How to construct Breather Solutions using Nonlinear ...scheider/media/scheider... · Breather solutions 2 16.01.2020-UniversityofStuttgart DominicScheider-The Sine-Gordon Breather

0 16.01.2020 - University of Stuttgart Dominic Scheider-

University of Stuttgart

How to construct Breather Solutionsusing Nonlinear Helmholtz Systems

Dominic Scheider

KIT – The Research University in the Helmholtz Association www.kit.edu

Outline


1 Breather solutions

2 Breather solutions for the cubic Klein-Gordon Equation

∂2tU − ∆U +m2U = U3 on R×R3

3 Breather solutions for the nonlinear Wave Equation

∂2tU − ∆U = |U |p−2U on R×RN

Breather solutions


The Sine-Gordon Breather

∂2tU − ∂2

xU + sin(U) = 0 on R×R

. Ablowitz et al. 1973:

U(t, x) = 4 arctan

(√1−ω2

ω

cos(ω(t − t0))

cosh(√1−ω2(x − x0))

).

. Kichenassamy 1991, Weinstein et al. 1994: Rigidity.

“breathing takes place only for isolated nonlinearities”

Breather solutions


The Sine-Gordon Breather

∂2tU − ∂2

xU + sin(U) = 0 on R×R

. Ablowitz et al. 1973:

U(t, x) = 4 arctan

(√1−ω2

ω

cos(ω(t − t0))

cosh(√1−ω2(x − x0))

).

. Kichenassamy 1991, Weinstein et al. 1994: Rigidity.

“breathing takes place only for isolated nonlinearities”

Breather solutions


Breathers in periodic structures (1/3)

V (x)∂2tU − ∂2

xU + q(x)U = U3 on R×R

. Blank, Chirilus-Bruckner, Lescarret, Schneider 2011:

Assumptions:V (x), q(x) = (q0 − ε2)V (x) explicit periodic step potentials.

Methods:Spatial dynamics, center manifold reduction, bifurcation.

Result: For 0 < ε < ε0 existence of a solution U(t, x) withI explicit period in t, I exponential decay in x .

. Hirsch, Reichel 2019.

Breather solutions



V (x)∂2tU − ∂2







Breather solutions



V (x)∂2tU − ∂2







Breather solutions



V (x)∂2tU − ∂2







Breather solutions



V (x)∂2tU − ∂2

xU + q(x)U = Γ(x)|U |p−2U on R×R

. Blank, Chirilus-Bruckner, Lescarret, Schneider 2011.

. Hirsch, Reichel 2019:

Assumptions:V (x) ∼ q(x) specific periodic delta / step / Hr potentials;2 < p < p∗(V ).

Methods:Variational approach (Nehari manifold).

Result:Existence of (possibly large) time-periodic ground state solutions.

Breather solutions



V (x)∂2tU − ∂2

xU + q(x)U = Γ(x)|U |p−2U on R×R


. Hirsch, Reichel 2019:

Assumptions:V (x) ∼ q(x) specific periodic delta / step / Hr potentials;2 < p < p∗(V ).

Methods:Variational approach (Nehari manifold).

Result:Existence of (possibly large) time-periodic ground state solutions.

Breather solutions



V (x)∂2tU − ∂2

xU +m2 V (x)U = f (x ,U) on R×R



Guiding principles:

U(t, x) = ∑k

eikωtuk (x) −u′′k + (m2 − k2ω2)V (x)uk = fk (x ,U).

Aim for 0 6∈ σ(− d2

dx2 + (m2 − k2ω2)V (x)). Problem: |k | → ∞ ...

Periodicity & roughness of V (x) yield uniformly open spectral gaps. “breathing takes place only for carefully designed potentials”

Breather solutions



V (x)∂2tU − ∂2




Guiding principles:

U(t, x) = ∑k





Breather solutions



V (x)∂2tU − ∂2




Guiding principles:

U(t, x) = ∑k





Breather solutions


Why not allow 0 in the spectra?

Then V (x) ≡ 1 is fine. Klein-Gordon equation:

∂2tU − ∆U +m2 U = f (x ,U) on R×RN

Cost: N = 1 not accessible, V (x) 6≡ const. hard, breathers decay slowly.

Gain: N > 1 accessible, V (x) ≡ const. accessible, many breathers.Here, breathing is not a rare phenomenon.

Breather solutions




∂2tU − ∆U +m2 U = f (x ,U) on R×RN



Breather solutions




∂2tU − ∆U +m2 U = f (x ,U) on R×RN


Gain: N > 1 accessible, V (x) ≡ const. accessible, many breathers.

Here, breathing is not a rare phenomenon.

Breather solutions




∂2tU − ∆U +m2 U = f (x ,U) on R×RN



Outline




∂2tU − ∆U +m2U = U3 on R×R3


∂2tU − ∆U = |U |p−2U on R×RN

Breathers for the Klein-Gordon Equation


∂2tU − ∆U +m2U = U3 on R×R3

(i) U(t, x) = ∑k eikωtuk (x) yields

−∆uk − (k2ω2 −m2) uk = ∑l+m+n=k

ul um un on R3.

Nonlinear Helmholtz System: 0 ∈ σ(−∆− (k2ω2 −m2)) for k 6= 0(ii) Study bifurcation from any stationary U0(t, x) = w0(x).

Family of breathers Uη(t, x) = ∑k eikωtuηk (x).

Bifurcation from simple eigenvalues:Need 1-dim. kernel of linearization

−∆vk − (k2ω2 −m2) vk = 3w20 vk on R3.

Key ideas: Radial symmetry, asymptotic phase condition.



∂2tU − ∆U +m2U = U3 on R×R3


−∆uk − (k2ω2 −m2) uk = ∑l+m+n=k

ul um un on R3.




−∆vk − (k2ω2 −m2) vk = 3w20 vk on R3.




∂2tU − ∆U +m2U = U3 on R×R3


−∆uk − (k2ω2 −m2) uk = ∑l+m+n=k

ul um un on R3.

Nonlinear Helmholtz System: 0 ∈ σ(−∆− (k2ω2 −m2)) for k 6= 0

(ii) Study bifurcation from any stationary U0(t, x) = w0(x).



−∆vk − (k2ω2 −m2) vk = 3w20 vk on R3.




∂2tU − ∆U +m2U = U3 on R×R3


−∆uk − (k2ω2 −m2) uk = ∑l+m+n=k

ul um un on R3.




−∆vk − (k2ω2 −m2) vk = 3w20 vk on R3.




∂2tU − ∆U +m2U = U3 on R×R3


−∆uk − (k2ω2 −m2) uk = ∑l+m+n=k

ul um un on R3.




−∆vk − (k2ω2 −m2) vk = 3w20 vk on R3.




∂2tU − ∆U +m2U = U3 on R×R3


−∆uk − (k2ω2 −m2) uk = ∑l+m+n=k

ul um un on R3.




−∆vk − (k2ω2 −m2) vk = 3w20 vk on R3.




∂2tU − ∆U +m2U = U3 on R×R3


−∆uk − (k2ω2 −m2) uk = ∑l+m+n=k

ul um un on R3.




−∆vk − (k2ω2 −m2) vk = 3w20 vk on R3.




Let X := {u ∈ Crad(R3,R) | sup(1+ |x |)|u(x)| < ∞}.

Let w0 ∈ X with −∆w0 +m2w0 = w30 on R3; choose ω > m and s ∈N.

Theorem 1 [S. 2019]There exist an interval I ⊆ R, 0 ∈ I and a family (Uη)η∈I ⊆ C2

per(R,X )

of real-valued, classical breather solutions Uη(t, x) = ∑k eikωtuηk (x) of

the Klein-Gordon equation

∂2tU − ∆U +m2U = U3 on R×R3

which is a continuous curve in C (R,X ) with

. U0(t, x) = w0(x),

. Uη is 2πω -periodic in time with ∞ many nonzero modes (η 6= 0),

. ddη

∣∣η=0u

ηk 6= 0 iff k = s (excitation of s-th mode).






per(R,X )



∂2tU − ∆U +m2U = U3 on R×R3


. U0(t, x) = w0(x),


. ddη

∣∣η=0u







per(R,X )



∂2tU − ∆U +m2U = U3 on R×R3


. U0(t, x) = w0(x),


. ddη

∣∣η=0u







per(R,X )



∂2tU − ∆U +m2U = U3 on R×R3


. U0(t, x) = w0(x),


. ddη

∣∣η=0u







per(R,X )



∂2tU − ∆U +m2U = U3 on R×R3


. U0(t, x) = w0(x),


. ddη

∣∣η=0u




Remarks

. Extension to

∂2tU − ∆U +m2U = Γ(x)U3 on R×R3

with Γ bounded, radial, continuously differentiable.

. Open: Other space dimensions / powers (easy?);non-constant potentials (hard!).

Aspects of the Proof

. Bifurcation from simple eigenvalues (in a nutshell),

. Linear Helmholtz equations in X (likewise),

. How to prove the Theorem.



Remarks

. Extension to

∂2tU − ∆U +m2U = Γ(x)U3 on R×R3









Remarks

. Extension to

∂2tU − ∆U +m2U = Γ(x)U3 on R×R3









Remarks

. Extension to

∂2tU − ∆U +m2U = Γ(x)U3 on R×R3









Aspects of the Proof 1/3: Bifurcation from simple eigenvalues.

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



Aspects of the Proof 1/3: Bifurcation from simple eigenvalues.Situation: Banach space E , u0 ∈ E and f ∈ C1(E ×R,E ) with































dim kerDuf (u0,λ0) = 1(and more)

Crandall-Rabinowitz Theorem:Bifurcationfrom a simple eigenvalue


















∂2tU − ∆U +m2U = U3 on R×R3

yields via U(t, x) = ∑k eikωtuk (x) the infinite system

−∆uk − (k2ω2 −m2) uk = ∑l+m+n=k

ul um un on R3.

. Reformulate as f ((uk )k ,λ) = 0.

. Introduce a bifurcation parameter λ.

. Ensure 1-dim. kernel of linearized problem

−∆vk − (k2ω2 −m2) vk = 3w20 (x) vk on R3.

. Verify remaining conditions of the CR Bifurcation Theorem(transversality).




∂2tU − ∆U +m2U = U3 on R×R3


−∆uk − (k2ω2 −m2) uk = ∑l+m+n=k

ul um un on R3.




−∆vk − (k2ω2 −m2) vk = 3w20 (x) vk on R3.





∂2tU − ∆U +m2U = U3 on R×R3


−∆uk − (k2ω2 −m2) uk = ∑l+m+n=k

ul um un on R3.




−∆vk − (k2ω2 −m2) vk = 3w20 (x) vk on R3.




Aspects of the Proof 2/3: Linear Helmholtz equations.

−∆u − µu = f on R3, µ > 0 (H)

. “Helmholtz” case: 0 ∈ σ(−∆− µ)

. Particular solution of (H):Limiting Absorption Principle,

u1 = <[limε→0

(−∆− µ− iε)−1f

]=

cos(| · |√µ)

4π| · | ∗ f .

. General solution of (H): u = u1 + u2

with any Herglotz wave −∆u2 − µu2 = 0, e.g. u2 =sin(|·|√µ)

4π|·| ∗ f .Summary: Multitude of (weakly) localized solutions.




−∆u − µu = f on R3, µ > 0 (H)



u1 = <[limε→0

(−∆− µ− iε)−1f

]=

cos(| · |√µ)

4π| · | ∗ f .







−∆u − µu = f on R3, µ > 0 (H)



u1 = <[limε→0

(−∆− µ− iε)−1f

]=

cos(| · |√µ)

4π| · | ∗ f .







−∆u − µu = f on R3, µ > 0 (H)



u1 = <[limε→0

(−∆− µ− iε)−1f

]=

cos(| · |√µ)

4π| · | ∗ f .



4π|·| ∗ f .

Summary: Multitude of (weakly) localized solutions.




−∆u − µu = f on R3, µ > 0 (H)



u1 = <[limε→0

(−∆− µ− iε)−1f

]=

cos(| · |√µ)

4π| · | ∗ f .







−∆u − µu = f (r) on R3, µ > 0 (H)


. Particular solution of (H):

−(ru1)′′ − µ(ru1) = rf (r), u1(0) = 1, u′1(0) = 0.

. General solution of (H):

u = u1 + c ·sin(| · |√µ)

4π| · |often=

cos(| · |√µ)

4π| · | ∗ f + c̃ ·sin(| · |√µ)

4π| · | ∗ f , c , c̃ ∈ R.

Radial symmetry 1-dim. solution spaces.




−∆u − µu = f (r) on R3, µ > 0 (H)



−(ru1)′′ − µ(ru1) = rf (r), u1(0) = 1, u′1(0) = 0.


u = u1 + c ·sin(| · |√µ)

4π| · |often=

cos(| · |√µ)

4π| · | ∗ f + c̃ ·sin(| · |√µ)

4π| · | ∗ f , c , c̃ ∈ R.





−∆u − µu = f (r) on R3, µ > 0 (H)



−(ru1)′′ − µ(ru1) = rf (r), u1(0) = 1, u′1(0) = 0.


u = u1 + c ·sin(| · |√µ)

4π| · |often=

cos(| · |√µ)

4π| · | ∗ f + c̃ ·sin(| · |√µ)

4π| · | ∗ f , c , c̃ ∈ R.





−∆u − µu = f (r) on R3, µ > 0 (H)



−(ru1)′′ − µ(ru1) = rf (r), u1(0) = 1, u′1(0) = 0.


u = u1 + c ·sin(| · |√µ)

4π| · |often=

cos(| · |√µ)

4π| · | ∗ f + c̃ ·sin(| · |√µ)

4π| · | ∗ f , c , c̃ ∈ R.





−∆u − µu = g(r) · u on R3, µ > 0 (H∗)

. Asymptotically, if g is localized:

−(ru)′′ − µ(ru) ≈ 0 u(r) ≈ $∞sin(r√

µ + τ∞)

r

. Lemma:(H∗) has a unique normalized solution in X . It satisfies

u(r) =sin(r√

µ + τ∞)

r+O

(1r2

)for some unique τ∞ ∈ [0,π).




−∆u − µu = g(r) · u on R3, µ > 0 (H∗)


−(ru)′′ − µ(ru) ≈ 0 u(r) ≈ $∞sin(r√

µ + τ∞)

r


u(r) =sin(r√

µ + τ∞)

r+O

(1r2





−∆u − µu = g(r) · u on R3, µ > 0 (H∗)


−(ru)′′ − µ(ru) ≈ 0 u(r) ≈ $∞sin(r√

µ + τ∞)

r


u(r) =sin(r√

µ + τ∞)

r+O

(1r2





−∆u − µu = g(r) · u on R3, µ > 0 (H∗)


−(ru)′′ − µ(ru) ≈ 0 u(r) ≈ $∞sin(r√

µ + τ∞)

r

. Lemma:(H∗) together with an asymptotic phase condition (far field condition)

u(r) ∼sin(r√

µ + τ)

r+O

(1r2

)(Aτ)

has a nontrivial solution in X iff τ = τ∞. (Unique up to constant multiple.)




− ∆u − µu = g(r) · u on R3, µ > 0 (H∗)

u(r) ∼sin(r√

µ + τ)

r+O

(1r2

)(Aτ)

. 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 and u ∈ X ,

(H∗), (Aτ) ⇔ u =sin(| · |√µ + τ)

4π| · | sin(τ) ∗ [g u] =: Ψτµ ∗ [g u].

Radial symmetry ⊕ “good” phase cond. 1-dim. solution spaces.Radial symmetry ⊕ “bad” phase cond. 0-dim. solution spaces.




− ∆u − µu = g(r) · u on R3, µ > 0 (H∗)

u(r) ∼sin(r√

µ + τ)

r+O

(1r2

)(Aτ)



(H∗), (Aτ) ⇔ u =sin(| · |√µ + τ)

4π| · | sin(τ) ∗ [g u] =: Ψτµ ∗ [g u].





− ∆u − µu = g(r) · u on R3, µ > 0 (H∗)

u(r) ∼sin(r√

µ + τ)

r+O

(1r2

)(Aτ)



(H∗), (Aτ) ⇔ u =sin(| · |√µ + τ)

4π| · | sin(τ) ∗ [g u] =: Ψτµ ∗ [g u].





− ∆u − µu = g(r) · u on R3, µ > 0 (H∗)

u(r) ∼sin(r√

µ + τ)

r+O

(1r2

)(Aτ)



(H∗), (Aτ) ⇔ u =sin(| · |√µ + τ)

4π| · | sin(τ) ∗ [g u] =: Ψτµ ∗ [g u].




Aspects of the Proof 3/3.

∂2tU − ∆U +m2U = U3 on R×R3


−∆uk − (k2ω2 −m2) uk = ∑l+m+n=k

ul um un on R3.




∂2tU − ∆U +m2U = U3 on R×R3


−∆uk − (k2ω2 −m2) uk = ∑l+m+n=k

ul um un on R3.




−∆uk − (k2ω2 −m2) uk = ∑l+m+n=k

ul um un on R3




−∆uk − (k2ω2 −m2) uk = (u ∗ u ∗ u)k on R3

where u = (uk )k ∈ `1(Z,X ).




−∆uk − µk uk = (u ∗ u ∗ u)k on R3

where u = (uk )k ∈ `1(Z,X ) and µk = k2ω2 −m2.

. Reformulate as f (u,λ) = 0.



−∆vk − µk vk = 3w20 (x) vk on R3.





−∆uk − µk uk = (u ∗ u ∗ u)k on R3

where u = (uk )k ∈ `1(Z,X ) and µk = k2ω2 −m2.

. Reformulate as f (u,λ) = 0.



−∆vk − µk vk = 3w20 (x) vk on R3.





uk −Ψτkµk∗ [(u ∗ u ∗ u)k ] = 0 (k 6= 0 !)

where u = (uk )k ∈ `1(Z,X ) and µk = k2ω2 −m2

with asymptotic conditions given by τk .

. X Reformulate as f (u,λ) = 0.



vk − 3Ψτkµk∗ [w2

0 vk ] = 0 (k 6= 0 !).





uk −Ψτkµk∗ [(u ∗ u ∗ u)k ] = 0 (k 6= 0 !)




. X Replace τs by τs + λ: “Invisible” bifurcation parameter.



0 vk ] = 0 (k 6= 0 !).





uk −Ψτkµk∗ [(u ∗ u ∗ u)k ] = 0 (k 6= 0 !)





. X Recall Lemma: Choose τs “good” and all other τk “bad” s.t.


0 vk ] = 0 ⇒ vk ≡ 0 holds iff k 6= s.

For k = 0, this is a result in the literature.. Verify remaining conditions of the CR Bifurcation Theorem

(transversality).




uk −Ψτkµk∗ [(u ∗ u ∗ u)k ] = 0 (k 6= 0 !)







0 vk ] = 0 ⇒ vk ≡ 0 holds iff k 6= s.

For k = 0, this is a result in the literature.. X Transversality condition: Direct computation using asymptotics.




uk −Ψτkµk∗ [(u ∗ u ∗ u)k ] = 0 (k 6= 0 !)







0 vk ] = 0 ⇒ vk ≡ 0 holds iff k 6= s.

For k = 0, this is a result in the literature.. X Transversality condition: Direct computation using asymptotics.. X Regularity of breathers: Scaling property of convolution with Ψτk

µk.






per(R,X )



∂2tU − ∆U +m2U = U3 on R×R3


. U0(t, x) = w0(x),


. ddη

∣∣η=0u


Outline




∂2tU − ∆U +m2U = U3 on R×R3


∂2tU − ∆U = |U |p−2U on R×RN

Breathers for the Wave Equation


This is work in progress.

Theorem 2 [sketch]Let Γ ∈ S(RN ), Γ > 0 and N ≥ 2, 2 < p < 2(N + 1)/(N − 1). Thenthe nonlinear wave equation

∂2tU − ∆U = Γ(x)|U |p−2U on [0, 2π]×RN

has a nontrivial dual ground state U : R×RN → R, which is 2π-periodic(π-antiperiodic) in time.



Remarks

. “Large” breathers via variational methods: dual ground states.

. Drawbacks: Need localized coefficient Γ(x) in the nonlinearity.N = 3, p = 4 not accessible (endpoint case).

. To do: Regularity, extension to Klein-Gordon Equation.


. A formal solution map for

∂2tU − ∆U = F on [0, 2π]×RN ,

. Dual variational techniques.



Remarks






∂2tU − ∆U = F on [0, 2π]×RN ,




Remarks






∂2tU − ∆U = F on [0, 2π]×RN ,




Remarks






∂2tU − ∆U = F on [0, 2π]×RN ,




Aspects of the Proof 1/2: Formal solution map.

∂2tU − ∆U = F on [0, 2π]×RN

with U(t, x) = ∑k eiktuk (x) and F (t, x) = ∑k eikt fk (x), this leads to

(−∆− k2)uk = fk on RN .

Restriction to odd k ∈ Z yields a nonlinear Helmholtz system. Idea:

uk = Ψk2 ∗ fk ,

U(t, x) = ∑k odd

eikt(Ψk2 ∗ fk )(x) with fk (x) =∫ 2π

0F (t, x)e−ikt dt

2π.




∂2tU − ∆U = F on [0, 2π]×RN


(−∆− k2)uk = fk on RN .


uk = Ψk2 ∗ fk ,

U(t, x) = ∑k odd


0F (t, x)e−ikt dt

2π.




∂2tU − ∆U = F on [0, 2π]×RN


(−∆− k2)uk = fk on RN .


uk = Ψk2 ∗ fk ,

U(t, x) = ∑k odd


0F (t, x)e−ikt dt

2π.




∂2tU − ∆U = F on [0, 2π]×RN


(−∆− k2)uk = fk on RN .


uk = Ψk2 ∗ fk ,

U(t, x) = ∑k odd


0F (t, x)e−ikt dt

2π.




∂2tU − ∆U = F on [0, 2π]×RN


(−∆− k2)uk = fk on RN .


uk = Ψk2 ∗ fk ,

U(t, x) = ∑k odd

eikt(Ψk2 ∗ fk )(x)

with fk (x) =∫ 2π

0F (t, x)e−ikt dt

2π.




∂2tU − ∆U = F on [0, 2π]×RN


(−∆− k2)uk = fk on RN .


uk = Ψk2 ∗ fk ,

U(t, x) = ∑k odd


0F (t, x)e−ikt dt

2π.



Aspects of the Proof 2/2: Dual variational techniques.

∂2tU − ∆U = Γ(x)|U |p−2U on [0, 2π]×RN

. Substitution V := Γ(x)1/p′ |U |p−2U yields

(∂2t − ∆)

[Γ(x)−1/p |V |p′−2V

]= Γ(x)1/pV on [0, 2π]×RN .

. Then, formally, solve

|V |p′−2V = LΓ[V ] on [0, 2π]×RN

where LΓ[V ](t, x) = ∑k odd

eiktΓ(x)1/p(Ψk2 ∗ [Γ1/pvk ])(x).

. LΓ : Lp′([0, 2π]×RN )→ Lp([0, 2π]×RN ) is symmetric, compact.




∂2tU − ∆U = Γ(x)|U |p−2U on [0, 2π]×RN


(∂2t − ∆)

[Γ(x)−1/p |V |p′−2V

]= Γ(x)1/pV on [0, 2π]×RN .


|V |p′−2V = LΓ[V ] on [0, 2π]×RN







∂2tU − ∆U = Γ(x)|U |p−2U on [0, 2π]×RN


(∂2t − ∆)

[Γ(x)−1/p |V |p′−2V

]= Γ(x)1/pV on [0, 2π]×RN .


|V |p′−2V = LΓ[V ] on [0, 2π]×RN







∂2tU − ∆U = Γ(x)|U |p−2U on [0, 2π]×RN


(∂2t − ∆)

[Γ(x)−1/p |V |p′−2V

]= Γ(x)1/pV on [0, 2π]×RN .


|V |p′−2V = LΓ[V ] on [0, 2π]×RN







∂2tU − ∆U = Γ(x)|U |p−2U on [0, 2π]×RN


(∂2t − ∆)

[Γ(x)−1/p |V |p′−2V

]= Γ(x)1/pV on [0, 2π]×RN .


|V |p′−2V = LΓ[V ] on [0, 2π]×RN







|V |p′−2V = LΓ[V ] on [0, 2π]×RN

with LΓ : Lp′([0, 2π]×RN )→ Lp([0, 2π]×RN ) symmetric and compact.

. Introduce J : Lp′([0, 2π]×RN )→ R via

J(V ) :=1p′

∫|V |p′ d(t, x)− 1

2

∫V LΓ[V ] d(t, x).

Following Evéquoz and Weth (2015, stationary case),. J has the mountain pass geometry,. J satisfies the Palais-Smale condition,. hence J has a nontrivial ground state V0 ∈ Lp

′([0, 2π]×RN ).

“Dual” ground state U0 = Γ(x)−1/p |V0|p′−2V0.




|V |p′−2V = LΓ[V ] on [0, 2π]×RN



J(V ) :=1p′

∫|V |p′ d(t, x)− 1

2

∫V LΓ[V ] d(t, x).


′([0, 2π]×RN ).





|V |p′−2V = LΓ[V ] on [0, 2π]×RN



J(V ) :=1p′

∫|V |p′ d(t, x)− 1

2

∫V LΓ[V ] d(t, x).

Following Evéquoz and Weth (2015, stationary case),. J has the mountain pass geometry,. J satisfies the Palais-Smale condition,

. hence J has a nontrivial ground state V0 ∈ Lp′([0, 2π]×RN ).





|V |p′−2V = LΓ[V ] on [0, 2π]×RN



J(V ) :=1p′

∫|V |p′ d(t, x)− 1

2

∫V LΓ[V ] d(t, x).


′([0, 2π]×RN ).



Thank you for your attention!

C. Blank, M. Chirilus-Bruckner, V. Lescarret and G. Schneider, BreatherSolutions in Periodic Media, Comm. Math. Phys. 3 (2011), pp. 815–841.A. Hirsch and W. Reichel, Real-valued, time-periodic localized weaksolutions for a semilinear wave equation with periodic potentials,Nonlinearity 32 (2019), no. 4, pp. 1408–1439.R. Mandel, E. Montefusco and B. Pellacci, Oscillating solutions fornonlinear Helmholtz equations, ZAMP 6 (2017), article 121.G. Evéquoz and T. Weth, Dual variational methods and nonvanishing forthe nonlinear Helmholtz equation Adv. in Math. 280 (2015), pp. 690–728.

The bifurcation result for the KG equation is part of my PhD thesis(KITopen, 2019).


Thank you for your attention!

C. Blank, M. Chirilus-Bruckner, V. Lescarret and G. Schneider, BreatherSolutions in Periodic Media, Comm. Math. Phys. 3 (2011), pp. 815–841.A. Hirsch and W. Reichel, Real-valued, time-periodic localized weaksolutions for a semilinear wave equation with periodic potentials,Nonlinearity 32 (2019), no. 4, pp. 1408–1439.R. Mandel, E. Montefusco and B. Pellacci, Oscillating solutions fornonlinear Helmholtz equations, ZAMP 6 (2017), article 121.G. Evéquoz and T. Weth, Dual variational methods and nonvanishing forthe nonlinear Helmholtz equation Adv. in Math. 280 (2015), pp. 690–728.

The bifurcation result for the KG equation is part of my PhD thesis(KITopen, 2019).

How to construct Breather Solutions using Nonlinear ...scheider/media/scheider... · Breather solutions 2 16.01.2020-UniversityofStuttgart DominicScheider-The Sine-Gordon Breather

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