Electronic Journal of Differential Equations, Vol. 2009(2009), No. 90, pp. 1–15.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
SECOND-ORDER BOUNDARY ESTIMATES FOR SOLUTIONS
TO SINGULAR ELLIPTIC EQUATIONS
CLAUDIA ANEDDA
Abstract. Let Ω ⊂ RN be a bounded smooth domain. We investigate the
effect of the mean curvature of the boundary ∂Ω in the behaviour of the solution to the homogeneous Dirichlet boundary value problem for the singular
semilinear equation ∆u + f (u) = 0. Under appropriate growth conditions on
f (t) as t approaches zero, we find an asymptotic expansion up to the second
order of the solution in terms of the distance from x to the boundary ∂Ω.
1. Introduction
In this article, we study the Dirichlet problem
∆u + f (u) = 0 in Ω
u = 0 on ∂Ω,
(1.1)
where Ω is a bounded smooth domain in RN , N ≥ 2, and f (t) is a decreasing and
positive smooth function in (0, ∞), which approaches infinity as t → 0. Equation
(1.1) arises in problems of heat conduction and in fluid mechanics.
Problems with singular data are discussed in many papers; see, for instance,
[8, 9, 10, 12, 16, 17] and references therein. Let f (t) = t−γ . For γ > 0, in [7] it is
shown that there exists a positive solution continuous up to the boundary ∂Ω. For
γ > 1, in [13] it is shown that the solution u satisfies
2
0 < c1 ≤ u(x)(δ(x))− 1+γ ≤ c2 ,
where δ = δ(x) denotes the distance of x from the boundary ∂Ω. Actually, in [7]
and in [13] the more general equation ∆u + p(x)u−γ = 0 with p(x) > 0 is discussed.
For equation (1.1) in case 1 < γ < 3, in [6] it is shown that there exists a constant
B > 0 such that
2 γ+1
1+γ
2γ
δ
u(x) − p
< Bδ γ+1 .
2(γ − 1)
For γ > 3, in [15] it is proved that
2 γ+1
1+γ
γ+3
δ
u(x) − p
< Bδ γ+1 .
2(γ − 1)
2000 Mathematics Subject Classification. 35B40, 35B05, 35J25.
Key words and phrases. Elliptic problems; singular equations;
second order boundary approximation.
c
2009
Texas State University - San Marcos.
Submitted June 17, 2009. Published July 30, 2009.
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C. ANEDDA
EJDE-2009/90
In [1], for γ > 3, it is proved that
2 h
γ+1
1+γ
i
N −1
u(x) = p
δ
1+
Kδ + o(δ) ,
3−γ
2(γ − 1)
where K = K(x) stands for the mean curvature of the surface {x ∈ Ω : δ(x) =
constant}.
In this article we extend the latter estimate to more general nonlinearities. More
precisely, assume
Z 1
f 0 (t)F (t)
γ
β
f (τ )dτ
(1.2)
=
+ O(1)t , F (t) =
f 2 (t)
1−γ
t
where γ > 3, β > 0 and O(1) denotes a bounded quantity as t → 0. In addition,
we suppose there is M finite such that for all θ ∈ (1/2, 2) and for t small we have
|f 00 (θt)|t2
≤ M.
f (t)
(1.3)
An example which satisfies these conditions is f (t) = t−γ + t−ν with 0 < ν < γ.
Here β = min[γ − ν, γ − 1].
When φ(δ) is defined as
Z φ(δ)
(2F (t))−1/2 dt = δ
(1.4)
0
and γ > 3, we prove that
i
h
N −1
Kδ + O(1)δ σ+1 ,
u(x) = φ(δ) 1 +
3−γ
(1.5)
2β
where σ is any number such that 0 < σ < min[ γ−3
γ+1 , γ+1 ]. Note that φ is a solution
to the one dimensional problem
φ00 + f (φ) = 0,
φ(0) = 0.
The estimate (1.5) shows that the expansion of u(x) in terms of δ has the first part
which is independent of the geometry of the domain, and the second part which
depends on the mean curvature of the boundary as well as on γ. For 1 < γ < 3,
the first part of the expansion is still independent of the geometry of the domain,
but it is not possible to find the second part in terms of the mean curvature even
in the special case f (t) = t−γ , see [1] for details.
We observe that similar results are known for boundary blow-up problems, see
[3, 4, 5]. In [3] the problem
∆u = f (u),
u → ∞ as x → ∂Ω
is discussed under the assumption that f (t) is positive, increasing and satisfying
Z t
p
f 0 (t)F(t)
−β
=
+
O(1)t
,
F(t)
=
f (τ )dτ,
f 2 (t)
1+p
0
where p > 1, β > 0 and O(1) is a bounded quantity as t → ∞. Under some
additional conditions for f , in [3] it is shown that
h
i
N −1
u(x) = Φ(δ) 1 +
Kδ + o(δ) ,
(1.6)
p+3
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SECOND-ORDER BOUNDARY ESTIMATES
where Φ is defined as
Z
3
∞
(2F (t))−1/2 dt = δ.
Φ(δ)
Note that Φ satisfies
Φ00 = f (Φ), Φ(0) = ∞.
We underline that (1.6) holds for p > 1, in contrast to (1.5) which holds when
γ > 3. Moreover, when f (t) = tp with p close to one, it is possible to find other
terms in the expansion (1.6). For example, the third term depends on the mean
curvature K as well as on its gradient ∇K (see [2]). Instead, the estimate (1.5)
cannot be improved for any value of γ because 0 < σ < 1.
2. Preliminary results
Let f (t) be a decreasing and positive smooth function in (0, ∞), which approaches infinity as t → 0. Assume condition (1.2) with γ > 1 and β > 0. Let us
rewrite (1.2) as
γ
(F (t)) γ−1
0
1
1−γ
(F (t))
= O(1)tβ .
(2.1)
f (t)
Since by [14, Lemma 2.1],
F (t)
lim
= 0,
(2.2)
t→0 f (t)
integration by parts on (0, t) of (2.1) yields
F (t)
1
=
+ O(1)tβ .
tf (t)
γ−1
(2.3)
Using the latter estimate and (1.2) again we find
tf 0 (t)
= −γ + O(1)tβ .
f (t)
(2.4)
Let us write (2.4) as
f 0 (t)
γ
= − + O(1)tβ−1 .
f (t)
t
Integration over (t, 1) yields
log
f (1)
= log tγ + O(1).
f (t)
Therefore, we can find two positive constants C1 , C2 such that
C1 t−γ < f (t) < C2 t−γ ,
Since F (t) =
R1
t
∀t ∈ (0, 1).
(2.5)
f (τ )dτ , using (2.5) we find two positive constants C3 , C4 such that
C3 t1−γ < F (t) < C4 t1−γ ,
N
∀t ∈ (0, 1/2).
(2.6)
Lemma 2.1. Let A(ρ, R) ⊂ R , N ≥ 2, be the annulus with radii ρ and R centered
at the origin, let f (t) > 0 smooth, decreasing for t > 0 and such that f (t) → ∞ as
t → 0. Assume condition (1.2) with γ > 3. If u(x) is a solution to problem (1.1)
in Ω = A(ρ, R) and v(r) = u(x) for r = |x|, then
R1
(F (t))1/2 dt
v(r) > φ(R − r) − C v
(R − r), r̃ < r < R,
(2.7)
(F (v))1/2
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C. ANEDDA
EJDE-2009/90
and
R1
(F (t))1/2 dt
(r − ρ), ρ < r < r,
(2.8)
F (v)
where φ is defined as in (1.4), ρ < r ≤ r̃ < R and C is a suitable positive constant.
0
v(r) < φ(r − ρ) + Cφ (r − ρ)
v
Proof. If Ω = A(ρ, R), the corresponding solution u(x) to problem (1.1) is radial.
With v(r) = u(x) for r = |x| we have
N −1 0
v + f (v) = 0, v(ρ) = v(R) = 0.
(2.9)
r
It is easy to show that there is r0 such that v(r) is increasing for ρ < r < r0 and
decreasing for r0 < r < R, with v 0 (r0 ) = 0. Multiplying (2.9) by v 0 and integrating
over (r0 , r) we find
Z r 0 2
(v )
(v 0 )2
+ (N − 1)
ds = F (v) − F (v0 ), v0 = v(r0 ).
(2.10)
2
s
r0
v 00 +
By (2.6), F (t) → ∞ as t → 0. Therefore, F (v(r)) → ∞ as r → R, and (2.10)
implies that
|v 0 | < 2(F (v))1/2 , r ∈ (r1 , R).
(2.11)
As a consequence we have
Z r 0 2
Z
Z
2 r
(v )
2 v0
ds ≤
(F (v))1/2 (−v 0 )ds =
(F (t))1/2 dt.
s
r
r
0
0
r0
r0
v
Since
(2.12)
v0
Z
(F (t))1/2 dt ≤ (F (v))1/2 v0 ,
(2.13)
v
using (2.12) we find
Rr
lim
r0
(v 0 )2
s ds
F (v)
r→R
R v0
v
= lim
r→R
(F (t))1/2 dt
= 0.
F (v)
On the other hand, by (2.10) we find
(N − 1)
(v 0 )2
=1−
2F (v)
Rr
r0
(v 0 )2
s ds
+ F (v0 )
F (v)
.
The above equation yields
−v 0
= 1 − Γ(r),
(2F (v))1/2
(2.14)
where
h
Γ(r) = 1 − 1 −
(N − 1)
Rr
r0
(v 0 )2
s ds
+ F (v0 ) i1/2
F (v)
Let us write equation (2.9) as
(rN −1 v 0 )0 (rN −1 v 0 ) + r2N −2 f (v)v 0 = 0.
Integration over (r0 , r) yields
(rN −1 v 0 )2
≥ r02N −2 [F (v) − F (v0 )],
2
.
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SECOND-ORDER BOUNDARY ESTIMATES
5
from which we find that |v 0 | > c(F (v))1/2 for some positive constant c. Hence,
Z r 0 2
Z r
Z v0
(v )
c
c
1/2
0
(F (v)) (−v )ds =
(F (t))1/2 dt.
ds >
s
r0 r 0
r0 v
r0
Rr 0 2
By using the estimate (2.6) for F (t), the latter inequality implies that r0 (vs) ds →
∞ as r → R. As a consequence, we have
Z r 0 2
(v )
(N − 1)
ds + F (v0 ) > 0
s
r0
for r close to R. Since for 0 < < 1, we have 1 − [1 − ]1/2 < . Using (2.12) we
find a constant M such that
R v0
2(N −1) R v0
(F (t))1/2 dt + F (v0 )
(F (t))1/2 dt
r0
v
≤M v
.
(2.15)
0 ≤ Γ(r) ≤
F (v)
F (v)
The inverse function of φ is
Z
s
1
dt.
(2F (t))1/2
ψ(s) =
0
Integration of (2.14) over (r, R) yields
R
Z
ψ(v) = R − r −
Γ(s)ds,
r
from which we find
Z
R
Γ(s)ds .
v(r) = φ R − r −
(2.16)
r
By (2.16) we have
Z
0
R
v(r) = φ(R − r) − φ (ω)
Γ(s)ds,
(2.17)
r
with
R
Z
R−r >ω >R−r−
Γ(s)ds.
r
Since
φ0 (ω) = (2F (φ(ω)))1/2 ,
and since the function t → F (φ(t)) is decreasing we have
Z R
1/2
φ0 (ω) < 2F φ(R − r −
Γ(s)ds)
= (2F (v))1/2 ,
r
where (2.16) has been used in the last step. Hence, by (2.17) we find
Z R
1/2
v(r) > φ(R − r) − (2F (v))
Γ(s)ds.
r
Using (2.15) we also have
1/2
v(r) > φ(R − r) − (2F (v))
Z
R
R v0
M
r
We claim that the function
R v0
r→
v(r)
(F (t))1/2 dt
F (v(r))
v
(F (t))1/2 dt
ds.
F (v)
(2.18)
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C. ANEDDA
EJDE-2009/90
is decreasing for r close to R. This is equivalent to show that the function
Z v0
(F (t))1/2 dt
s → (F (s))−1
s
is increasing for s close to 0. We have
Z v0
Z v0
i
dh
(F (s))−1
(F (t))1/2 dt = (F (s))−2 f (s)
(F (t))1/2 dt − (F (s))−1/2
ds
s
s
h R v0 (F (t))1/2 dt
i
−1/2
s
= (F (s))
−1 .
3
(F (s)) 2 (f (s))−1
R v0
3
By (2.5) and (2.6) it follows that s (F (t))1/2 dt and (F (s)) 2 (f (s))−1 tend to ∞
as s → 0. Therefore, we can use de l’Hôpital rule and condition (1.2) to find
R v0
(F (t))1/2 dt
2
γ+1
lim s
− 1 = lim
−1=
.
3
−2
0
−1
s→0 (F (s)) 2 (f (s))
s→0 3 + 2(F (s))(f (s))
f (s)
γ−3
Rv
Hence, since γ > 3, the function (F (s))−1/2 s 0 (F (t))1/2 dt is increasing for s close
to 0, and the claim follows. Using this fact, inequality (2.7) follows from (2.18).
To prove (2.8), let us write equation (2.10) as
Z r0 0 2
(v 0 )2
(v )
= F (v) − F (v0 ) + (N − 1)
ds,
(2.19)
2
s
r
with ρ < r < r0 . Note that, since (v 0 (r))2 → ∞ as r → ρ and v 00 > 0, we have [14,
Lemma 2.1]
R r0 (v0 )2
dt
lim r 0 t 2
= 0.
r→ρ
(v )
Hence, (2.19) implies 0 < v 0 < 2(F (v))1/2 for r near to ρ. As a consequence we
have
Z
Z
Z r0 0 2
2 r0
(v )
2 v0
1/2 0
ds ≤
(F (v)) (v )ds =
(F (t))1/2 dt.
s
ρ r
ρ v
r
Rv
Since v 0 (F (t))1/2 dt ≤ (F (v))1/2 v0 , the latter estimate implies that
R r0 (v0 )2
s ds
= 0.
lim r
r→ρ
F (v)
Using (2.10) again we find
(N − 1)
(v 0 )2
=1+
2F (v)
R r0
r
(v 0 )2
s ds
− F (v0 )
F (v)
.
The above equation yields
v0
= 1 + Γ(r),
(2F (v))1/2
where
h
Γ(r) = 1 +
(N − 1)
R r0
r
(v 0 )2
s ds
F (v)
− F (v0 ) i1/2
(2.20)
− 1.
Rr 0 2
Arguing as in the previous case one proves that r 0 (vs) ds → ∞ as r → ρ, and
Z r0 0 2
(v )
ds − F (v0 ) > 0
(N − 1)
s
r
EJDE-2009/90
SECOND-ORDER BOUNDARY ESTIMATES
7
for r close to ρ. Since for > 0 we have [1 + ]1/2 − 1 < , the function Γ(r) satisfies
(2.15) (possibly with a different constant M ). Integration of (2.20) over (ρ, r) yields
Z r
ψ(v) = r − ρ +
Γ(s)ds,
ρ
from which we find
r
Z
0
v(r) = φ(r − ρ) + φ (ω1 )
Γ(s)ds,
(2.21)
ρ
with
Z
r
r − ρ < ω1 < r − ρ +
Γ(s)ds.
ρ
Since φ0 (s) is decreasing, φ0 (ω1 ) < φ0 (r − ρ). This estimate and (2.21) imply
Z r
v(r) < φ(r − ρ) + φ0 (r − ρ)
Γ(s)ds.
(2.22)
ρ
−1/2
R v0
We have shown that the function (F (s))
(F (t))1/2 dt is increasing for s close
s
to 0. As a consequence, since v(r) is increasing for ρ < r, the function
R v0
(F (t))1/2 dt
v(r)
r→
F (v(r))
is increasing for r close to ρ. Hence, inequality (2.8) follows from (2.22) and (2.15).
The lemma is proved.
Corollary 2.2. Assume the same notation and assumptions of Lemma 2.1. Given
> 0 there are r1 and r2 such that
φ(R − r) > v(r) > (1 − )φ(R − r),
r1 < r < R,
(2.23)
φ(r − ρ) < v(r) < (1 + )φ(r − ρ),
ρ < r < r2 .
(2.24)
Proof. By (2.14) we have
−v 0
< 1.
(2F (v))1/2
Integrating over (r, R) we find ψ(v) < R − r, from which the left hand side of (2.23)
follows. By (2.7) we have
R1
h
(F (t))1/2 dt R − r i
v(r) > 1 − C v
φ(R − r).
(F (v))1/2 φ(R − r)
Since F (t) is decreasing we have
R1
v
(F (t))1/2 dt
≤ 1.
(F (v))1/2
Moreover, putting R − r = ψ(s) we have
R−r
ψ(s)
= lim
≤ lim (2F (s))−1/2 = 0.
s→0 s
s→0
r→R φ(R − r)
lim
The right hand side of (2.23) follows from these estimates. By (2.20) we have
v0
> 1.
(2F (v))1/2
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C. ANEDDA
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Integrating over (ρ, r) we find ψ(v) > r − ρ, from which the left hand side of (2.24)
follows. By (2.8) we have
R1
h
(F (t))1/2 dt r − ρ i
v(r) < 1 + Cφ0 (r − ρ) v
φ(r − ρ).
F (v)
φ(r − ρ)
We find
R1
(F (t))1/2 dt
1
= 0.
≤ lim
r→ρ
r→ρ (F (v))1/2
F (v)
Moreover, putting r − ρ = ψ(s) we have
lim
v
(r − ρ)φ0 (r − ρ)
ψ(s)(2F (s))1/2
=
≤ 1.
φ(r − ρ)
s
The right hand side of (2.24) follows from these estimates. The corollary is proved.
Lemma 2.3. If (1.2) holds with γ > 1 and if φ(δ) is defined as in (1.4), then
γ+1
φ0 (δ)
=
+ O(1)(φ(δ))β ,
δf (φ(δ))
γ−1
φ(δ)
γ+1
=
+ O(1)(φ(δ))β ,
δφ0 (δ)
2
φ(δ)
(γ + 1)2
=
+ O(1)(φ(δ))β ,
δ 2 f (φ(δ))
2(γ − 1)
2
φ(δ) = O(1)δ γ+1 ,
(2.25)
(2.26)
(2.27)
(2.28)
where O(1) is a bounded quantity as δ → 0.
Proof. By the relation
γ
γ+1
−1 − 2
+ O(1)tβ =
+ O(1)tβ ,
1−γ
γ−1
using (1.2) we have
−1 − 2F (t)f 0 (t)(f (t))−2 =
γ+1
+ O(1)tβ ,
γ−1
where O(1) is bounded as t → 0. Multiplying by (2F (t))−1/2 we find
γ+1
−(2F (t))−1/2 − (2F (t))1/2 f 0 (t)(f (t))−2 =
(2F (t))−1/2 + O(1)tβ (2F (t))−1/2 ,
γ−1
and
0
γ+1
(2F (t))1/2 (f (t))−1 =
(2F (t))−1/2 + O(1)tβ (2F (t))−1/2 .
(2.29)
γ−1
Using (2.5) and (2.6) we find that (2F (t))1/2 (f (t))−1 → 0 as t → 0. Hence,
integrating (2.29) on (0, s) we obtain
Z
Z s
−1/2
γ+1 s
1/2
−1
(2F (s)) (f (s)) =
2F (t)
dt + O(1)
tβ (2F (t))−1/2 dt, (2.30)
γ−1 0
0
where O(1) is bounded as s → 0. Since
Z s
Z s
β
−1/2
β
t (2F (t))
dt ≤ s
(2F (t))−1/2 dt,
0
0
EJDE-2009/90
SECOND-ORDER BOUNDARY ESTIMATES
9
equation (2.30) implies
(2F (s))1/2
γ+1
=
f (s)
γ−1
Z
s
−1/2
2F (t)
dt + O(1)sβ
s
Z
(2F (t))−1/2 dt.
0
0
0
1/2
Putting s = φ(δ) and recalling that φ (δ) = (2F (φ(δ)))
Recall that (1.2) implies (2.3). Hence, we have
, estimate (2.25) follows.
tf (t)
γ−1
=
+ O(1)tβ ,
2F (t)
2
γ+1
2F (t) + tf (t)
=
+ O(1)tβ ,
2F (t)
2
γ+1
(2F (t))−1/2 + tf (t)(2F (t))−3/2
=
+ O(1)tβ ,
−1/2
2
(2F (t))
0
γ+1
(2F (t))−1/2 + O(1)tβ (2F (t))−1/2 .
t(2F (t))−1/2 =
2
By (2.6), t(2F (t))−1/2 → 0 as t → 0. Hence, integrating over (0, s) we find
Z s
γ+1
γ+1
s(2F (s))−1/2 =
ψ(s) + O(1)
tβ (2F (t))−1/2 dt =
ψ(s) + O(1)sβ ψ(s),
2
2
0
where
Z
ψ(s) =
s
(2F (t))−1/2 dt.
0
Since ψ 0 (s) = (2F (s))−1/2 we find
sψ 0 (s)
γ+1
=
+ O(1)sβ .
ψ(s)
2
Putting s = φ(δ) and noting that φ is the inverse function of ψ we get (2.26). By
(2.25) we have
γ+1
β
φ(δ)
+
O(1)(φ(δ))
γ−1
φ(δ)
=
.
δ 2 f (φ(δ))
δφ0 (δ)
Using the latter estimate and (2.26) we get
γ + 1
γ + 1
φ(δ)
β
β
=
+
O(1)(φ(δ))
+
O(1)(φ(δ))
,
δ 2 f (φ(δ))
2
γ−1
from which (2.27) follows. Estimate (2.28) follows immediately from (2.6). The
lemma is proved.
3. Main result
N
Lemma 3.1. Let Ω ⊂ R , N ≥ 2 be a bounded smooth domain, and let f (t) > 0
smooth, decreasing for t > 0 and such that f (t) → ∞ as t → 0. Assume condition
(1.2) with γ > 3 and β > 0. If u(x) is a solution to problem (1.1), then
φ(δ) 1 − Cδ < u(x) < φ(δ) 1 + Cδ ,
(3.1)
where φ is defined as in (1.4), δ = δ(x) denotes the distance from x to ∂Ω and C
is a suitable positive constant.
10
C. ANEDDA
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Proof. If P ∈ ∂Ω we can consider a suitable annulus of radii ρ and R contained
in Ω and such that its external boundary is tangent to ∂Ω in P . If v(x) is the
solution of problem (1.1) in this annulus, by using the comparison principle for
elliptic equations we have u(x) ≥ v(x) for x belonging to the annulus. Choose the
origin in the center of the annulus and put v(x) = v(r) for r = |x|. By Lemma 2.1
we have
R1
(F (t))1/2 dt
(R − r), r̃ < r < R.
(3.2)
v(r) > φ(R − r) − C1 v
(F (v))1/2
R1
Since γ > 3, by (2.6) we find that t (F (τ ))1/2 dτ and t(F (t))1/2 approach infinite
as t approaches zero. Using de l’Hôpital rule and (2.3) (which follows from (1.2))
we find
R1
(F (τ ))1/2 dτ
−(F (t))1/2
1
2
lim t
=
lim
= lim
=
. (3.3)
tf (t)
t→0
t→0 (F (t))1/2 −
t→0 −1 + tf (t)
γ
−
3
t(F (t))1/2
1/2
2(F (t))
2F (t)
Therefore, (3.2) implies
v(r) > φ(R − r) − C2 v(r)(R − r).
The latter estimate and the left hand side of (2.23) yield
v(r) > φ(R − r) 1 − C2 (R − r) .
For x near ∂Ω we have δ = R − r; therefore, the latter estimate and the inequality
u(x) ≥ v(x) yield the left hand side of (3.1).
Consider a new annulus of radii ρ and R containing Ω and such that its internal
boundary is tangent to ∂Ω in P . If w(x) is the solution of problem (1.1) in this
annulus, by using the comparison principle for elliptic equations we have u(x) ≤
w(x) for x belonging to Ω. Choose the origin in the center of the annulus and put
w(x) = w(r) for r = |x|. By Lemma 2.1 with w in place of v we have
R1
(F (t))1/2 dt
0
w(r) < φ(r − ρ) + C1 (r − ρ)φ (r − ρ) w
,
ρ < r < r.
(3.4)
F (w)
Using (3.3) we can find a constant C2 such that
R1
(F (t))1/2 dt
w
w
≤ C2
.
F (w)
(F (w))1/2
Since φ0 = (2F (φ))1/2 , (3.4) and the previous inequality yield
F (φ) 1/2
w(r) < φ(r − ρ) + C3 (r − ρ)
w.
F (w)
(3.5)
By (2.24) with w in place of v and with = 1, and by (2.3) we find
F (φ) 1/2
F (φ) 1/2
w≤
2φ ≤ C4 φ.
F (w)
F (2φ)
Insertion of this estimate into (3.5) yields
w(r) < φ(r − ρ) 1 + C5 (r − ρ) .
(3.6)
For x near to ∂Ω we have δ = r − ρ; therefore, estimate (3.6) and the inequality
u(x) ≤ w(x) yield the right hand side of (3.1). The lemma is proved.
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SECOND-ORDER BOUNDARY ESTIMATES
11
Theorem 3.2. Let Ω ⊂ RN , N ≥ 2 be a bounded smooth domain, and let f (t) > 0
smooth, decreasing for t > 0 and such that f (t) → ∞ as t → 0. Assume condition
(1.2) with γ > 3 and β > 0. If u(x) is a solution to (1.1), then
i
h
i
h
N −1
N −1
(3.7)
φ(δ) 1 +
Kδ − Cδ 1+σ < u(x) < φ(δ) 1 +
Kδ + Cδ 1+σ ,
3−γ
3−γ
where φ is defined as in (1.4), K = K(x) is the mean curvature of the surface
2β
{x ∈ Ω : δ(x) = constant}, σ is a number such that 0 < σ < min[ γ−3
γ+1 , γ+1 ], and C
is a suitable constant.
Proof. We look for a super solution of the form
w(x) = φ(δ) 1 + Aδ + αδ 1+σ ,
where
H
, H = (N − 1)K
3−γ
and α is a positive constant to be determined. We have
wxi = φ0 δxi 1 + Aδ + αδ 1+σ + φ Axi δ + Aδxi + α(1 + σ)δ σ δxi .
A=
(3.8)
Recalling that
N
X
i=1
δxi δxi = 1,
N
X
δxi xi = −H,
i=1
we find that
∆w = φ00 1 + Aδ + αδ 1+σ − φ0 H 1 + Aδ + αδ 1+σ
+ 2φ0 ∇A · ∇δδ + A + α(1 + σ)δ σ
+ φ ∆A δ + 2∇A · ∇δ − AH + ασ(1 + σ)δ
σ−1
(3.9)
σ
− α(1 + σ)δ H .
By (1.4) we find φ00 = −f (φ). Using this equation together with (2.25) and (2.27),
by (3.9) we find that
h
γ + 1
+ O(1)φβ δH 1 + Aδ + αδ 1+σ
∆w = f (φ) −1 − Aδ − αδ 1+σ −
γ−1
γ + 1
+2
+ O(1)φβ δ ∇A · ∇δ δ + A + α(1 + σ)δ σ
γ−1
(γ + 1)2
+ O(1)φβ δ 2 ∆A δ + 2∇A · ∇δ − AH + ασ(1 + σ)δ σ−1
+
2(γ − 1)
i
− α(1 + σ)δ σ H .
(3.10)
This means that we can find suitable constants Ci such that
h
γ+1
∆w < f (φ) −1 − A +
(H − 2A) δ + C1 φβ δ + C2 δ 2
γ−1
i (3.11)
γ+1
(γ + 1)2
+ αδ 1+σ −1 + 2(1 + σ)
+ σ(1 + σ)
+ C3 φβ + C4 δ .
γ−1
2(γ − 1)
On the other hand, using Taylor’s expansion we have
h
i
f 0 (φ)
f 00 (φ)
f (w) = f (φ) 1 + φ
(Aδ + αδ 1+σ ) + φ2
(Aδ + αδ 1+σ )2 ,
f (φ)
f (φ)
(3.12)
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with φ between φ and φ(1 + Aδ + αδ 1+σ ). After α is fixed, we consider only points
x ∈ Ω such that
1
− < Aδ + αδ 1+σ < 1.
(3.13)
2
This means that 1/2 < 1 + Aδ + αδ 1+σ < 2. Therefore, the term φ which appears
in (3.12) satisfies φ = θφ, with 1/2 < θ < 2. Using (1.3) and (2.4) (which follows
from (1.2)), by (3.12) we find that
i
h
f (w) = f (φ) 1−γAδ−αγδ 1+σ +O(1)φβ δ+O(1)δ 2 +O(1)αφβ δ 1+σ +O(1)(αδ 1+σ )2 .
Using this equality, we can take suitable positive constants Ci such that
h
−f (w) > f (φ) −1 + γAδ + αγδ 1+σ − C5 φβ δ − C6 δ 2 − C7 αφβ δ 1+σ
i
− C8 (αδ 1+σ )2 .
(3.14)
Since by (3.8),
γ+1
− A+
(H − 2A) = γA,
γ−1
by (3.11) and (3.14), we have
∆w < −f (w)
(3.15)
provided that
(γ + 1)2
γ+1
C1 φβ δ + C2 δ 2 + αδ 1+σ −1 + 2(1 + σ)
+ σ(1 + σ)
+ C3 φβ + C4 δ
γ−1
2(γ − 1)
< αγδ 1+σ − C5 φβ δ − C6 δ 2 − C7 αφβ δ 1+σ − C8 (αδ 1+σ )2 .
Rearranging terms,
(C1 + C5 )φβ δ −σ + (C2 + C6 )δ 1−σ
γ+1
(γ + 1)2
< α γ + 1 − 2(1 + σ)
− σ(1 + σ)
− (C3 + C7 )φβ
γ−1
2(γ − 1)
− C4 δ − C8 αδ 1+σ .
(3.16)
Using (2.28) we find
2β
φβ δ −σ < Cδ γ+1 −σ .
2β
Since σ < γ+1
we have φβ δ −σ → 0 as δ → 0. Moreover, since σ < γ−3
γ+1 , we also
have δ 1−σ → 0 as δ → 0, and
γ − 3
γ+1
(γ + 1)2
(γ + 1)2
γ + 1 − 2(1 + σ)
− σ(1 + σ)
=
(σ + 2)
− σ > 0.
γ−1
2(γ − 1)
2(γ − 1)
γ+1
Hence, we can take α0 large and δ0 small so that (3.13) and (3.16) hold for α ≥ α0 ,
δ ≤ δ0 with αδ 1+σ ≤ α0 δ01+σ .
Let us show now that we can choose δ1 so that u(x) ≤ w(x) for δ(x) = δ1 . Using
Lemma 3.1 we have
h (N − 1)K
i
w(x) − u(x) ≥ φ(δ)
δ + αδ 1+σ − Cδ .
3−γ
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SECOND-ORDER BOUNDARY ESTIMATES
13
Take α0 and δ0 so that (3.13) and (3.16) hold, and put q = α0 δ01+σ . Decrease δ
and increasing α according to αδ 1+σ = q until
(N − 1)K
δ + q − Cδ > 0
3−γ
for δ(x) = δ1 . Then w(x) > u(x) for δ(x) = δ1 . By (3.15) and the comparison
principle [11, Theorem 10.1], it follows that u(x) ≤ w(x) in {x ∈ Ω : δ(x) < δ1 }.
We look for a sub solutions of the form
v(x) = φ(δ) 1 + Aδ − αδ 1+σ ,
where A and σ are the same as before and α is a positive constant to be determined.
Instead of (3.10) now we have
h
γ + 1
∆v = f (φ) −1 − Aδ + αδ 1+σ −
+ O(1)φβ δH 1 + Aδ − αδ 1+σ
γ−1
γ + 1
+2
+ O(1)φβ δ ∇A · ∇δδ + A − α(1 + σ)δ σ
γ−1
(γ + 1)2
+ O(1)φβ δ 2 ∆Aδ + 2∇A · ∇δ − AH − ασ(1 + σ)δ σ−1
+
2(γ − 1)
i
+ α(1 + σ)δ σ H .
(3.17)
This means that we can find suitable constants Ci (not necessarily the same as
before) such that
h
γ+1
(H − 2A) δ − C1 φβ δ − C2 δ 2
∆v > f (φ) −1 − A +
γ−1
i (3.18)
γ+1
(γ + 1)2
− αδ 1+σ −1 + 2(1 + σ)
+ σ(1 + σ)
+ C3 φβ + C4 δ .
γ−1
2(γ − 1)
After α is fixed, we consider only points x ∈ Ω such that
1
< Aδ − αδ 1+σ < 1.
(3.19)
2
Using Taylor’s expansion, now we find
h
i
−f (v) < f (φ) −1 + γAδ − αγδ 1+σ + C5 φβ δ + C6 δ 2 + C7 αφβ δ 1+σ + C8 (αδ 1+σ )2 .
(3.20)
Recalling that
γ+1
− A+
(H − 2A) = γA,
γ−1
by (3.18) and (3.20) we have
∆v > −f (v)
(3.21)
−
when
− C1 φβ δ − C2 δ 2
γ+1
(γ + 1)2
− αδ 1+σ −1 + 2(1 + σ)
+ σ(1 + σ)
+ C3 φβ + C4 δ
γ−1
2(γ − 1)
> −αγδ 1+σ + C5 φβ δ + C6 δ 2 + C7 αφβ δ 1+σ + C8 (αδ 1+σ )2 .
14
C. ANEDDA
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Rearranging terms,
(C1 + C5 )φβ δ −σ + (C2 + C6 )δ 1−σ
γ+1
(γ + 1)2
< α γ + 1 − 2(1 + σ)
− σ(1 + σ)
− (C3 + C7 )φβ
γ−1
2(γ − 1)
− C4 δ − C8 αδ 1+σ ,
(3.22)
which is the same as (3.16) (possibly with different constants). Therefore, we can
take δ small and α large in order to satisfy this inequality. Take α0 and δ0 so that
(3.19) and (3.22) hold, and put q = α0 δ01+σ . Using Lemma 3.1,
h (N − 1)K
i
v(x) − u(x) ≤ φ(δ)
δ − αδ 1+σ + Cδ .
3−γ
Decrease δ and increasing α according to αδ 1+σ = q until
(N − 1)K
δ − q + Cδ < 0
3−γ
for δ(x) = δ2 . Then v(x) < u(x) for δ(x) = δ2 . By (3.21) and the usual comparison
principle it follows that v(x) ≤ u(x) in {x ∈ Ω : δ(x) < δ2 }. The theorem follows.
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Claudia Anedda
Dipartimento di Matematica e Informatica, Universitá di Cagliari, Via Ospedale 72,
09124 Cagliari, Italy
E-mail address: canedda@unica.it