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Abstract and Applied Analysis
VolumeΒ 2012Β (2012), Article IDΒ 739156, 17 pages
http://dx.doi.org/10.1155/2012/739156
Research Article

On the Nonlinear Instability of Traveling Waves for a Sixth-Order Parabolic Equation

Department of Mathematics, Jilin University, Changchun 130012, China

Received 4 June 2012; Accepted 2 August 2012

Academic Editor: LjubisaΒ Kocinac

Copyright Β© 2012 Zhenbang Li and Changchun Liu. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

We study the instability of the traveling waves of a sixth-order parabolic equation which arises naturally as a continuum model for the formation of quantum dots and their faceting. We prove that some traveling wave solutions are nonlinear unstable under 𝐻4 perturbations. These traveling wave solutions converge to a constant as π‘₯β†’βˆž.

1. Introduction

In this paper, we consider the following sixth-order parabolic equation πœ•π‘’πœ•π‘‘=𝐷6𝑒+𝐷4ξ€·π‘’βˆ’π‘’3ξ€Έ+𝑔(𝑒),(π‘₯,𝑑)βˆˆπ‘…Γ—(0,𝑇),(1.1) where 𝑔(𝑒)=π‘Ž(1βˆ’π‘’2),π‘Ž>0.

Equation (1.1) arises naturally as a continuum model for the formation of quantum dots and their faceting; see [1]. Here 𝑒(π‘₯,𝑑) denotes the surface slope. The high-order derivatives are a result of the additional regularization energy which is required to form an edge between two-plane surfaces with different orientations.

During the past years, only a few works have been devoted to the sixth-order parabolic equation [2–7]. Barrett et al. [2] considered the above equation with π‘š=2. A finite element method is presented which proves to be well posed and convergent. Numerical experiments illustrate the theory.

Recently, JΓΌngel and MiliΕ‘iΔ‡ [5] studied the sixth-order nonlinear parabolic equation πœ•π‘’=𝑒1πœ•π‘‘π‘’ξ€·π‘’(ln𝑒)π‘₯π‘₯ξ€Έπ‘₯π‘₯+12ξ€·(ln𝑒)π‘₯π‘₯ξ€Έ2π‘₯ξ‚„π‘₯.(1.2) They proved the global-in-time existence of weak nonnegative solutions in one space dimension with periodic boundary conditions.

Evans et al. [3, 4] considered the sixth-order thin film equation containing an unstable (backward parabolic) second-order term πœ•π‘’ξ€Ίπœ•π‘‘=div|𝑒|π‘›βˆ‡Ξ”2π‘’ξ€»ξ€·βˆ’Ξ”|𝑒|π‘βˆ’1𝑒,𝑛>0,𝑝>1.(1.3) By a formal matched expansion technique, they show that, for the first critical exponent 𝑝=𝑝0=𝑛+1+4/𝑁 for π‘›βˆˆ(0,5/4), where 𝑁 is the space dimension, the free-boundary problem with zero-height, zero-contact-angle, zero-moment, and zero-flux conditions at the interface admits a countable set of continuous branches of radially symmetric self-similar blow-up solutions π‘’π‘˜(π‘₯,𝑑)=(π‘‡βˆ’π‘‘)(βˆ’π‘/(𝑛𝑁+6))π‘“π‘˜(𝑦), 𝑦=π‘₯/((π‘‡βˆ’π‘‘)(1/(𝑛𝑁+6))), where 𝑇>0 is the blow-up time.

Korzec et al. [8] considered the sixth-order equation π‘’π‘‘βˆ’πœˆπ‘’π‘’π‘₯βˆ’ξ€·π‘’βˆ’π‘’3+πœ€2𝑒π‘₯π‘₯ξ€Έπ‘₯π‘₯π‘₯π‘₯=0.(1.4) New type of stationary solutions is derived by an extension of the method of matched asymptotic expansion.

In this paper, we study instability of the traveling waves of (1.1). Our main result is as follows.

Theorem 1.1. All the traveling waves πœ‘(π‘₯βˆ’π‘π‘‘) of (1.1) satisfying πœ‘βˆˆπΏβˆž(𝑅), πœ‘(𝑛)∈𝐿∞(𝑅)∩𝐿2(𝑅)(𝑛=1,2,…,6) are nonlinearly unstable in the space 𝐻4(𝑅), where πœ‘(𝑛) denotes π‘›π‘‘β„Ž derivative of πœ‘.

The stability and instability of special solutions for the higher-order parabolic equation are very important in the applied fields. Carlen et al. [9] proved the nonlinear stability of fronts for the Cahn-Hilliard, under 𝐿1 perturbations. Gao and Liu [10] prove that it is nonlinearly unstable under 𝐻2 perturbations, for some traveling wave solution of the convective Cahn-Hilliard equation. The relevant equations have also been studied in [11, 12]. The main difficulties for treating (1.1) are caused by the principal part and the lack of the Lyapunov functional. Our proof is based on the principle of linearization. We invoke a general theorem that asserts that linearized instability implies nonlinear instability.

This paper is organized as follows. In the next section, we find an exact traveling wave solution for (1.1). In Section 3, we give the proof of our main result.

2. Exact Traveling Wave Solutions

In this section, we construct an exact traveling wave which satisfies all conditions of Theorem 1.1.

If πœ‘(π‘₯βˆ’π‘π‘‘)=πœ‘(𝑧) is a traveling wave solution of (1.1), then πœ‘ satisfies the ordinary differential equation βˆ’π‘πœ‘β€²=πœ‘(6)+ξ€·1βˆ’3πœ‘2ξ€Έπœ‘(4)βˆ’36πœ‘ξ…ž2πœ‘ξ…žξ…žβˆ’18πœ‘πœ‘2ξ…žξ…žβˆ’24πœ‘πœ‘β€²πœ‘ξ…žξ…žξ…žξ€·+π‘Ž1βˆ’πœ‘2ξ€Έ.(2.1) Let πœ‘β€²=πœ•πœ‘/πœ•π‘§=π‘˜(1βˆ’πœ‘2). Then πœ‘ξ…žξ…ž=πœ•ξ€·π‘˜ξ€·πœ•π‘§1βˆ’πœ‘2ξ€Έξ€Έ=βˆ’2π‘˜2πœ‘ξ€·1βˆ’πœ‘2ξ€Έ,πœ‘ξ…žξ…žξ…ž=πœ•ξ€·πœ•π‘§βˆ’2π‘˜2πœ‘ξ€·1βˆ’πœ‘2ξ€Έξ€Έ=2π‘˜3ξ€·βˆ’1+3πœ‘2ξ€Έξ€·1βˆ’πœ‘2ξ€Έ,πœ‘(4)=πœ•ξ€·πœ•π‘§2π‘˜3ξ€·βˆ’1+3πœ‘2ξ€Έξ€·1βˆ’πœ‘2ξ€Έξ€Έ=2π‘˜4ξ€·8πœ‘βˆ’12πœ‘3ξ€Έξ€·1βˆ’πœ‘2ξ€Έ,πœ‘(5)=πœ•ξ€·πœ•π‘§2π‘˜4ξ€·8πœ‘βˆ’12πœ‘3ξ€Έξ€·1βˆ’πœ‘2ξ€Έξ€Έ=8π‘˜5ξ€·1βˆ’πœ‘2ξ€Έξ€·2βˆ’15πœ‘2+15πœ‘4ξ€Έ,πœ‘(6)=πœ•ξ€·πœ•π‘§8π‘˜5ξ€·1βˆ’πœ‘2ξ€Έξ€·2βˆ’15πœ‘2+15πœ‘4ξ€Έξ€Έ=16π‘˜6πœ‘ξ€·1βˆ’πœ‘2ξ€Έξ€·60πœ‘2βˆ’45πœ‘4ξ€Έ.βˆ’17(2.2) Substituting the above equations into (2.1), we have ξ€·βˆ’π‘π‘˜βˆ’π‘Ž=360π‘˜4βˆ’720π‘˜6ξ€Έπœ‘5+ξ€·960π‘˜6βˆ’480π‘˜4ξ€Έπœ‘3+ξ€·136π‘˜4βˆ’272π‘˜6ξ€Έπœ‘.(2.3) Then comparing the order of πœ‘, we obtain βˆ’π‘π‘˜=π‘Ž,360π‘˜4βˆ’720π‘˜6=0,960π‘˜6βˆ’480π‘˜4=0,136π‘˜4βˆ’272π‘˜6=0.(2.4) A simple calculation shows that βˆšπ‘˜=1/2, βˆšπ‘=βˆ’2π‘Ž. Hence, we get 1πœ‘β€²=√2ξ€·1βˆ’πœ‘2ξ€Έ,(2.5) that is, 12ln1+πœ‘=11βˆ’πœ‘βˆš2𝑧,(2.6) that is, π‘’πœ‘(𝑧)=√(1/2)π‘§βˆ’π‘’βˆšβˆ’(1/2)π‘§π‘’βˆš(1/2)𝑧+π‘’βˆšβˆ’(1/2)𝑧1=tanh√2𝑧.(2.7) We easily proved that lim𝑧→+βˆžπœ‘(𝑧)=1,lim𝑧→+βˆžπœ‘(𝑧)=βˆ’1(2.8) and πœ‘(𝑧) satisfies the conditions of the Theorem 1.1.

3. Proof of The Result

To prove the Theorem 1.1, we first consider an evolution equation πœ•π‘’πœ•π‘‘=𝐿𝑒+𝐹(𝑒),(3.1) where 𝐿 is a linear operator that generates a strongly continuous semigroup 𝑒𝑑𝐿 on a Banach space 𝑋, and 𝐹 is a strongly continuous operator such that 𝐹(0)=0. In [13], authors considered the whole problem only on space X, that is, the nonlinear operator maps 𝑋 to 𝑋. However, many equations posses nonlinear terms that include derivatives and therefore, 𝐹 maps into a large Banach space 𝑍. Hence, they again got the following lemma.

Lemma 3.1 (see [14]). Assume the following.(i)X, Z are two Banach spaces with π‘‹βŠ‚π‘ and ‖𝑒‖𝑍≀𝑐1‖𝑒‖𝑋 for π‘’βˆˆπ‘‹.(ii)L generates a strongly continuous semigroup 𝑒𝑑𝐿 on the space Z, and the semigroup 𝑒𝑑𝐿 maps Z into X for 𝑑>0 and ∫10‖𝑒𝑑𝐿‖𝑍→𝑋𝑑𝑑=𝐢4<∞.(iii)The spectrum of 𝐿 on 𝑋 meets the right half-plane, {π‘…π‘’πœ†>0}.(iv)πΉβˆΆπ‘‹β†’π‘ is continuous and βˆƒβ€‰β€‰πœŒ0>0,𝐢3>0,𝛼>1 such that ‖𝐹(𝑒)‖𝑍<𝐢0‖𝑒‖𝛼𝑋, for ‖𝑒‖𝑋<𝜌0.
Then the zero solution of (3.1) is nonlinearly unstable in the space 𝑋.

In this paper, we are going to use Lemma 3.1 for the proof of Theorem 1.1.

Definition 3.2. A traveling wave solution πœ‘(π‘₯βˆ’π‘π‘‘) of (1.1) is said to be nonlinearly unstable in the space X, if there exist positive πœ€0 and 𝐢0, a sequence {𝑒𝑛} of solutions of (1.1) and a sequence of time 𝑑𝑛>0 such that ‖𝑒𝑛(0)βˆ’πœ‘(π‘₯)‖𝑋→0 but ‖𝑒𝑛(𝑑𝑛)βˆ’πœ‘(β‹…βˆ’π‘π‘‘π‘›)‖𝑋β‰₯πœ€0.

If πœ‘(π‘₯βˆ’π‘π‘‘)∈𝐻4(𝑅) is a traveling wave solution of (1.1), then letting πœ”(π‘₯,𝑑)=𝑒(π‘₯,𝑑)βˆ’πœ‘(π‘₯βˆ’π‘π‘‘), we have (πœ”+πœ‘)𝑑=πœ•6π‘₯(πœ”+πœ‘)+πœ•4π‘₯ξ€Ί(πœ”+πœ‘)βˆ’(πœ”+πœ‘)3ξ€»ξ€Ί+π‘Ž1βˆ’(πœ”+πœ‘)2ξ€»=πœ•6π‘₯πœ”+πœ‘(6)+πœ•4π‘₯ξ€·πœ”+πœ‘βˆ’πœ”3βˆ’πœ‘3βˆ’3πœ”2πœ‘βˆ’3πœ”πœ‘2ξ€Έξ€·+π‘Ž1βˆ’πœ”2βˆ’πœ‘2ξ€Έ,βˆ’2πœ”πœ‘(3.2) that is, πœ”π‘‘=πœ•6π‘₯πœ”+πœ•4π‘₯ξ€·πœ”βˆ’πœ”3βˆ’3πœ”2πœ‘βˆ’3πœ”πœ‘2ξ€Έξ€·+π‘Žβˆ’πœ”2ξ€Έ,βˆ’2πœ”πœ‘(3.3) that is, πœ”π‘‘βˆ’πœ•6π‘₯ξ€·πœ”βˆ’1βˆ’3πœ‘2ξ€Έπœ•4π‘₯πœ”+24πœ‘πœ‘β€²πœ•3π‘₯ξ‚€πœ”+36πœ‘πœ‘ξ…žξ…ž+36πœ‘ξ…ž2ξ‚πœ•2π‘₯πœ”+ξ€·24πœ‘πœ‘ξ…žξ…žξ…ž+72πœ‘ξ…žπœ‘ξ…žξ…žξ€Έπœ•π‘₯ξ‚€πœ”+18πœ‘2ξ…žξ…ž+24πœ‘ξ…žπœ‘ξ…žξ…žξ…ž+6πœ‘πœ‘(4)ξ‚πœ”+2π‘Žπœ‘=𝐹(πœ”),(3.4) where 𝐹(πœ”)=3πœ‘(4)ξ€Έπœ”βˆ’π‘Ž2+24πœ‘ξ…žξ…žξ…žπœ”πœ•π‘₯πœ”+36πœ‘ξ…žξ…ž(πœ•π‘₯πœ”)2+72πœ‘β€²πœ•π‘₯πœ”πœ•2π‘₯πœ”ξ€·πœ•+36π‘₯πœ”ξ€Έ2πœ•2π‘₯πœ”+(24πœ‘+24πœ”)πœ•π‘₯πœ”πœ•3π‘₯πœ”+36πœ‘ξ…žξ…žπœ”πœ•2π‘₯πœ”ξ€·πœ•+(18πœ‘+18πœ”)2π‘₯πœ”ξ€Έ2+24πœ‘β€²πœ”πœ•3π‘₯πœ”+6πœ‘πœ”πœ•4π‘₯πœ”+3πœ”2πœ•4π‘₯πœ”,(3.5) with the initial value πœ”(π‘₯,0)=πœ”0(π‘₯)≑𝑒0(π‘₯)βˆ’πœ‘(π‘₯).(3.6) So the stability of traveling wave solutions of (1.1) is translated into the stability of the zero solution of (3.4). In order to prove Theorem 1.1, taking 𝑍=𝐿2(𝑅), 𝑋=𝐻4(𝑅), we need to prove that the four conditions of Lemma 3.1 are satisfied by the associated equation (3.4). The condition (i) is satisfied by our choice of 𝑍 and 𝑋.

Denote the linear partial differential operator in (3.4) by 𝐿=(πœ•6π‘₯+πœ•4π‘₯)βˆ’[3πœ‘2πœ•4π‘₯+24πœ‘πœ‘β€²πœ•3π‘₯+(36πœ‘πœ‘ξ…žξ…ž+36πœ‘β€²2)πœ•2π‘₯+(24πœ‘πœ‘ξ…žξ…žξ…ž+72πœ‘β€²πœ‘ξ…žξ…ž)πœ•π‘₯+(18πœ‘2ξ…žξ…ž+24πœ‘β€²πœ‘ξ…žξ…žξ…ž+6πœ‘πœ‘(4)+2π‘Žπœ‘)] =𝐿0βˆ’[3πœ‘2πœ•4π‘₯+24πœ‘πœ‘β€²πœ•3π‘₯+(36πœ‘πœ‘ξ…žξ…ž+36πœ‘β€²2)πœ•2π‘₯+(24πœ‘πœ‘ξ…žξ…žξ…ž+72πœ‘β€²πœ‘ξ…žξ…ž)πœ•π‘₯+(18πœ‘2ξ…žξ…ž+24πœ‘β€²πœ‘ξ…žξ…žξ…ž+6πœ‘πœ‘(4)+2π‘Žπœ‘)] with 𝐿0=πœ•6π‘₯+πœ•4π‘₯. Then (3.4) may be rewritten in the form of (3.1) πœ”π‘‘=πΏπœ”+𝐹(πœ”).(3.7) Note the F maps 𝐻4(𝑅) into 𝐿2(𝑅), using the Sobolev embedding theorem, we have ‖‖𝐹(πœ”)𝐿2β‰€πΆβ€–πœ”β€–2𝐻4,𝐢>0,forβ€–πœ”β€–π»4<1.(3.8) So, the condition (iv) is satisfied.

To prove condition (ii) in Lemma 3.1, we need the following two lemmas.

Lemma 3.3. Let 𝐿0=πœ•6π‘₯+πœ•4π‘₯. Then ‖‖𝑒𝑑𝐿0β€–β€–π»π‘šβ†’π»π‘šβ‰€π‘’(4/27)𝑑,forπ‘šβˆˆπ‘…+‖‖𝑒,0≀𝑑<∞,(3.9)𝑑𝐿0‖‖𝐿2→𝐻4β‰€π‘Ž(𝑑)≑5π‘‘βˆ’2/3,for0<𝑑≀1.(3.10)

Proof. We write 𝑒(π‘₯,𝑑)=𝑒𝑑𝐿0𝑒0(π‘₯). By Fourier transformation ̂𝑒(πœ‰,𝑑)=π‘’βˆ’π‘‘(πœ‰6βˆ’πœ‰4)̂𝑒0(πœ‰),‖𝑒‖2π»π‘šβ‰‘ξ€œβˆžβˆ’βˆžξ€·1+πœ‰2ξ€Έπ‘š||||̂𝑒(πœ‰,𝑑)2=ξ€œπ‘‘πœ‰βˆžβˆ’βˆžξ€·1+πœ‰2ξ€Έπ‘šπ‘’βˆ’2𝑑(πœ‰6βˆ’πœ‰4)||̂𝑒0||(πœ‰)2π‘‘πœ‰β‰€supπœ‰βˆˆπ‘…π‘’βˆ’2𝑑(πœ‰6βˆ’πœ‰4)ξ€œβˆžβˆ’βˆžξ€·1+πœ‰2ξ€Έπ‘š||̂𝑒0||(πœ‰)2π‘‘πœ‰=𝑒(8/27)𝑑‖‖𝑒0β€–β€–2π»π‘š.(3.11) Hence, ‖‖𝑒𝑑𝐿0β€–β€–π»π‘šβ†’π»π‘šβ‰€π‘’(4/27)𝑑.(3.12) On the other hand, letting 𝑠=πœ‰2, we have ‖𝑒‖2𝐻4≀supπ‘ βˆˆπ‘…βˆ«π‘“(𝑠)βˆžβˆ’βˆž||̂𝑒0||(πœ‰)2π‘‘πœ‰,(3.13) with 𝑓(𝑠)=(1+𝑠)4π‘’βˆ’2𝑑(𝑠3βˆ’π‘ 2), 𝑑>0. Elementary computation shows that sup𝑠>0ξ‚€4𝑓(𝑠)≀3+16π‘‘βˆ’4/3𝑒(8/27)𝑑.(3.14) Thus, ‖𝑒(π‘₯,𝑑)‖𝐻4≀43+16π‘‘βˆ’4/31/2𝑒(4/27)𝑑‖‖𝑒0‖‖𝐿2,‖‖𝑒𝑑𝐿0‖‖𝐿2→𝐻4≀43+16π‘‘βˆ’4/31/2𝑒(4/27)𝑑≀5π‘‘βˆ’2/3,for0<𝑑≀1,(3.15) since 𝑒(4/27)𝑑≀𝑒4/27<2. Thus, Lemma 3.3 has been proved.

Lemma 3.4. Let 𝐿=(πœ•6π‘₯+πœ•4π‘₯)βˆ’[3πœ‘2πœ•4π‘₯+24πœ‘πœ‘ξ…žπœ•3π‘₯+(36πœ‘πœ‘ξ…žξ…ž+36πœ‘ξ…ž2)πœ•2π‘₯+(24πœ‘πœ‘ξ…žξ…žξ…ž+72πœ‘ξ…žπœ‘β€²β€²)πœ•π‘₯+(18πœ‘2ξ…žξ…ž+24πœ‘β€²πœ‘ξ…žξ…žξ…ž+6πœ‘πœ‘(4)+2π‘Žπœ‘)]=𝐿0βˆ’[3πœ‘2πœ•4π‘₯+24πœ‘πœ‘β€²πœ•3π‘₯+(36πœ‘πœ‘ξ…žξ…ž+36πœ‘β€²2)πœ•2π‘₯+(24πœ‘πœ‘ξ…žξ…žξ…ž+72πœ‘β€²πœ‘ξ…žξ…ž)πœ•π‘₯ + (18πœ‘2ξ…žξ…ž+24πœ‘β€²πœ‘ξ…žξ…žξ…ž+6πœ‘πœ‘(4)+2π‘Žπœ‘)] with 𝐿0=πœ•6π‘₯+πœ•4π‘₯, πœ‘(𝑖)∈𝐿∞(𝑅),𝑖=0,1,2,3,4. Then β€–β€–etLβ€–β€–L2β†’H4≀C1tβˆ’2/3β€–β€–e,for0<t≀1,(3.16)tLβ€–β€–H4β†’H4≀C2<∞,for0<t≀1.(3.17)

Proof. Consider the initial value problem 𝑒𝑑=𝐿𝑒=𝐿0π‘’βˆ’3πœ‘2πœ•4π‘₯π‘’βˆ’24πœ‘πœ‘β€²πœ•3π‘₯ξ‚€π‘’βˆ’36πœ‘πœ‘ξ…žξ…ž+36πœ‘ξ…ž2ξ‚πœ•2π‘₯π‘’βˆ’ξ€·24πœ‘πœ‘ξ…žξ…žξ…ž+72πœ‘ξ…žπœ‘ξ…žξ…žξ€Έπœ•π‘₯ξ‚€π‘’βˆ’18πœ‘2ξ…žξ…ž+24πœ‘ξ…žπœ‘ξ…žξ…žξ…ž+6πœ‘πœ‘(4)+2π‘Žπœ‘π‘’,𝑒(π‘₯,0)=𝑒0(π‘₯).(3.18) Then 𝑒(π‘₯,𝑑)=𝑒𝑑𝐿𝑒0(π‘₯), 𝑑β‰₯0, π‘₯βˆˆπ‘…, thus 𝑒(π‘₯,𝑑)=𝑒𝑑𝐿0𝑒0βˆ’ξ€œπ‘‘0𝑒(π‘‘βˆ’πœ)𝐿03πœ‘2πœ•4π‘₯𝑒+24πœ‘πœ‘β€²πœ•3π‘₯𝑒+36πœ‘πœ‘ξ…žξ…ž+36πœ‘ξ…ž2ξ‚πœ•2π‘₯𝑒+ξ€·24πœ‘πœ‘ξ…žξ…žξ…ž+72πœ‘ξ…žπœ‘ξ…žξ…žξ€Έπœ•π‘₯𝑒+ξ‚€18πœ‘2ξ…žξ…ž+24πœ‘ξ…žπœ‘ξ…žξ…žξ…ž+6πœ‘πœ‘(4)𝑒+2π‘Žπœ‘π‘‘πœ.(3.19) Denote 𝐴=β€–πœ‘β€–πΏβˆž, 𝐡=β€–πœ‘β€²β€–πΏβˆž, 𝐢=β€–πœ‘ξ…žξ…žβ€–πΏβˆž, 𝐷=β€–πœ‘ξ…žξ…žξ…žβ€–πΏβˆž, 𝐸=β€–πœ‘(4)β€–πΏβˆž and 𝑀=3𝐴2+24𝐴𝐡+36𝐴𝐢+36𝐡2+24𝐴𝐷+72𝐡𝐢+18𝐢2+24𝐡𝐷+6𝐴𝐸+2π‘Žπ΄.(3.20) Then, we have (‖𝑒𝑑)‖𝐻4≀‖‖𝑒𝑑𝐿0‖‖𝐻4→𝐻4‖‖𝑒0‖‖𝐻4+ξ€œπ‘‘0‖‖𝑒(π‘‘βˆ’πœ)𝐿0‖‖𝐻4→𝐻43β€–πœ‘β€–2πΏβˆžβ€–β€–πœ•4π‘₯𝑒‖‖𝐿2ξ€œπ‘‘πœ+24𝑑0‖‖𝑒(π‘‘βˆ’πœ)𝐿0‖‖𝐻4→𝐻4β€–πœ‘β€–πΏβˆžβ€–πœ‘β€²β€–πΏβˆžβ€–β€–πœ•3π‘₯𝑒‖‖𝐿2+ξ€œπ‘‘πœπ‘‘0‖‖𝑒(π‘‘βˆ’πœ)𝐿0‖‖𝐻4→𝐻4ξ‚€36β€–πœ‘β€–πΏβˆžβ€–β€–πœ‘ξ…žξ…žβ€–β€–πΏβˆžβ€–β€–πœ‘+36ξ…ž2β€–β€–πΏβˆžξ‚β€–β€–πœ•2π‘₯𝑒‖‖𝐿2+ξ€œπ‘‘πœπ‘‘0‖‖𝑒(π‘‘βˆ’πœ)𝐿0‖‖𝐻4→𝐻4ξ€·24β€–πœ‘β€–πΏβˆžβ€–β€–πœ‘ξ…žξ…žξ…žβ€–β€–πΏβˆž+72β€–πœ‘β€²β€–πΏβˆžβ€–β€–πœ‘ξ…žξ…žβ€–β€–πΏβˆžξ€Έβ€–β€–πœ•π‘₯𝑒‖‖𝐿2+ξ€œπ‘‘πœπ‘‘0‖‖𝑒(π‘‘βˆ’πœ)𝐿0‖‖𝐻4→𝐻4ξ‚€β€–β€–πœ‘18ξ…žξ…žβ€–β€–2𝐿∞+24β€–πœ‘β€²β€–πΏβˆžβ€–β€–πœ‘ξ…žξ…žξ…žβ€–β€–πΏβˆž+6β€–πœ‘β€–πΏβˆžβ€–β€–πœ‘(4)β€–β€–πΏβˆž+2π‘Žβ€–πœ‘β€–πΏβˆžξ‚β€–π‘’β€–πΏ2π‘‘πœβ‰€π‘’(4/27)𝑑‖‖𝑒0‖‖𝐻4ξ€œ+𝑀𝑑0𝑒(4/27)(π‘‘βˆ’πœ)‖𝑒(𝜏)‖𝐻4π‘‘πœ,(3.21) where we use 𝑒(𝑑) to denote 𝑒(β‹…,𝑑).
By iteration, (‖𝑒𝑑)‖𝐻4≀𝑒(4/27)𝑑‖‖𝑒0‖‖𝐻4ξ€œ+𝑀𝑑0𝑒(4/27)(π‘‘βˆ’πœ)𝑒(4/27)πœβ€–β€–π‘’0‖‖𝐻4ξ€œ+π‘€πœ0𝑒(4/27)(πœβˆ’π‘ )(‖𝑒𝑠)‖𝐻4ξ‚Ήπ‘‘π‘ π‘‘πœ=𝑒(4/27)𝑑‖‖𝑒0‖‖𝐻4ξ€œ+𝑀𝑑0𝑒(4/27)𝑑‖‖𝑒0‖‖𝐻4π‘‘πœ+𝑀2ξ€œπ‘‘0ξ€œπœ0𝑒(4/27)(π‘‘βˆ’π‘ )(‖𝑒𝑠)‖𝐻4π‘‘π‘ π‘‘πœβ‰€π‘’(4/27)𝑑‖‖𝑒0‖‖𝐻4+𝑀𝑑𝑒(4/27)𝑑‖‖𝑒0‖‖𝐻4+𝑀2ξ€œπ‘‘0ξ‚Έξ€œπ‘‘π‘ π‘’(4/27)(π‘‘βˆ’π‘ )‖𝑒(𝑠)‖𝐻4ξ‚Ήπ‘‘πœπ‘‘π‘ β‰€π‘’(4/27)𝑑‖‖𝑒0‖‖𝐻4+𝑀𝑑𝑒(4/27)𝑑‖‖𝑒0‖‖𝐻4+𝑀2𝑑𝑒(4/27)π‘‘ξ€œπ‘‘0‖𝑒(𝑠)‖𝐻4𝑑𝑠≀𝑒4/27‖‖𝑒0‖‖𝐻4+𝑀𝑒4/27‖‖𝑒0‖‖𝐻4+𝑒4/27𝑀2ξ€œπ‘‘0‖‖𝑒(𝑠)𝐻4𝑑𝑠,for0<π‘ β‰€πœβ‰€π‘‘β‰€1.(3.22) Let βˆ«π‘£(𝑑)=𝑑0‖𝑒(𝑠)‖𝐻4𝑑𝑠. Then 𝑑𝑣(𝑑)≀𝑒𝑑𝑑4/27+𝑒4/27𝑀‖‖𝑒0‖‖𝐻4+𝑒4/27𝑀2𝑣(𝑑),for0<𝑑≀1.(3.23) Multiplying both sides of the above inequality by π‘’βˆ’π‘’(4/27)𝑀2𝑑, we have π‘‘ξ‚€π‘’βˆ’π‘’4/27𝑀2𝑑𝑣(𝑑)π‘‘π‘‘β‰€π‘’βˆ’π‘’4/27𝑀2𝑑𝑒4/27‖‖𝑒(1+𝑀)0‖‖𝐻4,where0<𝑑≀1.(3.24) Integrating the above inequality with respect to 𝑑 over (0,𝑑), we obtain π‘’βˆ’π‘’4/27𝑀2π‘‘ξ€œπ‘£(𝑑)≀𝑑0π‘’βˆ’π‘’4/27𝑀2𝑠𝑒4/27(‖‖𝑒1+𝑀)0‖‖𝐻4𝑑𝑠,(3.25) that is, 𝑣(𝑑)≀𝑒𝑒4/27𝑀2π‘‘ξ€œπ‘‘0π‘’βˆ’π‘’4/27𝑀2𝑠𝑒4/27(‖‖𝑒1+𝑀)0‖‖𝐻4𝑑𝑠.(3.26) Observing that βˆ«π‘£(𝑑)=𝑑0‖𝑒(𝑠)‖𝐻4𝑑𝑠 is bounded and substituting the above inequality into (3.22), we get (‖𝑒𝑑)‖𝐻4≀𝑒4/27‖‖𝑒0‖‖𝐻4+𝑀𝑒4/27‖‖𝑒0‖‖𝐻4+𝑒4/27𝑀2ξ€œπ‘‘0(‖𝑒𝑠)‖𝐻4𝑑𝑠≀𝑐2<∞,for0<𝑑≀1,𝑐2>0,(3.27) thus (3.17) has been proven.
Next, we prove the inequality (3.16). Clearly, we have (‖𝑒𝑑)‖𝐻4≀‖‖𝑒𝑑𝐿0‖‖𝐿2→𝐻4‖‖𝑒0‖‖𝐿2+ξ€œπ‘‘0‖‖𝑒(π‘‘βˆ’πœ)𝐿0‖‖𝐿2→𝐻43β€–πœ‘β€–2πΏβˆžβ€–β€–πœ•4π‘₯𝑒‖‖𝐿2+ξ€œπ‘‘πœπ‘‘0‖‖𝑒(π‘‘βˆ’πœ)𝐿0‖‖𝐿2→𝐻424β€–πœ‘β€–πΏβˆžβ€–πœ‘β€²β€–πΏβˆžβ€–β€–πœ•3π‘₯𝑒‖‖𝐿2+ξ€œπ‘‘πœπ‘‘0‖‖𝑒(π‘‘βˆ’πœ)𝐿0‖‖𝐿2→𝐻4ξ‚€36β€–πœ‘β€–πΏβˆžβ€–β€–πœ‘ξ…žξ…žβ€–β€–πΏβˆžβ€–β€–πœ‘+36ξ…ž2β€–β€–πΏβˆžξ‚β€–β€–πœ•2π‘₯𝑒‖‖𝐿2+ξ€œπ‘‘πœπ‘‘0‖‖𝑒(π‘‘βˆ’πœ)𝐿0‖‖𝐿2→𝐻4ξ€·24β€–πœ‘β€–πΏβˆžβ€–β€–πœ‘ξ…žξ…žξ…žβ€–β€–πΏβˆž+72β€–πœ‘β€²β€–πΏβˆžβ€–β€–πœ‘ξ…žξ…žβ€–β€–πΏβˆžξ€Έβ€–β€–πœ•π‘₯𝑒‖‖𝐿2+ξ€œπ‘‘πœπ‘‘0‖‖𝑒(π‘‘βˆ’πœ)𝐿0‖‖𝐿2→𝐻4ξ‚€β€–β€–πœ‘18ξ…žξ…žβ€–β€–2𝐿∞+24β€–πœ‘β€²β€–πΏβˆžβ€–β€–πœ‘ξ…žξ…žξ…žβ€–β€–πΏβˆž+6β€–πœ‘β€–πΏβˆžβ€–β€–πœ‘(4)β€–β€–πΏβˆž+2π‘Žβ€–πœ‘β€–πΏβˆžξ‚β€–π‘’β€–πΏ2β€–β€–π‘’π‘‘πœβ‰€π‘Ž(𝑑)0‖‖𝐿2ξ€œ+𝑀𝑑0π‘Žβ€–(π‘‘βˆ’πœ)‖𝑒(𝜏)𝐻4π‘‘πœ,(3.28) where π‘Ž(𝑑) is defined in Lemma 3.3, and we use 𝑒(𝑑) to denote 𝑒(β‹…,𝑑).
By iteration, (‖𝑒𝑑)‖𝐻4β€–β€–π‘’β‰€π‘Ž(𝑑)0‖‖𝐿2ξ€œ+𝑀𝑑0ξ‚Έβ€–β€–π‘’π‘Ž(π‘‘βˆ’πœ)π‘Ž(𝜏)π‘œβ€–β€–πΏ2ξ€œ+π‘€πœ0π‘Ž(πœβˆ’π‘ )‖𝑒(𝑠)‖𝐻4ξ‚Ήβ€–β€–π‘’π‘‘π‘ π‘‘πœ=π‘Ž(𝑑)0‖‖𝐿2ξ€œ+𝑀𝑑0β€–β€–π‘’π‘Ž(π‘‘βˆ’πœ)π‘Ž(𝜏)0‖‖𝐿2π‘‘πœ+𝑀2ξ€œπ‘‘0ξ€œπœ0π‘Ž(π‘‘βˆ’πœ)π‘Ž(πœβˆ’π‘ )‖𝑒(𝑠)‖𝐻4π‘‘π‘ π‘‘πœ.(3.29) The second term on the right of (3.29) is π‘€ξ€œπ‘‘0β€–β€–π‘’π‘Ž(π‘‘βˆ’πœ)π‘Ž(𝜏)0‖‖𝐿2β€–β€–π‘’π‘‘πœ=𝑀0‖‖𝐿2ξ€œπ‘‘05(π‘‘βˆ’πœ)βˆ’2/35πœβˆ’2/3β€–β€–π‘’π‘‘πœ=25𝑀0‖‖𝐿2ξ€œπ‘‘0π‘‘βˆ’4/3ξ‚€πœ1βˆ’π‘‘ξ‚βˆ’2/3ξ‚€πœπ‘‘ξ‚βˆ’2/3π‘‘πœ=25𝑀𝐢3π‘‘βˆ’1/3‖‖𝑒0‖‖𝐿2,0<𝑑<1,(3.30) where 𝐢3=∫10(1βˆ’π‘Ÿ)βˆ’1/4π‘Ÿβˆ’1/4π‘‘π‘Ÿ. By exchanging the order of integration, we get from the third term on the right side of (3.29), ξ€œπ‘‘0ξ€œπœ0π‘Ž(π‘‘βˆ’πœ)π‘Ž(πœβˆ’π‘ )‖𝑒(𝑠)‖𝐻4ξ€œπ‘‘π‘ π‘‘πœ=𝑑0ξ‚Έξ€œπ‘‘π‘ ξ‚Ή(π‘Ž(π‘‘βˆ’πœ)π‘Ž(πœβˆ’π‘ )π‘‘πœβ€–π‘’π‘ )‖𝐻4𝑑𝑠,(3.31) then ξ€œπ‘‘π‘ ξ€œπ‘Ž(π‘‘βˆ’πœ)π‘Ž(πœβˆ’π‘ )π‘‘πœ=25𝑑𝑠(π‘‘βˆ’πœ)βˆ’2/3(πœβˆ’π‘ )βˆ’2/3π‘‘πœ=25𝐢3(π‘‘βˆ’π‘ )βˆ’1/3,0<𝑠≀𝑑≀1.(3.32) Therefore (3.28)–(3.32) imply ‖𝑒(𝑑)‖𝐻4β‰€ξ€Ίπ‘Ž(𝑑)+25𝐢3π‘€π‘‘βˆ’1/3‖‖𝑒0‖‖𝐿2+25𝐢3𝑀2ξ€œπ‘‘0(π‘‘βˆ’π‘ )βˆ’1/3‖𝑒(𝑠)‖𝐻4𝑑𝑠,0<𝑑≀1.(3.33) From (3.17), we know ‖𝑒(𝑑)‖𝐻4≀𝐢2‖𝑒0‖𝐻4,0<𝑑≀1. Then ‖𝑒(𝑑)‖𝐻4β‰€ξ€Ίπ‘Ž(𝑑)+25𝐢3π‘€π‘‘βˆ’1/3‖‖𝑒0‖‖𝐿2+752𝐢3𝐢2𝑀2‖‖𝑒0‖‖𝐻4𝑑2/3,0<𝑑≀1.(3.34) Therefore, there exists a π‘‘βˆ—, such that ‖‖𝑒(𝑑)𝐻4≀𝐢1π‘‘βˆ’2/3‖‖𝑒0‖‖𝐿2,0<π‘‘β‰€π‘‘βˆ—β‰€1,𝐢1>0.(3.35) So, we proved the inequality (3.16).
Hence (3.17) is proven and proof of Lemma 3.4 is finished.

By Lemma 3.4, the condition (ii) is proved.

We now proceed to verify condition (iii) of Lemma 3.1. Observing that if 𝑒(π‘₯,𝑑) satisfies πœ•π‘’(π‘₯,𝑑)=πœ•πœ•π‘‘6π‘’πœ•π‘₯6+πœ•4π‘’πœ•π‘₯4βˆ’3πœ‘2πœ•4π‘’πœ•π‘₯4πœ•βˆ’24πœ‘πœ‘β€²3π‘’πœ•π‘₯3βˆ’ξ‚€36πœ‘πœ‘ξ…žξ…ž+36πœ‘ξ…ž2ξ‚πœ•2π‘’πœ•π‘₯2βˆ’ξ€·24πœ‘πœ‘ξ…žξ…žξ…ž+72πœ‘ξ…žπœ‘ξ…žξ…žξ€Έπœ•π‘’βˆ’ξ‚€πœ•π‘₯18πœ‘2ξ…žξ…ž+24πœ‘ξ…žπœ‘ξ…žξ…žξ…ž+6πœ‘πœ‘(4)+2π‘Žπœ‘π‘’,(3.36) then 𝑒(π‘₯,𝑠+𝑑) also satisfies the above equation. By uniqueness of solution, we know that 𝐿 generates a strongly continuous semigroup on the Banach space 𝐻4(𝑅)(see [15] p.344). By Fourier transformation, the essential spectrum of 𝐿0 on 𝐻4(𝑅) is πœŽξ€·πΏ0ξ€ΈβŠƒξ€½βˆ’πœ‰6+πœ‰4ξ€Ύ.βˆ£πœ‰βˆˆπ‘…(3.37) The curve πœ†=βˆ’πœ‰6+πœ‰4 meets the vertical lines π‘…π‘’πœ†=𝛼 for βˆ’βˆž<𝛼≀4/27 because βˆ’βˆž<βˆ’πœ‰6+πœ‰4≀4/27.

We now prove that the same curve belongs to the essential spectrum of 𝐿.

Lemma 3.5. The essential spectrum of 𝐿 on 𝐻4(𝑅) contains that of 𝐿0.

Proof. Let πœ‰βˆˆπ‘… and let πœ†=𝑃(πœ‰)=βˆ’πœ‰6+πœ‰4. Following Schechter [16], πœ†βˆˆπœŽ(𝐿) if there exists a sequence {πœ‰π‘›}βŠ‚π»4(𝑅) with β€–β€–πœ‰π‘›β€–β€–π»4β€–β€–=1,(πΏβˆ’πœ†)πœ‰π‘›β€–β€–π»4⟢0,(3.38) and {πœ‰π‘›} does not have a strongly convergent subsequence in 𝐻4(𝑅). Here we use the definition πœ†βˆ‰πœŽ(𝐿) if and only if πΏβˆ’πœ† is Fredholm with index zero. Now let πœ‰0β‰’0 be a 𝐢∞ function with compact support in (0,∞). Define πœ‰π‘›π‘(π‘₯)=π‘›π‘’π‘–πœ‰π‘₯πœ‰0(π‘₯/𝑛)βˆšπ‘›,𝑛=1,2,…,(3.39) where 𝑐𝑛 is chosen so that β€–πœ‰π‘›β€–π»4=1. In fact, β€–β€–πœ‰π‘›β€–β€–πΏ2=π‘π‘›β€–β€–πœ‰0‖‖𝐿2β€–β€–πœ‰,1=𝑛‖‖𝐻4β‰€π‘˜π‘π‘›,(3.40) for some positive constant π‘˜. Hence 𝑐𝑛β‰₯1/π‘˜>0. Since β€–πœ‰π‘›β€–πΏβˆžβ†’0 but β€–πœ‰π‘›β€–πΏ2 is bounded away from zero, {πœ‰π‘›} can have no convergent subsequence in 𝐿2(𝑅).
It remains to show that β€–(πΏβˆ’πœ†)πœ‰π‘›β€–π»4β†’0. We write πΏβˆ’πœ†=𝐿0ξ‚ƒβˆ’πœ†βˆ’3πœ‘2πœ•4π‘₯+24πœ‘πœ‘β€²πœ•3π‘₯+ξ‚€36πœ‘πœ‘ξ…žξ…ž+36πœ‘ξ…ž2ξ‚πœ•2π‘₯+ξ€·24πœ‘πœ‘ξ…žξ…žξ…ž+72πœ‘ξ…žπœ‘ξ…žξ…žξ€Έπœ•π‘₯+ξ‚€18πœ‘2ξ…žξ…ž+24πœ‘ξ…žπœ‘ξ…žξ…žξ…ž+6πœ‘πœ‘(4).+2π‘Žπœ‘ξ‚ξ‚„(3.41) A simple calculation shows that 𝐿0ξ€Έπœ‰βˆ’πœ†π‘›(π‘₯)=π‘π‘›π‘’π‘–πœ‰π‘₯1≀𝑠≀6(βˆ’π‘–)𝑠𝑃(𝑠)(πœ‰)πœ‰0(𝑠)(π‘₯/𝑛)𝑠!𝑛(1/2)+𝑠,πœ•ξ€·πΏ0ξ€Έπœ‰βˆ’πœ†π‘›ξ€·πΏ(π‘₯)=π‘–πœ‰0ξ€Έπœ‰βˆ’πœ†π‘›(π‘₯)+π‘π‘›π‘’π‘–πœ‰π‘₯1≀𝑠≀6(βˆ’π‘–)𝑠𝑃(𝑠)(πœ‰)πœ‰0(𝑠+1)(π‘₯/𝑛)𝑠!𝑛(3/2)+𝑠,πœ•2𝐿0ξ€Έπœ‰βˆ’πœ†π‘›(π‘₯)=βˆ’πœ‰2𝐿0ξ€Έπœ‰βˆ’πœ†π‘›(π‘₯)+2π‘–πœ‰π‘π‘›π‘’π‘–πœ‰π‘₯1≀𝑠≀6(βˆ’π‘–)𝑠𝑃(𝑠)(πœ‰)πœ‰0(𝑠+1)(π‘₯/𝑛)𝑠!𝑛(3/2)+𝑠+π‘π‘›π‘’π‘–πœ‰π‘₯1≀𝑠≀6(βˆ’π‘–)𝑠𝑃(𝑠)(πœ‰)πœ‰0(𝑠+2)(π‘₯/𝑛)𝑠!𝑛(5/2)+𝑠,πœ•3𝐿0ξ€Έπœ‰βˆ’πœ†π‘›(π‘₯)=βˆ’π‘–πœ‰3𝐿0ξ€Έπœ‰βˆ’πœ†π‘›(π‘₯)βˆ’3πœ‰2π‘π‘›π‘’π‘–πœ‰π‘₯1≀𝑠≀6(βˆ’π‘–)𝑠𝑃(𝑠)(πœ‰)πœ‰0(𝑠+1)(π‘₯/𝑛)𝑠!𝑛(3/2)+𝑠+3π‘–πœ‰π‘π‘›π‘’π‘–πœ‰π‘₯1≀𝑠≀6(βˆ’π‘–)𝑠𝑃(𝑠)(πœ‰)πœ‰0(𝑠+2)(π‘₯/𝑛)𝑠!𝑛(5/2)+𝑠+π‘π‘›π‘’π‘–πœ‰π‘₯1≀𝑠≀6(βˆ’π‘–)𝑠𝑃(𝑠)(πœ‰)πœ‰0(𝑠+3)(π‘₯/𝑛)𝑠!𝑛(7/2)+𝑠,πœ•4𝐿0ξ€Έπœ‰βˆ’πœ†π‘›(π‘₯)=πœ‰4𝐿0ξ€Έπœ‰βˆ’πœ†π‘›(π‘₯)βˆ’4π‘–πœ‰3π‘π‘›π‘’π‘–πœ‰π‘₯1≀𝑠≀6(βˆ’π‘–)𝑠𝑃(𝑠)(πœ‰)πœ‰0(𝑠+1)(π‘₯/𝑛)𝑠!𝑛(3/2)+π‘ βˆ’6πœ‰2π‘π‘›π‘’π‘–πœ‰π‘₯1≀𝑠≀6(βˆ’π‘–)𝑠𝑃(𝑠)(πœ‰)πœ‰0(𝑠+2)(π‘₯/𝑛)𝑠!𝑛(5/2)+𝑠+4π‘–πœ‰π‘π‘›π‘’π‘–πœ‰π‘₯1≀𝑠≀6(βˆ’π‘–)𝑠𝑃(𝑠)(πœ‰)πœ‰0(𝑠+3)(π‘₯/𝑛)𝑠!𝑛(7/2)+𝑠+π‘π‘›π‘’π‘–πœ‰π‘₯1≀𝑠≀6(βˆ’π‘–)𝑠𝑃(𝑠)(πœ‰)πœ‰0(𝑠+4)(π‘₯/𝑛)𝑠!𝑛(9/2)+𝑠.(3.42) Thus, ‖‖𝐿0ξ€Έπœ‰βˆ’πœ†π‘›β€–β€–(π‘₯)𝐻4≀||πœ‰||+||πœ‰||1+2+||πœ‰||3+||πœ‰||41≀𝑠≀6||𝑃(𝑠)||𝑐(πœ‰)π‘›β€–β€–πœ‰0(𝑠)β€–β€–(π‘₯/𝑛)𝐿2𝑠!𝑛(1/2)+𝑠+ξ‚€||πœ‰||||πœ‰||1+2+32||πœ‰||+431≀𝑠≀6||𝑃(𝑠)(||π‘πœ‰)π‘›β€–β€–πœ‰0(𝑠+1)(β€–β€–π‘₯/𝑛)𝐿2𝑠!𝑛(3/2)+𝑠+ξ‚€||πœ‰||||πœ‰||1+3+621≀𝑠≀6||𝑃(𝑠)||𝑐(πœ‰)π‘›β€–β€–πœ‰0(𝑠+2)β€–β€–(π‘₯/𝑛)𝐿2𝑠!𝑛(5/2)+𝑠+ξ€·||πœ‰||1+41≀𝑠≀6||𝑃(𝑠)||𝑐(πœ‰)π‘›β€–β€–πœ‰0(𝑠+3)β€–β€–(π‘₯/𝑛)𝐿2𝑠!𝑛(7/2)+𝑠+1≀𝑠≀6||𝑃(𝑠)||𝑐(πœ‰)π‘›β€–β€–πœ‰0(𝑠+4)β€–β€–(π‘₯/𝑛)𝐿2𝑠!𝑛(9/2)+π‘ βŸΆ0,asπ‘›βŸΆβˆž.(3.43) Moreover, for any positive integer π‘š, β€–πœ•π‘šπ‘₯πœ‰π‘›β€–πΏβˆžβ†’0, as π‘›β†’βˆž, we have β€–β€–3πœ‘2πœ•4π‘₯πœ‰π‘›β€–β€–2𝐿2β‰€β€–β€–πœ•4π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–3πœ‘2β€–β€–2𝐿2β€–β€–πœ•βŸΆ0,π‘₯ξ€Ί3πœ‘2πœ•4π‘₯πœ‰π‘›ξ€»β€–β€–2𝐿2β‰€β€–β€–πœ•5π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–3πœ‘2β€–β€–2𝐿2+β€–β€–πœ•4π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–6πœ‘πœ‘β€²β€–2𝐿2β€–β€–πœ•βŸΆ0,2π‘₯ξ€Ί3πœ‘2πœ•4π‘₯πœ‰π‘›ξ€»β€–β€–2𝐿2β‰€β€–β€–πœ•6π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–3πœ‘2β€–β€–2𝐿2β€–β€–πœ•+25π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–6πœ‘πœ‘β€²β€–2𝐿2+β€–β€–πœ•4π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–6πœ‘β€²2+6πœ‘πœ‘ξ…žξ…žβ€–β€–2𝐿2β€–β€–πœ•βŸΆ0,3π‘₯ξ€Ί3πœ‘2πœ•4π‘₯πœ‰π‘›ξ€»β€–β€–2𝐿2β‰€β€–β€–πœ•7π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–3πœ‘2β€–β€–2𝐿2β€–β€–πœ•+36π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–6πœ‘πœ‘β€²β€–2𝐿2β€–β€–πœ•+35π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–6πœ‘β€²2+6πœ‘πœ‘ξ…žξ…žβ€–β€–2𝐿2+β€–β€–πœ•4π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–18πœ‘β€²πœ‘ξ…žξ…ž+6πœ‘πœ‘ξ…žξ…žξ…žβ€–β€–2𝐿2β€–β€–πœ•βŸΆ0,4π‘₯ξ€Ί3πœ‘2πœ•4π‘₯πœ‰π‘›ξ€»β€–β€–2𝐿2β‰€β€–β€–πœ•8π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–3πœ‘2β€–β€–2𝐿2β€–β€–πœ•+45π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–18πœ‘β€²πœ‘ξ…žξ…ž+6πœ‘πœ‘ξ…žξ…žξ…žβ€–β€–2𝐿2β€–β€–πœ•+47π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–6πœ‘πœ‘β€²β€–2𝐿2β€–β€–πœ•+66π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–6πœ‘β€²2+6πœ‘πœ‘ξ…žξ…žβ€–β€–2𝐿2+β€–β€–πœ•4π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–18πœ‘ξ…žξ…ž2+24πœ‘β€²πœ‘ξ…žξ…žξ…ž+6πœ‘πœ‘(4)β€–β€–2𝐿2⟢0.(3.44) From the assumptions on πœ‘, we obtain β€–β€–24πœ‘πœ‘β€²πœ•3π‘₯πœ‰π‘›β€–β€–2𝐿2β‰€β€–β€–πœ•3π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–24πœ‘πœ‘β€²β€–2𝐿2β€–β€–πœ•βŸΆ0,π‘₯ξ€Ί24πœ‘πœ‘ξ…žπœ•3π‘₯πœ‰π‘›ξ€»β€–β€–2𝐿2β‰€β€–β€–πœ•4π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–24πœ‘πœ‘β€²β€–2𝐿2+β€–β€–πœ•3π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–24πœ‘β€²2+24πœ‘πœ‘ξ…žξ…žβ€–β€–2𝐿2β€–β€–πœ•βŸΆ0,2π‘₯ξ€Ί24πœ‘πœ‘ξ…žπœ•3π‘₯πœ‰π‘›ξ€»β€–β€–2𝐿2β‰€β€–β€–πœ•5π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–24πœ‘πœ‘β€²β€–2𝐿2β€–β€–πœ•+24π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–24πœ‘β€²2+24πœ‘πœ‘ξ…žξ…žβ€–β€–2𝐿2+β€–β€–πœ•3π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–72πœ‘β€²πœ‘ξ…žξ…ž+24πœ‘πœ‘ξ…žξ…žξ…žβ€–β€–2𝐿2β€–β€–πœ•βŸΆ0,3π‘₯ξ€Ί24πœ‘πœ‘ξ…žπœ•3π‘₯πœ‰π‘›ξ€»β€–β€–2𝐿2β‰€β€–β€–πœ•6π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–24πœ‘πœ‘β€²β€–2𝐿2β€–β€–πœ•+35π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–24πœ‘β€²2+24πœ‘πœ‘ξ…žξ…žβ€–β€–2𝐿2β€–β€–πœ•+34π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–72πœ‘β€²πœ‘ξ…žξ…ž+24πœ‘πœ‘ξ…žξ…žξ…žβ€–β€–2𝐿2+β€–β€–πœ•3π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–72πœ‘2ξ…žξ…ž+96πœ‘β€²πœ‘ξ…žξ…žξ…ž+24πœ‘πœ‘(4)β€–β€–2𝐿2β€–β€–πœ•βŸΆ0,4π‘₯ξ€Ί24πœ‘πœ‘ξ…žπœ•3π‘₯πœ‰π‘›ξ€»β€–β€–2𝐿2β‰€β€–β€–πœ•7π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–24πœ‘πœ‘β€²β€–2𝐿2β€–β€–πœ•+46π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–24πœ‘β€²2+24πœ‘πœ‘ξ…žξ…žβ€–β€–2𝐿2β€–β€–πœ•+65π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–72πœ‘β€²πœ‘ξ…žξ…ž+24πœ‘πœ‘ξ…žξ…žξ…žβ€–β€–2𝐿2β€–β€–πœ•+44π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–72πœ‘2ξ…žξ…ž+96πœ‘β€²πœ‘ξ…žξ…žξ…ž+24πœ‘πœ‘(4)β€–β€–2𝐿2+β€–β€–πœ•3π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–240πœ‘ξ…žξ…žπœ‘ξ…žξ…žξ…ž+120πœ‘β€²πœ‘(4)+24πœ‘πœ‘(5)β€–β€–2𝐿2⟢0.(3.45) Similarly, we have β€–β€–ξ‚€36πœ‘πœ‘ξ…žξ…ž+36πœ‘ξ…ž2ξ‚πœ•2π‘₯πœ‰π‘›β€–β€–2𝐿2β‰€β€–β€–πœ•2π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–36πœ‘πœ‘ξ…žξ…ž+36πœ‘β€²2β€–β€–2𝐿2β€–β€–πœ•βŸΆ0,π‘₯36πœ‘πœ‘ξ…žξ…ž+36πœ‘ξ…ž2ξ‚πœ•2π‘₯πœ‰π‘›ξ‚„β€–β€–2𝐿2β‰€β€–β€–πœ•3π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–36πœ‘πœ‘ξ…žξ…ž+36πœ‘β€²2β€–β€–2𝐿2+β€–β€–πœ•2π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–108πœ‘β€²πœ‘ξ…žξ…ž+36πœ‘πœ‘ξ…žξ…žξ…žβ€–β€–2𝐿2β€–β€–πœ•βŸΆ0,2π‘₯36πœ‘πœ‘ξ…žξ…ž+36πœ‘ξ…ž2ξ‚πœ•2π‘₯πœ‰π‘›ξ‚„β€–β€–2𝐿2β‰€β€–β€–πœ•4π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–36πœ‘πœ‘ξ…žξ…ž+36πœ‘β€²2β€–β€–2𝐿2β€–β€–πœ•+23π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–108πœ‘β€²πœ‘ξ…žξ…ž+36πœ‘πœ‘ξ…žξ…žξ…žβ€–β€–2𝐿2+β€–β€–πœ•2π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–108πœ‘2ξ…žξ…ž+144πœ‘β€²πœ‘ξ…žξ…žξ…ž+36πœ‘πœ‘(4)β€–β€–2𝐿2β€–β€–πœ•βŸΆ0,3π‘₯36πœ‘πœ‘ξ…žξ…ž+36πœ‘ξ…ž2ξ‚πœ•2π‘₯πœ‰π‘›ξ‚„β€–β€–2𝐿2β‰€β€–β€–πœ•5π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–36πœ‘πœ‘ξ…žξ…ž+36πœ‘β€²2β€–β€–2𝐿2β€–β€–πœ•+34π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–108πœ‘β€²πœ‘ξ…žξ…ž+36πœ‘πœ‘ξ…žξ…žξ…žβ€–β€–2𝐿2β€–β€–πœ•+33π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–108πœ‘2ξ…žξ…ž+144πœ‘β€²πœ‘ξ…žξ…žξ…ž+36πœ‘πœ‘(4)β€–β€–2𝐿2+β€–β€–πœ•2π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–360πœ‘ξ…žξ…žπœ‘ξ…žξ…žξ…ž+180πœ‘β€²πœ‘(4)+36πœ‘πœ‘(5)β€–β€–2𝐿2β€–β€–πœ•βŸΆ0,4π‘₯36πœ‘πœ‘ξ…žξ…ž+36πœ‘ξ…ž2ξ‚πœ•2π‘₯πœ‰π‘›ξ‚„β€–β€–2𝐿2β‰€β€–β€–πœ•6π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–36πœ‘πœ‘ξ…žξ…ž+36πœ‘β€²2β€–β€–2𝐿2β€–β€–πœ•+45π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–108πœ‘β€²πœ‘ξ…žξ…ž+36πœ‘πœ‘ξ…žξ…žξ…žβ€–β€–2𝐿2β€–β€–πœ•+64π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–108πœ‘2ξ…žξ…ž+144πœ‘β€²πœ‘ξ…žξ…žξ…ž+36πœ‘πœ‘(4)β€–β€–2𝐿2β€–β€–πœ•+43π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–360πœ‘ξ…žξ…žπœ‘ξ…žξ…žξ…ž+180πœ‘β€²πœ‘(4)+36πœ‘πœ‘(5)β€–β€–2𝐿2+β€–β€–πœ•2π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–360πœ‘2ξ…žξ…žξ…ž+540πœ‘ξ…žξ…žπœ‘(4)+216πœ‘β€²πœ‘(5)+36πœ‘πœ‘(6)β€–β€–2L2β€–β€–ξ€·βŸΆ0,24πœ‘πœ‘ξ…žξ…žξ…ž+72πœ‘ξ…žπœ‘ξ…žξ…žξ€Έπœ•π‘₯πœ‰π‘›β€–β€–2𝐿2β‰€β€–β€–πœ•π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–24πœ‘πœ‘ξ…žξ…žξ…ž+72πœ‘β€²πœ‘ξ…žξ…žβ€–β€–2𝐿2β€–β€–πœ•βŸΆ0,π‘₯ξ€Ίξ€·24πœ‘πœ‘ξ…žξ…žξ…ž+72πœ‘ξ…žπœ‘ξ…žξ…žξ€Έπœ•π‘₯πœ‰π‘›ξ€»β€–β€–2𝐿2β‰€β€–β€–πœ•2π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–24πœ‘πœ‘ξ…žξ…žξ…ž+72πœ‘β€²πœ‘ξ…žξ…žβ€–β€–2𝐿2+β€–β€–πœ•π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–96πœ‘β€²πœ‘ξ…žξ…žξ…ž+72πœ‘2ξ…žξ…ž+24πœ‘πœ‘(4)β€–β€–2𝐿2β€–β€–πœ•βŸΆ0,2π‘₯ξ€Ίξ€·24πœ‘πœ‘ξ…žξ…žξ…ž+72πœ‘ξ…žπœ‘ξ…žξ…žξ€Έπœ•π‘₯πœ‰π‘›ξ€»β€–β€–2𝐿2β‰€β€–β€–πœ•3π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–24πœ‘πœ‘ξ…žξ…žξ…ž+72πœ‘β€²πœ‘ξ…žξ…žβ€–β€–2𝐿2β€–β€–πœ•+22π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–96πœ‘β€²πœ‘ξ…žξ…žξ…ž+72πœ‘2ξ…žξ…ž+24πœ‘πœ‘(4)β€–β€–2𝐿2+β€–β€–πœ•π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–240πœ‘ξ…žξ…žπœ‘ξ…žξ…žξ…ž+120πœ‘β€²πœ‘(4)+24πœ‘πœ‘(5)β€–β€–2𝐿2β€–β€–πœ•βŸΆ0,3π‘₯ξ€Ίξ€·24πœ‘πœ‘ξ…žξ…žξ…ž+72πœ‘ξ…žπœ‘ξ…žξ…žξ€Έπœ•π‘₯πœ‰π‘›ξ€»β€–β€–2𝐿2β‰€β€–β€–πœ•4π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–24πœ‘πœ‘ξ…žξ…žξ…ž+72πœ‘ξ…žπœ‘ξ…žξ…žβ€–β€–2𝐿2β€–β€–πœ•+33π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–96πœ‘β€²πœ‘ξ…žξ…žξ…ž+72πœ‘2ξ…žξ…ž+24πœ‘πœ‘(4)β€–β€–2𝐿2β€–β€–πœ•+32π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–240πœ‘ξ…žξ…žπœ‘ξ…žξ…žξ…ž+120πœ‘β€²πœ‘(4)+24πœ‘πœ‘(5)β€–β€–2𝐿2+β€–β€–πœ•π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–240πœ‘2ξ…žξ…žξ…ž+360πœ‘ξ…žξ…žπœ‘(4)+144πœ‘β€²πœ‘(5)+24πœ‘πœ‘(6)β€–β€–2𝐿2β€–β€–πœ•βŸΆ0,4π‘₯ξ€Ίξ€·24πœ‘πœ‘ξ…žξ…žξ…ž+72πœ‘ξ…žπœ‘ξ…žξ…žξ€Έπœ•π‘₯πœ‰π‘›ξ€»β€–β€–2𝐿2β‰€β€–β€–πœ•5π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–24πœ‘πœ‘ξ…žξ…žξ…ž+72πœ‘β€²πœ‘ξ…žξ…žβ€–β€–2𝐿2β€–β€–πœ•+44π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–96πœ‘β€²πœ‘ξ…žξ…žξ…ž+72πœ‘2ξ…žξ…ž+24πœ‘πœ‘(4)β€–β€–2𝐿2β€–β€–πœ•+63π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–240πœ‘ξ…žξ…žπœ‘ξ…žξ…žξ…ž+120πœ‘β€²πœ‘(4)+24πœ‘πœ‘(5)β€–β€–2𝐿2β€–β€–πœ•+42π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–240πœ‘2ξ…žξ…žξ…ž+360πœ‘ξ…žξ…žπœ‘(4)+144πœ‘β€²πœ‘(5)+24πœ‘πœ‘(6)β€–β€–2𝐿2+β€–β€–πœ•π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–840πœ‘ξ…žξ…žξ…žπœ‘(4)+504πœ‘ξ…žξ…žπœ‘(5)+168πœ‘β€²πœ‘(6)+24πœ‘πœ‘(7)β€–β€–2𝐿2⟢0.(3.46) In addition, β€–β€–ξ‚€18πœ‘2ξ…žξ…ž+24πœ‘ξ…žπœ‘ξ…žξ…žξ…ž+6πœ‘πœ‘(4)ξ‚πœ‰+2π‘Žπœ‘π‘›β€–β€–2𝐿2β‰€β€–β€–πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–18πœ‘2ξ…žξ…ž+24πœ‘β€²πœ‘ξ…žξ…žξ…ž+6πœ‘πœ‘(4)β€–β€–+2π‘Žπœ‘2𝐿2β€–β€–πœ•βŸΆ0,π‘₯18πœ‘2ξ…žξ…ž+24πœ‘ξ…žπœ‘ξ…žξ…žξ…ž+6πœ‘πœ‘(4)ξ‚πœ‰+2π‘Žπœ‘π‘›ξ‚„β€–β€–2𝐿2β‰€β€–β€–πœ•π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–18πœ‘2ξ…žξ…ž+24πœ‘β€²πœ‘ξ…žξ…žξ…ž+6πœ‘πœ‘(4)β€–β€–+2π‘Žπœ‘2𝐿2+β€–β€–πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–60πœ‘ξ…žξ…žπœ‘ξ…žξ…žξ…ž+30πœ‘β€²πœ‘(4)+6πœ‘πœ‘(5)β€–β€–+2π‘Žπœ‘β€²2𝐿2β€–β€–πœ•βŸΆ0,2π‘₯18πœ‘2ξ…žξ…ž+24πœ‘ξ…žπœ‘ξ…žξ…žξ…ž+6πœ‘πœ‘(4)ξ‚πœ‰+2π‘Žπœ‘π‘›ξ‚„β€–β€–2𝐿2β‰€β€–β€–πœ•2π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–18πœ‘2ξ…žξ…ž+24πœ‘β€²πœ‘ξ…žξ…žξ…ž+6πœ‘πœ‘(4)β€–β€–+2π‘Žπœ‘2𝐿2β€–β€–πœ•+2π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–60πœ‘ξ…žξ…žπœ‘ξ…žξ…žξ…ž+30πœ‘β€²πœ‘(4)+6πœ‘πœ‘(5)β€–β€–+2π‘Žπœ‘β€²2𝐿2+β€–β€–πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–60πœ‘2ξ…žξ…žξ…ž+90πœ‘ξ…žξ…žπœ‘(4)+36πœ‘β€²πœ‘(5)+6πœ‘πœ‘(6)+2π‘Žπœ‘ξ…žξ…žβ€–β€–2𝐿2β€–β€–πœ•βŸΆ0,3π‘₯18πœ‘2ξ…žξ…ž+24πœ‘ξ…žπœ‘ξ…žξ…žξ…ž+6πœ‘πœ‘(4)ξ‚πœ‰+2π‘Žπœ‘π‘›ξ‚„β€–β€–2𝐿2β‰€β€–β€–πœ•3π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–18πœ‘2ξ…žξ…ž+24πœ‘β€²πœ‘ξ…žξ…žξ…ž+6πœ‘πœ‘(4)β€–β€–+2π‘Žπœ‘2𝐿2β€–β€–πœ•+32π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–60πœ‘ξ…žξ…žπœ‘ξ…žξ…žξ…ž+30πœ‘β€²πœ‘(4)+6πœ‘πœ‘(5)β€–β€–+2π‘Žπœ‘β€²2𝐿2β€–β€–πœ•+3π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–60πœ‘2ξ…žξ…žξ…ž+90πœ‘ξ…žξ…žπœ‘(4)+36πœ‘β€²πœ‘(5)+6πœ‘πœ‘(6)+2π‘Žπœ‘ξ…žξ…žβ€–β€–2𝐿2+β€–β€–πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–210πœ‘ξ…žξ…žξ…žπœ‘(4)+126πœ‘ξ…žξ…žπœ‘(5)+42πœ‘β€²πœ‘(6)+6πœ‘πœ‘(7)+2π‘Žπœ‘ξ…žξ…žξ…žβ€–β€–2𝐿2β€–β€–πœ•βŸΆ0,4π‘₯18πœ‘2ξ…žξ…ž+24πœ‘ξ…žπœ‘ξ…žξ…žξ…ž+6πœ‘πœ‘(4)ξ‚πœ‰+2π‘Žπœ‘π‘›ξ‚„β€–β€–2𝐿2β‰€β€–β€–πœ•4π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–18πœ‘2ξ…žξ…ž+24πœ‘β€²πœ‘ξ…žξ…žξ…ž+6πœ‘πœ‘(4)β€–β€–+2π‘Žπœ‘2𝐿2β€–β€–πœ•+43π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–60πœ‘ξ…žξ…žπœ‘ξ…žξ…žξ…ž+30πœ‘β€²πœ‘(4)+6πœ‘πœ‘(5)β€–β€–+2π‘Žπœ‘β€²2𝐿2β€–β€–πœ•+62π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–60πœ‘2ξ…žξ…žξ…ž+90πœ‘ξ…žξ…žπœ‘(4)+36πœ‘β€²πœ‘(5)+6πœ‘πœ‘(6)+2π‘Žπœ‘ξ…žξ…žβ€–β€–2𝐿2β€–β€–πœ•+4π‘₯πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–210πœ‘ξ…žξ…žξ…žπœ‘(4)+126πœ‘ξ…žξ…žπœ‘(5)+42πœ‘β€²πœ‘(6)+6πœ‘πœ‘(7)+2π‘Žπœ‘ξ…žξ…žξ…žβ€–β€–2𝐿2+β€–β€–πœ‰π‘›β€–β€–2πΏβˆžβ€–β€–ξ€·πœ‘210(4)ξ€Έ2+12πœ‘ξ…žξ…žξ…žπœ‘(5)+168πœ‘ξ…žξ…žπœ‘(6)+48πœ‘β€²πœ‘(7)+6πœ‘πœ‘(8)+2π‘Žπœ‘(4)β€–β€–2𝐿2⟢0.(3.47) Thus, β€–β€–3πœ‘2πœ•4π‘₯πœ‰π‘›+24πœ‘πœ‘ξ…žπœ•3π‘₯πœ‰π‘›+ξ‚€36πœ‘πœ‘ξ…žξ…ž+36πœ‘ξ…ž2ξ‚πœ•2π‘₯πœ‰π‘›+ξ€·24πœ‘πœ‘ξ…žξ…žξ…ž+72πœ‘ξ…žπœ‘ξ…žξ…žξ€Έπœ•π‘₯πœ‰π‘›+ξ‚€18πœ‘2ξ…žξ…ž+24πœ‘ξ…žπœ‘ξ…žξ…žξ…ž+6πœ‘πœ‘(4)ξ‚πœ‰+2π‘Žπœ‘π‘›β€–β€–π»4⟢0,asπ‘›βŸΆβˆž.(3.48) So from the estimates above, β€–β€–(πΏβˆ’πœ†)πœ‰π‘›β€–β€–π»4⟢0,asπ‘›βŸΆβˆž.(3.49) The proof of Lemma 3.5 is completed.

Therefore all the four conditions of Lemma 3.1 are satisfied by the linearized equation (3.4) and Theorem 1.1 has been proved.

References

  1. T. V. Savina, A. A. Golovin, S. H. Davis, A. A. Nepomnyashchy, and P. W. Voorhees, β€œFaceting of a growing crystal surface by surface difusion,” Physical Review E, vol. 67, no. 2, part 1, Article ID 021606, 2003.
  2. J. W. Barrett, S. Langdon, and R. Nürnberg, β€œFinite element approximation of a sixth order nonlinear degenerate parabolic equation,” Numerische Mathematik, vol. 96, no. 3, pp. 401–434, 2004. View at Publisher Β· View at Google Scholar Β· View at Zentralblatt MATH
  3. J. D. Evans, V. A. Galaktionov, and J. R. King, β€œUnstable sixth-order thin film equation. I. Blow-up similarity solutions,” Nonlinearity, vol. 20, no. 8, pp. 1799–1841, 2007. View at Publisher Β· View at Google Scholar Β· View at Zentralblatt MATH
  4. J. D. Evans, V. A. Galaktionov, and J. R. King, β€œUnstable sixth-order thin film equation. II. Global similarity patterns,” Nonlinearity, vol. 20, no. 8, pp. 1843–1881, 2007. View at Publisher Β· View at Google Scholar Β· View at Zentralblatt MATH
  5. A. Jüngel and J.-P. Milišić, β€œA sixth-order nonlinear parabolic equation for quantum systems,” SIAM Journal on Mathematical Analysis, vol. 41, no. 4, pp. 1472–1490, 2009. View at Publisher Β· View at Google Scholar Β· View at Zentralblatt MATH
  6. C. Liu, β€œQualitative properties for a sixth-order thin film equation,” Mathematical Modelling and Analysis, vol. 15, no. 4, pp. 457–471, 2010. View at Publisher Β· View at Google Scholar Β· View at Zentralblatt MATH
  7. C. Liu, β€œA sixth order degenerate equation with the higher order P-Laplacian operator,” Mathematica Slovaca, vol. 60, no. 6, pp. 847–864, 2010. View at Publisher Β· View at Google Scholar
  8. M. D. Korzec, P. L. Evans, A. Münch, and B. Wagner, β€œStationary solutions of driven fourth- and sixth-order Cahn-Hilliard-type equations,” SIAM Journal on Applied Mathematics, vol. 69, no. 2, pp. 348–374, 2008. View at Publisher Β· View at Google Scholar Β· View at Zentralblatt MATH
  9. E. A. Carlen, M. C. Carvalho, and E. Orlandi, β€œA simple proof of stability of fronts for the Cahn-Hilliard equation,” Communications in Mathematical Physics, vol. 224, no. 1, pp. 323–340, 2001. View at Publisher Β· View at Google Scholar Β· View at Zentralblatt MATH
  10. H. Gao and C. Liu, β€œInstability of traveling waves of the convective-diffusive Cahn-Hilliard equation,” Chaos, Solitons and Fractals, vol. 20, no. 2, pp. 253–258, 2004. View at Publisher Β· View at Google Scholar Β· View at Zentralblatt MATH
  11. S. A. Gourley and N. F. Britton, β€œInstability of travelling wave solutions of a population model with nonlocal effects,” IMA Journal of Applied Mathematics, vol. 51, no. 3, pp. 299–310, 1993. View at Publisher Β· View at Google Scholar Β· View at Zentralblatt MATH
  12. C. Liu, β€œInstability of traveling waves for a generalized diffusion model in population problems,” Electronic Journal of Qualitative Theory of Differential Equations, no. 18, pp. 1–10, 2004. View at Zentralblatt MATH
  13. J. Shatah and W. Strauss, β€œSpectral condition for instability,” in Contemporary Mathematics, vol. 255, pp. 189–198, 2000. View at Publisher Β· View at Google Scholar Β· View at Zentralblatt MATH
  14. W. Strauss and G. Wang, β€œInstability of traveling waves of the Kuramoto-Sivashinsky equation,” Chinese Annals of Mathematics B, vol. 23, no. 2, pp. 267–276, 2002. View at Publisher Β· View at Google Scholar Β· View at Zentralblatt MATH
  15. Q. Ye and Z. Li, Theory of Reaction-Diffusion Equations, Science Press, Beijing, China, 1994.
  16. M. Schechter, Spectra of Partial Differential Operators, American Elsevier, New York, NY, USA, 1971.