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Universität Konstanz

The viscous model of quantum hydrodynamics in several dimensions

Li Chen M ichal Dreher

Konstanzer Schriften in Mathematik und Informatik Nr. 2 13 , Februar 2006

ISSN 1430-3558

© Fachbereich Mathematik und Statistik

© Fachbereich Informatik und Informationswissenschaft Universität Konstanz

Fach D 188, 78457 Konstanz, Germany E-Mail: preprints@informatik.uni-konstanz.de

Konstanzer Online-Publikations-System (KOPS) URL: http://www.ub.uni-konstanz.de/kops/volltexte/2007/2233/

URN: http://nbn-resolving.de/urn:nbn:de:bsz:352-opus-22338

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The Viscous Model of Quantum Hydrodynamics in Several Dimensions

Li Chen

and Michael Dreher

Abstract

We investigate the viscous model of quantum hydrodynamics in one and higher space dimensions. Exploiting the entropy dissipation method, we prove the exponential stability of the thermal equilibrium state in 1,2, and 3 dimensions, provided that the domain is a box.

Further, we show the local in time existence of a solution in the one dimensional case; and in the case of higher dimensions under the assumption of periodic boundary conditions. Finally, we discuss the semiclassical limit.

Keywords:quantum hydrodynamics; exponential decay; entropy dissipation method; local existence of solutions; semiclassical limit.

AMS Mathematics Subject Classification: 35B40, 35Q35, 76Y05

1 Introduction

The flow of charged particles in a semi-conductor can be simulated using different models. Typical examples are the quantum energy transport models (QET), the quantum drift diffusion model (QDD) or the quantum hydrodynamic model (QHD). Derivations of the quantum QET and QDD models can be found in [4]. The quantum hydrodynamic models can be derived directly from the Schr¨odinger–Poisson system by WKB wave functions [7]; or from the collision Wigner equa- tion by the momentum method and closing the system with the quantum thermal equilibrium distribution [5]; or by the entropy minimization method [9].

In this paper, we study the viscous QHD model, a model which is derived from the Wigner equation with the Fokker–Planck collision operator:

tn−divJ=ν0△n,

tJ−div

J⊗J n

−T∇n+n∇V +ε2 2n∇

△√

√nn

0△J−J τ, λ2△V =n−C(x),

n(0, x) =n0(x), J(0, x) =J0(x),













(1)

where (t, x)∈(0,∞)×Ω and Ω is a domain inRd.

Address: Department of Mathematical Sciences, Tsinghua University, Beijing, 100084, P.R. China, lchen@math.tsinghua.edu.cn

Address: Fachbereich Mathematik und Statistik, P.O.Box D187, Universit¨at Konstanz, 78457 Konstanz, Ger- many, michael.dreher@uni-konstanz.de

(3)

Our boundary conditions are either

νn(t, x) = 0, (t, x)∈(0,∞)×∂Ω, J(t, x) = 0, (t, x)∈(0,∞)×∂Ω,

νV(t, x) = 0, (t, x)∈(0,∞)×∂Ω,





(2)

where∂ν denotes the normal derivative, or we assume periodic boundary conditions, i.e.,

Ω =Td is ad–dimensional torus. (3)

Moreover, we suppose

x∈Ωinf n0(x)>0, (4)

Z

n0(x)−C(x)

dx= 0. (5)

The last condition is necessary; the Poisson equation forV would not be solvable otherwise.

The unknown functions in this system are the particle densityn=n(t, x) : [0,∞)×Ω→R+, the current densityJ =J(t, x) : [0,∞)×Ω→Rd, and the electrostatic potentialV =V(t, x) : [0,∞)× Ω→R. The given functionC=C(x) : Ω→Ris the doping profile of background charges.

The scaled physical constants are a viscosity constantν0describing the strength of the collisions, a temperature constantT, the Planck constantε, the momentum relaxation timeτ, and the Debye lengthλ. All these constants are assumed to be positive.

For quantum macroscopic models, some results on local or global existence or long time asymptotics are known. For the QDD model, the existence of weak solutions was shown in [2, 3, 10]; and the semiclassical limit and the long time behaviour were studied in [2, 3]. Concerning the QHD model without viscous terms, the existence of smooth solutions and their long time asymptotics for small initial data were investigated in [8, 13].

It seems that there are less mathematical results for the system (1). The authors are only aware of [6], where the exponential stability of a constant steady state to (1) in a one-dimensional setting with a certain boundary condition was proved, based on the entropy dissipation method [1]. Most of the difficulties arise from the Bohm potential term

B(n) = △√

√nn,

which introduces a third order perturbation to a system which could otherwise be interpreted as a parabolic system coupled to an elliptic equation. In this paper, we follow the approach of [6]

and generalize those results to domains of dimension two and three, see Theorem 2.1. Our key ingredient is an estimate on a certain term containing the Bohm potential, Proposition A.1.

Additionally, we are able to prove the local in time existence of sufficiently smooth solutions to (1).

The proof relies on the observation that the third-order perturbation term has a good sign, which makes standard energy estimates for parabolic systems possible after having introduced a fourth order viscous regularization.

Physically spoken, the periodic boundary conditions are of restricted interest; however, they enable us to prove the local in time existence of solutions in a one-dimensional setting with boundary conditions (2) immediately, see Theorem 2.5.

In the course of proving the local existence results of the Theorems 2.4 and 2.5, we will obtain a certaina prioriestimate of the solution, from which we then will be investigating the semiclassical limitε→+0, compare Theorem 2.7.

(4)

2 Main Results

Our notations are standard: Lp denote the usual Lebesgue spaces, andHk(Ω) :=W2k(Ω) are the L2–based Sobolev spaces, fork ∈N0. The brackets h·,·i stand for the scalar product inRd, and J⊗J is ad×dmatrix with entryJkJlat position (k, l).

We list our results:

Theorem 2.1 (Exponential stability). Let d= 1,2,3 andΩ =Qd

j=1(aj, bj)be a box. Let the triple (n, J, V)be a solution to (1)under the boundary conditions (2), and suppose that

n∈H1((0, t), H1(Ω))\

L2((0, t), H3(Ω)), J ∈H1((0, t), L2(Ω))\

L2((0, t), H2(Ω)), V ∈H1((0, t), H2(Ω)).

We assume thatC=C(x)≡C0>0 inΩand

(t,x)∈(0,tinf)×Ωn(t, x)>0.

Define an energy as follows:

E(t) = Z

ε2 2(∇√

n)2+T

n

ln n C0 −1

+C0

2

2 (∇V)2+|J|2 2n

dx. (6)

Let µ1 denote the first positive eigenvalue of − △ on Ω with Neumann boundary conditions, and fix

σ:= min 8T ν0

ε2 , µ1C0

µ1λ2T+C0

, 2 τ

. Then this energy satisfies the inequality

tE(t)≤ −σE(t)

−ε20

c1,d

Z

(△√

n)2dx+c2,d

Z

|∇n|4 n3 dx

−ν0

X

l

Z

n

∇ Jl

n 2

dx,

where the numbersc1,d andc2,d are given in the following table:

d 1 2 3

c1,d 2 13 19 c2,d 1

24 1 24

7 144

Remark 2.2. Physically spoken, the four terms in the energy (6) are the quantum energy, the thermodynamic entropy, the electric energy and the kinetic energy. We note that the energy of the steady state(n, J, V) = (C0,0,0)is zero, and that E can not become negative.

From the above differential inequality of the physical energy we then easily derive decay estimates:

(5)

Corollary 2.3. Assume that the solution mentioned in Theorem 2.1 exists up to t =∞. Then the function (n, J,∇V)decays to the steady state(C0,0,0) as follows:

ε2 2

∇√

n(t,·)−p C0

2

L2(Ω)≤e−σtE(0), (7)

T

√n(t,·)−p C0

2

L2(Ω)≤e−σtE(0), (8)

√n(t,·)−p C0

2

Lp(Ω)≤Cp,d,Ω2 2

ε2 + 1 T

e−σtE(0), (9) ε2

0c1,d

Z t=0

△√

n(t,·)−p C0

2

L2(Ω) dt≤E(0), (10)

1 2

J(t,·) pn(t,·)

2

L2(Ω)

≤e−σtE(0), (11)

λ2

2 k∇V(t,·)k2L2(Ω)≤e−σtE(0), (12) whereCp,d,Ω is the norm of the embedding H1(Ω)⊂Lp(Ω), with1≤p≤ ∞ford= 1,1≤p <∞ ford= 2, and1≤p≤6 ford= 3.

Theorem 2.4 (Local existence and uniqueness on a torus). Let dbe a positive integer and Ω =Td be a torus. Letb be the smallest integer greater than 12d, ands≥b be an integer. Suppose

n0∈Hs+1(Ω), J0∈Hs(Ω), C∈Hs−1(Ω)

and (4),(5). Then the problem(1)has a solution(n, J, V), local in a time interval[0, t], with the regularity properties

n∈H1((0, t), Hs(Ω))\

L2((0, t), Hs+2(Ω)), J ∈H1((0, t), Hs−1(Ω))\

L2((0, t), Hs+1(Ω)), V ∈H1((0, t), Hs+2(Ω))\

L2((0, t), Hs+4(Ω)), (n,∇n, J)∈C([0, t]×Ω).













(13)

The solution is unique and persists as long as n(t,·) and J(t,·) stay in L(Ω) and n remains positive. The life spant does not depend on the Sobolev regularitys.

Having shown the local existence in the periodic case, we can take advantage from geometric arguments and consider the case Ω⊂R1 effortlessly:

Theorem 2.5 (Local existence and uniqueness in one dimension). Let Ω⊂R1 be an open and bounded interval. Suppose

n0∈H2(Ω), J0∈H1(Ω), C∈L2(Ω)

and (4),(5). The initial functions are assumed to satisfy the compatibility conditions

νn0(x) = 0, J0(x) = 0, x∈∂Ω.

Then the problem (1)with the boundary conditions (2)has a local solution(n, J, V), with n∈H1((0, t), H1(Ω))\

L2((0, t), H3(Ω)), J ∈H1((0, t), L2(Ω))\

L2((0, t), H2(Ω)), V ∈H1((0, t), H3(Ω))\

L2((0, t), H5(Ω)), (n,∇n, J)∈C([0, t]×Ω).

(6)

This solution is unique and persists as long as n(t,·) and J(t,·) stay in L(Ω) and n remains positive.

Remark 2.6. The local in time existence in the one-dimensional case can also be proved for solutions of higher regularity provided that the usual compatibility conditions on the initial data are satisfied.

Theorem 2.7 (Semiclassical limit). Let Ω be either an open and bounded interval inR1 or a d-dimensional torus. Suppose that the given data(n0, J0, C)of (1) satisfy

n0∈Hb+1(Ω), J0∈Hb(Ω), C∈Hb−1(Ω)

and (4), (5). Let (nε, Jε, Vε), ε > 0, denote the solutions to (1) with boundary conditions (2) or (3), existing on a domain[0, t]×Ω.

As εtends to+0, a sub-sequence(nε, Jε, Vε)ε then converges to a limit(n, J, V), (nε, Jε,∇Vε)−→(n, J,∇V) inC([0, t]×Ω),

nε⇀ n inL2((0, t), Hb+2(Ω)), Jε⇀ J inL2((0, t), Hb+1(Ω)), (nε, Jε)⇀(n, J) inL((0, t), Hb(Ω)), which is a solution to the initial value problem













tn−divJ =ν0△n,

tJ−div

J⊗J n

−T∇n+n∇V =ν0△J−J τ, λ2△V =n−C(x),

n(0, x) =n0(x), J(0, x) =J0(x)

with boundary conditions (2) or (3), respectively. The regularity of (n, J, V)is given by n∈H1((0, t), Hb(Ω))\

L2((0, t), Hb+2(Ω)), J ∈H1((0, t), Hb−1(Ω))\

L2((0, t), Hb+1(Ω)), V ∈H1((0, t), Hb+2(Ω))\

L2((0, t), Hb+1(Ω)).

3 Exponential Stability

In this section, we prove Theorem 2.1, following an approach of [6]. Rewrite the energy from (6) as follows:

E(t) = Z

ε2 2(∇√

n)2+T

n

ln n C0 −1

+C0

2

2 (∇V)2+|J|2 2n

dx

=E1+E2+E3+E4.

We compute the time derivatives. In the sequel, the zeroes denote boundary integrals that vanish due to the boundary conditions:

tE12 Z

∂Ω

(∂ν√ n)(∂t

n) dσ−ε2 Z

(△√ n)(∂t

n) dx

= 0−ε2 2

Z

B(n)∂tndx=−ε2 2

Z

B(n) (divJ+ν0△n) dx.

(7)

Next, we have

tE2=T Z

ntln n C0

dx=T Z

(lnn) (divJ+ν0△n) dx

= 0−T Z

1

nh∇n, Jidx−T ν0

Z

|∇n|2 n dx

=−T Z

1

nh∇n, Jidx−8T ν0

ε2 E1. Further, we get

tE32 Z

(∇V)(∇Vt) dx=λ2 Z

∂Ω

V(∂νVt) dσ−λ2 Z

V(△Vt) dx

= 0− Z

V ntdx=− Z

V(divJ+ν0△n) dx

= 0 + Z

h∇V , Jidx+ν0

Z

h∇V ,∇nidx.

Finally, the identity

t

1

2n−1|J|2

0

1

nhJ,△Ji − 1

2n2(△n)|J|2

+ div 1

2n2J|J|2

− 1

nτ|J|2+T1

nh∇n, Ji − h∇V , Ji −ε2

2 (div(BJ)−BdivJ) implies

tE40

Z

1

nhJ,△Ji − 1

2n2(△n)|J|2

dx+ Z

∂Ω

1

2n2J|J|2d~σ−2 τE4

+T Z

1

nh∇n, Jidx− Z

h∇V , Jidx−ε2 2

Z

∂Ω

BJd~σ+ε2 2

Z

BdivJdx.

The two boundary integrals vanish, due to J = 0 on ∂Ω. Concerning the first integral, we can deduce, after partial integration, that

X

l

Z

1

nJl△Jl− 1

2n2Jl2△n

dx=−X

l

Z

n

∇ Jl

n 2

dx.

Summing up, we then find

tE=−ε20

Z

(△n)B(n) dx−8T ν0

ε2 E10

Z

h∇V ,∇nidx

−ν0

X

l

Z

n

∇ Jl

n 2

dx−2 τE4.

For the third term, we bring the constance of the doping profile into play:

Z

h∇V ,∇nidx= Z

h∇V ,∇(n−C0)idx

= Z

∂Ω

(∂νV)(n−C0) dσ− Z

(△V)(n−C0) dx= 0− 1 λ2

Z

(n−C0)2dx, Z

h∇V ,∇nidx=−λ2 Z

(△V)2dx

=−α0λ2 Z

(△V)2dx−1−α0

λ2 Z

(n−C0)2dx.

(8)

One easily checks thatx(lnx−1) + 1≤(x−1)2, forx >0, which implies C0−1(n−C0)2≥n

ln n

C0−1

+C0,

−1−α0

λ2 Z

(n−C0)2dx≤ −(1−α0)C0

λ2T E2, 0< α0<1.

Ifµ1 denotes the first positive eigenvalue of− △on Ω with Neumann boundary conditions, then k∇Vk2L2 ≤ 1

µ1k△Vk2L2, ∂νV = 0 on∂Ω.

As a consequence,

−α0λ2 Z

(△V)2dx≤ −α0λ2µ1k∇Vk2L2 =−α0µ1E3. Exploiting Proposition A.1, we then find

tE≤ −8T ν0

ε2 E1−(1−α0)C0

λ2T E2−α0µ1E3− 2 τE4

−ε2 2 ν0c1,d

Z

(△√

n)2dx−ε20c2,d

Z

|∇n|4

n3 dx−X

l

Z

n

∇ Jl

n 2

dx, where the numbersc1,dandc2,dare as in the theorem.

We chooseα0=µ1λ2CT0+C0, and obtain

tE≤ −8T ν0

ε2 E1− µ1C0

µ1λ2T+C0

(E2+E3)−2 τE4

−ε2 2 ν0c1,d

Z

(△√

n)2dx−ε20c2,d

Z

|∇n|4

n3 dx−X

l

Z

n

∇ Jl

n 2

dx.

This completes the proof of Theorem 2.1.

Proof of Corollary 2.3. We directly haveE(t)≤exp(−σt)E(0), which gives us (7) and (12) im- mediately. Next, it is easy to check that

(y−1)2≤y2(lny2−1) + 1, 0< y <∞, which then yields

(p

n(t, x)−p

C0)2≤n(t, x)

lnn(t, x) C0 −1

+C0,

and (8) follows quickly, as well as (9). The remaining estimates are proved similarly.

4 Existence on the Torus

The purpose of this section is to prove Theorem 2.4.

(9)

To this end, we choose numbersγ with 0< γ <1, and consider a family of parabolic initial value problems













tn−divJ =ν0△n−γ△2n,

tJ −div

J⊗J n

−T∇n+n∇V +ε2 2 n∇

△√n

√n

0△J−γ△2J−J τ, λ2△V =n−Cγ(x),

n(0, x) =n0γ(x), J(0, x) =Jγ0(x),

(14)

where (t, x)∈R×Ω. We assume that the functionsCγ,n0γ, andJγ0belong toC(Ω), satisfy the compatibility condition

Z

(n0γ(x)−Cγ(x)) dx= 0,

and converge toC,n0,J0in Sobolev norms as follows, forγtends to zero:

Cγ −→C inHs−1(Ω), n0γ −→n0 in Hs+1(Ω), Jγ0−→J0 inHs(Ω).

The system (14) is a fourth order nonlinear parabolic system with third order lower terms. It is standard to show that this problem has a unique solution

(nγ, Jγ, Vγ)∈C([0, tγ)×Ω)×C([0, tγ)×Ω)×C([0, tγ)×Ω),

for sometγ >0. The solution persists as long asnγ stays positive and (nγ, Jγ, Vγ) remain bounded.

A proof will be sketched in Lemma B.2.

Our approach is as follows:

• shrink the interval [0, tγ) to guarantee some boundedness assumptions on (nγ, Jγ, Vγ);

• derive uniform inγa priori estimates of the solutions (nγ, Jγ, Vγ);

• show that tγ can not go to zero forγ→0;

• prove convergence of a sub-sequence of (nγ, Jγ, Vγ)γ forγ→0;

• study the limit of that sub-sequence.

Fix a numberδ0 by the conditions 0< δ0<min

x∈Ωn0(x), max

x∈Ωn0(x)< δ0−1. In the following computations, we always assume that

δ0≤nγ(t, x)≤δ−10 . For a multi-indexα∈Nd0, define

nγ,α:=∂xαnγ, Jγ,α:=∂xαJγ, Vγ,α:=∂αxVγ.

(10)

Then we obtain

tJγ,α−ν0△Jγ,α+γ△2Jγ,α+1 τJγ,α

= div∂xα

Jγ⊗Jγ

nγ

+T∇nγ,α−∂xα(nγ∇Vγ)−ε2

2∂xα(nγ∇B(nγ)).

Multiplying this equation withJγ,α, integrating over Ω, performing partial integration, and taking advantage from the periodic boundary conditions, we find

1

2∂tkJγ,αk2L20k∇Jγ,αk2L2+γk△Jγ,αk2L2+1

τ kJγ,αk2L2

= Z

Jγ,αdiv∂xα

Jγ⊗Jγ

nγ

dx−T

Z

(divJγ,α)nγ,αdx

− Z

Jγ,αxα(nγ∇Vγ) dx−ε2 2

Z

Jγ,αxα(nγ∇B(nγ)) dx.

The integrals on the right-hand side are treated as follows:

Z

Jγ,αdiv∂xα

Jγ⊗Jγ

nγ

dx=−X

k,l

Z

(∂kJγ,α,l)∂xα

Jγ,lJγ,k

nγ

dx,

−T Z

(divJγ,α)nγ,αdx=−T Z

tnγ,α−ν0△nγ,α+γ△2nγ,α

nγ,αdx

=−T

2∂tknγ,αk2L2−T ν0k∇nγ,αk2L2−T γk△nγ,αk2L2, nγ∇B(nγ) = 1

2∇ △nγ−1 2

X

l

l

(∂lnγ)∇nγ

nγ

,

−ε2 2

Z

Jγ,αxα(nγ∇B(nγ)) dx

2 4

Z

(divJγ,α)△nγ,αdx−ε2 4

X

k,l

Z

(∂lJγ,α,k)∂xα

(∂lnγ)(∂knγ) nγ

dx

2 4

Z

(∂tnγ,α−ν0△nγ,α+γ△2nγ,α)△nγ,αdx−ε2 4

X

k,l

Z

. . .dx

=−ε2

8 ∂tk∇nγ,αk2L2−ε2

0k△nγ,αk2L2−ε2

4 γk∇ △nγ,αk2L2−ε2 4

X

k,l

Z

. . .dx.

Then it follows that T

2∂tknγ,αk2L2+T ν0k∇nγ,αk2L2+T γk△nγ,αk2L2 (15) +ε2

8∂tk∇nγ,αk2L22

4 ν0k△nγ,αk2L22

4γk∇ △nγ,αk2L2 +1

2∂tkJγ,αk2L20k∇Jγ,αk2L2+γk△Jγ,αk2L2+1

τkJγ,αk2L2

=−X

k,l

Z

(∂kJγ,α,l)∂xα

Jγ,lJγ,k

nγ

dx−

Z

Jγ,αxα(nγ∇Vγ) dx

−ε2 4

X

k,l

Z

(∂lJγ,α,k)∂xα

(∂lnγ)(∂knγ) nγ

dx

=I1,α+I2,α2 4 I3,α.

(11)

We define an energy:

Ek(t) = X

|α|=k

T

2 knγ,αk2L22

8 k∇nγ,αk2L2+1

2kJγ,αk2L2

, k≥0, (16)

E0,...,k:=

k

X

l=0

El. (17)

This energy is related to Sobolev space norms via knγk2L2≤ 2

TE0, X

|α|=k

knγ,αk2L2 ≤ C

ε2Ek−1, k≥1, knγk2Hk≤C

1 T + 1

ε2

E0,...,k−1. The identity (15) then yields

tEk+ X

|α|=k

T ν0k∇nγ,αk2L22

0k△nγ,αk2L20k∇Jγ,αk2L2+1

τkJγ,αk2L2

≤ X

|α|=k

|I1,α|+|I2,α|+ε2 4 |I3,α|

.

Next we estimate the integralsI1,α,I2,α,I3,αin terms ofE0,...,k. The constantsCin the following computations may change from one line to another, and can depend on the order of differentiation k=|α|, the space dimension dand the lower bound δ0 of n0, but are independent of ν0, γ, ε,τ, andλ. Recall the embeddingHb(Ω)⊂L(Ω). We will make free use of the estimates

kf gkHk≤C(kfkLkgkHk+kfkHkkfkL), k≥0, kf(u(·))kHk≤C(kukL)kukHk, k≥0, f(0) = 0.

Then we can conclude that

|I1,α| ≤X

l,m

k∂mJγ,α,lkL2

n−1γ Jγ,lJγ,m

H|α|

≤Ck∇Jγ,αkL2 n−1γ

H|α|kJγk2L+ n−1γ

LkJγkH|α|kJγkL

≤Ck∇Jγ,αkL2

knγkH|α|kJγk2L+kJγkH|α|kJγkL

≤Ck∇Jγ,αkL2 1 +T−1 E0,...,|α|

12

E0,...,b+p E0,...,b

≤ν0

4 k∇Jγ,αk2L2+ C

ν0 E0,...,b2 +E0,...,b

1 +T−1

E0,...,|α|. Concerning the second integral, we have

|I2,α| ≤ kJγ,αkL2knγ∇VγkH|α|

≤CkJγ,αkL2 knγkLk∇VγkH|α| +knγkH|α|k∇VγkL

≤CkJγ,αkL2 kVγkH|α|+1+knγkH|α|kVγkHb+1

≤ C

λ2kJγ,αkL2 knγ−CγkH|α|−1+knγkH|α|knγ−CγkHb−1

(12)

≤ C

λ2kJγ,αkL2 knγkH|α| +kCγkH|α|−1+knγkH|α| knγkHb−1+kCγkHb−1

≤ C λ2

kJγ,αk2L2+knγk2H|α|

1 +kCγk2Hb−1+knγk2Hb−1

+kCγk2H|α|−1

≤ C λ2

E0,...,|α|+T−1E0,...,|α|

1 +kCγk2Hb−1+T−1E0,...,b−1

+kCγk2H|α|−1

= C λ2

1 +T−1+T−1kCγk2Hb−1+T−2E0,...,b−1

E0,...,|α|+ C

λ2kCγk2H|α|−1. Concerning the last integral, we have

|I3,α| ≤X

k,l

k∂lJγ,α,kkL2

n−1γ (∂lnγ)(∂knγ) H|α|

≤X

k,l

k∂lJγ,α,kkL2k(∂llnnγ)(∂knγ)kH|α|

≤Ck∇Jγ,αkL2 klnnγkH|α|+1k∇nγkL+k∇lnnγkLknγkH|α|+1

≤Ck∇Jγ,αkL2knγkH|α|+1k∇nγkL.

Fix a numberθd withθd∈(1/2,1) for evendandθd∈(2/3,1) for oddd. Then we have k∇nγkL ≤CknγkθHdb+1knγk1−θLd,

by Lemma B.3. Together withknγkL ≤δ0−1, we then get the estimate

|I3,α| ≤Ck∇Jγ,αkL2knγkH|α|+1knγkθHdb+1. Now we distinguish two cases.

Case 1: b+ 1≤ |α|+ 2. Then we have the interpolation inequalities knγkHb+1 ≤Cknγk

1

|α|+2−b

H|α|+2 knγk

|α|+1−b

|α|+2−b

Hb ,

knγkH|α|+1 ≤Cknγk

|α|+1−b

|α|+2−b

H|α|+2 knγk

1

|α|+2−b

Hb ,

which imply

knγkH|α|+1knγkθHdb+1 ≤Cknγk1−̺H|α|+2knγkθHdb, where we have set̺=|α|+2−b1−θd ∈(0,1). Altogether, we get

|I3,α| ≤Ck∇Jγ,αkL2knγk1−̺H|α|+2knγkθHdb

≤Ck∇Jγ,αkL2

k△nγk2H|α| +knγk2H|α|1−̺2

knγkθHdb.

If we apply Young’s inequality with the exponents 2, 1−̺2 and 2̺ to the right-hand side, we deduce

(13)

that (puttingk=|α|) ε2

4 |I3,α| ≤

√ν0

2 k∇Jγ,αkL2·

ε2ν0

8M(d)(k△nγk2Hk+knγk2Hk) 1−̺2

×

×Cν−1+

̺ 2

0 ε1+̺knγkθHdb

√ν0

2 k∇Jγ,αkL2·

ε2ν0

8M(d)(k△nγk2Hk+knγk2Hk) 1−̺2

×

×Cν−1+

̺ 2

0 ε1−θd εknγkHb

θd

≤ ν0

8 k∇Jγ,αk2L2+ ε2ν0

8M(d)

k△nγk2Hk+knγk2Hk

+Cν1−

2

̺

0 ε2(1−̺θd)

ε2knγk2Hb

θd

̺ . Recalling that̺=̺(k) =k+2−b1−θd , we see that

ε2

4|I3,α| ≤ ν0

8 k∇Jγ,αk2L2+ ε2ν0

8M(d)k△nγk2Hk+Cν0 1 +ε2T−1

E0,...,k−1 +Cν1−

2

̺

0 ε2(k+2−b) 1 +ε2T−1

E0,...,b−1

1+θd̺

.

Case 2: b+1>|α|+2. In this case, we haveknγkH|α|+1 ≤ knγkHb−1. Applying Young’s inequality with the exponents 2, θ2

d and 1−θ2

d to the estimate

|I3,α| ≤Ck∇Jγ,αkL2knγkθHdb+1knγkHb−1, we get

ε2

4|I3,α| ≤

√ν0

2 k∇Jγ,αkL2·

ε2knγk2Hb+1θd2

·Cν012ε2−θdknγkHb−1

≤ν0

8 k∇Jγ,αk2L2+C 1 +ε2T−1 E0,...,b

+C

ν012ε1−θd1−2θd

ε2knγk2Hb−1

1−1θd

≤ν0

8 k∇Jγ,αk2L2+C 1 +ε2T−1 E0,...,b

+Cν

1 1−θd

0 ε2 1 +ε2T−1

E0,...,b−21−1θd .

Having now the estimates in both cases, we choose the numberM(d) sufficiently large. Then we can conclude that

tEk+ X

|α|=k

T ν0k∇nγ,αk2L22

8 ν0k△nγ,αk2L20

2 k∇Jγ,αk2L2+1

τ kJγ,αk2L2

≤ C

ν0 E0,...,b2 +E0,...,b

1 +T−1 E0,...,k + C

λ2

1 +T−1+T−1kCγk2Hb−1+T−2E0,...,b−1

E0,...,k+ C

λ2kCγk2Hk−1

+CR,

(14)

where the remainder termR equals ν0 1 +ε2T−1

E0,...,k−11−

2

̺(k)

0 ε2(k+2−b) 1 +ε2T−1

E0,...,b−11+̺(k)θd

in case ofk≥b−1; and fork < b−1 we have R= 1 +ε2T−1

E0,...,b

1 1−θd

0 ε2 1 +ε2T−1

E0,...,b−21−1θd .

For simplicity, we only discuss the cases=b. The other cases s > brun similarly. Summing up fork= 0, . . . , b, we find the energy estimate

tE0,...,b+T ν0k∇nγk2Hb2

0k△nγk2Hb0

2 k∇Jγk2Hb+1

τkJγk2Hb

≤C 1

ν0

+ 1 ν0T

E0,...,b2 +E0,...,b

+ (1 +ν0) 1 +ε2T−1 +1

λ2 + 1 λ2T + 1

λ2T kCγk2Hb−1+ 1

λ2T2E0,...,b−1

E0,...,b + C

λ2kCγk2Hb−1+Cν

1 1−θd

0 ε2 1 +ε2T−1

E0,...,b−2

1−1θd

+Cν

1+θd 1−θd

0 ε2 1 +ε2T−1

E0,...,b−11−1θd +Cν

3+θd 1−θd

0 ε4 1 +ε2T−1

E0,...,b−1

1+θd

1−θd. (18)

Observe that the right-hand side does not depend onγ.

It exists a numbert>0 such thatE0,...,b∈C1([0, t]) andE0,...,b(t)≤2E0,...,b(0) for 0≤t≤t. On the left-hand side, we have a termk∇Jγk2Hb. Shrinking the interval [0, t] if necessary, we can arrange thatt≤1 and

kdivJγkL2((0,t),Hb−1)≤1.

By Lemma B.1, we can show that knγ(t2,·)−nγ(t1,·)kL

≤C(Ω, α, ν0)|t2−t1|α2

kdivJγkL2([0,tγ],Hb−1)+ n0γ

Hb+1

≤C|t2−t1|α2, 0≤t1, t2≤tγ,

where 0< α < 12. Note that the right-hand side does not depend onγ, 0< γ <1. From this we can find a lower bound of the time the solution nγ needs to touch the boundary of the interval [δ0, δ−10 ].

We list the results obtained so far:

We have determined a numbert>0 with the property that, for eachγ with 0< γ <1, we have a solution

(nγ, Jγ, Vγ)∈C([0, t]×Ω)×C([0, t]×Ω)×C([0, t]×Ω), satisfyingδ0≤nγ(t, x)≤δ0−1 for all (t, x)∈[0, t]×Ω.

These functions satisfy thea prioriestimates

knγkL((0,t),Hb+1)≤C, kJγkL((0,t),Hb)≤C, k△nγkL2((0,t),Hb)≤C, k∇JγkL2((0,t),Hb)≤C.

(15)

These constantsC may depend onT,ν0,ε,λ, andδ0, but not onγ.

Because the functionCγ belongs toHb−1(Ω), we have k∇VγkL((0,t),Hb(Ω))≤C.

From the differential equations, we then obtain the uniform inγestimates on the time derivatives:

k∂tnγkL((0,t),Hb−4(Ω))+k∂tJγkL((0,t),Hb−4(Ω))≤C.

The embedding Hb+1(Ω) ⊂Hb(Ω) is compact. Therefore, the Aubin Lemma [11] implies that a sub-sequence (which we will not relabel) of (nγ)γconverges in the spaceC([0, t], Hb(Ω)) to a limit functionn. The sequence (nγ)γ is bounded inL2((0, t), Hb+2(Ω)). By interpolation, we get the strong convergences

nγ −→nin C([0, t], Hb+1−δ(Ω)), δ >0, nγ −→nin L2((0, t), Hb+2−δ(Ω)), δ >0.

And we have the weak convergences

nγ⇀ nin L2((0, t), Hb+2(Ω)), nγnin L((0, t), Hb+1(Ω)).

By a similar reasoning, we can show

Jγ −→J in C([0, t], Hb−δ(Ω)), δ >0, Jγ −→J in L2((0, t), Hb+1−δ(Ω)), δ >0, Jγ ⇀ J in L2((0, t), Hb+1(Ω)),

JγJ in L((0, t), Hb(Ω)).

Especially, we have the uniform convergences

(nγ,∇nγ, Jγ)−→(n,∇n, J) inC(Q),

where we have putQ:= (0, t)×Ω. In particular,nandJ satisfy the initial conditionsn(0, x) = n0(x) andJ(0, x) =J0(x).

The convergence of (nγ)γ yields the convergence of (∇Vγ)γ, too:

∇Vγ →V inC([0, t], Hb+2−δ), δ >0.

Finally, we show that (n, J, V) is a solution to (1). By the above reasoning, the identity λ2△V(t, x) =n(t, x)−C(x), (t, x)∈Q,

is obvious.

Take a functionϕ∈C0(Q). By the usual arguments, we find Z Z

Q −ϕtnγ0(∇nγ)(∇ϕ) +γnγ2ϕ

dxdt=− Z Z

Q

Jγ∇ϕdxdt.

We sendγto zero and find Z Z

Q

(−ϕtn+ν0(∇n)(∇ϕ)) dxdt=− Z Z

Q

J∇ϕdxdt, Z Z

Q

(−ϕtn+ϕ(−ν0△n−divJ)) dxdt= 0.

(16)

We conclude that the functionnhas distributional time derivative∂tn=ν0△n+ divJ. We study the terms of theJ–equation:

div

Jγ⊗Jγ

nγ

−→div

J⊗J n

in L2((0, t), L2(Ω)),

T∇nγ −→T∇n in C(Q),

nγ∇Vγ −→n∇V in C(Q),

nγ∇B(nγ)⇀ n∇B(n) in L2((0, t), L2(Ω)),

△Jγ ⇀ J in L2((0, t), L2(Ω)).

Similarly as for n, we can compute the distributional time derivative ofJ, and we will see that (n, J, V) solve (1).

To complete the proof of Theorem 2.4, we have to check the uniqueness of the solution: let (n1, J1, V1) and (n2, J2, V2) be two solutions with regularity as in (13). Put

n=n1−n2, J=J1−J2, V=V1−V2. Then we obtain the system

tn−ν0△n= divJ,

tJ−ν0△J+1

τJ−T∇n2

2∇ △n=R1−R2, Rj = div

Jj⊗Jj nj

−nj∇Vj+1 2

X

l

l

(∂lnj)(∇nj) nj

, j= 1,2, λ2△V=n,

with vanishing initial values fornandJ. Multiplying the second equation withJ, integrating over Ω, and performing partial integration gives

1

2∂tkJk2L20k∇Jk2L2+1 τ kJk2L2

+T Z

ndivJdx−ε2 2

Z

(△n) divJdx

= Z

J(R1−R2) dx, 1

2∂tkJk2L20k∇Jk2L2+1

τ kJk2L2+T

2∂tknk2L2+T ν0k∇nk2L2

2

2∂tk∇nk2L22

0k△nk2L2

= Z

J(R1−R2) dx.

Now it is standard to estimate

Z

J(R1−R2) dx ≤C

Jj L,

nj L,

∇nj L,

∇Vj L

×

×(k∇JkL2kJkL2+kJkL2(knkL2+k∇VkL2) +k∇JkL2knkH1). We apply Young’s inequality and find

T

2∂tknk2L22

2∂tk∇nk2L2+1

2∂tkJk2L2 ≤C

kJk2L2+knk2H1

.

An application of Gronwall’s lemma then yields n ≡ 0, J ≡0, which concludes the proof of Theorem 2.4.

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