Home Boundedness of Solutions to a Parabolic-Elliptic Keller–Segel Equation in ℝ2 with Critical Mass
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Boundedness of Solutions to a Parabolic-Elliptic Keller–Segel Equation in ℝ2 with Critical Mass

  • Toshitaka Nagai and Tetsuya Yamada EMAIL logo
Published/Copyright: June 21, 2017

Abstract

We consider the Cauchy problem for a parabolic-elliptic system in 2, called the parabolic-elliptic Keller–Segel equation, which appears in various fields in biology and physics. In the critical mass case where the total mass of the initial data is 8π, the unboundedness of nonnegative solutions to the Cauchy problem was shown by Blanchet, Carrillo and Masmoudi [7] under some conditions on the initial data, on the other hand, conditions for boundedness were given by Blanchet, Carlen and Carrillo [6] and López-Gómez, Nagai and Yamada [23]. In this paper, we investigate further the boundedness of nonnegative solutions.

MSC 2010: 35B45; 35K15; 35K55

1 Introduction

In this paper, we study the boundedness of nonnegative solutions to the Cauchy problem for the parabolic-elliptic system

(CPψ) { t u = Δ u - ( u ψ ) , t > 0 , x 2 , - Δ ψ = u , t > 0 , x 2 , u ( 0 , x ) = u 0 ( x ) , x 2 ,

where ψ(t,x) is defined by the convolution product of the Newtonian potential N(x) and u(t,x), namely,

ψ ( t , x ) := ( N u ) ( t , x ) = 2 N ( x - y ) u ( t , y ) 𝑑 y , N ( x ) = 1 2 π log 1 | x | .

This model is a simplified version of a chemotaxis system derived from the original Keller–Segel model [19] (see also Childress and Percus [12] and Hillen and Painter [16]), which is called the parabolic-elliptic Keller–Segel equation, and can be also regarded as a model of self-attracting particles in 2 (see Biler and Nadzieja [5] and Wolansky [38]). It is known that the decay property u(t)log(1+|x|)L1(2) is necessary and sufficient for ψ(t)Lloc1(2), and that if we assume the decay condition u0log(1+|x|)L1(2) for the nonnegative initial data u0L1(2), then u(t)log(1+|x|)L1(2) for t>0. In this paper, as the decay condition u0log(1+|x|)L1(2) is not assumed, we consider the following Cauchy problem in place of (CPψ):

(CP) { t u = Δ u - ( u ( N u ) ) , t > 0 , x 2 , u ( 0 , x ) = u 0 ( x ) , x 2 ,

where

( N u ) ( t , x ) := 2 N ( x - y ) u ( t , y ) 𝑑 y , N ( x ) = - 1 2 π x | x | 2 .

One of the most important properties for (CP) is the mass conservation, namely,

2 u ( t , x ) 𝑑 x = 2 u 0 ( x ) 𝑑 x , t > 0 ,

and the qualitative behavior of nonnegative solutions to (CP) depends on the initial mass of the system. It is known that in the subcritical case 2u0𝑑x<8π the nonnegative solutions exist globally in time and decay to zero as time goes to infinity (see, e.g., [4, 8, 9, 10, 30, 29, 31]), whereas in the supercritical case 2u0𝑑x>8π, there are blowing-up solutions in finite time (see, e.g., [3, 9, 21]). Some related results to finite-time blow-up are found in, e.g., [15, 17, 18, 25, 26, 28, 36, 37], and the references therein.

In this paper, we focus attention on the critical case 2u0𝑑x=8π. The global existence of nonnegative solutions with critical mass initial data was obtained by Biler, Karch, Laurençot, and Nadzieja [4] for radially symmetric solutions, and for non-radially symmetric solutions by Blanchet, Carrillo and Masmoudi [7] under the decay condition

(1.1) 2 | x | 2 u 0 ( x ) 𝑑 x < ,

and by Nagai and Ogawa [32] under the condition u0log(1+|x|)L1(2). It was also shown in [7] that the global nonnegative solution u under assumptions (1.1) is unbounded, more precisely,

lim t u ( t , x ) = 8 π δ x 0 ( x ) in the sense of measures ,

where x0=2xu0(x)𝑑x/(8π) is the center of mass of u0 and δx0(x) stands for the Dirac delta function at x0. The existence of another type of unbounded solution was established by Naito and Senba [35]: there is a radially symmetric nonnegative initial data u0 with 2|x|2u0(x)𝑑x= for which the nonnegative solution u satisfies

0 < lim inf t u ( t ) L < , lim sup t u ( t ) L = .

By the two results on the unbounded solutions, the infinite second moment of initial data is not sufficient for the boundedness of nonnegative solutions.

Problem (CP) admits stationary solutions θb,x0 (b>0, x02) given by

θ b , x 0 ( x ) = 8 b ( | x - x 0 | 2 + b ) 2 , x 2 ,

which belong to L1(2)L(2) and satisfy

2 θ b , x 0 𝑑 x = 8 π , 2 | x | 2 θ b , x 0 ( x ) 𝑑 x = .

When x0 is the origin, we denote θb,0 by θb for simplicity, namely,

(1.2) θ b ( x ) = 8 b ( | x | 2 + b ) 2 , x 2 .

The following result on boundedness was obtained by Blanchet, Carlen and Carrillo [6, Theorem 1.6]: let u0L1(2) be a nonnegative initial data with 2u0𝑑x=8π satisfying u0logu0,u0log(e+|x|2)L1(2). Assume

(1.3) b [ u 0 ] := 2 ( u 0 - θ b ) 2 θ b - 1 / 2 𝑑 x < for some  b > 0 .

Then suptτu(t)Lp< for any τ>0 and any 1<p<. The functional b was introduced in [6] to study the large-time behavior of nonnegative solutions to (CPψ) for the critical mass case. The next result on boundedness was given by López-Gómez, Nagai and Yamada [23, Theorem 4.1]: let u0L1(2)L(2) be a nonnegative initial data with 2u0𝑑x=8π satisfying

(1.4) lim inf R ( R 2 | x | > R u 0 𝑑 x ) > 0 .

Then there is b>0 such that u(t)LpθbLp on [0,) for any 1p. Here, u0 is the symmetric rearrangement of u0 (see Section 2.2 for the definition). As θb,x0=θb for every x02 and θb(x)=O(|x|-4) (|x|), it is apparent that condition (1.4) is satisfied for u0=θb,x0. We remark that if (1.4) is satisfied, then 2|x|2u0(x)𝑑x= (see [23, Lemma 4.1]), and that if b[u0]<, then there exists hL1(2) such that

(1.5) u 0 = θ b + h , lim R ( R 2 | x | > R | h | 𝑑 x ) = 0 ,

and (1.4) is satisfied (see [23, Lemma 4.3]).

For related studies about (CP) in the critical mass case, we refer to, e.g., [11, 24, 34], and the references therein.

In this paper, we first show that property (1.4) is inherited by the nonnegative solutions of (CP).

Theorem 1.1.

For a nonnegative initial data u0L1(R2) with R2u0𝑑x=8π, let u be the (unique) nonnegative mild solution of (CP) and let Tm be the maximal existence time of u. If (1.4) is satisfied, then

(1.6) lim inf R ( R 2 | x | > R u ( t , x ) 𝑑 x ) > 0 , 0 < t < T m ,

where u is the symmetric rearrangement of u with respect to x. Hence, Tm= and for any τ>0 there exists b>0 such that for all 1p,

(1.7) u ( t ) L p θ b L p , t τ .

The definition of mild solutions and their properties are stated in Section 2.1.

We next consider whether Theorem 1.1 is valid under the following condition in place of condition (1.4):

(1.8) lim inf R ( R 2 | x | > R u 0 𝑑 x ) > 0 .

Concerning the relation between (1.4) and (1.8), we have that (1.4) implies (1.8) because of

| x | > R f 𝑑 x | x | > R | f | 𝑑 x for all  f L 1 ( 2 )

(see (4.2)). If the statement

(1.9) lim inf R ( R 2 | x | > R f 𝑑 x ) > 0 lim inf R ( R 2 | x | > R f 𝑑 x ) > 0

for nonnegative functions fL1(2) is true, then condition (1.4) in Theorem 1.1 is replaced by condition (1.8) for boundedness. However, the statement is false in general and a counter example is given in Example 5.2.

For the boundedness of radially symmetric nonnegative solutions, u0 in (1.4) can be replaced by u0. As the equation in (CP) is invariant under rotations, the uniqueness of mild solutions ensures that the mild solution u(t,x) of (CP) is radially symmetric in x if the initial data u0 is radially symmetric.

Theorem 1.2.

Let u0L1(R2) be a nonnegative initial data with R2u0𝑑x=8π satisfying condition (1.8). If u0 is radially symmetric, then the radially symmetric nonnegative mild solution u of (CP) exists globally in time and

(1.10) sup t > τ u ( t ) L < for any  τ > 0 .

If u0L(R2) in addition, then (1.10) is valid for τ=0.

We study sufficient conditions for (1.4). Taking into account (1.5) and θb(x)=O(|x|-4) (|x|), we see that if a nonnegative initial data u0L1(2) with 2u0𝑑x=8π satisfies u0(x)C|x|-q (|x|1) for some 2<q<4 and a constant C>0, then

b [ u 0 ] = for all  b > 0 .

For such initial data, we give a comparison test to show the boundedness of the nonnegative solutions.

Theorem 1.3 (Comparison test).

For a nonnegative initial data u0L1(R2), let u be the nonnegative solution of (CP). Assume that there is a nonnegative function fL1(R2) such that

u 0 ( x ) f ( x ) for  | x | 1 , lim inf R ( R 2 | x | > R f 𝑑 x ) > 0 .

Then (1.4) holds. Hence, u(t)L is bounded on [τ,) for any τ>0, provided R2u0𝑑x=8π.

The following is a consequence of Theorem 1.3.

Corollary 1.4.

Let u0L1(R2) be a nonnegative initial data with R2u0𝑑x=8π satisfying

(1.11) u 0 ( x ) C | x | - 4 , | x | 1 ,

where C is a positive constant. Then u(t)L is bounded on [τ,) for any τ>0.

Remark 1.5.

If (1.11) in Corollary 1.4 is replaced by

u 0 ( x ) C | x | - q , | x | 1 ,

where q>4, then u(t)L as t because (1.1) is satisfied.

We consider statement (1.9) again. As mentioned above, (1.9) is false in general. We give some conditions in the following theorem under which (1.9) is true. We denote by |A| the Lebesgue measure of a measurable set A2.

Theorem 1.6.

Let fL1(R2) be a nonnegative function satisfying

  1. f ( x ) 0 as | x | ,

  2. μ ( t ) := | { x 2 : f ( x ) > t } | as t + 0 .

Define r(t)=sup{|x|:f(x)t} for t>0 and assume that

(1.12) lim inf t + 0 μ ( t ) r ( t ) 2 > 0 .

Then (1.9) holds.

This paper is organized as follows: In Section 2, we recall the definition of mild solutions to (CP) and their properties, and collect some fundamentals of rearrangements and related results. In Section 3, we prove Theorems 1.1 and 1.2. In Section 4, we begin with useful lemmas related to (1.4), and then prove Theorem 1.3, Corollary 1.4 and Theorem 1.6. In Section 5, we give an application of Theorem 1.6 in Example 5.1 and a counter example to statement (1.9) in Example 5.2.

2 Preliminaries

Throughout this paper, for 1p, Lp(d) stands for the usual Lebesgue space on d with norm Lp, and in the case of d=2, for simplicity, we write

L p := L p ( 2 ) , p := L p .

We denote by + the set of all integers n0, and for α=(α1,α2,,αd)+d, we set

x α := 1 α 1 2 α 2 d α d , j := x j .

We denote by Wm,p:=Wm,p(2) the usual Sobolev spaces.

2.1 Mild Solutions

The following definition introduces the concept of mild solutions for the Cauchy problem (CP).

Definition 2.1.

Given u0L1 and T(0,), a function u:[0,T]×2 is said to be a mild solution of (CP) in [0,T] if

  1. u C ( [ 0 , T ] ; L 1 ) C ( ( 0 , T ] ; L 4 / 3 ) , sup0<t<T(t1/4u(t)4/3)<,

  2. u satisfies the integral equation

    (2.1) u ( t ) = e t Δ u 0 - 0 t e ( t - s ) Δ ( u ( s ) ( N u ) ( s ) ) 𝑑 s , 0 < t < T ,

    where etΔ is the heat semigroup

    ( e t Δ f ) ( x ) = 2 G ( t , x - y ) f ( y ) 𝑑 y , G ( t , x ) = 1 4 π t exp ( - | x | 2 4 t ) .

A function u:[0,)×2 is said to be a global mild solution of (CP) with initial data u0 if u is a mild solution of (CP) in [0,T] for all T(0,).

The integral in (2.1) is well-defined by the Lp-Lq estimates for the heat semigroup etΔ in 2,

(2.2) t m x n e t Δ f p C t - 1 / q + 1 / p - m - n / 2 f q for all  f L q ,

where 1qp and m,n+, and the fact that, for every 4/3q<2,

(2.3) f ( N g ) 2 q / ( 4 - q ) C q f q g q for all  f , g L q ,

where Cq is a positive constant depending only on q. Estimate (2.3) follows easily by combining the Hölder inequality with the next inequality of Hardy–Littlewood–Sobolev in 2: for every 1<q<2,

1 | x | g 2 q / ( 2 - q ) C q g q for all  g L q ,

where Cq>0 is a constant depending only on q.

The next result was established in [30] by adapting methods used for the vorticity equation in 2.

Proposition 2.2.

Suppose u0L1. Then there exists T=T(u0)(0,) such that the Cauchy problem (CP) has a unique mild solution u in [0,T]. Moreover, u satisfies the following properties:

  1. u ( t ) u 0 in L 1 as t 0 .

  2. For every 1 p , there holds u C ( ( 0 , T ] ; L p ) and sup 0 < t < T ( t 1 - 1 / p u ( t ) p ) < .

  3. For every + , α+2 and 1<p<, there holds txαuC((0,T];Lp).

  4. u is a classical solution of t u = Δ u - ( u ( N u ) ) in ( 0 , T ) × 2 .

  5. 2 u ( t , x ) 𝑑 x = 2 u 0 𝑑 x for all 0 < t < T .

  6. If u 0 0 and u 0 0 , then u ( t , x ) > 0 for all ( t , x ) ( 0 , T ) × 2 .

  7. If u 0 log ( 1 + | x | ) L 1 , then u ( t ) log ( 1 + | x | ) L 1 for all 0 < t < T .

Remark 2.3.

For every u0L1 with u0log(1+|x|)L1, the Cauchy problem (CPψ) admits a solution (u,ψ) because, by u(t)L and u(t)log(1+|x|)L1, the function ψ(t):=(Nu)(t) is well-defined in Lloc1. Moreover, since u(t) is smooth in x, ψ(t) is also smooth in x and -Δψ=u.

We mention the behavior of solutions at the maximal existence time of the solution, which is used later. For the proof of the proposition, see, e.g., [7, 30].

Proposition 2.4.

Let u be a nonnegative mild solution of (CP) and let Tm be the maximal existence time of the solution u. If Tm<, then

lim sup t T m 2 ( 1 + u ( t , x ) ) log ( 1 + u ( t , x ) ) 𝑑 x = .

2.2 Rearrangements

For a measurable function f:d and t, we use the following notation for simplicity:

{ f > t } = { x d : f ( x ) > t } , | f > t | = | { x d : f ( x ) > t } | ,

where |A| is the Lebesgue measure of a Lebesgue measurable set A in d.

Let f:d be a measurable function. We assume that f vanishes at infinity in the sense that ||f|>t|< for all t>0. The distribution function μf of f is defined by

μ f ( t ) = | | f | > t | , t 0 ,

and the decreasing rearrangement f of f, the generalized inverse of μf, is defined by

f ( s ) = inf { t 0 : μ f ( t ) s } , s 0 .

The function f:d, called the symmetric rearrangement or the Schwarz symmetrization of f, is defined by

f ( x ) = f ( c d | x | d ) ,

where cd is the volume of the unit ball in d.

Some basic properties about rearrangements are as follows:

  1. | | f | > t | = | f > t | = | f > t | for t>0.

  2. f is non-increasing and right-continuous on [0,).

  3. f ( 0 ) = f L ( d ) , f()=0.

  4. If f is bounded and continuous on d, then f and f are bounded and continuous on [0,) and d, respectively.

  5. If |f||g|, then fg and fg.

  6. ( c f ) = | c | f , (cf)=|c|f for c.

  7. ( f + g ) ( s 1 + s 2 ) f ( s 1 ) + g ( s 2 ) for s1,s2>0.

  8. For s>0 and t>0 satisfying s=|f>t|,

    0 s f 𝑑 σ = f > t f 𝑑 x .

The following properties are well known.

Proposition 2.5.

  1. For every Borel measurable function Φ from to [0,),

    d Φ ( | f | ) 𝑑 x = d Φ ( f ) 𝑑 x = 0 Φ ( f ) 𝑑 s .

  2. Let f , g : d be integrable on d . If 0 s f 𝑑 σ 0 s g 𝑑 σ for all s > 0 , then

    d Φ ( | f | ) 𝑑 x d Φ ( | g | ) 𝑑 x

    for all convex functions Φ : [ 0 , ) [ 0 , ) with Φ ( 0 ) = 0 .

  3. (The Hardy–Littlewood inequality) Let 1p,q, 1/p+1/q=1. For fLp(d) and gLq(d),

    d | f | | g | 𝑑 x d f g 𝑑 x = 0 f g 𝑑 s .

  4. (Contraction property) Let 1p. For f,gLp(d),

    f - g L p ( 0 , ) f - g L p ( d ) , f - g L p ( d ) f - g L p ( d ) .

  5. (The Pólya-Szegö inequality) Let fW1,p(d) (1p). Then fW1,p(d) and

    f L p ( d ) f L p ( d ) .

For the properties on rearrangements mentioned above, we refer to, for example, [2, 22, 27, 33].

We now incorporate the following lemma [30, Lemma 5.1].

Lemma 2.6.

Let v(t,x) be a smooth function on (0,T)×R2, satisfying v(t)L1L, such that v is radially symmetric in x and satisfies

t v = Δ v - ( v ( N v ) ) in  ( 0 , T ) × 2 ,

where

( N v ) ( t , x ) := - 1 2 π 2 x - y | x - y | 2 v ( t , y ) 𝑑 y .

Then the functions φ(t,s)=v(t,x), s=π|x|2 and Φ(t,s):=0sφ(t,σ)𝑑σ satisfy

2 v ( t , x ) 𝑑 x = 0 φ ( t , s ) 𝑑 s , 0 < t < T ,
t Φ ( t , s ) = 4 π s s 2 Φ ( t , s ) + Φ ( t , s ) s Φ ( t , s ) , 0 < t < T , s > 0 .

Let u be a nonnegative mild solution of (CP) on [0,T) with nonnegative initial data u0L1. For the decreasing rearrangement u of the solution u with respect to the space variable x, define the function H(t,s) by

H ( t , s ) = 0 s u ( t , σ ) 𝑑 σ , 0 < t < T , s 0 .

Then we have the following proposition [13, 14, 30].

Proposition 2.7.

For almost all t(0,T),

t H - 4 π s s 2 H - H s H 0 , a.a.  s > 0 .

We give the following comparison principle for later uses (see [23, Proposition 3.3]).

Proposition 2.8.

For nonnegative initial data u0,v0L1, let u(t,x) and v(t,x) be nonnegative solutions of (CP) on [0,T) corresponding to the initial data u0 and v0, respectively. Assume that v0 and v are radially symmetric in x and

v ( t , x ) = φ ( t , s ) , v 0 ( x ) = φ 0 ( s ) , s = π | x | 2 .

If

0 s u 0 𝑑 σ 0 s φ 0 𝑑 σ for all  s > 0 ,

then

0 s u ( t , σ ) 𝑑 σ 0 s φ ( t , σ ) 𝑑 σ for all  s > 0 ,  0 < t < T .

3 Proofs of Theorems 1.1 and 1.2

In order to prove Theorems 1.1 and 1.2, we begin with the following lemma, which is a slight modification of [23, Lemma 4.2] with the decreasing rearrangement f of f replaced by f0 defined in Lemma 3.1. The proof of Lemma 3.1 is similar to that of [23, Lemma 4.2], and is omitted.

Lemma 3.1.

Let fL1 be a nonnegative radially symmetric function satisfying R2f𝑑x=8π. Define f0 by f(x)=f0(s), s=π|x|2.

  1. Assume that

    lim inf R ( R 2 | x | > R f 𝑑 x ) = 1 π lim inf s ( s s f 0 𝑑 σ ) > 0 .

    Then there exist b 0 > 0 and s 0 > 0 such that

    (3.1) 0 s f 0 𝑑 σ < 0 s θ b 0 𝑑 σ for all  s s 0 .

  2. If f L 1 L and ( 3.1 ) is satisfied, then there exists b ( 0 , b 0 ) such that

    0 s f 0 𝑑 σ < 0 s θ b 𝑑 σ for all  s > 0 .

We remark that as θb defined by (1.2) is radially symmetric and decreasing in |x|, we have the equalities θb(x)=θb(x)=θb(π|x|2) and

θ b ( s ) = 8 π 2 b ( s + π b ) 2 , 0 s θ b 𝑑 σ = 8 π s s + π b .

Proof of Theorem 1.1.

Let Tm be the maximal existence time of u. Recall that u0(x)=u0(s), s=π|x|2. Applying Lemma 3.1 (i) as f=u0 and f0=u0 yields that there exist b0>0 and s0>0 such that

(3.2) 0 s u 0 𝑑 σ < 0 s θ b 0 𝑑 σ for all  s s 0 .

Define H0(s), H(t,s) and Θb0(s) by

H 0 ( s ) = 0 s u 0 𝑑 σ , H ( t , s ) = 0 s u ( t , σ ) 𝑑 σ , 0 < t < T m , s 0 ,
Θ b 0 ( s ) = 0 s θ b 0 𝑑 σ = 8 π s s + π b 0 , s 0 .

Then

(3.3) { 4 π s Θ b 0 ′′ ( s ) + Θ b 0 ( s ) Θ b 0 ( s ) = 0 , s > 0 , Θ b 0 ( 0 ) = 0 , Θ b 0 ( ) = 8 π ,

where =d/ds, and by Proposition 2.7,

(3.4) { t H - 4 π s s 2 H - H s H 0 , 0 < t < T m , s > 0 , H ( t , ) = 8 π , 0 < t < T m .

Here we used the fact that by Proposition 2.5 (i),

H ( t , ) = 2 u ( t , x ) 𝑑 x = 2 u 0 𝑑 x = 8 π .

We observe that H(t,s0) is continuous in t[0,Tm) by Proposition 2.5 (iv) and uC([0,Tm);L1). As H(0,s0)=H0(s0)<Θb0(s0) by (3.2), there exists τ0>0 such that

(3.5) H ( t , s 0 ) < Θ b 0 ( s 0 ) , 0 < t τ 0 .

By (3.2)–(3.5), the comparison principle on the domain (0,τ0)×(s0,) ensures that

H ( t , s ) Θ b 0 ( s ) , 0 < t τ 0 , s s 0 ,

and thus b0 can be shortened to get

H ( t , s ) < Θ b 0 ( s ) , 0 < t τ 0 , s s 0 .

As u(τ)L1L (0<τ<Tm), we remark that u(τ)L1L by the basic property (iv) on rearrangements and Proposition 2.5 (i). Applying Lemma 3.1 (ii) as f=u(τ) and f0=u(τ) yields that for any 0<ττ0 there exists b(0,b0) such that

H ( τ , s ) < Θ b ( s ) , s > 0 .

Hence, by the comparison principle on the domain (τ,Tm)×(0,), we obtain that

H ( t , s ) Θ b ( s ) , τ t < T m , s 0 .

It follows from this inequality and H(t,)=Θb()=8π that

s s u ( t , σ ) 𝑑 σ = s ( 8 π - H ( t , s ) ) s ( 8 π - Θ b ( s ) ) = 8 π 2 b s s + π b ,

which proves (1.6).

As u(τ) is in L1L for 0<τ<Tm and (1.6) for t=τ is satisfied, taking τ and u(τ) as the initial time and the initial data, we obtain Tm= and (1.7) by the boundedness result of López-Gómez, Nagai and Yamada [23] mentioned in Section 1. ∎

To prove Theorem 1.2, we need the following lemma. This lemma was already proven by Moser’s technique (see, e.g., [1, 20]), but we give another proof.

Lemma 3.2.

For a nonnegative initial data u0L1L, let u be the nonnegative mild solution of (CP) on [0,T) for some 0<T. Assume that

A := sup 0 < t < T ( N u ) ( t ) < .

Then there exists a positive constant C(u01,u0,A) depending only on u01, u0 and A such that

(3.6) u ( t ) C ( u 0 1 , u 0 , A ) , 0 < t < T .

Proof.

Take any t0>0 with 4t0<T and fix it. By the definition of mild solutions, the following integral equation is satisfied: for τ0 with t0+τ<T,

u ( t 0 + τ ) = e t 0 Δ u ( τ ) - 0 t 0 e ( t 0 - s ) Δ ( u ( s + τ ) ( N u ) ( s + τ ) ) 𝑑 s .

Let 1qp and 1/q-1/p<1/2. By the Lp-Lq estimates (2.2) for etΔ and u(t)1=u01, we obtain that

u ( t 0 + τ ) p e t 0 Δ u ( τ ) p + C 0 t 0 ( t 0 - s ) - 1 / q + 1 / p - 1 / 2 u ( s + τ ) ( N u ) ( s + τ ) q 𝑑 s
C t 0 - 1 + 1 / p u ( τ ) 1 + C A 0 t 0 ( t 0 - s ) - 1 / q + 1 / p - 1 / 2 u ( s + τ ) q 𝑑 s
(3.7) C B 1 ( t 0 , p ) u 0 1 + C A B 2 ( t 0 , p , q ) sup 0 < s < t 0 u ( s + τ ) q ,

where

B 1 ( t 0 , p ) = t 0 - 1 + 1 / p , B 2 ( t 0 , p , q ) = t 0 - 1 / q + 1 / p + 1 / 2 0 1 ( 1 - s ) - 1 / q + 1 / p - 1 / 2 𝑑 s .

Take q=1 and p=4/3 in (3.7). Then

u ( t 0 + τ ) 4 / 3 C B 1 ( t 0 , 4 / 3 ) u 0 1 + C A B 2 ( t 0 , 4 / 3 , 1 ) u 0 1 .

Replacing t0+τ by t in this inequality, we have

(3.8) u ( t ) 4 / 3 C t 0 ( 1 + A ) u 0 1 , t 0 t < T .

Here, we denote by Ct0 a positive constant depending only on t0. Next, taking q=4/3 and p=3 in (3.7) for τt0 with t0+τ<T, we have

u ( t 0 + τ ) 3 C B 1 ( t 0 , 3 ) u 0 1 + C A B 2 ( t 0 , 3 , 4 / 3 ) sup 0 < s < t 0 u ( s + τ ) 4 / 3 .

Replacing t0+τ by t in this inequality yields that

(3.9) u ( t ) 3 C t 0 u 0 1 + C t 0 A sup t 0 < t < T u ( t ) 4 / 3 , 2 t 0 t < T .

Take q=3 and p= in (3.7) for τ2t0 with t0+τ<T. Then

u ( t 0 + τ ) C B 1 ( t 0 , ) u 0 1 + C A B 2 ( t 0 , , 3 ) sup 0 < s < t 0 u ( s + τ ) 3 ,

from which it follows that

(3.10) u ( t ) C t 0 u 0 1 + C t 0 A sup 2 t 0 < t < T u ( t ) 3 , 3 t 0 t < T .

Summing up (3.8)–(3.10), we obtain

(3.11) u ( t ) C t 0 ( 1 + A + A 2 + A 3 ) u 0 1 , 3 t 0 t < T .

We estimate u(t) for 0<t3t0, using the integral equation (2.1). Calculating similar to (3.7) and using

e t Δ u 0 p C u 0 p for  1 p

by (2.2), we have

u ( t ) 4 / 3 C u 0 4 / 3 + C t 0 A u 0 1 , 0 < t 3 t 0 .

Similarly,

u ( t ) 3 C u 0 3 + C t 0 A sup 0 < t 3 t 0 u ( t ) 4 / 3 , 0 < t 3 t 0 ,

and thus

u ( t ) C u 0 + C t 0 A sup 0 < t 3 t 0 u ( t ) 3 , 0 < t 3 t 0 .

Hence,

u ( t ) C u 0 + C t 0 ( A u 0 3 + A 2 u 0 4 / 3 + A 3 u 0 1 ) , 0 < t 3 t 0 .

Combining this estimate for 0<t3t0 with (3.11) concludes (3.6). ∎

The next lemma on the Poisson equation is used in the proof of Theorem 1.2.

Lemma 3.3.

Let f be in L1Lp for some 1<p and put

ψ ( x ) = 2 ( N ( x - y ) - N ( y ) ) f ( y ) 𝑑 y , x 2 ,

where N(x)=-1/(2π)log|x|. Then the following holds:

  1. ψ ( x ) is well-defined for all x 2 , and ψ C loc α ( 2 ) for every 0 < α < 1 .

  2. - Δ ψ = f in the sense of distributions in 2 .

  3. A weak derivative of ψ is given by

    (3.12) x j ψ ( x ) = 2 ( x j N ) ( x - y ) f ( y ) 𝑑 y for almost every  x 2 .

  4. If p > 2 , then ψ has a derivative for every x2 given by (3.12), and ψCloc1+α(2) for every 0<α<1-2/p.

Proof.

Take any x02 and R>1. Let φRC0 be such that 0φR1 and

φ R ( x ) = 1 for  | x - x 0 | 2 R , φ R ( x ) = 0 for  | x - x 0 | 3 R .

Put f1=φRf and f2=(1-φR)f, and define

ψ i ( x ) = - 1 2 π 2 ( log | x - y | - log | y | ) f i ( y ) 𝑑 y , i = 1 , 2 .

Then ψ=ψ1+ψ2. As the support of f1 is compact and f1L1Lp, we have

ψ 1 ( x ) = - 1 2 π 2 log | x - y | f 1 ( y ) 𝑑 y + 1 2 π 2 log | y | f 1 ( y ) 𝑑 y ,

and hence, by the book of Lieb and Loss [22, Theorems 6.21 and 10.2], we obtain (i)–(iv) in Lemma 3.3 with ψ and f replaced by ψ1 and f1, respectively. Next, as f2(x)=0 (|x-x0|2R) and f2L1, we derive that ψ2C(B3R/2(x0)) and for every α+2 with |α|1,

x α ψ 2 ( x ) = 2 ( x α N ) ( x - y ) f 2 ( y ) 𝑑 y for all  x B 3 R / 2 ( x 0 ) ,

where BR(x0)={x2:|x-x0|<R}. Hence, ψ2 is harmonic in B3R/2(x0). Therefore, the assertions of the lemma are obtained. ∎

Proof of Theorem 1.2.

Let Tm be the maximal existence time of the solution u. As u0(x) and u(t,x) are radially symmetric in x, we define φ(t,s) and φ0(s) for 0t<Tm, s0 by

u ( t , x ) = φ ( t , s ) , u 0 ( x ) = φ 0 ( s ) , s = π | x | 2 .

Put

Φ ( t , s ) = 0 s φ ( t , σ ) 𝑑 σ , Φ 0 ( s ) = 0 s φ 0 𝑑 σ , s 0 .

Then, by Lemma 2.6,

{ t Φ ( t , s ) = 4 π s s 2 Φ ( t , s ) + Φ ( t , s ) s Φ ( t , s ) , 0 < t < T m , s > 0 , Φ ( t , 0 ) = 0 , Φ ( t , ) = 8 π , 0 < t < T m , Φ ( 0 , s ) = Φ 0 ( s ) , s 0 .

As (1.8) is equivalent to

lim inf s ( s s φ 0 𝑑 σ ) > 0 ,

duplicating the argument in the proof of Theorem 1.1 with u0, u and H replaced by φ0, φ and Φ, respectively, we get that for any 0<τ<Tm there exists b>0 such that

(3.13) Φ ( t , s ) Θ b ( s ) = 8 π s s + π b , τ t < T m , s 0 .

For τt<Tm, define the function ψ~(t,x) by

ψ ~ ( t , x ) = - 1 2 π 2 ( log | x - y | - log | y | ) u ( t , y ) 𝑑 y , x 2 .

As u(t)L1L for any τt<Tm, by Lemma 3.3, ψ~(t,x) is well-defined for all x2, and it holds that

- Δ ψ ~ ( t ) = u ( t ) in the sense of distributions in  2 ,
ψ ~ ( t ) C loc 1 + α ( 2 ) for  0 < α < 1 , ψ ~ ( t , x ) = ( N u ) ( t , x ) .

Taking into account u(t)C(2), we see that ψ~(t)C(2) and

- Δ ψ ~ ( t , x ) = u ( t , x ) for all  x 2 .

As (u,ψ~) is radially symmetric with respect to x, we define u^(t,r) and ψ^(t,r) by

u ( t , x ) = u ^ ( t , r ) , ψ ~ ( t , x ) = ψ ^ ( t , r ) , r = | x | .

Then, by -Δψ~=u, we have

- 1 r r ( r r ψ ^ ) = u ^ ,

and hence

- r r ψ ^ ( t , r ) = 0 r u ^ ( t , η ) η 𝑑 η = 1 2 π 0 π r 2 φ ( t , σ ) 𝑑 σ = 1 2 π Φ ( t , π r 2 ) .

From this relation and (3.13) it follows that for τt<Tm and x2,

| ( N u ) ( t , x ) | = | ψ ~ ( t , x ) | = | r ψ ^ ( t , r ) | = 1 2 π r Φ ( t , π r 2 ) 4 r r 2 + b 2 b .

Applying Lemma 3.2 as T=Tm-τ for the solution u(t+τ,x), we have that

(3.14) sup 0 < t < T m - τ u ( t + τ ) C ( u ( τ ) 1 , u ( τ ) , A ) ,

where

A = sup τ t < T m ( N u ) ( t ) 2 b .

It follows from (3.14) that lim suptTmu(t)<. Therefore, Tm= by Proposition 2.4, that is, u exists globally in time, and u(t) is bounded on [τ,) by (3.14).

If u0L in addition, then Φ0(s)Θb(s) (s0) for some b>0 by Lemma 3.1, and hence (3.13) is valid for τ=0. Therefore, (1.10) is concluded for τ=0. ∎

4 Sufficient Conditions for (1.4)

In this section, we prove Theorems 1.3 and 1.6 and Corollary 1.4. We begin with two lemmas.

Lemma 4.1.

  1. Let f L 1 and assume that

    lim R ( R 2 | x | > R | f | 𝑑 x ) = 0 .

    Then

    lim R ( R 2 | x | > R f 𝑑 x ) = lim s ( s s f 𝑑 σ ) = 0 .

  2. Let f , g L 1 and assume that

    lim inf R ( R 2 | x | > R f 𝑑 x ) > 0 , lim R ( R 2 | x | > R | g | 𝑑 x ) = 0 .

    Then

    (4.1) lim inf R ( R 2 | x | > R ( f + g ) 𝑑 x ) > 0 .

Proof.

(i) We observe that by Proposition 2.5 (iii),

| x | < R | f | 𝑑 x = 2 χ B R | f | 𝑑 x 2 χ B R f 𝑑 x = | x | < R f 𝑑 x ,

where χBR is the characteristic function of BR={x2:|x|<R} and we used χBR=χBR. Hence, thanks to f1=f1 by Proposition 2.5 (i), we have

(4.2) | x | > R f 𝑑 x | x | > R | f | 𝑑 x ,

which yields the conclusion.

(ii) Putting h=f+g and applying the basic property (vii) on rearrangements, we observe that by f=h+(-g),

f ( 2 s ) h ( s ) + g ( s ) , s > 0 ,

and hence

s h ( σ ) 𝑑 σ s f ( 2 σ ) 𝑑 σ - s g ( σ ) 𝑑 σ = 1 2 2 s f ( σ ) 𝑑 σ - s g ( σ ) 𝑑 σ .

By (i), we have

lim s ( s s g 𝑑 σ ) = 0 .

Therefore,

lim inf s ( s s h 𝑑 σ ) 1 2 lim inf s ( s 2 s f 𝑑 σ ) > 0 ,

which proves (4.1). ∎

Lemma 4.2.

For fL1 and R0>0, define f0L1 by

f 0 ( x ) = 0 for  | x | R 0 , f 0 ( x ) = f ( x ) for  | x | > R 0 .

Then

f 0 ( x ) f ( 2 x ) , | x | R 0 ,
| x | > R f 0 𝑑 x 1 2 | x | > 2 R f 𝑑 x , R R 0 .

Proof.

Define f1L1 by

f 1 ( x ) = f ( x ) for  | x | R 0 , f 1 ( x ) = 0 for  | x | > R 0 .

Since ||f1|>t|πR02 for any t>0, we have

f 1 ( s ) = 0 , s π R 0 2 .

By f=f0+f1, we observe that f(2s)f0(s)+f1(s) for s>0 by the basic property (vii) on rearrangements, and thus

f 0 ( s ) f ( 2 s ) , s π R 0 2 ,

from which it follows that

s f 0 𝑑 σ 1 2 2 s f 𝑑 σ , s π R 0 2 .

Hence, the inequalities on f0 are obtained by virtue of f0(x)=f0(π|x|2). ∎

Proof of Theorem 1.3.

By u0(x)f(x) (|x|1), there is a number R0>1 such that u0(x)f(x) (|x|R0). Define f0L1 by

f 0 ( x ) = 0 for  | x | < R 0 , f 0 ( x ) = f ( x ) for  | x | R 0 .

As u0f00, we have u0f0 by the basic property (v) on rearrangements, and thus, by Lemma 4.2,

| x | > R u 0 𝑑 x | x | > R f 0 𝑑 x 1 2 | x | 2 R f 𝑑 x , R R 0 ,

from which it follows that

lim inf R ( R 2 | x | > R u 0 𝑑 x ) 1 2 lim inf R ( R 2 | x | > 2 R f 𝑑 x ) > 0 .

Hence, if 2u0𝑑x=8π, then Theorem 1.1 ensures the boundedness of u(t) on [τ,) for any τ>0. ∎

Proof of Corollary 1.4.

Taking b>0 such that C=8b, we have that

u 0 ( x ) C | x | 4 8 b ( | x | 2 + b ) 2 = θ b ( x ) , | x | R 0 ,

for a positive number R0>1. As θb=θb and

lim R ( R 2 | x | > R θ b 𝑑 x ) > 0 ,

the conclusion of the corollary is an immediate consequence of Theorem 1.3. ∎

Proof of Theorem 1.6.

We observe that by the definition of decreasing rearrangements,

| f > f ( s ) | s | f f ( s ) | for all  s > 0 ,

and by |f>t| (t+0),

f ( s ) > 0 for all  s > 0 , f ( s ) 0 as  s .

We claim that

(4.3) 0 s f 𝑑 σ | x | r ( t ) f 𝑑 x for  t := f ( s ) , s > 0 .

In fact, in the case of s=|f>t|, by the basic property (viii) on rearrangements and {ft}{|x|r(t)}, we have

0 s f 𝑑 σ = f > t f 𝑑 x | x | r ( t ) f 𝑑 x .

In the case of |f>t|<s|ft|, put s1=|f>t| and s2=|ft|. By the basic property (i) on rearrangements, we observe that

s 1 = | f > t | , s 2 = | f t | , s 2 - s 1 = | f = t | = | f = t | ,

and thus f(σ)=t for s1σs2. By these equalities, we have that

0 s f 𝑑 σ = 0 s 1 f 𝑑 σ + ( s - s 1 ) t
f > t f 𝑑 x + ( s 2 - s 1 ) t = f t f 𝑑 x | x | r ( t ) f 𝑑 x .

Hence, claim (4.3) is obtained.

As 0f𝑑s=2f𝑑x and μ(t):=|f>t|s for t=f(s), it follows from (4.3) that

s s f 𝑑 σ μ ( t ) | x | > r ( t ) f 𝑑 x = μ ( t ) r ( t ) 2 r ( t ) 2 | x | > r ( t ) f 𝑑 x .

Hence, as t=f(s)0 (s) and r(t) (t+0), by (1.12) we have that

lim inf s ( s s f 𝑑 σ ) C lim inf t 0 ( r ( t ) 2 | x | > r ( t ) f 𝑑 x )

for a constant C>0. Therefore, this inequality concludes that (1.9) holds. ∎

5 Examples

Example 5.1.

For given a1>0 and a2>0, define {rn}n=0 by

r 0 = 0 , r 1 = a 1 , r 2 n = r 2 n - 1 + a 2 , r 2 n + 1 = r 2 n + a 1 .

Put

A = n = 1 A 2 n - 1 , A n = { x 2 : r n | x | r n + 1 }

and define f(x) by

(5.1) f ( x ) = χ A ( x ) | x | - 4 , x 2 ,

where χA is the characteristic function of A. Then it holds that

(5.2) lim inf R ( R 2 | x | > R f 𝑑 x ) > 0 ,
(5.3) lim inf R ( R 2 | x | > R f 𝑑 x ) > 0 .

Let us prove (5.2). Observing that by r2n=r2n-1+a2 and r2n+1=r2n+a1,

r 2 n | x | r 2 n + 1 | x | - 4 𝑑 x = ( r 2 n + 2 r 2 n ) 2 2 a 1 r 2 n + a 1 2 2 a 2 r 2 n + 1 + a 2 2 r 2 n + 1 | x | r 2 n + 2 | x | - 4 𝑑 x

and

lim n ( r 2 n + 2 r 2 n ) 2 2 a 1 r 2 n + a 1 2 2 a 2 r 2 n + 1 + a 2 2 = a 1 a 2 ,

we see that there is n0 such that

r 2 n | x | r 2 n + 1 | x | - 4 𝑑 x 2 a 1 a 2 r 2 n + 1 | x | r 2 n + 2 | x | - 4 𝑑 x for all  n n 0 .

Then for r2n-1Rr2n+1 (nn0),

| x | > R | x | - 4 𝑑 x | x | > R χ A ( x ) | x | - 4 𝑑 x + | x | r 2 n ( 1 - χ A ( x ) ) | x | - 4 𝑑 x
= | x | > R f 𝑑 x + k = n r 2 k | x | r 2 k + 1 | x | - 4 𝑑 x
(5.4) ( 1 + 2 a 1 a 2 ) | x | > R f 𝑑 x .

Hence, (5.2) follows from (5.4).

Next, we prove (5.3) by applying Theorem 1.6. For (r2n)-4t<(r2n-1)-4, we have

μ ( t ) := | f > t | = π k = 1 n - 1 ( ( r 2 k ) 2 - ( r 2 k - 1 ) 2 ) + π ( t - 1 / 2 - ( r 2 n - 1 ) 2 ) ,

and for (r2n+1)-4t<(r2n)-4,

| f > t | = π k = 1 n ( ( r 2 k ) 2 - ( r 2 k - 1 ) 2 ) .

Put

s 2 n = | f > ( r 2 n ) - 4 | = π k = 1 n ( ( r 2 k ) 2 - ( r 2 k - 1 ) 2 ) .

Then

(5.5) μ ( t ) = { s 2 ( n - 1 ) + π ( t - 1 / 2 - ( r 2 n - 1 ) 2 ) if  ( r 2 n ) - 4 t < ( r 2 n - 1 ) - 4 , s 2 n if  ( r 2 n + 1 ) - 4 t < ( r 2 n ) - 4 .

We next observe that by r2k=r2k-1+a2 and r2k-1=(a1+a2)(k-1)+a1,

s 2 n = π k = 1 n { ( r 2 k ) 2 - ( r 2 k - 1 ) 2 } = π k = 1 n ( 2 a 2 r 2 k - 1 + a 2 2 )
(5.6) = π k = 1 n { 2 a 2 ( a 1 + a 2 ) ( k - 1 ) + 2 a 1 a 2 + a 2 2 } a 2 ( a 1 + a 2 ) π n 2 .

By (5.5) and (5.6),

μ ( t ) s 2 ( n - 1 ) a 2 ( a 1 + a 2 ) π ( n - 1 ) 2 , ( r 2 n + 1 ) - 4 t < ( r 2 n - 1 ) - 4 .

As r(t)=sup{|x|:f(x)t}, we observe that

r ( t ) r 2 n if  ( r 2 n ) - 4 t < ( r 2 n - 1 ) - 4 , r ( t ) = r 2 n if  ( r 2 n + 1 ) - 4 t < ( r 2 n ) - 4 .

Hence, by r2n=(a1+a2)n, we have that for (r2n+1)-4t<(r2n-1)-4,

μ ( t ) r ( t ) 2 a 2 ( a 1 + a 2 ) π ( n - 1 ) 2 ( a 1 + a 2 ) 2 n 2 ,

which yields that

lim inf t + 0 μ ( t ) r ( t ) 2 a 2 π a 1 + a 2 .

Therefore, by Theorem 1.6, inequality (5.3) is established.

With the function f defined by (5.1), we construct two kinds of nonnegative initial data u0L1L. We denote by u the nonnegative solution of (CP) corresponding to the initial data u0.

Case 1: Radial case. Define u0L1L by u0=cf, where c is taken as 2u0𝑑x=8π. Then u0 is radially symmetric and

0 u 0 C | x | - 4 for  | x | 1 , lim inf R ( R 2 | x | > R u 0 𝑑 x ) > 0 .

By Theorem 1.2, u(t) is bounded on [0,).

We remark that b[u0]= for all b>0, where b is the functional given by (1.3). In fact, we observe that by u0(x)=0 for xA,

b [ u 0 ] 2 / A θ b 1 / 2 𝑑 x
= 2 2 b k = 0 r 2 k | x | r 2 k + 1 1 | x | 2 + b 𝑑 x
= 4 π 2 b k = 0 r 2 k r 2 k + 1 r r 2 + b 𝑑 r
= 2 π 2 b k = 0 log ( ( r 2 k + 1 ) 2 + b ( r 2 k ) 2 + b ) .

As r2k+1=r2k+a1 and r2k=(a1+a2)k, we have

( r 2 k + 1 ) 2 + b ( r 2 k ) 2 + b = 1 + 2 a 1 r 2 k + a 1 2 ( r 2 k ) 2 + b 1 + 2 a 1 ( a 1 + a 2 ) k ( a 1 + a 2 ) 2 k 2 + b ,

and hence, by log(1+s)s/2 (0s1),

b [ u 0 ] 2 π 2 b a 1 ( a 1 + a 2 ) k = 2 k ( a 1 + a 2 ) 2 k 2 + b = .

Case 2: Non-radial case. Let cf be the same as in Case 1 and take gL1L satisfying

(5.7) c f + g 0 , 2 g 𝑑 x = 0 , lim R ( R 2 | x | > R | g | 𝑑 x ) = 0 .

Define u0L1L by u0=cf+g. We observe that by (5.3) and (cf)=cf,

lim inf R ( R 2 | x | > R ( c f ) 𝑑 x ) > 0 .

Hence, by Lemma 4.1 (ii),

lim inf R ( R 2 | x | > R u 0 𝑑 x ) > 0 .

Therefore, u(t) is bounded on [0,).

We remark that if |x|2g,|g|L1 in addition to (5.7), then b[u0]= for all b>0. In fact, observing that

( u 0 - θ b ) 2 θ b - 1 / 2 = ( c f - θ b ) 2 θ b - 1 / 2 + g θ b - 1 / 2 + 2 ( c f - u 0 )
( c f - θ b ) 2 θ b - 1 / 2 - | g | | x | 2 + b 2 2 b - 2 | g | ,

we have that

b [ u 0 ] b [ c f ] - 1 2 2 b 2 | g ( x ) | ( | x | 2 + b ) 𝑑 x - 2 2 | g | 𝑑 x .

Hence, as b[cf]=, we conclude b[u0]=.

Example 5.2.

Let a(t) and b(t) be such that

a ( t ) = t - α , b ( t ) = t - β    for  0 < t 1 ,  0 < β < α < 1 , α + β < 1 ,

and define the function f(x) (x=(x1,x2)2) by

f ( x ) = { 1 if  x 1 2 a ( 1 ) 2 + x 2 2 b ( 1 ) 2 1 , t if  x 1 2 a ( t ) 2 + x 2 2 b ( t ) 2 = 1 ,  0 < t < 1 .

Define the parameter regions of (α,β) as follows:

P 1 : 0 < β < α , 1 2 α + β < 1 ,
P 2 : 0 < β < α , α + β < 1 2 , 3 α + β 1 ,
P 3 : 0 < β < α , 3 α + β < 1 .

Put

A = lim R ( R 2 | x | > R f 𝑑 x ) , A = lim R ( R 2 | x | > R f 𝑑 x ) .

Then it holds that

(5.8) ( α , β ) P 1 A > 0 , A > 0 ,
(5.9) ( α , β ) P 2 A > 0 , A = 0 ,
(5.10) ( α , β ) P 3 A = A = 0 .

Statement (1.9) in Section 1 is false for (α,β)P2.

In what follows, we prove (5.8)–(5.10). As it is apparent that

{ f > t } = { , t 1 , { x 2 : x 1 2 a ( t ) 2 + x 2 2 b ( t ) 2 < 1 } , 0 < t < 1 ,

we have that

μ ( t ) := | f > t | = { 0 , t 1 , π a ( t ) b ( t ) , 0 < t < 1 ,
r ( t ) := sup { | x | : f ( x ) t } = a ( t ) = t - α , 0 < t < 1 ,
μ ( t ) r ( t ) 2 = π b ( t ) a ( t ) = π t α - β 0 as  t 0 .

We also see that

2 f ( x ) 𝑑 x = π 0 1 a ( t ) b ( t ) 𝑑 t < .

Let f(x)=t(0,1). Then, by b(t)|x|a(t),

| x | - 1 / β f ( x ) | x | - 1 / α , | x | > 1 .

As f on (π,) equals the inverse function of μ on (0,1), we have

f ( s ) = 1 for  0 s π , f ( s ) = ( s π ) - 1 / ( α + β ) for  s > π ,

and thus, for s>π,

s s f 𝑑 σ = π 1 / ( α + β ) α + β 1 - ( α + β ) s 2 - 1 / ( α + β ) .

Hence, we derive that

(5.11) lim R ( R 2 | x | > R f 𝑑 x ) = lim s ( s s f 𝑑 σ ) = { , α + β > 1 / 2 , π 2 , α + β = 1 / 2 , 0 , α + β < 1 / 2 .

It follows from (4.2) and (5.11) that

(5.12) α + β > 1 2 lim R ( R 2 | x | > R f 𝑑 x ) = .

We next study

(5.13) lim R ( R 2 | x | > R f 𝑑 x ) for  0 < β < α , α + β 1 2 .

For this aim, by {f>s}= (s1), we first note that for 0<t<1,

| x | > a ( t ) f ( x ) 𝑑 x = 0 1 ( | x | > a ( t ) χ { f > s } ( x ) 𝑑 x ) 𝑑 s = 0 1 g ( s , t ) 𝑑 s ,

where χ{f>s} is the characteristic function of {f>s} and

g ( s , t ) = | x | > a ( t ) χ { f > s } ( x ) d x = | { | x | > a ( t ) } { f > s } | .

Define γ(t) for 0<t<1 by b(γ(t))=a(t). Then

0 < γ ( t ) = t α / β < t .

We observe that

(5.14) g ( s , t ) = { 0 , t s 1 , π { a ( s ) b ( s ) - a ( t ) 2 } , 0 < s γ ( t ) .

Indeed, (5.14) follows from the following:

{ | x | > a ( t ) } { f > s } = if  t s 1 ,
{ | x | < a ( t ) } { f > s } if  0 < s γ ( t ) .

In the case of γ(t)<s<t<1, to describe g(s,t) by a(s) and b(s), we consider the intersection point (ξ(s,t),η(s,t)) in {x10,x20} of the two curves Γ1(s), Γ2(t):

Γ 1 ( s ) : x 1 2 a ( s ) 2 + x 2 2 b ( s ) 2 = 1 , Γ 2 ( t ) : x 1 2 + x 2 2 = a ( t ) 2 .

The point (ξ(s,t),η(s,t)) is given by

(5.15) ξ ( s , t ) = a ( s ) a ( t ) 2 - b ( s ) 2 a ( s ) 2 - b ( s ) 2 = t - α s 2 β - t 2 α s 2 β - s 2 α ,
(5.16) η ( s , t ) = b ( s ) 1 - ξ ( s , t ) 2 a ( s ) 2 = a ( t ) 2 - ξ ( s , t ) 2

and

0 < ξ ( s , t ) < a ( t ) for  γ ( t ) < s < t , ξ ( γ ( t ) , t ) = 0 , ξ ( t , t ) = a ( t ) .

We remark that there are no intersection points of the two curves for 0<s<γ(t). Then g(s,t) is expressed by the following: for γ(t)<s<t<1,

(5.17) g ( s , t ) = 4 { ξ ( s , t ) a ( s ) b ( s ) 1 - x 1 2 a ( s ) 2 𝑑 x 1 - ξ ( s , t ) a ( t ) a ( t ) 2 - x 1 2 𝑑 x 1 } .

It follows from (5.14) and (5.17) that

lim s t ± 0 g ( s , t ) = 0 ,
(5.18) lim s γ ( t ) ± 0 g ( s , t ) = π { a ( γ ( t ) ) b ( γ ( t ) ) - a ( t ) 2 } = π a ( t ) { a ( γ ( t ) ) - a ( t ) } .

Here we used ξ(γ(t),t)=0, ξ(t,t)=a(t) and b(γ(t))=a(t).

We claim that

(5.19) lim t 0 ( a ( t ) 2 | x | > a ( t ) f 𝑑 x ) = 2 lim t 0 { a ( t ) 4 γ ( t ) t ( π 2 - arcsin ξ ( s , t ) a ( t ) ) 𝑑 s } ,

provided that the limit on the right-hand side of (5.19) exists in the extended real values. To prove the claim, we first note that by g(s,t)=0 (0<ts1),

(5.20) | x | > a ( t ) f ( x ) 𝑑 x = 0 γ ( t ) g ( s , t ) 𝑑 s + γ ( t ) t g ( s , t ) 𝑑 s .

By (5.14),

(5.21) g t ( s , t ) = - 2 π a ( t ) a ( t ) for  0 < s < γ ( t ) .

For γ(t)<s<t, differentiating (5.17) with respect to t, we have that by (5.16),

g t ( s , t ) = - 4 { b ( s ) 1 - ξ ( s , t ) 2 a ( s ) 2 - a ( t ) 2 - ξ ( s , t ) 2 } ξ t ( s , t ) - 4 ξ ( s , t ) a ( t ) t a ( t ) 2 - x 1 2 𝑑 x 1
= - 4 ξ ( s , t ) a ( t ) a ( t ) a ( t ) a ( t ) 2 - x 1 2 𝑑 x 1 ,

from which it follows that

(5.22) g t ( s , t ) = - 4 a ( t ) a ( t ) ( π 2 - arcsin ξ ( s , t ) a ( t ) ) .

Hence, differentiating (5.20) with respect to t(0,1), by (5.18), (5.21) and (5.22), we derive that

d d t | x | > a ( t ) f ( x ) 𝑑 x = 0 γ ( t ) g t ( s , t ) 𝑑 s + γ ( t ) t g t ( s , t ) 𝑑 s
= - 2 π a ( t ) a ( t ) γ ( t ) - 4 a ( t ) a ( t ) γ ( t ) t ( π 2 - arcsin ξ ( s , t ) a ( t ) ) 𝑑 s .

Therefore, applying L’Hôpital’s rule and observing that by β<1/4,

lim t 0 a ( t ) 4 γ ( t ) = lim t 0 t - 4 α + α / β = 0 ,

we obtain claim (5.19).

By claim (5.19), it is shown that

(5.23) lim t 0 ( a ( t ) 2 | x | > a ( t ) f 𝑑 x ) = 1 2 0 1 ( 1 - s 2 α ) - 1 / 2 s - β 𝑑 s lim t 0 t - 3 α - β + 1 .

Thanks to (5.23) and a(t)=t-α, the question on (5.13) is concluded as follows:

(5.24) lim R ( R 2 | x | > R f 𝑑 x ) = { , 3 α + β > 1 , c , 3 α + β = 1 , 0 , 3 α + β < 1 ,

where

c = 1 2 0 1 ( 1 - s 2 α ) - 1 / 2 s - β 𝑑 s .

To prove (5.23) by (5.19), we recall γ(t)=tα/β by b(γ(t))=a(t), and observe that by ξ(t,t)=a(t) and ξ(γ(t),t)=0,

d d t γ ( t ) t ( π 2 - arcsin ξ ( s , t ) a ( t ) ) 𝑑 s
= ( π 2 - arcsin ξ ( t , t ) a ( t ) ) - ( π 2 - arcsin ξ ( γ ( t ) , t ) a ( t ) ) γ ( t ) - γ ( t ) t t ( arcsin ξ ( s , t ) a ( t ) ) 𝑑 s
= - π 2 γ ( t ) - γ ( t ) t t ( arcsin ξ ( s , t ) a ( t ) ) 𝑑 s .

By (5.15), we have ξ(s,t)/a(t)=(s2β-t2α)1/2(s2β-s2α)-1/2, and thus

t ( arcsin ξ ( s , t ) a ( t ) ) = - α t 2 α - 1 ( t 2 α - s 2 α ) - 1 / 2 ( s 2 β - t 2 α ) - 1 / 2 .

Hence,

d d t γ ( t ) t ( π 2 - arcsin ξ ( s , t ) a ( t ) ) 𝑑 s = - π α 2 β t α / β - 1 + α t 2 α - 1 γ ( t ) t ( t 2 α - s 2 α ) - 1 / 2 ( s 2 β - t 2 α ) - 1 / 2 𝑑 s ,

from which it follows that

(5.25) 1 ( a ( t ) - 4 ) d d t γ ( t ) t ( π 2 - arcsin ξ ( s , t ) a ( t ) ) 𝑑 s = - π 8 β t α / β - 4 α + 1 4 t - 2 α γ ( t ) t ( t 2 α - s 2 α ) - 1 / 2 ( s 2 β - t 2 α ) - 1 / 2 𝑑 s .

As α/β-4α>0 because of β<1/4, it is apparent that

(5.26) lim t 0 π 8 β t α / β - 4 α = 0 .

Next, putting s=tσ and γ1(t)=tα/β-1, we have that

t - 2 α γ ( t ) t ( t 2 α - s 2 α ) - 1 / 2 ( s 2 β - t 2 α ) - 1 / 2 𝑑 s = t - 3 α - β + 1 γ 1 ( t ) 1 ( 1 - σ 2 α ) - 1 / 2 ( σ 2 β - t 2 ( α - β ) ) - 1 / 2 𝑑 σ
(5.27) = t - 3 α - β + 1 I ( t ) ,

where

I ( t ) = γ 1 ( t ) 1 φ ( s , t ) 𝑑 s , φ ( s , t ) := ( 1 - s 2 α ) - 1 / 2 ( s 2 β - t 2 ( α - β ) ) - 1 / 2 .

We claim that

(5.28) lim t 0 I ( t ) = 0 1 ( 1 - s 2 α ) - 1 / 2 s - β 𝑑 s .

In fact, we first divide the integral into two parts:

I ( t ) = ( γ 1 ( t ) 2 γ 1 ( t ) + 2 γ 1 ( t ) 1 ) φ ( s , t ) d s .

Let γ1(t)<s<2γ1(t)<1/2. Then φ(s,t)(1-2-2α)-1/2(s2β-t2(α-β))-1/2, and, putting s=γ1(t)σ=tα/β-1σ, we obtain

γ 1 ( t ) 2 γ 1 ( t ) ( s 2 β - t 2 ( α - β ) ) - 1 / 2 𝑑 s = t - α + β + α / β - 1 1 2 ( σ 2 β - 1 ) - 1 / 2 𝑑 σ .

As -α+β+α/β-1=(α-β)(1-β)/β>0, we derive that

0 γ 1 ( t ) 2 γ 1 ( t ) φ ( s , t ) 𝑑 s C t - α + β + α / β - 1 0 as  t 0 .

Let 2γ1(t)<s<1. Then tα-β(s/2)β, and hence

φ ( s , t ) ( 1 - 2 - 2 β ) - 1 / 2 ( 1 - s 2 α ) - 1 / 2 s - β .

Putting A(t)=(2γ1(t),1), we have that

φ ( s , t ) χ A ( t ) ( s ) ( 1 - 2 - 2 β ) - 1 / 2 ( 1 - s 2 α ) - 1 / 2 s - β , 0 < s < 1 ,

where χA(t) is the characteristic function of A(t), and

lim t 0 φ ( s , t ) χ A ( t ) ( s ) = ( 1 - s 2 α ) - 1 / 2 s - β , 0 < s < 1 .

As the function (1-s2α)-1/2s-β is integrable on (0,1), Lebesgue’s convergence theorem ensures that

2 γ 1 ( t ) 1 φ ( s , t ) 𝑑 s = 0 1 φ ( s , t ) χ A ( t ) ( s ) 𝑑 s 0 1 ( 1 - s 2 α ) - 1 / 2 s - β 𝑑 s as  t 0 .

Hence, claim (5.28) is derived. Summing up (5.25)–(5.28) and applying L’Hôpital’s rule, we arrive at

lim t 0 { a ( t ) 4 γ ( t ) t ( π 2 - arcsin ξ ( s , t ) a ( t ) ) 𝑑 s } = lim t 0 1 ( a ( t ) - 4 ) d d t γ ( t ) t ( π 2 - arcsin ξ ( s , t ) a ( t ) ) 𝑑 s
(5.29) = 1 4 0 1 ( 1 - s 2 α ) - 1 / 2 s - β 𝑑 s lim t 0 t - 3 α - β + 1 .

Therefore, (5.23) follows from (5.19) and (5.29).

In conclusion, (5.8) follows from (5.11), (5.12) and (5.24), and (5.9) and (5.10) from (5.11) and (5.24).


Communicated by Julian Lopez Gomez


Award Identifier / Grant number: JP15KT0019

Funding statement: This work was supported by JSPS KAKENHI, Grant Number JP15KT0019.

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Received: 2017-05-09
Revised: 2017-05-25
Accepted: 2017-05-25
Published Online: 2017-06-21
Published in Print: 2018-04-01

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