Existence result for a singular nonlinear boundary value problem at resonance

Nonlinear Analysis 68 (2008) 671–680 www.elsevier.com/locate/na

Existence result for a singular nonlinear boundary value problem at resonance Ruyun Ma ∗ , Yunrui Yang Department of Mathematics, Northwest Normal University, Lanzhou 730070, PR China Received 15 April 2006; accepted 24 November 2006

Abstract We study the existence of solutions of the second-order boundary value problems u 00 (t) + π 2 u(t) + a(t)g(u(t)) = h(t),

a.e. t ∈ (0, 1),

u 0 ∈ AC loc (0, 1), u(0) = u(1) = 0, R where g : R → R is continuous, a, h ∈ {z ∈ L 1loc (0, 1) | 01 t|z(t)|dt < ∞}. The proof of the main result is based upon the Lyapunov–Schmidt procedure and the connectivity properties of the solution set of parametrized families of compact vector fields. c 2006 Elsevier Ltd. All rights reserved.

MSC: 34B15 Keywords: Connectivity; Solution set; Lyapunov–Schmidt procedure; Resonance

1. Introduction Singular nonlinear two-point boundary value problems at nonresonance have been studied by several authors; see [1,2,5,9,10,18] and the references therein. Also the existence and multiplicity of solutions of nonsingular twopoint boundary value problems at resonance have been extensively addressed in the literature; see [3,4,7,8,12–16] for references along these lines. However research into singular two-point boundary value problems at resonance has proceeded very slowly. To the best of our knowledge, only [6,17] used the Leray–Schauder principle to study the existence of solutions of singular nonlinear two-point boundary value problems at resonance. Following Ambrosetti and Mancini [4] who applied Lyapunov–Schmidt procedure to nonsingular b.v.p., we extend this method to singular b.v.p. obtaining multiplicity results. Let Z denote the Banach space defined by ( ) Z Z = z ∈ L 1loc (0, 1) |

1

t|z(t)|dt < ∞

0

∗ Corresponding author.

E-mail address: [email protected] (R. Ma). c 2006 Elsevier Ltd. All rights reserved. 0362-546X/$ - see front matter doi:10.1016/j.na.2006.11.030

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with the normal kzk Z =

R1 0

t|z(t)|dt. We will also define

Z + = {z ∈ Z | z(t) > 0 for a.e. t ∈ (0, 1)}. Define AC loc (0, 1) = {y ∈ Z | y ∈ AC([δ, 1 − δ]) for any δ ∈ (0, 1/2)}. Remark 1.1. It is worth remarking that in the study of the singular boundary value problems at resonance via the Lyapunov–Schmidt procedure, we cannot use the bigger space ) ( Z Z ∗ = z ∈ L 1loc (0, 1) |

1

t (1 − t)|z(t)|dt < ∞,

0

(which was used in [9,10] when some singular boundary value problems at nonresonance were considered). The R1 reason is that z ∈ Z ∗ does not guarantee 0 s| cos π s||z(s)|ds < ∞.  Let us consider the singular boundary value problem at resonance u 00 (t) + π 2 u + a(t)g(u) = h(t),

a.e. t ∈ (0, 1),

u ∈ AC loc (0, 1), 0

(1.1)

u(0) = u(1) = 0. We shall use the following assumptions: (H1) (H2) (H3) (H4)

a ∈ Z+; h ∈ Z; g : R → R is a continuous and bounded function; there exists r0 ≥ 0 such that g(s)s > 0,

(H40 )

|s| > r0 ; or

there exists r0 ≥ 0 such that g(s)s < 0,

|s| > r0 ;

(H5) lim|s|→∞ g(s) = 0. Remark 1.2. The extra restriction u 0 ∈ AC loc (0, 1) allows the b.v.p. u 00 (t) + π 2 u + a(t)g(u) = h(t),

a.e. t ∈ (0, 1),

u(0) = u(1) = 0, to be meaningful in the weak sense. In fact, the linear problem u 00 (t) = 0,

a.e. t ∈ (0, 1),

u(0) = u(1) = 0, has a nontrivial solution Z t ∗ ˆ u (t) = Θ(s)ds,

t ∈ (0, 1)

(1.2)

0

with  Θ(4s),    −Θ(4(s − 1/4)) + 1, ˆ Θ(s) = −Θ(4(s − 1/2)),    Θ(4(s − 3/4)) − 1,

0 ≤ s ≤ 1/4, 1/4 ≤ s ≤ 1/2, 1/4 ≤ s ≤ 1/2, 3/4 ≤ s ≤ 1,

(1.3)

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and Θ : [0, 1] → [0, 1] is a classical Cantor function, which is a nondecreasing function with Θ(0) = 0, Θ(1) > 0 and Θ 0 (t) = 0 for a.e. t ∈ [0, 1]; see Natanson [11, pp. 263–264]. Obviously (u ∗ )0 is a function of bounded variation over [0, 1] and is not a function in AC loc (0, 1). So in order to exclude such solutions, the restriction u 0 ∈ AC loc (0, 1) is needed in (1.1).  Finally we state the main tool which will be used in the proof of the main result. Theorem A ([7, Theorem 0]). Let C be a bounded closed convex set in a Banach space X and T : [α, β] × C → C, α < β, a continuous and compact mapping. Then the set Sα,β = {(s, w) ∈ [α, β] × C | T (s, x) = x} contains a component which connects {α} × C to {β} × C. 2. The preliminary results We begin by giving some elementary properties of functions in Z . Lemma 2.1. Suppose that z ∈ Z . Then Z t Z 1 sin(π s)z(s)ds ∈ L (0, 1), 0

1

cos(π s)z(s)ds ∈ L 1 (0, 1).

t

Proof. By a method similar to that used to prove [5, Lemma 2.1], we can get the desired result.



For each h ∈ Z , we have a unique decomposition h(t) = r sin(π t) + w,

(2.1)

where r sin(π t) ∈ V := {c sin(π t) | c ∈ R}, w ∈ V ⊥ and ) ( Z 1 w(s) sin(π s)ds = 0 . V⊥ = x ∈ Z | 0

Notice that w ∈ Z and the fact | sin(π s)| ≤ πs,

s ∈ (0, 1)

guarantee that V ⊥ is well defined. For v ∈ V, w ∈ V ⊥ , we define Z 1 v(s)w(s)ds. hv, wi = 2π 0

Let P : Z → Z and Q : Z → Z denote the orthogonal projections associated with h·, ·i. It is easy to check that φ(t) = √1π sin(π t), ψ(t) = √1π cos(π t), t ∈ [0, 1], are two linearly independently solutions of the equation u 00 (t) + π 2 u(t) = 0,

t ∈ [0, 1].

Then 1

Z

φ 2 (t)dt =

1

Z

0

0

ψ 2 (t)dt =

1 , 2π

and hφ, φi = 2π

1

Z 0

φ(s)φ(s)ds = 1.

(2.2)

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Let  G(t, s) =

ψ(t)φ(s), φ(t)ψ(s),

0 ≤ s ≤ t ≤ 1, 0 ≤ t ≤ s ≤ 1.

Lemma 2.2. Suppose that h¯ ∈ V ⊥ . Then (i) the function Z Z t ¯ φ(s)h(s)ds + φ(t) v(t) := ψ(t) 0

1

¯ ψ(s)h(s)ds

t ∈ (0, 1)

(2.3)

t

is well defined and v(0) = v(1) = 0; and then (ii) v ∈ AC[0, 1] ∩ C 1 (0, 1). ¯ ∈ L 1 (0, 1) and ψ(·)h(·) ¯ ∈ L 1 (δ, 1) for δ ∈ (0, 1), we have that Proof. (i) Since φ(·)h(·) Z t Z 1 ¯ ¯ v(t) = ψ(t) φ(s)h(s)ds + φ(t) ψ(s)h(s)ds. 0

t

1

Z

¯ G(t, s)h(s)ds

=

(2.4)

0

¯ ∈ L 1 (0, 1), it follows that is well defined. Again by the fact that φ(·)h(·) Z t ¯ lim ψ(t) φ(s)h(s)ds = 0. t→0+

(2.5)

0

¯

h Let h˜ = 1−s ; then Z 1 ˜ s(1 − s)|h(s)|ds < ∞. 0

¯ ˜ Combining this with the fact that |ψ(s)h(s)| ≤ (1 − s)|h(s)| and using Asakawa [5, Lemma 2.1(ii)], we conclude that Z 1 ¯ lim φ(t) ψ(s)h(s)ds = 0. (2.6) t→0+

t

The assumption

R1

lim ψ(t)

0

t

Z

t→1−

¯ φ(s)h(s)ds = 0 yields ¯ φ(s)h(s)ds = 0.

(2.7)

0

¯ It is easy to check that ψ(s)h(s) ∈ L 1 ( 12 , 1), and consequently, Z lim

t→1− t

1

¯ ψ(s)h(s)ds = 0.

(2.8)

Combining (2.4) with (2.5)–(2.8), it follows that v(0) = v(1) = 0. (ii) Integrating by parts or applying Fubini’s theorem we may write Z 1Z r Z tZ 1 0 ¯ ¯ v(t) = − ψ (r )φ(s)h(s)dsdr + φ 0 (r )ψ(s)h(s)dsdr +A 0

t

0

(2.9)

r

with A := ψ(1)

Z 0

1

¯ φ(s)h(s)ds −

1

Z 0

¯ φ(s)ψ(s)h(s)ds.

(2.10)

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R. Ma, Y. Yang / Nonlinear Analysis 68 (2008) 671–680

From Lemma 2.1 and the fact h¯ ∈ Z , it follows that the functions Z 1 Z r ¯ ¯ ψ(s)h(s)ds φ(s)h(s)ds, φ 0 (r ) ψ 0 (r ) 0

(2.11)

r

are in L 1 (0, 1) ∩ ACloc (0, 1). This together with (2.9) and (2.10) implies that v ∈ AC[0, 1] ∩ C 1 (0, 1).



¯ ∈ L 1 (δ, 1) for all δ ∈ (0, 1) is used to justify that Remark 2.1. In the proof of Lemma 2.2(i), the fact that ψ(·)h(·) R1 ¯ φ(t) t ψ(s)h(s)ds is well defined. The assertion is true for t > 0. For t = 0 we can use the following estimate: Z 1 φ(t) Z 1 t ¯ ¯ ψ(s)h(s)ds ≤ sψ h ds φ(t) t s t t which is bounded in t since φ(t)t −1 is bounded,

t s

≤ 1 and h¯ ∈ Z .

Remark 2.2. It is easy to see from the proof of Lemma 2.2(i) that if, more generally, h ∈ Z , then v(t) defined in (2.3) is AC([0, 1]) with v(0) = 0. Remark 2.3. It is worth remarking here that (2.3) is not sufficient to guarantee v ∈ C 1 (0, 1). In fact, we only get from (2.3) that Z 1 Z t ¯ ¯ v 0 (t) = ψ 0 (t) φ(s)h(s)ds + φ 0 (t) ψ(s)h(s)ds, a.e. t ∈ (0, 1) (2.12) 0

t

which does not mean that v ∈ difficulty.

C 1 (0, 1).

So we use the method used in the proof of Lemma 2.2(ii) to overcome this

Corollary 2.1. If h¯ ∈ V ⊥ and if v is the function defined in (2.3) (or, equivalently (2.9) and (2.10)), then v is a solution of the boundary value problem ¯ = 0, u 00 (t) + π 2 u(t) + h(t) u 0 ∈ AC loc (0, 1),

a.e. t ∈ (0, 1), (2.13)

u(0) = u(1) = 0. Proof. By Lemma 2.2(i), we only need to check that ¯ = 0, v 00 (t) + π 2 v(t) + h(t)

a.e. t ∈ (0, 1).

From (2.9), we obtain that Z Z t 0 0 0 ¯ φ(s)h(s)ds + φ (t) v (t) = ψ (t) 0

1

(2.14)

¯ ψ(s)h(s)ds,

t ∈ (0, 1)

(2.15)

t

and consequently v 0 ∈ ACloc (0, 1). Thus Z t Z 00 00 00 ¯ v (t) = ψ (t) φ(s)h(s)ds + φ (t) 0

= −π 2 ψ(t)

1

¯ ¯ − φ 0 (t)ψ(t)h(t) ¯ ψ(s)h(s)ds + ψ 0 (t)φ(t)h(t)

t t

Z

¯ φ(s)h(s)ds − π 2 φ(t)

1

Z

0

¯ ¯ − φ 0 (t)ψ(t)h(t) ¯ ψ(s)h(s)ds + ψ 0 (t)φ(t)h(t)

t

¯ = −π 2 v(t) − h(t) for a.e. t ∈ (0, 1).

(2.16)



From Lemma 2.2(i) and (2.2), we may define a linear operator K : V ⊥ → V ⊥ by * + Z Z 1

K (w)(t) = φ, −

0

1

G(t, s)w(s)ds φ(t) +

G(t, s)w(s)ds, 0

w ∈ V ⊥.

(2.17)

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R. Ma, Y. Yang / Nonlinear Analysis 68 (2008) 671–680

Obviously, for h¯ ∈ V , the solution set of (2.13) is ) ( Z 1 ¯ cφ(t) + G(t, s)h(s)ds | c ∈ R . 0

Although the restriction of the operator K to L 1 (0, 1) is compact, this is no longer true for K itself. The operator K has the following compactness property: Lemma 2.3. For every ζ ∈ Z + , let the subset A be defined by A = {z ∈ V ⊥ | |z(t)| ≤ ζ (t) a.e. t ∈ (0, 1)}. Then the set K (A) is precompact in C[0, 1]. Proof. Let K ∗ : Z −→ AC[0, 1] be a nonlinear operator defined by + * Z Z 1

K ∗ (y)(t) = φ,

1

G(t, s)|y(s)|ds φ(t) +

G(t, s)|y(s)|ds.

(2.18)

0

0

By Remark 2.2, K ∗ is well defined. Let ζ ∈ Z + . Then for every z ∈ A, we have that |K (z)(t)| ≤ kK ∗ (ζ )k. Hence K (A) is bounded in C[0, 1]. From Lemma 2.2(ii) and (2.17), we know that K (z) ∈ C 1 (0, 1) ∩ AC[0, 1]. This together with (2.15) implies that for every t1 , t2 ∈ [0, 1] with t1 < t2 , Z t 2 |K (z)(t2 ) − K (z)(t1 )| = (K (z))0 (r )dr t Z 1 Z 1 Z r t2 ψ 0 (r ) φ(s)z(s)ds + φ 0 (r ) ψ(s)z(s)ds = t1 0 r * + ! Z 1 + φ, − G(t, s)w(s)ds φ 0 (r ) dr 0 Z t2 Z r Z t2 Z 1 ≤ |ψ 0 (r )φ(s)||z(s)|dsdr + |φ 0 (r )ψ(s)||z(s)|dsdr t1 0 t1 r Z * + Z 1 t2 0 φ, − G(t, s)w(s)ds φ (r )dr + t1 0 Z t2 Z r Z t2 Z 1 ≤ |ψ 0 (r )φ(s)|ζ (s)dsdr + |φ 0 (r )ψ(s)|ζ (s)dsdr t1 0 t1 r Z * + Z 1 t2 0 φ, − G(t, s)w(s)ds φ (r )dr + t1 0 ! Z t2 Z 1 Z t2 Z r √ ≤ π φ(s)ζ (s)dsdr + ψ(s)ζ (s)dsdr t1

0

t1

r

Z * + Z 1 t2 0 φ, − G(t, s)w(s)ds φ (r )dr . + t1 0 Rr R1 This shows that K is equicontinuous, since by Lemma 2.1, the two functions 0 φ(s)ζ (s)ds and r ψ(s)ζ (s)ds are integrable on [0, 1]. Therefore, by the Arzela–Ascoli theorem, K (A) is precompact in C[0, 1]. 

R. Ma, Y. Yang / Nonlinear Analysis 68 (2008) 671–680

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3. Lyapunov–Schmidt procedure Define a linear operator L : D(L) ⊂ Z → Z by Lu = −u 00 − π 2 u with 1 D(L) := {u ∈ ACloc (0, 1) ∩ C[0, 1], | u(0) = u(1) = 0, u 00 ∈ Z }.

(3.1)

Then (1.1) can be rewritten as ¯ Lu = a(·)g(u) − (τ φ + h)

(3.2)

where h¯ ∈ V ⊥ . Now by projecting over V ⊥ and V , (3.2) is equivalent to the following system of equations: ¯ Lw = Qa(·)g(sφ + w) − h, Pa(·)g(sφ + w) − τ φ = 0,

a.e. t ∈ (0, 1)

(3.3)

(s ∈ R, w ∈ D(L) ∩ V ) ⊥

(3.4)

in V ⊥ and V , respectively. Set ¯ a.e. t ∈ (0, 1)}. Sh¯ = {(s, w) ∈ R × V ⊥ | w ∈ D(L), Lw = Qa(·)g(sφ + w) − h,

(3.5)

Then from Lemma 2.2(ii) and Corollary 2.1 and (2.17), we obtain ¯ Sh¯ = {(s, w) ∈ R × V ⊥ | w ∈ D(L), w = K [Qa(·)g(sφ + w) − h]}.

(3.6)

Define ¯ Ws = {w ∈ V ⊥ | w ∈ D(L), w = K [Qa(·)g(sφ + w) − h]}. Clearly Sh¯ = ∪s∈R ({s} × Ws ). From Lemma 2.3 and the continuity and uniform boundedness of g it follows that the operator Ts : (D(L) ∩ V ⊥ ) → (C[0, 1] ∩ V ⊥ ) defined by ¯ Ts := K [Qa(·)g(sφ + w) − h] is completely continuous and maps into the ball B¯ ρ (0) := {w ∈ V ⊥ | kwk∞ < ρ}, where ! Z 1 Z 1 ¯ G(t, s)a(s)ds + kQk G(t, s)|h(s)|ds ρ := sup |g(t)|kQk 0

t∈R

(3.7)

0

which is well defined since a and h¯ ∈ Z . Therefore, by Schauder’s fixed point theorem, each Ws is nonempty so that ¯ Now system (3.3) and (3.4) is equivalent to the equation projR Sh¯ = R and in fact, Sh¯ ⊂ R × B. Φ(s, w) = τ,

(s, w) ∈ Sh¯ ,

where Φ : R × V ⊥ → R is given by Z 1 a(r )g(sφ(r ) + w(r ))φ(r )dr = ha(·)g(sφ(·) + w(·)), φ(·)i. Φ(s, w) = 2π

(3.8)

(3.9)

0

It is clear that Φ is bounded and continuous. 4. The main results In this section, we will prove some multiplicity results on the singular boundary value problems at resonance via the Lyapunov–Schmidt procedure. Theorem 4.1. Let (H1)–(H3) hold. Then for each h¯ ∈ V ⊥ there exists a nonempty and bounded set Λh¯ ⊂ R such that (3.2) has a solution if and only if τ ∈ Λh¯ . Moreover if (H4) or (H40 ) hold, then τ = 0 is an interior point of Λh¯ .

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Proof. As we already observed in Section 3, Eq. (3.2) is equivalent to solving the equation Φ(s, w) = τ,

(s, w) ∈ Sh¯ .

(4.1)

Obviously Λh¯ in Theorem 4.1 is given by Λh¯ = Φ(Sh¯ ). Since Φ is continuous and bounded, Λh¯ is a bounded set of R. Now suppose h¯ ∈ V ⊥ and (H4) holds. Define W = {w | (s, w) ∈ Sh¯ }. Then we have from (3.7) that there exists a constant M > 0 such that kwk∞ ≤ M,

for all w ∈ W.

(4.2)

Let E s+ := {t ∈ [0, 1] | sφ(t) + w(t) > r0 for all w ∈ W }

(4.3)

E s− := {t ∈ [0, 1] | sφ(t) + w(t) < −r0 for all w ∈ W }.

(4.4)

and Since {t ∈ [0, 1] | sφ(t) − M > r0 } ⊂ lim

s→+∞

E s+

and

meas {t ∈ [0, 1] | sφ(t) − M > r0 } = 1,

(4.5)

it follows that lim

s→+∞

meas E s+ = 1.

(4.6)

meas E s− = 1.

(4.7)

Similarly lim

s→−∞

Using (4.6) and (H4) and the relation Z 1 Z a(r )g(sφ(r ) + w(r ))φ(r )dr = 0

E s+

a(r )g(sφ(r ) + w(r ))φ(r )dr

Z + [0,1]\E s+

a(r )g(sφ(r ) + w(r ))φ(r )dr

we conclude that there exists s2 > r0 > 0 such that Z 1 a(r )g(s2 φ(r ) + w(r ))φ(r )dr > 0

(4.8)

(4.9)

0

for all w ∈ W . Similarly, (4.7) and (4.8) and (H4) yield that there exists s1 < −r0 < 0 such that Z 1 a(r )g(s1 φ(r ) + w(r ))φ(r )dr < 0

(4.10)

0

for all w ∈ W . Thus Φ(s1 , w) < 0 < Φ(s2 , w),

for all w ∈ W.

(4.11)

From (4.11) and the fact that, by Theorem A, Sh¯ ⊂ R × B¯ ρ contains a component ζs1 ,s2 joining {s1 } × B¯ ρ to {s2 } × B¯ ρ , we conclude that Λh¯ contains an interval [−δ, δ], δ > 0. The case where (H40 ) holds can be treated in a similar way.  Theorem 4.2. Let (H1)–(H3) and (H4) or (H40 ) hold and let h¯ ∈ V ⊥ . Let Λh¯ be as in Theorem 4.1. If (H5) is satisfied, then there exists a set Λ∗h¯ ⊂ Λh¯ \ {0} with Λ∗h¯ ⊃ (γ , 0) ∪ (0, β) for some γ ∈ (−∞, 0) and β ∈ (0, ∞), such that (3.2) has at least two solutions if τ ∈ Λ∗h¯ .

R. Ma, Y. Yang / Nonlinear Analysis 68 (2008) 671–680

679

Proof. We only deal with the case where (H4) holds. The case (H40 ) is analogous. From the proof of Theorem 4.1, we have that there exists s2 > r0 > 0 such that Φ(s2 , w) > 0,

for all w ∈ W.

(4.12)

Let β := inf{Φ(s2 , w) | w ∈ W }.

(4.13)

Then β > 0. The existence of γ can be treated in a similar way. Now we only need to show that for any τ ∈ (0, β), (3.2) has at least two solutions. In fact, we have from the assumption (H5) that there exists r : r > s2 > 0 such that Sh¯ contains a component ζ−r,r joining {−r } × B¯ ρ to {r } × B¯ ρ , and max{Φ(s, w) | (s, w) ∈ ζ−r,r and s = ±r } ≤ max{Φ(s, w) | (s, w) ∈ {−r, r } × W } τ ≤ . 2

(4.14)

To see this, we have from (H3) and the fact that a(·)φ(·) ∈ L 1 (0, 1) that there exists δ > 0 such that for all s ∈ R, Z τ τ a(t)g(sφ(t) + w(t))φ(t)dt < , w ∈ W. (4.15) − < 2π 4 4 [0,δ]∪[1−δ,1] By (H5), there exists a positive number b > r0 , such that |g(x)| ≤

−1  Z τ , a(t)φ(t)dt 2π 4 [δ,1−δ]

(4.16)

|x| ≥ b.

Take r ∈ [s2 , ∞) so large that r φ(δ) − M > b (see (4.2) for the definition of M), then (4.16) and (H4) yield Z τ 0 < 2π a(t)g(r φ(t) + w(t))φ(t)dt ≤ , w ∈ W, 4 [δ,1−δ]

(4.17)

and 0 > 2π

Z

τ a(t)g(−r φ(t) + w(t))φ(t)dt ≥ − , 4 [δ,1−δ]

w ∈ W.

(4.18)

Combining (4.15) with (4.17) and (4.18) and using (3.9), it follows that max{Φ(s, w) | (s, w) ∈ {−r, r } × W } ≤

τ . 2

(4.19)

By the connectivity of ζ−r,r , there exist (s3 , w3 ) and (s4 , w4 ) in ζ−r,r with s3 ∈ (−r, s2 ) and s4 ∈ (s2 , r ), such that Φ(s3 , w3 ) = τ,

Φ(s4 , w4 ) = τ.

This implies that s3 φ + w3 and s4 φ + w4 are two distinct solutions of (3.2). Example 4.1. Let us consider the singular boundary value problem  u τ 00 2 −3  = √ sin(π t), a.e. t ∈ (0, 1), u (t) + π u(t) + t 2 2 1+u π u 0 ∈ AC loc (0, 1),   u(0) = u(1) = 0.

(4.20) 

(4.21)

It can be seen that the all conditions of Theorem 4.2 are satisfied. Therefore there exists a set Λ∗0 with int(Λ∗0 ) 6= ∅, such that (4.21) has at least two solutions if τ ∈ Λ∗0 .

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