Local properties of non–negative solutions to
some doubly non–linear degenerate parabolic
equations
S. Fornaro∗
U. Gianazza
∗
March 2, 2009
Abstract
In the present paper we study the local behavior of positive weak solutions of a wide
class of doubly non linear degenerate parabolic equations. We show, in particular, some
lower pointwise estimates of such solutions in terms of suitable sub-potentials (dictated
by the structure of the equation) and an alternative form of the Harnack inequality.
AMS subject classification (2000): 35K65, 35B65.
1
Introduction
In the present paper we are interested in local properties of non–negative solutions of
parabolic differential equations having the following form
ut − divA(x, t, u, Du) = 0,
(1.1)
where A is a measurable function defined in E × (0, T ] × RN +1 , for some open set E ⊂ RN
and T > 0, with values in RN and is subject to the structure conditions

 A(x, t, u, η) · η > C0 Φ(|u|)|η|p ,
(1.2)
 |A(x, t, u, η)| 6 C1 Φ(|u|)|η|p−1
for almost all (x, t) ∈ ET = E × (0, T ], for all u ∈ R and η ∈ RN , with p > 2 and C0 , C1
positive constants. The function Φ : [0, +∞[→ [0, +∞[ is assumed to be continuous and
bounded and such that c1 sα 6 Φ(s) 6 c2 sα , for every s ∈ [0, 1], for some α > 0 and
c1 , c2 > 0. In the literature these equations are classified as doubly non–linear and doubly
degenerate since their modulus of ellipticity vanishes whenever either u or Du tends to zero.
The prototype of (1.1)–(1.2) is
ut − div(|u|m−1 |Du|p−2 Du) = 0,
∗ Dipartimento
m > 1,
(1.3)
di Matematica “F. Casorati”, Università degli Studi di Pavia, via Ferrata, 1, 27100, Pavia,
Italy. e-mail: [email protected], [email protected]
which models the filtration of a polytropic non-newtonian fluid in a porous medium. Besides
its physical interpretation, equation (1.3) has been widely studied also from a theoretical
point of view, since it unifies the well–known p–Laplace equation (m = 1) with the classical
porous media equation (p = 2). Therefore, it is interesting to see how much of the regularity
properties of the solutions of the p–Laplace and the porous media equations is preserved in
this more general case.
Simple computations show that the Barenblatt–type functions
p−1
"
p # m+p−3
|x − x̄| p−1
kρν
1−b
,
Γ(x, t; x̄, t̄) = ν
1
γ λ (t)
γ λ (t)
(1.4)
+
with
λ = ν(m + p − 3) + p,
γ(t) = (kρν )m+p−3 (t − t̄) + ρλ ,
t > t̄,
(1.5)
1
m + p − 3 1−p
λ
,
b =
p
are sub-solutions of (1.3) in RN × (t̄, +∞), for every ν > N and k, ρ positive parameters. If
ν = N , then (1.4) turns out to be an exact solution for every choice of k, ρ > 0. As m → 1
and p → 2, Γ converges pointwise to a limit function Γ, which turns out to be a sub–solution
of the heat equation if ν > N and an exact solution if ν = N . For this reason, we may
consider the functions Γ as sub–potentials of (1.3).
It is worth observing that 0 6 Γ 6 k and that Γ has compact support with respect to
the space variables. More precisely, at each time t > t̄, the support of Γ(·, t; x̄, t̄) is the
p−1
1
ball centered at x̄ with radius r(t) = b− p γ λ (t). At t = t̄ the support is BR (x̄), where
p−1
R = b− p ρ. All these properties are useful to investigate the behavior of non–negative
local weak solutions of (1.3). In particular, if u is any such solution in Br(t) (x̄) × [t̄, t] such
that u(·, t̄) > k in BR (x̄), then, by the comparison principle, u(x, s) > Γ(x, s; x̄, t̄), for every
p−1
1
|x − x̄| 6 b− p γ λ (t), t̄ 6 s 6 t. This is the so–called expansion of positivity, which is one
of the main ingredients in proving a Harnack type inequality for u (see [7]).
Concerning the general doubly non–linear equations (1.1)–(1.2), there is no explicit analog of Γ, since only the structure of the equation is prescribed. Moreover, no comparison
principle holds. Nevertheless, one may ask if the functions Γ continue to drive, in some
sense, the local behavior of non–negative weak solutions of (1.1)–(1.2). This question can
be connected to a result of Moser ([6]), which shows that if u is a positive weak solution
of the linear equation
in RN × (0, T ), with measurable coefficients
P ut = div(a(x, t)Du)
−1
2
2
satisfying α |ξ| 6 aij (x, t)ξi ξj 6 α|ξ| , then
c
|x−x |2
t0
− t−t0
0
u(x, t) > γ u(x0 , t0 )e
,
(1.6)
t
for every x, x0 ∈ RN and 0 < t0 < t < T , with γ, c depending only on N and α. Notice that
the lower bound has the same structure as the fundamental solution of the heat equation.
Very recently the same issue has been successfully discussed by E. DiBenedetto, U. Gianazza,
V. Vespri in [3], where the authors prove that non–negative weak solutions of quasi–linear
degenerate parabolic equations of p–Laplace type are locally bounded below by suitable
Barenblatt–like sub–potentials. In this paper, we prove analogous results for equations
2
(1.1)–(1.2). We basically use the same ideas, but it must be remarked that the extension
to our situation is not straightforward, due to the double degeneracy exhibited by the
equations.
The main result, namely Proposition 2.1, is contained in Section 2. In Sections 3 and 4
we analyze some interesting consequences.
For the sake of simplicity, we restrict ourselves to proving all the results for the prototype
(1.3), stressing on the fact that equations with the full quasi–linear structure as (1.1)–(1.2)
can be handled in the same way, without any further effort.
In the sequel, by data we mean the quantities N, p, C0 , C1 , c1 , c2 , α, kΦk∞ , which reduce
to N, p, m for (1.3).
2
Time asymptotic decay
Throughout the section u denotes a non–negative local weak solution of (1.3), u ∈ C(0, T ;
m−1
L2loc (E)) with u p−1 |Du| ∈ Lploc (ET ) such that
u(·, t̄) > k,
in Bρ (x̄),
(2.1)
for some k > 0 and Bρ (x̄) ⊂ E. Set
M = µ(kρν )m+p−3 ,
S(t) = M (t − t̄) + ρλ ,
t > t̄,
ν
h(t) =
kρ
,
ν
S λ (t)
where ν and µ are positive constants to be chosen, λ is as in (1.5), and define
1
D(t) = {(x, s) ∈ RN × R : |x − x̄| < S λ (s), s ∈ (t̄, t)}.
(2.2)
Notice that the domain D(t) imitates the support of Γ. We are going to prove the following
result.
Proposition 2.1 Assume that D(t) ⊂ ET . Then one can choose the parameters λ, µ depending only on the data, such that
u(x, s) >
1
h(s),
2
|x − x̄| <
1 1
S λ (s), s ∈ (t̄, t).
2
For the proof of the previous proposition we need some preliminaries. In the sequel we
denote by C a constant depending only on p, m and that may change from line to line. For
every n = 0, 1, 2, . . ., t > t̄, we set
1
1
1
1
1
hn (t) =
1 + n h(t),
Rn (t) =
1 + n S λ (t)
2
2
2
2
and
Dn (t) = (x, s) ∈ RN × R : |x − x̄| < Rn (s), s ∈ (t̄, t) .
3
We also need the change of variables from (x, s) to (y, σ) given by
ν(m+p−3)
x − x̄
λ
λ
S
(s) − ρν(m+p−3)
y=
,
σ = σ(s) =
Ro (s)
ν(m + p − 3)
(2.3)
which transforms the domain Dn (t) into Bn × (0, τ ), where
ν(m+p−3)
λ
λ
Bn = B 1 (1+ 1n ) (0),
τ = σ(t) =
S
(t) − ρν(m+p−3) .
2
2
ν(m + p − 3)
en = B 1
Moreover, we set B
(1+
2
3
2n+2
) (0) and define
v(y, σ) = u(x, s),
e
hn (σ) = hn (s),
e
h(σ) = h(s).
(2.4)
The crucial point for Proposition 2.1 is represented by the energy estimates given by the
next lemma. Unlike the usual approach, we multiply equation (1.3) by −(u − h(t))q−1
− ζ,
with h(t) and ζ reflecting the structure of the Barenblatt–like functions Γ and q > 1 taking
the double non-linearity into account.
Lemma 2.2 Let q > 1. Then there exist two constants C, ω > 0 depending only on p, m, q
such that for every n ∈ N
Z τZ
Z
p+q−2 p
q
e
(v − hn+1 )− dy + ω
v m−1 Dy v − e
hn+1 − p dy dσ
sup M
06σ6τ
en
B
0
en
B
µ
6 C 2n
λ
Z τZ
+ C 2np
0
(v − e
hn+1 )q− e
hm+p−3 dy dσ
Bn
Z τZ
0
p+q−2
e
hm−1 (v − e
hn+1 )−
dy dσ.
(2.5)
Bn
Proof. We proceed formally, but the proof can be made rigorous by using the Steklov
averages of u (see Proposition 3.1 of Chapter II in [1]). Let ζn = ζn (y) be a cutoff function
having the following properties

1
3

 1
1
+
,
if
|y|
6



2
2n+2



1
3
1
1
1
1
ζn (y) =
2n+3
1 + n+2 6 |y| 6
1+ n ,
1 + n − |y|
if

2
2
2
2
2
2




1
1


if |y| >
1+ n .
 0
2
2
Let us multiply the equation us = div(um−1 |Du|p−2 Du) by
1
p x − x̄
−(u − hn+1 )q−1
(x,
s)
ζ
n
−
Ro (s) RoN (s)
and integrate on Dn (r), with t̄ < r < t. On the left hand side of the equation we change
the space variables from x into y, according to (2.3), getting
ZZ
1
q−1
p x − x̄
−
us (u − hn+1 )− (x, s) ζn
ds dx
N (s)
R
(s)
R
o
Dn (r)
o
Z Z r
=−
us (u − hn+1 )q−1
(Ro (s)y + x̄, s) ζnp (y) ds dy.
−
Bn t̄
4
Since
1 d − us (u − hn+1 )q−1
(u − hn+1 )q− (Ro (s)y + x̄, s)
(Ro (s)y + x̄, s) =
−
q ds
1
− Dx [(u − hn+1 )q− ](Ro (s)y + x̄, s) · yRo0 (s) − h0n+1 (s)(u − hn+1 )q−1
− (Ro (s)y + x̄, s)
q
and hn+1 is decreasing, we obtain
ZZ
1
p x − x̄
−
us (u − hn+1 )q−1
(x,
s)
ζ
ds dx
n
−
N (s)
R
(s)
R
o
Dn (r)
o
Z
1
>
(u − hn+1 )q− (Ro (r)y + x̄, r) ζnp (y) dy
q Bn
Z
1
−
(u − hn+1 )q− (Ro (t̄)y + x̄, t̄) ζnp (y) dy
q Bn
Z Z
1 r
R0 (s)
Dy (u − hn+1 )q− (Ro (s)y + x̄, s) · y ζnp (y) o
−
dy ds.
q t̄ Bn
Ro (s)
By (2.1), the second integral on the right hand side vanishes. By performing an integration
by parts with respect to the space variables, we find
ZZ
1
q−1
p x − x̄
ds dx
−
us (u − hn+1 )− (x, s) ζn
N (s)
R
(s)
R
o
Dn (r)
o
Z
1
>
(u − hn+1 )q− (Ro (r)y + x̄, r) ζnp (y) dy
q Bn
Z Z
M r
1
+
(u − hn+1 )q− (Ro (s)y + x̄, s)Dy ζnp (y) · y
dy ds.
qλ t̄ Bn
S(s)
Next, changing the time variable s into σ (see (2.3)) and recalling (2.4) we obtain
ZZ
1
p x − x̄
−
us (u − hn+1 )q−1
(x,
s)
ζ
ds dx
n
−
N
Ro (s) Ro (s)
Dn (r)
Z
1
(v − e
hn+1 )q− (y, σ(r)) ζnp (y) dy
>
q Bn
Z Z
2n µ τ
−C
(v − e
hn+1 )q− (y, σ) e
h(σ)m+p−3 dy dσ,
qM λ 0 Bn
(2.6)
where we have used the estimate |Dy ζnp · y| 6 C 2n .
Now, we deal with the right hand side of the equation. Integrating by parts, changing
the space variables from x into y and using Hölder and Young inequalities, we find
ZZ
1
q−1
m−1
p−2
p x − x̄
−
divx u
|Dx u| Dx u (u − hn+1 )− (x, s) ζn
ds dx
N (s)
R
(s)
R
o
Dn (r)
o
Z rZ
p+q−2 p
6 −ωp,q
um−1 Dx u − hn+1 − p ζnp (y) dy ds
t̄ B
Z rZ n
dy ds,
+ Cp,q
um−1 |Dx ζn |p (u − hn+1 )p+q−2
−
t̄
Bn
5
where Cp,q , ωp,q depends only on p, q. Finally, transforming the right hand side of the
previous inequality in terms of the function v, we obtain
ZZ
1
q−1
m−1
p−2
p x − x̄
−
divx u
|Dx u| Dx u (u − hn+1 )− (x, s) ζn
ds dx
N
Ro (s) Ro (s)
Dn (r)
Z
Z
p+q−2 p
ωp,q σ(r)
v m−1 Dy v − e
hn+1 − p ζnp dy dσ
6−
M 0
B
Z Zn
2np τ
p+q−2
+ Cp,q
v m−1 (v − e
hn+1 )−
dy dσ.
M 0 Bn
Combining this estimate with (2.6), taking the supremum over r and recalling that ζn ≡ 1
en , we finally get (2.5).
on B
Proof of Proposition 2.1. According to the notation introduced above, we consider the
auxiliary function
1e
w = max v, h .
2
Then it is easy to see that
(w − e
hn+1 )− 6 (v − e
hn+1 )− .
Moreover
2
1−m
Z τZ
0
p+q−2 p
e
hm−1 Dy w − e
hn+1 − p dy dσ
en
B
Z τZ
p+q−2 p
6
wm−1 Dy w − e
hn+1 − p dy dσ
e
0 B
Z τZ n
p+q−2 p
=
v m−1 Dy v − e
hn+1 − p dy dσ,
0
en
B
p+q−2
where the last equality follows from observing that Dy w − e
hn+1 − p vanishes on the set
where w differs from v. Hence, from (2.5) it follows that
Z
Z τZ
p+q−2 p
q
e
e
sup M
(w − hn+1 )− (y, σ) dy + ω
hm−1 Dy w − e
hn+1 − p dy dσ
06σ6τ
en
B
0
µ
6 C 2n
λ
en
B
Z τZ
0
+ C 2np
(v − e
hn+1 )q− e
hm+p−3 dy dσ
Bn
Z τZ
0
p+q−2
e
hm−1 (v − e
hn+1 )−
dy dσ.
Bn
Now, we observe that
(v − e
hn+1 )− 6 e
h χ{w<ehn+1 } 6 2(n+2) (w − e
hn )− χ{w<ehn+1 }
(2.7)
and so, taking also the inequality (w − e
hn+1 )− 6 (w − e
hn )− into account, we get
Z
Z τZ
p+q−2 p
e
sup M
(w − e
hn+1 )q− (y, σ) dy + ω
hm−1 Dy w − e
hn+1 − p dy dσ
06σ6τ
en
B
0
en
B
(m+p−3+q)n
6C2
µ
λ
6
+1
Z τZ
0
Bn
(w − e
hn )m+p−3+q
dy dσ. (2.8)
−
en which equals 1 on Bn+1 ,
Let ζen be a piecewise affine, non-negative, cutoff function in B
n
e
e
vanishes outside Bn and such that |Dζn | 6 C2 . Applying Hölder inequality and the
embedding result given in Proposition 3.1 of Chapter I of [1], with s > 1 to be chosen, we
have
Z τZ h
Z τZ
m+p−3+q ip
m+p−3+q
(w − e
hn+1 )−
dy dσ 6
w−e
hn+1 − p
ζen dy dσ
0
0
Bn+1
s
Z τZ
6 C τ N +s
0
6Cτ
s
N +s
en
B
en
B
N
m+p−3+q p NN+s N +s
hn+1 − p
ζen w−e
Z
w−e
hn+1
sup
06σ6τ
Z τZ
×
0
en
B
s m+p−3+q
p
en
B
−
ζens
Np+s
dy
NN+s
m+p−3+q
p
p
e
e
w
−
h
ζ
.
dy
dσ
D
n+1 −
n (2.9)
In order to estimate the last integral on the right hand side, we compute explicitly the
derivative
m+p−3+q ζen
D w−e
hn+1 − p
=
p+q−2
m+p−3+q
m−1
m+p−3+q e
hn+1 − p + w − e
hn+1 − p
Dζen
ζn w − e
hn+1 − p D w − e
p+q−2
so that
m+p−3+q p
m+p−3+q
p+q−2 p
hn+1 − p
hn+1 −
.
ζen 6 C e
hm−1 D w − e
hn+1 − p + C 2np w − e
D w − e
(2.10)
Choose
pq
m+p−3+q
s=
and note that s > 1 as soon as q satisfies q(p − 1) > m + p − 3. Thus, collecting estimates
(2.9), (2.10) and (2.8) we have proved that
Yn+1 6 C an
+p
N
τ Np+s µ
1+ p−s
N +s
+1
Yn N +s ,
M
λ
N +p
where a = 2(m+p−3+q) N +s and
1
Yn =
τ |Bn |
Set ε =
p−s
N +s .
w−e
hn
0
Bn
m+p−3+q
−
dy dσ.
According to Lemma 5.6 of Chapter II in [5], if
−
Yo 6 a
|
then
Z τZ
(N +s)2
(p−s)2
{zC
=δ
+s
−N
p−s
}
M
τ
n
Yn 6 θ a− ε ,
p
p−s
+p
µ
− Np−s
+1
=: θ,
λ
∀ n ∈ N.
7
(2.11)
From (2.7) it follows that
Z τZ
m+p−3+q
1
Zn+1 :=
v−e
hn+1 −
dy dσ
τ |Bn+1 | 0 Bn+1
Z Z
m+p−3+q
C 2(m+p−3+q)n τ
w−e
hn −
dy dσ = C 2(m+p−3+q)n Yn ,
6
τ |Bn |
0 Bn
n
N +s (m+p−3+q)n
1
. Thus, Zn
for every n ∈ N. Then Zn+1 6 C θ a− ε 2(m+p−3+q) = C θ 2− p−s
tends to 0 as n → +∞ and we obtain the statement, provided (2.11) holds. This requirement
is satisfied if
p
p−s
+p
µ
− Np−s
M
m+p−3+q
.
Yo 6 h(t)
6δ
+1
τ
λ
Recalling the definitions of M, τ , we have
M
τ
p
p−s
>
p
µ p−s
ν(m + p − 3)
λ
p
p−s
h(t)m+p−3+q
and so it suffices to impose that
N
1 6 δµ
p
p−s
λ−p
p
p−s
λ p−s
N +p
.
(µ + λ) p−s
This in turn holds true by choosing λ = µ and then µ large enough to get
2
N +p
p
6δ
p−s
p
(µ − p).
Note that δ → 0 and µ → +∞ as m + p − 3 → 0.
The following two corollaries are now immediate consequences of Proposition 2.1. The first
result concerns the expansion of positivity for the function u. It was already proved in [4] by
a different approach. The second result contains information on the time decay of u. More
ν
precisely, it states that u does not decay faster than t− λ , for t large.
Corollary 2.3 There exists θ > 0, depending only on the data, such that if D(t) ⊂ ET ,
with t = t̄ + θρp k −(m+p−3) , then
θρp
u x, t̄ + m+p−3 > η k,
|x − x̄| 6 2ρ,
k
where η =
1
ν .
2(µθ + 1) λ
Corollary 2.4 If k m+p−3
t − t̄
ρp
> 1 and D(t) ⊂ ET , then
p
1
kλ
u(x, t) >
ν
2 (µ + 1) λ
ρp
t − t̄
λν
8
,
|x − x̄| <
1 1
S λ (t).
2
3
Alternative Harnack inequality
Let θ, ρ > 0, y ∈ RN , s ∈ R and define the following cylinders
p
(y, s) + Q−
ρ (θ) = Bρ (y) × (s − θ ρ , s]
p
(y, s) + Q+
ρ (θ) = Bρ (y) × [s, s + θ ρ )
+
(y, s) + Qρ (θ) = (y, s) + Q−
ρ (θ) ∪ (y, s) + Qρ (θ) .
In [4] it has been proved the following intrinsic Harnack inequality.
Theorem 3.1 Let u be a non-negative, local weak solution to (1.3) in ET and let (x0 , t0 ) ∈
ET be such that u(x0 , t0 ) > 0. Then there exist constants c > 0 and κ > 1, depending only
upon the data, such that for all cylinders (x0 , t0 ) + Q4ρ (θ) ⊂ ET , where
m+p−3
c
,
(3.1)
θ=
u(x0 , t0 )
one has
u(x0 , t0 ) 6 κ inf u(x, t0 + θρp ).
(3.2)
Bρ (x0 )
The previous inequality is called intrinsic because the underlying geometry is dictated by
the solution u, as the choice of θ in (3.1) shows.
As a consequence of Proposition 2.1, we can state and prove an alternative form of
the Harnack inequality, where the geometry can be prescribed a priori independent of the
solution. Estimate (3.3) was first proved in [7] for ν = N and only for the model equation
(1.3).
Theorem 3.2 Let u be a non-negative, local weak solution to (1.3) in ET . Let (x0 , t0 ) ∈ ET
be such that u(x0 , t0 ) > 0 and assume that the cylinder (x0 , t0 ) + Q4ρ (θ), with θ given by
(3.1), is contained in ET . Let D(t) be the set defined in (2.2) with (x̄, t̄) = (x0 , t0 + θρp )
and k = κ−1 u(x0 , t0 ), where κ is given by Theorem 3.1. Suppose that D(t) ⊂ ET . Then
there exist two constants γ1 , γ2 > 0, depending only on the data, such that for every t > t0
1
p m+p−3
λp
ν ρ
t − t0 p
+ β2
.
(3.3)
inf
u(x,
t)
u(x0 , t0 ) 6 β1
t − t0
ρp
Bρ/2 (x0 )
Here λ is determined by Proposition 2.1 and ν is defined in (1.5).
Proof. Fix t > t0 . If t − t0 6 2γρp , where γ = (max{c, κ}/u(x0 , t0 ))m+p−3 then we have
1
p m+p−3
1
ρ
u(x0 , t0 ) 6 2 m+p−3 max{c, κ}
,
t − t0
1
which implies (3.3), with β1 = 2 m+p−3 max{c, κ}. Now assume that t − t0 > 2γρp . By (3.2)
u(x, t0 + θρp ) > κ−1 u(x0 , t0 ),
x ∈ Bρ (x0 ).
Then, applying Corollary 2.4 with t̄ = t0 + θρp , x̄ = x0 , k = κ−1 u(x0 , t0 ) we obtain for every
x ∈ Bρ/2 (x0 )
p p λν
λν
1 κ−1 u(x0 , t0 ) λ
ρp
1 κ−1 u(x0 , t0 ) λ
ρp
u(x, t) >
>
ν
ν
2
t − t0 − θρp
2
t − t0
(µ + 1) λ
(µ + 1) λ
λ
ν
which gives (3.3) again, with β2 = 2 p κ (µ + 1) p .
9
Remark 3.3 Recalling that µ → +∞ as m + p − 3 → 0, by the proof of Theorem 3.2, we
infer that the constants β1 , β2 in (3.3) tend to +∞, as m + p − 3 → 0, unlike the constants
c, κ in Theorem 3.1, which are stable.
Moreover, Theorem 3.2 can be the starting point to investigate the optimal growth of the
initial data as |x| → +∞ for the solvability of the Cauchy problem for the doubly non–linear
case (see Chapter XI in [1]).
Now, assuming Theorem 3.2 we can prove Theorem 3.1 and so the two formulations (3.2)
and (3.3) of the Harnack inequality turn out to be equivalent.
Theorem 3.4 Let u be a non-negative, local weak solution to (1.3) in ET . Let (x0 , t0 ) ∈ ET
be such that u(x0 , t0 ) > 0. Assume that there exist two constants β1 , β2 > 0, depending only
on the data, such that for every t > t0 , x ∈ Bρ (x0 ) ⊂ E
u(x0 , t0 ) 6 β1
ρp
t − t0
1
m+p−3
+ β2
t − t0
ρp
νp λp
inf u(x, t)
,
(3.4)
Bρ (x0 )
where λ and ν are as in (1.5). If (x0 , t0 ) + Q+
ρ (θ) ⊂ ET , with
θ=
2β1
u(x0 , t0 )
m+p−3
,
then (3.2) holds with
λp
(m+p−3)ν
p
κ = 2β2 (2β1 )
.
Proof. It suffices to apply (3.4) with t = t0 + θρp .
4
Sub–potential lower bounds
In this section we derive lower pointwise estimates for non-negative local weak solutions of
(1.3). In particular, Theorem 4.3 shows that any such solution u satisfying (2.1) is locally
bounded below by one of the Barenblatt type sub-solutions (1.4).
Theorem 4.1 Let u be a non-negative, local weak solution to (1.3) in ET . Let (x0 , t0 ) ∈ ET
be such that u(x0 , t0 ) > 0. Then for every (x, t) ∈ ET , with x 6= x0 , B4R (x), B4R (x0 ) ⊂ E,
1
R = |x − x0 |, and 0 < t − t0 < p t0
4
u(x, t) > γ0 u(x0 , t0 ) 1 − γ1
|x − x0 |p
m+p−3
u(x0 , t0 )
(t − t0 )
1 p−1
p−1
m+p−3
,
(4.1)
+
where γ0 , γ1 depend on c, κ, given by Theorem 3.1.
Proof. Let us consider (x, t) ∈ ET , with x 6= x0 and 0 < t − t0 < 4−p t0 . The line joining
(x0 , t0 ) with (x, t) is given by
y − x0 =
x − x0
(s − t0 ),
t − t0
10
s ∈ R.
Besides, we consider the intrinsic p–paraboloid with vertex at (x0 , t0 )
s − t0 = θ |y − x0 |p ,
with θ =
c
u(x0 , t0 )
m+p−3
given by (3.1). They intersect at the point (x1 , t1 ) with
m+p−3
u(x0 , t0 )
t − t0
|x1 − x0 |
=
,
c
|x − x0 |
m+p−3
c
t1 − t0 =
|x1 − x0 |p .
u(x0 , t0 )
p−1
Iterating this procedure, we obtain a finite number of points (xj , tj ), with j = 1, . . . , n, such
that t0 < t1 < · · · < tn 6 t and
m+p−3
u(xj , tj )
t − t0
|xj+1 − xj |p−1 =
,
c
|x − x0 |
(4.2)
m+p−3
c
|xj+1 − xj |p ,
tj+1 − tj =
u(xj , tj )
for every j = 1, . . . , n − 1. Notice that (xn , tn ) is the last point not overcoming (x, t).
According to Theorem 3.1, we can estimate
j = 0, . . . , n − 1,
u(xj , tj ) 6 κ u(xj+1 , tj+1 ),
m+p−3
c
and ρj = |xj+1 − xj |,
u(xj , tj )
is contained in ET . This is the case if tj − (4ρj )p θj > 0, which means, in view of (4.2),
tj −4p (tj+1 −tj ) > 0. This last inequality is true thanks to the assumption 0 < t−t0 < 4−p t0 .
A similar argument holds for the space variables. In particular, we infer that
provided the cylinder (xj , tj ) + Q4ρj (θj ), with θj =
u(xj , tj ) > κ−j u(x0 , t0 ).
(4.3)
It follows that
|x − x0 | >
n−1
X
|xj+1 − xj | =
j=0
t − t0
|x − x0 |
t − t0
|x − x0 |
>
=
where q = κ−
m+p−3
p−1
1
p−1
1
p−1
t − t0
m+p−3
c
|x − x0 |
1
1
p−1
n−1
X
u(xj , tj )
m+p−3
p−1
j=0
m+p−3
n−1
p−1
X
m+p−3 j
u(x0 , t0 )
κ− p−1
c
j=0
u(x0 , t0 )
c
m+p−3
p−1
1 − qn
,
1−q
(4.4)
. Then, combining (4.3) with j = n and (4.4), we get
u(xn , tn )
u(x0 , t0 )
m+p−3
p−1
> q n > 1 − γ1
11
|x − x0 |p
m+p−3
u(x0 , t0 )
(t − t0 )
1
p−1
,
whenever the right hand side is positive. Here γ1 =
explicitly, we have established that
u(xn , tn ) > u(x0 , t0 ) 1 − γ1
m+p−3
m+p−3
κ p−1 − 1 cκ−1 p−1 . More
|x − x0 |p
m+p−3
u(x0 , t0 )
(t − t0 )
1 p−1
p−1
m+p−3
.
+
Set γ0 = min{1, κ−1 }. If (xn , tn ) = (x, t), we are done. Otherwise, we apply once more the
Harnack inequality to get u(x, t) > κ−1 u(xn , tn ) and thus to complete the proof.
Remark 4.2 The proof above relies on the same idea used by Moser to prove (1.6). In our
case, however, the intrinsic paraboloids with vertices (xj , tj ) are not homothetic. Letting
m → 1, p → 2 in (4.1) one recovers Moser’s estimate.
Theorem 4.3 Let u be a non-negative, local weak solution of (1.3) in ET satisfying
u(·, t̄) > k,
in Bρ (x̄),
for some k > 0 and Bρ (x̄) ⊂ E. Then, for all (x, t) ∈ ET with x 6= x̄, B4R (x), B4R (x̄) ⊂ E,
R = |x − x̄| and 0 < 4p (t − t̄) 6 2t̄, µ (kρν )m+p−3 (t − t̄) > 2ρλ
γ0 kρν
u(x, t) >
ν
2 S λ (t)
(
1 − γ1
|x − x̄|
p−1
p )
p−1
m+p−3
,
1
S λ (t)
+
where ν, µ, λ, S(t) are given by Proposition 2.1 and γ1 is a suitable positive constant depending on the data.
Proof. Let (x, t) ∈ ET be as in the statement. Applying Proposition 2.1, we have u(x̄, t0 ) >
ν
1 kρ
, where t0 = (t + t̄)/2, and then, in particular
ν
2 λ
S
(t0 )
u(x̄, t0 ) >
1 kρν
.
ν
2 S λ (t)
(4.5)
On the other hand, by Theorem 4.1 we get
u(x, t) > γ0 u(x̄, t0 ) 1 − γ1
|x − x̄|p
m+p−3
u(x̄, t0 )
(t − t0 )
1 p−1
p−1
m+p−3
.
+
Now, taking (4.5) into account, the theorem is proved.
References
[1] E. DiBenedetto, Degenerate parabolic equations, Springer-Verlag, New York, (1993).
[2] E. DiBenedetto, Intrinsic Harnack type inequalities for solutions of certain degenerate parabolic equations, Arch. Ration. Mech. Anal., 100 (2), (1988),129-147.
12
[3] E. DiBenedetto, U. Gianazza, V. Vespri, Sub-potential lower bounds for nonnegative solutions to certain quasi-linear degenerate parabolic equations, Duke Math.
J., 143 Number 1, (2008), 1-15.
[4] S. Fornaro, M. Sosio, Intrinsic Harnack estimates for some doubly nonlinear degenerate parabolic equations, Adv. Differential Equations, 13 Numbers 1-2, (2008),
139-168.
[5] O.A. Ladizhenskaja, V.A. Solonnikov, N.N. Ural’ceva, Linear and quasilinear
equations of parabolic type, Nauka, Moskow 1967 (Russian). English transl.: American
Mathematical Society, Providence, 1968.
[6] J. Moser, A Harnack inequality for parabolic differential equations, Comm. Pure Appl.
Math., 17 (1964), 101–134.
[7] V. Vespri, Harnack type inequalities for solutions of certain doubly nonlinear parabolic
equations, J. Math. Anal. Appl., 181 (1994), 104–131.
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