Existence Criteria For Singular Boundary Value Problems With p-Laplacian Operators ∗

Existence Criteria For Singular Boundary Value Problems With p-Laplacian Operators ^∗

Yan-ping Guo

, Wei-hua Jiang

, Fang Xu

Received 13 September 2005

Abstract

The purpose of this paper is to prove some existence theorems for certain classes of singular nonlinear two point boundary value problems with p-Laplacian operators. In our study, we shall use nonlinear alternative theorem of Leray- Schauder type to prove a priori bound theorem, and from the theorem we get two existence results of differential equations without growth restrictions.

In recent years, singular two-point boundary value problems for ordinary differential equations has been studied by a number of authors. For the singular mixed boundary value problems for the second order differential equation, O’Regan discussed the exis- tence theorems in [1-3]. The aim of the paper is to generalize the O’Regan results in [2-3] to the p-Laplacian ordinary operator of the form

1

p pψn y =qf(t, y, pψn(y)),0< t <1,

whereψn:R→R is defined byψn(u) =|u|ⁿ⁻²uifu= 0,andψn(0) = 0 where n >1.

We begin to discuss the two point mixed boundary value problem

⎧⎪

⎪⎪

⎨

⎪⎪

⎪⎩

1

p pψn y =qf(t, y, pψn(y)),0< t <1,

t→lim0⁺pⁿ⁻¹¹ (t)y(t) = 0, ay(1) +b lim

t→1⁻pⁿ⁻¹¹ (t)y (t) =d, a >0, b≥0.

(1)

Wefirst assume throughout this paper that

(H1)f : [0,1]×R²→Ris continuous,q∈(0,1) with q >0 on (0,1);

(H2)p∈C[0,1]∩C¹(0,1) withp >0 on (0,1);

(H₃) ₀¹p(x)q(x)dx <∞and ₀¹[_p(s)¹ ]ⁿ⁻¹¹ [ ₀^sp(x)q(x)dx]ⁿ⁻¹¹ ds <∞.

∗Mathematics Subject Classifications: 34B15, 34B10.

†College of Sciences, Hebei University of Science and Technology, Shijiazhuang, Hebei 050018, P.

R. China

‡College of Sciences, Hebei University of Science and Technology, Shijiazhuang, Hebei 050018, P.

R. China

§College of Mathematics and Science of Information, Hebei Normal University, Shijiazhuang, Hebei 050016, P. R. China

276

LEMMA 1. Suppose (H1) and (H2) hold. In addition supposerm(x), r(x)∈C[0,1]

withrm→r. Then for eachε>0 there is N, independent of s (s∈[0,1]), such that

|ψ⁻_n¹(

s 0

p(x)q(x)r_m(x)dx)−ψ_n⁻¹(

s 0

p(x)q(x)r(x)dx)|≤εψ_n⁻¹(

s 0

p(x)q(x)dx)

whenm > N.

PROOF. Sincer(x)∈C[0,1], there is constantM >1 such that|r_m(x)|≤M −1 for allx∈[0,1]. Sinceψ_n⁻¹ is uniformly continuous in [−M , M ], for everyε∈(0,1), there existsδ>0 such that|ψ⁻_n¹(s)−ψ_n⁻¹(t)|<εwhen|s−t|≤δands, t∈[−M , M ].

Since r_m → r, for the above δ >0, there is N such that |r_m(x)−r(x)| < δ, for all x∈[0,1] whenm > N. From Integral Mean-value Theorem, there existsξ∈[0,1] such

that _s

0

p(x)q(x)r(x)dx=r(ξ)

s 0

p(x)q(x)dx).

Whenm > N, we have

|ψ⁻_n¹( ₀^sp(x)q(x)rm(x)dx)−ψ_n⁻¹( ₀^sp(x)q(x)r(x)dx)|

≤max{|ψ⁻_n¹( ₀^sp(x)q(x)[r(x)−δ]dx)−ψ_n⁻¹( ₀^sp(x)q(x)r(x)dx)|,

|ψ⁻_n¹( ₀^sp(x)q(x)[r(x) +δ]dx)−ψ_n⁻¹( ₀^sp(x)q(x)r(x)dx)|}

= max{|ψ⁻_n¹((r(ξ)−δ) ₀^sp(x)q(x)dx)−ψ⁻_n¹(r(ξ) ₀^sp(x)q(x)dx)|,

|ψ⁻_n¹((r(ξ) +δ) ₀^sp(x)q(x)dx)−ψ⁻_n¹(r(ξ) ₀^sp(x)q(x)dx)|}

= max{|ψ⁻_n¹(r(ξ)−δ)−ψ⁻_n¹(r(ξ))|,|ψ_n⁻¹(r(ξ) +δ)−ψ_n⁻¹(r(ξ))|}ψ⁻_n¹( ₀^sp(x)q(x)dx)

≤εψ_n⁻¹( ₀^sp(x)q(x)dx).

Associated with (1), we have a family of problems

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⎨

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1

p pψ_n y =λqf(t, y, pψ_n(y )),0< t <1,0<λ<1, lim

t→0⁺pⁿ⁻¹¹ (t)y (t) = 0, ay(1) +b lim

t→1⁻pⁿ⁻¹¹ (t)y(t) =d, a >0, b≥0.

(2)

THEOREM 1. Suppose (H1), (H2) and (H3) hold and assume there is a constant M, independent of λ, with |y|¹ ≤ M for each solution y to (2), for each λ ∈ (0,1).

Then (1) has a solution y ∈C[0,1]∩C²(0,1) with pψn(y) ∈ C[0,1]. Where we set

|y|¹= max{|y|⁰,|pψn(y )|⁰}= max{sup_[0,1]|y(t)|,sup_[0,1]|p(t)ψn(y (t))|}.

PROOF. Solving (2) is equivalent to finding a y ∈C[0,1] with pψ_n(y ) ∈C[0,1]

which satisfies

y(t) = ^d_a+λⁿ⁻¹¹ [ ₀^t(_p(s)¹ )ⁿ⁻¹¹ ψ_n⁻¹( ₀^sp(x)q(x)f(x, y(x), p(x)ψn(y (x)))dx)ds

−a^bψ⁻_n¹( ₀¹p(x)q(x)f(x, y(x), p(x)ψn(y (x)))dx)

− ₀¹ψ_n⁻¹(_p(s)¹ ₀^sp(x)q(x)f(x, y(x), p(x)ψ_n(y (x)))dx)ds].

(3)

Letµ=λⁿ⁻¹¹ . We can write (3) as

y(t) = (1−µ)d

a+µN y(t),

where N:K_B¹[0,1]→K_B¹[0,1] is given by

N y(t) = ^d_a+ ₀^t(_p(s)¹ )ⁿ⁻¹¹ ψ⁻_n¹( ₀^sp(x)q(x)f(x, y(x), p(x)ψ_n(y (x)))dx)ds

−^baψ⁻_n¹( ₀¹p(x)q(x)f(x, y(x), p(x)ψ_n(y (x)))dx)

− 0¹ψ_n⁻¹(_p(s)¹ ₀^sp(x)q(x)f(x, y(x), p(x)ψn(y(x)))dx)ds.

Let

K¹[0,1] ={u∈C[0,1], pψ_n(y )∈C[0,1] with norm|u|1}, which is a Banach space, and set

K_B¹[0,1] ={u∈K¹[0,1], lim

t→0⁺pⁿ⁻¹¹ (t)y (t) = 0, au(1) +b lim

t→1⁻pⁿ⁻¹¹ (t)y (t) =d}. Then K_B¹[0,1] is a convex subset of K¹[0,1]. Consequently (2) is equivalent to the fixed point problem

y= (1−µ)d

a+µN y.

Wefirst prove that N is continuous. Lety_m,y∈K_B¹[0,1], andy_m→y. So there is a constant M₀ so that|y_m|1≤M₀,|y|≤M₀. With lemma 1, for everyε>0 there is N, when m > N, we have

|N y−N y_m|0 = sup_[0,1]|N y(t)−N y_m(t)|

≤2 ₀¹(_p(s)¹ )ⁿ⁻¹¹ |ψ_n⁻¹( ₀^sp(x)q(x)f(x, y(x), p(x)ψn(y (x)))dx)

−ψ⁻_n¹( ₀^sp(x)q(x)f(x, ym(x), p(x)ψn(y_m(x)))dx)|ds +^b_a|ψ_n⁻¹( ₀¹p(x)q(x)f(x, y(x), p(x)ψ_n(y(x)))dx)

−ψ⁻_n¹( ₀¹p(x)q(x)f(x, ym(x), p(x)ψn(y_m(x)))dx)|

≤2ε ₀¹(_p(s)¹ )ⁿ⁻¹¹ ψ_n⁻¹( ₀^sp(x)q(x)dx)ds+_a^bε,

|pψn((N y) )−pψn((N ym) )|⁰= sup_[0,1]|p(t)ψn((N y) (t))−p(t)ψn(N ym) (t))|

≤ ₀¹p(x)q(x)|f(x, y(x), p(x)ψ_n(y(x)))−f(x, y_m(x), p(x)ψ_n(y_m(x)))|dx

≤ε ₀¹p(x)q(x)dx,

so (H₃) and the above inequalities implyN y_m→N y. Thus N is continuous.

Next we show N is completely continuous. LetΩ⊆K_B¹[0,1] be bounded i.e. there exists a constantM₁>0 with|y|1≤M₁ for eachy ∈Ω. For each y∈Ω, we have

|N y|0 ≤ ^d_a+ 2 ₀¹(_p(s)¹ )ⁿ⁻¹¹ |ψ_n⁻¹( ₀^sp(x)q(x)f(x, y(x), p(x)ψ_n(y(x)))dx)|ds +^b_a|ψ⁻_n¹( ₀¹p(x)q(x)f(x, y(x), p(x)ψn(y (x)))dx)|,

|p((N y) )|0 ≤ ₀¹p(x)q(x)|f(x, y(x), p(x)ψ_n(y (x)))|dx,

so there isM3such that|N y|¹≤M3 for eachy∈Ωi.e. NΩis completely bounded.

For eachy∈Ωands, t∈[0,1] withs < t, we have

|N y(t)−N y(s)| ≤ | s^t(_p(x)¹ )ⁿ⁻¹¹ ψ_n⁻¹( ₀^xp(z)q(z)f(z, y(z), p(z)ψn(y (z)))dz)dx|

≤ M4 t

s(_p(x)¹ )ⁿ⁻¹¹ ψ_n⁻¹( ₀^xp(z)q(z)dz)dx,

|p(t)ψn((N y) )(t)−p(s)ψn((N y) )(s)| ≤ s^tp(x)q(x)|f(x, y(x), p(x)ψn(y (x)))|dx

≤ M5 t

sp(x)q(x)dx,

where M₄ and M₅ are constant. (H₃) and above inequalities imply NΩis equicontin- uous. The Arzela-Ascoli theory in [4-5] guarantees that N is completely continuous.

Let U = {u ∈ K_B¹[0,1] : |u|1 < max{M + 1,|^da|+ 1}}, C = K_B¹[0,1] and E = K¹[0,1].

Setp^∗= ^d_a, with the choice of U,p^∗∈U and possibility (ii) of nonlinear alternative theorem of Leray-Schauder type in [6,7] is ruled out. By the theorem, we deduce that N has afixed point i.e. (1) has a solution y∈C[0,1] withpψn(y)∈C[0,1]. The fact that y∈C²(0,1) follows from (3) withλ= 1.

We consider the problem

⎧⎪

⎪⎪

⎨

⎪⎪

⎪⎩

1

p pψn y =qf(t, y, pψn(y)),0< t <1,

t→lim0⁺pⁿ⁻¹¹ (t)y(t) = 0, ay(1) +b lim

t→1⁻pⁿ⁻¹¹ (t)y (t) = 0, a >0, b≥0.

(4)

THEOREM 2. Suppose (H₁), (H₂) and (H₃) hold. In addition suppose (H₄) there is a constantM >0 withuf(t, u,0)>0 for|u|> M andt∈[0,1];

(H₅) there exists s₀ < r₀ with s₀ ≤ 0 ≤ r₀ and f(t, u, r₀) ≤ 0 ≤ f(t, u, s₀) for t∈[0,1] andu∈[−M, M].

Then (4) has a solutiony∈C[0,1]∩C²(0,1), withpψ_n(y)∈C[0,1].

PROOF. Lety be a solution to

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⎨

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1

p pψ_n y =λqf₁(t, y, pψ_n(y)),0< t <1,0<λ<1, lim

t→0⁺pⁿ⁻¹¹ (t)y(t) = 0, ay(1) +b lim

t→1⁻pⁿ⁻¹¹ (t)y (t) = 0, a >0, b≥0,

(5)λ

where

f1(t, u, v) =

⎧⎨

⎩

f(t, u, r0), v≥r0, f(t, u, v), s0≤v≤r0, f(t, u, s0), v≤s0. We noticef1: [0,1]×R²→R is continuous.

Wefirst show that

−M≤y(t)≤M fort∈[0,1]. (6) Suppose|y(t)|achieves a positive maximum att0∈[0,1]. First ift0 ∈(0,1), then y (t0) = 0. Assume |y(t0)| > M. Then (H4) together with s0 ≤ 0 ≤ r0 and the differential equation yields

y(t0)(p(t0)ψn(y(t0))) = λp(t0)q(t0)y(t0)f1(t0, y(t0),0)

= λp(t0)q(t0)y(t0)f(t0, y(t0),0)>0, i.e. y(t0)(ψn(y (t0))) >0. Then one of the following conditions occurs:

(a) ify(t0)>0, then (ψn(y(t0))) >0. Thus there isδ>0 such thatψn(y (t))<0 fort0−δ< t < t0, i.e. y (t)<0; andψn(y (t))>0 fort0< t < t0+δ, i.e. y(t)>0, a contradiction.

(b) ify(t0)<0, then (ψn(y (t0))) <0. Thus there isδ>0 such thatψn(y (t))>0 fort0−δ< t < t0, i.e. y (t)>0; andψn(y (t))<0 fort0< t < t0+δ, i.e. y(t)<0, a contradiction.

Next if t0 = 1, then b > 0. However y(1) lim

t→1⁻pⁿ⁻¹¹ (t)y (t) = −^aby²(1) < 0, i.e.

there is aδ>0 such thaty(1)y(t)<0 for t∈(1−δ,1),a contradiction.

It remains to consider the caset₀= 0. Suppose|y(0)|> M, then y(0)f1(0, y(0),0) =y(0)f(0, y(0),0)>0

together with the differential equation implies there exists δ > 0 with y(t)(p(t)ψn

(y (t))) >0 for t∈(0,δ).

(c) ify(t)>0 for t∈(0,δ), then (p(t)ψ_n(y (t))) >0. With lim

t→0⁺pⁿ⁻¹¹ (t)y (t) = 0, we reduce p(t)ψn(y (t))>0 for t∈(0,δ), i.e. y (t)>0, a contradiction.

(d) ify(t)<0 for t∈(0,δ), then (p(t)ψn(y(t))) <0. With lim

t→0⁺pⁿ⁻¹¹ (t)y(t) = 0, we reduce p(t)ψ_n(y (t))<0 for t∈(0,δ), i.e. y (t)<0, a contradiction.

Thus(6) is true.

Next we show

s0≤p(t)ψn(y (t))≤r0 fort∈[0,1]. (7) Suppose there existst∈(0,1) withp(t)ψn(y (t))> r0. Then since lim

t→0⁺pⁿ⁻¹¹ (t)y(t) = 0, lim

t→0⁺p(t)ψ_n(y (t)) = 0. So there existsµ∈[0, t) withp(s)ψ_n(y(s))> r₀fors∈(µ, t]

andp(µ)ψn(y (µ)) =r0. This together with (H5) yields

0< p(t)ψn(y (t))−p(µ)ψn(y (µ)) = _µ^t(p(s)ψn(y (s)))ds

= λ _µ^tp(s)q(s)f1(s, y(s), p(s)ψn(y (s)))ds

= λ _µ^tp(s)q(s)f(s, y(s), r₀)ds≤0,

a contradiction. Thus p(t)ψ_n(y (t))≤r₀ fort∈[0,1]. Similarly p(t)ψ_n(y(t))≥s₀ for t∈[0,1].

Thus we have shown that any solutionyof (5)_λsatisfied (6) and (7). Consequently theorem 1 implies that (5)_λ has solution y with −M ≤ y(t) ≤ M, t ∈ [0,1] and s₀≤p(t)ψ_n(y (t))≤r₀, t∈[0,1], so y is solution of (4) withy∈C[0,1]∩C²[0,1] and pψ_n(y)∈C[0,1].

THEOREM 3. Suppose (H1),(H2),(H3) and (H4) hold. In addition assume (H6) there exists s0 < r0 with s0 ≤ 0 ≤ r0 and uf(t, u, r0) ≥ 0 for t ∈ [0,1]

and u ∈ [−M0, M0], and uf(t, u, s0) ≥ 0 for t ∈ [0,1] and u ∈ [−M0, M0]; here M₀= max{M,^|d_a^|}andbs₀≤d≤br₀.

Then (1) has a solutiony∈C[0,1]∩C²[0,1] andpψn(y )∈C[0,1].

PROOF. Lety be a solution to

⎧⎪

⎪⎪

⎨

⎪⎪

⎪⎩

1

p pψn y =λqf1(t, y, pψn(y)),0< t <1,0<λ<1,

t→lim0⁺pⁿ⁻¹¹ (t)y(t) = 0, ay(1) +b lim

t→1⁻pⁿ⁻¹¹ (t)y (t) =d, a >0, b≥0,

(8)_λ

where f1 is as in theorem 2. Wefirst show that

|y(t)|≤M0 fort∈[0,1]. (9)

Suppose |y(t)| achieves a positive maximum at t0 ∈ [0,1]. If t0 ∈ [0,1), as in Theorem 2,|y(t0)|≤M. Ift0= 1, then one of the following conditions occurs:

(a) Ifb= 0, then |y(1)|= ^|d_a^|≤M0. (b) Ifb= 0, and suppose|y(1)|> ^|d_a^|, then by(1) lim

t→1⁻pⁿ⁻¹¹ (t)y(t) =y(1)d−a[y(1)]²≤|y(1)||d|−a[y(1)]²=|y(1)|(|d|−a|y(1)|)<0.

Thus there is δ >0 with y(t)pⁿ⁻¹¹ (t)y(t)<0 for t ∈ (1−δ,1), i.e. y(t)y (t) <0, a contradiction. Consequently (9) is true.

Next we show

s0≤p(t)ψn(y(t))≤r0 fort∈[0,1]. (10) Supposep(t)ψn(y (t))≤r0, then there ist1∈(0,1) such thatp(t1)ψn(y (t1))> r0. One of the following conditions occurs:

(c) Ify(t1)>0, letI1={t ∈[t1,1] :pψn(y )(s)> r0≥0,∀s∈[t1, t]}. ThenI1 is an interval, I1=φand y(t)>0 onI1. Suppose 1∈I1. Ifb= 0, then y1= ^d_a ≤0, a contradiction. Ifb= 0, then lim

t→1⁻p(t)ψ_n(y (t)) = ^d_b−^a_by(1)≤ ^d_b ≤r₀, a contradiction.

Thus 1∈/ I1. Then there ist1 < t2 ∈[0,1] such thaty(t)>0,p(t)ψn(y(t))> r0 for t∈(t1, t2) withp(t2)ψn(y (t2)) =r0.

(d) Ify(t1)<0, then there ist0< t1∈[0,1] such thaty(t)<0,p(t)ψn(y (t))> r0

fort∈(t0, t1) withp(t0)ψn(y (t0)) =r0. If (c) holds, then

0> p(t2)ψn(y (t2))−p(t1)ψn(y (t1)) = _t^t2

1(p(s)ψn(y(s)))ds

= λ _t^t2

1 p(s)q(s)f(s, y(s), r₀)ds≥0, a contradiction.

If (d) holds, then

0< p(t1)ψn(y (t1))−p(t0)ψn(y (t0)) = _t^t1

0(p(s)ψn(y(s)))ds

= λ _t^t1

0 p(s)q(s)f(s, y(s), r₀)ds≥0, a contradiction.

Thusp(t)ψn(y (t))≤r0 for t∈[0,1]. Similarly p(t)ψn(y (t))≥s0 fort ∈[0,1], so (10) is true.

Now (9), (10) and Theorem 1 imply (8)λ has a solution y with |y(t)| ≤ M0 for t∈[0,1] ands0≤p(t)ψn(y(t))≤r0 fort∈[0,1], so yis a solution of (1).

Acknowledgment. The project is supported by the Natural Science Foundation of China (10371030), and the Natural Science Foundation of Hebei Province (A2006000298) and the Doctoral Program Foundation of Hebei Province (B2004204)

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