亚洲免费av电影一区二区三区,日韩爱爱视频,51精品视频一区二区三区,91视频爱爱,日韩欧美在线播放视频,中文字幕少妇AV,亚洲电影中文字幕,久久久久亚洲av成人网址,久久综合视频网站,国产在线不卡免费播放

        ?

        Higher Derivative Estimates for a Linear Elliptic Equation

        2022-07-07 07:37:14WANGMing王明
        應(yīng)用數(shù)學 2022年3期

        WANG Ming (王明)

        (School of Mathematics and Physics, China University of Geosciences, Wuhan 430074, China)

        Abstract: It is well known that the solution of the elliptic equation ?u=V u is analytic if the potential V is analytic.But little is known on the ultra analyticity of the solution if V is ultra analytic.In this note, we show that the solution enjoys a log type ultra analyticity bound by an elegant and delicate induction argument.

        Key words: Elliptic equation; Analyticity; Higher derivative

        1.Introduction

        Let u : ? ?Rd→R be a smooth function.We say that u is analytic in ? if for every x ∈? there exists C,R > 0 so thatholds for all multi-index α =(α1,α2,··· ,αd) ∈Nd, where we adopt the convention |α| = α1+α2+···+αdand α! =α1!α2!···αd!.The quantity R is often referred to the analytic radius at x.The analyticity of solutions for elliptic equation is a classical topic studied in references.If u(x) is harmonic in?, namely ?u=0 in ?, then u is analytic in ?.[3]

        Moreover, let L be an elliptic differential operator with analytic coefficients and O be an open disk in Rd.Then any solution of Lu=0 in O is analytic.For the details of this result,we refer the readers to [2, 6, 8, 9].In particular, consider the linear elliptic equation

        If V is analytic on Rd, then the solution u of (1.1) is analytic on Rd.

        We ask the following question.

        Question:If V is ultra analytic, namely the analytic radius R = ∞, what can we say about the quantitative analytic bound for the solution of (1.1)?

        To our best knowledge, the problem has not been solved in the references.The question is motivated by the quantitative analytic bounds of solutions and its applications in the observability inequalities for heat equations[1,4,10]and KdV equations[5,7].

        To make the question more precise, we assume that there exists C0>0 so that

        Our main result reads as follows.

        Theorem 1.1Assume that (1.2) holds.Let u ∈L2(Rd) be a solution of (1.1).Then there exists a constant C >0 so that

        In our recent paper [10], we have established a similar bound as (1.3) for fractional heat equation on Rd.The main strategy in [10] is to derive some quantitative estimates in Gevrey spaces Gσ, endowed with the norm

        At this moment, we do not know whether the bound (1.3) is optimal or not.We hope to improve (1.3) to ∥u∥L∞(Rd)≤C|α|(α!)δ,?α ∈Ndfor some δ ∈(0,1).But this is also open.

        The notation in this paper is standard.We use AB to denote A ≤CB for some constant C >0.If both AB and BA hold, we write A ~B.

        2.Reduction Theorem 1.1 to Theorem 2.1

        In this section, we show that Theorem 1.1 follows from some higher derivative estimates in one variable, which may be easier to deal with.

        Theorem 2.1Let u ∈L2(Rd) be a solution of (1.1) with V satisfying

        Then there exists a constant C >0 so that

        Proof of Theorem 1.1 under Theorem 2.1Let u ∈L2(Rd) be a solution of

        with V satisfying

        Then, by (2.3)-(2.4), v ∈L2(Rd) is a solution of

        with W =a2V(ax) satisfying

        Suppose that Theorem 2.1 holds, then we apply Theorem 2.1 to v and conclude that for j =1,2,··· ,d

        On the other hand, we have the inequality

        which follows from the Plancherel theorem and the elementary inequality

        Combining (2.7) and (2.8) we find

        Note that u(x)=v(a?1x), we deduce from (2.9) that

        By the Sobolev embedding Hd(Rd)→L∞(Rd), we deduce from (2.10) that

        Combining the bound (2.11) and Lemma 2.2, we complete the proof of Theorem 1.1.

        In the proof above, we have used the following lemma, which is of independent interest.

        Lemma 2.2For all α ∈Nd, we have α!≥d?|α||α|!.

        ProofLet α=(α1,α2,··· ,αd).It suffices to show that

        where |α| = α1+α2+···+αd.We prove (2.12) by induction on |α|.Clearly, (2.12) holds for |α| = 0.Suppose (2.12) holds for all α satisfying |α| = k.Now we consider the case|α|=α1+α2+···+αd=k+1.Then in this case we have

        This, together with the induction hypothesis, gives that

        This shows that (2.12) holds for all α satisfying |α|=k+1.The proof is complete.

        3.Proof of Theorem 2.1

        Let

        We first study the monotonicity of aj.

        Lemma 3.1Let ajbe given by (3.1).Then when n →∞we have

        ProofThanks to the formula (3.1), we find that aj≤aj+1is equivalent to

        Rewrite

        and

        Note that

        here and below we use o(1/jk) to denote a quantity less than or equal some constant times 1/jkas j →∞.Similarly,

        Then we deduce from (3.6) that

        Similarly,

        Thanks to (3.7), we have

        This, together with (3.8), gives that

        Combining (3.4)-(3.5) and (3.9), we find aj≤aj+1is equivalent to

        Thus in order to prove(3.2),it suffices to show that if n/2

        Similarly, to prove (3.3), it suffices to show that if j ≥n ?log n+1 then

        In fact, if j ≥n ?log n+1, then j +log j ≥n ?log n+1+log(n ?log n+1) ≥n holds if 1 ≥log n ?log(n ?log n+1).But this holds clearly if n is large enough.Thus (3.11) holds and the proof is complete.

        The main estimate in this section is the following proposition, which is the key step in the proof of Theorem 2.1.

        Proposition 3.2When n →∞we have

        ProofLet ajbe given by (3.1).We split the sum as

        Then we find

        Thus for large enough n

        2) The case j > n ?2 log n.In this case, we have j > n/2 if n is large enough.So according to Lemma 3.1, we find

        Recall the definition (3.1), we infer that if n ?log n ?2 ≤j ≤n ?log n+1,

        It follows from (3.14)-(3.15) that

        This implies that

        holds if one can show that

        But this is equivalent to

        Thus we deduce that

        It follows that

        So (3.17) holds if n is large enough.

        3) The case n/2 ≤j ≤n ?2 log n.By the definition of aj, we have

        This, together with (3.5) and (3.9), gives

        Moreover, according to (3.14) and (3.15), we have

        Thanks to (3.18)-(3.19), similar to Step 2), we conclude that

        for n large enough.In fact, in 2) we have shown a stronger result: log n?RHS of (3.19) is less than or equal to RHS of (3.20).

        Finally, combining (3.13), (3.16) and (3.20), we complete the proof.

        Proof of Theorem 2.1Thanks to Proposition 3.2, there exists an integer n0> 0 so that for all n ≥n0

        On the other hand,let u ∈L2(Rd)be a solution of ?u=V(x)u.Then by the standard elliptic regularity and the smooth assumptions on V we find that, for every integer k, there exists a constant C(k)>0,∥u∥Hk+2(Rd)≤C(k)∥u∥Hk(Rd).Thus a bootstrapping argument shows that for every k ∈N, ∥u∥Hk(Rd)≤C(k).In particular, this implies that

        for some constant C >0, where n0is the same to that fulfilling (3.21).

        Now we claim that if

        holds for some n ≥n0, then

        In fact, by the Plancherel theorem and the equation ?u=V(x)u, we have

        Using the Leibniz rule, the assumption (2.1) and (3.23), we infer that

        Here in the last step we used (3.21).Combining (3.25)-(3.26) gives the claim.

        Finally, thanks to the bound (3.22) and the claim, we find that Theorem 2.1 holds.

        亚洲av色精品国产一区二区三区| www夜片内射视频在观看视频| 国产伦人人人人人人性| 粗大猛烈进出白浆视频| 18级成人毛片免费观看| 99热这里只有精品4| 日韩在线精品在线观看| 中文字幕人妻激情在线视频| 狠狠综合亚洲综合亚洲色| 久久青青草原精品国产app| a级毛片免费观看在线| 夜夜综合网| 亚洲天堂av另类在线播放| 虎白女粉嫩粉嫩的18在线观看| 国产国语亲子伦亲子| 农村欧美丰满熟妇xxxx| 国产精品国产三级国产av′| 91社区视频在线观看| 国产亚洲精品成人av在线| 日本一区二区在线高清观看| 无码人妻h动漫中文字幕| 中文字幕乱码免费视频| 久久久久亚洲精品美女| 无码中文字幕久久久久久| 日本加勒比精品一区二区视频| 午夜理论片yy6080私人影院| 亚洲人成网站77777在线观看| 波多野结衣一区二区三区视频| av永久天堂一区二区三区蜜桃 | 中文字幕无码精品亚洲资源网久久| 无码av永久免费大全| 亚洲最大一区二区在线观看| 超级碰碰色偷偷免费视频| a在线观看免费网站大全| 91精品国产色综合久久不卡蜜| 女同欲望一区二区三区| 在线观看特色大片免费视频| 国内精品久久久久久久影视麻豆| 丰满人妻中文字幕乱码| 日韩av一区二区网址| 日日澡夜夜澡人人高潮|