On quenching with logarithmic singularity

Nonlinear Analysis 52 (2003) 261 – 289

www.elsevier.com/locate/na

On quenching with logarithmic singularity Timo Salin ∗ Helsinki University of Technology, Institute of Mathematics, P.O. Box 1100, 02015 HUT, Finland Received 31 August 2001; accepted 2 January 2002

Abstract In this paper we study the quenching problem for the reaction di-usion equation ut −uxx =f(u) with Cauchy–Dirichlet data, in the case where we have only a logarithmic singularity, i.e., f(u) = ln(u),  ∈ (0; 1). We show that for su3ciently large domains of x quenching occurs, and that under certain assumptions on the initial function, the set of quenching points is 4nite. The main result of this paper concerns the asymptotic behavior of the solution in a neighborhood of a quenching point. This result gives the quenching rate for the problem. We also obtain new blow-up results for the equation vt − vxx = vev − vx2 . These concern the occurrence of blow-up, the blow-up set and the asymptotics in a neighborhood of a blow-up point. The analysis is based on the equivalence between the quenching and the blow-up for these two equations. ? 2002 Elsevier Science Ltd. All rights reserved. Keywords: Reaction–di-usion equation; Quenching; Quenching set; Quenching rate; Blow-up

1. Introduction Consider the nonlinear di-usion problem ut − uxx = −u−p ; u(x; 0) = u0 (x); u(±l; t) = 1;

x ∈ (−l; l);

t ∈ (0; T );

x ∈ [ − l; l]; t ∈ [0; T ):

(1.1)

Here p, l and T are positive constants, the initial function satis4es 0 ¡ u0 (x) 6 1 and u0 − u0−p 6 0. This type of reaction–di-usion equation with a singular reaction term arises in connection with the di-usion equation generated by a polarization phenomena in ionic conductors [16]. The problem can also be considered as a limiting case of ∗

Tel.: +358-9-451-3039; fax: +358-9-451-3016. E-mail address: [email protected] (T. Salin).

0362-546X/02/$ - see front matter ? 2002 Elsevier Science Ltd. All rights reserved. PII: S 0 3 6 2 - 5 4 6 X ( 0 2 ) 0 0 1 1 0 - 4

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T. Salin / Nonlinear Analysis 52 (2003) 261 – 289

models in chemical catalyst kinetics (Langmuir–Hinshelwood model) or of models in enzyme kinetics [8,23]. It is known, see [26, p. 34, Theorem 3.3], that Eq. (1.1) has a local unique solution in a set (0; t ) × (−l; l). This solution can be continued until t ¡ T , where T = inf  { ¿ 0 |lim supt↑; x∈(−l; l) (u(x; t)+(1=u(x; t)))=∞}. It is also known that [26, p. 41, Theorem 3.8] u(x; t) is a C ∞ -function with respect to x and t in (x; t) ∈ (−l; l)×(0; T ). Eq. (1.1) has been extensively studied under assumptions implying that the solution u(x; t) approaches zero in 4nite time. The reaction term then tends to in4nity and the smooth solution ceases to exist. This phenomenon is called quenching. We say that a is a quenching point and T is a quenching time for u(x; t), if there exists a sequence {(x n ; tn )} with x n → a and tn ↑ T , such that u(x n ; tn ) → 0 as n → ∞. For problem (1.1) it is well-known that for su3ciently large l quenching occurs in 4nite time [1,2,19]. It is also known that the set of quenching points is 4nite [14]. See also the review articles [17,22]. A primary interest for problem (1.1) has been the analysis of the local asymptotics of the solution as t ↑ T in the neighborhood of the quenching point. In particular, it has been shown that the quenching rate satis4es lim u(x; t)(T − t)−1=(1+p) = (1 + p)1=(1+p) ; t↑T

(1.2)

√ uniformly for |x − a| ¡ C T − t. This result was 4rst established by Guo [14] for p ¿ 3, and subsequently generalized to p ¿ 1 by Fila and Hulshof [6]. For the weaker singularity 0 ¡ p ¡ 1, (1.2) has been shown in [15]. Result (1.2) for higher dimensions has been obtained in [7] for the cases p ¿ 0. In this paper the quenching problem is studied for the equation ut − uxx = ln(u); u(x; 0) = u0 (x); u(±l; t) = 1;

x ∈ (−l; l);

t ∈ (0; T );

x ∈ [ − l; l]; t ∈ [0; T );

(1.3)

where  ∈ (0; 1) and u0 ∈ (0; 1]. The reaction term f(u) = −u−p in the Eq. (1.1) is thus replaced by a weaker logarithmic singularity f(u) = ln(u). In Eq. (1.3), an essential feature is the contest between the linear di-usion term uxx and the nonlinear reaction term f(u) = ln(u). If the dissipative di-usion term is dominant, then there is no quenching. Thus the nonlinear reaction term can achieve quenching. Because the reaction term is now much weaker than in Eq. (1.1), it is not obvious that quenching can happen. The 4rst problem is therefore to clear up, whether quenching may at all occur? It is assumed throughout this paper that the initial function satis4es u0 (x) + ln(u0 (x)) 6 0:

(1.4)

This technical assumption guarantees that u(x; t) is decreasing in time. In Section 2 we show that quenching is possible, i.e. Theorem 1.1. For l large enough; the solution u(x; t) of (1.3) quenches in 5nite time.

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The proof of this theorem is based on the fact that the stationary problem corresponding to Eq. (1.3) has no solution, if l is su3ciently large. In Section 3 it is studied how the weakening of the singularity a-ects the set of quenching points? It is known [9] that for a reaction term f(u)=−u−p the x-derivative of the 4nal pro4le u(x; T ) at the quenching point has a singularity when p ¿ 1 and is smooth (ux (a; T ) = 0), when p ∈ (0; 1). Can this regularity be strengthen in the logarithmic case so that ux (x; T ) = 0 for all x ∈ (c; b) ⊂ [ − l; l], in other words can quenching take place on a whole interval? The following result tells us that the situation does not qualitatively di-er from the situation, when we had the power-like singularity. Theorem 1.2. Suppose that u(x; t) satis5es (1.3) and that (1.4) holds. Then the set of quenching points is 5nite. The proof is based on the general method for certain parabolic equations developed by Angenent [3]. It is 4rst deduced, using this method, that ux cannot oscillate, when the quenching point is approached. Then it is shown that there is a time t ∗ such that there is a 4nite number of local minima with respect to x after t ∗ , and that this number is constant in time. Finally, one shows that quenching cannot occur on the boundary and that the set of quenching points is 4nite. The proof is essentially the same as Guo’s [14] (which is also based on [3]) for the stronger singularities f(u) = −u−p . In Section 4 the local asymptotics of the solution near the quenching point is studied. The main result of this paper is proven: Theorem 1.3. Let u(x; t) be the solution of the equation (1.3); where u0 is even; u0 (r) ¿ 0; u0 (x) ∈ (0; 1] and (1.4) holds (r = |x|). Assume that u(x; t) quenches at (0; T ) for some T ¡ ∞. Then 
 u(x; t) 1 d lim 1 + = 0; (1.5) t↑T T −t 0 ln() √ uniformly; when |x| ¡ C T − t; for every C ∈ (0; ∞). This theorem can also be proved in a somewhat stronger form Corollary 1.4. Let u(x; t) quench at (a; T ); with an initial function u0 that satis5es u0 (x) ∈ (0; 1] and (1.4). Then 
 u(x; t) 1 d = 0; lim 1 + t↑T T −t 0 ln() √ uniformly; when |x − a| ¡ C T − t; for every C ∈ (0; ∞). The content of Theorem 1.3 can be interpreted by comparing the quenching rate to a solution of the corresponding ordinary di-erential equation v = f(v) (where

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f(v) = ln(v), with ‘initial’ condition v(T ) = 0), √ and concluding that these solutions are asymptotically equal in the region |x − a| ¡ C T − t. Note that Theorem 1.3 is the same as the result (1.2) for the Eq. (1.1), if the term ln() in (1.5) is replaced by −−p . Our proof for a logarithmic singularity is not based on earlier results on quenching. The proof here uses similarity variables and energy estimates; in particular observe that our method is di-erent from the earlier versions to prove the corresponding quenching-rate estimate (1.2). (see [12,13,4,14]). Especially note that this is a consequence of the fact that (1.3) does not have the useful scaling property that the equation (1.1) has. The rather lengthy proof is further commented and given in Section 4. Consider now the blow-up problem for the equation vt − vxx = af(v) − bvxq

(1.6)

with Cauchy–Dirichlet data (v is given on boundary and v(x; 0) = v0 (x)), when t ¿ 0 and x ∈  (bounded). Here q, a and b are positive constants, and f(v)=vp or f(v)=erv (p and r are also positive constants). Blow-up means that a solution approaches in4nity in 4nite time. We say that b is a blow-up point and T is a blow-up time for v(x; t), if there exists a sequence {(x n ; tn )} with x n → b and tn ↑ T , such that v(x n ; tn ) → ∞ as n → ∞. The blow-up problem for Eq. (1.6) without the damping term—bvxq —has been studied extensively [21]. In problem (1.6), the key question is to 4nd out how this damping term can a-ect the possible occurrence of blow-up, the set of blow-up points and the asymptotics of blow-up. These questions have been studied in, e.g., [18,24,25]. The equivalence between the quenching problem and the blow-up problem is wellknown [18]. Putting u = e−v in Eq. (1.3), we get vt − vxx = vev − vx2 ;

x ∈ (−l; l);

v(x; 0) = −ln(u0 (x)); v(±l; t) = −ln();

t ∈ (0; T );

x ∈ [ − l; l];

t ∈ [0; T );

(1.7)

Note that quenching for Eq. (1.3) corresponds to blow-up in Eq. (1.7). Thus Theorems 1.1–1.3 yield the following new Corollaries Corollary 1.5. For su9ciently large l, the solution v(x; t) of (1.7) blows up in 5nite time. Corollary 1.6. The set of blow-up points for Eq. (1.7) is 5nite. Corollary 1.7. Let (0; T ) be the blow-up point for the Eq. (1.7). Then  ∞ 1 d lim = 1; t↑T T − t v(x; t) e √ uniformly, when |x| 6 C T − t.

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2. A sucient condition for quenching In this section the possibility of quenching in 4nite time is studied. There are two reasons why quenching may not happen. In the 4rst case, we may have u(x; t) ¿ c ¿ 0, for all t ¿ 0. This means, that there is a solution to the corresponding stationary equation, which is a subsolution of Eq. (1.3). On the other hand, it might happen, that u(x; t) ¿ 0 for all t and x, but that min u(x; t) → 0, as t → ∞. This second case is called quenching in in4nite time. It is known [20] for a general singularity, that u(x; t) quenches in 4nite time for l su3ciently large, provided that the similar stationary equation does not have a strong solution. We show in Lemma 2.2, that this last fact holds for the logarithmic singularity. Lemma 2.1. Let u(x; t) be the solution of (1.3), when x ∈ (−l; l) and t ∈ (0; T ). Suppose that (1.4) holds. Then (a) ut (x; t) ¡ 0, when x ∈ (−l; l) and t ∈ (0; T ). (b) 0 ¡ u(x; t) 6 1, when x ∈ [ − l; l] and t ∈ [0; T ). Proof. Let v(x; t) = ut (x; t), when x ∈ (−l; l) and t ∈ (0; T ). Di-erentiating Eq. (1.3), we get vt − vxx − (1=u)v = 0, where x ∈ (−l; l), t ∈ (0; T ) and 1=u is a locally bounded function. On the boundary it holds that: v(±l; t) = 0, t ∈ (0; T ), and v(x; 0) 6 0, by condition (1.4). Claim (a) follows now from the maximum principle. Claim (b) follows from (a) and from the fact that u0 (x) ∈ (0; 1]. Lemma 2.2. When l is su9ciently large, then the equation d 2 g(x) + ln(g(x)) = 0; d x2

g(±l) = 1;

(2.1)

does not have a solution g ∈ C 2 (−l; l)∩C[−l; l]; such that g(x) ∈ (0; 1] for −l 6 x 6 l. Proof. If g ∈ C 2 (−l; l) ∩ C[ − l; l] is a solution of Eq. (2.1) then g (x)2 + 2

 1

g(x)

ln() d = C;

x ∈ (−l; l):

(2.2)

If g(x) ∈ (0; 1]; and because  ∈ (0; 1); then g (x) = −ln(g(x)) ¿ 0. So g is strictly convex; and thus it has one minimum. Denote this minimum point by z and the minimum by m: g(z) = m and g (z) = 0. From formula (2.2); we can now see that  1

m

ln() d = C:

(2.3)

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T. Salin / Nonlinear Analysis 52 (2003) 261 – 289

From Eqs. (2.2) and (2.3) it follows that  m g (x)2 ln() d; = 2 g(x)   m 1=2  ln() d : g (x) = ± 2 g(x)

(2.4)

On the interval (−l; z); g(x) is decreasing; so an integration of the Eq. (2.4) yields −1=2  m  m 2 ln() d dr: (2.5) z+l=− 1

r

On the other hand; we can conclude from Eq. (2.4) that −1=2  m  m z−l= 2 ln() d dr: 1

r

(2.6)

By the addition of Eqs. (2.5) and (2.6) we see that z = 0. We have now obtained −1=2  1  m def l= 2 ln() d dr = I (m): m

r

r Because r ln() d = − m ln() d ¿ − ln()(r − m) and r; m ∈ (0; 1), then I (m) 6 M ¡ ∞. Choosing l ¿ M Lemma follows. m

Theorem 2.3. Suppose u(x; t) is the solution of Eq. (1.3). Then for l su9ciently large; u(x; t) quenches in 5nite time. Proof. Consider Eq. (1.3) with an initial function u0 = 1. The solution of this equation is a supersolution of (1.3). By Lemma 2.2 and [20] the claim follows. 3. There exist at most $nitely many quenching points In this section we investigate whether the weakening of the reaction term a-ects the size of the set of quenching points. Does the fact that f(u) = −u−p is replaced by the weaker singularity f(u) = ln(u) signi4cantly alter this size? Is it now possible to have quenching on an entire interval? The answer to these questions is provided by Theorems 3.7 and 3.8 below. In these results, we show that the qualitative properties of the quenching set do not change when a power singularity is replaced by a logarithmic one. We base our proof on the method developed by Angenent [3] for the analysis of the asymptotics of solutions of certain parabolic equations. More speci4cally, our proof is essentially that of Guo’s concerning the stronger singularity f(u) = −u−p [14]. To make use of that method possible, our quenching problem has to be rewritten as a blow-up problem, i.e., in form (3.2).

T. Salin / Nonlinear Analysis 52 (2003) 261 – 289

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We assume in this section that quenching happens, and that the assumption (1.4) holds. The quenching time is denoted by T as earlier. By a simple modi4cation, Eq. (1.3) can be written in the form ut − uxx = ln(u); u(x; 0) = u0 (x); u(±1; t) = 1;

x ∈ (−1; 1);

t ∈ (0; T );

x ∈ (−1; 1); t ∈ (0; T ):

(3.1)

Let v be de4ned by v = −ln(u). Then v satis4es the equation vt − vxx + vx2 − ev v = 0; v(x; 0) = −ln(u0 (x)); v(±1; t) = −ln();

x ∈ (−1; 1);

t ∈ (0; T );

x ∈ (−1; 1);

t ∈ (0; T ):

(3.2)

By Lemma 2.1, it holds that v ¿−ln() in the region A={(x; t)|x ∈ (−1; 1); t ∈ (0; T )}. Let also w = vt , then wt − wxx + 2vx wx − (1 + v)ev w = 0; w(x; 0) ¿ 0;

x ∈ (−1; 1);

w(±1; t) = 0;

t ∈ (0; T ):

x ∈ (−1; 1);

t ∈ (0; T );

It follows from Lemma 2.1, that w ¿ 0 in the region A. The purpose is now to prove the 4niteness of blow-up points for the Eq. (3.2). Let    +1 when b ¿ 0; sgn(b) =

 

0 when b = 0; −1 when b ¡ 0:

Lemma 3.1. For any a ∈ [ − 1; 1], the limit limt→T sgn(vx (a; t)) exists. Proof. Let a ∈ (−1; 0) and T0 ¡ T . Consider the function (x; t) = v(x; t) − v(2a − x; t); when x ∈ (−1; a) and t ∈ (0; T0 ). Then it follows from Eq. (3.2) that t

−

xx

+ b1

x

+ b2 = 0;

x ∈ (−1; a);

(−1; t) = −ln() − v(2a + 1; t) ¡ 0; (a; t) = v(a; t) − v(a; t) = 0;

satis4es

t ∈ (0; T0 );

t ∈ (0; T0 );

t ∈ (0; T0 ):

(3.3)

Here b1 = vx (x; t) − vx (2a − x; t) and b2 = b2 (v(x; t); v(2a − x; t)) are bounded functions. In addition the derivatives of b1 are bounded. Let also   1 y "(y; t) = exp − (b1 (x; t) d x) (y; t): 2 −1

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T. Salin / Nonlinear Analysis 52 (2003) 261 – 289

By Eq. (3.3) the function " satis4es "t − "yy + q" = 0;

y ∈ (−1; a);

"(−1; t) = (−1; t) ¡ 0; "(a; t) = 0;

t ∈ (0; T0 );

t ∈ (0; T0 );

t ∈ (0; T0 ):

(3.4)

Here q(y; t) is a bounded function. De4ne  "(y; t); −1 6 y 6 a; U (y; t) = −"(2a − y; t); a 6 y 6 2a + 1; and

 Q(y; t) =

q(y; t); −1 6 y 6 a; q(2a − y; t); a 6 y 6 2a + 1:

Then it can be seen by using Eq. (3.4), that U (y; t) satis4es the equation Ut − Uyy + QU = 0; U (−1; t) ¡ 0;

y ∈ (−1; 2a + 1);

t ∈ (0; T0 );

t ¿ 0;

U (2a + 1; t) ¿ 0;

t ¿ 0:

Here Q is a bounded function. Denote by z(t) the number of zeros of U on the interval [ − 1; 2a + 1], so z(t) = #{y ∈ [ − 1; 2a + 1]; U (y; t) = 0}. Concluding now by Ref. [3, Theorem D], we have that z(t) is nonincreasing and 4nite, when 0 ¡ t ¡ T0 . Therefore if there exists a point (y0 ; t0 ) such that U (y0 ; t0 ) = Uy (y0 ; t0 ) = 0, then z(t1 ) ¡ z(t2 ), when t2 ¡ t0 ¡ t1 . Because z(t) = 2m(t) − 1, where m(t) = #{x ∈ [ − 1; a]; (x; t) = 0}, and because T0 ¡ T , then m(t) is non-increasing and 4nite, when 0 ¡ t ¡ T . Furthermore, if there exists a point (x0 ; t0 ) such that (x0 ; t0 )= x (x0 ; t0 )=0, then m(t1 ) ¡ m(t2 ), when t2 ¡ t0 ¡ t1 . We will now deduce the claim by a contradiction. Suppose that vx (a; t) oscillates (or reaches zero without changing the sign), when t ↑ T . The same is then true for x , because x (a; t) = 2vx (a; t). Thus there exists a sequence tn ↑ T , when n → ∞ such that x (a; tn ) = 0 for all n, and because (a; t) = 0, then m(tn+1 ) ¡ m(tn ), and N

(m(ti ) − m(ti+1 )) + m(tN +1 ) ¿ N + m(tN +1 ); m(t1 ) = i=1

for arbitrary N ¿ 1. By this, m(t1 ) = ∞, which contradicts the 4niteness of m. Lemma is true, when a ∈ (−1; 0). The same proof applies in the case, where a ∈ (0; 1). Consider now the situation, when a = 0. Then, by Eq. (3.2), the function satis4es t

−

xx

+ b1

x

+ b2 = 0;

(−1; t) = (0; t) = 0;

x ∈ (−1; 0);

t ∈ (0; T0 );

t ¿ 0:

Applying [3, Theorem C], we deduce, that m(t) is nonincreasing and 4nite when 0 ¡ t ¡ T . Furthermore, if there exists a point (x0 ; t0 ) such that (x0 ; t0 )= x (x0 ; t0 )=0,

T. Salin / Nonlinear Analysis 52 (2003) 261 – 289

269

then m(t1 ) ¡ m(t2 ), when t2 ¡ t0 ¡ t1 . We can now conclude the claim by a contradiction as in the case a ∈ (−1; 0). When a = ±1, then vx (−1; t) ¿ 0 and vx (1; t) ¡ 0. Lemma 3.2. There exists a t∗ ∈ (0; T ) such that n(t) = #{a ∈ [ − 1; 1]; vx (a; t) = 0} is a positive constant for all t ∈ [t∗ ; T ). There also exist C 1 -functions s1 (t); : : : ; sm (t) from the interval [t∗ ; T ) to the interval [−1; 1], such that s1 (t) ¡ · · · ¡ sm (t), and {a ∈ [−1; 1]; vx (a; t)=0}={s1 (t); : : : ; sm (t)}, when t ¿ t∗ . Moreover, the limits si = limt↑T si (t) exist for all i = 1; : : : ; m. Proof. [14, p. 61]. Lemma 3.3. Let s1 and sm be de5ned as in the previous lemma. Then it holds that s1 ¿ − 1 and sm ¡ 1. Proof. [14, p. 62]. Lemma 3.4. The solution v cannot blow-up, when x = ±1. Proof. Let ) be an arbitrary positive constant. Consider the equation wt − wxx = 0; w(±1; t) = 1; w(x; t) = u(x; t);

x ∈ (−1; 1);

t ∈ (); ∞);

t ¿ ); x ∈ (−1; 1);

t = ):

(3.5)

By writing the equation for w − u; we can see by Eqs. (3.1) and (3.5); and by the maximum principle; that w ¿ u; when t ∈ (); T ); and x ∈ (−1; 1). It also holds that wx (−1; t) ¿ ux (−1; t); when t ∈ (); T ). Because vx (−1; t) = −ux (−1; t); and because there exists a constant c ¿ 0 such that wx (−1; t) 6 − c (see [11; p. 156]); when t ∈ (); T ); we obtain that vx (−1; t) ¿ c; for all t ∈ (); T ). Let a=(−1+s1 )=2. By Lemma 3.3; there exists a constant T0 ∈ (); T ); such that vx ¿ 0 in the set A = [ − 1; a] × [T0 ; T ). Consider now the function J (x; t) = vx (x; t) − n h(x)ev(x; t) ; on A; where h(x) = a − x and n ¿ 0 (a constant; determined later). Then we get from (3.2) for J that Jt − Jxx + b1 Jx + b2 J =  2 nh(x)e2v(x; t) ; where b1 =2vx and b2 =− ev (1+v). When n is chosen small enough; then J ¿ 0 on the parabolic boundary of A. By the maximum principle; J ¿ 0 in A. From this it follows that vx − n hev ¿ 0; in A. Dividing by ev and integrating this from x ∈ [ − 1; 12 (a − 1)] to a; when t ∈ (T0 ; T ); we get  v(a; t)  a 1 1 e− d ¿ n (a − ,) d, = n (a − x)2 ¿ n (1 + a)2 ¿ c ¿ 0; 2 8 x v(x; t)

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T. Salin / Nonlinear Analysis 52 (2003) 261 – 289

by Lemma 3.3. Thus e−v(x; t) ¿ c ¿ 0; when x ∈ [ − 1; 12 (a − 1)]. It follows that v is bounded in the set [ − 1; 12 (a − 1)] × (T0 ; T ); and v cannot blow-up; when x = −1. The case x = 1 is concluded similarly. Lemma 3.5. Let [a; b] ⊂ [ − 1; 1] \ {s1 ; : : : ; sm }. If v has a blow-up point c ∈ (a; b); then limt↑T v(x; t) = ∞; either for all x ∈ (c; b] or for all x ∈ [a; c). Proof. [14; p. 64]. Lemma 3.6. The function v does not have blow-up points in [ − 1; 1] \ {s1 ; : : : ; sm }. Proof. Let s0 =−1 and sm+1 =1. Suppose that there exists a blow-up point c ∈ (si ; si+1 ). Let a = 12 (si + c) and b = 12 (si+1 + c). Then there is T0 ∈ (0; T ); such that vx does not change the sign in the set [a; b] × [T0 ; T ) (denote this set by R0 ). We can now suppose that vx ¿ 0 in R0 . Consider the function; J (x; t) = vx (x; t) − n h(x)ev(x; t) ; when (x; t) ∈ R = [d; b] × [T0 ; T ); and h(x) = sin((x − d)/=(b − d)) and d ∈ (c; b); also n ¿ 0 (determined later). Then by (3.2); the function J satis4es   /2 v v Jt − Jxx + 2vx Jx + bJ = n e h ve − ; (b − d)2 where b = − ev (1 + v). By Lemma 3.5 there is a T1 ¿ T0 such that; if t ¿ T1 ; then

vev ¿ (/=(b − d))2 ; for all x ∈ (d; b). For this T1 we can choose n ¿ 0 so small that J ¿ 0; when t = T1 and x ∈ (d; b). Furthermore; J ¿ 0; when x = d or x = b; t ¿ T1 . By the maximum principle; we can see that J ¿ 0 in R. Hence vx e−v ¿ n h. Integrating this from d to b; when t ¿ T1 ; we get  b  x−d e−v(d; t) − e−v(b; t) ¿ n sin / d x: b−d d Letting here t ↑ T ; we can see that the left-hand side converges to zero; because of Lemma 3.5; but the right-hand side is strictly positive. This is a contradiction. Theorem 3.7. Let the initial function u0 in Eq. (1.3) satisfy the condition (1.4). Then the set of quenching points is 5nite. Proof. The claim follows directly from Lemma 3.6. Thus quenching cannot occur on the whole interval. Theorem 3.7 is of crucial importance in the proof of the main result 4.2 about the quenching rate. If it is also supposed, that u0 is symmetric (u0 =u0 (r) and u0 (r) ¿ 0), then it follows from the maximum principle and from Theorem 3.7, that the quenching point is (0; T ). We formulate this by the following corollary.

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Corollary 3.8. Let the assumptions of Theorem 3.7 hold. Let also the initial function u0 be even and u0 (r) ¿ 0. Then the quenching point is (0; T ). 4. Local asymptotics in the neighborhood of a quenching point The main result of this paper is Theorem 4.2. We 4rst prove the preliminary Lemma 4.1 concerning the asymptotics of the solution u(x; t). This lemma gives a lower bound as a function of t for u(x; t) (x ∈ (− ; )), when the quenching point is approached. It also gives an upper bound at a minimum point with respect to the x-variable. Theorem 4.2 improves√ this Lemma by giving a point-wise asymptotic behavior of u(x; t) in the region |x| ¡ C T − t, when the quenching point is approached. After this Lemma 4.1, the main Theorem 4.2 is formulated and the proof is commented on. Lemma 4.1. Suppose that the initial function u0 satis5es the condition (1.4); and that quenching occurs at t = T . Then there exist positive constants 0; l1 and t1 such that (a) ut −0 ln(u) 6 0; when x ∈ (−l1 ; l1 ) (the quenching points belong to this interval) and t ∈ [t1 ; T ). (b) ut blows up; when u quenches. (c) ut (x; t)−ln(u(x; t)) ¿ 0; when t ∈ (0; T ); and x is a local minimum point of u(x; t) with respect to x. Proof. Because we know that quenching happens (Theorem 2.3) and that the set of quenching points is 4nite (Theorem 3.7); we can apply [5; pp. 1053–1054] to conclude claim (a). Item (b) follows directly from (a). By the local existence theorem [26, p. 34] uxx (x; t) ¿ 0, where x is a local minimum point of u(x; t) with respect to x, and the claim (c) follows. Suppose in the following, that u0 is symmetric in the sense that u0 = u0 (r) and u0 (r) ¿ 0. Then it follows from Corollary 3.8 that the only possible quenching point is (0; T ). Suppose that l is su3ciently large, so that we know by Theorem 2.3 that quenching occurs. The local asymptotics of u(x; t) as the quenching point is approached, will be now studied. De4ne new variables: x y= √ ; T −t s = −ln(T − t); where x ∈ [ − l; l]; t ∈ [0; T ); y ∈ [ − le(1=2)s ; le−(1=2)s ] and s ∈ [ − ln T; ∞). Then the inverse functions x = x(y; s) and t = t(s) are well de4ned.

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The function w is de4ned in terms of these new variables:  u(x; t)  u(x(y; s); t(s)) d d 1 1 def w(y; s) = 1 + : =1+ T −t 0 ln() T − t(s) 0 ln()

(4.1)

Eq. (1.3) can now be written in the form ws = wyy − 12 ywy + w + F;

(4.2)

where F = ux2 =u(ln(u))2 , and (y; s) ∈ (−le(1=2)s ; le(1=2)s ) × (−ln T; ∞). Analogously to formula (4.1), we can write F = F(x(y; s); t(s)) = F(x; t). The boundary conditions are  1 d w(±le(1=2)s ; s) = 1 + es ; ln() 0 when s ¿ − ln T . The initial condition is now √  1 u0 (y T ) d w(y; −ln T ) = 1 + ; T 0 ln() where y ∈ [ − le−(1=2)s ; le−(1=2)s ]. Remark. Note that in the transformed equation (4.2) the nonlinear term F cannot be expressed explicitly as a function of y; s and w. For this reason in the following the variables x and s; or y and t might sometimes appear in a same equation. Another reason for this procedure is that in some cases it simpli4es notations. The goal is to prove: Theorem 4.2. Assume that u0 is even; u0 (r) ¿ 0 and u0 satis5es (1.4). Let u(x; t) be the corresponding solution of (1.3). Assume that u(x; t) quenches at (0; T ) for some T ¡ ∞. Then for any positive constant C; (a) w(y; s) − w(0; s)(1 − 12 y2 ) → 0; as s → ∞ uniformly with respect to |y| 6 C; and (b) w(0; s) → 0; when s → ∞. 4.1. Comment on the proof of the Theorem 4.2 The proof of (a) is built on Lemmas 4.3– 4.12 and Corollary 4.13. The statement (b) follows from (a) and from Lemmas 4.14 – 4.18. The proof of (a) is made more di3cult by the fact that in the transformed equation (4.2) the nonlinear term F cannot be expressed explicitly as a function of y, s and w. Furthermore, F depends on both y and s; therefore stationary equation cannot be de4ned. Finally note that on y-intervals we only get lower estimates on wyy . An upper bound can be obtained only at y = 0 (Lemma 4.4). At the beginning of the proof of (a) it is shown, that F → 0 uniformly on compact y-intervals for the Eq. (4.2) (Lemma 4.3). Therefore, on compact y-intervals the

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equation h − 12 yh + h = 0 can be considered as the stationary equation for (4.2), when s is large. A particular solution of this equation is h2 (y)=(1− 12 y2 ). Using Lemmas 4.3 and 4.4, we obtain 4rst the limit of Theorem 4.2(a) in a weak sense (Lemma 4.5(b)). The important Lemmas 4.10 and 4.12 show that w(y; s) − w(0; s)(1 − 12 y2 ) ¡ ( ¿ 0) for large s and for bounded y. The argument of Theorem 4.2(a) is based on this, weak convergence (Lemma 4.5(b)) and the estimates of w (Lemma 4.4). Lemmas 4.6 – 4.9 are needed in the proof of Lemma 4.10, and Lemma 4.11 is needed in the proof of Lemma 4.12. The idea of part (2) (the proof of Theorem 4.2(b)) is to conclude the claim from the nonlinear term F of Eq. (4.2), as s → ∞. It is known by part (1), that for large s the solution is formally w(y; s) ≈ w(0; s)h2 (y) (for bounded y), and then Lw = wyy − 12 wy + w ≈ 0. Concerning the reaction term F it is known, that it is zero only at the point y = 0 and otherwise positive. Thus for a large s∗ , the reaction term F has not any contribution on w(0; s), but for a large y it has a small increasing contribution on w(y; s). (In fact it is shown, that for large s, the reaction term is formally F ≈ f1 (y; s)f2 (s), where @=@yf1 (y; s) ¿ 0 (when y ¿ 0) and f2 (s) → 0). Therefore, somewhat later (s = s∗ + )), the pro4le of the solution w(y; s) is formally w(0; s)[h2 (y)+g(s) (y)], where the function (y) is nondecreasing (y ¿ 0) and (0)=0. Because the solution w(y; s) has to preserve the asymptotical form obtained in part (1), the only possibility is, that w(0; s) has to be decreasing, and that the limit value has to be zero. Eq. (4.2) is studied as a dynamical system in the space L24 (R). Then the eigenvalues and eigenfunctions of the operator L are well-known. The scaled Hermite polynomials form an orthonormal base on that space, and the eigenvalue of the second order polynomial h2 (y) is zero. By part (1), it is known that this polynomial is dominant, as s → ∞. So we obtain that the multiplier function a2 (s) of h2 in the Fourier expansion of the function w is asymptotically equal to w(0; s) (Lemma 4.16), and that a2 (s) → 0 (see (4.66)). The domain of the solution w(y; s) of Eq. (4.2) is not (with respect to y) the whole R. Therefore, the above properties of L24 and L cannot be applied directly to Eq. (4.2). This di3culty can be avoided by 4rst extending Eq. (1.3) to all x ∈ R, and observing that the solution of this equation in the region {(x; t) ∈ R2 |x ∈ (−l; l); t ∈ (0; T )} is the same as the solution of the original equation (1.3). Then the transformed solution w(y; ˜ s) corresponding to the extended solution u(x; ˜ t) is also de4ned for all y ∈ R. 4.2. Proof of Theorem 4.2(a) We begin by Lemma 4.3. Let u(x; t) be the solution of (1.3). Then √ (a) u(x; t) → 0 uniformly; when t ↑ T and |x| 6 C T − t. (b) F is uniformly bounded; when (x; t) ∈ [ − l;√l] × [0; T ). (c) F → 0 uniformly; when t ↑ T and |x| 6 C T − t.

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Proof. We show 4rst the inequality def 1 2 2 ux (x; t)

P(x; t) =

− u(x; t)(1 − ln(u(x; t)))

+ u(0; t)(1 − ln(u(0; t))) 6 0;

(4.3)

in the region A = {(x; t)|x ∈ (−l; l); t ∈ (0; T )}. Because the initial function u0 is symmetric, we can see that ux (0; t) = 0 for all t ∈ [0; T ), and thus P(0; t) = 0 for all t ∈ [0; T ). Di-erentiating the function P with respect to x, we get Px = ux uxx + ux ln(u) − ux + ux = ux ut ; by the Eq. (1.3). It follows from the symmetry and Lemma 2.1 that Pr 6 0 (r = |x|). So we have obtained inequality (4.3). By (4.3), we have that the function ux is uniformly bounded. Claim (a) follows now from Corollary 3.8. Claim (b) can be directly obtained from (4.3) by estimating: 2(1 − ln(u)) C 06F 6 6 6 M ¡ ∞; (4.4) 2 (ln(u)) |ln(u)| because u ∈ (0; 1] and  ∈ (0; 1). Claim (c) follows from inequality (4.4), the continuity of the logarithmic function and the item (a). Lemma 4.4. There exist positive constants c1 ; c2 and ) such that for all s ¿ − ln T ; (a) −c1 6 wyy (0; s) 6 0. (b) −c2 6 wyy (y; s); when −le1=2s 6 y 6 le1=2s . (c) −c2 y 6 wy (y; s) 6 0; when 0 6 y 6 le1=2s . Furthermore; 0 6 wy (y; s) 6 − c2 y; when −le1=2s 6 y 6 0. (d) 0 6 w(0; s) 6 1 − ). (e) − 12 c2 y2 6 w(y; s) 6 1 − ); when −le1=2s 6 y 6 le1=2s . Proof. Di-erentiating Eq. (4.1) with respect to y; we get 1 ux wy = √ ; ln(u) T −t and wyy =

uxx ut −F = − 1 − F; ln(u) ln(u)

(4.5)

(4.6)

by Eq. (1.3). Applying Lemma 4.1(c) and the fact that ux (0; t) = 0 to (4.6), we get wyy (0; s) 6 0. Recalling, in addition, Lemma 4.1(a), we get wyy (0; s) ¿ − c1 . This proves (a). Applying Lemmas 2.1(a,b) and (4.3)(b) to (4.6), we obtain claim (b). Item (c) follows by integrating (b), and noticing the symmetry of u. By Lemma 4.1(a), we get  0  T d d: (4.7) ¿0 u(0; t) ln() t

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275

From this it follows that w(0; s) = 1 +

1 T −t



u(0; t)

0

d 6 1 − 0: ln()

(4.8)

Similarly by Lemma 4.1(c), we get  0  T d d; 6 u(0; t) ln() t

(4.9)

and so 1 w(0; s) = 1 + T −t



u(0; t)

0

d ¿ 0: ln()

(4.10)

We get claim (d) from (4.8) and (4.10). An integration of item (c) yields claim (e) by item (d). Lemma 4.5. Let 4(y)=exp(−y2 =4); and let a(s) be a bounded function for s ¿−ln T . Let h2 (y) = (1 − 12 y2 ) be the second order (Hermite) polynomial. Then (a) (b)

 le(1=2)s 0

w(y; s)4(y) dy → 0; when s → ∞.

0

(w(y; s) − a(s)h2 (y))4(y) dy → 0; when s → ∞.

 le(1=2)s

Proof. (a) Multiply Eq. (4.2) by 4 to obtain (ws − w)4 = (wy 4)y + F4: Integrating this with respect to y from 0 to le(1=2)s ; we get 

le(1=2)s

0

(ws (y; s) − w(y; s))4(y) dy

= wy (le(1=2)s ; s)4(le(1=2)s ) +



le(1=2)s

0

F4(y) dy:

(4.11)

ux (l; t) → 0; ln()

(4.12)

Because ux (l; t) is bounded for all t; then wy (le(1=2)s ; s)4(le(1=2)s ) = e(1=2)s−(l as s → ∞. Writing 4.3(b;c); that  0

le(1=2)s

 le(1=2)s 0

F4 dy =

F4(y) dy → 0;

K 0

2 s

e =4)

F4 dy +

 le(1=2)s K

F4 dy; we can see; by Lemma

(4.13)

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T. Salin / Nonlinear Analysis 52 (2003) 261 – 289

as s → ∞. By formulas (4.12) and (4.13) the terms on the right-hand side of Eq. (4.11) converge to zero; as s → ∞. Furthermore; @ @s



le(1=2)s

0

1 w(y; s)4(y) dy = le(1=2)s w(le(1=2)s ; s)4(le(1=2)s ) 2  le(1=2)s + ws (y; s)4(y) dy: 0

(4.14)

De4ne I (s) =

@ @s



le(1=2)s

0

 w(y; s)4(y) dy −

0

le(1=2)s

ws (y; s)4(y) dy:

By (4.14); (4.15) and De4nition (4.1); we get 
   1 l 2 es d 1 1 s → 0: 1+e I (s) = l exp s − 2 2 4 0 ln() Let now J (s) = can see that

 le(1=2)s 0

(4.15)

(4.16)

w(y; s)4(y) dy. Using Eqs. (4.11) – (4.13); (4.15) and (4.16) we

J  (s) − J (s) = I (s) +

 0

le(1=2)s

(ws (y; s) − w(y; s))4(y) dy → 0;

(4.17)

as s → ∞. To obtain (a) from (4.17) we argue as follows. Suppose that there exists a sequence {si } such that |J (si )| ¿ . Then by (4.17) |J (si )| → ∞, as si → ∞. This contradicts Lemma 4.4. Claim (b) follows from item (a) and from the partial integration:  ∞  ∞  ∞ h2 (y)4(y) dy = 4(y) dy + (y(4(y)))(∞) − 4(y) dy = 0: 0

0

0

Lemma 4.6. There exist constants 7 ∈ (0; ∞); l1 ∈ (0; l) and t1 ∈ (0; T ) such that urt − 7ut ¿ 0; when x ∈ [ − l1 ; l1 ] and t ∈ [t1 ; T ). Proof. Let J (r; t) = urt (r; t) − 7ut (r; t). Di-erentiating this; we get Jt − Jrr − u1 J = −(ur ut )=u2 . The right-hand side of this equation is nonnegative; by the facts that ur ¿ 0 and ut ¡ 0. Because there is only one quenching point; we can choose 7 su3ciently big; so that the point (0; T ) has a neighborhood; for which J ¿ 0 on the parabolic boundary of that neighborhood. This follows from Lemma 4.1; and from boundedness of the functions urt (r; t) and ut (r; t). We obtain the claim now from the maximum principle. Lemma 4.7. The function (−ln(u(0; t))(T − t))=u(0; t) is bounded; when t ∈ [0; T ).

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277

Proof. The problem is the vanishing u(0; t); as t ↑ T . Using the inequality (4.7); we get  u(0; t) ln(u(0; t)) 0 d=ln() −ln(u(0; t))(T − t) 06 6 : u(0; t) 0u(0; t) Applying L‘Hospital’s rule; we obtain  u(0; t) d 1 ln() u(0;t) ln(u(0;t))

0

→

ln(u(0;t)) 1 1 ln(u(0;t)) − (ln(u(0;t)))2

→ 1;

(4.18)

when u(0; t) → 0. Lemma 4.8. The function (−ut (x; t)(T − t))=(u(x; t)) is bounded; for x ∈ [ − l1 ; l1 ] and t ∈ [t1 ; T ). Proof. Because u(x; t) is symmetric; it is su3cient to study the case x ¿ 0. By Lemma 4.6;  x  x ut (8; t) d8 utx (8; t) d8 ¿ 7 ut (x; t) − ut (0; t) = 0 0  x =7 (uxx (8; t) + ln(u(8; t))) d8 0

¿ 7(ux (x; t) + x ln(u(0; t))): Thus it follows from Lemma 4.1 that − ut (x; t) 6 − ut (0; t) − 7x ln(u(0; t)) 6 − ln(u(0; t))(1 + 7x);

(4.19)

and so

−ln((u(0; t)))(T − t) −ut (x; t)(T − t) 6 (1 + 7x): u(x; t) u(0; t) The claim now follows from Lemma 4.7. 06

√ √ Lemma 4.9. For |y| = |x=( T −t)|6C ¡∞; we have lim supt↑T T −turt =ln(u)60; uniformly. √ √ Proof. By Lemma 4.6; T − t(uxt =ln(u)) 6 7 T − t(ut =ln(u)); when x ¿ 0. So it su3ces to show that (T − t)ut2 =(ln(u))2 → 0; as t ↑ T . Lemma 4.8 implies ut2 [T − t=(ln(u))2 ] 6 Mu[ − ut =(ln(u))2 ]. Therefore it is su3cient, by Lemma 4.3(a), to obtain that −ut =(ln(u))2 is bounded. The Taylor expansions give: 1 ln(u(x; t)) = ln(u(0; t)) + (u(x; t) − u(0; t)) 8 6 ln(u(0; t)) +

1 (u(x; t) − u(0; t)) u(0; t)

6 ln(u(0; t)) +

u(x; t) ; u(0; t)

(4.20)

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where 8 = 8(x; t) ∈ (u(0; t); u(x; t)), and u(x; t) = u(0; t) + 12 uxx (,; t)x2

= u(0; t) + 12 (ut − ln(u))(,; t)x2 ;

(4.21)

where , ∈ (0; x). Using Lemma 2.1, and the fact that u(0; t) 6 u(,; t), in expansion (4.21), we get u(x; t) 6 u(0; t) − 21 ln(u(0; t))x2 :

(4.22)

By Lemma 2.1, and by (4.19) and (4.20), we have 06 6

−C ln(u(0; t)) −ut (x; t) 6 2 (ln(u(x; t))) (ln(u(x; t)))2 Cu(x; t) −C + : ln(u(x; t)) (ln(u(x; t)))2 u(0; t)

Making use of (4.22) and applying the de4nition of y yield 06

−C −ut (x; t) C 6 + (ln(u(x; t)))2 ln(u(x; t)) (ln(u(x; t)))2 (T − t) ln(u(0; t)) 1 − Cy2 : 2 u(0; t) ln(u(x; t))2

The claim now follows from Lemmas 4.3(a) and 4.7. Lemma 4.10. For 0 ¡ y ¡ C; we have lim sups→∞ wyyy (y; s) 6 0 uniformly. Proof. Di-erentiating Eq. (4.6) with respect to x; we get wyyy =

4

Gi ;

i=1

where G1 (x; t) =

√

T −t

uxt ; ln(u)

√ G2 (x; t) = − T − t G3 (x; t) =

√

u x ut ; u(ln(u))2

ux T −t 2 u

√ G4 (x; t) = − T − t



ux ln(u)

2ux u(ln(u))

2 ; 

uxx ux2 − ln(u) u(ln(u))2

:

By Lemma 4.9 it is su3cient to prove that Gi → 0 uniformly for bounded y; as s → ∞ and i = 2; 3; 4.

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Using Lemmas 2.1 and 4.8, we obtain 1 1 ux (T − t)(−ut ) : 0 6 G2 = − √ 6 Mwy ln(u) u −ln(u) T − t ln(u) Applying Lemmas 4.4 and 4.3, we get the claim limt↑T G2 = 0. From inequality (4.3), it follows that −ux 1 − ln(u) T −t ux2 T −t √ 0 6 G3 = 62 (−wy ) : u ( T − t ln(u)) (−u ln(u)) u −ln(u) By Lemmas 4.4 and 2.1, the two last term on the right-hand side are bounded. Using also the fact u(0; t) 6 u(x; t), Lemma 4.1 and applying L‘Hospital’s rule to the term T − t=u(0; t), we obtain the claim limt↑T G3 = 0. The function G4 can be written in the form T −t G4 = 2G2 + 2 (wy )(1 + F): u Applying Lemmas 4.4, 4.3 and 4.1, we conclude that the last term at the right-hand side converges to zero. Because limt↑T G2 = 0, then also limt↑T G4 = 0. Lemma 4.11. There exists a positive constant M such that (T − t)utt 6 M in some neighborhood N = (−a; a) × (T − ); T ) of (0; T ). Proof. Let H = (T − t)utt − M ; where the constant M ¿ 0 will be determined later. Therefore we get from Eq. (1.3) that  u 2 M  u t Ht − Hxx + bH = −(T − t) 1− ; + u u T −t where b = 1=(T − t) − 1=u is locally bounded function. We can see that u(x; t)=(T − t) ¿ u(0; t)=(T − t) → ∞; by Lemma 4.1. Hence there is a neighborhood N = (−a; a) × (T − ); T ) of (0; T ); which is independent of M ; such that Ht − Hxx + bH 6 0; on N . When M is chosen big enough; then H 6 0 on the parabolic boundary of N by Theorem 3.7. The claim now follows from the maximum principle. Lemma 4.12. Let w(y; s) be the solution of (4.2). Then lims→∞ ws (0; s) = 0. Proof. Using formulas (4.5) and (4.6); Eq. (4.2) can be written in the form xux 1 ut ws − w = wyy − ywy + F = −1− : 2 ln(u) 2(T − t) ln(u) Di-erentiating this with respect to t; we get   1 1 1 wss + ywys − ws − ywy T −t 2 2   ux 1 x utx u x ut =− + − 2 T − t (T − t) ln(u) ln(u) u(ln(u))2 +

utt ut2 : − ln(u) u(ln(u))2

(4.23)

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In (4.23); take y = 0 to obtain



wss (0; s) − ws (0; s) = (T − t)

utt (0; t) (ut (0; t))2 − ln(u(0; t)) u(0; t)(ln(u(0; t)))2

 :

Here the last term of the right-hand side converges to zero by Lemma 4.1(b;c). Thus; applying Lemma 4.11; we have lim inf {wss (0; s) − ws (0; s)} ¿ 0: s→∞

(4.24)

We prove next that lim inf s→∞ ws (0; s) ¿ 0. By Lemma 4.10, for every ¿ 0 and C ¿ 0, there exists s1 ¿ − ln T such that wyyy ¡ , when s ¿ s1 and 0 6 y 6 C ¡ ∞. Integrating this inequality three times with respect to y, we get wyy (y; s) − wyy (0; s) ¡ y;

(4.25)

wy (y; s) − ywyy (0; s) ¡ 12 y2 ;

(4.26)

w(y; s) − w(0; s) − 12 y2 wyy (0; s) ¡ 16 y3 ;

(4.27)

when y ∈ [0; C]. Because ws (0; s) = wyy (0; s) + w(0; s), it follows from the inequality (4.27) that    1 lim sup w(y; s) + wyy (0; s) 1 − y2 − ws (0; s) 6 0; (4.28) 2 s→∞ uniformly, when 0 6 y 6 C ¡ ∞. Let g(y; s)=w(y; s)+wyy (0; s)(1− 12 y2 ), and consider the function  y G(y; s) = (g(8; s) − ws (0; s))4(8) d8: 0

An application of (4.28) to this de4nition gives that for every ¿ 0 and K ∈ (0; ∞) there exists s2 ¿ − ln T such that G(y; s) ¡ ;

(4.29)

when s ¿ s2 and 0 ¡ y 6 K. Choosing y big enough, we can see, by Lemmas 4.4 and 4.5(b), that  y G(y; s) + ws (0; s) 4(8) d8 → 0; 0

(4.30)

as s → ∞. Using formulas (4.29) and (4.30), we get lim inf ws (0; s) ¿ 0: s→∞

(4.31)

Suppose now that there exists a sequence si → ∞ such that ws (0; si ) ¿ . From this it follows by (4.24) and (4.31), that ws (0; si ) → ∞. This contradicts Lemma 4.4. After these preliminary lemmas we turn to the proof of Theorem 4.2(a).

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281

Let g be as in the proof of Lemma 4.12. By this Lemma 4.12 and by (4.28), lim sup g(y; s) 6 0

(4.32)

s→∞

uniformly, when 0 6 y 6 C ¡ ∞. Furthermore, by Lemma 4.5(b) (where a(s) = −wyy (0; s))  le(1=2)s lim g(8; s)4(8) d8 = 0: (4.33) s→∞

0

By Lemma 4.4 it holds that |gy | 6 Cy. Therefore it follows from (4.32), (4.33) and the symmetry of the solution that lim g(y; s) = 0;

(4.34)

s→∞

uniformly, when |y| 6 C ¡ ∞. By Lemma 4.12, wyy (0; s) + w(0; s) → 0, as s → ∞, and so Theorem 4.2(a) follows from Eq. (4.34). Corollary 4.13. For |y| 6 C ¡ ∞; we have lims→∞ ws (y; s) = 0 uniformly. Proof. By Lemma 4.12; ws (0; s) = wyy (0; s) + w(0; s) → 0. Combining this with inequality (4.26); we have lim sup {wy (y; s) + yw(0; s)} 6 0; s→∞

(4.35)

uniformly; when 0 6 y 6 C ¡ ∞. Writing   y 1 2 w(y; s) − w(0; s) 1 − y = (w8 (8; s) + 8w(0; s)) d8; 2 0 we obtain; by Theorem 4.2(a); (4.35) and Lemma 4.4; that lim (wy (y; s) + yw(0; s)) = 0;

s→∞

uniformly for bounded y. Correspondingly; to conclude (4.38); we 4rst write  y wy (y; s) + yw(0; s) = (w88 (8; s) + w(0; s)) d8: 0

(4.36)

(4.37)

Using Inequality (4.25) and Lemma 4.12; we get lim (wyy (y; s) + w(0; s)) 6 0;

s→∞

uniformly for bounded y. Applying this together with (4.36) and Lemma 4.4 to Eq. (4.37); we can see that lim sup (wyy + w(0; s)) = 0; s→∞

(4.38)

uniformly for bounded y. Writing   1 1 2 ws = (wyy + w(0; s)) − y(wy + yw(0; s)) + w − w(0; s) 1 − y + F; 2 2 we obtain the claim from (4.38); (4.36); Theorem 4.2(a); and Lemma 4.3(c).

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T. Salin / Nonlinear Analysis 52 (2003) 261 – 289

4.3. Proof of Theorem 4.2(b) We will now replace Eq. (1.3) by an extended one, de4ned on the whole real line with respect to x. This equation of course admits the same solutions as (1.3) on the original interval (−l; l). The technical construction is done similarly as in [27] or in [9]. Without loss of generality, we may assume that l = 1 in Eq. (1.3). So let x ¿ 1, and de4ne the kernels:  2  2 x x x 1 ; W (x; t) = √ ; exp − V (x; t) = √ exp − 3 4t 4t /t 2 /t when |x| ¡ R and 0 ¡ t ¡ ∞. Di-erentiating these, we can see that Vx = −W , Vt = Vxx and Wt = Wxx . De4ne the extension uR of u(x; t), when x ¿ 1 and t ¿ 0 by  t u(x; t) = (x − 1) W (x − 1; t − )ux (1; ) d + 1: 0

(4.39)

Here ux (1; t) is obtained from Eq. (3.1) (ux (1; t) = limz↑1 ux (z; t)). Lemma 4.14. The function uR satis5es:



u t − u xx = 2ux (1; 0)V (x − 1; t) + 2

0

t

V (x − 1; t − )ux (1; ) d;

when x ¿ 1. Proof. The claim follows directly by di-erentiating (4.39). Correspondingly in the extension of u to the left of x = −1 the term ux (1; t) in Eq. (4.39) is replaced by the term ux (−1; t). An extended equation is now de4ned ˜ t)); u˜ t − u˜ xx = f(u(x; where

 u(x; ˜ t) =  f(u) ˜ =

x ∈ R \ {±1};

0 ¡ t ¡ T;

u(x; t) when |x| 6 1; u(x; t) when |x| ¿ 1;

ln(u) when |x| 6 1; g(x; R t) when |x| ¿ 1;

and

 g(x; R t) = 2ux (1; 0)V (x − 1; t) + 2

0

t

V (x − 1; t − )ux (1; ) d:

(4.40)

We can see that u˜ ∈ C 1 (R) (4xed t), but f is not continuous at x = ±1, and therefore u˜ is not twice continuously di-erentiable. Because u(x; t) cannot quench at x = ±1, then the functions ux (1; t) and uxt (1; t) are uniformly bounded.

T. Salin / Nonlinear Analysis 52 (2003) 261 – 289

283

Lemma 4.15. The functions u(x; R t) and g(x; R t) satisfy 1 6 u(x; t) ¡ c1 ¡ ∞

and 0 ¡ g(x; R t) ¡ c2 ¡ ∞;

when |x| ¿ 1 and 0 ¡ t ¡ T . Proof. Using inequality (4.3); we get   ux 6 2u(1 − ln(u)) → 2(1 − ln());

(4.41)

when x ↑ 1. Writing Eq. (4.39) in the form   t (x − 1)2 (x − 1)2  ux (1; ) d + 1; u= exp − 3 4(t − ) 0 2 /(t − ) and substituting (4.41); we can see that   2 u 6 C 2(1 − ln()) ,e−(, =4) d, + 1 ¡ ∞: R

(4.42)

The other part of Lemma follows from (4.40). From (4.42) we get the inequality  u 6 C 1 − ln() + ;

(4.43)

and from (4.39) there follows: |u x | 6 c4 ¡ ∞; when |x| ¿ 1 and 0 ¡ t ¡ T . √ De4ne, when y ∈ R (y = x=( T − t)) and s ¿ − ln T :  u(x; ˜ t) 1 d w(y; ˜ s) = 1 + ; T −t 0 f() where

 f() =

ln(); 0 ¡  6 1; ln() + g();  ¿ 1:

(4.44)

(4.45)

(4.46)

The function g is de4ned such that f() is smooth, increasing and negative, when  ¿ 0. Di-erentiating de4nition (4.45), we get ˜ w˜ s − w˜ yy + 12 yw˜ y − w˜ = F;

(4.47)

where F˜ =F, when y ∈ (−e(1=2)s ; e(1=2)s ), and at intervals (−∞; −e(1=2)s ) and (e(1=2)s ; ∞): g(x; R t) u2x −1+ F˜ = f (u); f(u) f(u)2

(4.48)

where gR is de4ned by Eq. (4.40) and f by Eq. (4.46). Using (4.43) – (4.46), (4.48) and Lemma 4.15, we obtain, when |y| ¿ le(1=2)s and s ¿ − ln T , |w(y; ˜ s)| 6 Cy2 + C; |w˜ y (y; s)| 6 C|y|;

(4.49)

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T. Salin / Nonlinear Analysis 52 (2003) 261 – 289

and ˜ 6 M ¡ ∞: |F|

(4.50)

Consider now the extended equation (4.47) as a dynamical system in the space    g(y)2 4(y) dy ¡ ∞ : L24 (R) = g ∈ L2loc (R): R

˜ Then ws − Lw = F, where We use from now on the notation w = w˜ and F = F. 1 Lw = wyy − 2 ywy + w, on the set R × [ − ln T; ∞). The space L24 is a Hilbert space with an inner product  f; gL24 = f(y)g(y)4(y) dy: R

Concerning the linear operator L it is known that (see [10]), it is self-adjoint i.e., that Lf; gL24 = f; LgL24 ;

(4.51)

with spectrum =k =1− 12 k; k =0; 1; 2; : : : : The corresponding eigenfunctions are h˜k (y)= k Hk ( 12 y), where Hk are the (standard) Hermite polynomials and k = (/1=2 2k+1 k!)−1=2 . The 4rst three eigenfunctions are  1 2 1 1 1 y −1 : h˜0 = √ /(−1=4) ; h˜1 = /(−1=4) y; h˜2 = /(−1=4) 2 2 2 2 The Fourier expansion of w with respect to this base is w(y; s) =

∞

a˜k (s)h˜k (y):

k=0

Then one has Lemma 4.16. Let a2 (s) = − 12 /(−1=4) a˜2 (s). Then a2 (s) − w(0; s) → 0; as s → ∞. Proof. Let "(y; s) = w(y; s) − w(0; s)h2 (y); where h2 is de4ned as in Lemma 4.5. Projecting the function " to the subspace generated by the function h˜2 ; we get

a˜k (s)h˜k ; h˜2 L24 − w(0; s)h2 ; h˜2 L24 = −2/(1=4) (a2 (s) − w(0; s)): "; h˜2 L24 = k

Applying HUolder’s inequality to this; it follows that  1=2 |a2 (s) − w(0; s)| 6 C (w(y; s) − w(0; s)h2 )2 4 h2 L1=2 2 → 0; R

4

by Theorem 4.2(a) and inequality (4.49). Lemma 4.17. The inequalities 0 6 (T − t)(−ln(T − t))=u(x; t) 6 M ¡ ∞ hold on the set [ − l; l] × [0; T ).

T. Salin / Nonlinear Analysis 52 (2003) 261 – 289

Proof. We 4rst prove that u u ( 0 [d= − ln()])(−ln(k 0 [d= − ln()])) =1 lim u↓0 u

285

(4.52)

holds for a positive constant k. To show this; we use L‘Hospital’s rule twice. We 4rst obtain u ln(k 0 [d= − ln()]) → 1: (4.53) ln(u) Claim (4.53) is true, because ln(k

u

d ) 0 −ln()

ln(u)

→

−ln(u)

1 u 0

d −ln()

1=u

=

ln(u)

u 0

u → 1; [d=ln()]

(4.54)

by (4.18). Claim (4.52), u u ( 0 [d= − ln()])(−ln(k 0 [d= − ln()])) u u [1=ln(u)] ln(k 0 [d= − ln()]) + 1=ln(u) → 1; → 1 now follows from Eq. (4.54). By inequalities (4.7) and (4.9) we can see that  u(0; t)  1 u(0; t) d d 6T − t6 ; −ln() 0 −ln() 0 0 when t ∈ [0; T ). Because u(0; t) 6 u(x; t), then  u(0; t)  u(0; t) [d= − ln()])(−ln( 0 [d= − ln()])) 1( 0 (T − t)(−ln(T − t)) 6 06 u(x; t) 0 u(0; t) 1 → ; 0 as u(0; t) → 0, by (4.52). Lemma 4.18. For the solution u(x; t) one has (T − t)(−ln(T − t)) 1 →0 − 1 − w(0; s)h2 (y) u(x; t) uniformly for bounded y; as t ↑ T (s → ∞). Proof. By Theorem 4.2(a);  u 1 d 1+ − w(0; s)h2 (y) → 0; T − t 0 ln()

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T. Salin / Nonlinear Analysis 52 (2003) 261 – 289

uniformly for bounded y; as t ↑ T . Dividing this by the function 1 − w(0; s)h2 (y) (= 0 by Lemma (4.4)); we get u [d= − ln()] 0 → 1; (4.55) (T − t)(1 − w(0; s)h2 (y)) uniformly for bounded y; as t ↑ T . From the properties of the logarithmic function it follows that   u d 1 − ln(T − t) → 0; ln 1 − w(0; s)h2 (y) 0 −ln() and

u ln([1=1 − w(0; s)h2 (y)] 0 [d= − ln()]) → 1; ln(T − t)

(4.56)

uniformly for bounded y; as t ↑ T . Let H=

(T − t)(−ln(T − t)) u u u [d= − ln()] −ln([1=1 − w(0; s)h2 (y)] 0 [d= − ln()]) 0 : − 1 − w(0; s)h2 (y) u

(4.57)

This can be written in the form  (T − t)(−ln(T − t)) H= u
 u u [d= −ln()] ln([1=1 −w(0; s)h2 (y)] 0 [d= −ln()]) 0 × 1− (T −t)(1−w(0; s)h2 (y)) ln(T −t) By Lemma 4.17, and by formulas (4.55) and (4.56), H → 0;

(4.58)

uniformly for bounded y, as t ↑ T . Writing Eq. (4.57) in the form H=

1 (T − t)(−ln(T − t)) 1 − + u 1 − w(0; s)h2 (y) 1 − w(0; s)h2 (y) 
u u ( 0 [d= − ln()])(−ln([1=1 − w(0; s)h2 ] 0 [d= − ln()])) ; × 1− u

we can see that the claim follows from (4.58) and (4.52), provided we recall that (1 − w(0; s)h2 (y)) ∈ [); C] for bounded y, by Lemma 4.4. After these preliminary lemmas we now turn to the proof of Theorem 4.2(b).

T. Salin / Nonlinear Analysis 52 (2003) 261 – 289

287

Proof. Projecting the equation ws = Lw + F to the subspace generated by the function h2 ; we obtain

a˜k (s)h˜k ; h2 L24 = Lw; h2 L24 + F; h2 L24 : k

 le(1=2)s  ∞ Note that F; h2 L24 = 2 0 + le(1=2)s Fh2 4; where the latter integral is less than C exp(− es ) by (4.50); and so only the 4rst one of these integral is essential in Eq. (4.59); as s → ∞. Factor 2 is included in the integrals below; because the solution is ˜ ∞ symmetric. √ We can conclude by (4.51) and the orthogonality of the base {hk }k=0 ; that (C = 4 /) as s → ∞  ∞  ∞ ux2 T −t 2 wy h2 (y)4(y) dy; (4.59) Ca2 (s) = 2 h (y)4(y) dy = 2 2 2 u(ln(u)) u 0 0 and Csa2 (s) = 2

 0

∞

(T − t)(−ln(T − t)) 2 wy h2 (y)4(y) dy: u

Write this in the form  ∞ (T − t)(−ln(T − t)) 2 Csa2 = 2 (wy − w(0; s)2 y2 )h2 (y)4(y) dy u 0  ∞ (T − t)(−ln(T − t)) 1 w(0; s)2 y2 h2 (y) +2 − u 1 − w(0; s)h (y) 2 0  ∞ 1 ×4(y) dy + 2 (w(0; s)2 − a2 (s)2 )y2 h2 (y)4(y) dy 1 − w(0; s)h2 (y) 0  ∞ 4

a2 (s)2 y2 +2 h2 (y)4(y) dy = Ij (s): (4.60) 1 − w(0; s)h2 (y) 0 j=1

Next we show that Ij (s) → 0, as s → ∞ and j = 1; 2; 3. When j = 1, then by Lemmas 4.4 and 4.17, we can apply the Lebesgue Dominated Convergence Theorem. Writing wy2 − w(0; s)2 y2 = (wy + w(0; s)y)(wy − w(0; s)y), we can conclude by Lemmas 4.4, 4.17 and formula (4.36), that lim I1 (s) = 0:

s→∞

(4.61)

Correspondingly, in the case j = 2, we obtain from Lemmas 4.4 and 4.18 that lim I2 (s) = 0:

s→∞

(4.62)

When j = 3, we obtain, by Lemmas 4.4 and 4.16, lim I3 (s) = 0:

s→∞

(4.63)

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T. Salin / Nonlinear Analysis 52 (2003) 261 – 289

Finally, we have that there exist positive constants c1 and c2 such that − c1 a2 (s)2 6 I4 (s) 6 − c2 a2 (s)2 ;

(4.64)

for all s ¿ − ln T . By relations (4.60) – (4.64) it follows that there exists a positive constant c3 such that lim sup (sa2 (s) + c3 a2 (s)2 ) 6 0: s→∞

(4.65)

We now conclude that (4.65) imply lim a2 (s) = 0:

s→∞

(4.66)

(1) If a2 has a non-zero limit a∗ (a∗ ¿ 0, because of Lemmas 4.4 and 4.16), then by (4.65) it holds that for every ¿ 0 there exists a s0 ¿ − ln T and C ¿ 0 such that sa2 6 − C, as s ¿ s0 . Integrating this, we obtain  s a2 (s) − a2 (s0 ) 6 − C ln → −∞; s0 which is a contradiction to Lemmas 4.4 and 4.16. (2) If a2 does not have a limit, then it it follows, by Lemmas 4.4 and 4.16, that there exists a sequence sj → ∞ such that a2 (sj ) ¿ 0, and a2 (sj ) ¿ ) ¿ 0, which is a contradiction to (4.65). Theorem 4.2(b) follows from (4.66) and Lemma 4.16. From Theorem 4.2 we can deduce Corollary 1.4: In the proof of Corollary 1.4 we can, by Lemma 3.2, replace the minimum point x = 0 of u by a function si (t), and note that the analysis in a neighborhood of quenching point is qualitatively similar as in the proof of Theorem 4.2. Acknowledgements The author wishes to thank for Professor S.-O. Londen for valuable discussions and encouragement. References [1] A. Acker, B. Kawohl, Remarks on quenching, Nonlinear Analysis TMA 13 (1989) 53–61. [2] A. Acker, W. Walter, The quenching problem for nonlinear parabolic di-erential equations, in: W.M. Everitt, B.D. Sleeman (Eds.), Ordinary and Partial Di-erential Equations, Lecture Notes in Mathematics, Vol. 564, Dundee, Springer, Berlin, 1976, pp. 1–12. [3] S. Angenent, The zero set of solutions of a parabolic equation, J. Reine Angew. Math. 390 (1988) 79–96. [4] J. Bebernes, D. Eberly, A description of self-similar blow-up for dimensions n ¿ 3, Ann. Inst. Henri PoincarXe 5 (1988) 1–21. [5] K. Deng, H.A. Levine, On blowup of ut at quenching, Proc. Amer. Math. Soc. 106 (1989) 1049–1056. [6] M. Fila, J. Hulshof, A note on quenching rate, Proc. Amer. Math. Soc. 112 (1991) 473–477.

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