Positive Solution of Fourth-Order Integral Boundary Value Problem with Two Parameters

Abstract

The author investigates the fourth-order integral boundary value problem with two parameters 𝑢⁽⁴⁾(𝑡)+𝛽𝑢^′′(𝑡)−𝛼𝑢(𝑡)=𝑓(𝑡,𝑢),𝑡∈(0,1),𝑢(0)=𝑢(1)=0,𝑢^′′(0)=∫¹₀𝑢(𝑠)𝜙₁(𝑠)𝑑𝑠,𝑢^′′∫(1)=₁₀𝑢(𝑠)𝜙₂(𝑠)𝑑𝑠, where nonlinear term function 𝑓 is allowed to change sign. Applying the fixed point index theorem on cone together with the operator spectrum theorem, some results on the existence of positive solution are obtained.

1. Introduction

The theory of boundary value problems with integral boundary conditions for ordinary differential equations arises in different areas of applied mathematics and physics. For example, heat conduction, chemical engineering, underground water flow, thermoelasticity, and plasma physics can all be reduced to nonlocal problems with integral boundary conditions (see, e.g., [1–3]). For boundary value problems with integral boundary conditions and comments on their importance, we refer the reader to the papers by Gallardo [4], Karakostas and Tsamatos [5], and Lomtatidze and Malaguti [6] and the references therein. For more information about the general theory of integral equations and their relation to boundary value problems, we refer to the books of Corduneanu [7] and Agarwal and O'Regan [8].

Moreover, boundary value problems with integral boundary conditions constitute a very interesting and important class of problems. They include two, three, multipoints and nonlocal boundary value problems as special cases. The existence and multiplicity of positive solutions for such problems have received a great deal of attention. To identify a few, we refer the reader to [9–15] and the references therein.

In the recent literature, several sorts of boundary value problems with integral boundary conditions have been studied further, see [16–20]. Especially, Ruyun Ma and Yulian An [18] investigated the global structure of positive solutions for nonlocal boundary value problems𝑢^′′(𝑡)(𝑡)𝑓(𝑢(𝑡𝑢(0)(1)=+𝜆ℎ))=0,0<𝑡<1,=0,𝑢¹₀𝑢(𝑠)𝑑𝐴(𝑠), (1.1) by using global bifurcation techniques, where 𝑓∈𝐶([0,∞),[0,∞)),ℎ∈𝐶((0,1),[0,∞)). In [19], Jiqiang Jiang et al. investigated the existence of positive solution for second-order singular Sturm-Liouville integral boundary value problems−𝑢^′′(𝑡)(𝑡)𝑓((𝑡(0)=𝜆ℎ𝑡,𝑢)),0<𝑡<1,𝛼𝑢−𝛽𝑢^′(0)=¹₀𝑎(𝑠)𝑢(𝑠)(1)𝑑𝑠,𝛾𝑢+𝛿𝑢^′(1)=¹₀𝑏(𝑠)𝑢(𝑠)𝑑𝑠, (1.2) by using the fixed point theory in cones, where 𝑓∈𝐶([0,1]×(0,∞),[0,∞)).

On the other hand, the fourth-order boundary value problem describe the deformations of an elastic beam in equilibrium state. Owing to its importance in physics, the existence of solutions to this problem has been studied by many authors; see, for example, [21–24] and references therein. Especially, in [22], Li studied existence of positive solution for fourth-order boundary value problem𝑢⁽⁴⁾(𝑡)+𝛽𝑢^′′(𝑡)(𝑡)()(),𝑢(0)(1)−𝛼𝑢=𝑓𝑡,𝑢,𝑡∈0,1=𝑢=0=𝑢^′′(0)==𝑢^′′(1)=0, (1.3) by using the fixed point index theorem, where 𝑓∈𝐶([0,∞),[0,∞)).

Motivated by the above-mentioned works [18, 19, 22], in this paper, we study the following fourth-order integral boundary value problem (for short BVP in the sequel) with two parameters:𝑢⁽⁴⁾(𝑡)+𝛽𝑢^′′(𝑡)(𝑡)()(),𝑢(0)(1)−𝛼𝑢=𝑓𝑡,𝑢,𝑡∈0,1=𝑢=0,𝑢^′′(0)=¹₀𝑢(𝑠)𝜙₁(𝑠)𝑑𝑠,𝑢^′′(1)=¹₀𝑢(𝑠)𝜙₂(𝑠)𝑑𝑠, (1.4) where nonlinear term function 𝑓 is allowed to change sign. To the best of our knowledge, BVP has not been investigated up to now. In the literature such as above-mentioned paper [18, 19, 22], the nonnegativity on 𝑓 is a usual assumption. In the present paper, since the function 𝑓 is not assumed to be nonnegative, the corresponding integral operator doesn't map the cone into cone, and so, there exists difficulty in applying the cone fixed point theorem. On the other hand, owing to the occurrence of parameter 𝛼,𝛽 in this boundary value problem including integral boundary conditions, it is not easy to transform the BVP (1.4) into an integral equation directly. To overcome these difficulties, we first introduce operator spectrum method combined with some analysis technique, next apply the fixed point index theorem, and establish existence of positive solution to BVP (1.4).

Let us begin with listing the following assumption conditions, which will be used in the sequel:

Let 𝐼=[0,1],𝐑=(−∞,+∞),𝐑₋=(−∞,0],𝐑₊=[0,+∞).(H1)𝑓∈𝐶[𝐼×𝐑₊,𝐑]andexists𝑀∈𝐿¹(0,1)∩𝐶[(0,1),𝐑₊]suchthat𝑓(𝑡,𝑢)+𝑀(𝑡)≥0,(𝑡,𝑢)∈(0,1)×𝐑₊. (1.5)(H2)𝛼,𝛽∈𝐑,𝛽<2𝜋²,𝛼≥−𝛽²/4,𝛼/𝜋⁴+𝛽/𝜋²<1.Let 𝜆₁,𝜆₂ be the roots of the polynomial 𝑝(𝜆)=𝜆²+𝛽𝜆−𝛼; namely,𝜆₁,𝜆₂=−𝛽±𝛽²+4𝛼2. (1.6) By (H2), it is to see that 𝜆₁≥𝜆₂>−𝜋².

Let Γ₀=𝜋⁴−𝛽𝜋²−𝛼. Then (H2) implies Γ₀>0. Let 𝑋=𝐶[0,1] be the real Banach space equipped with the norm ||𝑢||=max_0≤𝑡≤1|𝑢(𝑡)|. Denote by 𝑃 the set 𝑃={𝑢∈𝑋∶𝑢(t)≥0,𝑡∈𝐼} in 𝑋.

2. Preliminaries

In this section, we shall give some important preliminary lemmas, which will be used in proving of our main results.

Lemma 2.1 (see [22, 23]). Suppose that (H2) holds, then there exist unique 𝜑_𝑖,𝜓_𝑖,𝑖=1,2 satisfying −𝜑_𝑖^′′(𝑡)+𝜆_𝑖𝜑_𝑖(𝑡)[],𝜑=0,𝑡∈0,1_𝑖(0)=0,𝜑_𝑖(1)=1,−𝜓_𝑖^′′(𝑡)+𝜆_𝑖𝜓_𝑖(𝑡)[],𝜓=0,𝑡∈0,1_𝑖(0)=1,𝜓_𝑖(1)=0, (2.1) respectively, and 𝜑_𝑖≥0,𝜓_𝑖≥0 on [0,1], where 𝜆_𝑖 is as in (1.6). Moreover, 𝜑_𝑖,𝜓_𝑖 have the expression 𝜑_𝑖(𝑡)=sinh𝜔_𝑖𝑡sinh𝜔_𝑖,𝜆_𝑖>0,𝑡,𝜆_𝑖=0,sin𝜔_𝑖𝑡sin𝜔_𝑖,−𝜋²<𝜆_𝑖𝜓<0,^𝑖(𝑡)=sinh𝜔_𝑖(1−𝑡)sinh𝜔_𝑖,𝜆_𝑖>0,1−𝑡,𝜆_𝑖=0,sin𝜔_𝑖(1−𝑡)sin𝜔_𝑖,−𝜋²<𝜆_𝑖<0, (2.2) where 𝜔_𝑖=|𝜆_𝑖|,𝑖=1,2.

Let 𝐺_𝑖(𝑡,𝑠)(𝑖=1,2) be the Green function of the linear boundary value problem−𝑢^′′(𝑡)+𝜆_𝑖𝑢(𝑡)=0,𝑡∈[0,1],𝑢(0)=𝑢(1)=0. (2.3) By [22, 23], 𝐺_𝑖(𝑡,𝑠) can be expressed by the formula𝐺_𝑖()=1𝑡,𝑠𝜎_𝑖𝜑_𝑖(𝑡)𝜓_𝑖(𝑠)𝜓,0≤𝑡≤𝑠≤1,_𝑖(𝑡)𝜑_𝑖(𝑠),0≤𝑠≤𝑡≤1, (2.4) where 𝜎_𝑖=𝜔_𝑖sinh𝜔_𝑖,if𝜆_𝑖>0,1,if𝜆_𝑖𝜔=0,𝑖=1,2._𝑖sin𝜔_𝑖,if−𝜋²<𝜆_𝑖<0, (2.5)

Lemma 2.2 (see [22, 23]). 𝐺_𝑖=𝐺_𝑖(𝑡,𝑠)(𝑖=1,2) have the following properties: (i)𝐺_𝑖(𝑡,𝑠)>0,∀𝑡,𝑠∈(0,1). (ii)𝐺_𝑖(𝑡,𝑠)≤𝐶_𝑖𝐺_𝑖(𝑠,𝑠),∀𝑡,𝑠∈[0,1],𝜑_𝑖≤𝐶_𝑖,𝜓_𝑖≤𝐶_𝑖,𝑡∈[0,1]. (iii)𝐺_𝑖(𝑡,𝑠)≥𝛿_𝑖𝐺_𝑖(𝑡,𝑡)𝐺_𝑖(𝑠,𝑠),∀𝑡,𝑠∈[0,1],𝜑_𝑖(𝑡)≥𝛿_𝑖𝐺_𝑖(𝑡,𝑡),𝜓_𝑖(𝑡)≥𝛿_𝑖𝐺_𝑖(𝑡,𝑡),𝑡∈[0,1], where𝐶_𝑖=1,if𝜆_𝑖1≥0,sin𝜔_𝑖,if−𝜋²<𝜆_𝑖𝛿<0,^𝑖=𝜔_𝑖sinh𝜔_𝑖,if𝜆_𝑖>0,1,if𝜆_𝑖𝜔=0,_𝑖sin𝜔_𝑖,if−𝜋²<𝜆_𝑖<0. (2.6)

Put 𝐷_𝑖=max_𝑡∈𝐼∫¹₀𝐺_𝑖(𝑡,𝑠)𝑑𝑠,𝑖=1,2. Set 𝐸₂₁=𝐷₂𝐶₁,𝐸₁₂=𝐷₁𝐶₂, where 𝐶_𝑖 is described as before. We need also the following assumptions in the sequel.

(H3) Functions 𝜙_𝑖∈𝐶[𝐼,𝐑₋],𝑖=1,2, satisfy 𝐷=̇𝐸₁₂∫¹₀|𝜙₁(𝑠)|𝑑𝑠+𝐸₂₁∫¹₀|𝜙₂(𝑠)|𝑑𝑠<1.

Let ℎ∈𝐶(0,1)∩𝐿¹(0,1), consider the following BVP:𝑢⁽⁴⁾(𝑡)+𝛽𝑢^′′(𝑡)(𝑡)(𝑡)(),𝑢(0)(1)−𝛼𝑢=ℎ,𝑡∈0,1=𝑢=𝑢^′′(0)=𝑢^′′(1)=0. (2.7) By papers [22, 23], BVP (2.7) has a unique solution 𝑢=𝐾ℎ expressed by𝐾ℎ(𝑡)=¹₀¹₀𝐺₁(𝑡,𝑠)𝐺₂()ℎ(𝜏)=𝑠,𝜏𝑑𝜏𝑑𝑠₁₀₁₀𝐺₂(𝑡,𝑠)𝐺₁(𝑠,𝜏)ℎ(𝜏)𝑑𝜏𝑑𝑠,𝑡∈[0,1]. (2.8)

Let 𝑤=𝐾𝑀. Since 𝑀∈𝐿¹(0,1)∩𝐶[(0,1),𝑅₊], by Lemma 2.2, it is easy to verify that 𝑤∈𝑃.

Let𝑔₁(𝑡)=−¹₀𝐺₂(𝑡,𝑠)𝜑₁(𝑠)𝑑𝑠,𝑡∈[0,1], (2.9) where 𝜑₁ is as in (2. 1). By Lemmas 2.1 and 2.2, we have 𝑔₁∈𝐶²([0,1],𝑅₋) and−𝑔₁^′′(𝑡)+𝜆₂𝑔₁(𝑡)=−𝜑₁(𝑡)[],𝑔,𝑡∈0,1₁(0)=𝑔₁(1)=0. (2.10) On the other hand, 𝜑₁ satisfies the following relation:−𝜑₁^′′(𝑡)+𝜆₁𝜑₁(𝑡)[],𝜑=0,𝑡∈0,1₁(0)=0,𝜑₁(1)=1. (2.11) So, from (2.10)–(2.11), it follows that𝑔₁^′′(0)=𝜆₂𝑔₁(0)+𝜑₁(0)𝑔=0,_1′′(1)=𝜆₂𝑔₁(1)+𝜑₁(1)=1. (2.12) Now, we make the following decomposition:𝑔₁⁽⁴⁾+𝛽𝑔₁^′′−𝛼𝑔₁=−𝑑²𝑑𝑡²+𝜆₁−𝑑²𝑑𝑡²+𝜆₂𝑔₁=−𝑑²𝑑𝑡²+𝜆₁−𝑔₁^′′+𝜆₂𝑔₁=𝑑²𝜑₁𝑑𝑡²−𝜆₁𝑑𝜑₁𝑑𝑡=0. (2.13) So by (2.10), (2.12)-(2.13), it follows that𝑔₁⁽⁴⁾(𝑡)+𝛽𝑔₁^′′(𝑡)−𝛼𝑔₁(𝑡)[],𝑔=0,𝑡∈0,1₁(0)=𝑔₁(1)=0,𝑔₁^′′(0)=0,𝑔₁^′′(1)𝑔=1,₁(𝑡)≤0,𝑡∈[0,1]. (2.14) Similarly, by setting𝑔₂(𝑡)=−¹₀𝐺₁(𝑡,𝑠)𝜓₂(𝑠)𝑑𝑠,𝑡∈[0,1], (2.15) we have𝑔₂⁽⁴⁾(𝑡)+𝛽𝑔₂^′′(𝑡)−𝛼𝑔₂(𝑡)[],𝑔=0,𝑡∈0,1₂(0)=𝑔₂(1)=0,𝑔₂^′′(0)=1,𝑔₂^′′(1)𝑔=0,₂(𝑡)≤0,𝑡∈[0,1]. (2.16)

For any 𝑢∈𝑋, define 𝑢^∗ as𝑢^∗(𝑡)=𝑢(𝑡)(𝑡)(𝑡),if𝑢≥0,0,if𝑢<0. (2.17) Obviously, 𝑢^∗∈𝑃 for any 𝑢∈𝑋.

Let ℎ∈𝐿¹(0,1)∩𝐶(0,1); consider the BVP with integral boundary conditions𝑢⁽⁴⁾(𝑡)+𝛽𝑢^′′(𝑡)(𝑡)(𝑡)(),𝑢(0)(1)𝑢−𝛼𝑢=ℎ,𝑡∈0,1=𝑢=0,_′′(0)=¹₀[𝑢−𝑤]^∗(𝑠)𝜙₁(𝑠)𝑑𝑠,𝑢^′′(1)=¹₀[𝑢−𝑤]^∗(𝑠)𝜙₂(𝑠)𝑑𝑠. (2.18)

Denote operator B on 𝐶[0,1] by𝐵𝑢(𝑡)=𝑔₂(𝑡)¹₀[𝑢−𝑤]^∗(𝑠)𝜙₁(𝑠)𝑑𝑠+𝑔₁(𝑡)¹₀[𝑢−𝑤]^∗(𝑠)𝜙₂(𝑠)𝑑𝑠. (2.19) It is easy to see that 𝐵 maps 𝐶[0,1] into 𝐶[0,1].

Define operator L: 𝐶⁴(0,1)→𝐶(0,1) as follows:𝐿𝑢=𝑢⁽⁴⁾+𝛽𝑢^′′−𝛼𝑢. (2.20)

We need the following Lemma.

Lemma 2.3. Let (H2) holds. Assume that ℎ∈𝐿¹(0,1)∩𝐶(0,1) and 𝜙_𝑖∈𝐶[𝐈,𝐑₋],𝑖=1,2. Then 𝑢∈𝐶⁴(0,1)∩𝐶²[0,1] is a solution of (2.18) if and only if 𝑢 is a solution of operator equation 𝑢=𝐾ℎ+𝐵𝑢 in 𝐶[0,1].

Proof. (1) Assume 𝑢∈𝐶⁴(0,1)∩𝐶²[0,1] is a solution of (2.18). By (2.14)–(2.20), we have (𝐵𝑢)(0)=(𝐵𝑢)(1)=0,(𝐵𝑢)^′′(0)=¹₀[𝑢−𝑤]^∗(𝑠)𝜙₁(𝑠)(𝐵𝑑𝑠,𝑢)^′′(1)=¹₀[𝑢−𝑤]^∗(𝑠)𝜙₂(𝑠)𝐿(𝐵𝑑𝑠,𝑢)=𝐿𝑔₂¹₀[𝑢−𝑤]^∗(𝑠)𝜙₁(𝑠)𝑑𝑠+𝐿𝑔₁¹₀[𝑢−𝑤]^∗(𝑠)𝜙₂(𝑠)𝑑𝑠=0. (2.21) Let 𝑣=𝑢−𝐵𝑢. Then 𝐿𝑣(𝑡)=𝐿𝑢(𝑡)−𝐿𝐵𝑢(𝑡)=𝐿𝑢(𝑡)=ℎ(𝑡),𝑡∈(0,1);𝑣(0)=𝑢(0)−(𝐵𝑢)(0)=0,𝑣(1)=𝑢(1)−(𝐵𝑢)(1)=0;𝑣^′′(0)=𝑢^′′(0)−(𝐵𝑢)^′′(0)=0,𝑣^′′(1)=𝑢^′′(1)−(𝐵𝑢)^′′(1)=0. Thus, by (2.7)-(2.8), we have 𝑣=𝐾ℎ,𝑣∈𝐶[0,1], and so 𝑢=𝐾ℎ+𝐵𝑢,𝑢∈𝐶[0,1].
(2) Inversely, assume 𝑢∈𝐶[0,1] satisfies 𝑢=𝐾ℎ+𝐵𝑢. Then 𝑢∈𝐶⁴(0,1)∩𝐶²[0,1]. By (2.7), (2.8),(2.14)–(2.20), we have 𝐿𝐾ℎ=ℎ,𝐿𝐵𝑢=0,(𝐾ℎ)(0)=(𝐾ℎ)(1)=(𝐾ℎ)^′′(0)=(𝐾ℎ)^′′(1)(𝐵=0,𝑢)(0)=(𝐵𝑢)(1)=0,(𝐵𝑢)^′′(0)=¹₀[𝑢−𝑤]^∗(𝑠)𝜙₁(𝑠)(𝐵𝑑𝑠,𝑢)^′′(1)=¹₀[𝑢−𝑤]^∗(𝑠)𝜙₂(𝑠)𝑑𝑠. (2.22) Consequently, 𝐿𝑢=𝐿𝐾ℎ+𝐿𝐵𝑢=ℎ,𝑢(0)=(𝐾ℎ)(0)+(𝐵𝑢)(0)=0,𝑢(1)=(𝐾ℎ)(1)+(𝐵𝑢)(1)=0,𝑢^′′(0)=(𝐾ℎ)^′′(0)+(𝐵𝑢)^′′(0)=¹₀[𝑢−𝑤]^∗(𝑠)𝜙₁(𝑠)𝑑𝑠,𝑢^′′(1)=(𝐾ℎ)^′′(1)+(𝐵𝑢)^′′(1)=¹₀[𝑢−𝑤]^∗(𝑠)𝜙₂(𝑠)𝑑𝑠. (2.23) Hence, 𝑢 is a solution of (2.18). The proof is complete.
We have also the following lemma.

Lemma 2.4. Suppose (H3) holds. Then 𝐵∶𝑋→𝑋 is a bounded operator with ||𝐵||≤𝐷(<1) and 𝐵𝑋⊂𝑃.

Proof. In view of Lemma 2.2 (ii), by (2.9),(2.15),(2.19) and (H3), noticing that 𝑤∈𝑃, for any 𝑢∈𝑋 and 𝑡∈𝐼, we have |(𝐵𝑢)(𝑡)|≤𝑔₂(𝑡)¹₀[𝑢−𝑤]^∗𝜙₁(𝑠)𝑑𝑠+𝑔₁(𝑡)¹₀[𝑢−𝑤]^∗𝜙₂(𝑠)𝑑𝑠≤𝐸₁₂¹₀|𝑢(𝑠)|𝜙₁(𝑠)𝑑𝑠+𝐸₂₁¹₀|𝑢(𝑠)|𝜙₂(𝑠‖𝑢‖.)𝑑𝑠≤𝐷 (2.24) Thus, ||𝐵𝑢||≤𝐷||𝑢||, and so ||𝐵||≤𝐷(<1).
On the other hand, from 𝑔_𝑖(𝑡)≤0,𝜙_𝑖(𝑡)≤0,𝑡∈𝐼,𝑖=1,2, we have 𝐵𝑋⊂𝑃. So, Lemma 2.4 is true.
By (2.7)-(2.8), it follows from 𝑤=𝐾𝑀 that 𝑤⁽⁴⁾(𝑡)+𝛽𝑤^′′(𝑡)(𝑡)(𝑡)(),𝑤(0)(1)−𝛼𝑤=𝑀,𝑡∈0,1=𝑤=𝑤^′′(0)=𝑤^′′(1)=0. (2.25)
For any 𝑢∈𝑋, let 𝐟𝑢(𝑡)=𝑓(𝑡,[𝑢−𝑤]^∗(𝑡)),𝑡∈[0,1] and 𝐺𝑢(𝑡)=𝐟𝑢(𝑡)+𝑀(𝑡),𝑡∈(0,1). Under conditions (H1)–(H3), consider the following auxiliary BVP: 𝑢⁽⁴⁾(𝑡)+𝛽𝑢^′′(𝑡)(𝑡)(𝑡)(),𝑢(0)(1)𝑢−𝛼𝑢=𝐺𝑢,𝑡∈0,1=𝑢=0,_′′(0)=¹₀[𝑢−𝑤]^∗(𝑠)𝜙₁(𝑠)𝑑𝑠,𝑢^′′(1)=¹₀[𝑢−𝑤]^∗(𝑠)𝜙₂(𝑠)𝑑𝑠. (2.26) Notice that 𝑤(𝑡) satisfies (2.25), it is easy to see that 𝑢∈𝐶⁴(0,1)∩𝐶²[0,1] is a solution of (2.26) if and only if 𝑢−𝑤∈𝐶⁴(0,1)∩𝐶²[0,1] is a solution of the following BVP: 𝑢⁽⁴⁾(𝑡)+𝛽𝑢^′′(𝑡)−𝛼𝑢(𝑡)=𝑓𝑡,𝑢^∗(𝑡)(),𝑢(0)(1),𝑡∈0,1=𝑢=0,𝑢^′′(0)=¹₀𝑢^∗(𝑠)𝜙₁(𝑠)𝑑𝑠,𝑢^′′(1)=¹₀𝑢^∗(𝑠)𝜙₂(𝑠)𝑑𝑠. (2.27) Thus, if and only if 𝑢(𝑡)≥𝑤(𝑡),𝑡∈[0,1], then 𝑢−𝑤 is a solution of BVP (1.4).
Now, by Lemma 2.3, 𝑢∈𝐶⁴(0,1)∩𝐶²[0,1] is a solution of (2.26) if 𝑢∈𝑋 is a fixed point of the operator 𝐾𝐺+𝐵. So, we only need focusing our attention on the existence of the fixed point of 𝐾𝐺+𝐵.
For the remainder of this section, we give the definition of positive solution.
By a positive solution of BVP (1.4), we mean a function 𝑢∈𝐶⁴(0,1)∩𝐶²[0,1] such that 𝑢(𝑡)≥0,𝑡∈[0,1],  𝑢(𝑡)>0,𝑡∈(0,1), and 𝑢 satisfies (1.4).

3. Main Results

We introduce now some notations, which will be used in the sequel.

Let 𝐶₁, 𝛿₁, and 𝐷 be as described in Lemma 2.2 and (H3), respectively. We also set𝑑₁=¹₀¹₀𝐺₂(𝑠,𝜏)𝑀(𝜏)𝑑𝜏𝑑𝑠,𝑏₀=𝐶²₁𝑑₁𝛿₁(1−𝐷),𝑓₀=lim_𝑢→+0inf_𝑡∈𝐼𝑓(𝑡,𝑢)𝑢,𝑓^∞=lim_𝑢→+∞max_𝑡∈𝐼𝑓(𝑡,𝑢)𝑢. (3.1) We also need the following assumption.

(H4) There exists a number 𝑟₀∈(𝑏₀,+∞), and Γ₁≥𝑟₀/Γ₀(𝑟₀−𝑏₀) such that𝑓(𝑡,𝑢)+𝑀(𝑡)≥Γ₁𝑢,∀(𝑡,𝑢)∈(0,1)×0,𝑟₀. (3.2)

We are now in a position to state and prove our main results on the existence.

Theorem 3.1. Suppose that (H1)–(H4) hold. If 𝑓^∞=0, then BVP (1.4) has a positive solution.

Proof. By Lemma 2.4 together with (H3), we have ||𝐵||≤𝐷(<1). By operator spectrum theorem, we know that (𝐼−𝐵)⁻¹ exists and is bounded. Furthermore, by Neumann expression, (𝐼−𝐵)⁻¹can be expressed by (𝐼−𝐵)⁻¹=^∞_𝑛=0𝐵^𝑛. (3.3) Noticing that 𝐵𝑃⊂𝑃 and from (3.3), we have (𝐼−𝐵)⁻¹𝑢=^∞_𝑛=0𝐵^𝑛𝑢≥𝑢,∀𝑢∈𝑃, (3.4)(𝐼−𝐵)⁻¹≤^∞_𝑛=0‖𝐵‖^𝑛=1‖𝐵‖≤11−.1−𝐷 (3.5) Thus, from the reversibility of 𝐼−𝐵, we have 𝑢=𝐾G𝑢+𝐵𝑢,𝑢∈𝑋⟺𝑢=(𝐼−𝐵)⁻¹𝐾𝐺𝑢,𝑢∈𝑋. (3.6) The following proof will be divided into five steps.Step 1. We will show that (𝐼−𝐵)⁻¹𝐾𝐺∶𝑃→𝑃 is completely continuous.(1)𝐾𝐺 maps 𝑃 into 𝑃.For any 𝑢∈𝑃, it follows from (H1) that 𝑓(𝑡,[𝑢−𝑤]^∗(𝑡))∈𝑋, and so (𝐺𝑢)(𝑡)≥0,𝑡∈(0,1),𝐺𝑢∈𝐶(0,1)∩𝐿¹(0,1). By (H1)-(H2) together with Lemma 2.2, for any 𝑡∈[0,1], we have 0≤(𝐾𝐺𝑢)(𝑡)=¹₀¹₀𝐺₁(𝑡,𝑠)𝐺₂(𝑠,𝜏)𝑓𝜏,[𝑢−𝑤]^∗(𝜏)+𝑀(𝜏)𝑑𝜏𝑑𝑠≤𝜌𝜂+𝜌¹₀𝑀(𝜏)𝑑𝜏<+∞, (3.7) where 𝜂=max_𝑡∈[0,1]|𝑓(𝑡,[𝑢−𝑤]^∗(𝑡))|<∞,𝜌=max_{𝑡,𝑠,𝜏∈[0,1]}𝐺₁(𝑡,𝑠)𝐺₂(𝑠,𝜏)<∞.From the continuity of 𝐺₁(𝑡,𝑠), it is easy to see that 𝐾𝐺𝑢∈𝑋, and so 𝐾𝐺𝑢∈𝑃.(2)𝐾𝐺 is a compact operator on 𝑃.Assume that 𝑈 is a arbitrary bounded set in 𝑃. Then there exists a 𝐿₀>0 such that ‖𝑢‖≤𝐿₀ for all 𝑢∈𝑈. Also, we have ‖[𝑢−𝑤]^∗‖≤𝐿₀ for all 𝑢∈𝑈 since 𝑤∈𝑃. Consequently, 0≤(𝐾𝐺𝑢)(𝑡)≤𝜌𝑏+𝜌¹₀𝑀(𝜏)𝑑𝜏<+∞,∀𝑤∈𝑃, (3.8) where 𝑏=max_{(𝑡,𝑢)∈𝐼×[0,𝐿0]}|𝑓(𝑡,𝑢)|,𝜌=max_{𝑡,𝑠,𝜏∈[0,1]}𝐺₁(𝑡,𝑠)𝐺₂(𝑠,𝜏). That means {𝐾𝐺𝑢∣𝑢∈𝑈} is a uniformly bounded set in 𝑃.On the other hand, the continuity of 𝐺₁ on 𝐼×𝐼 yields that for every 𝜀>0, there exists 𝛿>0 such that for any 𝑡₁,𝑡₂∈𝐼 with |𝑡₁−𝑡₂|<𝛿, the following inequality𝐺₁𝑡₂,𝑠−𝐺₁𝑡₁,𝑠<𝜀 (3.9) holds for all 𝑠∈𝐼, and so, (𝐾𝐺𝑢)𝑡₂−(𝐾𝐺𝑢)𝑡₁≤𝜀¹₀¹₀𝐺₂(𝑠,𝜏)(𝐺𝑢)(𝜏)𝑑𝜏𝑑𝑠≤𝜀𝑏𝑒₂+𝑒₂¹₀𝑀(𝜏)𝑑𝜏, (3.10) for any 𝑢∈𝑈, where 𝑒₂=max_{𝑠,𝜏∈[0,1]}𝐺₂(𝑠,𝜏)<+∞,𝑏=max_{(𝑡,𝑢)∈𝐼×[0,𝐿0]}|𝑓(𝑡,𝑢)|<+∞. That is, {𝐾𝐺𝑢∣𝑢∈𝑈}is equicontinuous.
Hence, in view of Arzela-Ascoli theorem, we know that the operator 𝐾𝐺 is compact on 𝑃.(3)Now, we show that the operator 𝐾𝐺 is continuous.Indeed, for any sequence {𝑢_𝑛} in 𝑃 with 𝑢_𝑛→𝑢 and any 𝑡∈𝐼, we have 𝑢_𝑛−𝑤^∗(𝑡)−[𝑢−𝑤]^∗(𝑡=1)2𝑢_𝑛(𝑡)−𝑤(𝑡)+𝑢_𝑛(𝑡)(𝑡−𝑢(𝑡)(𝑡(𝑡)(𝑡=1−𝑤)[|−𝑤)|+𝑢−𝑤)]2𝑢_𝑛(𝑡)−𝑤(𝑡)−|𝑢(𝑡)−𝑤(𝑡)|+𝑢_𝑛(𝑡)(𝑡≤1−𝑢)2𝑢_𝑛(𝑡)−𝑤(𝑡)−|𝑢(𝑡)−𝑤(𝑡)|+𝑢_𝑛(𝑡)(𝑡≤𝑢−𝑢)_𝑛(𝑡)−𝑢(𝑡). (3.11) Thus, ‖[𝑢_𝑛−𝑤]^∗−[𝑢−𝑤]^∗‖→0, and, by Lemma 2.2, it follows from the continuity of 𝑓 that 𝐾𝐺𝑢_𝑛−𝐾𝐺𝑢≤𝐶₁¹₀¹₀𝐺₁(𝑠,𝑠)𝐺₂(𝑠,𝜏)𝑓𝜏,𝑢_𝑛−𝑤^∗(𝜏)−𝑓𝜏,[𝑢−𝑤]^∗(𝜏)𝑑𝜏𝑑𝑠⟶0. (3.12) By (1)–(3) we obtain that 𝐾𝐺∶𝑃→𝑃 is completely continuous.
Now, from (3.4), we have (𝐼−𝐵)⁻¹∶𝑃→𝑃 is continuous, and so, (𝐼−𝐵)⁻¹𝐾𝐺∶𝑃→𝑃 is completely continuous.
Now we set 𝑄=(𝐼−𝐵)⁻¹𝐾𝐺,𝑞₁(𝑡)=𝛿₁𝐶₁𝐺₁(𝑡,𝑡),𝑡∈𝐼, (3.13) where 𝛿₁,𝐶₁ are described in Lemma 2.2. Set 𝑃₀=𝑢∈𝑃∶𝑢(𝑡)≥(1−‖𝐵‖)𝑞₁(𝑡)‖𝑢‖,𝑡∈𝐼. (3.14) Obviously, 𝑃₀ is a cone in 𝑋.Step 2. 𝑄∶𝑃→𝑃₀.
In fact, for any 𝑢∈𝑃 and every 𝑡,𝜎 in 𝐼, by Lemma 2.2, we have (𝐾𝐺𝑢)(𝑡)=¹₀¹₀𝐺₁(𝑡,𝑠)𝐺₂(𝑠,𝜏)(𝐺𝑢)(𝜏)𝑑𝜏𝑑𝑠≥𝛿₁𝐺₁(𝑡,𝑡)¹₀¹₀𝐺₁(𝑠,𝑠)𝐺₂(𝜏)≥𝛿𝑠,𝜏)(𝐺𝑢)(𝑑𝜏𝑑𝑠₁𝐶₁𝐺₁(𝑡,𝑡)¹₀¹₀𝐺₁(𝜎,𝑠)𝐺₂(𝑠,𝜏)(𝐺𝑢)(𝜏)𝑑𝜏𝑑𝑠=𝑞₁(𝑡)(𝐾𝐺𝑢)(𝜎). (3.15) Thus, we have (𝐾𝐺𝑢)(𝑡)≥𝑞₁(𝑡)‖𝐾𝐺𝑢‖,𝑡∈𝐼. (3.16) Since 𝐾𝐺𝑃⊂𝑃, by (3.4) together with (3.16) for every 𝑡∈𝐼, we have (𝐼−𝐵)⁻¹𝐾𝐺𝑢(𝑡)≥(𝐾𝐺𝑢)(𝑡)≥𝑞₁(𝑡)‖𝐾𝐺𝑢‖. (3.17) On the other hand, since ‖(𝐼−𝐵)⁻¹(𝐾𝐺𝑢)‖≤‖(𝐼−𝐵)⁻¹‖⋅‖𝐾𝐺𝑢‖, by (3.5), we have ‖‖≥1𝐾𝐺𝑢(𝐼−𝐵)⁻¹(𝐼−𝐵)⁻¹≥(‖𝐵)𝐾𝐺𝑢1−‖)(𝐼−𝐵⁻¹𝐾𝐺𝑢. (3.18) Inequality (3.17) together with (3.18) implies for every 𝑡∈𝐼(𝐼−𝐵)⁻¹𝐾𝐺𝑢(𝑡)≥(1−‖𝐵‖)𝑞₁(𝑡)(𝐼−𝐵)⁻¹𝐾𝐺𝑢, (3.19) namely, (𝑄𝑢)(𝑡)≥(1−‖𝐵‖)𝑞₁(𝑡)‖𝑄𝑢‖,𝑡∈𝐼. Thus, we obtain that 𝑄 maps 𝑃 into 𝑃₀.Step 3. We shall deduce that for any 𝑢∈𝑃₀ and 𝑡∈𝐼, the following inequality holds: 𝑢(𝑡)(𝑡)≥𝑏−𝑤1−⁰‖𝑢‖𝑢(𝑡), (3.20) where 𝑏₀=𝐶²₁𝑑₁/𝛿₁(1−𝐷).
In fact, in view of Lemma 2.2 and the symmetry of 𝐺₁(𝑡,𝑠), we have 𝐺₁(𝑡,𝑠)=𝐺₁(𝑠,𝑡)≤𝐶₁𝐺₁(𝑡,𝑡),∀𝑡,𝑠∈𝐼. (3.21) Thus, keeping in mind that 𝑑₁=∫¹₀∫¹₀𝐺₂(𝑠,𝜏)𝑀(𝜏)𝑑𝜏𝑑𝑠, it follows from 𝑤=𝐾𝑀 that 𝑤(𝑡)=¹₀¹₀𝐺₁(𝑡,𝑠)𝐺₂(𝑠,𝜏)𝑀(𝜏)𝑑𝜏𝑑𝑠≤𝐶₁𝐺₁(𝑡,𝑡)¹₀¹₀𝐺₂()𝑀(𝜏)=𝐶𝑠,𝜏𝑑𝜏𝑑𝑠₂₁𝛿₁𝑑₁𝑞₁(𝑡),𝑡∈𝐼. (3.22) On the other hand, from 𝑢∈𝑃₀, it follows that 𝑢(𝑡)≥𝑞₁(𝑡)(1−‖𝐵‖)‖𝑢‖≥𝑞₁(𝑡)(1−𝐷)‖𝑢‖,𝑡∈𝐼. (3.23) Thus, by (3.22)-(3.23), we have 𝑤(𝑡)≤𝐶²₁𝑑₁𝛿₁(𝑢‖,1−𝐷)‖ (3.24) and so, 𝑢(𝑡)(𝑡)≥𝐶−𝑤1−²¹𝑑₁𝛿₁(𝑢‖𝑢(𝑡)=𝑏1−𝐷)‖1−⁰‖𝑢‖𝑢(𝑡),𝑡∈𝐼, (3.25) where 𝑏₀=𝐶²₁𝑑₁/(𝛿₁(1−𝐷)).Step 4. By (H4), we have 𝑓(𝑡,𝑢)+𝑀(𝑡)≥Γ₁𝑢,(𝑡,𝑢)∈(0,1)×0,𝑟₀. (3.26) Let 𝜙₀=sin𝜋𝑡. By (2.7)-(2.8), we easily know that sin𝜋𝑡 is a positive eigenfunction of operator 𝐾 with respect to positive eigenvalue Γ₀, that is, 𝐾𝜙₀=Γ₀𝜙₀.
Now, we show that 𝜙₀∈𝑃₀, that is, 𝜙₀(𝑡)≥(1−‖𝐵‖)𝑞₁(𝑡)‖𝜙₀‖,𝑡∈𝐼. We discuss it in three different cases.(1)𝜆₁=0. In this case, 𝐺₁(𝑡,𝑡)=𝑡(1−𝑡),𝑡∈𝐼, and 𝐶₁=𝛿₁=1.(i)If 𝑡∈[0,1/2], then 𝜋𝑡∈[0,𝜋/2]. By Jordan's inequality, we have 2sin𝜋𝑡≥𝜋1⋅𝜋𝑡=2𝑡,𝑡∈0,2. (3.27)(ii)If 𝑡∈[1/2,1], by setting 𝑥=1−𝑡, we have 𝑥∈[0,1/2]. Then from (3. 12), it follows that()()1sin𝜋𝑡=sin𝜋1−𝑥=sin𝜋𝑥≥2𝑥=21−𝑡,𝑡∈2,1. (3.28) Thus, by (i)-(ii) above, we have sin𝜋𝑡≥2𝑡(1−𝑡)=2𝐺₁(𝑡,𝑡)=2𝑞₁(𝑡),𝑡∈𝐼. (3.29)(2)𝜆₁>0. In this case, 𝐺₁(𝑡,𝑡)=(sinh𝜔₁𝑡⋅sinh𝜔₁(1−𝑡))/𝜔₁sinh𝜔₁,𝑡∈𝐼, and 𝐶₁=1,𝛿₁=𝜔₁/sinh𝜔₁.(i)If 𝑡∈[0,1/2], by setting 𝜑(𝑡)=(cosh(𝜔₁/2))𝑡−(sinh𝜔₁𝑡/𝜔₁),𝑡∈[0,1/2], we have𝜑^′(𝑡)𝜔=cosh¹2−cosh𝜔₁𝜔𝑡≥cosh¹2𝜔−cosh¹21=0,𝑡∈0,2. (3.30) From 𝜙(0)=0, it follows that 0≤sinh𝜔₁𝑡/𝜔₁≤(cosh(𝜔₁/2))𝑡,𝑡∈[0,1/2]. Keeping in mind that 0<sinh𝜔₁(1−𝑡)/sinh𝜔₁≤1 for all 𝑡∈[0,1/2], it follows immediately that 𝐺₁()≤𝜔𝑡,𝑡cosh¹21𝑡,𝑡∈0,2. (3.31)(ii)If 𝑡∈[1/2,1], by setting 𝑥=1−𝑡, we have 𝑥∈[0,1/2]. From (2)(i) above, it follows that𝐺₁(𝑡,𝑡)=𝐺₁(1−𝑥,1−𝑥)=𝐺₁(𝑥,𝑥)≤cosh𝜔₁/2𝑥=cosh𝜔₁()1/21−𝑡,𝑡∈2,1. (3.32) Hence, by (2)(i)-(ii) above, we have 𝐺()1𝑡,𝑡≤cosh𝜔₁⋅1𝑡,𝑡∈0,2,11−𝑡,𝑡∈2,1. (3.33) On the other hand, by (3.27)-(3.28), we have 1sin𝜋𝑡≥2⋅𝑡,𝑡∈0,2,11−𝑡,𝑡∈2,1. (3.34) Thus, we have immediately 2sin𝜋𝑡≥cosh𝜔₁𝐺/2¹(𝑡,𝑡)=2𝐶₁𝛿₁cosh𝜔₁𝑞/2¹(𝑡)=2sinh𝜔₁𝜔₁cosh𝜔₁𝑞/2¹(𝑡)=2sinh𝜔₁/2𝜔₁𝑞/2¹(𝑡),𝑡∈𝐼. (3.35) It is easy to verity that sinh(𝜔₁2)/𝜔₁/2≥1. Hence, sin𝜋𝑡≥2𝑞₁(𝑡),𝑡∈𝐼.
(3)−𝜋²<𝜆₁<0. In this case, 𝐺₁(𝑡,𝑡)=sin𝜔₁𝑡⋅sin𝜔₁(1−𝑡)/𝜔₁sin𝜔₁,𝑡∈𝐼, and 𝐶₁=1/sin𝜔₁, 𝛿₁=𝜔₁sin𝜔₁.(i)If 𝑡∈[0,1/2], then 0≤sin𝜔₁𝑡≤sin𝜋𝑡,0<sin𝜔₁(1−𝑡)≤1. Thus, 𝐺₁(𝑡,𝑡)≤sin𝜋𝑡𝜔₁sin𝜔₁1,𝑡∈0,2. (3.36)(ii)If 𝑡∈[1/2,1],  from (i), by letting 𝑥=1−𝑡, then we have 𝑥∈[0,1/2], and 𝐺₁(𝑡,𝑡)=𝐺₁(1−𝑥,1−𝑥)=𝐺₁()≤𝑥,𝑥sin𝜋𝑥𝜔₁sin𝜔₁=sin𝜋(1−𝑡)𝜔₁sin𝜔₁=sin𝜋𝑡𝜔₁sin𝜔₁1,𝑡∈2,1. (3.37)Thus, (3)(i)-(ii) above implies that sin𝜋𝑡≥𝜔₁sin𝜔₁𝐺₁(𝑡,𝑡)=𝜔₁sin𝜔₁𝐶₁𝛿₁𝑞₁(𝑡)=1sin𝜔₁𝑞₁(𝑡)≥𝑞₁(𝑡),𝑡∈𝐼. (3.38) Summing up (1)–(3) keeping in mind that ‖𝜙₀‖=1, we have 𝜙₀(t)=sin𝜋𝑡≥𝑞₁(𝑡)=𝑞₁(𝑡)𝜙₀≥(1−‖𝐵‖)𝑞₁(𝑡)𝜙₀,𝑡∈𝐼, (3.39) that is, 𝜙∈𝑃₀.
Now, set Ω_𝑟0={𝑢∈𝑃₀∶||𝑢₀||<𝑟₀}. We claim that 𝑢≠𝑄𝑢+𝜆𝜙₀,∀𝜆≥0,𝑢∈𝜕Ω_𝑟0. (3.40)Indeed, if not, then exists a 𝑢₀∈𝜕Ω_𝑟0 and 𝜆₀≥0 with 𝑢₀=𝑄𝑢₀+𝜆₀𝜙₀. Without loss of generality, assume that 𝜆₀>0 (otherwise, by proving later on, we will know that the theorem is true). By 𝑢₀∈𝜕Ω_𝑟0, we have ||𝑢₀||=𝑟₀, and so, it follows from (3.25) that 𝑟₀≥𝑢₀(𝑡)≥𝑢₀(𝑡)(𝑡)≥𝑏−𝑤1−⁰𝑟₀𝑢₀(𝑡)≥0,𝑡∈𝐼, (3.41) since 𝑟₀>𝑏₀.
Thus, by (3.26) and (3.41), we have 𝑓𝑡,𝑢₀−𝑤^∗(𝑡)+𝑀(𝑡)=𝑓𝑡,𝑢₀−𝑤(𝑡)+𝑀(𝑡)≥Γ₁𝑏1−⁰𝑟₀𝑢₀(𝑡),𝑡∈(0,1). (3.42) Therefore, by (3.4), (3.16), we have 𝑢₀=𝑄𝑢₀+𝜆₀𝜙₀=(𝐼−𝐵)⁻¹𝐾𝐺𝑢₀+𝜆₀𝜙₀≥𝐾𝐺𝑢₀+𝜆₀𝜙₀=¹₀¹₀𝐺₁(𝑡,𝑠)𝐺₂(𝑠,𝜏)𝑓𝜏,[𝑢−𝑤]^∗(𝜏)+𝑀(𝜏)𝑑𝜏𝑑𝑠+𝜆₀𝜙₀≥Γ₁𝑏1−⁰𝑟₀¹₀¹₀𝐺₁(𝑡,𝑠)𝐺₂(𝑠,𝜏)𝑢₀(𝜏)𝑑𝜏𝑑𝑠+𝜆₀𝜙₀=Γ₁𝑏1−⁰𝑟₀𝐾𝑢₀+𝜆₀𝜙₀. (3.43) Thus, 𝑢₀≥𝜆₀𝜙₀. Let 𝜆^∗=sup{𝜆∣𝑢₀≥𝜆𝜙₀}. Then 𝜆^∗≥𝜆₀, and 𝑢₀≥𝜆^∗𝜙₀. By 𝐾𝑃⊂𝑃 and 𝐾𝜙₀=Γ₀𝜙₀, it follows that 𝐾𝑢₀≥𝜆^∗𝐾𝜙₀=𝜆^∗Γ₀𝜙₀. (3.44) Thus, by (3.43), we have 𝑢₀≥Γ₁𝑏1−⁰𝑟₀Γ₀𝜆^∗𝜙₀+𝜆₀𝜙. (3.45) The hypothesis in (H4) yields Γ₁(1−𝑏₀/𝑟₀)Γ₀≥1, and so 𝑢₀≥(𝜆^∗+𝜆₀)𝜙₀, which contradicts to the definition of 𝜆^∗( noticing that 𝜆₀>0). This shows that (3.40) fulfils. Therefore, in terms of the fixed point index theorem on cone ([25]), we have 𝑖𝑄,Ω_𝑟0,𝑃₀=0. (3.46)
Step 5. Let 𝑑₂=max_𝑡∈[0,1]∫¹₀∫¹₀𝐺₁(𝑡,𝑠)𝐺₂(𝑠,𝜏)𝑑𝜏𝑑𝑠,Γ₂=(1−𝐷)/𝑑₂. By hypothesis 𝑓^∞=0, we have 𝑓^∞<(Γ₂−𝜀₀) for a fixed 𝜀₀∈(0,Γ₂), and so, there exists 𝑅₁>0 such that 𝑓(𝑡,𝑢)<Γ₂−𝜀₀𝑢,𝑡∈𝐼 (3.47) holds when 𝑢≥𝑅₁.Let 𝐶=max{|𝑓(𝑡,𝑢)|∶(𝑡,𝑢)∈𝐼×[0,𝑅₁]}. Then 𝑓(𝑡,𝑢)≤Γ₂−𝜀₀𝑢+𝐶,(𝑡,𝑢)∈𝐼×𝐑₊. (3.48) Let 𝑑₃=max_𝑡∈[0,1]∫¹₀∫¹₀𝐺₁(𝑡,𝑠)𝐺₂(𝑠,𝜏)𝑀(𝜏)𝑑𝜏𝑑𝑠,let𝐸₀=(𝐶𝑑₂+𝑑₃)/𝜀₀𝑑₂. Take 𝑅₀>𝑚𝑎𝑥{𝑟₀,𝐸₀}. Set Ω_𝑅0={𝑢∈𝑃₀∶‖𝑢‖<𝑅₀}. We shall show that 𝜆𝑢≠𝑄𝑢,𝑢∈𝜕Ω_𝑅0,𝜆≥1. (3.49) Suppose on the contradiction that there exist 𝑢₀∈𝜕Ω_𝑅0 and 𝜆₀≥1 with 𝜆₀𝑢₀=𝑄𝑢₀. Then ‖𝑢₀‖=𝑅₀>𝑟₀>𝑏₀. By (3.48), we have 𝑓𝑡,𝑢₀−𝑤^∗(𝑡)≤Γ₂−𝜀₀𝑢₀−𝑤^∗(𝑡)≤Γ+𝐶₂−𝜀₀𝑢₀(𝑡)+𝐶,𝑡∈𝐼. (3.50) Hence, 𝐾𝐺𝑢₀(𝑡)=¹₀