Electronic Journal of Differential Equations, Vol. 2007(2007), No. 15, pp. 1–6.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
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VISCOSITY SOLUTIONS TO DEGENERATE DIFFUSION
PROBLEMS
ZU-CHI CHEN, YAN-YAN ZHAO
Abstract. This paper concerns the weak solutions to a Cauchy problem in
RN for a degenerate nonlinear parabolic equation. We obtain the Hölder
regularity of the weak solutions to this problem.
1. Introduction
We consider the Cauchy problem
1
|∇u|2 , (x, t) ∈ RN × R+
2
u(x, 0) = u0 (x), x ∈ RN
ut = α1 uβ1 ∆u + α2 uβ2 w,
w=
(1.1)
where α1 , α2 , β1 , β2 are constants and u0 is a bounded continuous and nonnegative
function on RN , denote Ω = RN × R+ .
Problem (1.1) degenerates at the points where u vanishes. Therefore, in general,
it has no classical solutions and we have to consider its weak solutions. The weak
solution is defined as follows.
1
Definition 1.1. A function u ∈ L∞ (Ω) ∩ L2loc ([0, +∞); Hloc
(RN )) is called a weak
solution of (1.1) if u ≥ 0 a.e. in Ω and for all T > 0,
Z
Z
∂ψ
− α1 ∇u · ∇(uβ1 ψ) + α2 uβ2 |∇u|2 ψ dx dt = 0
u0 ψ(0)dx +
u
∂t
RN
RN ×(0,T )
for all ψ ∈ C 1,1 (RN × [0, T ]) with the compact support in RN × [0, T ).
Let u (x, t) ≥ 0 be the classical solution of the problem
1
|∇u |2 , (x, t) ∈ RN × R+
2
u (x, 0) = u0 (x) + , x ∈ RN
ut = α1 uβ 1 ∆u + α2 uβ 2 w ,
w =
By the maximum principle u (x, t) is decreasing with respect to , thus
u(x, t) = lim u (x, t)
→0
2000 Mathematics Subject Classification. 35K15, 35K55, 35K65.
Key words and phrases. Viscosity solution; degenerate; Hölder regularity.
c 2007 Texas State University - San Marcos.
Submitted November 4, 2006. Published January 23, 2007.
Supported by grant 10371116 from NNSF of China.
1
(1.2)
2
Z. CHEN, Y. ZHAO
EJDE-2007/15
is well defined in Ω̄. The function u is a weak solution of (1.1). Because u0 is
bounded, using the maximum principle in problem (1.2), u is bounded and {u }→0
is uniformly bounded.
Definition 1.2. The weak solution defined above is called a viscosity solution of
(1.1).
As its special cases, Bertsh, Passo, Ughi and Lu had considered the equation
ut = u∆u − γ|∇u|2 in [1-6]. When α1 = m, β1 = m − 1, α2 = 2m(m − 1),
β2 = β1 − 1, problem (1.1) is the porous medium equation, the well known case.
2. Main Result
Theorem 2.1. If α1 > 0, β2 = β1 − 1, there exists a constant s such that
2α2 β1 − 2α2 − sα2 + 2s(s + 1)α1 + N α1 β12 ≤ 0
and
1+ 2s
|∇(u0
)| ≤ M
for a nonnegative constant M . Then the viscosity solution u of (1.1) satisfies
s
|∇(u1+ 2 )| ≤ M in Ω̄.
Proof. In the definition of the viscosity solution, we let u > 0 be the classical
solution of (1.2). Then
u(x, t) = lim u (x, t)
→0
is the viscosity solution of (1.1). In the following we use the notation u,. to denote
the derivative of function u with respect to its independent variables. At first, we
have
N
X
1
w,t = ( |∇u |2 )t =
u,xi (u,xi )t
2
i=1
=
N
X
u,xi (α1 uβ 1 ∆u + α2 uβ 2 w,xi
i=1
=
N
X
u,xi (α1 β1 u,xi uβ1 −1 ∆u + α1 uβ 1 ∆u,xi
i=1
+ α2 β2 u,xi uβ 2 −1 w + α2 uβ 2 w,xi )
= 2α1 β1 uβ 1 −1 w ∆u + α1 uβ 1
N
X
u,xi ∆u,xi
i=1
+ 2α2 β2 uβ2 −1 w2 + α2 uβ 2
N
X
u,xi w,xi
i=1
= 2α1 β1 uβ 1 −1 w ∆u + α1 uβ 1 ∆w − α1 uβ 1
N
X
i,j=1
+ 2α2 β2 uβ2 −1 w2 + α2 uβ 2
N
X
i=1
u,xi w,xi .
u2,xi xj
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VISCOSITY SOLUTIONS
3
Let
z = us w ,
(2.1)
then
z,t = us,t w + us w,t
1 −1
1 −1
= sα1 us+β
w ∆u + sα2 us+β2 −1 w2 + 2α1 β1 us+β
w ∆u + α1 us+β1 ∆w
1
− α1 us+β
N
X
2
u2,xi xj + 2α2 β2 us+β2 −1 w2 + α2 us+β
i,j=1
N
X
u,xi w,xi .
i=1
(2.2)
From (2.1) and (2.2),
z,t = α1 uβ 1 ∆z + (α2 uβ 2 − 2sα1 uβ1 −1 )
N
X
u,xi z,xi
i=1
+ [(2α2 β2 − sα2 )uβ2 −s−1 + 2s(s + 1)α1 uβ1 −s−2 ]z2
1
+ 2α1 β1 uβ 1 −1 z ∆u − α1 us+β
N
X
(2.3)
u2,xi xj .
i,j=1
If β2 = β1 − 1, α1 > 0, then
z,t = α1 uβ 1 ∆z + (α2 − 2sα1 )uβ 1 −1
+ [(2α2 β2 − sα2 ) + 2s(s +
N
X
u,xi z,xi
i=1
1)α1 ]uβ1 −s−2 z2
+ 2α1 β1 uβ1 −1 z ∆u − α1 us+β1
N
X
u2,xi xj .
i,j=1
Since
N
X
u2,xi xj ≥
i,j=1
1
(∆u )2 ,
N
it follows that
z,t ≤α1 uβ 1 ∆z + (α2 − 2sα1 )uβ1 −1
N
X
u,xi z,xi
i=1
+ [(2α2 β1 − 2α2 − sα2 ) + 2s(s + 1)α1 ]uβ 1 −s−2 z2
α1 s+β1
u
(∆u)2
+ 2α1 β1 uβ 1 −1 z ∆u −
N N
X
= α1 uβ 1 ∆z + (α2 − 2sα1 )uβ 1 −1
u,xi z,xi
i=1
r
β1 −s−2
p
1
α1 s+β
u 2 ∆u − β1 N α1 u 2 z )2
−(
N
+ [(2α2 β1 − 2α2 − sα2 ) + 2s(s + 1)α1 + N α1 β12 ]uβ1 −s−2 z2
By the condition
2α2 β1 − 2α2 − sα2 + 2s(s + 1)α1 + N α1 β12 ≤ 0
(2.4)
4
Z. CHEN, Y. ZHAO
EJDE-2007/15
and (2.4), we obtain
z,t ≤ α1 uβ 1 ∆z + (α2 − 2sα1 )uβ 1 −1
N
X
u,xi z,xi .
i=1
Using the maximum principle, we obtain
kz k∞ ≤ kz0 k∞ .
Because z =
us w
=
1 s
2
2 u |∇u | ,
thus
kus |∇u |2 k∞ ≤ kus0 |∇u0 |2 k∞ ≤ M + .
1+ 2s
Since ∇(u
) is continuous,
1+ 2s
|∇(u
)| ≤ M + .
(2.5)
Because u(x, t) = lim→0 u (x, t), then
s
|∇(u1+ 2 )| ≤ M.
(2.6)
Theorem 2.2. Suppose α1 , α2 , β1 , β2 ,u0 are as in Theorem 2.1, if there exists a
nonpositive constant s 6= −2 satisfying
2α2 β1 − 2α2 − sα2 + 2s(s + 1)α1 + N α1 β12 ≤ 0,
then the viscosity solution u(x, t) of problem (1.1) is Lipschitz continuous in x and
Hölder continuous in t with exponent 1/2 in Ω.
Proof. Because {u }→0 is uniformly bounded, u(x, t) = lim→0 u , so there exists
s
a constant M1 such that |u| < M1 . By Theorem 2.1, |∇(u1+ 2 )| ≤ M , then
−s
−s
s
s
|∇u| ≤ |1 + |−1 M |u 2 | ≤ |1 + |−1 M M1 2 .
2
2
Therefore, u is Lipschitz continuous with respect to x. Hence, we get directly from
[7] that u is Hölder continuous in t with exponent 1/2 in Ω.
3. Examples
Example 3.1. Consider the problem
ut = u∆u − γ|∇u|2 ,
u(x, 0) = u0 (x),
If γ ≥
√
(x, t) ∈ Ω
x ∈ RN
(3.1)
N − 1 (N 6= 10), and there are constants
√
3− N −1
τ=
,
2
p
2γ 2 − 2N + 2 − [2τ − (3 − γ)]2
1 − γ − 2τ
s=
+
2τ
2τ
τ (1+ s )
satisfying |∇u0 2 | ≤ M , then the viscosity solution of (3.1) is Lipschitz continuous in x and Hölder continuous in t with exponent 1/2 in Ω.
EJDE-2007/15
VISCOSITY SOLUTIONS
5
Proof. Set v = uτ . From problem (3.1), we obtain
v,t = τ v∆u − τ γuτ −1 |∇u |2
= τ v
N
X
τ −1 1
1
1 1 −1
−1
( vτ v,xi )xi − τ γv τ | vτ ∇v |2
τ
τ
i=1
1
= vτ ∆v + v
1
= vτ ∆v +
N
X
1
γ 1 −1
1
−2 2
− v,τ |∇v |2
( − 1)vτ v,x
i
τ
τ
i=1
1 − γ − τ τ1 −1
v |∇v |2 .
τ
In problem (1.1), with α1 = 1, β1 = τ1 , α2 =
2−2γ−2τ
,
τ
β2 = β1 − 1, we have
2α2 β1 − 2α2 − sα2 + 2s(s + 1)α1 + N α1 β12
N
4(1 − γ − τ ) 4(1 − γ − τ ) 2s(1 − γ − τ )
−
−
+ 2s(s + 1) + 2
τ2
τ
τ
τ
1 − γ − 2τ 2 (1 − γ − 2τ )2
4(1 − γ − τ ) 4(1 − γ − τ )
N
= 2(s −
) −
+
−
+ 2
2τ
2τ 2
τ2
τ
τ
1 − γ − 2τ 2
1
= 2(s −
) + 2 [−γ 2 + (4τ − 6)γ + 4τ 2 − 12τ + 2N + 7]
2τ
2τ
= 0.
=
s
From Theorem 2.1 we get |∇(uτ (1+ 2 ) )| ≤ M . Because τ (1 + 2s ) − 1 ≤ 0, we have
s
s
|∇u| ≤ |τ (1 + )|−1 M |u−τ (1+ 2 )+1 | ≤ M2 .
2
We get the Hölder continuity of u with respect to t from [7] directly.
Remark 3.2. For the case N = 10, we take τ as a positive number, say δ, then
similar to the above arguments we can get the result that when γ ≥ 2δ − 3 +
p
δ(1+ s )
2(2δ − 3)2 + 2N − 2 and |∇u0 2 | ≤ M , then u is Lipschitz continuous in x
and Hölder continuous in t with exponent 1/2 in Ω. Since δ is any positive number
which√can be taken small enough so our conclusion for the parameter γ is that when
γ > N − 1 the solution is Lipschitz continuous in x and Hölder continuous in t
with exponent 1/2 in Ω. It is a improved result of the one in [9].
√
Remark 3.3. Let τ = 1 in Example 3.1, we could get the result as γ ≥ 2N − 1.
It is the main result in [3].
Example 3.4. The initial problem for the porous medium equation
ut = ∆(um ),
(x, t) ∈ Ω
u(x, 0) = u0 (x),
√
√
(3.2)
x ∈ RN
m+2
+ 16+2N
− 16+2N
If m > 0, 10+2N7+2N
≤ m ≤ 10+2N7+2N
, and |∇(u0 4 )| ≤ M , then the
viscosity solution u(x, t) is Lipschitz continuous in x and Hölder continuous in t
with exponent 1/2 in Ω.
Proof. In Theorem 2.2, let α1 = m, β1 = m − 1, α2 = 2m(m − 1), β2 = β1 − 1,
s = m−2
2 . Then the result follows.
6
Z. CHEN, Y. ZHAO
EJDE-2007/15
Example 3.5. Consider the initial-value problem for the singular equation
|∇u|2
, (x, t) ∈ Ω
um
u(x, 0) = u0 (x) x ∈ RN
ut = ∆u +
(3.3)
where m ≥ 0. If |∇u0 | ≤ M , then u(x, t) is Lipschitz continuous in x and Hölder
continuous in t with exponent 1/2 in Ω.
Proof. As the proof in Theorem 2.1, In problem (1.1), we take α1 = 1, β1 = 0,
α2 = 2, β2 = −m. From (2.3),
z,t ≤ ∆z + (2u−m
− 2su−1
)
N
X
u,xi z,xi + [(−4m − 2s)u−m+1
+ 2s(s + 1)]z2 u−s−2 .
i=1
Let s = 0, then
z,t ≤ ∆z + 2u−m
N
X
u,xi z,xi .
i=1
Thus kz k∞ ≤ kz0 k∞ and so |∇u| ≤ M . As in the proof in Theorem 2.2, u is
Lipschitz continuous in x and Hölder continuous in t with exponent 1/2 in Ω. Acknowledgements. The authors want to thank the anonymous referee for his/her
valuable comments and suggestions.
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parbolic equation, Trans. AMS 320 (1990), 2, 779-798.
[3] Yungguang Lu and Liwen Qian; Regularity of viscosity solutions of degenerate parabolic
equation, Proc. AMS. vol.130, No.4, 999-1004.
[4] M. Ughi; A dengenerate parabolic equation modelling the spread of an epidemic, Ann. Mat.
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[8] Yunguang Lu and Willi Jäger; On solutions to nonliner Reaction-diffusion-convection equations with dengenerate diffusion, J. Diff. Eqns 170, 1-21 (2001).
[9] Yan-Yan Zhao and Zu-Chi Chen; Regularity of viscosity solutions to a degenerate nonlinear
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Zu-Chi Chen
Department of mathematics, University of Science and Technology of China, Hefei
230026, China
E-mail address: chenzc@ustc.edu.cn
Yan-Yan Zhao
Department of mathematics, University of Science and Technology of China, Hefei
230026, China
E-mail address: yyzhao@mail.ustc.edu.cn