Quenching profile for a quasilinear parabolic equation

We consider the following first initial boundary value problem:ut -(u")xx ~x G (-/,/)> t > 0, (1.1) u(±l,t) = 1, t>0, (1.2)where 0<a<l,/?>0, Z>0, and uq(x) >0, Vx G [-1, Z].Without loss of generality, we may assume that uq(x) is smooth and bounded above by 1 such that uo(±Z) = 1.Since uo(x) is positive, the local (in time) existence and uniqueness of a classical solution of the problem (1.1)-(1.3)are trivial (see [8]).Many results in quenching, such as single point quenching and profiles, are similar to those blow-up results ([3], [5] and the references therein).The system (1.1)-(1.3) in the case a < 1 models fast diffusion and absorption (see [8]).The case a -1 was studied in [1], [2], [6], [7].To study the asymptotic behavior of quenching, a simple Lyapunov function can be constructed in the case a = 1.When a < 1, however, such an explicit Lyapunov function is not available.In this paper, we use the idea of [4] to construct a Lyapunov function, which is rather technical.We say that the solution u quenches if the minimum of u(-,t) reaches zero at some finite time.We shall always assume that u quenches at a quenching time T < oo.In this case the right-hand side of Eq. (1.1) becomes singular and we no longer have a classical solution after this quenching time.A point (c, T) is said to be a quenching point if there is a sequence {(xn,tn)} such that xn -> c, tn -> T, and u(xn,tn) -> 0 as n -> oo.It isshown in [8] that there can only be finitely many quenching points that stay a positive distance away from the boundary |x| = I for any positive initial data.The purpose of this paper is to study how the solution tends to zero.For simplicity we shall only consider the symmetric case, i.e., the case that uq is symmetric with respect to x = 0 and is monotone increasing in \x\.It has been shown that (0, T) is the only possible

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