The existence of solution for viscous Camassa–Holm equations on bounded domain in five dimensions

J. Math. Anal. Appl. 429 (2015) 849–872

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Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa

The existence of solution for viscous Camassa–Holm equations on bounded domain in five dimensions Yongjiang Yu a,b,1 a b

Department of Mathematics, Shanghai Jiaotong University, Shanghai 200240, China Department of Mathematics, University of Iowa, Iowa City 52242, USA

a r t i c l e

i n f o

Article history: Received 30 October 2014 Available online 15 April 2015 Submitted by J. Shi Keywords: Viscous Camassa–Holm equations Weak solution Strong solution

a b s t r a c t The existence of global weak solution and local existence of strong solution for five-dimensional viscous Camassa–Holm equations on bounded domain are proved in this note. The global existence of strong solution is also proved when small initial data is given. © 2015 Elsevier Inc. All rights reserved.

1. Background and the main result Assume L > 0. The five-dimensional viscous Camassa–Holm (abbreviated as VCH) equations on T = [0, L]5 considered in this note are ⎧ m ∂(u−α2 Δu) ⎪ − νΔ(u − α2 Δu) + (u · ∇)(u − α2 Δu) + j=1 (u − α2 Δu)j ∇uj + ∇ Pρ = f, ⎪ ∂t ⎪ ⎨ ∇ · u = 0, (1.1) ⎪ (x), u(x, 0) = u ⎪ 0 ⎪ ⎩ u is periodic on T , where u(x, t) = (u1 (x, t), · · · , u5 (x, t)) is the velocity of the fluid at point x = (x1 , · · · , x5 ) at time t, P π 2 2 ρ = ρ + |u| − α (u · Δu) is the modified pressure, while π is the pressure, ν > 0 is the constant viscosity and ρ is a constant density, α > 0 is scale parameter, at the limit α = 0 one obtains the Navier–Stokes equations, the function f is a given body forcing. The viscous Camassa–Holm equations first emerge in [7]. They average the motion of small scales of the Navier–Stokes equations. α > 0 is a scale at which the fluid motion is averaged. Specifically, for any fixed α, VCH equations are able to capture accurately the motion of the fluid at scales larger than α while averaging

1

E-mail address: [email protected]. Supported by Tianyuan Mathematical Foundation No. 10726055, NSFC No. 11171213 and NSFC No. BC0710057 in China.

http://dx.doi.org/10.1016/j.jmaa.2015.04.038 0022-247X/© 2015 Elsevier Inc. All rights reserved.

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Y. Yu / J. Math. Anal. Appl. 429 (2015) 849–872

out the motion of the fluid at scales smaller than α. The authors in [6] find the relationship between the Navier–Stokes equations (Eulerian formulation) and the VCH equations (Lagrangian formulation in some sense). Specifically, the unique weak solution of three-dimensional VCH equations converges to the weak solution of the three-dimensional Navier–Stokes equations when the scale converges to zero. The global existence of solution u ∈ L2 ([0, T ], H 3 ) ∩ L∞ ([0, T ]; H 2 ) ∩ W 1,∞ ([0, T ]; L2 ) has been proved for the equations on 2D periodic box in [10] with the assumption that f ∈ L2 , u0 ∈ H 2 . Assume 3 f ∈ L2 , u0 ∈ H 1 , the unique existence of regular solution u ∈ L∞ loc ((0, T ], H ) for the equations on 3D periodic box has been proved in [6]. Decay of solution and the unique existence of global weak solution  u ∈ L∞ ([0, T ], H 2 ) L2 ([0, T ], H 3 ) for the equations on bounded domain or in the whole space Rm has been obtained in [3] for m = 2, 3, 4 under the conditions that f = 0 and u0 ∈ H 2 . Assume f ∈ L2 ([0, T ]; H) u0 ∈ H 1 , the unique existence of weak solution for the equations on mD (m = 2, 3, 4) periodic box has been proved in [20]. Assume f = 0, u0 ∈ H 2+s (Rm ) with s > m 2 − 1, the regularity criteria for the equations has been considered in [21] for m = 2, 3, 4. Assume f = 0, u0 ∈ H s with s > m 2 + 1 and m = 2, 3, the uniqueness and smoothness of the global solution for VCH equations on m-dimensional Riemannian manifold with certain boundary conditions (Dirichlet, Neumann, and Mixed type boundary conditions) are proved in [15]. The higher dimensional cases are also considered in [15]. One can find in [2,8,9,16,18,19] for more research works about VCH equations. Note that VCH equations are formally similar to the Lagrangian averaged Navier–Stokes equations. The research history of Lagrangian averaged Navier–Stokes equations can be found in [1,5,11–14]. Above all, whether there exists a solution for the five-dimensional VCH equations is still an open problem. In this note the VCH equations on 5D periodic box are considered. The existence of global weak solution and unique existence of local strong solution for the equations are proved. The strong solution is proved globally when small initial data is given. The techniques of the proof in [3,6,10,21] are energy estimate and Sobolev imbedding. While additional interpolations between different Sobolev imbedding are used in [20] and this note. The proof idea for m = 2, 3, 4 case and m = 5 case is almost similar. Because of the higher dimension, some imbedding and interpolations used for m = 2, 3, 4 case cannot be used for m = 5 case. Also because of this, the uniqueness of weak solution for m = 5 case cannot be proved in this paper, though it is true for m = 2, 3, 4 case [20]. One needs some notations to state the result.    Assume T f dx = T u0 dx = 0 for simplicity. T udx = 0 will be obtained after integration by parts  ∞ ∞ from (1.1). Let V = {Φ ∈ Cper (T )5 ; ∇ · Φ = 0, T Φdx = 0}, where Cper (T )5 denotes the space of all ∞ T -periodic, C vector fields defined on T . H and V stand for the closure of V in L2 (T )5 and H 1 (T )5 respectively. (·, ·) and | · | will be used to denote the scalar product and norm in H. The scalar product and norm in V is denoted by ((·, ·)) and  · . Let A = P (−Δ) is the abstract Stokes operator with  domain D(A) = H 2 (T )5 V , where P is the Leray projector. Under space periodic boundary conditions A = −Δ|D(A) is a self-adjoint positive operator which is an isomorphism from V to V  (the dual of V ). Hence 1 1 2 2 A has eigenvalues {λj }∞ j=1 such that 0 < λ1 ≤ λ2 ≤ · · · ≤ λj → +∞. Moreover V = D(A ), || · || = |A · |, 1 1 ((·, ·)) = (A 2 ·, A 2 ·). By virtue of Poincaré inequality one can show that there is a constant c such that c−1 |Aw| ≤ ||w||H 2 ≤ c|Aw|,

∀ w ∈ D(A)

and c−1 |A 2 w| ≤ ||w||H 1 ≤ c|A 2 w|, 1

1

∀ w ∈ V.

Setting

B(φ, ψ) = P (φ · ∇)ψ ,

B(ψ)φ = B(φ, ψ),

where B(ψ) is a linear operator acting on φ for every fixed ψ.

∀ φ, ψ ∈ V,

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Moreover < B(φ, ψ), w >V  = − < B(φ, w), ψ >V  ,

∀ φ, ψ, w ∈ V.

(1.2)

One denotes m

 ˜ ψj ∇φj , ∀ φ, ψ ∈ V. B(φ, ψ) = P (φ · ∇)ψ + j=1

Then ˜ < B(φ, ψ), w >V  = < B(φ, ψ), w >V  − < B(w, ψ), φ >V  = < B(ψ)φ − B ∗ (ψ)φ, w >V  ,

∀ φ, ψ, w ∈ V,

where B ∗ (ψ) is the adjoint operator of B(ψ). As a result it reaches ˜ ψ) = B(ψ)φ − B ∗ (ψ)φ, B(φ,

∀ φ, ψ ∈ V.

Also one sets v = u + α2 Au, v0 = u0 + α2 Au0 .

(1.3)

The viscous Camassa–Holm equations are just written as operator equations ⎧ ⎨ ∂(u + α2 Au) ˜ u + α2 Au) = f, + νA(u + α2 Au) + B(u, ∂t ⎩ u(x, 0) = u0 (x),

(1.4)

or ⎧ ⎨ ∂v ˜ v) = f, + νAv + B(u, ∂t ⎩ u(x, 0) = u , v(x, 0) = v . 0 0

(1.5)

One starts with two definitions. Let T > 0. Definition 1.1. Let f ∈ L∞ ([0, T ]; V  ). A function u ∈ L∞ ([0, T ]; V ) ∩ L2 ([0, T ]; D(A)) ∩ C([0, T ]; V ) is said to be a weak solution to (1.4) on [0, T ] if it satisfies





d ˜ u + α2 Au), w (u + α2 Au), w + ν A(u + α2 Au), w + B(u, =< f, w >V  dt D(A) D(A) D(A) for every w ∈ D(A) and almost every t ∈ [0, T ]. Moreover u(0) = u0 in V . The above equality can be understood in the following sense: ∃ E ⊆ [0, T ] with L(E) = 0 and for t0 , t ∈ [0, T ] − E t (u(t) + α Au(t), w) − (u(t0 ) + α Au(t0 ), w) + ν 2

2

(u(s) + α2 Au(s), Aw)ds t0

t + t0

˜ B(u(s), u(s) + α2 Au(s)), w

t

D(A)

ds =

< f (s), w >V  ds. t0

(1.6)

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Definition 1.2. Let f ∈ L∞ ([0, T ]; V  ). A function u ∈ L∞ ([0, T ]; D(A)) ∩ L2 ([0, T ]; D(A 2 )) ∩ C([0, T ]; D(A)) is said to be a strong solution to (1.4) in [0, T ] if it satisfies 3





d ˜ u + α2 Au), w (u + α2 Au), w  + ν A(u + α2 Au), w  + B(u, =< f, w >V  dt V V V for every w ∈ V and almost every t ∈ [0, T ]. Moreover u(0) = u0 in D(A). The above equality can be understood in the following sense: ∃ E ⊆ [0, T ] with L(E) = 0 and for t0 , t ∈ [0, T ] − E t (u(t) + α Au(t), w) − (u(t0 ) + α Au(t0 ), w) + ν 2

2

1

1

(A 2 (u(s) + α2 Au(s)), A 2 w)ds t0

t +

˜ B(u(s), u(s) + α2 Au(s)), w

t

V

ds =

t0

< f (s), w >V  ds.

(1.7)

t0

Now the main result of this note is the following. Theorem 1.1. Assume u0 ∈ V , f ∈ L∞ ([0, T ]; V  ). Then there exists at least one global weak solution to (1.4) 4 3 on [0, T ] with du dt ∈ L ([0, T ]; H). Moreover, the energy inequality

1 |u(t)|2 + α2 ||u(t)||2 + ν 2

t

||u||2 + α2 |Au|2 ds

t0

1 ≤ |u(t0 )|2 + α2 ||u(t0 )||2 + 2

t < f (s), u(s) >V  ds

(1.8)

t0

holds for a.e. t0 , t ∈ [0, T ]. Theorem 1.2. Any two weak solutions belonging to L4 ([0, T ]; D(A)) must be equal almost everywhere. Theorem 1.3. Assume u0 ∈ D(A), f ∈ L∞ ([0, T ]; V  ). Then there exists a unique local strong solution    3  u ∈ L∞ [0, T ∗ ]; D(A) ∩ L2 [0, T ∗ ]; D(A 2 ) to (1.4) on [0, T ∗ ] with some constant T ∗ ≤ T .    3  Moreover, if f ∈ L∞ ([0, T ]; H), this solution also belongs to L∞ [τ, T ∗ ]; D(A 2 ) ∩ L2 [τ, T ∗ ]; D(A2 ) for any arbitrary but fixed τ .  3 3  Furthermore, if u0 ∈ D(A 2 ), f ∈ L∞ ([0, T ]; H), the unique local strong solution u ∈ L∞ [0, T ∗ ]; D(A 2 ) ∩ L2 ([0, T ∗ ]; D(A2 )). Theorem 1.4. Assume u0 ∈ D(A), f ∈ L∞ ([0, T ]; V  ). If there exists an absolute constant c such that 1

2T ν 2 λ12 α6 2 √ ||u0 || + α |Au0 | + ||f || . · ∞ ([0,T ];V  ) ≤ L 2 2 να2 2 c (λ−1 1 +α ) 2

2

2

(1.9)

   3 Then there exists a unique global strong solution u ∈ L∞ [0, T ]; D(A) ∩ L2 [0, T ]; D(A 2 ) to (1.4) on [0, T ], and the solution satisfies

sup t∈[0,T ]

 ||u(t)||2 + α2 |Au(t)|2  1

ν 2 λ12

T +

0

3

(|Au(s)|2 + α2 |A 2 u(s)|2 )ds 1

νλ12

2 α6 ≤ √ · −1 , c (λ1 + α2 )2

(1.10)

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moreover, if f ∈ L∞ ([0, T ]; H), the unique strong solution also belongs to    3  L∞ [τ, T ]; D(A 2 ) ∩ L2 [τ, T ]; D(A2 ) for any arbitrary but fixed τ,

(1.11)

and satisfies T   3 3 2 2 2 sup |Au(t)| + α |A 2 u(t)| + ν (|A 2 u(s)|2 + α2 |A2 u(s)|2 )ds t∈[τ,T ]

τ

4  5T ||f ||2 ∞ νλ α6 L ([0,T ];H) ≤ √ 1 −1 + 16νλ · . + 1 τ να2 c (λ1 + α2 )2 1 2

(1.12)

Furthermore, if u0 ∈ D(A 2 ), f ∈ L∞ ([0, T ]; H) satisfying (1.9), then there exists a unique global strong    3 solution u, not only belongs to L∞ [0, T ]; D(A) ∩ L2 [0, T ]; D(A 2 ) satisfying (1.10) but also belongs to   3  L∞ [0, T ]; D(A 2 ) ∩ L2 [0, T ]; D(A2 )). 3

2. Some preliminaries Several lemmas will be needed in this section. The first lemma is concern of the Stokes operator A defined in the space dimension five and its proof is similar to that in the case of space dimension three introduced in [17] and [4]. Lemma 2.1. (I) The operator A can be extended continuously to be defined on V = D(A 2 ) with values in V  = D(A− 2 ) such that 1

1

1

1

< Aφ, ψ >V  = (A 2 φ, A 2 ψ) for every φ, ψ ∈ V. (II) The operator A2 can be extended continuously to be defined on D(A) with values in D(A) (the dual space of the Hilbert space D(A)) such that < A2 φ, ψ >D(A) = (Aφ, Aψ) for every φ, ψ ∈ D(A). The second lemma can also be found in [17] and [4]. The proof of this lemma involves in Sobolev imbedding and interpolations. Lemma 2.2. Let T ⊆ Rm , m ≥ 2 be a bounded open piecewise C l -domain, l ≥ 1 and s1 , s2 , s3 are real numbers, 0 ≤ s1 ≤ l, 0 ≤ s2 ≤ l − 1, 0 ≤ s3 ≤ l. Assume that s1 + s2 + s3 ≥

m m m m with (s1 , s2 , s3 ) = (0, 0, ), (0, , 0), ( , 0, 0). 2 2 2 2

Then there exists a constant c depending on s1 , s2 , s3 , T , scale invariant, such that |(B(φ, ψ), w)| ≤ c|T |

s1 +s2 +s3 n

for all φ, ψ, w ∈ C ∞ (T )m .

− 12

1+[s ]−s1

||φ||[s1 ],T1

s −[s ]

1+[s ]−s

s −[s ]

1+[s ]−s3

1 2 2 2 ||φ||[s11 ]+1,T ||ψ||[s2 ]+1,T ||ψ||[s22 ]+2,T ||w||[s3 ],T3

s −[s ]

3 ||w||[s33 ]+1,T

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Since C ∞ (T )5 is dense in H, V and D(A) respectively, one takes (s1 , s2 , s3 ) = (2, 0, 12 ), ( 12 , 0, 2), (2, 12 , 0), (0, 12 , 2), (1, 12 , 1), ( 32 , 1, 0), ( 32 , 0, 1), (1, 0, 32 ), (2, 1, 0) to obtain 1

1

| < B(φ, ψ), w >V  | ≤ c|Aφ|||ψ|||w| 2 ||w|| 2 , ∀φ ∈ D(A), ψ, w ∈ V, 1 2

(2.1)

1 2

| < B(φ, ψ), w >D(A) | ≤ c|φ| ||φ|| ||ψ|||Aw|, ∀φ, ψ ∈ V, w ∈ D(A), 1

(2.2)

1

|(B(φ, ψ), w)| ≤ c|Aφ|||ψ|| 2 |Aψ| 2 |w|, ∀φ, ψ ∈ D(A), w ∈ H, 1 2

1 2

1 2

1 2

(2.3)

| < B(φ, ψ), w >D(A) | ≤ c|φ|||ψ|| |Aψ| |Aw|, ∀φ ∈ H, ψ, w ∈ D(A),

(2.4)

| < B(φ, ψ), w >V  | ≤ c||φ||||ψ|| |Aψ| ||w||, ∀φ, w ∈ V, ψ ∈ D(A), 1 2

(2.5)

1 2

|(B(φ, ψ), w)| ≤ c||φ|| |Aφ| |Aψ||w|, ∀φ, ψ ∈ D(A), w ∈ H, 1 2

(2.6)

1 2

| < B(φ, ψ), w >V  | ≤ c||φ|| |Aφ| ||ψ||||w||, ∀φ ∈ D(A), ψ, w ∈ V, 1 2

(2.7)

1 2

| < B(φ, ψ), w >D(A) | ≤ c||φ||||ψ||||w|| |Aw| , ∀φ, ψ ∈ V, w ∈ D(A),

(2.8)

|(B(φ, ψ), w)| ≤ c|Aφ||Aψ||w|, ∀φ, ψ ∈ D(A), w ∈ H.

(2.9)

Lemma 2.3. ˜ can be extended continuously from D(A) × H with values in V  , and it satisfies (III) The operator B 1

1

˜ | < B(φ, ψ), w >V  | ≤ c|Aφ|||ψ|||w| 2 ||w|| 2 , ∀φ ∈ D(A), ψ, w ∈ V.

(2.10)

Moreover, ˜ ψ), w >V  = − < B(w, ˜ < B(φ, ψ), φ >V  ,

˜ < B(φ, ψ), φ >V  = 0,

∀φ, ψ, w ∈ V.

(2.11)

˜ can be extended continuously from D(A) × D(A) with values in H, and it satisfies (IV) The operator B 1 1 ˜ | < B(φ, ψ), w >V  | ≤ c|Aφ|||ψ|| 2 |Aψ| 2 |w|,

∀φ, ψ ∈ D(A), w ∈ H.

(2.12)

˜ can be extended continuously from D(A) × V with values in V  , and it satisfies (V) The operator B 1 1 ˜ | < B(φ, ψ), w >V  | ≤ c||φ|| 2 |Aφ| 2 ||ψ||||w||,

∀φ, ∈ D(A), ψ, w ∈ V.

Proof of (III): By definition and (2.1), (2.2) ˜ | < B(φ, ψ), w >V  = | < B(φ, ψ), w >V  − < B(w, ψ), φ >V  | 1

1

1

1

1

1

≤ c|Aφ|||ψ|||w| 2 ||w|| 2 + c|w| 2 ||w|| 2 ||ψ|||Aφ| ≤ c|Aφ|||ψ|||w| 2 ||w|| 2 , ∀φ ∈ D(A), ψ, w ∈ V. ˜ ψ) and (1.7). This proves (2.10), and hence (III) holds. (2.11) is obvious by definition of B(φ, Proof of (IV): By definition and (2.3), (2.4) ˜ | < B(φ, ψ), w >V  | = | < B(φ, ψ), w >V  − < B(w, ψ), φ >V  | 1

1

1

1

1

1

≤ c|Aφ|||ψ|| 2 |Aψ| 2 |w| + c|w|||ψ|| 2 |Aψ| 2 |Aφ| ≤ c|Aφ|||ψ|| 2 |Aψ| 2 |w|, This proves (2.12), and hence (IV) holds.

∀φ, ψ ∈ D(A), w ∈ V.

(2.13)

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Proof of (V): By definition and (2.7), (2.8) ˜ ψ), w >V  | = | < B(φ, ψ), w >V  − < B(w, ψ), φ >V  | | < B(φ, 1

1

1

1

1

1

≤ c||φ|| 2 |Aφ| 2 ||ψ||||w|| + c||w||||ψ||||φ|| 2 |Aφ| 2 ≤ c||φ|| 2 |Aφ| 2 ||ψ||||w||,

∀φ ∈ D(A), ψ, w ∈ V.

This proves (2.13), and hence (V) holds. 3. Proof of Theorem 1.1 To prove the theorems of this note one needs some a priori estimates. Finite approximation of Galerkin procedure will be adopted to establish some a priori estimates by using lemmas in Section 2. Let u0 ∈ V . Assume the Stokes operator A has eigenfunctions wj , j = 1, 2, · · ·, which consist of an orthonormal basis of H. Let Hn = span{w1 , · · · , wn }. Pn denote the L2 -orthogonal projection from H onto Hn . The finite approximation of Galerkin procedure for (1.4) is ⎧ ∂(un + α2 Aun ) ⎪ ⎪ ˜ n , un + α2 Aun ) = Pn f, + νA(un + α2 Aun ) + B(u ⎨ ∂t un (x, 0) = Pn u0 . ⎪ ⎪ ⎩ vn = un + α2 Aun , vn0 = Pn (u0 + α2 Au0 ).

(3.1)

One takes the scalar product of the first equation in (3.1) with un in V  to obtain 1 d (|un |2 + α2 ||un ||2 ) + ν(||un ||2 + α2 |Aun |2 ) =< f, un >V  , 2 dt

(3.2)

˜ n , un + α2 Aun ), un >V  = 0 is used. Since where < B(u ν 1 ||un ||2 + ||f ||2V  2 2ν ν 1 ||f ||2V  . ≤ (||un ||2 + α2 |Aun |2 ) + 2 2ν

| < f, un >V  | ≤ ||f ||V  ||un ||V ≤

It follows that 1 d (|un |2 + α2 ||un ||2 ) + ν(||un ||2 + α2 |Aun |2 ) ≤ ||f ||2V  . dt ν

(3.3)

By Poincaré inequality 3

3

λ31 |φ|2 ≤ λ21 ||φ||2 ≤ λ1 |Aφ|2 ≤ |A 2 φ|2 , ∀φ ∈ D(A 2 )

(3.4)

1 d (|un |2 + α2 ||un ||2 ) + νλ1 (|un |2 + α2 ||un ||2 ) ≤ ||f ||2V  . dt ν

(3.5)

to reach

By Gronwall’s inequality one obtains −νλ1 t

|un (t)| + α ||un (t)|| ≤ e 2

2

2

1 (|un (0)| + α ||un (0)|| ) + ν 2

2

t

2

e−νλ1 (t−s) ||f (s)||2V  ds

0

≤ |u0 |2 + α2 ||u0 ||2 +

T ||f ||2L∞ ([0,T ];V  ) , ν

∀t ∈ [0, T ].

(3.6)

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Integrating (3.3) over [0, T ] T

T ν ||un ||2 + α2 |Aun |2 dt ≤ |un (0)|2 + α2 ||un (0)||2 + ||f ||2L∞ ([0,T ];V  ) . ν

(3.7)

0

By (3.6) and (3.7) T 0

||un ||2 + α2 |Aun |2 dt ≤ K1 , 

sup

 |un (t)|2 + α2 ||un (t)||2 ≤ K2 ,

(3.8)

t∈[0,T ]

where K1 =

 1 T |u0 |2 + α2 ||u0 ||2 + ||f ||2L2 ([0,T ];V  ) , ν ν

K2 = νK1 .

(3.9)

Note that K1 , K2 are two constants depending only on ν, α2 , |u0 |2 , ||u0 ||2 , ||f ||2L∞ ([0,T ];V  ) , but not on n. dvn n One also needs the uniform estimate with respect to n for du dt or dt . By the first equation of (3.1) dvn ˜ n , vn ) − νAvn + Pn f. = −Pn B(u dt

(3.10)

||Avn ||2L2 ([0,T ];D(A) ) ≤ cK1 .

(3.11)

By (3.8)

Calculation shows that ||Avn ||

4 L3

1

([0,T ];D(A) )

||f ||

1

1

≤ cT 4 ||Avn ||L2 ([0,T ];D(A) ) ≤ cT 4 K12 ,

(3.12)

4

4

L 3 ([0,T ];D(A) )

≤ cT ||f ||L3 ∞ ([0,T ];V  ) .

(3.13)

And ˜ n , vn ), w >D(A) = < B(un , vn ), w >D(A) − < B(w, vn ), un >D(A) < B(u = −(B(un , w), vn ) + (B(w, un ), vn ),

∀w ∈ D(A).

(3.14)

By (2.6) and (2.3) 1

1

1

1

˜ n , vn ), w >D(A) | ≤ c||un || 2 |Aun | 2 |Aw||vn | + c|Aw|||un || 2 |Aun | 2 |vn |, ∀w ∈ D(A). | < B(u By Poincare inequality (3.4) ˜ n , vn )||D(A) ≤ c||un || 2 |Aun | 2 |vn | ≤ c(λ−1 + α2 )||un || 2 |Aun | 2 . ||B(u 1 1

1

1

3

(3.15)

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By (3.8) ˜ n , vn )|| ||B(u

4 L3

([0,T ];D(A) )

≤

c(λ−1 1

+ α ) sup ||un || 2

1 2

t∈[0,T ]

T

34

0

K 14 K 38

2 ≤ c(λ−1 1 +α )

|Aun |2 dt

2

1

α2

α2

.

(3.16)

By (3.10), (3.12), (3.13) and (3.16) ||

dvn || 4 ≤ α2 · N < +∞, dt L 3 ([0,T ];D(A) ) dun || 4 || ≤ N < +∞, dt L 3 ([0,T ];H)

(3.17)

where N=

K 14 K 38  1 4 1 1  −1 2 1 2 4 K 2 + cT ||f || 3 + α ) + cT c(λ 1 1 L∞ ([0,T ];V  ) α2 α2 α2

is a constant dependent only on λ1 , ν, α2 , |u0 |2 , ||u0 ||2 , ||f ||L∞ ([0,T ];V  ) , T , but not on n. From (3.8) and (3.17), there exists a subsequence un , according to Aubin’s compactness theorem, such that un → u weakly in L2 ([0, T ]; D(A)),

(3.18)

un → u strongly in L2 ([0, T ]; V ),

(3.19)

u n → u

in C([0, T ]; H)

(3.20)

or equivalently vn → v weekly in L2 ([0, T ]; H),

(3.21)



vn → v strongly in L ([0, T ]; V ),

(3.22)

vn → v in C([0, T ]; D(A) ).

(3.23)

2

It is easy to conclude that v(t) = u(t) + α2 Au(t), a.e., t ∈ [0, T ]. By (3.8), (3.18) and (3.17) ||u||2L∞ ([0,T ];V ) ≤ lim inf ||un ||2L∞ ([0,T ];V ) ≤  n →∞

inf ||un ||2L2 ([0,T ];D(A)) ||u||2L2 ([0,T ];D(A)) ≤ lim  n →∞

||

K1 , α2 K2 ≤ 2, α

dun 2 du 2 || 4 || 4 ≤ lim inf || ≤ N. n →∞ dt L 3 ([0,T ];H) dt L 3 ([0,T ];H)

(3.24)

One relabels un and vn by un and vn respectively. From (3.1) t (vn (t), w) − (vn (t0 ), w) + ν

t (vn , Aw) +

t0

t ˜ n , vn ), Pn w >D(A) = < B(u

t0

< f, Pn w >V  t0

for all w ∈ D(A) and t, t0 ∈ [0, T ]. One wants to take limit about n on both sides of the above equality.

Y. Yu / J. Math. Anal. Appl. 429 (2015) 849–872

858

Since vn → v weekly in L2 ([0, T ]; H) then vn (s) → v(s) weakly in H for every s ∈ [0, T ] − E for some set E with L(E) = 0. Thus vn (τ ) → v(τ ) strongly in V  and D(A) for τ ∈ [0, T ] − E. Hence limn→+∞ (vn (t), w) = (v(t), w) limn→+∞ (vn (t0 ), w) = (v(t0 ), w) t

a.e., t0 , t ∈ [0, T ] − E, ∀w ∈ D(A). t (v(s), Aw)ds, ∀w ∈ D(A).

(vn (s), Aw)ds =

lim

n→+∞ t0

t0

And t

t ˜ n (s), vn (s)), Pn w >D(A) ds − < B(u

| t0

˜ < B(u(s), v(s)), w >D(A) ds| ≤ I1 + I2 + I3 . t0

By (2.9) and Poincare inequality (3.4) t ˜ n (s), vn (s)), Pn w − w >D(A) ds| < B(u

I1 = | t0

t

t (B(un (s), Pn w − w), vn (s))ds| + |

≤| t0

(B(Pn w − w, un (s)), vn (s))ds| t0

T ≤c

T |Aun (s)||Pn Aw − Aw||vn (s)|ds + c

0

|Pn Aw − Aw||Aun (s)||vn (s)|ds 0

 T  ≤ c|Pn Aw − Aw| |Aun (s)||vn (s)|ds 0

≤

c(λ−1 1

 T  + α )|Pn Aw − Aw| |Aun (s)|2 ds , 2

∀w ∈ D(A).

0

Hence lim I1 = 0

(3.25)

n→+∞

by the fact that limn→+∞ |Pn Aw − Aw| = 0 and un is bounded in L2 ([0, T ]; D(A)). By (2.6), (2.3) and Cauchy–Schwarz inequality t ˜ n (s) − u(s), vn (s)), w >D(A) ds| < B(u

I2 = | t0

t

t |(B(un (s) − u(s), w), vn (s))|ds +

≤ t0

|(B(w, un (s) − u(s)), vn (s))|ds t0

t

1 2

1 2

t

||un − u|| |Aun − Au| |Aw||vn |ds + c

≤c t0

1

1

|Aw|||un − u|| 2 |Aun − Au| 2 |vn |ds t0

Y. Yu / J. Math. Anal. Appl. 429 (2015) 849–872

t ≤ c|Aw|

1

859

1

||un − u|| 2 |Aun − Au| 2 |vn |ds t0

 t

12 t

12 ≤ c|Aw| ||un (s) − u(s)|||Aun (s) − Au(s)|ds |vn (s)|2 ds t0

t0

T

14 T

14 T

12 2 2 ≤ c|Aw| ||un (s) − u(s)|| ds |Aun (s) − Au(s)| ds |vn (s)|2 ds ≤ c|Aw|

0

0

T

T

14 ||un (s) − u(s)||2 ds

0

0

T

12

14 1 |vn (s)|2 ds · 2 4 (|Aun |2 + |Au|2 )ds

0

0

T

14 T

12

14 1 ≤ c|Aw| ||un − u||2 ds |vn |2 ds · 2 4 ||un ||2L2 ([0,T ],D(A)) + ||u||2L2 ([0,T ],D(A)) 0

0

T

14 1 K 14 1 1 ≤ c|Aw| ||un (s) − u(s)||2 ds K12 · 2 2 , α2

∀w ∈ D(A).

0

Hence lim I2 = 0

(3.26)

n→∞

by the fact that un −→ u strongly in L2 ([0, T ]; V ). t ˜ n , vn − v), w >D(A) |ds | < B(u

I3 = t0

t

t | < B(un , vn − v), w >D(A) |ds +

= t0

t0

t

t |(B(un , w), vn − v)|ds +

= t0

| < B(w, vn − v), un >D(A) |ds

|(B(w, un ), vn − v)|ds.

(3.27)

t0

By (2.9), |(B(un , w), ϕ)| ≤ c|Aun ||Aw||ϕ|,

∀ϕ ∈ H, w ∈ D(A),

|(B(w, un ), ϕ)| ≤ c|Aw||Aun ||ϕ|,

∀ϕ ∈ H, w ∈ D(A).

(3.28)

|B(w, un )| ≤ c|Aun ||Aw|.

(3.29)

It leads to |B(un , w)| ≤ c|Aun ||Aw|, By (3.8)

Y. Yu / J. Math. Anal. Appl. 429 (2015) 849–872

860

T ||B(un , w)||2L2 ([0,T ],H)

≤ c|Aw|

|Aun (s)|2 ds < ∞.

2 0

T ||B(w, un )||2L2 ([0,T ],H) ≤ c|Aw|2

|Aun (s)|ds < ∞. 0

Hence B(un , w), B(w, un ) ∈ L2 ([0, T ]; H) for any arbitrary but fixed w ∈ D(A).

(3.30)

By (3.27), (3.30) and (3.21) lim I3 = 0.

(3.31)

n→∞

As a result one gets for any t, t0 ∈ [0, T ] − E t (v(t), w) − (v(t0 ), w) + ν

t ˜ v), w >D(A) ds = < B(u,

(v, Aw)ds + t0

t

t0

< f, w >V  ds

(3.32)

t0

for every w ∈ D(A). Since v is bounded in L2 ([0, T ]; V  ). Moreover D(A) is dense in V  , (3.32) implies that v ∈ C([0, T ]; V  ) or equivalently u ∈ C([0, T ]; V ). So u(x, 0) = u0 in V is meaningful. This proves the existence of weak solution. As for the energy inequality, one obtains from (3.1) 1 (|un (t)|2 + α2 ||un (t)||2 ) + ν 2

t (||un (s)||2 + α2 |Aun (s)|2 )ds t0

1 ≤ (|un (t0 )|2 + α2 ||un (t0 )||2 ) + 2

t < f (s), un (s) >V  ds, a.e., t0 , t ∈ [0, T ].

(3.33)

t0

Since un → u strongly in L2 ([0, T ], V ), it leads to un → u strongly in L2 ([0, T ], H). Hence |un (t) − u(t)| → 0 in measure with respect to t

as n → ∞,

||un (t) − u(t)|| → 0 in measure with respect to t as n → ∞.

(3.34)

Therefore there exists a subsequence, relabeled as un (t), such that |un (t) − u(t)| → 0 a.e., t ∈ [0, T ] as n → ∞, ||un (t) − u(t)|| → 0 a.e., t ∈ [0, T ] as n → ∞.

(3.35)

Hence lim un (t) = u(t), lim un (t0 ) = u(t0 ), in H, a.e., t0 , t ∈ [0, T ],

n→∞

n→∞

lim un (t) = u(t), lim un (t0 ) = u(t0 ), in V, a.e., t0 , t ∈ [0, T ].

n→∞

Therefore

n→∞

(3.36)

Y. Yu / J. Math. Anal. Appl. 429 (2015) 849–872

861

lim |un (t)| = |u(t)|, lim |un (t0 )| = |u(t0 )|, a.e., t0 , t ∈ [0, T ],

n→∞

n→∞

lim ||un (t)|| = ||u(t)||, lim ||un (t0 )|| = ||u(t0 )||, a.e., t0 , t ∈ [0, T ].

n→∞

n→∞

(3.37)

Passing lim sup on both sides of (3.33), one gets 1 (|u(t0 )|2 + α2 ||u(t0 )||2 ) + 2

t < f (s), u(s) > ds t0

≥ lim sup

1 2

n→∞

t (|un (t)| + α ||un (t)|| ) + ν 2

2

2

 (||un (s)||2 + α2 |Aun (s)|2 )ds

t0

≥ lim sup

1

n→∞

2

t



(|un (t)| + α ||un (t)|| ) + ν lim inf 2

2

2

(||un (s)||2 + α2 |Aun (s)|2 )ds

n→∞

t0

1 = (|u(t)|2 + α2 ||u(t)||2 ) + ν lim inf n→∞ 2

t (||un (s)||2 + α2 |Aun (s)|2 )ds t0

1 ≥ (|u(t)|2 + α2 ||u(t)||2 ) + ν 2

t (||u(s)||2 + α2 |Au(s)|2 )ds.

(3.38)

t0

This proves the energy inequality. The proof of Theorem 1.1 is ended. 4. Proof of Theorem 1.2 Theorem 1.2 will be proved in this section. Let u1 , u2 be two weak solutions belonging to L4 ([0, T ]; D(A)) with initial data u1 (x, 0), u2 (x, 0) in V . Set u = u1 (x, t) − u2 (x, t), v = v1 − v2 = u + α2 Au,

u0 = u1 (x, 0) − u2 (x, 0), vi = ui + α2 Aui , i = 1, 2, du1 du2 du = − , dt dt dt

dv dv1 dv2 = − . dt dt dt

(4.1)

Then u0 ∈ V,

u1 , u2 , u ∈ L∞ ([0, T ]; V ) ∩ L2 ([0, T ]; D(A)) ∩ L4 ([0, T ]; D(A)).

(4.2)

And ⎧ ⎨ d(u + α2 Au) ˜ v1 ) + B(u ˜ 2 , u + α2 Au) = 0, + νA(u + α2 Au) + B(u, dt ⎩ u(x, 0) = u0 .

(4.3)

One takes the scalar product of the first equation in (4.3) with u to get 1 d ˜ 2 , v), u >D(A) = 0. (|u|2 + α2 ||u||2 ) + ν(||u||2 + α2 |Au|2 )+ < B(u 2 dt By (2.3) and (2.6)

(4.4)

Y. Yu / J. Math. Anal. Appl. 429 (2015) 849–872

862

˜ 2 , v), u >D(A) | = | − (B(u2 , u), v) + (B(u, u2 ), v)| | < B(u 1

1

1

1

≤ c|Au2 |||u|| 2 |Au| 2 |v| + c||u|| 2 |Au| 2 |Au2 ||v| 2 2 2 ≤ c(λ−1 1 + α )|Au2 |||u|| |Au| 1

3

≤

2 4 c(λ−1 να2 1 +α ) |Au|2 |Au2 |4 ||u||2 + 2 3 (να ) 2

≤

2 4 c(λ−1 ν 1 +α ) |Au2 |4 (|u|2 + α2 ||u||2 ) + (||u||2 + α2 |Au|2 ). 2 2 3 α (να ) 2

(4.5)

Hence

2 4 c(λ−1 d 4 2 2 2 1 +α ) (|u|2 + α2 ||u||2 ) + ν(||u||2 + α2 |Au|2 ) ≤ |Au | + α ||u|| |u| . 2 dt ν 3 α8

(4.6)

It leads to t |u(t)| + α ||u(t)|| ≤ (|u0 | + α ||u0 || ) exp 2

2

2

2

2

2

2 4 c(λ−1 1 +α ) |Au2 (τ )|4 dτ ν 3 α8

0

T ≤ (|u0 |2 + α2 ||u0 ||2 ) exp

2 4 c(λ−1 1 +α ) |Au2 (τ )|4 dτ. ν 3 α8

(4.7)

0

 T −1 2 )4 exp 0 c(λ1ν 3+α |Au2 (τ )|4 dτ < ∞, since u2 ∈ L4 ([0, T ]; D(A)). Hence |u(t)|2 + α2 ||u(t)||2 = 0, ∀t ∈ [0, T ]. α8 It deduces that u1 = u2 , a.e. in V . Hence any two weak solutions belonging to L4 ([0, T ]; D(A)) must be equal almost everywhere. This completes the proof of Theorem 1.2. 5. Proof of Theorem 1.3 One takes the scalar product of (3.1) with Aun in V  to obtain



3 1 d ˜ n , un + α2 Aun ), Aun >V  =< f, Aun >V  . ||un ||2 + α2 |Aun |2 + ν |Aun |2 + α2 |A 2 un |2 + < B(u 2 dt Notice that | < f, Aun >V  | ≤ ||f ||V  ||Aun ||V ≤

1 να2 3 |A 2 un |2 . ||f ||2V  + 2 να 4

This leads to

3ν

||f ||2 3 1 d V ˜ n , un + α2 Aun ), Aun >V  |. ||un ||2 + α2 |Aun |2 + |Aun |2 + α2 |A 2 un |2 ≤ + | < B(u 2 dt 4 να2 By (2.10), Poincaré inequality (3.4) and Young’s inequality 1

3

1

˜ n , un + α2 Aun ), Aun >V  | ≤ c|Aun |||un + α2 Aun |||Aun | 2 ||A 2 un | 2 | < B(u 2 2 2 2 ≤ c(λ−1 1 + α )|Aun | |A un | 3

≤

3

3

2 4 c(λ−1 να2 3 6 1 +α ) |A 2 un |2 |Au | + n (να2 )3 4

≤ k (||un ||2 + α2 |Aun |2 )3 +

να2 3 |A 2 un |2 , 4

Y. Yu / J. Math. Anal. Appl. 429 (2015) 849–872

863

where k = k (λ1 , ν, α) =

2 4 c(λ−1 1 +α ) . (να2 )3 α6

These lead to



2||f ||2 3 d V + k (||un ||2 + α2 |Aun |2 )3 ||un ||2 + α2 |Aun |2 + ν |Aun |2 + α2 |A 2 un |2 ≤ dt να2 ≤ k + k(||un ||2 + α2 |Aun |2 )3 ,

(5.1)

where  k = max k ,

 2 2 ||f || ∞ ([0,T ];V  ) L να2

depends only on λ1 , ν, α2 , ||f ||2L∞ ([0,T ];V  ) . Set z(t) = 1 + ||un (t)||2 + α2 |Aun (t)|2 ,

z0 = z(0) = 1 + ||u0 ||2 + α2 |Au0 |2 .

One gets dz(t) ≤ kz 3 (t). dt

(5.2)

Solving the inequality, one obtains z(t) ≤ 

z(0) 1−

2kz02

·t

,

as long as 0 ≤ t <

1 . 2kz02

(5.3)

It arrives at z(t) ≤ 2 · z(0),

as long as 0 ≤ t ≤

3 . 8kz02

(5.4)

Therefore ||un ||2 + α2 |Aun |2 ≤ k3 ,

0 ≤ t ≤ T ∗,

(5.5)

where K3 = 2(||u0 ||2 + α2 |Au0 |2 ) + 1,

 T ∗ = min T,

 3 . 8k(1 + ||u0 ||2 + α2 |Au0 |2 )2

(5.6)

Notice that K3 depends only on α2 , ||u0 ||2 , |Au0 |2 but not on n; while T ∗ depends only on λ1 , ν, α2 , ||u0 ||2 , |Au0 |2 , ||f ||2L∞ ([0,T ];V  ) , but not on n. Integrating (5.1) over [0, T ∗ ], one has T

∗ 3

(|Aun |2 + α2 |A 2 un |2 )ds ≤ ||u0 ||2 + α2 |Au0 |2

ν 0

2 + να2

T

||f ||2V  + k sup (||un (t)||2 + α2 |Aun (t)|2 )2

0

≤ ||u0 ||2 + α2 |Au0 |2 +

t∈[0,T ∗ ]

2T ||f ||2L∞ ([0,T ];V  ) + k K1 K32 . να2

T

∗

(||un (s)||2 + α2 |Aun (s)|2 )ds 0

(5.7)

Y. Yu / J. Math. Anal. Appl. 429 (2015) 849–872

864

Hence T

∗ 3

(|Aun |2 + α2 |A 2 un |2 )ds ≤ K4 ,

(5.8)

0

where K4 =

 1 2T 2  2 ||f || + k K K ||u0 ||2 + α2 |Au0 |2 + ∞  1 3 L ([0,T ];V ) ν να2

(5.9)

depends only on λ1 , ν, α2 , ||u0 ||2 , |Au0 |2 , ||f ||2L∞ ([0,T ];V  ) , but not on n. Passing to the limit, one obtains an local strong solution u ∈ L∞ ([0, T ∗ ]; D(A)) ∩ L2 ([0, T ∗ ]; D(A 2 )) on [0, T ∗ ]. From now on one assumes f ∈ L∞ ([0, T ]; H). Taking the scalar product of (3.1) with A2 un one has 3

3

3 1 d(|Aun |2 + α2 |A 2 un |2 ) ˜ n , un + α2 Aun ), A2 un ) = (f, A2 un ). + ν(|A 2 un |2 + α2 |A2 un |2 ) + (B(u 2 dt

By (2.12), Poincare inequality (3.4) and Young inequality ˜ n , un + α2 Aun ), A2 un )| |(B(u 1

1

≤ c|Aun |||un + α2 Aun || 2 |Aun + α2 A2 un | 2 |A2 un | 2 2 2 2 2 ≤ c(λ−1 1 + α )|Aun ||A un | |A un | 3

≤

1

3

2 4 3 c(λ−1 να2 2 2 4 1 +α ) 2 u |2 + |A un | |Au | |A n n (να2 )3 4

≤ k (||un ||2 + α2 |Aun |2 )2 (|Aun |2 + α2 |A 2 un |2 ) + 3

να2 2 2 |A un | . 4

(5.10)

Notice that |(f, A2 un )| ≤ |f ||A2 un | ≤

να2 2 2 1 |A un | + |f |2 . 4 να2

(5.11)

These lead to 3

3 d(|Aun |2 + α2 |A 2 un |2 ) + ν(|A 2 un |2 + α2 |A2 un |2 ) dt

≤ k (||un ||2 + α2 |Aun |2 )2 (|Aun |2 + α2 |A 2 un |2 ) + 3

2 |f |2 . να2

(5.12)

For any s, t with [s, t] ⊆ [0, T ∗ ] one integrates (5.12) on (s, t) to get 3 2

t

|Aun (t)| + α |A un (t)| + ν 2

2

3

3

(|A 2 un |2 + α2 |A2 un |2 )dτ ≤ |Aun (s)|2 + α2 |A 2 un (s)|2

2

s

+ k

t

3

(||un ||2 + α2 |Aun |2 )2 (|Aun |2 + α2 |A 2 un |2 )dτ +

2(t − s) ||f ||2L∞ ([0,T ];H) . να2

s

Taking (5.5) and (5.8) into account one has for a.e. s, t with 0 ≤ s < t ≤ T ∗

Y. Yu / J. Math. Anal. Appl. 429 (2015) 849–872

t

3 2

|Aun (t)| + α |A un (t)| + ν 2

2

865

3

(|A 2 un |2 + α2 |A2 un |2 )dτ

2

s

≤ |Aun (s)|2 + α2 |A 2 un (s)|2 + k K32 K4 + 3

2(t − s) ||f ||2L∞ ([0,T ];H) . να2

(5.13)

For any 0 < t ≤ T ∗ , integrating (5.13) over ( 2t , t) with respect to s, one obtains 3 t t2 t (|Aun (t)|2 + α2 |A 2 un (t)|2 ) ≤ K4 + · k  K32 K4 + ||f ||2L∞ ([0,T ];H) . 2 2 4να2

(5.14)

Hence 3 ˆ |Aun (t)|2 + α2 |A 2 un (t)|2 ≤ k(t), ∀0 < t ≤ T ∗ ,

(5.15)

ˆ = 2K4 + k K 2 K4 + t ||f ||2 ∞ k(t) 3 L ([0,T ];H) . t 2να2

(5.16)

where

ˆ satisfies the following properties: And k(t) ˆ (i) : k(t) is finite for all t > 0; ˆ (ii) : k(t) is independent of n; ˆ depends on ν, λ1 , α2 , |u0 |2 , ||u0 ||2 , |f |2 . Moreover, lim sup k(t) ˆ = +∞; (iii) : k(t) t→0+

∗

ˆ = kˆ = 2K4 + k K 2 K4 + T ||f ||2 ∞ (iv) : lim sup k(t) 3 L ([0,T ];H) < +∞. ∗ T∗ 2να2 t→T In particular, for any small but fixed positive τ one has |Aun (t)|2 + α2 |A 2 un (t)|2 ≤ K5 (τ ), ∀ τ ≤ t ≤ T ∗ , 3

(5.17)

where K5 (τ ) =

2K4 T + k  K32 K4 + ||f ||2L∞ ([0,T ];H) . τ 2να2

By (5.13) and (5.17) T

∗ 3

(|A 2 un (s)|2 + |A2 un (s)|2 )ds ≤ K6 (τ ),

(5.18)

τ

where K6 (τ ) =

1 2T 2 K5 (τ ) + k K32 K4 + ||f || ∞ ([0,T ];H) . L ν να2

(5.19)

By (5.12) 3

3 d(|Aun |2 + α2 |A 2 un |2 ) + ν(|A 2 un |2 + α2 |A2 un |2 ) dt 3 2 ≤ k K32 (|Aun |2 + α2 |A 2 un |2 ) + |f |2 , 0 ≤ t ≤ T ∗ . να2

(5.20)

Y. Yu / J. Math. Anal. Appl. 429 (2015) 849–872

866

One solves this inequality to obtain 3

|Aun |2 + α2 |A 2 un |2 ≤ K7 , T

(5.21)

3

(|A 2 un |2 + α2 |A2 un |2 )ds ≤ K8 ,

(5.22)

0

where  2T ||f ||2L∞ ([0,T ];H) )ek K3 T , 2 να  3 1 2T 2  K8 = ||f || + k K K T , |Au0 |2 + α2 |A 2 u0 |2 + ∞ 3 7 L ([0,T ];H) ν να2 3

K7 = (|Au0 |2 + α2 |A 2 u0 |2 +

(5.23)

3

if u0 ∈ D(A 2 ). Passing to the limit, one proves the existence of strong solutions. The existence of strong solutions can be made rigorous as in Section 2, but it will be long. For simplicity one omits the details. The strong solution must be unique as long as it exists. This will be stated as a lemma. Lemma 5.1. Let S > 0. Then any two strong solutions u ∈ L2 ([0, S]; D(A 2 )) ∩ L∞ ([0, S]; D(A)) coincide on [0, S]. 3

Proof of Lemma 5.1. Let u1 (x, t), u2 (x, t) be two strong solutions with the same initial value u1 (x, 0) = u2 (x, 0) in D(A). Set u = u1 (x, t) − u2 (x, t), v = v1 − v2 = u + α2 Au,

u0 = u1 (x, 0) − u2 (x, 0) = 0, vi = ui + α2 Aui , i = 1, 2, du1 du2 du = − , dt dt dt

dv dv1 dv2 = − . dt dt dt

(5.24)

Then u1 , u2 , u ∈ L∞ ([0, S]; D(A)) ∩ L2 ([0, S]; D(A 2 )), 3

v1 , v2 , v ∈ L∞ ([0, S]; H) ∩ L2 ([0, S]; V ).

(5.25)

And ⎧ ⎨ d(u + α2 Au) ˜ v1 ) + B(u ˜ 2 , u + α2 Au) = 0, + νA(u + α2 Au) + B(u, dt ⎩ u(x, 0) = u0 . One takes the scalar product of the first equation of (5.26) with Au to get 3 1 d (||u||2 + α2 |Au|2 ) + ν(|Au|2 + α2 |A 2 u|2 ) 2 dt ˜ v1 ), Au >V  | + | < B(u ˜ 2 , v), Au >V  |. ≤ | < B(u,

By (2.13) ˜ v1 ), Au >V  | ≤ c||u|| 12 |Au| 12 ||v1 |||A 32 u| | < B(u, ≤

να2 3 2 c |A 2 u| + ||u||||v1 ||2 |Au| 4 να2

(5.26)

Y. Yu / J. Math. Anal. Appl. 429 (2015) 849–872

≤

να2 3 2 c |A 2 u| + ||v1 ||2 |Au|2 1 4 2 2 νλ1 α

≤

να2 3 2 c |A 2 u| + ||v1 ||2 (||u||2 + α2 |Au|2 ). 1 4 2 4 νλ1 α

867

(5.27)

By (2.10) 1

3

1

˜ 2 , v), Au >V  | ≤ c|Au2 |||v|||Au| 2 |A 2 u| 2 | < B(u 2 2 2 2 ≤ c(λ−1 1 + α )|Au2 ||Au| |A u| 1

3

3

≤

2 4 να2 3 2 c(λ−1 1 +α ) |A 2 u| + |Au2 |4 |Au|2 4 (να2 )3

≤

2 4 να2 3 2 c(λ−1 1 +α ) |A 2 u| + |Au2 |4 (||u||2 + α2 |Au|2 ) 2 3 4 (να ) α2

≤

2 4 3 να2 3 2 c(λ−1 1 +α ) |A 2 u| + |Au2 |2 |A 2 u2 |2 (||u||2 + α2 |Au|2 ). 4 (να2 )3 α2 λ1

(5.28)

These lead to 3 d (||u||2 + α2 |Au|2 ) + ν(|Au|2 + α2 |A 2 u|2 ) dt c



2 4 3 c(λ−1 1 +α ) ||v1 ||2 + |Au2 |2 |A 2 u2 |2 ||u||2 + α2 |Au|2 . ≤ 1 2 3 2 (να ) α λ1 νλ12 α4

(5.29)

One solves the inequality to conclude that ||u(t)||2 + α2 |Au(t)|2

≤ ||u0 || + α |Au0 | 2

2

2

S

exp

c 1 2

0

νλ1 α4

||v1 (s)||2 +

2 4 3 c(λ−1 1 +α ) |Au2 |2 |A 2 u2 (s)|2 ds, 2 3 2 (να ) α λ1

(5.30)

for all t ∈ [0, S]. Since v1 ∈ L2 ([0, T ]; V ) and u2 ∈ L2 ([0, S]; D(A 2 )) ∩ L∞ ([0, S]; D(A)), it deduces that the exponential is finite. Hence ||u(t)||2 +α2 |Au(t)|2 = 0, ∀t ∈ [0, S]. This represents u1 = u2 in L∞ ([0, S]; D(A)). This completes the proof of the lemma. The proof of Theorem 1.3 is ended. 3

6. Proof of Theorem 1.4 Theorem 1.4 will be proved in this section. The estimates in this section is formal. The estimates can be made rigorous by considering the finite Galerkin approximation and then passing to the limit. One takes the scalar product of the first equation in (1.4) with u as in Section 2 to obtain d 1 (|u|2 + α2 ||u||2 ) + ν(||u||2 + α2 |Au|2 ) ≤ ||f ||2V  . dt ν

(6.1)

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And

|u(t)|2 + α2 ||u(t)||2 ≤ K2 ,

(6.2)

  ||u(s)||2 + α2 |Au(s)|2 ds ≤ K1 .

(6.3)

sup t∈[0,T ]

T 0

One takes the scalar product of (1.4) with Au as in Section 5 to have



3 d ||u||2 + α2 |Au|2 + ν |Au|2 + α2 |A 2 u|2 dt 2 4 2||f ||2V  c(λ−1 1 +α ) ≤ + (||u||2 + α2 |Au|2 )3 . 2 3 12 να ν α

(6.4)

It is an ordinary differential inequality with respect to ||u(t)||2 + α2 |Au(t)|2 . From (1.9) it deduces that there exists positive constant δ > 0 such that 1

||u(t)||2 + α2 |Au(t)|2 ≤ ν 2 λ12 · √

α6 , + α2 )2

c(λ−1 1

∀ t ∈ [0, δ]

(6.5)

by the smoothness of ||u(t)||2 + α2 |Au(t)|2 with respect to t. Therefore

c(λ−1 + α2 )4

3 3 2 2 2 1 ν |Au|2 + α2 |A 2 u|2 − + α |Au| ||u|| ν 3 α12

c(λ−1 + α2 )4

3 2 2 2 1 ≥ νλ1 ||u||2 + α2 |Au|2 − + α |Au| ||u|| ν 3 α12



2 4 c(λ−1 2 2 2 2 1 +α ) ≥ ||u||2 + α2 |Au|2 νλ1 − (||u|| + α |Au| ) ≥ 0, t ∈ [0, δ]. ν 3 α12

(6.6)

So

2||f ||2 d V , ∀ t ∈ [0, δ]. ||u||2 + α2 |Au|2 ≤ dt να2

(6.7)

It deduces that 2t ||f ||2L∞ ([0,T ];V  ) να2 2T ≤ ||u0 ||2 + α2 |Au0 |2 + ||f ||2L∞ ([0,T ];V  ) να2 1 1 α6 ≤ ν 2 λ12 · √ −1 , ∀ t ∈ [0, δ]. 2 c(λ1 + α2 )2

||u(t)||2 + α2 |Au(t)|2 ≤ ||u0 ||2 + α2 |Au0 |2 +

(6.8)

One has proved that as long as 1

||u(t)||2 + α2 |Au(t)|2 ≤ ν 2 λ12 · √

α6 , ∀t ∈ [0, δ], + α2 )2

c(λ−1 1

(6.9)

then ||u(t)||2 + α2 |Au(t)|2 ≤

1 2 12 α6 ν λ1 · √ −1 , ∀t ∈ [0, δ]. 2 c(λ1 + α2 )2

(6.10)

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Therefore for (6.4) the least upper bound of the set 

1

t ∈ [0, T ] : ||u(t)||2 + α2 |Au(t)|2 ≤ ν 2 λ12 · √

 α6 2 2 c(λ−1 1 +α )

(6.11)

must be T . By (6.4) and Poincare inequality (3.4)



3 d ||u||2 + α2 |Au|2 + ν |Au|2 + α2 |A 2 u|2 dt



2 2 4 3 2||f ||2V  c(λ−1 2 2 2 2 2 1 +α ) 2 u|2 ≤ + + α |A + α |Au| . |Au| ||u|| να2 ν 3 α12 λ1

(6.12)

Hence

d 2||f ||2V  2 2 2 ||u|| − + α |Au| να2 dt



2 4 3 c(λ−1 1 +α ) ≥ ν− (||u||2 + α2 |Au|2 )2 |Au|2 + α2 |A 2 u|2 3 12 ν λ1 α

2

2 4 3 α6 1 2 12 c(λ−1 2 2 1 +α ) 2 u|2 ν λ · + α |A |Au| ≥ ν− √ 1 −1 ν 3 λ1 α12 2 c(λ1 + α2 )2

3 3ν ≥ |Au|2 + α2 |A 2 u|2 . 4

(6.13)

One integrates (6.13) over [0, t] to obtain

||u(t)||2 + α2 |Au(t)|2 +

3ν 4

t

3

(|Au(s)|2 + α2 |A 2 u(s)|2 )ds 0

≤ ||u0 ||2 + α2 |Au0 |2 +

1 2T 1 α6 ||f ||2V  ≤ ν 2 λ12 · √ −1 . 2 να 2 c(λ1 + α2 )2

(6.14)

It deduces that ||u(t)||2 + α2 |Au(t)|2 1 2

ν 2 λ1

T +

0

3

(|Au(s)|2 + α2 |A 2 u(s)|2 )ds 1 2

νλ1

≤2· √

α6 , + α2 )2

c(λ−1 1

(6.15)

for all t ∈ [0, T ]. Hence u ∈ L∞ ([0, T ]; D(A)) ∩ L2 ([0, T ]; D(A 2 )). 3

(6.16)

One takes the scalar product of the first equation in (1.4) with A2 u as in Section 5 to obtain 3

3 d(|Au|2 + α2 |A 2 u|2 ) + ν(|A 2 u|2 + α2 |A2 u|2 ) dt

≤

2 4 3 c(λ−1 2 1 +α ) (||u||2 + α2 |Au|2 )2 (|Au|2 + α2 |A 2 u|2 ) + |f |2 . 3 12 ν α να2

(6.17)

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870

By (6.15) 3

3 3 2 d(|Au|2 + α2 |A 2 u|2 ) + ν(|A 2 u|2 + α2 |A2 u|2 ) ≤ 4νλ1 (|Au|2 + α2 |A 2 u|2 ) + |f |2 . dt να2

(6.18)

Integrating (6.18) over (s, t), one has t

3 2

|Au(t)| + α |A u(t)| + ν 2

2

3

(|A 2 u(r)|2 + α2 |A2 u(r)|2 )dr

2

s

t

3 2

≤ |Au(s)| + α |A u(s)| + 4νλ1 2

2

3

(|Au|2 + α2 |A 2 u|2 )dr +

2

2(t − s) ||f ||2L∞ ([0,T ];H) να2

s 3

8ν 2 λ 2 α6 2(t − s) ≤ |Au(s)| + α |A u(s)| + √ −1 1 2 2 + ||f ||2L∞ ([0,T ];H) . να2 c(λ1 + α ) 2

2

3 2

2

(6.19)

One integrates the above inequality over ( 2t , t) with respect to s to obtain 3 t · (|Au(t)|2 + α2 |A 2 u(t)|2 ) ≤ 2

t

3

4ν 2 λ 2 α6 · t t2 (|Au| + α |A u| ) + √ −11 + ||f ||2L∞ ([0,T ];H) 2 2 4να2 c(λ1 + α ) 2

t 2

2

3 2

2

1

3

2νλ 2 α6 4ν 2 λ 2 α6 · t t2 ≤ √ −11 + √ −11 + ||f ||2L∞ ([0,T ];H) . 2 2 2 2 4να2 c(λ1 + α ) c(λ1 + α )

(6.20)

Hence 1

3

4νλ12 α6 8ν 2 λ12 α6 T |Au(t)| + α |A u(t)| ≤ √ −1 + + ||f ||2L∞ ([0,T ];H) . √ −1 2 2 2 2 2 2να t · c(λ1 + α ) c(λ1 + α ) 2

3 2

2

2

(6.21)

For any arbitrary but fixed τ   3 sup |Au(t)|2 + α2 |A 2 u(t)|2 ≤ t∈[τ,T ]

1

3

8ν 2 λ12 α6 T 4νλ12 α6 + + ||f ||2L∞ ([0,T ];H) . √ −1 √ −1 2 2 2 2 2 2να τ · c(λ1 + α ) c(λ1 + α ) (6.22)

By (6.19) T ν τ

1

3

4νλ12 α6 16ν 2 λ 2 α6 5T (|A u(r)| + α |A u(r)| )dr ≤ + √ −1 1 2 2 + ||f ||2L∞ ([0,T ];H) . √ −1 2 2 να2 τ · c(λ1 + α ) c(λ1 + α ) 3 2

2

2

2

2

(6.23)

By (6.22) and (6.23) u ∈ L∞ ([τ, T ]; D(A 2 )) ∩ L2 ([τ, T ]; D(A2 )) for any arbitrary but fixed τ > 0. 3

(6.24)

One has from (6.18) and Poincare inequality (3.4) that 3

3 2 d(|Au|2 + α2 |A 2 u|2 ) ≤ 3νλ1 (|Au|2 + α2 |A 2 u|2 ) + |f |2 . dt να2

(6.25)

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871

By Gronwall’s Lemma 3

3

|Au(t)|2 + α2 |A 2 u(t)|2 ≤ (|Au0 |2 + α2 |A 2 u0 |2 )e3νλ1 t +

T 0

2||f ||2L∞ ([0,T ];H)

(e3νλ1 t − 1) 3ν 2 α2 λ1 ||f ||2L∞ ([0,T ];H) 3νλ T 3 ≤ |Au0 |2 + α2 |A 2 u0 |2 + e 1 , λ1 ν 2 α 2 3

3 2T ||f ||2L∞ ([0,T ];H) |Au0 |2 + α2 |A 2 u0 |2 8λ12 να6 + (|A u| + α |A u| )ds ≤ + , √ 2 2 ν ν 2 α2 c(λ−1 1 +α ) 3 2

2

2

2

2

3

if u0 ∈ D(A 2 ). This means u ∈ L∞ ([0, T ]; D(A 2 )) ∩ L2 ([0, T ]; D(A2 )) if u0 ∈ D(A 2 ) and f ∈ L∞ ([0, T ]; H). 3

3

(6.26)

This completes the proof of Theorem 1.4. Acknowledgments The author is grateful to the anonymous referee for her/his helpful comments which improve both the mathematical results and the way to present them. The author would also like to thank the support and encouragement of Professor Tong Li, Professor Lihe Wang and the support of Department of Mathematics in University of Iowa, during my visit in 2014. References 1

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