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Nonlinear and semilinear spectrum for asymptotically linear or positively homogeneous operators
Nonlinear Analysis 57 (2004) 891 – 903
www.elsevier.com/locate/na
Nonlinear and semilinear spectrum for asymptotically linear or positively homogeneous operators Wenying Feng∗ Computer Science/Studies Program, Trent University, 1600 West Bank Drive, Peterborough, Ontario, Canada K9J 7B8 Received 18 August 2003; accepted 9 March 2004
Abstract We prove some properties for the nonlinear spectrum (f) and semilinear spectrum (L; N ), when f and N are asymptotically linear or positively homogeneous, thus close to a linear operator. The results generalize a previous result which required N to be a linear operator and L to be the identity map. Applying the theorems, we prove a result on the existence of a positive eigenvalue and eigenvector for a compact, positive operator. Examples of applications to the study of a three-point boundary value problem are also given. ? 2004 Elsevier Ltd. All rights reserved. Keywords: Nonlinear spectrum; Semilinear spectrum; Asymptotically linear operators; Positively homogeneous operators; Boundary value problem
1. Introduction One of the important contributions in the area of nonlinear spectral theory was made by Furi et al. [11]. In [11], a spectrum for nonlinear operators, the so-called fmv-spectrum, was introduced and wide applications of the theory were given. The de:nition in [11] used two numbers (f), !(f) and the class of stably-solvable operators (see the de:nitions in Section 2). Papers related to [11] include [1–3,6–8,10,17], and others. In [7], by replacing two of the conditions of the de:nition of fmv-spectrum,
This research was supported by a grant from Natural Science and Engineering Research Council of Canada. ∗
Tel.: +1-7057481011-1249; fax: +1-705-748-1066. E-mail address: [email protected] (W. Feng).
0362-546X/$ - see front matter ? 2004 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2004.03.007
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a nonlinear spectrum that contains eigenvalues was de:ned. The de:nition of [7] used the class of -epi mappings [18]. In a recent paper [10], the nonlinear spectrum of [7] was generalized by introducing the notion of spectrum for a semilinear pair of operators (L; N ), where L is a linear Fredholm operator of index zero and N is a nonlinear operator. More recently, another de:nition of semilinear spectrum, which is based on the fmv-spectrum of [11], has been given by Appell et al. [2]. So far, although many results for nonlinear spectra have been obtained ([1,3,6,11,12,17]), few results are known for the semilinear spectrum except [2] and [10]. In this paper, we will prove some properties for the semilinear spectrum of [10] when N is asymptotically linear or positively homogeneous, thus when N shares many properties held by a linear operator. Some remarks on the fmv-spectrum for asymptotically linear operators were made in [6]. Recently, nonlinear spectrum for homogeneous operators were studied in [14]. Our results generalize some previous results which require N to be a linear operator and L to be the identity map. We apply our results to prove existence of a positive eigenvalue and eigenvector for a compact, positive operator. Examples of applications to the study of a three-point boundary value problem are given. Firstly, in Section 2, we will prove two properties of the nonlinear spectrum (f) de:ned by [7] when f is positively homogeneous. Similar results are known for the fmv-spectrum [11]. 2. Spectrum for positively homogeneous operators Suppose that E; F are complex Banach spaces, let A ⊂ E be bounded, and let (A) denote the measure of noncompactness of A, [4]. Let f : E → F be a continuous nonlinear operator. The number |f| = lim supx→∞ (f(x)=x) is called the quasinorm of f. (f) and !(f) are de:ned as in [11]: (f) = inf {k ¿ 0: (f(A)) 6 k(A) for any bounded A ⊂ E}; !(f) = sup{k ¿ 0: (f(A)) ¿ k(A) for any bounded A ⊂ E}: We also use the notations m(f) and (f) from [7]: m(f) = sup{k ¿ 0: f(x) ¿ kx for any x ∈ E}; r (f; 0) = inf {k ¿ 0; there exists a k-set contraction g : Br → E; with g ≡ 0 on @Br s:t: f(x) = g(x) has no solutions in Br }; (f) = inf {r (f; 0); r ¿ 0}; where Br = {x ∈ E: x 6 r} and @Br denotes the boundary of Br . (f) is the measure of solvability of f at 0 [7] and it is related to the following de:nition of (0; )-epi mappings [18]. De nition 2.1. A continuous map f : E → F is said to be 0-admissible if f(x) = 0 for x ∈ @. A 0-admissible mapping f : E → F is said to be (0; k)-epi if for each k-set
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contraction h : E → F with h(x) ≡ 0 on @ the equation f(x) = h(x) has a solution in . The class of (0; k)-epi mappings has similar properties to a class for which there is a topological degree theory [18]. We only give the ‘homotopy’ property here since it will be used in the sequel. Property 2.2 (Tarafdar and Thompson [18]): Let f : E → F be (0; k)-epi and h : [0; 1] × E → F be an -set contraction with 0 6 6 k ¡ 1 such that h(0; x) = 0 for E Further let f(x) + h(t; x) = 0 for all x ∈ @ and for all t ∈ [0; 1]. Then all x ∈ . f(x) + h(1; x) : E → F is (0; k − )-epi. (f) ¿ 0 if and only if there exists ¿ 0, such that f(x) is (0; )-epi on every Br with r ¿ 0 [7]. The following de:nition was given in [7]. De nition 2.3. Suppose that f : E → E is a continuous map, then f is said to be regular if !(f) ¿ 0; m(f) ¿ 0, and (f) ¿ 0. For each ∈ C, if I − f is regular, is said to be in the resolvent set, (f), of f. The spectrum of f is de:ned to be (f) = { ∈ C: I − f is not regular} = C\ (f): It was proved in [11] that if f; g : E → E are continuous maps and f − g = h, with h compact and with quasinorm |h| = 0, then fmv (f) = fmv (g). We have the following similar properties for the spectrum de:ned in De:nition 2.3 for positively homogeneous operators. Theorem 2.4. Suppose f and g are positively homogeneous operators and f − g = h, where h is a compact map with h(E) bounded. Then (f) = (g). The proof of the above theorem follows easily from Theorem 4.4 of [7] and the following lemma. Lemma 2.5. Suppose f; g : E → E are positively homogeneous. Suppose also that !(f) ¿ 0, f − g = h, and h is compact, h(E) is bounded. Then (f) ¿ 0 if and only if (g) ¿ 0. Proof. Assume that (f) ¿ 0. For 0 ¡ ¡ (f), f is (0; )-epi on B1 . Let h1 : [0; 1] × E → E be given by h1 (t; x) = −th(x), then h1 is a compact map and h1 (0; x) ≡ 0. Let S = {x ∈ E; f(x) − th(x) = 0; for some t ∈ (0; 1]}: Suppose there exist x n ∈ S with x n → ∞ as n → ∞. Then h(x n ) xn = tn f → 0 as n → ∞; x n x n ∞ ∞ and letting un = x n =x n , we obtain !(f)( n=1 un ) 6 n=1 f(un ). Since !(f) ¿ 0, ∞ we have ( n=1 un ) = 0. Hence, there exists unk → x0 , f(x0 ) = 0. This contradicts f
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being (0; )-epi on B1 . So, S is bounded. By Property 2.2, g is (0; 1 )-epi on B1 for every 0 ¡ 1 ¡ . Hence, (g) ¿ 0. Since !(g) ¿ !(f) − (h) = !(f) ¿ 0, by the same argument it can be proved that (f) ¿ 0 if (g) ¿ 0. Theorem 2.4 gives conditions such that f and g have the same spectrum when they are both positively homogeneous. When f is positively homogeneous and g is any continuous map, we can prove that if ∈ C is in the spectrum of f but not in the spectrum of g then must be an eigenvalue of f, that is, there exists a nonzero x0 ∈ E such that x0 − f(x0 ) = 0. Theorem 2.6. Let f : E → E be positively homogeneous and g : E → E be a continuous map. Assume that f − g = h and h is compact with |h| = 0. Then if ∈ (f)\(g), is an eigenvalue of f. Proof. Firstly, we have !(I − g) ¿ 0 since ∈ (g). Therefore !(I − f) = !(I − f) + (h) ¿ !(I − f + h) = !(I − g) ¿ 0: Assume that m(I − f) ¿ 0. Then there exists m ¿ 0, such that (I − f)x ¿ mx
for all x ∈ E:
Let S = {x ∈ E: x − f(x) + h(x) − th(x) = 0; t ∈ (0; 1]}: Then S is bounded. Otherwise there would exist x n ∈ S, with x n → ∞. Let tn ∈ (0; 1] be such that x n − f(x n ) + (1 − tn )h(x n ) = 0: Then (x n − f(x n ))=x n → 0 as n → ∞. This contradicts the assumption m(I − f) ¿ 0. By Property 2.2, I − f is (0; )-epi on every Br with r ¿ 0 since I − f + h = I − g is (0; )-epi on Br . So we obtain ∈ (f). This contradiction implies that m(I − f) = 0. Therefore, there exists x n ∈ E such that 1 x n − f(x n ) ¡ x n ; x n = 0: n Use the notation un = x n =x n , we have ∞ ∞ !(I − f) un 6 (I − f)un : So (
∞
n=1
n=1
n=1
un ) = 0. Let unk → x0 , then x0 = 1 and x0 − f(x0 ) = 0.
3. Semilinear spectrum for asymptotically linear or positively homogeneous operators Suppose L : dom(L) ⊂ E → F is a Fredholm operator of index zero, dom(L) is dense in E. E = ker(L) ⊕ E1 ; F = F0 ⊕ im(L). Let P : E → ker(L) and Q : F → F0
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be the respective projections. The operators LP and KP are de:ned as LP := L : dom(L) ∩ E1 → im(L); and KP = L−1 P : Let KPQ = KP (I − Q), and $ : F → F=im(L) be the quotient map. % : F=im(L) → ker(L) is the isomorphism. Let ⊂ E be a bounded open set with the property dom(L) ∩ = ∅. Let N : E → F be a nonlinear mapping. This is the same set up as used in de:ning coincidence degree, [13]. E is bounded. The nonlinSuppose that $N : E → coker(L) is continuous and $N () ear operator N is said to be L-compact if KPQ N : E → E is compact. N is said to be a L-k-set contraction [13] if KPQ N : E → E is a k-set contraction [4]. The semilinear spectrum for (L; N ) was de:ned in [10]. De nition 3.1. For ∈ C, let f (L; N )(x) = (I − P)x − (%$ + KPQ )Nx: L − N is regular if !(f ) ¿ 0; m(f ) ¿ 0 and (f ) ¿ 0. The resolvent set of (L; N ) is de:ned to be (L; N ) = { ∈ C: L − N is regular}. The spectrum is de:ned by (L; N ) = C\ (L; N ). To prove a property for the semilinear spectrum (L; N ) when the nonlinear part N is asymptotically linear, Theorem 3.4 below, we need a lemma, Lemma 3.3 below. For ∈ C, we use the following symbols. $(L; N ) = {: Lx − Nx ¿ mx for some m ¿ 0 and x ∈ dom L}: &+ (L; N ) = {: there exists a linear operator T : E → F such that L − N = T + R;
(3.1)
where ker(T ) = {0}; im(T ) is closed; and R(x)=x → 0 as x → ∞}. &0 (L; N ) = {: there exists a linear operator T : E → F such that L − N = T + R; where ker(T ) = {0}; im(T ) is closed; and R(x)=x → 0 as x → 0}. Remark 3.2. When T : D(T ) ⊂ H → H is a linear operator, in a Hilbert space H , the :eld of regularity $(T ) of T is de:ned (see [5]) to be the set of values ∈ C, for which there exists a positive constant k such that (I − T )(u) ¿ ku
for all u ∈ dom(T ):
Accordingly, $(L; N ) can be called the :eld of regularity of (L; N ). Lemma 3.3. (1) If N is an asymptotically linear operator, then $(L; N ) ⊂ &+ (L; N ): (2) If N is di:erentiable at 0, then $(L; N ) ⊂ &0 (L; N ).
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In the proof of Lemma 3.3, we will use the concept of reduced minimum modulus +(T ) of a linear operator T . It is known that T has a closed range if and only if +(T ) ¿ 0 (see [15, p. 231]). Proof of Lemma 3.3. (1) Suppose ∈ $(L; N ) and m ¿ 0 with (L − N )x ¿ mx for all x ∈ E. Since N is asymptotically linear, we can write L − N = T + R; where T is linear and R(x)=x → 0 as x → ∞. Assume that x0 ∈ E and Tx0 = 0. If x0 = 0, then R(nx0 )=nx0 ¿ m. This contradicts R(x)=x → 0 as x → ∞. Hence ker(T ) = 0. For every y0 ∈ E with y0 = 1, we have Tny0 + R(ny0 ) ¿ mn;
so Ty0 + R(ny0 )=n ¿ m:
Letting n → ∞, we obtain T (y0 ) ¿ m. Thus the reduced minimum modulus of T is positive, so im(T ) is closed. Hence ∈ &+ (L; N ). (2) Follows by similar arguments to case (1). Lemma 3.3 enables us to prove a property for the semilinear spectrum (L; N ) when the nonlinear part N is asymptotically linear. Theorem 3.4. Let L : E → F be a Fredholm operator of index 0. Suppose N is an asymptotically linear operator and is a L-k-set contraction. Assume that N = l − R, where l is linear, R is compact and R(x)=x → 0 as x → ∞. If ∈ (L; N ) and || ¿ k, then ∈ m (L; N ). Proof. Since || ¿ k, !(L − N ) ¿ 0, so ∈ ! (L; N ). Assume that ∈ m (L; N ), then ∈ $(L; N ). By the proof of (1) of Lemma 3.3, ∈ &+ (L; N ) and L − N = T + R; where T = L − l : E → F is a linear operator, ker(T ) = {0} and im(T ) is closed. Let H = L + JP, then %$ + KPQ = H −1 . By Lemma 4.4 of [10], (I − P) − H −1 N = H −1 T + H −1 R: H −1 T : E → E is a linear isomorphism, so H −1 T is (0; )-epi for some 0 ¡ ¡ 1= H −1 T (see [18]). Let U = {x: H −1 Tx + tH −1 Rx = 0; t ∈ [0; 1]}: Then U is bounded. Otherwise there would exist x n ∈ E with x n → ∞, such that H −1 T (x n =x n ) = −tn H −1 (R(x n )=x n ) → 0
as n → ∞:
Thus T (x n =x n ) → 0 as n → ∞. This contradicts the fact that im(T ) is closed. Hence for r1 suKciently large, if x ∈ E with x = r1 , we have H −1 Tx + tH −1 Rx = 0:
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This ensures that H −1 T + H −1 R is (0; )-epi on Br1 . Since (H −1 T + H −1 R)−1 {0} = f−1 {0} = {0}; we have H −1 T + H −1 R is (0; )-epi on every ball Br with r ¿ 0. By the de:nition, ∈ (L; N ). This contradiction shows that ∈ m (L; N ). The following corollary on linear operators extends previous result that required L to be the identity map [19]. Corollary 3.5. Suppose that N is linear and is an L-k-set contraction. If ∈ (L; N ) and || ¿ k, then is an eigenvalue of (L; N ). Our next result deals with the case that N is a positively homogeneous operator. Theorem 3.6. Suppose that T is a positively homogeneous L-k-set contraction. ∈ (L; T ) with || ¿ k. Then there exists t ∈ (0; 1], such that =t is an eigenvalue of (L; (I − Q)T ). Proof. Assume that m(f ) = 0. Then there exist {x n } ∈ E, such that (I − P)un − (%$ + KPQ )Tun ¡ 1=n → 0;
(n → ∞):
Here, un = x n =x n . Since || ¿ k, !(f ) ¿ 0. So un (n = 1; 2; : : :) has a convergent subsequence. Assume that un → x0 , then f (x0 ) = 0. Hence Lx0 = Tx0 = QTx0 + (I − Q)Tx0 : This implies that Lx0 = (I − Q)Tx0 . Thus is an eigenvalue of (L; (I − Q)T ). In this case t = 1. Now suppose that m(f ) ¿ 0. Let S = {x ∈ E: x − tPx − t(%$ + KPQ )Tx = 0; t ∈ [0; 1]}: We have the following two cases: Case (1): There exist x n ∈ S with x n → ∞. Let yn = x n =x n , then (yn ) − tn P(yn ) − tn (%$ + KPQ )T (yn ) = 0; where tn ∈ [0; 1]. Suppose that tn → t0 ∈ [0; 1], then (yn ) − t0 P(yn ) − t0 (%$ + KPQ )T (yn ) =(tn − t0 )P(yn ) + (tn − t0 )(%$ + KPQ )T (yn ): So ({yn }∞ n=1 )!(I − t0 P − t0 (%$ + KPQ )T ) 6 yn − t0 Pyn − t0 (%$ + KPQ )Tyn = 0: ∞ Since || ¿ k, !(I − t0 P − t0 (%$ + KPQ )T ) ¿ 0. Thus ({yn }∞ n=1 ) = 0. So {yn }n=1 has a convergent subsequence. Suppose that yn → x0 . Then
x0 − t0 Px0 − t0 (%$ + KPQ )Tx0 = 0:
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Since x0 = 1, t0 = 0. (I − P)x0 − t0 KPQ Tx0 = t0 Px0 − Px0 + t0 %$Tx0 ∈ ker(L): But (I − P)x0 − t0 KPQ Tx0 ∈ E1 ∩ dom(L), we have (I − P)x0 = t0 L−1 P (I − Q)Tx0 : Thus (=t0 )Lx0 − (I − Q)Tx0 = 0, and =t0 is an eigenvalue of (L; (I − Q)T ). Case (2): Assume that there exists R ¿ 0 such that S ⊂ BR \@BR . Then for every x ∈ @BR we have x − tP(x) − t(%$ + KPQ )T (x) = 0;
t ∈ [0; 1]:
By Property 2.2, I − P − (%$ + KPQ )T is (0; )-epi for some ¿ 0 on BR . By our assumption, m(f ) ¿ 0. Thus f is (0; )-epi on every Br with r ¿ 0. Hence (f ) ¿ 0. This shows that f is regular and ∈ (L; T ), which contradicts ∈ (L; T ). The above arguments complete the proof. When L is the identity, the projection Q = 0 and the semilinear spectrum (L; N ) reduces to the nonlinear spectrum (N ) [7]. So we have the following corollary. Corollary 3.7. Suppose that T is a positively homogeneous k-set contraction. ∈ (T ) with || ¿ k. Then there exists t ∈ (0; 1], such that =t is an eigenvalue of T . The following corollary is obvious. Corollary 3.8. Suppose T is a positively homogeneous L-k-set contraction. If the eigenvalues of (L; (I − Q)T ) are bounded, then (L; T ) is bounded. Proposition 3.9. Suppose T =B+R, where B is a linear operator, either R(x)=x → 0 as x → ∞ or as x → 0. If is an eigenvalue of (L; B), then ∈ (L; T ). Proof. Let x0 ∈ E, x0 = 0 be such that Lx0 −Bx0 =0. If R(x)=x → 0 as x → ∞, then (L(nx0 ) − T (nx0 )=n) → 0. Thus m(L; T ) = 0, and ∈ (L; T ). If R(x)=x → 0 as x → 0, then L(x0 =n) − T (x0 =n) → 0; 1=n
(n → ∞):
Hence ∈ (L; T ). 4. Applications In this section, we give two examples of the applications of our theorems. Firstly, we prove a theorem that gives a condition for a compact, positive operator to have a positive eigenvalue and eigenvector.
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Theorem 4.1. Suppose E is a Banach space and K is a cone of E. For r ¿ 0, let Kr = K ∩ Br , @Kr = {x ∈ K: x = r}. (1) Assume that f : K → K is a positively homogeneous, compact operator and inf {x=1; x∈K} f(x) ¿ 0. Then there exist r ¿ 0 and x0 ∈ K; x0 = 1 such that f(x0 ) = rx0 . (2) Assume that F : Kr → K is compact and inf {F(x): x = r} ¿ 0. Then F has a positive eigenvalue with an eigenvector x0 ∈ @Kr . Proof. (1) Since K is a cone of E; K is a closed, convex subset of E. By the Dugundji extension theorem, there exists an extension f1 with f1 (E) ⊂ K and with f1 compact map [4]. Let F : E → K be de:ned by xf1 (x=x) if x = 0; F(x) = 0 if x = 0: Then F : E → K is a positively homogeneous and compact operator. Also, F|K = f. Suppose (F) ∩ R+ = ∅ and u ∈ K, u = 0. Since (1=n) ∈ (F), (1=n)I − F is onto, and there exist {x n }∞ n=1 ∈ E satisfying (1=n)x n − F(x n ) = u: F(x n ) ∈ K and u ∈ K imply that x n ∈ K. So F(x n ) = f(x n ). Thus (1=n)x n − f(x n ) = u:
(4.1)
Assume that {x n ∞ n=1 } is unbounded and x nk → ∞ (k → ∞). Let unk = x nk =x nk , Then, (1=n)unk − f(unk ) = u=x nk → 0
(k → ∞):
Thus f(unk ) → 0. This contradicts inf {x∈K; x=1} f(x) ¿ 0. So we obtain that {x n ∞ n=1 } is bounded. Next, (4.1) implies that 0 6 u 6 (1=n)x n → 0
(n → ∞):
Thus u = 0. This contradiction shows that (F) ∩ R+ = ∅. Let r1 ¿ 0 and r1 ∈ (F). By Corollary 3.7, there exists r ¿ r1 such that r is an eigenvalue of F. Let x0 ∈ E with x0 = 1 be such that F(x0 ) = rx0 . Then x0 ∈ K since r ¿ 0 and F(x0 ) ∈ K. Thus f(x0 ) = F(x0 ) = rx0 . (2) Let FE : K → K be de:ned as follows: xF((rx)=x) for x = 0; E F(x) = 0 for x = 0: FE is a compact and positively homogeneous map since F is a compact map [16]. E Furthermore, inf x=1 (F)(x) = inf x=r F(x) ¿ 0. So, by (1), there exists x ∈ K E ) = x . Thus F(rx =x ) = x . Let x0 = rx with x = 1 and ¿ 0 such that F(x then x0 ∈ K and x0 = r. Also F(x0 ) = (=r)x0 .
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Now, consider the second order diLerential equation u (t) = f(t; u(t); u (t));
t ∈ (0; 1)
(4.2)
with one of the boundary conditions u (0) = 0;
u(1) = u(0);
(4.3)
u(0) = 0;
u(1) = u(0);
(4.4)
here ∈ R and 0 ∈ (0; 1). To prove the existence of a solution for these BVPs, various growth conditions are needed [9]. Theorem 3.6 enables us to prove the existence of an eigenvalue without any growth condition. We will use the classical spaces C[0; 1]; C 1 [0; 1]; C 2 [0; 1], and L1 [0; 1]. The Sobolev space is de:ned by W 2; 1 (0; 1) = {u : [0; 1] → R : u; u are absolutely continuous on [0; 1] with u ∈ L1 [0; 1]} with its usual norm. ∈ R is said to be an eigenvalue of the BVP (4.2), (4.3) if there exists a nonzero u ∈ C[0; 1] satisfying the boundary condition (4.3) and u = f(t; u(t); u (t));
t ∈ (0; 1):
(4.5)
If for any y(t) ∈ L1 [0; 1], there exists u ∈ C[0; 1] satisfying the boundary condition (4.3) and u − f(t; u(t); u (t)) = y(t);
t ∈ (0; 1);
(4.6)
the BVP (4.2), (4.3) is said to be surjective. Theorem 4.2. Suppose that f : [0; 1] × R2 → R is a continuous map and f(t; cx; cp) = cf(t; x; p) for any c ¿ 0;
(x; p) ∈ R2 :
(4.7)
Then, for = 1, either the BVP (4.2), (4.3) is surjective or has a positive eigenvalue r ¿ 1. Similar result holds for the BVP (4.2), (4.4) provided that = 1=0. Proof. Let E = C[0; 1] and F = L1 [0; 1]. L : D(L) ⊂ E → F be de:ned by L(u) = u and D(L) = {u ∈ W 2; 1 (0; 1): u (0) = 0; u(1) = u(0)}: The nonlinear map N : E → F is de:ned by (Nu)(t) = f(t; u; u );
t ∈ [0; 1]:
It was proved that L is invertible and N is L-compact [9]. By our assumptions, N is positively homogeneous. Suppose that 1 ∈ (L; N ). Then the BVP (4.2), (4.3) is surjective. Otherwise, by Theorem 3.6, there exists t ∈ (0; 1] such that r = 1=t is an eigenvalue of (L; N ). Let u ∈ D(L) be the corresponding eigenvector, then u(t) satis:es Eqs. (4.3), (4.5). A similar method proves the result for the BVP (4.2), (4.4).
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Theorem 4.2 gives an alternative condition for Eq. (4.7) to have a positive eigenvalue or being surjective. The following examples show that, in both boundary conditions (4.3) and (4.4), there are equations that are not surjective so that an eigenvalue exists. Example 4.3. Let f(t; x; p) = a|x| where a ¿ 0 is a constant. 1. If 1=02 ¿ ¿ 1 and a ¿ 2( − 1)=(1 − 02 ), then BVP (4.2), (4.3) has a positive eigenvalue. 2. If 1=02 ¿ ¿ 1=0 and a ¿ 2(0 − 1)=(0(1 − 0)), then the BVP (4.2), (4.4) has a positive eigenvalue. Proof. We will prove that in both cases 1 and 2, the BVPs are not surjective. 1. Suppose the BVP (4.2), (4.3) is surjective. For any c ¿ 0, c ∈ L[0; 1], there exist uc ∈ W 2; 1 (0; 1) such that uc − a|uc | = c;
(4.8)
uc (0) = 0;
(4.9)
uc (1) = uc (0):
Since Lu = u with these boundary conditions is invertible, let K = L−1 . For any y ∈ L1 [0; 1], by calculation, 0 t (0 − s)y(s) ds + (t − s)y(s) ds Ky(t) = 1− 0 0 1 1 − (1 − s)y(s) ds: 1− 0 From Luc − a|uc | = c we have uc = aK|uc | + Kc:
(4.10)
Since uc ¿ 0 and uc (0)=0, uc is positive and uc is increasing. Moreover, ¿ 1 ensures that both u(0) and u(1) are positive. So, uc (t) ¿ 0 for t ∈ [0; 1]. From (4.10), we can obtain that Kc 1 = aK1 + ; for t ∈ [0; 1]; (4.11) |uc | here
Kc(t) =
1 − 02 + 2( − 1)
0
t
(t − s) ds c:
Since 1=02 ¿ ¿ 1, it can be veri:ed that Kc(t) is increasing. So, for t ∈ [0; 1], 0 (1 − 02 )c 1 − 02 (0 − s) ds c = Kc(t) ¿ + : 2( − 1) 2( − 1) 0 Since a ¿ 2( − 1)=(1 − 02 ), the right side of (4.11) gives Kc 1 − 02 1 − 02 aK1 + ¿a + c ¿ 1: |uc | 2( − 1) 2|uc (1)|( − 1)
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This contradiction shows that (4.2), (4.3) is not surjective. By Theorem 4.2, it has a positive eigenvalue. 2. Suppose the BVP (4.2), (4.4) is surjective. For any c ¿ 0, there exist uc ∈ W 2; 1 (0; 1) such that uc − a|uc | = c;
(4.12)
uc (0) = 0;
(4.13)
uc (1) = uc (0):
With the above boundary condition, the linear operator K is: 0 t t (0 − s)y(s) ds + (t − s)y(s) ds Ky(t) = 1 − 0 0 0 1 t − (1 − s)y(s) ds: 1 − 0 0 Again, by calculation, we have 2 t (0 − 1)t + (t − s) ds c: (4.14) Kc(t) = 2(1 − 0) 0 We show that uc (0) = 0. Otherwise, uc (0) = 0 and uc (1) = 0 ensure that there exist 81 ∈ (0; 0) and 82 ∈ (0; 1) such that uc (81 ) = uc (82 ) = 0. This contradicts uc (t) ¿ 0 for t ∈ (0; 1). Next, if uc (0) ¡ 0, then uc (1) = uc (0) ¡ 0. uc (t) ¿ 0 implies that uc (t) ¡ 0 for any t ∈ [0; 1]. As in the proof of 1, we have Kc − 1 = aK1 + ; for t ∈ [0; 1]; (4.15) |uc | Since 1=02 ¿ ¿ 1=0, by (4.14), we can obtain that for any t ∈ [0; 1], (K1)(t) ¿ 0 and (Kc)(t) ¿ 0. Again, we reach a contradiction. Lastly, suppose that uc (1) ¿ uc (0) ¿ 0. Then uc (t) ¿ 0 implies that uc (t) ¿ 0 for any t ∈ [0; 1]. As in the proof of 1, we have Kc 1 = aK1 + ; for t ∈ [0; 1]; (4.16) |uc | By (4.14), we can see that Kc(t) ¿ 0 and is increasing. Thus, 2 0 (0 − 1)0 (t − s) ds a(K1)(t) ¿ a + 2(1 − 0) 0 0(1 − 0) ¿ 1: 2(0 − 1) This contradiction shows that the BVP (4.2), (4.4) is not surjective and the proof is completed. =a
Acknowledgements The author thanks Prof. J.R.L. Webb for valuable comments and suggestions, especially for his help with Example 4.3.
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References [1] J. Appell, E. De Pascale, A. Vignoli, A comparison of diLerent spectra for nonlinear operators, Nonlinear Anal. TMA 40 (2000) 73–90. [2] J. Appell, E. De Pascale, A. Vignoli, On a semilinear FMV-spectrum, Zeitschr. Anal. Anw. 20 (2001) 565–577. [3] J. Appell, M. DPorfner, Some spectral theory for nonlinear operators, Nonlinear Anal. TMA 28 (1997) 1955–1976. [4] K. Deimling, Nonlinear Functional Analysis, Springer, Berlin, Heidelberg, New York, 1985. [5] D.E. Edmunds, W.D. Evans, Spectral Theory and DiLerential Operators, Oxford Science Publications, Oxford University Press, New York, 1987. [6] D.E. Edmunds, J.R.L. Webb, Remarks on nonlinear spectral theory, Boll. Un. Mat. Ital. B (2) (1983) 377–390. [7] W. Feng, A new spectral theory for nonlinear operators and its applications, Abst. Appl. Anal. 2 (1997) 163–183. [8] W. Feng, J.R.L. Webb, Surjectivity results for nonlinear mappings without oddness conditions, Comment. Math. Univ. Carolin. 38 (1997) 15–28. [9] W. Feng, J.R.L. Webb, Solvability of the m-point boundary value problems with nonlinear growth, J. Math. Anal. Appl. 212 (1997) 467–480. [10] W. Feng, J.R.L. Webb, A spectral theory for semilinear operators and its applications, Progr. Nonlinear DiLerential Equations Appl. 40 (2000) 149–163. [11] M. Furi, M. Martelli, A. Vignoli, Contributions to the spectral theory for nonlinear operators in Banach spaces, Ann. Mat. Pura Appl. 118 (1978) 229–294. [12] M. Furi, M. Martelli, A. Vignoli, On the solvability of nonlinear operator equations in normed spaces, Ann. Mat. Pura Appl. 124 (1980) 321–343. [13] R.E. Gaines, J. Mawhin, Coincidence degree and nonlinear diLerential equations, in: Lecture Notes in Mathematics, Vol. 568, Springer, Berlin, 1977. [14] E. Giorgieri, J. Appell, M. Vath, Nonlinear spectral theory for homogeneous operators, Nonlinear Funct. Anal. Appl. 7 (4) (2002) 589–618. [15] T. Kato, Perturbation Theory of Linear Operators, Springer, Berlin, Heidelberg, New York, 1976. [16] M. Martelli, Positive eigenvectors of wedge maps, Ann. Mat. Pura Appl. 145 (1986) 1–32. [17] P. Santucci, M. Vath, Grasping the phantom a new approach to nonlinear spectral theory, Ann. Mat. 180 (2001) 255–284. [18] E.U. Tarafdar, H.B. Thompson, On the solvability of nonlinear noncompact operator equations, J. Austral. Math. Soc. (Ser. A) 43 (1987) 103–126. [19] W. Zhao, Geometrical coeKcients and measures of noncompactness, Ph.D Thesis, University of Glasgow, 1992.
www.elsevier.com/locate/na
Nonlinear and semilinear spectrum for asymptotically linear or positively homogeneous operators Wenying Feng∗ Computer Science/Studies Program, Trent University, 1600 West Bank Drive, Peterborough, Ontario, Canada K9J 7B8 Received 18 August 2003; accepted 9 March 2004
Abstract We prove some properties for the nonlinear spectrum (f) and semilinear spectrum (L; N ), when f and N are asymptotically linear or positively homogeneous, thus close to a linear operator. The results generalize a previous result which required N to be a linear operator and L to be the identity map. Applying the theorems, we prove a result on the existence of a positive eigenvalue and eigenvector for a compact, positive operator. Examples of applications to the study of a three-point boundary value problem are also given. ? 2004 Elsevier Ltd. All rights reserved. Keywords: Nonlinear spectrum; Semilinear spectrum; Asymptotically linear operators; Positively homogeneous operators; Boundary value problem
1. Introduction One of the important contributions in the area of nonlinear spectral theory was made by Furi et al. [11]. In [11], a spectrum for nonlinear operators, the so-called fmv-spectrum, was introduced and wide applications of the theory were given. The de:nition in [11] used two numbers (f), !(f) and the class of stably-solvable operators (see the de:nitions in Section 2). Papers related to [11] include [1–3,6–8,10,17], and others. In [7], by replacing two of the conditions of the de:nition of fmv-spectrum,
This research was supported by a grant from Natural Science and Engineering Research Council of Canada. ∗
Tel.: +1-7057481011-1249; fax: +1-705-748-1066. E-mail address: [email protected] (W. Feng).
0362-546X/$ - see front matter ? 2004 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2004.03.007
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a nonlinear spectrum that contains eigenvalues was de:ned. The de:nition of [7] used the class of -epi mappings [18]. In a recent paper [10], the nonlinear spectrum of [7] was generalized by introducing the notion of spectrum for a semilinear pair of operators (L; N ), where L is a linear Fredholm operator of index zero and N is a nonlinear operator. More recently, another de:nition of semilinear spectrum, which is based on the fmv-spectrum of [11], has been given by Appell et al. [2]. So far, although many results for nonlinear spectra have been obtained ([1,3,6,11,12,17]), few results are known for the semilinear spectrum except [2] and [10]. In this paper, we will prove some properties for the semilinear spectrum of [10] when N is asymptotically linear or positively homogeneous, thus when N shares many properties held by a linear operator. Some remarks on the fmv-spectrum for asymptotically linear operators were made in [6]. Recently, nonlinear spectrum for homogeneous operators were studied in [14]. Our results generalize some previous results which require N to be a linear operator and L to be the identity map. We apply our results to prove existence of a positive eigenvalue and eigenvector for a compact, positive operator. Examples of applications to the study of a three-point boundary value problem are given. Firstly, in Section 2, we will prove two properties of the nonlinear spectrum (f) de:ned by [7] when f is positively homogeneous. Similar results are known for the fmv-spectrum [11]. 2. Spectrum for positively homogeneous operators Suppose that E; F are complex Banach spaces, let A ⊂ E be bounded, and let (A) denote the measure of noncompactness of A, [4]. Let f : E → F be a continuous nonlinear operator. The number |f| = lim supx→∞ (f(x)=x) is called the quasinorm of f. (f) and !(f) are de:ned as in [11]: (f) = inf {k ¿ 0: (f(A)) 6 k(A) for any bounded A ⊂ E}; !(f) = sup{k ¿ 0: (f(A)) ¿ k(A) for any bounded A ⊂ E}: We also use the notations m(f) and (f) from [7]: m(f) = sup{k ¿ 0: f(x) ¿ kx for any x ∈ E}; r (f; 0) = inf {k ¿ 0; there exists a k-set contraction g : Br → E; with g ≡ 0 on @Br s:t: f(x) = g(x) has no solutions in Br }; (f) = inf {r (f; 0); r ¿ 0}; where Br = {x ∈ E: x 6 r} and @Br denotes the boundary of Br . (f) is the measure of solvability of f at 0 [7] and it is related to the following de:nition of (0; )-epi mappings [18]. De nition 2.1. A continuous map f : E → F is said to be 0-admissible if f(x) = 0 for x ∈ @. A 0-admissible mapping f : E → F is said to be (0; k)-epi if for each k-set
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contraction h : E → F with h(x) ≡ 0 on @ the equation f(x) = h(x) has a solution in . The class of (0; k)-epi mappings has similar properties to a class for which there is a topological degree theory [18]. We only give the ‘homotopy’ property here since it will be used in the sequel. Property 2.2 (Tarafdar and Thompson [18]): Let f : E → F be (0; k)-epi and h : [0; 1] × E → F be an -set contraction with 0 6 6 k ¡ 1 such that h(0; x) = 0 for E Further let f(x) + h(t; x) = 0 for all x ∈ @ and for all t ∈ [0; 1]. Then all x ∈ . f(x) + h(1; x) : E → F is (0; k − )-epi. (f) ¿ 0 if and only if there exists ¿ 0, such that f(x) is (0; )-epi on every Br with r ¿ 0 [7]. The following de:nition was given in [7]. De nition 2.3. Suppose that f : E → E is a continuous map, then f is said to be regular if !(f) ¿ 0; m(f) ¿ 0, and (f) ¿ 0. For each ∈ C, if I − f is regular, is said to be in the resolvent set, (f), of f. The spectrum of f is de:ned to be (f) = { ∈ C: I − f is not regular} = C\ (f): It was proved in [11] that if f; g : E → E are continuous maps and f − g = h, with h compact and with quasinorm |h| = 0, then fmv (f) = fmv (g). We have the following similar properties for the spectrum de:ned in De:nition 2.3 for positively homogeneous operators. Theorem 2.4. Suppose f and g are positively homogeneous operators and f − g = h, where h is a compact map with h(E) bounded. Then (f) = (g). The proof of the above theorem follows easily from Theorem 4.4 of [7] and the following lemma. Lemma 2.5. Suppose f; g : E → E are positively homogeneous. Suppose also that !(f) ¿ 0, f − g = h, and h is compact, h(E) is bounded. Then (f) ¿ 0 if and only if (g) ¿ 0. Proof. Assume that (f) ¿ 0. For 0 ¡ ¡ (f), f is (0; )-epi on B1 . Let h1 : [0; 1] × E → E be given by h1 (t; x) = −th(x), then h1 is a compact map and h1 (0; x) ≡ 0. Let S = {x ∈ E; f(x) − th(x) = 0; for some t ∈ (0; 1]}: Suppose there exist x n ∈ S with x n → ∞ as n → ∞. Then h(x n ) xn = tn f → 0 as n → ∞; x n x n ∞ ∞ and letting un = x n =x n , we obtain !(f)( n=1 un ) 6 n=1 f(un ). Since !(f) ¿ 0, ∞ we have ( n=1 un ) = 0. Hence, there exists unk → x0 , f(x0 ) = 0. This contradicts f
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being (0; )-epi on B1 . So, S is bounded. By Property 2.2, g is (0; 1 )-epi on B1 for every 0 ¡ 1 ¡ . Hence, (g) ¿ 0. Since !(g) ¿ !(f) − (h) = !(f) ¿ 0, by the same argument it can be proved that (f) ¿ 0 if (g) ¿ 0. Theorem 2.4 gives conditions such that f and g have the same spectrum when they are both positively homogeneous. When f is positively homogeneous and g is any continuous map, we can prove that if ∈ C is in the spectrum of f but not in the spectrum of g then must be an eigenvalue of f, that is, there exists a nonzero x0 ∈ E such that x0 − f(x0 ) = 0. Theorem 2.6. Let f : E → E be positively homogeneous and g : E → E be a continuous map. Assume that f − g = h and h is compact with |h| = 0. Then if ∈ (f)\(g), is an eigenvalue of f. Proof. Firstly, we have !(I − g) ¿ 0 since ∈ (g). Therefore !(I − f) = !(I − f) + (h) ¿ !(I − f + h) = !(I − g) ¿ 0: Assume that m(I − f) ¿ 0. Then there exists m ¿ 0, such that (I − f)x ¿ mx
for all x ∈ E:
Let S = {x ∈ E: x − f(x) + h(x) − th(x) = 0; t ∈ (0; 1]}: Then S is bounded. Otherwise there would exist x n ∈ S, with x n → ∞. Let tn ∈ (0; 1] be such that x n − f(x n ) + (1 − tn )h(x n ) = 0: Then (x n − f(x n ))=x n → 0 as n → ∞. This contradicts the assumption m(I − f) ¿ 0. By Property 2.2, I − f is (0; )-epi on every Br with r ¿ 0 since I − f + h = I − g is (0; )-epi on Br . So we obtain ∈ (f). This contradiction implies that m(I − f) = 0. Therefore, there exists x n ∈ E such that 1 x n − f(x n ) ¡ x n ; x n = 0: n Use the notation un = x n =x n , we have ∞ ∞ !(I − f) un 6 (I − f)un : So (
∞
n=1
n=1
n=1
un ) = 0. Let unk → x0 , then x0 = 1 and x0 − f(x0 ) = 0.
3. Semilinear spectrum for asymptotically linear or positively homogeneous operators Suppose L : dom(L) ⊂ E → F is a Fredholm operator of index zero, dom(L) is dense in E. E = ker(L) ⊕ E1 ; F = F0 ⊕ im(L). Let P : E → ker(L) and Q : F → F0
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be the respective projections. The operators LP and KP are de:ned as LP := L : dom(L) ∩ E1 → im(L); and KP = L−1 P : Let KPQ = KP (I − Q), and $ : F → F=im(L) be the quotient map. % : F=im(L) → ker(L) is the isomorphism. Let ⊂ E be a bounded open set with the property dom(L) ∩ = ∅. Let N : E → F be a nonlinear mapping. This is the same set up as used in de:ning coincidence degree, [13]. E is bounded. The nonlinSuppose that $N : E → coker(L) is continuous and $N () ear operator N is said to be L-compact if KPQ N : E → E is compact. N is said to be a L-k-set contraction [13] if KPQ N : E → E is a k-set contraction [4]. The semilinear spectrum for (L; N ) was de:ned in [10]. De nition 3.1. For ∈ C, let f (L; N )(x) = (I − P)x − (%$ + KPQ )Nx: L − N is regular if !(f ) ¿ 0; m(f ) ¿ 0 and (f ) ¿ 0. The resolvent set of (L; N ) is de:ned to be (L; N ) = { ∈ C: L − N is regular}. The spectrum is de:ned by (L; N ) = C\ (L; N ). To prove a property for the semilinear spectrum (L; N ) when the nonlinear part N is asymptotically linear, Theorem 3.4 below, we need a lemma, Lemma 3.3 below. For ∈ C, we use the following symbols. $(L; N ) = {: Lx − Nx ¿ mx for some m ¿ 0 and x ∈ dom L}: &+ (L; N ) = {: there exists a linear operator T : E → F such that L − N = T + R;
(3.1)
where ker(T ) = {0}; im(T ) is closed; and R(x)=x → 0 as x → ∞}. &0 (L; N ) = {: there exists a linear operator T : E → F such that L − N = T + R; where ker(T ) = {0}; im(T ) is closed; and R(x)=x → 0 as x → 0}. Remark 3.2. When T : D(T ) ⊂ H → H is a linear operator, in a Hilbert space H , the :eld of regularity $(T ) of T is de:ned (see [5]) to be the set of values ∈ C, for which there exists a positive constant k such that (I − T )(u) ¿ ku
for all u ∈ dom(T ):
Accordingly, $(L; N ) can be called the :eld of regularity of (L; N ). Lemma 3.3. (1) If N is an asymptotically linear operator, then $(L; N ) ⊂ &+ (L; N ): (2) If N is di:erentiable at 0, then $(L; N ) ⊂ &0 (L; N ).
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In the proof of Lemma 3.3, we will use the concept of reduced minimum modulus +(T ) of a linear operator T . It is known that T has a closed range if and only if +(T ) ¿ 0 (see [15, p. 231]). Proof of Lemma 3.3. (1) Suppose ∈ $(L; N ) and m ¿ 0 with (L − N )x ¿ mx for all x ∈ E. Since N is asymptotically linear, we can write L − N = T + R; where T is linear and R(x)=x → 0 as x → ∞. Assume that x0 ∈ E and Tx0 = 0. If x0 = 0, then R(nx0 )=nx0 ¿ m. This contradicts R(x)=x → 0 as x → ∞. Hence ker(T ) = 0. For every y0 ∈ E with y0 = 1, we have Tny0 + R(ny0 ) ¿ mn;
so Ty0 + R(ny0 )=n ¿ m:
Letting n → ∞, we obtain T (y0 ) ¿ m. Thus the reduced minimum modulus of T is positive, so im(T ) is closed. Hence ∈ &+ (L; N ). (2) Follows by similar arguments to case (1). Lemma 3.3 enables us to prove a property for the semilinear spectrum (L; N ) when the nonlinear part N is asymptotically linear. Theorem 3.4. Let L : E → F be a Fredholm operator of index 0. Suppose N is an asymptotically linear operator and is a L-k-set contraction. Assume that N = l − R, where l is linear, R is compact and R(x)=x → 0 as x → ∞. If ∈ (L; N ) and || ¿ k, then ∈ m (L; N ). Proof. Since || ¿ k, !(L − N ) ¿ 0, so ∈ ! (L; N ). Assume that ∈ m (L; N ), then ∈ $(L; N ). By the proof of (1) of Lemma 3.3, ∈ &+ (L; N ) and L − N = T + R; where T = L − l : E → F is a linear operator, ker(T ) = {0} and im(T ) is closed. Let H = L + JP, then %$ + KPQ = H −1 . By Lemma 4.4 of [10], (I − P) − H −1 N = H −1 T + H −1 R: H −1 T : E → E is a linear isomorphism, so H −1 T is (0; )-epi for some 0 ¡ ¡ 1= H −1 T (see [18]). Let U = {x: H −1 Tx + tH −1 Rx = 0; t ∈ [0; 1]}: Then U is bounded. Otherwise there would exist x n ∈ E with x n → ∞, such that H −1 T (x n =x n ) = −tn H −1 (R(x n )=x n ) → 0
as n → ∞:
Thus T (x n =x n ) → 0 as n → ∞. This contradicts the fact that im(T ) is closed. Hence for r1 suKciently large, if x ∈ E with x = r1 , we have H −1 Tx + tH −1 Rx = 0:
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This ensures that H −1 T + H −1 R is (0; )-epi on Br1 . Since (H −1 T + H −1 R)−1 {0} = f−1 {0} = {0}; we have H −1 T + H −1 R is (0; )-epi on every ball Br with r ¿ 0. By the de:nition, ∈ (L; N ). This contradiction shows that ∈ m (L; N ). The following corollary on linear operators extends previous result that required L to be the identity map [19]. Corollary 3.5. Suppose that N is linear and is an L-k-set contraction. If ∈ (L; N ) and || ¿ k, then is an eigenvalue of (L; N ). Our next result deals with the case that N is a positively homogeneous operator. Theorem 3.6. Suppose that T is a positively homogeneous L-k-set contraction. ∈ (L; T ) with || ¿ k. Then there exists t ∈ (0; 1], such that =t is an eigenvalue of (L; (I − Q)T ). Proof. Assume that m(f ) = 0. Then there exist {x n } ∈ E, such that (I − P)un − (%$ + KPQ )Tun ¡ 1=n → 0;
(n → ∞):
Here, un = x n =x n . Since || ¿ k, !(f ) ¿ 0. So un (n = 1; 2; : : :) has a convergent subsequence. Assume that un → x0 , then f (x0 ) = 0. Hence Lx0 = Tx0 = QTx0 + (I − Q)Tx0 : This implies that Lx0 = (I − Q)Tx0 . Thus is an eigenvalue of (L; (I − Q)T ). In this case t = 1. Now suppose that m(f ) ¿ 0. Let S = {x ∈ E: x − tPx − t(%$ + KPQ )Tx = 0; t ∈ [0; 1]}: We have the following two cases: Case (1): There exist x n ∈ S with x n → ∞. Let yn = x n =x n , then (yn ) − tn P(yn ) − tn (%$ + KPQ )T (yn ) = 0; where tn ∈ [0; 1]. Suppose that tn → t0 ∈ [0; 1], then (yn ) − t0 P(yn ) − t0 (%$ + KPQ )T (yn ) =(tn − t0 )P(yn ) + (tn − t0 )(%$ + KPQ )T (yn ): So ({yn }∞ n=1 )!(I − t0 P − t0 (%$ + KPQ )T ) 6 yn − t0 Pyn − t0 (%$ + KPQ )Tyn = 0: ∞ Since || ¿ k, !(I − t0 P − t0 (%$ + KPQ )T ) ¿ 0. Thus ({yn }∞ n=1 ) = 0. So {yn }n=1 has a convergent subsequence. Suppose that yn → x0 . Then
x0 − t0 Px0 − t0 (%$ + KPQ )Tx0 = 0:
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Since x0 = 1, t0 = 0. (I − P)x0 − t0 KPQ Tx0 = t0 Px0 − Px0 + t0 %$Tx0 ∈ ker(L): But (I − P)x0 − t0 KPQ Tx0 ∈ E1 ∩ dom(L), we have (I − P)x0 = t0 L−1 P (I − Q)Tx0 : Thus (=t0 )Lx0 − (I − Q)Tx0 = 0, and =t0 is an eigenvalue of (L; (I − Q)T ). Case (2): Assume that there exists R ¿ 0 such that S ⊂ BR \@BR . Then for every x ∈ @BR we have x − tP(x) − t(%$ + KPQ )T (x) = 0;
t ∈ [0; 1]:
By Property 2.2, I − P − (%$ + KPQ )T is (0; )-epi for some ¿ 0 on BR . By our assumption, m(f ) ¿ 0. Thus f is (0; )-epi on every Br with r ¿ 0. Hence (f ) ¿ 0. This shows that f is regular and ∈ (L; T ), which contradicts ∈ (L; T ). The above arguments complete the proof. When L is the identity, the projection Q = 0 and the semilinear spectrum (L; N ) reduces to the nonlinear spectrum (N ) [7]. So we have the following corollary. Corollary 3.7. Suppose that T is a positively homogeneous k-set contraction. ∈ (T ) with || ¿ k. Then there exists t ∈ (0; 1], such that =t is an eigenvalue of T . The following corollary is obvious. Corollary 3.8. Suppose T is a positively homogeneous L-k-set contraction. If the eigenvalues of (L; (I − Q)T ) are bounded, then (L; T ) is bounded. Proposition 3.9. Suppose T =B+R, where B is a linear operator, either R(x)=x → 0 as x → ∞ or as x → 0. If is an eigenvalue of (L; B), then ∈ (L; T ). Proof. Let x0 ∈ E, x0 = 0 be such that Lx0 −Bx0 =0. If R(x)=x → 0 as x → ∞, then (L(nx0 ) − T (nx0 )=n) → 0. Thus m(L; T ) = 0, and ∈ (L; T ). If R(x)=x → 0 as x → 0, then L(x0 =n) − T (x0 =n) → 0; 1=n
(n → ∞):
Hence ∈ (L; T ). 4. Applications In this section, we give two examples of the applications of our theorems. Firstly, we prove a theorem that gives a condition for a compact, positive operator to have a positive eigenvalue and eigenvector.
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Theorem 4.1. Suppose E is a Banach space and K is a cone of E. For r ¿ 0, let Kr = K ∩ Br , @Kr = {x ∈ K: x = r}. (1) Assume that f : K → K is a positively homogeneous, compact operator and inf {x=1; x∈K} f(x) ¿ 0. Then there exist r ¿ 0 and x0 ∈ K; x0 = 1 such that f(x0 ) = rx0 . (2) Assume that F : Kr → K is compact and inf {F(x): x = r} ¿ 0. Then F has a positive eigenvalue with an eigenvector x0 ∈ @Kr . Proof. (1) Since K is a cone of E; K is a closed, convex subset of E. By the Dugundji extension theorem, there exists an extension f1 with f1 (E) ⊂ K and with f1 compact map [4]. Let F : E → K be de:ned by xf1 (x=x) if x = 0; F(x) = 0 if x = 0: Then F : E → K is a positively homogeneous and compact operator. Also, F|K = f. Suppose (F) ∩ R+ = ∅ and u ∈ K, u = 0. Since (1=n) ∈ (F), (1=n)I − F is onto, and there exist {x n }∞ n=1 ∈ E satisfying (1=n)x n − F(x n ) = u: F(x n ) ∈ K and u ∈ K imply that x n ∈ K. So F(x n ) = f(x n ). Thus (1=n)x n − f(x n ) = u:
(4.1)
Assume that {x n ∞ n=1 } is unbounded and x nk → ∞ (k → ∞). Let unk = x nk =x nk , Then, (1=n)unk − f(unk ) = u=x nk → 0
(k → ∞):
Thus f(unk ) → 0. This contradicts inf {x∈K; x=1} f(x) ¿ 0. So we obtain that {x n ∞ n=1 } is bounded. Next, (4.1) implies that 0 6 u 6 (1=n)x n → 0
(n → ∞):
Thus u = 0. This contradiction shows that (F) ∩ R+ = ∅. Let r1 ¿ 0 and r1 ∈ (F). By Corollary 3.7, there exists r ¿ r1 such that r is an eigenvalue of F. Let x0 ∈ E with x0 = 1 be such that F(x0 ) = rx0 . Then x0 ∈ K since r ¿ 0 and F(x0 ) ∈ K. Thus f(x0 ) = F(x0 ) = rx0 . (2) Let FE : K → K be de:ned as follows: xF((rx)=x) for x = 0; E F(x) = 0 for x = 0: FE is a compact and positively homogeneous map since F is a compact map [16]. E Furthermore, inf x=1 (F)(x) = inf x=r F(x) ¿ 0. So, by (1), there exists x ∈ K E ) = x . Thus F(rx =x ) = x . Let x0 = rx with x = 1 and ¿ 0 such that F(x then x0 ∈ K and x0 = r. Also F(x0 ) = (=r)x0 .
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Now, consider the second order diLerential equation u (t) = f(t; u(t); u (t));
t ∈ (0; 1)
(4.2)
with one of the boundary conditions u (0) = 0;
u(1) = u(0);
(4.3)
u(0) = 0;
u(1) = u(0);
(4.4)
here ∈ R and 0 ∈ (0; 1). To prove the existence of a solution for these BVPs, various growth conditions are needed [9]. Theorem 3.6 enables us to prove the existence of an eigenvalue without any growth condition. We will use the classical spaces C[0; 1]; C 1 [0; 1]; C 2 [0; 1], and L1 [0; 1]. The Sobolev space is de:ned by W 2; 1 (0; 1) = {u : [0; 1] → R : u; u are absolutely continuous on [0; 1] with u ∈ L1 [0; 1]} with its usual norm. ∈ R is said to be an eigenvalue of the BVP (4.2), (4.3) if there exists a nonzero u ∈ C[0; 1] satisfying the boundary condition (4.3) and u = f(t; u(t); u (t));
t ∈ (0; 1):
(4.5)
If for any y(t) ∈ L1 [0; 1], there exists u ∈ C[0; 1] satisfying the boundary condition (4.3) and u − f(t; u(t); u (t)) = y(t);
t ∈ (0; 1);
(4.6)
the BVP (4.2), (4.3) is said to be surjective. Theorem 4.2. Suppose that f : [0; 1] × R2 → R is a continuous map and f(t; cx; cp) = cf(t; x; p) for any c ¿ 0;
(x; p) ∈ R2 :
(4.7)
Then, for = 1, either the BVP (4.2), (4.3) is surjective or has a positive eigenvalue r ¿ 1. Similar result holds for the BVP (4.2), (4.4) provided that = 1=0. Proof. Let E = C[0; 1] and F = L1 [0; 1]. L : D(L) ⊂ E → F be de:ned by L(u) = u and D(L) = {u ∈ W 2; 1 (0; 1): u (0) = 0; u(1) = u(0)}: The nonlinear map N : E → F is de:ned by (Nu)(t) = f(t; u; u );
t ∈ [0; 1]:
It was proved that L is invertible and N is L-compact [9]. By our assumptions, N is positively homogeneous. Suppose that 1 ∈ (L; N ). Then the BVP (4.2), (4.3) is surjective. Otherwise, by Theorem 3.6, there exists t ∈ (0; 1] such that r = 1=t is an eigenvalue of (L; N ). Let u ∈ D(L) be the corresponding eigenvector, then u(t) satis:es Eqs. (4.3), (4.5). A similar method proves the result for the BVP (4.2), (4.4).
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Theorem 4.2 gives an alternative condition for Eq. (4.7) to have a positive eigenvalue or being surjective. The following examples show that, in both boundary conditions (4.3) and (4.4), there are equations that are not surjective so that an eigenvalue exists. Example 4.3. Let f(t; x; p) = a|x| where a ¿ 0 is a constant. 1. If 1=02 ¿ ¿ 1 and a ¿ 2( − 1)=(1 − 02 ), then BVP (4.2), (4.3) has a positive eigenvalue. 2. If 1=02 ¿ ¿ 1=0 and a ¿ 2(0 − 1)=(0(1 − 0)), then the BVP (4.2), (4.4) has a positive eigenvalue. Proof. We will prove that in both cases 1 and 2, the BVPs are not surjective. 1. Suppose the BVP (4.2), (4.3) is surjective. For any c ¿ 0, c ∈ L[0; 1], there exist uc ∈ W 2; 1 (0; 1) such that uc − a|uc | = c;
(4.8)
uc (0) = 0;
(4.9)
uc (1) = uc (0):
Since Lu = u with these boundary conditions is invertible, let K = L−1 . For any y ∈ L1 [0; 1], by calculation, 0 t (0 − s)y(s) ds + (t − s)y(s) ds Ky(t) = 1− 0 0 1 1 − (1 − s)y(s) ds: 1− 0 From Luc − a|uc | = c we have uc = aK|uc | + Kc:
(4.10)
Since uc ¿ 0 and uc (0)=0, uc is positive and uc is increasing. Moreover, ¿ 1 ensures that both u(0) and u(1) are positive. So, uc (t) ¿ 0 for t ∈ [0; 1]. From (4.10), we can obtain that Kc 1 = aK1 + ; for t ∈ [0; 1]; (4.11) |uc | here
Kc(t) =
1 − 02 + 2( − 1)
0
t
(t − s) ds c:
Since 1=02 ¿ ¿ 1, it can be veri:ed that Kc(t) is increasing. So, for t ∈ [0; 1], 0 (1 − 02 )c 1 − 02 (0 − s) ds c = Kc(t) ¿ + : 2( − 1) 2( − 1) 0 Since a ¿ 2( − 1)=(1 − 02 ), the right side of (4.11) gives Kc 1 − 02 1 − 02 aK1 + ¿a + c ¿ 1: |uc | 2( − 1) 2|uc (1)|( − 1)
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This contradiction shows that (4.2), (4.3) is not surjective. By Theorem 4.2, it has a positive eigenvalue. 2. Suppose the BVP (4.2), (4.4) is surjective. For any c ¿ 0, there exist uc ∈ W 2; 1 (0; 1) such that uc − a|uc | = c;
(4.12)
uc (0) = 0;
(4.13)
uc (1) = uc (0):
With the above boundary condition, the linear operator K is: 0 t t (0 − s)y(s) ds + (t − s)y(s) ds Ky(t) = 1 − 0 0 0 1 t − (1 − s)y(s) ds: 1 − 0 0 Again, by calculation, we have 2 t (0 − 1)t + (t − s) ds c: (4.14) Kc(t) = 2(1 − 0) 0 We show that uc (0) = 0. Otherwise, uc (0) = 0 and uc (1) = 0 ensure that there exist 81 ∈ (0; 0) and 82 ∈ (0; 1) such that uc (81 ) = uc (82 ) = 0. This contradicts uc (t) ¿ 0 for t ∈ (0; 1). Next, if uc (0) ¡ 0, then uc (1) = uc (0) ¡ 0. uc (t) ¿ 0 implies that uc (t) ¡ 0 for any t ∈ [0; 1]. As in the proof of 1, we have Kc − 1 = aK1 + ; for t ∈ [0; 1]; (4.15) |uc | Since 1=02 ¿ ¿ 1=0, by (4.14), we can obtain that for any t ∈ [0; 1], (K1)(t) ¿ 0 and (Kc)(t) ¿ 0. Again, we reach a contradiction. Lastly, suppose that uc (1) ¿ uc (0) ¿ 0. Then uc (t) ¿ 0 implies that uc (t) ¿ 0 for any t ∈ [0; 1]. As in the proof of 1, we have Kc 1 = aK1 + ; for t ∈ [0; 1]; (4.16) |uc | By (4.14), we can see that Kc(t) ¿ 0 and is increasing. Thus, 2 0 (0 − 1)0 (t − s) ds a(K1)(t) ¿ a + 2(1 − 0) 0 0(1 − 0) ¿ 1: 2(0 − 1) This contradiction shows that the BVP (4.2), (4.4) is not surjective and the proof is completed. =a
Acknowledgements The author thanks Prof. J.R.L. Webb for valuable comments and suggestions, especially for his help with Example 4.3.
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