Behavior of solutions of nonlinear second-order elliptic inequalities

Nonlinear Analysis 42 (2000) 1253 – 1270

www.elsevier.nl/locate/na

Behavior of solutions of nonlinear second-order elliptic inequalities Andrej A. Kon’kov Department H -11, N.E. Bauman Moscow State Technical University, 2aya Baumanskaya ul. 5, 107005 Moscow, Russia Received 6 April 1998; accepted 20 November 1998

Keywords: Nonlinear elliptic inequalities; Unbounded domains

1. Introduction Suppose that 6= ∅ is an open subset of Rn ; n ≥ 2. We consider a dierential operator of the form L=

n X

n

aij (x)

i; j=1

X @2 @ + bi (x) ; @xi @xj @xi

(1)

i=1

where bi are locally bounded functions and aij satisfy the ellipticity condition 0¡

n X

aij (x)i j ≤ C||2

i; j=1

with some constant C ¿ 0 for all x = (x1 ; : : : ; x n ) ∈ and  = (1 ; : : : ; n ) ∈ Rn . The following notation is used: Qry is the open ball in Rn of radius r and center at a point y (Qry = {x: |x − y| ¡ r}); Sry is the sphere Sry = {x: |x − y| = r}; and Zry is the closed ball Zry = {x: |x − y| ≤ r}. If y = 0, we write Qr ; Sr , and Zr instead of Qr0 ; Sr0 , and Zr0 . Let F : × [0; ∞) → [0; ∞) be some function such that F(x; t1 ) ≥ F(x; t2 ) for all t1 ≥ t2 ¿ 0; x ∈ . We study nonnegative solutions of the inequality Lu ≥ F(x; u)

in ∩ QR \ ZR0 ;

 The research was supported by RFBR school grant No. 96-15-96177. E-mail address: [email protected]. (A.A. Kon’kov)

0362-546X/00/$ - see front matter ? 2000 Elsevier Science Ltd. All rights reserved. PII: S 0 3 6 2 - 5 4 6 X ( 9 9 ) 0 0 1 5 4 - 6

(2)

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A.A. Kon’kov / Nonlinear Analysis 42 (2000) 1253 – 1270

0 ≤ R0 ¡ R ≤ ∞, satisfying the condition u|(QR ∩@ )\ZR0 = 0:

(3)

By a solution of (2), (3) we mean a function u ∈ C 2 ( ∩ QR \ ZR0 ) ∩ C( ∩ QR \ ZR0 ) satisfying inequality (2) and condition (3) in the classical sense. Assume that r ∈ (R0 ; R) and Sr ∩ 6= ∅. We de ne M (r; u) by setting M (r; u) = max u:

∩Sr

The questions considered in this paper have been studied by a number of authors [1–3,5 –9]. Our aim is to estimate M (·; u) by solutions of the ordinary dierential equation that is the radial part of the Laplace operator. For L =
there exist two well-known ways to succeed in it: using the spherical coordinates in Rn , either to construct barrier functions [1,6,7] or to average solutions of inequality (2) on the unit sphere [5,8,9]. But both these methods are not admissible for the operator L of the general form (1). Diculties are overcome by Theorem 2.1 contained in this article. Similar result for divergent operators without lowest derivatives was earlier obtained in [4]. 2. Comparison theorems Theorem 2.1. Let  ¿ 1; 0 ≤ R0 ¡ R ≤ ∞; and SR0 ∩ 6= ∅. Also suppose that B : (R0 ; ∞) → [0; ∞) and f : (R0 ; ∞) × (0; ∞) → [0; ∞) are measurable functions such that n X |bi (x)| for any r ¿ R0 ; (4) B(r) ≥ sup x∈ ∩Qr \(Zr= ∪ZR0 ) i=1

f(r; t) ≤

inf

x∈ ∩Qr \(Zr= ∪ZR0 )

F(x; t)

for any r ¿ R0 ; t ¿ 0;

(5)

f(r; t − 0) = f(r; t) for any r ¿ R0 ; t ¿ 0

(6)

f(r; t1 ) ≥ f(r; t2 )

(7)

and for any r ¿ R0 ; t1 ≥ t2 ¿ 0:

Finally; let u be a nonnegative solution of (2); (3) satisfying conditions 0 ¡ M (R0 + 0; u)

(8)

and M (r1 ; u) ≤ M (r2 ; u)

for any R0 ¡ r1 ≤ r2 ¡ R:

(9)

Then for all real numbers k1 ¿ 0; k2 ¿ 0 there exist constants  ¿ 0; ¿ 0 that depend only on n; ; k1 ; k2 ; and the ellipticity constant C of the operator L such that

A.A. Kon’kov / Nonlinear Analysis 42 (2000) 1253 – 1270

the Cauchy problem   1 + k1 00 + k2 B(r) m0 = f(r; m); m + r m(R0 ) = M (R0 + 0; u);

m0 (R0 ) = 0

1255

(10) (11)

has a solution on [R0 ; R) satisfying the inequality M (r; u) ≥ m(r) for all r ∈ (R0 ; R). Remark 2.1. For n ≥ 3, by setting k1 = n − 2 and k2 = 1 in the condition of Theorem 2.1, we obtain the radial part of the operator
+ B(|x|)∇|x| · ∇ on the left-hand side in (10). The proof of Theorem 2.1 relies on the following assertion. Theorem 2.2. Under the assumptions of Theorem 2:1; for all real numbers k1 ¿ 0; k2 ¿ 0 there exist constants  ¿ 0; ¿ 0 that depend only on n; ; k1 ; k2 ; C such that Z r K(r; )f(; M (; u)) d (12) M (r; u) − M (R0 + 0; u) ≥  R0

for all r ∈ (R0 ; R); where  1+k1 Z r R  −k2 B(s) ds  d e : K(r; ) =  

(13)

Proof of Theorem 2.1. Let Theorem 2.2 be already proved. We construct a sequence of maps mi : [R0 ; R) → [0; ∞); i = 0; 1; 2; : : : , by setting m0 (r) ≡ M (R0 + 0; u) and Z r K(r; )f(; mi−1 ()) d; mi (r) = M (R0 + 0; u) +  R0

i = 1; 2; : : : . It can easily be seen from the de nition of the f that f(·; mi (·)) are measurable functions and mi−1 (r) ≤ mi (r) ≤ M (r; u) for any r ∈ (R0 ; R); i = 1; 2; : : : . Hence there exists a map m : [R0 ; R) → (0; ∞) such that mi → m almost everywhere in [R0 ; R) as i → ∞. We obviously have m(r) ≤ M (r; u) for every r ∈ (R0 ; R). Using Lebesque’s bounded convergence theorem, we also get Z r K(r; )f(; m()) d m(r) = M (R0 + 0; u) +  R0

for all r ∈ (R0 ; R). Thus m is a solution of the Cauchy problem (10), (11). 3. Proof of Theorem 2.2 Let  ¿ 1; k1 ¿ 0, and k2 ¿ 0 be some given real numbers. Next, let 0 ≤ R0 ¡ R ≤ ∞ and SR0 ∩ 6= ∅. Henceforward we assume that u is a solution of inequality (2)

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A.A. Kon’kov / Nonlinear Analysis 42 (2000) 1253 – 1270

satisfying conditions (3), (8), and (9). We also suppose that B : (R0 ; ∞) → [0; ∞) and f : (R0 ; ∞)×(0; ∞) → [0; ∞) are measurable functions such that relations (4) – (7) are valid. Without loss of generality, it can be assumed that B() ¿ 0 for all  ∈ (R0 ; ∞); otherwise, instead of B we consider a function B+, where  is a positive real number, and let  tend to zero afterwards. We put  = 1=3 , where  is the constant from the condition of Theorem 2.1. Finally, we de ne the function : (0; ∞) → (0; ∞) by setting (a) =

1 1 − 2 (1 − e−a ): a a

The map satis es the following properties. (a) If a1 ≥ a2 , then (a1 ) ≤ (a2 ). Proof. It is sucient to establish the validity of the equality Z 1 Z 1 d d e−a(−) : (a) = 0



(b) Suppose that 0 ≤ r1 ¡ r and b is a positive constant such that b ≤ B() for all  ∈ (r1 ; r); then Z r K(r; ) d ≤ (r − r1 )2 (bk2 (r − r1 )): (14) r1

Proof. Using (13), we get Z r 1 d e−bk2 (−) = (1 − e−bk2 (r−) ) K(r; ) ≤ bk2  for every  ∈ (r1 ; r), whence (14) follows at once. (c) For all 1 ¿ 0 and 2 ¿ 0 we have (1 ) ≥

1 2 e−2 : 6 1 + 2

Proof. If 1 ≥ 2, then Let 1 ¡ 2; then (1 ) =

(15)

(1 ) ≥ 1=(21 ) and 1 ≥ 2 e−2 . Therefore (15) is trivial.

1 1 1 (−1)i i−2 1 1 − 1 + · · · +  + ··· ≥ − 1 ≥ ; 2! 3! i! 1 2! 3! 6

whence we again derive (15). (d) For any  ¿ 0 1 − e− ≤ 6 ():

(16)

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Proof. Let  ≥ 2. It can easily be seen that (16) is equivalent to inequality   6 (1 − e− ) ≤ 6; 1+  which is evidently valid. Suppose that  ¡ 2; then  () =

2  2  i−1  − + · · · + (−1)i + ··· ≥ − ≥ : 2! 3! i! 2 6 6

Combining this with the inequality 1 − e− ≤ , we obtain (16). Lemma 3.1. Let R0 ¡ r1 ¡ r2 ¡ R; 0 ¡ ≤ 1=2; and M (r1 ; u) ≥ 2 M (r2 ; u). Also assume that b ¿ 0 is some real number such that n X |bi (x)|: b≥ sup x∈ ∩Qr2 \Zr1 i=1

Then M (r2 ; u) − M (r1 ; u) ≥ 1 (r1 − r2 )2 (bk2 (r2 − r1 ))

inf

∈ ∩Qr2 \Zr1

F(; M (r2 ; u));

(17)

where the constant 1 ¿ 0 depends only on n; k2 ; and the ellipticity constant C of the operator L. Proof. Take a point y ∈ S(r1 +r2 )=2 ∩ such that u(y) = M ((r1 + r2 )=2; u). Denote: v(x) = u(x) − M ((r1 + r2 )=2; u)=2 and ! = {x ∈ : v(x) ¿ 0}. Since y ∈ !, we have ! 6= ∅. Suppose that  ¿ 0 is some real number. Let us de ne the in nitely smooth function w : Rn → [0; ∞) by setting w (x) = 4 1 (|x − y|2 (bk2 |x − y| ) − 2 (bk2 )) Pn

where |x − y| = ( We get L(|x −

y|2

i=1 (xi

inf

∈ ∩Qr2 \Zr1

F(; M (r2 ; u));

− yi )2 + 2 )1=2 ; 1 = k2 =(8 + 4k2 (n + 1)C). n

X 1 (bk2 |x − y| )) = (1 − e−bk2 |x−y| ) aii (x) bk2 |x − y| i=1

n

X (xi − yi )(xj − yj ) 1 (1 − e−bk2 |x−y| ) aij (x) − bk2 |x − y| |x − y|2 i; j=1

n n X (xi − yi )(xj − yj ) 1 X xi − yi + bi (x) + e−bk2 |x−y| aij (x) bk2 |x − y| |x − y|2 i=1

−

1 −bk2 |x−y| e bk2

i; j=1

n X i=1

bi (x)

xi − yi 2 ≤ (n + 1)C + |x − y| k2

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A.A. Kon’kov / Nonlinear Analysis 42 (2000) 1253 – 1270

for any x ∈ ! ∩ Z(ry 2 −r1 )=2 . Therefore, Lw (x) ≤

inf

∈ ∩Qr2 \Zr1

F(; M (r2 ; u))

(18)

for all x ∈ ! ∩ Z(ry 2 −r1 )=2 . Denote =

(v − w );

sup

!∩Z(ry

2 −r1 )=2

then, by the relation 1 v(y) − w (y) = M 2 we obtain 1 ≥ M 2





 r1 + r2 ; u ¿ 0; 2

 r1 + r2 ; u ¿ 0: 2

(19)

On the other hand, since w depends only on |x − y|, there exists a constant A ¿ 0 such that w (x) = A for every x ∈ S(ry 2 −r1 )=2 . The following equality is valid: sup

!∩S(ry

v −  = A :

(20)

2 −r1 )=2

In fact, the inequality sup

!∩S(ry

v −  ≤ A

2 −r1 )=2

is given by the de nition of the . Let us prove that sup

!∩S(ry

v −  ≥ A :

(21)

2 −r1 )=2

Assume the converse. Then there exists a real number ˜ ∈ (0; ) such that ˜ sup v ¡ A + : !∩S(ry

(22)

2 −r1 )=2

˜ Taking into account (2) and (18), we obtain Consider a function w(x) ˜ = w (x) + . L(v(x) − w(x)) ˜ ≥0 y . At the same time, relation (22) implies the following inequality: for all x ∈ !∩Q(r 2 −r1 )=2

(v − w)| ˜ @(!∩Qy

(r2 −r1 )=2

)

≤ 0:

Thus, according to the maximum principle, v(x) − w(x) ˜ ≤0

A.A. Kon’kov / Nonlinear Analysis 42 (2000) 1253 – 1270

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for all x ∈ ! ∩ Z(ry 2 −r1 )=2 , or, in other words, ˜ ≥

sup

!∩Z(ry

(v − w ):

2 −r1 )=2

This contradiction proves (21). Combining (19), (20), and the evident inequality   1 r1 + r2 ; u ≥ sup v; M (r2 ; u) − M 2 2 !∩S y (r2 −r1 )=2

we have



M (r2 ; u) − M

r1 + r2 ;u 2

 ≥ A ;

whence, letting  tend to zero, we obtain   r1 + r2 ;u M (r2 ; u) − M 2   bk2 (r2 − r1 ) 2 inf F(; M (r2 ; u)): ≥ 1 (r1 − r2 ) 2 ∈ ∩Qr2 \Zr1 At last, (17) is given by the condition (9) and property (a) of the map

.

Lemma 3.2. Let R0 ¡ r1 ¡ r ≤ r0 ¡ R; r1 ≤ r; 0 ¡ ≤ 1=2; and M (r1 ; u) ≥ 2 M (r; u); then Z r K(r0 ; )f(; M (; u)) d; (23) M (r; u) − M (r1 ; u) ≥ 2 r1

where the constant 2 ¿ 0 depends only on n; C; k1 ; k2 ; and . Proof. Set i = r1 i ; i = 0; 1; 2; : : : . There is a natural number k such that k ≤ r and k+1 ¿ r. For every natural number i satisfying inequality 1 ≤ i ≤ k − 1, there exists ∗ ∈ (i−1 ; i ) such that Z i Z i K(r0 ; )f(; M (; u)) d ≤ f(∗ ; M (∗ ; u)) K(r0 ; ) d: (24) i−1

i−1

Suppose the constant b ¿ 0 is given by b=

inf

∈(i−1 ;i+1 )

B():

We have K(r0 ; ) = K(i+1 ; ) + for any  ∈ (i−1 ; i ).

Z

r0

i+1

d

 1+k1 R  −k2 B(s) ds  e 

(25)

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A.A. Kon’kov / Nonlinear Analysis 42 (2000) 1253 – 1270

For the second summand on the right in (25) the following estimate is valid:  1+k1  1+k1 Z r0 Z ∞ R   −k2 B(s) ds −bk2 (i+1 −i )  d e ≤e d   i+1  1 −bk2 (i+1 −i ) = e : k1 At the same time, by property (b) of the map , we get Z i+1 Z i K(i+1 ; ) d ≤ K(i+1 ; ) d ≤ (i+1 − i−1 )2 (bk2 (i+1 − i−1 )): i−1

i−1

Therefore, integrating (25) over , we obtain Z i K(r0 ; ) d ≤ (i+1 − i−1 )2 (bk2 (i+1 − i−1 )) i−1

+

2i − 2i−1 −bk2 (i+1 −i ) e : 2k1

It can easily be seen from property (c) of the map

(26) that

(bk2 (i+1 − i−1 )) ≥ 3 e−bk2 (i+1 −i ) ; where the constant 3 ¿ 0 depends only on . Hence (26) implies the following inequality: Z i K(r0 ; ) d ≤ (i+1 − i−1 )2 (bk2 (i+1 − i−1 ));

4 i−1

where the constant 4 ¿ 0 depends only on  and k1 . Combining this with (24), we obviously get Z i K(r0 ; )f(; M (; u)) d

4 i−1

≤ (i+1 − i−1 )2 (bk2 (i+1 − i−1 ))f(∗ ; M (∗ ; u)): On the other hand, from Lemma 3.1, it follows that M (i ; u) − M (i−1 ; u) ≥ 1 (i − i−1 )2 (bk2 (i − i−1 ))f(∗ ; M (∗ ; u)): Thus, M (i ; u) − M (i−1 ; u) Z i ≥ 5 K(r0 ; )f(; M (; u)) d; i−1

i = 1; : : : ; k − 1;

where 5 = 1 4 ( + 1)−2 . Further, there exists a real number ∗ ∈ (k−1 ; r) such that Z r Z r K(r0 ; )f(; M (; u)) d ≤ f(∗ ; M (∗ ; u)) K(r0 ; ) d: k−1

k−1

(27)

(28)

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1261

Now we set b=

inf

∈(k−1 ;)

B():

Assume that r0 ≤ r. Using property (b), we obtain Z r0 Z r K(r0 ; ) d ≤ K(r0 ; ) d ≤ (r0 − k−1 )2 (bk2 (r0 − k−1 )): k−1

k−1

Therefore, from (28), it follows that Z r K(r0 ; )f(; M (; u)) d k−1

≤ (r0 − k−1 )2 (bk2 (r0 − k−1 ))f(∗ ; M (∗ ; u)): At the same time, by Lemma 3.1, we have M (r; u) − M (k−1 ; u) ≥ 1 (r − k−1 )2 (bk2 (r − k−1 ))f(∗ ; M (∗ ; u)): Combining the last two inequalities, we obviously get Z r K(r0 ; )f(; M (; u)) d; M (r; u) − M (k−1 ; u) ≥ 6 k−1

(29)

(30)

where the constant 6 ¿ 0 depends only on 1 and . Now assume that r0 ¿ r. Repeating the previous arguments (see (26)), we readily obtain Z r K(r0 ; ) d ≤ (r − k−1 )2 (bk2 (r − k−1 )) k−1

+

r 2 − 2k−1 −bk2 (−1)r e : 2k1

(since we can always replace i−1 ; i , and i+1 by k−1 ; r, and r, respectively). The last inequality and property (c) of the map imply that Z r K(r0 ; ) d ≤ (r − k−1 )2 (bk2 (r − k−1 ));

7 k−1

where the constant 7 ¿ 0 depends only on  and k1 . Hence relation (28) can be rewritten as Z r K(r0 ; )f(; M (; u)) d

7 k−1

≤ (r − k−1 )2 (bk2 (r − k−1 ))f(∗ ; M (∗ ; u)): Combining this with (29), we again derive (30), where 6 = 1 7 (2 +  + 1)−2 . To complete the proof of the lemma it remains to notice that (23) is given by summing (27) and (30).

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A.A. Kon’kov / Nonlinear Analysis 42 (2000) 1253 – 1270

Lemma 3.3. Let R0 ¡ r1 ¡ r ¡ R; 0 ¡ ≤ 1=2; and M (r1 ; u) ≥ 2 M (r; u); then Z r K(r; )f(; M (; u)) d; M (r; u) − M (r1 ; u) ≥ 8 r1

where the constant 8 ¿ 0 depends only on n; C; k1 ; k2 ; and . Proof. If r1 ≤ r, then Lemma 3.3 follows from Lemma 3.2. Suppose that r1 ≥ r. There exists a real number ∗ ∈ (r1 ; r) such that Z r Z r K(r; )f(; M (; u)) d ≤ f(∗ ; M (∗ ; u)) K(r; ) d: r1

r1

Denote b = inf

∈(r1 ; r)

B():

Using Lemma 3.1 and property (b) of the map

, we obviously get

2

M (r; u) − M (r1 ; u) ≥ 1 (r − r1 )

(bk2 (r − r1 ))f(∗ ; M (∗ ; u)) Z r K(r; ) d: ≥ 1 f(∗ ; M (∗ ; u)) r1

Thus, Lemma 3.3 is proved. Lemma 3.4. Assume that R0 ¡ r1 ¡ r ≤ r0 ¡ R; r1 ≤ r; A ∈ [2; ∞); and AM (r1 ; u) = M (r; u). Also let  ∈ (0; 1=A] and ∈ (0; 1=(2A)] be some real numbers such that Z r1 K(r1 ; )f(; M (; u)) d: M (r1 ; u) ≥  R0

Then

Z M (r; u) − M (r1 ; u) ≥ 9 A

r1

R0

K(r0 ; )f(; M (; u)) d;

(31)

where the constant 9 ¿ 0 depends only on n; ; k1 , k2 ; C. Proof. At rst we suppose that R0 ≥ r1 . There exists ∗ ∈ (R0 ; r1 ) such that Z r1 Z r1 K(r0 ; )f(; M (; u)) d ≤ f(∗ ; M (∗ ; u)) K(r0 ; ) d: R0

R0

On the other hand, we have K(r0 ; ) = K(r1 ; ) + for all  ∈ (R0 ; r1 ). By de nition, put b=

inf

∈(R0 ;r1 )

B():

Z

r0

r1

 1+k1 R  −k2 B(s) ds  d e 

(32)

(33)

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For the second summand on the right in (33) the following estimate is valid:  1+k1  1+k1 Z ∞ Z r0 R   −k2 B(s) ds  d e ≤ e−bk2 (−1)r1 d   r1  1 −bk2 (−1)r1 = e : k1 Further, taking into account property (b) of the map , we obviously get Z r1 Z r1 K(r1 ; ) d ≤ K(r1 ; ) d ≤ (r1 − R0 )2 (bk2 (r1 − R0 )): R0

R0

Therefore, Z r1 r 2 − R20 −bk2 (−1)r1 K(r0 ; ) d ≤ (r1 − R0 )2 (bk2 (r1 − R0 )) + 1 e ; 2k1 R0 whence, by property (c), we have Z r1 K(r0 ; ) d ≤ r12 (bk2 (r1 − R0 ));

10 R0

where the constant 10 ¿ 0 depends only on  and k1 . The last estimate and inequality (32) imply that Z r1 K(r0 ; )f(; M (; u)) d ≤ r12 (bk2 (r1 − R0 ))f(∗ ; M (∗ ; u)):

10 R0

At the same time, according to Lemma 3.1, M (r1 ; u) − M (r1 ; u) ≥ 1 ( − 1)2 r12 (bk2 ( − 1)r1 )f(∗ ; M (∗ ; u)): Combining the last two inequalities, we evidently get Z r1 K(r0 ; )f(; M (; u)) d; M (r1 ; u) − M (r1 ; u) ≥ 11 R0

where 11 = 1 10 ( − 1)2 . Thus estimate (31) is valid. Now let R0 ¡ r1 and Z Z r1 1 r1 K(r0 ; )f(; M (; u)) d ≥ K(r0 ; )f(; M (; u)) d; 2 R0 r1 =

(34)

(35)

then repeating the argument given in the proof of (34), with R0 replaced by r1 =, we readily obtain Z r1 K(r0 ; )f(; M (; u)) d; M (r1 ; u) − M (r1 ; u) ≥ 11 r1 =

whence, taking (35) into account, we have (31). Finally, assume that R0 ¡ r1 and Z Z r1 = 1 r1 K(r0 ; )f(; M (; u)) d ≥ K(r0 ; )f(; M (; u)) d: 2 R0 R0

(36)

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A.A. Kon’kov / Nonlinear Analysis 42 (2000) 1253 – 1270

It follows from the conditions of the lemma that Z r1 = K(r1 ; )f(; M (; u)) d: M (r1 ; u) ≥  R0

(37)

On the other hand, for every  ∈ (R0 ; r1 =)  1+k1 Z r1 Rr Rr  1 −k2 1 B(s) ds −k2 1 B(s) ds −k1   d ≥ (1 −  )e : K(r1 ; ) ≥ e  k1  At the same time, it is obvious that  1+k1 Z r0 R  −k2 B(s) ds  d e K(r0 ; ) = K(r1 ; ) +  r1 for all  ∈ (R0 ; r1 =). For the second summand on the right in the last equality, the following estimate holds:  1+k1  1+k1 Z ∞ Z r0 R Rr   −k2 B(s) ds −k2 1 B(s) ds   d e ≤e d   r1  R 1 −k2 r1 B(s) ds = e : k1 Hence, K(r1 ; ) ≥ 12 K(r0 ; ) for all  ∈ (R0 ; r1 =), where the constant 12 ¿ 0 depends only on  and k1 . Combining this with (36) and (37), we conclude that Z r1 1 K(r0 ; )f(; M (; u)) d; M (r1 ; u) ≥  12 2 R0 whence, taking into account the evident inequality M (r; u) − M (r1 ; u) ≥

A M (r1 ; u); 2

we get (31). Thus Lemma 3.4 is completely proved. From now on we denote =

1 ; +1

and

  = min



A = max

1 8 1 2 ; ; ; A 3 12 2

2 2 ;  9

 ;

=

1 ; 2A2

 ;

where 1 ¿ 0; 2 ¿ 0; 8 ¿ 0, and 9 ¿ 0 are the constants of Lemmas 3.1, 3.2, 3.3, and 3.4, respectively.

A.A. Kon’kov / Nonlinear Analysis 42 (2000) 1253 – 1270

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Lemma 3.5. Let R0 ¡ r1 ¡ r ¡ R; r1 ≥ r; AM (r1 ; u) = M (r; u); and let Z  K(; )f(; M (; u)) d M (; u) ≥  R0

for any  ∈ (R0 ; r1 ); then M (r; u) − M (r1 ; u) ≥ 2

Z

r1

R0

(K(r; ) − K(r1 ; ))f(; M (; u)) d:

(38)

Proof. The proof is by induction over the minimal nonnegative integer N such that AN M (R0 + 0; u) ≥ M (r1 ; u). If N ≤ 1, then Lemma 3.3 implies that Z r K(r; )f(; M (; u)) d; M (r; u) − M (R0 + 0; u) ≥ 8 R0

whence, taking into account the inequality A (M (r; u) − M (R0 + 0; u)) A+1 ≥ 23 (M (r; u) − M (R0 + 0; u));

M (r; u) − M (r1 ; u) ≥

we obtain (38). Now suppose that N0 ≥ 2 and for every N ¡ N0 Lemma 3.5 is already proved. Let us show that the lemma is valid for N = N0 too. We construct a nite sequence of real numbers {ri }Ni=2 such that R0 ¡ ri ¡ ri−1 and i A M (ri ; u) = M (r; u) for any i = 2; : : : ; N . It is evident that M (rN ; u) ≤ AM (R0 + 0; u). By de nition, put rN +1 = R0 . By  we denote the set of natural numbers j such that 2 ≤ j ≤ N and for all natural numbers i ∈ {2; : : : ; j} the following conditions are valid: ri−1 − ri ≤ i−1 (r − r1 )

ri ≥ ri−1 ; and

Z

ri R0

(K(r; ) − K(r1 ; ))f(; M (; u)) d

1 ≥ 2

Z

ri−1

R0

(K(r; ) − K(r1 ; ))f(; M (; u)) d:

If  6= ∅, then we put k = max ; otherwise k = 1. Let us consider all possible cases. (1) Assume that rk+1 ¡ rk (possibly rk+1 = R0 ) and, moreover, the relation Z rk (K(r; ) − K(r1 ; ))f(; M (; u)) d rk+1

1 ≥ 2 is satis ed.

Z

rk

R0

(K(r; ) − K(r1 ; ))f(; M (; u)) d

(39)

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A.A. Kon’kov / Nonlinear Analysis 42 (2000) 1253 – 1270

Then, using Lemma 3.2, we obviously have Z rk K(r; )f(; M (; u)) d M (rk ; u) − M (rk+1 + 0; u) ≥ 2 ≥

2 2k

rk+1 r1

Z

R0

(K(r; ) − K(r1 ; ))f(; M (; u)) d:

Thus, taking into account the evident inequality M (r; u) − M (r1 ; u) ≥ Ak (M (rk ; u) − M (rk+1 + 0; u));

(40)

we get  k Z r1 A

2 (K(r; ) − K(r1 ; ))f(; M (; u)) d 2 R0 Z r1 (K(r; ) − K(r1 ; ))f(; M (; u)) d: ≥ 2

M (r; u) − M (r1 ; u) ≥

R0

(2) Suppose that (39) holds and, moreover, rk+1 ≥ rk and rk − rk+1 ¿ k (r − r1 ). Then there exists a real number ∗ ∈ (rk+1 ; rk ) such that Z rk (K(r; ) − K(r1 ; ))f(; M (; u)) d rk+1

Z

≤ f(∗ ; M (∗ ; u))

rk

rk+1

(K(r; ) − K(r1 ; )) d:

We de ne the constant b ¿ 0 by setting b=

inf

∈(rk+1 ;rk )

B():

For all  ∈ (rk+1 ; rk ) the following relation is valid:  1+k1 Z r R  −k2 B(s) ds  d e K(r; ) − K(r1 ; ) =  r1  1+k1 Z r  −bk2 (rk −) ≤e d ≤ (r − r1 )e−bk2 (rk −) :  r1 Therefore, Z rk rk+1

r − r1 (1 − e−bk2 (rk −rk+1 ) ) bk2 rk − rk+1 (1 − e−bk2 (rk −rk+1 ) ); ≤ k  bk2

(K(r; ) − K(r1 ; )) d ≤

whence, by property (d) of the map , we have Z rk (K(r; ) − K(r1 ; )) d ≤ 6−k (rk − rk+1 )2 (bk2 (rk − rk+1 )): rk+1

(41)

A.A. Kon’kov / Nonlinear Analysis 42 (2000) 1253 – 1270

1267

Hence estimate (41) can be rewritten as Z rk (K(r; ) − K(r1 ; ))f(; M (; u)) d rk+1

≤ 6−k (rk − rk+1 )2 (bk2 (rk − rk+1 ))f(∗ ; M (∗ ; u)): At the same time, according to Lemma 3.1, M (rk ; u) − M (rk+1 + 0; u) ≥ 1 (rk − rk+1 )2 (bk2 (rk − rk+1 ))f(∗ ; M (∗ ; u)): Combining the last two inequalities and (40), we conclude that Z rk

1 k (K(r; ) − K(r1 ; ))f(; M (; u)) d M (r; u) − M (r1 ; u) ≥ (A) 6 rk+1  k Z r1

1 A (K(r; ) − K(r1 ; ))f(; M (; u)) d: ≥ 6 2 R0 This implies relation (38). (3) Assume that rk+1 ≥ rk ; rk − rk+1 ¿ k (r − r1 ), and Z rk+1 (K(r; ) − K(r1 ; ))f(; M (; u)) d R0

1 ≥ 2

Z

rk

R0

(K(r; ) − K(r1 ; ))f(; M (; u)) d:

(42)

If rk+1 =R0 , then the right-hand side of (38) is equal to zero. Hence we may suppose that rk+1 ¿ R0 . By the induction hypothesis, Z rk+1 (K(rk ; ) − K(rk+1 ; ))f(; M (; u)) d: (43) M (rk ; u) − M (rk+1 ; u) ≥ 2 R0

At the same time, for any real number  ∈ (R0 ; rk+1 ) the following relation holds:  1+k1 Z rk R  −k2 B(s) ds  d e K(rk ; ) − K(rk+1 ; ) =  rk+1  1+k1 Z rk Rr  −k2 k B(s) ds  ≥e d  rk+1 R rk − rk+1 1+k1 −k2 rk B(s) ds ≥  e : rk1+k1 On the other hand, for every  ∈ (R0 ; rk+1 )  1+k1 Z r R  −k2 B(s) ds  d e K(r; ) − K(r1 ; ) =  r1  1+k1 Z r Rr  −k2 k B(s) ds  ≤e d  r1 R r − r1 1+k1 −k2 rk B(s) ds ≤ 1+k1  e : r1

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A.A. Kon’kov / Nonlinear Analysis 42 (2000) 1253 – 1270

Therefore we have K(rk ; ) − K(rk+1 ; ) ≥ k (K(r; ) − K(r1 ; )) for all  ∈ (R0 ; rk+1 ). Hence the estimate (43) implies that Z rk+1 k (K(r; ) − K(r1 ; ))f(; M (; u)) d: M (rk ; u) − M (rk+1 ; u) ≥ 2 R0

Combining this with (40), we get the inequality Z rk+1 (K(r; ) − K(r1 ; ))f(; M (; u)) d M (r; u) − M (r1 ; u) ≥ 2(A)k  ≥ 2

R0

A 2

k Z

r1

R0

(K(r; ) − K(r1 ; ))f(; M (; u)) d;

from which we obviously obtain (38). (4) Suppose that (39) is valid and, moreover, rk − rk+1 ≤ k (r − r1 ) and rk+1 ≥ rk . In this case, it can easily be seen that rk+1 ≥ r1 . There exists a real number ∗ ∈ (rk+1 ; rk ) such that Z rk (K(r; ) − K(r1 ; ))f(; M (; u)) d rk+1

Z

≤ f(∗ ; M (∗ ; u))

rk

rk+1

(K(r; ) − K(r1 ; )) d:

Denote b = inf

∈(r1 ;r)

B();

then we evidently have Z K(r; ) − K(r1 ; ) = ≤

r

r

Z 1r r1

 1+k1 R  −k2 B(s) ds  e  1 d e−bk2 (−r1 ) = (1 − e−bk2 (r−r1 ) ) bk2 d

for all  ∈ (rk+1 ; rk ). Therefore, the following estimate holds: Z rk rk − rk+1 (K(r; ) − K(r1 ; )) d ≤ (1 − e−bk2 (r−r1 ) ) bk2 rk+1 ≤

k (r − r1 ) (1 − e−bk2 (r−r1 ) ); bk2

whence, taking into account property (d) of the map , we get Z rk (K(r; ) − K(r1 ; )) d ≤ 6k (r − r1 )2 (bk2 (r − r1 )): rk+1

(44)

A.A. Kon’kov / Nonlinear Analysis 42 (2000) 1253 – 1270

1269

Combining this with (44), we conclude that Z rk (K(r; ) − K(r1 ; ))f(; M (; u)) d rk+1

≤ 6k (r − r1 )2 (bk2 (r − r1 ))f(∗ ; M (∗ ; u)): Further, according to Lemma 3.1, M (r; u) − M (r1 ; u) ≥ 1 (r − r1 )2 (bk2 (r − r1 ))f(∗ ; M (∗ ; u)): Combining the last two inequalities, we obviously obtain Z rk

1 (K(r; ) − K(r1 ; ))f(; M (; u)) d M (r; u) − M (r1 ; u) ≥ k 6 rk+1 Z r1

1 (K(r; ) − K(r1 ; ))f(; M (; u)) d; ≥ 6(2)k R0 whence (38) follows. (5) Assume that (42) is valid and rk+1 ¡ rk . We may also assume that rk+1 ¿ R0 ; if this is not so, the right-hand side of (38) is equal to zero. It follows from Lemma 3.4 that Z rk+1 K(r; )f(; M (; u)) d; M (rk ; u) − M (rk+1 ; u) ≥ 9 A R0

whence, by (40), we have M (r; u) − M (r1 ; u) ≥ Ak+1 9 A ≥

Z

rk+1

R0

K(r; )f(; M (; u)) d

 k Z r1 A A 9  K(r; )f(; M (; u)) d: 2 R0

This implies (38). Thus Lemma 3.5 is completely proved. Proof of Theorem 2.2. The proof is by induction over the minimal nonnegative integer N such that AN M (R0 + 0; u) ≥ M (r; u). If N ≤ 2, then Theorem 2.2 follows from Lemma 3.3. Now suppose that N0 ¿ 2 and Theorem 2.2 holds for all N ¡ N0 . Let us show that Theorem 2.2 is also valid for N = N0 . Take a real number r1 ∈ (R0 ; r) such that AM (r1 ; u) = M (r; u). By the induction hypothesis, Z r1 K(r1 ; )f(; M (; u)) d: (45) M (r1 ; u) − M (R0 + 0; u) ≥  R0

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A.A. Kon’kov / Nonlinear Analysis 42 (2000) 1253 – 1270

Hence the theorem will be proved if we succeed in establishing the validity of the inequality Z r1 (K(r; ) − K(r1 ; ))f(; M (; u)) d M (r; u) − M (r1 ; u) ≥  R0

Z

+

r

r1

K(r; )f(; M (; u)) d

since, summing (45) and (46), we readily obtain (12). On the other hand, (46) is given by the two following estimates: Z r1 (K(r; ) − K(r1 ; ))f(; M (; u)) d M (r; u) − M (r1 ; u) ≥ 2 R0

and

Z M (r; u) − M (r1 ; u) ≥ 2

r

r1

K(r; )f(; M (; u)) d:

(46)

(47)

(48)

It can easily be seen that Lemma 3.3 implies (48). At the same time, using Lemmas 3.4 and 3.5, we obviously get (47). Thus, Theorem 2.2 is completely proved. References [1] J.B. Keller, On solution of 4u = f(u), Comm. Pure Appl. Math. 10 (4) (1957) 503–510. [2] V.A. Kondratiev, E.M. Landis, Qualitative prorerties of the solutions of a second-order nonlinear equations, Mat. sb. 135 (3) (1988) 346–360. [3] V.A. Kondratiev, A.A. Kon’kov, On properties of solutions of a class of nonlinear second-order equations, Mat. sb. 185 (9) (1994) 81–94. [4] A.A. Kon’kov, Positive solutions of nonlinear second-order elliptic inequalities in unbounded domains, Russ. J. Math. Ph. 5 (1) (1997) 119–122. [5] I. Kuzin, S. Pohozaev S, Entire solutions of semilinear elliptic equations, Prog. Nonlinear Dierential Equations Appl. 33, 1991. [6] R. Osserman, On the inequality 4u ≥ f(u), Paci c J. Math. 7 (1957) 4. [7] J.L. Vazquez, An a priori interior estimate for the solutions of a nonlinear problem representing weak diusion, Nonlinear Anal. 5 (1981) 95–103. [8] L. Veron, Comportement asymptotique des solutions d’equations elliptiques semi-lineaires dans Rn , Ann. Math. Pure Appl. 127 (1981) 25–50. [9] L. Veron, Singularities of Solutions of Second Order Quasilinear Equations, Addison-Wesley, Reading, MA, 1996.