The boundary value
problems of a class of nth-order nonlinear integrodifferential equations of mixed type in Banach space are considered, and the
existence of three solutions is obtained by using the fixed-point
index theory.
Guo [1] considered the initial value problems of a class of integrodifferential equations of Volterra type and obtained the existence of maximal and minimal solutions by establishing a comparison result. In [2], the author and Qin investigated a first-order impulsive singular integrodifferential equation on the half line in a Banach space and proved the existence of two positive solutions by means of the fixed-point theorem of cone expansion and compression with norm type. For other results related to integrodifferential equations in Banach spaces please see also [3–6] and the references therein. It is worth pointing out that the nonlinear terms involved in the equations they considered are either sublinear or superlinear globally.
In this paper, by using fixed-point index theory (for details please see [7]), we consider the nth-order integrodifferential equations with nonlinear terms neither sublinear nor superlinear globally and prove the existence of three solutions.
Let E be a real Banach space and P a cone in E which defines a partial ordering in E by x≤y if and only if y-x∈P. P is said to be normal if there exists a positive constant N such that θ≤x≤y implies ∥x∥≤N∥y∥, where θ denotes the zero element of E and the smallest N is called the normal constant of P. If x≤y and x≠y, we write x<y. P is said to be solid if its interior is not empty; that is, int(P)≠ϕ. In case of y-x∈int(P), we write x≪y. For details on cone theory, please see [8].
We consider the following boundary value problem (BVP for short) in E:
(1)-u(n)(t)=f(t,u(t),u′(t),…,u(n-1)(t),(Tu)(t),(Su)(t)u(n-1)u′),∀t∈J,u(i)(0)=θ(i=0,1,…,n-2),u(n-1)(a)=θ,
where J:=[0,a](a>0), f∈C[J×P×P×⋯×P︸n+2,P], θ denotes the zero element of E, and
(2)(Tu)(t)=∫0tk(t,s)u(s)ds,(Su)(t)=∫0ah(t,s)u(s)ds,∀t∈J,
with k∈C[D,R+], h∈C[J×J,R+], D:={(t,s)∈J×J:t≥s}, and R+ the set of all nonnegative numbers. Let
(3)k0:=max(t,s)∈Dk(t,s),h0:=max(t,s)∈J×Jh(t,s),η:=2max{1,an}.
Denote that Cn-1[J,E]:={u:u is a map from J into E and u(n-1)(t) is continuous on J}. It is clear that Cn-1[J,E] is a Banach space with norm defined by
(4)∥u∥n-1:=maxi=0,1,…,n-1∥u(i)∥c,where∥u(i)∥c:=maxt∈J∥u(i)(t)∥.
Let
(5)Cn[J,P]≔{u∈Cn[J,E]:u(i)(t)≥θ(i=0,…,n-1),u(n)(t)≤θ},Cn-1[J,P]≔{u∈Cn-1[J,E]:u(i)(t)≥θ(i=0,…,n-1)[J,E]:u(i)}.
It is obvious that Cn[J,P] and Cn-1[J,P] are two cones in Cn[J,E] and Cn-1[J,E], respectively.
Lemma 1.
u∈Cn[J,P] is the solution of problem (1) if and only if u∈Cn-1[J,P] is the fixed point of operator A defined by
(6)(Au)(t)=1(n-1)![∫0atn-1f(s,u(s),u′(s),…,u(n-1)(s),(Tu)(s),(Su)(s)u′)ds-∫0t(t-s)n-1f((s)s,u(s),u′(s),…,u(n-1)(s),(Tu)(s),(Su)(s)u′)ds∫0a].
Proof.
For u∈Cn[J,E], Taylor's formula with the integral remainder term gives
(7)u(t)=∑k=0n-1(t-a)kk!u(k)(a)-1(n-1)!∫ta(t-s)n-1u(n)(s)ds,∀t∈J.
Taking a=0, we have
(8)u(t)=∑i=0n-1tii!u(i)(0)+1(n-1)!∫0t(t-s)n-1u(n)(s)ds,∀t∈J.
Substituting
(9)u(n-1)(0)=u(n-1)(a)-∫0au(n)(s)ds
into (8), we get
(10)u(t)=∑i=0n-2(t)ii!u(i)(0)+t(n-1)(n-1)!u(n-1)(a)-1(n-1)!×(∫0at(n-1)u(n)(s)ds-∫0t(t-s)n-1u(n)(s)ds),∀t∈J.
Let u∈Cn[J,P] be the solution of BVP (1). Then (10) implies
(11)u(t)=1(n-1)!×[∫0atn-1f(u(n-1)(s)s,u(s),u′(s),…,u(n-1)(s),(Tu)(s),(Su)(s)u′(s))ds-∫0t(t-s)n-1f(u(n-1)s,u(s),u′(s),…,u(n-1)(s),(Tu)(s),(Su)(s)u′(s))ds∫0a].
Comparing this with (6), we have u(t)=(Au)(t), which means that u(t) is the fixed point of the operator A in Cn-1[J,P].
On the other hand, let u(t)∈Cn-1[J,P] be the fixed point of the operator A. By (6),
(12)u(j)(t)=(Au)(j)(t)=1(n-1-j)!×[∫0atn-1-jf(s,u(s),u′(s),…,u(n-1)(s),(Tu)(s),(Su)(s)u′)ds-∫0t(t-s)n-1-jf(u(n-1)s,u(s),u′(s),…,u(n-1)(s),(Tu)(s),(Su)(s)u′)ds∫0a],
where j=1,2,…,n-1. It follows by taking t=0 and t=a in (12) that
(13)u(j)(0)=θ(j=0,1,…,n-2),u(n-1)(a)=θ,u(j)(t)≥θ(j=0,1,…,n-2),t∈J.
It is also clear from (12) that
(14)u(n-1)(t)=∫taf(u(n-1)s,u(s),u′(s),…,u(n-1)(s),(Tu)(s),(Su)(s)u′(s))ds,t∈J,u(n)(t)=-f(t,u(t),u′(t),…,u(n-1)(t),(Tu)(t),(Su)(t)u′(t)),t∈J.
Hence, u(n)(t)≤θ. Then (13)–(14) imply that u is the solution for BVP (1) in Cn[J,P].
To continue, let us formulate some conditions.
Let f(t,v0,v1,…,vn+1) be bounded and uniformly continuous in t on J×Br×Br×⋯×Br︸n+2, ∀r>0. There exist nonnegative constants ci(i=0,1,…,n+1)
such that
(15)ηk*(∑i=0,i≠n-1n+1ci+2cn-1)<1,(16)α(f(J,V0,V1,…,Vn-1,Vn,Vn+1))≤∑i=0n+1ciα(Vi),∀Vi⊂Br,
where k*:=max{1,k0a,h0a}, α denotes the Kuratowski measure of noncompactness, and Br={u∈E:∥u∥≤r}.
Assume that
(17)limr→∞¯M(r)r<η*k*,(18)limr→0+¯M(r)r<η*k*,
where
(19)M(r)≔sup{∥f(t,v0,v1,…,vn-1,vn,vn+1)∥:(t,v0,v1,…,vn-1,vn,vn+1)∈J×Pr×Pr×⋯×Pr×Pr×Pr},
Pr:={u∈P:∥u∥≤r}, η*:=η-1, and k* is defined by (H1).
There exist u*∈int(P), 0<t0<t1<a, and F(t)∈C[J,R+] such that
(20)f(t,v0,v1,…,vn-1,vn,vn+1)≥F(t)u*,∫t1aF(s)ds>max{1,1t0,2!t02,…,(n-1)!t0n-1},
for vi≥u*(i=0,1,…,n-1), vn≥θ, vn+1≥θ, and t∈[t0,t1].
Remark 2.
By (H2) and (H3), one can see that f is neither sublinear nor superlinear globally.
Lemma 3 (see [8]).
Let H be a bounded set of Cm[J,E]. Then
(21)αm(H)≥α(H(J)),αm(H)≥α(H′(J)),…,αm(H)≥α(H(m-1)(J)),αm(H)≥12α(H(m)(J)),
where H(i)(J):={u(i)(t):t∈J,u∈H}(i=0,1,2,…,m).
Lemma 4 (see [8]).
Let H be a bounded set of Cm[J,E]. Suppose that H(m):={u(m):u∈H} is equicontinuous. Then
(22)αm(H)=maxi=0,1,…,m{α(H(i)(J))}=maxi=0,1,…,m{maxt∈J{α(H(i)(t))}},
where H(i)(J)(i=0,1,2,…,m) is defined by Lemma 3 and H(i)(t):={u(i)(t):u∈H}(i=0,1,2,…,m).
Lemma 5.
Let (H1) hold. Then operator A defined by (6) is a strict set contraction from Cn-1[J,P] into Cn-1[J,P].
Proof.
It is easy to see that A:Cn-1[J,P]→Cn-1[J,P] and A is a bounded operator by (6), (12), and (H1).
Now we check that operator A is continuous from Cn-1[J,P] into Cn-1[J,P]. Let {um}m=1∞⊂Cn-1[J,P], u∈Cn-1[J,P], and
(23)∥um-u∥n-1⟶0(m⟶∞).
For any t∈J, by (6),
(24)∥(Aum)(t)-(Au)(t)∥≤1(n-1)!×[∫0atn-1∥f(um(n-2)s,um(s),um′(s),…,um(n-2)(s),(Tum)(s),(Sum)(s)um′(s))-f(s,u(s),u′(s),…,u(n-2)(s),(Tu)(s),(Su)(s)u′(s)u(n-2))∥ds+∫0t(t-s)n-1∥f(s,um(s),um′(s),…,um(n-2)(s),(Tum)(s),(Sum)(s)u′(s)um(n-2))-f(s,u(s),u′(s),…,u(n-1)(s),(Tu)(s),(Su)(s)u(n-1)u′(s))∥f(s,um(s),um′(s),…,um(n-2)(s),∥ds∫0a].
Then the Lebesgue dominated convergence theorem gives
(25)maxt∈J∥(Aum)(t)-(Au)(t)∥≤1(n-1)!×[∫0aan-1∥f(um(n-2)um(n-2)s,um(s),um′(s),…,um(n-2)(s),(Tum)(s),(Sum)(s)um′(s))-f(s,u(s),u′(s),…,u(n-2)(s),(Tu)(s),(Su)(s)u(n-2)um′(s))∥ds+∫0a(a-s)n-1∥f(um(n-2)s,um(s),um′(s),…,um(n-2)(s),(Tum)(s),(Sum)(s)um(n-2)um′(s))-f(s,u(s),u′(s),…,u(n-1)(s),(Tu)(s),(Su)(s)u(n-1)um′(s))∥ds∫0a]⟶0,(m⟶∞).
Hence,
(26)∥Aum-Au∥c=maxt∈J∥(Aum)(t)-(Au)(t)∥⟶0(m⟶∞).
Similarly, in view of (12), we get
(27)∥(Aum)(i)-(Au)(i)∥c⟶0,(m⟶∞);(i=1,2,…,n-1).
Then
(28)∥Aum-Au∥n-1=maxi=0,1,…,n-1∥(Aum)(i)-(Au)(i)∥c⟶0,(m⟶∞).
Consequently, the continuity of operator A is proved.
Let Q⊂Cn-1[J,P] be bounded. Then A(Q)⊂Cn[J,P] is bounded. We prove that (A(Q))(n-1) is equicontinuous on J. In fact, ∀(A(u))(n-1)∈(A(Q))(n-1), by (12),
(29)∥(A(u))(n-1)(t1)-(A(u))(n-1)(t2)∥≤∫t1t2∥f(s,u(s),u′(s),…,u(n-1)(s),(Tu)(s),(Su)(s)u(n-1)u′(s))∥ds.
According to the absolute continuity of Lebesgue integral, (A(Q))(n-1) is equicontinuous on J. Therefore, Lemma 4 implies that
(30)αn-1(A(Q))=maxi=0,1,…,n-1{maxt∈J{α((A(Q))(i)(t))}},
where α((A(Q))(i)(t))=α({(Au)(i)(t):u∈Q})(t is fixed, i=0,1,…,n-1). By (6), we see that
(31)α((Au)(t))≤ηα(Q(n-1)(J)),(TQ)(J),(SQ)(J)f(Q(n-1)(J)s,Q(J),Q′(J),…,Q(n-1)(J)Q′(J)),(TQ)(J),(SQ)(J)Q′(J)),
where Q(i)(J)={u(i)(s):s∈J,u∈Q}(i=0,1,…,n-1),
(32)(TQ)(J)={(Tu)(s):s∈J,u∈Q},(SQ)(J)={(Su)(s):s∈J,u∈Q}.
It follows from (31) and (H1) that
(33)α((Au)(t))≤η(∑i=0n-1ciα(Q(i)(J))+cnk0aα(Q(J))+cn+1h0aα(Q(J)))≤ηk*(∑i=0n-2ciα(Q(i)(J))+cn-1α(Q(n-1)(J))+cnα(Q(J))+cn+1α(Q(J))∑i=0n-2),
which implies, according to Lemma 3, that
(34)α((Au)(t))≤ηk*(∑i=0,i≠n-1n+1ci+2cn-1)αn-1(Q)=γαn-1(Q),
where γ=ηk*(∑i=0,i≠n-1n+1ci+2cn-1)<1 in view of (15).
Similarly, we have
(35)α((Au)(i)(t))≤γαn-1(Q)(i=1,2,…,n-1).
Thus, we get αn-1(A(Q))≤γαn-1(Q) by (34) and (35). Noticing that A is bounded and continuous, the conclusion follows.
Theorem 6.
Let P be a normal solid cone and let (H1), (H2), and (H3) hold. Then BVP (1) has at least three solutions in Cn[J,P].
Proof.
Condition (H2) implies that there exist ɛ′>0 and r′>0, such that, for r>k*r′,
(36)M(r)r<η*k*+ɛ′.
Choose r*>max{r′,2∥u*∥}. Let
(37)U:={u∈Cn-1[J,P]:∥u∥n-1<r*}.
For u∈U¯, we have ∥u(i)∥≤r*(i=0,1,…,n-1), ∥Tu∥≤k*r*, and ∥Su∥≤k*r*. So, it follows from (6), (12), and (36) that
(38)∥(Au)(i)∥≤ηM(k*r*)<ηη*k*k*+ɛ′r*<r*hhhhhhhhhhhhhhhhhh(i=0,1,…,n-1).
Hence, ∥Au∥n-1<r*. Thus, we have shown that
(39)A(U¯)⊂U.
Similarly, by (18), it is easy to get that there is a number r0 such that 0<r0<∥u*∥/N and
(40)A(U¯0)⊂U0,
where U0={u∈Cn-1[J,P]:∥u∥n-1<r0} and N is the normal constant of P.
Let
(41)U1≔{u∈Cn-1[J,P]:∥u∥n-1<r*,u(i)(t)≥λu*(i=0,1,…,n-1),t∈[t0,t1],λ>1dependingonuCn-1}.
It is easy to see that U, U0, and U1 are all nonempty bounded open convex sets of Cn-1[J,P], and
(42)Ui⊂U(i=0,1),U0∩U1=∅.
As the proof of (38), for u∈U¯1, by (H2),
(43)∥(Au)(i)∥<r*,(i=0,1,…,n-1).
On the other hand, according to (H3), for t∈[t0,t1], u(i)(t)≥u*(i=0,1,…,n-1), (Tu)(t)≥θ, and (Su)(t)≥θ, we get by (12) that
(44)(Au)(j-1)(t)≥1(n-j)!∫tatn-jF(s)u*ds≥1(n-j)!∫t1at0n-jF(s)dsu*,(j=1,2,…,n).
Condition (H3) also implies that
(45)1(n-j)!∫t1at0n-jF(s)ds>1(j=1,2,…,n).
Consequently, in view of (43) and (45), we have shown that
(46)A(U¯1)⊂U1.
It follows from (39), (40), (42), (46), and Lemma 5 that
(47)i(A,U,Cn-1[J,P])=1,i(A,U0,Cn-1[J,P])=1,i(A,U1,Cn-1[J,P])=1,i(A,U∖(U¯0⋃U¯1),Cn-1[J,P])=i(A,U,Cn-1[J,P])-i(A,U0,Cn-1[J,P])-i(A,U1,Cn-1[J,P])=-1,
where i(·,·,·,) denotes the fixed-point index [7]. Therefore, A has three fixed points u¯0∈U0, u¯1∈U1, andu¯3∈U∖(U¯0⋃U¯1). By Lemma 1, BVP (1) has at least three solutions in Cn[J,P].
Obviously, un(t)≡0 (n=1,2,…,m) is the trivial solution of BVP (48).
Conclusion. BVP (48) has at least two nontrivial nonnegative C4 solutions.
Proof.
Let E:={u=(u1,…,um)}, m-dimensional space, with norm ∥u∥:=supn=1,2,…,m|un| and
(49)P={u=(u1,…,um):un≥0,n=1,2,…,m}.
Then P is a normal and solid cone in E and (48) can be regarded as a BVP of the form (1), where
(50)a=2,k(t,s)=(2et+3t2s)-1,h(t,s)=lg(t+s2+1)cos2(t-s),u=(u1,…,um),v=(v1,…,vm),w=(w1,…,wm),x=(x1,…,xm),y=(y1,…,ym),z=(z1,…,zm),
and f=(f1,…,fm) with
(51)fn(t,u,v,w,x,y,z)=4tunln(1+5un+6vn+1+7wn-1+8xn+yn+1)+sin2(un+2vn-1+3wn+4xn+1)+133(vn)3/4(zn+1)1/4(n=1,2,…,m).
Obviously, f∈C[J×P×P×⋯×P︸6,P](J=[0,2]) and (H1) is satisfied for ci=0(i=0,1,…,5) since E is finite-dimensional.
One can see that
(52)|sin(un+2vn-1+3wn+4xn+1)|≤min{1,|un|+2|vn-1|+3|wn|+4|xn+1|}.
Then (51) implies that
(53)∥f(t,u,v,w,x,y,z)∥≤4t∥u∥ln(1+5∥u∥+6∥v∥+7∥w∥+8∥x∥+∥y∥)+min{1,(∥u∥+2∥v∥+3∥w∥+4∥x∥)2}+133∥v∥3/4∥z∥1/4,∀t∈J,u,v,w,x,y,z∈P.
Therefore,
(54)M(r)≤4rln(1+26r)+min{1,100r2}+133r.
Hence,
(55)limr→∞¯M(r)r<132,limr→0+¯M(r)r<132.
On the other hand, it is easy to see that
(56)η=32,η*=132,k*=1.
Thus, (55) and (56) imply that (H2) is satisfied.
Now, we check (H3). Let u*=(1,…,1), F(t)=4ln27 and t0=1, t1=3/2. Obviously, u*∈int(P) and, for t∈[t0,t1],u≥u*, v≥u*, w≥u*, x≥u*, y≥θ, and z≥θ (i.e., 1≤t≤3/2, un≥1, vn≥1, wn≥1, xn≥1, yn≥0, zn≥0, n=1,2,…,m). Then (51) implies that
(57)fn(t,u,v,w,x,y,z)≥4tunln(1+5un+6vn+1+7wn-1+8xn)≥4ln27,
where n=1,2,…,m. So, we have ∫3/224ln27ds>6. Hence, (H3) is satisfied. And, finally, the conclusion follows from Theorem 6.
Acknowledgment
The author is grateful to Professor Guo Dajun and two anonymous referees for their valuable suggestions and comments.
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