Electronic Journal of Differential Equations, Vol. 2003(2003), No. 63, pp. 1–11.
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
ftp ejde.math.swt.edu (login: ftp)
WEAK SOLUTIONS FOR A VISCOUS p-LAPLACIAN EQUATION
CHANGCHUN LIU
Abstract. In this paper, we consider the pseudo-parabolic equation ut −
k∆ut = div(|∇u|p−2 ∇u). By using the time-discrete method, we establish
the existence of weak solutions, and also discuss the uniqueness.
1. Introduction
This paper concerns the study of the viscous p-Laplacian equation
∂u
∂∆u
−k
= div(|∇u|p−2 ∇u),
∂t
∂t
with boundary condition
u = 0,
x ∈ Ω, p > 2,
∂Ω
(1.1)
(1.2)
and initial condition
u(x, 0) = u0 (x),
x ∈ Ω.
(1.3)
N
Here Ω is a bounded domain in R and k > 0 is the viscosity coefficient. The
term k ∂∆u
∂t in (1.1) is interpreted as due to viscous relaxation effects, or viscosity;
hence, the equation (1.1) is called “viscous p-Laplacian equations”. The well-known
p-Laplacian equation is obtained by setting k = 0.
Equation (1.1) arises as a regularization of the pseudo-parabolic equation
∂∆u
∂u
−k
= ∆u,
(1.4)
∂t
∂t
which arises in various physical phenomena. (1.4) can be assumed as a model
for diffusion of fluids in fractured porous media [1, 5, 4], or as a model for heat
conduction involving a thermodynamic temperature θ = u − k∆u and a conductive
temperature u [10, 3]. Equation (1.4) has been extensively studied, and there are
many outstanding results concerning existence, uniqueness, regularity, and special
properties of solutions, see for example [4, 5, 6, 7, 8, 9, 11].
To derive (1.4), B. D. Coleman, R. J. Duffin and V. J. Mizel considered a special
kinematical situation, of nonsteady simple shearing flow [4]. In fact, when the
influence of many factors, such as the molecular and ion effects, are considered,
one has the nonlinear relation div(|∇u|p−2 ∇u) in stead of ∆u in right-hand side of
(1.4). Hence, we obtain (1.1).
2000 Mathematics Subject Classification. 35G25, 35Q99, 35K55, 35K70.
Key words and phrases. Pseudo-parabolic equations, existence, uniqueness.
c
2003
Southwest Texas State University.
Submitted August 5, 2002. Published June 10, 2003.
1
2
CHANGCHUN LIU
EJDE–2003/63
Equation (1.1) is something like the p-Laplacian equation, but many methods
which are useful for the p-Laplacian equation are no longer valid for this equation.
Because of the degeneracy, problem (1.1)-(1.3) does not admit classical solutions in
general. So, we study weak solutions in the sense of following
Definition A function u is said to be a weak solution of (1.1)-(1.3), if the following
conditions are satisfied:
0
∞
−1,p
(1) u ∈ L∞ (0, T ; W01,p (Ω)) ∩ C(0, T ; H 1 (Ω)), ∂u
(Ω)), where
∂t ∈ L (0, T ; W
0
p is conjugate exponent of p.
(2) For ϕ ∈ C0∞ (QT ) and QT = Ω × (0, T ),
ZZ
ZZ
ZZ
∂ϕ
∂∇ϕ
∇u
dx dt −
|∇u|p−2 ∇u∇ϕdx dt = 0 .
u dx dt + k
∂t
∂t
QT
QT
QT
(3) u(x, 0) = u0 (x).
In this paper, we discuss first the existence of weak solutions. Most proofs of
existence for (1.4) are based on the Yoshida approximations [6], but these methods
do not apply to (1.1). Our method for proving the existence of weak solutions is
based on a time discrete method that constructs approximate solutions. Later on,
we discuss the uniqueness of a solution. For simplicity we set k = 1 in this paper.
2. Existence of weak solutions
Theorem 2.1. If u0 ∈ W01,p (Ω) with p > 2, then problem (1.1)-(1.3) has at least
one solution.
We use the a discrete method for constructing an approximate solution. First,
T
divide the interval (0, T ) in N equal segments and set h = N
. Then consider the
problem
1
1
(uk+1 − uk ) − (∆uk+1 − ∆uk ) = div(|∇uk+1 |p−2 ∇uk+1 ),
h
h
uk+1 |∂Ω = 0, k = 0, 1, . . . , N − 1,
(2.1)
(2.2)
where u0 is the initial value.
Lemma 2.2. For a fixed k, if uk ∈ H01 (Ω), problem (2.1)-(2.2) admits a weak
solution uk+1 ∈ W01,p (Ω), such that for any ϕ ∈ C0∞ (Ω), have
Z
Z
Z
1
1
(uk+1 − uk )ϕdx +
(∇uk+1 − ∇uk )∇ϕdx + |∇uk+1 |p−2 ∇uk+1 ∇ϕdx = 0.
h Ω
h Ω
Ω
(2.3)
Proof. On the space W01,p (Ω), we consider the functionals
Z
1
Φ1 [u] =
|∇u|p dx,
p Ω
Z
1
Φ2 [u] =
|u|2 dx,
2 Ω
Z
1
Φ3 [u] =
|∇u|2 dx,
2 Ω
Z
1
1
Ψ[u] = Φ1 [u] + Φ2 [u] + Φ3 [u] −
f udx,
h
h
Ω
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WEAK SOLUTIONS
3
where f ∈ H −1 (Ω) is a known function. Using Young’s inequality, there exist
constants C1 , C2 > 0, such that
Z
Z
Z
Z
1
1
1
Ψ[u] =
|∇u|p dx +
|u|2 dx +
|∇u|2 dx −
f u dx
p Ω
2h Ω
2h Ω
Ω
Z
≥ C1
|∇u|p dx − C2 kf k−1 ;
Ω
hence Ψ[u] → ∞, as kuk1,p → +∞. Here kuk1,p Rdenotes the norm of u in W01,p (Ω).
Since the norm is lower semi-continuous and Ω f udx is a continuous functional,
Ψ[u] is weakly lower semi-continuous on W01,p (Ω) and satisfying the coercive condition. From [2] we conclude that there exists u∗ ∈ W01,p (Ω), such that
Ψ[u∗ ] = inf Ψ[u],
and u∗ is the weak solutions of the Euler equation corresponding to Ψ[u],
1
1
u − ∆u − div(|∇u|p−2 ∇u) = f.
h
h
Taking f = (uk − ∆uk )/h, we obtain a weak solutions uk+1 of (2.1)–(2.2). The
proof is complete.
Now, we need to establish a priori estimates, for the weak solutions uk+1 of
(2.1)–(2.2). First, we define the weak solutions of (1.1)–(1.3) as follows:
uh (x, t) = uk (x),
kh < t ≤ (k + 1)h, k = 0, 1, . . . , N − 1,
uh (x, 0) = u0 (x).
Lemma 2.3. The weak solutions uk of (2.1)–(2.2) satisfy
h
N Z
X
k=1
Z
sup
0<t<T
|∇uk |p dx ≤ C,
(2.4)
|∇uh (x, t)|p dx ≤ C,
(2.5)
Ω
Ω
where C is a constant independent of h and k.
Proof. i) We take ϕ = uk+1 in the integral equality (2.3) (we can easily prove that
for ϕ ∈ W01,p (Ω), (2.3) also holds).
Z
Z
1
1
(uk+1 − uk )uk+1 dx +
(∇uk+1 − ∇uk )∇uk+1 dx
h Ω
h Ω
Z
+
|∇uk+1 |p−2 ∇uk+1 ∇uk+1 dx = 0,
Ω
i.e.,
1
h
Z
|uk+1 |2 dx +
Ω
−
1
h
1
h
Z
Z
|∇uk+1 |2 dx −
Ω
1
h
Z
Z
∇uk+1 ∇uk dx +
Ω
Ω
uk uk+1 dx
Ω
|∇uk+1 |p dx = 0.
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CHANGCHUN LIU
EJDE–2003/63
Thus
Z
Z
1
|uk+1 |2 dx +
|∇uk+1 |2 dx +
|∇uk+1 |p dx
h Ω
Ω
Ω
Z
Z
1
1
uk uk+1 dx +
∇uk+1 ∇uk dx.
=
h Ω
h Ω
1
h
Z
By Young’s inequality,
Z
Z
Z
1
1
2
2
|uk+1 | dx +
|∇uk+1 | dx +
|∇uk+1 |p dx
h Ω
h Ω
Ω
Z
Z
Z
Z
1
1
1
1
2
2
|uk | dx +
|uk+1 | dx +
|∇uk |2 dx +
|∇uk+1 |2 dx;
≤
2h Ω
2h Ω
2h Ω
2h Ω
that is,
Z
Z
1
|uk+1 |2 dx +
|∇uk+1 |2 dx + h
|∇uk+1 |p dx
2 Ω
Ω
Ω
Z
Z
1
1
2
≤
|uk | dx +
|∇uk |2 dx.
2 Ω
2 Ω
Adding these inequalities for k from 0 to N − 1, we have
Z
Z
N Z
X
1
1
h
|∇uk |p dx ≤
|u0 |2 dx +
|∇u0 |2 dx.
2
2
Ω
Ω
Ω
1
2
Z
(2.6)
k=1
Therefore, (2.4) holds.
ii) We take ϕ = uk+1 − uk in the integral equality (2.3) and have
Z
Z
1
1
(uk+1 − uk )(uk+1 − uk )dx +
(∇uk+1 − ∇uk )∇(uk+1 − uk )dx
h Ω
h Ω
Z
+
|∇uk+1 |p−2 ∇uk+1 ∇(uk+1 − uk )dx = 0 .
Ω
Since the first term and the second term of the left hand side of the above equality
is nonnegative, it follows that
Z
Z
|∇uk+1 |p dx ≤
|∇uk+1 |p−2 ∇uk+1 ∇uk dx
Ω
Ω
Z
Z
1
p−1
p
|∇uk+1 | dx +
|∇uk |p dx;
≤
p
p Ω
Ω
thus,
Z
p
Z
|∇uk |p dx.
|∇uk+1 | dx ≤
Ω
Ω
For any m, with 1 ≤ m ≤ N − 1, adding the above inequality for k from 0 to m − 1,
we have
Z
Z
|∇um |p dx ≤
Ω
|∇u0 |p dx.
Ω
Therefore, (2.5) holds.
Lemma 2.4. For a weak solutions uk+1 of (2.1)–(2.2), we have
Z
Z
Z
Z
−Ch ≤
|uk+1 |2 dx +
|∇uk+1 |2 dx −
|uk |2 dx −
|∇uk |2 dx ≤ 0,
Ω
Ω
where C is a constant independently of h.
Ω
Ω
(2.7)
EJDE–2003/63
WEAK SOLUTIONS
5
Proof. The second inequality in (2.7) is an immediate consequence of (2.6). To
prove the first inequality, we choose ϕ = uk in (2.3) and obtain
Z
Z
1
1
(uk+1 − uk )uk dx +
(∇uk+1 − ∇uk )∇uk dx
h Ω
h
Z Ω
+
|∇uk+1 |p−2 ∇uk+1 ∇uk dx = 0 .
Ω
Therefore,
Z
Z
Z
Z
|uk |2 dx +
|∇uk |2 dx −
uk+1 uk dx −
∇uk+1 ∇uk dx
Ω
Ω
Ω
Ω
Z
=h
|∇uk+1 |p−2 ∇uk+1 ∇uk dx
Ω
Z
(p−1)/p Z
1/p
≤h
|∇uk+1 |p dx
|∇uk |p dx
.
Ω
Ω
Here we have used Hölder inequality. By (2.5) again, we obtain
Z
Z
Z
Z
|uk |2 dx +
|∇uk |2 dx −
uk+1 uk dx −
∇uk+1 ∇uk dx ≤ Ch.
Ω
Ω
Ω
Ω
Therefore,
Z
Z
2
|uk | dx +
|∇uk |2 dx
Ω
Ω
Z
Z
≤ Ch +
uk+1 uk dx +
∇uk+1 ∇uk dx
Ω
Ω
Z
Z
Z
Z
1
1
1
1
2
2
2
|uk+1 | dx +
|uk | dx +
|∇uk+1 | dx +
|∇uk |2 dx.
≤ Ch +
2 Ω
2 Ω
2 Ω
2 Ω
i.e.,
Z
|uk |2 dx +
Ω
Z
|∇uk |2 dx −
Ω
Z
|uk+1 |2 dx −
Z
Ω
|∇uk+1 |2 dx ≤ Ch
Ω
which completes the proof.
Lemma 2.5.
sup
0<t<T
Z
|uh |2 dx +
Ω
Z
Ω
Z
Z
|∇uh |2 dx ≤
|u0 |2 dx +
|∇u0 |2 dx.
Ω
(2.8)
Ω
The proof follows by adding (2.4), for m with 1 ≤ m ≤ N − 1, for k from 0 to
m − 1.
Proof of Theorem 2.1. First, we define the operator At , At (∇uh ) = |∇uk |p−2 ∇uk ,
∆h uh = uk+1 − uk , where kh < t ≤ (k + 1)h, k = 0, 1, . . . , N − 1. By the dispersion
equation (2.1) and the (2.4) in Lemma 2.2, we know that
1
(uk+1 − uk )
h
0
in L∞ (0, T ; W −1,p (Ω))
is bounded.
(2.9)
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CHANGCHUN LIU
EJDE–2003/63
By (2.5), (2.7), (2.9) and (2.4) we known that exists a subsequence of {uh } (which
we denote as the original sequence) such that
uh → u
h
in L∞ (0, T ; W 1,p (Ω))
∇u → ∇u
∞
weak-?,
2
in L (0, T ; L (Ω)) weak-?,
0
1
∂u
(uk+1 − uk ) →
in L∞ (0, T ; W −1,p (Ω) weak-?,
h
∂t
0
|∇uh |p−2 ∇uh → w in L∞ (0, T ; Lp (Ω)) weak-?,
where p0 is conjugate exponent of p. From (2.3), we known, for any ϕ ∈ C0∞ (QT ),
ZZ
1
1
( ∆h uh ϕ + ∆h ∇uh ∇ϕ + |∇uh |p−2 ∇uh ∇ϕ)dx dt = 0,
h
h
QT
i.e.,
ZZ
QT
1
1
( ∆h uh ϕ − ∆h uh ∆ϕ + |∇uh |p−2 ∇uh ∇ϕ)dx dt = 0.
h
h
Letting h → 0, we obtain, in the sense of distributions,
∂u ∂∆u
−
− div(w) = 0.
∂t
∂t
(2.10)
Next, we prove that w = |∇u|p−2 ∇u a.e. in QT . Define
Z
Z
Z
Z
t − kh
fh (t) =
|uk+1 |2 dx +
|∇uk+1 |2 dx −
|uk |2 dx −
|∇uk |2 dx
2h
Ω
Ω
Ω
Ω
Z
Z
1
1
2
2
+
|uk | dx +
|∇uk | dx,
2 Ω
2 Ω
where kh < t ≤ (k + 1)h. by (2.7) we have
Z
Z
Z
Z
1
1
1
1
|∇uk |2 dx +
|uk |2 dx − Ch ≤ fh (t) ≤
|uk |2 dx +
|∇uk |2 dx,
2 Ω
2 Ω
2 Ω
2 Ω
−C ≤ fh0 (t) ≤ 0.
By Ascoli–Arzela theorem, there exists a function f (t) ∈ C([0, T ]), such that
lim fh (t) = f (t)
h→0
for t ∈ [0, T ] uniformly.
Using (2.7), we have
Z
1 Z
1
lim
|∇uh |2 dx +
|uh |2 dx = f (t)
h→0 2 Ω
2 Ω
for t ∈ [0, T ] uniformly.
(2.11)
By (2.6) again, we obtain
Z
Z
ZZ
Z
Z
1
1
1
1
|uN |2 dx +
|∇uN |2 dx +
|∇uh |p dx dt ≤
|u0 |2 dx +
|∇u0 |2 dx.
2 Ω
2 Ω
2
2
QT
Ω
Ω
EJDE–2003/63
WEAK SOLUTIONS
7
In the above inequality letting h → 0, and using (2.10) we have
ZZ
lim
|∇uh |p dx dt ≤ f (0) − f (T )
h→0
QT
1
= lim
ε→0 ε
Z
T −ε
(f (t) − f (t + ε))dt
0
h 1 Z T −ε Z
= lim lim
(|uh (x, t)|2 − |uh (x, t + ε)|2 )dx dt
ε→0 h→0 2ε 0
Ω
Z T −ε Z
i
1
+
(|∇uh (x, t)|2 − |∇uh (x, t + ε)|2 )dx dt .
2ε 0
Ω
R
R
Since Φ2 [u] = 12 Ω |u|2 dx and Φ3 [u] = 12 Ω |∇u|2 dx are convex functionals, and
δΦ2 [u]
= u,
δu
δΦ3 [u]
= −∆u,
δu
we have
Z
Z
1
|u (x, t)| dx −
|uh (x, t + ε)|2 dx
2
Ω
Ω
Z
Z
1
1
h
2
+
|∇u (x, t)| dx −
|∇uh (x, t + ε)|2 dx
2 Ω
2 Ω
Z
≤
uh (x, t)(uh (x, t) − uh (x, t + ε))dx
Ω
Z
+
∇uh (x, t)(∇uh (x, t) − ∇uh (x, t + ε))dx.
1
2
h
2
Ω
Therefore,
Z
Z
1 h T −ε
|uh (x, t)|2 − |uh (x, t + ε)|2 )dx dt
lim
h→0 2ε
0
Ω
Z T −ε Z
i
+
(|∇uh (x, t)|2 − |∇uh (x, t + ε)|2 )dx dt
0
1
ε
≤
Ω
T −ε Z
Z
(u(x, t) − u(x, t + ε))u dx dt
0
1
+
ε
Z
Ω
T −ε Z
(∇u(x, t) − ∇u(x, t + ε))∇u dx dt .
0
Ω
Hence, we obtain
ZZ
|∇uh |p dx dt ≤ −
lim
h→0
QT
Z
T
h
0
∂u
, uidt +
∂t
Z
T
h
0
∂u
, ∆uidt,
∂t
where h, i denotes the inner product. Form (2.10), we obtain
ZZ
ZZ
lim
|∇uh |p dx dt ≤
w∇udx dt .
h→0
QT
QT
(2.12)
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CHANGCHUN LIU
EJDE–2003/63
1 [u]
= − div(|∇u|p−2 ∇u) and the convexity of Φ1 [u], for any g ∈
Again by δΦδu
1,p
L∞ (0, T ; W0 (Ω)) we have
ZZ
ZZ
1
1
p
−
|∇g| dx dt +
|∇uh |p dx dt
p
p
QT
QT
ZZ
≤
− div(|∇uh |p−2 ∇uh )(uh − g)dx dt,
QT
that is
ZZ
ZZ
ZZ
1
1
|∇g|p dx dt −
|∇uh |p dx dt ≥
div(|∇uh |p−2 ∇uh )(uh − g)dx dt
p
p
QT
QT
QT
ZZ
=
(|∇uh |p−2 ∇uh )∇(g − uh )dx dt.
QT
By (2.11) and F (u) is weakly lower semicontinuous, in above equality letting h → 0,
we obtain
ZZ
ZZ
ZZ
1
1
|∇g|p dx dt −
|∇u|p dx dt ≤
w∇(g − u)dx dt.
(2.13)
p
p
QT
QT
QT
In (2.13), we take g = εg + u to obtain
ZZ
ZZ
i ZZ
1h1
1
|∇(εg + u)|p dx dt −
|∇u|p dx dt ≥
w∇g dx dt.
ε p
p
QT
QT
QT
Letting ε → 0,
ZZ
ZZ
ZZ
δΦ1 [u]
g dx dt =
|∇u|p−2 ∇u∇g dx dt ≥
w∇g dx dt .
δu
QT
QT
QT
Since g is arbitrary, taking g = −g, we get the opposite inequality above; hence
w = |∇u|p−2 ∇u.
The strong convergence of uh in C(0, T ; H 1 (Ω)) and the fact that uh (x, 0) = u0 (x)
completes the proof.
3. Uniqueness of solutions
In this section, we prove that the weak solution is unique. To this end we need
the following lemma.
Lemma 3.1. For ϕ ∈ L∞ (t1 , t2 ; W01,p (Ω)) with ϕt ∈ L2 (t1 , t2 ; H 1 (Ω)), the weak
solutions u of the problem (1.1)-(1.3) on QT satisfies
Z
Z
u(x, t1 )ϕ(x, t1 )dx +
∇u(x, t1 )∇ϕ(x, t1 )dx
Ω
Ω
Z t2 Z ∂ϕ
∂∇ϕ
p−2
+
u
+ ∇u
− |∇u| ∇u∇ϕ dx dt
∂t
∂t
Zt1 Ω
Z
=
u(x, t2 )ϕ(x, t2 )dx +
∇u(x, t2 )∇ϕ(x, t2 )dx.
Ω
Ω
EJDE–2003/63
WEAK SOLUTIONS
9
In particular, for ϕ ∈ W01,p (Ω), we have
Z
Z
(u(x, t1 ) − u(x, t2 ))ϕdx +
∇(u(x, t1 ) − u(x, t2 ))∇ϕdx
Ω
Ω
Z t2 Z
|∇u|p−2 ∇u∇ϕdx dt = 0 .
−
t1
(3.1)
Ω
Proof. From ϕ ∈ L∞ (t1 , t2 ; W01,p (Ω)) and ϕt ∈ L2 (t1 , t2 ; H 1 (Ω)), it follows that
there exists a sequence of functions {ϕk }, for fixed t ∈ (t1 , t2 ), ϕk (·, t) ∈ C0∞ (Ω),
and as k → ∞
kϕk − ϕkL∞ (t1 ,t2 ;W 1,p (Ω)) → 0.
kϕkt − ϕt kL2 (t1 ,t2 ;H 1 (Ω)) → 0,
0
C0∞ (R)
such that j(s) ≥ 0, for s ∈ R; j(s) = 0, for
Choose aR function j(s) ∈
∀|s| > 1; R j(s)ds = 1. For h > 0, define jh (s) = h1 j( hs ) and
Z t−t1 −2h
ηh (t) =
jh (s)ds.
t−t2 +2h
C0∞ (t1 , t2 ),
Clearly ηh (t) ∈
limh→0+ ηh (t) = 1, for all t ∈ (t1 , t2 ). In the definition
of weak solutions, choose ϕ = ϕk (x, t)ηh (t). We have
Z t2 Z
Z t2 Z
uϕk jh (t − t1 − 2h)dx dt −
uϕk jh (t − t2 + 2h)dx dt
t1
t2
Z
Ω
t1
Z
t2
Z
∇u∇ϕk jh (t − t1 − 2h)dx dt −
+
t1
Ω
∇u∇ϕk jh (t − t2 + 2h) dx dt
t1
Z
t2
Ω
Z
Ω
Z
+
Z
t2
Z
∇u∇ϕkt ηh dx dt
uϕkt ηh dx dt +
t1
Ω
t1
Z
t2
Z
−
t1
Ω
|∇u|p−2 ∇u∇ϕk ηh dx dt = 0.
Ω
Observe that
Z Z
Z
t2
uϕk jh (t − t1 − 2h)dx dt − (uϕk )|t=t1 dx
t1
=
≤
Ω
t1 +3h
Ω
Z t1 +3h Z
uϕk jh (t − t1 − 2h)dx dt −
(uϕk )|t=t2 jh (t − t1 − 2h)dx dt
t1 +h
Ω
t1 +h
Ω
Z
(uϕk )|t − (uϕk )|t1 dx,
sup
Z
Z
t1 +h<t<t1 +3h
Ω
and u ∈ C(0, T ; L2 (Ω)). We see that the right hand side tends to zero as h → 0.
Similarly,
Z
Z t2 Z
uϕk jh (t − t2 + 2h)dx dt − (uϕk )|t=t2 dx → 0, as h → 0,
t1
Z
t2
t1
Z t2
t1
Ω
Z
Ω
Z
∇u∇ϕk jh (t − t1 − 2h)dx dt −
Ω
Ω
Z
Z
∇u∇ϕk jh (t − t2 + 2h)dx dt −
Ω
Ω
(∇u∇ϕk )|t=t1 dx → 0,
as h → 0,
(∇u∇ϕk )|t=t2 dx → 0,
as h → 0.
10
CHANGCHUN LIU
EJDE–2003/63
Letting h → 0 and k → ∞, we obtain
Z
Z
u(x, t1 )ϕ(x, t1 )dx +
∇u(x, t1 )∇ϕ(x, t1 )dx
Ω
Ω
Z t2 Z ∂ϕ
∂∇ϕ
+
u
+ ∇u
− |∇u|p−2 ∇u∇ϕ dx dt
∂t
∂t
Zt1 Ω
Z
=
u(x, t2 )ϕ(x, t2 )dx +
∇u(x, t2 )∇ϕ(x, t2 )dx.
Ω
Ω
In particular for ϕ ∈ W01,p (Ω), we have
Z
Z
(u(x, t1 ) − u(x, t2 ))ϕdx + (∇u(x, t1 ) − ∇u(x, t2 ))∇ϕdx
Ω
Ω
Z t2 Z
−
|∇u|p−2 ∇u∇ϕ dx dt = 0
t1
Ω
which completes the proof.
For a fixed τ ∈ (0, T ), set h satisfying 0 < τ < τ + h < T . Letting t1 = τ ,
t2 = τ + h, then multiply (3.1) by h1 , for ϕ ∈ W01,p (Ω), we obtain
Z
Z
Z
(uh (x, τ ))τ ϕ(x)dx + ((∇u)h (x, τ ))τ ϕ(x)dx + (|∇u|p−2 ∇u)h (x, τ )∇ϕdx = 0,
Ω
Ω
Ω
(3.2)
where
( R t+h
1
uh (x, t) =
h
0,
t
u(·, τ )dτ, t ∈ (0, T − h),
t > T − h.
Theorem 3.2. Problem (1.1)-(1.3) admits only one weak solution.
Proof. Suppose u1 , u2 are two solutions of (1.1)-(1.3), then
Z
Z
(u1 (x, τ ) − u2 (x, τ ))hτ ϕ(x)dx + ((∇u1 − ∇u2 )h (x, τ ))τ ϕ(x)dx
Ω
Ω
Z
p−2
+ (|∇u1 | ∇u1 − |∇u2 |p−2 ∇u2 )h (x, τ )∇ϕdx = 0.
Ω
For a fixed τ , we take ϕ(x) = [u1 − u2 ]h ∈ W01,p (Ω), and hence
Z
(u1 (x, τ ) − u2 (x, τ ))hτ (u1 − u2 )h dx
Ω
Z
+
∇(u1 (x, τ ) − u2 (x, τ ))hτ ∇(u1 − u2 )h dx
Ω
Z
= − [(|∇u1 |p−2 ∇u1 − |∇u2 |p−2 ∇u2 )h ](x, τ )∇(u1 − u2 )h dx,
Ω
i.e.,
Z
(u1 (x, τ ) − u2 (x, τ ))hτ (u1 − u2 )h dx
Z
+
∇(u1 (x, τ ) − u2 (x, τ ))hτ ∇(u1 − u2 )h dx
Ω
Z
= − [(|∇u1 |p−2 ∇u1 − |∇u2 |p−2 ∇u2 )h ](x, τ )∇(u1 − u2 )h dx.
Ω
Ω
EJDE–2003/63
WEAK SOLUTIONS
Integrating the above equality with respect to τ over (0, t),
Z
Z
|(u1 − u2 )h |2 (x, t)dx +
|∇(u1 − u2 )h |2 (x, t)dx ≤ 0,
Ω
Ω
R
we have Ω |(u1 − u2 )h |2 dx = 0; therefore, u1 = u2 .
11
Acknowledgment. The author would like to thank referee for his/her valuable
suggestions and for providing the references E. Di Benedtto & M. Pierre [5] and E.
Di Benedetto & R. E. Showalter [6].
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Changchun Liu
Department of Mathematics, Nanjing Normal University, Nanjing 210097, China
Department of Mathematics, Jilin University, Changchun 130012, China
E-mail address: mathlcc@21cn.com