PICARD BOUNDARY VALUE PROBLEMS FOR SECOND ... - ANUBIH

SARAJEVO JOURNAL OF MATHEMATICS Vol.5 (18) (2009), 191–210

PICARD BOUNDARY VALUE PROBLEMS FOR SECOND ORDER NONLINEAR FUNCTIONAL INTEGRO-DIFFERENTIAL EQUATIONS YUJI LIU Abstract. Sufficient conditions for the existence of solutions of the Picard boundary value problem for the second order nonlinear integrodifferential equation are established. We allow G to grow linearly, superlinearly or sublinearly in our obtained results, see Theorem 2.1 and Theorem 2.2. Examples are presented to illustrate the efficiency of our theorems.

1. Introduction Recently, many papers have discussed the existence of solutions or positive solutions of two-pint boundary value problems for second order differential equations, one may see the text books [1,2] and references therein. In this paper, we study the solvability of Picard boundary value problems for second order of functional integro-differential equations  00 Rπ x (t) + lx0 (t) + kx(t) + G(t, 0 h(t, s)x(s)ds, x(t), x(t − τ1 (t)),     . . . , x(t − τm (t))) = p(t), t ∈ (0, π),  x(0) = x(π) = 0, (1)   x(t) = φ(t), t ∈ [−τ, 0],    x(t) = ψ(t), t ∈ [π, π + δ], where l, k ∈ R, G : [0, π] × Rm+2 → R is a continuous function, p ∈ C 0 [0, π], τi : [0, π] → R is continuous differentiable on [0, π], and τi0 (t) < 1 for all t ∈ [0, π] with its inverse function being denoted µi for i = 1, · · · , m, h(·, ·) : [0, π] × [0, π] → R+ is continuous, ψ : [π, π + δ] → R and φ : [−τ, 0] → R are 2000 Mathematics Subject Classification. 34B10, 34B15. Key words and phrases. Solutions, second order integro-differential equation, two-point boundary value problem, fixed-point theorem, growth condition. Supported by Natural Science Foundation of Hunan province, P.R.China (No:06JJ5008) and Natural Science Foundation of Guangdong province(No:7004569).

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continuous functions with φ(0) = ψ(π) = 0, τ and δ are defined by τ = max{ max {τi (t)} : i = 1, · · · , m}, t∈[0,π]

δ = − min{ min {τi (t)} : i = 1, · · · , m}. t∈[0,π]

This study is similar to the recent papers in which solvability of Picard boundary value problems for second order differential equations are studied. First, we find that BVP ½ 00 x (t) + lx0 (t) + x(t) = sin t, t ∈ (0, π), (2) x(0) = x(π) = 0 has no solution, see Remark 2 in Section 3. BVP(2) is a special case of BVP(1) when k = 1, G(t, x, y) ≡ 0 and p(t) = sin t. In [6], Kuo studied the boundary value problem ½ 00 u (t) + k 2 u + g(x, u) = h(x), x ∈ (0, π), (3) u(0) = u(π) = 0 under the assumption: (H1 ) there is a constant r0 > 0 and nonnegative functions p and b π such that ||p||L1 < 2k(k + 1) tan 2(k+1) , and for a.e x ∈ (0, π) and |u| ≥ r0 |g(x, u)| ≤ p(x)|u| + b(x), holds, where h ∈ L1 (0, π) and g : (0, π) × R → R is a Caratheodory function ( Condition (H1 ) admits g to grow linearly about u.) In [8,22], BVP(3) R π was also studied under the assumptions that (H1 ) holds and k = 1 and 0 h(x) sin xdx = 0. In [7,21], BVP(3) was studied under the assumptions that ||p||L1 ≤ 2k, and a Landesman-Lazer condition Z π Z Z h(x)v(x) < g+ (x)v(x)dx + g− (x)v(x)dx 0

v>0

v 0 and θ ≥ 1 such that g(t, x0 , y0 , x1 , . . . , xm )y0 ≤ −β|y0 |θ+1 , and there are continuous functions h1 , h2 , gi , q1 , q2 , pi , e such that |h(t, x0 , y0 , x1 , . . . , xm )| ≤ h1 (t, x0 ) + h2 (t, y0 ) +

m X

gi (t, xi ) + e(t)

i=1

with

|h1 (t, x)| = q1 (t) uniformly for t ∈ [0, π], x→∞ |x|θ lim

|h2 (t, x)| = q2 (t) uniformly for t ∈ [0, π], x→∞ |x|θ lim

and lim

x→∞

|gi (t, x)| = pi (t) uniformly for t ∈ [0, π], i = 1, . . . , m; |x|θ

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(H4 ) there are continuous functions h1 , h2 , gi , and functions q1 , q2 , pi , e ∈ L1 (0, π), and number 0 < θ < 1 such that |G(t, x0 , y0 , x1 , . . . , xm )| ≤ h1 (t, x0 ) + h2 (t, y0 ) +

m X

gi (t, xi ) + e(t)

i=1

and lim

|h1 (t, x)| = q1 (t) uniformly for t ∈ [0, π], |x|θ

lim

|h2 (t, x)| = q2 (t) uniformly for t ∈ [0, π], |x|θ

x→∞

x→∞

and lim

x→∞

|gi (t, x)| = pi (t) uniformly for t ∈ [0, π], i = 1, . . . , m |x|θ

hold.

¯ ¯ ¯ ¯ 1 Theorem 2.1. Let λi = maxt∈[0,π] ¯ 1−τ 0 (µ ¯, i = 1, . . . , m. Suppose that i (t)) i (H3 ) holds. Then BVP(1) has at least one solution provided k + ||q1 ||

||q1 ||

max

(t,s)∈[0,π]2

max

(t,s)∈[0,π]2

h(t, s)π + ||q2 || +

h(t, s)π θ + ||q2 || +

m X

||pi ||λi < β if θ = 1,

(8)

i=1 m X

||pi ||λθi < β if θ > 1.

(9)

i=1

Proof. To apply Lemma 2.1, we should define an open bounded subset Ω of X such that conditions Lemma 2.1 hold. Let Ω1 = {x ∈ Dom L, Lx = λN x for some λ ∈ (0, 1)}. We prove Ω1 is bounded. For y ∈ Ω1 , we get · µ Z π 00 0 y (t) + ly (t) = λ − ky(t) − G t, h(t, s)y(s)ds, y(t), y(t − τ1 (t)), 0 ¶ ¸ . . . , y(t − τm (t)) + p(t) (10) and y(t) = λφ(t), t ∈ [−τ, 0], y(t) = λψ(t), t ∈ [π, π + δ].

NONLINEAR FUNCTIONAL INTEGRO-DIFFERENTIAL EQUATIONS

197

Hence Z

π

0

[y 00 (t) + ly 0 (t)]y(t)dt µ Z Z π· =λ − ky(t) − G t, 0

0

π

h(t, s)y(s)ds, y(t), y(t − τ1 (t)), ¶ ¸ . . . , y(t − τm (t)) + p(t) y(t)dt.

Since y(0) = y(π) = 0, integrating (10) from 0 to π, we get Z

π

−

Z 0

2

π

Z 2

π

[y (t)] dt = −kλ y (t)dt + λ p(t)y(t)dt 0 0 0 ¶ Z π µ Z π −λ G t, h(t, s)y(s)ds, y(t), y(t − τ1 (t)), . . . , y(t − τm (t)) y(t)dt. 0

0

So Z k Z + 0

Z π y 2 (t)dt − p(t)y(t)dt 0 0 ¶ π µ Z π h(t, s)y(s)ds, y(t), y(t − τ1 (t)), . . . , y(t − τm (t)) y(t)dt ≥ 0. G t, π

0

We will prove that there is a constant B > 0 such that ||x|| ≤ B. We divide this into two steps. Rπ Step 1. Prove that there are constants M > 0 such that 0 |y(t)|θ+1 dt ≤ M . It follows from (H3 ) that Z

π

y 2 (t)dt 0 ¶ Z π µ Z π + g t, h(t, s)y(s)ds, y(t), y(t − τ1 (t)), . . . , y(t − τm (t)) y(t)dt 0 0 µ ¶ Z π Z π + h t, h(t, s)y(s)ds, y(t), y(t − τ1 (t)), . . . , y(t − τm (t)) y(t)dt 0 0 Z π − p(t)y(t)dt ≥ 0.

k

0

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Hence (H3 ) implies that Z π Z π θ+1 |y(t)| dt ≤ k y 2 (t)dt β 0 0 ¶ Z π µ Z π + h t, h(t, s)y(s)ds, y(t), y(t − τ1 (t)), . . . , y(t − τm (t)) y(t)dt 0 0 Z π Z π − p(t)y(t)dt ≤ k y 2 (t)dt 0 0 ¶¯ Z π¯ µ Z π ¯ ¯ ¯h t, ¯ |y(t)|dt + h(t, s)y(s)ds, y(t), y(t − τ (t)), . . . , y(t − τ (t)) 1 m ¯ ¯ 0 0 Z π Z π + |p(t)||y(t)|dt ≤ k y 2 (t)dt 0 0 ¶¯ Z π¯ µ Z π Z π ¯ ¯ ¯h1 t, ¯ |y(t)|dt + h(t, s)y(s)ds |h2 (t, y(t))||y(t)|dt + ¯ ¯ 0

+

m Z X i=1

0

0 π

Z |gi (t, y(t − τi (t))|y(t)|dt +

0 π

(|e(t)| + |p(t)|)|y(t)|dt. 0

From (8) and (9), choosing ² > 0 so that β−k−(||q1 ||+²)

max

(t,s)∈[0,π]2

h(t, s)π−(||q2 ||+²)−

m X

(||pi ||∞ +²)λi > 0 for θ = 1,

i=1

(11)

and β −(||q1 ||+²)

max

(t,s)∈[0,π]2

h(t, s)π θ −(||q2 ||+²)−

m X

(||pi ||∞ +²)λθi > 0 for θ > 1.

i=0

For such ², there is δ > 0 so that |gi (t, x)| ≤ (pi (t) + ²)|x|θ for |x| ≥ δ, t ∈ [0, π], i = 1, · · · , m and h1 (t, x) ≤ (q1 (t) + ²)|x|θ for |x| ≥ δ, t ∈ [0, π], h2 (t, x) ≤ (q2 (t) + ²)|x|θ for |x| ≥ δ, t ∈ [0, π]. Denote, gδ,i = max |gi (t, y)|, t∈[0,π] |y|≤δ

and hδ,1 = max |h1 (t, y)|, t∈[0,π] |y|≤δ

hδ,2 = max |h2 (t, y)|, t∈[0,π] |y|≤δ

(12)


199

and for i = 1, . . . , m and ∆1,i = {t : t ∈ [0, π], |y(t − τi (t))| ≤ δ}, ∆2,i = {t : t ∈ [0, π], |y(t − τi (t))| > δ}, and

¯Z π ¯ ¯ ¯ ¯ = {t ∈ [0, π], ¯ h(t, s)y(s)ds¯¯ ≤ δ}, 0 ¯Z π ¯ ¯ ¯ 0 ¯ ∆2 = {t ∈ [0, π], ¯ h(t, s)y(s)ds¯¯ > δ},

∆01

0

∆1 = {t ∈ [0, π], |y(t)| ≤ δ}, ∆2 = {t ∈ [0, π], |y(t)| > δ}. Hence Z ω Z π θ+1 β |y(s)| ds ≤ k y 2 (t)dt 0 0 ¶¯ Z ¯ µ Z π Z ¯ ¯ ¯h1 t, ¯ |y(t)|dt + h(t, s)y(s)ds + ¯ ¯ +

∆01 m Z X

0

i=1 ∆1,i m Z X

π

(|e(t)| + |p(t)|)|y(t)|dt 0

Z

+

i=1

Z

|gi (t, y(t − τi (t))||y(t)|dt +

|h2 (t, y(t))||y(t)|dt

∆1

∆2,i

|gi (t, y(t − τi (t))||y(t)|dt +

∆2

|h2 (t, y(t))||y(t)|dt

¯ µ Z π ¶¯ ¯ ¯ ¯ ¯ |y(t)|dt + h t, h(t, s)y(s)ds 1 ¯ ¯ ∆02 0 ¶¯ Z π Z ¯ µ Z π ¯ ¯ 2 ¯ ≤k y (t)dt + h(t, s)y(s)ds ¯¯ |y(t)|dt ¯h1 t, Z

0

Z + ∆2

|h2 (t, y(t))||y(t)|dt+ |y(t)|dt+

m X

∆1

0

∆2,i

i=1

Z +hδ,2

∆02 m XZ

Z

|gi (t, y(t−τi (t))|y(t)|dt+hδ,1

Z gδ,i

Z |y(t)|dt+(||e||+||p||)

π

Z

|y(t)|dt ≤ k 0

∆1,i

∆01 π

|y(t)|dt y 2 (t)dt

0

¯ µ Zi=1π ¶¯ Z ¯ ¯ ¯h1 t, ¯ |y(t)|dt + + h(t, s)y(s)ds |h2 (t, y(t))||y(t)|dt ¯ ¯ ∆02 0 ∆2 µ ¶ m Z m X X + |gi (t, y(t − τi (t))|y(t)|dt + hδ,2 + hδ,1 + gδ,i + ||e|| + ||p|| Z

Z

i=1 π

∆2,i

Z

|y(t)|dt ≤ k 0

π

2

y (t)dt + 0

µZ

Z ∆02

(q1 (t) + ²)

0

i=1

π

¶θ h(t, s)|y(s)|ds |y(t)|dt

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YUJI LIU

Z θ+1

+ ∆2

(q2 (t) + ²)|y(t)|

dt +

m Z X i=1

∆1,i

µ ¶Z m X + hδ,2 + hδ,1 + gδ,i + ||e|| + ||p|| i=1

max

(t,s)∈[0,π]2

+

π

0

µZ

(||pi ||∞ + ²)

i=1

0

Z

π

π

y 2 (t)dt + (||q1 || + ²)

|y(t)|dt ≤ k

0

Z

h(t, s)π θ

m X

(pi (t) + ²)|y(t − τi (t))|θ |y(t)|dt

0

Z

|y(s)|θ+1 ds + (||q2 || + ²)

π

|y(t)|θ+1 dt

0

¶ θ µZ π θ+1 θ+1 |y(s − τi (s))| ds

π

θ+1

||x(s)|

¶ θ θ+1 ds

0

µ ¶Z m X + hδ,2 + hδ,1 + gδ,i + ||e|| + ||p||

π

|y(t)|dt.

0

i=1

Since τi0 (t) < 1, and µi is the inverse function of τi , we see Z

π

0

¶θ+1 |y(s)| ds 0 s∈{t−τi (t), t∈[0,π]u[0,π] 1 − τi (µi (s)) ¶θ+1 µ Z |y(s)| + ds 0 s∈{t−τi (t), t∈[0,π]u[π,π+δ] 1 − τi (µi (s)) ¶θ+1 µ Z |y(s)| + ds 0 s∈{t−τi (t), t∈[0,π]u[−τ,0] 1 − τi (µi (s)) ¶θ+1 Z π+δ µ Z π |ψ(s)| θ+1 θ+1 |y(s)| ds + ≤ λi ds 1 − τi0 (µi (s)) π 0 ¶θ+1 Z π Z 0 µ φ(s)| θ+1 ds =: λ |x(s)|θ+1 ds + Qi . + i 0 (µ (s)) 1 − τ i 0 −τ i

Hence Z 1 Z θ+1 β |x(s)| ds ≤ k 0

π

y 2 (t)dt + (||q1 || + ²) Z π Z h(t, s)π θ |y(s)|θ+1 ds + (||q2 || + ²) 0

· +

µ

Z

|y(t − τi (t))|θ+1 dt =

max

(t,s)∈[0,π]2

µ Z θ+1 (||pi ||∞ + ²) λi

0

m X i=1

¶

π

θ+1

|y(s)|

0

ds + Qi

θ θ+1

|y(t)|θ+1 dt

0

µZ

π

θ+1

||x(s)|

¶ θ θ+1 ds

0

µ ¶ µZ m X + hδ,2 + hδ,1 + gδ,i + ||e|| + ||p|| π θ i=1

π

0

π

¶ θ+1

|y(t)|

dt

1 θ+1

.


201

y

−1 y Since limx→0+ (1+x) (1+y)x = 1+y < 1 for y > 0, then there is σ > 0 such that (1 + x)y ≤ 1 + (1 + y)x for 0 ≤ x ≤ σ. Let ) ( Z π Q i I1 = i ∈ {1, . . . , m} : |y(t)|θ+1 dt ≤ , σλθ+1 0 i ( ) Z π Q i I2 = i ∈ {1, . . . , m} : |y(t)|θ+1 dt > . σλθ+1 0 i

Then Z β

Z

1

θ+1

|x(s)|

0

·

max

(t,s)∈[0,π]2

+

π

y 2 (t)dt + (||q1 || + ²) 0 Z π Z θ θ+1 h(t, s)π |y(s)| ds + (||q2 || + ²)

ds ≤ k

X i∈I1

0

µ

Qi + Qi (||pi ||∞ + ²) σ

¶

θ θ+1

µZ

π

0

π

θ+1

|y(s)|

|y(t)|θ+1 dt ¶ θ θ+1 ds

0

¶ θ Z π θ+1 Qi + |y(s)|θ+1 ds R π θ+1 θ+1 λi dt 0 0 |y(t)| i∈I2 µ ¶ µZ π ¶ 1 m X θ+1 θ θ+1 + hδ,2 + hδ,1 + gδ,i + ||e|| + ||p|| π |y(t)| dt X

λθi (||pi ||∞

µ + ²) 1 +

i=1

·

max

(t,s)∈[0,π]2

0

Z

≤k Z θ h(t, s)π

π

0 π

y 2 (t)dt + (||q1 || + ²) Z θ+1

|y(s)|

0

µ

ds + (||q2 || + ²)

¶

|y(t)|θ+1 dt

0

¶ θ θ+1 Qi θ+1 + (||pi ||∞ + ²) + Qi |y(s)| ds σ 0 i∈I1 ¸Z π ¶ · µ X Qi θ θ |y(s)|θ+1 ds + λi (||pi ||∞ + ²) 1 + 1 + Rπ θ+1 dt θ + 1 λθ+1 |y(t)| 0 i 0 X

i∈I2

θ θ+1

µZ

π

π

µ ¶ µZ m X + hδ,2 + hδ,1 + gδ,i + ||e|| + ||p|| π θ i=1

·

max

(t,s)∈[0,π]2

0

θ+1

|y(t)|

dt

1 θ+1

0

Z

=k Z h(t, s)π θ

¶

π

π

0 π

y 2 (t)dt + (||q1 || + ²)

|y(s)|θ+1 ds + (||q2 || + ²)

Z 0

π

|y(t)|θ+1 dt

202

YUJI LIU

+

X

µ (||pi ||∞ + ²)

i∈I1

+

X

Z

λθi (||pi ||∞ + ²)

i∈I2

π

Qi + Qi σ

¶

θ θ+1

0

π

θ+1

|y(s)|

¶ θ θ+1 ds

0

X

|y(s)|θ+1 ds +

µZ

µ λθi (||pi ||∞ + ²) 1 +

i∈I2

µ ¶ µZ m X + hδ,2 + hδ,1 + gδ,i + ||e|| + ||p|| π θ

θ θ+1

¶

¶

π

θ+1

|y(t)|

dt

Qi λθ+1 i

1 θ+1

.

0

i=1

If θ = 1, then µ β − k − (||q1 || + ²) Z · 0

1

max

(t,s)∈[0,π]2

h(t, s)π − (||q2 || + ²) −

X

¶ (||pi ||∞ + ²)λi

i∈I2

µ

¶ 1 µZ π ¶1 2 2 Qi 2 |y(s)| ds ≤ (||pi ||∞ + ²) + Qi ||x(s)| ds σ 0 i∈I1 ¶ µ X 1 Qi + λθi (||pi ||∞ + ²) 1 + 2 λ2i X

2

i∈I2

µ ¶ µZ m X + hδ,2 + hδ,1 + gδ,i + ||e|| + ||p|| π Rπ

It follows from (11) that there is an M1 > 0 such that If θ > 1, then

β

1

θ+1

|x(s)|

ds ≤ |k|π

θ−1 θ+1

µZ

0 Z

+ (||q1 || + ²) +

max

(t,s)∈[0,π]2

X

+

i∈I2

λθi (||pi ||∞

h(t, s)π µ

(||pi ||∞ + ²)

i∈I1

X

π

Z + ²)

θ+1

|y(t)|

0

Qi + Qi σ θ+1

||x(s)| 0

.

|y(t)|2 dt ≤ M1 .

¶

ds +

θ θ+1

ds + (||q2 || + ²)

µZ

π

0

X

π

|y(t)|θ+1 dt

0

¶ θ θ+1 θ+1 ||x(s)| ds

λθi (||pi ||∞

µ + ²) 1 +

i∈I2

¶ µZ µ m X gδ,i + ||e|| + ||p|| π θ + hδ,2 + hδ,1 + i=1

2

¶ 2 θ+1 dt

θ+1

|y(s)| 0

π

0

Z

π

θ

¶1 |y(t)|2 dt

0

i=1

Z

π

0

π

θ θ+1

¶

¶ θ+1

|y(t)|

dt

Qi λθ+1 i

1 θ+1

.


Then µ β − (||q1 || + ²) Z

1

·

θ+1

|y(s)|

max

(t,s)∈[0,π]2

ds ≤

0

+ |k|π

µZ

θ

X

µ (||pi ||∞ + ²)

i∈I1 π

h(t, s)π − (||q2 || + ²) − Qi + Qi σ

¶

X

¶ (||pi ||∞ +

²)λθi

i∈I2

θ θ+1

203

µZ

π

θ+1

||x(s)|

¶ θ θ+1 ds

0

µ ¶ 2 X θ+1 θ λi (||pi ||∞ + ²) 1 + dt +

θ θ+1

¶

Qi λθ+1 0 i i∈I2 1 µ ¶ µ ¶ Z m π X θ+1 θ θ+1 + hδ,2 + hδ,1 + gδ,i + ||e|| + ||p|| π . |y(t)| dt

θ−1 θ+1

θ+1

|y(t)|

0

i=1

Rπ It follows from (12) that there is an M2 > 0 such that 0 |y(t)|θ+1 dt ≤ M2 . Hence above discussions imply that there is M > 0 such that Z π |y(s)|θ+1 ds ≤ M = max{M1 , M2 }. 0

Step 2. Prove that there is a constant B > 0 so that ||x|| ≤ B. Since Z π Z π Z π 0 2 2 [y (t)] dt = λk y (t)dt − λ p(t)y(t)dt 0 0 0 ¶ Z π µ Z 1 +λ G t, h(t, s)y(s)ds, y(t), y(t − τ1 (t)), . . . , y(t − τm (t)) y(t)dt 0 0 Z π Z π Z π 2 ≤ λk y (t)dt − λ p(t)y(t)dt − λβ |y(t)|θ+1 dt 0 0 0 ¶ Z π µ Z 1 +λ g t, h(t, s)y(s)ds, y(t), y(t − τ1 (t)), . . . , y(t − τm (t)) y(t)dt 0 0 Z π Z π 2 ≤k y (t)dt + |p(t)||y(t)|dt 0 0 ¶¯ Z π¯ µ Z 1 ¯ ¯ ¯g t, h(t, s)y(s)ds, y(t), y(t − τ1 (t)), . . . , y(t − τm (t)) ¯¯ |y(t)|dt ¯ 0 0 ¶¯ Z π Z π¯ µ Z π ¯ ¯ ¯ |y(t)|dt ¯h1 t, h(t, s)y(s)ds ≤k y 2 (t)dt + ¯ ¯ Z + 0

0

π

|h2 (t, y(t))||y(t)|dt+

m Z X

0 π

i=1 0

0

|gi (t, y(t − τi (t))|y(t)|dt+

Z

π

|e(t)||y(t)|dt. 0

204

YUJI LIU

Similar to the proof of that of Step 1, we get Z π Z π Z π 0 2 2 θ [y (t)] dt ≤ k y (t)dt + (||q1 || + ²) max h(t, s)π |y(s)|θ+1 ds 2 (t,s)∈[0,π] 0 0 0 Z π + (||q2 || + ²) |y(t)|θ+1 dt +

X

µ (||pi ||∞ + ²)

i∈I1

+

X

0

Qi + Qi σ

θ θ+1

Z λθi (||pi ||∞ + ²)

i∈I2

µ X θ + λi (||pi ||∞ + ²) 1 + i∈I2

¶

θ θ+1

¶

µZ

π

θ+1

|y(s)|

¶ θ θ+1 ds

0 π

|y(s)|θ+1 ds

0

Qi + ||e||π θ λθ+1 i

µZ

π

¶ θ+1

|y(t)|

dt

1 θ+1

.

0

Rπ from 0 |y(t)|θ+1 dt ≤ M that there is M3 > 0 such that R πIt 0 follows 2 0 [y (t)] dt ≤ M3 . For each t ∈ [0, π], we get ¯Z t ¯ Z π Z t ¯ ¯ 1 2 0 0 ¯ [y(t)] = y(s)y (s)ds = ¯ y(s)y (s)ds¯¯ ≤ |y(s)||y 0 (s)|ds 2 0 0 0 ¶ 1 µZ π ¶1 µZ π ¶ 1 µZ π 1 2 2 2 0 2 2 2 |y (t)| dt ≤ |y(t)| dt M32 ≤ |y(t)| dt 0 0 0  1 1  M 2 M 2 , θ = 1, 1 3 1 h θ−1 ¡R ≤ 1 2 i2  π θ+1 π |y(t)|θ+1 dt¢ θ+1 2 M 3 , θ >1 0  1 1   M12 M32 , θ = 1, · ¸1 2 ≤ 1 2 θ−1 θ+1   π θ+1 M2 M32 , θ > 1. It follows that there is a constant B > 0 such that supt∈[0,π] |y(t)| ≤ B. It follows that ½ ¾ ||y|| ≤ max B, max |φ(t)|, max |ψ(t)| for all x ∈ Ω1 . t∈[−τ,0]

t∈[π,π+δ]

Then Ω1 is bounded. Let Ω ⊇ Ω1 be a bounded open subset of X centered T at zero. It is easy to see that Lx 6= λN x for λ ∈ (0, 1] and x ∈ D(L) ∂Ω. It follows from Lemma 2.1 that Lx = N x has at least one solution x in Ω. Then x is a solution of BVP(1). ¤


205

Theorem 2.2. Suppose that (H4 ) holds. Then BVP(1) has at least one solution if |k| < 1. Proof. Similar to that of the proof of Theorem 2.1, we get (10). It follows that µ Z π Z π 0 2 [y (t)] dt = λ k [y(t)]2 dt 0 0 ¶ Z π µ Z 1 + G t, h(t, s)y(s)ds, y(t), y(t − τ1 (t)), . . . , y(t − τm (t)) y(t)dt 0 0 ¶ Z π − p(t)y(t)dt . 0

Let

( y(t), t ∈ [0, π], y1 (t) = −y(t), t ∈ −π, 0].

Suppose y1 (t) =

∞ X

an sin nt,

n=1

then y10 (t) =

∞ X

nan cos nt.

n=1

Rπ It is easy to see that −π y1 (t)dt = 0. It follows from the Parseval equality, Rπ Rπ Rπ P P [y(t)]2 dt = ∞ |an |2 and −π [y 0 (t)]2 dt = ∞ n2 |an |2 , that 0 [y(t)]2 dt n=1 n=1 −π Rπ ≤ 0 [y 0 (t)]2 dt, and the equality holds if and only if y(t) = c sin t, see [17]. Hence ¶¯ Z π¯ µ Z 1 Z π Z π ¯ ¯ 2 0 2 ¯ h(t, s)y(s)ds ¯¯ |y(t)|dt [y(t)] dt + [y (t)] dt ≤ |k| ¯h1 t, 0

0

Z +

0

π

|h2 (t, y(t))||y(t)dt +

m Z X i=1 π

Z

0

0

0

Z

π

|gi (t, y(t − τi (t))||y(t)dt + Z

π

|e(t)||y(t)|dt 0

π

+ ||p|| |y(t)|dt ≤ |k| [y 0 (t)]2 dt 0 0 ¶¯ Z π Z π¯ µ Z 1 ¯ ¯ ¯ |y(t)|dt + ¯h1 t, |h2 (t, y(t))||y(t)dt h(t, s)y(s)ds + ¯ ¯ 0

0

0

+

m Z X i=1

0

π

|gi (t, y(t − τi (t))||y(t)dt + (||e|| + ||p||)

Z

π

|y(t)|dt. 0

206

YUJI LIU

Then

Z

π

(1 − |k|) [y 0 (t)]2 dt 0 ¶¯ Z π¯ µ Z 1 Z ¯ ¯ ¯ ¯ h(t, s)y(s)ds ¯ |y(t)|dt + ≤ ¯h1 t, 0

+

0 m Z π X i=1

0

π

|h2 (t, y(t))||y(t)|dt

0

Z

|gi (t, y(t − τi (t))||y(t)|dt + (||e|| + ||p||)

π

|y(t)|dt. 0

Now, using the notations defined in the proof of Theorem 2.1, the methods used in the proof of Theorem 2.1, we get Z π (1 − |k|) [y 0 (t)]2 dt 0 Z π Z π θ θ+1 ≤ (||q1 || + ²) max h(t, s)π |y(s)| ds + (||q2 || + ²) |y(t)|θ+1 dt (t,s)∈[0,π]2

+

X

µ (||pi ||∞ + ²)

i∈I1

+

X

Z

λθi (||pi ||∞ +²)

π

0

Qi + Qi σ θ+1

|y(s)|

ds+

0

i∈I2

¶

θ θ+1

µZ

0

¶ θ θ+1 θ+1 |y(s)| ds

π

0

X

λθi (||pi ||∞ +²)

i∈I2

µ µZ m X θ + hδ,2 + hδ,1 + gδ,i + ||e|| + ||p||bigg)π max

θ

(t,s)∈[0,π]2

h(t, s)π π

+ (||q2 || + ²)π +

X

µ (||pi ||∞ + ²)

i∈I1

+

X

1−θ 2

Qi + Qi σ

λθi (||pi ||∞

i∈I2

µZ ¶

+ ²)π

1−θ 2

µZ

π

¶

π

θ+1

|y(t)|

dt

Qi λθ+1 i

1 θ+1

¶ θ+1 2 |y(s)| ds 2

¶ θ+1 2 |y(s)| ds 2

0 θ θ+1

θ(1−θ) 2(θ+1)

π

µZ

π

¶θ 2 |y(s)| ds 2

0 1−θ 2

µZ

π

¶ θ+1 2 |y(s)| ds

0

θ θ+1

2

¶

¶ µ m X 1−θ gδ,i + ||e|| + ||p|| π θ π 2(θ+1) + hδ,2 + hδ,1 + i=1

¶

0

π

µ X θ + λi (||pi ||∞ + ²) 1 + i∈I2

θ θ+1

0

i=1

≤ (||q1 || + ²)

µ 1+

Qi λθ+1 i µZ π 0

¶1 2 |y(s)| ds 2


≤ (||q1 || + ²)

θ

max

(t,s)∈[0,π]2

h(t, s)π π

+ (||q2 || + ²)π +

X

µ (||pi ||∞ + ²)

i∈I1

+

X

+

Qi + Qi σ

λθi (||pi ||∞

i∈I2

X

1−θ 2

+ ²)π

µZ

π

¶ θ+1 2 |y (s)| ds 0

2

0

µZ ¶

1−θ 2

207

π

¶ θ+1 2 |y (s)| ds 0

2

0 θ θ+1

θ(1−θ) 2(θ+1)

π

µZ

π

¶θ 2 |y (s)| ds 0

2

0 1−θ 2

µZ

π

¶ θ+1 2 |y (s)| ds 0

0

µ θ λi (||pi ||∞ + ²) 1 +

i∈I2

θ θ+1

2

¶

Qi λθ+1 i µZ

µ ¶ m X 1−θ + hδ,2 + hδ,1 + gδ,i + ||e|| + ||p|| π θ π 2(θ+1) i=1

Rπ

π

¶1 2 |y (s)| ds . 0

2

0

It follows from |k| < 1 that there is M > 0 such that 0 [y 0 (t)]2 dt ≤ M . One sees that ¯Z t ¯ Z π µZ π ¶1 2 ¯ ¯ 1 1 1 0 2 0 0 ¯ ¯ |y(t)| = ¯ y (s)ds¯ ≤ [y (t)] dt ≤ π2M 2. |y (t)|dt ≤ π 2 0

0

0

Hence there exists a constant B > 0 such that maxt∈[0,π] |y(t)| ≤ B. Then ½ ¾ ||y|| = max |y(t)| ≤ max B, max |φ(t)|, max |ψ(t)| t∈[−τ,π+δ]

t∈[−τ,0]

t∈[π,π+δ]

for all x ∈ Ω1 . Then Ω1 is bounded. The remainder of the proof is similar to that of the proof of Theorem 2.1 and is omitted. ¤ 3. Examples In this section, we present examples of equations, which can not be solved by known theorems in [4-8,10], to illustrate the main result in Section 2. Example 3.1. Consider the following problem for delay differential equation  00 P x (t) + lx0 (t) + kx(t) = 2n ai [x(t)]i + a0 [x(t)]2n+1   £ i=1 ¡ 2i−1 ¢¤2n+1 Pm  + p(t), t ∈ (0, π), + i=1 pi (t) x 2i t (13)  x(0) = x(π) = 0,   x(t) = φ(t), t ∈ [−1/2, 0], φ(0) = 0, where n ≥ 0 an integer, l, k ∈ R, ai ∈ R for i = 1, . . . , 2n, a0 < 0, t − τi (t) = 2i−1 2i t, and pi and p are continuous functions. Corresponding to BVP(1), we

208

YUJI LIU

get −G(t, x0 , y0 , x1 , . . . , xm ) = a0 y02n+1 + −g(t, x0 , y0 , x1 , . . . , xm ) = −h(t, x0 , y0 , x1 , . . . , xm ) =

2n X

ai y0i +

m X

i=1 i=1 2n+1 , a0 y0 2n m X X ai y0i + pi (t)x2k+1 i i=1 i=1

pi (t)x2k+1 + p(t), i

+ p(t),

h1 (t, x0 ) = 0, h2 (t, y0 ) =

2n X

ai y0i ,

i=1

gi (t, xi ) = pi (t)x2n+1 , ¯i ¯ ¯ ¯ 1 ¯ ¯ = 2i . λi = max ¯ 0 2i − 1 t∈[0,π] 1 − τi (µi (t)) ¯ We see that q2 (t) ≡ 0. It follows from Theorem 2.1 that, for each p, BVP(15) has at least one solution if n = 0 and m X 2i k+ ||pi || < −a0 , 2i − 1 i=1

and n > 0 and

¶2n+1 m µ X 2i ||pi || < −a0 . 2i − 1 i=1

Example 3.2. Consider the following problem for the delay differential equation  00 x (t) + lx0 (t) + kx(t) = p0 (t)[x(t)]θ   £ ¡ 2i−1 ¢¤θ P  + m + p(t), t ∈ (0, π), i=1 pi (t) x 2i t (14)  x(0) = x(π) = 0,   x(t) = φ(t), t ∈ [−1/2, 0], φ(0) = 0, where l, k ∈ R, t − τi (t) = 2i−1 2i t, and pi are continuous functions as in problem (1). It follows from Theorem 2.2 that problem (16) has at least one solution if θ ∈ [0, 1) and |k| < 1. Remark 3.1. In Example 3.1, G may grow superlinearly and linearly. In Example 3.2, G grows sublinearly. Remark 3.2. For BVP(2), we have k = 1. BVP(2) has no solution. In fact, if (2) has a solution x, then x00 (t) + lx0 (t) + x(t) = sin t, t ∈ (0, π), x(0) = x(π) = 0.


If l = 0, we get Z π Z 00 x (t) sin t dt + 0

Z

π

x(t) sin t dt =

0

π

[sin t]2 dt.

0

Rπ It follows that 0 [sin t]2 dt = 0, a contradiction. If l 6= 0, we get Z π Z π Z π Z 00 0 2 x (t)x(t)dt + l x (t)x(t)dt + [x(t)] dt = Z

0 π

0 π

Z 00

x (t) cos t dt + l 0

0 π

Z 0

x (t) cos t dt + 0

0

π

x(t) sin t dt, Z

π

x(t) cos t dt =

sin t cos t dt.

0

It follows that Z π Z π 0 2 − [x (t)] dt + [x(t)]2 dt 0 0 Z π Z = x(t) sin t dt, l 0

209

0

Z

π

x(t) sin t dt =

0

π

sin t cos t dt. 0

Rπ Rπ Rπ Since x(t) 6≡ c sin t we Rhave 0 [x0 (t)]2 dt > 0 [x(t)]2 dt, we get 0 x(t) sin tdt π < 0. l 6= 0 implies 0 sin t cos tdt 6= 0, a contradiction. One sees that conditions (H4 ) and |k| < 1 in Theorem 2.2 are un-improvable sufficient conditions.

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[11] P. K. Nagle and K. L. Pothoven, On a third order nonlineqar boundary value problem at resonance, J. Math. Anal. Appl., 195 (1995), 149–159. [12] C. P. Gupta, On a third order boundary value problem at resonance, Differ. Integral Equ., 2 (1989), 1–12. [13] C. J. Chyan and J. Henderson, Positive solutions of 2mth-order boundary value problems, Appl. Math. Lett., 15 (2002), 767–774. [14] Y. Liu and W. Ge, Double positive solutions of fourth-order nonlinear boundary value problems, Appl. Anal., 82(4)(2003), 369–380. [15] R. Ma and H. Y. Wang, On the existence of positive solutions of fourth-order ordinary differential equations, Appl. Anal. 59 (1995), 225–231. [16] C. P. Gupta, Existence and uniqueness theorem for a bending of an elastic beam equation, Appl. Anal., 26 (1988), 289–340. [17] E. F. Beckenbach and R. Bellman, Inequalities, Erg. Math. N. F. 30, Springer, 1961, p.p. 177–179. [18] D. S. Cohen, Multiple stable solutions nonlinear boundary value problems arising in chemical reactor theory, SIAM J. Appl. Math., 20 (1971), 1–13. [19] E. N. Dancer, On the structure of solutions of an equation in catalysis theory when a parameter is large, J. Differ. Equations, 37 (1980), 404–437. [20] Z. Han, Solvability of nonlinear ordinary differential equation when its associated linear equation has no nontrivial or sign-changing solution, Taiwanese J. Math. 8 (2004), 503–513. [21] A. Fonda and J. Mawhin, Quadratic forms, weighted eigenfunctions and boundary value problems for nonlinear second order ordinary differential equations, Proc. R. Soc. Edinb., 112A (1989), 145–153. [22] I. Iannacci and M. N. Nakashama, Nonlinear two-point boundary value problems at resonance without Landesman-Lazer conditions, Proc. Amer. Math. Soc., 106 (1989), 943–952. (Received: August 23, 2008) (Revised: January 18, 2009)

Department of Mathematics Hunan Institute of Science and Technology Yueyang 414000, P.R. China Department of Mathematics Guangdong University of Business Studies, Guangzhou 510000, P.R. China E–mail: [email protected]