0 and I;y/j < M. We then select n/r, large enough so that k 5 MrG(M), and take Mz = (1 - E)~(O)/G(M), in which case (1 - &)g(y) 2 MzG(y) for 1~1 < M, whereupon (5.7) holds. By the maximum principle, we then have that:
and since g(‘y) >> e-“‘G(y) as r + cc for any ;y E R”, follows by taking E > 0 arbitrarily small. Anmrle~ rlc I’lnstitrrt
Hcvzri Poinwrt:
the conclusion
n Analyw
non h&k
SEMILINEAR
PARABOLIC
85
PROBLEMS
2 5.2. The case N 5 u: = ~ p-l’
Under the assumptions above, it is no longer true that (3.44) and (5.6) have a positive solution. Instead, we have the following result: Let E > 0 be such that &
+ E - F # 0, 1,2,3,.
unique solution of (3.44) with d = &
. . Then there exists a
+ E such that G(0) = 1. G’(0) = 0.
as y+ cx), and G(y) N CEyp2(*+‘) where C, = I’(N/2)(I‘((N/2)*l/(p - 1) - E))-‘. Moreover G(y) has a finite number of zeroes (that depend on E).
(5.9)
Actually, G(y) = e-c M( $ - 5 - E, $; $), and the statements in (5.9) follow from classical results on Kummer’s functions (cf. [ 11, [l 11, . . .) Take now ~1, E:! such that 0 < 2~~ < ~2 for i = 1.2, E; satisfies the requirement made in (5.9) and I($ - & - E;) > 0. We now define: F(y, T) = 2e-E’TG,, (y) - e-C27GE2(y). We then have that: F(y,r)
> 0
for all y E R”
and r > 0.
(5.10)
Tocheck(5.10), wemerelyobserve that F(y,7) 2 e-E’r(2G,,(y)-G,,(y)) where function H(y) = 2G,,(y) - GEa(y) satisfies: fory>O,
H”(y)+(~+~)H’(y)=(~2-2~1)H(y)
H(0) = 1,
H’(0) = 0:
H(y) - DE1ym2(*+E1)
H(y) > 0
for y > 0.
for y -+ co; for some D,, > 0.
(5.11) (5.12) (5.13)
We now define: @+(Y>7) = Cl+ EMY/) + MIJYY, 7); @-(Y, ~1 = (ClEMyl) - M~JYY, 4), (compare with (5.4)-(5.5)).
One readily sees that:
For any positive numbers MI and M2, one has that @’ (resp. @- ) is a supersolution of (4.20) (resp. is a subsolution). (5.14) To check (5.14), we merely observe that CC,F = 0, where C, is the differential operator defined in the proof of Lemma 5.1. It then suffices to Vol. 16. no l-1999
86
L. HERRAIZ
repeat the argument there to derive the sought-for result. Since F(y. 7) is everywhere positive (cf. (5.10)), we now obtain (5.7) exactly as before. By the maximum principle, (5.8) continues to hold. The conclusion follows now by observing that eP1’(2G1 - G2) E He?lr << g as r + X. since there exists C > 0 such that g(:y) > CH(y) for all :y E I?“-. n 5.3. The proof of Theorem
3
A first step towards deriving such result is the following: LEMMA 5.2. - Let u(:c, t) be the solution One then has that: u(z: t) N .4lzl-^,
qf (1. I)-(1.3)
with &
< cv < N.
as t -+ 3cj, uniformly on sets
(5.15)
Proof. - Since u(z, t) 5 U(X, t), where ~(.7:, t) is the solution of (3.1)-
(3.3), it follows from (3.16) that U(X, t) satisfies in n an upper bound as that in the right of (5.15). Let now p be such that & = CX.Clearly. jZj< p. and therefore U(z: t) < U(IC, t) in i’l for t > 1, where ?I denotes the solution of (1 .l)-( 1.3) with p replaced by jY. By Theorem 2, we then have that ?i satisfies (5.15), and the proof is concluded. n End of the proof
qf Theorem
3:
Let E > 0 be given, and denote by g:(r) the solution of (5.1) with coefficient (--& + E) replaced there by 5, and where the second condition in (5.2) is replaced by G(s) - (A+E)s-” for s --+ 3~. Arguing as in part c) in Lemma 3.3, we readily see that $(r..t) = t-?gQ(r/v’?) is a supersolution for (1.1) when t >> 1. Consider now the auxiliary function:
u, = t-5 (g;(T/&) + e(g,(r/\/i))“t-V). where g;(s) denotes now the solution of (3.44) with d = 5, with boundary conditions (5.2) (with A replaced by (A - E) there). A routine computation reveals that u; satisfies all the required properties for a subsolution if 6’ and t are sufficiently large. Furthermore, for t >> 1 one also has that u; (T, t) 5 u; (T, t) for all T > 0. On the other hand, considering now f(t) such that ti < f(t) < tg, then by Lemma 5.2, (ui (r, t) - U(T, t,) + = 0 for T 2 f(t). This gives at once that: ug(r, t) 5 U(T, t) + C(t - to)-+
for t > t,.
87
SEMILINEAR PARABOLIC PROBLEMS
Hence: t-T (g;(r/Jt) and the result f(s) < SE,
follows
- C(t - to)-) by
6. THE PROOFS
observing
5 U(TJ)>
that g;(r/&)
OF THEOREMS
>
t+,
as n
4,s AND 6
For convenience, our discussion will be split into a number of subcases. 2 6.1. The case -
LEMMA
5
u(x, t) - AIx-”
and assume that
as t -+ 03, uniformly on sets
{1x/’ 2 kt log t} with k > 2(tr - N).
(6.1)
Proof. - It follows by adapting the argument already used in the proof of Lemma 5.2. We first notice that an upper bound alike to that in (6.1) holds in view of the corresponding result for the heat equation (cf. (3.14) in Lemma 3.2). We then select p exactly as in Lemma 5.1, and observe that the required lower bound holds for [XI 2 g(t) with g(t) >> t 3 when t >> 1, whence in particular in the region considered in (6.1). n We are now in a position to show: 6.2. - Let u(x, t) be the solution of (l.l)-(1.3), < N < cx. Then the following estimate holds:
LEMMA
5
u(x, t) = C,(4rt)-+
exp
and assume that
ast-+m!
uniformly on regions { 1x1’ 5 kt log t} with k < 2( cr - N), where
co =
.I’RN
Vol. 16, no I-1999.
u,(x)dx-
J7 0
IRNG(y,
s)dyds > 0.
(6.2)
88
L. HERRAIZ
- Recalling part a) in Lemma 3.3. the proof will follow as soon as we show that: Prmf
To derive (6.3), we proceed in several steps. the first of which is: Step 1. - Set:
where E > 0 is fixed and F’(x. t; y: s) = Then there holds: and
I L L,,
I,,,, = (1 + 0(~))Cr
To check (6.4), we observe that when s < fi
as t + CC. one has that:
Thus, when lz12 < Ct logt and (v.J)< E&-, I#
I:K-yj2
it turns out that:
fi
--4(t-41;
(6.4)
fort>>l
2
whereupon (6.4) follows. Write now:
14t) I I
IE
*0
Pu”dyds * m IN
+
where h(t) = tr-”
and E E (0,l)
I1=;I’h(t) .I
Pu”dyds
0
where M(s)
lYl
= (Mslogs)~
.t . I J
. h(t) R”’
Pu”dyds
E II f I,.
(6.5)
will be selected presently, +
h(t). I .!
. (1
Pu”dyds
lyl>M(s)
with M > 2(tr - IV).
E III + I12; (6.6)
SEMILINEAR
PARABOLIC
89
PROBLEMS
Step 2. - There holds:
To check (6.7), we notice that 12j2/4t - Iz - yj2/4(t Le-ir/;a; [zll91/2t. Therefore, for 1~1~ 5 Ctlogt 7
- s) 5 jz12/4t and ly/1 5 M(s),
N ~(tlogt)i(Mh(t)logt)t
< &~logI)*M(h(t))
E G(t),
where /ink - G(t) = 0. It then turns out that, as t -+ x :
whence (6.7). Let us examine now 112 in (6.6). We shall prove that: Step 3. - There holds:
To derive (6.8), we proceed as follows. Firstly, we observe that for t large enough and small E > 0 : 112
5
(I$-
jl(t) s I t; P(,,
0))
0
=
(1
5
ll2l
+
y, s)u"(yy;
s)dyds
* lyl>M(s)
41))
+
1122.
(6.9)
It is easy to see now that I 122 satisfies (6.8), since I~:l’/4t-lz-:yl~/4(ts) 5 lxjlyl/4t 5 ME. To estimate J1al, we take ~1 > 1 and cy2 > 0 such that al + o2 = y. Taking advantage of Lemma 6.1, we then have that:
Vol. 16, no I-1999
90
L. HERRAIZ
Since 1~1’ 2 &logt
with k < ‘L(tr - IV) , cxp
(
!$
>
5 t-.
whence:
Il.21 5 Ct--ylogt)c. for y = T((Y~ - 1) -
1 - r,-l (1 - E), and C = 9. A quick check ( c9 reveals that y = (F - 5) (~1 - ~)+(I)(E), which can be selected positive for u1 > 1 and 0 < E < 1. in which case I12r = o(l) as t -+ x and (6.8) follows. We have yet to deal with 12 in (6.5). Analysis of that integral will require of further splittings in its domain of integration. To this end, we shall write:
.t I I /I(t) +.I.tI.
I, =
I22
.f . I I
I%! + Pd’ E I, 1 + 12-J. . /l(t) . It,ll
=
.!.I
PUP
Ic,/>.lf(.~)./.y(>~~
PUP z I,,,
+ I,,,.
/l(t) . lY/>Af(s),ly/<~\
I222
=
I2221
+
(ci.10)
I22229
where integration in 122~~ (resp. in 12222) is performed in the region where Iz - yI 2 E (resp. where IX - y( < E). We then prove: Step 4. - The following 12
L
(1+
o(l)
+
inequality holds true:
.t . / 1,
O(E))
,,(t)
. ~~~
U”(YP
s)dYds
fort > 1.
(6.11)
To check (6.11) we first remark that: I2221
As q6
i
(l+o(l)+O(&))
a matter exp (pd)
lit,
iy,>Af,
s)dyds
,) ~L~(Y, ’ .5
of fact, (6.12) follows < t-4 exp (-!52$)
for
t 29
1.
(6.12)
from the fact that (t since IIC - yI # 0. One thus
has that P(x, t; y, s) 5 exp !$ - w , and the result is obtained by ( > repeating the argument already used to estimate 1122 in Step 3. We next show that: I2222
=
o(l) Amaks
ast+oo. de /‘hritut
(6.13) Henri
Poincarr’
- Analyse
non lint’aire
SEMILINEAR
PARABOLIC
91
PROBLEMS
To derive (6.13), we note that 191 < ~(t(logt)-‘)i and IIC- y[ < E give 5 C < co. Take now y1 at once 1x1 < e(t(logt)-I)$ , whence exp $ ( > and 72 positive and such that y1 < 1, 72 > ~1- 1, y1 + y2 = p. Denoting by C the region of variation of y, we then have that:
Since Iy/) 2 Ct-y(l-E)(logt)C. iv
12222 < ct 2
M(s)
> M(tl-E), it turns Therefore, for t >> 1 : .t
a71
2 (l-4(logt)C
v -y(‘---)
(,og t)C
(t _ S)-$S-*d
,I t’-’ .t s-s%ts
out
that
Iy-“Y1
2
S~exp(-~)dy N-‘rTl(‘--r) (log qc = ()(I). 2
provided that E > 0 is small enough. On the other hand, arguing as for Ilzl (cf. (6.9)) we easily see that: I221
=
o(l)
ast+co.
(6.14)
Let us write now Ci = ((9,s) : Iy( 5 M(s), ly12 5 2(ct - N)(l E)slogs}, c* = {(y,s) : 2(Q - N)(l - +ogs < Iyl2 5 (M(s))“}. We next set:
To bound 1ai2, we make use of the estimate: lLP(y,
s)
2
ct(
4(a-N)P-s)P(l-E)
5 s 5 t. (6.16) Inequality (6.16) is derived as follows. By (3.26), one certainly has that ~(y,s) 5 Cs-*:e-g at IyyI = (2(a - N)(l - ~)slog~)~, whence uP(y,s) 5 Clyl-“P 5 cs( w-s)p z fi (s) there. On the other hand, by (el), ~(y, s) 5 C(yl-” 5 C(slogs)-5 = fx(s) at /y/1 = M(s). Since 1, it follows from the maximum principle that f2(3) I fl(S) f or s > fi (s) provides a bound for uP(y, s) in all Ca, whereupon (6.16) follows by recalling that s > h(t) G t"+ under our current assumptions. Vol. 16, no I-1999
when (y,s) E C2 and
h(t)
92
L. HERRAIZ
A quick check reveals now that, if we set bl = $(c(o - N) - (~);o(1 - c) < tq for jz12 < 2(o - N)tlogf, then: and recall that t% exp
since 5 + bl + 1 = 1 - f(;u - 1) + O(E) for 0 < ,c << 1. To conclude with the proof, we have yet to show that:
To obtain (6.18), we take advantage of the fact that. for L!((Y- N)(l - ~)s log s with /1(t) 5 s 5 t. one has that:
I:(/[’ 5
We next take (I,, 0 positive and such that n + b = i. Clearly. exp
I:+
1:1:12 Ix: - y/2 - (
( 4t
((,,+ b)1.1: - !/I”
4(t - s) ) = cxl) ( ,4f
(f-s)
1
and: (f3.1’3)
We then have that: = o(1) whenever IyI 5 M(s)
as t -+ 3~.
and /z12 < 2(tr - N)t logt. with f’--E 5 s 2 t. (s.20)
To keep the main flow of the arguments here, we shall assume (6.20) for a moment to conclude the proof. From (6.19) and (6.20), it follows that:
as t B 1. .hznles
de I’hstitut
Henr-i Poincurt:
Analyse
non lint%irs
SEMILINEAR
PARABOLIC
93
PROBLEMS
Finally, to derive (6.20) we proceed as follows. Suppose first that IZ - yl 5 r(t) << A, as t -+ m. Then, since 1~1~I (1~ - yJ + lyl)2, we readily see that:
= o(l)
asticc.
On the other hand, if IX - y L: r(t) then IyI > r(t) - /XI: and
= o(1)
astico.
From Lemmata 6.1 and 6.2, it follows at once that Theorem 5 holds. n 2 6.2. The case N = N > ~ P-l We now consider the situation where the asymptotics is encoded in that of the corresponding linear heat equation. The behaviour in external regions is as follows: LEMMA 6.3. - Let U(Z, t) be the solution of (l.l)-(1.3), CY= N > 5. One then has that:
and assume that
u(x, t) - Al+” as t -+ cc, uniformly on sets { /ccl2 2 kt log t} with k > 4.
(6.21)
Proof. - It is entirely similar to that of Lemma 6.1, and will therefore be omitted. H Stabilization on internal regions is described in the following: 6.4. - Let u(z, t) be the solution of (].I)-(1.3), < N = a. Then the following estimate holds:
LEMMA
5
u(x, t) = +
log t(47rt)-+
exp
uniformly on sets /xl2 5 kt log(logt) Vol. 16. no l-1999
(
-$
)
(1 + o(l))
with 0 < k < 4.
and assume that
ast+c0, (6.22)
94
L. HERRAIZ
Proof. - Recalling the derivation of (3.36) in Lemma 3.3, it will suffice to show that:
is such that
(6.23)
I < 1ogt
ast--+m
0nC.
(6.24)
To this end we take up the argument in Lemma 6.2 and adapt it as follows. We first split I exactly as in (6.5), and subdivide again the term I1 thus obtained as in (6.6), but with a different choice of M(s). Namely, we now take M(s) = (MS log log s) 3 with A4 > 4. One then readily checks that: 111 = O(1)
astiocj
0nC.
(6.25)
To estimate 112, we split it in the form: ./l(t) I12
=
.I 0
. J R”
./c(t)
+
h(t)
P#dyd~
*I 0
=
.
I0
.I 21,
Pu”dyds (6.26)
.
I .\A
PuPdgds E Ilzl + Ilz2
where now 121 = {(y, s) : Iy/ > M(s) and IyI > ~(s(log(logs))-‘)i} and A2 = {(y,s) : 1y/1> M(s) and Iyl < ~(s(log(log.~))-l)~}. Just as in our previous Lemma, it is easy to check that: ~122
=
as t + 3~: in the region under consideration.
O(1)
(6.27)
To bound 1121, we select (~1 and a2 as when obtaining the corresponding estimate in Lemma 6.2. Arguing as in that result, and observing that now !$ 5 C log(log t), we now obtain:
I121
2
Cexp
5 CY(log
( ,> Id2 4f tp
&l-f+)(l-E)
t (
log log
=
o(l)
>
(log log t)C”?
where g = (1 - s) + $(l - (Y~) = (aI - l)(& that ~1 > 1. We have thus obtained that: 1121
t
+(l-nl)
ast-+cc
- 5)
< 0 provided
for (z, t) as in (4.36).
(6.28)
We next set out to control 12. This we shall do by splitting that integral as indicated in (6.10)-(6.10) with the modifications already introduced:
SEMILINEAR
PARABOLIC
95
PROBLEMS
namely, we take now M(s) = (MS logs) 4 with M > 4, and in (6.10) we replace IgyI 1 ~(t(logt)-‘)) (resp. IyI < ~(t(loglogt)-I)+) by 1~1> ~(t(loglogt)-l)i (resp. ly/1 < ~(t(loglogt)-l)$). As before, we readily check that IzzzI = o(l) in the region under consideration. The corresponding estimate for I,,,, is a minor modification of that obtained in the previous Lemma. Namely, we now derive: I2222
i
(2
~(l-r*(l-~))(loglogt)”
. p*
< Ct-x(loglogt)“. where x = $(n(l - E) - 1) - 1 + 3 = (kkl) (71 - 1) < 0. provided that y1 < 1 as in the previous case. The term I2221 is once again estimated as in Lemma 6.2, to yield that Izzl = o(1) for large t in the region under consideration. Finally, 121 is first splitted as in (6.15), with obvious modifications in the definition of sets C1 and C2. The contribution arising from 121~ is of similar order (and is analogously obtained) as in the previous Lemma. Finally, 1212 is easily shown to be o( 1) as f. + cc in the region considered by using the bound: ~(y,s) < .s-~(log~)~’
for some Ed, whenever (y,s) E X1,
which is derived in a similar way as (6.16). Summing up, we have obtained that: I2 = o(1)
as t -+ 00 in the region under consideration.
(6.29)
Putting together (6.25)-(6.29), (6.22) follows, and the proof of the Lemma is now complete. n Theorem 4 is now a consequence of Lemmata 6.3 and 6.4. 6.3. The proof of Theorem 6 We note that the result for the outer region follows by arguing as in the previous cases. To obtain the corresponding behaviour for the inner region, we shall proceed in several steps. To begin with, we obtain the following estimate: 6.5. - Let u(x, t) be the solution of (1.1)~(1.3), and assume that Q > N = 5. Then there exist positive constants S, Ikf such that for t large: LEMMA
s 5 u(z,t)(tlogt)W
Vol. 16. no l-1999.
5 A4, on sets {(x6, t) : 1x1 5 CA}
with C > 0. (6.30)
96
L. HERRAIZ
Proof. - We define for fixed positive T the functions: u,(n:, t) = b((t + T) log(t + T))-+e-*. and
for 6 small and positive, and for M: E large. We now apply the maximum principle in 1~ 5 Cv’fl for any C larger than a certain constant (this can be seen by using the technique developed in Section 4, together with Lemma 6.1), and hence we obtain the proof. n To proceed further. a suitable subsolution will be derived on regions of the type 121 2 C(T + t)i with C > 1. Namely. we will prove: 6.6. - Let T 2 1 be given. For any M > 0 and jar any B with 1 < H < 2, there exists R,, = R,(M. 0. N. p) > 0 such that the function t~(x,t) given by: LEMMA
‘(L(z,~) = M((t+T)
log(t+T))-+
exp
B + (t+T)”
’ 4(t+T)
(6.32) satisfies: (?L)t 5 A& - QP
,for
(6.33)
j:l:l 2 R,(t + T)f and t 2 0.
for any nonnegative value of the free parameter B. Pick now E such that 0 < E << 1: and let a be the function in ( 6.32) with MI = 15- E, where 5 is given in the previous Lemma. Let u(z, t) be the solution of(l.l)-(1.3). Then there exists t, >> 1 such that: g(z, to) 5 u(:c, to) for 1~1 2 R,(t, + T)$. G(:c, t) < u(:c, t)
(6.34)
at 1x1 = R,(t + T)* with t 1 t,.
(G.35)
Proof. - Checking that ~(2, t) is a subsolution in the region under consideration is made by means of a direct computation. To derive (6.34). we first select t, >> 1 so that: ‘dx,to)
2 (6 - ;)((to
+ T) log(t, + T))-+
exp (
$1
T)) ’ (6.36)
SEMILINEAR
PARABOLIC
97
PROBLEMS
for llcl = R,(t, + T)$ and t 2 t,, where T is fixed (this can be done by (6.30)). We then take B >> 1 such that for 1~12 &(t, + T); : + T) log(f, + T))-6 1 B
u(x. to) > (6 - c)((t,
4(t, + T) + (to + T)O
To obtain (6.37) we take advantage of the fact that U(Z. to) > 0 by the strong maximum principle. From this and the external Lemma 6.1, we merely have to observe that the quantity on the right in (6.37) decreases exponentially in B to derive the sought-for inequality. Finally, to derive (6.35) it suffices to check that: (6 - ;)((I
+ T) log(t + 7’))-$-&
> (6 - E)((t, + T) log(t, + T))-+e-(h+i6%i.‘i’, whenever IX/ = R,(t + T)f with t 2 t, > 1. Such inequality holds whenever cxp - ~(t+T)“-BR’ 1 5 1 5 2, and the proof is now comp1ete.m ( > As a natural complement to Lemma 6.6, a suitable supersolution in the same type of external zones is provided in the following: 6.7. - Let T > 1 be given. For any 1111> 0. S > 0 and y > f, there exists R,, > 0 and t, > 0 (both depending on y, S, Ml. N and y) such that the function u(:c, t) given by: LEMMA
?i(;r:, t) = (t + T)-+
(kf(log(t
+ T))+
+J,f2(log(t + M&r:(-yt
ILI2 ‘fr+‘ri(i+i&TJ)
+ T))-‘)e-
+ T)” 3z i&(x, t) + ;iz2(:r:,t) + ‘IL3(:c,t),
(6;.38)
satisjes: at 2 AE
for 1x1 2 R,(t + T)’ : and t 2 t,.
(6.39)
for any value of the free parameters Mz and MS. Take now E such that 0 < E < 1, and let E(x, t) be the function in (6.38) with Ml = M + E, where M is the constant of Lemma 6.5, and MS = 2A. Let u(z, t) be the solution of (l.l.L(l.3). Then there exists a constant C* such that: u(x,t,)
5 U(x, to) for I:cI 2 R,(t,
u(x, t) 5 qx, t)
Vol. 16. no I-1999.
+ T)*;
at 1x1 = R,(t + T);
with t 2 t,.
(6.40)
(6.41)
98
L. HERRAIZ
Proof. - As in our previous result, the fact that ;ii(~. f) given in (6.38)
satisfies (6.39) follows from a direct computation. Fix now T 2 1. As to (6.40) we observe that, by Lemma 6.5 for 1~1= R,,(f+T)i and t > t,, >> I: ‘Ib(Z. t) < (M + ;)((r
+ T) log(t + T))-+
exp ( - &).
(6.42)
By Lemma 6.1, (6.40) certainly holds for 1~1 2 S(t,,). where function S(f) is selected so that S(t) > (t log t)$ when t >> 1 : we just have to observe that w(:t.. to) 5 E;l(:c, to) in such regions. On the other hand, when rc,,(t, + T)f < 1x1 5 S(f,,) we take Mz > 1 in 712(3:,t) to derive that ‘IL(:~:to) < &(:c, to) there. Summing these results up, (6.40) follows. To check (6.41), we merely have to remark that ;%ir(z:,t) is larger than the term coming from the right of (6.30) for t >> 1. We shall omit further details. n End of the proof
of Theorem 6:
By the maximum principle, we now have that, for any E > 0:
where T > 1 is fixed, and R and t, are large enough. We note that ~(z, “) (h-~)(l;logt)-+r-* fort > 1 andE(z,t) - (M+~)(tlogt)-%e-g when R& 5 1x1 < 0,/w, provided that D < dw. The bound on the value of D is readily obtained by looking for the region where El N :u;j. We now return to the spectral analysis which was already introduced in Section 2. We recall the self-similar variables: ,u(x,t) = t-hP(y:T):
y = zt-).
7 = 1ogt.
(6.43)
With the new variables, equation (1.1) can be recast in the form: @,=A@+++-
p-l
l
Q,-@“=A,@+
I@
p-l
- cpp.
(6.44)
Spectral properties of operator A, were studied in Section 2 and will be used in the sequel. Because of the previous Lemmata, we have the bound: (6 - ,,,-+,-q
uniformly function:
< @(y,~) 5 (M + E)T-*e&,
in { IyI 5 Cfi}
for C < dw.
%Y> 7) = @(Y, T)f(Y, T);
(6.45)
We now define the (6.46)
SEMILINEAR
PARABOLIC
99
PROBLEMS
wheref(~d = f(lvl/~/-)7 .1s a
smooth and nonnegative cut-off function such that f(l) = 1 if [ < C1 and f(t) = 0 if t > C2, for some constants 0 < C1 < C, < dw. The previous bound (6.45) ensures that the new function &(y, P-) belongs to L:(IRN) (cf. Section 2) for large T. On the other hand, it can be easily seen that: 6, = A,& + LC P-1
where G(y,7)
= (a[f, - Af - +yVf]
- f&’
+ G(y, T)>
- 2VaVf.
6 = S(TMo(YY) +
We next define: (6.48)
WY, T)T
where g(T) = < 8?,ti0 > and 4,(y) is the first eigenfunction of the operator A,. Our goal now is to prove that 4 - g(r)$,(y) as time grows, uniformly in some region. Function R(y, T) satisfies the equation: R,-A,R-
-&I?
= -g’(+),(y)
- f&” + G(y, T) 3 H(y, T). (6.49)
From (6.45) it is easy to obtain the estimate < H, H >= O(T-%-~). Therefore, using the variation of constants formula and the orthogonality of R to $,, we obtain for some Ic (with llcl > 1): < R, R > = O(e-“br)
+
/-’ e-““(‘-“)O(s-+-l)&
= o(~-+‘), (6.50)
where vk = -XI, - & = !$, AS being the eigenvalue of the leading mode. Therefore by regularizing effects for parabolic equations we have that lR(y,~)j 5 C’7-4-l for T > 1 and IyI 5 n/r. In particular we obtain that @ - g(T)@,(y) as 7 >> 1 uniformly in compact sets 1~15 R. Substituting now (6.48) into (6.47) and taking the scalar product with $,(y), we obtain:
Vol. 16. no I-1999.
100
L,. HERRAIZ
The convergence + - ,y(r)li:O(y) on compact sets gives that:
We now have that by (6.45):
where Cr is a constant which does not depend on M. Finally, on using the bound (6.45) for each one for the terms in G(y,r), there holds:
IG(:y,~)ldy
= O(e?),
for some E > 0. For instance, we have that for some a, b > 0 such that a+b
= i:
From (6.51) we obtain, on taking the limit 111 --+ c~, that as 7 > 1:
As Go(y) = (47r-+e-q we have that y(r) N (CO(p - 1)~)~* for C, =< $$>I+!I,, >= (2,/%p%)-l. Back to our original variables we obtain that zl(z,t) N C,v(tlogt)-9 in regions IX/’ < Ct. where CN = (47r)-T(C,(p - l))-* = ($(l + $)u)%. To obtain the convergence in the regions 1~1’ < Ct log t for any C < 2(tu - N) one only needs to apply the maximum principle to functions (6.32) and (6.38). taking now constants M and MI equal to C,v f E, which completes the proof. n
SEMILINEAR
PARABOLIC
7. THE PROOF
101
PROBLEMS
OF THEOREM
7
In this Section we shall be concerned with the case when: (7.1)
Our first result is a non-optimal asymptotic estimate in an external region, namely: 7.1. - Let u(x, t) be the solution of (l.l)-(1.3), (7.1) holds. Then: LEMMA
u(x, t) N AIxl-”
and assume that
as t -+ 00 , uniformly on regions
{ (2, t) : /xl2 > kt log t}
(7.2)
with k > 2(c1 - N).
Proof. - As we have done in the previous Section, we just compare ~(5, t) with the solution of (1. I)-( 1.3) to derive a suitable upper bound, and then compare it from below with the solution I,(z, t) of (1 .l)-(1.3) with p replaced by ?, = 1 + :, this last function being estimated by means of n Lemma 6.1. A rather crude stabilization result in a parabolic inner region is provided by: LEMMA
7.2. - Let u(z: t) be as in the statement of Lemma 7.1. One then
has that: -1 u(z, t) = t p-ly(~:/&)(l
+ o(l)) as t -+ co,
uniformly on sets (1x1 5 Cfi} with C > 0, where g is a nontrivial
solution of (1.22).
(7.3)
Proof - It follows from the results obtained in [7]. As a matter of fact,
it has been proved there that under our assumptions: )iir
t+susuv
IU(zA) - t-*g(+/qj)
= 0.
whereupon (7.3) follows. n Let now y be a positive real number, and let F(s) be a solution of the following problem: [F”(S)+
I
F(R,)
(~+:)F’(s)+($--+y)F(s)=O
= C,
F(s) > 0 for s > R,,
F(s) - s 2( ~+44,4
Vol. 16. no I-1999.
fors>R,,
for s i
cc.
(7.4)
102
L. HERRAIZ
The existence of F(s) for some positive constants C, R, can be shown along the lines of (5.9) where the corresponding result for algebraically decreasing solutions was obtained. Recall now the change of variables given in (5.3). Then there holds: ‘7.3. - For any positive numbers M, y and E, the function
LEMMA
gqy; r) = (1 - c)g(y) - MF(y)e?‘,
(7.5)
is a subsolution of (4.20) in the region where @ is nonnegative. Moreover, for any R, large enough , there exist r,, > R,, M > 0 such that: 9(Y,
To) i WY> To)
TP(Y!r) 5 @(Y/,T)
,for R, 5 y < k7,: with k > 2(~ - N).
(7.6)
fory=krandr>r,,.
(7.8)
Proof. - Checking that (7.5) is a subsolution where positive follows from a straightforward computation. On the other hand, we have that TI!(Y,~) - (1 - c)g(y) for y = r >> 1, whence (7.8) follows by the external Lemma 7.1. Take now R, > 1 (in particular we need it to be greater than the largest zero of F). Then, by Lemma 7.2 one has that Q(R,,r,) > (1 - e)g(R,) > g(y;ro) which gives (7.7). It just remains to obtain (7.6). To this end, we observe that for fixed ru B 1, @(y,r,) 2 (1 - E)g(y) 2 9(y:rc,T,) for IYI 2 B = B(r,) in view of Lemma 7.2. By selecting R, sufficiently large, we may always assume that F(y) 2 Ci > 0 for R, < y < B. Inequality (f6) follows now n on taking M >> 1 in (7.5). A second subsolution that will used presently is provided by the following: LEMMA 7.4. - Let E > 0 be such that 0 < E < A. Then if cy > N there exists a constant R, = R,(A, N, p, CY,E, y) > 0 such that, for F(s) and y as before and M > 0 arbitrary, the function:
+,t)
= ((A-~)[~~-~~
- MI++h-T)+
(7.9)
satisjes: tL, 5 Au- - Id1 -
for [x1* > R,t, Anna/e.\
de l’lnsrirut
Henri
t > 1. PoinctrrL’
(7.10) Analyse
non lintaire
SEMILINEAR
Moreover, for 0 < ‘u(z,t,)
E
PARABOLIC
103
PROBLEMS
<< 1 and t, >> 1 one has that:
5 u(x,t,)
(7.11)
for [xl2 > R,t,,
3(x, t) 5 u(z, t)
for
[xl2 = R,t?
1:> t,.
(7.12)
Proof. - s in the case of our previous Lemma, a quick computation reveals that ~(r, t) given in (7.9) satisfies (7.10). To proceed further, we observe that there exists a constant R, large enough such that F is positive for /:c(~ > R,t, and such that (7.12) holds. To see the second statement we note that if 1~1~ = R,t, by the previous Lemma: a(~, t) < ct-4
< (1 - E)g(R,)t-p-l
5 us(x, t).
(7.13)
n Finally, once t, is chosen, we take M large enough to obtain (7.11). Let us summarize a bit. Putting together Lemmata 7.1-7.4 we have shown that we can obtain subsolutions that provide bounds for U(Z, t) exactly as required in Theorem 7. This can be seen, for instance, by checking that the second term in the right of (7.9) is negligible with respect to the first one there for T >> A. To conclude the proof of such result, we yet need to obtain supersolutions displaying the required behaviour. To this end, we consider the auxiliary function:
5(&T) = (1 ++I&)
+g&) + Fe-“;
where, as usual, E is a fixed (but small), positive number, gl(T) solution of (1.22), M > 0 is a free parameter and g2(T) solves:
(7.14) is the
We now prove: LEMMA 7.5. - Function q(y, -r) given in (7.14) is a supersolution of (7.15) for 1y( 2 R, and r 2 1, where R, is a positive number depending on N and p. Moreover, there exists I-~ >> 1 such that:
@(Y> To) I -(Y,
To>
Q(Y,T> 5 Q(Y,T)
for Y 2 R,,
(7.16)
fory=R,andr>r,.
(7.17)
Proof. - Since the sum of supersolutions is a new supersolution, the first statement in the Lemma is easily checked by noting that each Vol. 16, no I-1999.
104
L. HERRAIZ
one of the three terms on the right in (7.14) is itself a supersolution, which is straightforward. To obtain (7.16), we first select r,, >> 1 so that @(y, r) 5 (1+ $)$11(y) for y = R, and 7- > 70 (this can be done by Lemma 7.2). By redefining r,, if necessary, we next observe that the solution U(X. t) under consideration is such that u(:~:,t) < (A + s)I:cI-” for /:1:12> S(t). where S(t) = k:t log t with X: > 2( (I: - N). We next remark that for large enough t, t 2 (A+E)Ix-” for 1~1 > S(t) (this follows since di gz ( 3 > S(t) >> t for t >> 1). Denoting by 21(:1:.t) the function corresponding to q(y; T) under the self-similar resealing, we then have that: ?L(:l:.t,,) 5 ‘IL(:c.t,,)
for l:r:12> S(t,,).
To take care of the region Rz 5 ~x:~~5 S(t,): we make use of the third term on the right of (7.14). More precisely, we select &f >> 1 there. Finally the proof of (7.17) is straightforward. n End of the proof qf’ Theorem 7:
We deduce from Lemma 7.4 and the maximum principle that: u(z, t) < q:c. t)
for IX/ 2 R,&
and t 2 t,, >> 1.
(7.18)
Comparing the various behaviours of the functions appearing on the right of (7.14), we readily see that: q(y:~)
- (1 + E)gi(y) for 1 << y 5 6
q(W.7) - gn(y) for y > &
with k < 2(o - 5’.
with k > 2(o - 2). P-1
(7.1’3) (7.20)
It follows from (7.19)-(7.20) and Lemma 7.2 that, in order to obtain a supersolution exhibiting the required behaviour, the only region that needs of further analysis is: c = { (:r:,t) : Ct < 1~:/’ 5 k:t logt}
2 with C > 0 and k < 2(tr - =).
A (7.21) since gl (r) and g2(T) are of the same order in C. To conclude the argument, we now consider an auxiliary function z(y, T) given by &(y, T) in (7.14) without the term g2(y) included there. Clearly, @y, 7) is a supersolution. Moreover, the lower boundary corresponding to t = tl is taken care of by selecting M > 1 as before. As to the side conditions, 1~1~ = Ct is included in Lemma 7.4, whereas 1~1~ = ,kt log t follows from the result obtained with the previous supersolution. This concludes the proof. n
SEMILINEAR
PARABOLIC
105
PROBLEMS
ACKNOWLEDGMENTS
This work has been supported by a grant from Ministerio de Educaci6n y Ciencia of Spain, as well as by DGICYT Research Grant PB930438. The author is indebted to Professors Miguel A. Herrero and Juan J. L. Velazquez for many interesting discussions held during the preparation of the manuscript.
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Vol. 16, no I-1999.
received
November
1.5, 1996.)