEerential Equations 88 (1990) 347–395. [5] A. Capietto, J. Mawhin, F. Zanolin, A continuation approach to some forced superlinear Sturm–Liouville boundary value problems, Topol. Meth. Nonlinear Anal. 3 (1994) 81–100. [6] M. Henrard, Topological degree in boundary value problems: existence and multiplicity results for second-order diEerential equations, thesis, UniversitQe Catholique de Louvain, Louvain-la-Neuve, Belgium, 1995, 123pp. [7] M.A. Krasnosels’kiGi, Topological Methods in the Theory of Integral Equations, Pergamon Press, Oxford, 1964. [8] M.A. Krasnosels’kiGi, P.P. ZabreGiko, Geometrical methods of non-linear analysis, Springer, Berlin, 1994. [9] V.A. Il’in, E.I. Moiseev, Nonlocal boundary value problem of the Jrst kind for a Sturm–Liouville operator in its diEerential and Jnite diEerence aspects, DiEerential Equations 23 (7) (1987) 803–810. [10] V.A. Il’in, E.I. Moiseev, Nonlocal boundary value problem of the second kind for a Sturm–Liouville operator, DiEerential Equations 23 (8) (1987) 979–987.
www.elsevier.com/locate/na
Multiple solutions for a super-linear three-point boundary value problem Bruce Calverta , Chaitan P. Guptab; ∗ a Department
b Department
of Mathematics, University of Auckland, Auckland, New Zealand of Mathematics, 084, University of Nevada, Reno, NV 89557, USA Received 8 August 1999; accepted 3 February 2001
Keywords: Multiple solutions; Super-linear three-point boundary value problem; Leray–Schauder degree
1. Introduction Let g : R → R be a continuous function such that g(x)x ¿ 0, for all x = 0; g(x)=x is increasing and lim|x|→∞ (g(x)=x) = ∞. Let p : [0; 1] × R2 → R be a function satisfying the Caratheodory conditions. Let M1 : [0; 1] × [0; ∞) → [0; ∞) be such that for each s ∈ [0; ∞); M1 (·; s) is integrable on [0; 1] and for each t ∈ [0; 1]; M1 (t; ·) is increasing 1 on [0; ∞) with s−1 0 M1 (t; s) dt → 0 as s → ∞. Suppose, further, that |p(t; x; y)| 6 M1 (t; max(|x|; |y|)) for (t; x; y) ∈ [0; 1] × R2 . Also, let ∈ (0; 1), and ∈ (0; 1) ∪ (1; ∞) be given. The object of this paper is to show the existence of multiple solutions for the super-linear three point boundary value problem x (t) + g(x(t)) = p(t; x(t); x (t)); x(0) = 0;
x( ) = x(1):
(1)
This problem, with = 1, was studied earlier by M. Henrard in [6]. Henrard uses a variant of Leray–Schauder continuation theorem, (see [4,5]), for boundary value ∗ Corresponding author. E-mail addresses: [email protected] (B. Calvert), [email protected] (C.P. Gupta).
c 2002 Elsevier Science Ltd. All rights reserved. 0362-546X/02/$ - see front matter PII: S 0 3 6 2 - 5 4 6 X ( 0 1 ) 0 0 7 3 8 - 6
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problems when the set of solutions of a homotopic family is not necessarily a priori bounded. Our paper uses some of the ideas and results of Henrard’s thesis [6]. The study of multi-point boundary value problems for second-order ordinary diEerential equations was initiated by Il’in and Moiseev in [9,10], motivated by the work of Bitsadze and SamarskiGi on non-local elliptic boundary value problems in [1–3]. A study of second-order ordinary diEerential equations involving a superlinear term was done by Capieto et al. in [4] for periodic solutions and by Capieto et al. in [5] for the Sturm–Liouville conditions. 2. Preliminary lemmas Let g : R → R be a continuous increasing function with g(0) = 0; g(x)x ¿ 0 for x = 0 and g(x)=x → ∞ as |x| → ∞. For the super-linear three-point boundary value problem (1) we may assume that g(x)=x ¿ 1 for all non-zero x by adding something to g and x to p, without any loss of generality, and we do this from now on. Let G(x) = 0 g(s) ds for x ∈ R. Under our assumptions the function G is strictly increasing on [0; ∞) and is strictly decreasing on (−∞; 0]. Also, G(x) ¿ 0 for x ∈ R; x = 0 and −1 G(0) = 0. Let G+ : [0; ∞) → [0; ∞) denote the inverse of the function G=[0; ∞). Also, −1 let G− : [0; ∞) → (−∞; 0] denote the inverse of the function G=(−∞; 0]. Note that −1 −1 (y)=y → 0 as y → ∞, and G− (y)=y → 0 as y → ∞. We shall assume in the G+ following that ∈ R is given. We consider the system of diEerential equation x = y; y = − g(x);
(2)
where denotes diEerentiation with respect to t. If, now, (x(t); y(t)) is a solution of (2) then d 2 (y + 2G(x)) = 0: dt Accordingly, it follows that if (x(t); y(t)) is a solution of (2) with x(0) = 0; y(0) = , then y2 + 2G(x(t)) = 2
for all t:
(3)
We observe from Eq. (2) that when x(t) ¿ 0; y(t) is a decreasing function of t and when x(t) 6 0; y(t) is an increasing function of t. Also, we see from (3) that if (x(t); y(t)) is a solution of (2) with x(0) = 0; y(0) = , then the graph of Eq. (3) −1 2 −1 2 passes through the points (0; ); (G− ( =2); 0); (0; −); (G+ ( =2); 0) in the x–y plane and is oval shaped. Remark 1. We may remark that if (x(t); y(t)) is a solution of (2) with x(0) = 0; y(0) = then it is unique under our assumptions on the function g. This is because of the fact that (x(t); y(t)) lies on the curve (3). Indeed, let (t) denote the negative
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117
of the polar angle of (x(t); y(t)) in the x–y plane. Then d=dt = (y2 + xg(x))=(x2 + y2 ) ≡ f() ¿ 0; with f seen to be continuous. This implies that (t) is an invertthat t is a strictly increasing C 1 function of ible C 1 function of t and it follows . Also, dt=d = 1=f() and t = (1=f()) d. This formula Jxes (t) uniquely and hence (x(t); y(t)) is a unique function of t. Thus, we have obtained uniqueness of the Cauchy-problem associated with (2) without requiring that the function g be locally Lipschitz. Lemma 2. Let g : R → R be a continuous increasing function with g(0) = 0; g(x)=x ¿ 1 x for x = 0; g(x)=x → ∞ as |x| → ∞; and let G(x) = 0 g(s) ds. Also; let p : [0; 1] × R × R → R be a function satisfying the Caratheodory conditions. Let M1 : [0; 1]×[0; ∞) → [0; ∞) be such that for each s ∈ [0; ∞); M1 (·; s) is integrable on [0; 1] and for each 1 t ∈ [0; 1]; M1 (t; ·) is increasing on [0; ∞) with s−1 0 M1 (t; s) dt → 0 as s → ∞. Suppose; further; that |p(t; x; y)| 6 M1 (t; max(|x|; |y|)) for (t; x; y) ∈ [0; 1] × R2 : Suppose that for ∈ [0; 1]; (x (t); y (t)) is a solution for the system of di7erential equations x (t) = y(t);
(4)
y (t) + g(x(t)) = p(t; x(t); y(t));
0 ¡ t ¡ 1;
(5)
x(0) = 0;
(6)
y(0) = :
(7)
Let ¿ 0 be given. If ¿ 0 is large enough; then for t ∈ [0; 1] we have |y (t)| 6 (1 + );
(8)
2G(x (t)) 6 2 (1 + )
(9)
and there exists a solution (x(t); y(t)) on [0; 1] for the system of di7erential equations (4)–(7) for ∈ [0; 1]: Also; d (y (t)2 + 2G(x (t)) 6 2|y (t)|M1 (t; max(|x (t)|; |y (t)|)); dt 0 ¡ t ¡ 1; ∈ [0; 1]: Proof. We note from (4) and (5) that d ((y (t))2 + 2G(x (t))) = 2g(x (t))x (t) + 2y (t)y (t) dt = 2(y (t))(p(t; x (t); y (t)) 6 2|y (t)|M1 (t; max(|x (t)|; |y (t)|))
(10)
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giving (10). Let us write h (t) = (y (t))2 + 2G(x (t)). Then, d (h (t))2 6 2h (t)M1 (t; h (t)) (11) dt since |x| 6 2G(x) for all x, in view of the assumption g(x)=x ¿ 1 for x = 0. We then get by integrating from 0 to t in (11) that t 2 2 (h (t)) 6 + 2 h (s)M1 (s; h (s)) ds: (12) 0
Let, now, k (t) = max{h (s): 0 6 s 6 t}. For x ¿ 0, let M (x):= t ¿ 0; k (t) = h (u) for some u 6 t, giving u (k (t))2 = (h (u))2 6 2 + 2 h (s)M1 (s; h (s)) ds 6 2 + 2k (t) 2
6 + 2k (t)
0
0
M1 (t; x) dt. Given
0
u
0
1
1
M1 (s; k (t)) ds M1 (s; k (t)) ds
6 2 + 2k (t)M (k (t)): Now, take suKciently large such that for a given ¿ 0, to be chosen suKciently small, 2M (x) ¡ x
for x ¿ :
Since k (t) ¿ for all t ∈ [0; 1], we get (k (t))2 6 2 + (k (t))2
and
2 ; giving 1− √ h (t) 6 1 + ; choosing small: (k (t))2 6
This gives (8), and (9). This completes the proof of the lemma. Lemma 3. Let ¿ 0 be given and M1 be as in Lemma 2. Then; there exists an A ¿ 0 such that if for ∈ [0; 1]; (x (t); y (t)) is a solution for the system of di7erential equations (4)–(7) and a t0 ∈ [0; 1] is such that x (t0 ) ¿ 0; y (t0 ) = 0: then 1 x (t0 ) 61 + 6 −1 1+ G+ (2 =2)
if || ¿ A:
Similarly; if x (t0 ) ¡ 0; y (t0 ) = 0; then x (t ) 1 0 6 −1 6 1 + if || ¿ A: 2 1 + G− ( =2)
(13)
(14)
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119
Proof. We Jrst claim that G(x)=x is increasing on (0; ∞). Indeed, take an x ¿ 0 and a k ∈ (0; 1). Then kx g(s) ds G(kx) = 0
=k
0
6k
x
0
x
g(kz) d z;
where s = kz
g(z) d z
= kG(x): It follows that G(kx)=kx 6 G(x)=x, and accordingly we see that G(x)=x is increasing on (0; ∞). Similarly, one can see that −(G(x)=x) is increasing on (−∞; 0). To show the right inequality of (13), we use (9) and also G(x (t0 )) x (t0 ) 6 G : 1+ 1+ To show the left inequality of (13), or equivalently 2 6 G((1 + )x (t0 )); 2 it is enough to show that 2 6 (1 + )G(x (t0 )): 2 By integrating (10), using (8), we have, with h (t) as in Lemma 2, that for t ∈ [0; 1], t 2 2 M1 (s; max(|x (s)|; |y (s)|)) ds ¿ 0: (15) (h (t)) − + 2||(1 + ) 0
Take 1 so that 21 (1 + ) 6 min(; 12 ). Take A so that for ¿ A inequalities (8), (9) hold, and when 2 1 ; ; (16) max(|x (s)|; |y (s)|) ¿ min G −1 2 8 2 2
we have, with M as in Lemma 1, M (max(|x (s)|; |y (s)|)) ¡ 1 max(|x (s)|; |y (s)|):
(17)
Now for ¿ A, using g(x)=x ¿ 1, (8) and (9), (17) gives M (max(|x (s)|; |y (s)|)) ¡ 1 (1 + )
(18)
and hence (h (t))2 ¿ 2 (1 − 2(1 + )2 1 )
(19)
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if (16) holds for all s ∈ [0; t]. This implies (y (t))2 + 2G(x (t)) ¿
2 : 2
2 2 Therefore, either (y (t))2 ¿ 4 or |x (t)| ¿ G −1 8 and 2 −1 max(|x (t)|; |y (t)|) ¿ min G ; 8 2 holds and hence (16) holds for s near t by continuity. Thus, if I is the largest interval on which (16) holds, then we must have I = [0; 1]. Hence (19) holds for all t ∈ [0; 1]. Finally, if y (t0 ) = 0 then 2G(x (t0 )) ¿ 2 (1 − 2(1 + )2 1 ) and we see that (13) holds. A similar proof works for (14). This completes the proof of the lemma. For a solution (x(t); y(t)) of the system of equations x (t) = y(t); y (t) = − g(x); x(0) = 0; let
y(0) = ;
1 ()
denote the angle given by the integral 1 x(t)y (t) − y(t)x (t) () = − dt: 1 x2 (t) + y2 (t) 0
Lemma 4. Let g : R → R be a continuous increasing function with g(x)x ¿ 0 for every x ∈ R; g(x) |x| increasing on (−∞; 0); and on (0; ∞). Then 1 (); ∈ (0; ∞); is strictly increasing on (0; ∞). Proof. Let 0 ¡ ¡ be given and let (x (t); y (t)); (x (t); y (t)) be such that x (t) = y (t); y (t) = − g(x (t)); x (0) = 0;
y (0) = ;
x (t) = y (t); y (t) = − g(x (t)); x (0) = 0;
y (0) = :
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121
We note that (y (t))2 + 2G(x (t)) = 2 ; (y (t))2 + 2G(x (t)) = 2 : Let denote the polar angle of the point (x ; y ) in the x–y plane and let denote the polar angle of the point (x ; y ) in the x–y plane, so that x = r cos ; y = r sin with r = x2 + y2 ; x = r cos ;
y = r sin
with r =
x 2 + y 2 :
We note that since 0 ¡ ¡ the two curves y2 + 2G(x) = 2 ; y2 + 2G(x) = 2 , cannot intersect in the x–y plane and for the same polar angle the radial coordinate on the -curve is smaller than the radial coordinate of the corresponding point on the -curve. Let t1 ; t2 and k ¿ 1 be such that (x (t2 ); y (t2 )) = k(x (t1 ); y (t1 ))
and
(t1 ) = (t2 ):
(20)
Now, we see using our assumption, g(x)=x is increasing on (0; ∞), that g(x (t2 )) g(x (t1 )) : 6 x (t1 ) x (t2 ) It then follows from (20), (21) that (y (t2 ))2 (y (t1 ))2 = (x (t2 ))2 + (y (t2 ))2 (x (t1 ))2 + (y (t1 ))2
and
x (t2 )g(x (t2 )) (x (t2 ))2 g(x (t2 )) = 2 2 (x (t2 ))2 + (y (t2 ))2 x (t2 ) (x (t2 )) + (y (t2 )) =
g(x (t2 )) (x (t1 ))2 2 2 (x (t1 )) + (y (t1 )) x (t2 )
¿
g(x (t1 )) (x (t1 ))2 (x (t1 ))2 + (y (t1 ))2 x (t1 )
=
x (t1 )g(x (t1 )) : (x (t1 ))2 + (y (t1 ))2
Accordingly, we see that −
x g(x ) + y 2 x g(x ) + y2 (t ) 6 − (t1 ); 2 x 2 + y 2 x2 + y2
which implies that (t2 ) 6 (t1 ):
(21)
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Claim. (t) ¿ (t) for all t ¿ 0. Proof. We Jrst observe that (t) ¡ 0
for all t ¿ 0;
(t) ¡ 0
for all t ¿ 0 and
! (0) = (0) = : 2 Thus, (t) and (t) are strictly decreasing functions of t. We consider the functions ! ! f (t) = − (t) and f (t) = − (t): 2 2 Now, f (t) and f (t) are strictly increasing functions of t with f (0) = 0 and f (0) = 0. Also, we see from our work above that if f (t1 ) = f (t2 ) then f (t1 ) 6 f (t2 ). We observe that the claim is equivalent to showing that f (t) 6 f (t) for all t ¿ 0. We now assert that f−1 (s) ¿ f −1 (s) for all s ¿ 0. Indeed, let t1 = f−1 (s) and t2 = f −1 (s) so that f (t1 ) = f (t2 ) = s. We see that 1 1 (f−1 ) (s) = = f (f−1 (s)) f (t1 ) ¿
1 1 = = (f −1 ) (s): f (t2 ) f (f −1 (s))
This immediately implies the assertion since f (0) = 0 and f (0) = 0. Now, to prove that f (t) 6 f (t) for all t ¿ 0, let us suppose, on the other hand, that f (t) ¿ f (t). Then, there exists a t1 ¡ t such that f (t1 ) = f (t) = s. We now get from our assertion proved above that t1 ¿ t, a contradiction. This completes the proof of the claim. Finally, we note that f (1) = 1 () and accordingly we get that 1 () ¡ 1 ( ), proving thereby that 1 is strictly increasing on (0; ∞). This completes the proof of the lemma. Remark 5. We see from Lemma 4:3 of [6] that 1 () → ∞ as → ∞, and hence there exists an even natural number k0 such that for every k ¿ k0 , k ∈ {0; 1; 2 : : :}, there exists an k ¿ 0 such that 1 (k ) = (!=2) + k!, so that if (x(t); y(t)) is a solution for the system of equations x (t) = y(t); y (t) = − g(x); x(0) = 0;
y(0) = k :
then x(t) is either maximized or minimized at t = 1.
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123
De nition 6. For ∈ R, we deJne a bounded open subset of R2 by 2
2 2 y : G = (x; y) ∈ R + G(x) ¡ 2 2 Writing 0 := k0 as deJned in Remark 5 and ¿ 0 we deJne another bounded open subset of R2 by 0
G+ = {(x; y) ∈ G \GL |y ¿ 0}
and a bounded open subset of C([0; 1]; R2 ) by "+ }: = {u : [0; 1] → G |u(0) ∈ G+
Let ∈ (0; 1) and ¿ 1 be given. For = k , where k is as in Remark 5, we shall and G+ have a common core with respect to the system of diEerential see that "+ equations x (t) = y(t);
(22)
y (t) = − g(x(t));
(23)
x(0) = 0;
(24)
x( ) = x(1):
(25) 2
2
We recall the meaning of common core, from [8]. We deJne U : R → R as follows. For u = (x; y) ∈ R2 , letting (x(t); y(t)) denote the solution to the Cauchy-problem x (t) = y(t); y (t) = − g(x(t)); x(0) = x;
y(0) = y;
let U (u) = u + (x; x( ) − x(1)): We say "+ and G+ have common core with respect to the three-point boundary value , there are no Jxed point of U on problem above, to mean: it has no solutions on @"+ @G+ , and for a solution u; u ∈ "+ if any only if u(0) ∈ G+ . 2 We note that u ∈ C([0; 1]; R ) belongs to the boundary of "+ ; @"+ , if and only if u(t) ∈ GL for every t ∈ [0; 1]; u(0) ∈ GL + and either there exists a t0 ∈ [0; 1] such that . u(t0 ) ∈ @G or else u(0) ∈ @G+ We suppose that u(t) = (x(t); y(t)) is a solution of
x (t) = y(t); y (t) = − g(x(t)): 2
Now, if there exists a t0 ∈ [0; 1] with u(t0 ) ∈ @G then ((y(t))2 =2) + G(x(t)) = 2 for all t ∈ [0; 1].
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Suppose that u(t) = (x(t); y(t)) is a solution of x (t) = y(t); y (t) = − g(x(t)) with x(0) = 0 and y(0) = k ; so that 1 (k ) = !2 + k! and |x(t)| is maximized at t = 1. Then we must have x( ) − x(1) = 0, since ¿ 1 by assumption. Thus, u(t) cannot be a solution of (22) – (25). Suppose, next, u(t) = (x(t); y(t)) is a solution of x (t) = y(t); y (t) = − g(x(t)) k ; u(0) ∈ @G k and u(0) ∈ @G 0 . Then we must have y(0) = 0 with x(0) = 0; u(0) ∈ @G+ and u(t) = 0 for all t ∈ [0; 1]. Thus, we see that "+ and G+ have a common core with respect to the system of diEerential Eqs. (22) – (25) when = k . We deJne a nonlinear operator P : C([0; 1]; R2 ) → C([0; 1]; R2 ) by setting for u(t) = (x(t); y(t)) ∈ C([0; 1]; R2 ), t (y(s); −g(x(s))) ds: (Pu)(t) = u(0) + (x(0); x( ) − x(1)) + 0
k+1 L k Lemma 7. The Leray–Schauder degree deg(I − P; "+ \"+ ; 0) = (−1)k ; for k su
x (t) = y(t); y (t) = − g(x(t)); x(0) = x;
y(0) = y;
U (u) = u + (x; x( ) − x(1)): Next, we deJne for u = (x; y) ∈ GL + ; s ∈ [0; 1] a function h by
h(u; s) = (−x; y − U2 (sx; y); where U2 (sx; y) denotes the second coordinate of U (sx; y). Now, for u = (x; y) ∈ GL + ; s ∈ [0; 1]; h(u; s) = (0; 0)
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125
if and only if y = U2 (sx; y) = U2 (0; y)
x = 0; if and only if
h(u; 1) = (0; 0): Note that h(u; 1) = (−x; y − U2 (x; y)) = (I − U )(x; y): Hence, deg(I − U; G+ ; 0): ; 0) = deg(h(·; 1); G+ ; 0) = deg(h(·; 0); G+
(26)
Let, now, for u = (u1 ; u2 ) ∈ R2 ; &(t; u) denote the solution to the Cauchy-problem x (t) = y(t); y (t) = − g(x(t)); x(0) = u1 ;
y(0) = u2
and let us deJne a mapping ': R → R; for y ∈ R by '(y) = &( ; (0; y)) − &(1; (0; y)): We see from (26) that deg(I − U; G+ ; 0) = deg(h(·; 1); G+ ; 0) = deg(h(·; 0); G+ ; 0) = deg(−I × −'; G+ ; 0)
= deg(I × '; (−; ) × (0 ; ); 0)
for some ¿ 0
= deg('; (0 ; ); 0): Now, by the deJnition of 0 and k we have that 1 (0 ) = (!=2) + k0 ! and (!=2) + k!. Let (x(t); y(t)) be a solution for the system of equations
1 (k ) =
x (t) = y(t); y (t) = − g(x); x(0) = 0;
y(0) = k :
If k is even then x(t) is maximized at t = 1 and hence '(k ) ¡ 0. Similarly, '(0 ) ¡ 0. Thus, deg('; (0 ; k ); 0) = 0. Now, if k is odd then x(t) is minimized at t = 1 and hence '(k ) ¿ 0, which implies that deg('; (0 ; k )) = 1. k+1 L k k+1 L k Thus, if k is even deg(I −P; "+ \"+ ; 0) = 1 and if k is odd deg(I −P; "+ \"+ ; 0) = − 1. This completes the proof of the lemma.
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De nition 8. For u(t) = (x(t); y(t)) ∈ C 1 ([0; 1]; R2 \{(0; 0)}) we deJne 1 x(t)y (t) − y(t)x (t) ’1 (u) = − dt; x2 (t) + y2 (t) 0 as the angle traversed from u(0) to u(1). We also deJne for ∈ [0; 1] the family of nonlinear operators P(·; ) from C([0; 1]; R2 ) × [0; 1] to C([0; 1]; R2 ) by setting for u(t) = (x(t); y(t)) ∈ C([0; 1]; R2 ) t (P(u; ))(t) = u(0)+(x(0); x( ) − x(1))+ (y(s); −g(x(s))+p(s; x(s); y(s))) ds: 0
Let us, also, denote by the open subset of C([0; 1]; R2 ) deJned by 0 )k+ = u(t) = (x(t); y(t)) ∈ C([0; 1]; R2 \GL )|y(0) ¿ 0 ; )k+
’1 (u) ∈
! 2
+ k!;
! + (k + 1)! : 2
Lemma 9. For large k; there is a bounded open subset Uk of )k+ containing the >xed point set of P(·; 1); and for all such Uk ; the Leray–Schauder degree deg(I − P(·; 1); Uk ; 0) is de>ned; that is; P(·; 1) is continuous; cl{P(u; 1) : u ∈ Uk } is compact; and deg(I − P(·; 1); Uk ; 0) = (−1)k . Proof. Let , = {u(t) = (x(t); y(t)) ∈ C([0; 1]; R2 )|P(u(t); ) = u(t)
for some ∈ [0; 1]}:
That is, u(t) ∈ , if and only if u(t) satisJes (4), (5), (24), (25) for some ∈ [0; 1]. Claim. , ∩ @)k+ is empty. Proof. We Jrst note that u(t) ∈ @)k+ implies that one of the following holds: (i) u(t) ∈ R2 \G 0 for all t and there is some t0 ∈ [0; 1]; u(t0 ) ∈ @G 0 , (ii) u(t) ∈ R2 \G 0 for all t but y(0) = 0, (iii) ’1 (u) = (!=2) + k! or ’1 (u) = (!=2) + (k + 1)!. Suppose that u(t) ∈ , ∩ @)k+ satisJes (i). By a small extension of Lemma 2, u(t) is bounded. With the lower bound on (x(t))2 + (y(t))2 for t ∈ [0; 1] from u(t) ∈ G 0 , we have an upper bound on (y(t))2 + x(t)(g(x(t)(−p(t; x(t); y(t))) (x(t))2 + (y(t))2 1 and hence we have an upper bound on ’1 (u) = 0 − (t) dt. This contradicts ’1 (u) ¿ (!=2) + k! for k large. So u(t) ∈ , ∩ @)k+ cannot satisfy (i). Suppose, next, that u(t) ∈ , ∩@)k+ satisJes (ii). Then y(0) = 0 and so x(0) = 0 gives 0 = u(0) ∈ G 0 which contradicts that u(t) ∈ @)k+ . Hence u(t) ∈ , ∩ @)k+ cannot satisfy (ii) also. − (t) =
B. Calvert, C.P. Gupta / Nonlinear Analysis 50 (2002) 115 – 128
127
Finally suppose u(t) ∈ , ∩ @)k+ satisJes (iii). Now, let ¿ 0 be such that (1 + )2 ¡ and let A be as determined by Lemma 3, corresponding to this . Now, by the boundedness of ’1 (u) for u in a bounded subset of ,, as in (i), we may take k large enough such that if ’1 (u) ¿ (!=2) + k!, then = y(0) ¿ A. We then get by Lemma 3, −1 2 ( =2)=(1 + ). Let, now, x(t) supposing that x(1) ¿ 0 and y(1) = 0, that x(1) ¿ G+ be maximized at ∈ (0; 1]. Then we have, x( ) ¿ x( ) = x(1), and since y( ) = 0, −1 2 ( =2)(1 + ). It follows that 6 (1 + )2 , we get again by Lemma 3 that x( ) 6 G+ a contradiction. Similarly, we get a contradiction if x(1) ¡ 0. Thus, we must have that , ∩ @)k+ = - (empty set), and the claim is proved. Next, we note that , ∩ )k+ is bounded. Otherwise, take a sequence un in , with un → ∞, but then ’1 (un ) → ∞ by Lemma 4:3 of [6], contradicting ’1 (un ) ¡ (!=2)+ (k + 1)!. Next, we note that P : C([0; 1]; R2 )×[0; 1] → C([0; 1]; R2 ) is continuous and compact, that is, it maps bounded sets into relatively compact sets, since p(t; x; y) satisJes the Caratheodory conditions and we may apply a result of [7] for the NemytskiGi mappings. Now, take a bounded open subset Uk of )k+ containing , ∩ )k+ . By homotopy invariance we see that deg(I − P(·; 0); Uk ; 0) = deg(I − P(·; 1); Uk ; 1): However, using Lemma 4, deg(I − P(·; 0); Uk ; 0) = deg(I − P; Uk ; 0) k+1 L k = deg(I − P; "+ \"+ ; 0)
= (−1)k ; by Lemma 7. Finally, we note that this degree is independent of the choice of Uk by the excision property. This completes the proof of the lemma. 3. Main theorem Theorem 10. Let ∈ (0; 1) and ¿ 1 be given. Then; for each k su
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Proof. By Lemma 9, deg(I − P(·; 1); Uk ; 0) = (−1)k , for k suKciently large and so a 0 solution exists in )k+ . Similarly, if we deJne )k− = {u(t) ∈ C([0; 1]; R2 \GL )|y(0) ¡ 0, k ’1 (u) ∈ ((!=2) + k!; (!=2) + (k + 1)!)}, then a solution exists in )− . This completes the proof of the theorem. Remark 11. We note that the theorem above holds for 0 ¡ ¡ 1, for then we can use () and ’ (u). Remark 12. The above may be given in terms of continuation theorems of [4 – 6]. Also, one may use the degree dL (L − N (·; ); Uk ) of [6]. References [1] A.V. Bitsadze, On the theory of nonlocal boundary value problems, Soviet Math. Dokl. 30 (1) (1984) 8–10. [2] A.V. Bitsadze, On a class of conditionally solvable nonlocal boundary value problems for harmonic functions, Soviet Math. Dokl. 31 (1) (1985) 91–94. [3] A.V. Bitsadze, A.A. SamarskiGi, On some simple generalizations of linear elliptic boundary problems, Soviet. Math. Dokl. 10 (2) (1969) 398–400. [4] A. Capietto, J. Mawhin, F. Zanolin, A continuation approach to superlinear periodic boundary value problems, J. DiEerential Equations 88 (1990) 347–395. [5] A. Capietto, J. Mawhin, F. Zanolin, A continuation approach to some forced superlinear Sturm–Liouville boundary value problems, Topol. Meth. Nonlinear Anal. 3 (1994) 81–100. [6] M. Henrard, Topological degree in boundary value problems: existence and multiplicity results for second-order diEerential equations, thesis, UniversitQe Catholique de Louvain, Louvain-la-Neuve, Belgium, 1995, 123pp. [7] M.A. Krasnosels’kiGi, Topological Methods in the Theory of Integral Equations, Pergamon Press, Oxford, 1964. [8] M.A. Krasnosels’kiGi, P.P. ZabreGiko, Geometrical methods of non-linear analysis, Springer, Berlin, 1994. [9] V.A. Il’in, E.I. Moiseev, Nonlocal boundary value problem of the Jrst kind for a Sturm–Liouville operator in its diEerential and Jnite diEerence aspects, DiEerential Equations 23 (7) (1987) 803–810. [10] V.A. Il’in, E.I. Moiseev, Nonlocal boundary value problem of the second kind for a Sturm–Liouville operator, DiEerential Equations 23 (8) (1987) 979–987.