Citation: K. Ravindranadh, K. Durga Venkata Prasad, M.C. Rao. Spectroscopic and luminescent properties of Co2+ doped tin oxide thin films by spray pyrolysis[J]. AIMS Materials Science, 2016, 3(3): 796-807. doi: 10.3934/matersci.2016.3.796
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In this paper, we would like to investigate the initial value problem (IVP) associated with the (3+1)-dimensional modified Zakharov-Kuznetsov (mZK) equation
{∂tu+∂3xu+∂x∂2yu+∂x∂2zu+γu2∂xu=0,(x,y,z)∈R3t≥0,u(0)=u0, | (1.1) |
where u=u(x,y,z,t) is a real-valued function, u0=u0(x,y,z), and γ is a nonzero constant. Furthermore, it is proved that its solutions u(x,y,z,t) have the properties of exponential decay above the plane x+y+z=0.
Equation (1.1) was proposed by Zakharov and Kuznetsov [1] as a three-dimensional generalization of the Korteweg-de Vries (KdV) equation, which was derived from the Euler-Poisson system with magnetic field by Lannes et al. in [2]. This equation describes the unidirectional propagation of ionic-acoustic waves in magnetized plasma.
It is easy to see that the ZK equation can be regarded as a multidimensional generalization of the one-dimensional KdV equation
{∂tu+∂3xu+u∂xu=0,x∈R,t≥0,u(0)=u0. |
It is worth mentioning that two dimensional versions of the KdV equation and modified Korteweg-de Vries (mKdV) equation are the ZK equation and the mZK equation, respectively. Up to now, to the best of our knowledge, for the two-dimensional ZK equation, the well-posedness and uniqueness results have been studied extensively. For some related works, refer to [3,4,5,6,7,8] and references therein. At the same time, the Cauchy problem for the 2D mZK equation has also been discussed. The local well-posedness in H1(R2) was obtained by Biagioni and Linares in [9]. The local result was generalized to the data in Hs(R2), s>3/4, by Linares and Pastor in [10]. In terms of a smallness assumption on the L2-norm of the data [11], they proved the global well-posedness in Hs(R2), s>53/63. In [12], Ribaud and Vento investigated the local well-posedness in Hs(R2), s>1/4.
On the other hand, for the three-dimensional ZK equation, many interesting results have been obtained. For initial data in Hs(R3) with s>9/8, Linares and Saut [13] showed the local well-posedness of this initial problem. The local well-posedness theory of the Benjamin-Ono equation was established in [14] by utilizing similar techniques in [13]. For more discussions of the 3D ZK equation, see [15,16,17] and reference therein. Moreover, for the 3D mZK equation, in [18], Grünrock proved the local well-posedness of the Cauchy problem (1.1) for initial data in Hs(R3) with s>1/2. Kinoshita [19] established the well-posedness in the critical space H1/2(R3) for the Cauchy problem of the mZK equation. Ali et al. [20] developed the propagation of dispersive wave solutions for (3+1)-dimensional nonlinear mZK equation in plasma physics. Further analysis results can be also found in [21] and references therein.
The properties of decay preservation are of great interest. In an innovative paper, Isaza and León [22] studied the optimal exponential decay properties of solutions to the KdV equation. Larkin and Tronco [23] derived the decay properties of small solutions for the ZK equation posed on a half-strip. In [24], Larkin further established the exponential decay of the H1-norm for the 2D ZK equation. Recently, the decay properties for solutions of the ZK equation were also obtained in [25].
It is obvious that the decay properties are closely related to the aspect of unique continuation. It is noted that Bustamante et al. [15] derived the unique continuation property of the solutions of the 3D Zakharov-Kuznetsov equation. In recent years, the unique continuation principles of several models arising in nonlinear dispersive equations were investigated, see references [26,27,28,29,30] for example.
The well-posedness for the two dimensional generalized ZK equation in anisotropic weighted Sobolev spaces was discussed in [31]. In [32], Bustamante et al. established the well-posedness of the IVP for the 2D ZK equation in weighted Sobolev spaces Hs(R2)∩L2((1+x2+y2)rdxdy) for s,r∈R. Furthermore, they also showed in [15] that, for some small ε>0,
u1,u2∈C([0,1];H4(R3)∩L2((1+x2+y2+z2)85+εdxdydz))∩C1([0,1];L2(R3)), |
are solutions of the IVP for the three-dimensional ZK equation. Then, there exists a constant a0>0 such that if for some a>a0,
u1(0)−u2(0),u1(1)−u2(1)∈L2(ea(x2+y2+z2)3/4dxdydz), |
then u1≡u2.
The main goal of the present paper is to formally derive the decay properties of exponential type solutions u(x,y,z,t) to the IVP (1.1). In order to achieve this goal, we shall utilize Kato's approach to prove Kato's estimation in three-dimensional form.
Now, we are in the position to state our main results.
Theorem 1.1. Let a0 be a positive constant. For any given data
u0∈H2(R3)∩L2(ea0(x+y+z)3/2+dxdydz), | (1.2) |
the unique solution u(⋅,⋅,⋅) of the IVP (1.1) provided in [18]
u∈C([0,T];H2(R3)) |
satisfies
supt∈[0,T]∫R3ea(t)(x+y+z)3/2+|u(x,y,z,t)|2dxdydz≤c∗, | (1.3) |
where
c∗=c∗(a0;‖u0‖H1(R3);‖ea0(x+y+z)3/2+u0‖L2(R3);T), |
with
a(t)=a0√1+814a20t. |
Let us consider weighted spaces with symmetric weight, which take the form
L2(⟨x+y+z⟩bdxdydz)=L2((1+(x+y+z)2)b2dxdydz). |
Regardless of whether the time direction is forward t>0 or backward t<0, its persistent properties should hold.
Theorem 1.2. Let a0 be a positive constant. Let u1, u2 be solutions of the IVP (1.1) such that
u1∈C([0,T];H3(R3))∩L2(⟨x+y+z⟩2dxdydz),u2∈C([0,T];H3(R3)). |
If
Λ=∫R3ea0(x+y+z)3/2+|u0,1(x,y,z)−u0,2(x,y,z)|2dxdydz<∞, |
then
supt∈[0,T]∫R3ea(t)(x+y+z)3/2+|u1(x,y,z,t)−u2(x,y,z,t)|2dxdydz≤c∗∗, | (1.4) |
where c∗∗=c∗∗(a0;‖u0,1‖H4(R3);‖u0,2‖H4(R3);‖x2u0,1‖L2(R3);‖xu0,2‖L2(R3);Λ;T) and
a(t)=a0√1+814a20t. |
Let us denote the norm of the functional space Xs by
‖f‖Xs=‖Jsxf‖L2(R3)+‖f‖H1yzL2x, |
for s>1/2, where ^Jsx(ξ,η,γ)=(1+ξ2)s/2ˆf(ξ,η,γ).
In terms of the context, the previous result shows that it is necessary to have a similar property in a suitable Sobolev space Hs(R3) for a solution of the IVP (1.1) to satisfy the persistent property in L2(⟨x+y+z⟩bdxdydz).
Theorem 1.3. Let u0∈Xs, s>9/8. There exists T=T(‖u0‖Xs) and a unique solution of the IVP (1.1) such that u∈C([0,T];Xs) provided by Lemma 2.1 below. If there exist α>0 and two different instants of time t0,t1∈[0,T] such that
⟨x+y+z⟩αu(x,y,z,t0),⟨x+y+z⟩αu(x,y,z,t1)∈L2(R3), |
then for any t∈[0,T],
u(t)∈L2(⟨x+y+z⟩αdxdydz), |
(∂xu(t)+∂yu(t)),(∂xu(t)+∂zu(t))∈L2(⟨x+y+z⟩α−1/2dxdydz). |
The rest of this paper is organized as follows. In Section 2, some details on known results of the three-dimensional mZK equation will be introduced. In Section 3, the weights will be constructed to put forward the theory. Section 4 is devoted to proving Theorems 1.1 and 1.2. Finally, in Section 5, we demonstrate Theorem 1.3.
Attention in this section is now turned to prove some preliminary estimates which we often use in our analysis. We first give the following result.
Lemma 2.1. Given u0∈Xs, s>9/8, there exists T=T(‖u0‖Xs) and a unique solution of the IVP (1.1) such that u∈C([0,T];Xs), u,∂xu∈L1TL∞xyz. Moreover, the map u0↦u is continuous from a neighborhood of u0∈Xs into C([0,T];Xs).
The proof is similar to Theorem 3.9 in [13]. Meanwhile, using the assumptions of Theorem 1.3, we deduce that
∫T0‖u‖L∞xyzdt+∫T0‖∂xu‖L∞xyzdt≤cT, | (2.1) |
where cT is a constant.
Lemma 2.2. Let u∈C([0,T];H2(R3)) be a solution of the IVP (1.1), corresponding to data u0∈H2(R3)∩L2(eβ(x+y+z)dxdydz), β>0. Then,
eβ(x+y+z)u∈C([0,T];L2(R3)) |
and
‖eβ(x+y+z)u(t)‖L2(R3)≤c‖eβ(x+y+z)u0‖L2(R3),t∈[0,T]. |
Proof. Applying Kato's approach in [33], let us now prove this lemma. First, we consider the equation
∂tu+∂3xu+∂x∂2yu+∂x∂2zu+γu2∂xu=0,(x,y,z)∈R3,t≥0. | (2.2) |
Next, multiplying by uφδ on both sides of Eq (2.2) and integrating by parts, a direct computation gives rise to
ddt∫R3u2φδdxdydz+∫R3(∂xu+∂yu)2∂yφδdxdydz+∫R3(∂xu+∂zu)2∂zφδdxdydz+∫R3(∂xu)2(3∂xφδ−∂yφδ−∂zφδ)dxdydz+∫R3(∂yu)2(∂xφδ−∂yφδ)dxdydz+∫R3(∂zu)2(∂xφδ−∂zφδ)dxdydz=∫R3u2(∂3xφδ+∂xyyφδ+∂xzzφδ)dxdydz+γ2∫R3u4∂xφδdxdydz. |
For β>0, we define
φδ(x,y,z)=eβ(x+y+z)1+δeβ(x+y+z)forδ∈(0,1),δ≪1. |
Thus, we see that
φδ∈L∞(R3)and‖φδ‖L∞(R3)=1δ. | (2.3) |
0≤∂xφδ(x,y,z)=∂yφδ(x,y,z)=∂zφδ(x,y,z)=βeβ(x+y+z)(1+δeβ(x+y+z))2≤βφδ(x,y,z), |
∂2xφδ(x,y,z)=∂xyφδ(x,y,z)=∂xzφδ(x,y,z)=β2eβ(x+y+z)(1−δeβ(x+y+z))(1+δeβ(x+y+z))3, |
and then
|∂2xφδ(x,y,z)|=|∂xyφδ(x,y,z)|=|∂xzφδ(x,y,z)|≤β2eβ(x+y+z)(1+δeβ(x+y+z))2. |
∂3xφδ(x,y,z)=∂xyyφδ(x,y,z)=∂xzzφδ(x,y,z)=β3eβ(x+y+z)(1−4δeβ(x+y+z)+δ2e2β(x+y+z))(1+δeβ(x+y+z))4, |
hence
|∂3xφδ(x,y,z)|=|∂xyyφδ(x,y,z)|=|∂xzzφδ(x,y,z)|≤2β3eβ(x+y+z)(1+δeβ(x+y+z))2. |
Therefore,
∂3xφδ(x,y,z)+∂xyyφδ(x,y,z)+∂xzzφδ(x,y,z)≤c0β3φδ(x,y,z). | (2.4) |
Moreover,
φδ(x,y,z)≤φδ′(x,y,z),(x,y,z)∈R3if0<δ′<δ, |
and
limδ↓0φδ(x,y,z)=eβ(x+y+z). |
We apply properties (2.3) and (2.4) to obtain the estimate
ddt∫R3u2φδ(x,y,z)dxdydz≤c0β3∫R3u2φδ(x,y,z)dxdydz+γ2∫R3u4∂xφδ(x,y,z)dxdydz. | (2.5) |
In the case of u∈C([0,T];H2(R3)), there exists a positive constant c such that
‖u‖L∞(R3)≤c. |
Next, we consider the last term of (2.5). We write
∫R3u4∂xφδ(x,y,z)dxdydz≤β‖u‖2L∞(R3)∫R3u2φδ(x,y,z)dxdydz≤c∫R3u2φδ(x,y,z)dxdydz. |
Inserting this estimate into (2.5), one has
ddt∫R3u2φδ(x,y,z)dxdydz≤c∫R3u2φδ(x,y,z)dxdydz. | (2.6) |
Using Gronwall's lemma and integrating (2.6) in t∈[0,T], we deduce that
supt∈[0,T]∫R3u2(x,y,z,t)φδ(x,y,z)dxdydz≤c∫R3u20(x,y,z)φδ(x,y,z)dxdydz≤c∫R3u20(x,y,z)φ0(x,y,z)dxdydz, |
where c is a constant.
Letting δ↓0, this completes the proof of Lemma 2.2.
We multiply uϕN on both sides of (2.2). Then, for a fixed t∈[0,T], integrating over R3 with x, y, and z, and making use of integration by parts yields
ddt∫R3u2ϕNdxdydz+∫R3(∂xu+∂yu)2∂yϕNdxdydz+∫R3(∂xu+∂zu)2∂zϕNdxdydz+∫R3(∂xu)2(3∂xϕN−∂yϕN−∂zϕN)dxdydz+∫R3(∂yu)2(∂xϕN−∂yϕN)dxdydz+∫R3(∂zu)2(∂xϕN−∂zϕN)dxdydz=∫R3u2(∂3xϕN+∂xyyϕN+∂xzzϕN+∂tϕN)dxdydz+γ2∫R3u4∂xϕNdxdydz. | (3.1) |
A sequence of the weights {ϕN}∞N=1 will be constructed, which plays an important role in the proof of our main theorems.
Theorem 3.1. Given a0>0, there exists a sequence {ϕN}∞N=1 of functions with
ϕN:R3×[0,∞)→R |
satisfying for any N∈Z+:
(i) ϕN∈C2(R3×[0,∞)) with ∂3xϕN(⋅,⋅,⋅,t), ∂xyyϕN(⋅,⋅,⋅,t), ∂xzzϕN(⋅,⋅,⋅,t) having a jump discontinuity at x+y+z=N.
(ii) ϕN(x,y,z,t)>0 for all (x,y,z,t)∈R3×[0,∞).
(iii) ∂xϕN(x,y,z,t)=∂yϕN(x,y,z,t)=∂zϕN(x,y,z,t)>0 for all (x,y,z,t)∈R3×[0,∞).
(iv) There exist constants cN=c(N)>0 and c0=c0(a0)>0 such that
ϕN(x,y,z,t)≤cNc0⟨(x+y+z)+⟩2, |
with
(x+y+z)+=max{0;x+y+z},⟨x+y+z⟩=(1+(x+y+z)2)1/2. |
(v) For T>0, there is N0∈Z+ such that
ϕN(x,y,z,0)≤ea0(x+y+z)3/2+ifN>N0. |
Also,
limN↑∞ϕN(x,y,z,t)=ea(t)(x+y+z)3/2+, |
for any t>0 and x+y+z∈(−∞,0)∩(1,∞), where
a(t)=a0√1+814a20t. |
(vi) There exists a constant c0=c0(a0)>0 such that
∂tϕN+∂3xϕN+∂xyyϕN+∂xzzϕN≤c0ϕN, | (3.2) |
for any (x,y,z,t)∈R3×[0,∞).
(vii) There exists a constant c1=c1(a0)>0 such that
|∂xϕN(x,y,z,t)|≤c1⟨x+y+z⟩1/2ϕN(x,y,z,t), | (3.3) |
for any (x,y,z,t)∈R3×[0,∞).
Proof. For N∈Z+, given a0>0, let us first define
ϕN(x,y,z,t)={ea(t)φ(x,y,z),−∞<x+y+z≤1,ea(t)(x+y+z)3/2,1≤x+y+z≤N,PN(x,y,z,t),x+y+z≥N, |
where
a(t)=a0√1+814a20t∈(0,a0],t≥0, | (3.4) |
a0 being the initial parameter.
φ(x,y,z)=(1−η(x+y+z))(x+y+z)3++η(x+y+z)(x+y+z)3/2,(x+y+z)+=max{0;x+y+z}, |
for x+y+z∈(−∞,1] where η∈C∞(R3), ηx=ηy=ηz≥0, and
η(x+y+z)={0,x+y+z≤1/2,1,x+y+z≥3/4, | (3.5) |
i.e., for each x+y+z∈[0,1], φ(x,y,z) is a convex combination of (x+y+z)3 and (x+y+z)3/2. PN(x,y,z,t) is a polynomial of order 2 in (x+y+z), which matches the value of ea(t)(x+y+z)3/2 and its partial derivatives up to order 2 at x+y+z=N:
PN(x,y,z,t)=[1+32aN1/2(x+y+z−N)+(94a2N+34aN−1/2)(x+y+z−N)22]eaN3/2, |
with a=a(t) as in (3.4).
Thus, to prove Theorem 3.1, let us consider the regions x+y+z∈(−∞,0], [0,1], [1,N], and [N,∞), respectively.
In the first region x+y+z≤0, we get
ϕN(x,y,z,t)=ea(t)⋅0=1, |
which clearly satisfies Theorem 3.1.
In the region x+y+z∈[0,1], we deduce that
ϕN(x,y,z,t)=ea(t)φ(x,y,z), |
with
φ(x,y,z)=(1−η(x+y+z))(x+y+z)3+η(x+y+z)(x+y+z)3/2≥0,x+y+z∈[0,1], |
with η as in (3.5). Since in this region (x+y+z)3/2≥(x+y+z)3, it follows that
∂xφ(x,y,z)=(1−η(x+y+z))3(x+y+z)2+η(x+y+z)32(x+y+z)1/2+∂xη(x+y+z)((x+y+z)3/2−(x+y+z)3)≥(1−η(x+y+z))3(x+y+z)2+η(x+y+z)32(x+y+z)1/2≥0, |
likewise
∂yφ(x,y,z)≥0,∂zφ(x,y,z)≥0, |
with
∂xφ(x,y,z)=∂yφ(x,y,z)=∂zφ(x,y,z), | (3.6) |
and there exists c>0 such that
∂xφ(x,y,z),∂xxφ(x,y,z),∂xyφ(x,y,z),∂xzφ(x,y,z)≤c,∂xxxφ(x,y,z),∂xyyφ(x,y,z),∂xzzφ(x,y,z)≤c,x+y+z∈[0,1]. |
Note that for a′(t)≤0, it is found that
a(t)≤a0fort≥0, | (3.7) |
and it is deduced that
∂tϕN(x,y,z,t)=a′(t)φ(x,y,z)ϕN(x,y,z,t)≤0. |
Next, let us prove that there exists c0=c0(a0)>0 in this region such that
∂3xϕN+∂xyyϕN+∂xzzϕN≤c0ϕN. |
Since
∂xϕN=aφxϕN,∂yϕN=aφyϕN,∂zϕN=aφzϕN,∂2xϕN=(aφxx+(aφx)2)ϕN,∂xyϕN=(aφxy+a2φxφy)ϕN,∂xzϕN=(aφxz+a2φxφz)ϕN,∂3xϕN=(aφxxx+3a2φxxφx+(aφx)3)ϕN,∂xyyϕN=(aφxyy+2a2φxyφy+a2φxφyy+a3φxφ2y)ϕN,∂xzzϕN=(aφxzz+2a2φxzφz+a2φxφzz+a3φxφ2z)ϕN. |
From (3.6), we derive that
∂xϕN=∂yϕN=∂zϕN, |
thus, for x+y+z∼0 (x+y+z≥0),
∂xϕN=∂yϕN=∂zϕN∼3a(x+y+z)2ϕN,∂2xϕN=∂xyϕN=∂xzϕN∼(6a(x+y+z)+9a2(x+y+z)4)ϕN,∂3xϕN=∂xyyϕN=∂xzzϕN∼(6a+54a2(x+y+z)3+27a3(x+y+z)6)ϕN. |
Hence, for x+y+z∼0 (0≤x+y+z≤1),
∂3xϕN(x,y,z,t)≤c(a+a3)ϕN,∂xyyϕN(x,y,z,t)≤c(a+a3)ϕN,∂xzzϕN(x,y,z,t)≤c(a+a3)ϕN. |
Using (3.7) (i.e., a(t)≤a0 for t≥0), it is easy to see that there exist δ>0 and a universal constant c>0 such that
∂3xϕN+∂xyyϕN+∂xzzϕN≤c(a0+a30)ϕN,forx∈[0,δ],t≥0. | (3.8) |
In the region x+y+z∈[1,δ], we conclude that (3.8) still holds (with a possible large c>0). Applying the above estimates, it then follows that Theorem 3.1 holds in this region.
In the domain x+y+z∈[1,N], we observe that
ϕN(x,y,z,t)=ea(t)(x+y+z)3/2,x+y+z∈[1,N],t≥0. |
Then, a direct computation gives rise to
∂xϕN=32a(x+y+z)1/2ϕN>0,∂2xϕN=[94a2(x+y+z)+34a(x+y+z)−1/2]ϕN,∂xyϕN=[94a2(x+y+z)+34a(x+y+z)−1/2]ϕN,∂xzϕN=[94a2(x+y+z)+34a(x+y+z)−1/2]ϕN,∂3xϕN=[278a3(x+y+z)3/2+278a2−38a(x+y+z)−3/2]ϕN,∂xyyϕN=[278a3(x+y+z)3/2+278a2−38a(x+y+z)−3/2]ϕN,∂xzzϕN=[278a3(x+y+z)3/2+278a2−38a(x+y+z)−3/2]ϕN. | (3.9) |
Hence, ϕN>0 and
∂tϕN+∂3xϕN+∂xyyϕN+∂xzzϕN=[a′(x+y+z)32+818a3(x+y+z)32+818a2−98a(x+y+z)−32]ϕN. | (3.10) |
Taking advantage of
a′(t)+818a3(t)=0, |
we eliminate the terms with power 3/2 on the right-hand side of (3.10). Therefore,
a(t)=a0√1+814a20t. | (3.11) |
We show that
∂tϕN+∂3xϕN+∂xyyϕN+∂xzzϕN≤c0ϕN, |
with c0=c0(a0)>0, and it is easy to find that for 1≤x+y+z≤N,
818a2−98a(x+y+z)−32≤c0, |
since a(t)=a≤a0, −98a(x+y+z)−32≤0.
Next, it follows from (3.9) that
∂xϕN=∂yϕN=∂zϕN=32a(x+y+z)1/2ϕN≤ca0⟨x+y+z⟩1/2ϕN,∂2xϕN=∂xyϕN=∂xzϕN≤c(a20+a0)⟨x+y+z⟩ϕN,∂3xϕN=∂xyyϕN=∂xzzϕN≤c(a30+a0)⟨x+y+z⟩ϕN. |
Lastly, we remark that
ϕN(x,y,z,t)=ea(t)(x+y+z)3/2≤ea0N3/2fort≥0,x+y+z∈[1,N], |
which completes the proof of Theorem 3.1 in this region.
Finally, let us consider the last region x+y+z∈[N,∞]. In this domain,
ϕN(x,y,z,t)=PN(x,y,z,t)=[1+32aN1/2(x+y+z−N)+(94a2N+34aN−1/2)(x+y+z−N)22]eaN3/2, | (3.12) |
with a=a(t) as in (3.11). Hence, we obtain that there exists c>0 such that for x+y+z≥N,
PN(x,y,z,t)≥c[1+aN1/2(x+y+z−N)+(a2N+aN−1/2)(x+y+z−N)22]eaN3/2≥ceaN3/2>0, | (3.13) |
which proves (ⅱ) in Theorem 3.1 in this region. Furthermore, one gets
∂xPN(x,y,z,t)=∂yPN(x,y,z,t)=∂zPN(x,y,z,t)≥[32aN1/2+94a2N(x+y+z−N)]eaN3/2≥0, |
which proves (ⅲ) in Theorem 3.1 in this domain, and
∂tPN(x,y,z,t)=a′(t)SN(x,y,z,t)eaN3/2+a′(t)N3/2PN(x,y,z,t), |
where
SN(x,y,z,t)=32N1/2(x+y+z−N)+(94aN+38N−1/2)(x+y+z−N)2≥0. |
Next, we shall prove that if x+y+z≥N,
∂tϕN+∂3xϕN+∂xyyϕN+∂xzzϕN≤c0ϕN. | (3.14) |
Note that
∂3xϕN=∂xyyϕN=∂xzzϕN≡0,∂tϕN(x,y,z,t)=∂tPN(x,y,z,t)<0. |
Combining the above estimates completes the proof of (3.14). Then, (3.13) yields (3.3) in this region x+y+z≥N.
In order to complete the proof, it is necessary to prove (v) in the region x+y+z∈[N,∞). Taking advantage of (3.9) with t=0, we need only prove that for x+y+z≥N,
ϕN(x,y,z,0)=PN(x,y,z,0)≤ea0(x+y+z)3/2+. | (3.15) |
Let x+y+z=ω to prove (3.15). We need to prove that
ϕN(ω,0)=PN(ω,0)≤ea0ω3/2+. | (3.16) |
Considering PN(ω,0) and ea0ω3/2+, and their derivatives up to the second order, which coincide at ω=N, to prove (3.16), it is sufficient to prove that
∂2ωPN(ω,0)≤d2dω2ea0ω3/2,forω≥N. | (3.17) |
For this purpose, we deduce from (3.12) that the constant value of ∂2ωPN(ω,0) is given by
∂2ωPN(ω,0)=(34a0N−1/2+94a20N)ea0N3/2 |
and coincides at ω=N with
d2dω2ea0ω3/2=(34a0ω−1/2+94a20ω)ea0ω3/2. |
Let us observe that
d3dω3ea0ω3/2=(−38a0ω−3/2+98a20+94a20+278a30ω3/2)ea0ω3/2=278a30ω3/2(−19a20ω3+1a0ω3/2+1)ea0ω3/2>0, |
if ω>N and N>(9a20)−1/3≡N0(a0). Therefore, if N≥N0, d2dω2ea0ω3/2 is an increasing function with regard to the variable ω for ω≥N. According to this fact, we obtain (3.17). Hence, (3.16) and (3.15) hold, which gives the proof of (v) in this region.
Thus, the proof of Theorem 3.1 has been completed.
By virtue of Lemma 2.2, we deduce that the solution u of IVP (1.1) satisfies
u∈C([0,T];H2(R3)∩L2(eβ(x+y+z)dxdydz)),for any β>0. | (4.1) |
In general, if for some β>0, eβ(x+y+z)f,∂2xf∈L2(R3), then eβ(x+y+z)/2∂xf∈L2(R3) since
∫R3eβ(x+y+z)(∂xf)2dxdydz≤β2∫R3eβ(x+y+z)f2dxdydz+|∫R3eβ(x+y+z)f∂2xfdxdydz|. | (4.2) |
To prove (4.2), one initially assumes that f∈H2(R3) with compact support to obtain (4.2) by integration by parts, and then the density of this class is employed to achieve the desired result.
Therefore, applying the last argument and (4.1), it follows that
∂jxu,∂jyu,∂jzu∈C([0,T];H2−j(R3)∩L2(eβ(x+y+z)dxdydz)),j=0,1,2. | (4.3) |
In particular, for any k, u∈C([0,T];L2(⟨x+y+z⟩kdxdydz)), we assume that u is sufficiently regular, that is, u∈C([0,T];H3(R3)). Then, we derive energy estimates on u applying the weights {ϕN} (since ϕN≤c⟨x+y+z⟩2). Thus, multiplying by uϕN on both sides of (2.2), and integrating the result in R3 with x, y, z, we obtain
∫R3∂tuuϕNdxdydz+∫R3∂3xuuϕNdxdydz+∫R3∂xyyuuϕNdxdydz+∫R3∂xzzuuϕNdxdydz+γ∫R3u2∂xuuϕNdxdydz=0. |
Applying (3.1) and property (ⅲ) in Theorem 3.1, it is inferred that
ddt∫R3u2ϕNdxdydz≤∫R3u2(∂tϕN+∂3xϕN+∂xyyϕN+∂xzzϕN)dxdydz+γ2∫R3u4∂xϕNdxdydz. |
From (3.2), it follows that
ddt∫R3u2ϕNdxdydz≤c0∫R3u2ϕNdxdydz+γ2∫R3u4∂xϕNdxdydz. | (4.4) |
with c0=c0(a0).
Let us estimate the second term on the right-hand side of (4.4). To bound the contribution of the second term using (3.3), we write
0≤∂xϕN(x,y,z,t)≤c1⟨x+y+z⟩1/2ϕN(x,y,z,t)≤c(1+ex+y+z)ϕN(x,y,z,t), |
hence
∫R3u4∂xϕNdxdydz≤c(‖e12(x+y+z)u‖2L∞(R3)+‖u‖2L∞(R3))∫R3u2ϕNdxdydz. |
Combining the Sobolev embedding theorem and (4.3), one has
∫T0‖e12(x+y+z)u‖2L∞(R3)(t)dt<∞. |
Inserting the above estimates into (4.4), for any N∈Z+, we deduce that
ddt∫R3u2(x,y,z,t)ϕN(x,y,z,t)dxdydz≤L(t)∫R3u2(x,y,z,t)ϕN(x,y,z,t)dxdydz, |
with L(t)∈L∞([0,T]), where L(⋅)=L(a0;‖e12(x+y+z)u0‖L2(R3);‖u0‖H1(R3)). In view of property (v) in Theorem 3.1, and using Gronwall's lemma, it follows that for t∈[0,T],
∫R3u2(x,y,z,t)ϕN(x,y,z,t)dxdydz≤c(∫R3u20(x,y,z)ϕN(x,y,z,0)dxdydz)e∫T0L(t′)dt′≤c(a0;‖e14a0(x+y+z)3/2+u0‖L2(R3);‖u0‖H1(R3);T)∫R3u20(x,y,z)ea0(x+y+z)3/2+dxdydz. | (4.5) |
We are now in a position to establish (4.5) for our less regular solution u∈C([0,T];H2(R3)). For this purpose, let us consider IVP (1.1) with regularized initial data u0,δ:=ρδ∗u(⋅+δ,⋅+δ,⋅+δ,0), where δ>0, ρδ=1δ3ρ(⋅δ,⋅δ,⋅δ), ρ∈C∞(R3) is supported in (−1,1)×(−1,1)×(−1,1), and ∫R3ρdξdηdζ=1. Since
u0,δ→u0inH2(R3)asδ→0, | (4.6) |
for IVP (1.1) in H2(R3), according to the well-posedness result in [18], the corresponding solutions uδ satisfy uδ(t)→u(t) in H2(R3) uniformly for t∈[0,T] as δ→0. Furthermore, in terms of the Sobolev embedding theorem, for fixed t,
uδ(x,y,z,t)→u(x,y,z,t)for all(x,y,z)∈R3asδ→0. |
Meanwhile, by applying Minkowski's integral inequality, we prove that
‖e12a0(x+y+z)3/2+u0,δ‖L2(R3)≤‖e12a0(x+y+z)3/2+u0‖L2(R3). | (4.7) |
Note that uδ is sufficiently regular, and we obtain (4.5) with uδ and u0,δ instead of u and u0. For fixed t, taking account of (4.6), (4.7) and using Fatou's lemma, one can deduce that
∫R3u2(x,y,z,t)ϕN(x,y,z,t)dxdydz≤c(a0;‖e14a0(x+y+z)3/2+u0‖L2(R3);‖u0‖H1(R3);T)∫R3u20(x,y,z)ea0(x+y+z)3/2+dxdydz. |
Taking N↑∞, and making use of Fatou's lemma and the property (v) in Theorem 3.1, it follows that
supt∈[0,T]∫R3ea(t)(x+y+z)3/2+|u(x,y,z,t)|2dxdydz≤c∗, |
which completes the proof of Theorem 1.1.
Let us consider the difference of the two solutions to the equation
w(x,y,z,t)=(u1−u2)(x,y,z,t), |
that is,
∂tw+∂3xw+∂xyyw+∂xzzw+γ(u21∂xw+(u1+u2)w∂xu2)=0. | (4.8) |
Taking advantage of the argument developed in Theorem 1.1, we multiply (4.8) by wϕN, integrate the result in R3 with x,y,z, and formally use integration by parts to deduce that
∫R3u21∂xwwϕNdxdydz=−∫R3u1∂xu1w2ϕNdxdydz−12∫R3u21∂xϕNw2dxdydz, |
where
|∫R3u1∂xu1w2ϕNdxdydz|≤‖u1∂xu1‖L∞(R3)∫R3w2ϕNdxdydz. |
Using (3.3), it follows that
|∫R3u21∂xϕNw2dxdydz|≤c∫R3u21⟨x+y+z⟩12w2ϕNdxdydz≤c‖u1⟨x+y+z⟩14‖2L∞(R3)∫R3w2ϕNdxdydz. |
Altogether,
|∫R3∂xu2(u1+u2)w2ϕNdxdydz|≤‖u1∂xu2‖L∞(R3)∫R3w2ϕNdxdydz+‖u2∂xu2‖L∞(R3)∫R3w2ϕNdxdydz. |
Thus, combining the equality
∫R3∂twwϕNdxdydz+∫R3∂3xwwϕNdxdydz+∫R3∂xyywwϕNdxdydz+∫R3∂xzzwwϕNdxdydz+γ∫R3(u21∂xw+(u1+u2)w∂xu2)wϕNdxdydz=0, |
and the above estimates, one has
ddt∫R3w2(x,y,z,t)ϕN(x,y,z,t)dxdydz≤c0∫R3w2(x,y,z,t)ϕN(x,y,z,t)dxdydz+γ‖u1∂xu1‖L∞(R3)∫R3w2ϕNdxdydz+cγ‖u1⟨x+y+z⟩14‖2L∞(R3)∫R3w2ϕNdxdydz+γ‖u1∂xu2‖L∞(R3)∫R3w2ϕNdxdydz+γ‖u2∂xu2‖L∞(R3)∫R3w2ϕNdxdydz, |
i.e.,
ddt∫R3w2(x,y,z,t)ϕN(x,y,z,t)dxdydz≤G(t)∫R3w2(x,y,z,t)ϕN(x,y,z,t)dxdydz, |
where
G(t)=c(‖u1∂xu1‖L∞(R3)+‖u1⟨x+y+z⟩14‖2L∞(R3)+‖u1∂xu2‖L∞(R3)+‖u2∂xu2‖L∞(R3)), |
with G(t)∈L∞([0,T]). Therefore,
supt∈[0,T]∫R3w2(x,y,z,t)ϕN(x,y,z,t)dxdydz≤c(∫R3w2(x,y,z,0)ϕN(x,y,z,0)dxdydz)e∫T0G(t)dt, |
which completes the proof of Theorem 1.2.
Proof. We suppose t0=0 and 0<t1<T. First, let us consider α∈(0,1/2]. For x+y+z≥0,N∈Z+, and α>0, we are in a position to define
φN,α(x,y,z)={[1+(x+y+z)4]α2−1,x+y+z∈[0,N],(2N)2α,x+y+z≥10N, | (5.1) |
with φN,α(x,y,z)∈C3(x+y+z≥0), φN,α(x,y,z)≥0, and ∂xφN,α=∂yφN,α=∂zφN,α≥0, and for α∈(0,1/2],
|∂xφN,α(x,y,z)|=|∂yφN,α(x,y,z)|=|∂zφN,α(x,y,z)|≤C,|∂3xφN,α(x,y,z)|=|∂xyyφN,α(x,y,z)|=|∂xzzφN,α(x,y,z)|≤C, |
where C is independent of N.
Let θN,α be defined as the following:
θN,α(x,y,z)=θN(x,y,z)={φN,α(x,y,z),x+y+z≥0,−φN,α(−x,−y,−z),x+y+z≤0. | (5.2) |
Note that
∂xθN(x,y,z)=∂yθN(x,y,z)=∂zθN(x,y,z)≥0,∀(x,y,z)∈R3,θN∈C3(R3),‖θN‖L∞(R3)=(2N)2α. |
Then, let (u0,m)m∈Z+ be a sequence in C∞0(R3) such that
u0,m→u0inH2(R3)asm↑∞, | (5.3) |
and let um∈C([0,T];H∞(R3)) be the solution of Eq (1.1) corresponding to the initial data u0,m. We have
um→uinC([0,T];H2(R3)). | (5.4) |
From the continuous dependence of the solution upon the data (see Lemma 2.1), (5.3), and (5.4), there exists T>0 such that
(a)supt∈[0,T]‖u(t)−um(t)‖L2(R3)→0asm↑∞,(b)∫T0‖u(t)−um(t)‖L∞(R3)dt→0asm↑∞. | (5.5) |
Owing to um∈C([0,T],H∞(R3)) satisfying Eq (1.1), we multiply it by umθN. Then integrating the result and formally using integration by parts (justified since θN is bounded), we obtain
ddt∫R3u2mθNdxdydz+∫R3(∂xum+∂yum)2∂yθNdxdydz+∫R3(∂xum+∂zum)2∂zθNdxdydz+∫R3(∂xum)2(3∂xθN−∂yθN−∂zθN)dxdydz+∫R3(∂yum)2(∂xθN−∂yθN)dxdydz+∫R3(∂zum)2(∂xθN−∂zθN)dxdydz=∫R3u2m∂3xθNdxdydz+∫R3u2m∂xyyθNdxdydz+∫R3u2m∂xzzθNdxdydz+γ2∫R3u4m∂xθNdxdydz. | (5.6) |
Notice that ∂xθN=∂yθN=∂zθN, 3∂xθN−∂yθN−∂zθN>0, ∂xθN−∂yθN=0, and ∂xθN−∂zθN=0. We rewrite (5.6) as follows:
ddt∫R3u2mθNdxdydz+∫R3(∂xum+∂yum)2∂yθNdxdydz+∫R3(∂xum+∂zum)2∂zθNdxdydz≤∫R3u2m∂3xθNdxdydz+∫R3u2m∂xyyθNdxdydz+∫R3u2m∂xzzθNdxdydz+γ2∫R3u4m∂xθNdxdydz. | (5.7) |
For m large enough, thanks to the boundedness of the partial derivatives of θN, the L2−norm conservation law, and the convergence of the sequence {u0,m}, we deduce that
|∫R3(um)2∂3xθNdxdydz|≤C‖u0,m‖2L2(R3)≤2C‖u0‖2L2(R3),|∫R3(um)2∂xyyθNdxdydz|≤C‖u0,m‖2L2(R3)≤2C‖u0‖2L2(R3),|∫R3(um)2∂xzzθNdxdydz|≤C‖u0,m‖2L2(R3)≤2C‖u0‖2L2(R3),|∫R3(um)4∂xθNdxdydz|≤C‖um(t)‖2L∞(R3)‖u0,m‖2L2(R3)≤2C‖um(t)‖2L∞(R3)‖u0‖2L2(R3). | (5.8) |
Integrating (5.7) with regard to t in [0,t1] and using (5.8), it follows that
∫t10∫R3(∂xum+∂yum)2(x,y,z,t)∂yθN(x,y,z)dxdydzdt+∫t10∫R3(∂xum+∂zum)2(x,y,z,t)∂zθN(x,y,z)dxdydzdt≤‖u2m(t1)θN‖L1(R3)+‖u20,mθN‖L1(R3)+Ct1‖u0‖2L2(R3)+C‖u0‖2L2(R3)∫t10‖um(t)‖2L∞(R3)dt, |
where C represents a constant, and its value may change from line to line. Meanwhile, it does not depend on the initial parameters of the problem. Setting m↑∞ and making use of (5.5), Lemma 2.1, (2.1), and the assumptions of Theorem 1.3, we deduce that
¯limm↑∞∫t10∫R3(∂xum+∂yum)2(x,y,z,t)∂yθN(x,y,z)dxdydzdt+¯limm↑∞∫t10∫R3(∂xum+∂zum)2(x,y,z,t)∂zθN(x,y,z)dxdydzdt≤‖u2(t1)θN‖L1(R3)+‖u20θN‖L1(R3)+Ct1‖u0‖2L2(R3)+C‖u0‖2L2(R3)∫t10‖u(t)‖2L∞(R3)dt≤M, | (5.9) |
with M=M(‖⟨x+y+z⟩αu0‖L2(R3),‖⟨x+y+z⟩αu(t1)‖L2(R3)). Next, we use (5.4) and (5.5) to conclude that for any fixed ˉN∈Z+ and ˉN>10N,
∂xum→∂xu,∂yum→∂yu,∂zum→∂zu, | (5.10) |
in L2([0,t1]×{x+y+z∈[−¯N,¯N]}) as m↑∞.
Thanks to ∂yθN and ∂zθN having compact support, one has
∫t10∫R3(∂xu+∂yu)2(x,y,z,t)∂yθN(x,y,z)dxdydzdt≤M,∫t10∫R3(∂xu+∂zu)2(x,y,z,t)∂zθN(x,y,z)dxdydzdt≤M. | (5.11) |
At last, notice that ∂yθN=∂zθN≥0, and for x+y+z>1,
∂yθN=∂zθN→2α(x+y+z)3[1+(x+y+z)4]1−α2∼⟨x+y+z⟩2α−1. |
Using Fatou's lemma in (5.11), we obtain
∫t10∫|x+y+z|>1(∂xu+∂yu)2(x,y,z,t)⟨x+y+z⟩2α−1dxdydzdt≤M,∫t10∫|x+y+z|>1(∂xu+∂zu)2(x,y,z,t)⟨x+y+z⟩2α−1dxdydzdt≤M. | (5.12) |
From Lemma 2.1 and (2.1), it follows that
∫t10∫|x+y+z|≤1(∂xu+∂yu)2(x,y,z,t)dxdydzdt≤M,∫t10∫|x+y+z|≤1(∂xu+∂zu)2(x,y,z,t)dxdydzdt≤M. |
We derive that
∫t10∫R3(∂xu+∂yu)2(x,y,z,t)⟨x+y+z⟩2α−1dxdydzdt≤M,∫t10∫R3(∂xu+∂zu)2(x,y,z,t)⟨x+y+z⟩2α−1dxdydzdt≤M. | (5.13) |
With (5.13), we reapply the above argument with ψN,α(x,y,z)=ψN(x,y,z):
ψN,α(x,y,z)=ψN(x,y,z)={φN,α(x,y,z),x+y+z≥0,φN,α(−x,−y,−z),x+y+z≤0. | (5.14) |
Note that
|∂xψN(x,y,z)|=|∂yψN(x,y,z)|=|∂zψN(x,y,z)|≤C⟨x+y+z⟩2α−1. |
In Eq (5.6) with ψN(x,y,z) instead of θN(x,y,z), similar computations to that in (5.8) lead us to
¯limm↑∞∫t10∫R3(∂xum+∂yum)2(x,y,z,t)∂yψN(x,y,z)dxdydzdt=∫t10∫R3(∂xu+∂yu)2(x,y,z,t)∂yψN(x,y,z)dxdydzdt,¯limm↑∞∫t10∫R3(∂xum+∂zum)2(x,y,z,t)∂zψN(x,y,z)dxdydzdt=∫t10∫R3(∂xu+∂zu)2(x,y,z,t)∂zψN(x,y,z)dxdydzdt, |
and
|∫t10∫R3(∂xu+∂yu)2(x,y,z,t)∂yψN(x,y,z)dxdydzdt|≤∫t10∫R3(∂xu+∂yu)2(x,y,z,t)⟨x+y+z⟩2α−1dxdydzdt≤M,|∫t10∫R3(∂xu+∂zu)2(x,y,z,t)∂zψN(x,y,z)dxdydzdt|≤∫t10∫R3(∂xu+∂zu)2(x,y,z,t)⟨x+y+z⟩2α−1dxdydzdt≤M. |
Collecting all the above estimates and integrating in [0,t]⊂[0,t1] yields
⟨x+y+z⟩αu(t)∈L2(R3)t∈[0,t1]. | (5.15) |
Taking advantage of (5.13), for t∈[0,t1], it is inferred that
(∂xu(t)+∂yu(t))⟨x+y+z⟩α−1/2∈L2(R3),(∂xu(t)+∂zu(t))⟨x+y+z⟩α−1/2∈L2(R3). |
Hence, the desired result holds.
Next, let us consider the case α∈(1/2,1]. A direct computation gives rise to
∂xθN,α(x,y,z)+∂yθN,α(x,y,z)+∂zθN,α(x,y,z)+|∂xψN,α(x,y,z)|+|∂yψN,α(x,y,z)|+|∂zψN,α(x,y,z)|+|θN,α−12(x,y,z)|≤C⟨x+y+z⟩,|∂3xθN,α(x,y,z)|=|∂xyyθN,α(x,y,z)|=|∂xzzθN,α(x,y,z)|≤C. |
As before, we deduce that
ddt∫R3u2mθN,αdxdydz+∫R3(∂xum+∂yum)2∂yθN,αdxdydz+∫R3(∂xum+∂zum)2∂zθN,αdxdydz≤∫R3u2m∂3xθN,αdxdydz+∫R3u2m∂xyyθN,αdxdydz+∫R3u2m∂xzzθN,αdxdydz+γ2∫R3u4m∂xθN,αdxdydz, | (5.16) |
with um∈C([0,T];H∞(R3)).
In (5.16), we first utilize that
|∫R3(um)2∂3xθN,αdxdydz|≤C‖u0,m‖2L2(R3)≤2C‖u0‖2L2(R3),|∫R3(um)2∂xyyθN,αdxdydz|≤C‖u0,m‖2L2(R3)≤2C‖u0‖2L2(R3),|∫R3(um)2∂xzzθN,αdxdydz|≤C‖u0,m‖2L2(R3)≤2C‖u0‖2L2(R3). | (5.17) |
Next, for the last term on the right-hand side of (5.16), it follows that
|∂xθN,α(x,y,z)|=|∂yθN,α(x,y,z)|=|∂zθN,α(x,y,z)|≤C|θN,α−12(x,y,z)|2, |
with C independent of N. Accordingly,
|∫R3u4m∂xθN,αdxdydz|≤C‖um(t)‖2L∞(R3)‖um(t)θN,α−12‖2L2(R3). | (5.18) |
For each fixed N, the θN,α's are bounded, and
supt∈[0,t1]‖(u−um)(t)‖L2(R3)⟶0asm↑∞, |
we conclude that
supt∈[0,t1]‖(u−um)(t)θN,α−12‖L2(R3)⟶0asm↑∞. |
Hence,
supt∈[0,t1]‖um(t)θN,α−12‖L2(R3)≤2supt∈[0,t1]‖u(t)θN,α−12‖L2(R3)≤2supt∈[0,t1]‖⟨x+y+z⟩12u(t)‖L2(R3)≤M, | (5.19) |
with M=M(‖⟨x+y+z⟩1/2u0‖L2(R3),‖⟨x+y+z⟩1/2u(t1)‖L2(R3)) for m≫1.
Plugging the above estimates into (5.16) and applying the same argument in the previous case α∈(0,1/2], we obtain
∫t10∫R3(∂xum+∂yum)2(x,y,z,t)∂yθN(x,y,z)dxdydzdt≤(1+t1)M,∫t10∫R3(∂xum+∂zum)2(x,y,z,t)∂zθN(x,y,z)dxdydzdt≤(1+t1)M, | (5.20) |
for m≫1, and
∫t10∫R3(∂xu+∂yu)2(x,y,z,t)⟨x+y+z⟩2α−1dxdydzdt≤(1+t1)M,∫t10∫R3(∂xu+∂zu)2(x,y,z,t)⟨x+y+z⟩2α−1dxdydzdt≤(1+t1)M, | (5.21) |
where M=M(‖⟨x+y+z⟩1/2u0‖L2(R3),‖⟨x+y+z⟩1/2u(t1)‖L2(R3)).
In the following, we use ψN,α instead of θN,α. Then we obtain the similar formulation to (5.16). From (5.20) and (5.21), the desired result is valid.
For the case α∈(1,3/2] and higher α, we may apply a similar bootstrap technique to get the desired result.
The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.
All authors declare no conflicts of interest that may influence the publication of this paper.
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