Electronic Journal of Differential Equations, Vol. 2012 (2012), No. 200, pp. 1–21.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
HÖLDER REGULARITY FOR SIGNED SOLUTIONS TO
SINGULAR POROUS MEDIUM TYPE EQUATIONS
SIMONA PUGLISI
Abstract. We prove Hölder regularity for bounded signed solution to singular
porous medium type equations, whose prototype is
ut − div m|u|m−1 Du = 0
weakly in ET ,
with m ∈ (0, 1).
1. Introduction and statement of main result
Let E be an open set in RN , for T > 0 denote the cylindrical domain
ET = E × (0, T ]
and let Γ = ∂ET \ Ē × {T } be its parabolic boundary. We consider quasi-linear
homogeneous singular parabolic partial differential equation
ut − div A(x, t, u, Du) = 0
weakly in ET ,
(1.1)
where A : ET × RN +1 → RN is measurable and subject to the structure conditions
(
A(x, t, z, ξ) · ξ ≥ C0 m|z|m−1 |ξ|2
(1.2)
|A(x, t, z, ξ)| ≤ C1 m|z|m−1 |ξ|
for a.e. (x, t) ∈ ET , for every z ∈ R, ξ ∈ RN , where C0 , C1 are given positive
constants and 0 < m < 1.
The prototype of this class of parabolic equations is the porous medium equation
ut − div m|u|m−1 Du = 0
weakly in ET .
The modulus of ellipticity of this class of parabolic equations is m|u|m−1 . Whenever
m > 1, such a modulus vanishes when u vanishes, and for this reason we say that
the equation (1.1)-(1.2) is degenerate. Whenever 0 < m < 1, such a modulus
approaches infinity as u → 0, and for this reason we say that the equation (1.1)(1.2) is singular. One also speaks about slow, when m > 1, or fast diffusion, when
0 < m < 1 (see the monograph [8]).
2000 Mathematics Subject Classification. 35K67, 35B65, 35B45.
Key words and phrases. Singular parabolic equations; Hölder continuity.
c
2012
Texas State University - San Marcos.
Submitted July 17, 2012. Published November 15, 2012.
1
2
S. PUGLISI
EJDE-2012/200
We are interested only in local solutions to singular porous medium type equation. The parameters {N, m, C0 , C1 } are the data, and we say that a generic constant γ = γ(N, m, C0 , C1 ) depends upon the data, if it can be quantitatively determined a priori only in terms of the indicated parameters. As usual, in the following
the constant γ may change from line to line.
Let us give the notion of weak solution for this kind of equations as follows.
1
A function u ∈ Cloc (0, T ; L2loc (E)) with |u|m ∈ L2loc (0, T ; Hloc
(E)) is a local weak
sub(super)-solution to (1.1) if for every compact set K ⊂ E and every subinterval
[t1 , t2 ] ⊂ (0, T ]
Z
Z t2 Z
t2
uϕ dxt +
[−uϕt + A(x, t, u, Du) · Dϕ] dx dt ≤ (≥) 0,
1
K
t1
K
1
for all non-negative test functions ϕ ∈ Hloc
(0, T ; L2 (K)) ∩ L2loc (0, T ; H01 (K)).
Our aim is to show that locally bounded, local, weak solutions of variable sign
to our problem (1.1)-(1.2), with 0 < m < 1, are locally Hölder continuous.
Let us introduce the parabolic m-distance of a compact set K ⊂ ET from the
parabolic boundary Γ in the following way
1−m
m-dist(K, Γ) =
inf
kuk∞,2 ET |x − y| + |t − s|1/2 .
(x,t)∈K,(y,s)∈Γ
We can state the main result of this paper as follows.
Theorem 1.1. Let u be a bounded, local, weak solution to (1.1)-(1.2). Then u is
locally Hölder continuous in ET and there exist constants c > 1 and α ∈ (0, 1) such
that for every compact set K ⊂ ET
|u(x1 , t1 ) − u(x2 , t2 )| ≤ c kuk∞,ET
kuk
1−m
2
∞,ET |x1
− x2 | + |t1 − t2 |1/2 α
m-dist(K, Γ)
,
for every pair of points (x1 , t1 ), (x2 , t2 ) ∈ K.
The constant c depends only upon the data, the norm kuk∞,K and m-dist(K, Γ);
the constant α depends only upon the data and the norm kuk∞, K .
In some physical applications it is natural to consider positive solutions to quasilinear parabolic equations of the form (1.1), and it is also a very useful simplification
from the mathematical point of view. Therefore, most of the papers directly deal
with positive solutions.
A Hölder regularity result for signed solutions was obtained first by DiBenedetto
in [3] for degenerate (p > 2) p-laplacian type equations and then by Chen and
DiBenedetto in [1] for singular (1 < p < 2) p-laplacian type ones (see also [4]).
Later on, in 1993 Porzio and Vespri [7] considered the case of a degenerate doubly
non-linear equation, whose prototype is
ut − div |u|m−1 |Du|p−2 Du = 0,
for p ≥ 2 and m ≥ 1. Notice that this kind of equations admits as a particular
case both the degenerate p-laplacian type equations (for m = 1 and p > 2) and the
degenerate porous medium type equations (for p = 2 and m > 1). As a consequence,
it only remained open the case of the singular porous medium type equations.
We want to point out that the difficulty in our case is due to the presence of the
term |u|m−1 in the modulus of continuity; indeed, the fact that u changes sign plays
a crucial role here. In the p-laplacian case, the modulus of continuity is |Du|p−2 ,
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HÖLDER REGULARITY FOR SIGNED SOLUTIONS
3
thus the proof does not change if u is positive or if it changes sign. One could think
to follow the lines of [1] with minor modification, but at some point it will appear
|u|m−1 that one cannot control from above in a sublevel of the modulus of u, being
0 < m < 1.
An important point of our strategy is to work with a different equation, apparently more complicated, but instead easier to handle, to which we can reduce,
thanks to a change of variables introduced by Vespri in [9]. We will apply a technique due to DiBenedetto [3, 4] via an alternative argument; we will write energy
estimates for super(sub)-solutions and logarithmic estimates. We notice that, due
to the change of variables, our logarithmic function has to be different by the usual
one (see for instance [4]). Then we will use the so-called reduction of oscillation
procedure: the Hölder continuity of a solution u to the transformed equation (2.2)
will be heuristically a consequence of the following fact: for every (x0 , t0 ) ∈ ET
there exists a family of nested and shrinking cylinders in which the essential oscillation of u goes to zero in a way that can be quantitatively determined in terms
of the data. Since this result is well known for non-negative solutions (see [4, 5]),
it will suffice to consider the case in which the infimum of our solution is negative
and the supremum is positive.
2. Change of variables
To justify some of the following calculations, we assume u to be smooth. In no
way this is a restrictive assumption: indeed the modulus of continuity of u will play
no role in the forthcoming calculations.
Let us consider n ∈ N such that
1
,
n>
m
and define
|v|n−1 v = u,
which is equivalent to
1
v = |u| n −1 u.
Notice that
Du = n|v|n−1 Dv,
Dv =
1 1 −1
|u| n Du.
n
With this substitution equation (1.1) becomes
e t, v, Dv) = 0
|v|n−1 v t − div A(x,
weakly in ET ,
where
e t, v, Dv) = A(x, t, u, Du)
A(x,
.
u=|v|n−1 v
Now, let us see what the structure conditions become. We have
e t, v, Dv) · Dv = 1 |u| n1 −1 A(x, t, u, Du) · Du
A(x,
n
1
m
≥
C0 |u| n +m−2 |Du|2
n
= nmC0 |v|1+nm−2n |v|2(n−1) |Dv|2
= nmC0 |v|nm−1 |Dv|2 ;
since the exponent is nm − 1 > 0, the equation is “degenerate”.
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S. PUGLISI
EJDE-2012/200
In the same way
e t, v, Dv)| = |A(x, t, u, Du)| ≤ mC1 |u|m−1 |Du|
|A(x,
= mC1 |v|n(m−1) n|v|n−1 |Dv| = nmC1 |v|nm−1 |Dv|.
If we denote our variable with u again, we are then led to consider equations of the
type
e t, u, Du) = 0 weakly in ET ,
(|u|n−1 u)t − div A(x,
with structure conditions
(
e t, z, ξ) · ξ ≥ nmC0 |z|nm−1 |ξ|2
A(x,
(2.1)
e t, z, ξ)| ≤ nmC1 |z|nm−1 |ξ|,
|A(x,
for a.e. (x, t) ∈ ET and for every z ∈ R, ξ ∈ RN .
Without loss of generality, we can assume n to be odd; in this case
|u|n−1 u = un ,
and we can rewrite the equation as
e t, u, Du) = 0
(un )t − div A(x,
weakly in ET .
(2.2)
Hence we have reduced problem (1.1)-(1.2) to (2.2) with structure conditions (2.1).
Let us now see what the notion of weak solution becomes in this new setting. A
1
(E)) is a
function u such that un ∈ Cloc (0, T ; L2loc (E)) with |u|nm ∈ L2loc (0, T ; Hloc
local weak sub(super)-solution to (2.2) if for every compact set K ⊂ E and every
subinterval [t1 , t2 ] ⊂ (0, T ]
Z
Z t2 Z
t2
e t, u, Du) · Dϕ] dx dt ≤ (≥) 0,
un ϕ dxt1 +
[−un ϕt + A(x,
K
t1
K
1
(0, T ; L2 (K)) ∩ L2loc (0, T ; H01 (K)).
for all non-negative test functions ϕ ∈ Hloc
3. Preliminaries
Let r, s ≥ 1 and let us consider the Banach spaces
V r,s (ET ) = L∞ 0, T ; Lr (E) ∩ Ls 0, T ; W 1,s (E) ,
V0r,s (ET ) = L∞ 0, T ; Lr (E) ∩ Ls 0, T ; W01,s (E) ,
both equipped with the norm
kvkV r,s (ET ) = ess sup kv(·, t)kr,E + kDvks,ET ;
0<t<T
when r = s, let V (ET ) = V (ET ) and V0r,r (ET ) = V0r (ET ). Both spaces are
embedded in Lq (ET ), for some q > s (for a proof one can see [4]).
r,r
r
Proposition 3.1. If v ∈ V0r,s (ET ), then there exists a positive constant γ, depending only upon N, r, and s, such that
Z
ZZ
ZZ
s/N
|v|q dx dt ≤ γ q
|Dv|s dx dt ess sup
|v|r dx
ET
ET
0<t<T
E
with q = s NN+r . In particular
kvkq,ET ≤ γkvkV r,s (ET ) .
Note that, taking r = s in the previous proposition, and applying Hölder inequality, one obtains the following result.
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HÖLDER REGULARITY FOR SIGNED SOLUTIONS
5
Proposition 3.2. If v ∈ V0r (ET ), then there exists a positive constant γ depending
only upon N and r, such that
r
kvkrr,ET ≤ γ {|v| > 0} N +r kvkrV r (ET ) .
Given (y, s) ∈ ET , and λ, R > 0, we will denote by KR (y) the cube centered at
y with edge 2R; i.e.,
n
o
KR (y) = x ∈ RN : max |xi − yi | < R ,
1≤i≤N
and let ∂KR (y) be its boundary. Let (y, s) + QR (λ) be the generic cylinder
(y, s) + Qρ (λ) = Kρ (y) × [s − λ, s].
If k ∈ R, introduce the truncated functions
(u − k)± = max{±(u − k), 0}.
The following lemma, proved in [2], will be very useful in the sequel.
Lemma 3.3. Let v ∈ W 1,1 (Kρ (y)) and let k, l ∈ R, with k < l. There exists a
constant γ = γ(N, p) independent of k, l, v, y, ρ such that
Z
ρN +1
(l − k)|{v > l}| ≤ γ
|Dv| dx.
(3.1)
|{v < k}| {k<v<l}
Let us state now a lemma on fast geometric convergence one can find in [2]; for
a simple proof see again [4] and [6].
Lemma 3.4. Let {Yn }n∈N be a sequence of positive numbers satisfying
Yn+1 ≤ Cbn Yn1+α ,
being C, b > 1 and α > 0. If
1
1
Y0 ≤ C − α b− α2 ,
then Yn converges to 0, as n tends to +∞.
Let us prove energy estimates we will need later. We start with estimates for
super-solutions, then we will state the analogous ones for sub-solutions.
Proposition 3.5 (Energy estimates for super-solutions). Let u be a local, weak
super-solution to (2.1)-(2.2) in ET . There exists a positive constant γ, depending
only upon the data, such that for every cylinder (y, s) + QR (λ) ⊂ ET , every level
k ∈ R and every non-negative, piecewise smooth cutoff function ζ vanishing on
∂KR (y),
Z
Z k
ess sup
(k − s)+ sn−1 ds ζ 2 (x, t) dx
s−λ<t≤s KR (y)
u
ZZ
2
+
|u|nm−1 D[(u − k)− ζ] dx dτ
(y,s)+QR (λ)
≤γ
nZ
Z
KR (y)
k
(k − s)+ sn−1 ds ζ 2 (x, s − λ) dx
u
ZZ
+
(y,s)+QR (λ)
ZZ
+
(y,s)+QR (λ)
Z
k
(k − s)+ sn−1 ds ζ|ζτ | dx dτ
u
o
|u|nm−1 (u − k)2− |Dζ|2 dx dτ .
(3.2)
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S. PUGLISI
EJDE-2012/200
Proof. After a translation we may assume that (y, s) coincides with the origin and
it suffices to prove (3.2) for the cylinder QR (λ). In the weak formulation of (2.2),
take the test function
ϕ = −(u − k)− ζ 2
over Qt = KR × (−λ , t], where −λ < t ≤ 0.
Taking into account that
∂ ∂τ
Z
k
(k − s)+ sn−1 ds = −un−1 (u − k)− uτ ,
u
and estimating the various terms separately, we have first
Z
∂ k
(k − s)+ sn−1 ds ζ 2 dx dτ
(u )τ (u − k)− ζ dx dτ = n
u
Qt
Qt ∂τ
Z Z k
≥n
(k − s)+ sn−1 ds ζ 2 (x, t) dx
ZZ
−
n
2
ZZ
KR
u
Z
Z
−n
k
(k − s)+ sn−1 ds ζ 2 (x, −λ) dx
KR
u
ZZ
Z
− 2n
Qt
k
(k − s)+ sn−1 ds ζ|ζτ | dx dτ.
u
From the structure conditions (2.1) and Young’s inequality it follows that
ZZ
e τ, u, Du)D (u − k)− ζ 2 dx dτ
−
A(x,
Q
Z Zt
e τ, u, Du)D(u − k)− ζ 2 dx dτ
=−
A(x,
Q
Z Zt
e τ, u, Du)(u − k)− ζDζ dx dτ
−2
A(x,
Qt
ZZ
≥ nmC0
|u|nm−1 |D(u − k)− |2 ζ 2 dx dτ
Qt
ZZ
− 2nmC1
|u|nm−1 |D(u − k)− |(u − k)− ζ|Dζ| dx dτ
Qt
ZZ
2
C0
≥ nm
|u|nm−1 D[(u − k)− ζ] dx dτ
2
Qt
ZZ
C12
|u|nm−1 (u − k)2− |Dζ|2 dx dτ.
− 2nm
C0
Qt
Combining these estimates and taking the supremum over t ∈ (−λ, 0], completes
the proof.
Proposition 3.6 (Energy estimates for sub-solutions). Let u be a local, weak subsolution to (2.1)-(2.2) in ET . There exists a positive constant γ, depending only
upon the data, such that for every cylinder (y, s) + QR (λ) ⊂ ET , every level k ∈ R
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HÖLDER REGULARITY FOR SIGNED SOLUTIONS
7
and every non-negative, piecewise smooth cutoff function ζ vanishing on ∂KR (y),
Z
Z u
ess sup
(s − k)+ sn−1 ds ζ 2 (x, t) dx
s−λ<t≤s KR (y)
k
ZZ
2
+
|u|nm−1 D[(u − k)+ ζ] dx dτ
(y,s)+QR (λ)
≤γ
nZ
Z
KR (y)
u
(s − k)+ sn−1 ds ζ 2 (x, s − λ) dx
(3.3)
k
ZZ
Z
+
(y,s)+QR (λ)
ZZ
u
(s − k)+ sn−1 ds |ζτ | dx dτ
k
o
|u|nm−1 (u − k)2+ |Dζ|2 dx dτ .
+
(y,s)+QR (λ)
Proof. The proof is analogous to the previous one; we just need to take the test
function ϕ = (u − k)+ ζ 2 and observe that
Z
∂ u
(s − k)+ sn−1 ds = un−1 (u − k)+ uτ .
∂τ
k
Let us introduce the logarithmic function
ψ(H n , (un − k n )+ , ν n ) = log+
Hn
,
H n − (un − k n )+ + ν n
where
Hn =
ess sup (un − k n )+ ,
0 < ν n < min{1, H n },
(y,s)+QR (λ)
and for s > 0
log+ s = max{log s, 0}.
Proposition
(2.2) in ET .
such that for
non-negative,
3.7 (Logarithmic estimates). Let u be a local, weak solution to (2.1)There exists a positive constant γ, depending only upon the data,
every cylinder (y, s) + QR (λ) ⊂ ET , every level k ∈ R and every
piecewise smooth cutoff function ζ = ζ(x)
Z
ess sup
ψ 2 H n , (un − k n )+ , ν n (x, t) ζ 2 (x) dx
s−λ<t≤s KR (y)
Z
≤
ψ 2 (H n , (un − k n )+ , ν n ) (x, s − λ)ζ 2 (x) dx
(3.4)
KR (y)
ZZ
|u|n(m−1) ψ (H n , (un − k n )+ , ν n ) |Dζ|2 dx dτ.
+γ
(y,s)+QR (λ)
Proof. Again we assume that (y, s) coincides with the origin. Put v = un and, in
the weak formulation of (2.2), take the test function
∂ψ 2 2
ζ = 2ψψ 0 ζ 2 ,
∂v
over Qt = KR × (−λ, t], where −λ < t ≤ 0.
By direct calculation
00
ψ 2 = 2(1 + ψ)(ψ 0 )2 ∈ L∞
loc (ET ),
ϕ=
(3.5)
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S. PUGLISI
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which implies that such a ϕ is an admissible testing function. Estimating the
various terms separately, we have
ZZ
ZZ
∂ψ 2 2
∂ 2 2
vτ
ζ dx dτ =
ψ ζ dx dτ
∂v
Qt
Q ∂τ
Z t
Z
2
2
=
ψ (x, t)ζ (x) dx −
ψ 2 (x, −λ)ζ 2 (x) dx;
KR
KR
using (3.5) and the structure conditions (2.1)
ZZ
2 e τ, u, Du)D ∂ψ ζ 2 dx dτ
A(x,
∂v
Qt
ZZ
ZZ
2 00 2
e
e τ, u, Du)Dζ dx dτ
=
A(x, τ, u, Du)Dv(ψ ) ζ dx dτ + 2
(ψ 2 )0 ζ A(x,
Qt
Qt
ZZ
e τ, u, Du)Du(1 + ψ)(ψ 0 )2 ζ 2 dx dτ
= 2n
un−1 A(x,
Qt
ZZ
e τ, u, Du)Dζ dx dτ
+4
ψψ 0 ζ A(x,
Qt
ZZ
2
≥ 2n mC0
un−1 |u|nm−1 |D(u − k)+ |2 (1 + ψ)(ψ 0 )2 ζ 2 dx dτ
Qt
ZZ
− 4nmC1
|u|nm−1 |D(u − k)+ |ζ|Dζ|ψψ 0 dx dτ.
Qt
Applying Young’s inequality, we obtain
ZZ
2 e τ, u, Du)D ∂ψ ζ 2 dx dτ
A(x,
∂v
Qt
ZZ
≥ 2nm(nC0 − C1 ε2 )
|u|nm−1 |u|n−1 |D(u − k)+ |2 ψ(ψ 0 )2 ζ 2 dx dτ
Qt
ZZ
C1
− 2nm 2
|u|n(m−1) |Dζ|2 ψ dx dτ.
ε
Qt
Combining these estimates, discarding the term with the gradient on the left-hand
side, and taking the supremum over t ∈ (−λ , 0], proves the proposition.
4. Reduction of the oscillation
To obtain the Hölder regularity, we argue as usual with this kind of estimate by
a reduction-of-oscillation procedure. Let us state the basic result.
Theorem 4.1. Let (y, s) ∈ ET , and ρ, ω > 0 such that
(2ρ)2 (y, s) + Q2θρ nm−1 ⊂ ET ,
ess osc
u ≤ ω,
ω
(2ρ)2
(y,s)+Q2θρ
ω nm−1
where
θ=ω
1−n
2
.
Then, there exist η∗ , c0 ∈ (0, 1), depending only upon data, such that
ess osc
u ≤ η∗ ω ,
∗
Q
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HÖLDER REGULARITY FOR SIGNED SOLUTIONS
9
where
c0 1−nm
ω
.
2
As we show at the end, the local Hölder continuity of locally bounded solutions
is a straightforward consequence of Theorem 4.1. The proof of this theorem splits
into two alternatives.
Let ∈ (0, 1), R > 0, and (y, s) ∈ ET . Consider the cylinder
Q∗ = (y, s) + Qθρ θ∗ ρ2 ,
Q := K
R1−
n−1
2
θ∗ =
(y) × (s − R2−(nm−1) , s] ⊂ ET ,
and set
µ+ ≥ ess sup u ,
µ− ≤ ess inf u ,
(y,s)+Q
(y,s)+Q
ω = µ+ − µ− .
Let us recall that, without loss of generality, we can assume µ+ > 0, µ− < 0 and
µ+ ≥ |µ− |.
Indeed, otherwise just change the sign of u and work with the new function.
If we take 2ρ < R, and assume without loss of generality
ω > R ,
(4.1)
then we guarantee that
(y, s) + Q2θρ
(2ρ)2 ⊂ Q .
ω nm−1
5. The first alternative
We distinguish two alternatives; the first of them consists in assuming
n
(2ρ)2 (2ρ)2 o
ωo n
∩ (y, s) + Q2θρ nm−1 ≤ c0 Q2θρ nm−1 ,
u < µ− +
2
ω
ω
being c0 ∈ (0, 1) a constant to be determined later.
Let us prove now the following De Giorgi type lemma.
(5.1)
Lemma 5.1. There exists a number c0 ∈ (0, 1), depending only upon data, such
that if (5.1) holds, then
ρ2 ω
(5.2)
u ≥ µ− +
a.e. in (y, s) + Qθρ nm−1 .
4
ω
Proof. Without loss of generality we may assume (y, s) = (0, 0) and for k = 0, 1, . . .,
set
2
ρ
e k = Kθρ , Q
ek = K
e k × − ρk , 0 .
ρk = ρ + k , K
k
nm−1
2
ω
e k vanishing on the parabolic
Let ζk be a piecewise smooth cutoff function in Q
e k such that 0 ≤ ζk ≤ 1, ζk = 1 in Q
e k+1 and
boundary of Q
|Dζk | ≤
2k+2 n−1
ω 2 ,
ρ
0 ≤ ζk,t ≤
2k nm−1
ω
.
ρ2
We consider the following levels
ω
ω
+ k+2
4
2
ω
ω
hk = µ− + 5 + k+5
2
2
hk = µ− +
ω
if µ− ≥ − ,
8
ω
if µ− < − .
8
(5.3)
10
S. PUGLISI
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We first treat the least favorable case in which u might be close to zero; i.e., we
assume first that
ω
µ− ≥ − .
(5.4)
8
e k , to obtain
Write down the energy estimates (3.2) for (u − hk )− over the cylinder Q
Z Z hk
ess sup
(hk − s)+ sn−1 ds ζk2 (x, t) dx
−
ρ2
k
ω nm−1
<t≤0
ZZ
ek
K
u
2
|u|nm−1 D[(u − hk )− ζk ] dx dτ
+
ek
Q
≤γ
ek
Q
ZZ
+
hk
Z
n ZZ
(hk − s)+ sn−1 ds |ζk,τ | dx dτ
u
o
|u|nm−1 (u − hk )2− |Dζk |2 dx dτ .
ek
Q
Let us introduce the truncation
ω
v = max u, 4 ,
2
in order to estimate the terms with the integral over [u, hk ]; we have
Z hk
Z hk
(hk − s)+ sn−1 ds ≥
(hk − s)+ sn−1 ds
u
v
ω n−1 (v − h )2
− hk )2−
k −
≥ 4
.
≥v
2
2
2
On the other hand, as (u − hk )− ≤ ω and − ω8 ≤ µ− < 0, we have
Z hk
(u − hk )2−
ω n+1
≤
.
(hk − s)+ sn−1 ds ≤ hn−1
k
2
2
u
(5.5)
n−1 (v
By the definition of v, we obtain
ZZ
2
v nm−1 D[(v − hk )− ζk ] dx dτ
e
Qk
ZZ
2
=
|u|nm−1 D[(u − hk )− ζk ] dx dτ
ω
e
Qk ∩{u> 24 }
ZZ
ω nm−1 ω
2
+
−
h
|Dζk |2 dx dτ
k
4
2
e k ∩{u≤ ω } 24
−
Q
24
ZZ
2
22(k+1) n(m+1)
ω
|Ak | ,
≤
|u|nm−1 D[(u − hk )− ζk ] dx dτ +
ρ2
ek
Q
(5.6)
(5.7)
where
ek .
Ak = {u < hk } ∩ Q
Let us observe that
ek := {v < hk } ∩ Q
ek .
Ak = A
(5.8)
ek follows by the definition of v; let us now prove the
Indeed, the inclusion Ak ⊇ A
other one: if v = u there is nothing to prove; if v = 2ω4 , by (5.4) we have
ω
ω
ω
ω
ω
hk = µ− + + k+2 ≥ + k+2 ≥ 4 .
4
2
8
2
2
EJDE-2012/200
HÖLDER REGULARITY FOR SIGNED SOLUTIONS
11
Taking into account that |u| ≤ ω, (5.5)-(5.8) yield
ZZ
Z
ω n−1
2
nm−1 2 2
dx dτ
ess
sup
v
D[(v
−
h
)
ζ
]
(v
−
h
)
ζ
(x,
t)
dx
+
k
−
k
k
−
k
24
ek
ek
ρ2
Q
K
k
−
ω nm−1
<t≤0
2k
≤γ
2
ek |,
ω n(m+1) |A
ρ2
and again, thanks to the definition of v, it follows that
Z
ω n(m−1) Z Z 2 2
D[(v − hk )− ζk ]2 dx dτ
ess sup
(v − hk )− ζk (x, t) dx + 4
2
ek
ek
ρ2
Q
K
k
−
ω nm−1
≤γ
<t≤0
22k nm+1 e
ω
|Ak |.
ρ2
(5.9)
The change of variables
x̄ = x θ−1 ,
t̄ = ω nm−1 τ
e k into Kρ , and the cylinder Q
e k into Qk = Kρ × (−ρ2 , 0]. With
maps the cube K
k
k
k
(x̄, t̄) → u(x̄, t̄) denoting again the transformed function, the assumption (5.1) of
the lemma implies
n
ωo
∩ Q0 ≤ c0 |Q0 |.
(5.10)
v < µ− +
2
Performing such a change of variables in (5.9), we have
ZZ
Z
2 2
D[(v − hk )− ζk ]2 dx̄dt̄
(v − hk )− ζk (x̄, t) dx̄ +
ess sup
−ρ2k <t≤0
Kρk
Qk
2k
≤γ
2
ω 2 |Āk |,
ρ2
where
Āk = {v < hk } ∩ Qk .
This implies
22k 2
ω |Āk |.
k)
ρ2
Then from Proposition 3.2 with r = 2 and (5.11), one obtains
ZZ
ZZ
(v − hk )2− dx̄dt̄ ≤
(v − hk )2− ζk2 dx̄dt̄
(v − hk )− ζk 2 2
V (Q
Qk+1
≤γ
(5.11)
Qk
2
2
≤ γ|{v < hk } ∩ Qk | N +2 (v − hk )− ζk V 2 (Q
k)
2
22k
≤ γ 2 ω 2 |Āk |1+ N +2 ;
ρ
the left-hand side is estimated by
ZZ
ZZ
(v − hk )2− dx̄dt̄ =
(hk − v)2 dx̄dt̄
Qk+1 ∩{v<hk }
Qk+1
ZZ
≥
≥
(hk
Qk+1 ∩{v<hk+1 }
(hk − hk+1 )2 |Āk+1 |
− v)2 dx̄dt̄
12
S. PUGLISI
EJDE-2012/200
ω 2
|Āk+1 |.
2k+3
Combining the previous estimates yields
=
|Āk+1 | ≤ γ
2
24k
|Āk |1+ N +2 ,
ρ2
and setting
|Āk |
,
|Qk |
Yk =
it follows that
1+
2
Yk+1 ≤ γ 24k Yk N +2 .
Thanks to Lemma 3.4, we deduce that Yk tends to zero as k → ∞, provided
| v < µ− + ω2 ∩ Q0 |
N +2
2
|{v < h0 } ∩ Q0 |
Y0 =
=
≤ γ − 2 2−(N +2) ,
|Q0 |
|Q0 |
N +2
2
that is (5.10), with c0 := γ − 2 2−(N +2) .
Therefore,
ω
v ≥ µ− +
a.e. in Kρ × (−ρ2 , 0].
4
Returning to the variables x, t, we have
ρ2 ω
v ≥ µ− +
(5.12)
a.e. in Qθρ nm−1 ;
4
ω
ρ2
and, consequently, (5.2). In fact, by contrathis implies that u = v in Qθρ ωnm−1
ρ2
such that v(x, t) = 2ω4 , by (5.12)
diction, if there were a point (x, t) ∈ Qθρ ωnm−1
and (5.4), we would obtain
ω
ω
ω
≥ µ− + ≥ .
24
4
8
Assume now that (5.4) is violated; that is, µ− < − ω8 . Choosing the levels hk
according to (5.3), we have
ω
ω
ω
ω
ω
ω
hk = µ− + 5 + k+5 < − + 5 + k+5 ≤ − 5 .
2
2
8
2
2
2
Thus on the set {u ≤ hk }, one has
ω nm−1
|u|nm−1 ≥ 5
.
2
It follows that |u|nm−1 can be estimate above and below by ω nm−1 up to a constant,
depending only upon the data; the proof can be repeated as before, but in this case
there is no need to introduce the truncated function v.
Therefore under assumption (5.1),
−
ess inf
2
u ≤ −µ− −
ρ
Qθρ ( nm−1
ω
adding
ess sup
Qθρ (
ρ2
ω nm−1
)
ω
;
4
u, gives
)
ess osc
2
Qθρ (
ρ
ω nm−1
u≤
)
u − µ− −
ess sup
Qθρ (
ρ2
ω nm−1
)
ω
3
≤ ω.
4
4
EJDE-2012/200
HÖLDER REGULARITY FOR SIGNED SOLUTIONS
13
6. The second alternative
Let us recall the two fundamental hypotheses we assume, namely
µ+ > 0 ,
µ− < 0 ,
µ+ ≥ |µ− |.
Throughout this new section, let us assume that (5.1) does not hold; i.e.,
(2ρ)2 o
(2ρ)2 ω n
∩ (y, s) + Q2θρ nm−1 < (1 − c0 )Q2θρ nm−1 .
u ≥ µ− +
2
ω
ω
For simplicity in the following we assume (y, s) = (0, 0).
2
2 c0 (2ρ)
Lemma 6.1. There exists a time level t∗ in the interval − ω(2ρ)
nm−1 , − 2 ω nm−1
such that
c
ω
0
∩ K2θρ > |K2θρ |.
(6.1)
u(·, t∗ ) < µ− +
2
2
This in turn implies
c0 ω
∩ K2θρ ≤ 1 −
|K2θρ |.
(6.2)
u(·, t∗ ) ≥ µ+ −
4
2
Proof. By contradiction, suppose that (6.1) does not hold for any t∗ in the indicated
range; then
2
(2ρ)2 Z − c20 ω(2ρ)
nm−1 ω
ω
∗
∗
∩ Q2θρ nm−1 =
∩
K
u(·,
t
)
<
µ
+
dt
u < µ− +
2θρ
−
(2ρ)2
2
ω
2
− nm−1
ω
Z 0
ω
∗
∗
+
u(·,
t
)
<
µ
+
∩
K
dt
−
2θρ
c
(2ρ)2
2
− 20 nm−1
ω
c0
c0 (2ρ)2
c0 (2ρ)2
≤ |K2θρ | 1 −
+
|K
|
2θρ
2
2 ω nm−1
2 ω nm−1
2 (2ρ)
< c0 Q2θρ nm−1 .
ω
This proves (6.1); (6.2) follows by the fact that (6.1) is equivalent to
c0 ω
∩ K2θρ < 1 −
|K2θρ |,
u(·, t∗ ) ≥ µ− +
2
2
and µ− + ω2 ≤ µ+ − ω4 .
The next lemma asserts that a property similar to (6.2) continues to hold for all
time levels from t∗ up to zero.
Lemma 6.2. There exists a positive integer j ∗ , depending upon the data and c0 ,
such that
ω c2 u(·, t) > µ+ − j ∗ ∩ K2θρ < 1 − 0 |K2θρ |,
2
4
for all times t∗ < t < 0.
Proof. Consider the logarithmic estimates (3.4) written over the cylinder K2θρ ×
1/n
(t∗ , 0) for the function (un − k n )+ and for the level k = µn+ − ( ω4 )n
. Notice
that, thanks to our assumptions, µ+ > ω4 , so k > 0. The number ν in the definition
ω
of the logarithmic function is taken as ν = 2j+2
, where j is a positive number to
be chosen. Thus we have
Hn
,
ψ (H n , (un − k n )+ , ν n ) = log+
ωn
H n − (un − k n )+ + 2(j+2)n
14
S. PUGLISI
where
Hn =
h
ess sup
K2θρ ×(t∗ ,0)
EJDE-2012/200
ω i
un − µn+ − ( )n
.
4
+
The cutoff function x → ζ(x) is taken such that
on K(1−σ)2θρ for σ ∈ (0, 1),
ζ=1
|Dζ| ≤
1
.
σθρ
With these choices, inequality (3.4) yields
Z
ψ 2 (x, t) dx
K(1−σ)2θρ
Z
ψ 2 (x, t∗ ) dx + γ
≤
K2θρ
Z
0
Z
t∗
(6.3)
|u|n(m−1) ψ|Dζ|2 dx dτ,
K2θρ
∗
for all t ≤ t ≤ 0. Let us observe that
ωn 22n
ψ ≤ log
= jn log 2.
ωn
2(j+2)n
To estimate the first integral on the right-hand
side of (6.3),
observe that ψ vanishes
ω n
ω n
n
n
n
on the set {u < k } and that µ+ − 4 ≥ µ+ − 4 ; therefore by (6.2)
Z
c0 |K2θρ |.
ψ 2 (x, t∗ ) dx ≤ j 2 n2 log2 2 1 −
2
K2θρ
The remaining integral is estimated as follows
Z 0Z
γ
|u|n(m−1) ψ|Dζ|2 dx dτ
t∗
≤
K2θρ
γ
(2ρ)2 n(m−1)
γ
jn
log
2
ω
|K2θρ | = 2 jn|K2θρ |.
(σθρ)2
ω nm−1
σ
Combining the previous estimates,
Z
n
o
c0 γ
ψ 2 (x, t) dx ≤ j 2 n2 log2 2 1 −
+ 2 jn |K2θρ |
2
σ
K(1−σ)2θρ
(6.4)
for all t∗ ≤ t ≤ 0. The left-hand side of (6.4) is estimated below by integrating over
the smaller set
n
ωn o
un > µn+ − (j+2)n ;
2
on such a set, since ψ is a decreasing function of H n , we have
ωn 22n
ψ 2 ≥ log2
= (j − 1)2 n2 log2 2;
ωn
2(j+1)n
∗
hence, for all t ≤ t ≤ 0, we obtain
n
n j c0 γ o
ωn o
2
n
1−
+ 2 |K2θρ |.
u (·, t) > µn+ − (j+2)n ∩ K(1−σ)2θρ ≤
j−1
2
σ j
2
On the other hand,
n
ωn o
n
u (·, t) > µn+ − (j+2)n ∩ K2θρ 2
n
ωn o
≤ un (·, t) > µn+ − (j+2)n ∩ K(1−σ)2θρ + |K2θρ \ K(1−σ)2θρ |
2
EJDE-2012/200
HÖLDER REGULARITY FOR SIGNED SOLUTIONS
n
≤ un (·, t) > µn+ −
Then
n
n
u (·, t) > µn+ −
ωn
o
2(j+2)n
ωn
o
2(j+2)n
15
∩ K(1−σ)2θρ + N σ|K2θρ |.
n j 2 o
c0 γ
1−
+ 2 + N σ |K2θρ |.
∩ K2θρ ≤
j−1
2
σ j
for all t∗ ≤ t ≤ 0. Now choose σ so small and then j so large as to obtain
c2 ω n 1/n ∩ K2θρ ≤ 1 − 0 |K2θρ |
∀t∗ ≤ t ≤ 0.
u(·, t) > µn+ − (j+2)n
4
2
Notice that our hypotheses imply µ+ ≥ ω2 , µ+ < ω; therefore,
µn+ 1/n
ω n 1/n n
1 1/n
µn+ − (j+2)n
< µ+ − (j+2)n
= µ+ 1 − (j+2)n
2
2
2
1
ω
≤ µ+ 1 − (j+2)n
≤ µ+ − (j+2)n+1 .
2
n
2
n
∗
The proof is completed once we choose j as the smallest integer such that
ω
ω
µ+ − (j+2)n+1 ≤ µ+ − j ∗ .
2
2
n
2
Corollary 6.3. For all j ≥ j ∗ and for all times − c20 ω(2ρ)
nm−1 < t < 0,
n
c2 ωo
u(·, t) > µ+ − j ∩ K2θρ < 1 − 0 |K2θρ |.
2
4
Motivated by Corollary 6.3, introduce the cylinder
c0
Q∗ = K2θρ × − θ∗ (2ρ)2 , 0 , with θ∗ = ω 1−nm .
2
(6.5)
Lemma 6.4. For every ν∗ ∈ (0, 1), there exists a positive integer q∗ = q∗ (data, ν∗ )
such that
n
ω o
u ≥ µ+ − j∗ +q∗ ∩ Q∗ ≤ ν∗ |Q∗ |.
2
Proof. Write down the energy estimates (3.3) for the truncated functions (u − kj )+ ,
with kj = µ+ − 2ωj , for j = j∗ , . . . , j∗ + q∗ over the cylinder
i
2
e = K4θρ × − c0 (2ρ) , 0 ⊃ Q∗ ;
Q
ω nm−1
the cutoff function ζ is taken to be one on Q∗ , vanishing on the parabolic boundary
e and such that
of Q
1
ω nm−1
|Dζ| ≤
, 0 ≤ ζt ≤
.
θρ
c0 ρ2
Thanks to these choices, the energy estimates (3.3) take the form
ZZ
|u|nm−1 |D(u − kj )+ |2 ζ 2 dx dτ
Q̃
≤γ
n ω nm−1 Z Z Z
+
ω
c0 ρ2
n−1 Z Z
ρ2
Q̃
Q̃
u
(s − kj )+ sn−1 ds dx dτ
kj
o
|u|nm−1 (u − kj )2+ dx dτ .
16
S. PUGLISI
EJDE-2012/200
Estimating
Z
u
(s − kj )+ sn−1 ds ≤ un−1
kj
(u − kj )2+
(u − kj )2+
≤ ω n−1
,
2
2
and taking into account that (u − kj )+ ≤ 2ωj , yields
ZZ
ω 2
ω nm−1
|u|nm−1 |D(u − kj )+ |2 ζ 2 dx dτ ≤ γ j ω n−1
|Q∗ |.
2
c0 ρ2
Q̃
Note that u > kj ≥
ω
4:
indeed the second inequality is equivalent to
1
1 3
1 −1
,
µ+ ≥ |µ− |
+ j
− j
4 2
4 2
and this is implied by our assumptions. Thus we can estimate
ZZ
ZZ
|u|nm−1 |D(u − kj )+ |2 ζ 2 dx dτ ≥
|u|nm−1 |D(u − kj )+ |2 dx dτ
e
Q
Q∗
ω nm−1 Z Z
≥
|D(u − kj )+ |2 dx dτ ;
4
Q∗
it follows that
ZZ
|D(u − kj )+ |2 dx dτ ≤ γ
Q∗
ω 2
2j
ω n−1
1
|Q∗ |.
c0 ρ2
(6.6)
Next, apply the isoperimetric inequality (3.1) to the function u(·, t), for t in the
range (−θ∗ (2ρ)2 , 0], over the cube K2θρ , and for the levels
k = kj < l = kj+1 ;
ω
in this way (l − k) = 2j+1
.
Taking into account (6.5), this gives
ω
|{u(·, t) > kj+1 } ∩ K2θρ |
2j+1
Z
(2θρ)N +1
≤
|Du| dx
|{u(·, t) < kj } ∩ K2θρ | {kj <u(·,t)<kj+1 }∩K2θρ
Z
8θρ
≤ 2
|Du| dx;
c0 {kj <u(·,t)<kj+1 }∩K2θρ
integrating in dt over the indicated interval and applying the Hölder inequality, one
gets
ZZ
1/2
8θρ ω
2
|A
|
≤
|D(u
−
k
)
|
dx
dt
(|Aj | − |Aj+1 |)1/2 ,
j+1
j
+
2j+1
c20
Q∗
where
Aj = {u > kj } ∩ Q∗ .
Square both sides of this inequality and estimate above the term containing |D(u −
kj )+ | by inequality (6.6), to obtain
|Aj+1 |2 ≤
γ
|Q∗ | (|Aj | − |Aj+1 |) .
c50
EJDE-2012/200
HÖLDER REGULARITY FOR SIGNED SOLUTIONS
17
Add these recursive inequalities for j = j∗ + 1, . . . , j∗ + q∗ − 1, where q∗ is to be
chosen. Majorizing the right-hand side with the corresponding telescopic series,
gives
j∗ +q
∗ −1
X
γ
(q∗ − 2)|Aj∗ +q∗ |2 ≤
|Aj+1 |2 ≤ 5 |Q∗ |2 .
c
0
j=j +1
∗
From this
r
γ
1
|Q∗ |.
c50
q∗ − 2
The number ν∗ being fixed, choose q∗ from
r
1
γ
√
= ν∗ .
c50
q∗ − 2
|Aj∗ +q∗ | ≤ √
Now let ξ ∈ (0, 21 ), a ∈ (0, 1) be fixed numbers.
Lemma 6.5. There exists a number c∗ ∈ (0, 1), depending upon the data, ξ, and
a, such that if
| {u ≥ µ+ − ξω} ∩ Q∗ | ≤ c∗ |Q∗ |,
(6.7)
then
u ≤ µ+ − aξω a.e. in Qθρ (θ∗ ρ2 ).
Proof. For k = 0, 1, . . ., set
ρ
ρk = ρ + k , Kk = Kθρk , Qk = Kk × (−θ∗ ρ2k , 0].
2
Let ζ(x, t) = ζ1 (x)ζ2 (t) be a piecewise smooth cutoff function in Qk such that
(
1 in Kk+1
2k+2
,
ζ1 =
|Dζ1 | ≤
N
θρ
0 in R \ Kk ,
(
ρ2k+1
2k
1 if t ≥ − ωnm−1
.
ζ2 =
0 ≤ ζ2,t ≤
ρ2k
θ ∗ ρ2
0 if t < − nm−1 ,
ω
Choose the sequence of truncating levels
1−a
ξ,
2k
and write down the energy estimates (3.3) for (u − hk )+ over the cylinder Qk ,
Z Z u
ess sup
(s − hk )+ sn−1 ds ζ 2 (x, t) dx
hk = µ+ − ξk ω,
−
ρ2
k
ω nm−1
Kk
<t≤0
ZZ
where ξk = aξ +
hk
2
|u|nm−1 D[(u − hk )+ ζ] dx dτ
+
Qk
≤γ
n ZZ
Z
Qk
ZZ
u
(s − hk )+ sn−1 ds |ζt | dx dτ
hk
nm−1
|u|
+
o
(u − hk )2+ |Dζ|2 dx dτ .
Qk
Let us estimate
Z
u
hk
(s − hk )+ sn−1 ds ≥ hn−1
k
(u − hk )2+
,
2
18
S. PUGLISI
Z
EJDE-2012/200
u
(s − hk )+ sn−1 ds ≤ un−1
hk
(u − hk )2+
(u − hk )2+
≤ ω n−1
.
2
2
Taking into account that (u − hk )+ ≤ ξω and the definition of θ and θ∗ , one has
Z
ZZ
2
(u − hk )2+ 2
n−1
ess sup hk
ζ (x, t) dx +
|u|nm−1 D[(u − hk )+ ζ] dx dτ
2
ρ2
Kk
Qk
k
−
ω nm−1
<t≤0
n
2k o
2k
nm−1 2
|Ak |
+
ω
≤ γ(ξω)2 ω n−1
θ ∗ ρ2
(θρ)2
22k
= γ 2 (ξω)2 ω n−1 ω nm−1 |Ak |,
ρ
where
Ak = {u < hk } ∩ Qk .
Note that u > hk ≥
1
2
− ξ ω: indeed the last inequality is equivalent to
1
−1
1
,
µ+ ≥ |µ− |
− ξ + ξk
+ ξ − ξk
2
2
and this follows by our hypotheses. Therefore, we obtain
Z
1−n
22k 1
ess sup
(u − hk )2+ ζ 2 (x, t) dx ≤ γ 2
−ξ
(ξω)2 ω nm−1 |Ak |,
2
ρ
2
ρ
Kk
k
−
ω nm−1
<t≤0
(6.8)
1−nm
22k 1
−ξ
(ξω)2 ω n−1 |Ak |.
|D[(u − hk )+ ζ]|2 dx dτ ≤ γ 2
ρ 2
Qk
ZZ
By (u − hk )+ ≥
(6.8) yields
1−a
2k+1
ξω, applying the Hölder inequality, and then Proposition 3.1,
(1 − a)2
(ξω)2 |Ak+1 | ≤
22(k+1)
≤
ZZ
(u − hk )2+ dx dτ ≤
ZZ
Qk+1
ZZ
(u − hk )2+ ζ 2 dx dτ
Qk
[(u − hk )+ ζ]
2(N +2)
N
dx dτ
N/(N +2)
|Ak |2/(N +2)
Qk
≤γ
ZZ
N/(N +2)
D[(u − hk )+ ζ]2 dx dτ
Qk
×
Z
<t≤0
2/(N +2)
[(u − hk )+ ζ] dx
ess sup
ρ2
k
− nm−1
ω
≤γ
2
|Ak |2/(N +2)
Kk
1
2(nm−1)+N (n−1)
22k
N +2
(ξω)2 ω
−ξ
2
ρ
2
N (1−nm)+2(1−n)
N +2
2
|Ak |1+ N +2 .
It follows that
|Ak+1 | ≤ γ
1
2(nm−1)+N (n−1)
24k
N +2
ω
−
ξ
(1 − a)2 ρ2
2
Setting
Yk =
|Ak |
,
|Qk |
N (1−nm)+2(1−n)
N +2
2
|Ak |1+ N +2 .
EJDE-2012/200
HÖLDER REGULARITY FOR SIGNED SOLUTIONS
19
we obtain
1
N (1−nm)+2(1−n)
2(nm−1)+N (n−1)
2
24k
1+ N2+2
N +2
2 N
N +2
N +2 Y
ω
−
ξ
ρ
(θ
θ
)
∗
k
(1 − a)2 ρ2
2
N
(1−nm)+2(1−n)
1
24k
1+ 2
N +2
=γ
−ξ
Yk N +2 .
2
(1 − a) 2
Yk+1 ≤ γ
Applying Lemma 3.4, Yk tends to zero as k → ∞, provided
|{u > h0 } ∩ Q0 |
|{u > µ+ − ξω} ∩ Q0 |
=
|Q0 |
|Q0 |
N +2
N
(nm−1)+2(n−1)
1
2
γ− 2
2
≤
−ξ
2−(N +2) ,
−(N
+2)
2
(1 − a)
Y0 =
N +2
which is (6.7) with c∗ :=
the proof.
γ− 2
(1−a)−(N +2)
1
2 −ξ
N (nm−1)+2(n−1)
2
2
2−(N +2) . This completes
Thanks to Lemma 6.4, we can apply Lemma 6.5 with ξ =
getting
ω
u ≤ µ+ − j∗ +q∗ +1 a.e. in Qθρ θ∗ ρ2 ,
2
which implies
ω
ess sup u ≤ µ+ − j∗ +q∗ +1 .
2
Qθρ (θ∗ ρ2 )
1
2j∗ +q∗
and a =
1
2,
Hence
ess osc2 u ≤ µ+ − ess inf2 u −
Qθρ (θ∗ ρ )
Qθρ (θ∗ ρ )
ω
2j∗ +q∗ +1
≤ω 1−
1
2j∗ +q∗ +1
.
7. Conclusion
The two alternatives just discussed can be combined to prove Theorem 4.1.
Proof of Theorem 4.1. The concluding statement of the first alternative says that
ess osc
2
ρ
Qθρ ( nm−1
ω
u≤
)
3
ω;
4
analogously, the conclusion of the second alternative is that
1
ess osc
u
=
ess
osc
u
≤
ω
1
−
.
Q∗
2j∗ +q∗ +1
Qθρ (θ∗ ρ2 )
Recalling the definition of θ∗ , we observe that
Q∗ = Qθρ (θ∗ ρ2 ) ⊂ Qθρ
ρ2
ω nm−1
.
The thesis follows by defining
η∗ := 1 −
1
.
2j∗ +q∗ +1
We are now ready to prove the local Hölder regularity.
20
S. PUGLISI
EJDE-2012/200
Proof of Theorem 1.1. Let us remind that we fixed ∈ (0, 1), R > 0, (y, s) ∈ ET ,
and we considered the cylinder
Q = K
R1−
n−1
2
(y) × (s − R2−(nm−1) , s] ⊂ ET .
Let now β, δ ∈ (0, 1) to be chosen, and let us introduce the sequences
R2 1−n
k
,
Rk := β k R, ωk := δ k ω, θk := ωk 2 , Qk := (y, s) + Qθk Rk nm−1
ωk
for k ∈ N. The thesis follows by standard arguments once we prove that
Qk+1 ⊂ Qk ⊂ Q ⊂ ET ∀k ∈ N,
ess osc u ≤ ωk .
(7.1)
Qk
The inclusion Q0 ⊂ Q immediately follows by assumption (4.1), while Qk+1 ⊂ Qk
is equivalent to
n−1
nm−1
n−1
β ≤ min{δ 2 , δ 2 } = δ 2 .
To prove (7.1), we will argue by induction. The validity for k = 0 is true by
construction since
ess osc u ≤ ess osc u ≤ ω.
Q0
Q
Assume that (7.1) holds for k and apply Theorem 4.1 taking ρ = R2k and ω = ωk ;
thanks to these choices
(2ρ)2 θ = θk , (y, s) + Q2θρ nm−1 = Qk .
ω
The assumptions of Theorem 4.1 are satisfied because (7.1) holds for k; hence, we
have ess oscQ∗ u ≤ η∗ ωk , where in this setting
c
0 1−nm 2
Q∗ = (y, s) + Qθ Rk
ωk
Rk .
k 2
8
This leads us to choose δ = η∗ ∈ (0, 1), so that η∗ ωk = ωk+1 . It remains only to
check Qk+1 ⊂ Q∗ , which by a simple calculation is equivalent to
n 1 n−1 r c nm−1 o
0
δ 2 ,
δ 2
.
β ≤ min
2
8
We conclude by choosing β small enough.
Acknowledgments. The author wishes to warmly thank Ugo Gianazza for his
help and support while considering the problem. The author also would like to
thank the anonymous referee for the helpful comments.
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21
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Simona Puglisi
Dipartimento di Matematica e Informatica, University of Catania, Viale A. Doria 6,
95125 Catania, Italy
E-mail address: spuglisi@dmi.unict.it