Multiple Solutions for Boundary Value Problems of 𝑛th-Order Nonlinear Integrodifferential Equations in Banach Spaces

Abstract

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.

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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: −𝑢^(𝑛)(𝑡)=𝑓𝑡,𝑢(𝑡),𝑢^′(𝑡),…,𝑢^(𝑛−1)(𝑡),(𝑡),(𝑡𝑢𝑇𝑢)(𝑆𝑢)(),∀𝑡∈𝐽,_(𝑖)(0)(),𝑢=𝜃𝑖=0,1,…,𝑛−2_(𝑛−1)(𝑎)=𝜃, (1) where 𝐽∶=[0,𝑎](𝑎>0), 𝑓∈𝐶[𝐽×𝑃×𝑃×⋯×𝑃_𝑛+2,𝑃], 𝜃 denotes the zero element of 𝐸, and (𝑇𝑢)(𝑡)=^𝑡₀𝑘()𝑢(𝑠)(𝑡)=𝑡,𝑠𝑑𝑠,𝑆𝑢)(^𝑎₀ℎ(𝑡,𝑠)𝑢(𝑠)𝑑𝑠,∀𝑡∈𝐽, (2) 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 𝑘₀∶=max_{()𝑡,𝑠∈𝐷}𝑘(𝑡,𝑠),ℎ₀∶=max_{()𝑡,𝑠∈𝐽×𝐽}ℎ(𝑡,𝑠),𝜂∶=2max1,𝑎^𝑛. (3)

Denote that 𝐶^𝑛−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 𝐶^𝑛−1[𝐽,𝐸] is a Banach space with norm defined by ‖𝑢‖_𝑛−1∶=max_{𝑖=0,1,…,𝑛−1}𝑢^(𝑖)_𝑐,where𝑢^(𝑖)_𝑐∶=max_𝑡∈𝐽𝑢^(𝑖)(𝑡). (4) Let 𝐶^𝑛[𝐽,𝑃]∶=𝑢∈𝐶^𝑛[𝐽,𝐸]∶𝑢^(𝑖)(𝑡)(),𝑢≥𝜃𝑖=0,…,𝑛−1_(𝑛)(𝑡),𝐶≤𝜃_𝑛−1[𝐽,𝑃]∶=𝑢∈𝐶^𝑛−1[𝐽,𝐸]∶𝑢^(𝑖)(𝑡)≥𝜃(𝑖=0,…,𝑛−1). (5) It is obvious that 𝐶^𝑛[𝐽,𝑃] and 𝐶^𝑛−1[𝐽,𝑃] are two cones in 𝐶^𝑛[𝐽,𝐸] and 𝐶^𝑛−1[𝐽,𝐸], respectively.

Lemma 1. 𝑢∈𝐶^𝑛[𝐽,𝑃] is the solution of problem (1) if and only if 𝑢∈𝐶^𝑛−1[𝐽,𝑃] is the fixed point of operator 𝐴 defined by (𝑡)=1𝐴𝑢)((𝑛−1)!^𝑎₀𝑡^𝑛−1𝑓𝑠,𝑢(𝑠),𝑢^′(𝑠),…,𝑢^(𝑛−1)(𝑠),(𝑇𝑢)(𝑠),(𝑆𝑢)(𝑠)𝑑𝑠−^𝑡₀(𝑡−𝑠)^𝑛−1𝑓𝑠,𝑢(𝑠),𝑢^′(𝑠)𝑢,…,_(𝑛−1)(𝑠),(𝑠),(𝑠).𝑇𝑢)(𝑆𝑢)(𝑑𝑠 (6)

Proof. For 𝑢∈𝐶^𝑛[𝐽,𝐸], Taylor's formula with the integral remainder term gives 𝑢(𝑡)=^𝑛−1_𝑘=0(𝑡−𝑎)^𝑘𝑢𝑘!^(𝑘)(𝑎)−1(𝑛−1)!^𝑎_𝑡(𝑡−𝑠)^𝑛−1𝑢^(𝑛)(𝑠)𝑑𝑠,∀𝑡∈𝐽. (7) Taking 𝑎=0, we have 𝑢(𝑡)=^𝑛−1_𝑖=0𝑡^𝑖𝑢𝑖!^(𝑖)(0)+1(𝑛−1)!^𝑡₀(𝑡−𝑠)^𝑛−1𝑢^(𝑛)(𝑠)𝑑𝑠,∀𝑡∈𝐽. (8) Substituting 𝑢^(𝑛−1)(0)=𝑢^(𝑛−1)(𝑎)−^𝑎₀𝑢^(𝑛)(𝑠)𝑑𝑠 (9) into (8), we get 𝑢(𝑡)=^𝑛−2_𝑖=0(𝑡)^𝑖𝑢𝑖!^(𝑖)(0)+𝑡^(𝑛−1)()!𝑢𝑛−1^(𝑛−1)(𝑎)−1()!×𝑛−1_𝑎0𝑡^(𝑛−1)𝑢^(𝑛)(𝑠)−𝑑𝑠_𝑡0(𝑡−𝑠)^𝑛−1𝑢^(𝑛)(𝑠)𝑑𝑠,∀𝑡∈𝐽. (10) Let 𝑢∈𝐶^𝑛[𝐽,𝑃] be the solution of BVP (1). Then (10) implies 𝑢(𝑡)=1()!×𝑛−1_𝑎0𝑡^𝑛−1𝑓𝑠,𝑢(𝑠),𝑢^′(𝑠),…,𝑢^(𝑛−1)(𝑠),(𝑠),(𝑠)−𝑇𝑢)(𝑆𝑢)(𝑑𝑠_𝑡0(𝑡−𝑠)^𝑛−1𝑓𝑠,𝑢(𝑠),𝑢^′(𝑠),…,𝑢^(𝑛−1)(𝑠),(𝑇𝑢)(𝑠),(𝑆𝑢)(𝑠)𝑑𝑠. (11) Comparing this with (6), we have 𝑢(𝑡)=(𝐴𝑢)(𝑡), which means that 𝑢(𝑡) is the fixed point of the operator 𝐴 in 𝐶^𝑛−1[𝐽,𝑃].
On the other hand, let 𝑢(𝑡)∈𝐶^𝑛−1[𝐽,𝑃] be the fixed point of the operator 𝐴. By (6), 𝑢^(𝑗)(𝑡)=(𝐴𝑢)^(𝑗)(𝑡)=1!×𝑛−1−𝑗_𝑎0𝑡^{𝑛−1−𝑗}𝑓𝑠,𝑢(𝑠),𝑢^′(𝑠)𝑢,…,_(𝑛−1)(𝑠),(𝑠),(𝑠)−𝑇𝑢)(𝑆𝑢)(𝑑𝑠_𝑡0(𝑡−𝑠)^{𝑛−1−𝑗}𝑓𝑠,𝑢(𝑠),𝑢^′(𝑠),…,𝑢^(𝑛−1)(𝑠),(𝑇𝑢)(𝑠),(𝑆𝑢)(𝑠)𝑑𝑠, (12) where 𝑗=1,2,…,𝑛−1. It follows by taking 𝑡=0 and 𝑡=𝑎 in (12) that 𝑢^(𝑗)(0)=𝜃𝑗=0,1,…,𝑛−2,𝑢^(𝑛−1)(𝑎)𝑢=𝜃,_(𝑗)(𝑡)≥𝜃𝑗=0,1,…,𝑛−2,𝑡∈𝐽. (13) It is also clear from (12) that 𝑢^(𝑛−1)(𝑡)=^𝑎_𝑡𝑓𝑠,𝑢(𝑠),𝑢^′(𝑠)𝑢,…,_(𝑛−1)(𝑠),(𝑠),(𝑠)𝑢𝑇𝑢)(𝑆𝑢)(𝑑𝑠,𝑡∈𝐽,_(𝑛)(𝑡)=−𝑓𝑡,𝑢(𝑡),𝑢^′(𝑡),…,𝑢^(𝑛−1)(𝑡),(𝑇𝑢)(𝑡),(𝑆𝑢)(𝑡),𝑡∈𝐽. (14) Hence, 𝑢^(𝑛)(𝑡)≤𝜃. Then (13)–(14) imply that 𝑢 is the solution for BVP (1) in 𝐶^𝑛[𝐽,𝑃].

To continue, let us formulate some conditions.(𝐻₁) Let 𝑓(𝑡,𝑣₀,𝑣₁,…,𝑣_𝑛+1) be bounded and uniformly continuous in 𝑡 on 𝐽×𝐵_𝑟×𝐵_𝑟×⋯×𝐵_𝑟𝑛+2, ∀𝑟>0. There exist nonnegative constants 𝑐_𝑖(𝑖=0,1,…,𝑛+1) such that 𝜂𝑘^∗𝑛+1_{𝑖=0,𝑖≠𝑛−1}𝑐_𝑖+2𝑐_𝑛−1<1,𝛼𝑓𝐽,𝑉₀,𝑉₁,…,𝑉_𝑛−1,𝑉_𝑛,𝑉_𝑛+1≤^𝑛+1_𝑖=0𝑐_𝑖𝛼𝑉_𝑖,∀𝑉_𝑖⊂𝐵_𝑟, (15)(16) where 𝑘^∗∶=max{1,𝑘₀𝑎,ℎ₀𝑎}, 𝛼 denotes the Kuratowski measure of noncompactness, and 𝐵_𝑟={𝑢∈𝐸∶‖𝑢‖≤𝑟}.(𝐻₂) Assume that lim_𝑟→∞𝑀(𝑟)𝑟<𝜂^∗𝑘^∗,lim_𝑟→0+𝑀(𝑟)𝑟<𝜂^∗𝑘^∗, (17)(18) where 𝑀(𝑟)∶=sup𝑓𝑡,𝑣₀,𝑣₁,…,𝑣_𝑛−1,𝑣_𝑛,𝑣_𝑛+1∶𝑡,𝑣₀,𝑣₁,…,𝑣_𝑛−1,𝑣_𝑛,𝑣_𝑛+1∈𝐽×𝑃_𝑟×𝑃_𝑟×⋯×𝑃_𝑟×𝑃_𝑟×𝑃_𝑟, (19) 𝑃_𝑟∶={𝑢∈𝑃∶‖𝑢‖≤𝑟}, 𝜂^∗∶=𝜂⁻¹, and 𝑘^∗ is defined by (𝐻₁).(𝐻₃) There exist 𝑢^∗∈int(𝑃), 0<𝑡₀<𝑡₁<𝑎, and 𝐹(𝑡)∈𝐶[𝐽,𝑅₊] such that 𝑓𝑡,𝑣₀,𝑣₁,…,𝑣_𝑛−1,𝑣_𝑛,𝑣_𝑛+1≥𝐹(𝑡)𝑢^∗,_𝑎𝑡1𝐹(𝑠)1𝑑𝑠>max1,𝑡₀,2!𝑡²₀()!,…,𝑛−1𝑡₀^𝑛−1, (20) for 𝑣_𝑖≥𝑢^∗(𝑖=0,1,…,𝑛−1), 𝑣_𝑛≥𝜃, 𝑣_𝑛+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 𝐶^𝑚[𝐽,𝐸]. Then 𝛼_𝑚(𝐻)≥𝛼(𝐻(𝐽)),𝛼_𝑚(𝐻)≥𝛼𝐻^′(𝐽),…,𝛼_𝑚(𝐻)≥𝛼𝐻^(𝑚−1)(𝐽),𝛼_𝑚(𝐻)≥12𝛼𝐻^(𝑚)(𝐽), (21) 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 𝐶^𝑚[𝐽,𝐸]. Suppose that 𝐻^(𝑚)∶={𝑢^(𝑚)∶𝑢∈𝐻} is equicontinuous. Then 𝛼_𝑚(𝐻)=max_{𝑖=0,1,…,𝑚}𝛼𝐻^(𝑖)(𝐽)=max_{𝑖=0,1,…,𝑚}max_𝑡∈𝐽𝛼𝐻^(𝑖)(𝑡), (22) 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 𝐶^𝑛−1[𝐽,𝑃] into 𝐶^𝑛−1[𝐽,𝑃].

Proof. It is easy to see that 𝐴∶𝐶^𝑛−1[𝐽,𝑃]→𝐶^𝑛−1[𝐽,𝑃] and 𝐴 is a bounded operator by (6), (12), and (𝐻₁).
Now we check that operator 𝐴 is continuous from 𝐶^𝑛−1[𝐽,𝑃] into 𝐶^𝑛−1[𝐽,𝑃]. Let {𝑢_𝑚}^∞_𝑚=1⊂𝐶^𝑛−1[𝐽,𝑃], 𝑢∈𝐶^𝑛−1[𝐽,𝑃], and 𝑢_𝑚−𝑢_𝑛−1⟶0(𝑚⟶∞). (23) For any 𝑡∈𝐽, by (6), 𝐴𝑢_𝑚(𝑡)−(𝑡≤1𝐴𝑢)()()!×𝑛−1_𝑎0𝑡^𝑛−1𝑓𝑠,𝑢_𝑚(𝑠),𝑢^′_𝑚(𝑠)𝑢,…,_{𝑚(𝑛−2)}(𝑠),𝑇𝑢_𝑚(𝑠),𝑆𝑢_𝑚(𝑠)−𝑓𝑠,𝑢(𝑠),𝑢^′(𝑠),…,𝑢^(𝑛−2)(𝑠),(𝑇𝑢)(𝑠),(𝑆𝑢)(𝑠)𝑑𝑠+^𝑡₀(𝑡−𝑠)^𝑛−1𝑓𝑠,𝑢_𝑚(𝑠),𝑢^′_𝑚(𝑠),…,𝑢_𝑚^(𝑛−2)(𝑠),𝑇𝑢_𝑚(𝑠),𝑆𝑢_𝑚(𝑠)−𝑓𝑠,𝑢(𝑠),𝑢^′(𝑠),…,𝑢^(𝑛−1)(𝑠),(𝑇𝑢)(𝑠),(𝑆𝑢)(𝑠)𝑑𝑠. (24) Then the Lebesgue dominated convergence theorem gives max_𝑡∈𝐽𝐴𝑢_𝑚(𝑡)−(𝑡≤1𝐴𝑢)()()!×𝑛−1_𝑎0𝑎^𝑛−1𝑓𝑠,𝑢_𝑚(𝑠),𝑢^′_𝑚(𝑠)𝑢,…,_{𝑚(𝑛−2)}(𝑠),𝑇𝑢_𝑚(𝑠),𝑆𝑢_𝑚(𝑠)−𝑓𝑠,𝑢(𝑠),𝑢^′(𝑠),…,𝑢^(𝑛−2)(𝑠),(𝑇𝑢)(𝑠),(𝑆𝑢)(𝑠)𝑢^′_𝑚(𝑠)+𝑑𝑠_𝑎0(𝑎−𝑠)^𝑛−1𝑓𝑠,𝑢_𝑚(𝑠),𝑢^′_𝑚(𝑠),…,𝑢_𝑚^(𝑛−2)(𝑠),𝑇𝑢_𝑚(𝑠),𝑆𝑢_𝑚𝑠)(𝑠)(−𝑓𝑠,𝑢,𝑢^′(𝑠),…,𝑢^(𝑛−1)(𝑠),(𝑠),(𝑠)().𝑇𝑢)(𝑆𝑢)(𝑑𝑠⟶0,𝑚⟶∞ (25) Hence, 𝐴𝑢_𝑚−𝐴𝑢_𝑐=max_𝑡∈𝐽𝐴𝑢_𝑚(𝑡)−(𝐴𝑢)(𝑡)⟶0(𝑚⟶∞). (26) Similarly, in view of (12), we get 𝐴𝑢_𝑚^(𝑖)−(𝐴𝑢)^(𝑖)_𝑐();().⟶0,𝑚⟶∞𝑖=1,2,…,𝑛−1 (27) Then 𝐴𝑢_𝑚−𝐴𝑢_𝑛−1=max_{𝑖=0,1,…,𝑛−1}𝐴𝑢_𝑚^(𝑖)−(𝐴𝑢)^(𝑖)_𝑐⟶0,(𝑚⟶∞). (28) Consequently, the continuity of operator 𝐴 is proved.
Let 𝑄⊂𝐶^𝑛−1[𝐽,𝑃] be bounded. Then 𝐴(𝑄)⊂𝐶^𝑛[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), (𝐴(𝑢))^(𝑛−1)𝑡₁−(𝐴(𝑢))^(𝑛−1)𝑡₂≤^𝑡2_𝑡1𝑓𝑠,𝑢(𝑠),𝑢^′(𝑠),…,𝑢^(𝑛−1)(𝑠),(𝑇𝑢)(𝑠),(𝑆𝑢)(𝑠)𝑑𝑠. (29) According to the absolute continuity of Lebesgue integral, (𝐴(𝑄))^(𝑛−1) is equicontinuous on 𝐽. Therefore, Lemma 4 implies that 𝛼_𝑛−1(𝐴(𝑄))=max_{𝑖=0,1,…,𝑛−1}max_𝑡∈𝐽𝛼(𝐴(𝑄))^(𝑖)(𝑡), (30) 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 𝛼((𝐴𝑢)(𝑡))≤𝜂𝛼𝑓𝑠,𝑄(𝐽),𝑄^′(𝐽)𝑄,…,_(𝑛−1)(𝐽),(𝑇𝑄)(𝐽),(𝑆𝑄)(𝐽), (31) where 𝑄^(𝑖)(𝐽)={𝑢^(𝑖)(𝑠)∶𝑠∈𝐽,𝑢∈𝑄}(𝑖=0,1,…,𝑛−1), (𝐽)=𝑠)},(𝐽)=𝑠)}.𝑇𝑄)({(𝑇𝑢)(∶𝑠∈𝐽,𝑢∈𝑄𝑆𝑄)({(𝑆𝑢)(∶𝑠∈𝐽,𝑢∈𝑄 (32) It follows from (31) and (𝐻₁) that 𝛼((𝐴𝑢)(𝑡))≤𝜂^𝑛−1_𝑖=0𝑐_𝑖𝛼𝑄^(𝑖)(𝐽)+𝑐_𝑛𝑘₀𝑎𝛼(𝑄(𝐽))+𝑐_𝑛+1ℎ₀𝑎𝛼(𝑄(𝐽))≤𝜂𝑘^∗𝑛−2_𝑖=0𝑐_𝑖𝛼𝑄^(𝑖)(𝐽)+𝑐_𝑛−1𝛼𝑄^(𝑛−1)(𝐽)+𝑐_𝑛𝛼(𝑄(𝐽))+𝑐_𝑛+1𝛼(𝑄(𝐽)), (33) which implies, according to Lemma 3, that 𝛼((𝐴𝑢)(𝑡))≤𝜂𝑘^∗𝑛+1_{𝑖=0,𝑖≠𝑛−1}𝑐_𝑖+2𝑐_𝑛−1𝛼_𝑛−1(𝑄)=𝛾𝛼_𝑛−1(𝑄), (34) where 𝛾=𝜂𝑘^∗(∑^𝑛+1_{𝑖=0,𝑖≠𝑛−1}𝑐_𝑖+2𝑐_𝑛−1)<1 in view of (15).
Similarly, we have 𝛼(𝐴𝑢)^(𝑖)(𝑡)≤𝛾𝛼_𝑛−1(𝑄)(𝑖=1,2,…,𝑛−1). (35) Thus, we get 𝛼_𝑛−1(𝐴(𝑄))≤𝛾𝛼_𝑛−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 𝐶^𝑛[𝐽,𝑃].

Proof. Condition (𝐻₂) implies that there exist 𝜀^′>0 and 𝑟^′>0, such that, for 𝑟>𝑘^∗𝑟^′, 𝑀(𝑟)𝑟<𝜂^∗𝑘^∗+𝜀^′. (36) Choose 𝑟^∗>max{𝑟^′,2‖𝑢^∗‖}. Let 𝑈∶=𝑢∈𝐶^𝑛−1[𝐽,𝑃]∶‖𝑢‖_𝑛−1<𝑟^∗. (37) 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 (𝐴𝑢)^(𝑖)≤𝜂𝑀𝑘^∗𝑟^∗<𝜂𝜂^∗𝑘^∗𝑘^∗+𝜀^′𝑟^∗<𝑟^∗(𝑖=0,1,…,𝑛−1). (38) Hence, ‖𝐴𝑢‖_𝑛−1<𝑟^∗. Thus, we have shown that 𝐴𝑈⊂𝑈. (39) Similarly, by (18), it is easy to get that there is a number 𝑟₀ such that 0<𝑟₀<‖𝑢^∗‖/𝑁 and 𝐴𝑈₀⊂𝑈₀, (40) where 𝑈₀={𝑢∈𝐶^𝑛−1[𝐽,𝑃]∶‖𝑢‖_𝑛−1<𝑟₀}  and 𝑁 is the normal constant of 𝑃.
Let 𝑈₁∶=𝑢∈𝐶^𝑛−1[𝐽,𝑃]∶‖𝑢‖_𝑛−1<𝑟^∗,𝑢^(𝑖)(𝑡)≥𝜆𝑢^∗(𝑖=0,1,…,𝑛−1),𝑡∈𝑡₀,𝑡₁,𝜆>1dependingon𝑢. (41) It is easy to see that 𝑈, 𝑈₀, and 𝑈₁ are all nonempty bounded open convex sets of 𝐶^𝑛−1[𝐽,𝑃], and 𝑈_𝑖⊂𝑈(𝑖=0,1),𝑈₀∩𝑈₁=∅. (42) As the proof of (38), for 𝑢∈𝑈₁, by (𝐻₂), (𝐴𝑢)^(𝑖)<𝑟^∗,(𝑖=0,1,…,𝑛−1). (43) 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 (𝐴𝑢)^(𝑗−1)(𝑡)≥1𝑛−𝑗!^𝑎_𝑡𝑡^𝑛−𝑗𝐹(𝑠)𝑢^∗≥1𝑑𝑠𝑛−𝑗!^𝑎_𝑡1𝑡₀^𝑛−𝑗𝐹(𝑠)𝑑𝑠𝑢^∗,𝑗=1,2,…,𝑛. (44) Condition (𝐻₃) also implies that 1𝑛−𝑗!^𝑎_𝑡1𝑡₀^𝑛−𝑗𝐹(𝑠)𝑑𝑠>1𝑗=1,2,…,𝑛. (45) Consequently, in view of (43) and (45), we have shown that 𝐴𝑈₁⊂𝑈₁. (46) It follows from (39), (40), (42), (46), and Lemma 5 that 𝑖𝐴,𝑈,𝐶^𝑛−1[]𝑖𝐽,𝑃=1,𝐴,𝑈₀,𝐶^𝑛−1[]𝑖𝐽,𝑃=1,𝐴,𝑈₁,𝐶^𝑛−1[𝑖𝐽,𝑃]=1,𝐴,𝑈⧵𝑈₀𝑈₁,𝐶^𝑛−1[𝐽,𝑃]=𝑖𝐴,𝑈,𝐶^𝑛−1[𝐽,𝑃]−𝑖𝐴,𝑈₀,𝐶^𝑛−1[𝐽,𝑃]−𝑖𝐴,𝑈₁,𝐶^𝑛−1[𝐽,𝑃]=−1, (47) where 𝑖(⋅,⋅,⋅,) denotes the fixed-point index [7]. Therefore, 𝐴 has three fixed points 𝑢₀∈𝑈₀, 𝑢₁∈𝑈₁, and𝑢₃∈𝑈⧵(𝑈₀⋃𝑈₁). By Lemma 1, BVP (1) has at least three solutions in 𝐶^𝑛[𝐽,𝑃].

An application of Theorem 6 is as follows.

Example 7. Consider −𝑢_𝑛⁽⁴⁾(𝑡)=4𝑡𝑢_𝑛(𝑡)ln1+5𝑢_𝑛(𝑡)+6𝑢^′_𝑛+1(𝑡)+7𝑢^′′_𝑛−1(𝑡)+8𝑢_𝑛^′′′(𝑡)+^𝑡₀2𝑒^𝑡+3𝑡²𝑠⁻¹𝑢_𝑛+1(𝑠)𝑑𝑠+sin²𝑢_𝑛(𝑡)+2𝑢^′_𝑛−1(𝑡)+3𝑢_𝑛^′′(𝑡)+4𝑢^′′′_𝑛+1(𝑡)+1𝑢33^′𝑛(𝑡)^3/4ײ₀lg𝑡+𝑠2+1cos²(𝑡−𝑠)𝑢_𝑛+1(𝑠)𝑑𝑠^1/4,[],𝑢∀𝑡∈0,2_𝑛(𝑖)(0)(),𝑢=0𝑖=0,1,2_{𝑛′′′}(2)=0,(𝑛=1,2,…,𝑚), (48)
where 𝑢₀=𝑢_𝑚 and 𝑢_𝑚+1=𝑢₁.
Obviously, 𝑢_𝑛(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,…,𝑢_𝑚)}, 𝑚-dimensional space, with norm ‖𝑢‖∶=sup_{𝑛=1,2,…,𝑚}|𝑢_𝑛| and 𝑃=𝑢=𝑢₁,…,𝑢_𝑚∶𝑢_𝑛≥0,𝑛=1,2,…,𝑚. (49) Then 𝑃 is a normal and solid cone in 𝐸 and (48) can be regarded as a BVP of the form (1), where 𝑎=2,𝑘(𝑡,𝑠)=2𝑒^𝑡+3𝑡²𝑠⁻¹,ℎ(𝑡,𝑠)=lg𝑡+𝑠2+1cos²(),𝑢𝑡−𝑠𝑢=₁,…,𝑢_𝑚,𝑣=𝑣₁,…,𝑣_𝑚,𝑤=𝑤₁,…,𝑤_𝑚,𝑥=𝑥₁,…,𝑥_𝑚,𝑦=𝑦₁,…,𝑦_𝑚,𝑧=𝑧₁,…,𝑧_𝑚, (50) and 𝑓=(𝑓₁,…,𝑓_𝑚) with 𝑓_𝑛√𝑡,𝑢,𝑣,𝑤,𝑥,𝑦,𝑧=4𝑡𝑢_𝑛ln1+5𝑢_𝑛+6𝑣_𝑛+1+7𝑤_𝑛−1+8𝑥_𝑛+𝑦_𝑛+1+sin²𝑢_𝑛+2𝑣_𝑛−1+3𝑤_𝑛+4𝑥_𝑛+1+1𝑣33^𝑛3/4𝑧_𝑛+1^1/4(𝑛=1,2,…,𝑚). (51) Obviously, 𝑓∈𝐶[𝐽×𝑃×𝑃×⋯×𝑃₆,𝑃](𝐽=[0,2]) and (𝐻₁) is satisfied for 𝑐_𝑖=0(𝑖=0,1,…,5) since 𝐸 is finite-dimensional.
One can see that sin𝑢_𝑛+2𝑣_𝑛−1+3𝑤_𝑛+4𝑥_𝑛+1≤min1,𝑢_𝑛+2𝑣_𝑛−1+3𝑤_𝑛+4𝑥_𝑛+1. (52) Then (51) implies that 𝑓𝑡,𝑢,𝑣,𝑤,𝑥,𝑦,𝑧≤4𝑡‖𝑢‖‖𝑢‖‖𝑣‖‖𝑤‖‖𝑥‖+𝑦𝑢‖‖𝑣‖‖𝑤‖‖𝑥ln1+5+6+7+8+min1,(‖+2+3+4‖)²+1‖𝑣‖33^3/4‖𝑧‖^1/4,∀𝑡∈𝐽,𝑢,𝑣,𝑤,𝑥,𝑦,𝑧∈𝑃. (53) Therefore, 𝑀(𝑟)≤4√𝑟ln(1+26𝑟)+min1,100𝑟²+133𝑟. (54) Hence, lim_𝑟→∞𝑀(𝑟)𝑟<1,32lim_𝑟→0+𝑀(𝑟)𝑟<1.32 (55) On the other hand, it is easy to see that 𝜂=32,𝜂^∗=132,𝑘^∗=1. (56) 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, 𝑢_𝑛≥1, 𝑣_𝑛≥1, 𝑤_𝑛≥1, 𝑥_𝑛≥1, 𝑦_𝑛≥0, 𝑧_𝑛≥0, 𝑛=1,2,…,𝑚). Then (51) implies that 𝑓_𝑛√𝑡,𝑢,𝑣,𝑤,𝑥,𝑦,𝑧≥4𝑡𝑢_𝑛ln1+5𝑢_𝑛+6𝑣_𝑛+1+7𝑤_𝑛−1+8𝑥_𝑛≥4ln27, (57) where 𝑛=1,2,…,𝑚. So, we have ∫²_3/24ln27𝑑𝑠>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.