In this paper, we show the almost everywhere pointwise convergence of free Benjamin-Ono-Burgers equation in Hs(R) with s>0 with the aid of the maximal function estimate.
Citation: Fei Zuo, Junli Shen. Pointwise convergence problem of free Benjamin-Ono-Burgers equation[J]. AIMS Mathematics, 2022, 7(4): 5527-5533. doi: 10.3934/math.2022306
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In this paper, we show the almost everywhere pointwise convergence of free Benjamin-Ono-Burgers equation in Hs(R) with s>0 with the aid of the maximal function estimate.
In this paper, we investigate the pointwise convergence problem of the free Benjamin-Ono-Burgers equation
ut−∂2xu+H∂2xu=0, | (1.1) |
u(x,0)=f(x). | (1.2) |
Carleson [4] initiated the pointwise converge problem, more precisely, Carleson showed pointwise convergence problem of the one dimensional Schrödinger equation in Hs(R) with s≥14. Dahlberg and Kenig [8] showed that the pointwise convergence of the Schrödinger equation does not hold for s<14 in any dimension. The pointwise convergence problem of Schrödinger equations in higher dimension has been investigated by many authors, for example, see [1,2,6,7,9,14,16,17,20,21,22,23,24,25]. The results of Bourgain [3] and Du et al. [11,12] showed that s>n2(n+1) is the necessary condition for the pointwise convergence problem of n dimensional Schrödinger equation. Moreover, Miao et al. [19] studied the the maximal inequality for 2D fractional order Schrödinger operators.
The Benjamin-Ono-Burgers equation (1.1) was obtained by Ewdin and Roberts [13] in the study of intense magnetic flux tubes of the solar atmosphere. The dissipative effects ϵ∂2xu in that literature are due to weak thermal conduction, where ϵ is a measure of the importance of thermal conduction and is assumed small. Some people have studied the Cauchy problem for the Benjamin-Ono-Burgers equation [5,15,18]. In this paper, motivated by [10,25], we show the pointwise convergence of free Benjamin-Ono-Burgers equation in Hs(R) with s>0.
We present some notations before stating the main results. We always assume that 0<ϵ<10−8. |E| denotes by the Lebesgue measure of set E.
Fxf(ξ)=1√2π∫Re−ixξf(x)dx,F−1xf(ξ)=1√2π∫Reixξf(x)dx,U1(t)u0=1√2π∫Reixξ−itξ|ξ|−tξ2Fxu0(ξ)dξ,‖f‖LqxLpt=(∫R(∫R|f(x,t)|pdt)qpdx)1q. |
Hs(R)={f∈S′(R):‖f‖Hs(R)=‖⟨ξ⟩sFxf‖L2ξ(R)<∞}, where ⟨ξ⟩s=(1+ξ2)s2.
The main result is as follows.
Theorem 1.1. Let f∈Hs(R) with s>0. Then, we have
U1(t)f(x)⟶f(x) | (1.3) |
almost everywhere as t⟶0.
In this section, we present the proof of Lemma 2.1.
Lemma 2.1. For t≥0, we have
|∫Reixξ−itξ|ξ|−tξ|ξ|dξ|ξ|ϵ|≤C|x|−1+ϵ. | (2.1) |
Proof. Let xξ=η. Then, we have
∫Reixξ−itξ|ξ|−tξ|ξ|dξ|ξ|ϵ=Cx−1|x|ϵ∫Reiη−itx−1|x|−1η|η|−tx−2η2dη|η|ϵ. | (2.2) |
We define
I:=∫Reiη−tx−1|x|−1η|η|−tx−2η2dη|η|ϵ. | (2.3) |
We consider x≥0,x<0, respectively. When x>0, we have
I:=∫Reiη−iAη|η|−Aη2dη|η|ϵ=:I1+I2,. | (2.4) |
where I1:=∫∞0eiη−iAη2−Aη2dηηϵ,I2:=∫0−∞eiη+iAη2−Aη2dη(−η)ϵ. Here A=tx−2. For 12A≤2, then, we have I1:=I11+I12. Here, I11:=∫40eiη−iAη2−Aη2dηηϵ,I12:=∫∞4eiη−iAη2−Aη2dηηϵ. Obviously, we have
|I11|≤|∫40eiη−iAη2−Aη2dηηϵ|≤∫40η−ϵ=41−ϵ1−ϵ. | (2.5) |
By using the integration by parts, we have
I12:=∫∞4eiη−iAη2−Aη2dηηϵ=∫+∞41(1−2Aη)ηϵi+2Aη1+ϵdeiη−iAη2−Aη2=1(1−8A)4ϵi+2A41+ϵ−∫+∞4eiη−Aη2(1(1−2Aη)ηϵi+2Aη1+ϵ)′dη. | (2.6) |
Thus, from (2.6), we have
|(1u+iv)′|≤2|u′|u2+v2+2|v′|u2+v2. | (2.7) |
From (2.7), we have
(1(1−2Aη)ηϵi+2Aη1+ϵ)′≤2+3ϵη1+ϵ((1−2Aη)2+4A2η2)≤2+3ϵη1+ϵ≤3η1+ϵ. | (2.8) |
Combining (2.5) with (2.8), we have
|I12|≤|∫∞4eiη−iAη2−Aη2dηηϵ|≤|1(1−8A)4ϵi+2A41+ϵ|+∫+∞4|(1−2Aη1+ϵ+ηϵi)′|dη≤|1−8A|4ϵ(1−8A)242ϵ+4A242+2ϵ+4A242+2ϵ(1−8A)242ϵ+4A242+2ϵ+3∫+∞4η−1−ϵdη≤32+3ϵ≤6ϵ. | (2.9) |
When 12A≥2, we have
I1:=∫∞0eiη−iAη2−Aη2dηηϵ=∫10+∫14A1+∫1A14A+∫∞1A. | (2.10) |
We have
|∫10eiη−iAη2−Aη2dηηϵ|≤∫10η−ϵdη≤11−ϵ. | (2.11) |
Since |1−2Aη|≥12, we have
|∫14A1eiη−iAη2−Aη2(1(1−2Aη)ηϵi+2Aη1+ϵ)′|≤|∫14A12+3ϵη1+ϵ((1−2Aη)2+4A2η2)dη|≤12∫14A11η1+ϵdη≤12ϵ. | (2.12) |
Since 4A2η2≥14, we have
|∫1A14Aeiη−iAη2−Aη2(1(1−2Aη)ηϵi+2Aη1+ϵ)′|≤|∫1A14A2+3ϵη1+ϵ((1−2Aη)2+4A2η2)dη|≤12∫14A11η1+ϵdη≤12ϵ. | (2.13) |
Since Aη≥1, we have
|∫∞1Aeiη−iAη2−Aη2(1(1−2Aη)ηϵi+2Aη1+ϵ)′|≤|∫∞1A2+3ϵη1+ϵ((1−2Aη)2+4A2η2)dη|≤∫∞1A1η1+ϵdη≤1ϵ. | (2.14) |
For I2, let y=−η, we have
I2:=∫∞0e−iy+iAy2−Ay2dyyϵ. | (2.15) |
Thus, I2 can be proved similarly to I1. When x<0, we have
I:=∫Reiη+Aη|η|dη|η|ϵ=:I3+I4, | (2.16) |
where I3:=∫∞0eiη+Aη2dηηϵ,I4:=∫0−∞eiη−Aη2dη(−η)ϵ. Thus, I3,I4 can be proved similarly to I1.
This completes the proof of Lemma 2.1.
In this section, we establish the maximal function estimate.
Lemma 3.1. For f∈Hϵ(R), we have
‖U1(t)f‖L1xL∞t≤C‖f‖˙Hϵ(R). | (3.1) |
Proof. To prove Lemma 3.1, it suffices to prove
|∫1−1U1(t(x))f(x)dx|≤C(ϵ)‖f‖˙Hϵ(R) | (3.2) |
for all measurable functions t:R⟶R. By using the Fubini's theorem and the Cauchy-Schwarz inequality as well as Lemma 2.1, we have
∫1−1∫Reixξ−it(x)ξ|ξ|−t(x)ξ2Fxf(ξ)dξdx≤C∫RFxf(ξ)[∫1−1∫Reixξ−it(x)ξ|ξ|−t(x)ξ2dx]dξ≤C|∫R|Fxf(ξ)|2|ξ|ϵdξ|12∫R|∫1−1eixξ−it(x)ξ|ξ|−t(x)ξ2dx|2dξ|ξ|ϵ≤C‖f‖˙Hϵ(R)∫1−1∫1−1∫Rei(x−y)ξ−it(x)ξ|ξ|−(t(x)+t(y))ξ2dξ|ξ|ϵdxdy≤C‖f‖˙Hϵ(R)∫1−1∫1−1|x−y|−1+ϵdxdy≤Cϵ‖f‖˙Hϵ(R). | (3.3) |
This completes the proof of Lemma 3.1.
In this section, we apply Lemma 2.1 and the density theorem to prove Theorem 1.1.
If f is rapidly decreasing function, ∀ϵ>0, then, we have
|U1(t)f−f|≤|U1(t)f−U2(t)f|+|U2(t)f−f|≤C|t|∫R|ξ|2|Fxf(ξ)|dξ⟶0 | (4.1) |
as t⟶0. Here, U2(t)=12∫Reixξe−it|ξ|ξFxf(ξ)dξ.
When f∈Hs(R)(s≥ϵ), by using the density theorem which can be seen in Lemma 2.2 of [10], there exists a rapidly decreasing function g such that f=g+h, where ‖h‖Hs(R)<ϵ(s≥ϵ). Thus, we have
limt⟶0|U1(t)f−f|≤limt⟶0|U1(t)g−g|+limt⟶0|U1(t)h−h|. | (4.2) |
We define Eα={x∈R:limt⟶0|U1(t)f−f|>α}. Obviously, Eα⊂E1α∪E2α,
E1α={x∈R:limt⟶0|U1(t)g−g|>α2},E2α={x∈R:limt⟶0|U1(t)h−h|>α2}. | (4.3) |
Obviously,
Eα⊂E1α∪E2α. | (4.4) |
From (4.1), we have
|E1α|=0. | (4.5) |
Obviously,
E2α⊂E21α∪E22α, | (4.6) |
where
E21α={x∈R:supt>0|U1(t)h|>α4},E22α={x∈R:|h|>α4}. | (4.7) |
Thus, from Lemma 3.1, we have
|E21α|=∫E21αdx≤C∫E21αsupt>0|U1(t)h|αdx≤Cα−1‖h‖L1xL∞t≤Cα−1‖f‖˙Hϵ(R). | (4.8) |
Obviously, we have
|E22α|≤∫E22αdx≤C∫E22α|h|2α2dx≤Cα−2‖h‖2L2. | (4.9) |
From (4.5), (4.8) and (4.9), we have
|Eα|≤|E1α|+|E2α|≤|E1α|+|E21α|+|E22α|≤Cϵα+Cϵ2α2. | (4.10) |
Thus, for any α>0, from (4.10) we have
|Eα|=0. | (4.11) |
Thus, we have
U1(t)f−f⟶0 | (4.12) |
almost everywhere as t goes to zero.
This completes the proof of Theorem 1.1.
The authors declare there are no conflicts of interest.
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