SOME SIMPLE NONLINEAR PDE’S WITHOUT SOLUTIONSSOME SIMPLE NONLINEAR PDE’S WITHOUT SOLUTIONS Ha im...

SOME SIMPLE NONLINEAR PDE’S WITHOUT SOLUTIONS

Haim Brezis(1)(2)

and Xavier Cabre(1)

(1) Analyse Numerique, Universite Pierre et Marie Curie4 pl. Jussieu, 75252 Paris Cedex 05, France

(2) Department of Mathematics, Rutgers UniversityNew Brunswick, New Jersey 08903, USA

0. Introduction.

The original motivation of this work is the following. Consider the simple problem

(0.1) −∆u = a(x)u2 + f(x) in Ω

u = 0 on ∂Ω

where Ω is a smooth bounded domain in RN , N ≥ 3. If a(x) ∈ Lp(Ω) and p > N/2,then for any f ∈ Lp(Ω) with ‖f‖p small, problem (0.1) has a unique small solution uin W 2,p(Ω). This is an easy consequence of the Inverse Function Theorem applied toF (u) = −∆u − a(x)u2 which maps X = W 2,p(Ω) ∩W 1,p

0 (Ω) into Y = Lp(Ω) (recall thatW 2,p(Ω) ⊂ L∞(Ω) by the Sobolev imbedding theorem), since its differential at 0, DF (0) =−∆ is bijective.

As a special case suppose a(x) = |x|−α in a domain Ω containing 0, with 0 < α < 2.Then for any small constant c the problem

(0.2)

−∆u =u2

|x|α+ c in Ω

u = 0 on ∂Ω

has a unique small solution.

The case α = 2 is interesting since a(x) = |x|−2 does not belong to Lp(Ω) for p > N/2.One may then wonder what happens to the problem

(0.3)

−∆u =u2

|x|2+ c in Ω

u = 0 on ∂Ω.1

2 HAIM BREZIS AND XAVIER CABRE

On the one hand, a formal computation suggests that since the linearized operator at 0is −∆, which is bijective, problem (0.3) has a solution for small c. On the other hand, theF above does not map X = W 2,p(Ω) ∩W 1,p

0 (Ω) into Y = Lp(Ω) for any 1 < p <∞. Onemay then try to construct other function spaces, for example weighted spaces, where theInverse Function Theorem might apply. This is doomed to fail. In fact, our main resultsshow that for any constant c > 0 (no matter how small) problem (0.3) has no solution,even in a very weak sense. When c < 0, problem (0.3) does, however, have a solution (seeRemark 1.4).

In Sections 1 and 4 we propose several notions of weak solutions and establish nonexis-tence. A basic ingredient in Section 1 is the following:

Theorem 0.1. Assume 0 ∈ Ω. If u ∈ L1loc(Ω), u ≥ 0 a.e. with

u2

|x|2∈ L1

loc(Ω) is such that

(0.4) −∆u ≥ u2

|x|2in D′(Ω)

then u ≡ 0.

The proof of Theorem 0.1 uses an adaptation of a method introduced in [4]. In Section 4we prove a stronger result, namely:

Theorem 0.2. Assume 0 ∈ Ω. If u ∈ L2loc(Ω\0), u ≥ 0 a.e. is such that

(0.5) −|x|2∆u ≥ u2 in D′(Ω\0)

then u ≡ 0.

Theorem 0.2 is proved using appropriate powers of testing functions — an idea due toBaras-Pierre [2]. As a consequence we obtain the nonexistence of local solutions (i.e.,in any neighborhood of 0, without prescribing any boundary condition) for a very simplenonlinear equation:

Theorem 0.3. Assume 0 ∈ Ω and c > 0. There is no function u ∈ L2loc(Ω\0) satisfying

(0.6) −|x|2∆u = u2 + c in D′(Ω\0).

In Section 3 we examine what happens to a natural approximation procedure of (0.3).Consider for example the equation

(0.7)

−∆u =minu2, n|x|2 + (1/n)

+ c in Ω, c > 0,

u = 0 on ∂Ω.

SOME SIMPLE NONLINEAR PDE’S WITHOUT SOLUTIONS 3

For any n there is a minimal solution un. We prove that un(x) → +∞ for every x ∈ Ωas n → ∞, i.e., there is complete blow-up in the sense of Baras-Cohen [1]. Again, thisrules out any reasonable notion of weak solution for (0.3).

In Section 2 we extend the previous results to more general problems such as

−∆u = a(x)g(u) + b(x)

assuming only that g ≥ 0 on R, g is continuous and nondecreasing on [0,∞) and∫ ∞ ds

g(s)<∞,

with a ∈ L1loc(Ω), a ≥ 0 and ∫

a(x)|x|N−2

=∞.

The original motivation of our research came from observations made in [6] and [4].

For any N ≥ 3 the problem −∆u = 2(N − 2)eu in B1 = x ∈ RN ; |x| < 1u = 0 on ∂B1

admits the weak solution u(x) = log(1/|x|2). It was observed in [6] that when N ≥ 11 thelinearized operator at u namely

Lv = −∆v − 2(N − 2)euv

= −∆v − 2(N − 2)|x|2

v

is coercive and thus formally bijective; this is a simple consequence of Hardy’s inequality:∫|∇v|2 ≥ (N − 2)2

4

∫v2

|x|2∀v ∈ H1

0 (B1)

(note that (N − 2)2/4 > 2(N − 2) when N ≥ 11). On the other hand the results of [4]show that when N ≥ 10 the perturbed problem −∆u = 2(N − 2)eu + c in B1

u = 0 on ∂B1

has no solution even in a weak sense and no matter how small c is, provided c > 0.


This strange “failure” of the Inverse Function Theorem is only apparent. As was pointedout in [6] this just means that there is no functional setting in which it can be correctlyapplied. We have tried here, in the spirit of Open Problem 6 in [6], to analyze the samephenomenon for simple examples in low dimensions.

After our investigation was completed we learned about an interesting work of N. J.Kalton and I. E. Verbitsky [8] (which was carried out independently of ours). Consider forexample the problem

(0.8) −∆u = a(x)u2 + c in Ω

u = 0 on ∂Ω

with a ∈ L1loc(Ω), a ≥ 0 and c a positive constant.

Their result says that if (0.8) has a weak solution, then necessarily

(0.9) G(aδ2) ≤ Cδ in Ω

for some constant C, where G = (−∆)−1 (with zero boundary condition) and δ(x) =dist(x, ∂Ω). In particular, if 0 ∈ Ω and a(x) = 1/|x|2, then G(aδ2) ' | log |x|| as x → 0and thus (0.9) fails; hence (0.8) has no weak solution.

We present in Section 5 a very simple proof of the main result of [8] using a variant ofthe method developed in Section 1.

Finally, in Section 6 we present a parabolic analogue of Theorem 0.2. It extends, inparticular, a result of I. Peral and J. L. Vazquez [12]. Namely, the problem

ut −∆u = 2(N − 2)eu in B1 × (0, T )u = 0 on ∂B1 × (0, T )

u(x, 0) = u0

with u0 ≥ u = log(1/|x|2), u0 6≡ u, has no solution u ≥ u even for small time: instanta-neous and complete blow-up occurs.

The plan of the paper is the following:

(1) Proof of Theorem 0.1(2) General nonlinearities(3) Complete blow-up(4) Very weak solutions. Proofs of Theorems 0.2 and 0.3(5) Connection with a result of Kalton-Verbitsky(6) Evolution equations


Notation. Throughout this paper, Ω is a bounded smooth domain of RN , N ≥ 1, suchthat 0 ∈ Ω. We write

δ(x) = dist(x, ∂Ω)

and L1δ(Ω) = L1(Ω, δ(x)dx). We denote by C∞0 (Ω) the space of C∞ functions with compact

support in Ω, and by D′(Ω) the space of distributions in Ω. By C we denote a positiveconstant which may be different in each inequality.

1. Proof of Theorem 0.1.

In this section we prove Theorem 0.1 and its consequences. We first introduce someterminology about weak solutions.

Definition 1.1. Let h(x, u) be a Caratheodory function in Ω × R, that is, h(x, u) ismeasurable in x and continuous in u for a.e. x.

(a) We say that−∆u = h(x, u) in D′(Ω)

if u ∈ L1loc(Ω), h(x, u) ∈ L1

loc(Ω) and −∫u∆ϕ =

∫h(x, u)ϕ for any ϕ ∈ C∞0 (Ω).

(b) We say that−∆u ≥ h(x, u) in D′(Ω)

if u ∈ L1loc(Ω), h(x, u) ∈ L1

loc(Ω) and −∫u∆ϕ ≥

∫h(x, u)ϕ for any ϕ ∈ C∞0 (Ω)

with ϕ ≥ 0.(c) We say that u is a weak solution of −∆u = h(x, u) in Ω

u = 0 on ∂Ω

if u ∈ L1(Ω), h(x, u) ∈ L1δ(Ω) and −

∫Ωu∆ζ =

∫Ωh(x, u)ζ for any ζ ∈ C2(Ω) with

ζ = 0 on ∂Ω.

The following is the main result of this section; it is Theorem 0.1 of the Introduction.

Theorem 1.2. Let N ≥ 1 and u ∈ L1loc(Ω) satisfy u ≥ 0 a.e. in Ω,

u2

|x|2∈ L1

loc(Ω) and

(1.1) −∆u ≥ u2

|x|2in D′(Ω).

Then u ≡ 0.

This theorem easily implies two nonexistence results. The first one deals with thefollowing boundary value problem.


Corollary 1.3. Let N ≥ 1 and f ∈ L1δ(Ω) satisfy f ≥ 0 a.e. and f 6≡ 0. Then there is no

weak solution of

(1.2)

−∆u =u2

|x|2+ f(x) in Ω

u = 0 on ∂Ω,

in the sense of Definition 1.1.

Remark 1.4. When f ≡ 0, problem (1.2) has u ≡ 0 as the only weak solution; this followsimmediately from Theorem 1.2. When N ≥ 3 and f ∈ L1(Ω) satisfies f ≤ 0 and f 6≡ 0,(1.2) has a weak solution u ≤ 0, which is the unique solution among nonpositive functions.This fact is a consequence of a more general result of Gallouet and Morel [7], which extendswork of Brezis and Strauss [5].

As a consequence of Theorem 1.2, we may write down a simple PDE without localsolutions, i.e., no solution exists in any neighborhood of 0. Here, we do not impose anyboundary condition.

Corollary 1.5. Let N ≥ 3 and c > 0 be any positive constant. Then there is no function

u such thatu2

|x|2∈ L1

loc(Ω) and

(1.3) −∆u =u2 + c

|x|2in D′(Ω).

Remark 1.6. In contrast with (1.3), the equation

(1.4) −∆u =u2

|x|2+ c

has a weak solution in some neighborhood of 0, if N ≥ 3 and c > 0 is a constant. Thissolution is nonpositive and can be obtained from the results of [7] as follows.

We introduce the new unknown v = −u− c|x|2 so that (1.4) becomes

(1.5) −∆v +v2

|x|2+ 2cv = c(2N − 1)− c2|x|2 ≡ h(x).

We solve (1.5) on BR with the boundary condition v = 0 on ∂BR. Note that h ≥ 0 onBR provided R is sufficiently small (R2 ≤ (2N − 1)/c). The results of [7] give a uniquesolution v ≥ 0.

In Section 4 we will prove stronger nonexistence results for problems (1.2) and (1.3).

The proof of Theorem 1.2 is based on the following variant of Kato’s inequality [9].


Lemma 1.7. Let u ∈ L1loc(Ω) and f ∈ L1

loc(Ω) satisfy

−∆u ≥ f in D′(Ω).

Let φ : R→ R be a C1, concave function such that

0 ≤ φ′ ≤ C in R

for some constant C. Then φ(u) ∈ L1loc(Ω) and

−∆φ(u) ≥ φ′(u)f in D′(Ω).

The proof is standard, smoothing u, f and φ by convolution; see also Lemma 2 in [4].

The proof of Theorem 1.2 is a variant of a method introduced in [4]. Consider thefunction φ(s) = (1/ε) − (1/s). It is nonnegative, bounded, increasing and concave in theinterval [ε,∞), ε > 0. Note that if u ≥ ε satisfies

−∆u ≥ u2

|x|2

then

−∆φ(u) ≥ φ′(u)u2

|x|2=

1|x|2

.

This will lead to a contradiction with the fact that φ(u) is bounded. The details go asfollows.

Proof of Theorem 1.2. Suppose that u is as in the theorem, and that u 6≡ 0. Since u ≥ 0,u 6≡ 0, −∆u ≥ 0 in D′(Ω) and Ω is connected, we have that

u ≥ ε a.e. in Bη,

for some ε > 0 and Bη = Bη(0) with closure in Ω. If N ≤ 2, this is already a contradictionwith u2/|x|2 ∈ L1

loc(Ω). Suppose N ≥ 3. Let

φ(s) =1ε− 1s

for s ≥ ε,

and extend it by φ(s) = (s − ε)/ε2 for s ≤ ε. Note that φ : R → R is C1, concave and0 ≤ φ′ ≤ 1/ε2, so that φ satisfies all the conditions of Lemma 1.7. Recall that u ≥ ε inBη, and consider

v = φ(u) =1ε− 1u

in Bη.


It satisfies 0 ≤ v ≤ 1/ε in Bη and, by Lemma 1.7,

−∆v ≥ φ′(u)u2

|x|2=

1|x|2

in D′(Bη).

Hence v − 1N − 2

log1|x|∈ L1(Bη) and

−∆(v − 1

N − 2log

1|x|

)≥ 0 in D′(Bη),

which implies

v − 1N − 2

log1|x|≥ −C in Bη/2

for some constant C > 0. In particular, v(x) → +∞ as x → 0, which is a contradictionwith the fact that v ≤ 1/ε.

Remark 1.8. In the previous proof we could have used (in the spirit of [4]) the function

w =u

εu+ 1

for any ε > 0, instead of v =1ε− 1u

. Note that

ψ(s) =s

εs+ 1

satisfies ψ′(s) = (εs+ 1)−2 and hence

ψ′(s)s2 = ψ(s)2.

Moreover, ψ is bounded in [0,∞) and satisfies all the conditions (for φ) of Lemma 1.7 in[0,∞). In particular, if u ≥ 0, u 6≡ 0 and −∆u ≥ u2/|x|2 then

−∆w ≥ ψ′(u)u2

|x|2=

w2

|x|2,

and we can conclude as before, since w2/|x|2 ≥ ν/|x|2 in a subdomain of Ω, for someconstant ν > 0.

Analogous versions of the functions φ and ψ will appear, in Sections 2 and 5, when thenonlinearity u2 is replaced by more general nonlinearities g(u).


Finally, we use Theorem 1.2 to prove the nonexistence results of this section.

Proof of Corollary 1.3. Suppose that u is a weak solution of (1.2). Since∫

Ωu(−∆ζ) ≥ 0 for

any ζ ∈ C2(Ω) with ζ ≥ 0 in Ω and ζ = 0 on ∂Ω, we easily deduce u ≥ 0 in Ω. Moreover,−∆u ≥ u2/|x|2 in D′(Ω). We obtain, by Theorem 1.2, u ≡ 0. This is a contradiction with(1.2), since f 6≡ 0.

Proof of Corollary 1.5. Suppose that u is a solution of (1.3) in D′(Ω). Then −∆u ≥ c/|x|2in D′(Bη), for some ball Bη = Bη(0) with closure in Ω. As in the proof of Theorem 1.2,we deduce that

u− c

N − 2log

1|x|≥ −C in Bη/2,

for some constant C. In particular, u ≥ 0 in Bν for some small ν > 0.

We therefore have

u ≥ 0 in Bν

−∆u ≥ u2

|x|2in D′(Bν).

By Theorem 1.2, u ≡ 0 in Bν , which is a contradiction with equation (1.3).

2. General nonlinearities.

In this section we extend the previous nonexistence results to more general problems ofthe form

−∆u = a(x)g(u) + b(x).

We assume (here and throughout the rest of the paper) that g : R −→ [0,∞) is continuouson R, nondecreasing on [0,∞), g(s) > 0 if s > 0, and

(2.1)∫ ∞

1

ds

g(s)<∞.

Power functions g(u) = up for u ≥ 0, with p > 1 are examples of such nonlinearities. Wesuppose that N ≥ 3.

For the function a(x) we assume in this section that a ∈ L1loc(Ω), a ≥ 0 in Ω, and

(2.2)∫Bη(0)

a(x)|x|N−2

=∞

for some η > 0 small enough (or, equivalently, for any η > 0 small).


Theorem 2.1. Let g and a satisfy the assumptions above, with N ≥ 3.(a) Let u ≥ 0 a.e. in Ω satisfy

−∆u ≥ a(x)g(u) in D′(Ω).

Then u ≡ 0.(b) Let f ∈ L1

δ(Ω) satisfy f ≥ 0 a.e. and f 6≡ 0. Then there is no weak solution of −∆u = a(x)g(u) + f(x) in Ωu = 0 on ∂Ω.

(c) If b(x) satisfies the same conditions as a(x), then there is no weak solution of

−∆u = a(x)g(u) + b(x) in D′(Ω).

This theorem is proved with the same method as in the previous section. We only needto adapt two points. First, φ(s) = (1/ε)− (1/s) has to be replaced by a solution of

φ′(s) =1g(s)

if s ≥ ε,

where ε > 0 is a constant.

We therefore define

φ(s) =∫ s

ε

dt

g(t)if s ≥ ε,

which satisfies 0 ≤ φ ≤∫∞ε

dtg(t) < ∞ in [ε,∞) (by assumption (2.1)), φ(ε) = 0, φ′(ε) =

1g(ε) , φ is C1, concave (since φ′(s) = 1

g(s) is nonincreasing) and 0 ≤ φ′ ≤ 1g(ε) in [ε,∞).

Extending φ by φ(s) = 1g(ε) (s− ε) for s ≤ ε, we obtain a function φ on all of R, satisfying

the conditions of Lemma 1.7 and with φ bounded from above.

To complete the proof of Theorem 2.1, we only need to consider a solution w ∈ L1loc(Bη),

where Bη = Bη(0) with closure in Ω, of

−∆w = a(x) in D′(Bη)

(a solution always exists since a ∈ L1(Bη)) and show that

(2.3) ess infB1/n(0)

w −→ +∞ as n→ +∞.


For this purpose, we consider the convolution in Bη, w = a ∗ CN |x|2−N , where CN ischosen such that −∆(CN |x|2−N ) = δ0. Then w− w is harmonic in Bη and hence boundedin Bη/2. In particular, it suffices to show (2.3) for w. But this is true since, for |x| ≤ 1/n,

w(x) = CN

∫Bη

a(y)|y − x|N−2

dy

≥ C∫Bη

a(y)|y|N−2 + (1/n)N−2

dy → +∞

as n→∞, by (2.2).

3. Complete blow-up.

In Corollary 1.3 and Theorem 2.1 we have proved the nonexistence of weak solutions ofsome problems of the form

(3.1) −∆u = a(x)g(u) + f(x) in Ω

u = 0 on ∂Ω.

In this section we prove that, under the same assumptions on a(x), g(u) and f(x) madein Section 2, approximate solutions of (3.1) blow up everywhere in Ω, that is, there iscomplete blow-up. More precisely, we have the following.

Let g(u) and a(x) be as in Section 2. Suppose that f ∈ L1δ(Ω), f ≥ 0 a.e. and f 6≡ 0.

Let (gn) be a sequence of nonnegative, bounded, nondecreasing and continuous functionsin [0,∞) such that (gn) increases pointwise to g. Let an and fn be two sequences ofnonnegative bounded functions in Ω, increasing pointwise to a and f , respectively.

Theorem 3.1. Under the above assumptions, let un be the minimal nonnegative solutionof the approximate problem

(3.2)n

−∆u = an(x)gn(u) + fn(x) in Ωu = 0 on ∂Ω.

Then, as n→ +∞,un(x)δ(x)

→ +∞ uniformly in Ω.

In the proof of Theorem 3.1 we use two ingredients. First, the nonexistence result ofTheorem 2.1(b) (see also Corollary 1.3) and, second, the following estimate for the linearLaplace equation. It asserts that, for some positive constant c,

(3.3) G(x, y) ≥ cδ(x)δ(y) in Ω× Ω,

where G is the Green’s function of the Laplacian in Ω with zero Dirichlet condition. In anequivalent way, we can state this lower bound on G as follows.


Lemma 3.2. Suppose that h ≥ 0 belongs to L∞(Ω). Let v be the solution of −∆v = h in Ωv = 0 on ∂Ω.

Then

(3.4)v(x)δ(x)

≥ c∫

Ω

hδ ∀x ∈ Ω,

where c > 0 is a constant depending only on Ω.

Estimate (3.3) was proved by Morel and Oswald [11] (in unpublished work), and byZhao [13] (in a stronger form). For the convenience of the reader we give a simple proofof (3.3).

Proof of Lemma 3.2. We proceed in two steps.

Step 1. For any compact set K ⊂ Ω, we show

(3.5) v(x) ≥ c∫

Ω

hδ ∀x ∈ K,

where c is a positive constant depending only on K and Ω. To prove (3.5), let ρ =dist(K, ∂Ω)/2, and take m balls of radius ρ such that

K ⊂ Bρ(x1) ∪ . . . ∪Bρ(xm) ⊂ Ω.

Let ζ1, . . . , ζm be the solutions of −∆ζi = χBρ(xi) in Ωζi = 0 on ∂Ω,

where χA denotes the characteristic function of A. The Hopf boundary lemma impliesthat there is a constant c > 0 such that

ζi(x) ≥ cδ(x) ∀x ∈ Ω ∀1 ≤ i ≤ m.

Here and in the rest of the proof, c denotes various constants depending only on K andΩ. Let now x ∈ K, and take a ball Bρ(xi) containing x. Then Bρ(xi) ⊂ B2ρ(x) ⊂ Ω, andsince −∆v ≥ 0 in Ω, we conclude

v(x) ≥ ∫B2ρ(x)

v = c

∫B2ρ(x)

v ≥ c∫Bρ(xi)

v

= c

∫Ω

v(−∆ζi) = c

∫Ω

hζi

≥ c∫

Ω

hδ.


Step 2. Fix a smooth compact set K ⊂ Ω. By (3.5), v ≥ c∫

Ωhδ in K, so that it suffices

to prove (3.4) for x ∈ Ω\K.

Let w be the solution of −∆w = 0 in Ω\K

w = 0 on ∂Ωw = 1 on ∂K.

The Hopf boundary lemma gives again

w(x) ≥ cδ(x) ∀x ∈ Ω\K.

Since v is superharmonic and v ≥ c∫

Ωhδ on ∂K, the maximum principle implies

v(x) ≥ c(∫

Ω

hδ

)w(x)

≥ c(∫

Ω

hδ

)δ(x) ∀x ∈ Ω\K.

This completes the proof.

Proof of Theorem 3.1. Consider the approximate problem

(3.6)n

−∆u = an(x)gn(u) + fn(x) in Ωu = 0 on ∂Ω.

Since 0 ≤ an(x)gn(s) + fn(x) ≤ Cn in Ω× [0,∞) for some constant Cn, we have that Cnzis a supersolution of (3.6)n where

(3.7) −∆z = 1 in Ω

z = 0 on ∂Ω.

On the other hand, 0 is a subsolution of (3.6)n. We therefore obtain a minimal solutionun of (3.6)n by monotone iteration: −∆um+1 = an(x)gn(um) + fn(x) in Ω

um+1 = 0 on ∂Ω,

starting with u0 ≡ 0. In particular, since an(x)gn(s) + fn(x) increases with n, un+1 is asupersolution of (3.6)n, and hence

un ≤ un+1.


We claim that ∫Ω

an(x)gn(un)δ +∞ as n +∞.

Lemma 3.2 then givesun(x)δ(x)

+∞ uniformly in Ω

as n → +∞; this proves Theorem 3.1. Thus, it only remains to show the claim. Supposenot, that

(3.8)∫

Ω

an(x)gn(un)δ ≤ C ∀n.

Then, multiplying (3.6)n by the solution z of (3.7), we see that∫Ω

un ≤ C ∀n

(we have used that 0 ≤ fn ≤ f ∈ L1δ(Ω)). Hence, un u in L1(Ω), for some u, by

monotone convergence.

Since gn is a nondecreasing function, angn(un) + fn increases to ag(u) + f a.e. in Ω;(3.8) also gives

angn(un) + fn ag(u) + f in L1δ(Ω),

again by monotone convergence. We can now pass to the limit in the weak formulation of(3.6)n (recall Definition 1.1( c)), and obtain that u is a weak solution of −∆u = a(x)g(u) + f(x) in Ω

u = 0 on ∂Ω.

This is impossible by Theorem 2.1(b).

4. Very weak solutions. Proofs of Theorems 0.2 and 0.3.

In this section we return to the study of equation

−∆u =u2

|x|2+ f(x)

and its corresponding boundary value problem. We prove stronger versions of the nonex-istence results of Section 1 by considering a more general notion of solutions, that we callvery weak solutions. More precisely, we prove the following results.


Theorem 4.1. Let N ≥ 2 and u ∈ L2loc(Ω\0) satisfy u ≥ 0 a.e. in Ω and

(4.1) −|x|2∆u ≥ u2 in D′(Ω\0),

in the sense that −∫u∆(|x|2ϕ) ≥

∫u2ϕ for any ϕ ≥ 0, ϕ ∈ C∞0 (Ω\0).

Then u ≡ 0.

Note that now we are testing (4.1) only against functions with compact support in Ωand vanishing in a neighborhood of 0. As a consequence of the previous theorem, we willprove the following stronger version of Corollary 1.3.

Corollary 4.2. Let N ≥ 2 and f ∈ L1loc(Ω\0), fδ integrable near ∂Ω, f ≥ 0 a.e. in Ω,

f 6≡ 0. Then there is no very weak solution of

(4.2)

−∆u =u2

|x|2+ f(x) in Ω

u = 0 on ∂Ω,

in the sense that u ∈ L2loc(Ω\0), u and u2δ are integrable near ∂Ω and

−∫

Ω

u∆(|x|2ζ

)=∫

Ω

u2ζ +∫

Ω

f |x|2ζ

for any ζ ∈ C2(Ω), ζ = 0 on ∂Ω and ζ ≡ 0 in a neighborhood of 0.

Remark 4.3. Theorem 4.1 does not hold in dimension N = 1; a direct computation showsthat u(x) = |x|α, for any 0 < α < 1, satisfies (4.1) when Ω is a small interval containing 0.Corollary 4.2 is also false in dimension N = 1; in fact, for any p > 1 and f ∈ Lp(−1, 1),small in Lp, there exists a very weak solution u of (4.2) in Ω = (−1, 1) which is continuousin [−1, 1] and satisfies u(0) = 0. This solution u is defined in (0, 1) to be the solution of

(4.3)

−u′′ =u2

|x|2+ f(x) in (0, 1)

u(0) = u(1) = 0,

and similarly in (−1, 0); we obtain in this way a very weak solution u of (4.2) in (−1, 1).Note that if ‖f‖Lp(0,1) is small then there is a solution u ∈ C2(0, 1) ∩ C1([0, 1]) of

(4.3). It can be obtained through the Inverse Function Theorem applied to the operator−u′′ − u2/|x|2, which maps X = W 2,p(0, 1) ∩W 1,p

0 (0, 1) into Lp(0, 1); note that if u ∈ Xthen u/|x| ∈ L∞(0, 1) and u ∈ C1([0, 1]).

We also extend the local nonexistence result of Corollary 1.5, now for all N ≥ 1.


Corollary 4.4. Let N ≥ 1 and c > 0 be any positive constant. Then there is no veryweak solution of

(4.4) −|x|2∆u = u2 + c in D′(Ω\0),

in the sense that u ∈ L2loc(Ω\0), and −

∫u∆(|x|2ϕ) =

∫(u2 + c)ϕ for any function

ϕ ∈ C∞0 (Ω\0).

The proofs of the results of this section consist of using appropriate powers of testingfunctions; this is an idea due to Baras and Pierre [2] and employed in [2] for the studyof removable singularities of solutions of semilinear equations. In fact, the first step inour proof of Theorem 4.1 will be to use equation (4.1) to show that u/|x| ∈ L2

loc(Ω); thiscan be interpreted as a “removable singularity” result. The second step of the proof is toshow that −∆u ≥ u2/|x|2 is satisfied in D′(Ω). We may then conclude that u ≡ 0 withthe help of Theorem 1.2. We present here an alternative proof of Theorem 1.2 based onmultiplication by a sequence of appropriate testing functions.

Proof of Theorem 4.1.

Step 1. We prove that u/|x| ∈ L2loc(Ω). For this purpose, let ζn ∈ C∞0 (Ω\0) be such

that 0 ≤ ζn ≤ 1,

ζn =

0 if |x| ≤ 1/n1 if |x| ≥ 2/n, x ∈ ωζ if x ∈ Ω\ω,

where ω is open, 0 ∈ ω, ω ⊂ Ω and ζ is a fixed “tail” for all ζn. We take ζn such that if1/n < |x| < 2/n then

∇ζ4n = 4ζ3

n∇ζn∆ζ4

n = 4ζ3n∆ζn + 12ζ2

n|∇ζn|2

and

|∆ζ4n| ≤ Cn2ζ2

n,

for some constant C independent of n. Multiplying (4.1) byζ4n

|x|2yields

∫u2

|x|2ζ4n ≤ −

∫u∆ζ4

n

≤ C

(∫ 1n<|x|<

2nn2uζ2

n

)+ C.


Using the Cauchy-Schwarz inequality, we obtain∫u2

|x|2ζ4n ≤ Cn

(∫ 1n<|x|<

2n

u

|x|ζ2n

)+ C

≤ C

nN2 −1

(∫u2

|x|2ζ4n

)1/2

+ C.

Since N ≥ 2 we deduce that ∫u2

|x|2ζ4n ≤ C,

and hence u/|x| ∈ L2loc(Ω).

For later purposes, let us retain two more consequences of the previous proof. First, wedid not use u ≥ 0. Second, if carried out for N = 1, the proof gives

(4.5)∫ 3/n

2/n

u2

|x|2≤ Cn+ C, (N = 1).

Step 2. We show that

(4.6) −∆u ≥ u2

|x|2in D′(Ω).

Indeed, let ϕ ∈ C∞0 (Ω), ϕ ≥ 0, and ηn(x) = η1(n|x|) be such that 0 ≤ η1 ≤ 1 and

η1(x) =

0 if |x| ≤ 11 if |x| ≥ 2.

Multiplying (4.1) byϕ

|x|2ηn yields

∫u2

|x|2ϕηn ≤ −

∫u∆(ϕηn).

If we show that, as n→∞,

(4.7)∫u|∇ϕ| |∇ηn| → 0

and

(4.8)∫uϕ|∆ηn| → 0


then we obtain −∫u∆(ϕηn) → −

∫u∆ϕ and hence the statement of Step 2. To prove

(4.7) and (4.8), we use that u/|x| ∈ L2loc(Ω) — which we established in Step 1. Hence∫

u|∇ϕ| |∇ηn| ≤ Cn∫ 1n<|x|<

2nu

≤ C∫ 1n<|x|<

2n

u

|x|→ 0

and ∫uϕ|∆ηn| ≤ Cn2

∫ 1n<|x|<

2nu ≤ Cn

∫ 1n<|x|<

2n

u

|x|

≤ C

nN2 −1

(∫ 1n<|x|<

2n

u2

|x|2

)1/2

→ 0.

Note that, as in Step 1, we have not used u ≥ 0.

Step 3. We show that u ≡ 0 (it is only here where we use u ≥ 0). Let us suppose thatu 6≡ 0. Then, since u ≥ 0, −∆u ≥ 0 in D′(Ω) and Ω is connected, we have that

u ≥ ε a.e. in Bη,

for some ε > 0 and Bη = Bη(0) with closure in Ω. When N = 2 this is impossible sinceu/|x| ≥ ε/|x| near 0 and hence u/|x| 6∈ L2

loc(Ω) — a contradiction with Step 1.

When N ≥ 3, we use that

−∆u ≥ ε2

|x|2= −∆

(ε2

N − 2log

1|x|

)in D′(Bη)

and we conclude (as in the proof of Theorem 1.2) that

(4.9) u ≥ ε2

N − 2|log |x|| − C near 0

for some C > 0.

Let us now choose a sequence of functions χn(x) = χ1(n|x|) such that 0 ≤ χn ≤ 1 and

χn(x) =

1 if |x| ≤ 1/n0 if |x| ≥ 2/n.


Mutiplying (4.6) by χ4n yields∫u2

|x|2χ4n ≤

∫u|∆χ4

n| ≤ Cn2

∫ 1n<|x|<

2nuχ2

n

≤ Cn∫

u

|x|χ2n ≤

C

nN2 −1

(∫u2

|x|2χ4n

)1/2

,

and therefore ∫u2

|x|2χ4n ≤

C

nN−2.

But, using (4.9), we have∫u2

|x|2χ4n ≥ c

∫ 1

2n<|x|<1nn2| log n|2 ∼ | log n|2

nN−2

which contradicts the previous statement.

Finally we give the proofs of Corollaries 4.2 and 4.4.

Proof of Corollary 4.2. Recall that Steps 1 and 2 of the previous proof hold for any usatisfying (4.1) — without the assumption u ≥ 0. Therefore, since f ≥ 0, Step 1 of theproof of Theorem 4.1 gives

u

|x|∈ L2

loc(Ω).

Moreover, proceeding as in Step 2 of the same proof, we see that

−∫

Ω

u∆ζ =∫

Ω

u2

|x|2ζ +

∫Ω

fζ

for any ζ ∈ C2(Ω), ζ = 0 on ∂Ω. In particular, −∫u∆ζ ≥ 0 if, in addition, ζ ≥ 0 in Ω.

We conclude that u ≥ 0 in Ω. Theorem 4.1 implies that u ≡ 0, which contradicts (4.2)and f 6≡ 0.

Proof of Corollary 4.4. Suppose that u satisfies

−|x|2∆u = u2 + c in D′(Ω\0).

The proof of Step 1 in Theorem 4.1 gives

u2

|x|2+

c

|x|2∈ L1

loc(Ω) when N ≥ 2.


This is impossible when N = 2, since 1/|x|2 is not integrable near 0.

When N ≥ 3 we get (as in the proof of Theorem 4.1, Step 2) that

−∆u =u2

|x|2+

c

|x|2in D′(Ω).

We now proceed as in the proof of Corollary 1.5, i.e., we compare u with log(1/|x|). Weobtain that u ≥ 0 in a neighborhood of 0 and hence, by Theorem 4.1, u ≡ 0. This is acontradiction with equation (4.4).

We finally treat the case N = 1. We would have u ∈ L2loc(Ω\0) and

−x2u′′ = u2 + c in D′(Ω\0).

In particular, u belongs to C2(0, a) for some a > 0. Integrating the inequality −u′′ ≥ c/x2

in (s, a/2), we obtainu′(s) ≥ c

s− C

for some constant C. Integrating again yields

−u(s) ≥ −c log s− C.

Thus|u(s)| ≥ c| log s| − C near 0.

On the other hand, we recall (4.5):∫ 3/n

2/n

u2

s2ds ≤ Cn+ C,

which was proved in Step 1 of the proof of Theorem 4.1. Using |u(s)| ≥ c| log s| − C, thisinequality yields

c

6| log n|2n = c| log n|2

∫ 3/n

2/n

ds

s2

≤∫ 3/n

2/n

u2

s2ds ≤ Cn+ C,

which is a contradiction.


5. Connection with a result of Kalton-Verbitsky.

In this section we consider the problem (for u ≥ 0)

(5.1) −∆u = a(x)up + f(x) in Ω

u = 0 on ∂Ω,

where p > 1, a ≥ 0 and f ≥ 0 in Ω. Recently, Kalton and Verbitsky [8] have found aninteresting necessary condition for the existence of a weak solution of (5.1). Their resultstates that if (5.1) has a weak solution, then necessarily

(5.2) G(aG(f)p) ≤ CG(f) in Ω

for some constant C, where G = (−∆)−1 (with zero Dirichlet boundary condition). In [8]the authors also prove (5.2) for more general second-order elliptic operators.

In this section we give a simple proof of the necessary condition (5.2) (for the Laplacian)using a refinement of the method that we have developed in Section 1. Our proof gives(5.2) with constant C = 1

p−1 . Next, we replace f(x) by λf(x) in (5.1) (where λ > 0 is aparameter) and we study the problem of existence of solution depending on the value of λ.

As pointed out in the Introduction, condition (5.2) easily implies some of our nonexis-tence results. For instance, it gives the result of Theorem 2.1(b) when g(u) = up, for somep > 1, and f ∈ L∞, since in this case G(f) ∼ δ.

We recall that there is another necessary condition — due to Baras and Pierre [3] —for the existence of a weak solution of (5.1). Its proof consists of multiplying (5.1) by testfunctions and using Young’s inequality.

Throughout this section we assume that

(5.3) a ∈ L1loc(Ω), a ≥ 0 a.e., a 6≡ 0

and

(5.4) f ∈ L1δ(Ω), f ≥ 0 a.e., f 6≡ 0.

A function u ∈ L1(Ω), u ≥ 0 a.e. is a weak solution of (5.1) if aup ∈ L1δ(Ω) and (5.1) is

satisfied in the sense of Definition 1.1( c). Finally, for h ∈ L1δ(Ω) we denote by G(h) the

unique function in L1(Ω) satisfying −∆(G(h)) = h in ΩG(h) = 0 on ∂Ω

again in the sense of Definition 1.1( c) — see e.g. Lemma 1 of [4] for such result about thelinear Laplace equation. We now give a necessary condition and a sufficient condition forthe existence of a weak solution of (5.1).


Theorem 5.1. Assume (5.3) and (5.4).

(a) If

(5.5) −∆u = a(x)up + f(x) in Ω

u = 0 on ∂Ω

has a weak solution u ≥ 0, then aG(f)p ∈ L1δ(Ω) and

G(aG(f)p)G(f)

≤ 1p− 1

in Ω.

(b) If aG(f)p ∈ L1δ(Ω) and

G(aG(f)p)G(f)

≤(p− 1p

)p 1p− 1

in Ω,

then (5.5) has a weak solution u satisfying G(f) ≤ u ≤ CG(f) in Ω for some constant C.

The second result of this section is the following.

Theorem 5.2. Assume (5.3), (5.4) and

G(aG(f)p)G(f)

∈ L∞(Ω).

For λ a positive parameter, consider the problem

(5.6)λ

−∆u = a(x)up + λf(x) in Ωu = 0 on ∂Ω.

Then there exists λ? ∈ (0,∞) such that(i) if 0 < λ < λ?, then (5.6)λ has a weak solution uλ satisfying

λ ≤ uλG(f)

≤ C(λ) in Ω

for some constant C(λ) depending on λ.

(ii) if λ = λ?, then (5.6)λ has a weak solution.

(iii) if λ > λ?, then (5.6)λ has no weak solution.

Moreover,

(5.7)(p− 1p

)p 1p− 1

≤ (λ?)p−1

∥∥∥∥G(aG(f)p)G(f)

∥∥∥∥L∞(Ω)

≤ 1p− 1

.

The main ingredient in the proof of the above theorems is the following.


Lemma 5.3. Suppose u and v are C2 functions in Ω, and that v > 0. Let φ : R → R bea C2, concave function. Then

−∆[vφ(uv

)]≥ φ′

(uv

)(−∆u) +

[φ(uv

)− u

vφ′(uv

)](−∆v).

If, in addition, −∆v ≥ 0 in Ω, then

(5.8) −∆[vφ(uv

)]≥ φ′

(uv

)[−∆u+ ∆v] + φ(1)(−∆v).

Proof. We simply compute and use that φ′′ ≤ 0 and v > 0. We have (using the notation∂iv = vi)

−∆[vφ(uv

)]= −

N∑i=1

[φ′(uv

)v(uv

)i+ φ

(uv

)vi

]i

= −φ′′(uv

)v∣∣∇(u

v

) ∣∣2 − N∑i=1

φ′(uv

) [v(uv

)i

]i+[φ(uv

)vi

]i

≥ −N∑i=1

φ′(uv

) [ui −

u

vvi

]i+ φ

(uv

)vii + φ′

(uv

)(uv

)ivi

= φ′

(uv

)(−∆u) + φ′

(uv

)∇(uv

)∇v − u

vφ′(uv

)(−∆v)

+ φ(uv

)(−∆v)− φ′

(uv

)∇(uv

)∇v

= φ′(uv

)(−∆u) +

[φ(uv

)− u

vφ′(uv

)](−∆v),

which is the first inequality of the lemma. From this, we deduce the second inequality, asfollows. Since φ is concave, we have

φ(s) + (1− s)φ′(s) ≥ φ(1) ∀s ∈ R.

Thusφ(uv

)− u

vφ′(uv

)≥ −φ′

(uv

)+ φ(1);

multiplying this inequality by −∆v (which is nonnegative by assumption), we easily deduce(5.8).

To use the previous lemma, we will need (5.8) in its weak version for L1 functions.


Lemma 5.4. Let φ : R → R be a C1, concave function with φ′ bounded. Let h and kbelong to L1

δ(Ω), with k ≥ 0, k 6≡ 0, and let u and v be the L1(Ω) solutions of −∆u = h in Ωu = 0 on ∂Ω

and −∆v = k in Ωv = 0 on ∂Ω.

Then

(5.9) −∆[vφ(uv

)]≥ φ′

(uv

)(h− k) + φ(1)k,

in the sense that vφ(uv

)∈ L1(Ω), φ′

(uv

)(h− k) + φ(1)k ∈ L1

δ(Ω) and

−∫

Ω

vφ(uv

)∆ζ ≥

∫Ω

φ′(uv

)(h− k) + φ(1)k

ζ

for all ζ ∈ C2(Ω), ζ ≥ 0, with ζ = 0 on ∂Ω.

Proof. We first point out that (5.8) holds when φ is C1 and concave — not necessarily C2.This follows immediately from Lemma 5.3 convoluting φ with mollifiers.

We approximate h and k in L1δ(Ω) by sequences (hn) and (kn), respectively, of C∞0 (Ω)

functions and with kn ≥ 0, kn 6≡ 0. Let un, vn be the solutions of −∆un = hn in Ωun = 0 on ∂Ω

and −∆vn = kn in Ωvn = 0 on ∂Ω.

It follows that un → u and vn → v in L1(Ω) (for this, subtract the equations for un and u,multiply by G(1) and integrate). Moreover, un, vn ∈ C2(Ω) and vn > 0, −∆vn ≥ 0 in Ω.

By (5.8) we have

−∆[vnφ

(unvn

)]≥ φ′

(unvn

)(hn − kn) + φ(1)kn.


Moreover, using that φ′ is bounded, we see that∣∣∣∣vnφ(unvn)∣∣∣∣ =

∣∣∣∣vn(φ(unvn)− φ(0)

)+ φ(0)vn

∣∣∣∣≤ C(|un|+ |vn|)(5.10)

for some constant C. Hence, vnφ(un/vn) vanishes on ∂Ω and thus

(5.11) −∫

Ω

vnφ

(unvn

)∆ζ ≥

∫Ω

φ′(unvn

)(hn − kn) + φ(1)kn

ζ

for all ζ ∈ C2(Ω), ζ ≥ 0 in Ω and ζ = 0 on ∂Ω.

Note that v > 0 a.e. in Ω, so that vφ(u/v) is well defined a.e. Moreover, vnφ(un/vn)converges a.e. to vφ(u/v) — up to a subsequence. Since un and vn converge in L1(Ω),they are dominated (also up to a subsequence) by an L1(Ω) function. Thus, by (5.10),|vnφ(un/vn)| is also dominated by an L1 function (for a subsequence). We conclude that

vnφ

(unvn

)−→ vφ

(uv

)in L1(Ω).

Passing to the limit in (5.11), we finally obtain (5.9) and the lemma.

We write explicitly the concave functions φ that we use in this section. For Theorem 5.1we will use

(5.12) φ(s) =∫ s

1

dt

tp=

1p− 1

(1− 1

sp−1

)for s ≥ 1.

It satisfiesφ′(s)sp = 1 for s ≥ 1,

and hence φ is concave and φ′ is bounded in [1,∞). Moreover,

φ(1) = 0 and 0 ≤ φ(s) ≤ 1p− 1

for s ≥ 1.

Since φ′(1) = 1, we can extend φ by φ(s) = s − 1 in (−∞, 1] obtaining a function φ thatsatisfies the conditions of Lemma 5.4.

Note that the functions φ above are analogous versions of the ones that we employedin Sections 1 and 2, in the sense that they all satisfy φ′(s)g(s) = 1 (in an appropriateinterval) where g is the nonlinearity.


In Theorem 5.2 we will be led to take φ (that we denote now by ψ) satisfying anotherdifferential equation, namely: ψ′(s)sp = ψ(s)p. Precisely, we will take

(5.13) ψ(s) =s

(εsp−1 + 1)1p−1

for s ≥ 0.

It satisfiesψ′(s) =

1

(εsp−1 + 1)pp−1

for s ≥ 0

and henceψ′(s)sp = ψ(s)p for s ≥ 0.

Note that ψ is concave and ψ′ is bounded in [0,∞);

ψ(1) =(

11 + ε

) 1p−1

and 0 ≤ ψ(s) ≤(

1ε

) 1p−1

for s ≥ 0.

Extending ψ by ψ(s) = s in (−∞, 0], we obtain a function ψ which satisfies the conditions(for φ) of Lemma 5.4. Note that, for p = 2, ψ was already considered in Remark 1.8.

Proof of Theorem 5.1.

Part (a). Let u ≥ 0 be a weak solution of −∆u = a(x)up + f(x) in Ωu = 0 on ∂Ω.

We considerv = G(f);

note that, since a(x)up ≥ 0,u

v=

u

G(f)≥ 1 in Ω

by the weak maximum principle (which is an easy consequence of the weak formulation ofDefinition 1.1(c) used here). In particular aG(f)p ≤ aup and hence aG(f)p ∈ L1

δ(Ω).

We take φ, defined by (5.12), and apply Lemma 5.4 to obtain (we use u/v ≥ 1 and theproperties of φ(s) for s ≥ 1)

−∆[vφ(uv

)]≥ φ′

(uv

)(aup + f − f) + φ(1)f

= φ′(uv

)aup

= avp = aG(f)p


in the weak sense of the lemma. Hence (by the weak maximum principle)

G(aG(f)p) ≤ vφ(uv

)≤ 1p− 1

v =1

p− 1G(f).

Part (a) is now proved.

Part (b). We assume that aG(f)p ∈ L1δ(Ω) and

G(aG(f)p) ≤(p− 1p

)p 1p− 1

G(f).

It follows that the L1(Ω) function

u :=(

p

p− 1

)pG(aG(f)p) +G(f)

satisfiesu ≤ p

p− 1G(f).

Therefore

−∆u =(

p

p− 1

)paG(f)p + f

≥ aup + f

in the weak sense. That is, u is a weak supersolution of (5.5). On the other hand, 0is a subsolution of the problem. It is then easy to obtain a weak solution u of (5.5) bymonotone iteration (see e.g. Lemma 3 of [4]). Moreover

G(f) ≤ u ≤ u ≤ p

p− 1G(f).

This completes the proof of Theorem 5.1.

Proof of Theorem 5.2. We assume that

G(aG(f)p)G(f)

∈ L∞(Ω);

moreover, a 6≡ 0 and f 6≡ 0 and hence

0 < M∞ :=∥∥∥∥G(aG(f)p)

G(f)

∥∥∥∥L∞(Ω)

<∞.


Theorem 5.1 (applied with f replaced by λf) implies that if (5.6)λ has a weak solutionthen

λp−1M∞ ≤1

p− 1.

The theorem also gives that if

λp−1M∞ ≤(p− 1p

)p 1p− 1

then (5.6)λ has a weak solution. Hence, defining

λ? = supλ > 0; (5.6)λ has a weak solution,

we have 0 < λ? < ∞, and also estimate (5.7) of Theorem 5.2. Note that part (iii) of thetheorem is now obvious.

To prove part (i), we have to show that if 0 < λ < µ and (5.6)µ has a weak solution uthen (5.6)λ has a weak solution uλ satisfying

(5.14) λ ≤ uλG(f)

≤ C(λ) in Ω.

For this purpose, we considerv = G(µf) = µG(f),

and the function ψ defined by (5.13) with ε > 0 chosen small enough such that

λ ≤(

11 + ε

) 1p−1

µ = ψ(1)µ.

We apply Lemma 5.4 (with φ replaced by ψ) to obtain

−∆[vψ(uv

)]≥ ψ′

(uv

)(aup + µf − µf) + ψ(1)µf

= avp[ψ(uv

)]p+ ψ(1)µf

≥ a[vψ(uv

)]p+ λf

in the weak sense of the lemma. Hence vψ(u/v) is a weak supersolution of (5.6)λ. Againby monotone iteration we deduce that (5.6)λ has a weak solution uλ such that

uλ ≤ vψ(uv

)≤ Cv = CµG(f).


This, together with the trivial bound uλ ≥ G(λf), gives (5.14) and proves (i).

It remains to show part (ii). For λ < λ? we can take uλ to be the minimal solution of(5.6)λ, i.e., the solution obtained by monotone iteration starting from the function 0. Inthis manner, uλ ≤ uµ if 0 < λ < µ < λ?. Hence, in order to obtain a weak solution of(5.6)λ? , it suffices to show

(5.15)∫

Ω

[a(x)upλ + λf(x)] δ ≤ C

for some constant C independent of λ.

To prove (5.15) we proceed as follows. Since a ≥ 0 and a 6≡ 0, there exists a constantM ∈ (0,∞) such that

aM := χa≤Ma

satisfies 0 ≤ aM ≤ M and aM 6≡ 0 (here χa≤M denotes the characteristic function ofa ≤M).

It is well-known that the problem

(5.16)

−∆w = a

1/pM w1/p in Ω

w > 0 in Ωw = 0 on ∂Ω

has a unique solution w ∈W 2,r(Ω) (for any 1 < r <∞) with w 6≡ 0. This solution can beobtained for example by minimizing in H1

0 (Ω) the energy associated to (5.16):

E(v) =12

∫Ω

|∇v|2 − p

p+ 1

∫Ω

a1/pM (v+)(p+1)/p,

which is a coercive functional, bounded from below, in H10 (Ω) (note that a1/p

M ∈ L∞(Ω)and 1 < (p + 1)/p < 2). We have that E(tϕ1) < 0 for t small, where ϕ1 denotes the firsteigenfunction of −∆. In particular, the minimizer w of E satisfies E(w) < 0, and thusw 6≡ 0. The strong maximum principle then gives

(5.17) w ≥ cδfor some positive constant c.

Since ∆w ∈ L∞(Ω) we can multiply (5.6)λ by w and integrate by parts. We also useYoung’s inequality — in the spirit of the ideas of Baras and Pierre [3]. We obtain∫

Ω

aupλw +∫

Ω

λfw =∫

Ω

uλ(−∆w)

=∫

Ω

a1/pM uλw

1/p ≤∫

Ω

a1/puλw1/p

≤ 1p

∫Ω

aupλw +p− 1p

∫Ω

1,


and therefore ∫Ω

aupλw +∫

Ω

λfw ≤ C

for some constant C independent of λ. Using (5.17), we conclude (5.15), and hence theproof of part (ii).

Remark 5.5. An analogous version of Theorem 5.2 also holds for the problem −∆u = λ(a(x)up + f(x)) in Ωu = 0 on ∂Ω.

This follows from Theorem 5.2 rescaling the solution u, i.e., considering the problem forαu, for appropriate α.

Remark 5.6. In Theorem 5.2 we have shown that, for λ < λ?, (5.6)λ has a weak solutionuλ satisfying

(5.18)uλG(f)

∈ L∞(Ω).

We point out that this property may not be true for the minimal solution obtained forλ = λ?. To give an example of this, consider first the problem

(5.19)λ

−∆v = λ(v + 1)p in B1 = |x| < 1 ⊂ RN

v = 0 on ∂B1.

For N and p sufficiently large, (5.19)λ has

v(x) = |x|−2p−1 − 1

as weak solution for a certain parameter λ > 0; moreover v is the pointwise, increasinglimit (as λ λ) of the classical minimal solutions vλ of (5.19)λ for λ < λ (see e.g. [6]).Let us define

f = (v + 1)p − vp,

so that the problem

(5.20)λ

−∆u = λup + λf(x) in B1

u = 0 on ∂B1

has u = v as weak solution for λ = λ. Note that (5.20)λ is of the form (5.6)λ consideredin Theorem 5.2. We claim that v is the minimal weak solution of (5.20)λ. This is shownas follows: for λ < λ, let uλ be the minimal weak solution of (5.20)λ. Since uλ ≤ v, we


have (uλ + 1)p − upλ ≤ f , and hence uλ is a weak supersolution of (5.19)λ. Thus uλ ≥ vλ,and since vλ v as λ λ, we conclude that v is the minimal weak solution of (5.20)λ.

Finally, we have that

(5.21)v

G(f)6∈ L∞(B1),

since v ∼ |x|−2p−1 near 0 and G(f) ≤ C log(1/|x|) (note that f ≤ p(v + 1)p−1 = p|x|−2

in B1). This proves that (5.18) does not hold for λ = λ and, in particular, that λ is theextremal parameter λ? for problem (5.20)λ.

6. Evolution equations.

In Section 4 we proved that u ≡ 0 is the only nonnegative very weak supersolution of

−∆u ≥ u2

|x|2in D′(Ω\0).

Here we prove an analogous result for the equation

ut −∆u ≥ u2

|x|2in D′((Ω\0)× (0, T )).

More precisely, we have the following parabolic analogue of Theorem 4.1.

Theorem 6.1. Let N ≥ 2, T > 0 and u ∈ L2loc((Ω\0) × (0, T )) satisfy u ≥ 0 a.e. in

Ω× (0, T ) and

(6.1) |x|2(ut −∆u) ≥ u2 in D′((Ω\0)× (0, T )),

in the sense that−∫∫

u(|x|2ϕ)t + ∆(|x|2ϕ) ≥∫∫

u2ϕ

for any ϕ ≥ 0, ϕ ∈ C∞0 ((Ω\0)× (0, T )).Then u ≡ 0.

As an immediate consequence of the theorem we obtain an extension of a result of Peraland Vazquez (Theorem 7.1 of [12]). Here Ω = B1, the unit ball of RN , and N ≥ 3. Weconsider the function

u(x) = log1|x|2

,

which is a weak solution of −∆u = 2(N − 2)eu in B1

u = 0 on ∂B1.


Corollary 6.2. Let N ≥ 3, T > 0 and u ∈ L1loc((B1\0) × (0, T )) be a function such

that eu ∈ L1loc((B1\0)× (0, T )),

u(x, t) ≥ u(x) a.e. in B1 × (0, T ),

and

(6.2) ut −∆u = 2(N − 2)eu in D′((B1\0)× (0, T ))

in the sense that−∫∫

u(ϕt + ∆ϕ) = 2(N − 2)∫∫

euϕ

for any ϕ ∈ C∞0 ((B1\0)× (0, T )).Then u(x, t) ≡ u(x).

Note again that we only assume equation (6.2) to be satisfied in the distributional senseand away from x = 0 × [0, T ]. In particular, given any u0(x) ≥ u(x) a.e. in B1, u0 6≡ u,and any T > 0, there is no weak solution u, with

u(x, t) ≥ u(x) in B1 × (0, T ),

of the problem

(6.3)

ut −∆u = 2(N − 2)eu in B1 × (0, T )

u = 0 on ∂B1 × (0, T )u(x, 0) = u0 on B1

(as stated in [12]).A second consequence of Theorem 6.1 is the following nonexistence and complete blow-

up result. For any u0 ≥ 0, u0 6≡ 0 (say u0 ∈ C∞0 (Ω)) and for any T > 0, the problem

(6.4)

ut −∆u =

u2

|x|2in Ω× (0, T )

u = 0 on ∂Ω× (0, T )u(x, 0) = u0 on Ω

has no weak solution (by Theorem 6.1). Using similar ideas as in the elliptic case (seeSection 3), we prove that approximate solutions of (6.4) blow up completely. More pre-cisely, let gn be a sequence of nonnegative, nondecreasing and globally Lipschitz functionsin [0,∞) such that gn(u) increases pointwise to u2. Let an be a sequence of nonnegativebounded functions in Ω, increasing pointwise to 1/|x|2.


Theorem 6.3. Under the above assumptions, let un be the solution of

(6.5)n

∂un∂t−∆un = an(x)gn(un) in Ω× (0,+∞)

un = 0 on ∂Ω× (0,+∞)un(x, 0) = u0 on Ω.

Then, for any 0 < ε < T ,

(6.6)un(x, t)δ(x)

−→ +∞ uniformly in (x, t) ∈ Ω× [ε, T ]

as n→∞.

To prove Theorem 6.1 we adapt the method given in Section 4 for the elliptic case; itconsists of using appropriate powers of testing functions.

Proof of Theorem 6.1. We proceed as in the proof of Theorem 4.1; we use the same notationas there. We also fix a cut-off function in time:

ψ ∈ C∞0 ((0, T )), 0 ≤ ψ ≤ 1,

with ψ ≡ 1 in a given compact sub-interval of (0, T ).

Step 1. We prove that u/|x| ∈ L2loc(Ω × (0, T )). For this purpose, we multiply (6.1) by

ζ4n(x)ψ2(t)/|x|2; it yields∫∫

u2

|x|2ζ4nψ

2 ≤ −∫∫

u(∆ζ4n)ψ2 −

∫∫uζ4n(ψ2)t.

Let us denote by C different constants independent of n, but that may depend on Ω, Tand the cut-off ψ. We have

−∫∫

u(∆ζ4n)ψ2 ≤ Cn

(∫∫ 1n<|x|<

2n×(0,T )

u

|x|ζ2nψ

)+ C

and−∫∫

uζ4n(ψ2)t ≤ C

∫∫u

|x|ζ2nψ.

Hence we can conclude, as in Section 4, that∫∫u2

|x|2ζ4nψ

2 ≤ C.


Thus u/|x| ∈ L2loc(Ω× (0, T )).

Step 2. We show that

(6.7) ut −∆u ≥ u2

|x|2in D′(Ω× (0, T )).

This is done exactly as in the proof of Theorem 4.1, where now ϕ = ϕ(x, t) belongs toC∞0 (Ω× (0, T )).

Step 3. We finally prove u ≡ 0. We suppose u 6≡ 0. Since u ≥ 0, ut − ∆u ≥ 0 inD′(Ω× (0, T )) and Ω is connected, we have that

u ≥ ε a.e. in Bη × (τ, T ),

for some 0 < τ < T , ε > 0 and Bη = Bη(0) with closure in Ω. When N = 2 this is acontradiction with u/|x| ∈ L2

loc(Ω× (0, T )).

When N ≥ 3, (6.7) gives

ut −∆u ≥ ε2

|x|2= (∂t −∆)

(ε2

N − 2log

1|x|

)in D′(Bη × (τ, T )).

We deduce that

(6.8) u ≥ ε2

N − 2log

1|x|− C in Bη/2 × (τ ′, T )

for some C > 0 and τ < τ ′ < T .

Following the proof of Theorem 4.1, we now multiply (6.7) by χ4n(x)ψ2(t), with ψ a

cut-off as in the beginning of this proof, and with ψ ≡ 1 in (τ ′, T ′) for some T ′ withτ ′ < T ′ < T . We have∫∫

u2

|x|2χ4nψ

2 ≤∫∫

u|∆χ4n|ψ2 +

∫∫uχ4

n|(ψ2)t|

(where we are integrating on |x| < 2/n × (0, T ), since it contains the support of χ4nψ

2).Hence ∫∫

u2

|x|2χ4nψ

2 ≤ Cn∫∫

u

|x|χ2nψ,

where C is independent of n. From the Cauchy-Schwarz inequality, we deduce∫∫u2

|x|2χ4nψ

2 ≤ C

nN−2.


But using (6.8), we have ∫∫u2

|x|2χ4nψ

2 ≥ c | log n|2

nN−2,

which contradicts the previous statement. This proves the theorem.

Remark 6.4. Step 3 of the previous proof, (i.e., to show u ≡ 0 from u ≥ 0 and (6.7))could have been done using a parabolic analogue of the method of Section 1. That is, oneconsiders

v =1ε− 1u

in a subcylinder where u ≥ ε. Then (with the aid of the parabolic Kato’s inequality) vsatisfies

vt −∆v ≥ 1|x|2

,

which leads to contradiction since v is bounded (v ≤ 1/ε).

Corollary 6.2 follows immediately from Theorem 6.1:

Proof of Corollary 6.2. Let u be as in the corollary. Consider v(x, t) = u(x, t) − u(x). Itsatisfies, from our assumptions,

v ≥ 0 a.e. in B1 × (0, T ).

Moreover, in the distributional sense D′((B1\0)× (0, T )),

vt −∆v = ut −∆u+ ∆u

= 2(N − 2)(eu − eu) = 2(N − 2)eu(ev − 1)

=2(N − 2)|x|2

(ev − 1)

≥ (N − 2)|x|2

v2

since v ≥ 0. Hence (N − 2)v ≥ 0 satisfies (6.1). By Theorem 6.1 we deduce v ≡ 0, that isu ≡ u.

We finally give the proof of the complete blow-up result.

Proof of Theorem 6.3. We proceed in three steps. Recall that 0 ≤ un ≤ un+1, by themaximum principle.

Step 1. We prove that, for any τ > 0,∫ τ

0

∫Ω

angn(un)δ → +∞.


Suppose not, that∫ τ

0

∫Ωangn(un)δ ≤ C for some τ > 0. We multiply (6.5)n by the solution

of −∆z = 1 in Ωz = 0 on ∂Ω.

We obtain ∫ τ

0

∫Ω

un +∫

Ω

un(x, τ)z −∫

Ω

u0z =∫ τ

0

∫Ω

angn(un)z ≤ C

and, in particular ∫ τ

0

∫Ω

un ≤ C.

Hence un and angn(un)δ are bounded in L1(Ω × (0, τ)). By monotone convergence, weobtain that un → u in L1(Ω × (0, τ)) with u satisfying the assumptions of Theorem 6.1.By this theorem, u ≡ 0. Thus u1 ≡ 0 and u0(x) = u1(x, 0) ≡ 0, a contradiction.

Step 2. We show that∫Ω

un(x, t)δ(x)dx→ +∞ uniformly in t ∈ [ε/2, T ].

Indeed, let us multiply (6.5)n by eλ1tϕ1, where ϕ1 is the first eigenfunction of −∆ in Ωwith zero Dirichlet condition and λ1 its corresponding eigenvalue. We then integrate inspace and time, to obtain

eλ1τ

∫Ω

un(·, τ)ϕ1 −∫

Ω

u0ϕ1 =∫ τ

0

∫Ω

angn(un)ϕ1eλ1t.

Hence, if τ ∈ [ε/2, T ],∫Ω

un(x, τ)δ(x)dx ≥ ce−λ1T

∫ ε/2

0

∫Ω

angn(un)δ → +∞

by Step 1.

Step 3. We finally prove (6.6). For this purpose, we use a parabolic analogue of Lemma 3.2due to Martel (see Lemma 2 of [10]). It asserts that if ϕ ≥ 0 in Ω (say, ϕ ∈ L∞(Ω)) then

T (τ) · ϕ(x)δ(x)

≥ c(τ)∫

Ω

ϕδ ∀x ∈ Ω

for any τ > 0, where c(τ) > 0 is a constant depending on τ , and where T (τ) is the heatsemigroup at time τ . We apply this estimate with τ = ε/2, and obtain

un(x, t)δ(x)

≥ c (ε/2)∫

Ω

un(x, t− ε/2)δ(x)dx→ +∞


uniformly in t ∈ [ε, T ] by Step 2, since t− ε/2 ≥ ε/2.

Acknowledgments:

We thank Y. Martel, J. L. Vazquez and I. E. Verbitsky for useful discussions, as well asL. Nirenberg for a suggestion which led to the results of Section 4. The second author wassupported by CEE contract ERBCHRXCT940494. Part of this work was done while thesecond author was visiting Rutgers University. He thanks the Department of Mathematicsfor its hospitality.

References

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J. Funct. Anal. 71 (1987), 142–174.

[2] P. Baras and M. Pierre, Singularites eliminables pour des equations semi-lineaires, Ann. Inst.Fourier 34 (1984), 185–206.

[3] , Critere d’existence de solutions positives pour des equations semi-lineaires non mono-tones, Ann. Inst. Henri Poincare 2 (1985), 185–212.

[4] H. Brezis, T. Cazenave, Y. Martel and A. Ramiandrisoa, Blow-up for ut − ∆u = g(u) revisited,

Ad. Diff. Eq. 1 (1996), 73–90.

[5] H. Brezis and W. A. Strauss, Semi-linear second-order elliptic equations in L1, J. Math. Soc.

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[6] H. Brezis and J. L. Vazquez, Blow-up solutions of some nonlinear elliptic problems (to appear).

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