Universal bounds for global solutions of a forced porous medium equation

Nonlinear Analysis 57 (2004) 349–362www.elsevier.com/locate/na

Universal bounds for global solutions of a forcedporous medium equation

Michael Winkler∗

Department of Mathematics, Aachen University of Technology, Templergraben 55,Aachen D-52056, Germany

Abstract

We show that in a smooth bounded domain � ⊂ Rn, n¿ 2, all global nonnegative solutionsof ut − +um = up with zero boundary data are uniformly bounded in � × (�;∞) by a con-stant depending on �;p and � but not on u0, provided that 1¡m¡p¡ [(n + 1)=(n − 1)]m.Furthermore, we prove an a priori bound in L∞(� × (0;∞)) depending on ‖u0‖L∞(�) underthe optimal condition 1¡m¡p¡ [(n+ 2)=(n− 2)]m.? 2004 Elsevier Ltd. All rights reserved.

Keywords: Nonlinear di6usion; A priori estimates

1. Introduction

Consider the problem

ut −+um = up in � × (0; T );

u|@� = 0;

u|t=0 = u0 (1.1)

in a smooth bounded domain � ⊂ Rn, n¿ 1, with 1¡m¡p and u0 being a non-negative continuous function vanishing at @�. It is well-known [15] that for small u0,(1.1) has a global (in time) solution, while if u0 is su:ciently large then u blowsup in ;nite time. In the semilinear case m = 1, it has been shown in [3,7,13,14] thatif n6 4 and p¡ (n + 2)=(n − 2) or n¿ 4 and p¡ (n − 1)=(n − 3) then all globalsolutions, no matter how large u0 is, are a priori uniformly bounded in �× (�;∞) by

∗ Tel.: +49-516-661-3967; fax: +49-519-850-2459.E-mail address: [email protected] (M. Winkler).

0362-546X/$ - see front matter ? 2004 Elsevier Ltd. All rights reserved.doi:10.1016/j.na.2004.02.019

mailto:[email protected]

350 M. Winkler /Nonlinear Analysis 57 (2004) 349–362

a universal constant depending on �;p and � only; for a survey on further literatureon this topic, we refer to [14,6].In the degenerate cas m¿ 1, it is known (cf. [11,5]) that if p¡ [(n+2)=(n− 2)]m

then global solutions are bounded, and that this conclusion is false for p¿ [(n+2)=(n−2)]m (and n¿ 3). However, to the best of our knowledge it has not been clari;ed sofar whether in the subcritical case p¡ [(n+2)=(n− 2)]m a priori bounds of the form

‖u‖L∞(�×(0;∞))6C(‖u0‖L∞(�); p; �) (1.2)

or

‖u‖L∞(�×(�;∞))6C(�; p; �) (1.3)

are valid for all global solutions of (1.1). Recently, part of these questions wereanswered a:rmatively in [17], where (1.2) could be proved for p¡m + [(10m +2)=(3n − 4)] (which is less than (n + 2)=(n − 2)m) and (1.3) was established forp¡ [(n+ 2)=n]m.In this paper we aim at presenting entirely di6erent techniques which allow an

extension of these results up to the optimal exponent, that is, for all p¡ [(n+2)=(n−2)]m, in (1.2) (cf. Theorem 4.3) and up to p¡ [(n+1)=(n−1)]m in (1.3) (see Theorem4.4).On the way towards our main assertions, we make use of weighted Sobolev in-

equalities (Lemma 2.3) and an iterative method (Lemma 3.4) as well as a rescalingargument similar to that introduced in [8] for the semilinear case (Lemma 4.2).The author would like to thank Pavol Quittner and Philippe Souplet for fruitful

discussions on the subject of the present paper.

2. Approximation of solutions

Following [?], by a solution of (1.1) we mean a nonnegative function u∈C0([0; T ];L2(�))∩L∞(�×(0; T )) with um ∈L2((0; T );W 1;2

0 (�)) such that for all ’∈W 1;2((0; T );L2(�)) ∩ L∞((0; T );W 1;2

0 (�)), we have∫�u(t)’(t)−

∫ t

0

∫�(u’t −∇um · ∇’+ up’) =

∫�u0’(0) ∀t ∈ [0; T ]

u is said to be a global solution if T may be chosen arbitrarily large. Accordingto Theorem 12 in [2], weak solutions of (1.1) are unique.

Lemma 2.1. Suppose u is a global solution of (1.1) and let T ¿ 0 be given. Thenthere is �0 ∈ (0; 1) such that for all �∈ (0; �0) the regularized problem

u�t −+um� = up

� in � × (0; T );

u�|@� = �;

u�|t=0 = u0 + � (2.1)

has a unique classical solution u�¿ � with u� → u in C0( J�×[0; T ]) and in C2;1loc ({u¿ 0}

∩{t ¿ 0}) as � → 0.

M. Winkler /Nonlinear Analysis 57 (2004) 349–362 351

Proof. The claim follows easily by slightly modifying the corresponding proceduresin e.g. [2] or [18]. That the limit of the u� in fact coincides with u is a consequenceof the uniqueness of weak solutions.

In particular, Lemma 2.1 asserts that u is actually a continuous function being smoothwherever both u and t are positive.

3. Integral bounds

We start with the classical ‘eigenfunction test’ to derive weighted integral estimates.Throughout, we denote by � the ;rst eigenfunction of −� in � normalized such that∫� � = 1, corresponding to the ;rst eigenvalue �1 ¿ 0. It is well-known that due tothe smoothness of @�, c−1dist (x; @�)6�(x)6 cdist (x; @�) holds for some c¿ 0.

Lemma 3.1. There is a constant C ¿ 0 such that∫��u(t)6C1 ∀t ∈ (0;∞) (3.1)

and ∫ t

0

∫��up(t)6C2(1 + t) ∀t ∈ (0;∞): (3.2)

Proof. We multiply (2.1) by �, integrate over � × (0; t) and let � → 0 to see that∫� �u is absolutely continuous with respect to t and for a.e. t ¿ 0

@t

∫��u+ �1

∫��um =

∫��up:

By Young’s inequality we obtain �1um6 12u

p + cu, whence

@t

∫��u¿

12

∫��up − c

∫��u

¿12

(∫��u)p

− c∫��u; (3.3)

which implies, since u does not blow up in ;nite time, that∫��u(t)6 (2c)1=(p−1) ∀t ∈ (0;∞);

that is, (3.1). Now (3.2) follows upon integrating (3.3) over (0; t).

From (3.2) we immediately infer a weighted Lp bound for arbitrarily small times.Here and in the sequel, � denotes a small but ;xed positive constant.

Corollary 3.2. There is t1 ∈ (0; �=8) such that∫��up(t1)6

2C2

�: (3.4)


In a next step, we wish to get rid of the weight � in e.g. (3.2) and, as well, achievehigher integration powers. Both this will be done in Lemma 3.4 the proof of whichessentially requires the weighted Sobolev and interpolation inequalities presented in

Lemma 3.3. (i) Given �∈ (0; 1), let n?� := �+n=(�+n)=2−1. Then for all q∈ (1; n?

� ]there is a constant c¿ 0 such that

(∫��vq

)1=q6 c

(∫��|∇v|2

)1=2∀v∈C∞

0 (�):

(ii) In addition, suppose � and � are positive numbers satisfying �¡min{n?� ; �}

and �¡ 2 + (1 − 2=n?� )�. Then there is r ¿ 1 such that for any ¿ 0 there exists

c¿ 0 with

∫��v�6

∫��|∇v|2 + c

(∫��v�

)r∀v∈C∞

0 (�):

Proof. (i) is proved in Chapter 1.5 in [4].(ii) Let us split � = �1 + �2 with �1 := (� − �)=(1 − �=n?

� ) which is in (0; �) since�¡n?

� and �¡�. From �¡ 2 + (1− 2=n?� )� it follows at once that �1 ¡ 2¡n?

� andthus q1 := n?

� =�1 and q2 := 2=�1 satisfy 1¡q2 ¡q1, so that r := q′2=q′1 ¿ 1, where

1=qi+1=q′i =1; i=1; 2. Moreover, we easily calculate q′1�2=[(�−�1)=(n?� −�1)]n?

� =�.We now use HOolder’s and Young’s inequalities and part (i) to estimate

∫��v�6

(∫��vq1�1

)1=q1 (∫��vq

′1�2

)1=q′1

6 "(∫

��vq1�1

)q2=q1+ c(")

(∫��vq

′1�2

)q′2=q′1

6 c"∫��|∇v|2 + c

(∫��v�

)r

for any "¿ 0, from which the assertion follows.

The main result of this section is contained in

Lemma 3.4. For any q∈ [1;∞) there exist C3 ¿ 0, t2 ∈ (t1; �=2) and Jtq ¿ 0 such that

∫ t2+Jtq

t2

∫�uq6C3:

Proof. The proof is based upon an iteration method, using (3.4) as starting point. For�∈ [0; 1) and #¿ 1, multiply (2.1) by ��u#−1

� and integrate over �× (0; t) to obtain


for t ¿ 01#

∫��u#

� (t) +4m(#− 1)

(m+ #− 1)2

∫ t

0

∫��|∇u(m+#−1)=2

� |2

+m��1

m+ #− 1

∫ t

0

∫��um+#−1

� +m�(1− �)m+ #− 1

∫ t

0

∫��−2|∇�|2um+#−1

�

− �#−1∫ t

0

∫@�

��@Num� +

mm+ #− 1

�m+#−1∫ t

0

∫@�

@N��

61#

∫��(u0 + �)# +

∫ t

0

∫��up+#−1

� : (3.5)

Omitting positive terms and letting � → 0, we see that∫� ��u# is absolutely continuous

and, writing v := u(m+#−1)=2, that for a.e. t ¿ 01#@t

∫��v� + c#

∫��|∇v|26

∫��v� (3.6)

with � := 2#=(m+#−1)∈ (0; 2) and � := [2(p+#−1)=(m+#−1)]¿ 2. Now Lemma3.3(ii) applies to give∫

��v�6

c#2

∫��|∇v|2 + c

(∫��v�

)rfor some r ¿ 1, provided that �¡ 2 + (1 − 2=n?

� )� which is equivalent to #¿ (p −m)=(1− 2=n?

� ) or

#¿n+ �2

(p− m): (3.7)

In this case,

1#@t

∫��v� +

c#2

∫��|∇v|26 c0

(∫��v�

)rfor a:e: t ¿ 0: (3.8)

Assuming∫��v�(t?)6M ¡∞ (3.9)

for some t?¿ 0, we deduce from (3.8) that there exist C and Jt depending on r;Mand c0 such that∫ t?+Jt

t?

∫��|∇v|26C

and thus, by Lemma 3.3(i),∫ t?+Jt

t?

∫��vn?� 6C:

Concerning u, this means that if # and � are such that (3.7) holds as well as∫��u#(t?)6M; (3.10)


then ∫ t?+Jt

t?

∫��u(m+#−1)=2 n?� 6C: (3.11)

To make our choice of # and � concrete, observe that since p¡ [(n + 1)=(n − 1)]m,there is #1 ¿ 1 such that [(n + 1)=2](p − m)¡#1 ¡p. Choosing then �∈ (2#1=p; 1),we infer from Corollary 3.2 and HOolder’s inequality that∫

��u#1 (t1)6

(∫��up(t1)

)#1=p(∫��(p�−#1)=(p−#1)

)(p−#1)=p

6C;

since �" is integrable i6 "¿− 1. We conclude that (3.11) is valid with t? = t?1 := t1and #= #1. In particular, there exists t?2 ∈ (t?1 ; 38�) such that∫

��u#2 (t?2 )6C;

where

#2 =m+ #1 − 1

2n?� ¿

n+ 1n− 1

(m+ #1 − 1)¿n+ 1n− 1

#1 ¿#1:

We continue inductively to obtain a sequence of times t?k+1 ∈ (t?k ; (1 − 1=2k+1)�=2)and #k+1 = [(m + #k − 1)=2]n?

� ¿ [(n + 1)=(n − 1)]#k such that (3.11) is true (with Jtdepending on k). As #k → ∞, we conclude, using once again HOolder’s inequality, thatfor arbitrarily large q there are t? ∈ (t0; �=2), Jtq ¿ 0 and C ¡∞ with∫ t?+Jtq

t?

∫�uq6C;

which proves the lemma.

4. Pointwise bounds

Now a method familiar in the context of second-order parabolic equations (see [1])enables us to turn the integral estimate of Lemma 3.4 into an L∞ bound.

Lemma 4.1. There are t3 ∈ (0; �) and C4 ¡∞ such that

‖u(t3)‖L∞(�)6C4:

Proof. Fixing q¿ (n+ 2)=2, Lemma 3.4 allows us to rewrite (1.1) in the form

ut −+um = f(x; t) (4.1)

with ‖f‖Lq(�×(t2 ;t2+Jtq))6C(q; �). Note that due to the smoothness of u, (4.1) holdsclassically in {u¿ 0} ∩ {t ¿ 0}.


We choose Jt ∈ (0;max{�=4; Jtq=2}) and, following a standard iteration procedure,introduce for nonnegative integers j the parabolic cylinders

Qj := � × Ij with Ij := (Tj; T ]; Tj := t2 +(1− 1

2j

)Jt and T := t2 + 2Jt;

and the functions

zj := (u− kj)q=2+ with kj :=

(1− 1

2j+1

)k; k ¿ 0;

as well as nondecreasing smooth cut-o6 functions 06 ,j(t)∈C∞0 (Ij) with ,j ≡ 1 in

Ij+1 and ,jt6 2j+2= Jt. Multiplying (4.1) by (u− kj+1)q−1+ ,2j , we obtain

supt∈Ij

∫�,2j (u− kj+1)

q+ + m(q− 1)

∫ ∫Qj

,2j um−1(u− kj+1)

q−2+ |∇u|2

62j+3

Jt

∫ ∫Qj

(u− kj)q+ +

∫ ∫Qj

f(u− kj)q−1+ :

As ∫ ∫Qj

,2j um−1(u− kj+1)

q−2+ |∇u|2¿ ckm−1

∫ ∫Qj

,2j |∇(u− kj+1)q=2+ |2

and ∫ ∫Qj

f(u− kj)q−1+ 6

(∫ ∫Q0

fq)1=q(∫ ∫

Qj

(u− kj)q+

)1−1=q

;

we see that

supt∈Ij

∫�,2j z

2j+1 + ckm−1

∫ ∫Qj

,2j |∇zj+1|2

6 c2j

∫ ∫

Qj

z2j +

(∫ ∫Qj

z2j

)1−1=q :

Writing Aj+1 := {(x; t)∈Qj | u(x; t)¿kj+1}, we observe that

|Aj+1|6 1(kj+1 − kj)2

∫ ∫Qj

z2j 64j+1

k2

∫ ∫Qj

z2j :

Together with the interpolation inequality (cf. (4.5) and (4.2) in [10, pp. 74–75]) thisgives ∫ ∫

Qj+1

z2j+16∫ ∫

Qj

,2j z2j+1

6 |Aj+1|2=(n+2)

(∫ ∫Qj

(,jzj+1)2(n+2)=n

)n=(n+2)

6 c|Aj+1|2=(n+2)

(supt∈Ij

∫�,2j z

2j+1

)2=(n+2)(∫ ∫Qj

,2j |∇zj+1|2)n=(n+2)


and leads to∫ ∫Qj+1

z2j+16 c(2 · 42=(n+2))jk−(n(m−1)+4)=(n+2)

(∫ ∫

Qj

z2j

)1+n=(n+2)

+

(∫ ∫Qj

z2j

)1−1=q+2=(n+2) :

Employing a variant of Lemma 5.6 in [10], we conclude that if k is large enough suchthat ∫ ∫

Q0

uq6 ckn(m−1)+4=(n+2)(2=(n+2)−1=q)−2

then∫∫

Qjz2j → 0 as j → ∞, that is, u6 k in �× (t2 + Jt; t2 + 2Jt), which completes the

proof, since t2 + 2Jt ¡ �.

The next lemma will be a key to Theorem 4.3; it carries over the lemma in 9—proved for the semilinear case m = 1 there—to the present situation. For the reader’sconvenience, we adopt some of the notation used there.

Lemma 4.2. Suppose that a solution u of (1.1) ful7ls∫ T

t′

∫�um−1u2t 6N (4.2)

for some t′¿ 0 and assume that for a number t0 ¿t′,

sup�×(t′ ;T )

u is attained in � × (t0; T ]:

Then there is a constant M depending on �;p; N and t0 − t′ but not on u such that

‖u‖L∞(�×(t′ ;T ))6M:

Proof. If the assertion were false, (1.1) would have a sequence of solutions uk in� × (t′; T ] such that

Mk := ‖uk‖L∞(�×(t′ ;T )) = u(xk ; tk) → ∞with certain xk ∈� and tk ∈ (t0; T ], where we may assume xk → x∞ ∈ J� as k → ∞.Writing �k := M−(p−m)=2

k → 0, we rescale uk as well as the solutions uk� of (2.1)(approximating uk) by letting

vk(y; s) := �2=(p−m)k uk(xk + �ky; tk + �"

k s);

vk�(y; s) := �2=(p−m)k uk�(xk + �ky; tk + �"

k s)

with " := 2(p− 1)=(p− m). Then we have

vk(0; 0) = 1; (4.3)


vk 6 1 and vk�6 2 for s6 0 and su:ciently small �¡ �0(k); moreover,

vk�s −+vmk� = vpk�: (4.4)

(i) If x∞ ∈� then vk and vk� are de;ned in Q(d=�k), where Q(r) := {(y; s) | |y|¡r;−r" ¡ s6 0} and d := min{(t0 − t′)1="; dist(x∞; @�)}. By Lemma 2.1, vk� → v inC0( JQ(d=�k)) and in C2;1

loc ({vk ¿ 0}).Given K ⊂⊂ Q(∞) := Rn×(−∞; 0], we ;rst note that according to HOolder estimates

for degenerate equations of type (4.4) (cf. [12]),

‖vk�‖C�; �=2(K)6C (4.5)

for some �¿ 0, su:ciently large k and small �. Next, ;x ’∈C∞0 (Q(∞)) with ’ ≡ 1

in K and test (4.4) by v#−1k� ’2, #¿ 0. As a result, we obtain upon letting � → 0∫ ∫

K|∇v(m+#−1)=2

k |26C(#; K): (4.6)

Consequently, for a subsequence k = kj → ∞ we have, using standard parabolicSchauder estimates [10],

vk → v in C0loc(Q(∞)) and in C2;1

loc ({v¿ 0}) (4.7)

as well as ∇v�k * ∇v� in L2loc(Q(∞)) for all �¿ (m − 1)=2. We claim that in the

latter assertion we may replace weak convergence with strong convergence. Indeed, ;x�¿ (m−1)=2 and let �′ be any number in ((m−1)=2; �). Then we have for K ⊂⊂ Q(∞)and arbitrary �¿ 0∫ ∫

K|∇v�k −∇v�|2 =

(��′

)2 ∫ ∫K|v�−�′

k ∇v�′

k − v�−�′∇v�′ |2

6 c∫ ∫

K|v�−�′

k − v�−�′ |2|∇v�′

k |2

+ c∫ ∫

K∩{v¿�}v2(�−�′)|∇v�

′

k −∇v�′ |2

+ c�2(�−�′)∫ ∫

K∩{v6�}(|∇v�

′

k |2 + |∇v�′ |2)

=: I1 + I2 + I3:

Given �¿ 0, due to (4.6) and (4.7) we can choose ;rst � small such that I3 ¡�=2 forall k and then k0 large such that I1 + I2 ¡�=2 for all k ¿k0; hence,

∇v�k → ∇v� in L2(K): (4.8)

Now from (4.2) we gain∫ ∫Q(d=�k )

vm−1k v2ks6 �−n+2(p+m)=(p−m)

k

∫ T

t′

∫�um−1k u2kt → 0; (4.9)

since p¡ (n+ 2)=(n− 2)m.


If we multiply (3.4) by v(m−1)=2k ’, ’∈C∞

0 (Q(∞)) and let � → 0, we obtain for klarge enough∫ 0

−∞

∫Rn

v(m−1)=2k vks’+

∫ 0

−∞

∫Rn

∇vmk · ∇(v(m−1)=2k ’)

=∫ 0

−∞

∫Rn

v(m−1)=2+pk ’: (4.10)

Noting that∫ 0

−∞

∫Rn

∇vmk · ∇(v(m−1)=2k ’) =

8m(m− 1)(3m− 1)2

∫ 0

−∞

∫Rn

|∇v(3m−1)=4k |2’

+2m

3m− 1

∫ 0

−∞

∫Rn

∇v(3m−1)=2k · ∇’

and (3m − 1)=4 (and clearly (3m − 1)=2) is larger than (m − 1)=2, we infer from(4.7)–(4.10) that v does actually not depend on s and∫

Rn∇vm · ∇(v(m−1)=2’) =

∫Rn

v(m−1)=2+p’ ∀’∈C∞0 (Rn): (4.11)

We claim that w := vm ful;ls

+w + wp=m = 0 in Rn: (4.12)

For this purpose, we ;rst note that due to (4.11), w satis;es∫Rn

∇w · ∇(w(m−1)=2m’) =∫Rn

w(m−1)=2m+p=m’ (4.13)

for all ’∈W 1;∞0 (Rn), by completion, and for any K ⊂⊂ Rn and all 6¡ (m + 1)=m,

we have∫Kw−6|∇w|26C(K; 6): (4.14)

Fix ∈C∞0 (Rn) and insert ’� := w−(m−1)=(2m)−1(w − �)+ ∈W 1;∞

0 (Rn), �¿ 0, into(4.13) to obtain∫

Rn∇w · ∇(w−1(w − �)+ ) =

∫Rn

w(p=m)−1(w − �)+ :

Clearly,∫Rn

wp=m−1(w − �)+ →∫Rn

wp=m as � → 0:

On the other hand,∫Rn

∇w · ∇(w−1(w − �)+ ) =∫{w¿�}

∇w · ∇ − �∫{w¿�}

w−1∇w · ∇

+ �∫{w¿�}

w−2|∇w|2

=: J1 + J2 + J3;


where

|J3|6 c�6−1∫Kw−6|∇w|2

for any 6∈ (1; (m + 1)=m), with K := supp . Thus, by (4.14), J3 → 0 as � → 0.Similarly,

|J2|6 c�

(∫{w¿�}∩K

w−2|∇w|2)1=2

6 c�6=2(∫

Kw−6|∇w|2

)1=2→ 0:

Finally,∫{w¿�}

∇w · ∇ →∫{w¿0}

∇w · ∇ =∫Rn

∇w · ∇ ;

the latter equality sign being due to the fact that ∇w = 0 a.e. in {w = 0} (cf. [16]).Altogether,∫

Rn∇w · ∇ =

∫Rn

wp=m ∀ ∈C∞0 (Rn):

This implies by standard elliptic regularity theory that w is a classical solution of (4.12)which according to (4.3) is nontrivial. But in [9] it is proved that all nonnegativesolutions of (4.12) must vanish identically, a contradiction.(ii) The case x∞ ∈ @� runs similarly. Changing ;rst the spatial coordinates such that

@� ⊂ {xn = 0} and x∞ = 0, we consider instead of (1.1) the equation

ut −∑i; j

aij(x)(um)xixj −∑

i

bi(x)(um)xi = up

with smooth coe:cients aij(x) and bi(x) satisfying aij(0) = �ij. Now vk and vk� arede;ned in Q(�=�k)∩ {yn¿− dk=�k} for some �¿ 0 (with dk := dist(xk ; @�)) and vk�solves

vk�s −∑i; j

aij(xk + �ky)(vmk�)yiyj − �k

∑i

bi(xk + �ky)(vmk�)yi = vpk�;

and we have vk(0; 0) = 1, vk�6 2 for small � and vk� = ��2=(p−m)k on {yn = −dk=�k}.

Again, according to Remark 1.2 in [12],

‖vk�‖C�; �=2(K∩{yn¿−dk =�k})6C;

so that in particular

1 = vk(0; 0)− vk

(−dk

�k(0; : : : ; 0; 1); 0

)

6C(dk

�k

)�;

whence dk=�k ¿ c¿ 0 for all k. If dk=�k → ∞ for a subsequence, we are in a positionquite similar to that in part (i), while if dk=�k → c¿ 0 for a subsequence, we use


obvious variants of (4.6)–(4.14) (with K ⊂⊂ Q(∞) ∩ {yn ¿ − c} now) to achievevk → v(y) for a further subsequence, where v is such that w := vm solves

+w + wp=m = 0 in {yn ¿− c} and w|yn=−c = 0:

However, by Theorem 1.3 in [9], w ≡ 0 which contradicts the fact that v(0) = 1.

With the last lemma at hand, we may now invoke standard energy methods to derivea u0-dependent bound for all subcritical p in

Theorem 3.3. Let p¡ [(n+ 2)=(n− 2)]m. Then for all a¿ 0 there exists a constantC0(a; p; �) such that

‖u‖L∞(�×(0;∞))6C0(‖u0‖L∞(�); p; �)

whenever u is a global solution of (1.1).

Proof. First, parabolic comparison of u� with the solution y(t) of the intial-value prob-lem y′ = yp, y(0) = ‖u0‖L∞(�) + 1, yields in the limit � → 0 numbers Jt and JC ¡∞depending only on ‖u0‖L∞(�) and p such that

‖u‖L∞(�×(0; Jt))6 JC: (4.15)

Multiplying (2.1) by u#−1� − �#−1, #¿ 1, and integrating over � × (0; t), t ∈ (0;∞),

we obtain1#

[∫�u#� (t)−

∫�u#� (0)

]+

4m(#− 1)(m+ #− 1)2

∫ t

0

∫�|∇u

m+#−12

� |2

=∫ t

0

∫�up+#−1� + �#−1

[∫�u�(t)−

∫�u�(0)−

∫ t

0

∫�up�

]: (4.16)

We infer that∫ T0

∫� |∇u(m+#−1)=2

� |26 c(#; T; u) and thus (cf. the proof of Lemma 4.2)that ∇u�

� converges strongly in L2 for all �¿m=2; in particular, there is a sequence�= �j → 0 along which

∇um� → ∇um in L2(� × (0; T ))

and

∇um� (t) → ∇um(t) in L2(�) ∀t ∈ S; (4.17)

where S ⊂ (0;∞) is such that (0;∞)\S has measure zero. Thus, upon the specialchoice # := m+ 1 in (4.16) we may preserve the equality sign in the limit � → 0 andconclude that

∫� um+1(t) is absolutely continuous with

1m+ 1

@t

∫�um+1 =−

∫�|∇um|2 +

∫�up+m (4.18)

for a.e. t ¿ 0.To gain more information from this, let us take t0 ∈ S and multiply (2.1) by @tum

�with the result

m∫ t

t0

∫�um−1� u2�t + E(u�(t)) = E(u�(t0)) ∀t ¿ t0; (4.19)


having the ‘energy’ functional E for (1.1) de;ned by

E(v) :=12

∫�|∇vm|2 − m

p+ m

∫�vp+m:

By (4.17), E(u�(t0)) → E(u(t0)) as � → 0, so that Fatou’s lemma gives

m∫ t

t0

∫�um−1u2t + E(u(t))6E(u(t0)) ∀t0 ∈ S; ∀t ¿ t0; (4.20)

which shows that if E(u(t0)) were negative for some t0 ∈ S then so would be E(u(t))for all t ¿ t0. Hence, if we rewrite (4.18) in the form

1m+ 1

@t

∫�um+1 =−2E(u(t)) +

p− mp+ m

∫�up+m for a:e: t ¿ 0;

we see that in this case∫� um+1 should blow up in ;nite time, a contradiction. Therefore

E(u(t)) must be nonnegative for all t ∈ S. We can now ;nd t′ ∈ (Jt=4; 3Jt=4) ∩ S, andtogether with the last result, (4.20) yields for T ¿ t′

m∫ T

t′

∫�um−1u2t 6 E(u(t′))6 inf

t∈(0; t′)∩SE(u(t))

62Jt

∫ Jt

0

∫�|∇um|26C

due to (4.18) and (4.15). Now combining (4.15) with Lemma 4.2 shows that

‖u‖L∞(�×(0;T ))6max{ JC;M}for all T ¿ 0, which implies the claim.

As an immediate consequence of Theorem 4.3 and Lemma 4.1, we may state withoutfurther comment the ;nal

Theorem 4.4. Let p¡ [(n+1)=(n− 1)]m. Then for all �¿ 0 there exists a universalconstant C1(�; p; �) such that for any global solution u of (0.1), the estimate

‖u‖L∞(�×(�;∞))6C1(�; p; �)

holds.

Remark. Alternatively, Theorem 4.4 may be proved without Lemma 3.4 and Theorem4.3 if one involves more sophisticated test functions (rather than �) in Lemma 3.1and uses a result from [15]. However, much of the self-containedness of the abovepresentation would get lost in such a procedure.

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