On generalized Laplacian systems

Yong-Hoon LeePusan National University

(with X. Xu)

CMAC, Yonsei UniversityApril 1, 2016

1

Introduction1

1. Introduction

−ϕ(u′)′ = h(t)f(u), t ∈ (0, 1),

u(0) = 0 = (1)(P )

ϕ : R→ R is an increasing homeomorphism, ϕ(0) = 0

(H) h : (0, 1)→ R ; h ∈ Hϕ ⊂ L1loc(0, 1), h 6≡ 0 on any subinterval

Hϕ =

g : ϕ−1

(∫ 12

s|g(τ)|dτ

)∈ L1(0,+), ϕ−1

(∫ s

12

|g(τ)|dτ)∈ L1(+, 1)

2

Introduction1

1.1. ϕ ≡ identity

−u′′ = h(t)f(u), t ∈ (0, 1),

u(0) = 0 = (1),(P )

sh(s) ∈ L1(0,+) and (1− s)h(s) ∈ L1(+, 1)

h(t) = t−α, 1 ≤ α < 2

(P )⇔ u(t) =∫ 1

0G(t, s)h(s)f(u(s))ds := Tu(t)⇔ u = Tu

Operator Set-up

Compactness of the operator

Several nonlinear analytic approaches ⇒Fixed point theory, Topological degree theory,Bifurcation theory, Monotone method etc . . .

3

Introduction1

1.2. ϕ ≡ ϕp, ϕp(x) = |x|p−2x, p > 1

ϕp is an increasing homeomorphism.

−(|u′|p−2u′

)′ (≡ −ϕp(u′)′) = h(t)f(u), t ∈ (0, 1),

u(0) = 0 = (1)(P )

ϕ−1p

(∫ 12s|h(τ)|dτ

)∈ L1(0,+), ϕ−1

p

(∫ s12|h(τ)|dτ

)∈ L1(+, 1)

h(t) = t−α, 1 ≤ α < p

4

Introduction1

p-Laplacian : Operator Set-up

h ∈ L1(0, 1). Manasevich–Mawhin ’98 JDE, ’99 JKMS

(P ) ⇔ u(t) =

∫ t

0

ϕ−1p

(au +

∫ s

0

h(τ)f(u(τ))dτ

)ds := Tu(t)

where a = au is a unique solution of the equation∫ 1

0

ϕ−1p

(a+

∫ s

0

h(τ)f(u(τ))dτ

)dt = 0

h ∈ Hϕp . Sim-L ’12 AAA

(P ) ⇔ u(t) =

∫ t0ϕ−1p

(αu +

∫ 12sh(r)f(u(r))dr

)ds, 0 ≤ t ≤ 1

2∫ 1

tϕ−1p

(−αu +

∫ s12h(r)f(u(r))dr

)ds, 1

2≤ t ≤ 1,

where α = αu is a unique solution of the equation∫ 12

0ϕ−1p

(α+

∫ 12

sh(r)f(u(r))dr

)ds−

∫ 1

12

ϕ−1p

(−α+

∫ s

12

h(r)f(u(r))dr

)ds = 0

5

Introduction1

1.3. Motivation of our problem Hψ ⊂ Hϕ

Nice tools for p-Laplacian

(Prop 1) ϕ−1p (a+ b) ≤ Cp(ϕ−1

p (a) + ϕ−1p (b)), where a, b ≥ 0 and

Cp =

1, p > 2

22−pp−1 , 1 < p ≤ 2.

(Prop 2) ϕp(λx) = ϕp(λ)ϕp(x), ∀ λ, x ∈ RFor more general Laplcians lack of (Prop 1) or (Prop 2)???

ϕ : R→ R is an increasing homeomorphism, ϕ(0) = 0.

(A) there exists an increasing homeomorphism ψ : (0,∞)→ (0,∞) anda function γ : (0,∞)→ (0,∞) such that

ψ(σ) ≤ ϕ(σx)

ϕ(x)≤ γ(σ), for all σ > 0, x ∈ R

(H) h ∈ Hψ ⊂ L1loc(0, 1), h 6≡ 0 on any subinterval

6

Introduction1

Example 1

ϕ(x) = |x|p−2x+ |x|q−2x, x ∈ R, 1 < q < p

ϕ is an increasing homeomorphism

ψ(σ) =

σp−1, if 0 < σ ≤ 1,

σq−1, if σ > 1,

γ(σ) =

1, if 0 < σ ≤ 1,

σp−1, if σ > 1

ψ, γ : (0,∞)→ (0,∞) and ψ is an increasing homeomorphism

ψ(σ) ≤ ϕ(σx)ϕ(x)

≤ γ(σ), ∀σ > 0, x ∈ R

7

Introduction1

Example 1

h(t) = t−α, 1 < α < min2, q

1 < α < min2, q ⇒ ( 1α−1

)1

α−1 > s, s ∈ (0, 1) ⇒ 1α−1

s−(α−1) > 1

∫ 12

0

ψ−1

(∫ 12

s

τ−αdτ

)ds ≤

∫ 12

0

ψ−1

(1

α− 1s−(α−1)

)ds

=

∫ 12

0

(s−(α−1)

α− 1

) 1q−1

ds =q − 1

(α− 1)1q−1 (q − α)

sq−αq−1

∣∣∣ 120<∞

h(t) = t−α ∈ Hψ

ψ−1(σ) =

σ

1p−1 , if 0 < σ ≤ 1,

σ1q−1 , if σ > 1

8

Introduction1

Example 2

ϕ(x) = x13 , x ∈ R

ϕ is an increasing homeomorphism

ψ ≡ γ ≡ ϕ

∫ 12

0

ψ−1

(∫ 12

s

τ−54 dτ

)ds =

∫ 12

0

ψ−1(

4(s−14 − 2

14 ))ds

=

∫ 12

0

(4(s−

14 − 2

14 ))3ds ≤ 64

∫ 12

0

s−34 ds = 256s

14

∣∣∣ 120<∞

h(t) = t−54 ∈ Hψ

9

Introduction1

2. A solution operator2.1. g sign-changing g ∈ Hϕ : ϕ−1

(∫ 12s |g(τ)|dτ

)∈ L1(0, 1

2]

−ϕ(w′(t))′ = g(t), t ∈ (0, 1) (W )

w(0) = w(1) = 0 (D)

ϕ satisfies (A) and g ∈ Hϕ may change signs.

w ∈ C0[0, 1] ∩ C1(0, 1) with ϕ(w′) absolutely continuous satisfying (W ).

Integrating on (s, 12), ( 1

2, s) respectively,w

′(s) = ϕ−1(a+

∫ 12sg(τ)dτ

), w(0) = 0, s ∈ (0, 1

2],

w′(s) = ϕ−1(−a+

∫ s12g(τ)dτ

), w(1) = 0, s ∈ [ 1

2, 1)

(1)

a = ϕ(w′( 12))

Show ϕ−1(a+

∫ 12sg(τ)dτ

)∈ L1(0, 1

2]

10

Introduction1

Lemma 1

Assume (A) holds and g ∈ Hϕ. Then for any a ∈ R,

ϕ−1

(a+

∫ 12

s

g(τ)dτ

)∈ L1(0,

1

2], ϕ−1

(−a+

∫ s

12

g(τ)dτ

)∈ L1[

1

2, 1).

Proof. Case 1: g ∈ L1(0, 12] ⇒ ϕ−1

(a+

∫ 12sg(τ)dτ

)is continuous on [0, 1

2]

Case 2: g 6∈ L1(0, 12]

F (s) ,∫ 1

2s|g(τ)|dτ

F is continuous and F <∞ on (0, 12], F (0+) =∞, F ( 1

2) = 0

∃ s∗ ∈ (0, 12) satisfying |a| =

∫ 12s∗|g(τ)|dτ

11

Introduction1

ϕ−1[σϕ(x)] ≤ ψ−1(σ)x, ∀ σ, x > 0

Subcase 1. s ≤ s∗ : by (A), we get

ϕ−1

(|a|+

∫ 12

s

|g(τ)|dτ

)= ϕ−1

(∫ 12

s∗

|g(τ)|dτ +

∫ 12

s

|g(τ)|dτ

)

≤ ϕ−1

(2

∫ 12

s

|g(τ)|dτ

)≤ ψ−1(2)ϕ−1

(∫ 12

s

|g(τ)|dτ

)

g ∈ Hϕ ⇒ ϕ−1(∫ 1

2s|g(τ)|dτ

)is integrable near 0

Subcase 2. s > s∗

ϕ−1

(|a|+

∫ 12

s

|g(τ)|dτ

)= ϕ−1

(∫ 12

s∗

|g(τ)|dτ +

∫ 12

s

|g(τ)|dτ

)

≤ ϕ−1

(2

∫ 12

s∗

|g(τ)|dτ

)

g ∈ L1loc(0, 1) ⇒ ϕ−1

(2∫ 1

2s∗|g(τ)|dτ

)is a constant.

12

Introduction1

Similarly, we can show that ϕ−1(−a+

∫ s12g(τ)dτ

)∈ L1[ 1

2, 1).

(1) is equivalent to

w(t) =

∫ t0ϕ−1

(a+

∫ 12sg(τ)dτ

)ds, t ∈ [0, 1

2],∫ 1

tϕ−1

(−a+

∫ s12g(τ)dτ

)ds, t ∈ [ 1

2, 1]

Check w( 12

−) = w( 1

2

+): For a ∈ R, define

G(a) =

∫ 12

0

ϕ−1

(a+

∫ 12

s

g(τ)dτ

)ds−

∫ 1

12

ϕ−1

(−a+

∫ s

12

g(τ)dτ

)ds (2)

Lemma 2

For given g ∈ Hϕ, the function G defined in (2) has a unique zero a = a(g) in R.

13

Introduction1

Proof. Define

H(a) ,∫ 1

2

0

ϕ−1

(a+

∫ 12

s

g(τ)dτ

)ds,

W (a) ,∫ 1

12

ϕ−1

(−a+

∫ s

12

g(τ)dτ

)ds

Show H(a) and W (a) are continuous in a.

Show H(a) is strictly increasing in a and H(a)→ ±∞ as a→ ±∞.

Show W (a) is strictly decreasing in a and W (a)→ ∓∞ as a→ ±∞.

Show 1. H(a) and W (a) are continuous in a.Let an ⊂ R with an → a as n→∞. We need show

limn→∞

H(an) = H( limn→∞

an) = H(a),

limn→∞

W (an) = W ( limn→∞

an) = W (a)

14

Introduction1

For 0 < s ≤ 12

and a = supn|an|,

ϕ−1

(an +

∫ 12

s

g(τ)dτ

)≤ ϕ−1

(a+

∫ 12

s

|g(τ)|dτ

), ∀n

for s ∈ (0, 12], by Lebesgue Dominated Convergence Theorem,

limn→∞

∫ 12

0

ϕ−1

(an +

∫ 12

s

g(τ)dτ

)ds =

∫ 12

0

limn→∞

ϕ−1

(an +

∫ 12

s

g(τ)dτ

)ds

=

∫ 12

0

ϕ−1

(limn→∞

an +

∫ 12

s

g(τ)dτ

)ds =

∫ 12

0

ϕ−1

(a+

∫ 12

s

g(τ)dτ

)ds

limn→∞H(an) = H(a)

15

Introduction1

Show 2. H is strictly increasing in a and H(a)→ ±∞ as a→ ±∞.

H(a) ,∫ 1

20ϕ−1

(a+

∫ 12sg(τ)dτ

)ds : strictly increasing in a

a2

+∫ 1

2sg(τ)dτ > 0 for s ∈ [ 1

3, 12] and sufficiently large a > 0, we have

H(a) =

∫ 13

0ϕ−1

(a+

∫ 12

sg(τ)dτ

)ds+

∫ 12

13

ϕ−1

(a+

∫ 12

sg(τ)dτ

)ds

=

∫ 13

0ϕ−1

(a+

∫ 12

sg(τ)dτ

)ds+

∫ 12

13

ϕ−1

(a

2+a

2+

∫ 12

sg(τ)dτ

)ds

≥∫ 1

3

0ϕ−1

(∫ 12

sg(τ)dτ

)ds+

∫ 12

13

ϕ−1(a

2)ds

≥ −∫ 1

2

0ϕ−1

(∫ 12

s|g(τ)|dτ

)ds+

1

6ϕ−1(

a

2)→ +∞ as a→ +∞

16

Introduction1

a2

+∫ 1

2sg(τ)dτ < 0 on [ 1

3, 12] and for sufficiently negatively large a < 0,

H(a) =

∫ 13

0ϕ−1

(a+

∫ 12

sg(τ)dτ

)ds+

∫ 12

13

ϕ−1

(a+

∫ 12

sg(τ)dτ

)ds

=

∫ 13

0ϕ−1

(a+

∫ 12

sg(τ)dτ

)ds+

∫ 12

13

ϕ−1

(a

2+a

2+

∫ 12

sg(τ)dτ

)ds

≤∫ 1

3

0ϕ−1

(∫ 12

sg(τ)dτ

)ds+

∫ 12

13

ϕ−1(a

2)ds

≤∫ 1

2

0ϕ−1

(∫ 12

s|g(τ)|dτ

)ds+

1

6ϕ−1(

a

2)→ −∞ as a→ −∞

17

Introduction1

Show 3. W (a) is strictly decreasing in a and W (a)→ ∓∞ as a→ ±∞.

W (a) ,∫ 1

12ϕ−1

(−a+

∫ s12g(τ)dτ

)ds : strictly decreasing in a

−a2

+∫ s

12g(τ)dτ < 0 for s ∈ [ 1

2, 23] and sufficiently large a > 0, we have

W (a) =

∫ 23

12

ϕ−1

(−a+

∫ s

12

g(τ)dτ

)ds+

∫ 1

23

ϕ−1

(−a+

∫ s

12

g(τ)dτ

)ds

=

∫ 23

12

ϕ−1

(−a

2−a

2+

∫ s

12

g(τ)dτ

)ds+

∫ 1

23

ϕ−1

(−a+

∫ s

12

g(τ)dτ

)ds

≤∫ 2

3

12

ϕ−1(−a

2)ds+

∫ 1

23

ϕ−1

(∫ s

12

g(τ)dτ

)ds

≤1

6ϕ−1(−

a

2) +

∫ 1

12

ϕ−1

(∫ s

12

|g(τ)|dτ)ds→ −∞ as a→ +∞.

18

Introduction1

−a2

+∫ s

12g(τ)dτ > 0 on [ 1

2, 23] and for sufficiently negatively large a < 0,

W (a) =

∫ 23

12

ϕ−1

(−a+

∫ s

12

g(τ)dτ

)ds+

∫ 1

23

ϕ−1

(−a+

∫ s

12

g(τ)dτ

)ds

=

∫ 23

12

ϕ−1

(−a

2−a

2+

∫ s

12

g(τ)dτ

)ds+

∫ 1

23

ϕ−1

(−a+

∫ s

12

g(τ)dτ

)ds

≥∫ 2

3

12

ϕ−1(−a

2)ds+

∫ 1

23

ϕ−1

(∫ s

12

g(τ)dτ

)ds

≥1

6ϕ−1(−

a

2)−

∫ 1

12

ϕ−1

(∫ s

12

|g(τ)|dτ)ds→ +∞ as a→ −∞

G(a) is continuous in a and strictly increasing in a.

G(a)→ ±∞ as a→ ±∞Conclusion. There exists a unique a = a(g) ∈ R such that G(a) = 0.

19

Introduction1

−ϕ(w′(t))′ = g(t), w(0) = w(1) = 0

If ϕ satisfies (A) and g ∈ Hϕ,then the solution w of (W )+(D) can be represented by

w(t) =

∫ t0ϕ−1

(a(g) +

∫ 12sg(τ)dτ

)ds, t ∈ [0, 1

2],∫ 1

tϕ−1

(−a(g) +

∫ s12g(τ)dτ

)ds, t ∈ [ 1

2, 1]

(3)

a(g) ∈ R uniquely satisfies

∫ 12

0

ϕ−1

(a(g) +

∫ 12

s

g(τ)dτ

)ds =

∫ 1

12

ϕ−1

(−a(g) +

∫ s

12

g(τ)dτ

)ds

The function w defined in (3) satisfies w ∈ C0[0, 1] ∩ C1(0, 1), and ϕ(w′)absolutely continuous on (0, 1) and w is in turn a solution of (W )+(D).

20

Introduction1

3. Nonlinear ProblemsProblem set-up

−Φ(u′)

′= λh(t) · f(u(t)), t ∈ (0, 1),

u(0) = 0 = u(1),(Pλ)

u = (u1, · · · , uN ), Φ(u′) = (ϕ(u′1), · · · , ϕ(u′N ))

ϕ : R→ R is an increasing homeomorphism, ϕ(0) = 0, λ > 0 a parameter

h(t) · f(u(t)) := (h1(t)f1(u(t)), · · · , hN (t)fN (u(t)))−ϕ(u′1)′ = λh1(t)f1(u(t)),

...

−ϕ(u′N )′ = λhN (t)fN (u(t)), t ∈ (0, 1),

ui(0) = 0 = ui(1), i = 1, · · · , N

(Pλ)

21

Introduction1

Assumptions

(A) there exists an increasing homeomorphism ψ : (0,∞)→ (0,∞) anda function γ : (0,∞)→ (0,∞) such that

ψ(σ) ≤ ϕ(σx)

ϕ(x)≤ γ(σ), ∀ σ > 0, x ∈ R

(H) hi : (0, 1)→ R+, hi ∈ Hψ ⊂ L1loc(0, 1), hi 6≡ 0 on any subinterval

Hψ =

g :

∫ 12

0ψ−1

(∫ 12

sg(τ)dτ

)ds+

∫ 1

12

ψ−1

(∫ s

12

g(τ)dτ

)ds <∞

(F1) f i : RN+ → R+ is continuous for i = 1, · · · , N(F2) f i(u) > 0 for u ∈ RN+ with ‖u‖ > 0 for i = 1, · · · , N

22

Introduction1

4. Main results

Notation

f0 :=

N∑i=1

f i0, f∞ :=

N∑i=1

f i∞,

where

f i0 := lim‖x‖→0

f i(x)

ϕ(‖x‖) , f i∞ := lim‖x‖→∞

f i(x)

ϕ(‖x‖)

Theorem 1.

Assume that (A)(H) and (F1) hold.

(1) f0 = 0, f∞ =∞ or f0 =∞, f∞ = 0 =⇒(Pλ) has at least one positive solution for λ > 0.

(2) If f0 = 0, 0 < f∞ <∞ or 0 < f0 <∞, f∞ = 0 =⇒∃ λ0 > 0 such that (Pλ) has at least one positive solution for λ > λ0.

(3) If f0 =∞, 0 < f∞ <∞ or 0 < f0 <∞, f∞ =∞ =⇒∃ λ0 > 0 such that (Pλ) has at least one positive solution for λ ∈ (0, λ0).

23

Introduction1

Theorem 2.

Assume that (A)(H)(F1) and (F2) hold.

(1) If f0 = f∞ = 0 =⇒∃ λ0 > 0 such that (Pλ) has at least two positive solutions for λ > λ0.

(2) If f0 = f∞ =∞ =⇒∃ λ0 > 0 such that (Pλ) has at least two positive solutions for λ ∈ (0, λ0).

24

Introduction1

Example 3

ϕ(u′)′ + t−α[(u+ v)p−q + 1] = 0,

ϕ(v′)′ + t−βuq−1(1− e−v) = 0, t ∈ (0, 1),

u(0) = v(0) = u(1) = v(1) = 0

(E1)

ϕ(u′) = |u′|p−2u′ + |u′|q−2u′, 1 < q < p ⇒ condition (A)

h1(t) = t−α, h2(t) = t−β , 1 < α, β < min2, q ⇒ hi ∈ Hψf1(u, v) = (u+ v)p−q + 1, f2(u, v) = uq−1(1− e−v)

f10 = lim‖(u,v)‖→0

f1(u,v)ϕ(‖(u,v)‖) = lim‖(u,v)‖→0

(u+v)p−q+1

(u+v)q−1[(u+v)p−q+1]=∞

f1∞ = lim‖(u,v)‖→∞

f1(u,v)ϕ(‖(u,v)‖) = lim‖(u,v)‖→∞

1(u+v)q−1 = 0

f20 = lim‖(u,v)‖→0(1− e−v) · uq−1

(u+v)p−1+(u+v)q−1 ≤ lim‖(u,v)‖→0(1− e−v) = 0

25

Introduction1

Example 3

f2∞ = lim

‖(u,v)‖→∞(1− e−v) · uq−1

(u+ v)p−1 + (u+ v)q−1

≤ lim‖(u,v)‖→∞

(1− e−v) · (u+ v)q−1

(u+ v)p−1 + (u+ v)q−1

≤ lim‖(u,v)‖→∞

1

(u+ v)p−q + 1= 0.

f0 = f10 + f2

0 =∞, f∞ = f1∞ + f2

∞ = 0

Conclusion. (E1) has at least one positive solution by Theorem 1.

26

Introduction1

Example 4

ϕ(u′)′ + t−

54 (u+ v)

12 = 0,

ϕ(v′)′ + t−65 (1− e−(u+v))(u+ v)

13 = 0, t ∈ (0, 1),

u(0) = v(0) = u(1) = v(1) = 0

(E2)

ϕ(u′) = (u′)13 ⇒ condition (A)

h1(t) = t−54 , h2(t) = t−

65 ⇒ hi ∈ Hψ

f10 = lim‖(u,v)‖→0

f1(u,v)ϕ(‖(u,v)‖) = lim‖(u,v)‖→0

(u+v)12

(u+v)13

= lim‖(u,v)‖→0(u+ v)16 = 0

f1∞ = lim‖(u,v)‖→∞

f1(u,v)ϕ(‖(u,v)‖) = lim‖(u,v)‖→∞(u+ v)

16 =∞

f20 = lim‖(u,v)‖→0(1− e−(u+v)) · (u+v)

13

(u+v)13

= lim‖(u,v)‖→0(1− e−(u+v)) = 0

f2∞ = lim‖(u,v)‖→∞(1− e−(u+v)) = 1

f0 = 0, f∞ =∞, Theorem 1 ⇒ (E2) has at least one positive solution.

27

Introduction1

Fundamental theorem for existence

Fixed Point Theorem of Expansion/Compression on a Cone

(E, ‖ · ‖) a Banach space and K ⊂ E a cone in E.

For R > 0, define KR = u ∈ K : ‖u‖ < R.T : KR → K is completely continuous.

There exists 0 < r < R such that

‖Tu‖ ≤ ‖u‖, ∀ u ∈ ∂Kr, ‖Tu‖ ≥ ‖u‖, ∀u ∈ ∂KR, or

‖Tu‖ ≥ ‖u‖, ∀ u ∈ ∂Kr, ‖Tu‖ ≤ ‖u‖, ∀ u ∈ ∂KR.

Then T has a fixed point u ∈ K with r ≤ ‖u‖ ≤ R.

28

Introduction1

5. Sketch of proofsPreliminaries for proofs

Remark 1

From condition (A), we getσx ≤ ϕ−1[γ(σ)ϕ(x)],

ϕ−1[σϕ(x)] ≤ ψ−1(σ)x, for σ, x > 0

Lemma 2

w ∈ C0[0, 1] ∩ C1(0, 1), ϕ(w′)′ ≤ 0 on (0, 1) ⇒ w is concave on [0, 1],

mint∈[ 1

4, 34]w(t) ≥ 1

4‖w‖∞

Lemma 3

If h ∈ Hψ, then for given α ∈ C[0, 1], αh ∈ Hϕ.

29

Introduction1

E = C0[0, 1]× · · · × C0[0, 1]︸ ︷︷ ︸N

, ‖u‖∞ =∑Ni=1 ‖ui‖∞

K = u ∈ E : ui is concave on (0, 1), i = 1, · · · , NTλ = (T 1

λ , · · · , TNλ ) is defined by

T iλ(u)(t) =

∫ t0ϕ−1

(ai(λhif

i(u)) +∫ 1

2sλhi(τ)f i(u(τ))dτ

)ds, t ∈ [0, 1

2],∫ 1

tϕ−1

(−ai(λhif i(u)) +

∫ s12λhi(τ)f i(u(τ))dτ

)ds, t ∈ [ 1

2, 1]

where ∫ 12

0

ϕ−1

(ai(λhif

i(u)) +

∫ 12

s

λhi(τ)f i(u(τ)))dτ

)ds

=

∫ 1

12

ϕ−1

(−ai(λhif i(u)) +

∫ s

12

λhi(τ)f i(u(τ)))dτ

)ds.

(Pλ) ⇔ u = Tλ(u) on K

30

Introduction1

Lemma 4

ai sends any bounded set in (0,∞)×K into bounded set in R.

ai is continuous for i = 1, · · · , N .

Lemma 5

Tλ : K → K is completely continuous.

Tλ is continuous on K :Lebesgue Dominated Convergence Theorem and the continuity of ai.

Tλ(B) is relatively compact for any bounded subset B ⊂ K :Tλ(B) is bounded and equicontinuous (Arzela-Ascoli Theorem ).

31

Introduction1

Proof of Theorem 1 (1)

Step 1: show that ∃ r > 0 such that ‖Tλ(u)‖∞ ≤ ‖u‖∞, for u ∈ ∂Kr.

f i0 = 0, i = 1, · · · , N ⇒ ∀ ε > 0, ∃ ri(= ri(ε)) > 0 ; for x ∈ RN+ , ‖x‖ ≤ ri,

f i(x) ≤ εϕ(‖x‖), for i = 1, · · · , N.

Take r = minri : i = 1, · · · , N and let u ∈ ∂Kr.Tλ(u) ∈ K for u ∈ ∂Kr ⇒ ∃ unique σi ∈ (0, 1) such that

T iλ(u)(σi) = maxt∈[0,1]

T iλ(u)(t), T iλ(u)′(σi) = 0.

We first consider the case σi ∈ (0, 12].

0 = T iλ(u)′(σi) = ϕ−1

(aiλ,u +

∫ 12

σi

λhi(τ)f i(u(τ))dτ

).

32

Introduction1

Since ϕ is an odd homeomorphism, aiλ,u = −∫ 1

2σiλhi(τ)f i(u(τ))dτ .

By Remark 1, we have

‖T iλ(u)‖∞ = T iλ(u)(σi) =

∫ σi

0

ϕ−1

(aiλ,u +

∫ 12

s

λhi(τ)f i(u(τ))dτ

)ds

=

∫ σi

0

ϕ−1

(−∫ 1

2

σi

λhi(τ)f i(u(τ))dτ +

∫ 12

s

λhi(τ)f i(u(τ))dτ

)ds

=

∫ σi

0

ϕ−1

(∫ σi

s

λhi(τ)f i(u(τ))dτ

)ds ≤

∫ 12

0

ϕ−1

(∫ 12

s

λhi(τ)f i(u(τ))dτ

)ds

≤∫ 1

2

0

ϕ−1

(λεϕ(r)

∫ 12

s

hi(τ)dτ

)ds ≤ ψ−1(λε)

∫ 12

0

ϕ−1

(ϕ(r)

∫ 12

s

hi(τ)dτ

)ds

≤ ψ−1(λε) [

∫ 12

0

ψ−1

(∫ 12

s

hi(τ)dτ

)ds] r , ψ−1(λε)Hi

0r.

33

Introduction1

Similarly for the case σi ∈ [ 12, 1), we get

‖T iλ(u)‖∞ ≤ ψ−1(λε) [

∫ 1

12

ψ−1

(∫ 1

12

hi(τ)dτ

)ds] r , ψ−1(λε)Hi

1r.

Therefore combining the above two inequalities, we get

‖T iλ(u)‖∞ ≤ ψ−1(λε) maxHi0, H

i1r ≤ ψ−1(λε)Γr, for i = 1, · · · , N,

Γ = max

maxHi0, H

i1 : i = 1, · · · , N

Choose ε > 0 sufficiently small so that ψ−1(λε)Γ ≤ 1

N, thus

‖Tλ(u)‖∞ =N∑i=1

‖T iλ(u)‖∞ ≤ ‖u‖∞, for u ∈ ∂Kr.

34

Introduction1

Step 2: show that ∃ R > 0 such that ‖Tλ(u)‖∞ ≥ ‖u‖∞, for u ∈ ∂KR.

If f∞ =∞, then there exists an index i0 satisfying f i0∞ =∞.

∀ M > 0, ∃ RM > 0 such that for x ∈ RN+ with ‖x‖ ≥ RM ,

f i0(x) ≥Mϕ(‖x‖).

If u ∈ K with ‖u‖∞ ≥ 4RM , then by Lemma 2, for t ∈ [ 14, 34],

‖u(t)‖ =

N∑i=1

ui(t) ≥ mint∈[ 1

4, 34]

N∑i=1

ui(t) ≥1

4‖u‖∞ ≥ RM ,

and

f i0(u(t)) ≥Mϕ(‖u(t)‖) ≥Mϕ(1

4‖u‖∞).

Take R > maxr, 4RM. Then for u ∈ ∂KR, we get

35

Introduction1

2T i0λ (u)(1

2) =

∫ 12

0

ϕ−1

(ai0λ,u +

∫ 12

s

λhi0(τ)f i0(u(τ))dτ

)ds

+

∫ 1

12

ϕ−1

(−ai0λ,u +

∫ s

12

λhi0(τ)f i0(u(τ))dτ

)ds.

If ai0λ,u ≥ 0, then

∫ 12

0ϕ−1

(ai0λ,u +

∫ 12

sλhi0 (τ)f

i0 (u(τ))dτ

)ds ≥

∫ 12

0ϕ−1

(∫ 12

sλhi0 (τ)f

i0 (u(τ))dτ

)ds

∫ 1

12

ϕ−1

(−ai0λ,u +

∫ s

12

λhi0(τ)f i0(u(τ))dτ

)ds

=

∫ 12

0

ϕ−1

(ai0λ,u +

∫ 12

s

λhi0(τ)f i0(u(τ))dτ

)ds ≥ 0.

2T i0λ (u)(1

2) ≥

∫ 12

0

ϕ−1

(∫ 12

s

λhi0(τ)f i0(u(τ))dτ

)ds.

36

Introduction1

If ai0λ,u < 0, then −ai0λ,u > 0 and∫ 1

12

ϕ−1

(−ai0λ,u +

∫ s

12

λhi0(τ)f i0(u(τ))dτ

)ds

≥∫ 1

12

ϕ−1

(∫ s

12

λhi0(τ)f i0(u(τ))dτ

)ds

and by the same argument, we get

2T i0λ (u)(1

2) ≥

∫ 1

12

ϕ−1

(∫ s

12

λhi0(τ)f i0(u(τ))dτ

)ds

37

Introduction1

2‖T i0λ (u)‖∞ ≥ 2T i0λ (u)(1

2)

≥ min

∫ 12

0ϕ−1

(∫ 12

sλhi0 (τ)f

i0 (u(τ))dτ

)ds,

∫ 1

12

ϕ−1

(∫ s

12

λhi0 (τ)fi0 (u(τ))dτ

)ds

≥ min

∫ 14

0ϕ−1

(∫ 12

sλhi0 (τ)f

i0 (u(τ))dτ

)ds,

∫ 1

34

ϕ−1

(∫ s

12

λhi0 (τ)fi0 (u(τ))dτ

)ds

≥ min

∫ 14

0ϕ−1

(∫ 12

14

λhi0 (τ)fi0 (u(τ))dτ

)ds,

∫ 1

34

ϕ−1

(∫ 34

12

λhi0 (τ)fi0 (u(τ))dτ

)ds

≥ min∫ 1

4

0ϕ−1

(λMϕ(

1

4‖u‖∞)

∫ 12

14

hi0 (τ)dτ

)ds,

∫ 1

34

ϕ−1

(λMϕ(

1

4‖u‖∞)

∫ 34

12

hi0 (τ)dτ

)ds

=1

4ϕ−1

(λMϕ(

1

4‖u‖∞)min

∫ 12

14

hi0 (τ)dτ,

∫ 34

12

hi0 (τ)dτ

)

38

Introduction1

Choose M > 0 sufficiently large such that

M =γ(32)

min

∫ 1214

hi0(τ)dτ,∫ 3

412

hi0(τ)dτ

> 0,

where γ is the function appeared in condition (A). Then

2‖T i0λ (u)‖∞ ≥1

4ϕ−1

(γ(32)ϕ(

1

4‖u‖∞)

).

Applying Remark 1 with σ = 32 and x = 14‖u‖∞, we get

2‖T i0λ (u)‖∞ ≥1

4· 32 · 1

4‖u‖∞ = 2‖u‖∞.

Thus‖Tλ(u)‖∞ ≥ ‖T i0λ (u)‖∞ ≥ ‖u‖∞, for u ∈ ∂KR.

39

Introduction2

6. Strongly coupled p-Laplacian systems

ψp(u

′(t))′ + h(t) · f(u(t)) = 0, t ∈ (0, 1)

u(0) = u(1) = 0.(P )

• ψp : RN → RN ; ψp(x) = |x|p−2x, x ∈ RN , p > 1• h : (0, 1)→ RN , f : RN → RN continuous.• x · y := (x1y1, · · · , xNyN ) Hadamard product.

(|u′(t)|p−2u′1(t)

)′+ h1(t)f1(u1(t), · · · , uN (t))) = 0,

...(|u′(t)|p−2u′N (t)

)′+ hN (t)fN (u1(t), · · · , uN (t))) = 0, t ∈ (0, 1),

ui(0) = 0 = ui(1) = 0, i = 1, · · · , N.

40

Introduction2

Condition on h

Define H ⊂ L1loc((0, 1),RN ) given by

H :=h :

∫ 120 ϕ−1p

(∫ 12s |h(r)|dr

)ds+

∫ 112ϕ−1p

(∫ s12|h(r)|dr

)ds <∞

• ϕp : R→ R ; ϕp(s) = |s|p−2s

• L1(0, 1) ( H ( L1loc(0, 1)

• h(t) = (t−α, (1− t)−β), 1 ≤ α, β < p ⇒ h ∈ H \ L1(0, 1)

41

Operator Set-up g ∈ L1(0, 1)

7. Operator Set-up

7.1. g ∈ L1(0, 1) (’98 Manasevich-Mawhin, JDE)

−ψp(u′(t))′ = g(t), t ∈ (0, 1)

u(0) = u(1) = 0.(PP )

• g ∈ L1((0, 1),RN )∫ t0 ⇒ ψp(u

′(t)) = ψp(u′(0))−

∫ t0 g(r)dr

α , ϕp(u′(0))

u′(t) = ψ−1p (α−∫ t0 g(r)dr) ∈ C[0, 1]∫ t

0 ⇒ u(t) = u(0) +∫ t0 ψ−1p (α−

∫ t0 g(r)dr)ds

u(0) = 0 ⇒ u(t) =∫ t0 ψ−1p (α−

∫ t0 g(r)dr)ds

Need u(1) = 0 ⇐⇒∫ 10 ψ−1p (α−

∫ t0 g(r)dr)ds = 0

42

Operator Set-up g ∈ L1(0, 1)

Key Lemma

For g ∈ L1(0, 1),system

∫ 10 ψ−1p (α−

∫ t0 g(r)dr)ds = 0 has a unique zero α , α(g) in RN

Conclusion.The unique solution of (PP ) can be represented by

u(t) =

∫ t

0ψ−1p

(α(g)−

∫ s

0g(r)dr

)ds

43

Operator Set-up g ∈ L1(0, 1)

Some basic facts

1• ψ−1p (x) = ψp∗(x), p∗ = p

p−1

2• |ψ−1p (x+ y)| = |x+ y|p

∗−1 ≤ (|x|+ |y|)p∗−1 = ϕ−1

p (|x|+ |y|).

3• |∫ 1

2tg(r)dr| ≤ N

∫ 12t|g(r)|dr

4• ϕ−1p (a+ b) ≤ Cp

(ϕ−1p (a) + ϕ−1

p (b)), ∀a, b > 0 ; Cp =

1, p > 2

22−pp−1 , 1 < p ≤ 2

5• < ψ−1p (x)− ψ−1

p (y), x− y >> 0, ∀x, y ∈ RN , x 6= y

6• < ψp(x), x >= |x|p

7• < ψ−1p (x), x >= |ψ−1

p (x)|p = |x||x|p∗−1 = |x||ψ−1p (x)|.

44

Operator Set-up g ∈ H \ L1(0, 1)

7.2. g ∈ H \ L1(0, 1)

−ψp(u′(t))′ = g(t), t ∈ (0, 1)

u(0) = u(1) = 0.(PP )

t ∈(0, 12]

∫ 12

t⇒ ψp(u

′(t)) = ψp(u′( 12 )) +

∫ 12

tg(r)dr ; α , ψp(u

′( 12 ))

u′(t) = ψ−1p

(α+

∫ 12

tg(r)dr

)|ψ−1p

(α+

∫ 12

tg(r)dr

)| ≤ ϕ−1p

(|α|+ |

∫ 12

tg(r)dr|

)≤ Cp

(ϕ−1p (|α|) + ϕ−1p (|

∫ 12

tg(r)dr|)

)≤ Cpϕ−1p (|α|) + Cpϕ

−1p (N)ϕ−1p (

∫ 12

t|g(r)|dr) ∈ L1(0, 12 )

45

Operator Set-up g ∈ H \ L1(0, 1)∫ t0⇒ u(t) = u(0) +

∫ t0ψ−1p

(α+

∫ 12

tg(r)dr

)ds

u(0) = 0 ⇒ u(t) =∫ t0ψ−1p

(α+

∫ 12

tg(r)dr

)ds

• u satisfies (PP ) with u(0) = 0 only on[0, 12]

t ∈[12 , 1)

∫ t12⇒ ψp(u

′(t)) = ψp(u′( 12 ))−

∫ t12g(r)dr

u′(t) = ψ−1p

(α−

∫ t12g(r)dr

)∈ L1

(12 , 1)

∫ 1

t⇒ u(t) = u(1)−

∫ 1

tψ−1p

(α−

∫ s12g(r)dr

)ds

u(1) = 0 ⇒ u(t) =∫ 1

tψ−1p

(−α+

∫ s12g(r)dr

)ds

• u satisfies (PP ) with u(1) = 0 only on[12 , 1]

46

Operator Set-up g ∈ H \ L1(0, 1)

Need : u(12−) = u(12

+)

i.e.∫ 1

20ψ−1p

(α+

∫ 12sg(r)dr

)ds =

∫ 112ψ−1p

(−α+

∫ s12g(r)dr

)ds

H(α) :=∫ 1

2

0ψ−1p

(α+

∫ 12

sg(r)dr

)ds

W (α) :=∫ 1

12ψ−1p

(−α+

∫ s12g(r)dr

)ds, α ∈ RN

G(α) := H(α)−W (α) so that G : RN → RN .

Main Theorem 1

For given g ∈ H, G has a unique zero α = α(g), i.e. G(α(g)) = 0.

Conclusion.The unique solution of (PP ) can be represented by

u(t) :=

∫ t0 ψ−1p (α(g) +

∫ 12s g(r)dr)ds, 0 ≤ t ≤ 1

2 ,∫ 1t ψ−1p (−α(g) +

∫ s12g(r)dr)ds, 1

2 ≤ t ≤ 1

47

Operator Set-up Proof of Main Theorem 1

7.3. Proof of Main Theorem 1

Uniqueness:

< G(α1)−G(α2), α1 − α2 >=< H(α1)−W (α1)−H(α2) +W (α2), α1 − α2 >=< H(α1)−H(α2), α1 − α2 > + < W (α2)−W (α1), α1 − α2 >

=∫ 1

20 < ψ−1p (α1 +

∫ 12s g(τ)dτ)− ψ−1p (α2 +

∫ 12s g(τ)dτ), α1 − α2 > ds

+∫ 1

12< ψ−1p (−α2 +

∫ s12g(τ)dτ)− ψ−1p (−α1 +

∫ s12g(τ)dτ), α1 − α2 > ds

> 0, ∀α1, α2 ∈ RN , α1 6= α2.

5• < ψ−1p (x)− ψ−1

p (y), x− y >> 0, ∀x, y ∈ RN , x 6= y

48

Operator Set-up Proof of Main Theorem 1

Existence:

Claim 1. ∃ r > 0 such that < G(α), α >> 0 for all α ∈ ∂Br(0) ⊂ RN .

Construct a homotopy

h(λ, α) = λα+ (1− λ)G(α), ∀λ ∈ [0, 1].

< h(λ, α), α >=< λα+ (1− λ)G(α), α >= λ|α|2 + (1− λ) < G(α), α >> 0,∀α ∈ ∂Br(0), ∀λ ∈ [0, 1].

Take Ω = Br(0), then < h(λ, α), α >> 0, for all α ∈ ∂Ωh(λ, ·) 6= 0 on ∂Ω, for all λ ∈ [0, 1].

Brouwer degree dB(h(λ, α),Ω, 0) is well-defined and by the homotopy invariance,

dB(G(·),Ω, 0) = dB(h(0, ·),Ω, 0) = dB(h(1, ·),Ω, 0) = dB(id,Ω, 0) = 1

Thus there exists a solution α , α(g) for G(α(g)) = 0.

49

Operator Set-up Proof of Main Theorem 1

Proof of Claim 1:∃ r > 0 such that < G(α), α >> 0 for all α ∈ ∂Br(0) ⊂ RN .

Show : < H(α), α >> 0 and < W (α), α >≤ 0, for sufficiently large |α|.

< H(α), α >=∫ 1

2

0< ψ−1p (α+

∫ 12

sg(τ)dτ), α > ds

=∫ δ0< ψ−1p (α+

∫ 12

sg(τ)dτ), α > ds+

∫ 12

δ< ψ−1p (α+

∫ 12

sg(τ)dτ), α > ds

A ,∫ δ0< ψ−1p (α+

∫ 12

sg(τ)dτ), α > ds,

B ,∫ 1

2

δ< ψ−1p (α+

∫ 12

sg(τ)dτ), α > ds

50

Operator Set-up Proof of Main Theorem 1

B =∫ 1

2

δ< ψ−1p (α+

∫ 12

sg(τ)dτ), α > ds

=∫ 1

2

δ< ψ−1p (α+

∫ 12

sg(τ)dτ), α+

∫ 12

sg(τ)dτ > ds

−∫ 1

2

δ< ψ−1p (α+

∫ 12

sg(τ)dτ),

∫ 12

sg(τ)dτ > ds

≥∫ 1

2

δ|α+

∫ 12

sg(τ)dτ ||ψ−1p (α+

∫ 12

sg(τ)dτ)|ds−Mδ

∫ 12

δ|ψ−1p (α+

∫ 12

sg(τ)dτ)|ds

=∫ 1

2

δ|ψ−1p (α+

∫ 12

sg(τ)dτ)|(|α+

∫ 12

sg(τ)dτ | −Mδ)ds

• |∫ 1

2

sg(τ)dτ |2 ≤

∑Ni=1(

∫ 12

δ|gi(τ)|dτ)2 ,M2

δ .

7• < ψ−1p (x), x >= |ψ−1p (x)|p = |x||x|p∗−1 = |x||ψ−1p (x)|.

51

Operator Set-up Proof of Main Theorem 1

Aside.

B ≥∫ 1

2

δ|ψ−1p (α+

∫ 12

sg(τ)dτ)|(|α+

∫ 12

sg(τ)dτ | −Mδ)ds

|ψ−1p (α+∫ 1

2

sg(τ)dτ)| = |α+

∫ 12

sg(τ)dτ |p∗−1 ≥ (|α| − |

∫ 12

sg(τ)dτ |)p∗−1

≥ (|α| −Mδ)p∗−1 →∞, as |α| → ∞.

|α+∫ 1

2

sg(τ)dτ | −Mδ ≥ |α| − |

∫ 12

sg(τ)dτ | −Mδ

≥ |α| − 2Mδ →∞, as |α| → ∞.

52

Operator Set-up Proof of Main Theorem 1

A =∫ δ0< ψ−1p (α+

∫ 12

sg(τ)dτ), α > ds.

| < ψ−1p (α+

∫ 12sg(τ)dτ), α > | ≤ |ψ−1

p (α+∫ 1

2sg(τ)dτ)||α|

≤ ϕ−1p (|α|+ |

∫ 12sg(τ)dτ |)|α|

≤ Cpϕ−1p (|α|)|α|+ Cpϕ

−1p (N

∫ 12s|g(τ)|dτ)|α|

= Cp|α|p∗

+ Cpϕ−1p (N)ϕ−1

p (∫ 1

2s|g(τ)|dτ)|α|

A ≥∫ δ0−Cp|α|p

∗− Cpϕ−1

p (N)ϕ−1p (∫ 1

2s|g(τ)|dτ)|α|

= −Cpδ|α|p∗− Cpϕ−1

p (N)∫ δ0ϕ−1p (∫ 1

2s|g(τ)|dτ)|α|.

2• |ψ−1p (x+ y)| ≤ ϕ−1

p (|x|+ |y|).

3• |∫ 1

2sg(τ)dτ | ≤ N

∫ 12s|g(τ)|dτ.

53

Operator Set-up Proof of Main Theorem 1

< H(α), α >= A+B

≥ −Cpδ|α|p∗ − Cpϕ−1p (N)

∫ δ0ϕ−1p (

∫ 12

s|g(τ)|dτ)ds|α|

−Mδ

∫ 12

δ|ψ−1p (α+

∫ 12

sg(τ)dτ)|ds

+∫ 1

2

δ|ψ−1p (α+

∫ 12

sg(τ)dτ)||α+

∫ 12

sg(τ)dτ |ds

• |ψ−1p (α+∫ 1

2

sg(τ)dτ)||α+

∫ 12

sg(τ)dτ | = |α+

∫ 12

sg(τ)dτ |p∗

≥ (|α| −Mδ)p∗ = |α|p∗(1− M(δ)

|α| )p∗

• −Mδ

∫ 12

δ|ψ−1p (α+

∫ 12

sg(τ)dτ)|ds = −Mδ

∫ 12

δ|α+

∫ 12

sg(τ)dτ |p∗−1ds

≥ −Mδ(|α|+Mδ)p∗−1 · 12 = −Mδ

2 ·(|α|+Mδ)

p∗−1

|α|p∗ · |α|p∗

= −Mδ

2 · (1 +Mδ

|α| )p∗−1 · 1

|α| · |α|p∗ .

54

Operator Set-up Proof of Main Theorem 1

< H(α), α > ≥ |α|p∗[(1− Mδ

|α| )p∗

− Cpδ

− Cpϕ−1p (N)

∫ δ0ϕ−1p (

∫ 12

s|g(τ)|dτ)ds · 1

|α|p∗−1

− Mδ

2 · (1 +Mδ

|α| )p∗−1 · 1

|α|].

Taking big |α| and small δ, get < H(α), α >> 0.

Similarly, we can show< W (α), α >≤ 0 for all α ∈ ∂Br(0).

55

Compactness of solution operators Main result

8. Compactness of solution operators

8.1. Main result

ψp(u

′(t))′ + h(t) · f(u(t)) = 0, t ∈ (0, 1)

u(0) = u(1) = 0.(P )

• h ∈ H \ L1((0, 1),RN ), f ∈ C(RN ,RN )

T (u)(t) :=

∫ t0ψ−1p

(α(hf(u)) +

∫ 12

sh(r) · f(u(r))dr

)ds, 0 ≤ t ≤ 1

2∫ 1

tψ−1p

(−α(hf(u)) +

∫ s12h(r) · f(u(r))dr

)ds, 1

2 ≤ t ≤ 1,

• α , α(hf(u)) ∈ RN is the unique solution of∫ 12

0ψ−1p

(α+

∫ 12

sh(r) · f(u(r))dr

)ds

=∫ 1

12ψ−1p

(−α+

∫ s12h(r) · f(u(r))dr

)ds

56

Compactness of solution operators Main result

• T : C[0, 1]→ C[0, 1]• (P )⇐⇒ u = T (u)• α : C[0, 1]→ RN defined by α(u) = α(hf(u)).

Lemma

α is continuous on C[0, 1]α maps bounded sets in C[0, 1] into bounded sets in RN

• T is continuous on C[0, 1].• For B ∈ C[0, 1] bounded, T (B) is bounded and equicontinuous.

Main Theorem 2.

T : C[0, 1]→ C[0, 1] is completely continuous.

57

Compactness of solution operators Proof of Lemma

8.2. Proof of Lemma

Claim 2. α sends any bounded sets in C[0, 1] into bounded sets in RN .

Assume that un is bounded in C[0, 1], αn , α(hf(un)).Suppose αn is unbounded in RN ,∃ αnk such that |αnk | → ∞ as k →∞.We know G(αnk) = 0, i.e. < G(αnk), αnk >= 0, ∀k.By Claim 1 in the proof of Main Theorem 1,Some < G(αnk), αnk >> 0 when |αnk | → ∞.

This is a contradiction.

58

Compactness of solution operators Proof of Lemma

Claim 3. α : C[0, 1]→ RN is continuous.

Assume that un → u in C[0, 1]Show α(hf(un))→ α(hf(u)) as n→∞.

Claim 2 ⇒ α(hf(un)) is bounded ⇒ ∃ α(hf(unk))→ α

•∫ 1

2

0ψ−1p

(α(hf(unk)) +

∫ 12

sh(τ) · f(unk(τ))dτ

)ds

=∫ 1

12ψ−1p

(−α(hf(unk)) +

∫ s12h(τ) · f(unk(τ))dτ

)ds.∣∣∣ψ−1p (

α(hf(unk)) +∫ 1

2

sh(τ) · f(unk(τ))dτ

) ∣∣∣≤ ϕ−1p

(∣∣∣α(hf(unk))∣∣∣+ ∣∣∣ ∫ 12

sh(τ) · f(unk(τ))dτ)

∣∣∣)≤ ϕ−1p

(K ′ +NM ′

∫ 12

s|h(τ)|dτ

)≤ Cpϕ−1p (K ′) + Cpϕ

−1p (NM ′)ϕ−1p

(∫ 12

s|h(τ)|dτ

)• K ′ , maxk |α(hf(unk))|, M ′ , maxk ‖f(unk)‖∞

59

Compactness of solution operators Proof of Main Theorem 2

8.3. Proof of Main Theorem 2

Claim 4. T (B) is bounded for B a bounded subset in C[0, 1]For 0 ≤ t ≤ 1

2,

|T (u)(t)| =∣∣∣ ∫ t

0ψ−1p (α(hf(u)) +

∫ 12

sh(τ) · f(u(τ))dτ)ds

∣∣∣≤ N

∫ t0

∣∣∣ψ−1p

(α(hf(u)) +

∫ 12

sh(τ) · f(u(τ))dτ

) ∣∣∣ds≤ N

∫ t0ϕ−1p

(|α(hf(u))| + |

∫ 12

sh(τ) · f(u(τ))dτ)|

)ds

≤ N

∫ t0ϕ−1p

(K +NM

∫ 12

s|h(τ)|dτ

)ds

≤ NCpϕ−1p (K) ·

1

2+NCpϕ

−1p (NM)

∫ 12

0ϕ−1p

(∫ 12

s|h(τ)|dτ

)ds,

• K = maxu∈B |α(hf(u))|, M = maxu∈B ‖f(u)‖∞.

Similarly, for 12≤ t ≤ 1, we have

|T (u)(t)| ≤ NCpϕ−1p (K) ·

1

2+NCpϕ

−1p (NM)

∫ 1

12

ϕ−1p

(∫ s12

|h(τ)|dτ)ds.

60

Compactness of solution operators Proof of Main Theorem 2

Claim 5. T (B) is equicontinuous.

There are four cases : Case 0 ≤ t1, t2 ≤ 12

(or 12≤ t1, t2 ≤ 1),∣∣∣T (u)(t1)− T (u)(t2)

∣∣∣=∣∣∣ ∫ t2

t1

ψ−1p

(α(hf(u)) +

∫ 12

s

h(τ) · f(u(τ))dτ

)ds∣∣∣

≤ N∣∣∣ ∫ t2

t1

∣∣∣ψ−1p

(α(hf(u)) +

∫ 12

s

h(τ) · f(u(τ))dτ

)∣∣∣ds∣∣∣≤ N

∣∣∣ ∫ t2

t1

ϕ−1p

(|α(hf(u))|+ |

∫ 12

s

h(τ) · f(u(τ))dτ)|

)ds∣∣∣

≤ N∣∣∣ ∫ t2

t1

ϕ−1p

(K +NM

∫ 12

s

|h(τ)|dτ

)ds∣∣∣

≤ NCpϕ−1p (K)|t2 − t1|+NCpϕ

−1p (NM)

∣∣∣ ∫ t2

t1

ϕ−1p

(∫ 12

s

|h(τ)|dτ

)ds∣∣∣.

h ∈ H ⇒ the above upper bound tends to 0 as |t2 − t1| does.

61

Compactness of solution operators Proof of Main Theorem 2

Case 0 ≤ t1 ≤ 12≤ t2 ≤ 1 (or 0 ≤ t2 ≤ 1

2≤ t1 ≤ 1),

|T (u)(t1)− T (u)(t2)|

=∣∣∣ ∫ t1

0ψ−1p

(α(hf(u)) +

∫ 12

sh(τ) · f(u(τ))dτ

)ds−

∫ 1

t2

ψ−1p

(−α(hf(u)) +

∫ s12

h(τ) · f(u(τ))dτ)ds∣∣∣

=∣∣∣ ∫ 1

2

0ψ−1p

(α(hf(u)) +

∫ 12

sh(τ) · f(u(τ))dτ

)ds−

∫ 12

t1

ψ−1p

(α(hf(u)) +

∫ 12

sh(τ) · f(u(τ))dτ

)ds

−[ ∫ 1

12

ψ−1p

(−α(hf(u)) +

∫ s12

h(τ) · f(u(τ))dτ)ds−

∫ t212

ψ−1p

(−α(hf(u)) +

∫ s12

h(τ) · f(u(τ))dτ)ds]∣∣∣

=∣∣∣ ∫ t2

12

ψ−1p

(−α(hf(u)) +

∫ s12

h(τ) · f(u(τ))dτ)ds−

∫ 12

t1

ψ−1p

(α(hf(u)) +

∫ 12

sh(τ) · f(u(τ))dτ

)ds∣∣∣

≤∣∣∣ ∫ t2

12

ψ−1p

(−α(hf(u)) +

∫ s12

h(τ) · f(u(τ))dτ)ds∣∣∣ + ∣∣∣ ∫ 1

2

t1

ψ−1p

(α(hf(u)) +

∫ 12

sh(τ) · f(u(τ))dτ

)ds∣∣∣

≤ N∫ t2

12

∣∣∣ψ−1p

(−α(hf(u)) +

∫ s12

h(τ) · f(u(τ))dτ) ∣∣∣ds +N

∫ 12

t1

∣∣∣ψ−1p

(α(hf(u)) +

∫ 12

sh(τ) · f(u(τ))dτ

) ∣∣∣ds≤ N

∫ t212

ϕ−1p

(K +NM

∫ s12

|h(τ)|dτ)ds +N

∫ 12

t1

ϕ−1p

(K +NM

∫ s12

|h(τ)|dτ)ds

≤ 2NCpϕ−1p (K)|t2 − t1|

+NCpϕ−1p (NM)

[ ∫ t212

ϕ−1p

(∫ s12

|h(τ)|dτ)ds +

∫ 12

t1

ϕ−1p

(∫ 12

s|h(τ)|dτ

)ds].

62

Compactness of solution operators Proof of Main Theorem 2

Claim 6. T is continuous.

Assume un → u in C[0, 1].The continuity of α and Lebesgue Dominated Convergence Theorem ⇒

‖T (un)− T (u)‖∞

= sup0≤t≤1

∣∣∣T (un)(t)− T (u)(t)∣∣∣

≤ sup0≤t≤ 1

2

∣∣∣T (un)(t)− T (u)(t)∣∣∣ + sup

12≤t≤1

∣∣∣T (un)(t)− T (u)(t)∣∣∣

≤ sup0≤t≤ 1

2

∣∣∣ ∫ t0ψ−1p

(α(hf(un)) +

∫ 12

sh(τ) · f(un(τ))dτ

)

− ψ−1p

(α(hf(u)) +

∫ 12

sh(τ) · f(u(τ))dτ

)ds∣∣∣

+ sup12≤t≤1

∣∣∣ ∫ 1

tψ−1p

(−α(hf(un)) +

∫ s12

h(τ) · f(un(τ))dτ)

− ψ−1p

(−α(hf(u)) +

∫ s12

h(τ) · f(u(τ))dτ)ds∣∣∣

→ 0, as n→∞.

63

Compactness of solution operators Proof of Main Theorem 2

Thank you for your attention...

64