Hindawi Publishing Corporation
Boundary Value Problems
Volume 2007, Article ID 35825, 28 pages
doi:10.1155/2007/35825
Research Article
Interior Gradient Estimates for Nonuniformly Parabolic
Equations II
Gary M. Lieberman
Received 31 May 2006; Revised 6 November 2006; Accepted 9 November 2006
Recommended by Vincenzo Vespri
We prove interior gradient estimates for a large class of parabolic equations in divergence
form. Using some simple ideas, we prove these estimates for several types of equations
that are not amenable to previous methods. In particular, we have no restrictions on the
maximum eigenvalue of the coefficient matrix and we obtain interior gradient estimates
for so-called false mean curvature equation.
Copyright © 2007 Gary M. Lieberman. This is an open access article distributed under
the Creative Commons Attribution License, which permits unrestricted use, distribution,
and reproduction in any medium, provided the original work is properly cited.
1. Introduction
A key step in the study of second-order quasilinear parabolic equations is establishing
suitable a priori estimates for any solution of the equation. This fact is the theme of many
books on the subject [1–5] and our focus here is on one particular such estimate: a local
pointwise gradient estimate for solutions of equations in divergence form:
ut = div A(X,u,Du) + B(X,u,Du).
(1.1)
The role of this divergence structure has been noted many times under varying hypotheses on the functions A and B (see, in particular [6, Sections VIII.4 and VIII.5], [3, Section
V.4], [5, Section 11.5]). Our current interest is deriving this estimate using a surprising
variant (detailed below) of standard methods. Although this variant seems, at first, to be
a purely technical modification, we mention here two quite different types of estimates
which follow from this variant and which appear to be new. First, we derive a local gradient estimate for a class of equations which includes the parabolic false mean curvature
2
Boundary Value Problems
equation, that is, the equation with
A(X,z, p) = exp
1
1 + | p|2 p
2
(1.2)
and some conditions on B. Such an operator does not fall under the hypotheses from,
for example, [3], and the present author has, previously, given an incorrect proof of this
estimate [7, page 569] (we will point out the error later), and then in [5, Section 11.5,
page 281] a correct but weaker version of the estimate. Second, we estimate the gradient
of a solution to a large class of equations only in terms of the structure of the equation
and a bound for the gradient of the initial function. (Ordinarily, a gradient estimate is
given in terms of a maximum estimate for the solution, which, in turn, depends on some
estimate on the boundary and initial data.) Such an estimate was first proved by Ecker for
the parabolic prescribed mean curvature equation [8, Theorem 3.1], but we also show
that such an estimate is valid for the parabolic p-Laplacian if p < 2, and this fact seems
to be new. (In [9], a corresponding estimate was given for the Lq norm of the solution in
terms of the Lq norm of the initial data, and this estimate can be used to infer a gradient
estimate, but our goal here is to give an estimate directly.) This gradient estimate provides
an interesting counterpoint to known results on these equations (see [6, Chapter XII] for
a detailed description of these results). In particular, it is known that for p > 2n/(n + 1),
solutions of this equation are bounded (and have Hölder continuous spatial derivatives)
at any positive time for quite general initial data, in particular for L1 initial data. On the
other hand, [6, Section XII.13-(i)] provides an initial datum in L1 for which the solution
is unbounded for all sufficiently small positive time. Although the counterexample is described in all of Rn × (0, ∞), it should be noted that it satisfies the boundary condition
u = 0 on {|x| = 1,t > 0}, so the regularity of the solution is affected only by that of the
initial datum. An important point for our comparison is that the solution becomes infinite only at x = 0 (for t > 0 as well) and the initial function is smooth except at x = 0. Our
result shows that this is the only configuration in which the solution can be unbounded
since we obtain a gradient estimate at any x = 0. Of course, the additional surprise is that
our gradient estimate also applies to some equations with p > 2n/(n + 1).
The basic plan is to modify the Moser iteration technique [10] along the lines of Simon’s estimate for elliptic equations [11]. Of course, this is the plan followed by the author before (especially [7]) but we add two important new twists. As in [12], we obtain
an estimate that does not use an upper bound on the maximum eigenvalue of the matrix
∂A/∂p. Such an approach is also useful in studying anisotropic problems (see [13, 14])
and we present the calculations for this case in [15]. In addition, we use a modified version
of the Sobolev inequality from [11]. This inequality will allow us to prove some unusual
estimates (in particular the estimates for parabolic p-Laplace equations) and also to use
some more standard notations, in particular, we will use ai j to denote the components
of the matrix ∂A/∂p; in [7, 11, 16], ai j denoted the components of a slightly different
matrix.
Following [11], we break the estimate into several steps. After giving some notation in
Section 2, we prove an energy-type inequality in Section 3. We then present the Sobolev
Gary M. Lieberman 3
inequality in Section 4, and we use the energy inequality along with the Sobolev inequality in Section 5 to bound the maximum of the gradient in terms of an integral:
q
w |Du| Du · AdX
(1.3)
for some function w and some exponent q, which we will detail in that section. This integral is estimated in Section 6 in terms of the integral of Du · A, and this final integral is
easily estimated; we will quote [5, Lemma 11.13]. Section 7 contains some examples, especially the false mean curvature equation, to illustrate our structure conditions. We also
discuss some interesting variants of our estimate. In Section 8, we examine the application of our Sobolev inequality to some equations satisfying structure conditions depending on the maximum eigenvalue of ∂A/∂p; the most important of such equations are the
parabolic prescribed mean curvature equation and parabolic p-Laplacian with p < 2 described above. Finally, we look at parabolic equations with faster than exponential growth
in Section 9; our method is only partially successful in dealing with such problems.
2. Notation
For the most part, we follow the notation in [5], so X = (x,t) denotes a point in Rn+1 with
|X | =
n
i 2
x
1/2
+ |t |
,
(2.1)
i =1
and, for R > 0, we write
Q(R) = X ∈ Rn+1 : |x| < R, −R2 < t < 0 ,
B(R) = x ∈ Rn : |x| < R .
(2.2)
We also use ᏼQ(R) to denote the parabolic boundary of Q(R), that is, the set of X such
that either
|x| = R,
−R2 ≤ t ≤ 0,
(2.3)
t = −R2 .
(2.4)
or
|x| < R,
Moreover, we use N to denote n if n > 2 and an arbitrary constant greater than 2 if n = 2.
We always assume that u ∈ C 2,1 (Q(R)) for some R > 0 and we set
v = 1 + |Du|2
1/2
,
ν=
Du
,
v
g i j = δ i j − νi ν j .
(2.5)
We will also use this notation, without further comment, with p in place of Du to describe
structural conditions on the functions A and B (and their derivatives). We also set
ai j =
∂Ai
,
∂p j
Ꮿ2 = ai j g km Dik uD jm u,
Ᏹ = ai j Di vD j v,
(2.6)
4
Boundary Value Problems
where we use the Einstein summation convention that repeated indices are summed
from 1 to n. (Note that ai j , Ꮿ2 , and Ᏹ are not quite the same as in [7, 11, 16].)
We also define the oscillation of u over a set S by
osc u = sup u − inf u.
S
(2.7)
S
S
In addition, for parameters τ > 1 and r ∈ (0,R], we write Qτ (r) and qτ (r,t) for the
subsets of Q(r) and B(r) × {t }, respectively, on which v > τ.
3. The energy inequality
In this section, we prove an energy inequality, that is, an inequality which estimates integrals involving second spatial derivatives of u in terms of integrals involving only first
derivatives. Before stating this inequality, we present some preliminary structure conditions. We suppose that there are matrices [Cki ] and [Dki ] such that Dki is differentiable with
respect to (x,z, p) and
Cki + Dki =
∂Ai
∂Ai
pk + k + Bδki .
∂z
∂x
(3.1)
For simplicity, we set
Ᏸi j = νk
∂Dki
,
∂p j
Ᏺ = pi
∂Dki ∂Dki k
+ i ν.
∂z
∂x
(3.2)
Our structure conditions are stated in terms of these expressions. We assume that there
are nonnegative constants τ0 ≥ 1, β1 , and β2 along with positive functions Λ0 , Λ1 , and Λ2
such that
ai j ηik η jk
Cki g jk ηi j ≤ β1 Λ1/2
0
1/2 Cki νk ξi ≤ β1 Λ0
1/2 vᏰi j ηi j ≤ β1 Λ0
ai j ξ i ξ j
1/2
ai j ηik η jk
vᏲ ≤ β12 Λ0 ,
1/2
,
,
1/2
(3.3a)
(3.3b)
,
vνk Dki − νi B ≤ β1 Λ1 ,
ij
1/2
|A|g i j ηi j ≤ β2 Λ1/2
a ηik η jk ,
2
ij
1/2
|A|ν · ξ ≤ β2 Λ1/2
a ξi ξ j ,
2
(3.3c)
(3.3d)
(3.3e)
(3.3f)
(3.3g)
for all n × n matrices η, all n-vectors ξ, and all (X,z, p) ∈ Q(R) × R × Rn such that z =
u(X) and v > τ0 . Note that conditions (3.3a)–(3.3d) are exactly the same as [5, (11.41a–d)]
(except for a slight variation in notation).
Our energy estimate is then a variant of [5, Lemma 11.10] (which in turn comes from
[11, (2.11)]).
Gary M. Lieberman 5
Lemma 3.1. Let χ be an increasing, nonnegative Lipschitz function defined on [τ, ∞) for
some τ ≥ τ0 and let ζ be a nonnegative C 2,1 (Q(R)) function which vanishes in a neighborhood of ᏼQ(R). Suppose conditions (3.3) hold, and define
Ξ(σ) =
σ
τ
(ξ − τ)χ(ξ)dξ.
(3.4)
Then
qτ (R,s)
Ξ(v)ζ 2 dx +
≤ 20β12
+4
Qτ (R)
Qτ (R)
Qτ (R)
1−
τ 2
Ꮿ + Ᏹ χζ 2 dX
v
Λ0 (v − τ)χ + χ ζ dX + 4β1
2
Qτ (R)
|A|χ D2 ζ ζ + |Dζ |2 v dX + 32β22
Qτ (R)
Qτ (R)
+4
Λ1 χζ |Dζ |dX
Λ2 (v − τ)χ + χ |Dζ |2 dX
Ξζζt dX
(3.5)
for any s ∈ (−R2 ,0). (Here, and in what follows, the argument v from χ and Ξ is suppressed.)
Proof. We begin just as in [5, Lemma 11.10]. Let θ be a vector-valued C 2 function which
vanishes in a neighborhood of ᏼQ(R), and set Q = B(R) × (−R2 ,s). If we multiply the
differential equation by div θ and then integrate by parts, we obtain
Q
− ut Dk θ k + Dk Ai Di θ k + BDk θ k dX = 0.
(3.6)
An easy approximation argument shows that this identity holds for any θ which is only
Lipschitz (with respect to x only); in particular, we take
θ = (v − τ)+ χ(v)ζ 2 ν.
(3.7)
Just as in [5, pages 270-271], we see that
k
Q
−ut Dk θ dX =
Ξζ dx − 2
2
qτ (R,s)
Q
Ξζζt dX.
(3.8)
Next, we have
Q
Dk Ai Di θ k + BDk θ k dX =
Q
+
Q
+
Dk Ai + Bδki Di (v − τ)+ χνk ζ 2 dX
Q
Dk Ai (v − τ)+ χνk Di ζ 2 dX
B(v − τ)+ χν · D ζ 2 dX.
(3.9)
6
Boundary Value Problems
The first integral is handled as usual. We set
⎧
⎨(v − τ)χ + χ
if v > τ,
if v ≤ τ,
Ψ=⎩
0
(3.10)
and we note that
Di (v − τ)χνk = ΨDi vνk + 1 −
τ
v
+
χg k j Di j u.
(3.11)
It follows that
τ
χ Ꮿ2 + Cki g k j Di j u
v +
+ Ψ Ᏹ + νk Cki Di v + Dki Di (v − τ)+ χνk .
Dk Ai + Bδki Di (v − τ)+ χνk = 1 −
(3.12)
An integration by parts then yields
Q
Dki Di (v − τ)+ χνk ζ 2 dX = −
Q
Ᏸi j Di j u + Ᏺ (v − τ)+ χ ζ 2 dX
−2
(3.13)
Q
Dki (v − τ)+ χνk ζDi ζ dX.
For the second integral, we integrate by parts again (cf. the proof of [12, Lemma 2.3]):
Q
Dk Ai (v − τ)+ χνk Di ζ 2 dX
= −2
Q
Ai Dk (v − τ)+ χνk ζDi ζ dX − 2
Q
Ai χ(v − τ)+ νk ζDik ζ + Di ζDk ζ dX.
(3.14)
To simplify the notation, we now set
I1 =
I2 =
Qτ (R)
qτ (R,s)
Ξζ 2 dx,
τ
1 − Ꮿ2 χ + Ᏹ χ (v − τ) + χ ζ 2 dX.
v
(3.15)
Then
I1 + I2 = 2
Q
Ξζζt dX +
10
j =3
Ij,
(3.16)
Gary M. Lieberman 7
where
I3 = −
Q
I4 = −
Q
I5 =
Q
ΨCki νk Di vζ 2 dX,
Ᏺ(v − τ)+ χζ 2 dX,
Q
(3.17)
Dki − Bδki (v − τ)χνk ζDi ζ dX,
Q
I8 = 2
Cki χg k j Di j udX,
Q
1−
I9 = 2
Q
I10 = 2
+
Ᏸi j Di j u(v − τ)+ χζ 2 dX,
I6 =
I7 = 2
τ
v
1−
τ
v
+
Ai Di ζg k j Dk j udX,
Ai (v − τ)+ χν · DvζDi ζ dX,
Q
Ai χ(v − τ)+ νk ζDik ζ + Di ζDk ζ dX.
These terms are estimated as in [5, Lemma 11.10] using (3.3) and Cauchy’s inequality.
For the reader’s convenience, we give a brief estimate of each integral.
First, from (3.3a), we have
I3 ≤ β1
Q
Λ1/2
ai j Dik uD jk u
0
1/2
1−
τ
v
χζ 2 dX.
(3.18)
+
Since
ai j Dik uD jk u = Ꮿ2 + Ᏹ
(3.19)
and χ ≥ 0, we have
ai j Dik uD jk u 1 −
τ
v
+
χ ≤ 1−
τ
v
+
χᏯ2 + ΨᏱ.
(3.20)
Therefore, by Cauchy’s inequality,
I3 ≤ 3β12
Q
Λ0 Ψζ 2 dX +
1
I2 .
12
(3.21)
Similarly, since Ꮿ2 ≥ 0, we see from (3.3b) and Cauchy’s inequality that
I4 ≤ 3β12
Q
Λ0 Ψζ 2 dX +
1
I2 .
12
(3.22)
Next, we use (3.3c), (3.20), and Cauchy’s inequality to obtain
I5 ≤ 3β12
Q
Λ0 Ψζ 2 dX +
1
I2 .
12
(3.23)
8
Boundary Value Problems
Moreover, (3.3d) gives
Λ0 Ψζ 2 dX,
(3.24)
Λ1 χ(v − τ)+ ζ |Dζ |dX.
(3.25)
I6 ≤ β12
Q
and (3.3e) gives
I7 ≤ 2β1
Q
From (3.3f) and Cauchy’s inequality, we infer that
I8 ≤ 8β22
Q
1
Λ2 Ψ|Dζ |2 dX + I2 ,
8
(3.26)
and, finally, (3.3g), (3.20), and Cauchy’s inequality imply that
I9 ≤ 8β22
It follows that
I1 + I2 ≤ 2
Q
Q
1
Λ2 Ψ|Dζ |2 dX + I2 .
8
Ξζζt dX
+ 16β2
+ 10β12
Q
Λ0 Ψζ dX + 2β1
2
Q
(3.27)
Λ2 Ψ|Dζ |2 dX + 2
Q
Q
Λ1 χζ |Dζ |dX
1
|A|χ D2 ζ ζ + |Dζ |2 dX + I2 .
(3.28)
2
Then (3.5) follows from this inequality by simple algebra.
In Section 6, we will need a sharper version of this lemma. To obtain this version, we
note that (3.3d) is only needed to estimate the positive part of Ᏺ, so (3.5) also holds with
an additional term of
−
Q(R)
Ᏺ− χ(v − τ)+ ζ 2 dX
(3.29)
on the right-hand side.
4. The Sobolev inequality
We now present our modified Sobolev inequality, which is an easy consequence of [17,
Theorem 2.1]; however, for notational reasons (in particular the use of n and m), we
quote a consequence of this theorem (see [5, Corollary 11.9]).
Lemma 4.1. Let n ≥ 2, and let g ∈ L∞ (Q(R)) be nonnegative. Set H i = D j (g i j ) and κ =
(N + 2)/N. Then
2κ 2/N
Q(R)
|h| g
dX ≤ C(N)
×
sup
s∈(−R2 ,0) B(R)
Q(R)
ij
h(x,s)2 g(x,s)dx
2
2
g Di hD j h + h |H | dX
2/N
n/N 2
Q(R)
h dX
(N −n)/N
(4.1)
Gary M. Lieberman 9
for any h ∈ C(Q(R)) that vanishes on {|x| = R} and which is uniformly Lipschitz with respect to x.
Proof. Let us set m = n + 1 and U = B(R). We define νn+1 = −1/v and extend the definition g i j = δ i j − νi ν j for i and j in {1,...,m}. With dμ = dx, it is easy to check that all the
hypotheses of [5, Corollary 11.9] are satisfied, and this corollary gives
U
|h|2κ g 2/N dx ≤ C(N)
×
U
2/N
U
|h|2 g dx
ij
2
2
n/N g Di hD j h + h |H | dx
2
U
(N −n)/N
(4.2)
h dx
for each t ∈ (−R2 ,0). (In this equation, all functions are evaluated at (x,t).) The proof is
completed as in [5, Theorem 6.9]: note that
U
h2 g dx ≤ sup
s∈(−R2 ,0)
U
h(x,s)2 g(x,s)dx,
(4.3)
integrate the resulting inequality with respect to t, and then apply Hölder’s inequality if
n = 2.
Note that the vector H is not quite the usual mean curvature vector. For later reference,
we observe that
v2 |H |2 ≤ C(n) g i j Dik ug km D jm u + g i j Di vD j vv .
(4.4)
5. Estimate of the maximum in terms of an integral
From our energy inequality and the Sobolev inequality, we can now reduce our pointwise
estimate of |Du| to an integral estimate of a suitable quantity. For this reduction, we
introduce three positive C 1 [τ0 , ∞) functions w, λ, and Λ. In addition to their smoothness,
the functions w, λ, and Λ obey the following monotonicity properties:
w is increasing,
(5.1a)
ξ w(ξ) is a decreasing function of ξ,
(5.1b)
−β
Λ(ξ)/λ(ξ)
ξ2
w(ξ)−β
ξ
−β
Λ(ξ)
λ(ξ)
N/2 is an increasing function of ξ,
(5.1c)
N/2
is a decreasing function of ξ
(5.1d)
for some nonnegative constant β. We also assume that
Λ0 ≤ vΛ,
Λ1 ≤ vΛ,
Λ2 ≤ vΛ,
(5.2a)
(5.2b)
λ ≤ Λ,
(5.2c)
1 ≤ Λ,
(5.2d)
10
Boundary Value Problems
and that
|A| ≤ β2 Λ.
(5.3)
Finally, we assume that
vλ
λ 1+
λ
2 g i j ξi ξ j ≤ vai j ξi ξ j ,
(5.4)
where (as before) we suppress the argument v from λ, Λ, and their derivatives. These
hypotheses imply a pointwise estimate for the gradient in terms of an integral.
Lemma 5.1. Suppose that conditions (3.3), (5.1), (5.2), (5.3), and (5.4) hold. Then there is
a constant c1 (n,β,β1 R,β2 ) such that
1−
sup
Qτ (R/2)
τ
v
N+2
w ≤ c1 R−n−2
Qτ (R)
w
Λ
λ
N/2
Λ
dX.
v
(5.5)
Proof. The proof is essentially the same as that of [5, Lemma 11.11], so we only give a
sketch.
First, for q ≥ 1 + β a parameter at our disposal, we set
χ=
Λ
λ
N/2
wq
1−
τ
v
(N+2)(q−1)
v −2 .
(5.6)
+
Then conditions (5.1a), (5.1c) imply that χ is increasing while conditions (5.1b), (5.1d)
imply that Ψ ≤ C(β)q2 χ. Now let ζ be as in Lemma 3.1, and note that we can take ζ so that
|Dζ | ≤ C/R, |D2 ζ | + |ζt | ≤ C/R2 , and 0 ≤ ζ ≤ 1 in Q(R). It then follows from Lemma 3.1
with ζ (N+2)q−N in place of ζ 2 that
sup
t ∈(−R2 ,0) qτ (R,t)
Ξ(v)ζ 2 dx +
≤ C β,β1 R,β2
q2
Qτ (R)
R2
Qτ (R)
1−
τ 2
Ꮿ + Ᏹ χζ 2 dX
v
(5.7)
χΛζ (N+2)(q−1) v dX
by taking (5.2a), (5.2b), (5.2c), and (5.3) into account and observing that
1
Ξ(v) ≤ χ(v)(v − τ)2
2
(5.8)
(because χ is increasing).
Now we define h by the equation
h2 = χλ 1 −
so
ij
2 (N+2)(q−1) χλ
g Di hD j h ≤ Cq ζ
v
τ
v
ζ
2
2
vζ (N+2)q−N ,
vλ
1+
λ
2 (5.9)
v2
g Di vD j v + 2 .
R
ij
(5.10)
Gary M. Lieberman 11
In addition, from conditions (5.1b), (5.1d), we infer that
Ξ(v) ≥
1
χ(v)(v − τ)2 .
2 + C(β)q
(5.11)
It then follows from (4.4) and (5.4) that
sup
t ∈(−R2 ,0)
v
h2 dx+
λ
B(R)×{t }
Q(R)
g i j Di hD j h+h2 |H |2 dX ≤ Cq4 R−2
Qτ (R)
χΛvζ (N+2)(q−1) dX.
(5.12)
Lemma 4.1, with g = v/λ, then yields
1/κ
κq
Σ
w dμ
≤ Cq
4
Σ
w q dμ
(5.13)
for w = ζ(1 − τ/v)+ w,
Σ = X ∈ Qτ (R) : ζ(X) > 0, v > τ ,
Λ
λ
dμ =
N/2
Λ
v
1−
−N −2
τ
ζ
v
(5.14)
dX.
A standard iteration argument (based on [10]) completes the proof.
If we assume further that there are nonnegative constants β3 and β4 such that
Λ
w
λ
N/2
Λ ≤ β3 wβ4 +1 Du · A
(5.15)
(see [11, (1.5)] or [5, (11.50)]), then we have reduced the pointwise estimate to an estimate of the integral
v q Du · AdX
(5.16)
for q = β4 , and we estimate this integral in the next section. (Note that if β4 = 0, this
estimate is particularly simple.)
6. Estimate of the integral
We now examine the integral (5.16), and we provide an estimate specifically for the case
w = v. To this end, we make some basic assumptions relating the sizes of A, B, and Du · A:
v|A| ≤ β5 Du · A,
(6.1a)
B ≤ β6 Du · A.
(6.1b)
We also use a variant of (3.3e): we assume that there is a decreasing function ε such that
−νk Dki νi ≤ ε(v)Du · A.
(6.2)
12
Boundary Value Problems
Next, we suppose that the functions Λ0 , Λ1 , and Λ2 can be estimated suitably in terms of
Du · A:
Λ0 ≤ ε(v)2 v2 Du · A,
(6.3a)
for the same function ε as in (6.2),
Λ1 ≤ vDu · A,
(6.3b)
Λ2 ≤ Du · A.
(6.3c)
v ≤ β7 Du · A.
(6.4)
Finally, we assume that
Under these hypotheses, we obtain an estimate for (5.16) provided that ε can be made
sufficiently small when v is large.
Lemma 6.1. Suppose conditions (3.3), (6.1), (6.2), (6.3), and (6.4) are satisfied. Let q >
0 and set ω = oscQ(R) u, E = exp(β6 ω), Σ = 1 + β7 ω/R, and q∗ = max{q,2}. If there is a
constant τ1 greater than max{τ,2} such that
2
8ωε τ1 + 10β2 q∗3 + 640β2 q∗4 + 1280 β1 q∗ ωε τ1
2
+ 80q∗2 Eβ12 q∗ ω2 ε τ1
≤ 1, (6.5)
then there is a constant C determined only by β1 ωε(τ1 ), β2 , β5 , and q such that
Qτ (R)
v q Du · AdX ≤ C τ1 + ΣE
ω
R
q Qτ (2R)
Du · AdX.
(6.6)
Proof. Suppose first that q ≥ 2. Our proof in this case is a modification of the proof of
[11, Lemma 2]. First, we set
⎧
⎨σ q − qτ q−1 σ + (q − 1)τ q
G(σ) = ⎩
0
if σ > τ,
if σ ≤ τ,
(6.7)
and we observe that G (σ) = q(σ q−1 − τ q−1 )+ . Hence
1
0 ≤ G(σ) ≤ G (σ)σ,
2
τ
q −1
G (σ) ≤ qσ
1−
,
σ +
τ
,
G(σ) ≤ σ q 1 −
σ +
0 ≤ G (σ) ≤ q2 σ q−2 .
(6.8a)
(6.8b)
(6.8c)
(6.8d)
In what follows, we suppress the argument v from G and its derivatives. Next, we set
F(z) =
1
2 β6 z exp β6 z + 1 − exp β6 z
β6
(6.9)
Gary M. Lieberman 13
and F1 (z) = F(z)exp(−β6 z). We note that F1 (0) = F1 (0) = 0 and F1 (z) ≤ 1 for z ≥ 0, so
F1 (z) ≤ (1/2)z2 for z ≥ 0. It follows that for z replaced by u = u − inf Q(R) u, F satisfies the
properties
1
1
0 ≤ F ≤ ω2 E ≤ ω2 E,
2
2
0 ≤ F ≤ ωE ≤ ωE,
(6.10a)
(6.10b)
0 ≤ F ≤ 1 + β6 ω E,
(6.10c)
F − β6 F = E,
(6.10d)
where E = exp(β6 u).
We also define
ζ(X) =
|x|2
1−
4R2
2 1+
+
t
,
4R2
(6.11)
so that
|Dζ | ≤
2 D ζ ≤ 1 ,
2
1
,
2R
R
ζt ≤ 1 .
2
4R
(6.12)
Now, we set
I = −
Q(2R)
ζ 2q F GDu · AdX = −
Q(2R)
ζ 2q GA · D(F )dX,
(6.13)
and an integration by parts gives
I = I1 + I2 + I3
(6.14)
with
I1 =
Q(2R)
I2 =
Q(2R)
ζ 2q F G A · Dv dX,
I3 = 2q
ζ 2q F Gdiv AdX,
Q(2R)
(6.15)
ζ 2q−1 F GA · Dζ dX.
The estimate for I1 is, in the present situation, the most complex. First, we use the differential equation to see that I1 = I4 + I5 with
I4 =
Q(2R)
ζ 2q F Gut dX =
I5 = −
Q(2R)
2q Q(2R)
ζ 2q Ft GdX,
ζ F GB dX.
(6.16)
14
Boundary Value Problems
To estimate I4 , we need some further integration by parts which is easily justified if A,
B, and u are smoother than we have assumed. The justification under our current hypotheses is to let (um ) be a sequence of C ∞ functions which converge in C 2,1 to u. Writing
vm = (1 + |Dum |2 )1/2 and Gm for G(vm ), we have
Q(2R)
ζ 2q Ft Gm dX = −
Q(2R)
= −2q
Q(2R)
B(R)×{0}
ζ 2q FGm dx
ζ 2q−1 Gm Fζt dX
Q(2R)
−
ζ 2q Gm t F dX +
(6.17)
ζ 2q Gm vm t F dX +
B(R)×{0}
ζ 2q FGm dx.
But (vm )t = νkm Dk (um )t , so
−
Q(2R)
ζ 2q Gm vm t F dX =
Q(2R)
Dk ζ 2q Gm νkm F um t dX.
(6.18)
Sending m → ∞ then gives
I4 = −2q
Q(2R)
ζ 2q−1 FGζt dX +
Q(2R)
Dk ζ 2q FG νk ut dX +
B(2R)×{0}
ζ 2q FGdx. (6.19)
Then we use the differential equation again to conclude that
I4 = −2q
Q(2R)
ζ
2q−1
FGζt dX +
Q(2R)
Dk ζ FG νk Di A +B dX +
2q
i
B(2R)×{0}
ζ 2q FGdx,
(6.20)
and another integration by parts (as in Lemma 3.1) gives us
Q(2R)
Dk ζ 2q FG νk Di Ai dX =
Q(2R)
Di ζ 2q FG νk Dk Ai dX.
(6.21)
It follows that I4 = I6 + I7 + I8 + I9 with
I6 =
Q(2R)
I7 =
Q(2R)
I8 = −2q
I9 =
Di ζ 2q FG νk Dk Ai dX,
Di ζ 2q FG νk δki BdX,
Q(2R)
(6.22)
ζ 2q−1 FGζt dX,
B(2R)×{0}
ζ 2q FGdx.
Gary M. Lieberman 15
Next, we write
I6 =
Q(2R)
Di ζ 2q F G νk Dk Ai dX +
Q(2R)
ζ 2q FDi G νk Dk Ai dX,
(6.23)
and another integration by parts yields
Q(2R)
Di ζ 2q F G νk Dk Ai dX = −
Q(2R)
−
Dik ζ 2q F G νk Ai dX
Q(2R)
k
(6.24)
Di ζ F Dk G ν A dX,
2q
i
so
I6 =
19
Ij,
(6.25)
j =10
with
I10 = −2q
Q(2R)
I11 = −2q
ζ 2q−2 FG νk ζDik ζ + (2q − 1)Di ζDk ζ Ai dX,
Q(2R)
ζ 2q−1 F G (ν · Dζ)(Du · A) + (ν · Du)(Dζ · A) dX,
I12 = −I2 ,
I13 = −
Q(2R)
ζ 2q F G ν · DuDu · AdX,
I14 = −2q
I15 = −2q
Q(2R)
1
ζ 2q−1 FG g k j D jk uDζ · AdX,
v
Q(2R)
I16 = −
I17 = −
I18 =
Q(2R)
ζ 2q−1 FG ν · DvDζ · AdX,
Q(2R)
Q(2R)
ζ 2q F G ν · Dv Du · AdX,
1
ζ 2q F G g k j D jk uDu · AdX,
v
ζ 2q FG νk ai j D jk uDi v dX +
I19 =
Q(2R)
+
ζ 2q FG νk Di v
Q(2R)
ζ 2q F
G km
g Dim uai j D jk udX,
v
∂Ai
∂Ai
Dk u + k dX
∂z
∂x
∂Ai
1
∂Ai
ζ 2q FG g k j Di j u
Dk u + k dX.
v
∂z
∂x
Q(2R)
(6.26)
16
Boundary Value Problems
We now combine some of these integrals:
I5 + I7 + I19 =
Q(2R)
+
ζ 2q FDi G νk Cki + Dki dX
Dk ζ 2q F G νk B dX −
Q(2R)
(6.27)
ζ 2q F GB dX
Q(2R)
= I20 + I21 + I22 + I23 + I24 + I25 + I26 ,
with
I20 =
1
ζ 2q FG Cki g k j Di j udX,
v
Q(2R)
I21 =
ζ 2q FG νk Cki Di v dX,
Q(2R)
I22 = −
I23 = −2q
Q(2R)
ζ 2q F G νk Dki Di udX,
Q(2R)
I24 = −
Q(2R)
I25 = −
I26 =
Q(2R)
(6.28)
ζ 2q FG Ᏸi j Di j udX,
ζ 2q−1 FG νk Dki − νi B Di ζ dX,
Q(2R)
ζ 2q FG Ᏺ dX,
ζ 2q F (G ν · Du − G)B dX.
It follows that
Q(2R)
ζ 2q F (G ν · Du − G)Du · AdX = I − I13 = I3 +
11
Ij +
j =8
18
Ij +
j =14
26
Ij.
(6.29)
j =20
If we assume now that τ ≥ 2, we have ν · Du ≥ (3/4)v for v ≥ τ, and hence (6.8a) implies
that
q
1
G ν · Du − G ≥ G v = v q − τ q−1 v + .
4
4
(6.30)
From (6.1b) and (6.10d), we then conclude that
q
4
Q(2R)
ζ 2q E v q − τ q−1 v + Du · AdX ≤
Q(2R)
≤ I3 +
ζ 2q E[G ν · Du − G]Du · AdX
11
j =8
Ij +
18
j =14
Ij +
25
j =20
(6.31)
Ij.
Gary M. Lieberman 17
We are now ready to estimate the right-hand side of this inequality, one term at a time.
First, we define the measure μ by
μ(S) =
Qτ (2R)∩S
Du · AdX,
(6.32)
so, for any function f , we have
f dμ =
S
Qτ (2R)∩S
then
ω
R
I3 ≤ β5 qE
f (X)Du · AdX,
ζ 2v
Q(2R)
q −1
(6.33)
dμ
(6.34)
by (6.1a), (6.8c), and (6.10b);
ω2
1
I8 ≤ β7 qE 2
2
R
ζ 2v
Q(2R)
q −1
dμ
(6.35)
by (6.4), (6.8c), and (6.10a);
1
I9 ≤ Eω2
2
B(R)×{0}
ζ 2q G(v)dx
(6.36)
by (6.10a);
I10 ≤ q3 β5 E
ω2
R2
Q(2R)
ζ 2v
q −2
dμ
(6.37)
dμ
(6.38)
by (6.1a), (6.8b), and (6.10a);
I11 ≤ 2β5 q2 E
ω
R
Q(2R)
ζ 2v
q −1
by (6.1a), (6.8b), (6.10b), and the observation that β5 ≥ 1. Next
ω2
1
I14 ≤ β2 q3 E
2
R
2q q−2
Qτ (2R)
ζ v
1/2 ᏱdX
Q(2R)
2
ζ v
q −2
1/2
dμ
(6.39)
by (3.3g), (6.3c), (6.8d), and (6.10a);
ω2
1
I15 ≤ β2 q2 E
2
R
2q q−2
Qτ (2R)
ζ v
τ
1 − Ꮿ2 + Ᏹ dX
v
1/2 Q(2R)
ζ 2v
q −2
1/2
dμ
(6.40)
by (3.3f), (6.3c), (6.8b), and (6.10a);
2
I16 ≤ β2 q ω E
Qτ (2R)
2q q−2
ζ v
1/2 ᏱdX
Q(2R)
2
q
E ζ v dμ
1/2
(6.41)
18
Boundary Value Problems
by (3.3g), (6.3c), (6.8d), and (6.10b);
I17 ≤ β2 qω E
2q q−2
Qτ (2R)
τ
1 − Ꮿ2 + Ᏹ dX
v
ζ v
1/2 2
1/2
q
E ζ v dμ
Q(2R)
(6.42)
by (3.3f), (6.3c), (6.8b), and (6.10b);
1
I18 ≤ q2 Eω2
2
Qτ (2R)
ζ 2q v q−2
τ 2
Ꮿ + Ᏹ dX
v
1−
(6.43)
by (6.8b), (6.8d), and (6.10b);
1
I20 ≤ β1 qω2 ε(τ) E
2
2q q−2
ζ v
Qτ (2R)
τ
1 − Ꮿ2 + Ᏹ dX
v
1/2 Q(2R)
2
q
1/2
E ζ v dμ
(6.44)
by (3.3a), (6.3a), (6.8a), and (6.10b);
1
I21 ≤ β1 q2 ω2 ε(τ) E
2
2q q−2
Qτ (2R)
ζ v
1/2 ᏱdX
Q(2R)
q
2
1/2
E ζ v dμ
(6.45)
by (3.3b), (6.3a), (6.8d), and (6.10a);
I22 ≤ ωε(τ)q
Q(2R)
q
E ζ 2 v dμ
(6.46)
by (6.2), (6.8b), and (6.10b);
ω2
1
I23 ≤ β1 q2 E
2
R
Q(2R)
ζ 2v
q −1
dμ
(6.47)
by (3.3e), (6.3b), (6.8b), and (6.10a);
1
I24 ≤ β1 qω2 ε(τ) E
2
2q q−2
Qτ (2R)
ζ v
τ
1 − Ꮿ2 + Ᏹ dX
v
1/2 Q(2R)
2
q
1/2
E ζ v dμ
(6.48)
by (3.3c), (6.3a), (6.8b), and (6.10a);
1
I25 ≤ qEω2
2
by (6.8b) and (6.10a).
Q(2R)
Ᏺ− ζ 2q v q−2 (v − τ)+ dX
(6.49)
Gary M. Lieberman 19
Combining all these estimates and using Cauchy’s inequality, we find that
q
4
Eζ 2q v q − τ q−1 v dμ
Q(2R)
ω
≤ K1 ΣE
R
+ K3 Eω
+
2
ζ v
Q(2R)
q −1
ω2
dμ + K2 E2 2
R
2q q−2
2
Qτ (2R)
1
+ ωε(τ)q
4
ζ 2v
Q(2R)
q −2
τ
1 − Ꮿ2 + Ᏹ dX + Eω2
v
ζ v
q
1
E ζ 2 v dμ + qEω2
2
Q(2R)
Q(2R)
dμ
(6.50)
2q
B(R)×{0}
ζ Gdx
Ᏺ− ζ 2q v q−2 (v − τ)+ dX
with
1
1
K2 = β 5 q 3 + β 2 q 3 ,
K1 = 4β5 q2 + q2 β1 ω2 + q,
2
4
2 1
1
K3 = β2 q3 + 4β22 q4 + 8 β1 ωε(τ)q + q2 .
8
2
(6.51)
We now use the remark after Lemma 3.1 with χ = v q−2 and ζ q in place of ζ. Since
G ≤ qΞ and K3 ≥ q, we infer from (6.1), (6.3), and (6.4) that
ω2
B(R)×{0}
ζ 2q Gdx + K3 ω2
≤ K4 Σ
ω
R
Q(2R)
ζ 2v
q −1
+ K6 ω2 ε(τ)2
Qτ (2R)
ζ 2q v q−2
dμ + K5
ω2
R2
q
q
ζ 2 v dμ + ω2
2
τ 2
Ꮿ + Ᏹ dX
v
Q(2R)
Q(2R)
1−
ζ 2v
Q(2R)
q −2
(6.52)
dμ
Ᏺ− ζ 2q v q−2 (v − τ)+ dX
with
K4 = 2 + 2β1 ω K3 ,
K5 = 4β5 q2 + 8β2 q2 K3 ,
(6.53)
K6 = 20q2 β12 K3 .
Since K3 ≥ q/2, it follows that
q
4
Q(2R)
q
E ζ 2 v dμ
ω
≤ K1 + K4 ΣE
R
+
Q(2R)
ζ 2v
q −1
1
+ qωε(τ) + K6 Eω2 ε(τ)2
4
dμ + K2 + K5 E2
Q(2R)
q
ω2
R2
Q(2R)
E ζ 2 v dμ + qτ q−1
ζ 2v
q −2
dμ
Q(2R)
ζ 2q v dμ.
(6.54)
20
Boundary Value Problems
If we now replace τ by τ1 and write μ1 for the measure defined by replacing τ by τ1 in
(6.32), we infer that
Q(2R)
q
q −1
ζ 2 v dμ1 ≤ Cτ1
Q(2R)
ω
+ CΣE
R
ζ 2 v dμ1
Q(2R)
2
ζ v
q −1
ω2
dμ1 + CE2 2
R
2
ζ v
Q(2R)
q −2
(6.55)
dμ1 .
Applying Young’s inequality yields
q
ζ 2 v dμ1 ≤ C τ1 + ΣE
Q(2R)
ω
R
q Q(2R)
dμ1 ,
(6.56)
so
Qτ1 (2R)
q
ζ 2 v Du · AdX ≤ C τ1 + ΣE
ω
R
q Qτ1 (2R)
Du · AdX,
(6.57)
and it is clear that
Qτ (2R)\Qτ1 (2R)
q
q
ζ 2 v Du · AdX ≤ τ1
Qτ (2R)\Qτ1 (2R)
Du · AdX.
Adding these last two inequalities gives the desired result.
The case q < 2 follows from this one via Hölder’s inequality.
(6.58)
Note that we can take ε to be a constant provided that a modulus of continuity is
known for u; all we need
is to take R small enough that (6.5) holds.
The estimate of Du · AdX is given in [5, Lemma 11.13], so we give the estimate
without proof.
Lemma 6.2. Suppose conditions (6.1) hold and set ω = oscQ(R) u. Also set τ2 = max{τ0 ,
8β5 ω/R} and
Δ = sup
v<τ2
B − β6 Du · A
+ + (Du · A)+ +
ω
|A| .
R
(6.59)
Then
Qτ (R/2)
Du · AdX ≤ C(n)exp β6 ω Rn ω2 + ΔR2 .
(6.60)
We can combine all of these results into a single estimate although we will see in the
next section that sometimes a different combination is more useful.
Theorem 6.3. Suppose there are functions w, Λ0 , Λ1 , Λ2 , Λ, λ, and ε such that conditions
(3.3), (5.1), (5.2), (5.3), (5.4), (5.15), (6.1), (6.2), (6.3), and (6.4) hold for some nonnegative
constants β, β1 ,...,β7 , and τ1 ≥ max{2,τ0 } with ω = oscQ(R) u and q∗ = max{β4 ,2}. Set
τ2 = max{τ0 ,8β5 ω/R}, E = exp(β6 ω), and Σ = 1 + β7 ω/R, and define Δ by (6.59). Then
Gary M. Lieberman 21
there is a constant C, determined only by n, β, β1 R, β1 ωε(τ1 ), β2 , β3 , β4 , β5 such that
sup w(v) ≤ max w 2τ0 ,C τ1 + ΣE
Q(R/8)
ω
R
β4 E
ω2
+Δ
R2
.
(6.61)
7. Examples
We start by assuming that the functions A and B satisfy the conditions
|A| ≤ γ2 Ψ(v),
Du · A ≥ γ0 vΨ(v) − γ1 ,
Ψ(v) 2
|ξ | ≤ ai j ξi ξ j ,
v
vAz + Ax + |B | ≤ ε1 (v)vΨ(v)
γ0
(7.1)
for a positive constant γ0 , nonnegative constants γ1 and γ2 , an increasing function Ψ ∈
C 1 ([1, ∞)) such that Ψ(1) = 1 and
Ψ (v) ≤ ψ0 vα−1 Ψ(v)
(7.2)
for some nonnegative constants ψ0 and α and a decreasing, positive function ε1 such that
limτ →∞ ε1 (τ) = 0. Then conditions (3.3) are satisfied with Dki = 0,
Λ0 =
γ0 ε1 (v) 3
v Ψ(v),
2 ε1 (1)
γ0 ε1 (v) 2
γ0
Λ2 = vΨ(v),
v Ψ(v),
2 ε1 (1)
2
γ2
ε (1)
β2 = n1/2 .
β1 = (2n)1/2 1 ,
γ0
γ0
Λ1 =
(7.3)
In addition, we can take w = v, Λ = (1/2)γ0 v2 Ψ(v), and
λ=
γ0
v−2α Ψ(v)
1 + 4α2 + ψ02
(7.4)
to satisfy (5.1)–(5.4) with β = (1 + α)N. Condition (5.15) holds with
β3 = 2 1 + 4α2 + ψ02
N/2 γ0
(7.5)
2
and β4 = (1 + α)N + 1 if τ0 is sufficiently large. Finally, conditions (6.1)–(6.4) hold with
β5 = 2
γ2
,
γ0
β6 = 2
ε(v) =
ε1 (1)
,
γ0
β7 =
1
,
γ0
ε1 (v)
.
ε1 (1)
(7.6)
Since ε(v) → 0 as v → ∞, we have a gradient estimate under these hypotheses.
In particular, the equation
ut = div exp
1 2
v Du + B(X,u,Du)
2
(7.7)
22
Boundary Value Problems
is included under these hypotheses if |B | = o(v2 exp((1/2)v2 )) as | p| → ∞: we take Ψ(v) =
v exp((1/2)[v2 − 1]), and note that (7.2) is satisfied with ψ0 = 2 and α = 2. It would be of
interest to know if a gradient estimate can be obtained for |B | = O(v2 exp((1/2)v2 )).
The difficulty with [7, Lemma 5.4] is easy to explain in terms of the notation here. We
write div A = ai j Di j u since, in this case, A is independent of z and x. Moreover, under the
hypotheses of that lemma, one needs to estimate the integral
I=
Q(R)
wq 1 −
τ
v
+
ζ 2q ai j Di j udX
(7.8)
for some function w, which was claimed to equal v2 in [7]. The structure of the function
A shows that
ai j Di j u ≤ Ꮿ2 + Ᏹ
1/2
(Du · A)1/2 ,
(7.9)
so
I≤
Qτ (R)
(ζw)
q
τ
1 − Ꮿ2 + Ᏹ dX
v
1/2 1/2
q
Q(R)
(ζw) dμ
,
(7.10)
and the integral
Qτ (R)
(ζw)
q
τ
1 − Ꮿ2 + Ᏹ dX
v
(7.11)
cannot be estimated by a small multiple of
Q(R)
(ζw)q dμ.
(7.12)
(Note that this estimation does not arise in the proof of [13, Lemma 2.3], so the latter
result is correct.)
Note also that when Ψ satisfies (7.2) with α = 0, we have the uniformly parabolic equations described in [16, Example 4] but without any assumptions on the maximum eigenvalue of the matrix [ai j ]. In particular, we reproduce the usual gradient estimate for parabolic p-Laplacian equations once we observe that the condition Ψ(1) = 1 can be replaced
by Ψ(τ ∗ ) = 1 for some τ ∗ ≥ 1. If we further assume that ε1 (v) = γ3 /v for some positive
constant γ3 and that Ψ(v) ≥ v (which is the case if vΨ (v) ≥ Ψ(v)), then we can take as
structure functions
γ0
(7.13)
Λ0 = Λ1 = Λ2 = vΨ(v),
2
and hence Λ = λ = (1/2)γ0 Ψ(v). With w = v2 , (5.5) reads
sup v2 ≤ C τ 2 + R−n−2
Qτ (R/2)
Qτ (R)
vΨ(v)dX .
(7.14)
The integral here can be estimated directly via Lemma 6.2 and our estimate has the same
form as [6, Equation VIII.5.1] although we have used the choices σ = 1/2, θ = ρ2 = R2 for
Gary M. Lieberman 23
the parameters in [6]. Moreover, if Ψ(v) = (v2 − 1)(m−1)/2 with m ∈ (1,2), then we choose
r ≥ 2 so that n[m − 2] + 2r > 2, and we take Λ = v, λ = vm−1 , and w = vr+N(2−m)/2 . In this
way, we also reproduce [6, Equation VIII.5.3] (with the same choice of parameters).
On the other hand, when A = ν and B ≡ 0, our method does not apply. To see why, we
examine (3.3g) and (6.3c) with ξ = ν. First, |A|ν · ξ ≥ 1/8 for v sufficiently large, while
ai j ξi ξ j ≤ v−3 , so the structure function Λ2 needs to be at least (some multiple of) v3 and
this choice of Λ2 clearly does not satisfy (6.3c). This example is important because it is the
motivating case for the structure described in [11]. Moreover, the hypotheses for gradient
estimates in [11] and [5] are clearly satisfied for this choice of A and B.
8. Gradient estimates without boundary data
In [8], Ecker showed that the gradient of a solution to a prescribed mean curvature equation can be estimated, locally in space, just in terms of its initial data. Here, we show
how that result follows from a simple modification of our estimates. In fact, we obtain a
corresponding estimate for a larger class of equations.
To this end, we need to adjust our notation slightly. First, for any R > 0 and T > 0, we
set
Q(R,T) = X ∈ Rn+1 : |x| < R, 0 < t < T ,
(8.1)
and we write Qτ (R,T) for the subset of Q(R,T) on which v > τ. We then have the following form of the energy inequality.
Lemma 8.1. Let χ be a nonnegative Lipschitz function defined on [τ, ∞) for some τ ≥ τ0 and
let ζ be a nonnegative C 2 (B(R)) function which vanishes on ∂B(R). Suppose conditions (3.3)
hold, and define Ξ by (3.4). If v(x,0) ≤ τ for all x ∈ B(R), then
qτ (R,s)
Ξ(v)ζ 2 dx +
≤ 20β12
+4
Qτ (R,T)
Qτ (R,T)
Qτ (R,T)
1−
τ 2
Ꮿ + Ᏹ χζ 2 dX
v
Λ0 (v − τ)χ + χ ζ dX + 4β1
2
Qτ (R,T)
|A|χ D2 ζ ζ + |Dζ |2 vdX + 32β2
Λ1 χζ |Dζ |dX
Qτ (R,T)
Λ2 (v − τ)χ + χ |Dζ |2 dX
(8.2)
for any s ∈ (0,T).
Proof. We proceed exactly as in Lemma 3.1 except that the integral involving ζt is not
present.
Next, we note (see, e.g., [5, Corollary 6.9]) that our Sobolev inequality (4.1) holds if
we replace Q(R) by Q(R,T) and (−R2 ,0) by (0,T). Then the proof of Lemma 5.1 gives the
following gradient bound.
Lemma 8.2. Suppose that all the hypotheses of Lemma 5.1 hold except for (5.2d), which
is replaced by the assumption that v(x,0) ≤ τ for all x ∈ B(R). Then there is a constant
24
Boundary Value Problems
c3 (n,β,β1 R,β2 ) such that
N+2
τ
v
1−
sup
Qτ (R/2,T)
w ≤ c3 R−n−2
Qτ (R,T)
w
Λ
λ
N/2
Λ
dX.
v
(8.3)
Note that if Λ = λ = vθ for some constant θ < 1, then we can take w = v1−θ to infer
that the integrand in (8.3) is identically one, and hence we obtain a gradient bound directly which depends only on a gradient bound for the initial function and on data of the
equation. In particular, we have the following result for p-Laplacian equations.
Corollary 8.3. Let m ∈ (1,2), and suppose u is a solution of the equation
−ut + div |Du|m−2 Du = 0
(8.4)
in some cylinder Q(R,T) with |Du| bounded on B(R) × {0}. Then
T
sup |Du| ≤ C(m,n) 1 + sup |Du| +
2
R
B(R/2)×(0,T)
B(R)×{0}
1/(2−m) .
(8.5)
To include the mean curvature equations, we must modify our structure conditions to
include a condition on the maximum eigenvalue of the matrix ∂A/∂p. Following [11], we
assume that there is a positive function μ such that
ai j ψi ξ j ≤ μ|ψ |2
1/2 ai j ξ i ξ j
1/2
(8.6)
for all vectors ξ and ψ. Of course if [ai j ] is symmetric, then we can take μ to be the
maximum eigenvalue of this matrix. With this hypothesis in hand, we have the following
version of the energy inequality.
Lemma 8.4. Let χ be a nonnegative Lipschitz function defined on [τ, ∞) for some τ ≥ τ0
and let ζ be a nonnegative C 1 (B(R)) function which vanishes on ∂B(R). Suppose conditions
(3.3a)–(3.3d) and (8.6) hold, and define Ξ by (3.4). If v(x,0) ≤ τ for all x ∈ B(R), then
qτ (R,s)
Ξ(v)ζ 2 dx +
≤ 12β12
Qτ (R,T)
Qτ (R,T)
1−
τ 2
Ꮿ + Ᏹ χζ 2 dX
v
Λ0 (v − τ)χ + χ ζ 2 dX + 4
Qτ (R,T)
(8.7)
v2 μ|Dζ |2 dX
for any s ∈ (0,T).
Proof. This is an easy modification of the proof of Lemma 5.1. See [5, Lemma 11.10]
for details but note the differences in notation between that reference and the current
paper.
From this energy inequality, we obtain the following gradient estimate.
Lemma 8.5. Suppose that conditions (3.3a)–(3.3d), (5.1), (5.2a), (5.2d), (5.3), (8.6) and
vμ ≤ Λ
(8.8)
Gary M. Lieberman 25
are satisfied and that v(x,0) ≤ τ for all x ∈ B(R), then there is a constant c3 (n,β,β1 R,β2 )
such that (8.3) holds.
Note that Corollary 8.3 also follows from this lemma. Furthermore, in case A(X,z, p) =
ν and B depends only on X and z, we suppose that B is nonincreasing as a function of z
and Lipschitz with respect to x, all the conditions of this lemma are satisfied (cf. [5, pages
279-280]) with Cki = 0,
Λ0 = v,
β = 2,
1
μ= ,
w = v,
v
1/2
β1 = sup Bx ,
λ = Λ = 1,
τ0 = sup v,
(8.9)
B(R)×{0}
and hence we infer that
sup v ≤ C n, R sup Bx |
2
Q(R/2,T)
sup v + TR
−2
,
(8.10)
B(R)×{0}
which is a sharper form of [8, Theorem 3.1] in case the constant κ there is zero. To infer
the estimate for general κ, we perform a simple transformation. In our notation, the
assumption involving κ is that Bz ≤ κ, so let us note that u = exp(κt)u is a solution of the
equation
−ut + div A(X,u,Du) + B(X,u,Du) = 0
(8.11)
with
A(X,z, p) = exp(−κt)A X, exp(κt)z, exp(κt)p ,
B(X,z, p) = exp(−κt)B x,exp(κt)z − κz.
(8.12)
Now the hypotheses of Lemma 8.5 are satisfied for u, A, and B with Cki = 0,
Λ0 = v,
β = 2,
1
μ= ,
λ = exp(−2κT),
Λ = 1,
w = v,
v
1/2
β1 = exp(κT) sup Bx ,
τ0 = sup v.
(8.13)
B(R)×{0}
The corresponding estimate for v(Du) then implies that
−2
sup v ≤ C n, exp(κT), R sup Bx
sup v + TR
,
2
Q(R/2,T)
(8.14)
B(R)×{0}
which is a sharper version of the full force of [8, Theorem 3.1].
9. Equations with faster than exponential growth
An important element in the theory of a priori estimates is the question of what classes of
operators are encompassed. As we have already seen, if A(p) = Ψ(v)ν for some increasing
scalar function Ψ, then our method provides a gradient estimate for some choices of Ψ
26
Boundary Value Problems
but not others. In particular, if Ψ grows too slowly (e.g., if Ψ is a constant), then our
method does not supply a gradient estimate. In this section, we examine this structure
when Ψ grows more rapidly than any exponential function.
Our first step is a positive one, reducing the pointwise gradient estimate to an integral
estimate. We only need a slight variant of the argument in Section 5.
We assume that there is a positive increasing function y with y(τ0 ) ≥ 1 such that
w (v) ≤
d Λ(v)
dv λ(v)
v y
y
2
+
N/2
vλ
λ
y(v)
w(v),
v
y(v) Λ(v)
≤
v
λ(v)
2
N/2
(9.1)
.
+ y 2 yλg i j ξi ξ j ≤ vai j ξi ξ j
(9.2)
for all ξ ∈ Rn . In addition, we assume that
Λ0 y ≤ vΛ,
Λ2 y ≤ vΛ.
(9.3)
Theorem 9.1. Suppose u is a solution of (1.1), and suppose that w, Λ, and λ satisfy conditions (3.3), (5.1a), (5.1c), (5.2), (5.3), (9.1), (9.2), and (9.3). Then (5.5) holds with c1
determined by n, β, β1 R, and β2 .
Proof. Take χ as in Lemma 5.1 so that χ is increasing. It is not hard to see that
(v − τ)χ (v) ≤ Y (v)χ(v)
(9.4)
for Y (v) = (n + 2)qy(v), and hence
v
v
(σ − τ)2 χ (σ)dσ
Y (σ)
τ
τ
v
1
1 ≥
(σ − τ)2 χ (σ)dσ =
(v − τ)2 χ(v) − 2Ξ(v) .
Y (v) τ
Y (v)
Ξ(v) =
(σ − τ)χ(σ)dσ ≥
(9.5)
Simple rearrangement gives
Ξ(v) ≥
1
(v − τ)2 χ(v)
Y (v) + 2
(9.6)
and a simple calculation (cf. [5, Lemma 6.15]) gives
1
Ξ(v) ≤ (v − τ)2 χ(v).
2
(9.7)
We now define h by
h2 = χ yλ 1 −
τ
v
2
vζ (N+2)q−N
to infer (5.12). From this inequality, the proof is exactly the same as for Lemma 5.1.
(9.8)
Gary M. Lieberman 27
As a specific example, we suppose that
A(X,z, p) = exp exp
1 2
v
2
B(X,z, p) ≡ 0.
p,
(9.9)
Then (3.3) is easily checked with
Λ0 = Λ1 = 0,
Λ2 = exp exp
1 2
v
2
β1 = 0,
,
(9.10)
and suitable β2 . The remainder of the hypotheses are satisfied with w = vK for any K > 0
and
1
Λ = v exp exp v2
2
3
1
,
λ = v exp − v exp exp v2
2
2
y = (N + 1)v .
−5
2
,
(9.11)
In particular, for K = 1, we thus obtain
supv ≤ C 1 + R
−n−2
Q
v4N+3 exp
N 2
1
v exp exp v2
2
2
dX .
(9.12)
To see why we cannot infer a complete gradient estimate for this example, we note that
(9.2) immediately implies that
λ ≤ exp exp
1 2
v
2
exp − v2 v−2
(9.13)
while (9.3) implies that Λ ≥ exp(exp((1/2)v2 )), so we must take y no less than some
constant times v2 . Hence, the integral in (5.5) is at least
w
1
exp exp v2
v
2
exp
N 2
v dX,
2
(9.14)
and this integrand cannot be estimated by an expression of the form v q Du · A for any
power q, so Lemma 6.1 does not apply to this example. If we note that the integrand can
q
be estimated by an expression of the form w1 Du · A with w1 = exp(v2 ), then it would
seem that the proof of that lemma could be modified. If we try to imitate the proof of
Lemma 6.1 but with G(w1 ) in place of G(v) (as was done in [11]), the integral I3 causes
problems since the integrand has the form
q
ζ 2q−1 w1 |A|,
(9.15)
q −θ
and this cannot be estimated by ζ 2q−1 w1 Du · A for any positive constant θ. Hence it is
not possible to adapt the proof of Lemma 6.1 to this situation.
Acknowledgments
Some of this research was performed while the author was on a Faculty Professional Development Assignment from Iowa State University at the Centre for Mathematics and its
28
Boundary Value Problems
Applications at the Australian National University. The author thanks both institutions
for their support.
References
[1] A. Friedman, Partial Differential Equations of Parabolic Type, Krieger, Malabar, Fla, USA, 1983.
[2] N. V. Krylov, Nonlinear Elliptic and Parabolic Equations of the Second Order, vol. 7 of Mathematics
and Its Applications, D. Reidel, Dordrecht, The Netherlands, 1987.
[3] O. A. Ladyzhenskaya, V. S. Solonnikov, and N. N. Ural’tseva, Linear and Quasilinear Differential
Equations of Parabolic Type, American Mathematical Society, Providence, RI, USA, 1968.
[4] E. M. Landis, Second Order Equations of Elliptic and Parabolic Type, vol. 171 of Translations of
Mathematical Monographs, American Mathematical Society, Providence, RI, USA, 1998.
[5] G. M. Lieberman, Second Order Parabolic Differential Equations, World Scientific, River Edge,
NJ, USA, 1996.
[6] E. DiBenedetto, Degenerate Parabolic Equations, Universitext, Springer, New York, NY, USA,
1993.
[7] G. M. Lieberman, “Maximum estimates for solutions of degenerate parabolic equations in divergence form,” Journal of Differential Equations, vol. 113, no. 2, pp. 543–571, 1994.
[8] K. Ecker, “Estimates for evolutionary surfaces of prescribed mean curvature,” Mathematische
Zeitschrift, vol. 180, no. 2, pp. 179–192, 1982.
[9] G. M. Lieberman, “A new regularity estimate for solutions of singular parabolic equations,”
Discrete and Continuous Dynamical Systems. Series A, vol. 2005, supplement, pp. 605–610, 2005.
[10] J. Moser, “A new proof of De Giorgi’s theorem concerning the regularity problem for elliptic
differential equations,” Communications on Pure and Applied Mathematics, vol. 13, pp. 457–468,
1960.
[11] L. Simon, “Interior gradient bounds for non-uniformly elliptic equations,” Indiana University
Mathematics Journal, vol. 25, no. 9, pp. 821–855, 1976.
[12] I. Fonseca and N. Fusco, “Regularity results for anisotropic image segmentation models,” Annali
della Scuola Normale Superiore di Pisa. Classe di Scienze. Serie IV, vol. 24, no. 3, pp. 463–499,
1997.
[13] G. M. Lieberman, “Gradient estimates for a new class of degenerate elliptic and parabolic equations,” Annali della Scuola Normale Superiore di Pisa. Classe di Scienze. Serie IV, vol. 21, no. 4,
pp. 497–522, 1994.
[14] F. Siepe, “On the Lipschitz regularity of minimizers of anisotropic functionals,” Journal of Mathematical Analysis and Applications, vol. 263, no. 1, pp. 69–94, 2001.
[15] G. M. Lieberman, “Gradient estimates for anisotropic elliptic equations,” Advances in Differential Equations, vol. 10, no. 7, pp. 767–812, 2005.
[16] G. M. Lieberman, “Interior gradient bounds for nonuniformly parabolic equations,” Indiana
University Mathematics Journal, vol. 32, no. 4, pp. 579–601, 1983.
[17] J. H. Michael and L. M. Simon, “Sobolev and mean-value inequalities on generalized submanifolds of Rn ,” Communications on Pure and Applied Mathematics, vol. 26, pp. 361–379, 1973.
Gary M. Lieberman: Department of Mathematics, Iowa State University, Ames, IA 50011, USA
Email address: lieb@iastate.edu