4 pl. Jussieu, Paris Cedex 05, France (2) Department of Mathematics, Rutgers University. New Brunswick, New Jersey 08903, USA

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1 SOME SIMPLE NONLINEAR PDE S WITHOUT SOLUTIONS Haïm Brezis (1( and Xaier Cabré (1 (1 Analyse Numérique, Uniersité Pierre et Marie Curie 4 pl. Jussieu, 755 Paris Cedex 05, France ( Department of Mathematics, Rutgers Uniersity New Brunswick, New Jersey 08903, USA 0. Introduction. The original motiation of this work is the following. Consider the simple problem (0.1 { u = a(xu + f(x in u = 0 on where is a smooth bounded domain in R N, N 3. If a(x L p ( and p > N/, then for any f L p ( with f p small, problem (0.1 has a unique small solution u in W,p (. This is an easy consequence of the Inerse Function Theorem applied to F (u = u a(xu which maps X = W,p ( W 1,p 0 ( into Y = L p ( (recall that W,p ( L ( by the Sobole imbedding theorem, since its differential at 0, DF (0 = is bijectie. As a special case suppose a(x = x α in a domain containing 0, with 0 < α <. Then for any small constant c the problem u = u (0. x α + c in u = 0 on has a unique small solution. The case α = is interesting since a(x = x does not belong to L p ( for p > N/. One may then wonder what happens to the problem u = u (0.3 x + c in u = 0 on. 1

2 HAÏM BREZIS AND XAVIER CABRÉ On the one hand, a formal computation suggests that since the linearized operator at 0 is, which is bijectie, problem (0.3 has a solution for small c. On the other hand, the F aboe does not map X = W,p ( W 1,p 0 ( into Y = L p ( for any 1 < p <. One may then try to construct other function spaces, for example weighted spaces, where the Inerse Function Theorem might apply. This is doomed to fail. In fact, our main results show that for any constant c > 0 (no matter how small problem (0.3 has no solution, een in a ery weak sense. When c < 0, problem (0.3 does, howeer, hae a solution (see Remark 1.4. In Sections 1 and 4 we propose seeral notions of weak solutions and establish nonexistence. A basic ingredient in Section 1 is the following: Theorem 0.1. Assume 0. If u L 1 u loc (, u 0 a.e. with x L1 loc ( is such that (0.4 u u x in D ( then u 0. The proof of Theorem 0.1 uses an adaptation of a method introduced in [4]. In Section 4 we proe a stronger result, namely: Theorem 0.. Assume 0. If u L loc (\{0}, u 0 a.e. is such that (0.5 x u u in D (\{0} then u 0. Theorem 0. is proed using appropriate powers of testing functions an idea due to Baras-Pierre []. As a consequence we obtain the nonexistence of local solutions (i.e., in any neighborhood of 0, without prescribing any boundary condition for a ery simple nonlinear equation: Theorem 0.3. Assume 0 and c > 0. There is no function u L loc (\{0} satisfying (0.6 x u = u + c in D (\{0}. In Section 3 we examine what happens to a natural approximation procedure of (0.3. Consider for example the equation (0.7 u = min{u, n} x + c in, c > 0, + (1/n u = 0 on.

3 SOME SIMPLE NONLINEAR PDE S WITHOUT SOLUTIONS 3 For any n there is a minimal solution u n. We proe that u n (x + for eery x as n, i.e., there is complete blow-up in the sense of Baras-Cohen [1]. Again, this rules out any reasonable notion of weak solution for (0.3. In Section we extend the preious results to more general problems such as u = a(xg(u + b(x assuming only that g 0 on R, g is continuous and nondecreasing on [0, and ds g(s <, with a L 1 loc (, a 0 and a(x =. x N The original motiation of our research came from obserations made in [6] and [4]. For any N 3 the problem { u = (N e u in B 1 = {x R N ; x < 1} u = 0 on B 1 admits the weak solution u(x = log(1/ x. It was obsered in [6] that when N 11 the linearized operator at u namely L = (N e u = (N x is coercie and thus formally bijectie; this is a simple consequence of Hardy s inequality: (N 4 x H 1 0 (B 1 (note that (N /4 > (N when N 11. On the other hand the results of [4] show that when N 10 the perturbed problem { u = (N e u + c in B 1 u = 0 on B 1 has no solution een in a weak sense and no matter how small c is, proided c > 0.

4 4 HAÏM BREZIS AND XAVIER CABRÉ This strange failure of the Inerse Function Theorem is only apparent. As was pointed out in [6] this just means that there is no functional setting in which it can be correctly applied. We hae tried here, in the spirit of Open Problem 6 in [6], to analyze the same phenomenon for simple examples in low dimensions. After our inestigation 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 for example the problem (0.8 { u = a(xu + c in u = 0 on with a L 1 loc (, a 0 and c a positie constant. Their result says that if (0.8 has a weak solution, then necessarily (0.9 G(aδ Cδ in for some constant C, where G = ( 1 (with zero boundary condition and δ(x = dist(x,. In particular, if 0 and a(x = 1/ x, then G(aδ log x as x 0 and thus (0.9 fails; hence (0.8 has no weak solution. We present in Section 5 a ery simple proof of the main result of [8] using a ariant of the method deeloped in Section 1. Finally, in Section 6 we present a parabolic analogue of Theorem 0.. It extends, in particular, a result of I. Peral and J. L. Vázquez [1]. Namely, the problem u t u = (N e u in B 1 (0, T u = 0 on B 1 (0, T u(x, 0 = u 0 with u 0 u = log(1/ x, u 0 u, has no solution u u een for small time: instantaneous and complete blow-up occurs. The plan of the paper is the following: (1 Proof of Theorem 0.1 ( General nonlinearities (3 Complete blow-up (4 Very weak solutions. Proofs of Theorems 0. and 0.3 (5 Connection with a result of Kalton-Verbitsky (6 Eolution equations

5 SOME SIMPLE NONLINEAR PDE S WITHOUT SOLUTIONS 5 Notation. Throughout this paper, is a bounded smooth domain of R N, N 1, such that 0. We write δ(x = dist(x, and L 1 δ ( = L1 (, δ(xdx. We denote by C 0 ( the space of C functions with compact support in, and by D ( the space of distributions in. By C we denote a positie constant which may be different in each inequality. 1. Proof of Theorem 0.1. In this section we proe Theorem 0.1 and its consequences. We first introduce some terminology about weak solutions. Definition 1.1. Let h(x, u be a Caratheodory function in R, that is, h(x, u is measurable in x and continuous in u for a.e. x. (a We say that u = h(x, u in D ( if u L 1 loc (, h(x, u L1 loc ( and u ϕ = h(x, uϕ for any ϕ C 0 (. (b We say that u h(x, u in D ( if u L 1 loc (, h(x, u L1 loc ( and u ϕ h(x, uϕ for any ϕ C 0 ( with ϕ 0. (c We say that u is a weak solution of { u = h(x, u in u = 0 on if u L 1 (, h(x, u L 1 δ ( and u ζ = h(x, uζ for any ζ C ( with ζ = 0 on. The following is the main result of this section; it is Theorem 0.1 of the Introduction. Theorem 1.. Let N 1 and u L 1 loc ( satisfy u 0 a.e. in, u x L1 loc ( and (1.1 u u x in D (. Then u 0. This theorem easily implies two nonexistence results. following boundary alue problem. The first one deals with the

6 6 HAÏM BREZIS AND XAVIER CABRÉ Corollary 1.3. Let N 1 and f L 1 δ ( satisfy f 0 a.e. and f 0. Then there is no weak solution of (1. in the sense of Definition 1.1. u = u x + f(x u = 0 in on, Remark 1.4. When f 0, problem (1. has u 0 as the only weak solution; this follows immediately from Theorem 1.. When N 3 and f L 1 ( satisfies f 0 and f 0, (1. has a weak solution u 0, which is the unique solution among nonpositie functions. This fact is a consequence of a more general result of Gallouët and Morel [7], which extends work of Brezis and Strauss [5]. As a consequence of Theorem 1., we may write down a simple PDE without local solutions, i.e., no solution exists in any neighborhood of 0. Here, we do not impose any boundary condition. Corollary 1.5. Let N 3 and c > 0 be any positie constant. Then there is no function u u such that x L1 loc ( and (1.3 u = u + c x in D (. Remark 1.6. In contrast with (1.3, the equation (1.4 u = u x + c has a weak solution in some neighborhood of 0, if N 3 and c > 0 is a constant. This solution is nonpositie and can be obtained from the results of [7] as follows. We introduce the new unknown = u c x so that (1.4 becomes (1.5 + x + c = c(n 1 c x h(x. We sole (1.5 on B R with the boundary condition = 0 on B R. Note that h 0 on B R proided R is sufficiently small (R (N 1/c. The results of [7] gie a unique solution 0. In Section 4 we will proe stronger nonexistence results for problems (1. and (1.3. The proof of Theorem 1. is based on the following ariant of Kato s inequality [9].

7 SOME SIMPLE NONLINEAR PDE S WITHOUT SOLUTIONS 7 Lemma 1.7. Let u L 1 loc ( and f L1 loc ( satisfy u f in D (. Let φ : R R be a C 1, concae function such that 0 φ C in R for some constant C. Then φ(u L 1 loc ( and φ(u φ (uf in D (. The proof is standard, smoothing u, f and φ by conolution; see also Lemma in [4]. The proof of Theorem 1. is a ariant of a method introduced in [4]. Consider the function φ(s = (1/ε (1/s. It is nonnegatie, bounded, increasing and concae in the interal [ε,, ε > 0. Note that if u ε satisfies then u u x φ(u φ (u u x = 1 x. This will lead to a contradiction with the fact that φ(u is bounded. The details go as follows. Proof of Theorem 1.. Suppose that u is as in the theorem, and that u 0. Since u 0, u 0, u 0 in D ( and is connected, we hae that u ε a.e. in B η, for some ε > 0 and B η = B η (0 with closure in. If N, this is already a contradiction with u / x L 1 loc (. Suppose N 3. Let φ(s = 1 ε 1 s for s ε, and extend it by φ(s = (s ε/ε for s ε. Note that φ : R R is C 1, concae and 0 φ 1/ε, so that φ satisfies all the conditions of Lemma 1.7. Recall that u ε in B η, and consider = φ(u = 1 ε 1 u in B η.

8 8 HAÏM BREZIS AND XAVIER CABRÉ It satisfies 0 1/ε in B η and, by Lemma 1.7, Hence 1 N log 1 x L1 (B η and which implies φ (u u x = 1 x in D (B η. ( 1 N log 1 0 in D (B η, x 1 N log 1 x C in B η/ for some constant C > 0. In particular, (x + as x 0, which is a contradiction with the fact that 1/ε. Remark 1.8. In the preious proof we could hae used (in the spirit of [4] the function w = u εu + 1 for any ε > 0, instead of = 1 ε 1. Note that u satisfies ψ (s = (εs + 1 and hence ψ(s = s εs + 1 ψ (ss = ψ(s. Moreoer, ψ is bounded in [0, and satisfies all the conditions (for φ of Lemma 1.7 in [0,. In particular, if u 0, u 0 and u u / x then w ψ (u u x = w x, and we can conclude as before, since w / x ν/ x in a subdomain of, for some constant ν > 0. Analogous ersions of the functions φ and ψ will appear, in Sections and 5, when the nonlinearity u is replaced by more general nonlinearities g(u.

9 SOME SIMPLE NONLINEAR PDE S WITHOUT SOLUTIONS 9 Finally, we use Theorem 1. to proe the nonexistence results of this section. Proof of Corollary 1.3. Suppose that u is a weak solution of (1.. Since u( ζ 0 for any ζ C ( with ζ 0 in and ζ = 0 on, we easily deduce u 0 in. Moreoer, u u / x in D (. We obtain, by Theorem 1., u 0. This is a contradiction with (1., since f 0. Proof of Corollary 1.5. Suppose that u is a solution of (1.3 in D (. Then u c/ x in D (B η, for some ball B η = B η (0 with closure in. As in the proof of Theorem 1., we deduce that u c N log 1 x C in B η/, for some constant C. In particular, u 0 in B ν for some small ν > 0. We therefore hae u 0 in B ν u u x in D (B ν. By Theorem 1., u 0 in B ν, which is a contradiction with equation (1.3.. General nonlinearities. In this section we extend the preious nonexistence results to more general problems of the form u = a(xg(u + b(x. We assume (here and throughout the rest of the paper that g : R [0, is continuous on R, nondecreasing on [0,, g(s > 0 if s > 0, and (.1 1 ds g(s <. Power functions g(u = u p for u 0, with p > 1 are examples of such nonlinearities. We suppose that N 3. For the function a(x we assume in this section that a L 1 loc (, a 0 in, and a(x (. x N = B η (0 for some η > 0 small enough (or, equialently, for any η > 0 small.

10 10 HAÏM BREZIS AND XAVIER CABRÉ Theorem.1. Let g and a satisfy the assumptions aboe, with N 3. (a Let u 0 a.e. in satisfy u a(xg(u in D (. Then u 0. (b Let f L 1 δ ( satisfy f 0 a.e. and f 0. Then there is no weak solution of { u = a(xg(u + f(x in u = 0 on. (c If b(x satisfies the same conditions as a(x, then there is no weak solution of u = a(xg(u + b(x in D (. This theorem is proed with the same method as in the preious section. We only need to adapt two points. First, φ(s = (1/ε (1/s has to be replaced by a solution of φ (s = 1 g(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 dt ε g(t < in [ε, (by assumption (.1, φ(ε = 0, φ (ε = 1 g(ε, φ is C1, concae (since φ (s = 1 g(s is nonincreasing and 0 φ 1 g(ε in [ε,. Extending φ by φ(s = 1 g(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 aboe. To complete the proof of Theorem.1, we only need to consider a solution w L 1 loc (B η, where B η = B η (0 with closure in, of w = a(x in D (B η (a solution always exists since a L 1 (B η and show that (.3 ess inf w + B 1/n (0 as n +.

11 SOME SIMPLE NONLINEAR PDE S WITHOUT SOLUTIONS 11 For this purpose, we consider the conolution in B η, w = a C N x N, where C N is chosen such that (C N x N = δ 0. Then w w is harmonic in B η and hence bounded in B η/. In particular, it suffices to show (.3 for w. But this is true since, for x 1/n, a(y w(x = C N dy B η y x N a(y C y N dy + + (1/n N as n, by (.. 3. Complete blow-up. B η In Corollary 1.3 and Theorem.1 we hae proed the nonexistence of weak solutions of some problems of the form { u = a(xg(u + f(x in (3.1 u = 0 on. In this section we proe that, under the same assumptions on a(x, g(u and f(x made in Section, approximate solutions of (3.1 blow up eerywhere in, that is, there is complete blow-up. More precisely, we hae the following. Let g(u and a(x be as in Section. Suppose that f L 1 δ (, f 0 a.e. and f 0. Let (g n be a sequence of nonnegatie, bounded, nondecreasing and continuous functions in [0, such that (g n increases pointwise to g. Let a n and f n be two sequences of nonnegatie bounded functions in, increasing pointwise to a and f, respectiely. Theorem 3.1. Under the aboe assumptions, let u n be the minimal nonnegatie solution of the approximate problem { u = an (xg n (u + f n (x in (3. n u = 0 on. Then, as n +, u n (x δ(x + uniformly in. In the proof of Theorem 3.1 we use two ingredients. First, the nonexistence result of Theorem.1(b (see also Corollary 1.3 and, second, the following estimate for the linear Laplace equation. It asserts that, for some positie constant c, (3.3 G(x, y cδ(xδ(y in, where G is the Green s function of the Laplacian in with zero Dirichlet condition. In an equialent way, we can state this lower bound on G as follows.

12 1 HAÏM BREZIS AND XAVIER CABRÉ Lemma 3.. Suppose that h 0 belongs to L (. Let be the solution of { = h in Then (3.4 = 0 on. (x δ(x c hδ x, where c > 0 is a constant depending only on. Estimate (3.3 was proed by Morel and Oswald [11] (in unpublished work, and by Zhao [13] (in a stronger form. For the conenience of the reader we gie a simple proof of (3.3. Proof of Lemma 3.. We proceed in two steps. Step 1. For any compact set K, we show (3.5 (x c hδ x K, where c is a positie constant depending only on K and. To proe (3.5, let ρ = dist(k, /, and take m balls of radius ρ such that K B ρ (x 1... B ρ (x m. Let ζ 1,..., ζ m be the solutions of { ζi = χ Bρ (x i in ζ i = 0 on, where χ A denotes the characteristic function of A. The Hopf boundary lemma implies that there is a constant c > 0 such that ζ i (x cδ(x x 1 i m. Here and in the rest of the proof, c denotes arious constants depending only on K and. Let now x K, and take a ball B ρ (x i containing x. Then B ρ (x i B ρ (x, and since 0 in, we conclude (x = c c B ρ (x = c B ρ (x ( ζ i = c hδ. c B ρ (x i hζ i

13 SOME SIMPLE NONLINEAR PDE S WITHOUT SOLUTIONS 13 Step. Fix a smooth compact set K. By (3.5, c hδ in K, so that it suffices to proe (3.4 for x \K. Let w be the solution of w = 0 in \K w = 0 on w = 1 on K. The Hopf boundary lemma gies again w(x cδ(x x \K. Since is superharmonic and c hδ on K, the maximum principle implies ( (x c ( c hδ w(x hδ δ(x x \K. This completes the proof. Proof of Theorem 3.1. Consider the approximate problem { u = an (xg n (u + f n (x in (3.6 n u = 0 on. Since 0 a n (xg n (s + f n (x C n in [0, for some constant C n, we hae that C n z is a supersolution of (3.6 n where (3.7 { z = 1 in z = 0 on. On the other hand, 0 is a subsolution of (3.6 n. We therefore obtain a minimal solution u n of (3.6 n by monotone iteration: { um+1 = a n (xg n (u m + f n (x in u m+1 = 0 on, starting with u 0 0. In particular, since a n (xg n (s + f n (x increases with n, u n+1 is a supersolution of (3.6 n, and hence u n u n+1.

14 14 HAÏM BREZIS AND XAVIER CABRÉ We claim that Lemma 3. then gies a n (xg n (u n δ + u n (x δ(x + as n +. uniformly in as n + ; this proes Theorem 3.1. Thus, it only remains to show the claim. Suppose not, that (3.8 a n (xg n (u n δ C n. Then, multiplying (3.6 n by the solution z of (3.7, we see that u n C n (we hae used that 0 f n f L 1 δ (. monotone conergence. Hence, u n u in L1 (, for some u, by Since g n is a nondecreasing function, a n g n (u n + f n increases to ag(u + f a.e. in ; (3.8 also gies a n g n (u n + f n ag(u + f in L 1 δ(, again by monotone conergence. 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(xg(u + f(x in u = 0 on. This is impossible by Theorem.1(b. 4. Very weak solutions. Proofs of Theorems 0. and 0.3. In this section we return to the study of equation u = u x + f(x and its corresponding boundary alue problem. We proe stronger ersions of the nonexistence results of Section 1 by considering a more general notion of solutions, that we call ery weak solutions. More precisely, we proe the following results.

15 SOME SIMPLE NONLINEAR PDE S WITHOUT SOLUTIONS 15 Theorem 4.1. Let N and u L loc (\{0} satisfy u 0 a.e. in and (4.1 x u u in D (\{0}, in the sense that u ( x ϕ u ϕ for any ϕ 0, ϕ C 0 (\{0}. Then u 0. Note that now we are testing (4.1 only against functions with compact support in and anishing in a neighborhood of 0. As a consequence of the preious theorem, we will proe the following stronger ersion of Corollary 1.3. Corollary 4.. Let N and f L 1 loc (\{0}, fδ integrable near, f 0 a.e. in, f 0. Then there is no ery weak solution of (4. u = u x + f(x u = 0 in on, in the sense that u L loc (\{0}, u and u δ are integrable near and u ( x ζ = u ζ + f x ζ for any ζ C (, ζ = 0 on and ζ 0 in a neighborhood of 0. Remark 4.3. Theorem 4.1 does not hold in dimension N = 1; a direct computation shows that u(x = x α, for any 0 < α < 1, satisfies (4.1 when is a small interal containing 0. Corollary 4. is also false in dimension N = 1; in fact, for any p > 1 and f L p ( 1, 1, small in L p, there exists a ery weak solution u of (4. in = ( 1, 1 which is continuous in [ 1, 1] and satisfies u(0 = 0. This solution u is defined in (0, 1 to be the solution of (4.3 u = u + f(x in (0, 1 x u(0 = u(1 = 0, and similarly in ( 1, 0; we obtain in this way a ery weak solution u of (4. in ( 1, 1. Note that if f Lp (0,1 is small then there is a solution u C (0, 1 C 1 ([0, 1] of (4.3. It can be obtained through the Inerse Function Theorem applied to the operator u u / x, which maps X = W,p (0, 1 W 1,p 0 (0, 1 into L p (0, 1; note that if u X then u/ x L (0, 1 and u C 1 ([0, 1]. We also extend the local nonexistence result of Corollary 1.5, now for all N 1.

16 16 HAÏM BREZIS AND XAVIER CABRÉ Corollary 4.4. Let N 1 and c > 0 be any positie constant. Then there is no ery weak solution of (4.4 x u = u + c in D (\{0}, in the sense that u L loc (\{0}, and u ( x ϕ = (u + cϕ for any function ϕ C 0 (\{0}. The proofs of the results of this section consist of using appropriate powers of testing functions; this is an idea due to Baras and Pierre [] and employed in [] for the study of remoable singularities of solutions of semilinear equations. In fact, the first step in our proof of Theorem 4.1 will be to use equation (4.1 to show that u/ x L loc (; this can be interpreted as a remoable singularity result. The second step of the proof is to show that u u / x is satisfied in D (. We may then conclude that u 0 with the help of Theorem 1.. We present here an alternatie proof of Theorem 1. based on multiplication by a sequence of appropriate testing functions. Proof of Theorem 4.1. Step 1. We proe that u/ x L loc (. For this purpose, let ζ n C0 (\{0} be such that 0 ζ n 1, 0 if x 1/n ζ n = 1 if x /n, x ω ζ if x \ω, where ω is open, 0 ω, ω and ζ is a fixed tail for all ζ n. We take ζ n such that if 1/n < x < /n then and ζ 4 n = 4ζ 3 n ζ n ζ 4 n = 4ζ 3 n ζ n + 1ζ n ζ n ζ 4 n Cn ζ n, for some constant C independent of n. Multiplying (4.1 by ζ4 n x yields u x ζ4 n u ζn 4 ( C n uζ { 1 n < x < n } n + C.

17 SOME SIMPLE NONLINEAR PDE S WITHOUT SOLUTIONS 17 Using the Cauchy-Schwarz inequality, we obtain ( u x ζ4 n Cn u { 1 n < x < n } x ζ n + C C n N 1 ( u x ζ4 n 1/ + C. Since N we deduce that u and hence u/ x L loc (. x ζ4 n C, For later purposes, let us retain two more consequences of the preious proof. First, we did not use u 0. Second, if carried out for N = 1, the proof gies (4.5 3/n /n u Cn + C, (N = 1. x Step. We show that (4.6 u u x in D (. Indeed, let ϕ C0 (, ϕ 0, and η n (x = η 1 (n x be such that 0 η 1 1 and { 0 if x 1 η 1 (x = 1 if x. Multiplying (4.1 by ϕ x η n yields u x ϕη n u (ϕη n. If we show that, as n, (4.7 u ϕ η n 0 and (4.8 uϕ η n 0

18 18 HAÏM BREZIS AND XAVIER CABRÉ then we obtain u (ϕη n u ϕ and hence the statement of Step. To proe (4.7 and (4.8, we use that u/ x L loc ( which we established in Step 1. Hence u ϕ η n Cn { 1 n < x < n } u u C { 1 n < x < n } x 0 and uϕ η n Cn { u Cn 1n < x < n } C n N 1 ( { 1 n < x < n } u x x { 1 n < x < n } u 1/ 0. Note that, as in Step 1, we hae not used u 0. Step 3. We show that u 0 (it is only here where we use u 0. Let us suppose that u 0. Then, since u 0, u 0 in D ( and is connected, we hae that u ε a.e. in B η, for some ε > 0 and B η = B η (0 with closure in. When N = this is impossible since u/ x ε/ x near 0 and hence u/ x L loc ( a contradiction with Step 1. When N 3, we use that ( u ε ε x = N log 1 x and we conclude (as in the proof of Theorem 1. that (4.9 u ε log x C near 0 N for some C > 0. in D (B η Let us now choose a sequence of functions χ n (x = χ 1 (n x such that 0 χ n 1 and { 1 if x 1/n χ n (x = 0 if x /n.

19 Mutiplying (4.6 by χ 4 n yields SOME SIMPLE NONLINEAR PDE S WITHOUT SOLUTIONS 19 u x χ4 n Cn u χ 4 n Cn { 1n < x < n } uχ n u x χ n C n N 1 ( u x χ4 n 1/, and therefore u But, using (4.9, we hae x χ4 n C n N. u x χ4 n c n log n { 1 n < x < 1 n } log n n N which contradicts the preious statement. Finally we gie the proofs of Corollaries 4. and 4.4. Proof of Corollary 4.. Recall that Steps 1 and of the preious proof hold for any u satisfying (4.1 without the assumption u 0. Therefore, since f 0, Step 1 of the proof of Theorem 4.1 gies u x L loc(. Moreoer, proceeding as in Step of the same proof, we see that u ζ = u x ζ + fζ for any ζ C (, ζ = 0 on. In particular, u ζ 0 if, in addition, ζ 0 in. We conclude that u 0 in. Theorem 4.1 implies that u 0, which contradicts (4. and f 0. Proof of Corollary 4.4. Suppose that u satisfies x u = u + c in D (\{0}. The proof of Step 1 in Theorem 4.1 gies u x + c x L1 loc( when N.

20 0 HAÏM BREZIS AND XAVIER CABRÉ This is impossible when N =, since 1/ x is not integrable near 0. When N 3 we get (as in the proof of Theorem 4.1, Step that u = u x + c x in D (. We now proceed as in the proof of Corollary 1.5, i.e., we compare u with log(1/ x. We obtain that u 0 in a neighborhood of 0 and hence, by Theorem 4.1, u 0. This is a contradiction with equation (4.4. We finally treat the case N = 1. We would hae u L loc (\{0} and x u = u + c in D (\{0}. In particular, u belongs to C (0, a for some a > 0. Integrating the inequality u c/x in (s, a/, we obtain u (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 /n u ds Cn + C, s which was proed in Step 1 of the proof of Theorem 4.1. Using u(s c log s C, this inequality yields c 3/n 6 log n n = c log n 3/n /n /n ds s u ds Cn + C, s which is a contradiction.

21 SOME SIMPLE NONLINEAR PDE S WITHOUT SOLUTIONS 1 5. Connection with a result of Kalton-Verbitsky. In this section we consider the problem (for u 0 { u = a(xu p + f(x in (5.1 u = 0 on, where p > 1, a 0 and f 0 in. Recently, Kalton and Verbitsky [8] hae found an interesting necessary condition for the existence of a weak solution of (5.1. Their result states that if (5.1 has a weak solution, then necessarily (5. G(aG(f p CG(f in for some constant C, where G = ( 1 (with zero Dirichlet boundary condition. In [8] the authors also proe (5. for more general second-order elliptic operators. In this section we gie a simple proof of the necessary condition (5. (for the Laplacian using a refinement of the method that we hae deeloped in Section 1. Our proof gies (5. with constant C = 1 p 1. Next, we replace f(x by λf(x in (5.1 (where λ > 0 is a parameter and we study the problem of existence of solution depending on the alue of λ. As pointed out in the Introduction, condition (5. easily implies some of our nonexistence results. For instance, it gies the result of Theorem.1(b when g(u = u p, for some p > 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 test functions and using Young s inequality. Throughout this section we assume that (5.3 a L 1 loc(, a 0 a.e., a 0 and (5.4 f L 1 δ(, f 0 a.e., f 0. A function u L 1 (, u 0 a.e. is a weak solution of (5.1 if au p L 1 δ ( and (5.1 is satisfied in the sense of Definition 1.1( c. Finally, for h L 1 δ ( we denote by G(h the unique function in L 1 ( satisfying { (G(h = h in G(h = 0 on again in the sense of Definition 1.1( c see e.g. Lemma 1 of [4] for such result about the linear Laplace equation. We now gie a necessary condition and a sufficient condition for the existence of a weak solution of (5.1.

22 HAÏM BREZIS AND XAVIER CABRÉ Theorem 5.1. Assume (5.3 and (5.4. (a If (5.5 { u = a(xu p + f(x in u = 0 has a weak solution u 0, then ag(f p L 1 δ ( and G(aG(f p G(f 1 p 1 on in. (b If ag(f p L 1 δ ( and G(aG(f p G(f ( p p 1 1 p p 1 in, then (5.5 has a weak solution u satisfying G(f u CG(f in for some constant C. The second result of this section is the following. Theorem 5.. Assume (5.3, (5.4 and G(aG(f p G(f For λ a positie parameter, consider the problem L (. { u = a(xu p + λf(x in (5.6 λ u = 0 on. Then there exists λ (0, such that (i if 0 < λ < λ, then (5.6 λ has a weak solution u λ satisfying λ u λ G(f C(λ 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. Moreoer, (5.7 ( p 1 p p 1 p 1 (λ p 1 in G(aG(f p G(f L ( 1 p 1. The main ingredient in the proof of the aboe theorems is the following.

23 SOME SIMPLE NONLINEAR PDE S WITHOUT SOLUTIONS 3 Lemma 5.3. Suppose u and are C functions in, and that > 0. Let φ : R R be a C, concae function. Then [ ( u ] φ If, in addition, 0 in, then [ ( u ] (5.8 φ ( φ u [ ( u ( u + φ u ( u ] φ (. ( φ u [ u + ] + φ(1(. Proof. We simply compute and use that φ 0 and > 0. We hae (using the notation i = i [ ( u ] φ = N [ i=1 ( = φ u ( u N { i=1 φ ( u φ ( u N [ u i u i ( u ( u ] + φ i i i { i=1 ] i φ ( u ( = φ u ( ( u + φ u ( u ( u ( + φ ( φ u ( u ( = φ u [ ( u ( u + φ u ( u φ [ ( u ] [ ( u ] + φ i i i i ( u ( + φ ii + φ u u φ ( u ] (, ( u i ( which is the first inequality of the lemma. From this, we deduce the second inequality, as follows. Since φ is concae, we hae Thus φ(s + (1 sφ (s φ(1 s R. ( u φ u ( u φ ( φ u + φ(1; multiplying this inequality by (which is nonnegatie by assumption, we easily deduce (5.8. To use the preious lemma, we will need (5.8 in its weak ersion for L 1 functions. i } }

24 4 HAÏM BREZIS AND XAVIER CABRÉ Lemma 5.4. Let φ : R R be a C 1, concae function with φ bounded. Let h and k belong to L 1 δ (, with k 0, k 0, and let u and be the L1 ( solutions of { u = h in u = 0 on and { = k in = 0 on. Then [ ( u ] (5.9 φ ( φ u (h k + φ(1k, ( u ( u in the sense that φ L 1 (, φ (h k + φ(1k L 1 δ ( and ( u { φ ζ for all ζ C (, ζ 0, with ζ = 0 on. φ ( u } (h k + φ(1k ζ Proof. We first point out that (5.8 holds when φ is C 1 and concae not necessarily C. This follows immediately from Lemma 5.3 conoluting φ with mollifiers. We approximate h and k in L 1 δ ( by sequences (h n and (k n, respectiely, of C 0 ( functions and with k n 0, k n 0. Let u n, n be the solutions of { un = h n in u n = 0 on and { n = k n in n = 0 on. It follows that u n u and n in L 1 ( (for this, subtract the equations for u n and u, multiply by G(1 and integrate. Moreoer, u n, n C ( and n > 0, n 0 in. By (5.8 we hae [ n φ ( un n ] ( φ un (h n k n + φ(1k n. n

25 SOME SIMPLE NONLINEAR PDE S WITHOUT SOLUTIONS 5 Moreoer, using that φ is bounded, we see that (5.10 nφ ( un n = n ( φ ( un n C( u n + n φ(0 + φ(0 n for some constant C. Hence, n φ(u n / n anishes on and thus (5.11 n φ ( un for all ζ C (, ζ 0 in and ζ = 0 on. n { ( } ζ φ un (h n k n + φ(1k n ζ n Note that > 0 a.e. in, so that φ(u/ is well defined a.e. Moreoer, n φ(u n / n conerges a.e. to φ(u/ up to a subsequence. Since u n and n conerge in L 1 (, they are dominated (also up to a subsequence by an L 1 ( function. Thus, by (5.10, n φ(u n / n is also dominated by an L 1 function (for a subsequence. We conclude that n φ ( un n ( u φ in L 1 (. Passing to the limit in (5.11, we finally obtain (5.9 and the lemma. We write explicitly the concae functions φ that we use in this section. For Theorem 5.1 we will use s dt (5.1 φ(s = t p = 1 ( 1 1 p 1 s p 1 for s 1. It satisfies 1 φ (ss p = 1 for s 1, and hence φ is concae and φ is bounded in [1,. Moreoer, φ(1 = 0 and 0 φ(s 1 p 1 for s 1. Since φ (1 = 1, we can extend φ by φ(s = s 1 in (, 1] obtaining a function φ that satisfies the conditions of Lemma 5.4. Note that the functions φ aboe are analogous ersions of the ones that we employed in Sections 1 and, in the sense that they all satisfy φ (sg(s = 1 (in an appropriate interal where g is the nonlinearity.

26 6 HAÏM BREZIS AND XAVIER CABRÉ In Theorem 5. we will be led to take φ (that we denote now by ψ satisfying another differential equation, namely: ψ (ss p = ψ(s p. Precisely, we will take (5.13 ψ(s = s (εs p p 1 for s 0. It satisfies and hence ψ (s = 1 (εs p p p 1 for s 0 ψ (ss p = ψ(s p for s 0. Note that ψ is concae and ψ is bounded in [0, ; ψ(1 = ( 1 1 p ε and 0 ψ(s ( 1 1 p 1 ε 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 =, 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(xu p + f(x in u = 0 on. We consider note that, since a(xu p 0, u = = G(f; u G(f 1 by the weak maximum principle (which is an easy consequence of the weak formulation of Definition 1.1(c used here. In particular ag(f p au p and hence ag(f p L 1 δ (. We take φ, defined by (5.1, and apply Lemma 5.4 to obtain (we use u/ 1 and the properties of φ(s for s 1 in [ ( u ] ( φ φ u (au p + f f + φ(1f ( = φ u au p = a p = ag(f p

27 SOME SIMPLE NONLINEAR PDE S WITHOUT SOLUTIONS 7 in the weak sense of the lemma. Hence (by the weak maximum principle ( u G(aG(f p φ Part (a is now proed. 1 p 1 = 1 p 1 G(f. Part (b. We assume that ag(f p L 1 δ ( and ( p p 1 G(aG(f p 1 p p 1 G(f. It follows that the L 1 ( function ( p p ũ := G(aG(f p + G(f p 1 satisfies Therefore ũ p p 1 G(f. ( p p ũ = ag(f p + f p 1 aũ p + f in the weak sense. That is, ũ is a weak supersolution of (5.5. On the other hand, 0 is a subsolution of the problem. It is then easy to obtain a weak solution u of (5.5 by monotone iteration (see e.g. Lemma 3 of [4]. Moreoer G(f u ũ This completes the proof of Theorem 5.1. Proof of Theorem 5.. We assume that p p 1 G(f. G(aG(f p G(f L (; moreoer, a 0 and f 0 and hence 0 < M := G(aG(f p G(f <. L (

28 8 HAÏM BREZIS AND XAVIER CABRÉ Theorem 5.1 (applied with f replaced by λf implies that if (5.6 λ has a weak solution then λ p 1 M 1 p 1. The theorem also gies that if ( p p 1 λ p 1 1 M p p 1 then (5.6 λ has a weak solution. Hence, defining λ = sup{λ > 0; (5.6 λ has a weak solution}, we hae 0 < λ <, and also estimate (5.7 of Theorem 5.. Note that part (iii of the theorem is now obious. To proe part (i, we hae to show that if 0 < λ < µ and (5.6 µ has a weak solution u then (5.6 λ has a weak solution u λ satisfying (5.14 λ u λ G(f For this purpose, we consider C(λ in. = G(µf = µg(f, and the function ψ defined by (5.13 with ε > 0 chosen small enough such that λ ( 1 1 p 1 µ = ψ(1µ. 1 + ε We apply Lemma 5.4 (with φ replaced by ψ to obtain [ ( u ] ψ ( ψ u (au p + µf µf + ψ(1µf [ ( u ] p = a p ψ + ψ(1µf [ ( u ] p a ψ + λf in the weak sense of the lemma. Hence ψ(u/ is a weak supersolution of (5.6 λ. Again by monotone iteration we deduce that (5.6 λ has a weak solution u λ such that ( u u λ ψ C = CµG(f.

29 SOME SIMPLE NONLINEAR PDE S WITHOUT SOLUTIONS 9 This, together with the triial bound u λ G(λf, gies (5.14 and proes (i. It remains to show part (ii. For λ < λ we can take u λ to be the minimal solution of (5.6 λ, i.e., the solution obtained by monotone iteration starting from the function 0. In this manner, u λ u µ if 0 < λ < µ < λ. Hence, in order to obtain a weak solution of (5.6 λ, it suffices to show (5.15 [a(xu p λ + λf(x] δ C for some constant C independent of λ. To proe (5.15 we proceed as follows. Since a 0 and a 0, there exists a constant M (0, such that a M := χ {a M} a satisfies 0 a M M and a M 0 (here χ {a M} denotes the characteristic function of {a M}. It is well-known that the problem w = a 1/p M w1/p (5.16 w > 0 w = 0 in in on has a unique solution w W,r ( (for any 1 < r < with w 0. This solution can be obtained for example by minimizing in H0 1 ( the energy associated to (5.16: E( = 1 p a 1/p M p + 1 (+ (p+1/p, which is a coercie functional, bounded from below, in H0 1 ( (note that a 1/p M L ( and 1 < (p + 1/p <. We hae that E(tϕ 1 < 0 for t small, where ϕ 1 denotes the first eigenfunction of. In particular, the minimizer w of E satisfies E(w < 0, and thus w 0. The strong maximum principle then gies (5.17 w cδ for some positie constant c. Since w L ( we can multiply (5.6 λ by w and integrate by parts. We also use Young s inequality in the spirit of the ideas of Baras and Pierre [3]. We obtain au p λ w + λfw = u λ ( w = a 1/p M u λw 1/p a 1/p u λ w 1/p 1 au p λ p w + p 1 1, p

30 30 HAÏM BREZIS AND XAVIER CABRÉ and therefore au p λ w + λfw C for some constant C independent of λ. Using (5.17, we conclude (5.15, and hence the proof of part (ii. Remark 5.5. An analogous ersion of Theorem 5. also holds for the problem { u = λ(a(xu p + f(x in u = 0 on. This follows from Theorem 5. rescaling the solution u, i.e., considering the problem for αu, for appropriate α. Remark 5.6. In Theorem 5. we hae shown that, for λ < λ, (5.6 λ has a weak solution u λ satisfying (5.18 u λ G(f L (. We point out that this property may not be true for the minimal solution obtained for λ = λ. To gie an example of this, consider first the problem (5.19 λ { = λ( + 1 p in B 1 = { x < 1} R N = 0 on B 1. For N and p sufficiently large, (5.19 λ has (x = x p 1 1 as weak solution for a certain parameter λ > 0; moreoer is the pointwise, increasing limit (as λ λ of the classical minimal solutions λ of (5.19 λ for λ < λ (see e.g. [6]. Let us define f = ( + 1 p p, so that the problem (5.0 λ { u = λu p + λf(x in B 1 u = 0 on B 1 has u = as weak solution for λ = λ. Note that (5.0 λ is of the form (5.6 λ considered in Theorem 5.. We claim that is the minimal weak solution of (5.0 λ. This is shown as follows: for λ < λ, let u λ be the minimal weak solution of (5.0 λ. Since u λ, we

31 SOME SIMPLE NONLINEAR PDE S WITHOUT SOLUTIONS 31 hae (u λ + 1 p u p λ f, and hence u λ is a weak supersolution of (5.19 λ. Thus u λ λ, and since λ as λ λ, we conclude that is the minimal weak solution of (5.0 λ. Finally, we hae that (5.1 G(f L (B 1, since x p 1 near 0 and G(f C log(1/ x (note that f p( + 1 p 1 = p x in B 1. This proes that (5.18 does not hold for λ = λ and, in particular, that λ is the extremal parameter λ for problem (5.0 λ. 6. Eolution equations. In Section 4 we proed that u 0 is the only nonnegatie ery weak supersolution of u u x Here we proe an analogous result for the equation in D (\{0}. u t u u x in D ((\{0} (0, T. More precisely, we hae the following parabolic analogue of Theorem 4.1. Theorem 6.1. Let N, T > 0 and u L loc ((\{0} (0, T satisfy u 0 a.e. in (0, T and (6.1 x (u t u u in D ((\{0} (0, T, in the sense that u{( x ϕ t + ( x ϕ} u ϕ for any ϕ 0, ϕ C 0 ((\{0} (0, T. Then u 0. As an immediate consequence of the theorem we obtain an extension of a result of Peral and Vázquez (Theorem 7.1 of [1]. Here = B 1, the unit ball of R N, and N 3. We consider the function u(x = log 1 x, which is a weak solution of { u = (N e u in B 1 u = 0 on B 1.

32 3 HAÏM BREZIS AND XAVIER CABRÉ Corollary 6.. Let N 3, T > 0 and u L 1 loc ((B 1\{0} (0, T be a function such that e u L 1 loc ((B 1\{0} (0, T, and u(x, t u(x a.e. in B 1 (0, T, (6. u t u = (N e u in D ((B 1 \{0} (0, T in the sense that u(ϕ t + ϕ = (N e u ϕ for any ϕ C 0 ((B 1 \{0} (0, T. Then u(x, t u(x. Note again that we only assume equation (6. to be satisfied in the distributional sense and away from {x = 0} [0, T ]. In particular, gien any u 0 (x u(x a.e. in B 1, u 0 u, and any T > 0, there is no weak solution u, with u(x, t u(x in B 1 (0, T, of the problem (6.3 u t u = (N e u in B 1 (0, T u = 0 on B 1 (0, T u(x, 0 = u 0 on B 1 (as stated in [1]. A second consequence of Theorem 6.1 is the following nonexistence and complete blowup result. For any u 0 0, u 0 0 (say u 0 C 0 ( and for any T > 0, the problem (6.4 u t u = u x in (0, T u = 0 on (0, T u(x, 0 = u 0 on has no weak solution (by Theorem 6.1. Using similar ideas as in the elliptic case (see Section 3, we proe that approximate solutions of (6.4 blow up completely. More precisely, let g n be a sequence of nonnegatie, nondecreasing and globally Lipschitz functions in [0, such that g n (u increases pointwise to u. Let a n be a sequence of nonnegatie bounded functions in, increasing pointwise to 1/ x.

33 SOME SIMPLE NONLINEAR PDE S WITHOUT SOLUTIONS 33 Theorem 6.3. Under the aboe assumptions, let u n be the solution of (6.5 n Then, for any 0 < ε < T, u n t u n = a n (xg n (u n u n = 0 u n (x, 0 = u 0 on. in (0, + on (0, + (6.6 u n (x, t δ(x + uniformly in (x, t [ε, T ] as n. To proe Theorem 6.1 we adapt the method gien in Section 4 for the elliptic case; it consists 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 notation as there. We also fix a cut-off function in time: ψ C 0 ((0, T, 0 ψ 1, with ψ 1 in a gien compact sub-interal of (0, T. Step 1. We proe that u/ x L loc ( (0, T. For this purpose, we multiply (6.1 by ζn(xψ 4 (t/ x ; it yields u x ζ4 nψ u( ζnψ 4 uζ 4 n(ψ t. Let us denote by C different constants independent of n, but that may depend on, T and the cut-off ψ. We hae u( ζ 4 nψ Cn ( u { 1 n < x < n } (0,T x ζ nψ + C and uζn(ψ 4 t C Hence we can conclude, as in Section 4, that u x ζ nψ. u x ζ4 nψ C.

34 34 HAÏM BREZIS AND XAVIER CABRÉ Thus u/ x L loc ( (0, T. Step. We show that (6.7 u t u u x in D ( (0, T. This is done exactly as in the proof of Theorem 4.1, where now ϕ = ϕ(x, t belongs to C 0 ( (0, T. Step 3. We finally proe u 0. We suppose u 0. Since u 0, u t u 0 in D ( (0, T and is connected, we hae that u ε a.e. in B η (τ, T, for some 0 < τ < T, ε > 0 and B η = B η (0 with closure in. When N = this is a contradiction with u/ x L loc ( (0, T. When N 3, (6.7 gies We deduce that ( u t u ε ε x = ( t N log 1 x (6.8 u ε N log 1 x C in B η/ (τ, T for some C > 0 and τ < τ < T. in D (B η (τ, T. Following the proof of Theorem 4.1, we now multiply (6.7 by χ 4 n(xψ (t, with ψ a cut-off as in the beginning of this proof, and with ψ 1 in (τ, T for some T with τ < T < T. We hae u x χ4 nψ u χ 4 n ψ + uχ 4 n (ψ t (where we are integrating on { x < /n} (0, T, since it contains the support of χ 4 nψ. Hence u u x χ4 nψ Cn x χ nψ, where C is independent of n. From the Cauchy-Schwarz inequality, we deduce u x χ4 nψ C n N.

35 But using (6.8, we hae SOME SIMPLE NONLINEAR PDE S WITHOUT SOLUTIONS 35 u x χ4 nψ c log n n N, which contradicts the preious statement. This proes the theorem. Remark 6.4. Step 3 of the preious proof, (i.e., to show u 0 from u 0 and (6.7 could hae been done using a parabolic analogue of the method of Section 1. That is, one considers = 1 ε 1 u in a subcylinder where u ε. Then (with the aid of the parabolic Kato s inequality satisfies t 1 x, which leads to contradiction since is bounded ( 1/ε. Corollary 6. follows immediately from Theorem 6.1: Proof of Corollary 6.. Let u be as in the corollary. Consider (x, t = u(x, t u(x. It satisfies, from our assumptions, 0 a.e. in B 1 (0, T. Moreoer, in the distributional sense D ((B 1 \{0} (0, T, t = u t u + u = (N (e u e u = (N e u (e 1 = (N x (e 1 (N x since 0. Hence (N 0 satisfies (6.1. By Theorem 6.1 we deduce 0, that is u u. We finally gie the proof of the complete blow-up result. Proof of Theorem 6.3. We proceed in three steps. Recall that 0 u n u n+1, by the maximum principle. Step 1. We proe that, for any τ > 0, τ a n g n (u n δ +. 0

36 36 HAÏM BREZIS AND XAVIER CABRÉ Suppose not, that τ 0 a ng n (u n δ C for some τ > 0. We multiply (6.5 n by the solution of { z = 1 in We obtain τ 0 and, in particular u n + z = 0 u n (x, τz τ 0 u 0 z = u n C. on. τ 0 a n g n (u n z C Hence u n and a n g n (u n δ are bounded in L 1 ( (0, τ. By monotone conergence, we obtain that u n u in L 1 ( (0, τ with u satisfying the assumptions of Theorem 6.1. By this theorem, u 0. Thus u 1 0 and u 0 (x = u 1 (x, 0 0, a contradiction. Step. We show that u n (x, tδ(xdx + uniformly in t [ε/, T ]. Indeed, let us multiply (6.5 n by e λ1t ϕ 1, where ϕ 1 is the first eigenfunction of in with zero Dirichlet condition and λ 1 its corresponding eigenalue. We then integrate in space and time, to obtain τ e λ 1τ u n (, τϕ 1 u 0 ϕ 1 = a n g n (u n ϕ 1 e λ1t. Hence, if τ [ε/, T ], by Step 1. u n (x, τδ(xdx ce λ 1T ε/ 0 0 a n g n (u n δ + Step 3. We finally proe (6.6. For this purpose, we use a parabolic analogue of Lemma 3. due to Martel (see Lemma of [10]. It asserts that if ϕ 0 in (say, ϕ L ( then T (τ ϕ(x c(τ ϕδ x δ(x for any τ > 0, where c(τ > 0 is a constant depending on τ, and where T (τ is the heat semigroup at time τ. We apply this estimate with τ = ε/, and obtain u n (x, t c (ε/ u n (x, t ε/δ(xdx + δ(x

37 SOME SIMPLE NONLINEAR PDE S WITHOUT SOLUTIONS 37 uniformly in t [ε, T ] by Step, since t ε/ ε/. Acknowledgments: We thank Y. Martel, J. L. Vázquez and I. E. Verbitsky for useful discussions, as well as L. Nirenberg for a suggestion which led to the results of Section 4. The second author was supported by CEE contract ERBCHRXCT Part of this work was done while the second author was isiting Rutgers Uniersity. He thanks the Department of Mathematics for its hospitality. References [1] P. Baras and L. Cohen, Complete blow-up after T max for the solution of a semilinear heat equation, J. Funct. Anal. 71 (1987, [] P. Baras and M. Pierre, Singularités éliminables pour des équations semi-linéaires, Ann. Inst. Fourier 34 (1984, [3], Critère d existence de solutions posities pour des équations semi-linéaires non monotones, Ann. Inst. Henri Poincaré (1985, [4] H. Brezis, T. Cazenae, Y. Martel and A. Ramiandrisoa, Blow-up for u t u = g(u reisited, Ad. Diff. Eq. 1 (1996, [5] H. Brezis and W. A. Strauss, Semi-linear second-order elliptic equations in L 1, J. Math. Soc. Japan 5 (1973, [6] H. Brezis and J. L. Vázquez, Blow-up solutions of some nonlinear elliptic problems (to appear. [7] T. Gallouët and J.-M. Morel, Resolution of a semilinear equation in L 1, Proc. Royal Soc. Edinburgh 96A (1984, [8] N. J. Kalton and I. E. Verbitsky, Nonlinear equations and weighted norm inequalities (to appear. [9] T. Kato, Schröedinger operators with singular potentials, Israel J. Math. 13 (197, [10] Y. Martel, Complete blow-up and global behaiour of solutions of u t u = g(u (to appear. [11] J.-M. Morel and L. Oswald, A uniform formulation for Hopf maximum principle, (1985, unpublished preprint. [1] I. Peral and J. L. Vázquez, On the stability or instability of the singular solution of the semilinear heat equation with exponential reaction term, Arch. Rational Mech. Anal. 19 (1995, [13] Z. Zhao, Green function for Schrödinger operator and conditional Feynman-Kac gauge, J. Math. Anal. Appl. 116 (1986,