ISSN 0965-5425, Computational Mathematics and Mathematical Physics, 2008, Vol. 48, No. 10, pp. 1857–1874. © Pleiades Publishing, Ltd., 2008. Original Russian Text © M.O. Korpusov, A.G. Sveshnikov, 2008, published in Zhurnal Vychislitel’noi Matematiki i Matematicheskoi Fiziki, 2008, Vol. 48, No. 10, pp. 1860–1877.
An Initial–Boundary Value Problem for a Sobolev-Type Strongly Nonlinear Dissipative Equation M. O. Korpusov and A. G. Sveshnikov Faculty of Physics, Moscow State University, Moscow, 119992 Russia e-mail:
[email protected] Received November 1, 2007
Abstract—An initial–boundary value problem is considered for a model equation governing waves in crystalline semiconductors with allowance for strong spatial dispersion, linear dissipation, and sources of free charges. The weak generalized local-in-time solvability of the problem is proved. Sufficient conditions are obtained for the blowup of the solution and for global-in-time solvability. Two-sided estimates for the blowup time are derived. DOI: 10.1134/S0965542508100096 Keywords: Sobolev-type nonlinear dissipative equations, waves in crystalline semiconductors, local solvability conditions, sufficient blowup conditions.
1. INTRODUCTION The goal of this paper is to derive sufficient conditions for the blowup of the solution to the following initial–boundary value problem for a strongly nonlinear pseudoparabolic dissipative wave equation: ∂⎛ ----- ⎜ ∆u + ∂t ⎝
N
∂
∂u ------- ⎛ -----∑ ∂x ⎝ ∂x-
i=1
i
p–2
i
u ( x, t )
∂u ⎞ 3 -------⎞ ⎟ – u + u x1 + uu x1 + u = 0, ∂x i⎠ ⎠
∂Ω
= 0,
(1.1)
t 0,
(1.2)
x ∈ Ω,
(1.3)
u ( x, 0 ) = u 0 ( x ),
N
where p > 2, (x, t) ∈ Q = Ω × (0, T), and x = (x1, x2, …, xN) ∈ Ω ⊂ . 2. WEAK GENERALIZED LOCAL UNIQUE SOLVABILITY OF THE PROBLEM (2, δ)
Let ∂Ω ∈ be the boundary of the bounded domain Ω, δ ∈ (0, 1], and s 2 be a positive integer such that (s – 1)/N 1/2 – 1/p. Consider the eigenvalue problem s
s
( – 1 ) ∆ w k – λ k w k = 0,
n
∂ wk ----------n∂x i
n = 1, s – 1 .
= 0,
(2.1)
∂Ω
By the definition of s, we have s
1, p
1
2
0 ( Ω ) ⊂ 0 ( Ω ) ⊂ 0 ( Ω ) ⊂ ( Ω ),
p > 2. s
Since the eigenfunctions of problem (2.1) form a Galerkin basis in 0 (Ω), it follows from these inclu1, p
sions that the eigenfunctions of problem (2.1) form a Galerkin basis in 0 (Ω). Definition 1. The weak generalized solution to problem (1.1)–(1.3) is defined as the solution to the problem T
∫ dtϕ ( t )
d 3 ---- ( 〈 ∆u, w〉 0 + 〈 ∆ p u, w〉 ) + ( u , w ) – ( u, w ) + 〈 u x1, w〉 0 + 〈 uu x1, w〉 0 = 0, dt
0
1857
1858
KORPUSOV, SVESHNIKOV 1, p
1, p
2
u ( 0 ) = u0 ∈ 0 ( Ω )
∀ϕ ( t ) ∈ ( 0, T ),
∀w ∈ 0 ( Ω ), 1
–1
where 〈·, ·〉0 denotes the duality brackets between the Hilbert spaces 0 (Ω) and (Ω), 〈·, ·〉 stands for the 1, p
–1, p'
duality brackets between the Banach spaces 0 (Ω) and (Ω), and the symbol (·, ·) denotes the inner 2 product in (Ω). A solution to the given problem is sought in the following class of functions: ∞
1, p
2
u ∈ ( 0, T ; 0 ( Ω ) ), ∞
∆ p u ∈ ( 0, T ; 1
∆u + ∆ p u ∈ ( 0, T ;
– 1, p'
1
u' ∈ ( 0, T ; 0 ( Ω ) ), – 1, p'
( Ω ) ),
(1)
( Ω ) ) ∩ w ( [ 0, T ];
– 1, p'
( Ω ) ).
In this class of smoothness, the problem in question is equivalent to the problem T
∫ dtϕ ( t )
d 3 ---- ( ∆u + ∆ p u ) – u + u x1 + uu x1 + u , w = 0 dt
0 1, p
2
∀ϕ ( t ) ∈ ( 0, T ),
∀w ∈ 0 ( Ω ),
which, in turn, is equivalent to the problem T
∫ dt
d 3 ---- ( ∆u + ∆ p u ) – u + u x1 + uu x1 + u , v = 0 dt
0 1, p
2
∀v ∈ ( 0, T ; 0 ( Ω ) ). 1, p
Remark 1. As w ∈ 0 (Ω), we can use elements wj ∈ 0 (Ω), j = 1, +∞ , since their linear combinations s
1, p
are dense in 0 (Ω). 1, p
Theorem 1. Let u0(x) ∈ 0 (Ω). If N p or if N > p for p 4N/(N + 4), then there is T0 = T0(u0, q, p) > 0 such that, for any T ∈ (0, T0), we have and there is a unique function u(x, t) : (0, T0) × Ω 1, p
∞
∂ ∂u ----- ⎛ -----∂t ⎝ ∂x-i ∂ ∂u ----- ⎛ -----∂t ⎝ ∂x-i
( p – 2 )/2
p–2
2 ∂u -------⎞ ∈ ( Q ), ⎠ ∂x i
1
∆u + ∆ p u ∈ ( 0, T ;
– 1, p'
u' ∈ ( 0, T ; 0 ( Ω ) ),
∂u ------∂x i
2 p' ∂u -------⎞ ∈ ( 0, T ; ( Ω ) ), ⎠ ∂x i
1
2
u ∈ ( 0, T ; 0 ( Ω ) ),
( p – 2 )/2
∂u ------∂x i
∞ 2 ∂u ------- ∈ ( 0, T ; ( Ω ) ), ∂x i
p–2
∞ p' ∂u ------- ∈ ( 0, T ; ( Ω ) ), ∂x i
(1)
( Ω ) ) ∩ w ( [ 0, T ];
– 1, p'
( Ω ) ),
p' = p/ ( p – 1 ).
Moreover, u(0) = u0 almost everywhere in Ω and ∂ 3 2 – 1, p' ----- ( ∆u + ∆ p u ) – u + u x1 + uu x1 + u = 0 in ( 0, T ; ( Ω ) ), ∂t N
∆ pu ≡
j=1
∞
Remark 2. Since u ∈ (0, T;
∂
⎛ ∂u ∑ ------∂x ⎝ ------∂x j
1, p 0 (Ω))
the result of [2] implies that u(t) : [0, T]
j
p–2
∂u -------⎞ , ∂x j⎠ 2
–1
p + p'
–1
= 1.
1
and u' ∈ (0, T; 0 (Ω)) in the assumptions of Theorem 1, 1
0 (Ω) is a strongly continuous function.
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Prof. To prove the local solvability of the problem, we use the Galerkin method in conjunction with the monotonicity and compactness methods (see [2]). An approximate solution to problem (1.1)–(1.3) is sought in the form T
∫ dtϕ ( t )
d 3 ---- ( 〈 ∆u m, w〉 0 + 〈 ∆ p u m, w〉 ) + ( u m , w ) – ( u m, w ) + 〈 u m x1, w〉 0 + 〈 u m u m x1, w〉 0 = 0 dt
0 2
∀ϕ ( t ) ∈ ( 0, T ),
(2.2)
j = 1, m ,
m
um =
∑c
mk ( t )w k ,
k=1
(2.3)
m
∑c
u 0m ≡ u m ( 0 ) =
mk ( 0 )w k
u0
strongly in
1, p 0 ( Ω ).
k=1
(1)
As before, in the class cmk(t) ∈ [0, Tm0), problem (2.2) implies the pointwise equality N
( ∇u 'm, ∇w j ) + ( p – 1 )
--------∑ ⎛⎝ ∂u ∂x
p–2
m i
i=1
∂u 'm ∂w j⎞ ∂w 1 2 ∂w 2 ---------, --------- + ( u m, w j ) – ⎛ u m, ---------j⎞ – --- ⎛ u m , ---------j⎞ = ( u m u m, w j ), (2.4) ⎝ ∂x i ∂x i ⎠ ∂x 1 ⎠ 2 ⎝ ∂x 1 ⎠
j = 1, m . It follows from (2.4) that m
N
p–2
⎛ 1 + ( p – 1 ) ∂u m --------⎝ ∂x i k = 1i = 1
∑∑
m
+
m
N
∑ ∑w w c k
j mk
∂w j
-⎛c ∑ w -------∂x ⎝
–
k
k = 1i = 1
mk
1
k=1
∂w ∂w ' ---------k, ---------j⎞ c mk ∂x i ∂x i ⎠
1 2 q + --- c mk ⎞ = ( u m u m, w j ), ⎠ 2
j = 1, m .
1, p
1
Now, as a Galerkin basis in 0 (Ω) and 0 (Ω), we use the eigenfunctions of problem (2.1). Define N
a kj ≡
p–2
⎛ 1 + ( p – 1 ) ∂u m --------⎝ ∂x i i=1
∑
∂w ∂w ---------k, ---------j⎞ . ∂x i ∂x i ⎠
It is true that m, m
∑
N
a kj ξ k ξ j = ∇η
2 2
+ ( p – 1)
k, j = 1, 1 m
ηi =
∑ k=1
The norm ∇η
2 2
⎛ ∂u m ⎝ -------∂x i i=1
∑
p–2
, η i ⎞ ∇η 2 , ⎠ 2
2
m
∂w k --------- ξ k , ∂x i
η =
∑ξ w . k
k
k=1
1
vanishes in 0 (Ω) if and only if m
η =
∑ξ w k
k
= 0.
k=1 1, p
Since { w k } k = 1 is a Galerkin basis in 0 (Ω), we conclude that +∞
m
η =
∑ξ w k
k
= 0
k=1
COMPUTATIONAL MATHEMATICS AND MATHEMATICAL PHYSICS
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KORPUSOV, SVESHNIKOV m, m
m
if and only if { ξ k } k = 1 = 0. Therefore, the Sylvester criterion implies that det { a kj } k, j = 1, 1 > 0. The general results concerning nonlinear systems of ordinary differential equations guarantee that problem (2.4) has a (1) solution on some interval [0, tm] (where tm > 0) in the sense of ckm ∈ [0, tm]. Below, we obtain a priori (1) (1) estimates that yield tm = T, where T is independent of m. Therefore, ckm ∈ [0, T] and um ∈ ([0, T]; 1, p
0 (Ω)). Let us derive a priori estimates. Multiplying both sides of (2.4) by cmj and summing the result over j = 1, m gives 1d --- ----- ∇u m 2 dt
2 2
p–1d + ------------ ----p dt
N
p
∂u m --------∂x i
∑ i=1
2 2
+ um
4
= um 4 .
(2.5)
p
Note that the norms p ⎛ ∂v ⎞ ⎜ ------- ⎟ ⎝ i = 1 ∂x i p⎠ N
∑
1/ p
⎛ ⎜ v ⎝
,
p ∂v ⎞ ------- ⎟ ∂x i p⎠
N
p p
∑
+
i=1
1/ p
1, p
1, p
are equivalent for v ∈ 0 (Ω). For this reason, we consider the Banach space 0 (Ω) equipped with the norm p ⎛ ∂v ⎞ ⎜ ------- ⎟ ⎝ i = 1 ∂x i p⎠ N
∑
1/ p
.
It follows from (2.5) that 1 --- ∇u m 2
2 2
p–1 + -----------p
N
∑ i=1
∂u m --------∂x i
p p
1 --- ∇u 0m 2
2 2
p–1 + -----------p
N
∑ i=1
∂u 0m ----------∂x i
p
t
∫
4
+ ds u m 4 ( s ).
p
(2.6)
0
By virtue of the assumptions on p and N formulated in Theorem 1 and by the Sobolev embedding theorems, 1, p
4
we have 0 (Ω) ⊂ (Ω). Therefore, p ⎛ ∂u m ⎞ B1 ⎜ --------- ⎟ ⎝ i = 1 ∂x i p⎠ N
um
1 --- ∇u m 2
2 2
p–1 + -----------p
N
∑ i=1
∂u m --------∂x i
p p
4
1 --- ∇u 0m 2
∑
2 2
p–1 + -----------p
N
∑ i=1
1/ p
,
∂u 0m ----------∂x i
t
p
+
4 B1
p
p ⎛ ∂u m ⎞ ds ⎜ --------- ⎟ ⎝ i = 1 ∂x i p⎠ 0 N
∫ ∑
4/ p
( s ).
Thus, t α
∫
E m ( t ) E 0m + B 2 dsE m ( s ),
p 4/ p ------------⎞ , ⎝ p – 1⎠
q + 2⎛
4 α = --- , p
B2 ≡ B1
0
1 E m ( t ) ≡ --- ∇u m 2
2 2
p–1 + -----------p
N
∑ i=1
(2.7) p
∂u m , --------∂x i p
E 0m ≡ E m ( 0 ).
By applying the Gronwall–Bellman and Bihary lemmas, (2.7) yields 1–α
E m ( t ) [ E 0m + ( 1 – α )B 2 t ) ]
1/ ( 1 – α )
E m ( t ) E 0m exp { B 2 t } (α – 1)
E m ( t ) E 0m [ 1 – ( α – 1 )B 2 E 0m
if t]
if
0 < α < 1,
(2.8)
α = 1,
– 1/ ( 1 – α )
if
(2.9) α > 1.
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' over j = 1, m and integrating the result with respect to s ∈ [0, t] gives Summing Eq. (2.4) times c mj t
N
2 4( p – 1) ∂ ∂u m ds ∇u 'm 2 ( s ) + ------------------d x ⎛ ----- ⎛ -------2 ⎝ ⎝ ∂x ∂t p i i = 1Ω 0
∑∫
∫
( p – 2 )/2
∂u m⎞ ⎞ 2 --------∂x i ⎠ ⎠ (2.11)
t
1 + --- u m 2
2 2
1 + --- u 0m 4
4 4
1 1 2 ∂u ' + ds ⎛ u m + --- u m⎞ --------m- = --- u m ⎝ ⎠ 4 2 ∂x 1
∫
4 4
1 2 + --- u m0 2 . 2
0
We have the auxiliary estimates
∫
Ω
∂u ' ε dsu m --------m- --- ∇u 'm 2 ∂x 1
∫ dsu
Ω
2 ∂u 'm m --------∂x 1
ε --- ∇u 'm 2
2 2
1 2 + ----- u m 2 , 2ε
2 2
1 4 + ----- u m 4 , 2ε
p ⎛ ∂u m ⎞ B3 ⎜ --------- ⎟ ⎝ i = 1 ∂x i p⎠ N
um
B2 um
2
4
∑
1/ p
.
By virtue of the assumptions on p and N in Theorem 1 and by the Sobolev embedding theorems, it follows from (2.11) that t
∫ 0
2 4( p – 1) 3ε ds 1 – ----- ∇u 'm 2 ( s ) + ------------------2 4 p t
ε + ds --- u m 4
∫
4 4
N
∑∫
i = 1Ω
∂ ∂u m d x ⎛ ----- ⎛ -------⎝ ∂t ⎝ ∂x i
ε + --- u m 2
2 2
( p – 2 )/2
1 2 + --- u 0m 2 , 2
∂u m⎞ ⎞ 2 1 4⎛ -------- --- B 1 ⎜ 4 ⎝ ∂x i ⎠ ⎠
N
∑ i=1
p ∂u m ⎞ --------- ⎟ ∂x i p⎠
4/ p
(2.12)
ε ∈ ( 0, 4/3 ).
0 1, p
u0 strongly in 0 (Ω) and E0m A, where A is independent of m. This result In view of (2.3), u0m and relations (2.8)–(2.10) imply that there is T0 > 0 such that Em(t) C for t ∈ [0, T] and any ∀T ∈ (0, T0), where C is independent of m or t. Thus, (2.8)–(2.12) imply that there is T0 ≡ T0(u0, q, p) such that, for any T ∈ (0, T0), um
is bounded in
u m' ∂ ∂u m ----- ⎛ -------∂t ⎝ ∂x -i
is bounded in
( p – 2 )/2
3
um
∂u m⎞ --------∂x i ⎠
1, p
∞
( 0, T ; 0 ( Ω ) ),
is bounded in
is bounded in
1
2
( 0, T ; 0 ( Ω ) ), 2
(2.13)
2
( 0, T ; ( Ω ) ),
∞
4/3
( 0, T ; ( Ω ) ).
Finally, it is clear that ∂u m --------∂x i
( p – 2 )/2
∂u m --------∂x i
is bounded in
∞
2
( 0, T ; ( Ω ) ).
(2.14)
By virtue of (2.13) and (2.14), we have tm = T, where T is independent of m ∈ . Therefore, m
um =
∑c
km ( t )w k
(1)
1, p
∈ ( [ 0, T ]; 0 ( Ω ) ).
k=1
COMPUTATIONAL MATHEMATICS AND MATHEMATICAL PHYSICS
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KORPUSOV, SVESHNIKOV
Based on (2.13), (2.14), and the results of [2], we conclude that there is a subsequence of the sequence um (again denoted by um) such that um
u
u m'
∞
u'
1, p
( 0, T ; 0 ( Ω ) ),
*-weakly in
1
2
( 0, T ; 0 ( Ω ) ),
weakly in
∞
–1
( 0, T ; ( Ω ) ),
∆u m
∆u
–∆ p u m
χ
*-weakly in
( 0, T ;
w
*-weakly in
( 0, T ; ( Ω ) ),
3
um ∂u --------m∂x 1
*-weakly in
∂u -------∂x 1
∞
∞
– 1, p'
(2.15)
(Ω))
4/3
∞
p
( 0, T ; ( Ω ) ).
*-weakly in
Note that the first relation in (2.15) implies that u 'm
u'
1, p
' ( 0, T ; 0 ( Ω ) ).
in 1
2
Because of the weak convergence in (0, T; 0 (Ω)), we then have η.
u 'm
1
Therefore, η(t) = u'(t) for almost all t ∈ (0, T), which yields um u weakly in (Q), where Q = (0, T) × Ω. 1 2 2 u strongly in (Q) and, hence, Since the embedding (Q) ⊂ (Q) is compact, we conclude that um ∞
∂u m --------∂x i
( p – 2 )/2
∂u m 1 2 --------- ∈ ( 0, T ; ( Ω ) ), ∂x i
Taking into account that –∆pum
4/3
u3 *-weakly in (0, T; (Ω)). Thus, w = u3.
3
almost everywhere. Then, by Lemma 1.3 in [2], u m From (2.13) and (2.14), it follows that
1
1
u m ∈ ( 0, T ; 0 ( Ω ) ). 1, p'
∞
χ *-weakly in (0, T; 0
(Ω)), we prove that
χ = – ∆ p u.
(2.16)
Let ϕ(t) = 1 in (2.2). Then, t
〈 ∆u m + ∆ p u m, w j〉 +
t
∫ dsu
3 m (s),
wj –
0
Since ∆um
∫ dsu
m(s),
wj +
0
∞
∫ 0
1 2 ∂ ⎛ u m + --- u m⎞ ⎝ 2 ⎠ ds ------------------------------, w j = 〈 ∆u 0m + ∆ p u 0m, w j〉 . (2.17) ∂x 1
–1
∆u *-weakly in (0, T; (Ω)), we conclude that 〈∆um, wj 〉
j = 1, m . Moreover, –∆pum wj 〉
t
∞
χ *-weakly in (0, T;
∞
∞
〈∆u, wj 〉*-weakly in (0, T),
–1, p'
(Ω)), where p' = p/(p – 1). Therefore, –〈∆pum,
〈χ, wj 〉 *-weakly in (0, T), j = 1, m . Finally, t
∫
t 3 ds 〈 u m ,
w j〉 =
0
∫
t 3 ds 〈 u m
∫
3
0
∞
3
– u , w j〉 + ds 〈 u , w j〉 , 0
4/3
On the other hand, u m , u3 ∈ (0, T; (Ω)). Moreover, u m fore, the Lebesgue theorem implies that 3
j = 1, m .
3
∞
4/3
u3 *-weakly in (0, T; (Ω)). There-
t
∫ ds 〈 u
3 m
3
– u , w j〉
+0
0
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AN INITIAL–BOUNDARY VALUE PROBLEM 1, p
1863
4
as m +∞ for all t ∈ (0, T). Since 0 (Ω) ⊂ (Ω), we see that the right-hand side of the last inequality tends to zero as m +∞. 1, p
1
u0 strongly in 0 (Ω) ⊂ 0 (Ω), we have 〈∆u0m, wj 〉
Since u0m
〈∆u0, wj 〉 and 〈∆pu0m, wj 〉
〈∆pu0,
wj 〉, j = 1, m . Passage to the limit as m
+∞ in (2.17) yields t
t
t
∫
∫
∫
0
0
0
3
∆u – χ = ∆u 0 + ∆ p u 0 + dsu – ds ( u x1 + uu x1 ) – dsu ( s ).
(2.18)
Define (v) = –∆pv. It follows from (2.17) that t
〈 ( u m ), u m〉 = 〈 ∆u m, u m〉 +
∫ dsu
3 m ( s ),
um ( t )
0 t
∫ ds ( u
+
m
– u m x1 – u m u m x1 ) ( s ), u m ( t ) – 〈 ∆u 0m + ∆ p u 0m, u m〉
0
(2.19)
t
= – ∇u m
2 2
+
∫ dsu
3 m ( s ),
u m ( t ) – 〈 ∆u 0m + ∆ p u 0m, u m〉
0 t
+
t
∫ dsu
m ( s ),
∫ ds ( u
um ( t ) +
0
Since um
m x1 ( s )
+ u m u m x1 ( s ) ), u m ( t ) .
0 1, p
1
u weakly in 0 (Ω) ⊂ 0 (Ω) for almost all t ∈ (0, T), we have 2
2
lim inf ∇u m 2 ( t ) ∇u 2 ( t ). m → +∞
2
u strongly in (Q), we have
Moreover, since um
t
t
∫ dsu
m ( s ),
∫ dsu ( s ), u ( t )
um ( t )
0
t m x1 ( s )
+ u m u m x1 ( s ) ), u m ( t )
0
as m
+∞,
0
t
∫ ds ( u
as m
∫ ds ( u
x1 ( s )
+ uu x1 ( s ) ), u ( t )
0
+∞. Moreover, T
t
∫ dt ∫ 0
T 3 dsu m ( s ),
um ( t ) =
0
t
∫ dt ∫ 0
T 3 dsu m ( s ),
t
∫ ∫ dsu
u m ( t ) – u ( t ) + dt
0
0
3 m ( s ),
u(t ) .
0
1
It was previously proved that the sequence {um} converges weakly in (Q). Therefore, there is a subse4 ∞ 4 quence of {um} that converges strongly in (Q). Moreover, the sequence is bounded in (0, T; (Ω)). Then, T
t
∫ dt ∫ dsu 0
3 m ( s ),
um ( t ) – u ( t )
+0 as m
+∞.
0
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Indeed, we have 3 〈 u m ( s ),
T
⎛ 4⎞ u m ( t ) – u ( t )〉 ⎜ d x u m ( s ) ⎟ ⎝Ω ⎠
∫
t
∫ dt ∫ 0
3/4
⎛ 4⎞ ⎜ dx um ( t ) – u ( t ) ⎟ ⎝Ω ⎠
∫
T
⎛ 4⎞ u m ( t ) – u ( t ) C ( T ) ⎜ dt u m ( t ) – u ( t ) 4 ⎟ ⎝0 ⎠
0
,
1/4
∫
3 dsu m ( s ),
1/4
.
Now, consider T
t
∫ dt ∫ dsu 0
∞
Since u(t) ∈ (0, T;
1, p 0 (Ω))
3 m ( s ),
u(t ) .
0
∞
4
⊂ (0, T; (Ω)), t
∫ ds u
q m
∞
4/3
u m ( s ) ∈ ( 0, T ; ( Ω ) ),
0
and
3 um
∞
4/3
u3 *-weakly in (0, T; (Ω)), we conclude that T
t
∫ dt ∫
lim
m → +∞
0
T 3 dsu m ( s ),
u(t ) =
0
t
∫ dt ∫ dsu ( s ), u ( t ) . 3
0
0
Let T
m = ∫ dt 〈 ( u m ) – ( v ), u m – v 〉 0 1, p
r
r'
∀v ∈ ( 0, T ; 0 ( Ω ) )
∀( v ) ∈ ( 0, T ;
r ∈ ( 1, ∞ ),
– 1, p'
( Ω ) ),
r r' = ----------- . r–1
It follows from (2.19) that T
T
m = – ∫ dt
2 ∇u m 2
0 t
T 2 dsu m u m ( s ),
∫ ∫ dsu
m ( s ),
0
∫
u m ( t ) – dt 〈 ∆u 0m + ∆ p u 0m, u m〉
0
0 T
– dt 0
∫ ∫
+ dt
0 T
t
t
∫ ∫ ds ( u
u m ( t ) – dt 0
m x1 ( s )
+ u m u m x 1 ( s ) ), u m ( t )
(2.20)
0
T
T
∫
∫
0
0
– dt 〈 ( u m ), v 〉 – dt 〈 ( v ), u m – v 〉 . In this expression, it remains to consider the term m: T
T
∫ dt 〈 ∆u 0
0m
+ ∆ p u 0m, u m〉 =
∫ dt 〈 ∆u
T 0m,
0
∫
u m〉 + dt 〈 ∆ p u 0m, u m〉 .
(2.21)
0
∞
Consider the first term in (2.21). The functions um are uniformly bounded with respect to m in (0, T; 1, p
∞
1
0 (Ω)) ⊂ (0, T; 0 (Ω)). On the other hand, u0m
1, p
1
u0 strongly in 0 (Ω) ⊂ 0 (Ω) and, since
COMPUTATIONAL MATHEMATICS AND MATHEMATICAL PHYSICS
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AN INITIAL–BOUNDARY VALUE PROBLEM 1
1865
–1
∆ ∈ ( 0 (Ω)); (Ω) (see [3]) is a set of linear continuous (bounded because ∆ is linear) operators from 1
–1
–1
0 (Ω) to (Ω), we conclude that ∆u0m
∆u0 strongly in (Ω) and
〈 ∆u 0m – ∆u 0, u m〉 C ∆u 0m – ∆u 0 Since um
∞
1
as m
0
–1
+∞.
–1
u *-weakly in (0, T; 0 (Ω)) and ∆u0 ∈ (Ω), we have T
T
∫ dt 〈 ∆u , u 0
∫ dt 〈 ∆u , u〉
m〉
as m
0
0
+∞.
0
In this case, T
T
∫ dt 〈 ∆u
lim
m → +∞
0m,
u m〉 =
0
∫ dt 〈 ∆u , u〉 . 0
0
∞
Consider the second term in (2.21). The functions um are uniformly bounded with respect to m in (0, T; 1, p
1, p
0 (Ω)). On the other hand, u0m
1, p
p
u0 strongly in 0 (Ω). Since ∂/∂xi ∈ ( 0 (Ω); (Ω)), i = 1, N , p
∂u0 /∂xi strongly in (Ω) for all i = 1, N . It is true that
we conclude that ∂u0m /∂xi
p–2
w
p–2
w– v
v
p/ ( p – 1 )
( p – 1) f (w – v )
p/ ( p – 1 ) ,
where f = max(|w|p – 2, |v |p – 2), f (w – v )
p1
f
p–2
w– v f
p 1 = p/ ( p – 2 ),
q2 p1 ,
q 1 = ( p – 1 )/ p – 2,
q 2 = p – 1, w
w–v
q1 p1
p–2
v
p/ ( p – 2 )
p/ ( p – 1 )
q 2 p 1 = p,
q 1 p 1 = p/ ( p – 2 ),
( p – 1) f
p/ ( p – 2 )
2 ( max { w p, v p } )
p–2
w – v p,
.
If v = ∂u0m /∂xi and w = ∂u0 /∂xi , we obtain ∂u 0m ----------∂x i p'
p–2
Since ∂/∂xi ∈ ( (Ω),
∂u 0m ----------∂x i
∂u 0 -------∂x i
p–2
∂u 0 -------∂x i
–1, p'
Since um
∆pu0 strongly in
(Ω)), it follows that ∆pu0m
〈 ∆ p u 0m – ∆ p u 0, u m〉 C ∆ p u 0m – ∆ p u 0 ∞
0
– 1, p'
1
u *-weakly in (0, T; 0 (Ω)) and ∆pu0 ∈ T
p p' = ------------ . p–1
p'
( Ω ),
strongly in
–1, p'
as m
(Ω) and
+∞.
–1, p'
(Ω), we conclude that
T
∫ dt 〈 ∆ u , u p 0
∫ dt 〈 ∆ u , u〉 as
m〉
m
p 0
0
+∞.
0
Therefore, T
lim
T
∫ dt 〈 ∆ u
p 0m,
m → +∞
0
u m〉 =
∫ dt 〈 ∆ u , u 〉 . p 0
0
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Passage to the upper limit in (2.20) as m
+∞ gives
T
T
∫
0 lim sup m – dt ∇u m → +∞
2 2
t
T
∫ ∫ dsu u ( s ), u ( t ) – ∫ dt 〈 ∆u 0
∫ ∫ dsu ( s ), u ( t ) – ∫ dt ∫ ds ( u 0
0
0
x1 ( s )
0
+ ∆ p u 0, u〉
0
t
– dt 0
T 2
+ dt
0 T
t
T
T
∫
∫
0
0
+ uu x1 ( s ) ), u ( t ) – dt 〈 χ, v 〉 – dt 〈 ( v ), u – v 〉 .
0
In view of (2.18), we conclude that T
T
∫ dt 〈 χ, u〉
∫
T 2 2
= – dt ∇u
0
T
∫ ∫ dsu u ( s ), u ( t ) – ∫ dt 〈 ∆u 2
+ dt
0 T
t
0
0
t
t
∫ ∫ dsu ( s ), u ( t ) – ∫ dt ∫ ds ( u 0
+ ∆ p u 0, u〉
0
T
– dt 0
0
0
x1 ( s )
+ uu x1 ( s ) ), u ( t ) .
0
Therefore, T
∫ dt 〈 χ – ( v ), u – v 〉
= 0
0 1, p
r
∀v ∈ ( 0, T ; 0 ( Ω ) ),
1, p
r
w ∈ ( 0, T ; 0 ( Ω ) ),
where v = u – λw,
r > 1.
Following the standard procedure yields T
∫ dt 〈 χ – ( u ), w〉
1, p
r
∀w ∈ ( 0, T ; 0 ( Ω ) ),
= 0
0 ∞
χ, ( u ) ∈ ( 0 , T ;
– 1, p'
r'
( Ω ) ) ⊂ ( 0, T ;
– 1, p'
( Ω ) ).
Hence, (2.16) holds true. Now, we prove that ∂u ------∂x i
( p – 2 )/2
1 2 ∂u ------- ∈ ( 0, T ; ( Ω ) ), ∂x i
i = 1, N .
Let ∂ ∂x i ∂x i
∂v i ( v ) ≡ – ------- ⎛⎝ ------
p–2
∂v -------⎞ , ∂x i⎠
p
∂v , 〈 i ( v ), v 〉 = -----∂x i p
i = 1, N .
1, p
p
u weakly in (0, T; 0 (Ω)), we have
Since um
T
∫
lim sup dt 〈 i ( u m ), u m〉 – m → +∞ T
T
∫
∫
lim sup 〈 ( u m ), u m〉 – dt ∇u 0 T
t
∫ ∫
– dt 0
0
∑ ∫
j = 1, j ≠ i 0
0
m → +∞
T
N
0 T
t
∫ ∫
dsu ( s ), u ( t ) – dt 0
0
T 2 2
∂u dt ------∂x j
p p
T
∫
+ lim sup dt 〈 ( u m ), u m〉 , m → +∞
0
t
T
∫ ∫ dsu u ( s ), u ( t ) – ∫ dt 〈 ∆u 2
+ dt 0
0
0
+ ∆ p u 0, u〉
0 T
T
N
∫
∫ ∑
0
0
ds ( u x1 ( s ) + uu x1 ( s ) ), u ( t ) – dt 〈 ∆ p u, u〉 =
COMPUTATIONAL MATHEMATICS AND MATHEMATICAL PHYSICS
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j=1
Vol. 48
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∂u ------- . ∂x j p No. 10
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AN INITIAL–BOUNDARY VALUE PROBLEM
1867
Thus, we obtain T
lim sup dt ∂u --------mm → +∞ ∂x i
∫ 0
∞
p p
T
p
∂u . dt -----∂x i p
∫
(2.22)
0
p
p
On the other hand, ∂um /∂xi is bounded in (0, T; (Ω)) ⊂ (Q), where Q = (0, T) × Ω. Therefore, since p (Q) is reflexive, there is a subsequence (again denoted by um) such that ∂u m --------∂x i
∂u ------∂x i
p
weakly in (Q), i = 1, N .
(2.23)
Hence, we obtain T
lim inf dt ∂u --------mm → +∞ ∂x i
∫ 0
p p
T
p
∂u , dt -----∂x i p
∫
i = 1, N .
(2.24)
0
Combining (2.22) with (2.24) gives T
∫
lim
m → +∞
0
∂u m dt -------∂x i
p p
T
p
∂u , = dt -----∂x i p
∫
i = 1, N .
(2.25)
0
From (2.23) and (2.25), it follows that ∂u m --------∂x i
∂u ------∂x i
p
strongly in (Q), i = 1, N ,
(2.26)
which means that ∂u m --------∂x i
∂u ------∂x i
i = 1, N .
almost everywhere in Q ≡ (0, T) × Ω,
(2.27)
Passing to a subsequence, we obtain ∂u m --------∂x i
∂u m --------∂x i
χi *-weakly in (0, T; (Ω)), i = 1, N .
∂u m --------∂x i
∂u ------∂x i
( p – 2 )/2
∞
2
By virtue of (2.27), ∂u m --------∂x i
( p – 2 )/2
( p – 2 )/2
∂u ------∂x i
∞
2
*-weakly in (0, T; (Ω)).
Therefore, ∂ ∂u m ----- ⎛ -------∂t ⎝ ∂x -i
( p – 2 )/2
∂u m⎞ --------∂x i ⎠
∂ ∂u ----- ⎛ -----∂t ⎝ ∂x-i
( p – 2 )/2
∂u -------⎞ ∂x i⎠
2
in '(0, T; (Ω)).
By virtue of (2.13), there is a subsequence such that ∂ ∂u m ----- ⎛ -------∂t ⎝ ∂x -i
( p – 2 )/2
∂u m⎞ --------∂x i ⎠
2
ψ weakly in (Q), Q = (0, T) × Ω,
which means that ∂ ∂u ψ = ----- ⎛ -----∂t ⎝ ∂x-i
( p – 2 )/2
2 ∂u -------⎞ ∈ ( Q ), ∂x i⎠
Q = ( 0, T ) × Ω,
COMPUTATIONAL MATHEMATICS AND MATHEMATICAL PHYSICS
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Thus, ∂ ∂u ----- ⎛ -----∂t ⎝ ∂x-i
( p – 2 )/2
2 ∂u -------⎞ ∈ ( Q ), ⎠ ∂x i
∂u ------∂x i
( p – 2 )/2
∂u ------∂x i
( p – 2 )/2
∞ 2 ∂u ------- ∈ ( 0, T ; ( Ω ) ), ∂x i
(2.28)
1 2 ∂u ------- ∈ ( 0, T ; ( Ω ) ). ∂x i
In a similar manner, with the help of (2.27), we prove that ∂u m --------∂x i
p–2
∂u m --------∂x i
p–2
∂u ------∂x i
∂u ------∂x i
∞
p'
*-weakly in (0, T; (Ω)).
From (2.28) and the embedding theorems, it follows that ∂ ∂u ----- ⎛ -----∂t ⎝ ∂x-i ∂u ------∂x i
p–2
p–2
2 p' ∂u -------⎞ ∈ ( 0, T ; ( Ω ) ), ⎠ ∂x i
(2.29)
∞ p' ∂u ------- ∈ ( 0, T ; ( Ω ) ). ∂x i
Let us prove (2.29). Let ψ ≡ |∂u/∂xi |(p – 2)/2∂u/∂xi . Then, |ψ|αψ = |∂u/∂xi |p – 2∂u/∂xi, where α = (p – 2)/p and i = 1, N . Note that ∞ p' |ψ|αψ ∈ (0, T; (Ω)). 2 p We prove that |ψ|αψ ) t' ∈ (0, T; (Ω)). Indeed, T
⎛ dt ⎜ d x ψ ⎝Ω 0
∫ ∫
p'⎞
T
2/ p'
ψ 't ⎟ ⎠
⎛ dt ⎜ d x ψ ⎝Ω 0
p' p 2 = 2,
p 1 + p 2 = 1,
α p'
∫ ∫
α p' p 1⎞
2/ ( p' p 1 )
⎟ ⎠
⎛ ⎜ d x ψ 't ⎝Ω
∫
p' p 2⎞
2/ ( p' p 2 )
⎟ ⎠
,
where α p' p 1 = 2, ∞
–1
2
–1
2
2( p – 1) p 1 = -------------------- , p–2
2( p – 1) p 2 = -------------------- . p
2
Since ψ ∈ (0, T; (Ω)) and ψ 't ∈ (0, T; (Ω)), we have T
⎛ dt ⎜ d x ψ ⎝Ω 0
∫ ∫
α p'
p'⎞
2/ p'
ψ t' ⎟ ⎠
sup ψ t ∈ ( 0, T )
2α 2
∫ d x dt ψ '
2
t
< +∞.
Q
+∞ in Eq. (2.2). Indeed, setting ϕ(t) = 1 in Eq. (2.2) gives the
Now, we can pass to the limit as m integral equality
t
∫
3
∆u m + ∆ p u m = ∆u m0 + ∆ p u m0 – ds [ u m x1 + u m u m x1 + u m – u ], 0
which is understood in the sense of sage to the limit as m +∞ yields
–1, p'
(Ω) for almost all t ∈ (0, T). In view of the previous results, past
∫
3
∆u + ∆ p u = ∆u 0 + ∆ p u 0 – ds [ u + u x1 + uu x1 – u ] 0
in the sense of
–1, p'
(Ω) for almost all t ∈ (0, T). This relation implies that 1
∆u + ∆ p u ∈ ( 0, T ;
– 1, p'
(1)
( Ω ) ) ∩ w ( [ 0, T ];
– 1, p'
COMPUTATIONAL MATHEMATICS AND MATHEMATICAL PHYSICS
( Ω ) ). Vol. 48
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AN INITIAL–BOUNDARY VALUE PROBLEM
1869
Note the pointwise equality d 3 ----- [ ∆u m + ∆ p u m ] = –u m – u m x1 – u m u m x1 + u m . dt Since ∆u m + ∆ p u m 2
weakly in (0, T;
∆u + ∆ p u
–1, p'
(Ω)), we have d ----- ( ∆u + ∆ p u ) dt
d ----- ( ∆u m + ∆ p u m ) dt in the sense of '(0, T;
–1, p'
(Ω)). On the other hand,
3
3
um 2
weakly in (0, T;
u ,
u m x1
u x1 ,
u m u m x1
uu x1 ,
–1, p'
(Ω)). Therefore, d/dt ( ∆u m + ∆ p u m )
2
weakly in (0, T;
2
ξ
–1, p'
(Ω)). Thus, d ----- ( ∆u m + ∆ p u m ) dt
weakly in (0, T;
u
um
d ----- ( ∆u + ∆ p u ) dt
–1, p'
(Ω)).
Now, passage to the limit in (2.2) produces T
∫ dtϕ ( t )
d 3 ---- 〈 ∆u + ∆ p u, w j〉 + 〈 –u + u x1 + uu x1 + u , w j〉 = 0 dt
0
2
∀ϕ(t) ∈ (0, T), j = 1, +∞ . This relation is equivalent to T
∫ dt
d 3 ---- [ ∆u + ∆ p u ] + u – u + u x1 + uu x1, v dt
1, p
2
∀v ∈ ( 0, T ; 0 ( Ω ) ).
0
Now, we prove the uniqueness of a weak generalized solution. Assume that ui (i = 1, 2) are two weak generalized solutions to the problem T
∫ dtϕ ( t )
d 3 ---- ( 〈 ∆u, w〉 0 + 〈 ∆ p u, w〉 ) + ( u , w ) – ( u, w ) + 〈 u x1, w〉 0 + 〈 uu x1, w〉 0 = 0 dt
0
2
∀ϕ ( t ) ∈ ( 0, T ),
1, p
1, p
∀w ∈ 0 ( Ω ),
u ( 0 ) = u 0 ∈ 0 ( Ω ),
in the classes defined in the conditions of the theorem with the same initial data. If ϕ(t) = 1, we find that t
∆u i + ∆ p u i = ∆u 0 + ∆ p u 0 –
∫ 0
3 dsu i ( s )
t
t
∫
∫
0
0
+ dsu i – ds ( u i x1 + u i u i x1 ),
COMPUTATIONAL MATHEMATICS AND MATHEMATICAL PHYSICS
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KORPUSOV, SVESHNIKOV
which yields – 〈 ∆u 1 – ∆u 2, u 1 – u 2〉 – 〈 ∆ p u 1 – ∆ p u 2, u 1 – u 2〉 t
=
∫ ds ∫ dx ( u ( s ) – u ( s ) ) ( u ( t ) – u ( t ) ) 3 1
3 2
1
2
Ω
0 t
∫ ∫
– ds d x ( u 1 ( s ) – u 2 ( s ) ) ( u 1 ( t ) – u 2 ( t ) ) Ω
0 t
∫ ∫
+ ds d x ( u 1 x 1 ( s ) – u 2 x 1 ( s ) ) ( u 1 ( t ) – u 2 ( t ) ) Ω
0 t
1 2 2 + ds d x --- ( u 1 ( s ) – u 2 ( s ) ) ( u 1 x1 ( t ) – u 2 x1 ( t ) ). 2
∫ ∫ 0
Ω
Since the operator ∆p is monotone, applying the embedding theorems gives the inequality t
∇u 1 –
2 ∇u 2 2
∫ ∫
2
2
3 ds d x max { u 1 ( s ), u 2 ( s ) } u 1 ( s ) – u 2 ( s ) u 1 ( t ) – u 2 ( t ) 0
Ω
t
t
∫ ∫
∫ ∫
+ ds d x u 1 ( s ) – u 2 ( s ) u 1 ( t ) – u 2 ( t ) + ds d x ∇u 1 ( s ) – ∇u 2 ( s ) u 1 ( t ) – u 2 ( t ) 0
Ω
Ω
0 t
∫ ∫
+ ds d x max { u 1 ( s ) , u 2 ( s ) } u 1 ( s ) – u 2 ( s ) ∇u 1 ( t ) – ∇u 2 ( t ) 0
Ω t
∫
C ( T ) ∇u 1 – ∇u 2 2 ( t ) ds ∇u 1 – ∇u 2 2 ( s ), 0
which implies that u1 = u2 almost everywhere in (0, T) × Ω. The theorem is proved. 3. UNIQUE SOLVABILITY OF THE PROBLEM IN ANY FINITE CYLINDER AND FINITE-TIME BLOWUP OF ITS SOLUTION We introduce the following notation: 1 E m ( t ) ≡ --- ∇u m 2
2 2
p–1 + -----------p
1 E ( t ) ≡ --- ∇u 2
2 2
N
∑ i=1
p
∂u m , --------∂x i p
p–1 + -----------p
N
∑ i=1
E 0m ≡ E m ( 0 ), p
∂u ------- , ∂x i p
p 4 B 2 ≡ B 1 ⎛ ------------⎞ ⎝ p – 1⎠ 1, p
4 α = --- , p
E 0 ≡ E ( 0 ),
4/ p
, 4
where B1 is the best constant for the embedding 0 (Ω) ⊂ (Ω). Theorem 2. Let all the assumptions of Theorem 1 be satisfied. Then, the following assertions hold true: (i) If p > 4, then T0 = +∞ and 1–α
E ( t ) [ E0
+ ( 1 – α )B 2 t ) ]
1/ ( 1 – α )
,
α = 4/ p.
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AN INITIAL–BOUNDARY VALUE PROBLEM
1871
(ii) If p = 4, then T0 = +∞ and E ( t ) E 0 exp { B 2 t }. (iii) If p ∈ (2, 4) and γ E' ( 0 ) > ⎛ --------------⎞ ⎝ α 1 – 1⎠
1/2
β E 0 + --------------E 0 , α1 – 1
then T0 ∈ [T1, T2], where E' ( 0 ) = u 0
4 4
4+ p α 1 = ------------, 2p
2
– u0 2 ,
1 E 0 = E ( 0 ) = --- ∇u 0 2
2 2 ( B3
+ 3) β = -----------------------, 4– p
+ 3) γ = --------------------------------- , 4– p
1 – α1
2
– 2α 1
N
∑ i=1
p
∂u ------- , ∂x i p
p ∈ ( 2, 4 ),
p 4/ p ------------⎞ , ⎝ p – 1⎠
q + 2⎛
B2 ≡ B1
T 2 = E0 2
p–1 + -----------p
2 2 16B 3 ( B 3
1 –1 1 – α T 1 = ------------ B 2 E 0 , α–1
A ≡ ( 1 – α1 ) E
2 2
–1
A ,
2 β γ 2 ( 0 ) ⎛ E' ( 0 ) – --------------E 0⎞ – -------------- ( E ( 0 ) ) , ⎝ α1 – 1 ⎠ α1 – 1 1
2
and B3 is the best constant for the embedding 0 (Ω) ⊂ (Ω). 1, p
u0 strongly in 0 (Ω)
Proof. Let um be the Galerkin approximations defined by (2.4). Since u0m 1, p
u weakly in 0 (Ω) for almost all t ∈ (0, T), passing to the limit as m and um (2.9) and taking into account E(t) lim inf E m ( t ) , we obtain
+∞ in (2.8) and
m → +∞
E(t )
1–α [ E0
+ ( 1 – α )B 2 t ]
1 -----------1–α
E ( t ) E 0 exp { B 2 t }
for α < 1,
for α = 1.
Consider the case α > 1. 1, p
1
u0 strongly in 0 (Ω) ⊂ 0 (Ω), we have E0m E0. In this case, the number sequence Since u0m E0m contains either a monotonically nondecreasing or monotonically nonincreasing subsequence. We retain the same notation for the desired subsequence of E0m and for the corresponding subsequences of u0m and um. {n}
Consider inequality (2.10). Assume that E0m is a monotonically nonincreasing sequence. Let { E 0m } be the sequence obtained from E0m by deleting its first n ∈ members. Then, (2.10) holds uniformly with respect {n}
{n}
–1
1–α
to m for all t ∈ [0, T 1 ), where T 1 ≡ (α – 1)–1 B 2 E 0n . Since um passage to the limit as m +∞ in (2.10) for all such t yields α–1
E ( t ) E 0 [ 1 – ( α – 1 )B 2 E 0
t]
– 1/ ( α – 1 )
for
∞
1, p
u *-weakly in (0, T; 0 (Ω)), {n}
t ∈ [ 0, T 1 ).
(3.1)
{n}
Since n ∈ is arbitrary and T 1 ↑ T1 as n +∞, we conclude that (3.1) holds for t ∈ [0, T1); moreover, T0 T1. Now, let E0m be a monotonically nondecreasing sequence. Then, repeating the previous argument, we find that (3.1) holds for all t ∈ [0, T1) and T0 T1. For the further analysis, we need to prove that Em(t)
E(t) for almost all t ∈ [0, T] ∀T ∈ (0, T0).
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For this purpose, we note that (2.26) was derived in the proof of Theorem 1. Therefore, taking into p 2 account (Ω) ⊂ (Q), where Q = (0, T) × Ω ∀T ∈ [0, T0), we have t
t
∫ dsE
∫ dsE ( s )
m(s)
0
∀t ∈ [ 0, T ],
∀T ∈ ( 0, T 0 ).
(3.2)
0
1
4
On the other hand, under the conditions of Theorem 1, we have the compact embedding (Q) ⊂ (Q) and 1 4 u strongly in (Q), where Q = (0, T) × Ω ∀T ∈ the sequence um converges weakly in (Q). Hence, um (0, T0). Since the right-hand side of the expression for Em(t) converges to a function η(t) for any t ∈ [0, T] with T ∈ (0, T0) and since the proof of Theorem 1 implies that Em(t) C < +∞, where C is independent of m or t, the Lebesgue theorem on passage to the limit under the sign of the Lebesgue integral gives η(t) ∈ 1 (0, T) ∀T ∈ (0, T0) and t
t
∫ dsE
∫ dsη ( s )
m(s)
0
∀t ∈ [ 0, T ],
∀T ∈ ( 0, T 0 ).
(3.3)
0
A comparison of (3.2) with (3.3) suggests that t
∫ ds [ E ( s ) – η ( s ) ] = 0
∀t ∈ [ 0, T ],
∀T ∈ ( 0, T 0 ).
(3.4)
0
1
1
Moreover, E(t) ∈ (0, T) and η(t) ∈ (0, T). Therefore, η(t ) = E(t )
∀t ∈ [ 0, T ],
∀T ∈ ( 0, T 0 ).
Thus, by virtue of (3.4), we conclude that, under additional conditions on q > 0, the sequence Em(t) converges to E(t) almost everywhere on the set of t ∈ [0, T] ∀T ∈ (0, T0). 1, p
(1)
Taking into account (2.6), (2.11), and um(t) ∈ ([0, T]; 0 (Ω)), we have 1d --- ----- ∇u m 2 dt
2 2
p–1d + ------------ ----p dt
N
∑ i=1
N
2 ∇u 'm 2
+ ( p – 1)
∑∫
i = 1Ω
1d = --- ----- u m 4 dt
4 4
∂u m --------∂x i
d x ∂u --------m∂x i
p
2 2
+ um
4
= u m 4 ( s ),
(3.5)
p p–2
2
d ∂u 'm + 1--- ---- u --------2 dt m ∂x i
2 2
(3.6)
1 2 + d xu 'm x1 u + --- d xu m u 'm x1 . 2
∫
∫
Ω
Ω
It follows from (3.5) and (3.6) that Em(t), E m' (t) ∈ [0, T]. It holds that 2
∫ dx ( ∇u' , ∇u m
m)
∇u m'
2 2
2
∇u m 2 ,
(3.7)
Ω
∫
Ω
d x ∂u --------m∂x i
p–2
∂u m ∂u 'm --------- --------- d x ∂u --------m∂x i ∂x i ∂x i
∫
Ω
p–1
∂u m' ∂u m ----------------∂x i ∂x i
p/2 ⎛ p
∂u ⎜ d x --------m∂x i ⎝Ω
∫
p–2
2 ∂u m' ⎞ --------- ⎟ ∂x i ⎠
1/2
.
(3.8)
In view of the Schwarz inequality, we have N
∑ i=1
∂u m --------∂x i
p/2 ⎛ p
∂u ⎜ d x --------m∂x i ⎝Ω
∫
p–2
2 ∂u 'm ⎞ --------- ⎟ ∂x i ⎠
1/2
p ⎛ ∂u m ⎞ ⎜ --------- ⎟ ⎝ i = 1 ∂x i p⎠ N
∑
1/2
⎛ ∂u m d x -------⎜ ∂x i ⎝i = 1 Ω N
∑∫
p–2
COMPUTATIONAL MATHEMATICS AND MATHEMATICAL PHYSICS
2 ∂u 'm ⎞ --------- ⎟ ∂x i ⎠
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2008
AN INITIAL–BOUNDARY VALUE PROBLEM
1873
By virtue of (3.5)–(3.9), it is true that N
2
2
〈 u m, [ –∆u m – ∆ p u m ] 't〉 〈 ∇u 'm, ∇u m〉 + ( p – 1 )
∑∫
2
i = 1Ω N
d x ∂u --------m∂x i
∑∫
+ 2 ( p – 1 ) ( ∇u 'm, ∇u m )
i = 1Ω N
∇u 'm
2 2
2 ∇u m 2
+ ( p – 1)
2
∑
p ⎛ ∂u m ⎞ ∇u m 2 ⎜ --------- ⎟ ⎝ i = 1 ∂x i p⎠ N
+ 2 ( p – 1 ) ∇u 'm ⎛ ⎜ ∇u m ⎝
2
∑
N
2 2
+ ( p – 1)
∑ i=1
⎛ p ⎜ ∇u 'm ⎝
p ∂u m ⎞ ⎛ ∇u ' --------- ⎟ ⎜ m ∂x i p⎠ ⎝
N
2 2
+ ( p – 1)
∑∫
i = 1Ω
d x ∂u --------m∂x i
p–2
p N
∂u m --------∂x i
i=1
p–2
2
p–2
∂u 'm --------∂x i
⎛ d x ∂u ⎜ --------m∂x i ⎝i = 1 Ω
p–2
2 ∂u 'm ⎞ --------- ⎟ ∂x i ⎠
∑∫ N
∑∫
+ ( p – 1)
∑∫
i = 1Ω
2 ∂u 'm ⎞ ⎛ 1--- ∇u --------- ⎟ ⎜ 2 m ∂x i ⎠ ⎝
2
∂u m ∂u 'm --------- --------∂x i ∂x i
N
2 2
∂u m ∂u m' --------- --------∂x i ∂x i
d x ∂u --------m∂x i
pi = 1 Ω
1/2
p–2
d x ∂u --------m∂x i
2 2
p–2
d x ∂u --------m∂x i p–1 + -----------p
N
∑ i=1
1/2
2 ∂u 'm ⎞ --------- ⎟ ∂x i ⎠ p ∂u m ⎞ . --------- ⎟ ∂x i p⎠
Thus, we obtain the inequality 2
( E m' ) pE m ( t )J m ( t ),
(3.10)
where N
J m ( t ) = ∇u 'm
2 2
+ ( p – 1)
∑∫
i = 1Ω
d x ∂u --------m∂x i
p–2
2
∂u 'm . --------∂x i
Now, we use (3.5) and (3.6) to derive an upper bound for Jm. Specifically, 1 1d J m = --- E m'' + --- ----- u m 4 4 dt
2 2
1 2 + d xu 'm x1 u + --- d xu 'm x1 u 2
∫
∫
Ω
Ω
1 1 2 ε 1 2 ε 1 ε 2 --- E m'' + --- B 3 J m + ----- B 3 E m + --- J m + --- B 3 E m + --- J m + ----- u m 4 2ε 2 ε 4 4ε 4
(3.11)
4 4
2
2B 1 1 ε 2 --- E m'' + --- ( B 3 + 3 )J m + --------3-E m + -----E m' . 4 4ε ε 4 Combining (3.10) with (3.11) yields [1] 2
2
E m E m'' – α 1 ( E m' ) + βE m' E m + γ E m 0, where 4+ p α 1 = ------------, 2p
2
2 ( B3 + 3 ) -, β = ---------------------4– p
2
2
16B 3 ( B 3 + 3 ) γ = -------------------------------4– p
for p < 4. Then, α1 > 1 and β, γ > 0. COMPUTATIONAL MATHEMATICS AND MATHEMATICAL PHYSICS
Vol. 48
No. 10
2008
1874
KORPUSOV, SVESHNIKOV
Proceeding as in [4], we complete the proof of the theorem. ACKNOWLEDGMENTS This work was supported by the Russian Foundation for Basic Research (project no. 08-01-00376-a) and by the Russian Federation President’s Program for the Support of Young Doctors of Sciences (project no. MD-1006.2007.1). REFERENCES 1. V. K. Kalantarov and O. A. Ladyzhenskaya, “Blowup Formation in Parabolic and Hyperbolic Quasilinear Equations,” Zap. LOMI 69, 77–102 (1977). 2. J.-L. Lions, Quelques méthodes de résolution des problémes aux limites non linéires (Dunod, Paris, 1969; Mir, Moscow, 1972). 3. J.-L. Lions and E. Magenes, Problemes aux limites non homogénes et applications (Dunod, Paris, 1968; Mir, Moscow, 1971). 4. A. G. Sveshnikov, A. B. Al’shin, M. O. Korpusov, and Yu. D. Pletner, Linear and Nonlinear Equations of Sobolev Type (Fizmatlit, Moscow, 2006) [in Russian].
COMPUTATIONAL MATHEMATICS AND MATHEMATICAL PHYSICS
Vol. 48
No. 10
2008
