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A complete characterization of sharp thresholds to spherically symmetric multidimensional pressureless Euler-Poisson systems

Manas Bhatnagar and Hailiang Liu Department of Mathematics and Statistics, University of Massachusetts Amherst, Amherst, Massachusetts 01003 [email protected] Department of Mathematics, Iowa State University, Ames, Iowa 50010 [email protected]
Abstract.

The Euler-Poisson system describes the dynamic behavior of many important physical flows including charge transport, plasma with collision and cosmological waves. We prove sharp threshold conditions for the global existence/finite-time-breakdown of solutions to the multidimensional pressureless Euler-Poisson (EP) system with or without background and general initial data. In particular, the initial data could include points where velocity is negative, that is, the flow is directed towards the origin. Obtaining threshold conditions for such systems is extremely hard due to the coupling of various local/nonlocal forces. Remarkably, we are able to achieve a sharp threshold for the zero background case and most importantly, the positive background case, which is quite delicate due to the oscillations present in the solutions. We discover a completely novel nonlinear quantity that helps to analyze the system. In the case of positive background, if the initial data results in a global-in-time solution, then we show that the density is periodic along any single characteristic path. We use the Floquet Theorem to prove periodicity.

Key words and phrases:
Critical thresholds, global regularity, shock formation, Euler-Poisson system
2020 Mathematics Subject Classification:
35A01; 35B30; 35B44; 35L45

1. Introduction

A general system of pressureless Euler-Poisson (EP) equations has the following form,

(1.1a) ρt+(ρ𝐮)=0,t>0,𝐱N,\displaystyle\rho_{t}+\nabla\cdot(\rho\mathbf{u})=0,\quad t>0,\mathbf{x}\in\mathbb{R}^{N},
(1.1b) 𝐮t+(𝐮)𝐮=kϕ,\displaystyle\mathbf{u}_{t}+(\mathbf{u}\cdot\nabla)\mathbf{u}=-k\nabla\phi,
(1.1c) Δϕ=ρc,\displaystyle-\Delta\phi=\rho-c,

with smooth initial data (ρ00,𝐮𝟎)(\rho_{0}\geq 0,\mathbf{u_{0}}). The constant parameters k,c0k,c\geq 0 are the forcing coefficient and background state respectively. The sign of the forcing coefficient kk signifies the type of particles being modeled and its magnitude gives a measure of the strength between them. When the force in-between particles is repulsive, for example in the case of charge flow, then k>0k>0. Within the pressureless setup (1.1), k<0k<0 is relevant in the case of interstellar clouds where the pressure gradient becomes negligible compared to the gravitation forces, see [11]. In the pressureless setup with same charge particles (k>0k>0), the background state is, in practicality, a profile, that is, a function of the spatial variable, c=c(x)c=c(x), see [13]. The background models the doping profile for charge flow in semiconductors. However, the sheer complicacy of the system has restricted researchers to consider the background as a constant. To our knowledge, [3] is the only work that considers background as a profile in obtaining critical thresholds.

A locally well-posed PDE system exhibiting critical threshold phenomena is the one wherein existence of global-in-time solutions is dependent on whether the initial data crosses a certain threshold manifold. This critical threshold manifold divides the phase space of initial data into two mutually exclusive regions or sets. If the initial data lies completely in one of the sets (also called subcritical region), there is global solution. However, if some part of initial data lies outside this set, or in other words, in the supercritical region, a breakdown occurs and solution loses its smoothness in finite time.

As one can expect, the question of proving the existence and subsequently finding the critical threshold manifold is simpler in one dimension (N=1N=1). In this case, the threshold is a curve on the (u0x,ρ0)(u_{0x},\rho_{0}) plane and the subcritical region is given by,

|u0x|<k(2ρ0c).|u_{0x}|<\sqrt{k(2\rho_{0}-c)}.

A vast amount of literature exists for critical thresholds to (1.1) and other similar systems. The existence of such curve was first identified and analyzed in [9] for EP systems. The authors analyzed the one dimensional and multidimensional (with spherical symmetry) cases. A series of works then followed for EP as well as other systems, [2, 3, 4, 5, 7, 12, 16, 17, 18, 19, 20, 21, 22, 24, 25, 26, 27, 28, 29] among many others. As is known for general hyperbolic conservation laws that for any initial data, singularities develop as one moves forward in time, [15]. This is due to the convective forces present. Addition of source terms however, can result in a set of initial data that leads to global solutions. To extract a set of initial data (subcritical region) for which there is global well-posedness, is an interesting territory to be explored. The ‘goodness’ of the external forces can balance or even outweigh the ‘bad’ convective forces and result in a ‘large’ subcritical region. An example where the external forces completely obliterate the convective forces is the strongly singular Euler-Poisson-alignment (EPA) system. In [14], the authors show that the system (1.1) with N=1N=1 and an additional nonlocal alignment force in the momentum equation results in global-in-time solutions for any initial data. This result is remarkable and in stark opposition to the conventional results for EP systems.

For threshold results in one dimensional EP and EPA systems, one can refer to [2, 3, 4, 5, 8, 14, 26, 28]. In [26], the authors include pressure and exclude background. It is quite difficult to obtain thresholds for the full EP system (with pressure) due to the strict hyperbolicity resulting in two characteristic flow paths. Critical threshold for EP systems with pressure and c>0c>0 is largely an open problem.

For pressureless EP systems, the key is to obtain bounds on the gradient of velocity. Owing to the local existence result, global-in-time solutions to (1.1) can be ‘extended’ upon a priori smooth local solutions. The local existence for such systems is well-known, see [23] for a proof. We state the result here.

Theorem 1.1 (Local wellposedness).

Consider (1.1) with smooth initial data, ρ0cHs(N)\rho_{0}-c\in H^{s}(\mathbb{R}^{N}) and 𝐮𝟎(Hs+1(N))N\mathbf{u_{0}}\in\left(H^{s+1}(\mathbb{R}^{N})\right)^{N} with s>N/2s>N/2. Then there exists a time T>0T>0 and functions ρ,𝐮\rho,\mathbf{u} such that,

ρcC([0,T];Hs(N)),𝐮(C([0,T];Hs+1(N)))N,\rho-c\in C([0,T];H^{s}(\mathbb{R}^{N})),\quad\mathbf{u}\in\left(C([0,T];H^{s+1}(\mathbb{R}^{N}))\right)^{N},

are unique smooth solutions to (1.1). In addition, the time TT can be extended as long as,

0T𝐮(t,)𝑑t<.\int_{0}^{T}||\nabla\mathbf{u}(t,\cdot)||_{\infty}dt<\infty.

As a result, in 1D, one needs to bound |ux||u_{x}| to guarantee global well-posedness. However, things are different in higher dimensions where the gradient of velocity is in fact an N×NN\times N matrix. One finds out that an ODE along the characteristic path can be obtained for the divergence of the gradient matrix. However, analyzing the divergence is not enough. One needs to control the spectral gap (sum of all the absolute differences of eigenvalues), see for example [12], in order to guarantee smooth solutions for all time. This is the tricky part. Given the inherent property of hyperbolic balance laws, it is in general easier to obtain sufficient conditions for breakdown of solutions as compared to conditions for global existence. In [6, 7], the authors obtain bounds on the supercritical region for (1.1) for k<0k<0.

To obtain bounds on subcritical region is more involved. Several workarounds have been used by various researchers by simplifying the EP system. In [16, 21], the authors study and obtain thresholds for the restricted EP equations in two and three dimensions respectively.

Another derivative of the EP system that has been extensively studied is (1.1) with spherical symmetry. It can be shown that if the initial data to (1.1) is such that ρ(0,𝐱)=f(|𝐱|),𝐮(0,𝐱)=g(|𝐱|)𝐱|𝐱|\rho(0,\mathbf{x})=f(|\mathbf{x}|),\mathbf{u}(0,\mathbf{x})=g(|\mathbf{x}|)\frac{\mathbf{x}}{|\mathbf{x}|} for smooth functions g,fg,f, then for any fixed time and as long as the solution is smooth, the symmetry extends, that is, the solutions are ρ(t,|𝐱|),u(t,|𝐱|)𝐱|𝐱|\rho(t,|\mathbf{x}|),u(t,|\mathbf{x}|)\frac{\mathbf{x}}{|\mathbf{x}|}. Applying this simplification to (1.1), we obtain the following system of equations,

(1.2a) ρt+(ρurN1)rrN1=0,t>0,r>0,\displaystyle\rho_{t}+\frac{(\rho ur^{N-1})_{r}}{r^{N-1}}=0,\qquad t>0,\quad r>0,
(1.2b) ut+uur=kϕr,\displaystyle u_{t}+uu_{r}=-k\phi_{r},
(1.2c) (rN1ϕr)r=rN1(ρc),\displaystyle-(r^{N-1}\phi_{r})_{r}=r^{N-1}(\rho-c),

with r=|𝐱|>0,N2r=|\mathbf{x}|>0,N\geq 2 and subject to smooth initial density and velocity

.(ρ(t,),u(t,))|t=0=(ρ00,u0).\big{.}(\rho(t,\cdot),u(t,\cdot))\big{|}_{t=0}=(\rho_{0}\geq 0,u_{0}).

Even though the system is now simplified as compared to the general EP system, it turns out that it is still tricky to analyze for thresholds, especially when k>0k>0. (1.2) was first studied in [9] with c=0c=0 and expanding flows (u0>0u_{0}>0). Sufficient conditions on global existence and finite-time-breakdown were obtained for N=2,3N=2,3 and a sharp condition was obtained for N=4N=4. Later, a sharp condition was derived in [29] however, once again, only for zero background and expanding flows. A sufficient blow-up condition was derived in [30] for c=0c=0, however, no conclusion was made with regards to global existence.

The dynamics of (1.2) is quite different for c>0c>0 compared with c=0c=0. One major difference is that in the former, the density is a perturbation around the constant background. The mass is infinite and the following holds,

0(ρ(t,r)c)𝑑r=0,\int_{0}^{\infty}(\rho(t,r)-c)dr=0,

which is different from the zero background state where the mass is finite and conserved. A major step forward to mitigate the restrictiveness of the past results was taken in [28]. The author there reduced (1.2) to a 4×44\times 4 ODE system along a characteristic path, and subsequently proved the existence of critical thresholds. This 4×44\times 4 ODE system derived by the author is crucial in our analysis as well. An intriguing discovery by the author was that the Poisson forcing in (1.2b) is enough to avoid concentration at the origin, even if there are points where the initial velocity points towards the origin. To our knowledge, it is the first work with regards to thresholds of spherically symmetric EP system wherein the assumption of expanding flows was dropped. It was also noted that N=2N=2 is the critical case and the analysis needs to be done fairly differently as compared to the case when N3N\geq 3. The author obtained partial results for thresholds for c=0c=0, that is, bounds on the subcritical and supercritical regions were presented. However, no results were presented for c>0c>0 citing the highly chaotic dynamics in such a scenario.

In this paper, we present sharp thresholds for both c>0c>0 as well as c=0c=0. Several steps are needed to arrive at the precise thresholds. Special care has to be taken with regards to finding the subcritical region for c>0c>0. We essentially start out by characterizing the supercritical region and narrowing down the possible initial configurations that might allow for global solutions, eventually extracting out the subcritical region.

Our main results can be stated non-technically as follows:

  • For the nonzero background case (c0c\not\equiv 0), we show that the EP system admits a global-in-time smooth solution if and only if the initial data is smooth, lying in a subcritical region, ΘN\Theta_{N}. Theorem 2.1 contains such critical threshold result. The explicit definition of ΘN\Theta_{N} is stated after the result.

  • For the zero background case (c=0c=0), we show that the EP system admits a global-in-time smooth solution if and only if the initial data is smooth, and lying in certain subcritical region, ΣN\Sigma_{N}. Theorem 2.6 contains the precise thresholds for dimensions greater than or equal to three. Theorem 2.8 contains the threshold results in dimension two. The explicit definition of ΣN\Sigma_{N} is stated after each of the results.

We discover a completely novel nonlinear quantity that helps to analyze the system. At t=0t=0, this is given by

(1.3) A0(r):=u0(r)u0r(r)+kϕ0r(r)rρ0(r),ρ0(r)>0.\displaystyle A_{0}(r):=\frac{u_{0}(r)u_{0r}(r)+k\phi_{0r}(r)}{r\rho_{0}(r)},\quad\rho_{0}(r)>0.

In fact, such quantity is very crucial in analyzing and simplifying the representation of subcritical/supercritical regions. The full motivation and usage of this quantity will be thoroughly discussed in the Sections to come. It is found that a simpler breakdown condition can be obtained using this expression, the corresponding result for dimensions greater than or equal to three is provided in Theorem 2.4.

1.1. A roadmap of the c>0c>0 case

Before stating our main results, we give a short roadmap of how threshold regions are identified. Following is a list of key points:

  • The full dynamics of (1.2) can be reduced to a weakly coupled system of four equations (along characteristic path {(t,X):dX/dt=u(t,X),X(0)=β,β>0}\{(t,X):dX/dt=u(t,X),\ X(0)=\beta,\beta>0\}) as,

    ρ=(N1)ρqpρ,\displaystyle\rho^{\prime}=-(N-1)\rho q-p\rho,
    p=p2k(N1)s+k(ρc),\displaystyle p^{\prime}=-p^{2}-k(N-1)s+k(\rho-c),
    q=ksq2,\displaystyle q^{\prime}=ks-q^{2},
    s=q(c+Ns)\displaystyle s^{\prime}=-q(c+Ns)

    with

    p:=ur,q:=ur,s:=ϕrr.\displaystyle p:=u_{r},\quad q:=\frac{u}{r},\quad s:=-\frac{\phi_{r}}{r}.

    The initial data being ρ0,p0,q0,s0\rho_{0},p_{0},q_{0},s_{0} respectively, which depends on β\beta, but we do not specify it explicitly as we will be analyzing one characteristic path at a time. A thorough analysis of this ODE system becomes the main task. We restrict to the case k>0k>0, N2N\geq 2 and c0c\geq 0.

  • The qsq-s system is decoupled and admits a closed form of trajectory curves, expressed as,

    RN(q(t),s~(t))=RN(q0,s0+c/N),t0,s~=s+c/NR_{N}(q(t),\tilde{s}(t))=R_{N}(q_{0},s_{0}+c/N),\quad t\geq 0,\quad\tilde{s}=s+c/N

    with,

    (1.6) RN(q,s~)={s~1(q2+kc2+ks~ln(s~)),N=2,s~2N(q2+kcN+2ks~N2),N3.\displaystyle R_{N}(q,\tilde{s})=\left\{\begin{array}[]{c}\tilde{s}^{-1}\left(q^{2}+\frac{kc}{2}+k\tilde{s}\ln\left(\tilde{s}\right)\right),\quad N=2,\\ \tilde{s}^{-\frac{2}{N}}\left(q^{2}+\frac{kc}{N}+\frac{2k\tilde{s}}{N-2}\right),\quad N\geq 3.\end{array}\right.

    This ensures that the qsq-s system admits a global bounded solution if and only if s0>c/Ns_{0}>-c/N. Moreover, the solutions are periodic and the trajectories rotate clockwise on the (q,s)(q,s) plane as time progresses. We also have

    smin<0<smax,s_{\rm min}<0<s_{\rm max},

    where smin,smaxs_{min},s_{max} are the minimum/maximum values of ss attained. In addition, 0tq(τ)𝑑τ\int_{0}^{t}q(\tau)\,d\tau is shown to be bounded along with,

    Γ(t):=e0tq(τ)𝑑τ=(s(t)+c/Ns0+c/N)1N.\Gamma(t):=e^{-\int_{0}^{t}q(\tau)d\tau}=\left(\frac{s(t)+c/N}{s_{0}+c/N}\right)^{\frac{1}{N}}.
  • Another key transformation of form,

    η:=1ρΓN1,w=pρΓN1\displaystyle\eta:=\frac{1}{\rho}\Gamma^{N-1},\qquad w=\frac{p}{\rho}\Gamma^{N-1}

    leads to a new system,

    η=w,\displaystyle\eta^{\prime}=w,
    w=kη(c+s(N1))+kΓ(N1),\displaystyle w^{\prime}=-k\eta(c+s(N-1))+k\Gamma^{(N-1)},

    This is a linear system with time-dependent yet bounded coefficients. This way, the key to existence of global solution to (ρ,p)(\rho,p) is equivalent to ensuring η(t)>0\eta(t)>0 for all t>0t>0.

  • A nonlinear quantity of the form,

    A=qwkηs,A=qw-k\eta s,

    is shown to be bounded. A0:=q0w0kη0s0A_{0}:=q_{0}w_{0}-k\eta_{0}s_{0} and A0A_{0} in (1.3) are essentially the same and we will see more about this in Section 5. More precisely, for N3N\geq 3, we obtain

    A(Γ)=(A0+kN2)ΓkN2ΓN1.A(\Gamma)=\left(A_{0}+\frac{k}{N-2}\right)\Gamma-\frac{k}{N-2}\Gamma^{N-1}.

    We are able to conclude that η(t)\eta(t) will surely be zero at some positive time if AA does not change sign. This is in fact the case if either 1+A0(N2)k01+\frac{A_{0}(N-2)}{k}\leq 0, or 1+A0(N2)k>01+\frac{A_{0}(N-2)}{k}>0 with

    ΓmaxκorΓminκ,\Gamma_{\rm max}\leq\kappa\quad\text{or}\;\Gamma_{\rm min}\geq\kappa,

    where κ\kappa is the unique positive root of AA,

    κ:=(1+A0(N2)k)1N2.\displaystyle\kappa:=\left(1+\frac{A_{0}(N-2)}{k}\right)^{\frac{1}{N-2}}.
  • Therefore, to precisely identify the possible initial configurations for global existence, it is necessary to require

    κ(Γmin,Γmax).\kappa\in(\Gamma_{\rm min},\Gamma_{\rm max}).

    The key idea is to construct two nonnegative functions to precisely demarcate the subcritical region. We will show that if a solution, η\eta starts in between these functions at t=0t=0, then it remains so for all except discrete times (where it is positive), thereby providing a condition to ensure positivity of η\eta. The converse also holds. The two constructed functions form beads and the strictly positive solutions are contained within those beads, see Figure 5. More precisely, we show that if κ\kappa satisfies the above inclusion, then there exists two nonnegative functions ηi=ηi(t;q0,s0,A0),i=1,2\eta_{i}=\eta_{i}(t;q_{0},s_{0},A_{0}),i=1,2 so that the following holds. For the case q00q_{0}\neq 0, we show that if

    min{η1(0),η2(0)}<η0<max{η1(0),η2(0)},\min\{\eta_{1}(0),\eta_{2}(0)\}<\eta_{0}<\max\{\eta_{1}(0),\eta_{2}(0)\},

    then,

    min{η1(t),η2(t)}η(t)max{η1(t),η2(t)},t>0.\min\{\eta_{1}(t),\eta_{2}(t)\}\leq\eta(t)\leq\max\{\eta_{1}(t),\eta_{2}(t)\},\quad t>0.

    The converse also holds. Quite similarly, for q0=0q_{0}=0, we have that if

    dη1dt(0)<w0<dη2dt(0),\frac{d\eta_{1}}{dt}(0)<w_{0}<\frac{d\eta_{2}}{dt}(0),

    then,

    min{η1(t),η2(t)}η(t)max{η1(t),η2(t)},t>0.\min\{\eta_{1}(t),\eta_{2}(t)\}\leq\eta(t)\leq\max\{\eta_{1}(t),\eta_{2}(t)\},\quad t>0.

    The functions η1,η2\eta_{1},\eta_{2} are the key components through which the threshold curves will be defined.

2. Main Results

This section is devoted to stating precise versions of our main results. We start by specifying that for (1.2), the finite-time-breakdown is manifested in the following form,

limttc|ur(t,rc)|=,limttcρ(t,rc)=(or 0),\lim_{t\to t_{c}^{-}}|u_{r}(t,r_{c})|=\infty,\quad\lim_{t\to t_{c}^{-}}\rho(t,r_{c})=\infty\ (or\,0),

for some rc>0r_{c}>0 and finite tct_{c}. The first blowup (shock formation) can be concluded from Theorem 1.1. The density blowing up at the same time/position is a consequence of weak hyperbolicity of the pressureless Euler-Poisson systems and is a well-known phenomena for such systems. In the next section, our analysis will allow us to conclude the same. Now we state our results and follow it up with an interpretation. As mentioned in the previous section that the N=2N=2 case is critical and needs to be handled separately. Consequently, the representation of subcritical regions is also different for N=2N=2 than those for N3N\geq 3. In this section, we also provide the subcritical region sets.

Theorem 2.1 (Sharp threshold condition).

Suppose c>0c>0 and N2N\geq 2 in (1.2). If for all r>0r>0, the set of points

(r,u0(r),ϕ0r(r),u0r(r),ρ0(r))ΘN,(r,u_{0}(r),\phi_{0r}(r),u_{0r}(r),\rho_{0}(r))\in\Theta_{N},

then there is global solution. Moreover, if there exists an rc>0r_{c}>0 such that,

(rc,u0(rc),ϕ0r(rc),u0r(rc),ρ0(rc))ΘN,(r_{c},u_{0}(r_{c}),\phi_{0r}(r_{c}),u_{0r}(r_{c}),\rho_{0}(r_{c}))\notin\Theta_{N},

then there is finite-time-breakdown. Here,

ΘN{(α,x,y,z,ω):y/α<c/N,ω>0},\Theta_{N}\subseteq\{(\alpha,x,y,z,\omega):y/\alpha<c/N,\omega>0\},

is as in Definition 2.3 (for N=2N=2) and in Definition 2.2 (for N3N\geq 3).

We will describe the subcritical region set with the help of the quantity (1.3). Motivated by this, we set, for a point (α,x,y,z,ω)(\alpha,x,y,z,\omega),

(2.1) a:=xz+kyαω.\displaystyle a:=\frac{xz+ky}{\alpha\omega}.
Definition 2.2.

A point (α,x,y,z,ω)ΘN(\alpha,x,y,z,\omega)\in\Theta_{N} (N3N\geq 3) if and only if the following holds,

  • For akN2((ym,NcNyα)N2N1,(yM,NcNyα)N2N1),a\in\frac{k}{N-2}\left(\left(\frac{y_{m,N}}{\frac{c}{N}-\frac{y}{\alpha}}\right)^{\frac{N-2}{N}}-1,\left(\frac{y_{M,N}}{\frac{c}{N}-\frac{y}{\alpha}}\right)^{\frac{N-2}{N}}-1\right),

    (2.2) 1max{η1(0),η2(0)}<ω<1min{η1(0),η2(0)},for x0,zkyαa(dη1dt(0),dη2dt(0)),for x=0.\displaystyle\begin{aligned} &\frac{1}{\max\{\eta_{1}(0),\eta_{2}(0)\}}<\omega<\frac{1}{\min\{\eta_{1}(0),\eta_{2}(0)\}},\quad\text{for }x\neq 0,\\ &z\in\frac{ky}{\alpha a}\left(\frac{d\eta_{1}}{dt}(0),\frac{d\eta_{2}}{dt}(0)\right),\quad\text{for }x=0.\end{aligned}
Definition 2.3.

A point (α,x,y,z,ω)Θ2(\alpha,x,y,z,\omega)\in\Theta_{2} if and only if the following holds,

  • For ak2(ln(ym,2c2yα),ln(yM,2c2yα)),a\in\frac{k}{2}\left(\ln\left(\frac{y_{m,2}}{\frac{c}{2}-\frac{y}{\alpha}}\right),\ln\left(\frac{y_{M,2}}{\frac{c}{2}-\frac{y}{\alpha}}\right)\right),

    (2.3) 1max{η1(0),η2(0)}<ω<1min{η1(0),η2(0)},for x0,zkyαa(dη1dt(0),dη2dt(0)),for x=0.\displaystyle\begin{aligned} &\frac{1}{\max\{\eta_{1}(0),\eta_{2}(0)\}}<\omega<\frac{1}{\min\{\eta_{1}(0),\eta_{2}(0)\}},\quad\text{for }x\neq 0,\\ &z\in\frac{ky}{\alpha a}\left(\frac{d\eta_{1}}{dt}(0),\frac{d\eta_{2}}{dt}(0)\right),\quad\text{for }x=0.\end{aligned}

Here, 0<ym,N<cNyα<yM,N0<y_{m,N}<\frac{c}{N}-\frac{y}{\alpha}<y_{M,N} are the roots of the equation,

kln(y)+kc2y=R2,k\ln(y)+\frac{kc}{2y}=R_{2},

and

2kN2y12N+kcNy2N=RN,N3,\frac{2k}{N-2}y^{1-\frac{2}{N}}+\frac{kc}{N}y^{\frac{2}{N}}=R_{N},\quad N\geq 3,

with

RN=RN(u0(α)α,cNϕ0r(α)α).R_{N}=R_{N}\left(\frac{u_{0}(\alpha)}{\alpha},\frac{c}{N}-\frac{\phi_{0r}(\alpha)}{\alpha}\right).

The explicit expression for RN(,)R_{N}(\cdot,\cdot) is in (1.6), also in (4.7). The functions

ηi=ηi(t;u0(α)α,ϕ0r(α)α,A0(α)),i=1,2,\eta_{i}=\eta_{i}\left(t;\frac{u_{0}(\alpha)}{\alpha},\frac{\phi_{0r}(\alpha)}{\alpha},A_{0}(\alpha)\right),\quad i=1,2,

are defined via a second order linear IVP and are positive for all except at infinitely many discrete times.

We now give an interpretation of Theorem 2.1. Similar to existing threshold results, Theorem 2.1 enables us to check pointwise whether or not the given initial data lies in the subcritical region. Not only this, we can in fact construct the subcritical region piece by piece. Our result enables us to divide the 5\mathbb{R}^{5} space of (r,u0,ϕ0r,u0r,ρ0)(r,u_{0},\phi_{0r},u_{0r},\rho_{0}) into a 3D3D space and a plane, that is, (r,u0,ϕ0r)(r,u_{0},\phi_{0r}) (say P1P_{1}) and (u0r,ρ0)(u_{0r},\rho_{0}) (say P2P_{2}). For each point in P1P_{1}, we can construct the subricitical region on P2P_{2} through A0A_{0}. Let (α,x,y):=(r,u0,ϕ0r)(\alpha,x,y):=(r,u_{0},\phi_{0r}) be any point in P1P_{1}. Note that ym,N,yM,Ny_{m,N},y_{M,N} can now be calculated using α,x,y\alpha,x,y and hence, one can now explicitly find the interval given in Definition 2.2. For any aa in that interval, (2.1) outputs a line in P2(z,ω)P_{2}(z,\omega). In other words, by fixing an aa, we have a linear relation between (u0r,ρ0)(u_{0r},\rho_{0}). For this value of aa, we can find the two constants, η1(0),η2(0)\eta_{1}(0),\eta_{2}(0). Indeed for fixed α,x,y,a\alpha,x,y,a, the functions ηi\eta_{i} are known functions. Finally, (2.2) gives the desired portion of the line on P2P_{2} that forms part of the subcritical region, as and according to whether xx is zero or not. We can do this for all aa in the above interval and obtain the complete part of the subcritical region for the fixed point (α,x,y)(\alpha,x,y) in P1P_{1}. Carrying out this procedure for each point in P1P_{1} gives the entire subcritical region.

The next two results give a complete picture of the zero background case with dimension greater than or equal to three.

Theorem 2.4 (Sufficient condition for blow-up for N3N\geq 3).

Suppose c=0c=0 and N3N\geq 3 in (1.2). If there exists an rc>0r_{c}>0 such that,

A0(rc)kN2,A_{0}(r_{c})\leq-\frac{k}{N-2},

then there is finite-time-breakdown.

Remark 2.5.

This result gives an easy criteria to check breakdown. Moreover, it shows the criticality at N=2N=2. For N3N\geq 3, there is a flat cutoff that bounds the subcritical region from one side. However, this flat cutoff is absent in N=2N=2 where A0(r)A_{0}(r) can take any negative value and still possibly lie in the subcritical region.

In the next theorem, we give a full picture of what happens for initial data which satisfies A0(r)>kN2A_{0}(r)>-\frac{k}{N-2} for all r>0r>0.

Theorem 2.6 (Global solution for N3N\geq 3 with zero background).

Suppose c=0c=0 and N3N\geq 3 in (1.2). If for all r>0r>0, the set of points

(r,u0(r),ϕ0r(r),u0r(r),ρ0(r))ΣN{(α,x,y,z,0):x>0,zkyx},(r,u_{0}(r),\phi_{0r}(r),u_{0r}(r),\rho_{0}(r))\in\Sigma_{N}\cup\left\{(\alpha,x,y,z,0):x>0,z\geq-\frac{ky}{x}\right\},

where ΣN{(α,x,y,z,ω):y<0,ω>0,xz+kyαω>kN2}\Sigma_{N}\subseteq\left\{(\alpha,x,y,z,\omega):y<0,\omega>0,\ \frac{xz+ky}{\alpha\omega}>-\frac{k}{N-2}\right\} is as in Definition 2.7, then there is global solution.

Moreover, if there exists an rc>0r_{c}>0 such that

(rc,u0(rc),ϕ0r(rc),u0r(rc),ρ0(rc))ΣN{(α,x,y,z,0):x>0,zkyx},(r_{c},u_{0}(r_{c}),\phi_{0r}(r_{c}),u_{0r}(r_{c}),\rho_{0}(r_{c}))\notin\Sigma_{N}\cup\left\{(\alpha,x,y,z,0):x>0,z\geq-\frac{ky}{x}\right\},

then there is finite-time-breakdown.

The quantity aa in the Definitions 2.7, 2.9 below is the same as in (2.1).

Definition 2.7.

A point (α,x,y,z,ω)ΣN(\alpha,x,y,z,\omega)\in\Sigma_{N} (for N3N\geq 3) if and only if one of the following holds,

  • For a(kN2,0)a\in\left(-\frac{k}{N-2},0\right),

    (2.4) z>kyx+αaxη1(0),for x0,z>kyαadη1dt(0),for x=0.\displaystyle\begin{aligned} &z>-\frac{ky}{x}+\frac{\alpha a}{x\eta_{1}(0)},\quad\text{for }x\neq 0,\\ &z>\frac{ky}{\alpha a}\frac{d\eta_{1}}{dt}(0),\quad\text{for }x=0.\end{aligned}
  • For a=0a=0,

    (2.5) z=kyx,ω>1η2(0),for x<0,z=kyx,ω>0,for x>0.\displaystyle\begin{aligned} &z=-\frac{ky}{x},\ \omega>\frac{1}{\eta_{2}(0)},\quad\text{for }x<0,\\ &z=-\frac{ky}{x},\ \omega>0,\quad\text{for }x>0.\end{aligned}
  • For a(0,kN2((αyNy)12N1))a\in\left(0,\frac{k}{N-2}\left(\left(\frac{-\alpha y_{N}}{y}\right)^{1-\frac{2}{N}}-1\right)\right),

    (2.6) kyx+αaxη1(0)<z<kyx+αaxη2(0),for x<0,ω>0,for x>0.\displaystyle\begin{aligned} &-\frac{ky}{x}+\frac{\alpha a}{x\eta_{1}(0)}<z<-\frac{ky}{x}+\frac{\alpha a}{x\eta_{2}(0)},\quad\text{for }x<0,\\ &\omega>0,\quad\text{for }x>0.\end{aligned}
  • For a[kN2((αyNy)12N1),)a\in\left[\frac{k}{N-2}\left(\left(\frac{-\alpha y_{N}}{y}\right)^{1-\frac{2}{N}}-1\right),\infty\right),

    (2.7) ω>0,for x>0.\displaystyle\begin{aligned} &\omega>0,\quad\text{for }x>0.\end{aligned}

Here, 0<yα<yN=yN(u0(α)α,ϕ0α(α)α)=((N2)2k)NN2(RN(u0(α)α,ϕ0α(α)α))NN20<-\frac{y}{\alpha}<y_{N}=y_{N}\left(\frac{u_{0}(\alpha)}{\alpha},\frac{\phi_{0\alpha}(\alpha)}{\alpha}\right)=\left(\frac{(N-2)}{2k}\right)^{\frac{N}{N-2}}\left(R_{N}\left(\frac{u_{0}(\alpha)}{\alpha},\frac{-\phi_{0\alpha}(\alpha)}{\alpha}\right)\right)^{\frac{N}{N-2}}. The explicit expression for RN(,)R_{N}(\cdot,\cdot) is in (4.7) (with c=0c=0). The functions

ηi=ηi(t;u0(α)α,ϕ0r(α)α,A0(α)),i=1,2,\eta_{i}=\eta_{i}\left(t;\frac{u_{0}(\alpha)}{\alpha},\frac{\phi_{0r}(\alpha)}{\alpha},A_{0}(\alpha)\right),\quad i=1,2,

are defined later via linear IVPs.

Theorem 2.8 (Global solution for N=2N=2 with zero background).

Suppose c=0c=0 and N=2N=2 in (1.2). If for all r>0r>0, the set of points

(r,u0(r),ϕ0r(r),u0r(r),ρ0(r))Σ2{(α,x,y,z,0):x>0,zkyx},(r,u_{0}(r),\phi_{0r}(r),u_{0r}(r),\rho_{0}(r))\in\Sigma_{2}\cup\left\{(\alpha,x,y,z,0):x>0,z\geq-\frac{ky}{x}\right\},

where Σ2{(α,x,y,z,ω):y<0,ω>0}\Sigma_{2}\subseteq\left\{(\alpha,x,y,z,\omega):y<0,\omega>0\right\} is as in Definition 2.9, then there is global solution.

Moreover, if there exists an rc>0r_{c}>0 such that

(rc,u0(rc),ϕ0r(rc),u0r(rc),ρ0(rc))Σ2{(α,x,y,z,0):x>0,zkyx},(r_{c},u_{0}(r_{c}),\phi_{0r}(r_{c}),u_{0r}(r_{c}),\rho_{0}(r_{c}))\notin\Sigma_{2}\cup\left\{(\alpha,x,y,z,0):x>0,z\geq-\frac{ky}{x}\right\},

then there is finite time breakdown.

Definition 2.9.

A point (α,x,y,z,ω)Σ2(\alpha,x,y,z,\omega)\in\Sigma_{2} if and only if one of the following holds,

  • For a(,0)a\in\left(-\infty,0\right),

    (2.8) z>kyx+αaxη1(0),for x0,z>kyαadη1(0)dt,for x=0.\displaystyle\begin{aligned} &z>-\frac{ky}{x}+\frac{\alpha a}{x\eta_{1}(0)},\quad\text{for }x\neq 0,\\ &z>\frac{ky}{\alpha a}\frac{d\eta_{1}(0)}{dt},\quad\text{for }x=0.\end{aligned}
  • For a=0a=0,

    (2.9) z=kyx,ω>1η2(0),for x<0,z=kyx,ω>0,for x>0.\displaystyle\begin{aligned} &z=-\frac{ky}{x},\ \omega>\frac{1}{\eta_{2}(0)},\quad\text{for }x<0,\\ &z=-\frac{ky}{x},\ \omega>0,\quad\text{for }x>0.\end{aligned}
  • For a(0,k2ln(αy2y))a\in\left(0,\frac{k}{2}\ln\left(\frac{-\alpha y_{2}}{y}\right)\right),

    (2.10) kyx+αaxη1(0)<z<kyx+αaxη2(0),for x<0,ω>0,for x>0.\displaystyle\begin{aligned} &-\frac{ky}{x}+\frac{\alpha a}{x\eta_{1}(0)}<z<-\frac{ky}{x}+\frac{\alpha a}{x\eta_{2}(0)},\quad\text{for }x<0,\\ &\omega>0,\quad\text{for }x>0.\end{aligned}
  • For a[k2ln(αy2y),)a\in\left[\frac{k}{2}\ln\left(\frac{-\alpha y_{2}}{y}\right),\infty\right),

    (2.11) ω>0,for x>0.\displaystyle\begin{aligned} &\omega>0,\quad\text{for }x>0.\end{aligned}

Here, 0<yα<y2=y2(u0(α)α,ϕ0α(α)α)=eR2(u0(α)α,ϕ0α(α)α)k0<-\frac{y}{\alpha}<y_{2}=y_{2}\left(\frac{u_{0}(\alpha)}{\alpha},\frac{\phi_{0\alpha}(\alpha)}{\alpha}\right)=e^{\frac{R_{2}\left(\frac{u_{0}(\alpha)}{\alpha},\frac{-\phi_{0\alpha}(\alpha)}{\alpha}\right)}{k}}. The explicit expression for R2(,)R_{2}(\cdot,\cdot) is in (4.7) (with c=0c=0).

This paper is arranged as follows. Section 3 entails some preliminary calculations showing how to reduce the full dynamics to a weakly coupled system of four ODEs along characteristics. Section 4 is devoted to the analysis of the system of two of the four ODEs, that are decoupled from the other two, obtaining a closed form of trajectory curves with related properties established for later use. Section 5 entails conditions under which the velocity gradient/density blow up. Section 6 is devoted to constructions of the precise subcritical region leading to global solutions, proving Theorem 2.1. The zero-background case is analysed in Section 7. Finally concluding remarks are given in Section 8.

3. Preliminary calculations

From Theorem 1.1, we note that to extend the local solutions, we need to obtain bounds on the gradient of velocity. The following result, that was derived by the authors in [25], designates the quantities that need to be bounded for (1.2).

Lemma 3.1.

Suppose f(𝐱)=g(r)f(\mathbf{x})=g(r) be a radially symmetric function. Then for 𝐱0\mathbf{x}\neq 0, 2f\nabla^{2}f has exactly two eigenvalues: grrg_{rr} and gr/rg_{r}/r. Also, xx is an eigenvector corresponding to grrg_{rr} and xx^{\perp} is the eigenspace for the other eigenvalue.

Though the proof has already been given by the authors in [25], we include it here for the sake of completion.

Proof.

Let r=|x|r=|x|. Taking gradient of ff, we have

f=grxr.\nabla f=g_{r}\frac{x}{r}.

Taking gradient again,

2f=(grrgrr)xxr2+grr𝕀.\nabla^{2}f=\left(g_{rr}-\frac{g_{r}}{r}\right)\frac{x\otimes x}{r^{2}}+\frac{g_{r}}{r}\mathbb{I}.

Note that (2f)x=grrx(\nabla^{2}f)x=g_{rr}x and (2f)v=grrv(\nabla^{2}f)v=\frac{g_{r}}{r}v for any vxv\in x^{\perp}. This completes the proof of the Lemma. ∎

Through this Lemma, we conclude that the Hessians of radial functions are diagonalizable using the same matrix of eigenvectors. From Theorem 1.1, we know that for solutions to persist for all time, the gradient matrix of velocity should be bounded. Given the velocity is radial vector field, we have the following relation for 𝐮\mathbf{u}, uu in (1.1b), (1.2b) respectively,

𝐮=u𝐱r.\displaystyle\mathbf{u}=u\frac{\mathbf{x}}{r}.

Hence, the two eigenvalues of 𝐮\nabla\mathbf{u} will be uru_{r} and u/ru/r. Therefore, to ensure that the gradient of velocity is bounded, we need to control the two quantities: ur,u/ru_{r},u/r.

Inspired by this, we will obtain ODE along the characteristic path,

(3.1) {(t,X(t)):dXdt=u(t,X),X(0)=β},\displaystyle\left\{(t,X(t)):\frac{dX}{dt}=u(t,X),\ X(0)=\beta\right\},

for the desired quantities. Rearranging (1.2a), we obtain,

(3.2) ρt+(ρu)r=(N1)ρur.\displaystyle\rho_{t}+(\rho u)_{r}=-(N-1)\rho\frac{u}{r}.

Note that by (1.2c),

ϕrr=(N1)ϕrr+ρc.-\phi_{rr}=(N-1)\frac{\phi_{r}}{r}+\rho-c.

Taking spatial derivative of (1.2b) and using the above equation we obtain,

urt+uurr+ur2\displaystyle u_{rt}+uu_{rr}+u_{r}^{2} =kϕrr\displaystyle=-k\phi_{rr}
(3.3) =k(N1)ϕrr+k(ρc).\displaystyle=k(N-1)\frac{\phi_{r}}{r}+k(\rho-c).

Next using (1.2b),

(ur)t+u(ur)r\displaystyle\left(\frac{u}{r}\right)_{t}+u\left(\frac{u}{r}\right)_{r} =ut+uurru2r2\displaystyle=\frac{u_{t}+uu_{r}}{r}-\frac{u^{2}}{r^{2}}
(3.4) =kϕrru2r2.\displaystyle=-k\frac{\phi_{r}}{r}-\frac{u^{2}}{r^{2}}.

Upon integrating (1.2c) from 0 to rr,

(3.5) rN1ϕr+limr0+rN1ϕr\displaystyle-r^{N-1}\phi_{r}+\lim_{r\to 0^{+}}r^{N-1}\phi_{r} =0r(ρc)yN1𝑑y.\displaystyle=\int_{0}^{r}(\rho-c)y^{N-1}\,dy.

Local well-posedness requires a boundary condition at the origin. The convention is to assume zero boundary conditions, that is, the above limit is zero. Similarly the density flux (ρurN1\rho ur^{N-1}) approaches zero, which signifies that there is no loss of material at the origin. Taking time derivative of (3.5),

rN1ϕrt\displaystyle-r^{N-1}\phi_{rt} =0rρtyN1𝑑y\displaystyle=\int_{0}^{r}\rho_{t}y^{N-1}\,dy
=0r(ρuyN1)y𝑑yfrom (1.2a)\displaystyle=-\int_{0}^{r}(\rho uy^{N-1})_{y}\,dy\qquad\text{from \eqref{mainsysmass}}
=ρurN1+limr0+ρurN1\displaystyle=-\rho ur^{N-1}+\lim_{r\to 0^{+}}\rho ur^{N-1}
=ρurN1.\displaystyle=-\rho ur^{N-1}.

As a consequence, we get ϕrt=ρu\phi_{rt}=\rho u. We use this and the previously obtained expression for ϕrr\phi_{rr} in the below calculation.

(ϕrr)t+u(ϕrr)r\displaystyle\left(\frac{\phi_{r}}{r}\right)_{t}+u\left(\frac{\phi_{r}}{r}\right)_{r} =ϕrtr+uϕrrrurϕrr\displaystyle=\frac{\phi_{rt}}{r}+u\frac{\phi_{rr}}{r}-\frac{u}{r}\frac{\phi_{r}}{r}
=ρur(N1)urϕrr(ρc)ururϕrr\displaystyle=\frac{\rho u}{r}-(N-1)\frac{u}{r}\frac{\phi_{r}}{r}-(\rho-c)\frac{u}{r}-\frac{u}{r}\frac{\phi_{r}}{r}
(3.6) =Nurϕrr+cur.\displaystyle=-N\frac{u}{r}\frac{\phi_{r}}{r}+c\frac{u}{r}.

Set

(3.7) p:=ur,q:=ur,s:=ϕrr.\displaystyle p:=u_{r},\quad q:=\frac{u}{r},\quad s:=-\frac{\phi_{r}}{r}.

Equations (3.2), (3.3), (3.4) and (3.6) can be used to obtain an ODE system,

(3.8a) ρ=(N1)ρqpρ,\displaystyle\rho^{\prime}=-(N-1)\rho q-p\rho,
(3.8b) p=p2k(N1)s+k(ρc),\displaystyle p^{\prime}=-p^{2}-k(N-1)s+k(\rho-c),
(3.8c) q=ksq2,\displaystyle q^{\prime}=ks-q^{2},
(3.8d) s=q(c+Ns),\displaystyle s^{\prime}=-q(c+Ns),

with initial data ρ0:=ρ(0,β),p0:=p(0,β),q0:=q(0,β),s0:=s(0,β)\rho_{0}:=\rho(0,\beta),p_{0}:=p(0,\beta),q_{0}:=q(0,\beta),s_{0}:=s(0,\beta) respectively. Here, indicates differentiation along the path (3.1). Note that we use the notation ρ\rho for the density as in (1.2a), which is a function of time and the spatial variable, as well as for the solution to the ODE (3.8a), which is another function of time only (for a fixed parameter β\beta). Similarly for the notation ρ0\rho_{0}. However, it will be clear from context whether we are referring to ρ\rho as the solution of the main PDE system (1.2) or as the unknown in (3.8a).

4. The qsq-s system

Note that (3.8c), (3.8d) are decoupled from (3.8a), (3.8b). In fact, a closed form for the trajectory curve for equations (3.8c), (3.8d) can be obtained. We have the following Lemma which describes the behavior of the system (3.8c),(3.8d).

Lemma 4.1.

Global solution to (3.8c), (3.8d):

  1. (1)

    s,qs,q exist for all time and are uniformly bounded if and only if s0>cNs_{0}>-\frac{c}{N}. In particular, if s0cNs_{0}\leq-\frac{c}{N}, then qq\to-\infty in finite time.

  2. (2)

    When s0>cNs_{0}>-\frac{c}{N}, the solutions lie on bounded trajectory curves satisfying,

    (4.1) 1(s+c2)(q2+kc2+k(s+c2)ln(s+c2))=constant,N=2,\displaystyle\frac{1}{\left(s+\frac{c}{2}\right)}\left(q^{2}+\frac{kc}{2}+k\left(s+\frac{c}{2}\right)\ln\left(s+\frac{c}{2}\right)\right)=\text{constant},\qquad N=2,
    (4.2) (s+cN)2N(q2+kcN+2k(s+cN)N2)=constant,N>2.\displaystyle\left(s+\frac{c}{N}\right)^{-\frac{2}{N}}\left(q^{2}+\frac{kc}{N}+\frac{2k\left(s+\frac{c}{N}\right)}{N-2}\right)=\text{constant},\qquad N>2.
  3. (3)

    When s0>cNs_{0}>-\frac{c}{N}, the solutions are periodic and the trajectories rotate clockwise on the (q,s)(q,s) plane as time progresses.

We will time and again use the variable,

(4.3) s~:=s+cN,s~0:=s0+cN,\displaystyle\tilde{s}:=s+\frac{c}{N},\quad\tilde{s}_{0}:=s_{0}+\frac{c}{N},

and (4.4b) instead of ss and (3.8d) as the situation demands since this will make calculations simpler and easy to understand. The equivalent q,s~q,\tilde{s} system is,

(4.4a) q=ks~kcNq2,\displaystyle q^{\prime}=k\tilde{s}-\frac{kc}{N}-q^{2},
(4.4b) s~=qNs~.\displaystyle\tilde{s}^{\prime}=-qN\tilde{s}.

Proof of Lemma 4.1: It seems intuitive to use s~\tilde{s} instead of ss. We first prove the only if part of the first assertion through a contradiction argument. From (4.4b), one can see that s~\tilde{s} maintains sign as long as qq exists. Now suppose s~00\tilde{s}_{0}\leq 0. This implies s~0\tilde{s}\leq 0 as long as qq exists. For the sake of contradiction, suppose qq remains bounded for all time. From (4.4a) we have,

qkcNq2min{kcN,q2}.q^{\prime}\leq-\frac{kc}{N}-q^{2}\leq\min\left\{-\frac{kc}{N},-q^{2}\right\}.

Since qq^{\prime} is bounded above by a strictly negative constant, there exists some finite time, t0t_{-}\geq 0, such that q(t)<0q(t_{-})<0. Once qq is negative, the dynamics of qq is that of a Ricatti equation,

q<q2,q(t)<0,q^{\prime}<-q^{2},\qquad q(t_{-})<0,

and therefore, limttcq=\lim_{t\to t_{c}^{-}}q=-\infty for some tc<t1/q(t)t_{c}<t_{-}-1/q(t_{-}). This is a contradiction.

Now we prove the if part of the first assertion along with the second assertion. We derive the equation of trajectory assuming s~0>0\tilde{s}_{0}>0, which in turn implies s~>0\tilde{s}>0 for all t>0t>0 as long as the solutions to (4.4) exist. Dividing (4.4a) by (4.4b) we have the following,

qdqds~=kN+kcN2s~+q2Ns~,\displaystyle q\frac{dq}{d\tilde{s}}=-\frac{k}{N}+\frac{kc}{N^{2}\tilde{s}}+\frac{q^{2}}{N\tilde{s}},
d(q2)ds~2q2Ns~=2kN2(cs~N).\displaystyle\frac{d(q^{2})}{d\tilde{s}}-\frac{2q^{2}}{N\tilde{s}}=\frac{2k}{N^{2}}\left(\frac{c}{\tilde{s}}-N\right).

Using the integrating factor s~2N\tilde{s}^{-\frac{2}{N}},

dds~(q2s~2N)=2kN2(cs~12NNs~2N).\displaystyle\frac{d}{d\tilde{s}}\left(q^{2}\tilde{s}^{-\frac{2}{N}}\right)=\frac{2k}{N^{2}}\left(c\tilde{s}^{-1-\frac{2}{N}}-N\tilde{s}^{-\frac{2}{N}}\right).

Owing to the last term, we see that N=2N=2 case has to be handled separately before integrating the above equation. First, we move on with N>2N>2 case. Integrating the equation above, we obtain

RN(q,s~):=s~2/N(q2+2ks~N2+kcN)=constant.\displaystyle R_{N}(q,\tilde{s}):=\tilde{s}^{-2/N}\left(q^{2}+\frac{2k\tilde{s}}{N-2}+\frac{kc}{N}\right)=\text{constant}.

For N=2N=2, upon integration,

R2(q,s~):=q2s~+kln(s~)+kc2s~=constant.\displaystyle R_{2}(q,\tilde{s}):=\frac{q^{2}}{\tilde{s}}+k\ln(\tilde{s})+\frac{kc}{2\tilde{s}}=\text{constant}.

From the trajectory equation for N>2N>2, we have for s~0>0\tilde{s}_{0}>0,

s~<(RN(N2)2k)NN2,\displaystyle\tilde{s}<\left(\frac{R_{N}(N-2)}{2k}\right)^{\frac{N}{N-2}},
|q|<s~1/NRN<RNN2N4(N22k)NN2.\displaystyle|q|<\tilde{s}^{1/N}\sqrt{R_{N}}<R_{N}^{\frac{N}{2N-4}}\left(\frac{N-2}{2k}\right)^{\frac{N}{N-2}}.

Hence, solutions are uniformly bounded. Very similarly, bounds on q,s~q,\tilde{s} can be derived for N=2N=2 as well. This completes the proof to the first and second assertions.

Note that trajectory equation is invariant under the transformation qqq\to-q. Putting this together with the fact that linearized system around the only critical point, (0,c/N)(0,c/N), has imaginary eigenvalues, we obtain that the solution trajectories starting at any s~0>0\tilde{s}_{0}>0 are closed curves around the critical point. Hence, the solutions are periodic. The direction of motion of trajectory as time progresses is clear from (4.4b) since s~<0\tilde{s}^{\prime}<0 for q>0q>0 and s~>0\tilde{s}^{\prime}>0 for q<0q<0. ∎

Refer to caption
Figure 1. qsq-s phase diagram for k=c=1,N=4k=c=1,N=4. Trajectories move clockwise with increasing time.
Corollary 4.2.

q,sq,s in system (3.8c), (3.8d) exist for all time. In particular, assertions 22 and 33 of Lemma 4.1 apply.

Proof.

Using the expression of ss from (3.7) in (3.5) at t=0t=0, we have,

s0\displaystyle s_{0} =ϕ0r(β)β=1βN0β(ρ0(ξ)c)ξN1𝑑ξ\displaystyle=-\frac{\phi_{0r}(\beta)}{\beta}=\frac{1}{\beta^{N}}\int_{0}^{\beta}(\rho_{0}(\xi)-c)\xi^{N-1}\,d\xi
>cN.\displaystyle>-\frac{c}{N}.

Therefore, by Lemma 4.1, we conclude the result. ∎

Remark 4.3.

This result simply asserts that in multidimension, Poisson forcing is enough to avoid concentrations at the origin, irrespective of how negative the initial velocity is. This is not the case in one dimension. A more detailed discussion comparing 1D and multi-D cases is included in Section 8 (conclusion).

At this point, we set up the notation for the trajectory curve as obtained in Lemma 4.1. It will be used in this as well as in later sections.

(4.7) RN(q,s~)={s~2/N(q2+2ks~N2+kcN),N>2,q2s~+kln(s~)+kc2s~,N=2.\displaystyle R_{N}(q,\tilde{s})=\left\{\begin{array}[]{c}\tilde{s}^{-2/N}\left(q^{2}+\frac{2k\tilde{s}}{N-2}+\frac{kc}{N}\right),\quad N>2,\\ \frac{q^{2}}{\tilde{s}}+k\ln(\tilde{s})+\frac{kc}{2\tilde{s}},\quad N=2.\end{array}\right.

From Lemma 4.1,

RN(q,s~)=RN(q0,s~0).R_{N}(q,\tilde{s})=R_{N}(q_{0},\tilde{s}_{0}).

Since RNR_{N} is a constant, we will use the notation RNR_{N} to denote it as a single constant or RN(q,s~)R_{N}(q,\tilde{s}) for a function of (q,s~)(q,\tilde{s}). The next Lemma pertaining to (4.4) will be useful in proving the blowup/global existence results in Section 6.

Lemma 4.4.

Consider the system (3.8c) and (3.8d). Denote the two coordinates at which the trajectory curve intersects the ss-axis by (0,smin)(0,s_{min}) and (0,smax)(0,s_{max}). Then

0<smin<smax.0<-s_{min}<s_{max}.

Moreover, the algebraic equations,

RN(0,y)=RN(q0,s~0),y>0,R_{N}(0,y)=R_{N}(q_{0},\tilde{s}_{0}),\ y>0,

have exactly two roots, which are smin+cNs_{min}+\frac{c}{N} and smax+cNs_{max}+\frac{c}{N}.

Proof.

We give a proof for N>2N>2 only. Very similar arguments apply for N=2N=2. Note that for s~\tilde{s}, the statement is equivalent to saying,

(4.8) 0<cNs~min<s~maxcN,\displaystyle 0<\frac{c}{N}-\tilde{s}_{min}<\tilde{s}_{max}-\frac{c}{N},

where s~max:=smax+c/N\tilde{s}_{max}:=s_{max}+c/N, s~min:=smin+c/N\tilde{s}_{min}:=s_{min}+c/N.

Since the solution trajectories, (q,s~)(q,\tilde{s}), are bounded periodic orbits around (0,c/N)(0,c/N), (0,s~min)(0,\tilde{s}_{min}) and (0,s~max)(0,\tilde{s}_{max}) lie on either side of (0,c/N)(0,c/N) implying that c/Ns~min>0c/N-\tilde{s}_{min}>0. We now analyze the following function,

g(s~):=2kN2s~12N+kcNs~2N.g(\tilde{s}):=\frac{2k}{N-2}\tilde{s}^{1-\frac{2}{N}}+\frac{kc}{N}\tilde{s}^{-\frac{2}{N}}.

The function gg is essentially the left-hand-side of (4.2) with q=0q=0. The aim is to show that g(s~)=RNg(\tilde{s})=R_{N} has exactly two roots.

gg goes to infinity as s~0\tilde{s}\to 0 or s~\tilde{s}\to\infty. We also have,

dgds~=2kNs~2N2kcN2s~12N.\frac{dg}{d\tilde{s}}=\frac{2k}{N}\tilde{s}^{-\frac{2}{N}}-\frac{2kc}{N^{2}}\tilde{s}^{-1-\frac{2}{N}}.

Setting the derivative as zero, we obtain that the minimum is unique and is obtained at s~=cN\tilde{s}=\frac{c}{N}. Hence, g:(0,)(g(c/N),)g:(0,\infty)\longrightarrow(g(c/N),\infty).

Given the structure of gg, we have that the algebraic equation,

g(s~)=RN,g(\tilde{s})=R_{N},

has exactly two roots if the constant, RN>g(cN)R_{N}>g\left(\frac{c}{N}\right). These roots correspond to s~min\tilde{s}_{min} and s~max\tilde{s}_{max}. Clearly, if s~max2cN\tilde{s}_{max}\geq\frac{2c}{N}, the second inequality in (4.8) stands true since nonpositive reals are not in the domain of gg. In other words, we only need to prove the inequality for RN<g(2cN)R_{N}<g\left(\frac{2c}{N}\right) or equivalently, when s~max<2c/N\tilde{s}_{max}<2c/N. We achieve this by making use of the sign of the third derivative of gg,

d3gds~3=4k(N+2)N4(s~cN(N+1))s~32N.\frac{d^{3}g}{d\tilde{s}^{3}}=\frac{4k(N+2)}{N^{4}}\left(\tilde{s}-\frac{c}{N}\left(N+1\right)\right)\tilde{s}^{-3-\frac{2}{N}}.

Since,

cN(N+1)>2cN,\frac{c}{N}(N+1)>\frac{2c}{N},

we have that for all s~(0,2c/N)\tilde{s}\in(0,2c/N), d3gds~3<0\frac{d^{3}g}{d\tilde{s}^{3}}<0. Hence,

d2gds~2(s~1)>d2gds~2(c/N)>d2gds~2(s~2),s~1(0,cN),s~2(cN,2cN).\displaystyle\frac{d^{2}g}{d\tilde{s}^{2}}(\tilde{s}_{1})>\frac{d^{2}g}{d\tilde{s}^{2}}(c/N)>\frac{d^{2}g}{d\tilde{s}^{2}}(\tilde{s}_{2}),\quad\tilde{s}_{1}\in\left(0,\frac{c}{N}\right),\ \tilde{s}_{2}\in\left(\frac{c}{N},\frac{2c}{N}\right).

Consequently, for any δ<c/N\delta<c/N,

c/Nδc/Nd2gds~2(s~1)𝑑s~1>δd2gds~2(c/N)>c/Nc/N+δd2gds~2(s~2)𝑑s~2\displaystyle\int_{c/N-\delta}^{c/N}\frac{d^{2}g}{d\tilde{s}^{2}}(\tilde{s}_{1})d\tilde{s}_{1}>\delta\frac{d^{2}g}{d\tilde{s}^{2}}(c/N)>\int_{c/N}^{c/N+\delta}\frac{d^{2}g}{d\tilde{s}^{2}}(\tilde{s}_{2})d\tilde{s}_{2}
dgds~(c/Nδ)>dgds~(c/N+δ).\displaystyle-\frac{dg}{d\tilde{s}}(c/N-\delta)>\frac{dg}{d\tilde{s}}(c/N+\delta).

The second inequality is a result of the fact that dgds~(c/N)=0\frac{dg}{d\tilde{s}}(c/N)=0. Integrating the obtained inequality with respect to δ\delta with zero as the lower limit,

g(c/Nδ)>g(c/N+δ).g(c/N-\delta)>g(c/N+\delta).

Consequently, for the same shift from s~=c/N\tilde{s}=c/N the function attains a higher value on the left than on the right. Therefore, the two points, s~min,s~max\tilde{s}_{min},\tilde{s}_{max}, in the level set {s~:g(s~)=RN}\{\tilde{s}:g(\tilde{s})=R_{N}\} will be such that

cNs~min<s~maxcN.\frac{c}{N}-\tilde{s}_{min}<\tilde{s}_{max}-\frac{c}{N}.

This completes the proof. ∎

We also include a short lemma to derive a different relationship between s~\tilde{s} and qq which will be helpful in the later sections. To this end, we first define the following useful quantity,

(4.9) Γ(t)=e0tq(τ)𝑑τ.\displaystyle\Gamma(t)=e^{-\int_{0}^{t}q(\tau)\,d\tau}.
Lemma 4.5.

We have the following,

(4.10) Γ(t)=(s~(t)s~0)1N,\displaystyle\Gamma(t)=\left(\frac{\tilde{s}(t)}{\tilde{s}_{0}}\right)^{\frac{1}{N}},

for all time. In particular, 0Tq(t)𝑑t=0\int_{0}^{T}q(t)dt=0 and Γ(t)\Gamma(t) is uniformly bounded and periodic. Here, TT is the period of the system (4.4) (equivalently (3.8c),(3.8d)).

Proof.

From Corollary 4.2, we have the all time existence of s~,q\tilde{s},q and strict positivity of s~\tilde{s}. Note that Γ=qΓ\Gamma^{\prime}=-q\Gamma. We can divide this by (4.4b) to obtain,

dΓds~=ΓNs~.\frac{d\Gamma}{d\tilde{s}}=\frac{\Gamma}{N\tilde{s}}.

Upon integrating, we conclude (4.10). Uniform boundedness of e0tqe^{-\int_{0}^{t}q} follows immediately from the uniform boundedness of s~\tilde{s}. Moreover, since s~\tilde{s} is periodic with period TT, we have that for any tt\in\mathbb{R},

e0tq=e0t+Tq.e^{-\int_{0}^{t}q}=e^{-\int_{0}^{t+T}q}.

Therefore,

0\displaystyle 0 =tt+Tq\displaystyle=\int_{t}^{t+T}q
=t0q+0Tq+Tt+Tq\displaystyle=\int_{t}^{0}q+\int_{0}^{T}q+\int_{T}^{t+T}q
=0Tq.\displaystyle=\int_{0}^{T}q.

To obtain the last equality, we used the fact that qq has period TT. ∎

5. Blow up of solutions

Now we move onto the analysis of (3.8a) and (3.8b). In this section, our aim is to entail conditions under which p,ρp,\rho blow up. From Corollary 4.2, we already know that q,sq,s are uniformly bounded, periodic. Hence, the quantities in (3.8) that can blowup are pp or ρ\rho.

If ρ0=0\rho_{0}=0, then from (3.8a), ρ0\rho\equiv 0 as long as ρ,p\rho,p exist. From (3.8b),

pp2kc,p^{\prime}\leq-p^{2}-kc,

leading to Ricatti-type blow up of pp. Consequently, zero density always leads to blowup. Hence, further onwards we will assume ρ0>0\rho_{0}>0. This in turn implies ρ(t)>0\rho(t)>0 for t>0t>0 as long as ρ,p\rho,p exist. With this, we move onto the simplification of the system (3.8a), (3.8b). This simplification is along the lines of [28]. From (3.8a), (3.8b), we obtain that,

(1ρ)\displaystyle\left(\frac{1}{\rho}\right)^{\prime} =q(N1)1ρ+pρ,\displaystyle=q(N-1)\frac{1}{\rho}+\frac{p}{\rho},
(pρ)\displaystyle\left(\frac{p}{\rho}\right)^{\prime} =k(c+s(N1))1ρ+q(N1)pρ+k.\displaystyle=-k(c+s(N-1))\frac{1}{\rho}+q(N-1)\frac{p}{\rho}+k.

Noticing that the coefficients of 1/ρ,p/ρ1/\rho,p/\rho in their respective ODEs are the same, we can multiply the above ODEs by integrating factor e(N1)0tqe^{-(N-1)\int_{0}^{t}q} to obtain,

(1ρe(N1)0tq)=pρe(N1)0tq,\displaystyle\left(\frac{1}{\rho}e^{-(N-1)\int_{0}^{t}q}\right)^{\prime}=\frac{p}{\rho}e^{-(N-1)\int_{0}^{t}q},
(pρe(N1)0tq)=k(c+s(N1))1ρe(N1)0tq+ke(N1)0tq.\displaystyle\left(\frac{p}{\rho}e^{-(N-1)\int_{0}^{t}q}\right)^{\prime}=-k(c+s(N-1))\frac{1}{\rho}e^{-(N-1)\int_{0}^{t}q}+ke^{-(N-1)\int_{0}^{t}q}.

Setting

(5.1) η:=1ρΓN1,w=pρΓN1,\displaystyle\eta:=\frac{1}{\rho}\Gamma^{N-1},\qquad w=\frac{p}{\rho}\Gamma^{N-1},

we obtain a new system,

(5.2a) η=w,\displaystyle\eta^{\prime}=w,
(5.2b) w=kη(c+s(N1))+kΓN1.\displaystyle w^{\prime}=-k\eta(c+s(N-1))+k\Gamma^{N-1}.

We label the initial data as η0,w0\eta_{0},w_{0}.

Remark 5.1.

Owing to Corollary 4.2 and Lemma 4.5, the coefficients in the linear ODE system (5.2) are uniformly bounded. Hence, η,w\eta,w remain bounded and well-defined for all t(,)t\in(-\infty,\infty). Consequently, the key to existence of global solution is to ensure η(t)>0\eta(t)>0 for all t>0t>0. From (5.1), this means that p,ρp,\rho are both bounded for all t>0t>0. Conversely, if there is a finite time, tct_{c}, at which η\eta becomes zero, then

limttcρ(t)=limttc1η(t)=.\lim_{t\to t_{c}^{-}}\rho(t)=\lim_{t\to t_{c}^{-}}\frac{1}{\eta(t)}=\infty.

Moreover, since ww is bounded for all times,

limttc|p(t)|=limttc|w(t)|η(t)=.\lim_{t\to t_{c}^{-}}|p(t)|=\lim_{t\to t_{c}^{-}}\frac{|w(t)|}{\eta(t)}=\infty.

Therefore, at the time of breakdown, ρ,|p|\rho,|p| blow up together.

The above remark results in the following key proposition.

Proposition 5.2.

Suppose ρ00.\rho_{0}\neq 0. ρ,p,q,s\rho,p,q,s in (3.8) are well-defined for all t>0t>0 if and only if η(t)>0\eta(t)>0 for all t>0t>0 in (5.2). In particular, if there is a tc>0t_{c}>0 such that η(tc)=0\eta(t_{c})=0, then limttc|p(t)|=limttcρ(t)=\lim_{t\to t_{c}^{-}}|p(t)|=\lim_{t\to t_{c}^{-}}\rho(t)=\infty.

Next, we have one of the key contributions of this paper in the form of a nonlinear quantity. This quantity will be instrumental in analyzing the system (3.8a),(3.8b). Set,

(5.3) A:=qwkηs.\displaystyle A:=qw-k\eta s.

Using (3.8c),(3.8d) and (5.2),

A\displaystyle A^{\prime} =qw+qwkηskηs\displaystyle=q^{\prime}w+qw^{\prime}-k\eta^{\prime}s-k\eta s^{\prime}
=(ksq2)w+q(kη(c+s(N1))+kΓN1)kws+kqη(c+Ns)\displaystyle=(ks-q^{2})w+q\left(-k\eta(c+s(N-1))+k\Gamma^{N-1}\right)-kws+kq\eta(c+Ns)
=q2w+kqηs+kqΓN1\displaystyle=-q^{2}w+kq\eta s+kq\Gamma^{N-1}
=q(qwkηs)+kqΓN1\displaystyle=-q(qw-k\eta s)+kq\Gamma^{N-1}
(5.4) =qA+kqΓN1.\displaystyle=-qA+kq\Gamma^{N-1}.

We have,

A+qA=(kN1ΓN1).A^{\prime}+qA=-\left(\frac{k}{N-1}\Gamma^{N-1}\right)^{\prime}.

We assume N>2N>2. As evident in the calculations below, the N=2N=2 case has to be handled separately and we will tackle it at the end of the section. Upon integration and setting A0:=q0w0kη0s0A_{0}:=q_{0}w_{0}-k\eta_{0}s_{0},

AΓ1A0\displaystyle A\Gamma^{-1}-A_{0} =kN10t(Γ(s))1(ΓN1(s))𝑑s\displaystyle=-\frac{k}{N-1}\int_{0}^{t}(\Gamma(s))^{-1}\left(\Gamma^{N-1}(s)\right)^{\prime}ds
=kN1[ΓN210tq(s)(Γ(s))N2𝑑s]\displaystyle=-\frac{k}{N-1}\left[\Gamma^{N-2}-1-\int_{0}^{t}q(s)(\Gamma(s))^{N-2}ds\right]
=kN1[ΓN21+1N2(ΓN21)]\displaystyle=-\frac{k}{N-1}\left[\Gamma^{N-2}-1+\frac{1}{N-2}\left(\Gamma^{N-2}-1\right)\right]
=kN2[1ΓN2].\displaystyle=\frac{k}{N-2}\left[1-\Gamma^{N-2}\right].

Finally,

(5.5) A(Γ)=(A0+kN2)ΓkN2ΓN1,Γ(t):=e0tq>0,A(t):=A(Γ(t)).\displaystyle\begin{aligned} &A(\Gamma)=\left(A_{0}+\frac{k}{N-2}\right)\Gamma-\frac{k}{N-2}\Gamma^{N-1},\quad\Gamma(t):=e^{-\int_{0}^{t}q}>0,\\ &A(t):=A(\Gamma(t)).\end{aligned}

Here, we have abused the notation AA and Γ\Gamma by assigning both of them to two functions. A(t)A(t) is a function of time and A(Γ)A(\Gamma) a function of Γ\Gamma as in (5.5). Γ\Gamma is the argument of the function A(Γ)A(\Gamma) as well as a function itself as in (4.9) and takes positive values. It will, however, be clear from context which functions we are referring to.

Remark 5.3.

It is important to note that A(t)A(t) is connected to η,w\eta,w only through the initial data, η0,w0\eta_{0},w_{0}. In other words, (3.8c), (3.8d), (5.4) with initial data q0,s0,A0q_{0},s_{0},A_{0} constitute an IVP independent from the dynamics of (5.2). Moreover, A(t)A(t) can be explicitly solved for. Also, from Lemma 4.5, A(t)A(t) has the same period as q,sq,s.

Next, we state a sufficient condition for η\eta taking zero value in finite time.

Proposition 5.4.

If A(t)A(t) is nonnegative (or nonpositive) for all time, then there exists a time tc>0t_{c}>0 such that η(tc)=0\eta(t_{c})=0.

Proof.

Suppose A(t)0A(t)\geq 0 for all time. At the time t0t_{0} where ss is maximum, from (3.8d) we have q(t0)=0q(t_{0})=0 and,

0A(t0)=q(t0)w(t0)kη(t0)s(t0)=kη(t0)s(t0).\displaystyle 0\leq A(t_{0})=q(t_{0})w(t_{0})-k\eta(t_{0})s(t_{0})=-k\eta(t_{0})s(t_{0}).

From Lemma 4.4, s(t0)>0s(t_{0})>0. Hence, η(t0)0\eta(t_{0})\leq 0. This gives us the existence of a time tc>0t_{c}>0 with tct0t_{c}\leq t_{0} such that η(tc)=0\eta(t_{c})=0.

The proof is similar for A(t)0A(t)\leq 0 situation. ∎

Remark 5.5.

At the times where ss achieves its extrema, η\eta is a priori known and essentially depends only on A,q,sA,q,s. Remarkably, a simple computation gives η\eta at these times. Indeed when ss achieves max/min (say at t=tt=t_{\ast}), q(t)=0q(t_{\ast})=0. Hence, from (5.3), η(t)=A(t)ks(t)\eta(t_{\ast})=-\frac{A(t_{\ast})}{ks(t_{\ast})}. Moreover, since A,sA,s are periodic with same period TT, η(t)=η(t+lT),l=0,1,2,\eta(t_{\ast})=\eta(t_{\ast}+lT),l=0,1,2,\ldots.

Lemma 5.6.

Suppose N3N\geq 3.

  • If 1+A0(N2)k01+\frac{A_{0}(N-2)}{k}\leq 0, then A(t)<0A(t)<0 for all t>0t>0.

  • If 0tq1N2ln(1+A0(N2)k)\int_{0}^{t}q\leq-\frac{1}{N-2}\ln\left(1+\frac{A_{0}(N-2)}{k}\right) for all t>0t>0, then A(t)0A(t)\leq 0 for all t>0t>0.

  • If 0tq1N2ln(1+A0(N2)k)\int_{0}^{t}q\geq-\frac{1}{N-2}\ln\left(1+\frac{A_{0}(N-2)}{k}\right) for all t>0t>0, then A(t)0A(t)\geq 0 for all t>0t>0.

Proof.

The first assertion follows from the fact that if 1+A0(N2)k01+\frac{A_{0}(N-2)}{k}\leq 0, then from (5.5), A(t)<0A(t)<0 for all time.

Note that if 1+A0(N2)k>01+\frac{A_{0}(N-2)}{k}>0, then it can be readily seen from (5.5) that A(Γ)A(\Gamma) has exactly two real roots, 0 and

(5.6) κ:=(1+A0(N2)k)1N2>0.\displaystyle\kappa:=\left(1+\frac{A_{0}(N-2)}{k}\right)^{\frac{1}{N-2}}>0.

Moreover, for nonnegative arguments Γ\Gamma in (5.5), A(Γ)>0A(\Gamma)>0 for Γ(0,κ)\Gamma\in(0,\kappa) and A(Γ)<0A(\Gamma)<0 for Γ>κ\Gamma>\kappa.

Now suppose the hypothesis of the second assertion holds. Straightforward calculations then imply Γ(t)κ\Gamma(t)\geq\kappa for all tt. Therefore, A(Γ)0A(\Gamma)\leq 0 for all attainable values of Γ\Gamma. This proves the second assertion.

The third assertion is similar, only that here the hypothesis implies Γ(t)κ\Gamma(t)\leq\kappa. Therefore, A(t)0A(t)\geq 0. ∎

Corollary 5.7.

Suppose N3N\geq 3. If one of the following is true,

  • 1+A0(N2)k01+\frac{A_{0}(N-2)}{k}\leq 0,

  • 1+A0(N2)k>01+\frac{A_{0}(N-2)}{k}>0 and the two roots, y2>y1>0y_{2}>y_{1}>0, of the equation,

    (5.7) 2kyN2+kcNRNy2/N=0,(equivalently RN(0,y)=RN),\displaystyle\frac{2ky}{N-2}+\frac{kc}{N}-R_{N}y^{2/N}=0,\qquad(\text{equivalently }R_{N}(0,y)=R_{N}),

    are such that,

    κ((y1s~0)1/N,(y2s~0)1/N),\kappa\notin\left(\left(\frac{y_{1}}{\tilde{s}_{0}}\right)^{1/N},\left(\frac{y_{2}}{\tilde{s}_{0}}\right)^{1/N}\right),

then there is a time tc>0t_{c}>0 such that η(tc)=0\eta(t_{c})=0. Here, RNR_{N} is the constant in (4.7) and κ\kappa is as in (5.6).

Remark 5.8.

Note that all the conditions in the hypothesis depend only on the initial data, thereby, will lead to provide a characterization of the supercritical region for (1.2).

Proof.

If the first condition holds, then by the first assertion of Lemma 5.6, A(t)<0A(t)<0 for all t>0t>0. Proposition 5.4 leads to the conclusion.

As mentioned, (5.7) is equivalent to RN(0,y)=RN(q0,s~0)R_{N}(0,y)=R_{N}(q_{0},\tilde{s}_{0}). Hence, from Lemma 4.4, the two roots of (5.7) are the maximum and minimum attainable values of s~\tilde{s}. Now suppose the hypothesis of the second assertion holds. There could be two situations:

  1. (1)

    κ(y1/s~0)1/N(s~(t)/s~0)1/N\kappa\leq(y_{1}/\tilde{s}_{0})^{1/N}\leq(\tilde{s}(t)/\tilde{s}_{0})^{1/N} or,

  2. (2)

    κ(y2/s~0)1/N(s~(t)/s~0)1/N\kappa\geq(y_{2}/\tilde{s}_{0})^{1/N}\geq(\tilde{s}(t)/\tilde{s}_{0})^{1/N},

for all t>0t>0. Suppose (1)(1) holds. Using Lemma 4.5,

e0tq=(s~(t)s~0)1Nκ.e^{-\int_{0}^{t}q}=\left(\frac{\tilde{s}(t)}{\tilde{s}_{0}}\right)^{\frac{1}{N}}\geq\kappa.

Hence, 0tqln(κ)\int_{0}^{t}q\leq-\ln(\kappa) for all t>0t>0. Using the second assertion of Lemma 5.6, we obtain that A(t)0A(t)\leq 0 for all time. Then by Proposition 5.4, we conclude the result. Very similar arguments hold for (2) as well. ∎

We complete this section by presenting the N=2N=2 case. Upon integrating (5.4) with N=2N=2, we have,

(5.8) B(Γ):=(A0klnΓ)Γ,B(t):=B(Γ(t)),Γ(t)=e0tq.\displaystyle\begin{aligned} &B(\Gamma):=(A_{0}-k\ln\Gamma)\Gamma,\\ &B(t):=B(\Gamma(t)),\quad\Gamma(t)=e^{-\int_{0}^{t}q}.\end{aligned}

For the initial data to BB, we use the same notation, A0A_{0}. Proposition 5.4 holds as it is for BB in place of AA. Analogous to Lemma 5.6, we have the following result.

Lemma 5.9.

Suppose N=2N=2.

  • If 0tqA0k\int_{0}^{t}q\leq-\frac{A_{0}}{k} for all t>0t>0, then B(t)0B(t)\leq 0 for all t>0t>0.

  • If 0tqA0k\int_{0}^{t}q\geq-\frac{A_{0}}{k} for all t>0t>0, then B(t)0B(t)\geq 0 for all t>0t>0.

The proof is very similar to that of Lemma 5.6 and fairly straightforward given the simplicity of (5.8).

Remark 5.10.

Interestingly, the result of Lemma 5.9 can be related to that of Lemma 5.6 by replacing,

1N2ln(1+A0(N2)k),-\frac{1}{N-2}\ln\left(1+\frac{A_{0}(N-2)}{k}\right),

with

limβ2+1β2ln(1+A0(β2)k).-\lim_{\beta\to 2^{+}}\frac{1}{\beta-2}\ln\left(1+\frac{A_{0}(\beta-2)}{k}\right).

This shows that N=2N=2 is indeed a critical case.

Finally, we have the result analogous to Corollary 5.7.

Corollary 5.11.

Suppose N=2N=2. If the two roots, y2>y1>0y_{2}>y_{1}>0, of the equation,

(5.9) kyln(y)+kc2R2y=0,\displaystyle ky\ln(y)+\frac{kc}{2}-R_{2}y=0,

are such that,

eA0k(y1s~0,y2s~0),e^{\frac{A_{0}}{k}}\notin\left(\sqrt{\frac{y_{1}}{\tilde{s}_{0}}},\sqrt{\frac{y_{2}}{\tilde{s}_{0}}}\right),

then there is a finite time tc>0t_{c}>0 such that η(tc)=0\eta(t_{c})=0. Here, R2R_{2} is the constant right-hand-side in (4.7).

Once again, the proof is very similar.

Before we begin the new section, we would like to introduce some notations. tq,ts,tAt_{q},t_{s},t_{A} will be used to denote a time when q,s,Aq,s,A are zero respectively. Owing to their periodicity, there are infinitely many such times. In view of this, we will also use the notations tqi,tsi,tAit_{q}^{i},t_{s}^{i},t_{A}^{i} for i=1,2,i=1,2,\ldots to refer to more than one such times as and when it is required.

6. Global Solution

Up to this point, we have narrowed down the class of possible initial data for (3.8) that could ensure η(t)>0\eta(t)>0, or equivalently, ρ(t)<\rho(t)<\infty for all t>0t>0. This section is devoted to precisely identifying the possible initial configurations, from the set remaining from Section 5, that lead to the all-time positivity of η\eta. Therefore, from this point onwards, we will assume the negation of the hypothesis in Corollary 5.7. Hence, for this section, we assume the following for N>2N>2,

(6.1) 1+A0(N2)k>0,κ((s~mins~0)1/N,(s~maxs~0)1/N),\displaystyle 1+\frac{A_{0}(N-2)}{k}>0,\qquad\kappa\in\left(\left(\frac{\tilde{s}_{min}}{\tilde{s}_{0}}\right)^{1/N},\left(\frac{\tilde{s}_{max}}{\tilde{s}_{0}}\right)^{1/N}\right),

with κ\kappa as in (5.6) and s~min,s~max\tilde{s}_{min},\tilde{s}_{max} are the minimum and maximum attainable values of s~(t)\tilde{s}(t), which are also the roots of (5.7). For N=2N=2, we set κ=eA0k\kappa=e^{\frac{A_{0}}{k}} and s~min,s~max\tilde{s}_{min},\tilde{s}_{max} being the minimum and maximum attainable values of s~(t)\tilde{s}(t), which are the roots of (5.9). The first assumption in (6.1) is a vacuous statement for N=2N=2. Under these notational assumptions, the analysis and results of this section for the N>2N>2 case is the same as that of N=2N=2 with AA replaced by BB. Hence, in this section, we do not differentiate between the N>2N>2 and N=2N=2 cases as there is no critical behaviour with respect to the system (5.2).

Owing to the first assumption in (6.1), A(Γ)A(\Gamma) has a unique positive root equal to κ\kappa. From (4.10), the second assumption in (6.1) implies,

κ{Γ(t):t>0},\kappa\in\{\Gamma(t):t>0\}^{\circ},

where denotes the interior of the set. Since κ\kappa is the root of A(Γ)A(\Gamma), from (5.5) we conclude that A(t)A(t) changes sign. Recall that this is essentially the negation of the hypothesis of Proposition 5.4 and, therefore, AA must necessarily change sign to hope for a situation wherein η(t)>0\eta(t)>0 for all t>0t>0. See Figure 2 for a visualization of this situation.

Refer to caption
Figure 2. AA vs Γ\Gamma

From the expression for Γ(t)\Gamma(t) in (5.5), we see that the leftmost point on the green line in Figure 2 corresponds to the time when s~\tilde{s} attains its minimum and the rightmost point is when it attains the maximum. Using this fact in (4.10) and (5.5), we make an important conclusion that will be quite helpful throughout this section. For a t>0t>0,

(6.2) s~(t)=s~maxA(t)<0,s~(t)=s~minA(t)>0.\displaystyle\begin{aligned} &\tilde{s}(t)=\tilde{s}_{max}\implies A(t)<0,\\ &\tilde{s}(t)=\tilde{s}_{min}\implies A(t)>0.\end{aligned}

In most parts of this section, we will be analyzing the system (5.2). Our aim will be to obtain a necessary and sufficient condition to ensure η(t)>0\eta(t)>0 for all t>0t>0. Recall from Remark 5.5 that at the times when ss (or equivalently s~\tilde{s}) attains extrema, the values of η\eta are a priori known. That is to say, if tqt_{q} is such that q(tq)=0q(t_{q})=0, then from (5.3),

(6.3) η(tq)=A(tq)ks(tq).\displaystyle\eta(t_{q})=-\frac{A(t_{q})}{ks(t_{q})}.

In particular, when s(tq)=smins(t_{q})=s_{min}, we have η(tq)=A(tq)ksmin\eta(t_{q})=-\frac{A(t_{q})}{ks_{min}}. From Lemma 4.4, we have smin<0s_{min}<0. Using this in (6.2), we have A(tq)>0A(t_{q})>0. From (6.3), this implies η(tq)>0\eta(t_{q})>0. Likewise, at the time when ss achieves maximum, ss and AA switch signs ensuring that η\eta is again positive.

Keeping this in mind, we need to ensure that η\eta remains strictly positive at other times apart from the ones where ss attains its extreme values. To this end, we will construct two functions that form a cloud around η\eta. This will enable us to prove the positivity of η\eta.

We ensure this as follows. Suppose q0,s0,A0q_{0},s_{0},A_{0} are given. This fixes the functions q,s,Aq,s,A, which are unknowns of a closed ODE system, whose properties have been analyzed. Fixing A0A_{0} also establishes a linear relation between η0,w0\eta_{0},w_{0}. Given these three functions, we will derive a condition for exactly one of either η0\eta_{0} or w0w_{0}. Each of these conditions will ensure the positivity of η\eta. Once a condition on η0\eta_{0} (or w0w_{0}) is imposed, it automatically implies a condition on w0w_{0} (or η0\eta_{0}) through A0A_{0} as in (5.3). For example, if δ1<η0<δ2\delta_{1}<\eta_{0}<\delta_{2}, then for given q0,s0,A0q_{0},s_{0},A_{0}, and using (5.3) we have,

δ1<q0w0A0ks0<δ2,\delta_{1}<\frac{q_{0}w_{0}-A_{0}}{ks_{0}}<\delta_{2},

which gives the subsequent bounds on w0w_{0} as well. Therefore, it remains to find the appropriate conditions.

From Sections 4 and 5, we have that if the initial data q0,s0,A0q_{0},s_{0},A_{0} is given, then functions s,q,As,q,A are all known, periodic and uniformly bounded. Then (5.2) is an inhomogeneous, linear 2×22\times 2 system with uniformly bounded coefficients and, therefore, solutions η,w\eta,w exist for all t(,)t\in(-\infty,\infty). We now state and prove a few lemmas.

Lemma 6.1.

Consider the first order linear ODE,

(6.4) η~=A+kη~sq.\displaystyle\tilde{\eta}^{\prime}=\frac{A+k\tilde{\eta}s}{q}.

Let 𝔻={t(,):q(t)=0}\mathbb{D}=\{t\in(-\infty,\infty):q(t)=0\}. For all intervals II with I𝔻cI\subset\mathbb{D}^{c}, suppose η~1\tilde{\eta}_{1} and η~2\tilde{\eta}_{2} satisfy (6.4) and η~1(t)η~2(t)\tilde{\eta}_{1}(t_{\ast})\neq\tilde{\eta}_{2}(t_{\ast}) for some t𝔻ct_{\ast}\in\mathbb{D}^{c}. Then following statements hold,

  1. (1)

    η~1,η~2\tilde{\eta}_{1},\tilde{\eta}_{2} satisfy

    (6.5) η~′′+kη~(c+(N1)s)=kΓN1,\displaystyle\tilde{\eta}^{\prime\prime}+k\tilde{\eta}(c+(N-1)s)=k\Gamma^{N-1},

    for all t(,)t\in(-\infty,\infty),

  2. (2)

    η~1(t)=η~2(t)=A(t)ks(t)>0\tilde{\eta}_{1}(t)=\tilde{\eta}_{2}(t)=-\frac{A(t)}{ks(t)}>0 for all t𝔻t\in\mathbb{D},

  3. (3)

    η~1(t)η~2(t)\tilde{\eta}_{1}(t)\neq\tilde{\eta}_{2}(t) for all t𝔻ct\in\mathbb{D}^{c},

  4. (4)

    η~1(t)η~2(t)\tilde{\eta}^{\prime}_{1}(t)\neq\tilde{\eta}_{2}^{\prime}(t) for all t𝔻t\in\mathbb{D}. In particular, η~1η~2\tilde{\eta}_{1}-\tilde{\eta}_{2} changes sign at only and all t𝔻t\in\mathbb{D}.

Proof.

Taking derivative of (6.4), and using (3.8c), (3.8d), (5.4) results in (6.5), which is linear with bounded coefficients. Note that since 𝔻\mathbb{D} is discrete, η~1,η~2\tilde{\eta}_{1},\tilde{\eta}_{2} satisfy (6.5) for all tt by continuity. Hence, the first statement holds.

Consequently, η~1,η~2\tilde{\eta}_{1},\tilde{\eta}_{2} are well-defined for all times. Therefore, the limit in (6.4),

limtt0η~i(t),tq𝔻,i=1,2,\lim_{t\to t_{0}}\tilde{\eta}_{i}^{\prime}(t),\quad t_{q}\in\mathbb{D},\ i=1,2,

must exist. Hence, A(tq)+kη~i(tq)s(tq)=0A(t_{q})+k\tilde{\eta}_{i}(t_{q})s(t_{q})=0 for any tq𝔻t_{q}\in\mathbb{D}. The arguments to the fact that η~i(tq)>0\tilde{\eta}_{i}(t_{q})>0 are the same as were used in the paragraph following (6.3). This completes the proof to the second assertion.

We will prove the third assertion by contradiction. To that end, assume η~1(τ)=η~2(τ)\tilde{\eta}_{1}(\tau_{\ast})=\tilde{\eta}_{2}(\tau_{\ast}) for some τ𝔻c\tau_{\ast}\in\mathbb{D}^{c}. Note that η~1,η~2\tilde{\eta}_{1},\tilde{\eta}_{2} satisfy (6.4) as well as (6.5). Consider the IVP (6.5) along with initial data η~(τ)=η~1(τ),η~(τ)=A(τ)+kη~1(τ)s(τ)q(τ)\tilde{\eta}(\tau_{\ast})=\tilde{\eta}_{1}(\tau_{\ast}),\tilde{\eta}^{\prime}(\tau_{\ast})=\frac{A(\tau_{\ast})+k\tilde{\eta}_{1}(\tau_{\ast})s(\tau_{\ast})}{q(\tau_{\ast})}. This IVP has a unique solution. Consequently, η~1η~2\tilde{\eta}_{1}\equiv\tilde{\eta}_{2}. However, this is a contradiction since η~1(t)η~2(t)\tilde{\eta}_{1}(t_{\ast})\neq\tilde{\eta}_{2}(t_{\ast}). Hence, the assertion stands.

For tq𝔻t_{q}\in\mathbb{D}, η~1(tq)η~2(tq)\tilde{\eta}_{1}^{\prime}(t_{q})\neq\tilde{\eta}_{2}^{\prime}(t_{q}) follows by the second assertion and a uniqueness of ODE argument as above. Consequently, if η~1(t)η~2(t)<0\tilde{\eta}_{1}(t)-\tilde{\eta}_{2}(t)<0 for t(tqϵ,tq)t\in(t_{q}-\epsilon,t_{q}) for ϵ\epsilon sufficiently small, then from the second and third assertions, we have η~1(tq)η~2(tq)>0\tilde{\eta}_{1}^{\prime}(t_{q})-\tilde{\eta}_{2}^{\prime}(t_{q})>0. To see this,

η~1(t)>η~2(t),t(tqϵ,tq),\displaystyle-\tilde{\eta}_{1}(t)>-\tilde{\eta}_{2}(t),\quad t\in(t_{q}-\epsilon,t_{q}),
η~1(tq)η~1(t)tqt>η~2(tq)η~2(t)tqt,by Assertion 2,\displaystyle\frac{\tilde{\eta}_{1}(t_{q})-\tilde{\eta}_{1}(t)}{t_{q}-t}>\frac{\tilde{\eta}_{2}(t_{q})-\tilde{\eta}_{2}(t)}{t_{q}-t},\quad\text{by Assertion }2,
η~1(tq)η~2(tq),\displaystyle\tilde{\eta}_{1}^{\prime}(t_{q})\geq\tilde{\eta}_{2}^{\prime}(t_{q}),
η~1(tq)>η~2(tq),by Assertion 3.\displaystyle\tilde{\eta}_{1}^{\prime}(t_{q})>\tilde{\eta}_{2}^{\prime}(t_{q}),\quad\text{by Assertion }3.

Hence, η~1η~2\tilde{\eta}_{1}-\tilde{\eta}_{2} changes sign at tqt_{q}. This completes the proof. ∎

Figure 3 below gives an illustration of the Lemma.

Refer to caption
Figure 3. Visualization of Lemma 6.1. k=c=1,N=4,q0=0.1,s0=0.1,A0=0.15k=c=1,N=4,q_{0}=0.1,s_{0}=-0.1,A_{0}=0.15
Lemma 6.2.

Suppose η~\tilde{\eta} satisfies (6.4) on the intervals as specified in Lemma 6.1 and η~(0)>0\tilde{\eta}(0)>0. Let tc>0t_{c}>0 be the first time (if it exists) when η~(tc)=0\tilde{\eta}(t_{c})=0. Let J:=(tq1,tq2)J:=(t_{q}^{1},t_{q}^{2}) be the smallest interval such that tcJt_{c}\in J and q(tqi)=0q(t_{q}^{i})=0, i=1,2i=1,2. t1qt_{1}^{q} could possibly be negative. Let tAJt_{A}\in J be the unique time when A(tA)=0A(t_{A})=0. Then it must be that tc(tq1,tA]t_{c}\in(t_{q}^{1},t_{A}] and η~(t)<0\tilde{\eta}(t)<0 for t(tc,tA]t\in(t_{c},t_{A}].

In particular, if η~(tA)>0\tilde{\eta}(t_{A})>0, then there is no such tct_{c} and η~(t)>0\tilde{\eta}(t)>0 for all t[tq1,tq2]t\in[t_{q}^{1},t_{q}^{2}]. Or if η~(tA)=0\tilde{\eta}(t_{A})=0, then η~(t)>0\tilde{\eta}(t)>0 for all t[tq1,tq2]\{tA}t\in[t_{q}^{1},t_{q}^{2}]\backslash\{t_{A}\}.

Note that from (4.4b) and (6.2), AA and qq cannot be zero at the same time. Moreover, from (4.10) and (5.5) there is a unique time tA(tq1,tq2)t_{A}\in(t_{q}^{1},t_{q}^{2}) with A(tA)=0A(t_{A})=0. Therefore, tAJt_{A}\in J exists and is unique. Also, tAt_{A} could be nonpositive. In that case, η~(t)>0\tilde{\eta}(t)>0 for t[0,tq2][tA,tq2]t\in[0,t_{q}^{2}]\subset[t_{A},t_{q}^{2}].

Proof.

From Lemma 6.1, η~(tq1)>0\tilde{\eta}(t_{q}^{1})>0. Hence, for some ϵ>0\epsilon>0 small enough such that tq1+ϵ<tAt_{q}^{1}+\epsilon<t_{A}, we have η~(t)>0\tilde{\eta}(t)>0 for t[tq1,tq1+ϵ]t\in[t_{q}^{1},t_{q}^{1}+\epsilon]. η~\tilde{\eta} satisfies (6.4) in the interval [tq1+ϵ,tq2)[t_{q}^{1}+\epsilon,t_{q}^{2}). Multiplying (6.4) with the appropriate integrating factor, we have,

(η~ektsq)=Aqektsq.\left(\tilde{\eta}e^{-k\int^{t}\frac{s}{q}}\right)^{\prime}=\frac{A}{q}e^{-k\int^{t}\frac{s}{q}}.

Note that, A(t)A(t) and q(t)q(t) have opposite signs in the interval [tq1+ϵ,tA)[t_{q}^{1}+\epsilon,t_{A}) and same sign in (tA,tq2)(t_{A},t_{q}^{2}). Indeed, if tq1t_{q}^{1} is such that s(tq1)=smaxs(t_{q}^{1})=s_{max}, then from (4.10) and (5.5), A(t)<0A(t)<0 for t[tq1+ϵ,tA)t\in[t_{q}^{1}+\epsilon,t_{A}) and strictly positive in (tA,tq2)(t_{A},t_{q}^{2}). From assertion three of Lemma 4.1, q(t)>0q(t)>0 in JJ and hence, signs of A(t),q(t)A(t),q(t) are opposite. Same holds if s(tq1)=smins(t_{q}^{1})=s_{min}.

Therefore, from the ODE above, the quantity η~ektsq\tilde{\eta}e^{-k\int^{t}\frac{s}{q}} is decreasing in the interval [tq1+ϵ,tA)[t_{q}^{1}+\epsilon,t_{A}) and increasing in the interval (tA,tq2)(t_{A},t_{q}^{2}). Since η~ektsq|tq1+ϵ>0\left.\tilde{\eta}e^{-k\int^{t}\frac{s}{q}}\right|_{t_{q}^{1}+\epsilon}>0, we conclude that tc(tq1+ϵ,tA](tq1,tA]t_{c}\in(t_{q}^{1}+\epsilon,t_{A}]\subset(t_{q}^{1},t_{A}]. η~(t)<0\tilde{\eta}(t)<0 for t(tc,tA]t\in(t_{c},t_{A}] follows directly from the above ODE.

Lastly, if η~(tA)>0\tilde{\eta}(t_{A})>0, then there is no tct_{c} because if it is so, then η(tA)<0\eta(t_{A})<0 which is a contradiction. Hence, η~(t)>0\tilde{\eta}(t)>0 for all t(tq1,tq2)t\in(t_{q}^{1},t_{q}^{2}) and η~\tilde{\eta} is positive at tq1,tq2t_{q}^{1},t_{q}^{2} by Lemma 6.1. Also, if tc=tAt_{c}=t_{A}, then from (6.4) we have η~(tA)=0\tilde{\eta}^{\prime}(t_{A})=0 and (6.5) then implies that η~′′(tA)>0\tilde{\eta}^{\prime\prime}(t_{A})>0. Hence, tc=tAt_{c}=t_{A} is a local minima and η~\tilde{\eta} is positive in a neighbourhood of tct_{c}. However, on the right, A,qA,q have same sign and the ODE above implies η~>0\tilde{\eta}>0 after tAt_{A}. ∎

We will now move on to define two functions through an IVP using the linear differential equation (6.5) along with appropriate initial conditions. These functions will form a cloud around the solution, η\eta.

Let tAi,i=1,2t_{A}^{i},i=1,2 with 0tA1<tA20\leq t_{A}^{1}<t_{A}^{2} be the first two times when A(tAi)=0A(t_{A}^{i})=0. Define the two functions ηi,i=1,2\eta_{i},i=1,2 as follows,

(6.6) ηi′′+kηi(c+(N1)s)=kΓN1,ηi(tAi)=0,ηi(tAi)=0.\displaystyle\begin{aligned} &\eta_{i}^{\prime\prime}+k\eta_{i}(c+(N-1)s)=k\Gamma^{N-1},\\ &\eta_{i}(t_{A}^{i})=0,\qquad\eta_{i}^{\prime}(t_{A}^{i})=0.\end{aligned}

Now we state an important result for these functions as well as η\eta.

Proposition 6.3.

The functions η,ηi,i=1,2\eta,\eta_{i},i=1,2 are periodic with the same period as that of the system (3.8c),(3.8d).

Proof.

Part 1: The homogeneous system.
To prove this, we will use the Floquet Theorem, [10]. To use the result, we will first study the homogeneous form of (5.2) as follows,

(6.13) ΘH:=[ηHwH]=[01k(c+(N1)s)0][ηHwH]=:EΘH.\displaystyle\Theta_{H}^{\prime}:=\left[\begin{array}[]{c}\eta_{H}\\ w_{H}\end{array}\right]^{\prime}=\left[\begin{array}[]{cc}0&1\\ -k(c+(N-1)s)&0\end{array}\right]\left[\begin{array}[]{c}\eta_{H}\\ w_{H}\end{array}\right]=:E\Theta_{H}.

Let TT be the period of ss and therefore, the period of the matrix EE. By Floquet Theorem, a fundamental matrix to (6.13) is of the form F(t)etGF(t)e^{tG}, where FF is a TT-periodic, 2×22\times 2 matrix which is nonsingular for all times and GG is a 2×22\times 2 constant matrix. F,GF,G could possibly be complex but if one of them is real, then the other will be since EE is real.

The substitution ΦH:=F1ΘH\Phi_{H}:=F^{-1}\Theta_{H} reduces (6.13) to

(6.14) ΦH=GΦH.\displaystyle\Phi_{H}^{\prime}=G\Phi_{H}.

To see this, note that

EFΦH=EΘH=ΘH=FΦH+FΦH.EF\Phi_{H}=E\Theta_{H}=\Theta_{H}^{\prime}=F^{\prime}\Phi_{H}+F\Phi_{H}^{\prime}.

Also, since F(t)etGF(t)e^{tG} is a fundamental matrix of (6.13), we have,

(FetG)=EFetG,\displaystyle(Fe^{tG})^{\prime}=EFe^{tG},
F+FG=EF.\displaystyle F^{\prime}+FG=EF.

Substituting this for FF^{\prime} above and multiplying by F1F^{-1}, we obtain (6.14).

Next, we define a quantity for (6.13) analogous to (5.3). Define AH:=qwHkηHsA_{H}:=qw_{H}-k\eta_{H}s. Using (6.13), (3.8c), (3.8d),

AH\displaystyle A_{H}^{\prime} =(ksq2)wHkq(c+(N1)s)ηHkwHs+kηHq(c+Ns)\displaystyle=(ks-q^{2})w_{H}-kq(c+(N-1)s)\eta_{H}-kw_{H}s+k\eta_{H}q(c+Ns)
=q(qwHkηHs)\displaystyle=-q(qw_{H}-k\eta_{H}s)
=qAH.\displaystyle=-qA_{H}.

Therefore, AH(t)=AH(0)e0tq=AH(0)(s~(t)/s~0)1/NA_{H}(t)=A_{H}(0)e^{-\int_{0}^{t}q}=A_{H}(0)(\tilde{s}(t)/\tilde{s}_{0})^{1/N}. We used (4.10) for the last equality. If tqt_{q} is a time where s~\tilde{s} (or equivalently ss) achieves maximum, then for l=0,1,2,l=0,1,2,\ldots,

(6.15) β:=ηH(tq+lT)=AH(0)ksmax(s~maxs~0)1N=constant.\displaystyle\beta:=\eta_{H}(t_{q}+lT)=-\frac{A_{H}(0)}{ks_{max}}\left(\frac{\tilde{s}_{max}}{\tilde{s}_{0}}\right)^{\frac{1}{N}}=\text{constant}.

Since BB is a constant matrix, the general solution form of (6.14) is known. Consequently, from (6.14) and that ΘH=FΦH\Theta_{H}=F\Phi_{H}, we obtain that a general solution to (6.13) depending on whether the eigenvalues of BB are distinct or repeated,

ΘH=c1eλ1t𝐟𝟏(t)+c2eλ2t𝐟𝟐(t),λ1λ2.\Theta_{H}=c_{1}e^{\lambda_{1}t}\mathbf{f_{1}}(t)+c_{2}e^{\lambda_{2}t}\mathbf{f_{2}}(t),\quad\lambda_{1}\neq\lambda_{2}.

or,

ΘH=c1eλ1t𝐟𝟏(t)+c2teλ1t𝐟𝟐(t),\Theta_{H}=c_{1}e^{\lambda_{1}t}\mathbf{f_{1}}(t)+c_{2}te^{\lambda_{1}t}\mathbf{f_{2}}(t),

where λ1,λ2\lambda_{1},\lambda_{2} are the eigenvalues (possibly complex) of BB, 𝐟𝟏,𝐟𝟐\mathbf{f_{1}},\mathbf{f_{2}} are linearly independent, periodic vectors and c1,c2c_{1},c_{2}\in\mathbb{R} are arbitrary constants. We will show that ηH\eta_{H} is periodic. Suppose the eigenvalues are repeated and latter formula is the general solution, then we have,

ηH(t)=eλ1t(c1f1(t)+c2tf2(t)),\eta_{H}(t)=e^{\lambda_{1}t}\left(c_{1}f_{1}(t)+c_{2}tf_{2}(t)\right),

where f1,f2f_{1},f_{2} are the first elements of 𝐟𝟏,𝐟𝟐\mathbf{f_{1}},\mathbf{f_{2}} respectively. This should be a general solution form for any constant β\beta. Using (6.15),

βeλ1(tq+lT)=c1f1(tq)+c2(tq+lT)f2(tq),l=0,1,2,\beta e^{-\lambda_{1}(t_{q}+lT)}=c_{1}f_{1}(t_{q})+c_{2}(t_{q}+lT)f_{2}(t_{q}),\quad l=0,1,2,\ldots

Assume β0\beta\neq 0. If Re(λ1)>0(\lambda_{1})>0, then left-hand-side tends to zero as ll\to\infty but right-hand-side does not. If Re(λ1)<0(\lambda_{1})<0, then left-hand-side is an exponential function of ll while the right-hand-side is linear and therefore, the equation cannot hold for all ll. Lastly, if λ1\lambda_{1} is purely imaginary, then the left-hand-side changes periodically with ll but the right-hand-side does not. This leaves us with the only possibility that the repeated eigenvalue must be zero. In this case,

ΘH=c1𝐟𝟏(t)+c2t𝐟𝟐(t).\Theta_{H}=c_{1}\mathbf{f_{1}}(t)+c_{2}t\mathbf{f_{2}}(t).

However, note that since ηH=wH\eta_{H}^{\prime}=w_{H} and the vectors 𝐟𝟏,𝐟𝟐\mathbf{f_{1}},\mathbf{f_{2}} are periodic, it must be that,

(tf2)=tf2+f2=tg2,(tf_{2})^{\prime}=tf_{2}^{\prime}+f_{2}=tg_{2},

for some periodic function g2g_{2}. This could only hold if f2f20f_{2}\equiv f_{2}^{\prime}\equiv 0, which is a contradiction to FF being nonsingular. Consequently,

ηH(t)=c1eλ1tf1(t)+c2eλ2tf2(t),\eta_{H}(t)=c_{1}e^{\lambda_{1}t}f_{1}(t)+c_{2}e^{\lambda_{2}t}f_{2}(t),

for λ1λ2\lambda_{1}\neq\lambda_{2}. Again, using (6.15),

β=c1eλ1(tq+lT)f1(tq)+c2eλ2(tq+lT)f2(tq),l=0,1,.\beta=c_{1}e^{\lambda_{1}(t_{q}+lT)}f_{1}(t_{q})+c_{2}e^{\lambda_{2}(t_{q}+lT)}f_{2}(t_{q}),\quad l=0,1,\ldots.

It is clear that the real part of both eigenvalues is zero because if not, then a contradiction is obtained as ll\to\infty. Consequently, λ1,λ2\lambda_{1},\lambda_{2} are purely imaginary. Taking difference with l=0l=0 in (6.15) with itself,

0=c1f1(tq)eλ1tq(eλ1lT1)+c2f2(tq)eλ2tq(eλ2lT1).0=c_{1}f_{1}(t_{q})e^{\lambda_{1}t_{q}}\left(e^{\lambda_{1}lT}-1\right)+c_{2}f_{2}(t_{q})e^{\lambda_{2}t_{q}}\left(e^{\lambda_{2}lT}-1\right).

Since c1,c2c_{1},c_{2} are arbitrary,

f1(tq)(eλ1lT1)=f2(tq)(eλ2lT1)=0,l=1,2,f_{1}(t_{q})\left(e^{\lambda_{1}lT}-1\right)=f_{2}(t_{q})\left(e^{\lambda_{2}lT}-1\right)=0,\quad l=1,2,\ldots

As argued before, both f1(tq),f2(tq)f_{1}(t_{q}),f_{2}(t_{q}) cannot be zero together. So, at least one of the eλiT1=0e^{\lambda_{i}T}-1=0. In fact, both these terms are zero, because if eλ1T=ω1e^{\lambda_{1}T}=\omega\neq 1 (WLOG), then f1(tq)=0f_{1}(t_{q})=0, in which case c2c_{2} is no longer arbitrary, hence, a contradiction. Therefore,

ηH(t)=c1f1(t)+c2f2(t),\eta_{H}(t)=c_{1}f_{1}(t)+c_{2}f_{2}(t),

where f1,f2f_{1},f_{2} are new TT-periodic functions obtained after dissolving the exponential terms, which had the same period.

Part 2: Periodicity.
Owing to the fact that ηH=wH\eta_{H}^{\prime}=w_{H}, we have eventually proved that a fundamental matrix to (6.13), is of the form,

F¯(t)=[f1(t)f2(t)f1(t)f2(t)].\bar{F}(t)=\left[\begin{array}[]{cc}f_{1}(t)&f_{2}(t)\\ f_{1}^{\prime}(t)&f_{2}^{\prime}(t)\end{array}\right].

We will now show that η1\eta_{1} is periodic. Very similar arguments apply for η2\eta_{2} as well as any solution to (5.2a), η\eta. The important fact is that all of these three functions satisfy (6.4) in the intervals as mentioned in the hypothesis of Lemma 6.1 and hence, the assertions of the Lemma hold.

[η1η1]T[\eta_{1}\ \eta_{1}^{\prime}]^{T} has the following closed form expression,

[η1(t)η1(t)]=F¯(t)tA1tF¯1(τ)[0k(Γ(τ))N1]𝑑τ.\displaystyle\left[\begin{array}[]{c}\eta_{1}(t)\\ \eta_{1}^{\prime}(t)\end{array}\right]=\bar{F}(t)\int_{t_{A}^{1}}^{t}\bar{F}^{-1}(\tau)\left[\begin{array}[]{c}0\\ k(\Gamma(\tau))^{N-1}\end{array}\right]d\tau.

Through routine calculations, one can check that η1\eta_{1} in the above expression indeed satisfies (6.6) for i=1i=1. We can further evaluate this expression as follows,

(6.20) [η1(t)η1(t)]\displaystyle\left[\begin{array}[]{c}\eta_{1}(t)\\ \eta_{1}^{\prime}(t)\end{array}\right] =F¯(t)tA1tF¯1(τ)[0k(Γ(τ))N1]𝑑τ\displaystyle=\bar{F}(t)\int_{t_{A}^{1}}^{t}\bar{F}^{-1}(\tau)\left[\begin{array}[]{c}0\\ k(\Gamma(\tau))^{N-1}\end{array}\right]d\tau
(6.25) =F¯(t)tA1t1det(F¯(τ))[f2(τ)f2(τ)f1(τ)f1(τ)][0k(Γ(τ))N1]𝑑τ\displaystyle=\bar{F}(t)\int_{t_{A}^{1}}^{t}\frac{1}{\text{det}(\bar{F}(\tau))}\left[\begin{array}[]{cc}f_{2}^{\prime}(\tau)&-f_{2}(\tau)\\ -f_{1}^{\prime}(\tau)&f_{1}(\tau)\end{array}\right]\left[\begin{array}[]{c}0\\ k(\Gamma(\tau))^{N-1}\end{array}\right]d\tau
(6.30) =[f1(t)f2(t)f1(t)f2(t)]tA1tk(Γ(τ))N1det(F¯(τ))[f2(τ)f1(τ)]𝑑τ\displaystyle=\left[\begin{array}[]{cc}f_{1}(t)&f_{2}(t)\\ f_{1}^{\prime}(t)&f_{2}^{\prime}(t)\end{array}\right]\int_{t_{A}^{1}}^{t}\frac{k(\Gamma(\tau))^{N-1}}{\text{det}(\bar{F}(\tau))}\left[\begin{array}[]{c}-f_{2}(\tau)\\ f_{1}(\tau)\end{array}\right]d\tau
(6.33) =[f1(t)tA1tg1(τ)𝑑τ+f2(t)tA1tg2(τ)𝑑τf1(t)tA1tg1(τ)𝑑τ+f2(t)tA1tg2(τ)𝑑τ],\displaystyle=\left[\begin{array}[]{c}f_{1}(t)\int_{t_{A}^{1}}^{t}g_{1}(\tau)d\tau+f_{2}(t)\int_{t_{A}^{1}}^{t}g_{2}(\tau)d\tau\\ f_{1}^{\prime}(t)\int_{t_{A}^{1}}^{t}g_{1}(\tau)d\tau+f_{2}^{\prime}(t)\int_{t_{A}^{1}}^{t}g_{2}(\tau)d\tau\end{array}\right],

where g1,g2g_{1},g_{2} are T-periodic functions defined by,

g1(t):=f2(t)kΓN1det(F¯(t)),g2(t):=f1(t)kΓN1det(F¯(t)).g_{1}(t):=-f_{2}(t)\frac{k\Gamma^{N-1}}{\text{det}(\bar{F}(t))},\quad g_{2}(t):=f_{1}(t)\frac{k\Gamma^{N-1}}{\text{det}(\bar{F}(t))}.

Periodicity of g1,g2g_{1},g_{2} follows from (4.10) as 0tq\int_{0}^{t}q also has period TT. Since [η1η1]T[\eta_{1}\quad\eta_{1}^{\prime}]^{T} is a solution to a 2×22\times 2 ODE system, proving periodicity is equivalent to proving equality at a single point. In particular, if we prove the following for some tt^{\ast}, then we obtain periodicity.

[η1(t+T)η1(t+T)]=[η1(t)η1(t)].\left[\begin{array}[]{c}\eta_{1}(t^{\ast}+T)\\ \eta_{1}^{\prime}(t^{\ast}+T)\end{array}\right]=\left[\begin{array}[]{c}\eta_{1}(t^{\ast})\\ \eta_{1}^{\prime}(t^{\ast})\end{array}\right].

From (6.20) and periodicity of g1,g2g_{1},g_{2},

[η1(t+T)η1(t+T)]\displaystyle\left[\begin{array}[]{c}\eta_{1}(t^{\ast}+T)\\ \eta_{1}^{\prime}(t^{\ast}+T)\end{array}\right] =[f1(t)tA1t+Tg1(τ)𝑑τ+f2(t)tA1t+Tg2(τ)𝑑τf1(t)tA1t+Tg1(τ)𝑑τ+f2(t)tA1t+Tg2(τ)𝑑τ]\displaystyle=\left[\begin{array}[]{c}f_{1}(t^{\ast})\int_{t_{A}^{1}}^{t^{\ast}+T}g_{1}(\tau)d\tau+f_{2}(t^{\ast})\int_{t_{A}^{1}}^{t^{\ast}+T}g_{2}(\tau)d\tau\\ f_{1}^{\prime}(t^{\ast})\int_{t_{A}^{1}}^{t^{\ast}+T}g_{1}(\tau)d\tau+f_{2}^{\prime}(t^{\ast})\int_{t_{A}^{1}}^{t^{\ast}+T}g_{2}(\tau)d\tau\end{array}\right]
=[η1(t)η1(t)]+[f1(t)0Tg1(τ)𝑑τ+f2(t)0Tg2(τ)𝑑τf1(t)0Tg1(τ)𝑑τ+f2(t)0Tg2(τ)𝑑τ].\displaystyle=\left[\begin{array}[]{c}\eta_{1}(t^{\ast})\\ \eta_{1}^{\prime}(t^{\ast})\end{array}\right]+\left[\begin{array}[]{c}f_{1}(t^{\ast})\int_{0}^{T}g_{1}(\tau)d\tau+f_{2}(t^{\ast})\int_{0}^{T}g_{2}(\tau)d\tau\\ f_{1}^{\prime}(t^{\ast})\int_{0}^{T}g_{1}(\tau)d\tau+f_{2}^{\prime}(t^{\ast})\int_{0}^{T}g_{2}(\tau)d\tau\end{array}\right].

Set

h(t):=f1(t)0Tg1(τ)𝑑τ+f2(t)0Tg2(τ)𝑑τ.h(t):=f_{1}(t)\int_{0}^{T}g_{1}(\tau)d\tau+f_{2}(t)\int_{0}^{T}g_{2}(\tau)d\tau.

If h(t)=0=h(t)h(t^{\ast})=0=h^{\prime}(t^{\ast}), then hh is identically zero. Indeed, if h,hh,h^{\prime} are both zero together at the same time then from nonsingularity of FF, we have that 0Tgi(τ)𝑑τ=0,i=1,2\int_{0}^{T}g_{i}(\tau)d\tau=0,i=1,2, and hence, h(t)=h(t)0h(t)=h^{\prime}(t)\equiv 0, thereby proving that η1\eta_{1} is periodic. Our aim is to find such a tt^{\ast}. We begin with some calculations. For l=1,2,l=1,2,\ldots and t[0,T)t\in[0,T), and owing to periodicity of gi,i=1,2g_{i},i=1,2,

tA1t+lTgi(τ)𝑑τ\displaystyle\int_{t_{A}^{1}}^{t+lT}g_{i}(\tau)d\tau =tA1tgi(τ)𝑑τ+t0gi(τ)𝑑τ+0lTgi(τ)𝑑τ+lTt+lTgi(τ)𝑑τ\displaystyle=\int_{t_{A}^{1}}^{t}g_{i}(\tau)d\tau+\int_{t}^{0}g_{i}(\tau)d\tau+\int_{0}^{lT}g_{i}(\tau)d\tau+\int_{lT}^{t+lT}g_{i}(\tau)d\tau
=tA1tgi(τ)𝑑τ+l0Tgi(τ)𝑑τ.\displaystyle=\int_{t_{A}^{1}}^{t}g_{i}(\tau)d\tau+l\int_{0}^{T}g_{i}(\tau)d\tau.

Using this, we can obtain,

η1(t+lT)\displaystyle\eta_{1}(t+lT) =f1(t)tA1t+lTg1(τ)𝑑τ+f2(t)tA1t+lTg2(τ)𝑑τ\displaystyle=f_{1}(t)\int_{t_{A}^{1}}^{t+lT}g_{1}(\tau)d\tau+f_{2}(t)\int_{t_{A}^{1}}^{t+lT}g_{2}(\tau)d\tau
=η1(t)+l(f1(t)0Tg1(τ)𝑑τ+f2(t)0Tg2(τ)𝑑τ)\displaystyle=\eta_{1}(t)+l\left(f_{1}(t)\int_{0}^{T}g_{1}(\tau)d\tau+f_{2}(t)\int_{0}^{T}g_{2}(\tau)d\tau\right)
(6.34) =η1(t)+lh(t).\displaystyle=\eta_{1}(t)+lh(t).

Similar calculation for η1\eta_{1}^{\prime} leads to

η1(t+lT)\displaystyle\eta_{1}^{\prime}(t+lT) =η1(t)+l(f1(t)0Tg1(τ)𝑑τ+f2(t)0Tg2(τ)𝑑τ)\displaystyle=\eta_{1}^{\prime}(t)+l\left(f_{1}^{\prime}(t)\int_{0}^{T}g_{1}(\tau)d\tau+f_{2}^{\prime}(t)\int_{0}^{T}g_{2}(\tau)d\tau\right)
(6.35) =η1(t)+lh(t).\displaystyle=\eta_{1}^{\prime}(t)+lh^{\prime}(t).

Note that from the second assertion of Lemma 6.1, we have η1(tq)=0=η1(tq+T)\eta_{1}(t_{q})=0=\eta_{1}(t_{q}+T), where tqt_{q} is such that q(tq)=0q(t_{q})=0. From (6.34), h(tq)=0h(t_{q})=0. However, h(tq)h^{\prime}(t_{q}) may not be equal to zero. In a similar way, observing (6.4) and applying it to (6.35), we also have that h(ts)=0h^{\prime}(t_{s})=0, where tst_{s} is such that s(ts)=0s(t_{s})=0, whereas h(ts)h(t_{s}) may not be zero. The key is to find a tt^{\ast} such that both h(t),h(t)h(t^{\ast}),h^{\prime}(t^{\ast}) are zero.

We will consider two cases. The first is when s(t)s(t) and A(t)A(t) are zero at the same times and the other when they are never zero together. The two situations are exhaustive, in the sense that there can never be a scenario when A,sA,s are zero together at some times and there is the existence of some other times when one of them is zero and the other is not. A,sA,s sharing zeros is only dependent on the initial data s0,A0s_{0},A_{0}. Indeed from (4.10), ss is zero at the times when Γ(t)=(c/Ns~0)1/N\Gamma(t)=(c/N\tilde{s}_{0})^{1/N} and from (5.5), AA is zero at times Γ(t)=κ\Gamma(t)=\kappa. Hence, s,As,A are zero at the same times if and only if κN=cNs~0\kappa^{N}=\frac{c}{N\tilde{s}_{0}}, which only depends on the s0,A0s_{0},A_{0}. Conversely, if κNcNs~0\kappa^{N}\neq\frac{c}{N\tilde{s}_{0}}, then there is no time when s,As,A are both zero.

Taking note of this discussion, we first prove periodicity of η1\eta_{1} for the situation when A(ts)0A(t_{s})\neq 0 for any ts{t:s(t)=0}t_{s}\in\{t:s(t)=0\}. Since hh is periodic, it is enough to consider the restricted function h:[tq1,tq1+T]h:[t_{q}^{1},t_{q}^{1}+T]\to\mathbb{R}, where tq1t_{q}^{1} is such that q(tq1)=0q(t_{q}^{1})=0. From second assertion of Lemma 6.1 and (6.34), we have h(tq1)=0=h(tq2)h(t_{q}^{1})=0=h(t_{q}^{2}), where tq2(tq1,tq1+T)t_{q}^{2}\in(t_{q}^{1},t_{q}^{1}+T) such that q(tq2)=0q(t_{q}^{2})=0. In a similar way, observing (6.4) and applying it to (6.35), we also have that h(tsi)=0h^{\prime}(t_{s}^{i})=0, where tsi,i=1,2t_{s}^{i},i=1,2 are the two unique times in the interval (tq1,tq1+T)(t_{q}^{1},t_{q}^{1}+T) where s(tsi)=0s(t_{s}^{i})=0. Now suppose there is a point t[tq1,ts1)(ts1,ts2)(ts2,tq1+T]t^{\ast}\in[t_{q}^{1},t_{s}^{1})\cup(t_{s}^{1},t_{s}^{2})\cup(t_{s}^{2},t_{q}^{1}+T] such that h(t)=0h^{\prime}(t^{\ast})=0. Then firstly from (6.35), η1(t+lT)=η1(t)\eta_{1}^{\prime}(t^{\ast}+lT)=\eta_{1}^{\prime}(t^{\ast}). Secondly, from (6.4) and (6.34),

A(t)η1(t)q(t)ks(t)=η1(t+lT)=η1(t)+lh(t),l=1,2,.-\frac{A(t^{\ast})-\eta_{1}^{\prime}(t^{\ast})q(t^{\ast})}{ks(t^{\ast})}=\eta_{1}(t^{\ast}+lT)=\eta_{1}(t^{\ast})+lh(t^{\ast}),\quad l=1,2,\ldots.

Therefore, h(t)=0h(t^{\ast})=0. Consequently, h(t)=h(t)=0h(t^{\ast})=h^{\prime}(t^{\ast})=0. Hence, if hh is not identically zero, then it must be that it is, without loss of generality, strictly increasing in the intervals [tq1,ts1][ts2,tq1+T][t_{q}^{1},t_{s}^{1}]\cup[t_{s}^{2},t_{q}^{1}+T] and strictly decreasing in [ts1,ts2][t_{s}^{1},t_{s}^{2}]. This implies that hh has exactly one maximum and it occurs at ts1t_{s}^{1}. Therefore, from (6.34), we have the existence of a natural number LL large enough so that,

max[t0+(l1)T,t0+lT]η1(t)=η1(ts+lT),lL,\max_{[t_{0}+(l-1)T,t_{0}+lT]}\eta_{1}(t)=\eta_{1}(t_{s}+lT),\quad l\geq L,

implying that η1(ts+lT)=0\eta_{1}^{\prime}(t_{s}+lT)=0 for all lLl\geq L. However, this is a contradiction since from (6.4),

η1(ts+lT)=A(ts)q(ts)0,l=1,2,.\eta_{1}^{\prime}(t_{s}+lT)=\frac{A(t_{s})}{q(t_{s})}\neq 0,\quad l=1,2,\ldots.

Therefore, hh is identically zero.

Now we prove periodicity for the scenario when A,sA,s are zero at the same times. Firstly, we argue that η1\eta_{1} being uniformly bounded implies periodicity. This can be seen by plugging in a special sequence of times in (6.34) and (6.35). To this end, let {tl}l=1\{t_{l}^{\ast}\}_{l=1}^{\infty} be a sequence of times such that η1(tl)=0\eta_{1}^{\prime}(t_{l}^{\ast})=0. We obtained this sequence by applying Rolle’s Theorem to η1\eta_{1} in the intervals [t0+(l1)T,t0+lT][t_{0}+(l-1)T,t_{0}+lT]. Rewrite tlt_{l}^{\ast} as,

tl=t1+(l1)T+αlT,αl(1,1).\displaystyle t_{l}^{\ast}=t_{1}^{\ast}+(l-1)T+\alpha_{l}T,\quad\alpha_{l}\in(-1,1).

By compactness, there exists a convergent subsequence αlm\alpha_{l_{m}}. Let t1+αlmTtt_{1}^{\ast}+\alpha_{l_{m}}T\to t^{\ast} as mm\to\infty. Plugging in tlt_{l}^{\ast} in (6.35), we obtain,

0=η1(t1+αlmT)+(lm1)h(t1+αlmT).0=\eta_{1}^{\prime}(t_{1}^{\ast}+\alpha_{l_{m}}T)+(l_{m}-1)h^{\prime}(t_{1}^{\ast}+\alpha_{l_{m}}T).

Dividing by lm1l_{m}-1 and letting mm\to\infty, we have that

h(t)=0.h^{\prime}(t^{\ast})=0.

Similarly using (6.34),

η1(tl)=η1(t1+αlmT)+(lm1)h(t1+αlmT).\eta_{1}(t_{l}^{\ast})=\eta_{1}(t_{1}^{\ast}+\alpha_{l_{m}}T)+(l_{m}-1)h(t_{1}^{\ast}+\alpha_{l_{m}}T).

If η1\eta_{1} is uniformly bounded in time, then dividing by lm1l_{m}-1 and letting mm\to\infty results in,

h(t)=0.h(t^{\ast})=0.

As a result of this, it is enough to show that if A,sA,s are zero at the same times, then η1\eta_{1} is uniformly bounded. Without loss of generality, assume it is not bounded above and attains a positive maximum value in the interval [0,T][0,T]. Let,

M:=max[0,T]η1(t).M:=\max_{[0,T]}\eta_{1}(t).

Let tM>Tt_{M}>T be a time when η1(tM)=3M\eta_{1}(t_{M})=3M. By continuous dependence of ODE solutions on initial data, we can choose a solution to (6.5), η~\tilde{\eta}, with initial data such that,

q0η~(0)kη~(0)s0A0,q_{0}\tilde{\eta}^{\prime}(0)-k\tilde{\eta}(0)s_{0}\neq A_{0},

and |η1(0)η~(0)|,|η1(0)η~(0)||\eta_{1}(0)-\tilde{\eta}(0)|,|\eta_{1}^{\prime}(0)-\tilde{\eta}^{\prime}(0)| small enough so that,

max[0,tM]|η~(t)η1(t)|<M.\max_{[0,t_{M}]}|\tilde{\eta}(t)-\eta_{1}(t)|<M.

However, by previous discussion, η~\tilde{\eta} is periodic. Therefore, it must be that for a tm[0,T]t_{m}\in[0,T],

M<η~(tM)M=η~(tm)M<η1(tm)M,M<\tilde{\eta}(t_{M})-M=\tilde{\eta}(t_{m})-M<\eta_{1}(t_{m})\leq M,

hence, a contradiction. This completes the full proof. ∎

Refer to caption
(a) η1η2\eta_{1}\equiv\eta_{2}.
Refer to caption
(b) η1,η2\eta_{1},\eta_{2} are distinct.
Figure 4.
Proposition 6.4.

For each i=1,2i=1,2,

ηi(t)>0,t[0,T]\{tAi}.\eta_{i}(t)>0,\quad t\in[0,T]\backslash\{t_{A}^{i}\}.

In particular, ηi(0)>0\eta_{i}(0)>0 and ηi\eta_{i}’s are distinct.

Proof.

We will show that for any A0A_{0} satisfying the condition (6.1), there exists a strictly positive solution to (6.4). The proposition statement can then be proved through the following argument. From Lemma 6.2, η1(t)>0\eta_{1}(t)>0 for t[tq1,tq2]\{tA1}t\in[t_{q}^{1},t_{q}^{2}]\backslash\{t_{A}^{1}\}. Hence, it can only be zero in (tq2,tq3)(t_{q}^{2},t_{q}^{3}). If η1\eta_{1} just touches zero in (tq2,tq3)(t_{q}^{2},t_{q}^{3}), then by (6.4), that time must be tA2t_{A}^{2} and by (6.5) and uniqueness of ODE, η1η2\eta_{1}\equiv\eta_{2}. If η1\eta_{1} crosses zero, then by Lemma 6.2, tA2t_{A}^{2} must be in between the two roots. Similar statement holds for η2\eta_{2} as well. Both these cases are illustrated in Figure 4. In any case, Lemma 6.1 implies that there can never be a strictly positive solution. Hence, it must be that for each i=1,2i=1,2 ηi\eta_{i} is positive everywhere on [0,T][0,T] except at tAit_{A}^{i}.

In view of the above discussion, we prove the statement mentioned at the beginning of the proof. Let,

𝒞:={a:A0=a,(6.4) has a strictly positive solution}\mathcal{C}:=\{a:A_{0}=a,\eqref{firstordereta}\text{ has a strictly positive solution}\}

Claim 1: 𝒞\mathcal{C} is non-empty.
We check that g(t)=s~1NNs~011Ng(t)=\frac{\tilde{s}^{-\frac{1}{N}}}{N\tilde{s}_{0}^{1-\frac{1}{N}}} is a positive solution to (6.5). gg is positive since s~\tilde{s} is. Note from (4.4),

g′′+kg(c+s(N1))\displaystyle g^{\prime\prime}+kg(c+s(N-1)) =1Ns~011N[(s~1Nq)+ks~1N(cN+(N1)s~)]\displaystyle=\frac{1}{N\tilde{s}_{0}^{1-\frac{1}{N}}}\left[(\tilde{s}^{-\frac{1}{N}}q)^{\prime}+k\tilde{s}^{-\frac{1}{N}}\left(\frac{c}{N}+(N-1)\tilde{s}\right)\right]
=1Ns~011N[q2s~1N+s~1N(ks~kcNq2)+ks~1N(cN+(N1)s~)]\displaystyle=\frac{1}{N\tilde{s}_{0}^{1-\frac{1}{N}}}\left[q^{2}\tilde{s}^{-\frac{1}{N}}+\tilde{s}^{-\frac{1}{N}}(k\tilde{s}-\frac{kc}{N}-q^{2})+k\tilde{s}^{-\frac{1}{N}}\left(\frac{c}{N}+(N-1)\tilde{s}\right)\right]
=k(s~s~0)11N\displaystyle=k\left(\frac{\tilde{s}}{\tilde{s}_{0}}\right)^{1-\frac{1}{N}}
=kΓN1.\displaystyle=k\Gamma^{N-1}.

The last equality follows from (4.10). Consequently a=q0g(0)kg(0)s0=q02ks0Ns~0a=q_{0}g^{\prime}(0)-kg(0)s_{0}=\frac{q_{0}^{2}-ks_{0}}{N\tilde{s}_{0}} belongs to 𝒞\mathcal{C}.
Claim 2: 𝒞\mathcal{C} is open.
If a𝒞a\in\mathcal{C}, then for A0=aA_{0}=a, (6.4) has a strictly positive solution. By continuous dependence of solutions on initial data, there is a neighbourhood around aa for which there is strictly positive solution.
Claim 3: 𝒞\mathcal{C} is closed in the set where (6.1) holds.
Let {ai}i=1𝒞\{a_{i}\}_{i=1}^{\infty}\subset\mathcal{C} be a sequence with aiaa_{i}\to a and aa belongs to the set where (6.1) holds. By well-posedness of ODEs, there must be solution η~\tilde{\eta} to (6.4) for A0=aA_{0}=a satisfying η~(t)0\tilde{\eta}(t)\geq 0. If η~\tilde{\eta} is strictly positive then a𝒞a\in\mathcal{C}. If not, then from Lemma 6.2 and (6.4), it is positive everywhere except, η~(tA1)=η~(tA1)=0\tilde{\eta}(t_{A}^{1})=\tilde{\eta}^{\prime}(t_{A}^{1})=0 or η~(tA2)=η~(tA2)=0\tilde{\eta}(t_{A}^{2})=\tilde{\eta}^{\prime}(t_{A}^{2})=0 or both. If η~(tAi)=η~(tAi)=0\tilde{\eta}(t_{A}^{i})=\tilde{\eta}^{\prime}(t_{A}^{i})=0 for both i=1,2i=1,2, then by uniqueness, η~\tilde{\eta} is (tA2tA1)(t_{A}^{2}-t_{A}^{1})-periodic. Note that tA2tA1<Tt_{A}^{2}-t_{A}^{1}<T, the period of q,s,Aq,s,A. If this is true, then by (6.5),

(N1)η~(t)[s~(t+tA2tA1)s~(t)]=1s~011N[(s~(t+tA2tA1))11N(s~(t))11N],(N-1)\tilde{\eta}(t)\left[\tilde{s}(t+t_{A}^{2}-t_{A}^{1})-\tilde{s}(t)\right]=\frac{1}{\tilde{s}_{0}^{1-\frac{1}{N}}}\left[\left(\tilde{s}(t+t_{A}^{2}-t_{A}^{1})\right)^{1-\frac{1}{N}}-\left(\tilde{s}(t)\right)^{1-\frac{1}{N}}\right],

for all tt. Since tA2tA1<Tt_{A}^{2}-t_{A}^{1}<T, we can obtain an interval on which a closed form of η~\tilde{\eta} is obtained given by above formula. This is a contradiction as it can be readily checked that it does not satisfy (6.5).

Therefore, η~(tA1)=η~(tA1)=0\tilde{\eta}(t_{A}^{1})=\tilde{\eta}^{\prime}(t_{A}^{1})=0 (WLOG) and η~(t)>0\tilde{\eta}(t)>0, t[0,T]\{tA1}t\in[0,T]\backslash\{t_{A}^{1}\}. But by uniqueness, η~=η1\tilde{\eta}=\eta_{1}. By definition of η2\eta_{2} and Lemma 6.1, it must be such that η2(t)>0\eta_{2}(t)>0, t[0,T]\{tA2}t\in[0,T]\backslash\{t_{A}^{2}\}. However, if this is the case, then from Lemma 6.1 there has to be a strictly positive solution to (6.4) squeezed between η1\eta_{1} and η2\eta_{2}. Hence, a𝒞a\in\mathcal{C} and consequently, 𝒞\mathcal{C} is closed.

This finishes the proof. ∎

The key results are as follows,

Proposition 6.5.

Suppose q00q_{0}\neq 0. If

min{η1(0),η2(0)}<η0<max{η1(0),η2(0)},\min\{\eta_{1}(0),\eta_{2}(0)\}<\eta_{0}<\max\{\eta_{1}(0),\eta_{2}(0)\},

then,

min{η1(t),η2(t)}<η(t)<max{η1(t),η2(t)},t𝔻c.\min\{\eta_{1}(t),\eta_{2}(t)\}<\eta(t)<\max\{\eta_{1}(t),\eta_{2}(t)\},\quad t\in\mathbb{D}^{c}.

Conversely, if

η0(min{η1(0),η2(0)},max{η1(0),η2(0)}),\eta_{0}\notin\left(\min\{\eta_{1}(0),\eta_{2}(0)\},\max\{\eta_{1}(0),\eta_{2}(0)\}\right),

then

η(t)(min{η1(t),η2(t)},max{η1(t),η2(t)}),t𝔻c.\eta(t)\notin\left(\min\{\eta_{1}(t),\eta_{2}(t)\},\max\{\eta_{1}(t),\eta_{2}(t)\}\right),\quad t\in\mathbb{D}^{c}.

Here, 𝔻\mathbb{D} is the same as in the statement of Lemma 6.1.

Proof.

Firstly, note that ηis\eta_{i}^{\prime}s satisfy (6.4) along with η\eta. Indeed, if a function satisfies (6.4) with η~(tA1)=0\tilde{\eta}(t_{A}^{1})=0, then

η~(tA1)=A(tA1)+kη~(tA1)s(tA1)q(tA1)=0,\tilde{\eta}^{\prime}(t_{A}^{1})=\frac{A(t_{A}^{1})+k\tilde{\eta}(t_{A}^{1})s(t_{A}^{1})}{q(t_{A}^{1})}=0,

and hence, by first assertion of Lemma 6.1 and uniqueness of ODE, η~η1\tilde{\eta}\equiv\eta_{1}. Similarly for η2\eta_{2}. Also, the solution to (5.2a), η\eta, satisfies (6.4). This is true directly from the definition of AA as in (5.3) and (5.2a). Consequently, all the three functions, η,η1,η2\eta,\eta_{1},\eta_{2} pairwise satisfy hypothesis of Lemma 6.1 with t=0t_{\ast}=0. The result follows from Lemma 6.1. Indeed at each t𝔻t\in\mathbb{D}, all η1,η2,η\eta_{1},\eta_{2},\eta cross each other, hence, maintaining that η\eta will be contained in between η1,η2\eta_{1},\eta_{2} if and only if it was so initially. ∎

An illustration of the situation is provided in Figure 5.

Refer to caption
Figure 5. A representation of a plausible solution squeezed between ηi\eta_{i}’s. k=c=1,N=4,q0=0.1,s0=0.1,A0=0.15k=c=1,N=4,q_{0}=0.1,s_{0}=-0.1,A_{0}=0.15
Proposition 6.6.

Suppose q0=0q_{0}=0. If

η1(0)<w0<η2(0),\eta_{1}^{\prime}(0)<w_{0}<\eta_{2}^{\prime}(0),

then for any t>0t>0,

min{η1(t),η2(t)}<η(t)<max{η1(t),η2(t)},t𝔻c.\min\{\eta_{1}(t),\eta_{2}(t)\}<\eta(t)<\max\{\eta_{1}(t),\eta_{2}(t)\},\quad t\in\mathbb{D}^{c}.

Conversely, if

w0(η1(0),η2(0)),w_{0}\notin\left(\eta_{1}^{\prime}(0),\eta_{2}^{\prime}(0)\right),

then

η(t)(min{η1(t),η2(t)},max{η1(t),η2(t)}),t𝔻c.\eta(t)\notin\left(\min\{\eta_{1}(t),\eta_{2}(t)\},\max\{\eta_{1}(t),\eta_{2}(t)\}\right),\quad t\in\mathbb{D}^{c}.
Proof.

The proof is very similar to that of Proposition 6.5 and follows from Lemma 6.1, only that here the starting time t=0t=0 is when the functions η1,η2,η\eta_{1},\eta_{2},\eta cross each other. In other words, 0𝔻0\in\mathbb{D}. ∎

Corollary 6.7.

η(t)>0\eta(t)>0 for all t>0t>0 if and only if one of the following holds,

  • If q00q_{0}\neq 0 then,

    min{η1(0),η2(0)}<η0<max{η1(0),η2(0)}.\min\{\eta_{1}(0),\eta_{2}(0)\}<\eta_{0}<\max\{\eta_{1}(0),\eta_{2}(0)\}.
  • If q0=0q_{0}=0 then,

    η1(0)<w0<η2(0).\eta_{1}^{\prime}(0)<w_{0}<\eta_{2}^{\prime}(0).
Proof.

The result follows from Propositions 6.5 or 6.6, followed by an application of Proposition 6.4. ∎

Using the results developed above, we move on to proving Theorem 2.1.

Proof of Theorem 2.1: We will prove for N3N\geq 3 as N=2N=2 is very similar. Suppose initial data satisfies the hypothesis of the Theorem. Along the characteristic path (3.1), this translates to the condition that for all β>0\beta>0,

(β,u0(β),ϕ0r(β),u0r(β),ρ0(β))ΘN.(\beta,u_{0}(\beta),\phi_{0r}(\beta),u_{0r}(\beta),\rho_{0}(\beta))\in\Theta_{N}.

We will now analyze a single characteristic path and replace the initial data notations with (β,u0,ϕ0r,u0r,ρ0)(\beta,u_{0},\phi_{0r},u_{0r},\rho_{0}). Under the transformation (3.7), we now turn to the unknowns of the ODE system (3.8), (q,s,p,ρ)(q,s,p,\rho) with initial data (q(0),s(0),p(0),ρ(0))=(u0β,ϕ0rβ,u0r,ρ0)(q(0),s(0),p(0),\rho(0))=\left(\frac{u_{0}}{\beta},-\frac{\phi_{0r}}{\beta},u_{0r},\rho_{0}\right). Global-in-time existence of these variables is equivalent to the global-in-time existence of the original variables. Note that if ρ(0)=0\rho(0)=0, then as argued, through simple calculations, at the beginning of Section 5, there is blowup of density, ρ\rho. Hence, we can safely assume ρ(0)>0\rho(0)>0. Using the transformations (5.1), we work with the unknowns (q,s,η,w)(q,s,\eta,w) with initial data (q(0),s(0),η(0),w(0))=(q(0),s(0),p(0)/ρ(0),1/ρ(0))(q(0),s(0),\eta(0),w(0))=(q(0),s(0),p(0)/\rho(0),1/\rho(0)). Through (3.7) and (5.1), we see that indeed a=q(0)w(0)kη(0)s(0)a=q(0)w(0)-k\eta(0)s(0). Set A(0)=aA(0)=a to obtain A(t)A(t) as in (5.3) satisfying (5.4). Turning to the Definition 2.2, we use (4.3) and ym,N=y1,yM,N=y2y_{m,N}=y_{1},y_{M,N}=y_{2} as in (5.7) to get,

A(0)kN2((y1s~0)N2N1,(y2s~0)N2N1).A(0)\in\frac{k}{N-2}\left(\left(\frac{y_{1}}{\tilde{s}_{0}}\right)^{\frac{N-2}{N}}-1,\left(\frac{y_{2}}{\tilde{s}_{0}}\right)^{\frac{N-2}{N}}-1\right).

Rearranging this result in condition (6.1). Also, u0=0u_{0}=0 if and only if q(0)=0q(0)=0. In (2.2), this is equivalent to whether xx is zero or not. If x0x\neq 0 (equivalently q(0)0q(0)\neq 0), then through the transformation (5.1), (2.2) is equivalent to,

min{η1(0),η2(0)}<η(0)<max{η1(0),η2(0)},\min\{\eta_{1}(0),\eta_{2}(0)\}<\eta(0)<\max\{\eta_{1}(0),\eta_{2}(0)\},

which is exactly the hypothesis of Proposition 6.5. On the other hand, if x=0x=0 (equivalently q(0)=0q(0)=0), then using (3.7) and (5.1), (2.2) reduces to

w(0)ks(0)η(0)A(0)(η1(0),η2(0)).w(0)\in-\frac{ks(0)\eta(0)}{A(0)}\ (\eta_{1}^{\prime}(0),\eta_{2}^{\prime}(0)).

From q(0)=0q(0)=0 and (5.3), the above inclusion becomes,

w(0)(η1(0),η2(0)),w(0)\in(\eta_{1}^{\prime}(0),\eta_{2}^{\prime}(0)),

which is exactly the hypothesis of Proposition 6.6. Note that on a single characteristic path, η1,η2\eta_{1},\eta_{2} are known functions because A(0),q(0),s(0)A(0),q(0),s(0) are fixed. In particular, η1,η2\eta_{1},\eta_{2} can first be evaluated and then it can be checked that one of the above hypothesis is satisfied depending on whether q(0)=0q(0)=0 or not. By Corollary 6.7, we obtain the all-time positivity of η\eta. By Proposition 5.2, all unknowns (q,s,p,ρ)(q,s,p,\rho) in (3.8) exist for all time. Since the above analysis holds for all characteristic paths, then by Lemma 3.1 and Theorem 1.1 we have global-in-time solution to (1.2).

Conversely, suppose there is a characteristic path corresponding to some parameter β>0\beta^{\ast}>0 such that,

(β,u0,ϕ0r,u0r,ρ0):=(β,u0(β),ϕ0r(β),u0r(β),ρ0(β))ΘN.(\beta^{\ast},u_{0}^{\ast},\phi_{0r}^{\ast},u_{0r}^{\ast},\rho_{0}^{\ast}):=(\beta^{\ast},u_{0}(\beta^{\ast}),\phi_{0r}(\beta^{\ast}),u_{0r}(\beta^{\ast}),\rho_{0}(\beta^{\ast}))\notin\Theta_{N}.

Without loss of generality, we assume ρ0>0\rho_{0}^{\ast}>0. Then there could be two situations. Either the inclusion in Definition 2.2 does not hold or the inclusion holds but (β,u0,ϕ0r,u0r,ρ0)(\beta^{\ast},u_{0}^{\ast},\phi_{0r}^{\ast},u_{0r}^{\ast},\rho_{0}^{\ast}) violates (2.2). Suppose the first case is true. Similar to how we argued above, we have,

A0kN2((y1s~0)N2N1,(y2s~0)N2N1).A_{0}^{\ast}\notin\frac{k}{N-2}\left(\left(\frac{y_{1}}{\tilde{s}_{0}}\right)^{\frac{N-2}{N}}-1,\left(\frac{y_{2}}{\tilde{s}_{0}}\right)^{\frac{N-2}{N}}-1\right).

Using (5.6) in the above expression, and subsequently with the help of Corollary 5.7, we obtain that there is blow up of density. On the other hand suppose the above inclusion holds and (2.2) does not. Then very similar to the way as we argued for the global existence result, we have that for q(0)0q(0)\neq 0,

η(0)(min{η1(0),η2(0)},max{η1(0),η2(0)}),\eta(0)\notin\left(\min\{\eta_{1}(0),\eta_{2}(0)\},\max\{\eta_{1}(0),\eta_{2}(0)\}\right),

and for q(0)=0q(0)=0,

w(0)(η1(0),η2(0)).w(0)\notin(\eta_{1}^{\prime}(0),\eta_{2}^{\prime}(0)).

From Corollary 6.7, we obtain that there is finite time tct_{c} with η(tc)=0\eta(t_{c})=0. From Proposition 5.2, there is breakdown at t=tct=t_{c} and the solution ceases to be smooth. This completes the proof of the Theorem. ∎

7. The zero background case

The zero background case has been analyzed by several researchers. Most notably, the authors in [29] give a sharp threshold condition, however, assuming that the flow is expanding (u0>0u_{0}>0). A more refined analysis was done by the author in [28], however, the threshold condition was not sharp. In this section, we present a sharp characterization of the subcritical and supercritical regions for general velocity. To this end, we consider (3.8) with c=0c=0,

(7.1a) ρ=(N1)ρqpρ,\displaystyle\rho^{\prime}=-(N-1)\rho q-p\rho,
(7.1b) p=p2k(N1)s+kρ,\displaystyle p^{\prime}=-p^{2}-k(N-1)s+k\rho,
(7.1c) q=ksq2,\displaystyle q^{\prime}=ks-q^{2},
(7.1d) s=Nqs,\displaystyle s^{\prime}=-Nqs,

with the same notation for initial data. Also recall the system (5.2). With zero background, it reduces to,

(7.2a) η=w,\displaystyle\eta^{\prime}=w,
(7.2b) w=ks(N1)η+kΓN1.\displaystyle w^{\prime}=-ks(N-1)\eta+k\Gamma^{N-1}.

We will make use of the same quantity AA as in (5.3). Using similar computations, one finds that the expression for AA is exactly the same as in (5.5) for N>2N>2. Also for N=2N=2, the expression for the corresponding quantity BB is the same as in (5.8).

A robust analysis of the q,sq,s system in this case has been carried out by the author in [28]. We only state the important results which will be used directly.

Proposition 7.1 ([28][Theorem 3.5, Lemma 3.4, 3.7).

] Consider the 2×22\times 2 ODE system (7.1c), (7.1d). Suppose s0>0s_{0}>0. Then

  • (q(t),s(t))(0,0)(q(t),s(t))\to(0,0) as tt\to\infty. Moreover, if q0<0q_{0}<0 there exists a unique time tqt_{q} such that q(tq)=0q(t_{q})=0, s(tq)=max[0,)s(t)s(t_{q})=\max_{[0,\infty)}s(t) and q(t)>0q(t)>0 for all t>tqt>t_{q}.

  • There is a TT large enough so that,

    (7.3a) Cmq(t+1)1\displaystyle C^{q}_{m}(t+1)^{-1}\leq q(t)CMq(t+1)1,\displaystyle q(t)\leq C^{q}_{M}(t+1)^{-1},
    (7.3b) Cms(t+1)N\displaystyle C_{m}^{s}(t+1)^{-N}\leq s(t)CMs(t+1)N,N3,\displaystyle s(t)\leq C_{M}^{s}(t+1)^{-N},\qquad N\geq 3,
    (7.3c) Cms(t+1)2(1+ln(t+1))1\displaystyle C_{m}^{s}(t+1)^{-2}(1+\ln(t+1))^{-1}\leq s(t)CMs(t+1)2(1+ln(t+1))1,N=2,\displaystyle s(t)\leq C_{M}^{s}(t+1)^{-2}(1+\ln(t+1))^{-1},N=2,

    for all tTt\geq T. Cmq,CMq,CMs,CmsC^{q}_{m},C^{q}_{M},C_{M}^{s},C_{m}^{s} are all positive constants depending on N,q0,s0N,q_{0},s_{0}.

Once again, the Poisson forcing is enough to avoid concentrations at the origin. In particular, Corollary 4.2 holds and s0>0s_{0}>0. We state it as a Lemma below.

Lemma 7.2.

q,sq,s in system (7.1c), (7.1d) exist for all time.

We now move on to some key results. We first give a sufficient condition for η\eta achieving zero in finite time.

Proposition 7.3.

Suppose N3N\geq 3. If

A0+kN2<0,A_{0}+\frac{k}{N-2}<0,

then there is a time tc>0t_{c}>0 such that η(tc)=0\eta(t_{c})=0.

Proof.

Since η\eta satisfies (6.4), we have,

(7.4) (ηektsq)=Aqektsq.\displaystyle\left(\eta e^{-k\int^{t}\frac{s}{q}}\right)^{\prime}=\frac{A}{q}e^{-k\int^{t}\frac{s}{q}}.

We will use the convergence estimates from Proposition 7.1. Suppose t1>0t_{1}>0 is a time so that the convergence rates are valid for all tt1t\geq t_{1}. Throughout our calculations 0<Cm<CM0<C_{m}<C_{M} are constants that may change from step to step but only depend on k,N,q0,s0,η0,w0,t1k,N,q_{0},s_{0},\eta_{0},w_{0},t_{1}. From (7.3), we have for t1τ<tt_{1}\leq\tau<t,

kτtsqCMτt(ξ+1)(N1)𝑑ξ,\displaystyle k\int_{\tau}^{t}\frac{s}{q}\leq C_{M}\int_{\tau}^{t}(\xi+1)^{-(N-1)}d\xi,
(7.5) kτtsqCM.\displaystyle k\int_{\tau}^{t}\frac{s}{q}\leq C_{M}.

Integrating (7.4), we have,

η(t)\displaystyle\eta(t) =η(t1)ekt1tsq+t1tA(τ)q(τ)ekτtsq𝑑τ\displaystyle=\eta(t_{1})e^{k\int_{t_{1}}^{t}\frac{s}{q}}+\int_{t_{1}}^{t}\frac{A(\tau)}{q(\tau)}e^{k\int_{\tau}^{t}\frac{s}{q}}d\tau
=:I+II.\displaystyle=:\textup{I}+\textup{II}.

From (7.5), I is uniformly bounded. Owing to (5.5) and (4.10), we can find convergence rates for AA as well. Using (7.3b) in (5.5) along with (4.10), we have for tt1t\geq t_{1},

A(t)\displaystyle A(t) =(A0+kN2)ΓkN2ΓN1\displaystyle=\left(A_{0}+\frac{k}{N-2}\right)\Gamma-\frac{k}{N-2}\Gamma^{N-1}
Cm(t+1)1.\displaystyle\leq-C_{m}(t+1)^{-1}.

Consequently, from (7.3a),

A(t)q(t)Cm,tt1.\frac{A(t)}{q(t)}\leq-C_{m},\quad t\geq t_{1}.

Since q,sq,s are positive and AA is negative, we have for t>t1t>t_{1},

II =t1tA(τ)q(τ)ekτtsq𝑑τ\displaystyle=\int_{t_{1}}^{t}\frac{A(\tau)}{q(\tau)}e^{k\int_{\tau}^{t}\frac{s}{q}}d\tau
t1tA(τ)q(τ)𝑑τ\displaystyle\leq\int_{t_{1}}^{t}\frac{A(\tau)}{q(\tau)}d\tau
Cmt1t𝑑τ.\displaystyle\leq-C_{m}\int_{t_{1}}^{t}d\tau.

Finally, for t>t1t>t_{1}, we have,

η(t)\displaystyle\eta(t) =I+II\displaystyle=\textup{I}+\textup{II}
CMCmt1t𝑑τ\displaystyle\leq C_{M}-C_{m}\int_{t_{1}}^{t}d\tau

Therefore, there exists a tct_{c} such that η(tc)=0\eta(t_{c})=0. This completes the proof. ∎

We now move on to characterizing the subcritical region. We first define a function η1\eta_{1} through an IVP using (6.4),

(7.6) η1=Aq+ksqη1,η1(tA)=0,\displaystyle\eta_{1}^{\prime}=\frac{A}{q}+\frac{ks}{q}\eta_{1},\quad\eta_{1}(t_{A})=0,

where tA>0t_{A}>0 is a time (if it exists) such that A(tA)=0A(t_{A})=0.

Proposition 7.4.

Consider the N3N\geq 3 case. Suppose s0>0,q0>0s_{0}>0,q_{0}>0. Given η0>0\eta_{0}>0, we have that η(t)>0\eta(t)>0 for all t>0t>0 if one of the following conditions is satisfied,

  • A00A_{0}\geq 0,

  • kN2<A0<0-\frac{k}{N-2}<A_{0}<0 and η0>η1(0)\eta_{0}>\eta_{1}(0).

Additionally, if kN2<A0<0-\frac{k}{N-2}<A_{0}<0 and η0η1(0)\eta_{0}\leq\eta_{1}(0), then there is a time tc>0t_{c}>0 such that η(tc)=0\eta(t_{c})=0.

Proof.

Firstly, note that the hypothesis and Proposition 7.1 imply s,qs,q are positive for all times. Hence, from (7.1d), ss is monotonically decreasing to zero. Owing to (4.10), Γ(t)=e0tq\Gamma(t)=e^{-\int_{0}^{t}q} is strictly decreasing and tends to zero as tt\to\infty. Moreover, if A00A_{0}\geq 0, then,

κ1,\kappa\geq 1,

where κ\kappa is the root of A(Γ)A(\Gamma) as in (5.6). From (5.5), A(t)>0A(t)>0 for all t>0t>0. Consequently, from (7.4), η\eta can never be zero in finite time if it was initially positive.

Now suppose the second hypothesis holds. Here, we have that κ<1\kappa<1. Owing to the properties of s,q,Γ,As,q,\Gamma,A as listed above, we obtain that there is a unique time tA>0t_{A}>0 such that A(tA)=0A(t_{A})=0. Moreover, A(t)<0A(t)<0 for t<tAt<t_{A} and A(t)>0A(t)>0 for t>tAt>t_{A}. Owing to (7.4) once again, η\eta remains greater than zero if it is so at t=tAt=t_{A}. In fact, η1(t)\eta_{1}(t) serves as a lower bound for η(t)\eta(t) at each time, see Figure 6. Since η,η1\eta,\eta_{1} satisfy the same first order ODE we have a comparison principle. Indeed on taking difference, we obtain,

(ηη1)=ksq(ηη1).(\eta-\eta_{1})^{\prime}=\frac{ks}{q}(\eta-\eta_{1}).

From (7.6), we have η1(tA)=0\eta_{1}^{\prime}(t_{A})=0. Also, upon taking derivative of (7.6), one can check that η1′′(tA)>0\eta_{1}^{\prime\prime}(t_{A})>0 and, therefore, tAt_{A} is indeed the unique time where the minimum of η1\eta_{1} is attained. Hence, η(t)>0\eta(t)>0 for all t>0t>0. Conversely, if η0η1(0)\eta_{0}\leq\eta_{1}(0), then since η\eta remains below η1\eta_{1}, there is a time tctAt_{c}\leq t_{A}, such that η(tc)=0\eta(t_{c})=0. ∎

Refer to caption
Figure 6. η1\eta_{1} and a plausible solution η\eta. k=1,N=4,q0=0.1,s0=1,A0=0.4k=1,N=4,q_{0}=0.1,s_{0}=1,A_{0}=-0.4
Proposition 7.5.

Consider the N3N\geq 3 case. Suppose s0>0,q0=0s_{0}>0,q_{0}=0 and A0>kN2A_{0}>-\frac{k}{N-2}. Given η0>0\eta_{0}>0, we have that η(t)>0\eta(t)>0 for all t>0t>0 if and only if w0>η1(0)w_{0}>\eta_{1}^{\prime}(0).

Note that η1(0)\eta_{1}^{\prime}(0) is in the limit sense in (7.6) since q(0)=0q(0)=0. We know that this limit exists because η1\eta_{1} satisfies (6.5) (with c=0c=0), which is an inhomogeneous second order linear ODE with bounded coefficients. Also, the assumption q0=0q_{0}=0 implies A0<0A_{0}<0.

Proof.

The proof of Proposition 7.5 is very similar to that of the second assertion of Proposition 7.4. The comparison principle holds because of uniqueness of ODE. Since η,η1\eta,\eta_{1} both satisfy (6.5) and η0=η1(0)=A0ks0\eta_{0}=\eta_{1}(0)=-\frac{A_{0}}{ks_{0}}, we have that if η(0)=w0>η1(0)\eta^{\prime}(0)=w_{0}>\eta_{1}^{\prime}(0), then η(t)>η1(t)\eta(t)>\eta_{1}(t) for all time, thereby maintaining positivity, since η1(t)0\eta_{1}(t)\geq 0. Converse also holds since η1(tA)=0\eta_{1}(t_{A})=0. ∎

Now we present the result for q0<0q_{0}<0. We take derivative of (7.6) and state two second order IVPs for i=1,2i=1,2,

(7.7) ηi′′+k(N1)sηi=kΓN1,ηi(tAi)=0,ηi(tAi)=0,\displaystyle\begin{aligned} &\eta_{i}^{\prime\prime}+k(N-1)s\eta_{i}=k\Gamma^{N-1},\\ &\eta_{i}(t_{A}^{i})=0,\qquad\eta_{i}^{\prime}(t_{A}^{i})=0,\end{aligned}

where tAi0t_{A}^{i}\geq 0 is such that A(tAi)=0A(t_{A}^{i})=0. There need not be two such times. In that case, we only consider η1\eta_{1}. Note that by uniqueness, η1\eta_{1} in (7.6) is the same function as η1\eta_{1} above. Also, similar to (4.2), we obtain the trajectory for this case as,

(7.8) q2s2N+2kN2s12N=RN,\displaystyle q^{2}s^{-\frac{2}{N}}+\frac{2k}{N-2}s^{1-\frac{2}{N}}=R_{N},

From this, one can directly note the maximum attained value of ss, which is when q=0q=0, is

(7.9) smax:=(RN(N2)2k)NN2.\displaystyle s_{max}:=\left(\frac{R_{N}(N-2)}{2k}\right)^{\frac{N}{N-2}}.
Proposition 7.6.

Consider the N3N\geq 3 case. Suppose s0>0,q0<0s_{0}>0,q_{0}<0. Given η0>0\eta_{0}>0, we have the following.

  • Suppose A00A_{0}\geq 0. Then η(t)>0\eta(t)>0 for all t>0t>0 if and only if κ<(smax/s0)1/N\kappa<(s_{max}/s_{0})^{1/N} and η1(0)<η0<η2(0)\eta_{1}(0)<\eta_{0}<\eta_{2}(0).

  • Suppose kN2<A0<0-\frac{k}{N-2}<A_{0}<0. Then η(t)>0\eta(t)>0 for all t>0t>0 if and only if η0<η1(0)\eta_{0}<\eta_{1}(0).

Proof.

Suppose A00A_{0}\geq 0. From the first assertion of Proposition 7.1 and (5.3), we conclude that at t=tqt=t_{q},

η(tq)=A(tq)ks(tq)=A(tq)ksmax.\eta(t_{q})=-\frac{A(t_{q})}{ks(t_{q})}=-\frac{A(t_{q})}{ks_{max}}.

Therefore, a necessary condition for η\eta to be positive is that A(tq)<0A(t_{q})<0. From (4.10) and (5.5), the condition is equivalent to,

κ<(smaxs0)1N.\kappa<\left(\frac{s_{max}}{s_{0}}\right)^{\frac{1}{N}}.

Here, we must keep in mind that the dynamics of ss of is such that increases until t=tqt=t_{q} then decreases monotonically and approaches zero as tt\to\infty. If the above inequality holds, then there are two positive times, tAi,i=1,2t_{A}^{i},i=1,2, when A(tAi)=0A(t_{A}^{i})=0. Also, tq(tA1,tA2)t_{q}\in(t_{A}^{1},t_{A}^{2}). The very same arguments as in Proposition 6.4 allow us to conclude that ηi(t)>0\eta_{i}(t)>0 for t[0,)\{tAi}t\in[0,\infty)\backslash\{t_{A}^{i}\}. Since ηi\eta_{i}’s also satisfy (7.6), the all-time-positivity of η\eta is guaranteed once again by arguments as in Lemma 6.1 if it was in between η1\eta_{1} and η2\eta_{2} at t=0t=0. In particular, assertions 1-4 of Lemma 6.1 are valid with a slight modification that the set 𝔻={tq}\mathbb{D}=\{t_{q}\} has only one element, see Figure 7 for a visualization. Conversely, if η0(η1(0),η2(0))\eta_{0}\notin(\eta_{1}(0),\eta_{2}(0)), then from Lemma 6.1 η\eta becomes zero in finite time.

If kN2<A0<0-\frac{k}{N-2}<A_{0}<0, then κ<1\kappa<1. From Proposition 7.1, ss increases to a maximum and then decreases to zero. Once again, making use of the relation (4.10) and (5.5), we have that AA is zero only once, at t=tAt=t_{A}. From (7.6), we have,

(η1ektsq)=Aqektsq.\left(\eta_{1}e^{-k\int^{t}\frac{s}{q}}\right)^{\prime}=\frac{A}{q}e^{-k\int^{t}\frac{s}{q}}.

We conclude that for t<tqt<t_{q}, η(t)>0\eta(t)>0 since A,qA,q are both negative in this interval. Since from Lemma 6.1, η(t)>η1(t)\eta(t)>\eta_{1}(t) for t>tqt>t_{q}, we conclude that η(t)>0\eta(t)>0 for all t>tqt>t_{q} since η1\eta_{1} is nonnegative and serves as a lower bound for η\eta in this domain. Conversely, if η0>η1(0)\eta_{0}>\eta_{1}(0), then η(t)<η1(t)\eta(t)<\eta_{1}(t) for all t>tqt>t_{q} and hence, it must be zero in a finite time, t<tAt<t_{A}, since η1(tA)=0\eta_{1}(t_{A})=0. ∎

Refer to caption
Figure 7. η1,η2\eta_{1},\eta_{2} and a plausible solution, η\eta. k=1,N=4,q0=5,s0=1,A0=1k=1,N=4,q_{0}=-5,s_{0}=1,A_{0}=1
Proposition 7.7.

Consider the N3N\geq 3 case. Suppose s0>0,q0s_{0}>0,q_{0} be given. If

A0=kN2,A_{0}=-\frac{k}{N-2},

then there is a time tc>0t_{c}>0 such that η(tc)=0\eta(t_{c})=0.

We state this proposition and its proof separately because the technique used in the proof of Proposition 7.3 does not apply. In fact, it turns out that it is much easier to work in the p,ρp,\rho variables instead of the η,w\eta,w variables.

Proof of Proposition 7.7: Using (5.3) and A0=k/(N2)A_{0}=-k/(N-2) in the expression of AA as in (5.5), we obtain,

q(t)w(t)kη(t)s(t)=A(t)=kN2ΓN1.q(t)w(t)-k\eta(t)s(t)=A(t)=-\frac{k}{N-2}\Gamma^{N-1}.

Substituting for η,w\eta,w using (5.1), we obtain,

qpksρ=kN2.\frac{qp-ks}{\rho}=-\frac{k}{N-2}.

As a result, we can find pp in terms of the other variables as,

p=ksqkρ(N2)q.p=\frac{ks}{q}-\frac{k\rho}{(N-2)q}.

We can divide by qq because from (7.3a), qq is eventually positive and decays accordingly. We will consider sufficiently large times so that this holds. Plugging this in the ODE of ρ\rho, (7.1a), we get,

ρ=kq(N2)ρ2ρ(q(N1)+ksq).\rho^{\prime}=\frac{k}{q(N-2)}\rho^{2}-\rho\left(q(N-1)+\frac{ks}{q}\right).

For all sufficiently large times, the rates in Proposition 7.1 hold. We can assume ρ\rho has not already blown up, because if it has then the proof is done. Using these rates, we conclude that ks/qks/q, qq are bounded. Therefore, there occurs a Riccati-type blow up of density, and the blow up is aggravated by the qq in the denominator. ∎

Propositions 7.3, 7.4, 7.5, 7.6 and 7.7 enable us to put together the picture for N3N\geq 3. Now, we move onto the critical case N=2N=2. We will omit the repetitive parts in the proofs of these results. Note the quantity BB in (5.8) analogous to AA. Also note that there always exist a positive root of B(Γ)B(\Gamma), κ=eA0k\kappa=e^{\frac{A_{0}}{k}}, no matter the sign of A0A_{0}, which is unlike the case for AA. κ\kappa could lie on or either side of 11 depending on the sign of A0A_{0}. For the N=2N=2 case, we will consider the function η1\eta_{1} as in (7.6) and the functions ηi\eta_{i}’s as in (7.7) with AA replaced by BB in the definition.

Proposition 7.8.

Suppose N=2N=2 and consider η1\eta_{1} as in (7.6). Suppose s0>0,q0>0s_{0}>0,q_{0}>0. Given η0>0\eta_{0}>0, we have that η(t)>0\eta(t)>0 for all t>0t>0 if one of the following conditions is satisfied,

  • A00A_{0}\geq 0,

  • A0<0A_{0}<0 and η0>η1(0)\eta_{0}>\eta_{1}(0).

Additionally, if A0<0A_{0}<0 and η0η1(0)\eta_{0}\leq\eta_{1}(0), then there is a time tc>0t_{c}>0 such that η(tc)=0\eta(t_{c})=0.

The proof is very similar to that of Proposition 7.4.

Proposition 7.9.

Consider the N=2N=2 case. Suppose s0>0,q0=0s_{0}>0,q_{0}=0. Given η0>0\eta_{0}>0, we have that η(t)>0\eta(t)>0 for all t>0t>0 if and only if w0>η1(0)w_{0}>\eta_{1}^{\prime}(0).

The proof is very similar to that of Proposition 7.5.

Proposition 7.10.

Consider the N=2N=2 case. Suppose s0>0,q0<0s_{0}>0,q_{0}<0. Given η0>0\eta_{0}>0, we have the following.

  • Suppose A00A_{0}\geq 0. Then η(t)>0\eta(t)>0 for all t>0t>0 if and only if eA0k<(s2,max/s0)e^{\frac{A_{0}}{k}}<\sqrt{(s_{2,max}/s_{0})} and η1(0)<η0<η2(0)\eta_{1}(0)<\eta_{0}<\eta_{2}(0).

  • Suppose A0<0A_{0}<0. Then η(t)>0\eta(t)>0 for all t>0t>0 if and only if η0<η1(0)\eta_{0}<\eta_{1}(0).

Once again, the proof is very similar to that of Proposition 7.6. The expression for s2,maxs_{2,max} is different from smaxs_{max} since the trajectory equation in the critical case is different. Similar to the way we obtained (7.8), we can obtain the trajectory for the critical case N=2N=2,

(7.10) q2s+kln(s)=R2,\displaystyle\frac{q^{2}}{s}+k\ln(s)=R_{2},

where R2=q02s0+kln(s0)R_{2}=\frac{q_{0}^{2}}{s_{0}}+k\ln(s_{0}). From this, one concludes,

s2,max=eR2k.s_{2,max}=e^{\frac{R_{2}}{k}}.

Finally, we analyze the case when the initial density is zero, that is, ρ0=0\rho_{0}=0. Firstly, from (7.1a), ρ0=0\rho_{0}=0 is equivalent to ρ0\rho\equiv 0, as long as pp in (7.1b) exists. Therefore, as long as pp exists, (7.1b) reduces to,

(7.11) p=p2k(N1)s.\displaystyle p^{\prime}=-p^{2}-k(N-1)s.
Proposition 7.11.

Consider (7.1) and suppose ρ0=0\rho_{0}=0 in (7.1a). Then pp is bounded for all times if and only if q0>0q_{0}>0 and p0ks0q0p_{0}\geq\frac{ks_{0}}{q_{0}}. Moreover, if q0>0q_{0}>0 and p0ks0q0p_{0}\geq\frac{ks_{0}}{q_{0}} then

p(t)ks(t)q(t),t>0,p(t)\geq\frac{ks(t)}{q(t)},\quad t>0,

and if q00q_{0}\leq 0 or p0<ks0q0p_{0}<\frac{ks_{0}}{q_{0}}, then there exists a time, tc>0t_{c}>0, such that,

limttcp(t)=.\lim_{t\to t_{c}^{-}}p(t)=-\infty.
Proof.

Consider the quantity qpksqp-ks. From (7.11), (7.1c) and (7.1d), we have,

(qpks)\displaystyle(qp-ks)^{\prime} =p(ksq2)q(p2+k(N1)s)+kNsq\displaystyle=p(ks-q^{2})-q(p^{2}+k(N-1)s)+kNsq
=kpspq2qp2+kqs\displaystyle=kps-pq^{2}-qp^{2}+kqs
=(p+q)(qpks).\displaystyle=-(p+q)(qp-ks).

Consequently, as long as pp exists, qpksqp-ks maintains sign and we have,

(7.12) qpks=(q0p0ks0)e0tqe0tp.\displaystyle qp-ks=(q_{0}p_{0}-ks_{0})e^{-\int_{0}^{t}q}e^{-\int_{0}^{t}p}.

Suppose q0>0q_{0}>0 and p0ks0q0p_{0}\geq\frac{ks_{0}}{q_{0}}. From the first assertion of Proposition 7.1, we have that q(t)>0q(t)>0 for all time. For the sake of contradiction, suppose that pp breaks down in finite time. Since, ss is uniformly bounded, it is clear from (7.11) that at the time of breakdown, pp\to-\infty. Since q,s,e0tq,s,e^{-\int_{0}^{t}} are uniformly bounded quantities, it must be that at a certain time before breakdown, the left-hand-side in (7.12) is negative but the right-hand-side is non-negative. This is a contradiction and hence, p(t)p(t) is finite for all times. In particular, it decays with a lower bound, p(t)ks(t)/q(t)p(t)\geq ks(t)/q(t), the decay rate for which can be obtained directly from Proposition 7.1.

Now suppose q0>0q_{0}>0 but p0<ks0q0p_{0}<\frac{ks_{0}}{q_{0}}. We first assume N3N\geq 3. The critical case when N=2N=2 needs separate treatment in this regard and will be analyzed at the end. Note that if pp has not broken down, the following holds,

k(N1)ts(τ)𝑑τp(t)ks(t)q(t).k(N-1)\int_{t}^{\infty}s(\tau)d\tau\leq p(t)\leq\frac{ks(t)}{q(t)}.

From Proposition 7.1, the integral on the left-hand-side is well-defined. The second inequality is a direct result of (7.12). The first inequality can be shown as follows. Suppose for the sake of contradiction, there is a t1t_{1} such that k(N1)t1s(τ)𝑑τ>p(t1)k(N-1)\int_{t_{1}}^{\infty}s(\tau)d\tau>p(t_{1}). Then using (7.11), we obtain that for some t2>t1t_{2}>t_{1},

p(t2)<p(t1)k(N1)t1t2s(τ)𝑑τ<0.p(t_{2})<p(t_{1})-k(N-1)\int_{t_{1}}^{t_{2}}s(\tau)d\tau<0.

Consequently, from (7.11), we conclude that a Riccati-type blowup occurs and pp\to-\infty at some time greater than t2t_{2}.

Therefore, we can assume that k(N1)ts(τ)𝑑τp(t)k(N-1)\int_{t}^{\infty}s(\tau)d\tau\leq p(t) for all t>0t>0. Choose tt_{\ast} large enough so that the convergence estimates of Proposition 7.1 hold. Using these estimates, we can rewrite the above bounds on pp as follows,

(7.13) 0<C1(1+t)(N1)p(t)C2(1+t)(N1),tt\displaystyle 0<C_{1}(1+t)^{-(N-1)}\leq p(t)\leq C_{2}(1+t)^{-(N-1)},\quad t\geq t_{\ast}

CisC_{i}^{\prime}s are appropriate positive constants whose values may change through the proof but they depend only on s0,q0,t,ks_{0},q_{0},t_{\ast},k. Once again from Proposition 7.1, and (7.13), we obtain

0>qpksC1(1+t)NC2(1+t)NC1(1+t)N.0>qp-ks\geq C_{1}(1+t)^{-N}-C_{2}(1+t)^{-N}\geq-C_{1}(1+t)^{-N}.

Now we analyze the right-hand-side of (7.12).

(q0p0ks0)e0tqe0tp\displaystyle(q_{0}p_{0}-ks_{0})e^{-\int_{0}^{t}q}e^{-\int_{0}^{t}p} =C2s1Ne0tp\displaystyle=-C_{2}s^{\frac{1}{N}}e^{-\int_{0}^{t}p}
C2(1+t)1.\displaystyle\leq-C_{2}(1+t)^{-1}.

We used Lemma 4.5 to obtain the equality. We used (7.13) to conclude that pp is integrable and hence, the inequality holds. Combining the above two inequalities for the two sides of the equation (7.12), we have for sufficiently large times,

qpksC1(1+t)N>C2(1+t)1(q0p0ks0)e0tqe0tp,qp-ks\geq-C_{1}(1+t)^{-N}>-C_{2}(1+t)^{-1}\geq(q_{0}p_{0}-ks_{0})e^{-\int_{0}^{t}q}e^{-\int_{0}^{t}p},

which is a contradiction. Therefore, pp must blow up in finite time, that is, limttcp(t)=\lim_{t\to t_{c}^{-}}p(t)=-\infty for some tc>0t_{c}>0.

Now suppose q00q_{0}\leq 0. Firstly, note that p0>0p_{0}>0 because if not, then a Riccati-type blowup occurs. This implies

q0p0ks0<0.q_{0}p_{0}-ks_{0}<0.

Hence, qpks<0qp-ks<0 for all time. Now from Proposition 7.1, we know that after a sufficiently large time q>0q>0, and q,sq,s follow the convergence rates. Hence, the same arguments as above apply which lead to a finite-time-breakdown of pp.

We now analyze the N=2N=2 case. Just as for N3N\geq 3 case, we need to prove a contradiction when q0>0q_{0}>0 and p0<ks0q0p_{0}<\frac{ks_{0}}{q_{0}}. All the other arguments are the same. To this end, we assume q0>0q_{0}>0 and p0<ks0q0p_{0}<\frac{ks_{0}}{q_{0}}. Assuming pp exists for all times, we have the following,

kts(τ)𝑑τp(t)ks(t)q(t).k\int_{t}^{\infty}s(\tau)d\tau\leq p(t)\leq\frac{ks(t)}{q(t)}.

For ttt\geq t_{\ast} (tt_{\ast} such that the rates of Proposition 7.1 hold), we have

(7.14) C1t(τ+1)2(1+ln(1+τ))1𝑑τp(t)C2(t+1)1(1+ln(1+t))1,\displaystyle C_{1}\int_{t}^{\infty}(\tau+1)^{-2}(1+\ln(1+\tau))^{-1}d\tau\leq p(t)\leq C_{2}(t+1)^{-1}(1+\ln(1+t))^{-1},

for all ttt\geq t_{\ast}. We focus on the integral above. A substitution changes it into,

C11+ln(1+t))eττ𝑑τ.C_{1}\int_{1+\ln(1+t))}^{\infty}\frac{e^{-\tau}}{\tau}d\tau.

This expression can be represented using a well-known exponential integral function given by

E1(t)=teττ𝑑τ.E_{1}(t)=\int_{t}^{\infty}\frac{e^{-\tau}}{\tau}d\tau.

Moreover, it has the following bounds, see [1, Page 229],

et2ln(1+2t)<E1(t)<etln(1+1t).\frac{e^{-t}}{2}\ln\left(1+\frac{2}{t}\right)<E_{1}(t)<e^{-t}\ln\left(1+\frac{1}{t}\right).

Using the above lower bound in (7.14), we obtain the following bounds,

(7.15) C11+tln(1+21+ln(1+t))p(t)C21+t(1+ln(1+t))1,tt.\displaystyle\frac{C_{1}}{1+t}\ln\left(1+\frac{2}{1+\ln(1+t)}\right)\leq p(t)\leq\frac{C_{2}}{1+t}(1+\ln(1+t))^{-1},\quad t\geq t_{\ast}.

Therefore, we have,

0>qpks\displaystyle 0>qp-ks C1(1+t)2ln(1+21+ln(1+t))C2(1+t)21(1+ln(1+t))\displaystyle\geq\frac{C_{1}}{(1+t)^{2}}\ln\left(1+\frac{2}{1+\ln(1+t)}\right)-\frac{C_{2}}{(1+t)^{2}}\frac{1}{(1+\ln(1+t))}
C1(1+t)21(1+ln(1+t)),\displaystyle\geq-\frac{C_{1}}{(1+t)^{2}}\frac{1}{(1+\ln(1+t))},

for all times sufficiently large. On the other hand, if we analyze the right-hand-side of (7.12), we see that for ttt\geq t_{\ast},

(q0p0ks0)s0se0tp\displaystyle\frac{(q_{0}p_{0}-ks_{0})}{\sqrt{s_{0}}}\sqrt{s}e^{-\int_{0}^{t}p} C1(1+t)1+ln(1+t)e0tp\displaystyle\leq-\frac{C_{1}}{(1+t)\sqrt{1+\ln(1+t)}}e^{-\int_{0}^{t}p}
=C1(1+t)1+ln(1+t)ettp\displaystyle=-\frac{C_{1}}{(1+t)\sqrt{1+\ln(1+t)}}e^{-\int_{t_{\ast}}^{t}p}
C1(1+t)1+ln(1+t)ettC21+τ(1+ln(1+τ))1𝑑τ\displaystyle\leq-\frac{C_{1}}{(1+t)\sqrt{1+\ln(1+t)}}e^{-\int_{t_{\ast}}^{t}\frac{C_{2}}{1+\tau}(1+\ln(1+\tau))^{-1}d\tau}
=C1(1+t)1+ln(1+t)eC2ln(1+ln(1+t))\displaystyle=-\frac{C_{1}}{(1+t)\sqrt{1+\ln(1+t)}}e^{-C_{2}\ln(1+\ln(1+t))}
=C1(1+t)[1+ln(1+t)]C2.\displaystyle=-\frac{C_{1}}{(1+t)[1+\ln(1+t)]^{C_{2}}}.

Once again, this contradicts (7.12) since for sufficiently large times,

qpksC1(1+t)2(1+ln(1+t))>C1(1+t)[1+ln(1+t)]C2(q0p0ks0)e0tqe0tp.qp-ks\geq-\frac{C_{1}}{(1+t)^{2}(1+\ln(1+t))}>-\frac{C_{1}}{(1+t)[1+\ln(1+t)]^{C_{2}}}\geq(q_{0}p_{0}-ks_{0})e^{-\int_{0}^{t}q}e^{-\int_{0}^{t}p}.

This completes the proof. ∎

Proof of Theorem 2.4: Suppose the hypothesis holds. Then by using (5.1) in Proposition 7.3, one immediately obtains the blow up of density in finite time if A0<k/(N2)A_{0}<-k/(N-2).

If A0=k/(N2)A_{0}=-k/(N-2), then the result follows from Proposition 7.7. ∎

Proof of Theorem 2.6: Suppose initial data satisfies the hypothesis of the Theorem. Along the characteristic path (3.1), this translates to the condition that for all β>0\beta>0,

(β,u0(β),ϕ0r(β),u0r(β),ρ0(β))ΣN{(β,x,y,z,0):x>0,zky/x}.(\beta,u_{0}(\beta),\phi_{0r}(\beta),u_{0r}(\beta),\rho_{0}(\beta))\in\Sigma_{N}\cup\{(\beta,x,y,z,0):x>0,z\geq-ky/x\}.

We will now analyze a single characteristic path and replace the initial data notations with (β,u0,ϕ0r,u0r,ρ0)(\beta,u_{0},\phi_{0r},u_{0r},\rho_{0}). Under the transformation (3.7), we now turn to the unknowns of system (7.1), (q,s,p,ρ)(q,s,p,\rho). Global-in-time existence of these variables is equivalent to the global-in-time existence of the original variables. If ρ(0)=0\rho(0)=0, then Proposition 7.11 gives the all-time existence of pp and hence, the solution is global.

Next, we prove global existence for the case when ρ(0)>0\rho(0)>0. Turning to the Definition 2.7, we will use the equivalence of aa and (5.3), and analyze the conditions (2.4), (2.5), (2.6), (2.7) one by one. To this end, let it be such that,

A0(kN2,0).A_{0}\in\left(-\frac{k}{N-2},0\right).

We thus have fulfilment of (2.4). Rearranging (2.4), and noting the transformations (3.7), (5.1), we obtain for x0x\neq 0,

η0βq0<η1(0)βq0,\frac{\eta_{0}}{\beta q_{0}}<\frac{\eta_{1}(0)}{\beta q_{0}},

and for x=0x=0,

w0>dη1(0)dt.w_{0}>\frac{d\eta_{1}(0)}{dt}.

Also, x=0x=0 if and only if q0=0q_{0}=0. The first inequality then is the hypothesis to the second assertion of Proposition 7.4 and the second assertion of Proposition 7.6 (depending on whether q0>0q_{0}>0 or q0<0q_{0}<0). The second inequality is the hypothesis to Proposition 7.5. As a result, through (5.1), we obtain the all time existence of ρ,p\rho,p in (7.1a), (7.1b).

Next, we suppose that A0=0A_{0}=0. Then (2.5) reduces to η0<η2(0)\eta_{0}<\eta_{2}(0) for q0<0q_{0}<0 and there are no extra conditions if q0>0q_{0}>0. Note that q0q_{0} cannot be equal to zero because that would be a violation to A0=0A_{0}=0. These two scenarios form the hypothesis of the first assertion of Proposition 7.4 and the first assertion of Proposition 7.6. Therefore, we have the all time existence of ρ,p\rho,p.

Now, suppose

A0(0,kN2((βy𝔐,Ny)12N1)).A_{0}\in\left(0,\frac{k}{N-2}\left(\left(\frac{-\beta y_{\mathfrak{M},N}}{y}\right)^{1-\frac{2}{N}}-1\right)\right).

Then using (7.9) with y𝔐,N=smaxy_{\mathfrak{M},N}=s_{max} and rearranging this gives,

κ(1,(smaxs0)1N),\kappa\in\left(1,\left(\frac{s_{max}}{s_{0}}\right)^{\frac{1}{N}}\right),

with κ\kappa as in (5.6). Note that yMy^{M}, which in the definition of ΣN\Sigma_{N} initially, was not given explicitly is now given explicitly by (7.9). Clearly it only depends on q0,s0q_{0},s_{0}. Also note that since A0>0A_{0}>0, κ>1\kappa>1 directly from its formula, (5.6). From (2.6), we obtain,

η1(0)<η0<η2(0),\eta_{1}(0)<\eta_{0}<\eta_{2}(0),

if q0<0q_{0}<0, and no extra conditions if q0>0q_{0}>0. Note that q0q_{0} cannot be equal to zero because that would be a violation to A0>0A_{0}>0. Therefore, all time existence of ρ,p\rho,p follows from the first assertion of Proposition 7.6 (if q0<0q_{0}<0) and first assertion of Proposition 7.4 (if q0>0q_{0}>0).

Lastly, we assume

A0kN2((βy𝔐,Ny)12N1).A_{0}\geq\frac{k}{N-2}\left(\left(\frac{-\beta y_{\mathfrak{M},N}}{y}\right)^{1-\frac{2}{N}}-1\right).

Similar to how we argued in the previous case, here the above inequality is equivalent to saying that κ(smax/s0)1N\kappa\geq\left(s_{max}/s_{0}\right)^{\frac{1}{N}}. The conditions (2.7) imply that q0>0q_{0}>0. We can then apply the first assertion of Proposition 7.4 to obtain the all time existence of ρ,p\rho,p.

Putting all the cases together and applying Proposition 5.2, we have obtained that solutions to (7.1) exist for all time if any single characteristic lies in any of the above situations. In particular, if all the characteristics satisfy the above conditions, then an application of Lemma 3.1 and Theorem 1.1 gives the existence of global-in-time solutions to (1.2) with c=0c=0.

Conversely, suppose there is a characteristic path corresponding to some parameter β>0\beta^{\ast}>0 such that,

(β,u0,ϕ0r,u0r,ρ0)\displaystyle(\beta^{\ast},u_{0}^{\ast},\phi_{0r}^{\ast},u_{0r}^{\ast},\rho_{0}^{\ast}) :=(β,u0(β),ϕ0r(β),u0r(β),ρ0(β))\displaystyle:=(\beta^{\ast},u_{0}(\beta^{\ast}),\phi_{0r}(\beta^{\ast}),u_{0r}(\beta^{\ast}),\rho_{0}(\beta^{\ast}))
ΣN{(β,x,y,z,0):x>0,zky/x}.\displaystyle\notin\Sigma_{N}\cup\{(\beta,x,y,z,0):x>0,z\geq-ky/x\}.

If ρ0=0\rho_{0}^{\ast}=0, then a direct application of Proposition 7.11 gives the finite-time-breakdown of pp.

Now, we suppose ρ0>0\rho_{0}^{\ast}>0. Then it could be that A0k/(N2)A_{0}^{\ast}\leq-k/(N-2). Finite time breakdown is then a direct result of Propositions 7.3 or 7.7.

If A0>k/(N2)A_{0}^{\ast}>-k/(N-2), then negation of one of the conditions among (2.4), (2.5), (2.6), (2.7) has to be true as and according to the value of A0A_{0}^{\ast}. Once again, we can check each condition one by one. All the analysis is a repetition of above except from the fact that, instead of the all-time existence results, we use the finite-time-breakdown results of Propositions 7.4, 7.5 and 7.6. Consequently, solutions to (1.2) cease to be smooth.

This completes the proof of the Theorem. ∎

Theorem 2.8 can be proved in a very similar way to that of Theorem 2.6 only that instead of Propositions 7.4, 7.5 and 7.6, we use Propositions 7.8, 7.9 and 7.10.

8. Conclusion

The techniques developed in this paper work for the one-dimensional case as well. However, there is one main difference in the 1D and multi-dimensional scenarios. As pointed out by the author in [28], that in multi-D, the Poisson forcing is enough to avoid flow concentration at the origin. In particular, by Corollary 4.2, we know that no matter how large (in absolute value) the initial velocity is, there are no concentrations at the origin. Let’s do similar calculations for the 1D case. In 1D, system (3.8c), (3.8d) for N=1N=1 reduce to,

q=ks~kcq2,s~=qs~,q^{\prime}=k\tilde{s}-kc-q^{2},\qquad\tilde{s}^{\prime}=-q\tilde{s},

with s~:=s+c\tilde{s}:=s+c. Using a transformation, a=q/s~,b=1/s~a=q/\tilde{s},b=1/\tilde{s}, we obtain a simple linear ODE system,

a=kkcb,b=a.a^{\prime}=k-kcb,\qquad b^{\prime}=a.

We can analytically solve this to obtain,

b(t)=1c+(b(0)1c)cos(kct)+a(0)kcsin(kct).b(t)=\frac{1}{c}+\left(b(0)-\frac{1}{c}\right)\cos(\sqrt{kc}t)+\frac{a(0)}{\sqrt{kc}}\sin(\sqrt{kc}t).

From this, one can conclude that b(t)=0b(t_{\ast})=0 (or equivalently, limtts~(t)=\lim_{t\to t_{\ast}^{-}}\tilde{s}(t)=\infty) for some t>0t_{\ast}>0 if

a(0)22kb(0)kc(b(0))2,a(0)^{2}\geq 2kb(0)-kc(b(0))^{2},

which in the original variables is equivalent to,

|q|>k(2s~(0)c).|q|>\sqrt{k(2\tilde{s}(0)-c)}.

Hence, for a large initial velocity (absolute value), there are concentrations at the origin. For c=0c=0, the result is one sided, that is, a sufficiently large negative initial velocity (flow pointing towards origin) would lead to concentrations at the origin. This concentration is completely avoided for N2N\geq 2, wherein the Poisson forcing turns out to be sufficient.

In this work, we make a subsequent important discovery that, in multi-D, the subcritical region can contain arbitrarily large initial velocities. In other words, no matter how large the initial velocity is (positive or negative), one can have a region in the phase plane of initial density and gradient of velocity, that corresponds to all-time-existence of the solution.

Moreover, we are hopeful that our techniques can be applied to other systems such as Euler-Poisson-alignment and Euler-Poisson with swirl. However, we leave that for future studies.

Acknowledgments

This research was partially supported by the National Science Foundation under Grant DMS1812666.


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