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Research article Special Issues

Effect of Excess Food Nutrient on Producer-Grazer Model under Stoichiometric and Toxicological Constraints

  • Received: 10 September 2018 Accepted: 24 September 2018 Published: 11 December 2018
  • Accurately assessing the risks of contaminants requires more than an understanding of the effects of contaminants on individual organism, but requires further understanding of complex ecological interactions, elemental cycling, and the interactive effects of natural stressors, such as resource limitations, and contaminant stressors. There is increasing evidence that organisms experience interactive effects of contaminant stressors and food conditions, such as resource stoichiometry, availability and excess of nutrient. Here, we develop and analyze the first producer-grazer population model that incorporates the effects of excess nutrients, as well as nutrient limitations on grazer exposed to toxicants. We use analytical, numerical and bifurcation analysis to reduce and explore model parameterized for an aquatic system of algae and zooplankton exposed to methylmercury under varying phosphorus conditions. Under certain environmental conditions, our models predict higher toxicity than previous models that neglect the consequences excess nutrient conditions can have on grazer populations.

    Citation: Md Nazmul Hassan, Kelsey Thompson, Gregory Mayer, Angela Peace. Effect of Excess Food Nutrient on Producer-Grazer Model under Stoichiometric and Toxicological Constraints[J]. Mathematical Biosciences and Engineering, 2019, 16(1): 150-167. doi: 10.3934/mbe.2019008

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  • Accurately assessing the risks of contaminants requires more than an understanding of the effects of contaminants on individual organism, but requires further understanding of complex ecological interactions, elemental cycling, and the interactive effects of natural stressors, such as resource limitations, and contaminant stressors. There is increasing evidence that organisms experience interactive effects of contaminant stressors and food conditions, such as resource stoichiometry, availability and excess of nutrient. Here, we develop and analyze the first producer-grazer population model that incorporates the effects of excess nutrients, as well as nutrient limitations on grazer exposed to toxicants. We use analytical, numerical and bifurcation analysis to reduce and explore model parameterized for an aquatic system of algae and zooplankton exposed to methylmercury under varying phosphorus conditions. Under certain environmental conditions, our models predict higher toxicity than previous models that neglect the consequences excess nutrient conditions can have on grazer populations.


    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+x2yu+x2zu+γu2xu=0,(x,y,z)R3t0,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+uxu=0,xR,t0,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,rR. Furthermore, they also showed in [15] that, for some small ε>0,

    u1,u2C([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 u1u2.

    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

    u0H2(R3)L2(ea0(x+y+z)3/2+dxdydz), (1.2)

    the unique solution u(,,) of the IVP (1.1) provided in [18]

    uC([0,T];H2(R3))

    satisfies

    supt[0,T]R3ea(t)(x+y+z)3/2+|u(x,y,z,t)|2dxdydzc, (1.3)

    where

    c=c(a0;u0H1(R3);ea0(x+y+z)3/2+u0L2(R3);T),

    with

    a(t)=a01+814a20t.

    Let us consider weighted spaces with symmetric weight, which take the form

    L2(x+y+zbdxdydz)=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

    u1C([0,T];H3(R3))L2(x+y+z2dxdydz),u2C([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)|2dxdydzc, (1.4)

    where c=c(a0;u0,1H4(R3);u0,2H4(R3);x2u0,1L2(R3);xu0,2L2(R3);Λ;T) and

    a(t)=a01+814a20t.

    Let us denote the norm of the functional space Xs by

    fXs=JsxfL2(R3)+fH1yzL2x,

    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+zbdxdydz).

    Theorem 1.3. Let u0Xs, s>9/8. There exists T=T(u0Xs) and a unique solution of the IVP (1.1) such that uC([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+zαu(x,y,z,t0),x+y+zαu(x,y,z,t1)L2(R3),

    then for any t[0,T],

    u(t)L2(x+y+zαdxdydz),
    (xu(t)+yu(t)),(xu(t)+zu(t))L2(x+y+zα1/2dxdydz).

    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 u0Xs, s>9/8, there exists T=T(u0Xs) and a unique solution of the IVP (1.1) such that uC([0,T];Xs), u,xuL1TLxyz. Moreover, the map u0u is continuous from a neighborhood of u0Xs 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

    T0uLxyzdt+T0xuLxyzdtcT, (2.1)

    where cT is a constant.

    Lemma 2.2. Let uC([0,T];H2(R3)) be a solution of the IVP (1.1), corresponding to data u0H2(R3)L2(eβ(x+y+z)dxdydz), β>0. Then,

    eβ(x+y+z)uC([0,T];L2(R3))

    and

    eβ(x+y+z)u(t)L2(R3)ceβ(x+y+z)u0L2(R3),t[0,T].

    Proof. Applying Kato's approach in [33], let us now prove this lemma. First, we consider the equation

    tu+3xu+x2yu+x2zu+γu2xu=0,(x,y,z)R3,t0. (2.2)

    Next, multiplying by uφδ on both sides of Eq (2.2) and integrating by parts, a direct computation gives rise to

    ddtR3u2φδdxdydz+R3(xu+yu)2yφδdxdydz+R3(xu+zu)2zφδdxdydz+R3(xu)2(3xφδyφδzφδ)dxdydz+R3(yu)2(xφδyφδ)dxdydz+R3(zu)2(xφδzφδ)dxdydz=R3u2(3xφδ+xyyφδ+xzzφδ)dxdydz+γ2R3u4xφδdxdydz.

    For β>0, we define

    φδ(x,y,z)=eβ(x+y+z)1+δeβ(x+y+z)forδ(0,1),δ1.

    Thus, we see that

    φδL(R3)andφδL(R3)=1δ. (2.3)
    0xφδ(x,y,z)=yφδ(x,y,z)=zφδ(x,y,z)=βeβ(x+y+z)(1+δeβ(x+y+z))2βφδ(x,y,z),
    2xφδ(x,y,z)=xyφδ(x,y,z)=xzφδ(x,y,z)=β2eβ(x+y+z)(1δeβ(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)(14δeβ(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

    limδ0φδ(x,y,z)=eβ(x+y+z).

    We apply properties (2.3) and (2.4) to obtain the estimate

    ddtR3u2φδ(x,y,z)dxdydzc0β3R3u2φδ(x,y,z)dxdydz+γ2R3u4xφδ(x,y,z)dxdydz. (2.5)

    In the case of uC([0,T];H2(R3)), there exists a positive constant c such that

    uL(R3)c.

    Next, we consider the last term of (2.5). We write

    R3u4xφδ(x,y,z)dxdydzβu2L(R3)R3u2φδ(x,y,z)dxdydzcR3u2φδ(x,y,z)dxdydz.

    Inserting this estimate into (2.5), one has

    ddtR3u2φδ(x,y,z)dxdydzcR3u2φδ(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)dxdydzcR3u20(x,y,z)φδ(x,y,z)dxdydzcR3u20(x,y,z)φ0(x,y,z)dxdydz,

    where c is a constant.

    Letting δ0, 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

    ddtR3u2ϕNdxdydz+R3(xu+yu)2yϕNdxdydz+R3(xu+zu)2zϕNdxdydz+R3(xu)2(3xϕNyϕNzϕN)dxdydz+R3(yu)2(xϕNyϕN)dxdydz+R3(zu)2(xϕNzϕN)dxdydz=R3u2(3xϕN+xyyϕN+xzzϕN+tϕN)dxdydz+γ2R3u4xϕNdxdydz. (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 NZ+:

    (i) ϕNC2(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 N0Z+ such that

    ϕN(x,y,z,0)ea0(x+y+z)3/2+ifN>N0.

    Also,

    limNϕN(x,y,z,t)=ea(t)(x+y+z)3/2+,

    for any t>0 and x+y+z(,0)(1,), where

    a(t)=a01+814a20t.

    (vi) There exists a constant c0=c0(a0)>0 such that

    tϕN+3xϕN+xyyϕN+xzzϕNc0ϕN, (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)|c1x+y+z1/2ϕN(x,y,z,t), (3.3)

    for any (x,y,z,t)R3×[0,).

    Proof. For NZ+, given a0>0, let us first define

    ϕN(x,y,z,t)={ea(t)φ(x,y,z),<x+y+z1,ea(t)(x+y+z)3/2,1x+y+zN,PN(x,y,z,t),x+y+zN,

    where

    a(t)=a01+814a20t(0,a0],t0, (3.4)

    a0 being the initial parameter.

    φ(x,y,z)=(1η(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 ηC(R3), ηx=ηy=ηz0, and

    η(x+y+z)={0,x+y+z1/2,1,x+y+z3/4, (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+zN)+(94a2N+34aN1/2)(x+y+zN)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+z0, 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)=(1η(x+y+z))(x+y+z)3+η(x+y+z)(x+y+z)3/20,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)=(1η(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)(1η(x+y+z))3(x+y+z)2+η(x+y+z)32(x+y+z)1/20,

    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)a0fort0, (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+xzzϕNc0ϕN.

    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+z0 (x+y+z0),

    xϕN=yϕN=zϕN3a(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+z0 (0x+y+z1),

    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 t0), it is easy to see that there exist δ>0 and a universal constant c>0 such that

    3xϕN+xyyϕN+xzzϕNc(a0+a30)ϕN,forx[0,δ],t0. (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],t0.

    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+278a238a(x+y+z)3/2]ϕN,xyyϕN=[278a3(x+y+z)3/2+278a238a(x+y+z)3/2]ϕN,xzzϕN=[278a3(x+y+z)3/2+278a238a(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+818a298a(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)=a01+814a20t. (3.11)

    We show that

    tϕN+3xϕN+xyyϕN+xzzϕNc0ϕN,

    with c0=c0(a0)>0, and it is easy to find that for 1x+y+zN,

    818a298a(x+y+z)32c0,

    since a(t)=aa0, 98a(x+y+z)320.

    Next, it follows from (3.9) that

    xϕN=yϕN=zϕN=32a(x+y+z)1/2ϕNca0x+y+z1/2ϕN,2xϕN=xyϕN=xzϕNc(a20+a0)x+y+zϕN,3xϕN=xyyϕN=xzzϕNc(a30+a0)x+y+zϕN.

    Lastly, we remark that

    ϕN(x,y,z,t)=ea(t)(x+y+z)3/2ea0N3/2fort0,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+zN)+(94a2N+34aN1/2)(x+y+zN)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+zN,

    PN(x,y,z,t)c[1+aN1/2(x+y+zN)+(a2N+aN1/2)(x+y+zN)22]eaN3/2ceaN3/2>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+zN)]eaN3/20,

    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+zN)+(94aN+38N1/2)(x+y+zN)20.

    Next, we shall prove that if x+y+zN,

    tϕN+3xϕN+xyyϕN+xzzϕNc0ϕN. (3.14)

    Note that

    3xϕN=xyyϕN=xzzϕN0,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+zN.

    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+zN,

    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,forωN. (3.17)

    For this purpose, we deduce from (3.12) that the constant value of 2ωPN(ω,0) is given by

    2ωPN(ω,0)=(34a0N1/2+94a20N)ea0N3/2

    and coincides at ω=N with

    d2dω2ea0ω3/2=(34a0ω1/2+94a20ω)ea0ω3/2.

    Let us observe that

    d3dω3ea0ω3/2=(38a0ω3/2+98a20+94a20+278a30ω3/2)ea0ω3/2=278a30ω3/2(19a20ω3+1a0ω3/2+1)ea0ω3/2>0,

    if ω>N and N>(9a20)1/3N0(a0). Therefore, if NN0, d2dω2ea0ω3/2 is an increasing function with regard to the variable ω for ωN. 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

    uC([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,2xfL2(R3), then eβ(x+y+z)/2xfL2(R3) since

    R3eβ(x+y+z)(xf)2dxdydzβ2R3eβ(x+y+z)f2dxdydz+|R3eβ(x+y+z)f2xfdxdydz|. (4.2)

    To prove (4.2), one initially assumes that fH2(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,jzuC([0,T];H2j(R3)L2(eβ(x+y+z)dxdydz)),j=0,1,2. (4.3)

    In particular, for any k, uC([0,T];L2(x+y+zkdxdydz)), we assume that u is sufficiently regular, that is, uC([0,T];H3(R3)). Then, we derive energy estimates on u applying the weights {ϕN} (since ϕNcx+y+z2). Thus, multiplying by uϕN on both sides of (2.2), and integrating the result in R3 with x, y, z, we obtain

    R3tuuϕNdxdydz+R33xuuϕNdxdydz+R3xyyuuϕNdxdydz+R3xzzuuϕNdxdydz+γR3u2xuuϕNdxdydz=0.

    Applying (3.1) and property (ⅲ) in Theorem 3.1, it is inferred that

    ddtR3u2ϕNdxdydzR3u2(tϕN+3xϕN+xyyϕN+xzzϕN)dxdydz+γ2R3u4xϕNdxdydz.

    From (3.2), it follows that

    ddtR3u2ϕNdxdydzc0R3u2ϕNdxdydz+γ2R3u4xϕNdxdydz. (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

    0xϕN(x,y,z,t)c1x+y+z1/2ϕN(x,y,z,t)c(1+ex+y+z)ϕN(x,y,z,t),

    hence

    R3u4xϕNdxdydzc(e12(x+y+z)u2L(R3)+u2L(R3))R3u2ϕNdxdydz.

    Combining the Sobolev embedding theorem and (4.3), one has

    T0e12(x+y+z)u2L(R3)(t)dt<.

    Inserting the above estimates into (4.4), for any NZ+, we deduce that

    ddtR3u2(x,y,z,t)ϕN(x,y,z,t)dxdydzL(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)u0L2(R3);u0H1(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)dxdydzc(R3u20(x,y,z)ϕN(x,y,z,0)dxdydz)eT0L(t)dtc(a0;e14a0(x+y+z)3/2+u0L2(R3);u0H1(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 uC([0,T];H2(R3)). For this purpose, let us consider IVP (1.1) with regularized initial data u0,δ:=ρδu(+δ,+δ,+δ,0), where δ>0, ρδ=1δ3ρ(δ,δ,δ), ρC(R3) is supported in (1,1)×(1,1)×(1,1), and R3ρdξdηdζ=1. Since

    u0,δu0inH2(R3)asδ0, (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 δ0. 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)R3asδ0.

    Meanwhile, by applying Minkowski's integral inequality, we prove that

    e12a0(x+y+z)3/2+u0,δL2(R3)e12a0(x+y+z)3/2+u0L2(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)dxdydzc(a0;e14a0(x+y+z)3/2+u0L2(R3);u0H1(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)|2dxdydzc,

    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)=(u1u2)(x,y,z,t),

    that is,

    tw+3xw+xyyw+xzzw+γ(u21xw+(u1+u2)wxu2)=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

    R3u21xwwϕNdxdydz=R3u1xu1w2ϕNdxdydz12R3u21xϕNw2dxdydz,

    where

    |R3u1xu1w2ϕNdxdydz|u1xu1L(R3)R3w2ϕNdxdydz.

    Using (3.3), it follows that

    |R3u21xϕNw2dxdydz|cR3u21x+y+z12w2ϕNdxdydzcu1x+y+z142L(R3)R3w2ϕNdxdydz.

    Altogether,

    |R3xu2(u1+u2)w2ϕNdxdydz|u1xu2L(R3)R3w2ϕNdxdydz+u2xu2L(R3)R3w2ϕNdxdydz.

    Thus, combining the equality

    R3twwϕNdxdydz+R33xwwϕNdxdydz+R3xyywwϕNdxdydz+R3xzzwwϕNdxdydz+γR3(u21xw+(u1+u2)wxu2)wϕNdxdydz=0,

    and the above estimates, one has

    ddtR3w2(x,y,z,t)ϕN(x,y,z,t)dxdydzc0R3w2(x,y,z,t)ϕN(x,y,z,t)dxdydz+γu1xu1L(R3)R3w2ϕNdxdydz+cγu1x+y+z142L(R3)R3w2ϕNdxdydz+γu1xu2L(R3)R3w2ϕNdxdydz+γu2xu2L(R3)R3w2ϕNdxdydz,

    i.e.,

    ddtR3w2(x,y,z,t)ϕN(x,y,z,t)dxdydzG(t)R3w2(x,y,z,t)ϕN(x,y,z,t)dxdydz,

    where

    G(t)=c(u1xu1L(R3)+u1x+y+z142L(R3)+u1xu2L(R3)+u2xu2L(R3)),

    with G(t)L([0,T]). Therefore,

    supt[0,T]R3w2(x,y,z,t)ϕN(x,y,z,t)dxdydzc(R3w2(x,y,z,0)ϕN(x,y,z,0)dxdydz)eT0G(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+z0,NZ+, and α>0, we are in a position to define

    φN,α(x,y,z)={[1+(x+y+z)4]α21,x+y+z[0,N],(2N)2α,x+y+z10N, (5.1)

    with φN,α(x,y,z)C3(x+y+z0), φN,α(x,y,z)0, and xφN,α=yφN,α=zφN,α0, 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+z0,φN,α(x,y,z),x+y+z0. (5.2)

    Note that

    xθN(x,y,z)=yθN(x,y,z)=zθN(x,y,z)0,(x,y,z)R3,θNC3(R3),θNL(R3)=(2N)2α.

    Then, let (u0,m)mZ+ be a sequence in C0(R3) such that

    u0,mu0inH2(R3)asm, (5.3)

    and let umC([0,T];H(R3)) be the solution of Eq (1.1) corresponding to the initial data u0,m. We have

    umuinC([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)T0u(t)um(t)L(R3)dt0asm. (5.5)

    Owing to umC([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

    ddtR3u2mθNdxdydz+R3(xum+yum)2yθNdxdydz+R3(xum+zum)2zθNdxdydz+R3(xum)2(3xθNyθNzθN)dxdydz+R3(yum)2(xθNyθN)dxdydz+R3(zum)2(xθNzθN)dxdydz=R3u2m3xθNdxdydz+R3u2mxyyθNdxdydz+R3u2mxzzθNdxdydz+γ2R3u4mxθNdxdydz. (5.6)

    Notice that xθN=yθN=zθN, 3xθNyθNzθN>0, xθNyθN=0, and xθNzθN=0. We rewrite (5.6) as follows:

    ddtR3u2mθNdxdydz+R3(xum+yum)2yθNdxdydz+R3(xum+zum)2zθNdxdydzR3u2m3xθNdxdydz+R3u2mxyyθNdxdydz+R3u2mxzzθNdxdydz+γ2R3u4mxθNdxdydz. (5.7)

    For m large enough, thanks to the boundedness of the partial derivatives of θN, the L2norm conservation law, and the convergence of the sequence {u0,m}, we deduce that

    |R3(um)23xθNdxdydz|Cu0,m2L2(R3)2Cu02L2(R3),|R3(um)2xyyθNdxdydz|Cu0,m2L2(R3)2Cu02L2(R3),|R3(um)2xzzθNdxdydz|Cu0,m2L2(R3)2Cu02L2(R3),|R3(um)4xθNdxdydz|Cum(t)2L(R3)u0,m2L2(R3)2Cum(t)2L(R3)u02L2(R3). (5.8)

    Integrating (5.7) with regard to t in [0,t1] and using (5.8), it follows that

    t10R3(xum+yum)2(x,y,z,t)yθN(x,y,z)dxdydzdt+t10R3(xum+zum)2(x,y,z,t)zθN(x,y,z)dxdydzdtu2m(t1)θNL1(R3)+u20,mθNL1(R3)+Ct1u02L2(R3)+Cu02L2(R3)t10um(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

    ¯limmt10R3(xum+yum)2(x,y,z,t)yθN(x,y,z)dxdydzdt+¯limmt10R3(xum+zum)2(x,y,z,t)zθN(x,y,z)dxdydzdtu2(t1)θNL1(R3)+u20θNL1(R3)+Ct1u02L2(R3)+Cu02L2(R3)t10u(t)2L(R3)dtM, (5.9)

    with M=M(x+y+zαu0L2(R3),x+y+zαu(t1)L2(R3)). Next, we use (5.4) and (5.5) to conclude that for any fixed ˉNZ+ and ˉN>10N,

    xumxu,yumyu,zumzu, (5.10)

    in L2([0,t1]×{x+y+z[¯N,¯N]}) as m.

    Thanks to yθN and zθN having compact support, one has

    t10R3(xu+yu)2(x,y,z,t)yθN(x,y,z)dxdydzdtM,t10R3(xu+zu)2(x,y,z,t)zθN(x,y,z)dxdydzdtM. (5.11)

    At last, notice that yθN=zθN0, and for x+y+z>1,

    yθN=zθN2α(x+y+z)3[1+(x+y+z)4]1α2x+y+z2α1.

    Using Fatou's lemma in (5.11), we obtain

    t10|x+y+z|>1(xu+yu)2(x,y,z,t)x+y+z2α1dxdydzdtM,t10|x+y+z|>1(xu+zu)2(x,y,z,t)x+y+z2α1dxdydzdtM. (5.12)

    From Lemma 2.1 and (2.1), it follows that

    t10|x+y+z|1(xu+yu)2(x,y,z,t)dxdydzdtM,t10|x+y+z|1(xu+zu)2(x,y,z,t)dxdydzdtM.

    We derive that

    t10R3(xu+yu)2(x,y,z,t)x+y+z2α1dxdydzdtM,t10R3(xu+zu)2(x,y,z,t)x+y+z2α1dxdydzdtM. (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+z0,φN,α(x,y,z),x+y+z0. (5.14)

    Note that

    |xψN(x,y,z)|=|yψN(x,y,z)|=|zψN(x,y,z)|Cx+y+z2α1.

    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

    ¯limmt10R3(xum+yum)2(x,y,z,t)yψN(x,y,z)dxdydzdt=t10R3(xu+yu)2(x,y,z,t)yψN(x,y,z)dxdydzdt,¯limmt10R3(xum+zum)2(x,y,z,t)zψN(x,y,z)dxdydzdt=t10R3(xu+zu)2(x,y,z,t)zψN(x,y,z)dxdydzdt,

    and

    |t10R3(xu+yu)2(x,y,z,t)yψN(x,y,z)dxdydzdt|t10R3(xu+yu)2(x,y,z,t)x+y+z2α1dxdydzdtM,|t10R3(xu+zu)2(x,y,z,t)zψN(x,y,z)dxdydzdt|t10R3(xu+zu)2(x,y,z,t)x+y+z2α1dxdydzdtM.

    Collecting all the above estimates and integrating in [0,t][0,t1] yields

    x+y+zαu(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+zα1/2L2(R3),(xu(t)+zu(t))x+y+zα1/2L2(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,α12(x,y,z)|Cx+y+z,|3xθN,α(x,y,z)|=|xyyθN,α(x,y,z)|=|xzzθN,α(x,y,z)|C.

    As before, we deduce that

    ddtR3u2mθN,αdxdydz+R3(xum+yum)2yθN,αdxdydz+R3(xum+zum)2zθN,αdxdydzR3u2m3xθN,αdxdydz+R3u2mxyyθN,αdxdydz+R3u2mxzzθN,αdxdydz+γ2R3u4mxθN,αdxdydz, (5.16)

    with umC([0,T];H(R3)).

    In (5.16), we first utilize that

    |R3(um)23xθN,αdxdydz|Cu0,m2L2(R3)2Cu02L2(R3),|R3(um)2xyyθN,αdxdydz|Cu0,m2L2(R3)2Cu02L2(R3),|R3(um)2xzzθN,αdxdydz|Cu0,m2L2(R3)2Cu02L2(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,α12(x,y,z)|2,

    with C independent of N. Accordingly,

    |R3u4mxθN,αdxdydz|Cum(t)2L(R3)um(t)θN,α122L2(R3). (5.18)

    For each fixed N, the θN,α's are bounded, and

    supt[0,t1](uum)(t)L2(R3)0asm,

    we conclude that

    supt[0,t1](uum)(t)θN,α12L2(R3)0asm.

    Hence,

    supt[0,t1]um(t)θN,α12L2(R3)2supt[0,t1]u(t)θN,α12L2(R3)2supt[0,t1]x+y+z12u(t)L2(R3)M, (5.19)

    with M=M(x+y+z1/2u0L2(R3),x+y+z1/2u(t1)L2(R3)) for m1.

    Plugging the above estimates into (5.16) and applying the same argument in the previous case α(0,1/2], we obtain

    t10R3(xum+yum)2(x,y,z,t)yθN(x,y,z)dxdydzdt(1+t1)M,t10R3(xum+zum)2(x,y,z,t)zθN(x,y,z)dxdydzdt(1+t1)M, (5.20)

    for m1, and

    t10R3(xu+yu)2(x,y,z,t)x+y+z2α1dxdydzdt(1+t1)M,t10R3(xu+zu)2(x,y,z,t)x+y+z2α1dxdydzdt(1+t1)M, (5.21)

    where M=M(x+y+z1/2u0L2(R3),x+y+z1/2u(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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