Smoothing effect and Fredholm property for first-order hyperbolic - - PowerPoint PPT Presentation

Smoothing effect and Fredholm property for first-order hyperbolic PDEs

Irina Kmit

Humboldt University of Berlin, Germany

Padova, 28 June 2012

The problem under consideration

∂tuj + aj(x, t)∂xuj +

n

X

k=1

bjk(x, t)u = fj(x, t), j ≤ n, (x, t) ∈ (0, 1) × R uj(x, 0) = ϕj(x), j ≤ n, x ∈ [0, 1] uj(0, t) = (Ru)j(t), 1 ≤ j ≤ m, t ∈ R+ uj(1, t) = (Ru)j(t), m < j ≤ n, t ∈ R+ Motivation: traveling-wave models from

• laser dynamics
• population dynamics
• chemical kinetics

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Smoothing, fredholmness and hyperbolic PDEs

• Motivation

– bifurcation analysis of traveling-wave models – local investigations of semilinear and quasilinear hyperbolic problems via ∗ Implicit Function Theorem ∗ Lyapunov-Schmidt reduction – periodic synhronizations via ∗ averaging procedure

• Complications and related topics

– bad regularity properties of hyperbolic operator: ∗ no smoothing effect in general ∗ lost-of-smoothness effect – small denominators – non-strict hyperbolicity and constraints on lower-order terms

2

Examples

• 1. Smoothing effect. Heat equation:

∂u ∂t − ∆u = 0 u(x, 0) = ϕ(x) u(x, t) = 1 (4πt)n/2 Z

Rn exp

„−|x − y|2 4t « ϕ(y) dy

3

Examples

• 1. Smoothing effect. Heat equation:

∂u ∂t − ∆u = 0 u(x, 0) = ϕ(x) u(x, t) = 1 (4πt)n/2 Z

Rn exp

„−|x − y|2 4t « ϕ(y) dy

• 2. No smoothing effect. Wave equation:

∂2u ∂t2 − ∂2u ∂x2 = 0 u = ϕ(x − t) + ϕ(x + t) 2 + 1 2 Z x+t

x−t

ψ(s) ds

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Smoothing property (definition)

Definition. A solution u is called smoothing if for any k there exists T > 0 such that u is Ck-smooth above the line t = T. Remark. Any t-periodic and smoothing solution immediately meets the C∞- regularity in the entire domain. Remark. The domain of influence of the initial conditions is in general infinite and is determined by the boundary operator as well as the lower-order terms .

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Smoothing effect I: Classical boundary conditions

∂tuj + aj(x, t)∂xuj +

n

X

k=1

bjk(x, t)u = fj(x, t), j ≤ n, (x, t) ∈ (0, 1) × R uj(x, 0) = ϕj(x), j ≤ n, x ∈ [0, 1] uj(0, t) = hj(t), 1 ≤ j ≤ m, t ∈ R+ uj(1, t) = hj(t), m < j ≤ n, t ∈ R+ Assumptions: aj > 0 for all j ≤ m and aj < 0 for all j > m, inf

x,t |aj| > 0 for all j ≤ n,

for all 1 ≤ j = k ≤ n there exists pjk ∈ C1([0, 1] × R) such that bjk = pjk(ak − aj) and pjk = 0 in the interior of the domain {(x, t) : aj(x, t) = ak(x, t)}.

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Smoothing effect I: main result

• Theorem. Assume that aj, bjk, fj, hj ∈ C∞ and ϕj ∈ C. Then any

continuous solution is smoothing. Operator representation of the problem: u = CBu + Du + Ff, where (Cu)j(x, t) = cj(xj, x, t)uj (xj, ωj(xj; x, t)) , cj(ξ, x, t) = exp Z ξ

x

„bjj aj « (η, ωj(η; x, t)) dη, (Bu)j(x, t) = ( hj(t) if t > 0, ϕj(x) if t = 0, (Du)j (x, t) = Z xj

x

cj(ξ, x, t) aj(ξ, ωj(ξ; x, t)) X

k=j

“ bjkuk ” (ξ, ωj(ξ; x, t))dξ

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Proof idea

• 1-st step: u = CBu + Du + Ff.

Hence u = (CBu + DCBu) + D2u + (I + D)Ff. If T1 > 0 is sufficiently large, then u|t>T1 = (C + DC)h + D2u + (I + D)Ff. The operator D2 maps C(Π)n continuously into C1

t (Π ∩ t > T1)n.

• 2-nd step: Set v = ∂tu. Then

(∂t + aj∂x)vj +

n

X

k=1

bjkvk − ∂taj aj vj = Gj(fj, ∂tfj, u) and we have the representation v|ΠT2 = ˜ Ch′ + ˜ Dv + ˜ FG(f, ∂tf, u).

• Further we use induction on the order of regularity.

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Smoothing effect II: Integral boundary conditions in population models

(∂t + ∂x + µ)u = 0, (x, t) ∈ (0, 1) × R, u(x, 0) = ϕ(x), x ∈ [0, 1], u(0, t) = h „Z 1 γ(x)u(x, t) dx « , t ∈ R, Theorem. Assume that h, γ ∈ C∞ and ϕ ∈ C. Then any continuous solution is smoothing.

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Proof idea

1-st step: Any continuous solution for large enough T1 is given by the formula u|t≥T1 = CBu = CBCu = e−µxh „Z 1 γ(ξ)e−µξu(0, t − x − ξ) dξ « = e−µxh „Z t−x

t−x−1

γ(t − x − τ)eµ(x−t+τ)u(0, τ) dτ « , 2-nd step: Set v = ut. Then for large enough T2 we have v|t≥T2 = e−µxh′ „Z 1 γ(ξ)u(ξ, t − x) dξ « Z 1 γ(ξ)e−µξv(0, t − x − ξ) dξ = e−µxh′ „Z 1 γ(ξ)u(ξ, t − x) dξ « Z t−x

t−x−1

γ(t − x − τ)eµ(x−t+τ)v(0, τ) dτ.

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Smoothing effect III: Dissipative boundary conditions and periodic problems

∂tu + a(x, t)∂xu + b(x, t)u = f(x, t), (x, t) ∈ (0, 1) × R u(x, t + 2π) = u(x, t), x ∈ [0, 1] uj(0, t) = hj(z(t)), 1 ≤ j ≤ m, t ∈ R+ uj(1, t) = hj(z(t)), m < j ≤ n, t ∈ R+ with z(t) = (u1(1, t), . . . , um(1, t), um+1(0, t), . . . , un(0, t)) .

• Theorem. Assume that aj, bjk, fj, hj ∈ C∞ and aj, bjk, fj are 2π-

periodic in t. If ˛ ˛∂kh′

j(z)

˛ ˛ exp (Z xj

x

bjj aj − l∂taj a2

j

! (η, ωj(η; x, t)) dη ) < 1 for all j, k ≤ n, x ∈ [0, 1], t ∈ R, z ∈ Rn, and l = 0, 1, . . . , r, then any continuous solution belongs to Cr.

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Proof idea

• u = Cu + Du + Ff,

u = CBu + Du + Ff ⇒ u = CBu + (DC + D2)u + (I + D)Ff

• The operator I − CB ∈ L (C1

t )n is bijective

⇒ u = (I − CB)−1h (DC + D2)u + (I + D)Ff i

• The operator DC + D2 is smoothing (maps C into C1

t )

11

Fredholmness: a problem from laser dynamics

∂tuj + aj(x)∂xuj +

n

X

k=1

bjk(x)uk = fj(x, t), j ≤ n, (x, t) ∈ (0, 1) × R uj(x, t + 2π) = uj(x, t), j ≤ n, (x, t) ∈ [0, 1] × R uj(0, t) =

n

X

k=m+1

r0

jkuk(0, t),

j ≤ m, t ∈ R uj(1, t) =

m

X

k=1

r1

jkuk(1, t),

m + 1 ≤ j ≤ n, t ∈ R.

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Choice of function spaces

(u, f) ∈ (V γ, W γ) Spaces for the right-hand sides: W γ = L2 ` 0, 1; Hγ

per(R)

´n Spaces for the solutions: U γ(a) = n u ∈ W γ : ∂xu ∈ W γ−1, [∂tuj + aj∂xuj]n

j=1 ∈ W γo

u2

Uγ(a) = u2 W γ +

‚ ‚ ‚[∂tuj + aj∂xuj]n

j=1

‚ ‚ ‚

2 W γ .

V γ(a) = n u ∈ U γ(a) : u fulfills refl. bound. conditions

• 13

Properties of the solution spaces

Lemma. (i) The space U γ(ω, a) is complete. (ii) If γ ≥ 1, then for any x ∈ [0, 1] there exists a continuous trace map u ∈ U γ(ω, a) → u(x, ·) ∈ L2 ((0, 2π); Rn) . (iii) If γ > 3/2, then U γ(ω, a) is continuously embedded into C([0, 1]× [0, 2π]; Rn).

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Operator representation of the problem

A(a, b0)u + B(b1)u = f, where A(a, b0) ∈ L(V γ(a); W γ) is defined by A(a, b0)u = [∂tu + aj∂xuj + bjjuj]n

j=1 ,

and B(b1) ∈ L(W γ) is defined by B(b1)u = 2 4X

j=k

bjkuk 3 5

n j=1

15

Fredholm property

• Theorem. (K., Recke, 2012) Assume that

aj, bjj ∈ L∞(0, 1), inf

0≤x≤1 |aj(x)| > 0 for all j ≤ n,

for all j = k there is cjk ∈ BV (0, 1) such that bjk(x) = cjk(x)(aj(x) − ak(x)) for a.a. x ∈ [0, 1], ff R(a, b0)

n

X

j,l=m+1 m

X

k=1

|r1

jkr0 kl| < 1.

Then (i) The operator A+B is a Fredholm operator with index zero from V γ(a) into W γ for all γ ≥ 1. (ii) (smoothing effect) The subspaces ker(A+B) and ker( ˜ A+ ˜ B) do not depend on γ, ker(A + B)∗ = ker( ˜ A + ˜ B), and im(A + B) = n f ∈ W γ : f, uL2 = 0 for all u ∈ ker “ ˜ A + ˜ B ”o .

16

Fredholmness and non-strict hyperbolicity

Fredholmness for non-strict hyperbolicity necessarily entails constraints on zero-order terms: ∂tu1 + ∂xu1 = ∂tu2 + ∂xu2 + bu1 = 0, u1(x, t + 2π) − u1(x, t) = u2(x, t + 2π) − u2(x, t) = 0, u1(0, t) − r0u2(0, t) = u2(1, t) − r1u1(1, t) = 0. No small denominators: r0r1 < 1. Assume that b = r0r1 − 1 r0 , covering arbitrarily large |b|. Then u1(x, t) = sin l(t − x), u2(x, t) = b „ 1 1 − r0r1 − x « sin l(t − x), l ∈ N, gives infinitely many linearly independent solutions. Hence, the Fredholmness is destroyed.

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Fredholmness: proof idea

Theorem (Fredholmness criterion). Let W be a Banach space, I the identity in W, and C ∈ L(W) with Ck being compact for some k ∈ N. Then I + C is a Fredholm operator of index zero.

• A ∈ L(V γ; W γ) is an isomorphism, hence A + B ∈ L(V γ; W γ) is

Fredholm iff I + BA−1 ∈ L(W γ) is Fredholm.

• (BA−1)2 ∈ L(W γ) is compact.
• (right) parametrix or (right) regularizer: A−1 (I − BA−1):

(A + B) ˆ A−1 ` I − BA−1´˜ = I − ` BA−1´2 .

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Thank you!

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