Global Existence and Asymptotic Behavior for a Compressible Energy - - PowerPoint PPT Presentation

Global Existence and Asymptotic Behavior for a Compressible Energy Transport Model

Yong LI

Department of Mathematical Sciences, Tsinghua University BEIJING,

• P. R. CHINA

Joint with Ling Hsiao

The energy transport model is a degenerate quasi-linear cross diffusion parabolic system with principal part in divergence form, the common form is governed by

              

∂ ∂tn(µ, T) − divJn = 0, ∂ ∂tU(µ, T) − divJw = −∇V · Jn + W(µ, T), λ2△V = n − b(x). (1) Where µ, T are chemical potential of the electrons and the electron temperature respectively, V is the electrostatic po- tential, n(µ, T) is the electron density, U(µ, T) is the density of the internal energy, λ is the scaled Debye length, b(x) is the doping profile which represents the background of the device,

2

W(µ, T) is the energy relaxation term satisfying W(µ, T)(T − T0) ≤ 0, the positive constant T0 is the lattice temperature, Jn is the carrier flux density and Jw is the energy flux density, which are given by

          

Jn = L11

• ∇

µ

T

• + ∇V

T

• + L12∇
• − 1

T

• ,

Jw = L21

• ∇

µ

T

• + ∇V

T

• + L22∇
• − 1

T

• ,

L is the diffusion matrix, L =

L11

L12 L21 L22

• .

3

We consider the following relations for n(µ, T) and U(µ, T), which derived from the Boltzmann statistics, n(µ, T) = T

1 γ−1 exp

µ

T

• ,

U(µ, T) = 1 γ − 1T

γ γ−1 exp

µ

T

• .

The energy relaxation term W(µ, T) is given by W(µ, T) = 1 γ − 1T

1 γ−1 exp

µ

T

T0 − T

ε , the diffusion matrix L is a positive semidefinite matrix, L = (Lij) = εT

γ γ−1 exp

µ

T

• 

  

1

γT γ−1 γT γ−1 γ2T 2 (γ−1)2

    .

4

We could rewritten the above energy transport model as the following form:

                    

nt − ε∇ · [∇(nT) − n E] = 0,

nT

γ − 1

• t

− ε∇ ·

γT

γ − 1[∇(nT) − n E]

• = ε[n

E − ∇(nT)] · E − n(T − T0) ε(γ − 1) , λ2∆V = n − b(x),

• E = ∇V.

(2) Where ε is the energy relaxation time, and the adiabatic ex- ponent γ > 1. The diffusion matrix L satisfying L = (Lij) = εnT

   

1

γT γ−1 γT γ−1 γ2T 2 (γ−1)2

    .

Eigenvalues: µ1 = 0, µ2 = 1 + γ2 (γ − 1)2T 2.

5

◮ Derivation

The model (2) can be derived from the non-isentropic Euler- Poisson system in semiconductors or plasma,

                

nt + ∇ · (n u) = 0, (n u)t + ∇ · (n u u) + ∇P = n E − n

u τp ,

Wt + ∇ · ( uW + uP) = n u · E − W−W0

τw

, λ2∆V = n − b(x),

• E = ∇V,

with P = nT, W = n| u|2 + nT γ − 1, W0 = nT0

γ−1;

τp, τw are the momentum and energy relaxation times. This model was introduced in

♦ K. Bl¨

• tekjær, IEEE Trans. Electron. Devices, ED-17 (1970), 38-47.

♦ Markowich, Ringhofer and Schmeiser, Semiconductors Equations, 1990.

6

Define τ2 = τp τw << 1, and consider the following variables transformations, t → t

τ ,

• u → τ

u, τp = ετ, τw = ε

τ

(n, T, E) → (n, T, E). We rewrite the system as

                            

nτ

t + ∇ · (nτ

uτ) = 0, (τ2nτ uτ)t + ∇ · (τ2nτ uτ ⊗ uτ) + ∇(nτT τ) = nτ Eτ − nτ

uτ ε

,

τ2

2 nτ| uτ|2 + nτT τ γ − 1

• t + ∇ ·

τ2

2 nτ| uτ|2 uτ + γnτT τ

γ−1

uτ = nτ Eτ · uτ −

τ2

2 nτ| uτ|2 + nτ(T τ−T0)

ε(γ−1)

• ,

λ2∇ · Eτ = nτ − b(x).

7

Taking the formal limit as τ → 0, we obtain the compressible energy transport model,

                    

nt − ε∇ · [∇(nT) − n E] = 0,

nT

γ − 1

• t

− ε∇ ·

γT

γ − 1[∇(nT) − n E]

• = ε[n

E − ∇(nT)] · E − n(T − T0) ε(γ − 1) , λ2∇ · E = n − b(x). (3) This formal limit has been rigorously justified by

♦ I. Gasser and R. Natalini, Quart.Appl.Math., LVII, No.2(1999), 269-282. ♦ Y. Li, Acta Math. Sci., (in press, 2007).

8

◮Some Previous Results:

(The diffusion matrix L is (uniformly) positive definite)

• Numerical results

– Chen-Shu-Dutton, Degond-J¨ ungel-Pietra, Jerome-Shu, Souissi-Gnudi, Holst-J¨ ungel-Pietra,...

• Steady state

–Degond-G´ enieys-J¨ ungel, Griepentrog, Xie, Chen-Hsiao

• Global existence

–The diffusion matrix L is uniformly positive definite Degond-G´ enieys-J¨ ungel, J¨ ungel, Al ` ı etc. – The diffusion matrix L just positive definite Chen-Hsiao, Chen-Hsiao-Li

9

Consider the one-dimensional case of model (2) (ε = λ = T0 = 1)

              

nt − [(nT)x − nE]x = 0, J(x, t) = nE − (nT)x, (nT)t − {γT[(nT)x − nE]}x = (γ − 1)[nE − (nT)x]E − n(T − 1), Ex = n − b(x). (4) The insulated boundary conditions and initial values J(x, t)|x=0,1 = 0, Tx(x, t)|x=0,1 = 0, E(0, t) = 0, n(x, 0) = nI(x), T(x, 0) = TI(x), x ∈ Ω, (5) where the initial density nI(x) is chosen to satisfy

• Ω(nI − b)(x)dx = 0.

Then the boundary conditions are nx(x, t)|x=0,1 = 0, Tx(x, t)|x=0,1 = 0, E(x, t)|x=0,1 = 0. (6)

10

Introduce the flux as nu = nE − (nT)x. So (4) can be rewritten as

      

nt + (nu)x = 0, (nT)t + (γnTu)x = (γ − 1)nuE − n(T − 1), Ex = n − b(x). (7) The model (7) can be understood as the ”full” compressible Navier-Stokes equations (with a temperature equation),

                    

nt + (nu)x = 0, (nu)t + (γnu2)x + (nT)x = γ(nTux)x − [2 + b(x)]nu + 2nE − nx Tt + uTx + (γ − 1)Tux = (γ − 1)u2 − (T − 1), Ex = n − b(x).

11

◮ The isothermal stationary solution

We consider a typical stationary solution (N, 1, V).

    

∇N − N∇V = 0, x ∈ Ω, ∆V = N − b(x), x ∈ Ω, ∇V · γ|∂Ω = 0, γ : the unit outer normal vector Theorem 1 Assume that 0 < C ≤ b(x) ≤ C and b ∈ L∞(Ω), then the problem has a solution (N, V), and satisfing 0 < C ≤ N(x) ≤ C, x ∈ Ω, c ≤ V(x) ≤ c, x ∈ Ω, |∆V(x)|, |∇V(x)|, |∇N(x)|, |∆N(x)| ≤ C, x ∈ Ω, where C is a positive constant and c, c are constants.

12

◮ The GlobalExistence and exponential decay

Theorem 2 Suppose 0 < C ≤ b(x) ≤ C, b(x) ∈ C2(Ω) and (nI(x), TI(x)) ∈ H3(Ω). There exists a positive constant δ1, such that if nI(x) − N (x)H3 + TI(x) − 1H3 ≤ δ1, then the problem (4)-(5) has a unique global solution (n, T, E) in Ω × [0, ∞) satisfying n(·, t) − N (·)H3 + T(·, t) − 1H3 + E(·, t) − Vx(·)H3 ≤ c0(n(·, 0) − N (·)H3 + T(·, 0) − 1H3) exp (−ηt) for some positive constants c0 and η. Remark 3 In our theorem, the doping profile b(x) can be roughly large.

13

◮ A-priori estimates

The 1-D compressible Energy Transport Model (7),

      

nt + (nu)x = 0, (nT)t + (γnTu)x = (γ − 1)nuE − n(T − 1), Ex = n − b(x), with nu = nE − (nT)x. Setting n = N + Nρ, T = 1 + θ, E = E + ϕ, E = Vx, where the stationary solution is (N, 1, E). Introduce the entropy function s as follows θ = (1 + ρ)γ−1(1 + s) − 1.

14

Then the model (7) can be rewritten as

                    

ρt + [(1 + ρ)u]x = −E(1 + ρ)u, st + usx = −(1 + ρ)γ−1(1 + s) − 1 (1 + ρ)γ−1 + γ − 1 (1 + ρ)γ−1|u|2 + (γ − 1)E(1 + s)u, ϕx = Nρ, (8) u = ϕ − 1 1 + ρ[(1 + ρ)γ(1 + s)]x − E[(1 + ρ)γ−1(1 + s) − 1]. The corresponding initial data and the boundary conditions, ρ(x, 0) = N(x)−1nI(x) − 1, s(x, 0) = N(x)γ−1TI(x)nI(x)−(γ−1) − 1, (9) ρx|x=0,1 = 0, sx|x=0,1 = 0, ϕ|x=0,1 = 0. (10)

15

For the solutions of the above problem, we are able to get the following a-priori estimates. A-priori estimates: There exist δ1 > 0, such that, for any T1 > 0, if the solution (ρ, s, ϕ) exists on Ω × [0, T1] satisfying A(t) = sup

0≤t≤T1

(ρ, s)(·, t)2

H3 ≤ δ2 1,

then d dt

• 3
• i=0

(|∂i

xρ|2 + |∂i xs|2 + |∂i xϕ|2)

+ d dt

• 1
• i=0

(|∂i

xρt|2 + |∂i xst|2 + |∂i xϕt|2)

+C

• 3
• i=0

(|∂i

xρ|2 + |∂i xs|2 + |∂i xϕ|2)

+C

• 1
• i=0

(|∂i

xut|2 + |∂i xρt|2 + |∂i xst|2 + |∂i xϕt|2) ≤ 0

for any t ∈ [0, T1], and some generic positive constants C.

16

Lemma 1 Denote H0(t) =

• N

γ − 1[(1 + ρ)γ − 1 − γρ](1 + s) + Nρs + N 2(γ − 1)(1 + ρ)s2 + 1 2(1 + λ0 N )ϕ2, G0(t) =

• N{u2 + (γ − 1 + γλ0)ρ2 + (2 + λ0)ρs +

1 γ − 1s2} +λ0ϕ2 − λ0Eϕ[(γ − 1)ρ + s], where λ0 is a positive constant to be determined later. We

• btain the following estimates for small δ1,

d dtH0(t) + G0(t) ≤ 0. From (17), it holds that ϕt + N(1 + ρ)u = 0.

17

Lemma 2 Denote Hi(t) =

γ

2N(1 + s)(1 + ρ)γ−2|∂i

xρ|2 + N(1 + ρ)γ−1∂i xρ∂i xs

+ N(1 + ρ)γ 2(γ − 1)(1 + s)|∂i

xs|2 + 1

2(1 + λi N )|∂i

xϕ|2,

Gi(t) =

• |∂i

xu|2 + (γ − 1 + γλi)|∂i xρ|2 + (2 + λi)∂i xρ∂i xs

+ 1 γ − 1|∂i

xs|2 + λi|ϕ|2,

i = 1, 2, where λi are positive constants to be determined. We have the following estimates for small δ1, d dtHi(t) + Gi(t) ≤ diECi

• |∂i−1

x

U||∂i

xU|,

where di are positive constants and U = (ρ, s, u, E).

18

It can be shown that there exist λ0 and λi, satisfying 0 < λ0 < 4 γ − 1 inf

• N

N + E2

• ,

0 < λi < 4 γ − 1, i = 1, 2, such that (G0 + G1 + G2)(t) ≥ c1

• 2
• i=0
• |∂i

xu|2 + |∂i xρ|2 + |∂i xs|2 + |∂i xϕ|2

• .

Also it is easy to show that c2(ρ, s, ϕ)2

H2 ≤ (H0 + H1 + H2)(t) ≤ c3(ρ, s, ϕ)2 H2,

where ci (i = 1, 2, 3) are positive constants. Then we obtain d dt

• 2
• i=0

(|∂i

xρ|2 + |∂i xs|2 + |∂i xϕ|2)

+C

• 2
• i=0

(|∂i

xu|2 + |∂i xρ|2 + |∂i xs|2 + |∂i xϕ|2) ≤ 0.

(11)

19

Since we could’t directly deal with the estimates of ∂3

xρ, we use

the following energy density functionals to obtain the deriva- tives estimates with respect to the time variable. Ht

i(t) =

γ

2N(1 + s)(1 + ρ)γ−2|∂i

xρt|2 + N(1 + ρ)γ−1∂i xρt∂i xst

+ N(1 + ρ)γ 2(γ − 1)(1 + s)|∂i

xst|2 + 1

2(1 + Λi N )|∂i

xϕt|2,

i = 0, 1. We have d dt

• 1
• i=0

(|∂i

xρt|2 + |∂i xst|2 + |∂i xϕt|2)

+C

• 1
• i=0

(|∂i

xut|2 + |∂i xρt|2 + |∂i xst|2 + |∂i xϕt|2) ≤ 0.

(12) Is OK? Not enough. ∂xρt ∼ ∂2

xu

∼ ∂3

xρ + ∂3 xs.

20

Noting the Poisson equation and the damping term in (17)2, st+usx = −(1 + ρ)γ−1(1 + s) − 1 (1 + ρ)γ−1 + γ − 1 (1 + ρ)γ−1|u|2+(γ−1)E(1+s)u. Lemma 3 Denote H3(t) =

• λ3

2N |∂3

xϕ|2 + N

2 |∂3

xs|2,

G3(t) =

• N[γλ3|∂3

xρ|2 + (λ3 + γ − 1)∂3 xρ∂3 xs + |∂3 xs|2] + λ3|∂3 xϕ|2,

where λ3 is a positive constant to be determined. We have the following estimates for small δ1, d dtH3(t) + G3(t) ≤ d4EC2

• |Uxx + Uxt||∂3

xρ + ∂3 xs + ∂3 xϕ|,

where d4 is a positive constant.

21

With help of the estimates (11), (12) and Lemma 3, we have d dt

• 2
• i=0

(|∂i

xρ|2 + |∂i xs|2 + |∂i xϕ|2) + d

dt

• (|∂3

xs|2 + |∂3 xϕ|2)

+ d dt

• 1
• i=0

(|∂i

xρt|2 + |∂i xst|2 + |∂i xϕt|2)

+c8

• 3
• i=0

(|∂i

xρ|2 + |∂i xs|2 + |∂i xϕ|2) + c8

• 2
• i=0

|∂i

xu|2

+c8

• 1
• i=0

(|∂i

xut|2 + |∂i xρt|2 + |∂i xst|2 + |∂i xϕt|2) ≤ 0,

where c8 is a positive constant. With the help of (17)1 and the flux, it is easy to get

• |∂3

xρ|2 ≤ O(1)

• (|ρxt|2+|∂3

xs|2+|∂2 xρ|2+|∂2 xs|2+|ρx|2+|sx|2).

22

♠ For the multi-dimensional case, the model is

      

nt + ∇ · (nu) = 0, (nT)t + γ∇ · (nTu) = (γ − 1)nu · ∇V − n(T − 1), ∆V = n − b(x), (13) with the flux nu is nu = n∇V − ∇(nT). The initial values n(x, 0) = nI(x), T(x, 0) = TI(x), x ∈ Ω, (14) and the Neumann boundary conditions ∇n · γ|∂Ω = 0, ∇T · γ|∂Ω = 0, ∇V · γ|∂Ω = 0, (15) where γ is the unit outer normal vector to ∂Ω.

23

Theorem 4 Suppose 0 < C ≤ b(x) ≤ C, b(x) ∈ C3(Ω) and (nI(x), TI(x)) ∈ H4(Ω). There exists a positive constant δ1 such that if C − C < c∗C

3 2,

(16) where c∗ is a positive constant, and nI(x) − N (x)H4 + TI(x) − 1H4 ≤ δ1, then the problem (13)-(15) has a unique solution (n, T, E) inΩ × [0, ∞) satisfying n(·, t) − N (·)H4 + T(·, t) − 1H4 + E(·, t) − ∇V (·)H4 ≤ c0(n(·, 0) − N(·)H4 + T(·, 0) − 1H4) exp (−ηt) for some positive constants c0 and η.

24

The system for ρ, s, E

                    

ρt + ∇ · [(1 + ρ)u] = −(1 + ρ)E · u, st + u · ∇s = −(1 + ρ)γ−1(1 + s) − 1 (1 + ρ)γ−1 + γ − 1 (1 + ρ)γ−1|u|2 + (γ − 1)(1 + s)E · u, divE = Nρ, (17)

u = E −

1 1 + ρ∇[(1 + ρ)γ(1 + s)] − E[(1 + ρ)γ−1(1 + s) − 1]. Different: For 1-dimensional case, we have the following equation, Et + N(1 + ρ)u = 0. For multi-dimensional case, we just have,

Et + ∇∆−1∇ · [N(1 + ρ)u)] = 0.

25

For the zero order estimate, we have

• H0(t) =
• N

γ − 1[(1 + ρ)γ − 1 − γρ](1 + s) + Nρs +

N 2(γ−1)(1 + ρ)s2 + 1 2(1 + λ0)|E|2,

• G0(t) =
• N[|u|2 + (γ − 1 + γλ0N)ρ2 + (2 + λ0N)ρs

+ 1

γ−1s2 + λ0|E|2 + λ0E · Eρ],

We obtain the following estimates for small δ1, d dt

• H0(t) +

G0(t) ≤ 0. It can be shown that there exist λ0,

• G0(t) ≥ c1
• (|u|2 + ρ2 + s2 + |E|2),

0 < λ0 < 1 γ − 1 inf

4N − E2

N 2

• ,

Need (16).

26

THANK YOU !!

27

◮ The Local Existence

Introduce the entropy function s as T = nγ−1s. So (7) can be changed to the following system,

    

nt + (nu)x = 0, st + usx = n1−γ(γ − 1)u2 − s + n1−γ, Ex = n − b(x), with u = E − nγ−2(nsx + γsnx). Define a mapping F : (n, s) ∈ M → (v, w), M :=

• sup

0≤t≤t1

((n, s)2

H3 + (nt, st)2 H1) +

t1

0 nt2 H2dt ≤ M,

n, s ≥ D

• ,

where M, D are positive constants.

28

Then we solve the following problem to get v and w,

                          

vt − (γnγ−1svx + nγwx − Ev)x = 0, wt − Et n wx = n−(γ+1)[(γ − 1)E2

t + n2] − s,

Ex = n − b(x), vx(x, t)|x=0,1 = wx(x, t)|x=0,1 = E(0, t) = 0. v(x, 0) = nI(x), w(x, 0) = sI(x) = n1−γ

I

(x)TI(x). We could show that the mapping F exists exactly one fixed point (n, s) with (n, s) = F(n, s) in the corresponding M.

29

Chen Model

The diffusion matrix and energy relaxation term satisfying L = (Lij) = µ0ρ

    

1 3 2T 3 2T 15 4 T 2

     ,

W(µ, T) = 3 2ρT0 − T τ . So the Chen Model can be rewritten as:

                      

ρt + divj1 = 0, (ρT)t + divj2 = 2 3E · j1 + ρ(T0 − T) τ , j1 = ρ T E − ∇ρ, j2 = ρE − T∇ρ − ρ∇T, λ2∆V = ρ − C(x), E = ∇V. (18)

30

Lyumkis Model

The diffusion matrix and energy relaxation term satisfying L = (Lij) = 2µ0 √π ρT

1 2

  1

2T 2T 6T 2

  ,

W(ρ, T) = 2 √π ρ(T0 − T) τT

1 2

. So the Lyumkis Model can be rewritten as:

                                  

ρt + divj1 = 0, 3 2(ρT)t + divj2 = E · j1 + 2 √π n(T0 − T) τT

1 2

, j1 = 2µ0 √π

ρ

T

1 2

E − ∇(ρT

1 2)

• ,

j2 = 4µ0 √π

• ρT

1 2E − ∇(ρT 3 2)

• ,

λ2∆V = ρ − C(x), E = ∇V. (19)

31