Sep 8, 2017 - value condition. 1. Introduction. Consider the polytropic filtration equation with a convection term ut = div(a(x)|u|α|âu|pâ2âu) +. âbi(um). âxi.
Electronic Journal of Differential Equations, Vol. 2017 (2017), No. 207, pp. 1–10. ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
SOLUTIONS TO POLYTROPIC FILTRATION EQUATIONS WITH A CONVECTION TERM HUASHUI ZHAN
Abstract. We introduce a new type of the weak solution of the polytropic filtration equations with a convection term, ut = div(a(x)|u|α |∇u|p−2 ∇u) +
∂bi (um ) . ∂xi
Here, Ω ⊂ RN is a domain with a C 2 smooth boundary ∂Ω, a(x) ∈ C 1 (Ω), α p > 1, m = 1 + p−1 , α > 0, a(x) > 0 when x ∈ Ω and a(x) = 0 when x ∈ ∂Ω. Since the equation is degenerate on the boundary, its weak solutions may lack the needed regularity to have a trace on the boundary. The main aim of the paper is to establish the stability of the weak solution without any boundary value condition.
1. Introduction Consider the polytropic filtration equation with a convection term ut = div(a(x)|u|α |∇u|p−2 ∇u) +
∂bi (um ) , ∂xi
(x, t) ∈ QT = Ω × (0, T ),
(1.1)
α where p > 1, m = 1 + p−1 , α > 0, Ω ⊂ RN is with a C 2 smooth boundary ∂Ω, a(x) ∈ C 1 (Ω), a(x) > 0. The equations like (1.1) arise from a variety of diffusion phenomena, such as soil physics, fluid dynamics, combustion theory, reaction chemistry, one can see [1, 10] and the references therein. In particular, when α > 0, a(x) ≡ 1, the well-posedness of equation (1.1) with the usual initial-boundary value conditions
u|t=0 = u0 (x), x ∈ Ω, u(x, t) = 0,
(x, t) ∈ ΓT = ∂Ω × (0, T ),
(1.2) (1.3)
has been studied thoroughly, one can refer to [2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 15, 16]. In this article, we assume that a(x) > 0, x ∈ Ω, a(x) = 0, x ∈ ∂Ω. 2010 Mathematics Subject Classification. 35L65, 35K85, 35R35. Key words and phrases. Polytropic filtration equation; convection term; stability; boundary value condition. c
2017 Texas State University. Submitted February 16, 2017. Published September 8, 2017. 1
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Consequently, equation (1.1) is always degenerate on the boundary. Not only the degeneracy comes from the physics quantity u itself, but also comes from the diffusion coefficient a(x). Now, let us introduce some basic definitions and the main results. For every fixed t ∈ [0, T ], the Banach space Vt (Ω) = u(x, t) : u(x, t) ∈ L2 (Ω) ∩ W01,1 (Ω), |∇u(x, t)|p ∈ L1 (Ω) , is with the norm kukVt (Ω) = kuk2,Ω + k∇ukp,Ω , and we denote its dual space as Vt0 (Ω). By W (QT ) we denote the Banach space W (QT ) = {u : [0, T ] → Vt (Ω)|u ∈ L2 (QT ), |∇u|p ∈ L1 (QT ), u = 0
on ∂Ω},
kukW (QT ) = k∇ukp,QT + kuk2,QT . 0
Here W (QT ) is the dual of W (QT ) (the space of linear functionals over W (QT ), w ∈ W 0 (QT ) if n X 0 Di wi , w0 ∈ L2 (QT ), wi ∈ Lp (QT ), w = w0 + i=1
ZZ
∀φ ∈ W (QT ), hw, φi =
w0 φ +
X
QT
wi Di φ dx dt.
i
0
The norm in W (QT ) is defined by kvkW 0 (QT ) = sup{hv, φi : φ ∈ W(QT ), kφkW (QT ) 6 1}. Definition 1.1. A nonnegative function u(x, t) is said to be a weak solution of (1.1) with the initial value (1.2), if u satisfies ∂u ∈ W 0 (QT ), a(x)|u|α |∇u|p ∈ L1 (QT ), (1.4) u ∈ L∞ (QT ), ∂t and for any function ϕ1 ∈ L1 (0, T ; C01 (Ω)), ϕ2 ∈ L∞ (QT ) such that for any given 1,p t ∈ [0, T ), ϕ2 (x, ·) ∈ Wloc (Ω), we have ZZ ∂u (ϕ1 ϕ2 ) + a(x)|u|α |∇u|p−2 ∇u · ∇(ϕ1 ϕ2 ) QT ∂t (1.5) i m + b (u )(ϕ1 ϕ2 )xi dx dt = 0. The initial value (1.2) is satisfied in the sense that Z Z lim u(x, t)φ(x) dx = u0 (x)φ(x) dx, ∀φ(x) ∈ C0∞ (Ω). t→0
Ω 1,γ
(1.6)
Ω
If u ∈ L∞ (0, T ; W (Ω)) for some constant γ > 1, the boundary value condition (1.3) is satisfied in the sense of the trace, then we say u is a weak solution of the initial-boundary problem of equation (1.1). Clearly, if noticing m = 1 +
α p−1 ,
by (1.4), then
a(x)|∇um |p ∈ L1 (QT ), and (1.5) is equivalent to ZZ ∂u 1 (ϕ1 ϕ2 ) + p−1 a(x)|∇um |p−2 ∇um · ∇(ϕ1 ϕ2 ) m QT ∂t i m + b (u )(ϕ1 ϕ2 )xi dx dt = 0.
(1.7)
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In general, since (1.1) is always degenerate on the boundary, instead of u(x, t) ∈ 1.p (Ω)). Thus, we can not L∞ (0, T ; W01,p (Ω)), we only have u(x, t) ∈ L∞ (0, T ; Wloc define the trace of the weak solution u on the boundary. If u, v are two weak solutions of equation (1.1), to prove the stability (or uniqueness) of the weak solutions, one generally must choose a test function with the form f (x, t, u − v) which involves the boundary value condition u(x, t) = v(x, t) = 0,
(x, t) ∈ ΓT = ∂Ω × (0, T ).
(1.8)
However, the weak solution defined in this paper can not guarantee this condition. This is the main reason that we need to choose the test function ϕ1 ϕ2 in Definition 1.1. If α = 0, m = 1, bi ≡ 0, the existence of the weak solutions had been proved in our previous paper [14]. In this paper, we mainly concern with the stability of the weak solutions of equation (1.1). Theorem 1.2. Let u, v be two nonnegative solutions of (1.1) with the same homogeneous boundary value condition (1.3) and with the different initial values u0 , v0 respectively. Then Z Z |u(x, t) − v(x, t)| dx 6 |u0 − v0 | dx. (1.9) Ω
Ω
Theorem 1.3. Let u, v be two nonnegative solutions of equation (1.1) with the initial values u0 , v0 respectively. If 1 < p 6 2, and Z 1 (1.10) a− p−1 (x)dx < ∞, Ω
then the stability of the weak solutions is true in the sense of (1.9). Theorem 1.4. Let u, v be two nonnegative solutions of (1.1) with the initial values u0 , v0 respectively. If p > 1 and for small enough λ > 0, u(x) and v(x) satisfy 1 λ
Z
p−1 p a(x)|∇u | dx 6 c,
Ω\Ωλ
m p
1 λ
Z
p−1 p a(x)|∇v m |p dx 6 c,
(1.11)
Ω\Ωλ
then (1.9) is true. Here Ωλ = {x ∈ Ω : a(x) > λ} Theorem 1.5. Let u, v be two weak solutions of problem (1.1) with the initial values u0 (x), v0 (x) respectively. If p > 1, m > 0, Z Z |∇a| m |∇a| m |u |dx 6 c, |v |dx 6 c, (1.12) Ω a Ω a then (1.9) is true. At the end, we suggest that not any boundary value condition is required in Theorems 1.3–1.5. However, from my own perspective, the condition (1.12) in Theorem 1.5 makes a substitute of the boundary value condition. Moreover, if bi ≡ 0, i.e. equation (1.1) has no convection term, Theorem 1.5 is true without the condition (1.12).
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2. Proof of Theorem 1.2 Let u, v are two nonnegative solutions of equation (1.1) with the same homogeneous boundary value and with the different initial values u0 , v0 respectively. From the definition of the weak solution, we let ϕ1 = ϕ ∈ L1 (0, T ; C01 (Ω)), ϕ2 ≡ 1. Then Z Z ∂(u − v) ϕ dx + a(x)(uα |∇u|p−2 ∇u − v α |∇v|p−2 ∇v) · ∇ϕ dx ∂t Ω Ω Z (2.1) + [bi (um ) − bi (v m )]ϕxi dx = 0, Ω
or equivalently Z Z 1 ∂(u − v) dx + p−1 a(x)(|∇um |p−2 ∇um − |∇v m |p−2 ∇v m ) · ∇ϕ dx ϕ ∂t m Ω Ω Z (2.2) + [bi (um ) − bi (v m )]ϕxi dx = 0. Ω
For small η > 0, let Z Sη (s) =
s
hη (τ )dτ,
hη (s) =
0
|s| 2 1− . η η +
(2.3)
Obviously hη (s) ∈ C(R), and |shη (s)| 6 1,
hη (s) > 0,
|Sη (s)| 6 1,
lim sSη0 (s) = 0.
lim Sη (s) = sgn s,
η→0
(2.4)
η→0
We can choose ϕ = Sη (um − v m ) as the test function, then Z Z ∂(u − v) 1 Sη (um − v m ) dx + p−1 a(x)(|∇um |p−2 ∇um − |∇v m |p−2 ∇v m ) ∂t m Ω Ω · ∇(um − v m )Sη0 (um − v m )dx Z = − [bi (um ) − bi (v m )](um − v m )xi Sη0 (um − v m )dx. Ω
(2.5) Clearly, Z lim
η→0
and Z
Sη (um − v m )
Ω
∂(u − v) dx ∂t Ω Z ∂(u − v) dx = sgn(u − v) ∂t Ω d = ku − vkL1 (Ω) , dt
∂(u − v) dx = ∂t
Z
sgn(um − v m )
(2.6)
a(x)|(|∇um |p−2 ∇u − |∇v m |p−2 ∇v m ) · ∇(um − v m )Sη0 (um − v m )dx > 0. (2.7)
Ω
At the same time, Z
−1
a p−1 (x)dx < ∞, Ω
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using Lebesgue dominated convergence theorem, by (2.4), we have Z lim [bi (um ) − bi (v m )](um − v m )xi Sη0 (um − v m )dx η→0
Ω
6 lim
η→0
×
p−1 p 1 p |[bi (um ) − bi (v m )]Sη0 (um − v m )a− p | p−1 dx
Z
(2.8)
Ω
Z
a(x)(|∇um |p + |∇v m |p )dx
1/p
= 0.
Ω
Let η → 0 in (2.2). Then d ku − vkL1 (Ω) 6 0. dt
(2.9)
This implies Z
Z |u(x, t) − v(x, t)| dx 6
Ω
|u0 − v0 | dx,
∀t ∈ [0, T ).
Ω
3. Proofs of Theorem 1.3 and 1.4 Proof of Theorem 1.3. By Definition 1.1, for any function ϕ1 ∈ L1 (0, T ; C01 (Ω)), 1,p ϕ2 ∈ L∞ (QT ) such that for any given t ∈ [0, T ), ϕ2 (x, ·) ∈ Wloc (Ω), we have ZZ ∂(u − v) 1 (ϕ1 ϕ2 ) + p−1 a(x)(|∇um |p−2 ∇um ∂t m (3.1) QT − |∇v m |p−2 ∇v m ) · ∇(ϕ1 ϕ2 ) + (bi (um ) − bi (v m ))(ϕ1 ϕ2 )xi ] dx dt = 0. For a small positive constant λ > 0, let Ωλ = {x ∈ Ω : a(x) > λ}, ( 1, if x ∈ Ωλ , φλ (x) = 1 a(x), if x ∈ Ω \ Ωλ . λ
(3.2)
Now, we choose ϕ1 = φλ (x)χ[τ,s] , ϕ2 = Sη (um − v m ), and then integrate it over Ω, to have Z sZ Z sZ ∂(u − v) 1 φλ (x)Sη (um − v m ) dx dt + p−1 φλ (x)a(x) ∂t m τ Ω τ Ω × |∇um |p−2 ∇um − |∇v m |2 ∇v m · ∇(um − v m )Sη0 (um − v m ) dx dt Z sZ 1 + p−1 a(x)(|∇um |p−2 ∇um − |∇v m |p−2 ∇v m ) m (3.3) τ Ω · ∇φλ (x)Sη (um − v m ) dx dt Z sZ + [bi (um ) − bi (v m )][φλ (x)Sη0 (um − v m )(um − v m )xi τ
Ω
+ Sη (um − v m )φλxi (x)] dx dt = 0. Clearly, Z
φλ (x)a(x)(|∇um |p−2 ∇um − |∇v m |2 ∇v m )
Ω
· ∇(um − v m )Sη0 (um − v m ) dx > 0.
(3.4)
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Z
a(x)(|∇um |p−2 ∇um − |∇v m |2 ∇v m ) · ∇φλ (x)Sη (um − v m ) dx Ω Z 6 a(x)|(|∇um |p−2 ∇um − |∇v m |2 ∇v m ) · ∇φλ (x)Sη (um − v m )|dx Ω\Ωλ Z 6 a(x)|(|∇um |p−2 ∇um − |∇v m |2 ∇v m )||∇φλ (x)|dx
(3.5)
Ω\Ωλ
ch 6 λ
Z
m p−1
a(x)|∇u |
Z
s
Z
|∇a| dx + τ
Ω\Ωλ
Since 1 < p 6 2, |∇a| 6 c and Z
i a(x)|∇v m |p−1 |∇a|dx .
Ω\Ωλ
|∇a|p dx 6 cλ 6 cλp−1 ,
Ω\Ωλ
it follows that Z Z 1/p 1/p c c 6 6 c. a(x)|∇a|p dx λ |∇a|p dx λ Ω\Ωλ λ Ω\Ωλ By (3.5)-(3.6), using the H¨ older inequality, Z a(x)(|∇um |p−2 ∇um − |∇v m |2 ∇v m ) · ∇φλ (x)Sη (um − v m ) dx Ω Z Z sZ i ch 6 a(x)|∇um |p−1 |∇a| dx + a(x)|∇v m |p−1 |∇a|dx λ Ω\Ωλ τ Ω\Ωλ Z Z p−1 1/p c p 6 a|∇a|p dx a(x)|∇um |p dx λ Ω\Ωλ Ω\Ωλ Z Z p−1 1/p c p + a(x)|∇a|p dx a(x)|∇v m |p dx λ Ω\Ωλ Ω\Ωλ Z p−1 Z p−1 p p 6c a(x)|∇um |p dx +c a(x)|∇v m |p dx . Ω\Ωλ
(3.6)
(3.7)
Ω\Ωλ
Then, we have lim
λ→0
Z
a(x)(|∇um |p−2 ∇um − |∇v m |p−2 ∇v m )
Ω m
(3.8)
m
· ∇φλ (x)Sη (u − v ) dx| = 0. R 1 At the same time, by that Ω a− p−1 (x)dx < c, using (2.4) and the Lebesgue dominated convergence theorem, we also have Z lim φλ [bi (um ) − bi (v m )]Sη0 (um − v m )(um − v m )xi dx = 0, (3.9) η→0
Ω
and Z
φλxi [bi (um ) − bi (v m )]Sη (um − v m )dx| Z c 6 lim |∇a|dx λ→0 λ Ω\Ω λ Z p−1 1/p Z 1 c p 6 lim a(x)|∇a|p dx a− p−1 (x)dx = 0, λ→0 λ Ω\Ωλ Ω\Ωλ lim |
λ→0
Ω
(3.10)
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a− p−1 (x)dx < c. At last, Z sZ ∂(u − v) lim lim φλ (x)Sη (um − v m ) dx dt η→0 λ→0 τ ∂t Ω Z sZ ∂(u − v) Sη (um − v m ) = lim dx dt η→0 τ ∂t Ω Z sZ ∂(u − v) dx dt = sgn(um − v m ) ∂t Zτ s ZΩ ∂(u − v) = sgn(u − v) dx dt ∂t Zτ s Ω d = ku − vkL1 (Ω) dt. dt τ
R
by (3.6) and
7
Ω
(3.11)
Now, after letting λ → 0, let η → 0 in (3.2). Then by (3.4), (3.8)-(3.11), Z Z |u(x, t) − v(x, t)|dx 6 |u0 − v0 |dx. Ω
Ω
Proof of Theorem 1.4. As in the proof of Theorem 1.3, we have (3.3)- (3.5). Since u(x) and v(x) satisfy (1.11) by (3.6)-(3.7), using the H¨older inequality, we have Z a(x)(|∇um |p−2 ∇um − |∇v m |2 ∇v m ) · ∇φλ (x)Sη (um − v m ) dx Ω
6
c λ +
Z Ω\Ωλ
Z c
λ Z 6c
p−1 p a(x)|∇um |p dx
1/p Z a|∇a|p dx Ω\Ωλ p
a(x)|∇a| dx
(3.12)
p−1 p a(x)|∇v | dx
1/p Z
m p
Ω\Ωλ
Ω\Ωλ
1/p Z a|∇a|p dx +c
Ω\Ωλ
1/p a(x)|∇a|p dx ,
Ω\Ωλ
which approaches zero as λ → 0 since that a(x) ∈ C 1 (Ω), we have (3.8). At last, R 1 since Ω a− p−1 (x)dx < ∞, similar as the proof of Theorem 1.3, we have (3.9)-(3.10). So, as the proof of Theorem 1.3, we know that the stability (1.9) is true. 4. Proof Theorem 1.5 It is not difficult to show that the following definition is equivalent to Definition 1.1. Definition 4.1. A function u(x, t) is said to be a weak solution of (1.1) with initial value (1.2), if u ∈ L∞ (QT ),
ut ∈ L2 (QT ), 1
a(x)|∇u|p ∈ L1 (QT ), C01 (Ω),
∞
(4.1)
1,p (0, T ; Wloc (Ω)),
and for any function g(s) ∈ C (R), g(0) = 0, ϕ1 ∈ ϕ2 ∈ L ZZ [ut g(ϕ1 ϕ2 ) + a(x)|∇u|p−2 ∇u · ∇g(ϕ1 ϕ2 ) + u bixi (x)g(ϕ1 ϕ2 ) QT + bi (x)gxi (ϕ1 ϕ2 ) − c(x, t)ug(ϕ1 ϕ2 ) + f (x, t)g(ϕ1 ϕ2 )] dx dt = 0.
(4.2)
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The initial value is satisfied in the sense that Z Z lim u(x, t)φ(x) dx = u0 (x)φ(x) dx, ∀φ(x) ∈ C0∞ (Ω). t→0
Ω
(4.3)
Ω
Proof of Theorem 1.5. Let u, v be two solutions of equation (1.1) with the initial values u0 (x), v0 (x). We can choose Sη (aβ (um − v m )) as the test function. Then Z Z 1 ∂(u − v) dx + p−1 aβ+1 (x) |∇um |p−2 ∇um Sη (aβ (um − v m )) ∂t m Ω Ω − |∇v m |p−2 ∇v m · ∇(um − v m )Sη0 (aβ (um − v m )) dx Z + a(x)(|∇um |p−2 ∇um − |∇v m |p−2 ∇v m ) (4.4) Ω β m m 0 β m m · ∇a (u − v )Sη (a (u − v )) dx Z + [bi (um ) − bi (v m )][Sη0 (aβ (um − v m )) Ω β (axi (um
− v m ) + aβ (um − v m )xi dx = 0.
Thus Z lim
η→0
Sη (aβ (um − v m ))
d ∂(u − v) dx = ku − vk1 , ∂t dt
ZΩ aβ+1 (x)(|∇um |p−2 ∇um − |∇v m |p−2 ∇v m ) Ω m
· ∇(u − v
m
)Sη0 (aβ (um
(4.5)
(4.6)
m
− v )) dx > 0.
From |∇a(x)| 6 c in Ω, we have Z a(x)(um − v m )Sη0 (aβ (um − v m ))(|∇um |p−2 ∇um − |∇v m |p−2 ∇v m ) Ω · ∇aβ dx Z 6 c aβ (um − v m )Sη0 (aβ (um − v m ))
(4.7)
Ω
× (|∇um |p−2 ∇um − |∇v m |p−2 ∇v m ) dx|, Z aβ (um − v m )Sη0 (aβ (um − v m )) Ω × (|∇um |p−2 ∇um − |∇v m |p−2 ∇v m ) dx Z p−1 = a− p aβ (um − v m )Sη0 Ω:aβ |um −v m |
