Annales Academiæ Scientiarum Fennicæ
Mathematica
Volumen 35, 2010, 633–640
A POTENTIAL THEORY APPROACH
TO THE EQUATION −∆u = |∇u|2
Anna Tuhola-Kujanpää
University of Jyväskylä, Department of Mathematics and Statistics
P.O. Box 35 (MaD), FI-40014 University of Jyväskylä, Finland; [email protected]
Abstract. In this paper we show that if
(
−∆u = |∇u|2 in Ω,
(1)
u ∈ W01,2 (Ω),
then
−∆(eu − 1) = µ in Ω,
(2)
when µ ⊥ cap2 , and conversely.
1. Introduction
In this paper we show that by a change of variable we can transform a Laplace
equation with quadratic growth in the gradient to one with a singular measure on
the right hand side. More precisely we have:
1.1. Theorem. Let Ω ⊂ Rn be a bounded domain. Then u is a solution of
(
−∆u = |∇u|2 in Ω,
(3)
u ∈ W01,2 (Ω),
if and only if eαu/2 − 1 ∈ W01,2 (Ω) for all 0 < α < 1 and there exists a positive Radon
measure µ such that eu − 1 is a weak solution of
−∆(eu − 1) = µ in Ω
(4)
and µ ⊥ cap2 .
The equation (3) is an analytic equation that does not allow any other bounded
solutions but the constant 0. Here we characterize all possible solutions.
A similar result can be found in [1] and its corrigendum [2]. However, our proof
extends to a case where µ is an arbitrary Radon measure, not necessarily bounded.
Our approach is based on a very different technique, namely potential theory and it
relies in partial on the Riesz decomposition theorem. We also employ renormalized
solutions discussed in [10].
In the proof of Theorem 1.1 we also need the uniqueness of harmonic functions in
W01,p (Ω). This is an interesting result of its own and Section 2 is devoted to its proof
and comments. To show that our assumptions on the domain Ω are relevant, we
include a counterexample by Hajłasz [13] and construct another counterexample in
a domain with a very irregular boundary. Recently Brezis [4] and Jin et al. [15] have
studied similar problems locally without considering the regularity of the domain.
doi:10.5186/aasfm.2010.3540
2000 Mathematics Subject Classification: Primary 35J05, 35J25, 31B20, 31B05.
Key words: Elliptic equations, singular solutions, uniqueness.
634
Anna Tuhola-Kujanpää
Problems with equations similar to (3) have been widely studied. See for instance
[6],[8], [7], [9], [11] and [16].
2. Uniqueness of harmonic functions in W 1,p (Ω)
In W 1,2 (Ω) the uniqueness of harmonic functions is a familiar fact: for fixed
v ∈ W 1,2 (Ω) there is a unique harmonic function u such that u − v ∈ W01,2 (Ω). In
W 1,p (Ω), when 1 < p < 2, it is difficult to locate the corresponding fact from the
literature.
In Theorem 2.1 we find a sufficient condition for the uniqueness to hold in a
domain Ω: the smoothness of the domain is expressed in terms of the integrability of
the gradient of the Green function. (For more information of the Green function, see
for instance [5].) This Theorem 2.1 will be later applied in the proof of Theorem 1.1
in a smooth domain. However, a bounded domain with C 1,α boundary for some
α > 0 is regular enough to satisfy the assumptions.
By the space W01,p (Ω) we mean the closure of C0∞ (Ω) in W 1,p (Ω).
2.1. Theorem. Let Ω ⊂ Rn be a bounded domain and G the Green function
associated with Ω. Suppose v ≥ 0 is superharmonic W01,p (Ω)-function for some p ≥ 1.
If for some x0 ∈ Ω there exists K ⊂⊂ Ω such that ∇y G(x0 , y) ∈ Lp/(p−1) (Ω \ K),
when p > 1, or ∇y G(x0 , y) ∈ L∞ (Ω \ K), when p = 1, then the greatest harmonic
minorant h of v is 0.
Proof. Notice first that by the minimum principle either h < v in Ω or v itself is
harmonic: if h(x) = v(x) for some x ∈ Ω, then for the non-negative superharmonic
function v − h we have (v − h)(x) = 0 and so v − h attains its minimum in Ω. Hence
v − h is a constant function and it follows that v is harmonic.
Assume first that h < v. Take a sequence ϕj ∈ C0∞ (Ω) such that ϕj → v in
1,p
W (Ω) and that for every compact S ⊂ Ω there exists J ∈ N such that ϕj ≥ h in
S when j > J.
Fix x0 ∈ Ω and denote g(y) = G(x0 , y). Define Ωt = {y ∈ Ω : g(y) > t} for each
t > 0. Denote the Green function of Ωt by Gt and gt (y) = Gt (x0 , y). Observe that
gt (y) = g(y) − t, and hence |∇gt (y)| = |∇g(y)|.
Denote the greatest harmonic minorant of ϕj in Ωt by ht,j . We have h(x0 ) ≤
ht,j (x0 ) for all large j. Functions ϕj have compact supports and so it is justified to
use the Green formula in Ω \ Ωt for ϕj and g. By the harmonicity of g near the
boundary we have
Z
Z
∂gt
∂g
ht,j (x0 ) =
ϕj
dS =
ϕj
dS
∂ν
∂ν
∂Ωt
∂Ωt
Z
Z
Z
∂g
=
∇ϕj · ∇g dy +
ϕj ∆g dy −
ϕj
dS
(5)
∂ν
Ω\Ωt
Ω\Ωt
∂Ω
Z
Z
=
∇ϕj · ∇g dy →
∇v · ∇g dy,
Ω\Ωt
Ω\Ωt
for when t is small enough, Ω \ Ωt ⊂ Ω \ K and hence by Hölder’s inequality
µZ
¶ p1 µZ
¶ p−1
Z
p
p
p
(∇ϕj − ∇v) · ∇g dy ≤
|∇ϕj − ∇v| dy
→ 0,
|∇g| p−1 dy
Ω\Ωt
Ω\Ωt
Ω\Ωt
A potential theory approach to the equation −∆u = |∇u|2
635
when j → ∞. This implies
³
¢
lim lim ht,j (x0 ) = lim lim
¡
t→0
j→∞
t→0
j→∞
Z
Z
´
∇ϕj · ∇g dy = lim
t→0
Ω\Ωt
∇v · ∇g dy = 0,
Ω\Ωt
since ∇v · ∇g is integrable in Ω \ Ωt when t is small. Since h(x) ≤ ht,j (x), it follows
that h(x) = 0.
In the case v is harmonic, we can find a sequence ϕj ∈ C0∞ (Ω) such that ϕj → v in
W 1,p (Ω) and ϕj converge to v locally uniformly. If we denote the greatest harmonic
minorants of ϕj and v in Ωt by ht,j and ht respectively, we have by the uniform
convergence on Ω̄t that ht,j (y) → ht (y) for all y ∈ Ωt .On the other hand, we know
by the previous calculation (5) which is also valid in this case, that
Z
ht,j (x0 ) →
∇v · ∇g dy,
Ω\Ωt
when j → ∞. Hence
Z
ht (x0 ) =
∇v · ∇g dy
Ω\Ωt
and by the integrability of ∇v · ∇g in Ω \ Ωt when t is small, we obtain ht (x0 ) → 0,
when t → 0, since |Ω \ Ωt | → 0. The result follows from the fact that ht (x0 ) ≥
h(x0 ).
¤
In the proof above it is explicitely shown the following.
2.2. Corollary. If Ω is as in Theorem 2.1 and p ≥ 1, then the only harmonic
function in W01,p (Ω) is the zero function.
As an immediate consequence we get the next corollary.
2.3. Corollary. Suppose that p ≥ 1. If Ω is as in Theorem 2.1 and v ∈ W 1,p (Ω),
then there exists at most one harmonic function u such that u − v ∈ W01,p (Ω).
2.4. Remark. When p ≥ 2 the previous Theorem is trivial, but also the assumptions of the Theorem are apparent: Let x ∈ Ω and denote Gx (y) = G(x, y).
Then the zero extension of Gx is subharmonic in Rn \ B(x, r) for all r > 0, for Gx is
1,2
harmonic in Ω \ B(x, r). Subharmonic functions belong to Wloc
(Rn \ B(x, r)). Since
2 ≥ p/(p − 1), when p ≥ 2, we have ∇G ∈ Lp/(p−1) (Ω \ B(x, r)) for all r > 0.
2.5. Remark. Theorem 2.1 is not completely trivial. In [13] Hajłasz gives a
counterexample in the case 1 < p < 34 : There exists a domain Ω ⊂ R2 and a non-zero
harmonic function u ∈ W01,p (Ω). Here Ω is the image of set D = {z ∈ C : |z − i| < 1}
under mapping z 7→ z 2 . The domain Ω has one inward cusp and it satisfies the cone
property. In the following we construct a counterexample in Rn for all 1 < p < 2
with a domain far from simply connected.
2.6. Example. Let 1 < p < 2. We can find a Cantor set E ⊂ Rn such that
cap2 (E) > 0 and dimH (E) < n − p [3, Section 5.3]. Then capp (E) = 0. Take a ball
1
B ⊂ Rn containing E and denote Ω = 2B \ E. Now the 2-potential R̂E
of E in 2B is
harmonic in Ω, but not the zero function, since cap2 (E) > 0 [5, Theorem 5.3.4.(iii)
1
and Lemma 5.3.3]. Clearly R̂E
∈ W01,p (Ω) because capp (E) = 0 [14, Theorem 8.6].
636
Anna Tuhola-Kujanpää
3. Proof of Theorem 1.1 with some preparatory results
For the proof of the main theorem we need few auxiliary results. For them,
denote
Z
vµ,Ω (x) =
G(x, y)dµ(y),
Ω
where G is the Green function of Ω.
The p-capacity capp (A) for 1 < p ≤ N is defined in the following classical way:
The p-capacity of a compact set K ⊂ Ω is first defined as
Z
capp (K) = inf{ |∇ϕ|p dx : ϕ ∈ C0∞ (Ω), ϕ(x) ≥ 1 for all x ∈ K}.
Ω
The p-capacity of any open subset U ⊂ Ω is then defined by
capp (U ) = sup{capp (K) : K compact, K ⊂ U }
Finally, the p-capacity of an arbitrary subset A ⊂ Ω is defined by
capp (A) = inf{capp (U ) : U open, A ⊂ U }.
For the properties of the p-capacity, see [3].
3.1. Lemma. For every Radon measure µ in Ω there exist unique Radon measures µ0 and µs in Ω such that µ = µ0 + µs , µ0 << cap2 and µs ⊥ cap2 .
Proof. See [12], Lemma 2.1.
¤
3.2. Lemma. Suppose µ is a positive Radon measure in Ω and µ = µ0 + µs as
above in Lemma 3.1. Then µs {vµ,Ω (x) < ∞} = 0.
Proof. Let A ⊂ Ω such that cap2 (A) = 0 and µs (Ω \ A) = 0. If µs {vµ,Ω (x) <
∞} > 0, then µs {vµ,Ω (x) < k} > 0 for some k > 0. Let K ⊂ {vµ,Ω (x) < k} ∩ A be
compact. By an alternative definition of capacity [5, Theorem 5.5.5] we have
0 = cap2 (K)
= sup{ν(K) : ν positive measure, spt(ν) ⊂ K, vµ,Ω (x) < 1 for all x ∈ K}
1
1
≥ µbK (K) ≥ µs (K).
k
k
It follows that µs {vµ,Ω (x) < k} = 0. This is a contradiction.
¤
Denote Tk (f ) = min{k, max{f, −k}} for all k ≥ 0.
α
3.3. Lemma. If u ∈ W01,2 (Ω) such that −∆u = |∇u|2 , then e 2 u − 1 ∈ W01,2 (Ω)
for all α < 1.
Proof. In the view of the Sobolev inequality is enough to show the integrability of
|∇e
| , since the zero boundary values follow immediately from the zero boundary
values of u.
Fix k > 0 and 0 < α < 1. Function eTk (u) − 1 ∈ W01,2 (Ω) and therefore it can be
chosen as a test function in equation (3). So
Z
Z
Z
2 u
Tk (u)
|∇u| e dx =
∇u · ∇e
dx =
|∇u|2 (eTk (u) − 1)dx
{u≤k}
Ω
Ω
Z
Z
Z
2 u
2 k
=
|∇u| e dx +
|∇u| e dx − |∇u|2 dx,
αu/2 2
{u≤k}
{u>k}
Ω
A potential theory approach to the equation −∆u = |∇u|2
which gives
Z
(6)
e
637
Z
−k
2
|∇u|2 dx.
|∇u| dx =
Ω
{u>k}
Also eαTk (u) − 1 ∈ W01,2 (Ω), so it is a valid test function in equation (3). Hence
we have
Z
Z
2 αTk (u)
|∇u| (e
− 1)dx =
∇u · ∇(eαTk (u) − 1)dx
Ω
Ω
Z
= α eαTk (u) ∇u · ∇Tk (u)dx,
Ω
which together with equation (6) yields
Z
Z
αu
2
αk
(α − 1)
e |∇u| dx = e
{u≤k}
Z
2
|∇u|2 dx
|∇u| dx −
{u>k}
Z
Z
Ω
|∇u|2 dx.
|∇u|2 dx −
= eαk e−k
Ω
Ω
By letting k → ∞ we have
Z
³ α ´2 Z
αu/2
2
(7)
(1 − α) |∇(e
− 1)| dx =
|∇u|2 dx < ∞.
2
Ω
Ω
α
Hence by the Sobolev inequality we have e 2 u − 1 ∈ W01,2 (Ω).
¤
u
3.4. Remark. Lemma 3.3 is sharp: Function e 2 − 1 6∈ W01,2 (Ω) unless u ≡ 0.
u
This can be seen by letting α → 1 in equation (7). If e 2 − 1 ∈ W01,2 (Ω), then the left
hand side of the equation tends to zero making ∇u the zero function.
3.5. Lemma. If u ∈ W01,2 (Ω) such that −∆u = |∇u|2 , then eαu − 1 ∈ W 1,1 (Ω)
and eαu − 1 is superharmonic for all α < 1.
Proof. To see that eαu − 1 ∈ W 1,1 (Ω) we need to notice only, that by denoting
ν = eαu dx, a bounded Radon measure, we have
¶1/2
µZ
Z
Z
2
αu
|∇u| dν
|∇(e − 1)|dx = α |∇u|dν ≤ c
Ω
Ω
Ω
µZ
αu/2
=c
|∇(e
¶1/2
2
− 1)| dx
Ω
which is finite by Lemma 3.3.
Function eαu − 1 ∈ W 1,1 (Ω) is a supersolution for the equation −∆v = 0 in
Ω for every 0 < α < 1: Let ϕ ∈ C0∞ (Ω), ϕ ≥ 0. Now eαTk (u) ϕ ∈ W01,2 (Ω) and
eαTk (u) ϕ ≥ 0 for every k > 0. By the dominated convergence theorem, valid here
because of Lemma 3.3, we have
Z
Z
Z
2 αu
2 αTk (u)
|∇u| e ϕ dx = lim
|∇u| e
ϕ dx = lim
∇u · ∇(eαTk (u) ϕ) dx
k→∞
k→∞
Ω
µΩZ
¶
ZΩ
αTk (u)
αTk (u)
2
= lim
e
∇u · ∇ϕ dx + α e
ϕ|∇Tk (u)| dx
k→∞
Ω
Ω
638
Anna Tuhola-Kujanpää
Z
Z
e ∇u · ∇ϕ dx + α eαu ϕ|∇u|2 dx
Ω
Z
ZΩ
1
=
∇eαu · ∇ϕ dx + α eαu ϕ|∇u|2 dx
α Ω
Ω
=
which implies
αu
Z
Z
∇(e
αu
eαu ϕ|∇u|2 dx ≥ 0.
− 1) · ∇ϕ dx = α(1 − α)
Ω
Ω
So eαu − 1 is superharmonic.
¤
Now we have all the ingredients for the proof of the main result.
Proof of Theorem 1.1. Assume first that u is a solution of equation (3). Since
eαu − 1 is superharmonic for all 0 < α < 1 (Lemma 3.5), we have by letting α → 1
that eu − 1 is superharmonic [14, Lemma 7.3]. Consequently [14, Theorem 7.45]
∇(eu − 1) ∈ Lq (Ω) for all q < n/(n − 1) and hence eu − 1 ∈ W01,q (Ω).
Denote by µ the Riesz measure of function eu − 1. Let ϕ ∈ C0∞ (Ω). Choose a
∞
C -set D ⊂⊂ Ω such that spt(ϕ) ⊂⊂ D. Then µ(D) < ∞ and there is a positive
function w, that solves the equation
(
−∆w = µ in D,
w = 0 on ∂D
in the renormalized sense [10, Theorem 3.1]. By the Riesz decomposition theorem
there exist harmonic minorants of eu − 1 and w in D, h and hw respectively, such
that eu − 1 = vµ,D + h and ω = vµ,D + hw . However, by Theorem 2.1 we know that
hw = 0. Hence w = vµ,D and in D we have eu − 1 = w + h, where h is a non-negative
harmonic function.
Let k > 0. We have e−Tk (u) ϕ ∈ W01,2 (D) ∩ L∞ (D) , spt(e−Tk (u) ϕ) ⊂⊂ D and
e−k ϕ = e−Tk (u) ϕ in {eu > k + 1}, e−k ϕ ∈ C0∞ (D). Since w is a solution in the
renormalized sense and {w > k} ⊂ {eu − 1 > k}, we have
Z
Z
³ ϕ ´
e−Tk (u) ϕ dµ.
∇w · ∇ T (u) dx =
k
e
Ω
Ω
1,2
By harmonicity h ∈ Wloc
(Ω), and hence we have
Z
³ ϕ ´
∇h · ∇ T (u) dx = 0.
e k
Ω
Thus
Z
Z
Z
³ ϕ ´
³ ϕ ´
u
∇(e − 1) · ∇ T (u) dx =
∇(w + h) · ∇ T (u) dx =
e−Tk (u) ϕ dµ.
k
k
e
e
Ω
Ω
Ω
This implies
Z
Z
Z
∇ϕ
2
|∇u| ϕ dx =
∇u · ∇ϕ dx =
∇(eu − 1) · u dx
e
Ω
Ω
Ω
Z
∇ϕ
= lim
∇(eu − 1) · T (u) dx
k→∞ Ω
e k
µZ
¶
Z
³ ϕ ´
ϕ
u
u
= lim
∇(e − 1) · ∇ T (u) dx + ∇(e − 1) · ∇Tk (u) T (u) dx
k→∞
e k
e k
Ω
Ω
A potential theory approach to the equation −∆u = |∇u|2
639
µZ
(8)
¶
Z
³ ϕ ´
ϕ
Tk (u)
∇(e − 1) · ∇ T (u) dx + ∇e
= lim
· ∇Tk (u) T (u) dx
k→∞
e k
e k
Ω
Ω
µZ
¶
Z
= lim
e−Tk (u) ϕ dµ + ϕ ∇Tk (u) · ∇Tk (u) dx
k→∞
Ω
µ ZΩ
¶
Z
−Tk (u)
2
= lim
e
ϕ dµ + |∇Tk (u)| ϕ dx
k→∞
Ω
Ω
Z
Z
=
e−u ϕ dµ + |∇u|2 ϕ dx,
u
Ω
and hence
Ω
Z
e−u ϕ dµ = 0
Ω
C0∞ (Ω).
for all ϕ ∈
We obtain that µ({u < ∞}) = 0. Since u is superharmonic, we
have cap2 ({u = ∞}) = 0. Thus µ ⊥ cap2 .
Next prove the converse. Clearly u ∈ W01,2 (Ω). Function eu − 1 is superharmonic
and by the Riesz decomposition theorem
Z
u(x)
e
−1=
G(x, y)dµ(y) + h(x),
Ω
where h is harmonic.
Since h cannot take values ±∞ in Ω, we have eu(x) = ∞ if and
R
only if vµ,Ω (x) = Ω G(x, y)dµ(y) = ∞. By lemma 3.2 we have µ({vµ,Ω (x) < ∞}) = 0.
This implies
(9)
µ({eu(x) < ∞}) = 0.
Let k > 0, ϕ ∈ C ∞ (Ω) and D ⊂⊂ Ω smooth such that spt(ϕ) ⊂⊂ D. As in
the proof of the first part, we find a renormalized solution w in D and by lemma 2.1
w = vµ,D . So we have eu − 1 = w + h in D. Now, e−Tk (u) ϕ is a valid test function for
w + h and we find, calculating as earlier in (8), that
Z
Z
Z
Z
Z
u ∇ϕ
−u
2
∇u · ∇ϕ dx =
∇e · u dx =
e ϕ dµ + |∇u| ϕ dx =
|∇u|2 ϕ dx,
e
0
Ω
Ω
Ω
Ω
R −u
since by equation (9) we have Ω e ϕ dµ = 0.
¤
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Received 29 January 2010