35 Acta Math. Univ. Comenianae Vol. LX, 1(1991), pp. 35–103 STATIONARY SOLUTIONS, BLOW UP AND CONVERGENCE TO STATIONARY SOLUTIONS FOR SEMILINEAR PARABOLIC EQUATIONS WITH NONLINEAR BOUNDARY CONDITIONS M. CHIPOT, M. FILA AND P. QUITTNER 1. Introduction Consider the problem  ut = △u − aup    ∂u (1.1) = uq  ∂n   u(x, 0) = uo (x) ≥ 0 x ∈ Ω, t > 0 x ∈ ∂Ω, t > 0 x ∈ Ω, with p, q > 1, a > 0, Ω – bounded domain in RN , uo 6≡ 0. If a = 0 then it follows from [F] that any solution blows up in finite time. The starting point of our investigations was the question whether the damping term in the equation can prevent blow up if a > 0. For N = 1 we give the following complete answer: (i) If p < 2q − 1 or p = 2q − 1, a < q then there are initial data for which blow up occurs. (ii) If p > 2q − 1 or p = 2q − 1, a > q then any solution exists globally and stays uniformly bounded. (iii) If p = 2q −1, a = q then any solution exists globally but it is not uniformly bounded. More precisely, any solution tends pointwise (as t → ∞) to the unique function v which satisfies vxx − qv 2q−1 = 0 v=∞ in Ω on ∂Ω. For N > 1 and Ω a ball we also show that (i), (ii) hold. For general domains the answer is far from being complete. We show global existence and boundedness only for N − q(N − 2) N +1 , p> (q + 1) − 1 q< N −1 N + 1 − q(N − 1) Received February 1, 1991. 1980 Mathematics Subject Classification (1985 Revision). Primary 35K60, 35B40, 35B35, 35B32, 35J65. 36 M. CHIPOT, M. FILA and P. QUITTNER and blow up of solutions starting from initial functions with negative energy for p ≤ q. For p ≤ q and q subcritical (q < N/(N − 2) if N > 2) we give also another sufficient condition for blow up. Namely, u blows up provided uo ≥ v, uo 6≡ v, v is any positive stationary solution. Positive stationary solutions exist if p < q or |∂Ω| . If p = q, q is subcritical and a < aΩ then any solution p = q, a > aΩ := |Ω| blows up. If Ω is a ball and p, q, a are as in (i) then we prove blow up of solutions which emanate from radial subsolutions that are sufficiently large on ∂Ω. For N = 1 and p, q, a as in (i), a sufficient condition for blow up is that uo lies above an arbitrary maximal stationary solution. If q ≤ p ≤ 2q − 1 then we shall see below that for any interval Ω there exists ao = ao (Ω, p, q) > 0 such that for a < ao the maximal stationary solution is 0, which means that any solution blows up. For N = 1 we also show that for suitable initial functions blow up occurs only on the boundary of the interval Ω. Since we are interested in all possible types of behavior of solutions, we are led to the question if there are global unbounded solutions for p, q, a as in (i). For N = 1 or p ≤ q, q subcritical, the answer is no. Therefore, there are only two possibilities in this case: blow up in finite time or global existence and boundedness. The latter possibility means that the ω–limit set is nonempty and consists of stationary solutions. Let us now give a sketch of our results concerning the stationary solutions. For N = 1 (Ω = (−l, l)) our description of the set of (positive) stationary solutions is almost complete. Denote the set of positive stationary solutions by E and the subset of symmetric positive stationary solutions by Es . For fixed l > 0 we distinguish five cases: (i) If p > 2q − 1 then card E = 1, E = Es for any a > 0. (ii) If p = 2q − 1 then E = ∅ for 0 < a ≤ q, card E = 1, E = Es for a > q. (iii) If q < p < 2q − 1 then there are 0 < ao < a1 such that E = ∅ for 0 < a < ao , card E = 1, E = Es for a = ao , card E = 2, E = Es for ao < a ≤ a1 , card E ≥ 4, card E is even, card Es = 2 for a > a1 . 1 , then If, in addition, p ≤ 4 or p > 4, q ≥ p − 1 − p−2 card E = 4 for a > a1 . (iv) If p = q then there is an a1 > 0 such that E = ∅ for 0 < a ≤ 1/l, card E = 1, E = Es for 1/l < a ≤ a1 , card E = 3, card Es = 1 for a > a1 . (v) If p < q then there is an a1 > 0 such that card E = 1, E = Es for 0 < a ≤ a1 , 37 BLOW UP FOR PARABOLIC EQUATIONS card E = 3, card Es = 1 for a > a1 . Our results are summarized in the following bifurcation diagrams: u(l; a) u(l; a) ao = 0 a1 a 0 u(l; a) ao = 1/l Fig.1: p < q a Fig.2: p = q u(l; a) 0 a1 0 ao a1 a Fig.3: 2q−1 > p > q u(l; a) ao = q Fig.4: p = 2q − 1 a ao = 0 a Fig.5: p > 2q − 1 In higher space dimension we have also some existence, nonexistence and multiplicity results for the stationary problem on general domains and more precise results for the radially symmetric problem on a ball. These results confirm that several facts indicated in Figures 1–5 hold also for N > 1. See Theorems 2.1, 2.2 for more details. We mentioned above that for N = 1 a sufficient condition for blow up is that uo lies above an arbitrary maximal stationary solution. This leads to the question how are the stationary solutions ordered. We show that for N = 1 any positive stationary solution is maximal except for the case q < p < 2q − 1, a > ao , when there is a v ∈ Es such that v < w for any w ∈ E, w 6= v. Any w ∈ E, w 6= v is maximal. 38 M. CHIPOT, M. FILA and P. QUITTNER To give a description of the local semiflow generated by the problem (1.1) we determine the stability properties of stationary solutions. For N = 1 we show that positive stationary solutions which do not correspond to a = ao or a = a1 are hyperbolic, i.e. zero is not an eigenvalue of the linearization (if q < p < 2q − 1 then also the smaller solution corresponding to a = a1 is hyperbolic). Then we compute the Morse indices of the hyperbolic stationary solutions. This will be used to draw the picture of the flow, more precisely, to find orbits which connect the stationary solutions. For N = 1, p < q, a > a1 the flow is depicted in the following figure. v2 v1 v3 0 Figure 6. The flow for N = 1, p < q, a > a1 . In Figure 6, the function v1 is the symmetric positive stationary solution, v2 and v3 are nonsymmetric stationary solutions. The zero solution is stable, the unstable manifolds of v2 , v3 are one–dimensional, the unstable manifold of v1 is two–dimensional. Any positive stationary solution is connected by an orbit to 0, v1 is connected to v2 and v3 . Moreover, if N = 1, p < q, then for any uo there is a λo > 0 such that the solution u(t, λuo ) starting from λuo tends to 0 in W 1,2 (Ω) as t → ∞ if λ < λo ; u(t, λo uo ) tends to a positive stationary solution; while u(t, λuo ) blows up in finite time if λ > λo . A weaker result is proved in a more general situation. Denote the set of initial nonnegative data for which the solutions exist globally by G. Then G is star– shaped with respect to zero and closed in C + := {v ∈ W 1,2 (Ω) ; v ≥ 0 a.e. } provided N > 1, p < q < (N + 1)/(N − 1) or p = q < min(2, (N + 2)/N ). The paper is organized as follows. Section 2 contains results on the N –dimensional stationary problem. The bifurcation diagrams for the 1–dimensional stationary problem are established in Section 3. In Section 3 also the Morse BLOW UP FOR PARABOLIC EQUATIONS 39 indices of the stationary solutions for N = 1 are computed. In Section 4 we give sufficient conditions for blow up and global existence. In Section 5 we establish the connecting orbits and study the behavior of u(t, λuo ), λ > 0. 2. Stationary solutions for N ≥ 1 Throughout this section1 we shall suppose that Ω ⊂ RN is a bounded domain N +2 with the smooth boundary ∂Ω, a > 0 and p, q > 1 are subcritical, i.e. p < N −2 N if N > 2. Then we have the compact imbedding of the Sobolev and q < N −2 1,2 space W (Ω) into Lp+1 (Ω) and the trace operator Tr : W 1,2 (Ω) → Lq+1 (∂Ω) is also compact. We shall look for (weak) solutions of the problem   △u = a |u|p−1 u  ∂u = |u|q−1 u ∂n (2.1) in Ω on ∂Ω By standard Lp regularity theory (see e.g. [A1, Theorem 3.2]) we get that any solution of (2.1) is in C 1 (Ω) ∩ C 2 (Ω). Moreover, the maximum principle (see [GT, Theorem 3.5, Lemma 3.4]) implies that any nonnegative solution u 6≡ 0 of (2.1) is positive in Ω. In what follows, by |∂Ω| we denote the (N -1)–dimensional measure of ∂Ω, by |Ω| we mean the N –dimensional measure of Ω. Finally, we put |∂Ω| and cΩ = |Ω|−1/2 . aΩ = |Ω| The main result of this section are the following two theorems. Theorem 2.1. (i) Let p ≤ q and let a > ao , where ao := 0 if p < q and ao := aΩ if p = q. Then there exists a positive solution of (2.1). The zero solution is stable, any positive solution is unstable (both from above and from below) in W 1,2 (Ω) in the Lyapunov sense. The graphs of any two positive solutions intersect. (ii) Let p = q and a < aΩ . Then (2.1) does not have positive solutions. The zero solution is unstable. (iii) Let p > q. Then the zero solution is unstable and there exists ao ∈ [0, ∞) such that (2.1) has a positive stable solution for a > ao and (2.1) does not have positive solutions for 0 < a < ao .  p + 1 (p−1)/(q−1) (iv) Let q < p < 2q − 1 and put ζ = . If ã > 0 is sufficiently 2 large, then there exists a ∈ (ã, ãζ) such that (2.1) has at least two positive solutions. 1 except of Remark 2.6 where supercritical p, q are considered 40 M. CHIPOT, M. FILA and P. QUITTNER (v) N − q(N − 2) N +1 and p + 1 > (q + 1)q ∗ , where q ∗ = . N −1 N + 1 − q(N − 1) Then p > 2q − 1 and ao = 0, i.e. (2.1) has a positive stable solution for any a > 0. 2 Let q < Theorem 2.2. Let Ω be a ball in RN . (i) If p < q or p = q and a > aΩ , then there exists a positive symmetric solution of (2.1). This solution is unique among positive symmetric functions. (ii) If q < p < 2q − 1 then there exists aso > 0 such that (2.1) has a symmetric positive solution iff a ≥ aso . If a > aso , then (2.1) has et least 2 symmetric positive solutions. (iii) Let p = 2q−1. If a > q then (2.1) has a symmetric positive stable solution. If a ≤ q then (2.1) does not have symmetric positive solutions. (iv) If p > 2q − 1 then there exists a symmetric positive stable solution of (2.1) for any a > 0. We shall use the variational formulation of (2.1), i.e. we shall look for critical points of the C 2 functional Φ : X → R : u 7→ I(u) + aP(u) − Q(u), where X = W 1,2 (Ω) is endowed with the scalar product Z Z hu, vi = ∇u ∇v dx + uv dx, Ω Ω Z 1 I(u) = |∇u(x)|2 dx, 2 Ω Z 1 |u(x)|p+1 dx and P(u) = p+1 Ω Z 1 Q(u) = |u(x)|q+1 dS. q + 1 ∂Ω Z 1 Hence kuk2 := hu, ui = 2I(u) + 2K(u), where K(u) = u(x)2 dx. By F , 2 Ω P , Q and K we denote the Fréchet derivatives of Φ, P, Q and K, respectively. Notice that K, P and Q are compact C 1 operators in X and the problem (2.1) is equivalent to the problem (2.2) F (u) = 0, where F = Fa : X → X : u 7→ u − K(u) + aP (u) − Q(u). If u is an isolated solution of (2.2), we shall denote by d(u) or da (u) the local Leray–Schauder degree of F at u with respect to 0, i.e. d(u) = deg(F, 0, Bε (u)) for ε sufficiently small (where Bε (u) = {v ∈ X ; kv − uk ≤ ε}). 2 cf. also Remark 2.5(i) BLOW UP FOR PARABOLIC EQUATIONS 41 If C is a closed convex set in X, we denote by P C the orthogonal projection in
X onto C and we put F C (u) = u − P C K(u) − aP (u) + Q(u) i.e. the solutions of F C (u) = 0 correspond to the solutions of the variational inequality (2.3) u∈C: hF (u), ϕ − ui ≥ 0 for any ϕ ∈ C which are the critical points of Φ with respect to C. If C = C + := {u ∈ X ; + u ≥ 0 a.e.}, then we write briefly F + instead of F C and we denote by d+ (u) the + local Leray–Schauder degree of F at u with respect to 0. We call u a subsolution of (2.2) if Φ′ (u)ϕ ≤ 0 for any ϕ ∈ C + . Analogously we define a supersolution of (2.2). Following [H2], we call an operator T : X → X E–regular, if there exists a finite sequence {Ei }n+1 i=0 of real Banach spaces such that E = E0 ⊂ E1 ⊂ . . . ⊂ En ⊂ En+1 = X and T induces continuous operators Ti ∈ C(Ei , Ei−1 ) for i = 1, . . . , n + 1. The Lp regularity for (2.1) implies that the operators K, P and Q are W 1,r (Ω)–regular for any r ≥ 2. Moreover, one can easily prove the following Lemma (cf. [H2, Lemma 2]). Lemma 2.1. Let Tj : X → X be E–regular operators for j = 1, . . . , m and let the corresponding Ei spaces in the definition of E–regularity be independent of j. (j) (j) Let {αk }∞ for j = 1, . . . , m. k=1 be a sequence of real numbers converging to α m P (j) αk Tj (vk ). Then vk ∈ E and Let vk ∈ X, vk → v ∈ E in X and let vk = j=1 vk → v in E. In the following two lemmas we study solutions which are close to zero. Lemma 2.2. Let p, q > 1 be fixed, ∞ > A ≥ ak ≥ 0 (k = 1, 2, . . . ), Fak (uk ) = 0, 0 6= kuk k → 0. Then one of the following assertions is true (i) p < q, ak → 0, ak > 0 for k large enough. (ii) p = q, ak → aΩ . |uk | Moreover, → cΩ in X ∩ C(Ω). kuk k |uk | we may suppose that vk converges weakly in X to kuk k some element v ∈ X (otherwise we choose a suitable subsequence). Dividing the equation Fak (uk ) = 0 by kuk k we obtain Proof. Putting vk = (2.4) vk = Kvk − ak P (vk )kuk kp−1 + Q(vk )kuk kq−1 . Passing to the limit in (2.4) and using the compactness of K we get vk → v = Kv (strong convergence), kvk = 1, which implies v ≡ ±cΩ . Lemma 2.1 and the imbedding W 1,r (Ω) ⊂ C(Ω) for r > N imply vk → v in C(Ω), hence vk > 0 (or vk < 0) for k large enough. Without loss of generality we may suppose 42 M. CHIPOT, M. FILA and P. QUITTNER vk > 0. Integrating the equation △uk = ak upk over Ω and multiplying the resulting equation by kuk k−q we get Z Z p p−q vkq dS → |∂Ω| |Ω|−q/2 , vk dx kuk k = (2.5) ak ∂Ω Ω which implies p ≤ q, ak > 0, ak → 0 if p < q and ak → aΩ if p = q.  Remark 2.1. By Theorem 2.1(ii) it will follow that ak ≥ aΩ for k large enough in the case of Lemma 2.2(ii). If Ω is a ball, then using (2.5) one can even prove ak > aΩ , since △(upk ) > 0. Lemma 2.3. (i) If p < q and a > 0 or if p = q and a > aΩ , then u = 0 is a strict local minimum of Φ, d(0) = 1. (ii) If a = 0 or p = q and 0 ≤ a < aΩ or if p > q and a ≥ 0, then d(0) = −1. Proof. (i) We shall argue by contradiction. Suppose there exist 0 6= uk → 0 (in X) such that Φ(uk ) ≤ 0. Since Φ is bounded on bounded sets and weakly lower semicontinuous, there exists 0 6= uk such that Φ(uk ) = min Φ(v) ≤ 0. Hence, kvk≤1/k there exists a Lagrange multiplier λk ≥ 0 such that F (uk ) + λk uk = 0, i.e. (2.6) uk =
1 Kuk − aP (uk ) + Q(uk ) . 1 + λk 1 uk ⇀ v, → µ ∈ [0, 1]. Dividing (2.6) by kuk k 1 + λk kuk k and passing to the limit we get vk → v = µKv, kvk = 1, which yields µ = 1, |v| ≡ cΩ . By Lemma 2.1 we get vk → v in C(Ω). Now We may suppose that vk := (2.7) Φ(uk ) ≥ aP(vk ) − Q(vk ) kuk kq−p , kuk kp+1 where the right-hand side converges to aP(v) for q > p or to aP(v) − Q(v) for p = q. Since in both cases the limit is positive, we have a contradiction. Hence u = 0 is a (strict) local minimizer for Φ and by [A2] d(0) = 1. (ii) Using the homotopies H t (u) = Fta (u), t ∈ [0, 1], and Hα (u) = u − Q(u) − (1 + α)Ku, α ∈ [0, αo ], we obtain d(0) = deg(Hα , 0, Bε (u)) = −1, since the operator Hα′ (0) is regular and has exactly one negative eigenvalue for α > 0 small. We have to verify H t (u) 6= 0 and Hα (u) 6= 0 for kuk = ε small and α ≥ 0 small. The condition H t (u) 6= 0 and H0 (u) 6= 0 follows from Lemma 2.2. Hence suppose uk we Hαk (uk ) = 0 for 0 6= uk → 0 and αk > 0, αk → 0. Putting vk = kuk k get similarly as in Lemma 2.2 vk → v ≡ ±cΩ in X ∩ C(Ω) and we may assume ∂uk = uqk > 0, which yields a vk > 0 for k large. Then △uk = −αk uk < 0, ∂n contradiction.  BLOW UP FOR PARABOLIC EQUATIONS 43 Remark 2.2 It can be shown that in the situation of Lemma 2.3(ii) the critical point u = 0 is of mountain–pass type in the sense of [H1]. Lemma 2.4. Let 0 ≤ u ≤ u ≤ M < ∞, where u and u are a subsolution and a supersolution of (2.2), respectively. Then there exists a solution u of (2.2) with u ≤ u ≤ u. Moreover, if u, u ∈ C 1 (Ω) ∩ C 2 (Ω) are not local minimizers of Φ with respect to C := {v ∈ X ; u ≤ v ≤ u} and u < u in Ω, then there exists a solution u lying strictly between u and u and being a local minimizer of Φ. Proof. In the first part of the proof we shall proceed similarly as in [St, Theorem I.2.4.]. The set C is convex and (weakly) closed and Φ : C → R is lower bounded and weakly lower semicontinuous, hence there exists u ∈ C such that Φ(u) = min Φ(v). Consequently, u solves (2.3). v∈C Choose ϕ ∈ C 1 (Ω), ε > 0 and put vε = min{u, max{u, u + εϕ}} = u + εϕ − ϕε + ϕε ∈ C, where ϕε = max{0, u + εϕ − u} ≥ 0 and ϕε = − min{0, u + εϕ − u} ≥ 0. We have 0 ≤ hΦ′ (u), vε − ui = εhΦ′ (u), ϕi − hΦ′ (u), ϕε i + hΦ′ (u), ϕε i, so that (2.8) hΦ′ (u), ϕi ≥  1 ′ hΦ (u), ϕε i − hΦ′ (u), ϕε i . ε Since u is a supersolution, we have hΦ′ (u), ϕε i ≥ hΦ′ (u) − Φ′ (u), ϕε i Z
∇(u − u) ∇(u + εϕ − u) + a(up − up )(u + εϕ − u) dx = Ωε Z (uq − uq )(u + εϕ − u) dS − ε Γ Z Z
∇(u − u) ∇ϕ + a(up − up )ϕ dx − ε ≥ε |uq − uq | |ϕ| dS, Ωε Γε where Ωε or Γε are the sets of all x ∈ Ω or x ∈ ∂Ω, for which u(x)+εϕ(x) ≥ u(x) > u(x), respectively. Since |Ωε | → 0 and |Γε | → 0 as ε → 0, we get hΦ′ (u), ϕε i ≥ o(ε). Analogously we get hΦ′ (u), ϕε i ≤ o(ε), hence (2.8) implies hΦ′ (u), ϕi ≥ 0 for all ϕ ∈ C 1 (Ω), so that Φ′ (u) = 0. Suppose now the additional assumptions on u and u and let u be as above. ∂u ≥ uq . Putting w = u − u one obtains w 6≡ 0, w ≥ 0, Then △u ≤ aup , ∂n △w ≤ a(up − up ) ≤ cw, where c = apM p−1 . By [GT, Theorem 3.5] w > 0 in ∂w (xo ) < 0. Ω. If w(xo ) = 0 for some xo ∈ ∂Ω, then [GT, Lemma 3.4] implies ∂n ∂w (xo ) ≥ uq (xo ) − uq (xo ) ≥ 0, a contradiction. Hence, w = u − u > 0 However, ∂n in Ω. Similarly one gets also u − u > 0 in Ω. 44 M. CHIPOT, M. FILA and P. QUITTNER Now suppose that u is not local minimum of Φ. Similarly as in the proof of Lemma 2.3 we find uk → u such that Φ(u) > Φ(uk ) = min Φ(v), F (uk ) + kv−uk≤1/k λk (uk − u) = 0 for some λk ≥ 0. The last equation is equivalent to uk = λk 1 (K − aP + Q)(uk ) + u, 1 + λk 1 + λk which together with Lemma 2.1 implies uk → u in C(Ω). However, this is a  contradiction with Φ(uk ) < Φ(u) = min Φ(v). u≤v≤u Lemma 2.5. Any solution of the variational inequality (2.3) with C = C + solves also the problem (2.2). Proof. Proof is based on the same arguments as the first part of the proof of Lemma 2.4. Choosing ϕ ∈ C 1 (Ω) and putting vε = max{0, u + εϕ} one gets 0≤ hence Φ′ (u) = 0. 1 ′ hΦ (u), vε − ui = hΦ′ (u), ϕi + o(1), ε  Lemma 2.6. If u ∈ C + is an isolated solution of (2.2), then the degree d+ (u) is well defined. If N = 1 and u 6= 0, then d+ (u) = d(u). Moreover, except for the case p = q, a = aΩ , we have (for any N ) (i) d+ (0) = 1 if d(0) = 1, (ii) d+ (0) = 0 if d(0) = −1. Proof. If u ∈ C + is an isolated solution of (2.2) then u is an isolated solution of (2.3) by Lemma 2.5. If N = 1, then u lies in the interior of C + ⊂ X, hence F + = F in a neighbourhood of u. If p < q and a > 0 or p = q and a > aΩ , then 0 is a strict local minimum of Φ by Lemma 2.3, hence it is a (strict) local minimum of Φ with respect to C + . Now [Q2] implies d+ (0) = 1. Now it is sufficient to show d+ (0) = 0 for a = 0, since then (ii) follows from the homotopy invariance property of the degree. Hence suppose a = 0.
Then we may use the homotopies Hαt (u) = u−P + (1+α)Ku+tQ(u) , α ∈ [0, αo ], t ∈ [0, 1], to derive d+ (0) = deg(Hα1 o , 0, Bε (0)) = deg(Hα0 0 , 0, Bε (0)) = 0, where the last equality follows from [Q1, Theorem 2(i)]. The admissibility of Hαt follows from the fact the the solutions of Hαt (u) = 0 correspond to the solutions of the ∂u ≥ tuq .  inequality u ≥ 0, −△u ≥ αu, ∂n Proof of Theorem 2.1(i). Suppose p < q and a > 0 or p = q and a > aΩ . Then 0 is a strict local minimum of Φ by Lemma 2.3(i). Choosing u > 0 such that aP(u) < Q(u) we simply get Φ(tu) < 0 for t > 0 sufficiently large. Put Φ+ (u) = Φ(u) for u ∈ C + , Φ+ (u) = +∞ for u ∈ / C + . We show that the functional + Φ fulfils the Palais–Smale condition introduced by Szulkin [Sz], hence by the BLOW UP FOR PARABOLIC EQUATIONS 45 corresponding mountain–pass theorem [Sz, Theorem 3.2] there exists a nontrivial solution u of the variational inequality (2.3) with C = C + . By Lemma 2.5 u is a positive solution of (2.2). Thus suppose uk ∈ C + , εk ↓ 0, Φ(uk ) → d and (2.9) hΦ′ (uk ), v − uk i ≥ −εk kv − uk k for any v ∈ C + . Put wk := P + (uk − F (uk )), then (2.10) huk − F (uk ) − wk , wk − vi ≥ 0 for any v ∈ C + . To prove the relative compactness of the sequence {uk } it is sufficient to show its boundedness, since then {wk } is relatively compact and putting v = wk in (2.9), v = uk in (2.10) and adding the resulting inequalities one simply gets kuk − wk k ≤ εk . Now using (2.9) with v = 2uk we get for k sufficiently large (2.11) (q +1)(d+1)+εk kuk k ≥ (q +1)Φ(uk )−hΦ′ (uk ), uk i = (q −1)I(uk )+a(q −p)P(uk ). If q > p, then the right-hand side in (2.11) can be estimated below by ckuk k2 for some c > 0, hence the assertion follows.R Let p = q and suppose kuk k → ∞. Using ⊥ the decomposition uk = ck +u⊥ k , where Ω uk dx = 0 and ck is constant, (2.11) and ⊥ ⊥ [N, Theorem 7.1] yield kuk k ≤ M I(uk ) = o(ck ) for some M > 0, which implies uk /kuk k → cΩ . Therefore, cp+1 Φ(uk ) Ω → aP(c ) − Q(c ) = (a|Ω| − |∂Ω|) > 0, Ω Ω kuk kp p+1 which gives a contradiction with the assumption Φ(uk ) → d. To see that any positive solution u is unstable (both from above and from below) notice that (2.12) Φ′′ (u)(u, u) = qhΦ′ (u), ui + (1 − q)2I(u) + a(p − q)(p + 1)P(u) < 0 and suppose e.g. that u is stable from above. Choosing ε > 0 we may find δ > 0 such that the solution uδ of (1.1) starting from (1 + δ)u fulfils kuδ (t)
− uk < ε for any t > 0. Moreover, choosing δ sufficiently small we have Φ uδ (0) < Φ(u) and due to the compactness and monotonicity of the flow (see Proposition 5.1) we get uδ (t) → uδ as t → +∞, where uδ is a stationary solution fulfilling kuδ − uk ≤ ε, uδ ≥ u and
Φ(uδ ) < Φ(u); the last inequality follows from the fact that the function Φ uδ (·) is nonincreasing. The maximum principle implies uδ > u in Ω and Lemma 2.4 together with (2.12) (used both for u and for uδ ) yield a contradiction. The last argument shows also the nonexistence of two positive solutions u1 , u2 with u1 ≤ u2 .  Remarks 2.3. Let us briefly mention some other possibilities how to prove Theorem 2.1(i). 46 M. CHIPOT, M. FILA and P. QUITTNER (i) One can use the standard mountain–pass theorem for the functional Φ to get a critical point u which is either a local minimum or of mountain–pass type (see [H1, Theorem]). If u changes sign in Ω, one gets similarly as in (2.12) Φ′′ (u)(w, w) < 0 for any 0 6= w ∈ span{u+ , u− } (where u+ (x) = max{u(x), 0}, u− (x) = − min{u(x), 0}) and using this information it is not difficult to show that u is neither local minimum nor of mountain–pass type. (ii) If one is able to prove suitable apriori estimates for the positive solutions of (2.2), then one can use the degree theory: if kuk < R for any solution u of (2.2) with 0 ≤ a ≤ A, then + + + 0 = d+ 0 (0) = deg(F0 , 0, BR (0)) = deg(FA , 0, BR (0)) 6= dA (0) = 1, hence there exists a nontrivial solution for a = A. The apriori estimates can be easily found e.g. for symmetric solutions on a ball (see the proof of Theorem 2.2). For a general domain we have the following assertion: N −1 if N > 2. Then for any A > 0 there exists R > 0 such Let p < q and let q < N −2 that any positive solution u of (2.2) with 0 ≤ a ≤ A fulfils kuk < R. Moreover, the solutions tend to zero if a → 0+. Proof. Denote by k · kr or ||| · |||r the norm in Lr (Ω) or Lr (∂Ω), respectively. By R we denote various constants, which may vary from step to step. We have kuk2 ≤ R(I(u) + Q(u)) + η for any u ∈ X, where η > 0 and R = R(η). If u is a solution, then obviously 2I(u) ≤ (q + 1)Q(u). Choosing ε > 0 such that 2−ε the trace operator Tr : X → Lr (∂Ω), where r = q , is continuos, we obtain 1−ε using Hölder inequality |||u|||qq−1+ε ≤ Rkuk2−ε|||u|||qq−1+ε , −η + kuk2 ≤ RQ(u) ≤ R|||u|||2−ε r hence (2.13) kukε ≤ η ′ + R|||u|||qq−1+ε , where η ′ → 0 as η → 0. Now △(up ) ≥ 0, hence ||up ||1 ≤ R|||up |||1 , where R does not depend on u. Using this inequality, Hölder inequality and the equation △u = aup integrated over Ω, we obtain ||u||qp ≤ R|||u|||qp ≤ R|||u|||qq = Ra||u||pp , hence ||u||p ≤ Ra1/(q−p) and |||u|||q ≤ Ra1/(q−p) . Now (2.13) implies kuk ≤ R and kuk → 0 if a → 0+.  Let us also note that using the degree theory and Lemmas 2.3, 2.6 one can easily prove (without apriori estimates) the following assertion: (2.14) (∀ε > 0)(∃δ > 0)(∀η ∈ (0, δ))(∃a ∈ (ao , ao + ε))(∃u ∈ X) u is a positive solution of (2.1) and kuk = η. 47 BLOW UP FOR PARABOLIC EQUATIONS (iii) In Section 4 we show that under the assumptions of Theorem 2.1(i) there exists a positive bounded initial condition uo , for which the solution of the parabolic problem (1.1) blows up in a finite time, and that any global solution of (1.1) with bounded initial condition is globally bounded. Since zero is a stable stationary solution, we may use Theorem 5.1 to show the existence of α ∈ (0, 1) such that the solution with the initial condition αuo tends to a positive stationary solution as time tends to infinity. However, this dynamical proof of the existence of stationary solution has (similarly as in the case (ii)) one disadvantage: we have to impose some additional assumptions on p and q (see Theorem 5.1). Proof of Theorem 2.1(ii). Let p = q, a < aΩ , and suppose there exists a positive solution u of (2.2). Choose ã ∈ (a, aΩ ). Then u is a supersolution for the operator Fã , 0 is a solution of Fã (v) = 0 and neither u nor 0 is a minimizer of Φ = Φã with respect to C = {v ∈ X ; 0 ≤ v ≤ u}. By Lemma 2.4 the equation Fã (v) = 0 has a solution ũ ∈ C, which is a local minimizer of Φã . However, this a contradiction with the estimate (2.12).  Proof of Theorem 2.1(iii). Let p > q, a > 0. If there exists a positive solution u of (2.2) and ã > a, then similarly as in the proof of Theorem 2.1(ii) we get a positive solution ũ of Fã (v) = 0, which is a local minimizer of Φã and fulfils 0 < ũ < u in Ω. Hence to prove the assertion (iii), it is sufficient to prove the existence of a positive solution for some a > 0. Choose p̃ ∈ (1, q), ã > 0 and let ũ be a positive solution of (2.1) with p and a replaced by p̃ and ã, respectively (its existence follows from Theorem 2.1(i)). It is easily seen that ũ is a supersolution for our problem if a is sufficiently large, since then aũp > ãũp̃ . Hence Lemma 2.4 yields the assertion.  Proof of Theorem 2.1(iv). Choose b > 0 and put Λb (u) = I(u) + bP(u), M = {u ∈ X ; Q(u) = 1}. Due to the compactness of the trace operator Tr : X → Lq+1 (∂Ω), the set M is weakly closed. The C 1 functional Λb : X → R is convex and coercive, hence there exists ub ∈ M such that Λb (ub ) = inf Λb (u). We may suppose 0 6≡ ub ≥ 0 u∈M (otherwise we put ũb = |ub |). The minimizer ub fulfils the equation Λ′b (ub ) = νb Q′ (ub ), where (2.15) νb = 2I(ub ) + b(p + 1)P(ub ) hΛ′b (ub ), ub i = >0 ′ hQ (ub ), ub i q+1 1/(q−1) is the corresponding Lagrange multiplier. Putting tb = νb can easily show that u is a positive solution of (2.2) with (2.16) a= b −(p−1)/(q−1) = bνb =: f (b), tp−1 b and u = tb ub one 48 M. CHIPOT, M. FILA and P. QUITTNER where the function f depends not only on b but also on ub . It is easily seen that the function g : b 7→ Λb (ub ) is continuous (and does not depend on ub , of course). Moreover, (2.15) implies 2g(b) ≤ (q + 1)νb ≤ (p + 1)g(b), (2.17) so that (2.16) yields the estimate (2.18)  q + 1 (p−1)/(q−1) 2 h(b) > f (b) >  q + 1 (p−1)/(q−1) p+1 h(b),
−(p−1)/(q−1) where h(b) := b g(b) is continuous. Now (2.18) and the continuity of h will imply our assertion if we show lim f (b) = +∞ and kuk = ktb ub k → ∞ b→+∞ for the corresponding solutions, since the solutions that we found in the proof of (iii) were bounded (in L∞ and, consequently, in X). Hence, suppose b → +∞. If  q + 1 1/(q+1) √ , we have we put vb (x) = d max{0, 1 − b dist(x, ∂Ω)}, where d = |∂Ω| Q(vb ) = 1, hence √ (2.19) g(b) ≤ Λb (vb ) ≤ c b, where c is some constant independent of b. This implies √ h(b) ≥ b(c b)−(p−1)/(q−1) = c̃ b(2q−p−1)/(2q−2) → ∞, hence ∞. Now (2.15), (2.17) and (2.19) imply P(ub ) ≤ √ by (2.18) also f (b) → 2 c/ b, so that ub → 0 in L (Ω). Now choose ξ < 1 such that the trace operator Tr : W ξ,2 (Ω) → Lq+1 (∂Ω) is continuous. Using an interpolation inequality we obtain (2.20) (1−ξ)(q+1) ξ(q+1) 1 = Q(ub ) ≤ ckub kq+1 kub k2 ξ,2 ≤ ckub k , where k · kξ,2 and k · k2 is the norm in W ξ,2 (Ω) and L2 (Ω), respectively. Since kub k2 → 0, (2.20) implies kub k → ∞.  Remark 2.4. If we could choose ub such that f (b) became continuous, then this would imply in the case of Theorem 2.1(iv) the existence of two positive solutions for any a large. If one could prove Palais-Smale condition in this case, this would also lead to the proof of two positive solutions for a large. Another way how to prove this existence is to prove corresponding apriori estimates and to use the degree theory – this will be done for the symmetric solutions on the ball. In the proof of Theorem 2.1(v) we will need the following lemma from [FK]. Lemma 2.7. Let q, q ∗ be as in Theorem 2.1(v), let ε > 0 and r > q ∗ . Then there exists a constant c = c(ε, r) such that Z Z r q+1 2 |u| dS ≤ εkuk + c (2.21) |u|q+1 dx ∂Ω Ω BLOW UP FOR PARABOLIC EQUATIONS 49 for any u ∈ X. Proof. Proof is based on the continuity of the trace operator Tr : W θz,q+1 (Ω) → L (∂Ω), on an interpolation inequality and the continuity of the imbedding X ⊂ W z,q+1 (Ω) for suitable z, θ ∈ (0, 1). A detailed proof can be found in [FK]. q+1 Proof of Theorem 2.1(v). Let a > 0 be fixed. Our assumptions imply p + 1 > (q + 1)r for suitable r > q ∗ . Choosing ε > 0 and using Lemma 2.7 and Hölder inequality we obtain for any u ∈ X (and suitable c > 0 varying from step to step) Z |u|q+1 dS hQ(u), ui = ∂Ω Z  r(q+1) Z  2 p+1 p+1 p+1 p+1 +c |u| dx |∇u| dx + c ≤ε |u| dx Ω ZΩ ZΩ p+1 2 |u| dx + c |∇u| dx + εa ≤ε Z Ω 2 Ω = εhu − Ku + aP (u), ui + c which implies a uniform apriori bound for the solutions t ∈ [0, 1], u ∈ C + of the inequality (2.22) Consequently, denoting Ht (u) = u − P + (2.23) ∀v ∈ C + .
Ku − aP (u) + tQ(u) we get hu − Ku + aP (u) − tQ(u), v − ui ≥ 0 deg(F + , 0, Bc (0)) = deg(H1 , 0, Bc (0)) = deg(H0 , 0, Bc (0)) = 1, where the last equality follows from [Q2, Corollary 1], since the functional Λa (u) = I(u)+aP(u) corresponding to H0 is coercive. On the other hand, Lemma 2.6 yields (2.24) deg(F + , 0, Bε (0)) = d+ (0) = 0. The existence of a positive solution follows from (2.23), (2.24) and Lemma 2.5.  Remarks 2.5. (i) According to the results for Ω being a ball, the condition on p, q in Theorem 2.1(v) does not seem to be optimal. In fact, a finer apriori estimate can lead to weaker assumptions. Suppose e.g. that p, q fulfil the following assumptions: p−1 N +1 , p ≥ q + 1 and p + 1 + > (q + 1)q ∗ (so that p, q need not fulfil the q< N −1 p+1 condition from Theorem 2.1(v)). We show that this condition is also sufficient for the apriori bound and, consequently, also for the existence. ∂u Let u be a solution of (2.22), i.e. it solves the problem △u = aup , = tuq . ∂n Choosing a test function ϕd (x) = min{1, d1 dist(x, ∂Ω)} for d > 0 small and putting Ωd = {x ∈ Ω ; ϕd (x) = 1} we get Z Z Z c ∇u∇ϕ dx ≤ kuk kϕk ≤ √ kuk up dx ≤ a up ϕ dx = − (2.25) a d Ω Ω Ωd 50 M. CHIPOT, M. FILA and P. QUITTNER and using Hölder inequality we obtain Z Z  p 1 p+1 p+1 . up dx ≤ c (2.26) up+1 dx d Ω\Ωd Ω
−ν Choosing d = Ω up+1 dx , where ν = (p − 1)/(p + 3), and using (2.25) and (2.26) in Lemma 2.7 we get the desired apriori estimate for u. Similar improvements can be made also for p < q + 1. (ii) In order to prove Theorem 2.1(v) one can use also the function f (b) introduced in the proof of Theorem 2.1(iv) and show lim inf f (b) = 0. However, this R b→∞ leads to estimates which are close to those already used in the proof of Theorem 2.1(v). (iii) The investigation of the function f (b) gives an information for the existence of solutions also in other cases; however, in these cases other methods turned out to be more powerfull. Nevertheless, the likely behaviour of f (indicated in the figures below) gives us a good insight on the stationary solutions. To support the figures below, let us only mention that it is easy to show that f (b) → ∞ if p > q, b → 0, or if p < q, b → ∞. In both cases one can use a simple estimate νb ≤ cb. f f f b b b p=q p<q 2q − 1 > p > q f f b b p = 2q − 1 p > 2q − 1 Figure 7. The graphs of f . BLOW UP FOR PARABOLIC EQUATIONS 51 Proof of Theorem 2.2(i). Let Ω = BR (0). The existence of a positive symmetric solution to (2.1) follows by the same way as in Theorem 2.1(i); we have only to restrict ourselves to the space Xs of all radially symmetric functions in X = W 1,2 (Ω). Hence it suffices to prove the uniqueness. Denote r = |x|. Any positive symmetric solution of (2.1) fulfils the O.D.E. (2.27) urr + N −1 ur = aup , r r ∈ (0, R) together with the boundary conditions (2.28) ur (0) = 0, ur (R) = uq (R). If u1 , u2 are two different positive symmetric solutions, then the uniqueness of the solution of the initial problem for (2.27) implies u1 (0) 6= u2 (0). Hence we may suppose u1 (0) < u2 (0). Since w := u2 − u1 fulfils wrr + N −1 wr = a(up2 − up1 ), r wr (0) = 0, w(0) > 0, it is easily seen that w(r) > 0 for any r ∈ [0, R], so that u2 > u1 in Ω. By (2.12) neither u1 nor u2 is a local minimum of Φ with respect to C := {u ; u1 ≤ u ≤ u2 }, hence Lemma 2.4 implies the existence of a local minimizer of Φ between u1 and u2 , which contradicts (2.12).  Proof of Theorem 2.2(ii). Let Ω = BR (0). Considering only the space Xs of symmetric functions we get similarly as in the proof of Theorem 2.1(iii) the existence of aso ≥ 0 such that the problem (2.1) has a stable symmetric positive solution if a > aso and (2.1) does not have symmetric positive solution if a < aso . To show the rest of the assertion we need some apriori estimates for symmetric positive solutions. Hence suppose that u is such solution. Multiplying (2.27) by ur and integrating resulting equation over (0, R) we get using (2.28) Z R 1 2q 1 1 N −1 2 u (R) = u2r (R) ≤ u2r (R) + ur (r) dr 2 2 2 r 0
a a u(R)p+1 − u(0)p+1 < = u(R)p+1 p+1 p+1 which implies (2.29) u(R)2q−p−1 < 2a . p+1 Moreover, (2.27) implies urr > 0 whenever ur ≤ 0, hence ur ≥ 0 and (2.29) yields an apriori bound for u, which is independent of a ∈ [0, A] for any A < ∞ fixed. Denoting by ds+ the local degree corresponding to F + /Xs and using apriori estimates (2.29) we obtain for R > 0 sufficiently large

s+ (2.30) deg Fa+ /Xs , 0, BR (0) = deg F0+ /Xs , 0, BR (0) = ds+ 0 (0) = 0 = da (0), 52 M. CHIPOT, M. FILA and P. QUITTNER where the last two equalities follow analogously as the corresponding equality in Lemma 2.6(ii). Now if a > aso , then we have a positive symmetric solution u1 which is a local minimizer of Φ in Xs (cf. the proof of Theorem 2.1(iii)), hence [Q2] implies ds+ (u1 ) = 1. If this were the only positive symmetric solution, (2.30) would imply
s+ 0 = deg Fa+ /Xs , 0, BR (0) = ds+ a (u1 ) + da (0) = 1, a contradiction. Hence there exist at least two symmetric positive solutions for a > aso . Now we show the existence of a positive symmetric solution for a = aso and this will also imply aso > 0, since the equation F0 (u) = 0 does not have positive solutions. Thus let un be positive symmetric solutions of (2.1) with a = an ↓ aso . Then (2.31) un = Kun − an P (un ) + Q(un ) and the boundedness of un implies that we may suppose un ⇀ u (weak convergence). Now (2.31) implies un → u = Ku − aso P (u) + Q(u), hence u is a nonnegative symmetric solution for a = aso . It is now sufficient to notice that u 6= 0 by Lemma 2.2.  Proof of Theorem 2.2(iii), (iv). If p > 2q − 1 or p = 2q − 1 and a > q, then the proof of Theorem 4.1 yields a positive symmetric supersolution to our problem, hence the existence follows from Lemma 2.4 (used for the space Xs ). If p = 2q − 1, a ≤ q and u were a positive symmetric solution, then (2.29) yields a simple contradiction.  Remark 2.6. If p > 1 or q > 1 is not subcritical, then one can still expect similar results as in Theorems 2.1, 2.2. More precisely, (i) if p > q, then there exists ao ∈ [0, ∞) such that (2.1) has a classical positive solution for a > ao and (2.1) does not have classical positive solutions for 0 < a < ao . If Ω is a ball and p > 2q − 1, then ao = 0. (ii) If Ω is a ball, p ≤ q and a > ao (where ao is defined in Theorem 2.1(i)), then (2.1) has a classical positive symmetric solution. If Ω is a ball and q < p < 2q − 1, then the conclusions of Theorem 2.2(ii) are true. Proof. (i) Let p > q > 1, let u be a classical positive solution of (2.1) and let ã > a. Then u is a supersolution of (2.1) in which a is replaced by ã and the nonlinearities v p and v q are suitably modified for v > max u (so that the corresponding functional is well defined and differentiable). An obvious modification of Lemma 2.4 implies now the existence of a solution ũ for the problem (2.1) with a replaced by ã. Hence the existence of ao ∈ [0, ∞] follows. BLOW UP FOR PARABOLIC EQUATIONS 53 To see that ao < ∞, choose subcritical p̃, q̃ > 1 such that q̃ < min(p̃, q). If ã > 0 is large enough, we have a positive solution ũ of (2.1) with p, q and a replaced by p̃, q̃ and ã, respectively. The proof of Theorem 2.1(iii) shows that we may suppose 0 < ũ < 1 in Ω, hence ũq̃ > ũq . Moreover, choosing a > 0 large enough we have aũp > ãũp̃ , so that ũ is a supersolution for the problem (2.1) (with the nonlinearities v p , v q modified for v > 1), which implies the existence of a solution for a large. If Ω is a ball and p > 2q − 1, we may use the supersolution from Theorem 4.1. (ii) Replacing the nonlinearities up and uq by m(u) = u min(u, C)p−1 and n(u) = u1+ε min(u, C)q−1−ε , respectively (where ε > 0 is small and C > 0 is large) we obtain similarly as in (2.29) the following apriori bound for the positive symmetric solutions of the modified problem:
n2 u(R)
< 2a, (2.32) M u(R) where M (u) = 2a > Ru o m(v) dv. If u(R) > C, then (2.32) yields u(R)2+2ε C 2q−2−2ε u(R)2 −C 2 2 C p−1 + C p+1 p+1 > u(R)2+2ε C 2q−2−2ε > C 2q−p−1 , u(R)2 C p−1 which is a contradiction for C large. Consequently, any positive symmetric solution of the modified problem is a solution of our original problem for C large enough. The existence of a positive symmetric solution for the modified problem for p ≤ q and a > ao follows from the mountain pass theorem similarly as in Theorems 2.1(i), 2.2(i) or from the degree theory (see Remark 2.3(ii)). The existence of aso (as in Theorem 2.2(ii)) for q < p < 2q − 1 follows from an obvious modification of the proof of Theorem 2.2(ii).  Finally let us note, that if Ω is a general domain in RN and p ≤ q, then one can easily show that (2.14) is true also for supercritical p, q. 3. Stationary solutions for N = 1 Consider the O.D.E. (3.1) uxx = aup for x > 0, with the initial conditions (3.2) u(0) = m > 0, ux (0) = 0. We are looking for L > 0 such that (3.3) ux (L) = uq (L). 54 M. CHIPOT, M. FILA and P. QUITTNER This will provide a symmetric solution to  on (−l, l),  uxx = aup (3.4)  ∂u = uq at − l, l, ∂n with l = L. If for given m there are two values L1 , L2 such that (3.3) is satisfied, then by shift and reflection we obtain a pair of nonsymmetric solutions u1 , u2 to the problem (3.4) with l = (L1 + L2 )/2, u1 (x) = u2 (−x). Multiplying (3.1) by ux and integrating we see that a a 1 2 ux − up+1 = const = − mp+1 . 2 p+1 p+1 (3.4a) Note that uxx ≥ 0, hence ux is nondecreasing and since ux (0) = 0 we have that ux ≥ 0. Therefore r 2a p p+1 u − mp+1 (3.4b) ux = p+1 and integrating this equation we obtain Z (3.5) u(x) m dv √ = p+1 v − mp+1 r 2a x. p+1 For m given, the solvability of (3.1)–(3.3) is equivalent to finding L such that Z u(L) m dv √ = v p+1 − mp+1 r 2a L, p+1 r 2a p p+1 u (L) − mp+1 . uq (L) = p+1 The last equation may be written in the form p + 1 2q u (L) − up+1 (L) + mp+1 = 0. 2a If we now denote by R(m) a root of the equation (3.6) p + 1 2q x − xp+1 + mp+1 = 0 2a and assume that
R(m) > m, then (3.5) gives us a solution to (3.4) on the interval −L(m), L(m) with L(m) = r p+1 2a Z R(m) m dv √ . p+1 v − mp+1 BLOW UP FOR PARABOLIC EQUATIONS Setting V = (3.7) 55 v we get m L(m) = r p + 1 −(p−1)/2 m 2a Z R(m) m 1 dV √ . p+1 V −1 Theorem 3.1. Assume that p > 2q − 1. Then for any l the problem (3.4) has a unique nontrivial solution. This solution is symmetric. Proof. Consider the function (3.8) One has p + 1 2q x − xp+1 + mp+1 . 2a F(x) =  q F ′ (x) = (p + 1) x2q−1 − xp . a Hence F ′ vanishes only for (3.9) x=  a 1/(2q−p−1) q . Thus F is increasing up to this value and decreasing next. Hence (3.6) has only one root  a 1/(2q−p−1) R(m) ≥ , q in particular (3.10) lim m→0 R(m) = +∞. m Since (3.11) 0< Z +∞ 1 dV √ < +∞, V p+1 − 1 we deduce from (3.7), (3.10) that (3.12) lim L(m) = +∞. m→0 Combining (3.7), (3.11) we have also (3.13) lim L(m) = 0 m→∞ 56 M. CHIPOT, M. FILA and P. QUITTNER and the range of L is (0, +∞). Now we show that L is a decreasing function. Indeed, from (3.7) we have r Z R(m) m p − 1 p + 1 −(p+1)/2 dV √ m L (m) = − p+1 2 2a V −1 1 r ′  p + 1 −(p−1)/2 1 R(m)  q + . m 2a m R(m)
p+1 −1 m ′ (3.14) But since R(m) is the only root to (3.6), it follows from the implicit function theorem that R is differentiable and by differentiation one gets R′ (m) = (3.15) R(m)p mp . − aq R(m)2q−1 It follows that  R(m) ′ m  mp+1 1 1 ′ 1  R(m) + R(m) + R (m) = − q 2q−1 − R(m)p m2 m m2 a R(m)     −1 q 1 q =− 2 R(m)2q−1 − R(m)p R(m)2q − R(m)p+1 + mp+1 m a a −1  p + 1  1 q 2q−1 p <− 2 R(m) − R(m) R(m)2q − R(m)p+1 + mp+1 m a 2a = 0, =− the last inequality follows from the fact, that
q R(m)2q−1 − R(m)p = F ′ R(m) < 0. a Recalling (3.14) we obtain that (3.16) L′ (m) < 0. (3.12), (3.13) and (3.16) yield the assertion.  Theorem 3.2. Assume that p = 2q − 1. (i) If a ≤ q then the problem (3.4) cannot have nontrivial solutions. (ii) If a > q then for any l the problem (3.4) has a unique nontrivial solution. This solution is symmetric. Proof. (i) The boundary value u(L) must be a solution to (3.6). But (3.6) reduces to  q mp+1 = xp+1 1 − ≤ 0. a 57 BLOW UP FOR PARABOLIC EQUATIONS (ii) In this case (3.6) has a unique root R(m) which is given by the explicit formula 1  q − 2q R(m) = m 1 − . a  R(m) ′ = 0 and it is easily seen from (3.14) that L′ (m) < 0. (3.7) Hence m immediately yields (3.12) and (3.13).  Next we turn to the case p < 2q − 1. Considering F given by (3.8) we see that F has an absolute minimum given by (3.9). So, in order for (3.6) to have a root we need  a  1  2q−p−1 F ≤0 q which reads also 1 m ≤ c(a) := a 2q−p−1 c(p, q), (3.17) where c(p, q)p+1 = 1 q 1  2q−p−1 2q 1 − p − 1  p+1 . 2q Then for m satisfying (3.17), the graph of F looks like F mp+1 2q−p−1 (a q) 1 0 R1 (m) R2 (m) x Figure 8. The graph of F. and (3.6) has two roots R1 (m), R2 (m) which are equal to m = c(a) . Note that if m satisfies (3.17) then m≤ a q 1 2q−p−1 . a q 1 2q−p−1 when 58 M. CHIPOT, M. FILA and P. QUITTNER Since F(m) ≥ 0, one has m ≤ R1 (m) ≤ R2 (m). Let us now study the two curves (3.18) Li (m) = r p + 1 −(p−1)/2 m 2a Z Ri (m) m 1 dV √ V p+1 − 1 i = 1, 2 on the interval (0, c(a) ). Lemma 3.1. Assume that p < 2q − 1. Then we have L1 (m) ≤ L2 (m), (3.19) and L2 (m) is decreasing for m ∈ (0, c(a) ). Moreover, r Z d(p,q) p + 1 − p−1 dV 2 √ c(a) =: L(a) , (3.20) lim Li (m) = p+1 m→c(a) 2a V −1 1 where d(p, q) = 1 1  1  2q−p−1 . c(p, q) q Proof. (3.19) and (3.20) are obvious. In order to show that L′2 < 0 it is sufficient  R (m) ′ 2 < 0 (see (3.14)). From (3.15) we get to prove that m q  R (m) ′ 1  Ri (m)2q − Ri (m)p+1 + mp+1  i = 2 a . m m Ri (m)p − aq Ri (m)2q−1 According to (3.6), the last equality implies that (3.21)  R (m) ′ Ri (m)2q 1  p + 1 i = q− . 2 p m am 2 Ri (m) − aq Ri (m)2q−1 Since R2 (m) ≥ a q 1 2q−p−1 , R2 (m) is in the region where 1 q F ′ (x) = x2q−1 − xp > 0. p+1 a Hence, the right hand side of (3.21) is negative for i = 2.  Lemma 3.2. Assume that p ≤ q. Then L1 (m) is increasing. Proof. To prove that L′1 > 0 means (see (3.14)) to prove that ′ p−1 m− 2 q R1 (m)
m R1 (m)
p+1 m −1 p − 1 − p+1 > m 2 2 Z 1 R1 (m) m dV √ . p+1 V −1 BLOW UP FOR PARABOLIC EQUATIONS 59 The last inequality is equivalent to the following one: (3.22) s (m) Z R1m  R (m)  p+1  R (m) ′ p − 1 dV 1 1 √ =: ΨR (m). −1 > ΨL (m) := m p+1 m 2 m V −1 1 Using (3.21) we get ΨL (m) =  p + 1 1 1 R1 (m) R1 (m)2q−p−1 q − . a m 2 1 − aq R1 (m)2q−p−1 But  R (m) −(p+1) p+1 q 1 , R1 (m)2q−p−1 = 0 < 1 − R1 (m)2q−p−1 < 1 − a 2a m the last equality follows from (3.6). Hence (3.23) ΨL (m) > On the other hand, 1 p + 1  R1 (m) p+2 q− R1 (m)2q−p−1 . a 2 m s (m) Z R1m p+1 V p dV p − 1  R1 (m)  √ −1 ΨR (m) < 2 m V p+1 − 1   1  p−1 R1 (m) p+1 = −1 . p+1 m According to (3.6) we have   p − 1  R1 (m) p+1 p − 1  R1 (m) p+1 −1 = R1 (m)2q−p−1 . p+1 m 2a m Our assumption on p, q implies now that ΨR (m) < 1 p + 1  R1 (m) p+1 q− R1 (m)2q−p−1 . a 2 m Recalling the inequality R1 (m) ≥ m, we obtain (3.24) ΨR (m) < p + 1  R1 (m) p+2 1 q− R1 (m)2q−p−1 . a 2 m (3.23) and (3.24) yield (3.22). Lemma 3.3. Assume that p < 2q − 1. Then (3.25) (3.26) lim L2 (m) = + ∞, m→0 lim m→0 aL1 (m) = 1. mq−p  60 M. CHIPOT, M. FILA and P. QUITTNER Proof. First remark that if R is a limit point (as m → 0) of R1 (m) or R2 (m) one must have (see (3.6)) p + 1 2q R − Rp+1 = 0, 2a which means that R = 0 or R = R1 (m) ≤  2a  1 2q−p−1 . Since p+1 a 1 2q−p−1 q and a <  2a  1 2q−p−1 p+1 1 2q−p−1 < R2 (m), q  2a  1 2q−p−1 the only limit point of R2 (m) is and the only limit point of R1 (m) p+1 is 0. Thus we have R2 (m) = +∞. lim m→0 m One concludes like in (3.12) that (3.25) holds. Since R1 (m) → 0 and (3.6) implies that   p+1 R1 (m)2q−p−1 , mp+1 = R1 (m)p+1 1 − 2a we have R1 (m) = 1. m→0 m (3.27) lim In the sequel it will be convenient for us to use the following notation: “f (x) ∼ g(x) f (x) = 1. When h → 0+, we have when x → xo ” means that lim x→xo g(x) Z 1 1+h dV √ = V p+1 − 1 Z 0 h dV q ∼ (V + 1)p+1 − 1 Z 0 h dV 2 √ p =√ h. p+1 (p + 1)V Using (3.18), (3.27) it follows that r r r
2 p + 1 − p−1 R1 (m) 2 (3.28) L1 (m) ∼ R1 (m) − m . m 2 √ −1= p 2a m am p+1 From (3.6) we deduce  m p+1  p+1 m  (p + 1), ∼ 1− R1 (m)2q−p−1 = 1 − 2a R1 (m) R1 (m) BLOW UP FOR PARABOLIC EQUATIONS hence R1 (m) − m ∼ 61 1 2q−p 1 R1 (m)2q−p ∼ m . 2a 2a Going back to (3.28) we get r L1 (m) ∼ p 1 2 −p 1 m 2 √ mq− 2 = mq−p a a 2a and (3.26) is shown.  Lemma 3.4. Assume that p < 2q − 1. Then lim L′1 (m) = + ∞, m→c(a) lim L′2 (m) = − ∞. m→c(a) Proof. From (3.14), (3.7) we have L′i (m) = − (3.29) where Gi (m) := r p−1 Li (m) + Gi (m), 2m  − 1  2 Ri (m) ′ p + 1 − p−1  Ri (m) p+1 2 −1 . m 2a m m According to (3.6) we get (3.30) − 1 r  2 p+1 Ri (m) p+1 2a −1 Ri (m)−q m 2 . = m p+1 (3.21) and (3.30) imply that (3.31) Gi (m) = −1  p + 1 1  q . q− Ri (m)q−p 1 − Ri (m)2q−p−1 am 2 a The first term on the right hand side of (3.29) tends to a finite limit as m → c(a) (see (3.20)), while G1 (m) → +∞, G2 (m) → −∞ since R1 (m) ≤ Ri (m) → a 1 2q−p−1 q a q 1 2q−p−1 as m → c(a) ≤ R2 (m), as m → c(a) .  62 M. CHIPOT, M. FILA and P. QUITTNER Lemma 3.5. Assume that q < p < 2q − 1. Then L1 has a unique minimum in (0, c(a) ). Proof. It suffices to prove that L′′1 (m) > 0 at any point m where L′1 (m) = 0. We first rewrite (3.14) in the following form: r (3.32) p+1 p−1 2a m 2 L′1 (m) = − I(m) + J(m), p+1 2 where I(m) := Z ̺ 1 dV √ , p+1 V −1 m̺′ , J(m) := p ̺p+1 − 1 ̺ := R1 (m) . m Differentiating (3.32) and multiplying the result by m, we get (3.33) r  p+3 m̺′′ (p + 1)̺p m̺′  2a p2 − 1 . m 2 L′′1 (m) = I(m)+J(m) 1−p+ ′ − p+1 4 ̺ 2(̺p+1 − 1) Let us now compute m̺′ , m̺′′ . From (3.6) we obtain that ̺′ R1 (m)2q−p−1 = 2a ̺p+1 − 1 . p + 1 ̺p+1 (3.15) and the last equality yield (3.34) m̺′ = R1′ (m) − = 1 1 R1 (m) = p −̺ q m ̺ 1 − a R1 (m)2q−p−1 ̺(̺p+1 − 1) , k − ̺p+1 k := 2q . 2q − p − 1 Further (m̺′ )′ = ̺′ + m̺′′ , hence (3.35) m̺′′ (p + 2)̺p+1 − 1 (p + 1)̺p+1 (̺p+1 − 1) = −1 + + . ̺′ k − ̺p+1 (k − ̺p+1 )2 If L′1 (m) = 0, then (3.36) J(m) = p−1 I(m). 2 Using (3.34)–(3.36) we obtain from (3.33) that r  1 − p (p + 3)σ − 2 (p + 1)σ(σ − 1)  p+3 2a , σ := ̺p+1 . m 2 L′′1 (m) = J(m) + + p+1 2 2(k − σ) (k − σ)2 63 BLOW UP FOR PARABOLIC EQUATIONS We will be done if we show that the expression in big brackets is positive. It can be easily seen that this holds if and only if σ(3kp − 2p + k) > k 2 (p − 1) + 2k. (3.37) To prove (3.37) we need the following lower bound for σ: σ≥ (3.38) p−1 k 2p if L′1 (m) ≥ 0. To derive (3.38) we use successively (3.34), the nonnegativity of L′ and an obvious inequality: p Z dV ̺(̺p+1 − 1) m̺′ p−1 ̺ p − 1 ̺p+1 − 1 √ p . =p ≥ ≥ 2 p+1 ̺p V p+1 − 1 (k − ̺p+1 ) ̺p+1 − 1 ̺p+1 − 1 1 Now an easy calculation yields (3.38). According to (3.38) it is sufficient to prove that
p−1 2kp − 2p + k(p + 1) > k(p − 1) + 2. 2p Writing this inequality in the form k(p − 1) − (p − 1) + k we see that it holds if k > q < p. p2 − 1 > k(p − 1) + 2 2p 1 2p 2p 1 > = . But k = , since p+1 p+1 p−1 p−1 1 − 2q 1 − 2p  The results of Lemmas 3.1–3.5 are summarized in the following figures. L L2 (m) L1 (m) L 1 a L2 (m) L1 (m) c(a) m 0 0 p<q L L∗ c(a) m 0 p=q Figure 9. The graphs of Li (m). L2 (m) L1 (m) c(a) m q < p < 2q − 1 64 M. CHIPOT, M. FILA and P. QUITTNER Concerning symmetric solutions to (3.4) we have the following theorem. Theorem 3.3. (i) If p < q, then for any l > 0 the problem (3.4) has a unique positive symmetric solution. (ii) If p = q, then for any l > 1/a the problem (3.4) has a unique positive symmetric solution, while for l ≤ 1/a there are no positive solutions. (iii) If q < p < 2q − 1, then there is a number L∗ (depending on (a, p, q) such that for l > L∗ there are exactly two positive symmetric solutions, for l = L∗ there is a unique positive symmetric solution and for l < L∗ there are no positive solutions. Proof. It is an immediate consequence of Lemmas 3.1–3.5. We only remark that the nonexistence results in (ii), (iii) hold also for nonsymmetric solutions (recall the observations at the beginning of this section).  Now we turn to the study of nonsymmetric solutions. From the fact that for p < 2q − 1 and 0 < m < c(a) there are two values L1 (m), L2 (m) such that (3.3) holds it follows that there is at least one pair of nonsymmetric solutions for l = 12 (L1 (m) + L2 (m)). The following lemma is motivated by the question, whether this pair is unique. Lemma 3.6. Assume that p < 2q − 1 and either p ≤ 4 or p > 4, q ≥ 1 . Then p−1− p−2 (3.39) L′1 (m) + L′2 (m) < 0 for m ∈ (0, c(a) ). Proof. According to (3.29) a sufficient condition for (3.39) is that G1 (m) + G2 (m) ≤ 0. (3.40) By (3.31) this is equivalent to

F ′ R2 (m) F ′ R1 (m) + ≤ 0, R1 (m)q R2 (m)q where F is defined by (3.8). Setting H(y) := p+1 2q p + 1 q+1 − y q+1 + mp+1 , y 2a we obtain that (q + 1)H ′ (y) = x−q F ′ (x) if y = xq+1 , hence (3.40) holds if and only if (3.41) H ′ (y1 ) + H ′ (y2 ) ≤ 0, yi := Ri (m)q+1 , i = 1, 2. BLOW UP FOR PARABOLIC EQUATIONS 65 Now we show that if H ′′′ (y) < 0 for y ∈ [y1 , y2 ], then (3.41) holds. To do this we first observe that yo − y1 ≤ y2 − yo , (3.42) where yo is the unique point where H ′ (yo ) = 0. Indeed, from Taylor’s theorem we have 1 0 = H(yi ) = H(yo ) + H ′′ (θi )(yi − yo )2 , 2 (3.43) θi lies between yi and yo . From (3.43) it is easily seen that yo − y1 = s H ′′ (θ2 ) (y2 − y1 ) H ′′ (θ1 ) and (3.42) follows from the assumption on H ′′′ . Suppose now that H ′ (y2 ) > −H ′ (y1 ). Then (3.44) H ′ (y2 − η) > −H ′ (y1 + η) for η ∈ [0, yo − y1 ] since H ′′ (y2 − η) < H ′′ (y1 + η). But (3.44) leads to
H y2 − (yo − y1 ) < H(yo ) what is a contradiction. Suppose now that there is a point m such that L′1 (m) + L′2 (m) ≥ 0 (hence  R (m) ′ 2 < 0 that L′1 (m) ≥ 0). For such m we get using (3.34) and the fact that m p (m) Z ̺  Z R2m ̺ ̺p+1 − 1 p−1 dV m̺′ dV √ √ > =p + k − ̺p+1 2 V p+1 − 1 V p+1 − 1 ̺p+1 − 1 1 1 p Z ̺ p − 1 ̺p+1 − 1 dV √ . ≥2 ≥ (p − 1) p+1 ̺p V p+1 − 1 1 This implies that ̺p+1 ≥ 2(p − 1) 4(p − 1)q k= . 3p − 1 (3p − 1)(2q − p − 1) By (3.6) we have   p+1 ̺p+1 1 − R1 (m)2q−p−1 = 1, 2a 66 M. CHIPOT, M. FILA and P. QUITTNER hence (3.45) 2q−p−1 q+1 y1 = R1 (m)2q−p−1 ≥ 2a  (3p − 1)(2q − p − 1)  1− . p+1 4q(p − 1) If we show that H ′′′ (y) < 0 for y ∈ [y1 , y2 ] then we arrive at a contradiction. A straightforward calculation yields that (3.46) 2q−p−1 q+3 (q + 1)3 q+1 2q − H ′′′ (y) = − (q − 1) + (p − q)(2q − p + 1)y q+1 . y p+1 a Taking (3.45) into account we see, that we need only to consider y − 2q−p−1 q+1 ≤ 2q(p − 1) . a(3p − 2q − 1) The right hand side of (3.46) is then nonpositive if (p − q)(2q − p + 1) p−1 ≤ q − 1. 3p − 2q − 1 By straightforward calculations it can be shown that the last inequality holds if and only if (2q − p − 1)(p2 − pq + 2q − 3p + 1) ≤ 0. The first term is positive and the second one is nonpositive if and only if (3.47) q(p − 2) ≥ p2 − 3p + 1. 1 If p > 4 and q ≥ p − 1 − p−2 , then (3.47) is easily seen to hold. Consider now p ≤ 4. If p ≤ 2, then q(p − 2) ≥ p(p − 2) > p2 − 3p + 1. If 2 < p ≤ 4, then  q(p − 2) > 21 (p + 1)(p − 2) ≥ p2 − 3p + 1. Remark 3.1. The method of proof of Lemma 3.6 does not work for any p < 2q − 1, since for q > 3 there exists p ∈ (q, 2q − 1) such that H ′′′ (yo ) > 0. Theorem 3.4. Assume that p < 2q − 1. Then the following holds: (i) There is a number L∗∗ ∈ (0, L(a) ] (which depends on a, p, q) such that for any l > L∗∗ the problem (3.4) has at least one pair of positive nonsymmetric solutions u1 , u2 , u1 (x) = u2 (−x) for x ∈ [−l, l], while for l < L∗∗ there are no positive nonsymmetric solutions. 1 , then L∗∗ = L(a) and the pair of (ii) If p ≤ 4 or p > 4, q ≥ p − 1 − p−2 nonsymmetric positive solutions is unique. Proof. In order to prove (i) we need only to show that the range of 21 (L1 + L2 ) contains the interval (L(a) , ∞). This follows from (3.20), (3.25). Lemma 3.6 implies (ii).  BLOW UP FOR PARABOLIC EQUATIONS 67 In sections 4 and 5 it will be important to know how are the stationary solutions ordered. For p ≥ 2q − 1 we have shown that there is at most one positive solution, for p ≤ q it follows from Theorem 2.1(i) that any two positive solutions cross each other. Concerning the remaining case q < p < 2q − 1 we have the following result. Proposition 3.1. Assume that q < p < 2q − 1, l > L∗ . Let u1 , u2 be the two symmetric solutions from Theorem 3.3(iii), m1 = u1 (0) < u2 (0) = m2 . Then (i) u1 < v (i.e. u1 (x) < v(x), x ∈ [−l, l]) for any positive solution v, v 6≡ u1 , (ii) any nonsymmetric positive solution crosses u2 , (iii) any two nonsymmetric positive solutions cross each other. Proof. (i) We show first that u1 < u2 . Suppose there is a point xo ∈ (0, l] such that u1 (xo ) = u2 (xo ), u1 (x) < u2 (x) for x ∈ [0, xo ). Set w := u2 − u1 . Then wx (0) = 0 and wxx (x) > 0 for x ∈ [0, xo ), hence wx (x) > 0 for x ∈ (0, xo ). But then w(xo ) > w(0) > 0, a contradiction. Let now v be an arbitrary nonsymmetric solution. If v ≥ u1 then v > u by the maximum principle. Suppose
there is a point xo ∈ [−l, l] such that u1 (xo ) > v(xo ). Set w(x) := min v(x), u1 (x) . Then (3.47a) w ≤ u1 , w 6≡ u1 , w is a supersolution. The problem (1.1) generates a strongly monotone compact local semiflow in C + := {v ∈ W 1,2 (Ω) ; v ≥ 0} (see Proposition 5.1) and it is easily seen that the subset Cs+ := {v ∈ C + ; v(x) = v(−x)} is invariant. The zero solution is unstable from above (Theorem 2.1(iii)), the W 1,2 –norm of any orbit can be estimated in terms of its sup–norm (see (4.10)), therefore there is an orbit lying in Cs+ which connects 0 to u1 ([M, Theorem 8]), a contradiction to (3.47a). (It is not difficult to see that Theorem 8 from [M] is applicable in our case, although it was formulated in [M] only for semiflows on whole Banach spaces.) (ii) Let v be an arbitrary nonsymmetric solution. Suppose v does not cross u2 , i.e. either v ≥ u2 or v ≤ u2 . In both cases we arrive at a contradiction, because according to [M, Theorem 8] there are orbits (in Cs+ ) which connect u2 to u1 and to ∞. (iii) Let v1 , v2 be nonsymmetric solutions, v1 6≡ v2 . If v1 (x) = v2 (−x), then they cross at x = 0. Assume now that there is a point xo such that v1 (xo ) 6= v2 (−xo ). Then v1 , v2 lie on two different trajectories of the planar system u′ = w, w′ = aup . The phase portrait for this problem is depicted in Figure 10. Trajectories going through the points (m, 0), m < c(a) (cf. (3.17)), cross both of the curves u′ = uq , u′ = −uq exactly twice. The trajectory going through (c(a) , 0) hits any of the curves u′ = uq , u′ = −uq exactly once. Trajectories going through the points (m, 0), m > c(a) , cannot yield solutions to (3.4). 68 M. CHIPOT, M. FILA and P. QUITTNER u′ u′ = uq c(a) 0 u u′ = −uq Figure 10. The phase portrait for u′ = v, v ′ = aup . If a pair of nonsymmetric solutions is not unique, then there are two trajectories that need the same “time” to go from the first intersection with u′ = −uq to the first intersection with u′ = uq . It is easy to see that it is sufficient to consider v1 , v2 as depicted in the following Figure 11. u′ u′ v2 u v1 v2 u v1 Figure 11. Two nonsymmetric solutions v1 , v2 . In both cases v1 (−l) > v2 (−l) and v1 (l) < v2 (l), i.e. v1 , v2 cross each other.  Now we turn to the investigation of the Morse indices of the stationary solutions. These results will be used in Section 5. BLOW UP FOR PARABOLIC EQUATIONS 69 Theorem 3.5 (i) The symmetric stationary solutions are hyperbolic except of the cases p < 2q − 1, m = c(a) q < p < 2q − 1, l = L∗ . or (ii) The nonsymmetric stationary solutions are hyperbolic if they correspond to m such that L′1 (m) + L′2 (m) 6= 0 (cf. Lemma 3.6). Proof. Let v(x; µ) be a solution of (3.1) with v(−l, µ) = µ > 0, vx (−l, µ) = −µq . We have to show that the linearized problem w′′ = apv p−1 w, (3.48) (3.49) (3.50) ′ x ∈ (−l, l) w (−l) = − qv(−l)q−1 w(−l) w′ (l) = qv(l)q−1 w(l) cannot have a nontrivial solution, if µ is such that vx (l; µ) = v q (l, µ). Obviously, w(x) = vµ (x; µ) satisfies (3.48), (3.49). Since (3.48) is a linear second order equation, any solution of (3.48), (3.49) must be a scalar multiple of vµ (x; µ). Assume now that v corresponds to a symmetric solution u, hence l = Li (m) for i = 1 or 2 (or l = L(m)). The numbers m = v(0; µ) and µ are related (cf. (3.6)) by the equation p + 1 2q µ . mp+1 (µ) = µp+1 − 2a Differentiating the equality vx (Li (m); µ) = v q (Li (m); µ) with respect to µ, we obtain that (3.51) vxµ − qv q−1 vµ = −1 ′ p q 2q−1 L (µ − µ )(vxx − qv q−1 vx ). mp i a Since vxx (Li (m); µ) − qv q−1 vx (Li (m); µ) = avp (Li (m); µ) − qv 2q−1 (Li (m); µ), we see that the right hand side of (3.51) is nonzero under the assumptions of the first part of the theorem (L′1 vanishes if and only if q < p < 2q − 1, l = L∗ ; aµp − qµ2q−1 = avp (Li (m); µ) − qv 2q−1 (Li (m); µ) vanishes if and only if m = c(a) ). Hence, (3.48)–(3.50) has no nontrivial solution. 70 M. CHIPOT, M. FILA and P. QUITTNER To prove (ii) we can argue exactly as before with the only difference that now 2l = L1 (m) + L2 (m).  In the remainder of this section we shall work with a fixed length l, we shall vary the parameter a and we shall use the notations introduced in Section 2. We want to use the bifurcation diagrams shown in Section 1; their correctness is shown 1 )) by Theorems 3.1–3.4 and (except of the case 2q − 1 > p > max(4, q + 1 + p−2 by the following lemma: Lemma 3.7. Let ua be (any) positive solution of (3.4) and let L(a) be as in (3.20) (if p < 2q − 1). Then we have d L(a) < 0, lim L(a) = 0, lim L(a) = +∞. (i) If p < 2q − 1, then a→∞ a→0+ da (ii) If p < q and a → 0+ or if p = q and a → 1l +, then kua k → 0. d (iii) If q < p < 2q − 1 and m ∈ (0, c(a) ) is fixed, then L1 (m) < 0, da   min L1 (m) = +∞. lim 0<m<c(a) a→0+ (iv) If p > 2q − 1 (or if p = 2q − 1 and a > q), then lim kua k = +∞ (or a→0+ lim kua k = +∞) and a→q+ lim kua k = 0. a→∞ (v) If p < 2q − 1, then ua → ua1 in W 1,2 (Ω) as a → a1 +, where ua1 is the maximal positive solution of (3.4) and a1 is as in Figs. 1–3. Proof. (i) Follows immediately from (3.10) and (3.17). (ii) By the same way as in (2.29) we obtain (3.52) u2q−p−1 (l) ≤ 2a , p+1 hence kua k → 0 if a → 0+ and kua k is bounded if a → 1l +. If p = q, a → 1l + and ua 6→ 0, choose a sequence an ↓ 1l such that kun k → c > 0 (where un := uan ). We may suppose un ⇀ u (weak convergence) and passing to the limit in the equality un = Kun − an P (un ) + Q(un ) we get un → u = Ku + 1l P (u) + Q(u), which contradicts Theorem 3.3(ii). (iii) Using (3.7) we obtain r p + 1 −(p+1)/2 1 1 d d q L1 (m) = − L1 (m) + m R1 (m)
p+1 da 2a 2a da R1 (m) − 1 m and differentiating (3.6) we get p + 1 R1 (m)2q d
< 0, R1 (m) = da 2a2 F ′ R1 (m) 71 BLOW UP FOR PARABOLIC EQUATIONS d L1 (m) < 0. da Now suppose ak → 0 and hence min L1 (m) < l < +∞, where ck = cak . Then 0<m<ck ck → 0 by (3.17), hence there exists a sequence uk of positive solutions of (3.4) with a = ak and uk (0) = mk → 0. Since ak → 0, this implies kuk k → 0, which contradicts Lemma 2.2. (iv) The estimate (3.52) implies kua k → +∞ for a → 0+ and p > 2q − 1. If p = 2q − 1, ak → q+ and kuk k < c (where uk = uak and c is a constant), then choosing a weakly convergent subsequence we get (as in the proof of (ii)) uk → u, where u is a positive solution corresponding to a = q. Now Theorem 3.2(i) and Lemma 2.2 yield a contradiction. Finally, choose ε > 0 and choose any positive function u : [−l, l] → (0, ε) fulfilling the boundary conditions in (3.4). Then uxx ≤ aup for sufficiently large a, hence ua ≤ u < ε, which implies kua k → 0 for a → ∞. (v) This follows from the continuous dependence of Li (m), L(a) and c(a) on a and from the continuous dependence of the solution of (3.1)–(3.2) on m.  If u is a solution of (3.4) (or, equivalently, (2.2)), then the number of the negative or zero eigenvalues of the operator F ′ (u) = I − K + aP ′ (u) − Q′ (u) (where I denotes the identity), will be denoted by M − (u) or M o (u), respectively. Recall that any eigenvalue λ 6= 1 of F ′ (u) is simple since the corresponding eigenvector is a solution of a second order linear differential equation with a fixed boundary condition. Moreover, the variational characterization of eigenvalues of F ′ (u) gives us immediately the continuous dependence of these eigenvalues on the solution u, which implies (3.53) (3.54) M − (un ) → M − (u)
0 ≤ lim M − (un ) − M − (u) ≤ 1 n→∞ if un → u and M o (u) = 0 if un → u and M o (u) 6= 0 Finally, if M o (u) = 0, then the degree d(u) is well defined and d(u) = (−1)M − (u) . Theorem 3.6. Let ao and a1 be as in Figs. 1–5. (i) Let p ≤ q. If u is a positive symmetric solution of (3.4), then M − (u) = 1 for a ≤ a1 and M − (u) = 2 for a > a1 . Moreover, M o (u) = 0 if a 6= a1 . If u is a positive nonsymmetric solution of (3.4), then M − (u) = 1, M o (u) = 0. (ii) Let q < p < 2q − 1 and let the assumptions of Theorem 3.4(ii) be fulfilled. Let a > ao and let u1 < u2 be the two corresponding symmetric positive solutions of (3.4). Then M − (u1 ) = M o (u1 ) = 0, M − (u2 ) = 1 for a < a1 , M − (u2 ) = 2 for a > a1 and M o (u) = 0 if a 6= a1 . If u is a positive nonsymmetric solution of (3.4), then M − (u) = 1, M o (u) = 0. (iii) Let p ≥ 2q − 1 and let u be a positive solution of (3.4). Then M − (u) = M o (u) = 0. (iv) M − (0) = 0 and M o (0) = 1 for any p, q. 72 M. CHIPOT, M. FILA and P. QUITTNER Proof. (i) Choose a > ao sufficiently close to ao . Using Lemma 2.6, Lemma 2.3 and the homotopy invariance property of the degree one easily gets + + + 0 = d+ ao (0) = da (u) + da (0) = da (u) + 1, where u is the unique positive solution of (3.4), hence d+ a (u) = da (u) = −1. Since u → 0 for a → ao and M − (0) = 0 6= M o (0), we get using (3.54) (which is easily seen to hold also for varying a) and Theorem 3.5 that M − (u) = 1 and M o (u) = 0. By Theorem 3.5 this will hold for any a < a1 . Now choose a− < a1 < a+ close to a1 and let u− , u+ be the corresponding + positive symmetric solutions and u+ 1 , u2 the corresponding positive nonsymmetric solutions (for a = a+ ). Using the homotopy invariance of the degree we get (3.55) + + + −1 = da− (u− ) = da+ (u+ ) + da+ (u+ 1 ) + da+ (u2 ) = da+ (u ) + 2da+ (u1 ) + due to the symmetry u+ 1 (x) = u2 (−x). Theorem 3.5 implies |da+ (u+ )| = |da+ (u+ 1 )| = 1, so that (3.55) yields da+ (u+ ) = 1, da+ (u+ 1 ) = −1. Repeating our considerations with a close to ao for F/Xs , where Xs is the space of symmetric functions from X, we get dsa (u) = −1 and (M s )− (u) = 1 for any a > ao and any positive symmetric solution u (where ds and M s is the degree and the Morse index corresponding to F/Xs , respectively), hence M − (u+ ) ≥ (M s )− (u+ ) = 1. Since da+ (u+ ) = 1 and M − (u+ ) ≤ 2 by (3.54), + we have M − (u+ ) = 2. Finally, (3.54) and da+ (u+ 1 ) = da+ (u2 ) = −1 imply − + − + M (u1 ) = M (u2 ) = 1. (ii) We have M o (u1 ) = 0 by Theorem 3.5. If M − (u1 ) > 0, then this would imply the existence of a positive (symmetric) solution lying between 0 and u1 (see Lemma 2.4 and the proof of Theorem 2.2(ii)). Hence M − (u1 ) = 0, d(u1 ) = da (u1 ) = 1. Let uo be the unique positive solution for a = ao and choose a− < + ao < a+ sufficiently close to ao , ε > 0 small. If u+ 1 < u2 are the positive solutions + corresponding to a , we have + 0 = deg(Fa− , 0, Bε (uo )) = deg(Fa+ , 0, Bε (uo )) = da+ (u+ 1 ) + da+ (u2 ) = 1 + da+ (u+ 2 ), − + hence da+ (u+ 2 ) = −1 and using (3.54) and Theorem 3.5 we get M (u2 ) = 1. The rest of the assertion can be proved analogously as in (i). (iii) The assertion can be proved by the same way as the equality M − (u1 ) = o M (u1 ) = 0 in the proof of (ii). (iv) This is trivial.  Corollary. If p < 2q − 1 and u is a maximal non-negative solution, then u is unstable from above. BLOW UP FOR PARABOLIC EQUATIONS 73 Proof. If M − (u) > 0, then the instability from above of u follows by the same way as in the proof of Theorem 2.1(i), since the eigenvector corresponding to the first eigenvalue of F ′ (u) is positive. If M − (u) = 0, then p ≥ q, a ≤ ao and either u = 0 or p > q, a = ao . In both cases d(u) 6= 1, hence u is unstable (see [Q3]). Assume now that u is stable both from above and from below and choose ε > 0. Then there exists δ > 0 such that the solution u(t, uo ) of (1.1) stays in Bε (u) = {v ∈ W 1,2 (−l, l) ; kv − uk < ε} whenever uo ∈ Bδ (u) and either uo ≥ u or uo ≤ u. Choosing u− , u+ ∈ Bδ (u) such that u− < u < u+ in [−l, l] we can find ν > 0 such that u− < v < u+ for any v ∈ Bν (u), since W 1,2 (−l, l) ⊂ C([−l, l]). The monotonicity of the flow (see the proof of Proposition 5.1) implies ku(t, v)−ukL∞ (−l,l) ≤ ku(t, u+ )−u(t, u−)kL∞ (−l,l) ≤ cku(t, u+ )−u(t, u−)k ≤ 2εc for any v ∈ Bν (u), and the variation-of-constants formula from [A1] implies (3.56) ku(t, v) − uk ≤ C(̺) sup t−̺≤τ ≤t ku(τ, v) − ukL∞ (−l,l) ≤ 2εcC(̺) for any v ∈ Bν (u), where C(̺) → +∞ as ̺ → 0+. Fix ̺ > 0. Taking ν > 0 smaller, if necessary, we may suppose u(t, v) ∈ Bε (u) for v ∈ Bν (u) and t ∈ [0, ̺] (see Proposition 5.1). This estimate together with (3.56) imply u(t, v) ∈ Bε̃ (u) for v ∈ Bν (u) and any t > 0 (where ε̃ = ε max(1, 2cC(̺))), which gives us a contradiction with the instability of u. Consequently, u is unstable from above or from below. If u = 0 then the instability from above follows from the fact that the functional Φ corresponding to (2.1) is even. If p > q, a = ao , then u is stable from below by [M, Theorem 8], hence it is unstable from above.  4. Blow up and global existence In this section we consider the problem (4.1) ut = △u − aup x ∈ Ω, t > 0, ∂u = uq x ∈ ∂Ω, t > 0,  ∂n   u(x, 0) = uo (x) ≥ 0 x ∈ Ω,     with p, q > 1, a > 0, Ω ⊂ RN is a smoothly bounded domain. We assume that ∂uo = uqo , x ∈ ∂Ω uo ≥ 0 is smooth enough and that the compatibility condition ∂n is satisfied. By a solution we mean a nonnegative classical solution. Theorem 4.1. If Ω is the unit ball B1 (0) and p > 2q − 1 or p = 2q − 1 and a>q 74 M. CHIPOT, M. FILA and P. QUITTNER then the solution u exists globally and stays uniformly bounded for any uo . Proof. For any uo there is a smooth function vo satisfying the compatibility condition and such that uo (x) ≤ vo (x) = Vo (|x|) for |x| ≤ 1. We shall construct a sequence {wn (r)} such that for n large enough wn (r) ≥ Vo (r) (4.2) and N −1 ′ wn (r) − awnp (r) ≤ 0, r wn′ (0) = 0, wn′ (1) = wnq (1). wn′′ (r) + (4.3) (4.4) r ∈ (0, 1), The maximum principle implies then that the solution emanating from uo stays below wn for t > 0.  q − 1 n 1/(1−q) q−1 Put wn (r) := Cn − r , Cn := + εn , εn > 0, εn → 0 as n n 1/(1−q) n → ∞. Then wn (r) ≥ Cn , hence (4.2) holds for n large enough. q−1 n r . Then Set ϕn (r) := Cn − n wn′ (r) = ϕn (r)q/(1−q) rn−1 and (4.4) follows. Since wn′′ (r) = qϕn (r)(2q−1)/(1−q) r2n−2 + (n − 1)ϕn (r)q/(1−q) rn−2 it suffices to show that (4.5) qϕn (r)(2q−1)/(1−q) r2n−2 + (n + N − 2)ϕn (r)q/(1−q) rn−2 ≤ aϕn (r)p/(1−q) . Multiplying (4.5) by ϕn (r)p/(q−1) , we obtain (4.6) qϕn (r)(p−2q+1)/(q−1) r2n−2 + (n + N − 2)ϕn (r)(p−q)/(q−1) rn−2 ≤ a. If p > 2q − 1, then the left hand side of (4.6) is easily seen converge to zero as n → ∞. If p = 2q − 1, a > q then (4.6) has the form (4.7)  1−   N −2 N − 2  n−2 (q − 1) r2n−2 + nCn 1 + r ≤ a. n n BLOW UP FOR PARABOLIC EQUATIONS 75 It is obvious that it is sufficient to prove (4.7) for r = 1. But for r = 1, (4.7) reduces to q + εn (n + N − 2) < a, a−q and we are done.  n+N −2 In what follows we show that Theorem 3.1 is sharp. More precisely, we prove that for Ω = B1 (0), p < 2q − 1 or p = 2q − 1, a < q blow up occurs, while for N = 1, p = 2q − 1, a = q all solutions are global but unbounded. We are also interested in all possible types of behaviour of solutions to (4.1). Three possibilities are conceivable: (i) global existence and boundedness, (ii) blow up in finite time, (iii) global existence without uniform boundedness. In several cases we will be able to prove that the third possibility cannot occur. therefore we only need to choose εn < Theorem 4.2. Assume that N = 1 (Ω = (−l, l)) and p < 2q − 1 or p = 2q − 1, a < q. Then (i) any global solution is uniformly bounded in X = W 1,2 , i.e. sup ku(·, t)kX < ∞. t>0 (ii) If uo ≥ v, uo 6≡ v, where v is any maximal stationary solution, then u blows up in a finite time. Remark 4.1. Under the assumptions of Theorem 4.2 we have the following list of maximal stationary solutions. The trivial solution is maximal if 1 (Theorem 3.3(ii)), a q < p < 2q − 1, a > 0, l < L∗ (Theorem 3.3(iii)), p = q, a > 0, l ≤ or or p = 2q − 1, a < q, l > 0 (Theorem 3.2(i)). Any positive solution is maximal if or or p < q, a > 0, l > 0 (Theorem 3.3(i)), 1 p = q, a > 0, l > (Theorem 3.3(ii)), a q < p < 2q − 1, a > 0, l = L∗ (Theorem 3.3(iii)). Except of the minimal symmetric solution, any nontrivial solution is maximal if q < p < 2q − 1, a > 0, l > L∗ (Proposition 3.1). 76 M. CHIPOT, M. FILA and P. QUITTNER Proof of Theorem 4.2. (i) We proceed by contradiction. Suppose that u is a global solution which is unbounded in the W 1,2 –norm (which we denote similarly as in Section 2 by k · k). Then one of the following possibilities must occur: lim ku(·, t)k = ∞ (4.8) t→∞ or lim sup ku(·, t)k = ∞, lim inf ku(·, t)k < ∞. (4.9) t→∞ t→∞ Exactly in the same manner as in [F], the variation of constants formula from [A1] can be used to prove that for any constant C large enough, (4.9) implies the existence of a positive stationary solution v with kvk = C. This is impossible, since under our assumptions we have an apriori bound for stationary solutions due to (2.29), (3.52) and Theorem 3.2(i). Suppose now that (4.8) holds. The solution u satisfies the well known energy identity Z tZ (4.10) 0 l −l
u2t dx dt + Φ u(·, t) = Φ(uo ), t > 0, where Φ is the energy functional introduced in Section 2, i.e. Φ(v) = Z l  1  a p+1  1 2 v(l)q+1 + v(−l)q+1 . dx − vx + v p+1 q+1 −l 2 If we set F(v) := v(l)2q + v(−l)2q for v ∈ C([−l, l]), then we get from (4.10) and (4.8) that
F u(·, t) → ∞ as t → ∞. Our next aim is to show that there is a constant co > 0 such that
F u(·, t) ≤ co (4.11) Z l −l u2t dx for t large enough. To do this we first choose for any t > 0 a point xo = xo (t) such that ux (xo , t) = 0 and observe that Z (4.12) l xo since ut ux = 1 2 2 ux − ut ux dx ≥ a p+1 . p+1 u x
1 a u(l, t)2q − u(l, t)p+1 , 2 p+1 77 BLOW UP FOR PARABOLIC EQUATIONS Analogously, (4.13) − Z xo −l a 1 u(−l, t)2q − u(−l, t)p+1 . 2 p+1 ut ux dx ≥ Adding (4.12) to (4.13) and using the assumption on p, q, a we get, that for t large enough (t > 0 if p = 2q − 1, a < q) there exists an ε > 0 such that (4.14)
εF u(·, t) ≤ ≤ l Z |ut ux | dx ≤ −l ε 8l Z Z l ux (l, t) − ux (x, t) = l −l But Z x hence Z l x (4.15) Z l −l −l l −l l −l 1/2 u2x dx u2t dx. l −l  Z ux (x, t) ≤ u(l, t) + 2l Z l 1/2 Z dx  Z ut dx ≥ − 2l q and u2t −l u2x dx + Cε uxx dx ≥ l 1/2 u2t dx , 1/2 u2t dx Z  u2x dx ≤ 4l u(l, t)2q + 2l l −l  u2t . Using (4.14) and (4.15) we obtain (4.11). If we now set f (t) := Z 0 t
F u(·, s) ds, then (4.11) and (4.10) yield that f (t) ≤ co Z tZ 0 l −l 
 u2t dx dt = co Φ(uo ) − Φ u(·, t)
 q+1 2q ≤ c1 + c2 F u(·, t)  for some positive constants c1 , c2 . But this means that there is a constant c3 > 0 such that 2q f ′ (t) ≥ c3 f (t) q+1 for t large enough. Hence f blows up in a finite time what is a contradiction. (ii) We show that (ii) is a consequence of (i). We recall from Section 3 (Corollary of Theorem 3.6) that under our assumptions any maximal solution v is unstable 78 M. CHIPOT, M. FILA and P. QUITTNER from above. By [A1], (4.1) defines a local semiflow in X = W 1,2 (−l, l). The maximum principle implies the strong monotonicity of this semiflow. Hence, according to [M, Theorem 5] there is a function w defined on [−l, l] × (−∞, T ), T > 0, with the following properties: w satisfies the equation wt = wxx − awp in (−l, l) × (−∞, T ), together with the boundary condition wx (±l, t) = ±wq (±l, t) for t ∈ (−∞, T ), further (4.16) for x ∈ [−l, l], −∞ < t1 < t2 < T, w(x, t1 ) < w(x, t2 ) and (4.17) w(·, t) → v in X as t → −∞. Suppose that T = ∞. Then sup kw(·, t)k < ∞, t>0 hence w tends to a stationary solution which is by (4.16), (4.17) greater than v — a contradiction. By the maximum principle u(·, t) > v for t > 0, therefore (4.17) implies the assertion.  Concerning the localization of blow up points we have the following result. Theorem 4.3. Assume that N = 1, Ω = (−1, 1) and p < 2q − 1 or p = 2q − 1, a < q. Let uo 6≡ 0 satisfy the conditions : uo (x) = uo (−x) u′o (x) ≥0 for x ∈ [−1, 1], for x ∈ [0, 1]. If p > q assume further that uo is a subsolution, i.e. u′′o − aupo ≥ 0 u′o (±1) = ± for x ∈ (−1, 1), uqo (±1). Let u blow up. Then u blows up only at the points −1, 1. Proof. We shall use an idea from [FML]. We set J(x, t) := ux (x, t) − ϕ(x)uq (x, t) BLOW UP FOR PARABOLIC EQUATIONS 79 and show that J ≥ 0 in [0, 1] × [0, T ) for a suitably chosen function ϕ; T is the blow up time of u. For given uo we choose ϕ smooth as follows: on [0, 1 − ε], 0 < ε < 1, ϕ=0 ′ ′′ ≥ ϕuqo ϕ, ϕ , ϕ > 0 u′o on (1 − ε, 1], ϕ(1) = ε, on (1 − ε, 1]. With this choice of ϕ we have J(x, 0) ≥ 0 for x ∈ [0, 1], J(0, t) = 0, J(1, t) > 0 for t ∈ [0, T ). If we derive for J a linear parabolic inequality such that the maximum principle enables us to conclude that J ≥ 0 in [0, 1] × [0, T ) then we are done, since then ε ux (x, t) ≥ ϕ(1 − )uq (x, t) 2 ε for x ∈ [1 − , 1) 2 and integrating this, we obtain ε for x ∈ [1 − , 1), 2 u(x, t) ≤ k(1 − x)1/(1−q) Obvious calculations yield ε
1/(1−q) k := (q − 1)ϕ(1 − ) . 2 Jt −Jxx = uxt − uxxx − qϕuq−1 (ut − uxx ) + ϕ′′ uq + 2qϕ′ uq−1 ux + q(q − 1)ϕuq−2 u2x = − apup−1 ux + aqϕup+q−1 + ϕ′′ uq + 2qϕ′ uq−1 ux + q(q − 1)ϕuq−2 u2x . From the definition of J we have −apup−1 ux + aqϕup+q−1 = −aqup−1 J − a(p − q)up−1 ux . If p ≤ q, then Jt − Jxx + aqup−1 J ≥ 0 and we are done. If p > q, then we obtain −a(p−q)up−1 ux = −(p−q) ux ux Jx +(p−q) ut −(p−q)ϕ′ uq−1 ux −(p−q)qϕuq−2 u2x . u u Since ut ≥ 0 by the maximum principle, we arrive at ux Jx + aqup−1 J ≥ u ≥ (3q − p)ϕ′ uq−1 ux + q(2q − 1 − p)ϕuq−2 u2x ≥ 0 Jt − Jxx + (p − q) 80 M. CHIPOT, M. FILA and P. QUITTNER and the proof is finished.  Remark 4.2. If a = 0, Ω = (−1, 1) and uo ≥ 0 fulfils some additional assumptions, then one can use the similarity variables 1 1−x , s = − log(T − t), λ = w(y, s) = (T − t)λ u(x, t), y = √ 2(q − 1) T −t (where T is the blow up time) in order to show that for any y ≥ 0, √ (T − t)λ u(1 − y T − t, t) → wo (y) as t → T, where wo is the unique positive bounded solution of y in (0, ∞), w′′ = w′ + λw 2 w′ (0) = − wq (0), see [FQ]. Repeating formally these considerations also for a > 0, we get the same result if p < 2q − 1, while for p = 2q − 1 we obtain √ as t → T, (T − t)λ u(1 − y T − t, t) → wa (y) where wa is a positive solution of y w′′ = w′ + λw + awp 2 w′ (0) = − wq (0). in (0, ∞), The existence of such a solution for small a > 0 can be shown e.g. by investigation of critical points of the functional Z ∞  1 q+1 a 1 2 λ 2 ̺vy + ̺v + ̺v p+1 dy − v (0), E(v) := 2 2 p + 1 q + 1 0 2 where ̺(y) = e−y /4 (cf. [FQ]). Notice also that this problem does not have positive solution for a > 0 large, since in this case we have E ′ (v)v > 0 for any v 6= 0. Next we turn to the higher dimensional radially symmetric case. Theorem 4.4. Assume that N > 1, Ω = B1 (0) and p < 2q − 1 or p = 2q − 1, a < q. (i) If uo (=uo (r)) is such that (4.18) (4.19) (4.20) N −1 ′ uo − aupo ≥ 0 for r ∈ (0, 1), r u′o (0) = 0, u′o (1) = uqo (1),  2a  1 2q−p−1 uo (1) > 1 and uo (1) > if p < 2q − 1, p+1 u′′o + then u blows up in a finite time. (ii) Initial data uo with the properties required in (i) exist. 81 BLOW UP FOR PARABOLIC EQUATIONS Proof. (i) According to the maximum principle it follows from (4.18),(4.19) that ut ≥ 0. On the other hand, (4.18) yields that (rN −1 u′o )′ ≥ arN −1 up ≥ 0, hence u′o ≥ 0 on [0, 1] because u′o (0) = 0. The maximum principle implies then that ur ≥ 0. R1 We want to show that U (t) := 0 u(r, t) dr satisfies an O.D.E. which has the property that all its positive solutions blow up in finite time. To do this we first derive some estimates. Z 1 Z 1
ut ur dr ≥ urr ur − aup ur dr 0 0 Z 1  a 1 2 ur − up+1 dr = 2 p+1 r 0 a 1 2q up+1 (1, t). ≥ u (1, t) − 2 p+1 If p = 2q − 1, a < q, then a 1 2q u (1, t) − up+1 (1, t) = δu2q (1, t), 2 p+1 δ := 1 a 1− . 2 q If p < 2q − 1, then we use (4.20) to obtain 1 2q a u (1, t) − up+1 (1, t) ≥ δu2q (1, t) 2 p+1 for some δ > 0. In both cases Z (4.21) 1 0 ut ur dr ≥ δu2q (1, t). On the other hand (4.22) Z Z ε Z r Z ε Z r Z 1   uq (1, t) 1 ut dr + ut ut dr dr + ut aup dr dr ut ur dr ≤ N −1 ε ε 0 0 0 0 0 =: I1 + I2 + I3 , where we used the facts that ur ≤ ur (1, t) , rN −1 and that ur = since Z rN −1 ur r 0 urr dr ≤ Z 0
r = (ut + aup )rN −1 ≥ 0, r (ut + aup ) dr. 82 M. CHIPOT, M. FILA and P. QUITTNER Put now ε := u−α (1, t), Then I1 ≤ u2q−1−η (1, t) I2 ≤ Z 1 ut dr 0 I3 ≤ aεup (1, t) 2 Z Z 1 ut dr, , 0 ε ut dr ≤ aup−α (1, t) Z 1 1 ut dr 0 ut dr, Z 1 ut dr 0 ξ := min{α, η}. For t such that 2Au2q−1−ξ (1, t) ≤ we obtain I1 + I2 + I3 ≤ 2 If Z 0 I1 + I2 + I3 ≤ 2Au2q−1−ξ (1, t) + A := max{1, a}, 0 < η < q − 1. 0 ≤ au2q−1−α (1, t) hence q−1−η , N −1 α= Z then Z 1 ut dr, 0 1 Z ut dr 0 1 ut dr 0 2Au2q−1−ξ (1, t) ≥
Z 2 . 1 ut dr, 0 I1 + I2 + I3 ≤ 4Au2q−1−ξ (1, t) In both cases we get from (4.21), (4.22) that Z 1 ut dr. 0 d U (t) ≥ λU (t)µ dt for some λ > 0, µ > 1 which means that u must blow up in a finite time. (ii) We show that there is a number ã > a such that a solution of the equation (4.23) u′′o = ãupo , x ∈ (0, 1) satisfies (4.19), (4.20). Consider first the case p < 2q − 1. By Lemma 3.7 we have L(a) → 0 as a → ∞. Hence, for any ã large enough the equation L2 (m; ã) = 1 BLOW UP FOR PARABOLIC EQUATIONS 83 has a solution m̃. The value at 1 of the solution to (4.23), (4.19) which corresponds to m̃ is equal to R2 (m̃). But R2 (m̃) >  ã  1 2q−p−1 q and the assertion follows. If p = 2q − 1, a < q, then the assertion follows from Lemma 3.7(iv).  For general domains we have the following result. Theorem 4.5. Assume that p ≤ q. Then (i) u blows up if Φ(uo ) < 0. (ii) If u is global, then (4.24) sup ku(·, t)kC(Ω) < ∞, (4.25) sup ku(·, t)kX < ∞, t≥0 t≥0 provided (4.26) N ≤2 or N > 2, q < N . N −2 In the case p = q we assume in addition that a 6= aΩ . (iii) Assume that (4.26) holds. If p = q and a < aΩ then any solution blows up. (iv) Assume that (4.26) holds. If p = q assume in addition that a > aΩ . Then u blows up provided uo ≥ v, uo 6≡ v, v is any positive stationary solution. Remark 4.3. Under the assumption (4.26), the assertion (i) was proved already in [E, Theorem 1.1(a)]. The proof there is similar to ours, it is based on the classical concavity method (see [L]). Proof of Theorem 4.5. We prove first the assertion (ii). Observe that (4.25) implies (4.24), since the trace operator Tr : W 1,2 (Ω) → Lq+1 (∂Ω) is continuous and according to [Fo] ku(·, t)kC(Ω) ≤ c(kuo kC(Ω) , sup ku(·, s)kLr (∂Ω) ) 0≤s≤t if r > (q − 1)(N − 1), N > 1. Exactly by the same reasoning as at the beginning of the proof of Theorem 4.2, it can be seen that we have only to prove that (4.8) leads to a contradiction. To do this we proceed similarly as in [F, Lemma 1]. Put M (t) := Z tZ 0 u2 dx dt. Ω 84 M. CHIPOT, M. FILA and P. QUITTNER Then M ′ (t) = Z u2 dx = Ω Z tZ 0 (u2 )t dx dt + Ω Z Ω u2o dx. Assuming that p < q and setting ε := p − 1 we obtain that (4.27) ε 1 ′′ M (t) = −(2 + ε)Φ(u) + 2 2 Z Ω |∇u|2 dx + q−1−ε q+1 Z uq+1 dS. ∂Ω In what follows, positive constants depending only on a, p, q, Ω, uo will be denoted by ci (i = 1, 2, . . . ). From (4.27) we get M ′′ (t) ≥ c1 ku(·, t)k2X − c2 , (4.28) therefore M ′ (t) → ∞ (4.29) as t → ∞. On the other hand, using (4.10) we obtain from (4.27) that Z tZ   u2t dx dt + c3 M ′ (t) − c4 , M ′′ (t) ≥ 2 (2 + ε) 0 Ω hence 
2 ε M M ′′ − 1 + M′ ≥ 2 Z t Z Z t Z  Z t Z 2  2 2 ≥ 2(2 + ε) + u dx dt (ut ) dx dt − uut dx dt 0 Ω 0 Ω 0 Ω + 2M (c3 M ′ − c4 ) − c5 M ′ . The first term on the right hand side is nonnegative according to the Cauchy inequality and the second one tends to infinity as t → ∞. Thus, there is a to ≥ 0 such that the right hand side is positive for t ≥ to . This implies that (4.30) M −ε/2
′′ <0 for t ≥ to . Since M −ε/2 is decreasing, it must have a root – a contradiction. This proves (ii) for p < q. Let us now prove (i) for p < q. From (4.27) and (4.10) we obtain that M ′′ (t) ≥ −2(2 + ε)Φ(uo ). This again yields (4.29), hence also (4.30) and (i) follows. BLOW UP FOR PARABOLIC EQUATIONS 85 Now we turn to the case p = q. We again prove first (ii) showing that (4.8) 2+ε . Then leads to a contradiction. Choose 0 < ε < p − 1 and set ε̃ = 1 − p+1 Z Z Z ε 1 ′′ M (t) = − (2 + ε)Φ(u) + |∇u|2 dx − aε̃ up+1 dx + ε̃ up+1 dS 2 2 Ω Ω ∂Ω ε Z ε̃ = − (2 + ε)Φ(u) + + ε̃ |∇u|2 dx + M ′′ (t), 2 2 Ω hence (4.31) ε  1 − ε̃ ′′ M (t) ≥ −(2 + ε)Φ(u) + + ε̃ 2 2 Z Ω |∇u|2 dx. Using (4.10) we get (4.32) Z tZ ε Z 1 − ε̃ ′′ 2 (ut ) dx dt + M (t) ≥ −(2 + ε)Φ(uo ) + (2 + ε) + ε̃ |∇u|2 dx. 2 2 0 Ω Ω Our next aim is to show by contradiction that M ′′ (t) → ∞ as t → ∞. Suppose that there exist a c > 0 and a sequence {tn }, tn → ∞, such that (4.33) M ′′ (tn ) ≤ c for n ∈ IN. Z 1 u(x, t) dx and write u in the form u = d+u⊥ , where u⊥ belongs |Ω| Ω to the subspace of functions Rin X which are orthogonal to constants. From [N, Theorem 7.1] it follows that ( Ω |∇v|2 dx)1/2 is an equivalent norm for v from this subspace. Thus, (4.32) yields that ku⊥ (·, tn )kX is bounded because Set d(t) := Z Ω ⊥ 2 |∇u | dx = Z Ω |∇u|2 dx. Therefore (4.8) implies that d(tn ) → ∞. Now (4.10) yields that Φ(u(·, tn )) is bounded from above and according to (4.31) it is also bounded from below. Thus (4.34) Setting vn :=
Φ u(·, tn ) → 0. d(tn )p+1 u(·, tn ) we have that vn → 1 in X since d(tn ) kvn − 1kX = ku⊥ (·, tn )kX . d(tn ) 86 M. CHIPOT, M. FILA and P. QUITTNER Under our assumption on p, X = W 1,2 (Ω) is continuously embedded into Lp+1 (Ω) and the trace operator Tr : X → Lp+1 (∂Ω) is also continuous. Hence, (4.34) implies that Z Z
p+1 vnp+1 dS = a|Ω| − |∂Ω| 0 = lim a vn dx − n→∞ ∂Ω Ω what is a contradiction with our assumption on a. Since M ′′ (t) → ∞ as t → ∞, it is possible to find for any C > 0 a τ ≥ 0 such that M ′ (t) ≥ Ct, (4.35) M (t) ≥ Ct for t ≥ τ. From (4.32) it follows that M ′′ (t) ≥ 2(2 + ε) + c6 thus
Z tZ 0 Ω (ut )2 dx dt − c7 , Z tZ 
2
ε M ′ ≥ M c6 (ut )2 dx dt − c7 − c8 M ′ . M M ′′ − 1 + 2 0 Ω (4.36) We want to show that the right hand side of (4.36) is positive for t large enough. To do this, we use the estimate ′ (4.37) M (t) = Z 2 Ω u dx ≤ 2 Z Ω u2o dx +t Z tZ 0 Ω  (ut )2 dx dt which follows from the next simple observation: u(x, t) = uo (x) + Z 0 t ut (x, s) ds ≤ uo (x) + According to (4.37) we obtain M c6 Z tZ 0 Z √ t 0 t
2  12 ut (x, s) ds .
c9 (ut )2 − c7 − c8 M ′ ≥ M M ′ − c10 M − c8 M ′ t Ω  c  c 9 9 M − c8 + M M ′ − c10 . = M′ 2t 2t Now (4.35) yields that the right hand side of (4.36) is positive for t large enough and the proof of (ii) for p = q can be finished in the same way as in the case p < q. To prove (i) for p = q we recall (4.32). It implies now that ′′ M (t) ≥ 2(2 + ε) + c6
Z tZ 0 Ω (ut )2 dx dt + c7 . BLOW UP FOR PARABOLIC EQUATIONS 87 Hence, for any C > 0 there is a τ ≥ 0 such that M (t) ≥ Ct (4.38) for t ≥ τ. The right hand side of (4.36) reads now  Z tZ  M c6 (ut )2 dx dt + c7 − c8 M ′ 0 Ω and using (4.37) we get Z c      Z tZ 1 6 u2o dx + M ′ M − c8 . M c6 (ut )2 dx dt + c7 − c8 M ′ ≥ M c7 − c6 t 2t Ω 0 Ω The estimate (4.38) ensures positivity of the last expression for large t. This completes the proof of (i) and (ii). Proof of (iii). Suppose there is a global solution u. According to (ii) it is bounded, hence its ω–limit set consist of stationary solutions. But the only stationary solution is the unstable trivial solution (see Theorem 2.1(ii)), a contradiction. To prove (iv) we argue similarly. If u were global then its ω–limit set would consist of stationary solutions. But according to Theorem 2.1(i) there is no stationary solution larger than v and v is unstable from above.  Remark 4.4. If p ≤ q then initial functions uo with Φ(uo ) < 0 always exist. If p < q then for any v (Tr v 6≡ 0), Φ(λv) < 0 provided λ is large enough. If p = q then we choose v such that Z Z v p+1 dS. a v p+1 dx < Ω ∂Ω Concerning global existence for general domains we have the following result. Theorem 4.6. Let p, q be such as in Theorem 2.1(v). Then all solutions are global and bounded. Proof. Using the energy identity (4.10), we derive an apriori estimate for ku(·, t)kX . This implies (by the same argument as at the beginning of the proof of Theorem 4.5) that ku(·, t)kC(Ω) is bounded, too. By Lemma 2.7 and Hölder inequality we have that for any ε > 0 small there exists a constant Cε such that Z uq+1 dS ≤ εkuk2 + Cε kukp+1−ε ≤ εkuk2 + εkukp+1 (4.39) q+1 p+1 + Cε , ∂Ω where k · kr denotes the norm in Lr (Ω). Since Hölder inequality implies also kuk22 ≤ εkukp+1 p+1 + Cε , 88 M. CHIPOT, M. FILA and P. QUITTNER we get using (4.39) (4.40) Φ(u) ≥ 1 2  − ε kuk2 − Cε . Since Φ(u) ≤ Φ(uo ), the estimate (4.40) proves our assertion.  Let us now turn to the interesting case p = 2q − 1, a = q. Theorem 4.7. Assume that p = 2q − 1, a = q, N = 1, Ω = (−1, 1). Then there exists a unique function w which satisfies the equation (4.41) w′′ − qw2q−1 = 0 in (−1, 1) together with the boundary condition w(±1) = ∞. (4.42) All nontrivial solutions of (4.1) are global and tend pointwise to w as t → ∞. Remark 4.5. It is known (see [KN] and the references there), that positive solutions of the problem (4.43) (4.44) △u = aup u=∞ x ∈ Ω, x ∈ ∂Ω exist for a > 0, p > 1. In [KN] it is shown that u(x) behaves near ∂Ω like (4.45)  a(p − 1)2  1 1−p 2(p + 1)
2 dist(x, ∂Ω) 1−p . From this it follows that solutions to (4.43), (4.44) are not singular stationary ∂u solutions to the problem (1.1), except of the case a = q, p = 2q − 1, when ∂n behaves like uq near ∂Ω. In [KN] also uniqueness of solutions to (4.43), (4.44) is shown for p ≥ 3. We prove Theorem 4.7 in the following series of lemmas. Lemma 4.1. There is a unique function w which satisfies (4.41), (4.42). Moreover, (4.46) w(x) = w(−x) for x ∈ [−1, 1]. Proof. Denote by ϕα the solution of the initial value problem ϕ′′ = qϕ2q−1 ϕ(0) = α > 0 ϕ′ (0) = 0. 89 BLOW UP FOR PARABOLIC EQUATIONS Then ϕα is given by the formula (cf. (3.5)) ϕα (x) Z (4.47) dv √ = |x|. 2q v − α2q α The function ϕα exists for |x| ≤ Lmax (α) < ∞, Lmax (α) is given by the formula Z Lmax (α) = ∞ α dv 1 √ = q−1 α v 2q − α2q Z ∞ √ 1 dz . z 2q − 1 Lmax (α) is decreasing, it tends to zero as α → ∞ and there is a unique αo such that Lmax (αo ) = 1, namely Z αo = ∞ 0 √ dz z 2q − 1  1 q−1 . Hence, w = ϕαo is the unique solution to (4.41), (4.42) with w′ (0) = 0. The function ϕαo obviously satisfies (4.46). Suppose there is a nonsymmetric solution. Let m be its minimum attained at 0 6= xo ∈ (−1, 1). Instead of (4.47) we obtain now the formula Z ϕm (x) m dv √ = |x − xo |. v 2q − m2q (4.42) implies that Z ∞ m √ dv = |1 − xo | = | − 1 − xo | − m2q v 2q what is a contradiction.  Lemma 4.2. Let p, q, a, N be as in Theorem 4.7 and let αo , ϕα be as in Lemma 4.1. If 0 ≤ uo ≤ w, uo 6≡ 0, then there exist α1 , α2 ∈ (0, αo ) and functions g1 , g2 such that ψi = ϕαi + gi satisfy the conditions (4.48) (4.49) (4.50) (4.51) (4.52) ψ ′′i ψ 1 (x) ≤ u(x, to ) ≤ ψ2 (x) − qψi2q−1 ψ ′i (±1) ≥0 in for some to > 0, (−1, 1), ψiq (±1), = ± ψ i (x) = ψi (−x) ψ ′i (x) ≥ 0 for for |x| ≤ 1, x ∈ [0, 1]. Proof. By the maximum principle 0 < u(x, to ) < w(x)
for any to ∈ 0, tmax (uo ) , 90 M. CHIPOT, M. FILA and P. QUITTNER where tmax (uo ) is the maximal existence time. From (4.47) it is easily seen that ϕα (1) → 0 as α → 0. Therefore, there is a α1 > 0 such that ϕα1 < u(·, to ). On the other hand ϕα (1) → ∞ as α → αo and ϕα (x) → ϕαo (x) pointwise in (−1, 1), hence, there is a α2 > 0 such that ϕα2 > u(·, to ). The functions ϕαi satisfy all conditions except of (4.50). We shall show that it is possible to find functions gi such that ψi = ϕαi + gi satisfies (4.48)–(4.52). Observe first that ϕ′α (1) < ϕqα (1) for any α ∈ (0, αo ). This is easy to see if α is small, because then ϕα (1) is small, let us say ϕα (1) = ε and ϕ′α (1) = Z 1 0 ϕ′′α (x) dx < qε2q−1 < εq = ϕqα (1). But the mapping α 7→ ϕ′α (1) − ϕqα (1) is continuous, therefore its values must be negative for all α ∈ (0, αo ) since there is no β with ϕ′β (1) − ϕqβ (1) = 0. Set ( 0 for |x| ≤ 1 − η, 0 < η < 1 gη,n (x) =
n |x| − 1 + η for 1 − η < |x| ≤ 1 Taking gi = gηi ,ni with ni sufficiently large and suitable ηi (ηi small), it is not difficult to check that the conditions (4.48)–(4.52) are satisfied.  Lemma 4.3. Assume that p, q, a, Ω are as in Theorem 4.7. Let uo (x) = uo (−x) u′o (x) ≥0 for for |x| ≤ 1, x ∈ [0, 1]. Assume further that u(0, t) ≤ K on [0, tmax (uo )) for some K >0 and either (i) ut ≥ 0 in [−1, 1] × [0, tmax ) or (ii) for any t ∈ [0, tmax ) there is a unique point y(t) ∈ (0, 1) such that ut (x, t) < 0 for 0 ≤ x < y(t), ut (x, t) > 0 for y(t) < x ≤ 1. Then tmax = ∞. Proof. Consider the case (ii). Since ux (0, t) = 0 and ux (x, t) ≥ 0 for x ∈ (0, 1], we have uxx (0, t) ≥ 0, hence ut (0, t) ≥ −qu2q−1 (0, t) ≥ −qK 2q−1 . By the maximum principle, there is a constant c1 > 0 such that ut (x, t) ≥ −c1 for |x| ≤ 1, t ∈ [0, tmax ). 91 BLOW UP FOR PARABOLIC EQUATIONS Therefore, for z ∈ (0, 1) we get Z Z z (4.53) ut ux dx ≥ −c1 0 z ux dx = −c1 u(z, t). 0 Further, (4.54) 1 Z ut ux dx = 0 1 2 Z 1 0 u2x − u2q
Z z and (4.55) Z 1 ut ux dx = z Z 1 0 ut ux dx − 0 x dx = 1 1 2q u (0, t) ≤ K 2q 2 2
ut ux dx ≤ c2 1 + u(z, t) . The inequalities (4.53)–(4.55) will be used to derive an apriori estimate of u(1, t). Using (4.55) we get u2x (z, t) 2q = u (z, t) − 2 Z 1 z ut ux dx ≥ u2q (z, t) − 2c2 u(z, t) + 1 1 ≥ u2q (z, t) − c3 . 2
Hence ux (z, t) ≥ (4.56) 1 q u (z, t) − c4 . 2 Using (4.54), (4.53) we get (4.57) 1 2q K ≥ 2 Z 0 1
ut ux dx ≥ −c1 u y(t), t + Z 1 ut ux dx. y(t) By (4.56) we have Z 1 y(t) ut ux dx ≥ Z 1 y(t) Z 1 Z y(t)     1 q 1 q uq −c4 ut dx = u −c4 ut dx− u −c4 ut dx. 2 2 2 0 0 1 Since −c1 ≤ ut (x, t) ≤ 0 for x ∈ [0, y(t)), we obtain Z 0 y(t)   1 q u − c4 ut dx ≤ c1 c4 2 thus (4.58) Z 1 y(t) ut ux dx ≥ d dt Z 1  q+1  u − c4 u dx − c5 . 2(q + 1) 0 92 M. CHIPOT, M. FILA and P. QUITTNER Combining (4.57), (4.58) we get d dt (4.59) Z 1  q+1 
u − c4 u dx ≤ c6 1 + u(1, t) . 2(q + 1) 0 Integrating from 0 to T < tmax and taking into account that ut (1, t) > 0, it follows that (4.60) 1 2(q + 1) Z 1 u q+1 0 (x, T ) dx − c4 Z 1 0
u(x, T ) dx ≤ c6 1 + u(1, T ) T + c7 . By Hölder and Young inequalities c4 Z 0 0 1 u(x, T ) dx ≤ η Z 1 uq+1 (x, T ) dx + cη , η > 0. 0 1 , then (4.60) yields 2(q + 1) If we take η < Z 1
uq+1 (x, T ) dx ≤ c8 1 + u(1, T ) T + c9 for T ∈ (0, tmax ). Suppose that tmax < ∞. Then u(1, t) → ∞ as t → tmax . Therefore, there is a τ ∈ (0, tmax ) such that (4.61) Z 0 1 uq+1 (x, T ) dx ≤ c10 T u(1, T ) for T ∈ (τ, tmax ). Using (4.53), let us now estimate ux from above in the following way: u2x (z, t) = u2q (z, t) − 2 2q Z 1 z ≤ 2u (1, t) + c11 ,
ut ux dx ≤ u2q (1, t) + 2c2 1 + u(z, t) hence ux (z, T ) ≤ 2uq (1, T ) if u(1, T ) is large enough. By the mean value theorem u(1, T ) − u(1 − ε, T ) = εux (ξ, T ) ≤ 2εuq (1, T ), thus (4.62) u(1 − ε, T ) ≥ u(1, T ) − 2εuq (1, T ) ≥ 1 u(1, T ) 2 BLOW UP FOR PARABOLIC EQUATIONS 93 if ε := 41 u1−q (1, T ). Using (4.61), (4.62) we obtain c10 T u(1, T ) ≥ ≥ 1 Z u q+1 0 1 Z 1−ε (x, T ) dx ≥  u(1, T ) q+1 2 Z 1 uq+1 (x, T ) dx 1−ε dx = u2 (1, T ) . 2q+3 This means that u(1, T ) ≤ c11 T, c11 does not depend on T . This is a contradiction. In the case (i), the proof is slightly simpler. We only mention that c1 = 0, hence (4.57), (4.58) are not needed to derive (4.59). The estimate d dt Z 1  q+1  u − c4 u dx ≤ c6 2(q + 1) 0 follows from (4.54), (4.56) in the following way: 1 2q K ≥ 2 Z 0 1 ut ux dx ≥ Z 1 ut 0 Z   d 1  uq+1 uq − c4 = − c4 u dx. 2 dt 0 2(q + 1) 1  Lemma 4.4 Let p, q, a, Ω be as in Theorem 4.7. Then u is global and u(·, t) → w pointwise as t → ∞, provided 0 ≤ uo ≤ w, uo 6≡ 0. Proof. According to Lemma 4.2 we need only to prove that u is global and tends to w if uo = ϕα + gη,n , α ∈ (0, αo ) (with suitable η, n). We first show that the assumptions of Lemma 4.3 are fulfilled. We have u(0, t) ≤ αo for t ∈ (0, tmax ) since (4.63) u(x, t) ≤ w(x) for |x| ≤ 1, t ∈ [0, tmax ) according to the maximum principle. We have also (4.64) because ut ≥ 0 in [−1, 1] × [0, tmax ), u′′o − qu2q−1 ≥0 o u′o (±1) Hence, u is global. = ± in (−1, 1), uqo (±1). 94 M. CHIPOT, M. FILA and P. QUITTNER From the fact that there are no stationary solutions it follows that u cannot be bounded. By (4.63), (4.64) the pointwise limit V exists in (−1, 1). But V must satisfy (4.41), (4.42), hence V = w.  Lemma 4.5. Let p, q, a, Ω be as in Theorem 4.7. Then u is global and u(·, t) → w pointwise as t → ∞, provided uo = k + gη,n , where k is any positive constant and gη,n is from Lemma 4.2, η, n are suitably chosen. Proof. According to Lemma 4.4, we need only to consider k > αo . Obviously, u′o (x) ≥ 0 for x ∈ [0, 1] and it is not difficult to verify that there are η, n such that u′′o − quo2q−1 has exactly one sign change in [0, 1] and uo satisfies the compatibility condition. From the maximum principle it follows that there is at most one sign change of ut (·, t) in (0, 1) for t ∈ (0, tmax ). Take any t1 ∈ (0, tmax ). If ut (x, t1 ) ≥ 0 for x ∈ [0, 1], then u(·, t1 ) ≤ w by the maximum principle and Lemma 4.4 yields the assertion. If ut (x, t1 ) ≤ 0 for x ∈ [0, 1], then ut ≤ 0 in [0, 1] × [t1 , tmax ), hence tmax = ∞ and u tends to a stationary solution from above. But the only stationary solution is 0 and 0 is unstable from above, a contradiction. We only need to consider the case when there is a function y(t) such that ut (x, t) < 0 for x < y(t), ut (x, t) > 0 for x > y(t), t ∈ [0, tmax ). Lemma 4.3(ii) can be applied if we show that u(0, t) ≤ K on [0, tmax ) for some K > 0. Take any to < tmax and choose xo such that w(xo ) > u(1, t) for t ≤ to . By the maximum principle max (u − w) = max (u − w). x=0 0≤x≤xo 0≤t≤to 0≤t≤to
If u(0, t)−w(0) = max u(0, τ )−w(0) , then we use the fact that ux (0, t)−wx (0) = 0 τ ≤t which implies uxx (0, t) − wxx (0) ≤ 0 and uxx (0, t) ≤ wxx (0) = qw2q−1 (0). This yields
ut (0, t) = uxx (0, t) − qu2q−1 (0, t) ≤ q w2q−1 (0) − u2q−1 (0, t) < 0.
This means that max u(0, t)−w(0) = uo (0)−w(0) = k−αo , therefore u(0, t) ≤ k t≤to for t ∈ [0, tmax ).  Proof of Theorem 4.7. Theorem 4.7 is an immediate consequence of Lemma 4.5, since for any uo ≥ 0, uo 6≡ 0 and any to small, there are constants k1 , k2 > 0 and functions g1 , g2 as in Lemma 4.2 such that k1 + g1 ≤ u(·, to ) ≤ k2 + g2 .  BLOW UP FOR PARABOLIC EQUATIONS 95 5. Convergence to equilibria The aim of this section is to study the problem (1.1) from the point of view of dynamical systems. The solution starting from uo will be often denoted by u(t, uo ), by k · k we mean the norm in X = W 1,2 (Ω). Proposition 5.1. The problem (1.1) defines a compact local semiflow in C + = {v ∈ W 1,2 (Ω) ; v ≥ 0 a.e.} provided N = 1, 2 or N > 2, q < N/(N − 2), p < (N + 2)/(N − 2). This local semiflow is monotone in the following sense: if uo ≤ ũo a.e., uo 6≡ ũo , then u(t, uo ) < u(t, ũo ) in Ω for any t ∈ (0, tmax (ũo )). If N = 1, this is the strong monotonicity (u(t, ũo ) − u(t, uo ) lies in the interior of C + ). Proof. A straightforward modification of [A1, Lemma 14.3] in virtue of [A1, Remark 14.7(b)] shows that [A1, Theorem 12.3] is applicable for WBβ = W 1,2 (Ω), i.e. (1.1) (where up := |u|p−1 u and uq := |u|q−1 u) defines a local semiflow in W 1,2 (Ω). Moreover, a repeated use of the variation-of-constants formula [A1, Corollary 12.2] with suitable WBβ = W 1,ri (Ω), 2 =: ro < r1 < · · · < rm =: r > N shows the continuity and boundedness on bounded sets of u(t, ·) : W 1,ri (Ω) → W 1+ε,ri (Ω) ⊂ W 1,ri+1 (Ω), hence u(t, ·) : W 1,2 (Ω) → W 1,r (Ω) is continuous and the flow in W 1,2 (Ω) is compact. Suppose now uo ≤ ũo . It can be easily shown that we may find un , ũn ∈ C 2 (Ω) such that un ≤ ũn , un → uo , ũn → ũo in W 1,2 (Ω) and un , ũn fulfil the boundary ∂u = |u|q−1 u. Using the maximum principle we get u(t, un ) ≤ u(t, ũn ) condition ∂n for t small enough. The continuous dependence on initial values implies now u(t, uo ) ≤ u(t, ũo ). If uo ≥ 0, uo 6≡ 0, then u1 := u(τ, uo ) ∈ W 1,r (Ω) ⊂ C(Ω) is nonnegative and u1 6≡ 0 for τ > 0 small enough, hence we may find ϕ1 ∈ D(Ω) (a smooth function with compact support in Ω) such that ϕ1 6≡ 0, 0 ≤ ϕ1 ≤ u1 . The maximum principle implies u(t, ϕ1 ) > 0 for t ∈ (0, tmax (ϕ1 )), hence u(t + τ, uo ) = u(t, u1 ) ≥ u(t, ϕ1 ) > 0. Consequently, u(t, uo ) is positive for t ∈ (0, tmax (uo )) and [A3, Corollary 9.3 or 9.4] implies that u(·, uo ) is a classical solution of (1.1) for t ∈ (0, tmax (uo )). Finally, let 0 ≤ uo ≤ ũo , uo 6≡ ũo . Then we have u(τ, uo ) ≤ u(τ, ũo ) and u(τ, uo ) 6≡ u(τ, ũo ) for τ small enough. Moreover, u(·, uo ), u(·, ũo ) are classical solutions, hence the maximum principle implies u(t + τ, uo ) < u(t + τ, ũo ) on the time interval where both solutions exist. (4.10) implies then that tmax (uo ) ≥ tmax (ũo ) and the assertion follows.  Remark 5.1. The problem (1.1) defines a strongly monotone compact local semiflow in C + ∩ W 1,r (Ω) for any r > N . 96 M. CHIPOT, M. FILA and P. QUITTNER Theorem 5.1.Denote the set of initial nonnegative data for which the solutions exist globally by G. Then G is star-shaped with respect to zero. Moreover, G is closed in C + provided one of the following conditions holds: N +1 (i) p < q and q < if N > 1, N  N−+12  . (ii) p = q < min 2, N Proof. The fact that G is star-shaped follows from Proposition 5.1. To prove that G is closed we proceed by contradiction. Suppose that u(t, uo ) blows up in a finite time T and that there is a sequence {un } ⊂ G, un → uo in X. By continuous dependence on initial values, it is possible to choose for any K > 0 a t1 < T and an no such that ku(t1 , uno )k = max ku(t, uno )k > K. (5.1) t≤t1 Differentiating the equation with respect to t, multiplying it by ut and integrating, we obtain (5.2)Z Z Z Z Z 1 d uq−1 u2t dS. |∇ut |2 dx − ap up−1 u2t dx + q utt ut dx = − u2t dx = 2 dt Ω ∂Ω Ω Ω Ω By Hölder inequality we have (5.3) Z Z  1s Z uq−1 u2t dS ≤ u2s dS t ∂Ω ∂Ω ∂Ω ′ us (q−1) dS  1′ s for s, s′ > 1, 1 1 + = 1. s s′ Consider first the case N > 1 or N = 1 and p = q. N −1 If N > 1, set s := , θ := 21 + ε. Then the trace operator Tr : W θ,2 (Ω) → N − 2θ N −1 and the trace operator Tr : W 1,2 (Ω) → L2s (∂Ω) is continuous, s′ = 2θ − 1 N −1 ′ Ls (q−1) (∂Ω) is continuous provided N = 2 or N > 2, s′ (q − 1) ≤ 2 , i.e. N −2 (N − 2)(q − 1) 1 (N − 2)(q − 1) θ≥ + . Hence, if N > 2, then we take ε = . 2 4 4 If N = 1, then we choose arbitrary s > 1. With this choice of s, ε, we obtain Z 2(1−θ) , uq−1 u2t dS ≤ c1 kukq−1 kut k2θ,2 ≤ c2 kukq−1 kut k2θ kut k2 ∂Ω where k · kθ,2 or k · k2 denotes the norm in W θ,2 (Ω) or L2 (Ω), respectively. Using Young inequality we obtain Z Z   − θ 2 q−1 2 q−1 1−θ ηkut k + η u ut dS ≤ c2 kuk u2t dx . ∂Ω Ω BLOW UP FOR PARABOLIC EQUATIONS 97 Now (5.2) yields that (5.4) Z Z Z   1 d − θ u2t dx u2t dx + c2 kukq−1 ηkut k2 + η 1−θ u2t dx ≤ − kut k2 + 2 dt Ω Ω Ω Z q−1 1 ≤ − kut k2 + c3 kuk 1−θ u2t dx 2 Ω
q−1 −1 if we put η := . Since u is global, we know from Theorem 4.5(i)
2c2 kuk that Φ u(·, t) ≥ 0 for t ≥ 0, hence Z tZ (5.5) 0 Ω


u2t dx dt = Φ u(·, 0) − Φ u(·, t) ≤ Φ u(·, 0) . Therefore, integrating (5.4) we get (5.6) Z τ 2 0 kut (·, t)k dt + Estimating both R Ωu 2 Z dx and Ω   q−1 u2t (x, τ ) dx ≤ c4 max ku(·, t)k 1−θ + 1 . t≤τ 2 Ω |∇u| dx R as in (4.37) we see that Z  ku(·, τ )k2 ≤ 2 ku(·, 0)k2 + τ τ 0  kut (·, t)k2 dt . Hence, (5.6) yields   q−1 ku(·, τ )k2 ≤ 2τ c4 max ku(·, t)k 1−θ + 1 + c5 . (5.7) t≤τ In the case p = q it is easy to check that 2−q q−1 < 2 if N = 1, 2 and ε < 1−θ 2 (N − 2)(q − 1) . Therefore (5.1), (5.7) yield a contradiction if we 4 choose K large enough and τ = t1 , u(·, 0) = uno . In the case N > 1, p < q, we proceed slightly differently. According to (4.28) or if N > 2, ε = kuk2 ≤ c6 M ′′ + c7 = 2c6 (5.8) Z uut dx + c7 . Ω By Hölder inequality, (4.37) and (5.5), we obtain Z Ω Z tZ 1/2 Z 1/2 √ Z 2 2 uo dx + t u2t dx dt u2t dx Ω 0 Ω Ω Z 1/2 ≤ (c8 + c9 t)1/2 u2t dx . uut dx ≤ Ω 98 M. CHIPOT, M. FILA and P. QUITTNER Now (5.8) implies that 2 ku(·, τ )k ≤ c10 (5.9) Z Ω 1/2 u2t (x, τ ) dx + c7 , where c10 depends on τ . But taking u(·, 0) = uno , τ = t1 we obtain from (5.6), (5.9) that   q−1 ku(t1 , uno )k2 ≤ c11 ku(t1 , uno )k 2(1−θ) + 1 (5.10) what is a contradiction to (5.1), since q−1 < 2 under the assumption N > 1, 2(1 − θ) N +1 . N −1 Consider now the case N = 1, p < q. From (4.10) it follows that Z Z Z a 1 1 |∇u|2 dx + up+1 dx ≤ Φ(uo ) + uq+1 dS c12 kuk ≤ 2 Ω p+1 Ω q + 1 ∂Ω p<q< hence   ku(·, t)k ≤ c13 sup u(x, t) + 1 . x∈Ω This means that for K large enough max ku(t, uno )kC(Ω) cannot be attained for 0≤t≤t1 t = 0. Therefore, there is a t2 ∈ (0, t1 ] for which max u(x, t) = max u(x, t2 ) = max u(x, t2 ) =: U. x∈∂Ω x∈Ω x∈Ω 0≤t≤t1  q+1  1 q−p If U > a |Ω| then there is a δ ∈ (0, 1) such that p+1 Z Z a δ (5.10a) up+1 (x, t2 ) dx ≤ uq+1 (x, t2 ) dS. p+1 Ω q + 1 ∂Ω Since Φ(u) ≥ 0 by Theorem 4.5(i), we obtain from (5.10a) that Z Z 1 1−δ q+1 u (x, t2 ) dS ≤ |∇u(x, t2 )|2 dx. (5.11) q + 1 ∂Ω 2 Ω Taking s′ = Z q+1 , (5.11) yields q−1 ∂Ω ′ us (q−1) (x, t) dS  1′ s ≤ c14 Z Ω  q−1 q+1 |∇u(x, t2 )|2 dx ≤ c14 max ku(·, t)k t≤t1 q−1 2 q+1 for t ≤ t1 . 99 BLOW UP FOR PARABOLIC EQUATIONS By (5.3) we have Z ∂Ω uq−1 u2t dS ≤ c14 Z ∂Ω uq+1 dS t  2 q+1 max ku(·, t)k t≤t1 ≤ c15 kut k2θ,2 max ku(·, t)k t≤t1 q−1 2 q+1 q−1 2 q+1 , θ= 1 + ε. 2 Instead of (5.10) we obtain now by the same arguments as before that   1 q−1 ku(t1 , uno )k2 ≤ c16 ku(t1 , uno )k 1−θ q+1 + 1 , 1 q−1 1 < 2 if ε < and we arrive at a contradiction. 1−θq+1 q+1  q+1  1 q−p The proof is trivial if U ≤ a |Ω| , since then p+1 Z ′ us (q−1) dS ≤ c17 . where ∂Ω  Theorem 5.2. Assume that N = 1 and p < q. Then for any uo ∈ C + , uo 6≡ 0 on ∂Ω, there is a λo > 0 such that u(t, λuo ) → 0 in X as t → ∞ for λ < λo ; u(t, λo uo ) tends to a positive stationary solution as t → ∞, while u(t, λuo ) blows up in finite time for λ > λo . Proof. Set λo = sup{λ > 0 ; u(t, λuo ) exists globally}. From Theorem 4.5(i) and Remark 4.2 it follows that λo < ∞. Set λ1 = sup{λ > 0 ; u(t, λuo ) → 0 in X as t → ∞}. According to Theorem 2.1(i), zero is a stable stationary solution, therefore λ1 > 0. Obviously, λ1 ≤ λo . The domain of attraction of 0 is open in X, hence u(t, λ1 uo ) cannot tend to zero. By Theorem 5.1, u(t, λ1 uo ) is global and according to Theorem 4.5(ii), it is bounded in X. Therefore the ω–limit set ω(λ1 uo ) is nonempty and consists of positive stationary solutions. Since ω(λ1 uo ) is connected and the positive stationary solutions are isolated (Theorems 3.3(i) and 3.4(ii)), ω(λ1 uo ) = {v}, where v is a positive stationary solution. The proof will be finished if we show that λ1 = λo . Suppose λ1 < λo . By Theorem 5.1, u(t, λo uo ) is global, hence u(t, λo uo ) → w as t → ∞, where w is a positive stationary solution. Since (5.13) u(t, λo uo ) > u(t, λ1 uo ) for t > 0, 100 M. CHIPOT, M. FILA and P. QUITTNER we have that w = v, because any two positive stationary solutions must intersect (Theorem 2.1(i)). The stationary solution v is hyperbolic (Theorem 3.5) and unstable (Theorem 3.6(i)). Its stable manifold W s (v) is an immersed submanifold of X with codimension ≥ 1, therefore it cannot contain an open set – a contradiction to (5.13). (For the existence of the local (un)stable manifold see [S, Theorem 5.2], for the globalization see [H, Theorem 6.1.9].)  Let v1 and v2 be stationary solutions. We say that v1 connects to v2 , iff there is an orbit {u(t) ; t ∈ R} such that u(t) → v1 in X as t → −∞, u(t) → v2 in X as t → +∞. For semilinear parabolic equations with homogeneous Dirichlet or Neumann boundary conditions, the connecting orbits problem was solved completely in [BF1],[BF2]. But nonlinear boundary conditions were not considered there. Theorem 5.3. Assume that N = 1, p ≤ q and a > a1 . Let v1 denote the symmetric positive stationary solution and v2 , v3 the nonsymmetric positive stationary solutions. Then (i) vi connects to 0 for i = 1, 2, 3; (ii) v1 connects to v2 and v3 . Proof. (i) follows from [M, Theorem 8] and Theorem 2.1(i). To prove (ii) we first recall that M − (v1 ) = 2 (Theorem 3.6(i)) and that there exists an orbit w lying in the unstable manifold W u (v1 ) such that w blows up in a finite time T (see the proof of Theorem 4.2(ii)). Let Γ = {γ(s) ; s ∈ [0, 1]} be a Jordan curve in W u (v1 ) around v1 such that γ(0) = γ(1) =: γo lies on the orbit which connects v1 to 0. Set so := sup{s ; u(t, γ(σ)) → 0 for σ ∈ [0, s]}. Then so > 0, because the domain of attraction of 0 is open. There is a s1 ∈ (0, 1) such that γ(s1 ) ∈ w, hence so < s1 . By Theorem 5.1, u(t, γ(so )) exists for t ≥ 0 and u(t, γ(so )) converges to a stationary solution vo . The semiflow is gradient like with respect to the functional Φ, therefore vo 6= v1 . This means that v1 connects to v2 or v3 . But if {u(x, t) ; t ∈ R} is an orbit connecting v1 to v2 , then {u(−x, t) ; t ∈ R} connects v1 to v3 .  Theorem 5.4. Assume that N = 1, q < p < 2q − 1, a > ao . Let u1 denote the smaller symmetric positive stationary solution. Then the following holds. (i) Any positive stationary solution v (v 6≡ u1 ) connects to u1 . (ii) Let v be a stationary solution, v > u1 . If 0 ≤ uo ≤ v, uo 6≡ 0, uo 6≡ v, then u(t, uo ) → u1 . (iii) Let all nonsymmetric positive stationary solutions be hyperbolic. Then for any uo ∈ C + , uo > 0, there is a λo > 0 such that u(t, λuo ) → u1 as t → ∞ for λ < λo ; while u(t, λuo ) blows up in finite time for λ > λo . Remark 5.2. The nonsymmetric positive stationary solutions are hyperbolic 1 (see Theorem 3.5 and Lemma 3.6). if e.g. p ≤ 4 or p > 4, q ≥ p − 1 − p−2 BLOW UP FOR PARABOLIC EQUATIONS 101 Proof of Theorem 5.4. (i) Since u1 is stable (see the proof of Theorem 3.6(i)), the assertion follows from Proposition 3.1 and [M, Theorem 8]. (ii) is an immediate consequence of (i). To prove (iii) we define λo , λ1 as in the proof of Theorem 5.2. Since the set of stationary solutions is bounded in L∞ (Ω) (cf. (2.29), (3.52)), Theorem 4.2(ii) implies that λo < ∞. Obviously, λ1 ≤ λo and according to (ii), λ1 > 0. Suppose that λ1 < λo . Take λ2 ∈ (λ1 , λo ). Then u(t, λ1 uo ) and u(t, λ2 uo ) converge to the same stationary solution by Proposition 3.1. But this stationary solution is unstable and hyperbolic. Hence, we arrive at a contradiction as in the proof of Theorem 5.2.  We formulate our next (and last) result as a remark, since we only indicate some possible proofs. To give a complete proof is out of the scope of this paper. Remark 5.3. If N = 1, p, q are as in Lemma 3.6 and a > a1 then the larger symmetric stationary solution u2 connects to both of the nonsymmetric stationary solutions. There are several possibilities to prove this fact. We shall sketch two of them. (i) ∂W s (u1 ) (the boundary of the domain of attraction of the smaller symmetric stationary solution u1 ) is an invariant Lipschitz manifold with codimension one. (This might be shown in the same way as Theorem 5.5 in [P], see also [T, Propositions 1.2 and 1.3].) From Theorem 5.4(i) it follows that u2 ∈ ∂W s (u1 ). Since W u (u2 ) contains initial data for which blow up occurs and also initial data for which the corresponding solutions tend to u1 and dim W u (u2 ) = 2, there is a uo ∈ W u (u2 ) ∩ ∂W s (u1 ), uo 6≡ u2 . Analogously as in [P, Theorem 5.4(v)] it could be shown that any two functions in ∂W s (u1 ) cross each other. Hence, if we denote by z(f ) the number of sign changes (zero number) of f ∈ C([−l, l]), then z(uo − u2 ) ≥ 1. But we also know that
lim z u(t, uo ) − u2 = 1, t→−∞ since lim t→−∞ u(t, uo ) − u2 = ±ϕ2 , ku(t, uo ) − u2 k ϕ2 being the second eigenfunction of the linearized (at u2 ) stationary problem
(which is a Sturm–Liouville problem). It is well known that z u(t, uo ) − u2 is nonincreasing in t for t ∈ (−∞, T ), T ≤ ∞ being the maximal existence time. Therefore (5.14) Assume that (5.15)
z u(t, uo ) − u2 = 1 u(−l, t; uo) > u2 (−l), for t ∈ (−∞, T ). u(l, t; uo) < u2 (l) 102 M. CHIPOT, M. FILA and P. QUITTNER for t close to −∞. From (5.14) it follows that (5.16) u(l, t; uo ) ≤ u2 (l) for t ∈ (−∞, T ). Let v1 be the nonsymmetric stationary solution with v1 (−l) > u2 (−l), v1 (l) < u2 (l).
Since z(v1 − u2 ) = 1, we have z u(t, uo ) − v1 = 1 for t close to −∞. But then (5.17) u(−l, t; uo) ≤ v1 (−l) for t ∈ (−∞, T ) according to the nonincrease of z. (5.16), (5.17) and (4.10) imply that ku(t, uo)k is bounded for t ∈ (−∞, T ) which means that T = ∞ and u(t, uo ) → v1 as t → ∞. If the inequalities in (5.15) are reversed, we can argue exactly as before to prove that u2 connects to v2 , v2 (x) = v1 (−x). (ii) Another possibility to prove the connections is to use the y–map (see [BF1, Section 2]). This tool enables us to show the
existence of an orbit {w(t) ; t ∈ (−∞, T )} lying in W u (u2 ) with z w(t) − u2 = 1 for t ∈ (−∞, T ). As before, we can conclude that T = ∞ and w(t) converges to a nonsymmetric stationary solution as t → ∞. Acknowledgement. The authors are indebted to Peter Poláčik for his suggestions concerning Theorem 3.5 and Section 5 and to Pavol Brunovský for his suggestions concerning Remark 5.3. References [A1] Amann H., Parabolic evolution equations and nonlinear boundary conditions, J. Diff. Equations 72 (1988), 201–269. , A note on degree theory for gradient mappings, Proc. Amer. Math. Soc. 85 [A2] (1982), 591–595. [A3] , Dynamic theory of quasilinear parabolic equations: II. Reaction-diffusion systems, Diff. Integral Equs. 3 (1990), 13–75. [BF1] Brunovský P. and Fiedler B., Connecting orbits in scalar reaction diffusion equations, Dynamics Reported, Vol. 1 (U. Kirchgraber & H.O. Walther, eds.), 1988, pp. 57–89. , Connecting orbits in scalar reaction diffusion equations II. The complete solution, [BF2] J. Diff. Equations 81 (1989), 106–135. [E] Escher J., Global existence and nonexistence for semilinear parabolic systems with nonlinear boundary conditions, Math. Ann. 284 (1989), 285–305. [F] Fila M., Boundedness of global solutions for the heat equation with nonlinear boundary conditions, Comment. Math. Univ. Carol. 30 (1989), 479–484. [FQ] Fila M. and Quittner P., The blow-up rate for the heat equation with a nonlinear boundary condition, Math. Meth. Appl. Sci. (to appear). [Fo] Filo J., Comment. Math. Univ. Carol., Uniform bounds of solutions to a degenerate diffusion equation with nonlinear boundary conditions 30 (1989), 485–495. [FK] Filo J. and Kačur J. (to appear). [FML] Friedman A. and McLeod B., Blow up of positive solutions of semilinear heat equations, Indiana Univ. Math. J. 34 (1985), 425–447. BLOW UP FOR PARABOLIC EQUATIONS [GT] [H] [H1] [H2] [KN] [L] [M] [N] [P] [Q1] [Q2] [Q3] [S] [St] [Sz] [T] 103 Gilbarg D. and Trudinger N. S., Elliptic partial differential equations of second order, Springer, Berlin–Heidelberg–New York–Tokyo, 1983. Henry D., Geometric theory of semilinear parabolic equations, LNM 840, Springer, Berlin–Heidelberg, 1981. Hofer H., A geometric description of the neighbourhood of a critical point given by the mountain-pass theorem, J. London Math. Soc. (2) 31 (1985), 566–570. , Variational and topological methods in partially ordered Hilbert spaces, Math. Ann. 261 (1982), 493–514. and , , 26, 3 (1990), 465–468; English transl.:, Diff. Equations 26, 3 (1990), 345–348. Levine H. A., Some nonexistence and instability theorems for solutions of formally parabolic equations of the form P ut = −Au + F (u), Arch. Rat. Mech. Anal. 51 (1973), 371–386. Matano H., Existence of nontrivial unstable sets for equilibriums of strongly order-preserving systems, J. Fac. Sci. Univ. Tokyo, Sect. 1A Math. 30 (1983), 645–673. Nečas J., Les méthodes directes en théorie des équations elliptiques, Academia, Praha, 1967. Poláčik P., Domains of attraction of equilibria and monotonicity properties of convergent trajectories in parabolic systems admitting strong comparison principle, J. reine angew. Math. 400 (1989), 32–56. Quittner P., Solvability and multiplicity results for variational inequalities, Comment. Math. Univ. Carol. 30 (1989), 281–302. , A note on the degree theory for variational inequalities involving gradient operators, Preprint. , An instability criterion for variational inequalities, Nonl. Analysis TM&A. 15 (1990), 1167–1180. Shub M., Global stability of dynamical systems, Springer, New York, 1987. Struwe M., Variational methods and their applications to nonlinear partial differential equations and Hamiltonian systems, Springer, Berlin–Heidelberg, 1990. Szulkin A., Minimax principles for lower semicontinuous functions and applications to nonlinear boundary value problems, Ann. Inst. Henri Poincaré 3 (1986), 77–109. Takáč P., Domains of attraction of generic ≤–limit sets for strongly monotone semiflows, Z. Anal. Anwendungen (1991) (to appear). M. Chipot, Département de Mathématiques et Informatique, Université de Metz, Ile du Saulcy, 57045 Metz Cedex 01, France M. Fila, Department of Mathematical Analysis, Comenius University, 842 15 Bratislava, Czechoslovakia P. Quittner, Institute of Applied Mathematics, Comenius University, 842 15 Bratislava, Czechoslovakia