time-periodic solutions of a cubic wave equation - dominic...
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0 29.08.2019 - Waves 2019 Dominic Scheider-
WAVES 2019, Vienna
Time-Periodic Solutions of a Cubic Wave Equation
Dominic Scheider
KIT – The Research University in the Helmholtz Association www.kit.edu
Outline
1 29.08.2019 - Waves 2019 Dominic Scheider-
1 The main result
2 The strategy
3 Conclusion
The main result
2 29.08.2019 - Waves 2019 Dominic Scheider-
Aim:Find solutions U = U(t, x) of the wave-type equation
∂2tU − ∆U − U = Γ(x)U3 on R×R3 (W)
which are real-valued and .... periodic in t, . localized in x , . radially symmetric in x .
Method:(i) Polychromatic ansatz
U(t, x) = ∑k∈N0
cos(kt) uk (x). (P)
(ii) Bifurcation from a stationary solution U0(t, x) = w0(x).
The main result
2 29.08.2019 - Waves 2019 Dominic Scheider-
Aim:Find solutions U = U(t, x) of the wave-type equation
∂2tU − ∆U − U = Γ(x)U3 on R×R3 (W)
which are real-valued and .... periodic in t, . localized in x , . radially symmetric in x .
Method:(i) Polychromatic ansatz
U(t, x) = ∑k∈N0
cos(kt) uk (x). (P)
(ii) Bifurcation from a stationary solution U0(t, x) = w0(x).
The main result
3 29.08.2019 - Waves 2019 Dominic Scheider-
Aim:Find solutions U = U(t, x) of the wave-type equation
∂2tU − ∆U − U = Γ(x)U3 on R×R3 (W)
which are real-valued and .... periodic in t, . localized in x , . radially symmetric in x .
Method:(i) Polychromatic ansatz
U(t, x) = ∑k∈N0
cos(kt) uk (x), (P)
uk ∈ X := {u ∈ Crad(R3,R) | sup(1+ |x |)|u(x)| < ∞}.
(ii) Bifurcation from a stationary solution U0(t, x) = w0(x).
The main result
4 29.08.2019 - Waves 2019 Dominic Scheider-
Aim:Find solutions U ∈ C2
per(R,X ) of the wave-type equation
∂2tU − ∆U − U = Γ(x)U3 on R×R3. (W)
Method:(i) Polychromatic ansatz
U(t, x) = ∑k∈N0
cos(kt) uk (x), uk ∈ X . (P)
(ii) Bifurcation from a stationary solution U0(t, x) = w0(x).
The main result
5 29.08.2019 - Waves 2019 Dominic Scheider-
Let Γ ∈ L∞(R3,R) be radial and continuously differentiable.Let w0 ∈ X with −∆w0 − w0 = Γ(x)w3
0 on R3.
Theorem: (S, 2019)There exist an interval I ⊆ R, 0 ∈ I and a family (Uη)η∈I of real-valued,classical solutions Uη = Uη(t, x) ∈ C2
per(R,X ) of the wave equation
∂2tU − ∆U − U = Γ(x)U3 on R×R3 (W)
which is a continuous curve in C (R,X ) with. U0(t, x) = w0(x),. Uη non-stationary and 2π-periodic in time (η 6= 0).
The main result
5 29.08.2019 - Waves 2019 Dominic Scheider-
Let Γ ∈ L∞(R3,R) be radial and continuously differentiable.Let w0 ∈ X with −∆w0 − w0 = Γ(x)w3
0 on R3.
Theorem: (S, 2019)There exist an interval I ⊆ R, 0 ∈ I and a family (Uη)η∈I of real-valued,classical solutions Uη = Uη(t, x) ∈ C2
per(R,X ) of the wave equation
∂2tU − ∆U − U = Γ(x)U3 on R×R3 (W)
which is a continuous curve in C (R,X ) with
. U0(t, x) = w0(x),
. Uη non-stationary and 2π-periodic in time (η 6= 0).
The main result
5 29.08.2019 - Waves 2019 Dominic Scheider-
Let Γ ∈ L∞(R3,R) be radial and continuously differentiable.Let w0 ∈ X with −∆w0 − w0 = Γ(x)w3
0 on R3.
Theorem: (S, 2019)There exist an interval I ⊆ R, 0 ∈ I and a family (Uη)η∈I of real-valued,classical solutions Uη = Uη(t, x) ∈ C2
per(R,X ) of the wave equation
∂2tU − ∆U − U = Γ(x)U3 on R×R3 (W)
which is a continuous curve in C (R,X ) with. U0(t, x) = w0(x),. Uη non-stationary and 2π-periodic in time (η 6= 0).
The main result
6 29.08.2019 - Waves 2019 Dominic Scheider-
Remarks:Family of polychromatic solutions (“breathers”)
Uη(t, x) = ∑k∈N0
cos(kt) uηk (x).
. (Excitation of s-th mode)
Given s ∈N, find Uη with ddη
∣∣η=0u
ηk 6= 0 iff k = s.
. Extension to the Klein-Gordon equation
∂2tU − ∆U +m2 U = Γ(x)U3 on R×R3.
. Open:Other space dimensions / powers (easy?); non-constant potentials (hard!).
The main result
6 29.08.2019 - Waves 2019 Dominic Scheider-
Remarks:Family of polychromatic solutions (“breathers”)
Uη(t, x) = ∑k∈N0
cos(kt) uηk (x).
. (Excitation of s-th mode)
Given s ∈N, find Uη with ddη
∣∣η=0u
ηk 6= 0 iff k = s.
. Extension to the Klein-Gordon equation
∂2tU − ∆U +m2 U = Γ(x)U3 on R×R3.
. Open:Other space dimensions / powers (easy?); non-constant potentials (hard!).
The main result
6 29.08.2019 - Waves 2019 Dominic Scheider-
Remarks:Family of polychromatic solutions (“breathers”)
Uη(t, x) = ∑k∈N0
cos(kt) uηk (x).
. (Excitation of s-th mode)
Given s ∈N, find Uη with ddη
∣∣η=0u
ηk 6= 0 iff k = s.
. Extension to the Klein-Gordon equation
∂2tU − ∆U +m2 U = Γ(x)U3 on R×R3.
. Open:Other space dimensions / powers (easy?); non-constant potentials (hard!).
The main result
6 29.08.2019 - Waves 2019 Dominic Scheider-
Remarks:Family of polychromatic solutions (“breathers”)
Uη(t, x) = ∑k∈N0
cos(kt) uηk (x).
. (Excitation of s-th mode)
Given s ∈N, find Uη with ddη
∣∣η=0u
ηk 6= 0 iff k = s.
. Extension to the Klein-Gordon equation
∂2tU − ∆U +m2 U = Γ(x)U3 on R×R3.
. Open:Other space dimensions / powers (easy?); non-constant potentials (hard!).
Outline
7 29.08.2019 - Waves 2019 Dominic Scheider-
1 The main result
2 The strategy
3 Conclusion
The strategy
8 29.08.2019 - Waves 2019 Dominic Scheider-
Aim: Find solutions U ∈ C2per(R,X ) of the wave-type equation
∂2tU − ∆U − U = Γ(x)U3 on R×R3. (W)
Method:(i) Polychromatic ansatz
U(t, x) = ∑k∈N0
cos(kt) uk (x), uk ∈ X . (P)
(ii) Bifurcation from a stationary solution U0(t, x) = w0(x).
The strategy
9 29.08.2019 - Waves 2019 Dominic Scheider-
Aim: Find solutions U ∈ C2per(R,X ) of the wave-type equation
∂2tU − ∆U − U = Γ(x)U3 on R×R3. (W)
Method:(i) Polychromatic ansatz
U(t, x) = ∑k∈Z
eikt uk (x), uk= u−k ∈ X . (P)
(ii) Bifurcation from a stationary solution U0(t, x) = w0(x).
The strategy
10 29.08.2019 - Waves 2019 Dominic Scheider-
Aim: Find solutions U ∈ C2per(R,X ) of the wave-type equation
∂2tU − ∆U − U = Γ(x)U3 on R×R3. (W)
Method:(i) Polychromatic ansatz
U(t, x) = ∑k∈Z
eikt uk (x), uk = u−k ∈ X . (P)
(ii) Bifurcation from a stationary solution U0(t, x) = w0(x).
The strategy
11 29.08.2019 - Waves 2019 Dominic Scheider-
Aim: Find solutions U ∈ C2per(R,X ) of the wave-type equation
∂2tU − ∆U − U = Γ(x)U3 on R×R3. (W)
Method:(i) Polychromatic ansatz
U(t, x) = ∑k∈Z
eikt uk (x), uk = u−k ∈ X . (P)
(infinite) Helmholtz system
−∆uk − (k2 + 1)uk = ∑l+m+n=k
Γ(x) ulumun. (W ∗)
(ii) Bifurcation from a stationary solution U0(t, x) = w0(x).
The strategy
12 29.08.2019 - Waves 2019 Dominic Scheider-
Aim: Find solutions U ∈ C2per(R,X ) of the wave-type equation
∂2tU − ∆U − U = Γ(x)U3 on R×R3. (W)
Method:(i) Polychromatic ansatz
U(t, x) = ∑k∈Z
eikt uk (x), uk = u−k ∈ X . (P)
(infinite) Helmholtz system for u = (uk )k ∈ `1(N0,X )
−∆uk − (k2 + 1)uk = Γ(x) (u ∗ u ∗ u)k . (W ∗)
(ii) Bifurcation from w = (w0, 0, 0, ...).
The strategy
13 29.08.2019 - Waves 2019 Dominic Scheider-
Aim: Find solutions U ∈ C2per(R,X ) of the wave-type equation
∂2tU − ∆U − U = Γ(x)U3 on R×R3. (W)
Method:(i) Polychromatic ansatz
U(t, x) = ∑k∈Z
eikt uk (x), uk = u−k ∈ X . (P)
(infinite) Helmholtz system for u = (uk )k ∈ `1(N0,X )
−∆uk − (k2 + 1)uk = Γ(x) (u ∗ u ∗ u)k . (W ∗)
(ii) Bifurcation from w = (w0, 0, 0, ...).
The strategy
14 29.08.2019 - Waves 2019 Dominic Scheider-
Questions:
(1) What does “Helmholtz” mean?
(2) What is bifurcation?
(3) Why are we talking about this?
The strategy
15 29.08.2019 - Waves 2019 Dominic Scheider-
Question 1: What does “Helmholtz” mean?
−∆u − λu = f on R3, λ > 0 (H)
. “Helmholtz” case: 0 ∈ σ(−∆− λ)
. Particular solution of (H):Limiting Absorption Principle,
u1 = <[limε→0
(−∆− λ− iε)−1f
]=
cos(| · |√
λ)
4π| · | ∗ f .
. General solution of (H):
u = u1 + u2 with any Herglotz wave − ∆u2 − λu2 = 0.
Summary: Multitude of (weakly) localized solutions.
The strategy
15 29.08.2019 - Waves 2019 Dominic Scheider-
Question 1: What does “Helmholtz” mean?
−∆u − λu = f on R3, λ > 0 (H)
. “Helmholtz” case: 0 ∈ σ(−∆− λ)
. Particular solution of (H):Limiting Absorption Principle,
u1 = <[limε→0
(−∆− λ− iε)−1f
]=
cos(| · |√
λ)
4π| · | ∗ f .
. General solution of (H):
u = u1 + u2 with any Herglotz wave − ∆u2 − λu2 = 0.
Summary: Multitude of (weakly) localized solutions.
The strategy
15 29.08.2019 - Waves 2019 Dominic Scheider-
Question 1: What does “Helmholtz” mean?
−∆u − λu = f on R3, λ > 0 (H)
. “Helmholtz” case: 0 ∈ σ(−∆− λ)
. Particular solution of (H):Limiting Absorption Principle,
u1 = <[limε→0
(−∆− λ− iε)−1f
]=
cos(| · |√
λ)
4π| · | ∗ f .
. General solution of (H):
u = u1 + u2 with any Herglotz wave − ∆u2 − λu2 = 0.
Summary: Multitude of (weakly) localized solutions.
The strategy
15 29.08.2019 - Waves 2019 Dominic Scheider-
Question 1: What does “Helmholtz” mean?
−∆u − λu = f on R3, λ > 0 (H)
. “Helmholtz” case: 0 ∈ σ(−∆− λ)
. Particular solution of (H):Limiting Absorption Principle,
u1 = <[limε→0
(−∆− λ− iε)−1f
]=
cos(| · |√
λ)
4π| · | ∗ f .
. General solution of (H):
u = u1 + u2 with any Herglotz wave − ∆u2 − λu2 = 0.
Summary: Multitude of (weakly) localized solutions.
The strategy
15 29.08.2019 - Waves 2019 Dominic Scheider-
Question 1: What does “Helmholtz” mean?
−∆u − λu = f on R3, λ > 0 (H)
. “Helmholtz” case: 0 ∈ σ(−∆− λ)
. Particular solution of (H):Limiting Absorption Principle,
u1 = <[limε→0
(−∆− λ− iε)−1f
]=
cos(| · |√
λ)
4π| · | ∗ f .
. General solution of (H):
u = u1 + u2 with any Herglotz wave − ∆u2 − λu2 = 0.
Summary: Multitude of (weakly) localized solutions.
The strategy
16 29.08.2019 - Waves 2019 Dominic Scheider-
Question 1: What does “Helmholtz” mean?
−∆u − λu = f (r) on R3, λ > 0 (H)
. “Helmholtz” case: 0 ∈ σ(−∆− λ)
. Particular solution of (H):
−(ru1)′′ − λ(ru1) = rf (r), u1(0) = 1, u′1(0) = 0.
. General solution of (H):
u = u1 + c · sin(| · |√
λ)
4π| · | , c ∈ R.
Radial symmetry 1-dim. solution spaces.
The strategy
16 29.08.2019 - Waves 2019 Dominic Scheider-
Question 1: What does “Helmholtz” mean?
−∆u − λu = f (r) on R3, λ > 0 (H)
. “Helmholtz” case: 0 ∈ σ(−∆− λ)
. Particular solution of (H):
−(ru1)′′ − λ(ru1) = rf (r), u1(0) = 1, u′1(0) = 0.
. General solution of (H):
u = u1 + c · sin(| · |√
λ)
4π| · | , c ∈ R.
Radial symmetry 1-dim. solution spaces.
The strategy
17 29.08.2019 - Waves 2019 Dominic Scheider-
Question 1: What does “Helmholtz” mean?
−∆u − λu = g(r) · u on R3, λ > 0 (H∗)
. Asymptotically, if g is localized:
−(ru)′′ − λ(ru) ≈ 0 u(r) ≈ $∞sin(r√
λ + τ∞)
r
Lemma:(H∗) has a unique normalized solution in X . It satisfies
u(r) =sin(r√
λ + τ∞)
r+O
(1r2
)for some unique τ∞ ∈ [0,π).
The strategy
17 29.08.2019 - Waves 2019 Dominic Scheider-
Question 1: What does “Helmholtz” mean?
−∆u − λu = g(r) · u on R3, λ > 0 (H∗)
. Asymptotically, if g is localized:
−(ru)′′ − λ(ru) ≈ 0 u(r) ≈ $∞sin(r√
λ + τ∞)
r
Lemma:(H∗) has a unique normalized solution in X . It satisfies
u(r) =sin(r√
λ + τ∞)
r+O
(1r2
)for some unique τ∞ ∈ [0,π).
The strategy
17 29.08.2019 - Waves 2019 Dominic Scheider-
Question 1: What does “Helmholtz” mean?
−∆u − λu = g(r) · u on R3, λ > 0 (H∗)
. Asymptotically, if g is localized:
−(ru)′′ − λ(ru) ≈ 0 u(r) ≈ $∞sin(r√
λ + τ∞)
r
Lemma:(H∗) has a unique normalized solution in X . It satisfies
u(r) =sin(r√
λ + τ∞)
r+O
(1r2
)for some unique τ∞ ∈ [0,π).
The strategy
18 29.08.2019 - Waves 2019 Dominic Scheider-
Question 1: What does “Helmholtz” mean?
−∆u − λu = g(r) · u on R3, λ > 0 (H∗)
. Asymptotically, if g is localized:
−(ru)′′ − λ(ru) ≈ 0 u(r) ≈ $∞sin(r√
λ + τ∞)
r
Lemma:(H∗) together with an asymptotic phase condition (far field condition)
u(r) ∼ sin(r√
λ + τ)
r+O
(1r2
)(Aτ)
has a nontrivial solution in X iff τ = τ∞. (Unique up to constant multiple.)
The strategy
19 29.08.2019 - Waves 2019 Dominic Scheider-
Question 1: What does “Helmholtz” mean?
Lemma:(H∗) together with the asymptotic phase condition (Aτ) has a nontrivialsolution in X iff τ = τ∞. (Unique up to constant multiple.)
Remark: For τ 6= 0,
(H∗), (Aτ) ⇔ u = Rτλ[g u] =
sin(| · |√
λ + τ)
4π| · | sin(τ) ∗ [g u].
Radial symmetry ⊕ “good” phase cond. 1-dim. solution spaces.Radial symmetry ⊕ “bad” phase cond. 0-dim. solution spaces.
The strategy
19 29.08.2019 - Waves 2019 Dominic Scheider-
Question 1: What does “Helmholtz” mean?
Lemma:(H∗) together with the asymptotic phase condition (Aτ) has a nontrivialsolution in X iff τ = τ∞. (Unique up to constant multiple.)
Remark: For τ 6= 0,
(H∗), (Aτ) ⇔ u = Rτλ[g u] =
sin(| · |√
λ + τ)
4π| · | sin(τ) ∗ [g u].
Radial symmetry ⊕ “good” phase cond. 1-dim. solution spaces.Radial symmetry ⊕ “bad” phase cond. 0-dim. solution spaces.
The strategy
20 29.08.2019 - Waves 2019 Dominic Scheider-
Question 2: What is bifurcation?
Situation: Banach space E , u0 ∈ E and f ∈ C1(E ×R,E ) with
f (u0,λ) = 0 for all λ ∈ R.
Question: Solutions of f (u,λ) = 0 with (u,λ) ≈ (u0,λ0) but u 6= u0?
dim kerDuf (u0,λ0) = 0
Implicit Function Theorem
The strategy
20 29.08.2019 - Waves 2019 Dominic Scheider-
Question 2: What is bifurcation?Situation: Banach space E , u0 ∈ E and f ∈ C1(E ×R,E ) with
f (u0,λ) = 0 for all λ ∈ R.
Question: Solutions of f (u,λ) = 0 with (u,λ) ≈ (u0,λ0) but u 6= u0?
dim kerDuf (u0,λ0) = 0
Implicit Function Theorem
The strategy
20 29.08.2019 - Waves 2019 Dominic Scheider-
Question 2: What is bifurcation?Situation: Banach space E , u0 ∈ E and f ∈ C1(E ×R,E ) with
f (u0,λ) = 0 for all λ ∈ R.
Question: Solutions of f (u,λ) = 0 with (u,λ) ≈ (u0,λ0) but u 6= u0?
dim kerDuf (u0,λ0) = 0
Implicit Function Theorem
The strategy
20 29.08.2019 - Waves 2019 Dominic Scheider-
Question 2: What is bifurcation?Situation: Banach space E , u0 ∈ E and f ∈ C1(E ×R,E ) with
f (u0,λ) = 0 for all λ ∈ R.
Question: Solutions of f (u,λ) = 0 with (u,λ) ≈ (u0,λ0) but u 6= u0?
dim kerDuf (u0,λ0) = 0
Implicit Function Theorem
The strategy
21 29.08.2019 - Waves 2019 Dominic Scheider-
Question 2: What is bifurcation?Situation: Banach space E , u0 ∈ E and f ∈ C1(E ×R,E ) with
f (u0,λ) = 0 for all λ ∈ R.
Question: Solutions of f (u,λ) = 0 with (u,λ) ≈ (u0,λ0) but u 6= u0?
dim kerDuf (u0,λ0) = 1(and more)
Crandall-Rabinowitz Theorem:Bifurcationfrom a simple eigenvalue
The strategy
21 29.08.2019 - Waves 2019 Dominic Scheider-
Question 2: What is bifurcation?Situation: Banach space E , u0 ∈ E and f ∈ C1(E ×R,E ) with
f (u0,λ) = 0 for all λ ∈ R.
Question: Solutions of f (u,λ) = 0 with (u,λ) ≈ (u0,λ0) but u 6= u0?
dim kerDuf (u0,λ0) = 1(and more)
Crandall-Rabinowitz Theorem:Bifurcationfrom a simple eigenvalue
The strategy
22 29.08.2019 - Waves 2019 Dominic Scheider-
Question 3: Why are we talking about this? - The strategy, finally.
Aim: Solve ∂2tU − ∆U − U = Γ(x)U3, U ∈ C2
per(R3,X ).
Method:(i) Polychromatic ansatz U(t, x) = ∑k∈Z eikt uk (x)
(infinite) Helmholtz system for u = (uk )k ∈ `1(N0,X )
−∆uk − (k2 + 1)uk = Γ(x) (u ∗ u ∗ u)k . (W ∗)
(ii) Bifurcation from w = (w0, 0, 0, ...).
The strategy
22 29.08.2019 - Waves 2019 Dominic Scheider-
Question 3: Why are we talking about this? - The strategy, finally.Aim: Solve ∂2
tU − ∆U − U = Γ(x)U3, U ∈ C2per(R
3,X ).Method:
(i) Polychromatic ansatz U(t, x) = ∑k∈Z eikt uk (x)
(infinite) Helmholtz system for u = (uk )k ∈ `1(N0,X )
−∆uk − (k2 + 1)uk = Γ(x) (u ∗ u ∗ u)k . (W ∗)
(ii) Bifurcation from w = (w0, 0, 0, ...).
The strategy
23 29.08.2019 - Waves 2019 Dominic Scheider-
Question 3: Why are we talking about this? - The strategy, finally.Aim: Solve ∂2
tU − ∆U − U = Γ(x)U3, U ∈ C2per(R
3,X ).Method:
(i) Polychromatic ansatz U(t, x) = ∑k∈Z eikt uk (x)
(infinite) Helmholtz system for u = (uk )k ∈ `1(N0,X )
with asymptotic conditions
uk = Rτkk2+1[Γ(x) (u ∗ u ∗ u)k ]. (W ∗∗)
(ii) Bifurcation from w = (w0, 0, 0, ...).
The strategy
24 29.08.2019 - Waves 2019 Dominic Scheider-
Question 3: Why are we talking about this? - The strategy, finally.Aim: Solve ∂2
tU − ∆U − U = Γ(x)U3, U ∈ C2per(R
3,X ).Method:
(i) Polychromatic ansatz U(t, x) = ∑k∈Z eikt uk (x)
(infinite) Helmholtz system for u = (uk )k ∈ `1(N0,X )
with asymptotic conditions
uk = Rτkk2+1[Γ(x) (u ∗ u ∗ u)k ]. (W ∗∗)
(ii) Bifurcation from w = (w0, 0, 0, ...).
. “Invisible” bifurcation parameter: τs τs + λ.
. Study linearization of (W ∗∗) at u = w, λ = 0:
ψk = 3Rτkk2+1[Γ(x)w
20 (x)ψk ]. (W ∗∗∗)
Lemma!
The strategy
24 29.08.2019 - Waves 2019 Dominic Scheider-
Question 3: Why are we talking about this? - The strategy, finally.Aim: Solve ∂2
tU − ∆U − U = Γ(x)U3, U ∈ C2per(R
3,X ).Method:
(i) Polychromatic ansatz U(t, x) = ∑k∈Z eikt uk (x)
(infinite) Helmholtz system for u = (uk )k ∈ `1(N0,X )
with asymptotic conditions
uk = Rτkk2+1[Γ(x) (u ∗ u ∗ u)k ]. (W ∗∗)
(ii) Bifurcation from w = (w0, 0, 0, ...).. “Invisible” bifurcation parameter: τs τs + λ.
. Study linearization of (W ∗∗) at u = w, λ = 0:
ψk = 3Rτkk2+1[Γ(x)w
20 (x)ψk ]. (W ∗∗∗)
Lemma!
The strategy
24 29.08.2019 - Waves 2019 Dominic Scheider-
Question 3: Why are we talking about this? - The strategy, finally.Aim: Solve ∂2
tU − ∆U − U = Γ(x)U3, U ∈ C2per(R
3,X ).Method:
(i) Polychromatic ansatz U(t, x) = ∑k∈Z eikt uk (x)
(infinite) Helmholtz system for u = (uk )k ∈ `1(N0,X )
with asymptotic conditions
uk = Rτkk2+1[Γ(x) (u ∗ u ∗ u)k ]. (W ∗∗)
(ii) Bifurcation from w = (w0, 0, 0, ...).. “Invisible” bifurcation parameter: τs τs + λ.. Study linearization of (W ∗∗) at u = w, λ = 0:
ψk = 3Rτkk2+1[Γ(x)w
20 (x)ψk ]. (W ∗∗∗)
Lemma!
The strategy
24 29.08.2019 - Waves 2019 Dominic Scheider-
Question 3: Why are we talking about this? - The strategy, finally.Aim: Solve ∂2
tU − ∆U − U = Γ(x)U3, U ∈ C2per(R
3,X ).Method:
(i) Polychromatic ansatz U(t, x) = ∑k∈Z eikt uk (x)
(infinite) Helmholtz system for u = (uk )k ∈ `1(N0,X )
with asymptotic conditions
uk = Rτkk2+1[Γ(x) (u ∗ u ∗ u)k ]. (W ∗∗)
(ii) Bifurcation from w = (w0, 0, 0, ...).. “Invisible” bifurcation parameter: τs τs + λ.. Study linearization of (W ∗∗) at u = w, λ = 0:
ψk = 3Rτkk2+1[Γ(x)w
20 (x)ψk ]. (W ∗∗∗)
Lemma!
The strategy
25 29.08.2019 - Waves 2019 Dominic Scheider-
Question 3: Why are we talking about this? - The strategy, finally.
ψk = 3Rτkk2+1[Γ(x)w
20 (x)ψk ] (W ∗∗∗)
is equivalent to
− ∆ψk − (k2 + 1)ψk = 3 Γ(x)w20 (x)ψk on R3, (H∗)
ψk (r) ∼sin(r√
λ + τk )
r+O
(1r2
). (Aτk )
Radial symmetry ⊕ “good” phase cond. 1-dim. solution spaces.Radial symmetry ⊕ “bad” phase cond. 0-dim. solution spaces.
The strategy
25 29.08.2019 - Waves 2019 Dominic Scheider-
Question 3: Why are we talking about this? - The strategy, finally.
ψk = 3Rτkk2+1[Γ(x)w
20 (x)ψk ] (W ∗∗∗)
is equivalent to
− ∆ψk − (k2 + 1)ψk = 3 Γ(x)w20 (x)ψk on R3, (H∗)
ψk (r) ∼sin(r√
λ + τk )
r+O
(1r2
). (Aτk )
Radial symmetry ⊕ “good” phase cond. 1-dim. solution spaces.Radial symmetry ⊕ “bad” phase cond. 0-dim. solution spaces.
The strategy
25 29.08.2019 - Waves 2019 Dominic Scheider-
Question 3: Why are we talking about this? - The strategy, finally.
ψk = 3Rτkk2+1[Γ(x)w
20 (x)ψk ] (W ∗∗∗)
is equivalent to
− ∆ψk − (k2 + 1)ψk = 3 Γ(x)w20 (x)ψk on R3, (H∗)
ψk (r) ∼sin(r√
λ + τk )
r+O
(1r2
). (Aτk )
Radial symmetry ⊕ “good” phase cond. 1-dim. solution spaces.Radial symmetry ⊕ “bad” phase cond. 0-dim. solution spaces.
The strategy
26 29.08.2019 - Waves 2019 Dominic Scheider-
Question 3: Why are we talking about this? - The strategy, finally.Aim: Solve ∂2
tU − ∆U − U = Γ(x)U3, U ∈ C2per(R
3,X ).Method:
(i) Polychromatic ansatz U(t, x) = ∑k∈Z eikt uk (x)
(infinite) Helmholtz system for u = (uk )k ∈ `1(N0,X )
with asymptotic conditions
uk = Rτkk2+1[Γ(x) (u ∗ u ∗ u)k ]. (W ∗∗)
(ii) Bifurcation from w = (w0, 0, 0, ...).. “Invisible” bifurcation parameter: τs τs + λ.. Study linearization of (W ∗∗) at u = w, λ = 0:
ψk = 3Rτkk2+1[Γ(x)w
20 (x)ψk ]. (W ∗∗∗)
Choice of τk such that ψk ≡ 0 (k 6= s), ψs 6≡ 0.
Outline
27 29.08.2019 - Waves 2019 Dominic Scheider-
1 The main result
2 The strategy
3 Conclusion
Conclusion
28 29.08.2019 - Waves 2019 Dominic Scheider-
Theorem: (S, 2019)There exist an interval I ⊆ R, 0 ∈ I and a family (Uη)η∈I of real-valued,classical solutions Uη = Uη(t, x) ∈ C2
per(R,X ) of the wave equation
∂2tU − ∆U − U = Γ(x)U3 on R×R3 (W)
which is a continuous curve in C (R,X ) with U0(t, x) = w0(x) and (forη 6= 0) Uη nonstationary and 2π-periodic.
Key ideas:
Conclusion
28 29.08.2019 - Waves 2019 Dominic Scheider-
Theorem: (S, 2019)There exist an interval I ⊆ R, 0 ∈ I and a family (Uη)η∈I of real-valued,classical solutions Uη = Uη(t, x) ∈ C2
per(R,X ) of the wave equation
∂2tU − ∆U − U = Γ(x)U3 on R×R3 (W)
which is a continuous curve in C (R,X ) with U0(t, x) = w0(x) and (forη 6= 0) Uη nonstationary and 2π-periodic.
Key ideas:
Conclusion
29 29.08.2019 - Waves 2019 Dominic Scheider-
Theorem: (S, 2019)There exist an interval I ⊆ R, 0 ∈ I and a family (Uη)η∈I of real-valued,classical solutions Uη = Uη(t, x) ∈ C2
per(R,X ) of the wave equation
∂2tU − ∆U − U = Γ(x)U3 on R×R3 (W)
which is a continuous curve in C (R,X ) with U0(t, x) = w0(x) and (forη 6= 0) Uη nonstationary and 2π-periodic.
Key ideas:
Conclusion
30 29.08.2019 - Waves 2019 Dominic Scheider-
Theorem: (S, 2019)There exist an interval I ⊆ R, 0 ∈ I and a family (Uη)η∈I of real-valued,classical solutions Uη = Uη(t, x) ∈ C2
per(R,X ) of the wave equation
∂2tU − ∆U − U = Γ(x)U3 on R×R3 (W)
which is a continuous curve in C (R,X ) with U0(t, x) = w0(x) and (forη 6= 0) Uη nonstationary and 2π-periodic.
Key ideas:
Conclusion
31 29.08.2019 - Waves 2019 Dominic Scheider-
Related results:Breather solutions for the wave equation
V (x)∂2tU − ∂2
xU + c V (x)U = Γ(x)U3 on R×R.
. Blank, Chirilus-Bruckner, Lescarret, Schneider 2011,
. Hirsch, Reichel 2019.
Common feature: polychromatic ansatzDifferences:. specific, rough periodic potentials (↔ constant potentials). analysis in spectral gaps (↔ exploitation of kernel elements). strongly localized solutions (↔ power decay). large solutions (↔ close-to-stationary solutions)
Conclusion
31 29.08.2019 - Waves 2019 Dominic Scheider-
Related results:Breather solutions for the wave equation
V (x)∂2tU − ∂2
xU + c V (x)U = Γ(x)U3 on R×R.
. Blank, Chirilus-Bruckner, Lescarret, Schneider 2011,
. Hirsch, Reichel 2019.
Common feature: polychromatic ansatz
Differences:. specific, rough periodic potentials (↔ constant potentials). analysis in spectral gaps (↔ exploitation of kernel elements). strongly localized solutions (↔ power decay). large solutions (↔ close-to-stationary solutions)
Conclusion
31 29.08.2019 - Waves 2019 Dominic Scheider-
Related results:Breather solutions for the wave equation
V (x)∂2tU − ∂2
xU + c V (x)U = Γ(x)U3 on R×R.
. Blank, Chirilus-Bruckner, Lescarret, Schneider 2011,
. Hirsch, Reichel 2019.
Common feature: polychromatic ansatzDifferences:. specific, rough periodic potentials (↔ constant potentials). analysis in spectral gaps (↔ exploitation of kernel elements). strongly localized solutions (↔ power decay). large solutions (↔ close-to-stationary solutions)
Conclusion
32 29.08.2019 - Waves 2019 Dominic Scheider-
Thank you for your attention!... Questions?
Theorem: (S, 2019)There exist an interval I ⊆ R, 0 ∈ I and a family (Uη)η∈I of real-valued,classical solutions Uη = Uη(t, x) ∈ C2
per(R,X ) of the wave equation
∂2tU − ∆U − U = Γ(x)U3 on R×R3 (W)
which is a continuous curve in C (R,X ) with U0(t, x) = w0(x) and (forη 6= 0) Uη nonstationary and 2π-periodic.
Conclusion
32 29.08.2019 - Waves 2019 Dominic Scheider-
Thank you for your attention!... Questions?
Theorem: (S, 2019)There exist an interval I ⊆ R, 0 ∈ I and a family (Uη)η∈I of real-valued,classical solutions Uη = Uη(t, x) ∈ C2
per(R,X ) of the wave equation
∂2tU − ∆U − U = Γ(x)U3 on R×R3 (W)
which is a continuous curve in C (R,X ) with U0(t, x) = w0(x) and (forη 6= 0) Uη nonstationary and 2π-periodic.
References
33 29.08.2019 - Waves 2019 Dominic Scheider-
C. Blank, M. Chirilus-Bruckner, V. Lescarret and G. Schneider, BreatherSolutions in Periodic Media, Comm. Math. Phys. 3 (2011), pp. 815–841A. Hirsch and W. Reichel, Real-valued, time-periodic localized weaksolutions for a semilinear wave equation with periodic potentials,Nonlinearity (2019), to appearR. Mandel, E. Montefusco and B. Pellacci, Oscillating solutions fornonlinear Helmholtz equations, ZAMP 6 (2017), article 121R. Mandel and D. Scheider, Bifurcations of nontrivial solutions of a cubicHelmholtz system, to appear in ANONA. URL: http://www.waves.kit.edu/downloads/CRC1173_Preprint_2018-32.pdf
The main result will be part of my PhD thesis (KIT, Oct 2019).