Resumen:
|
Consider the Cauchy problem ut=uxx+up, x?R, t>0, u(x,0)=u0(x), x?R, where p>1 and u0(x) is continuous, nonnegative and bounded. Assume that u(x,t) blows up at x=0, t=T and set u(x,t)=(T?t)?1/(p?1)?(y,?), y=x/T?t?????, ?=?ln(T?t). Here we show that there exist initial values u0(x) for which the corresponding solution is such that two maxima collapse at x=0, t=T. One then has that ?(y,?)=(p?1)1/(p?1)?C1e??H4(y)+o(e??)as???,(1) with C1>0, H4(y)=c4H˜4(y/2), where c4=(23(4?)1/4)?1, H˜4(s) is the standard 4th Hermite polynomial, and convergence in (1) takes place in Ck,?loc for any k?1 and some ??(0,1). We also show that in this case, limt?T(T?t)1/(p?1)u(?(T?t)1/4,t)=(p?1)(1+C1c4?n)?1/(p?1),(2) where the convergence is uniform on sets |?|?R with R>0. This asymptotic behaviour is different (and flatter) than that corresponding to solutions spreading from data u0(x) having a single maximum, in which case (3)?(y,?)=(p?1)?1/(p?1)?(4?)1/4(p?1)?1/(p?1)2?p?H2(y)?+o(1?)
as ???, and limt?T(T?t)1/(p?1)u(?(T?t)1/2|ln(T?t)|1/2,t)=(p?1)?1/(p?1)(1+(p?1)4p?2)?1/(p?1).(4)
|