3.3 Local Existence and Uniqueness
3.3.2 Proof of Theorem 3.3.1
Local Existence
To prove local existence in Theorem 3.3.1 we will use the iteration method and compactness arguments. The main task is to construct a sequence of ap-proximate solutions which is uniformly bounded in a certain Sobolev space in a fixed time interval. Compactness arguments then imply that there exists a limit which proves to be a local-in-time solution of (3.0.1). The first step is to linearize the system (3.0.1) around its initial state (n0, J0, V0), where V0 solves the Dirichlet problem
( λ24V0 =n0(x)− C(x), V0(x)|∂Ω =VΓ.
(3.3.1) We study the equations for the perturbation P := (P0, Pd) := (n−n0, J− J0), and first construct approximate solutions Pk := (Pk0, Pkd) for Pk0 := nk− n0, Pkd:=Jk−J0 (k≥1) from a fixed-point procedure, which are expected to converge to a solutionP of the perturbed problem ask→ ∞. For this, we shall derive uniform bounds in certain Sobolev spaces on a uniform time interval and apply standard compactness arguments. A further analysis will show that (n, J) = (P0+n0, Pd+J0) withn >0 is the expected local (in time) solution of
the original problem (3.0.1)(V is via (3.3.1) uniquely determinate). For given (Pk−1, Vk−1)2 we obtain the following linearized problems forPk (k≥1)
∂tPk+A(∂x)Pk =Fk−1, Pk(0, x) = 0,
Pk(t, x) = 0, on∂Ω for a.e. 0≤t≤T,
(3.3.2)
where
Fk−1:=
0 S(Pk−1)
!
−A(∂x) n0
J0
!
(k≥2),
S(Pk−1) := div (Pk−1d +J0)⊗(Pk−1d +J0) Pk−10 +n0
!
−(Pk−10 +n0)∇Vk−1 +2div
∇q
Pk−10 +n0
⊗
∇q
Pk−10 +n0
(k≥2), λ24Vk−1 =Pk−10 +n0− C(x), Vk−1(t, x)|∂Ω =VΓ(x).
In the next lemma we will show that fork≥2 and any time interval [0, T], ifPk−1 satisfies (3.2.101) then Fk−1 satisfies (3.2.100).
Lemma 3.3.1. Let T >0, k≥1, ifPk−1= (Pk−10 , Pk−1d ) satisfies
Pk−10 ∈L∞(0, T;H3(Ω))∩C([0, T], C1(Ω))∩C([0, T], H2(Ω)), Pk−1d ∈L∞(0, T;H2(Ω))∩C([0, T], C(Ω))∩C([0, T], H01(Ω)),
∂tPk−10 ∈L∞(0, T;H01(Ω))∩C([0, T], L2(Ω))∩L2(0, T;H2(Ω)),
∂tPk−1d ∈L∞(0, T;L2(Ω))∩C([0, T], L2(Ω))∩L2(0, T;H01(Ω)),
(3.3.3)
and
Pk−1(0, x) = (Pk−10 , Pk−1d )(0, x) = 0, (3.3.4) then Fk−1 satisfies
Fk−10 ∈L∞(0, T;H1(Ω)), Fk−1d ∈L∞(0, T; (L2(Ω))d);
F˙k−10 ∈L∞(0, T;L2(Ω)), F˙k−1d ∈L∞(0, T; (H−1(Ω))d);
Fk−10 (0, x)∈H01(Ω), Fk−1d (0, x)∈L2(Ω).
(3.3.5)
2Actually it is sufficient to givePk−1as the known function since (3.3.1) holds, whereVk−1
can be considered as the unique solution of (3.3.1) for a givenPk−10 .
Proof. Since Fk−10 =ν04n0+ divJ0 and thereby ˙Fk−10 = 0, then from (3.0.4)
where Since Ω satisfies the cone condition under our assumptions then from the Sobolev embedding theorem the following embedding
H1(Ω),→L6(Ω),→L4(Ω) (3.3.9) holds. From (3.3.9) we estimate K1 and K3 as follows.
K1≤C
K3 ≤C
d
X
l=1 d
X
k=1
n−10
L∞(Ω)k∂xk∂xln0kH1(Ω)k∂xkn0kH1(Ω)
+ n−10
L∞(Ω)
∂x2kn0
H1(Ω)k∂xln0kH1(Ω)
+ n−20
L∞(Ω)k∂xkn0kH1(Ω)k∂xln0kH1(Ω)k∂xkn0kH1(Ω)
! , from which we obtain
K3≤C n−10
L∞(Ω)kn0kH3(Ω)kn0kH2(Ω)+ n−10
L∞(Ω)kn0kH3(Ω)kn0kH2(Ω)
+ n−20
L∞(Ω)kn0k3H1(Ω)
!
<∞.
To estimateK2 we go back to (3.3.1) then with (3.3.9) to find K2 ≤
d
X
l=1
kn0kH1(Ω)k∂xlV0kH1(Ω)
≤Ckn0kH1(Ω)kV0kH2(Ω)
≤Ckn0kH1(Ω)
kn0kL2(Ω)+kC(x)kL2(Ω)+kVΓkH3/2(Ω)
<∞.
(3.3.10)
(3.3.6), (3.3.7) and (3.3.8) together with the estimations ofK1,K2,K3and our assumptions thatn0 ∈H3(Ω), J0∈(H2(Ω))d yield that
Fk−1d (0, x)∈L2(Ω).
In order to proveFk−1d ∈L∞(0, T; (L2(Ω))d), it is sufficient to show that S(Pk−1)∈L∞(0, T; (L2(Ω))d).
To see this we use our assumptions onPk−1 then find for a.e. t∈[0, T] Z
Ω
div (Pk−1d +J0)⊗(Pk−1d +J0) Pk−10 +n0
!!2
dx
≤C sup
[0,T]×Ω
(Pk−1d +J0)4 inf
[0,T]×Ω
(Pk−10 +n0)4kPk−10 +n0k2H1(Ω)
+C sup
[0,T]×Ω
(Pk−1d +J0)2 inf
[0,T]×Ω
(Pk−10 +n0)2kPk−1d +J0k2H1(Ω)<∞,
from which it follows
div (Pk−1d +J0)⊗(Pk−1d +J0) Pk−10 +n0
!
∈L∞(0, T; (L2(Ω))d). (3.3.11) Concerning the second term ofS(Pk−1) we have
Z It remains to give a estimation of the third term of S(Pk−1). For this we find
Z
Notice that in this equation we use the notation of (2.1.1). More precisely,
from which we infer
By a similar calculation, which is equal to
2 Combining (3.3.14), (3.3.15) and (3.3.16) we conclude
F˙k−1d ∈L∞(0, T; (H−1(Ω))d).
Now select two positive numbersδ0, M >0 such that δ0< inf
x∈Ωn0 (3.3.17) max k∇n0kL∞(Ω),kn0kL∞(Ω),kJ0kL∞(Ω)
< δ0−1, (3.3.18) and
kn0kH3(Ω) < M, kJ0kH2(Ω) < M. (3.3.19) Next we shall use iteration method to obtain a sequence of approximate solu-tions.
Lemma 3.3.2. Let k≥0 and P0 = 0, there exist t0 >0, C0 >0, C∗ >0 and C0 >0 independent ofk such that in the interval [0, t0] Pk satisfies
Pk0∈L∞(0, t0;H3(Ω))∩C([0, t0], C1(Ω))∩C([0, t0], H2(Ω)), Pkd∈L∞(0, t0;H2(Ω))∩C([0, t0], C(Ω))∩C([0, t0], H01(Ω)),
∂tPk0∈L∞(0, t0;H01(Ω))∩C([0, t0], L2(Ω))∩L2(0, t0;H2(Ω)),
∂tPkd∈L∞(0, t0;L2(Ω))∩C([0, t0], L2(Ω))∩L2(0, t0;H01(Ω)),
(3.3.20)
and the following uniform bounds
kPk0kL∞(0,t0;H3(Ω))≤C0, kPkdkL∞(0,t0;H2(Ω))≤C0, kPk0kL∞(0,t0;H2(Ω))≤C∗, kPkdkL∞(0,t0;H10(Ω))≤C∗, kPk0kL∞(0,t0;H01(Ω))≤C0, kPkdkL∞(0,t0;L2(Ω))≤C0, kP˙k0kL∞(0,t0;H01(Ω))≤C0, kP˙kdkL∞(0,t0;L2(Ω))≤C0, kP˙k0kL2(0,t0;H2(Ω))≤C0, kP˙kdkL2(0,t0;H10(Ω))≤C0,
(3.3.21)
and
inf
[0,t0]inf
x∈Ω
(Pk0+n0)> δ0, sup
[0,t0]
max
k∇(Pk0+n0)kL∞(Ω),kPk0+n0kL∞(Ω), kPkd+J0kL∞(Ω)
< δ0−1,
(3.3.22)
hold.
Remark 3.3.1. Lemma 3.3.2 interprets a uniform bounds of Pk in given re-flexive spaces, that guarantees the existence of a convergent subsequence of Pk in certain weak senses.
Proof. Throughout the proof we will use C as a generic constant which may change from line to line and depend onδ0, Ω and all physical constants, but is independent of k.
Obviously, forP0 = 0,F0satisfies (3.2.100). Starting withP0= 0, by theorem 3.2.13 we obtain a solutionP1 of (3.3.2) on any time interval [0, T],T >0, i.e.,
Since obviously P0 = 0 satisfies (3.3.3) and (3.3.4) then F0 satisfies(3.3.5). By Theorem 3.2.13 P1 satisfies (3.2.101).Thus we can shrink the interval [0, T] to [0, t1] such that on this interval
Then we use mathematical induction to complete the proof. Let c∗ >0,T >0 assume Pk−1(k≥1) satisfies
From lemma 3.3.1Fk−1 satisfies (3.3.5). Then by theorem 3.2.13 we obtainPk
in [0, T) as the solution of (3.3.2) which satisfies the regularity mentioned in (3.2.101) withu replaced byPk. We expand (3.3.2) then obtain two equations forPk0 and Pkd for a.e. t∈[0, T]
where the following calculations
(Qk, Pk0) =−ν0(∇n0,∇Pk0) + (divJ0, Pk0), (3.3.30)
Taking theL2(Ω) scalar product of the first equation in (3.3.27) with−2 44Pk0,
since∂tPk0 ∈L∞(0, T;H01(Ω)) from theorem 3.2.13.
Summing up (3.3.28), (3.3.29) and (3.3.32) yields 1 where I1,I3 are defined as (2.1.1). The itemsIk can be estimated as follows.
First by H¨older’s inequality
|I1|= We use the assumptions (3.3.25) and (3.3.26) and Cauchy’s inequality then infer for a.e. t∈[0, T]
|I1| ≤d2δ−20 (c∗+M)k∇PkdkL2(Ω) ≤1k∇Pkdk2L2(Ω)+d4δ−40 (c∗+M)2
41 ,
(3.3.38)
here
Using the assumptions 3.3.25 and (3.3.26) again we obtain
|I3| ≤ 2
4d2δ−20 (c∗+M)k∇PkdkL2(Ω)
≤3k∇Pkdk2L2(Ω)+ 4
16d4δ0−4(c∗+M)2 43
.
(3.3.43)
In the following, we will prove uniform in k estimates of Pk, which will imply that the life span can not tend to zero for k going to infinity. Then we will have a uniform existence interval as well as uniform estimates and can prove the convergence of a subsequence Pki by compactness argument.
Recall (3.3.30), we have T0|(Qk, Pk0)|
≤T0ν0k∇n0kL2(Ω)k∇Pk0kL2(Ω)+T0kJ0kL2(Ω)k∇Pk0kL2(Ω)
≤T0M(ν0+ 1)k∇Pk0kL2(Ω)
≤4k∇Pk0k2L2(Ω)+ T02M2(ν0+ 1)2
44 .
(3.3.44)
Recall (3.3.34):
|(Rk, Pkd)| ≤ν0|(4J0, Pkd)|+1
τ|(J0, Pkd)|+T0|(∇n0, Pkd)|
+ 2
4|(∇4n0, Pkd)|+|I1|+|I2|+|I3|.
Using H¨older’s inequality and (3.3.19)
|(Rk, Pkd)| ≤ν0k4J0kL2(Ω)kPkdkL2(Ω)+1
τkJ0kL2(Ω)kPkdkL2(Ω)
+T0k∇n0kL2(Ω)kPkdkL2(Ω)+2
4k∇4n0kL2(Ω)kPkdkL2(Ω)
+|I1|+|I2|+|I3|
≤
ν0+ 1
τ +T0+ 2 4
MkPkdkL2(Ω)+|I1|+|I2|+|I3|.
Then from (3.3.38), (3.3.42), (3.3.43) and Cauchy’s inequality we conclude
|(Rk, Pkd)| ≤5kPkdk2L2(Ω)+
ν0+ 1
τ +T0+2 4
2
M2 45
+1k∇Pkdk2L2(Ω)+d4δ−40 (c∗+M)2 41
+2kPkdk2L2(Ω)+(c∗+M)2Cλ,Ω,c2 ∗,M,C(x),VΓ
42
+3k∇Pkdk2L2(Ω)+ 4
16d4δ−40 (c∗+M)2 43
.
(3.3.45)
It is easy to verify 2
4|(Qk,4Pk0)| ≤ 2
4ν0|(4n0,4Pk0)|+ 2
4|(divJ0,4Pk0)|
≤ 2
4ν0k4n0kL2(Ω)k4Pk0kL2(Ω)+2
4kdivJ0kL2(Ω)k4Pk0kL2(Ω).
(3.3.46)
By (3.3.19):
2
4|(Qk,4Pk0)| ≤ 2
4M(ν0+ 1)k4Pk0kL2(Ω)
≤6k4Pk0k2L2(Ω)+ 4
16M2(ν0+ 1)2 46
.
(3.3.47)
Combining (3.3.33), (3.3.44), (3.3.45) and (3.3.47) we obtain the following in-equality
1
2∂t T0kPk0k2+2
4k∇Pk0k2+kPkdk2
+ν0T0k∇Pk0k2 +2
4ν0k4Pk0k2+ν0k∇Pkdk2+1 τkPkdk2
≤4k∇Pk0k2+ (2+5)kPkdk2+ (1+3)k∇Pkdk2 +6k4Pk0k2+C1,···,6,ν0,T0,τ,,λ,VΓ,C(x),Ω,δ0,M,c∗,C14,
(3.3.48)
C1,···,6,ν0,T0,τ,,λ,VΓ,C(x),Ω,δ0,M,c∗ >0 depends only upon the constants that oc-cur in the subscript. Notice that C1,···,6,ν0,T0,τ,,λ,VΓ,C(x),Ω,δ0,M,c∗ > 0 doesn’t depend onk.
Selecti, i= 1,· · ·,6 sufficiently small such that 4< ν0T0, (2+5)< 1
τ, (1+3)< ν0, 6< 2 4ν0.
Then we obtain
∂t T0kPk0k2+2
4k∇Pk0k2+kPkdk2 +C1
T0kPk0k2+2
4k∇Pk0k2+k4Pk0k2+k∇Pkdk2+kPkdk2
≤ C2, (3.3.49) where C1 > 0 is independent of c∗, C2 depends on all physical known quanti-ties, the imbedding constants and c∗, δ0 and M. Via solving the differential inequality (3.3.49) we infer the following estimation.
T0kPk0k2L∞(0,T;L2(Ω))+2
4k∇Pk0k2L∞(0,T;L2(Ω))+kPkdk2L∞(0,T;L2(Ω))
≤ C2 Z T
0
eC1sds.
(3.3.50)
Integrating (3.3.49) from 0 toT yields
kPk0k2L2(0,T;H2(Ω))+kPkdk2L2(0,T;H1(Ω))≤ C3T, (3.3.51) C3>0 depends on all physical known quantities andc∗,δ0 and M.
Next we shall estimate ˙Pk0 and ˙Pkd. Since Pk−1 satisfies the assumptions (3.3.23) and (3.3.24), then from Lemma 3.3.1 Fk−1 satisfies (3.3.5), thus the time derivatives of (Pk0, Pkd) satisfy the same boundary conditions as (Pk0, Pkd) from Theorem 3.2.13. Differentiating formally (3.3.27) with respect to 0 ≤ t≤T, then we obtain for a.e. 0≤t≤T
∂tP˙k0−ν04P˙k0−div ˙Pkd = 0,
∂tP˙kd−ν04P˙kd+ 1
τP˙kd−T0∇P˙k0+2
4∇4P˙k0 =−S(P˙ k−1).
(3.3.52)
Notice that − Z
Ω
∂tP˙k04P˙k0dx = 1
2∂tk∇P˙k0k2 by Theorem 3.2.11. Then we deduce, by a similar computation as before(see (3.3.33)), that
1
2∂t T0kP˙k0k2+2
4k∇P˙k0k2+kP˙kdk2
+ν0T0k∇P˙k0k2 +2
4ν0k4P˙k0k2+ν0k∇P˙kdk2+ 1 τkP˙kdk2
=( ˙S(Pk−1),P˙kd).
(3.3.53)
The scalar product −( ˙S(Pk−1),P˙kd) has the representation
−( ˙S(Pk−1),P˙kd) =I10 +I20 +I30, (3.3.54)
where Then by Cauchy’s inequality
|Z1| ≤01
Furthermore
Combining (3.3.61), (3.3.62) and (3.3.63) we obtain
|I10| ≤(01+02+03) where the constant C0
1,02,03,Ω,δ0,c∗>0 depends only on01, 02, 03,Ω, δ0, c∗.
Recall (3.3.40), (3.3.41) and use the imbedding
H1(Ω),→L4(Ω) (3.3.65)
where let C14 denote the imbedding constant of (3.3.65), then under the as-sumptions (3.3.25) we find
|I20| ≤C142 kP˙kdkL2(Ω)kVk−1kH2(Ω)kP˙k−10 kH1(Ω)
+C142 kP˙kdkL2(Ω)kPk−10 +n0kH1(Ω)kV˙k−1kH2(Ω)
≤c∗C142 Cλ,Ω,c∗,M,C(x),VΓkP˙kdkL2(Ω)
+c∗(c∗+M)C142 CΩ,λkP˙kdkL2(Ω), which implies
|I20| ≤04kP˙kdk2L2(Ω)+(c∗)2C144 Cλ,Ω,c2 ∗,M,C(x),VΓ
404
+05kP˙kdk2L2(Ω)+(c∗(c∗+M))2C144 CΩ,λ2 405
≤(04+05)kP˙kdk2L2(Ω)+C0
4,05,λ,Ω,c∗,M,C(x),VΓ,C14.
(3.3.66)
Now we considerI30. By a similar computation asI10 we obtain
I30 =2 4
X
i,j=1,···,d
∂i(Pk−10 +n0)∂j(Pk−10 +n0) Pk−10 +n0
!0
, ∂xj
P˙kd
i
!
=2
4(Z1+Z2+Z3), where we define
Z1: = X
i,j=1,···,d
(∂xiPk−10 )0∂xj(Pk−10 +n0) Pk−10 +n0
, ∂xj P˙kd
i
!
, (3.3.67)
Z2: = X
i,j=1,···,d
∂xi(Pk−10 +n0)(∂xjPk−10 )0 Pk−10 +n0 , ∂xj
P˙kd
i
!
, (3.3.68)
Z3: =− X
i,j=1,···,d
(Pk−10 )0∂xi(Pk−10 +n0)∂xj(Pk−10 +n0) (Pk−10 +n0)2 , ∂xj
P˙kd
i
! . (3.3.69) Under the assumptions (3.3.25) and (3.3.26) the items Zl, l = 1,2,3 can be estimated as follows.
|Z1| ≤ X Then by Cauchy’s inequality
|Z1| ≤06 Using Cauchy’s inequality yields
|Z3| ≤08
Combining (3.3.70), (3.3.71) and (3.3.72) we obtain
|I30| ≤ 2
where the constant C0
6,07,08,,Ω,δ0,c∗ > 0 depends only on 06, 07, 08, ,Ω, δ0, c∗. Recall (3.3.53), (3.3.54), (3.3.64), (3.3.66) and (3.3.73) then
1 imbedding constants and all physical known quantities.
Select0l,(l= 1,· · · ,8) sufficiently small such that quanti-ties, the imbedding constants and c∗, δ0 and M. Via solving the differential inequality (3.3.75) we infer the following estimation.
T0kP˙k0k2L∞(0,T;L2(Ω))+2
and
hold. kS(Pk−1)(0)k can be estimated via a same procedure as (3.3.7)-(3.3.10) from which we obtain that kS(Pk−1)(0)k is bounded by a constant depending only upon δ0, M, C14, C16 where C1s, s = 4,6 denotes the imbedding constant of H1(Ω),→Ls(Ω) respectively. Thus
kP˙kd(0,·)k2 ≤Cν0,T0,τ,,δ0,M,C14,C16. (3.3.78) Together with (3.3.76) and (3.3.77) we obtain
T0kP˙k0k2L∞(0,T;L2(Ω))+2
Integrate (3.3.75) from 0 to T:
C10 T0kP˙k0k2L2(0,T;L2(Ω))+2
RecallTheorem3.2.9
Fix now a positive numberγ >0 and define
C0 := max
We recall (3.3.23)-(3.3.26) and conclude that ifPk−1 defined in [0, t∗) satisfies
then first from (3.3.85) and (3.3.86) together withLemma3.3.1 andTheorem 3.2.13 the solution Pk of (3.3.2) in [0, t∗) satisfies
second from (3.3.50), (3.3.80), (3.3.81), (3.3.83) and (3.3.84)Pk satisfies
which implies
kPk0kL∞(0,t∗;H2(Ω))≤C∗, kPkdkL∞(0,t∗;H01(Ω))≤C∗. (3.3.97) To get more estimates first from (3.3.2) we conclude that for a.e. t∈[0, t∗]Pk
solves
(A(∂x)Pk=Fk−1−P˙k, Pk(x) = 0, on ∂Ω,
(3.3.98) where Fk−1 −P˙k ∈ H01(Ω)×(L2(Ω))d because of the assumptions (3.3.85), (3.3.86) together with Lemma 3.3.1 and (3.3.89). Using Theorem 3.2.6 we have the following a priori estimate forPk:
kPk0kL∞(0,t∗;H3(Ω))+kPkdkL∞(0,t∗;(H2(Ω))d)
≤CA(∂x) kFk−10 kL∞(0,t∗;H01(Ω))+kFk−1d kL∞(0,t∗;L2(Ω))
+kP˙k0kL∞(0,t∗;H01(Ω))+kP˙kdkL∞(0,t∗;(L2(Ω))d)
,
(3.3.99)
the constantCA(∂x)>0 depends only on the coefficients of operatorA(∂x).
Since Fk−10 =ν04n0+ divJ0 then
kFk−10 kL∞(0,t∗;H01(Ω))≤ν0kn0kH3(Ω)+kJ0kH2(Ω) ≤M(ν0+ 1). (3.3.100) Moreover
Fk−1d =S(Pk−1) +ν04J0−1
τJ0+T0∇n0−2
4∇4n0, which can be estimated as
kFk−1d kL∞(0,t∗;L2(Ω))
≤ kS(Pk−1)kL∞(0,t∗;L2(Ω))+ν0kJ0kH2(Ω)+ 1
τkJ0kL2(Ω)
+T0kn0kH1(Ω)+2
4kn0kH3(Ω)
≤M
ν0+1
τ +T0+2 4
+kS(Pk−1)kL∞(0,t∗;L2(Ω)).
(3.3.101)
For a.e. 0≤t≤t∗, kS(Pk−1)kL2(Ω)≤
div (Pk−1d +J0)⊗(Pk−1d +J0) Pk−10 +n0
! L2(Ω)
+
(Pk−10 +n0)∇Vk−1 L2(Ω)
+
2div
∇q
Pk−10 +n0
⊗
∇q
Pk−10 +n0 L2(Ω)
=:M1+M2+M3.
• Estimates of M1. Combining (3.3.19), (3.3.87), (3.3.88) and (3.3.102) we decuce
M1 ≤2dδ−20 (C∗+M) +dδ−40 (C0+M). (3.3.103) From (3.3.87) and (3.3.41) we infer
M2 ≤C142 (C0+M)Cλ,Ω,C0,M,C(x),VΓ. (3.3.104)
• Estimates of M3.
we use a similar reasoning asM1 to find 4
Thus combining (3.3.103)-(3.3.105) we conclude that for a.e. t∈[0, t∗]
kS(Pk−1)kL2(Ω)≤Cλ,,Ω,C0,C∗,δ0,M,C(x),VΓ,C14. (3.3.106) Together with (3.3.91), (3.3.99), (3.3.100) and (3.3.101) we deduce the following estimation.
kPk0kL∞(0,t∗;H3(Ω))+kPkdkL∞(0,t∗;(H2(Ω))d)
≤Cν0,λ,,τ,Ω,C0,C∗,δ0,M,C(x),VΓ,C14 =:C0,
(3.3.107) C0 >0 depends upon all known physical quantities, the imbedding constant of H1(Ω),→L4(Ω),δ0, M and C0, C∗, but is independent of kand t∗.
Now we make a summary. Given a function Pk−1; three positive constants C0, C∗, C0 which are defined in (3.3.83), (3.3.96), (3.3.107) respectively and depend only upon all physical constants, the corresponding imbedding constants C14, C16, the initial functions n0, J0 and the boundary conditions (nΓ, JΓ, VΓ);
a time interval [0, t∗) which satisfies (3.3.84). IfPk−1 satisfies (3.3.85), (3.3.86), (3.3.87), (3.3.88), (3.3.102) and
then we obtainPk via solving (3.3.2), andPk satisfies
Pk0 ∈L∞(0, t∗;H3(Ω))∩C([0, t∗], C1(Ω))∩C([0, t∗], H2(Ω)), Pkd ∈L∞(0, t∗;H2(Ω))∩C([0, t∗], C(Ω))∩C([0, t∗], H01(Ω)),
∂tPk0 ∈L∞(0, t∗;H01(Ω))∩C([0, t∗], L2(Ω))∩L2(0, t∗;H2(Ω)),
∂tPkd ∈L∞(0, t∗;L2(Ω))∩C([0, t∗], L2(Ω))∩L2(0, t∗;H01(Ω)),
(3.3.109)
and
kPk0kL∞(0,t∗;H3(Ω))≤C0, kPkdkL∞(0,t∗;H2(Ω)) ≤C0, kPk0kL∞(0,t∗;H2(Ω))≤C∗, kPkdkL∞(0,t∗;H01(Ω)) ≤C∗, kPk0kL∞(0,t∗;H01(Ω))≤C0, kPkdkL∞(0,t∗;L2(Ω)) ≤C0, kP˙k0kL∞(0,t∗;H01(Ω))≤C0, kP˙kdkL∞(0,t∗;L2(Ω)) ≤C0, kP˙k0kL2(0,t∗;H2(Ω))≤C0, kP˙kdkL2(0,t∗;H01(Ω)) ≤C0.
(3.3.110)
It is pointed out that up to now we have not derived (3.3.22) yet. In order to let Pk satisfy (3.3.22) we can shrink the time interval [0, t∗) into [0, tk) such that
inf
[0,tk]inf
x∈Ω
(Pk0+n0)> δ0, sup
[0,tk]
max
k∇(Pk0+n0)kL∞(Ω),kPk0+n0kL∞(Ω), kPkd+J0kL∞(Ω)
< δ0−1.
(3.3.111)
This process can be realized because of (3.3.17), (3.3.18) and (3.3.109). Under the conditions (3.3.109), (3.3.110), (3.3.111) and via shrinking the time interval into [0, tk+1) the solutionPk+1satisfies the same estimates asPkbut in [0, tk+1).
More precisely, we have derived a sequence{Pk}∞k=1 in [0, tk) which satisfies the uniform bounds (3.3.110) and (3.3.111).
Next we are able to show that to guarantee (3.3.111) tk will not converge to zero for k→ ∞. For 0≤t1 ≤t2 ≤tk, recallnk =Pk0+n0, Jk =Pkd+J0, we deduce the H¨older estimates
knk(t1,·)−nk(t2,·)kH2(Ω)≤ Z t2
t1
kn0k(s,·)kH2(Ω)ds
≤ |t1−t2|1/2kP˙k0kL2(0,tk;H2(Ω)),
which allows us to estimate Pk in C1/2([0, tk], C(Ω)). Fix a number β with 0< β < 1
2 then by Sobolev’s embedding theorem
kPk(t1)−Pk(t2)kC(Ω)≤CβkPk(t1)−Pk(t2)kH2−β(Ω), (3.3.112)
here letCβ denote the imbedding constant depending only on Ω. Interpolation yields
kPk(t1)−Pk(t2)kH2−β(Ω)
≤CinterkPk(t1)−Pk(t2)kβ/2L2(Ω)kPk(t1)−Pk(t2)k(2−β)/2H2(Ω) .
(3.3.113)
Finally
kPk(t1)−Pk(t2)kL2(Ω)≤ Z t2
t1
k∂tPkkL2(Ω)dt
≤ |t1−t2|k∂tPkkL∞(0,tk;L2(Ω))
(3.3.114)
together with (3.3.112) and (3.3.113) yields kPk(t1)−Pk(t2)kC(Ω)
≤2(2−β)/2CβCinterk∂tPkkβ/2L∞(0,tk;L2(Ω))kPkk(2−β)/2L∞(0,tk;H2(Ω))|t1−t2|β/2. (3.3.115) Thus under (3.3.110)
knk(t1,·)−nk(t2,·)kC(Ω)
≤2(2−β)/2CβCinterkP˙k0kβ/2L∞(0,tk;L2(Ω))knkk(2−β)/2L∞(0,tk;H2(Ω))|t1−t2|β/2
≤2(2−β)/2CβCinterC0β/2(C∗+M)(2−β)/2|t1−t2|β/2.
(3.3.116)
kJk(t1,·)−Jk(t2,·)kC(Ω)
≤2(2−β)/2CβCinterkP˙kdkβ/2L∞(0,tk;L2(Ω))kJkk(2−β)/2L∞(0,tk;H2(Ω))|t1−t2|β/2
≤2(2−β)/2CβCinterC0β/2(C0+M)(2−β)/2|t1−t2|β/2.
(3.3.117)
By a similar reasoning
k∇nk(t1,·)− ∇nk(t2,·)kC(Ω)
≤2(2−β)/2CβCinterk∇P˙k0kβ/2L∞(0,tk;L2(Ω))knkk(2−β)/2L∞(0,tk;H3(Ω))|t1−t2|β/2
≤2(2−β)/2CβCinterC0β/2(C0+M)(2−β)/2|t1−t2|β/2.
(3.3.118)
Since the right-hand sides of (3.3.116), (3.3.117) and (3.3.118) are uniformly bounded with respect to k, there is a time interval [0, t0] with 0< t0 ≤t∗ such that for all k = 1,2, ..., (3.3.20), (3.3.21) and (3.3.22) hold. We have finished the proof of Lemma 3.3.2.
To obtain a local-in-time solution the convergence property of the total sequence {Pk}∞k=0 needs to be considered. For this we have
Lemma 3.3.3. There exits t∗∈(0, t0)such that {Pk}∞k=0 is a Cauchy sequence inL∞(0, t∗;H1(Ω)×(L2(Ω))d). Moreover there exists aP∗∈L∞(0, t∗;H1(Ω)×
(L2(Ω))d) such that Pk converges to P∗ in L∞(0, t∗;H1(Ω)×(L2(Ω))d).
Proof. Recall (3.3.2) the following system
∂t(Pk+1−Pk) +A(∂x)(Pk+1−Pk) =Fk−Fk−1, (Pk+1−Pk)(0, x) = 0,
(Pk+1−Pk)(t, x) = 0, on ∂Ω for a.e. 0≤t≤t0, holds in the time interval [0, t0]. By a similar calculations as (3.3.28)-(3.3.33):
1
2∂t T0kPk+10 −Pk0k2+2
4k∇(Pk+10 −Pk0)k2+kPk+1d −Pkdk2 +ν0T0k∇(Pk+10 −Pk0)k2+2
4ν0k4(Pk+10 −Pk0)k2 +ν0k∇(Pk+1d −Pkd)k2+1
τkPk+1d −Pkdk2
=T0(Fk0−Fk−10 , Pk+10 −Pk0) + (Fkd−Fk−1d , Pk+1d −Pkd)
−2
4(Fk0−Fk−10 ,4(Pk+10 −Pk0)).
By H¨older’s inequality 1
2∂t T0kPk+10 −Pk0k2+2
4k∇(Pk+10 −Pk0)k2+kPk+1d −Pkdk2 +ν0T0k∇(Pk+10 −Pk0)k2+2
4ν0k4(Pk+10 −Pk0)k2 +ν0k∇(Pk+1d −Pkd)k2+ 1
τkPk+1d −Pkdk2
≤T0kFk0−Fk−10 kL2(Ω)kPk+10 −Pk0kL2(Ω)+kFkd−Fk−1d kH−1(Ω)×
× kPk+1d −PkdkH1
0(Ω)+2
4kFk0−Fk−10 kL2(Ω)k4(Pk+10 −Pk0)kL2(Ω), then by Cauchy’s inequality
∂t T0kPk+10 −Pk0k2+2
4k∇(Pk+10 −Pk0)k2+kPk+1d −Pkdk2
≤C
kFk0−Fk−10 k2L2(Ω)+kFkd−Fk−1d k2H−1(Ω)
for someC >0 depending only on the coefficients of operatorA(∂x). Then we
To estimate L1 we selectg∈(H01(Ω))d withkgk(H1
Now considerL2. Set43 :=Vk−Vk−1, then
nk∇Vk−nk−1∇Vk−1=nk∇Vk−nk∇Vk−1+nk∇Vk−1−nk−1∇Vk−1
=nk∇43+∇Vk−142, which yields
L2≤ knk∇43kL2(Ω)+k∇Vk−142kL2(Ω)
≤ knkkL∞(Ω)k43kH1(Ω)+k∇Vk−1kL4(Ω)k42kL4(Ω)
≤δ−10 k43kH1(Ω)+C142 kVk−1kH2(Ω)k42kH1(Ω). Since43 solves
(λ2443=42, 43|∂Ω= 0 recall (3.3.41) we obtain
L2 ≤Ck42kH1(Ω), (3.3.124) C > 0 is independent of k, t0, but depends upon C0. The representation and estimates of L3 follows from a similar process as L1. Precisely, let ∂xlnk (∂xsnk,∂xlnk−1,∂xsnk−1) replace (Jk)l ((Jk)s,(Jk−1)l,(Jk−1)s) in (3.3.122) then from (3.3.123)
L3 ≤2δ0−4
k∇Pk0− ∇Pk−10 kL2(Ω)+kPk0−Pk−10 kL2(Ω)
≤4δ0−4kPk0−Pk−10 kH1(Ω).
(3.3.125) Combining (3.3.123), (3.3.124) and (3.3.125) we conclude
kS(Pk)−S(Pk−1)k2L∞(0,t0;H−1(Ω))
≤C
k41k2L2(Ω)+k42k2H1(Ω)
≤C
kPk0−Pk−10 k2L∞(0,t0;H1(Ω))+kPkd−Pk−1d k2L∞(0,t0;L2(Ω))
,
(3.3.126)
whereC >0 is independent ofk andt0. Recall (3.3.119), (3.3.120) and (3.3.121):
kPk+10 −Pk0k2L∞(0,t0;H1(Ω))+kPk+1d −Pkdk2L∞(0,t0;L2(Ω))
≤Ct0
kPk0−Pk−10 k2L∞(0,t0;H1(Ω))+kPkd−Pk−1d k2L∞(0,t0;L2(Ω))
. Define a Banach spaceB:=L∞(0, t0;H1(Ω))×L∞(0, t0; (L2(Ω))d) with norm
k(x1, x2)kB =q
kx1k2L∞(0,t0;H1(Ω))+kx2k2
L∞(0,t0;(L2(Ω))d,
then
kPk+1−PkkB ≤C(t0)1/2kPk−Pk−1kB.
We shrink [0, t0] into [0, t∗] such thatC(t∗)1/2 <1 sinceC is independent of the time interval. Consequently let k > l,
kPk−PlkB ≤ kP1−P0kB
k−1
X
j=l
(C(t∗)1/2)j.
Hence {Pk = (Pk0, Pkd)}∞k=0 is a Cauchy sequence in B, and therefore there exists a point P∗ = ((P∗)0,(P∗)d) ∈ L∞(0, t∗;H1(Ω))×L∞(0, t∗; (L2(Ω))d) withPk→P∗ in L∞(0, t∗;H1(Ω))×L∞(0, t∗; (L2(Ω))d).
Recall the uniform bounds of {Pk0}∞k=0 it follows
Lemma 3.3.4. The limit P∗ in Lemma 3.3.3 satisfies P∗∈A∗1∩A∗2∩A∗3
with
∂tP∗∈L2(0, t∗;H2(Ω))×L2(0, t∗; (H1(Ω))d) where
A∗1 : =C([0, t∗];H2(Ω))×C([0, t∗]; (H1(Ω))d), A∗2 : =L∞(0, t∗;H3(Ω))×L∞(0, t∗; (H2(Ω))d), A∗3 : =C([0, t∗];C1(Ω))×C([0, t∗]; (C(Ω))d).
Proof. Since the embeddings H3(Ω) ,→ H2(Ω), H2(Ω) ,→ H1(Ω) are com-pact, (Pk0, Pkd) (k = 0,1,· · ·) are bounded in L∞(0, t∗;H3(Ω)× (H2(Ω))d) and ( ˙Pk0,P˙kd) (k = 0,1,· · ·) are bounded in L2(0, t∗;H1(Ω)×(L2(Ω))d) from Lemma 3.3.2, then Aubin’s Lemma (Corollary 4 in [72]) yields {Pk0}∞k=0 and {Pkd}∞k=0 are relatively compact in C([0, t∗], H2(Ω)) and C([0, t∗], H1(Ω)) re-spectively, i.e., there are a subsequence
{(Pk0s, Pkds)}∞s=0⊆ {(Pk0, Pkd)}∞k=0 and a function
PS= (PS0, PSd)∈C([0, t∗];H2(Ω))×C([0, t∗]; (H1(Ω))d) (3.3.127) such that
(Pk0s, Pkds)→(PS0, PSd) in C([0, t∗];H2(Ω))×C([0, t∗]; (H1(Ω))d). (3.3.128)
Furthermore from the uniform bounds ofPk(see Lemma 3.3.2) there exists a subsequence{(Pk0
sm, Pkd sequence of simple functions{fj} such that
j→∞lim
where {Aj1,· · · , Ajk} is a finite collection of mutually disjoint subsets of (0, t∗) Recall (3.3.131) it follows
j→∞lim Combining (3.3.133) and (3.3.134) we obtain
ω
From a same reasoning ω
From (3.3.130) then it follows ω from a similar reasoning.
As derived above
Pk0sm −→PS0, inC([0, t∗];H2(Ω)) as m→ ∞, Pkdsm −→PSd, inC([0, t∗]; (H1(Ω))d) as m→ ∞,
we infer
Pk0sm *∗PS0, inL∞(0, t∗;H2(Ω)) asm→ ∞, Pkdsm *∗PSd, inL∞(0, t∗; (H1(Ω))d) as m→ ∞.
Since (H1(Ω))∗⊂(H2(Ω))∗⊂(H3(Ω))∗, then from (3.3.129) Pk0sm *∗ PM0 , inL∞(0, t∗;H2(Ω)) asm→ ∞, Pkdsm *∗ PMd, inL∞(0, t∗; (H1(Ω))d) as m→ ∞.
This implies
PS0=PM0 , PSd=PMd. (3.3.139) Fix 0< γ < 1
2 and anyt∈(0, t∗) we deduce from the interpolation inequality (C.0.1) that
kPk0
sm −PM0 kH3−γ(Ω)≤CinkPk0
sm −PM0 k(3−γ)/3H3(Ω) kPk0
sm −PM0 kγ/3L2(Ω), then by Lemma 3.3.2
kPk0
sm −PM0 kH3−γ(Ω)≤Cin2(3−γ)/3C0(3−γ)/3kPk0
sm −PM0 kγ/3L2(Ω). (3.3.140) Similarly
kPkdsm −PMdkH2−γ(Ω)≤Cin2(2−γ)/2C0(2−γ)/2kPkdsm −PMdkγ/2L2(Ω). (3.3.141) From (3.3.140), (3.3.141) and Sobolev embedding theorem we infer
(Pk0sm,∇Pk0
sm, Pkdsm)→(PM0 ,∇PM0 , PMd) in C([0, t∗]×Ω). (3.3.142) FromLemma 3.3.3 the subsequence {Pksm}∞m=1 converges toP∗ in
L∞(0, t∗;H1(Ω))×L∞(0, t∗; (L2(Ω))d).
This implies
Pksm *∗ P∗ inL∞(0, t∗;H1(Ω))×L∞(0, t∗; (L2(Ω))d).
From (3.3.129) and the embeddingsL2(Ω)⊂(H2(Ω))∗, (H1(Ω))∗ ⊂(H3(Ω))∗ we conclude
Pksm *∗ (PM0 , PMd) in L∞(0, t∗;H1(Ω))×L∞(0, t∗; (L2(Ω))d).
By the uniqueness of the limit
P∗ = (PM0 , PMd).
Furthermore using (3.3.127), (3.3.139), (3.3.129) and (3.3.142) we complete the proof.
Remark 3.3.2. This lemma ensures that the limit function (n, J) := P∗ + (n0, J0) satisfies the initial conditions (3.0.1) and the boundary conditions from (3.0.3).
It is also obvious that the sequence {Vk}∞m=0 converges to a limit V in the space C([0, t∗];H2(Ω)) which solves the Possion equation λ24V = n− C(x)
Then we find
Going back to (3.3.144) we find
C(ϕ) >0 is a constant depending upon ϕ. Substitute (3.3.145) into (3.3.146), UsingLemma 3.3.3 we conclude
By a similar calculation it is easy to verify that 2
Thus we conclude thatP∗ is the solution we seek.
Uniqueness
Let (n1, J1, V1) and (n2, J2, V2) be two solutions of (3.0.1)-(3.0.4). Put n4=n1−n2, J4 =J1−J2, V4 =V1−V2, then we obtain the system
Similarly as (3.3.28)-(3.3.33) we obtain 1
2∂t
T0kn4k2+2
4k∇n4k2+kJ4k2
+ν0T0k∇n4k2 +2
4ν0k4n4k2+ν0k∇J4k2+1 τkJ4k2
=(F1−F2, J4)≤ kF1−F2kH−1(Ω)kJ4kH1 0(Ω).
(3.3.147)
We go back to the proof ofLemma3.3.3. By a similar calculations as (3.3.121)-(3.3.126) we deduce
kF1−F2k2H−1(Ω) ≤C(kn4k2H1(Ω)+kJ4k2L2(Ω)). (3.3.148) (3.3.147) and (3.3.148) together with Cauchy’s inequality yield
1 2∂t
T0kn4k2+2
4k∇n4k2+kJ4k2
≤C(kn4k2H1(Ω)+kJ4k2L2(Ω)).
(3.3.149)
Applying Gronwall’s lemma on (3.3.149) then n4≡J4 ≡0.
Global Existence and
Exponential Decay on a Torus
We assume that the doping profile of background charges is constant, denoted by C0. Then it is easy to verify that (C0,0,0) is a steady state of (3.0.1) on a torus Tdor on a bounded domain Ω under insulating boundary conditions. In order to extend the local solution of (3.0.1)-(3.0.3) globally in time we need to establish uniform bounds. Here we only consider the situation when the initial data is close to the steady state (C0,0,0).
4.1 Reformulation
We consider the viscous QHD model (3.0.1) on ad-dimensional torus Td. The local existence of solutions to the viscous model of quantum hydrodynamics on a torus is proved in [17]. We will study the situation when the initial data are assumed in a small neighborhood of the steady state (C0,0,0). The idea of extending the local classical solution globally in time is first to establish uniform estimates, and then to use the usual continuity argument.
Suppose
x∈Ωinf n0(x)>0, (4.1.1) Z
Ω
(n0− C0)dx= 0. (4.1.2)
The second condition (4.1.2) is necessary, otherwise the Poisson equation for V would not be solvable from Green’s formulas. Note that (4.1.1) and (4.1.2) imply C0 >0.
In the following, we use the abbreviation
u=n− C0, u0 =n0− C0. (4.1.3) Now we reformulate the original problem (3.0.1) into an equivalent one with respect to the classical solution (u, J, V) which describes the perturbation to
105
the steady state (C0,0,0).
Differentiate (4.1.4) with respect to t, take divergence of (4.1.5), add the resul-tants, then substitute (4.1.8) (4.1.9) into it, we obtain a nonlinear fourth-order wave equation for u
utt+ 1
Introduce the perturbations of the steady-state (C0,0,0) u=n− C0, J =J−0, V =V −0.
Then, using (4.1.4)-(4.1.6) and (4.1.10), the evolution equations for (u, J, V) read as follows
where
H:=G+T0∇u−(u+C0)∇V −2 4∇4u.
The initial data are given:
u(0, x) =u0(x), ut(0, x) =u1(x), J(0, x) =J0(x) (4.1.12) with
u1(x) := divJ0+ν04n0.