Yong-Hoon Lee Pusan National University (with X. Xu)cmac.yonsei.ac.kr/lecture/lee-20160401.pdf ·...
Transcript of Yong-Hoon Lee Pusan National University (with X. Xu)cmac.yonsei.ac.kr/lecture/lee-20160401.pdf ·...
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)
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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 . . .
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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
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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
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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
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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
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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
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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
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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]
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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
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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.
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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.
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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)
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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)
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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→ +∞
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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→ −∞
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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→ +∞.
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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.
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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).
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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)
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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).
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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).
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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
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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.
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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.
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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