Nonlinear Schrodinger equations with randompotentials existence and uniqueness

Leandro Cioletti - UnBcioletti@mat.unb.br

XVII - EBP

Mambucaba

August 2013

Paper

On nonlinear Schrodinger equations with random potentials:existence and probabilistic properties

Joint work with:

• Marcelo Furtado - UnB

• Lucas C. Ferreira - UNICAMP

Nonlinear Schrodinger Equation

We consider the nonlinear Schrodinger equation

ih∂ψ

∂t= −h2∆ψ + V (x)ψ − |ψ|p−1ψ, x ∈ Rn (1)

where t ∈ R, n ≥ 3, p > 1, h is the Planck constant and i is theimaginary unit.

Stading wave solutions

ih∂ψ

∂t= −h2∆ψ + V (x)ψ − |ψ|p−1ψ, x ∈ Rn.

Standing wave solutions

ψ(x , t) := e−iEhtu(x).

Elliptic Problem

Standing wave solutions

ψ(x , t) := e−iEhtu(x).

−∆u + V (x)u = |u|p−1u, x ∈ RN .

Choosing the domain and boundary conditions.−∆u + V (x)u = b(x)u|u|p−1 + g(x), if x ∈ U;

u = 0, if x ∈ ∂U,(2)

Introducing the Randomness

−∆u + V (x)u = |u|p−1u, x ∈ RN .

Dirichlet boundary condition−∆u + Vω(x)u = b(x)u|u|p−1 + g(x), if x ∈ U;

u = 0, if x ∈ ∂U,(3)

Basic Notation

• (Ω,F ,P) is a complete probability space.

• If (E , E) is a measurable space, any (F , E)-measurablefunction X : Ω→ E will be called a E -valued random variable.

• M(U) – set of all Random measures over U having finitevariation.

• BC (U) – bounded continuous functions .

The Random PotentialGiven a continuous function f : RN → R we consider

Vω(x) :=

∫Uf (x − y) dµω(y) (4)

where ω → µω is a M(U)-valued random variable.

Examples:

• Unordered alloy (on the square lattice Zd)

Vω(x) =∑i∈Zd

qi (ω)f (x − i)

• Discrete models for glass or rubber

Vω(x) =∑i∈Zd

f (x − ηi (ω))

Discrete models were studied by

J.-M. Combes and P.D. Hislop: Localization for Some Continuous,Random Hamiltonians in d-dimensions. 1994.

J. Bourgain and C. Kenig : On localization in the continuousAnderson-Bernoulli model in higher dimension. 2005.

F. Germinet, A. Klein and J. Schenker: Dynamical delocalization inrandom Landau Hamiltonians. 2007.

O. Safronov: Absolutely continuous spectrum of one randomelliptic operator. 2008.

Basic questions

−∆u + Vω(x)u = b(x)u|u|p−1 + g(x), if x ∈ U;

u = 0, if x ∈ ∂U,(5)

1- Is the set

S = ω ∈ Ω : the problem (5) has a unique solution in L∞(U)

an event (F-measurable)?

2- What is the P(S)?

Integral formulation

• Let G be the Green function of the laplacian operator (−∆)in the bounded domain U ⊂ Rn with n ≥ 3.

0 ≤ G (x , y) ≤ 1

nαn(n − 2)

1

|x − y |n−2,

where αn is the volume of the unit ball in Rn.∫UG (x , y)dy ≤ 1

nαn(n − 2)

∫BdU

(x)

1

|x − y |n−2dy

=1

nαn(n − 2)

nαnd2U

2=

d2U

2(n − 2),

(6)

• definition of l0(n,U) ≡ l0

l0 :=d2U

2(n − 2). (7)

• We consider the mapping H : L∞(U)→ L∞(U) given by

H(ϕ)(x) :=

∫UG (x , y)ϕ(y)dy , x ∈ U.

• H is a bounded linear operator

‖H(ϕ)‖∞ ≤ l0‖ϕ‖∞. (8)

• Equivalent integral equation

u(x) = H(g) + H(Vωu) + H(bu|u|p−1). (9)

Key estimatesLet X := L∞(U) and define the linear operator T : X → X by

T (u) := H(Vωu), ∀ u ∈ X .previous estimates and the definition of the potential ⇒

‖T‖∞ ≤ l0‖f ‖∞ µω .

For the nonlinear term we define B : X → X by

B(u) := H(b|u|p−1u)

If x , y ∈ R there holds∣∣x |x |p−1 − y |y |p−1∣∣ ≤ p|x − y |

(|x |p−1 − |y |p−1

),

and therefore it follows that

‖b(·)(u|u|p−1 − u|u|p−1

)‖∞ ≤ ‖b‖∞‖u−u‖∞

(‖u‖p−1

∞ − ‖u‖p−1∞).

From previous inequality

‖B(u)− B(u)‖∞ ≤ l0p‖b‖∞‖u − u‖∞(‖u‖p−1

∞ − ‖u‖p−1∞), (10)

for any u, u ∈ L∞(U).

TheoremGiven f , b, g ∈ L∞(U) and ω ∈ Ω, we consider the potential Vω

induced by the random measure µω := X (ω).

τω := l0‖f ‖∞ µω and K := l0p‖b‖∞. (11)

If ε > 0 and ω ∈ Ω are such that

0 ≤ τω < 1,2pKεp−1

(1− τω)p−1+ τω < 1, (12)

and ‖g‖∞ ≤ ε/l0, then the equation (5) has an unique integralsolution

uω = u(·, ω) ∈ L∞(U) such that ‖uω‖∞ ≤2ε

1− τω. (13)

For each ω ∈ Ω, we consider the closed ball

Bε =

u ∈ L∞(U); ‖u‖∞ ≤

2ε

(1− τω)

Now we want to prove that Φ is a contraction on (Bε, d).

Φ(u) :=H(g) + H(Vωu) + H(bu |u|p−1) (14)

= H(g) + T (u) + B(u) (15)

using again the previous estimates

‖Φ(u)‖∞ ≤ ‖H(g)‖∞ + ‖T (u)‖∞ + ‖B(u)‖∞

≤ l0 ‖g‖∞ + τω‖u‖∞ + K‖u‖p∞

≤ 2ε

1− τω. (16)

The contraction ...

‖Φ(u)− Φ(u)‖∞ = ‖T (u − u)‖∞ + ‖B(u)− B(u)‖∞

≤ τω‖u − u‖∞ + K‖u − u‖∞(‖u‖p−1

∞ + ‖u‖p−1∞)

≤ γω‖u − u‖∞.

TheoremLet ν be the probability measure induced on R by the randomvariable ω 7→ µω . Let g ∈ L∞(U) be such that ‖g‖∞ is smallenough.Let A be the set of ω ∈ Ω such that (5) has a unique solutionu(·, ω) given by the last theorem. The set A is called theadmissible for the random variable X .

(i) The set A is F-measurable and

P(A) = ν

([0,

1

l0‖f ‖∞

)).

(ii) Let uω, uω be two solutions of (5) corresponding,respectively, to µω, g ,A and µω, g , A. We have that

‖u(·, ω)− u(·, ω)‖∞ = O (‖g − g‖∞ + ‖f ‖∞ µω − µω )

for all ω ∈ A∩A.

(iii) The map U : A → L∞(U) given by U(ω) := u(·, ω) is arandom variable and there holds

‖u(·, ω)‖∞ ≤2ε0

1− τω, (17)

for all ω ∈ A.

a.s. existence and uniqueness

TheoremLet µjj∈N be a sequence inM(U) and let aj(ω)j∈N be asequence of random variables from Ω to R. Assume that thefollowing series is convergent inM(U)

µω =∞∑j=1

aj(ω)µj .

Denote by Lk = ω ∈ Ω : Sk ≥ c, with 0 < c < 1/(l0‖f ‖∞). If

∞∑k=1

P(Lk) <∞

then there is an integral solution for (5) a.s.

Let (X , ‖ · ‖X ) be a Banach space and (Ω,F ,P) be a probabilityspace.If X : Ω→ X is a X -valued Borel random variable∫

Ω‖X (ω)‖X dP(ω) <∞,

then there exist a unique element E[X ] ∈ X with the property

`(E[X ]) =

∫Ω`(X (ω)) dP(ω)

for all ` ∈ X ∗, where X ∗ is the dual of X .Notation

E[X ] =

∫ΩX (ω) dP(ω).

We call E[X ] the Bochner integral of X with respect to P.

TheoremUnder hypotheses of Theorem 2 let us denote byuω(x) = u(x , ω) ∈ A the solution of (5). Let m ∈ N and supposethat

∞∑j=1

(m + j − 1)!

(m − 1)!j!(l0‖f ‖∞)j EA[ µω

j ] < +∞. (18)

Then EA[|u|m(x , ω)] ∈ L∞(U) and

EA[‖|u|m(·, ω)‖L∞(U)

]<∞. (19)

In particular, EA[u(x , ω)] ∈ L∞(U).

Theorem (Central Limit Theorem)

Let Xjj∈N be an independent identically distributed (i.i.d.)sequence of random variables Xj : Ω→M(U). Assume that theadmissible set Aj = Ω for all j , and let uj(·, ω) ∈ L∞(U) be thesolution given by Theorem 2 with respect to Xj(ω) = µω,j and g .We have that Zjj∈N given by Zj(ω) := ‖uj(·, ω)‖∞ is a i.i.d.sequence of random variables, and if m = E[‖uj(·, ω)‖∞] <∞ andσ2 := Var Zj <∞ then following holds as k → +∞

k∑j=1

(Zj −m)

σ√k→ N(0, 1).

Theorem (Weak Law of Large Numbers)

Let Xjj∈N be an independent sequence of random variablesXj : Ω→M(U). Assume that the admissible set Aj = Ω for all j ,and let uj(·, ω) ∈ L∞(U) be the solution given by Theorem 2 withrespect to Xj(ω) = µω,j and g . If Xj → X a.s. and

L = supj∈N

(ess sup

ω∈Ωµω,j

)<

1

l0‖f ‖∞, (20)

thenk∑

j=1

‖uj(·, ω)‖∞ − EΩ[‖uj(·, ω)‖∞]

k→ 0, (21)

when k →∞.

References

A. Ambrosetti, A. Malchiodi and S. Secchi, Multiplicity results for somenonlinear Schrodinger equations with potentials, Arch. Rational Mech.Anal. 159 (2001), 253–271.

A. Ambrosetti, M. Badiale and S. Cingolani, Semiclassical states ofnonlinear Schrodinger equations, Arch. Rational Mech. Anal. 140 (1997),285-300.

G. Bal, T. Komorowski and L. Ryzhik: Asymptotic of the Solutions of theRandom Schrodinger Equation. Arch. Rational Mech. Anal. 200, 613-664(2011).

J. Bourgain: Nonlinear Schrodinger Equation With a Random Potential.Illinois J. math. 50, 183-188 (2006).

The End