Global Existence and Finite-Time Blow-Up for a Nonlinear Nonlocal Evolution Equation

Abstract

We study the Cauchy problem for a nonlinear nonlocal evolution equation arising in the study of oceanic flows in equatorial regions. We present a well-posedness result and show that while some initial data develop into solutions that exist for all times, others lead to blow-up in finite time.

1 Introduction

We consider the Cauchy problem for spatially periodic solutions of the evolution equation

$$\begin{aligned} u_t + u u_x + (\mathcal {H}u) (\mathcal {H}u) _x + \beta \,{\mathcal {H}} u - \mu \, u_{xx} = k(t,x) \end{aligned}$$

(1.1)

with initial data

$$\begin{aligned} u(0,\cdot ) =u_0, \end{aligned}$$

(1.2)

where $u=u(t,x)\in {\mathbb {R}}$ for $(t,x) \in (0,T) \times {\mathbb {T}}$, $\beta \in {{\mathbb {R}}}$ and $\mu >0$ are constants, and $k: [0,T) \times {\mathbb {T}}\rightarrow {\mathbb {R}}$ is a known smooth forcing with zero mean at every $t \in [0,T)$; here ${\mathbb {T}}={\mathbb {R}}/{\mathbb {Z}}$ is the one-dimensional torus. Equation (1.1) arises in the context of equatorial ocean flows, describing the evolution of the restriction of the horizontal fluid velocity u to a fixed depth, with the forcing term k due to the corresponding horizontal pressure gradient (see Sect. 2). The study of (1.1) is not only of theoretical interest. Indeed, while submerged velocimeters and pressure transducers are commonly used devices to assess the state of ocean flows, the investigations that explore the global implications for the bulk of the fluid flow are typically pursued within the simpler setting of travelling waves (see the discussion in [4]). Equation (1.1) opens up a wider range of physically relevant possibilities.

In Sect. 2 we derive (1.1) from the governing equations for equatorial ocean flow (the Navier–Stokes equation with Coriolis effects accounted for, and the equation of mass conservation for an incompressible homogeneous fluid). In Sect. 3 we discuss the well-posedness of (1.1) and in Sect. 4 we prove global existence for small initial data. The decomposition in Fourier modes performed in Sect. 5 permits us to obtain a global existence result for suitable large initial data. In Sect. 6 we show that some large initial data develop into solutions that blow-up in finite time.

2 Derivation of the Model Equation

We now present a derivation of the model equation (1.1) from the governing equations for ocean flow in equatorial regions. The limited latitudinal extent of equatorial flows makes the effects of the Earth’s sphericity unimportant and we can therefore use the Cartesian f-plane representation of the equations of motion (see [17]). We choose a coordinate system with the origin at a point on the Equator, with the x-axis pointing horizontally due East, the y-axis horizontally due North and the z-axis upward. We denote by u, v, w, the corresponding fluid velocity components in the direction of increasing azimuth, latitude and elevation. If $\Omega$ is the (constant) rotational speed of the Earth round the polar axis toward the East, t stands for time, g is the (constant) gravitational acceleration at the Earth’s surface, $\mu _1$ and $\mu _2$ are the (constant) horizontal and vertical eddy viscosity coefficients, respectively, $\rho$ is the constant water density, and P is the pressure, the governing equations in the f-plane approximation are the Navier–Stokes equations (see [16])

$$\begin{aligned} u_t + uu_x + vu_y + wu_z +2 \Omega w= & {} \,-\tfrac{1}{\rho }\,P_{x} + \mu _1( u_{xx}+u_{yy}) + \mu _2 u_{zz} , \end{aligned}$$

(2.1)

$$\begin{aligned} v_t + uv_x + vv_y + wv_z= & {} \,-\tfrac{1}{\rho }\,P_{y} + \mu _1( v_{xx}+v_{yy}) + \mu _2 v_{zz}, \end{aligned}$$

(2.2)

$$\begin{aligned} w_t + uw_x + vw_y + ww_z - 2 \Omega u= & {} \,-\tfrac{1}{\rho }\,P_{z} - g + \mu _1( w_{xx}+w_{yy}) + \mu _2 w_{zz}, \end{aligned}$$

(2.3)

coupled with the equation of mass conservation

$$\begin{aligned} u_x+v_y+w_z=0, \end{aligned}$$

(2.4)

and with the constraint

$$\begin{aligned} (u,v,w) \rightarrow 0 \quad \text {for}\quad z \rightarrow -\infty ,\quad \text {uniformly in}\ (x,y), \end{aligned}$$

(2.5)

that ensures that the water is practically at rest at great depths. A significant feature of equatorial air dynamics is that the change of sign of the Coriolis force across the Equator produces an effective waveguide: the Equator acts as a (fictitious) natural boundary that facilitates azimuthal flow propagation, with a negligible meridional velocity component (see [5, 6]). Imposing a vanishing meridional velocity ($v'\equiv 0$), for flows periodic in the x-variable (with normalized wavelength $\lambda =2\pi$) and independent of the y-variable the equations (2.1)–(2.4) simplify to

$$\begin{aligned} u_t + uu_x + wu_z +2 \Omega w= & {} -\tfrac{1}{\rho }\,P_{x} + \mu _1 u_{xx} + \mu _2 u_{zz} , \end{aligned}$$

(2.6)

$$\begin{aligned} w_t + uw_x + ww_z - 2 \Omega u= & {} -\tfrac{1}{\rho }\,P_{z} - g + \mu _1 w_{xx} + \mu _2 w_{zz}, \end{aligned}$$

(2.7)

$$\begin{aligned} u_x+w_z= & {} 0. \end{aligned}$$

(2.8)

note that (2.8) yields the existence of a stream function $\psi (t,x,z)$, satisfying

$$\begin{aligned} \psi _z=u,\qquad \psi _x=- w. \end{aligned}$$

(2.9)

Assuming $\psi$ to be harmonic ensures that w and u are the real and imaginary parts of an analytic function of the complex variable $x + \textrm{i}z$. The Cauchy–Riemann equations would then permit us to formulate the restriction of the first equation in (2.6) to a submerged level $z=z_0$ as the evolution equation (1.1) in the single horizontal space-variable x, with

$$\begin{aligned} \mu = \mu _1-\mu _2,\qquad \beta = -2\Omega ,\qquad k(t,x)=-\tfrac{1}{\rho }\,P_x(t,x,z_0), \end{aligned}$$

given that

$$\begin{aligned} u_z= w_x \quad \text {and}\quad w=-(\mathcal {H}u) \quad \text {on}\quad z=z_0, \end{aligned}$$

(2.10)

where $\mathcal {H}$ is the Hilbert transform acting on square integrable functions of mean zero on $[0,2\pi ]$ as the Fourier multiplier

$$\begin{aligned} {{\mathcal {H}}}u= -\textrm{i} \sum _{k \in {{\mathbb {Z}}}\setminus \{0\}} \text {sgn}(k)\, \eta _k\,\textrm{e}^{\textrm{i}k x} \quad \text {for}\quad u=\sum _{k \in {{\mathbb {Z}}}\setminus \{0\}} \eta _k\,\textrm{e}^{\textrm{i}k x}, \end{aligned}$$

(2.11)

where

$$\begin{aligned} \text {sgn}(x)=\left\{ \begin{array}{ll} -1 &{}\quad \text {if}\;\; x<0,\\ 0 &{}\quad \text {if}\;\; x=0,\\ 1 &{}\quad \text {if}\;\; x>0. \end{array} \right. \end{aligned}$$

The second relation in (2.10) follows because (2.8) and the periodicity yield

$$\begin{aligned} \partial _z \int _0^{2\pi } w(t,x,z)\,\textrm{d}x = \int _0^{2\pi } u_x(t,x,z)\,\textrm{d}x =0,\qquad z \le z_0, \end{aligned}$$

and now (2.5) ensures that w has zero horizontal mean at every fixed time:

$$\begin{aligned} \int _0^{2\pi } w(t,x,z)\,\textrm{d}x =0,\qquad z \le z_0. \end{aligned}$$

We have $\mu >0$ since the vertical eddy viscosity coefficient is several orders of magnitude smaller than the horizontal one (see [16]). Note that reversing the flow direction ($x \mapsto -x$ and $u \mapsto -u$) has the effect of changing $\beta$ to $-\beta$ in (1.1) and a solution to (1.1) determines the deep-water flow beneath the level $z=z_0$.

The harmonic hypothesis is convenient from the point of view of analytical tractability and presents also the advantage of ensuring that we investigate the propagation of large-amplitude oscillations along the submerged level. Indeed, (2.13) and the fact that the Hilbert transform is an isomorphism on the space of $L^2({\mathbb T})$-functions of mean zero ensure that at the level $z=z_0$ the horizontal and vertical kinetic energies are of the same order of magnitude. Note that the harmonicity of $\psi$ beneath the submerged level $z=z_0^*$ for some $z_0^*>z_0$ captures the physically realistic assumption of irrotational flow below $z=z_0^*$. Indeed, equatorial currents are confined to a near-surface layer (see the data in [6]) and the assumption of zero vorticity $\Gamma =0$, where

$$\begin{aligned} \Gamma =u_z-w_x, \end{aligned}$$

is consistent with the equations (2.6)–(2.8) since using (2.8) and taking the curl of (2.6)–(2.7) yields

$$\begin{aligned} \Gamma _t + u\Gamma _x + w\Gamma _z -\mu _1 \Gamma _{xx} - \mu _2 \Gamma _{zz}=0. \end{aligned}$$

(2.12)

Zero initial data ($\Gamma =0$ at $t=0$) ensures $\Gamma =0$ at all subsequent times for the solution to the linear parabolic equation (2.12), for any bounded and continuously differentiable velocity field (u, w). Let us also point out that (1.1) captures the entire dynamics throughout the region of irrotational flow beneath $z=z_0$. To see, this, it is convenient to introduce the velocity potential $\varphi (t,x,z)$ as the harmonic conjugate of the stream function at every fixed time, uniquely determined up to an additive function of time by

$$\begin{aligned} \varphi _x=u,\qquad \varphi _z= w. \end{aligned}$$

(2.13)

In terms of $\varphi$, we can recast the validity of the equations (2.6)–(2.7) in the region $z<z_0^*$ as

$$\begin{aligned} \varphi _t + \tfrac{1}{2}\,(\varphi _x^2 + \varphi _z^2) - 2 \Omega \psi + \tfrac{1}{\rho }\,P + gz - \mu \varphi _{xx} =0 \quad \text {for}\quad z <z_0^*, \end{aligned}$$

(2.14)

which is the viscous geophysical analogue of the Bernoulli equation for irrotational inviscid two-dimensional flow (see [3] for the relevance of the latter to water flows). Equation (1.1) is obtained by differentiating the restriction of (2.14) to $z=z_0$ with respect to the x-variable. This derivation forces $k(t,\cdot )$ to have mean zero for all $t\in [0,T)$.

3 Well-Posedness

In the sequel we denote by $L^2_0({\mathbb {T}})$ and $H^s_0({\mathbb {T}})$, $s\in {\mathbb {R}}$, the closed subspaces of zero-mean functions in $L^2({\mathbb {T}})$ and $H^s({\mathbb {T}})$, respectively. The local well-posedness of (1.1)–(1.2) can be obtained by adapting the standard approaches that were developed for the heat and the (viscous) Burgers equation (see for instance [14, 15, 18, 19] and also [13] for heat equations and [2, 8] for the Burgers equation). Here we follow mainly the point of view used by Dix in [8]. We choose to present two different results. In both cases we work with the integral equation

$$\begin{aligned} u(t) = e^{t \mu \partial _x^2} u_0 - \int _0^t e^{(t-t') \mu \partial _x^2} \Bigl [ \frac{1}{2}\partial _x \big (u^2+ (\mathcal {H}u)^2\big ) +\beta \mathcal {H}u(t') -k(t')\Bigr ] dt' \;. \end{aligned}$$

(3.1)

that is equivalent to (1.1)–(1.2) for solutions $u \in L^\infty (]0,T[; L^2({\mathbb {T}}))$ such that $u_t\in L^\infty (]0,T[; H^{-2}({\mathbb {T}}))$; here $\{e^{t \mu \partial _x^2}\}_{t>0}$ denotes the heat semigroup generated on $L^2({\mathbb {T}})$ by $\mu \partial _x^2$. Our aim is to prove that the associated map

$$\begin{aligned} \Lambda _{u_0}: u \mapsto e^{t \mu \partial _x^2} u_0 - \int _0^t e^{(t-t') \mu \partial _x^2} \Bigl [ \frac{1}{2}\partial _x \big (u^2+ (\mathcal {H}u)^2\big ) + \beta \mathcal {H}u(t') -k(t')\Bigr ] dt' \; \end{aligned}$$

is 1/2-contractive in a ball of a suitable metric function space.

In our approach we will make use of the two following technical lemmas. The first one deals with classical product estimates in Sobolev spaces (see for instance [1] or [12]).

Lemma 3.1

Let $(t,r_1,r_2)\in {\mathbb {R}}^3$ with $r_1+r_2>t+1/2$, $r_1+r_2\ge 0$ and $r_1,r_2\ge t$. Then for any $u\in H^{r_1}({\mathbb {T}})$ and $v\in H^{r_2}({\mathbb {T}})$, we have $uv \in H^t({\mathbb {T}})$ with

$$\begin{aligned} \Vert u v \Vert _{H^{t}({\mathbb {T}})}\le C\Vert u\Vert _{H^{r_1}({\mathbb {T}})} \Vert v\Vert _{H^{r_2}({\mathbb {T}})}, \end{aligned}$$

(3.2)

for some $C=C(r_1,r_2,t)>0$.

The second lemma concerns the regularizing effect of the heat kernel in Sobolev spaces.

Lemma 3.2

Let $\mu >0$, $\theta \ge 0$ and $f\in L^2({\mathbb {T}})$. Then for $0<t<1$ we have

$$\begin{aligned} \Bigl \Vert e^{t\mu \partial _x^2} f \Bigr \Vert _{H^{\theta }({\mathbb {T}})} \le C_{\theta ,\mu } t^{-\theta /2}\Vert f\Vert _{L^2({\mathbb {T}})} \quad . \end{aligned}$$

(3.3)

Proof

Relation (3.3) follows directly from the exact formula

$$\begin{aligned} e^{t\mu \partial _x^2} f= \displaystyle \sum _{k\in {\mathbb {Z}}} e^{-\mu k^2 t} {\hat{f}}(k) e^{ikx} \;. \end{aligned}$$

(3.4)

Indeed, the Plancherel identity yields

$$\begin{aligned} \Vert e^{t\mu \partial _x^2} f\Vert _{H^\theta ({\mathbb {T}})}^2 =\sum _{k\in {\mathbb {Z}}} \langle k\rangle ^{2\theta } e^{-2\mu k^2 t} |{\hat{f}}(k)|^2 \le \sup _{\xi \in {\mathbb {R}}_+} \Bigl ( \langle \xi \rangle ^{2\theta } e^{-2\mu \xi ^2 t}\Bigr ) \Vert f\Vert _{L^2({\mathbb {T}})}^2 , \end{aligned}$$

and this leads to (3.3) since for $0\le t\le 1$ we have

$$\begin{aligned} \sup _{\xi \in {\mathbb {R}}} \Bigl ( \langle \xi \rangle ^{2\theta } e^{-2\mu \xi ^2 t}\Bigr ){} & {} \le \sup _{\zeta \in {\mathbb {R}}} \frac{1}{(\mu t)^\theta }\Bigl ( (\mu t + \zeta ^2)^{\theta } e^{-2\zeta ^2}\Bigr ) \\{} & {} \le \frac{\max (1,\mu ^\theta )}{(\mu t)^\theta } \sup _{\zeta \in {\mathbb {R}}} \Bigl ( (1 + \zeta ^2)^{\theta } e^{-2\zeta ^2}\Bigr )\;. \end{aligned}$$

$\square$

3.1 The case $s\ge 0$

We start by an approach that enables us to get the unconditional local-wellposedness of (1.1)–(1.2) in $H^s({\mathbb {T}})$ for $s\ge 0$. The notion of unconditional local-wellposedness, introduced by Kato in [11], means that uniqueness holds in $L^\infty (0,T; H^s({\mathbb {T}}))$, which, roughly speaking, ensures the uniqueness of weak solutions.

Proposition 3.3

Let $\mu >0$, $\beta \in {\mathbb {R}}^*$, and let $s'\ge s\ge 0$. For any $u_0\in H^s_0({\mathbb {T}})$ and any $k\in L^\infty ({\mathbb {R}}_+; H^{s'}_0({\mathbb {T}}))$ there exists $T=T(\Vert u_0\Vert _{L^2({\mathbb {T}})}, \Vert k\Vert _{L^\infty ({\mathbb {R}}^*_+;H^{-1}({\mathbb {T}}))} )>0$ and a solution $u\in C([0,T];H^s_0({\mathbb {T}}))\cap C(]0,T];H^{s'+1}({\mathbb {T}}))$ of (1.1)–(1.2). This solution is the unique solution to (1.1)–(1.2) that belongs to $L^\infty (0,T;H^s({\mathbb {T}}))$. Finally, for any $M>0$, the map $u_0\mapsto u$ is continuous from the ball of $H_0^s({\mathbb {T}})$ centered at the origin and with radius M into $C([0,T(M)]; H^s_0({\mathbb {T}}))$.

Proof

We first notice that since $u_0$ has zero mean-value and $k \in L^\infty ({\mathbb {R}}_+; H^{s'}_0({\mathbb {T}}))$, $\Lambda _{u_0}$ maps zero mean-value functions to zero mean-value functions. Now, a direct application of (3.3) with $\theta =\frac{13}{8}$, in combination with (3.2) for $r_1=r_2=s \ge 0$ and $t=s-\frac{5}{8}$, leads us to

$$\begin{aligned}&\Bigl \Vert \int _0^t e^{\mu (t-t')\partial _x^2} \partial _x (u v + \mathcal {H}u \mathcal {H}v )(t') \, dt' \Bigr \Vert _{H^{s}({\mathbb {T}})} \nonumber \\&\quad \le \int _0^t \Bigl \Vert e^{\mu (t-t')\partial _x^2} \partial _x (u v + \mathcal {H}u \mathcal {H}v(t') \Bigr \Vert _{H^{s}({\mathbb {T}})} \, dt' \nonumber \\&\quad \le C_{\mu } \int _0^t (t-t')^{-\frac{13}{16} }\Bigl ( \Vert \partial _x (u(t')v(t'))\Vert _{H^{s-13/8} ({\mathbb {T}})}+ \Vert \partial _x (\mathcal {H}u(t') \mathcal {H}v(t'))\Vert _{H^{s-13/8}({\mathbb {T}})}\Bigr ) \, dt' \nonumber \\&\quad \le C_{\mu } \int _0^t (t-t')^{-\frac{13}{16} }\Bigl ( \Vert u(t')v(t')\Vert _{H^{s-5/8} ({\mathbb {T}})}+ \Vert \mathcal {H}u(t') \mathcal {H}v(t')\Vert _{H^{s-5/8}({\mathbb {T}})}\Bigr ) \, dt' \nonumber \\&\quad \le C_{\mu } \int _0^t (t-t')^{-\frac{13}{16}} \Vert u(t')\Vert _{H^{s}({\mathbb {T}})} \Vert v(t')\Vert _{H^{s}({\mathbb {T}})} \, dt' \nonumber \\&\quad \le C_{\mu } \, \Vert u\Vert _{L^\infty _T H^s} \Vert v\Vert _{L^\infty _T H^s} \int _0^t (t-t')^{-\frac{13}{16}} \, dt' \nonumber \\&\quad \le C_{\mu } \, T^\frac{3}{16} \Vert u\Vert _{L^\infty _T H^s} \Vert v\Vert _{L^\infty _T H^s}\,,\qquad 0< t<T <1\,, \end{aligned}$$

(3.5)

where in the fourth step we used that the Hilbert transform is a non-expansive map in $H^s({\mathbb {T}})$ for any $s\in {\mathbb {R}}$; here the constant $C_\mu$ may change from line to line.

On the other hand, using the contractivity of the semigroup $e^{\mu t\partial _x^2}$, $t>0$, in Sobolev spaces and Lemma 3.2 with $\theta =1$, it is straightforward to check that

$$\begin{aligned} \Vert e^{\mu t\partial _x^2} u_0\Vert _{H^s({\mathbb {T}})}\le \Vert u_0\Vert _{H^s({\mathbb {T}})},\qquad t\ge 0, \end{aligned}$$

and that

$$\begin{aligned} \sup _{t\in ]0,T[} \Bigl \Vert \int _0^t e^{\mu (t-t')\partial _x^2} ( \beta \mathcal {H}u(t') -k(t')) \, dt' \Bigr \Vert _{H^{s}({\mathbb {T}})} \le T |\beta | \Vert u\Vert _{L^\infty _T H^s} + C_\mu \, T^{1\over 2} \Vert k\Vert _{L^\infty _T H^{s-1}} \;. \end{aligned}$$

Using these two estimates we obtain

$$\begin{aligned} \Vert \Lambda _{u_0}(u)\Vert _{L^\infty _T H^s} \le \Vert u_0\Vert _{H^s} +C_\mu T^\frac{3}{16} \Vert u\Vert _{L^\infty _T H^s}^2 +T|\beta |\Vert u\Vert _{L^\infty _T H^s} + C_\mu T^{1\over 2} \Vert k\Vert _{L^\infty _T H^{s-1}}\nonumber \\ \end{aligned}$$

(3.6)

and

$$\begin{aligned} \Vert \Lambda _{u_0}(v_1)- \Lambda _{u_0}(v_2)\Vert _{L^\infty _T H^s}\le & {} C_\mu \, T^\frac{3}{16} \Vert v_1+v_2\Vert _{L^\infty _T H^s} \Vert v_1-v_2\Vert _{L^\infty _T H^s}\nonumber \\{} & {} \quad + T |\beta | \Vert v_1-v_2\Vert _{L^\infty _T H^s} \ \nonumber \\\le & {} C_\mu \, T^\frac{3}{16} \Vert v_1-v_2\Vert _{L^\infty _T H^s} (T^{\frac{13}{16}}|\beta |+ \Vert v_1+v_2\Vert _{L^\infty _T H^s}).\nonumber \\ \end{aligned}$$

(3.7)

This proves that for

$$\begin{aligned} T_s= \min \Bigl \{\frac{1}{8(|\beta |+1)}, (8 C_\mu \, \Vert u_0\Vert _{H^s})^{-\frac{16}{3}}, \Vert u_0\Vert _{H^s}^2 (8 C_\mu \Vert k\Vert _{L^\infty _1 H^{s-1}})^{-2}\Bigr \} \end{aligned}$$

the map $\Lambda _{u_0}$ is 1/2 contractive on the ball centered at the origin of $C([0,T_s]; H^s_0 ({\mathbb {T}}))$, with radius $2 \Vert u_0\Vert _{H^s}$. Consequently, we showed the unconditional well-posedness of (1.1)–(1.2) in $H^s_0({\mathbb {T}})$.

To prove that the maximal time of existence only depends on $\Vert u_0\Vert _{L^2({\mathbb {T}})}$ we proceed by induction on s. Let us denote by $T^*(s)>0$, $s\ge 0$, the maximal time of existence of u in $H^s_0({\mathbb {T}})$. For $s\in ]0,5/8]$, we notice that Lemma 3.1 ensures

$$\begin{aligned} \Vert u v \Vert _{H^{s-\frac{5}{8}}({\mathbb {T}})} \le C\, \Vert u\Vert _{H^s({\mathbb {T}})} \Vert v\Vert _{L^2({\mathbb {T}})}. \end{aligned}$$

Inserting this inequality in the third step of (3.5) we eventually get

$$\begin{aligned}&\Bigl \Vert \int _0^t e^{\mu (t-t')\partial _x^2} \partial _x (u v + \mathcal {H}u \mathcal {H}v )(t') \, dt' \Bigr \Vert _{H^{s}({\mathbb {T}})} \le C_{\mu } \, T^\frac{3}{16} \Vert u\Vert _{L^\infty _T H^s} \Vert v\Vert _{L^\infty _T L^2}\,,\nonumber \\&\qquad 0< t<T <1\,. \end{aligned}$$

(3.8)

Translating this estimate in time and setting $I=]t_0,T^*(s)[$ ensures that for any $t_0\in ]\min \{T^*(s)-1,0\},T^*(s)[$ the solution u satisfies

$$\begin{aligned}{} & {} \Vert u\Vert _{L^\infty (I; H^s)} \le \Vert u(t_0)\Vert _{H^s} +C_\mu |I|^\frac{3}{16} \Vert u\Vert _{L^\infty (I; H^s)} \Vert u\Vert _{L^\infty (I; L^2)}\nonumber \\{} & {} \quad +|I||\beta | \Vert u\Vert _{L^\infty (I; H^s)}+ |I|^{1\over 2}\Vert k\Vert _{L^\infty (I; H^{s-1})} \;. \end{aligned}$$

(3.9)

Assuming that $T^*(s)<T^*(0)$ and taking $t_0< T^*(s)$ close enough to $T^*(s)$ would lead us to

$$\begin{aligned} \Vert u\Vert _{L^\infty (]t_0,T^*(s)[;H^s)}<\infty , \end{aligned}$$

which contradicts the definition of $T^*(s)$. It thus follows that $T^*(s) =T^*(0)$ for $0<s\le \frac{5}{8}$. Since according to Lemma 3.1,

$$\begin{aligned} \Vert u v \Vert _{H^{s-\frac{5}{8}}({\mathbb {T}})} \le C\, \Vert u\Vert _{H^s({\mathbb {T}})} \Vert v\Vert _{{s-\frac{5}{8}}({\mathbb {T}})},\qquad s\ge \frac{5}{8}, \end{aligned}$$

we see that we can reach any $s>0$ by repeating this argument a finite number of times.

It remains to prove that $u\in C(]0,T^*(0)[;H^{s'+1}({\mathbb {T}}))$. For this we show a smoothing effect for our solution by making use of a space of weighted continuous function in time introduced by Kato and Fujita in [10]. More precisely, we define the Banach space

$$\begin{aligned} Y_T=\{ u\in C(]0,T];H^{s+\frac{1}{16}}): \, \Vert u\Vert _{Y_T}=\sup _{t\in ]0,T]} t^\frac{1}{32} \Vert u(t)\Vert _{H^{s+\frac{1}{16}}} <\infty \}\;. \end{aligned}$$

We first notice that Lemma 3.2 ensures

$$\begin{aligned} t^\frac{1}{32} \Vert e^{\mu \partial _x^2 t} u_0 \Vert _{H^{s+\frac{1}{16}}({\mathbb {T}})}\le C_\mu \Vert u_0\Vert _{H^s({\mathbb {T}})}. \end{aligned}$$

Then proceeding exactly as for (3.5), applying (3.3) with $\theta =\frac{13}{8}$ but this time (3.2) with $r_1=r_2=s \ge 0$ and $t=s+\frac{1}{16}-\frac{5}{8}=s-\frac{9}{16}$, we obtain

$$\begin{aligned} \qquad&t^{-\frac{1}{32}} \Bigl \Vert \int _0^t e^{\mu (t-t')\partial _x^2} \partial _x (u v + \mathcal {H}u \mathcal {H}v )(t') \, dt' \Bigr \Vert _{H^{s+\frac{1}{16}}({\mathbb {T}})} \nonumber \\&\qquad \le t^{-\frac{1}{32}} \int _0^t \Bigl \Vert e^{\mu (t-t')\partial _x^2} \partial _x (u v + \mathcal {H}u \mathcal {H}v(t') \Bigr \Vert _{H^{s+\frac{1}{16}}({\mathbb {T}})} \, dt' \nonumber \\&\qquad \le C_{\mu } t^{-\frac{1}{32}} \int _0^t (t-t')^{-\frac{13}{16} }\Bigl ( \Vert \partial _x (u(t')v(t'))\Vert _{H^{s-\frac{25}{16}} ({\mathbb {T}})} \nonumber \\&\qquad + \Vert \partial _x (\mathcal {H}u(t') \mathcal {H}v(t'))\Vert _{H^{s-\frac{25}{16}}({\mathbb {T}})}\Bigr ) \, dt' \nonumber \\&\qquad \le C_{\mu } t^{-\frac{1}{32}} \int _0^t (t-t')^{-\frac{13}{16} }\Bigl ( \Vert u(t')v(t')\Vert _{H^{s-\frac{9}{16}} ({\mathbb {T}})}+ \Vert \mathcal {H}u(t') \mathcal {H}v(t')\Vert _{H^{s-\frac{9}{16}}({\mathbb {T}})}\Bigr ) \, dt' \nonumber \\&\qquad \le C_{\mu } t^{-\frac{1}{32}} \int _0^t (t-t')^{-\frac{13}{16}} \Vert u(t')\Vert _{H^{s}({\mathbb {T}})} \Vert v(t')\Vert _{H^{s}({\mathbb {T}})} \, dt' \nonumber \\&\qquad \le C_{\mu } t^{-\frac{1}{32}} \, \Vert u\Vert _{L^\infty _T H^s} \Vert v\Vert _{L^\infty _T H^s} \int _0^t (t-t')^{-\frac{13}{16}} \, dt' \nonumber \\&\qquad \le C_{\mu } \, T^\frac{5}{32} \Vert u\Vert _{L^\infty _T H^s} \Vert v\Vert _{L^\infty _T H^s}\,,\qquad 0< t<T <1\,. \end{aligned}$$

(3.10)

Finally, applying (3.3) with $\theta =\frac{1}{16}$ and $\theta = \frac{9}{8}$, we get

$$\begin{aligned} \qquad&t^{-\frac{1}{32}} \Bigl \Vert \int _0^t e^{\mu (t-t')\partial _x^2} ( \beta \mathcal {H}u(t'))+k(t')) \, dt' \Bigr \Vert _{H^{s+\frac{1}{16}}({\mathbb {T}})} \nonumber \\&\quad \le C_\mu t^{-\frac{1}{32}} \int _0^t (t-t')^{-\frac{1}{32}} |\beta | \Vert u(t')\Vert _{H^s} +(t-t')^{-\frac{9}{16}}\Vert k(t')\Vert _{H^{s-\frac{17}{16}}}\, dt'\nonumber \\&\quad \le C_\mu T^\frac{13}{32} \Bigl ( |\beta | \Vert u\Vert _{L^\infty _T H^s} + \Vert k\Vert _{L^\infty _T H^{s-\frac{17}{16}}} \Bigr )\,. \end{aligned}$$

(3.11)

The estimates (3.10)–(3.11) ensure that, starting with $e^{\mu t \partial _x^2} u_0\in C([0,T];H^s_0({\mathbb {T}}))\cap Y_T$, the function sequence constructed by the Picard iterative scheme associated with $\Lambda _{u_0}$ that, according to the contraction theorem, converges to the solution u in $C([0,T_s];H^s({\mathbb {R}}))$, is a Cauchy sequence in $Y_{T_s}$ and thus $u\in Y_{T_s}$. Hence $u\in C(]0,T_s]; H^{s+\frac{1}{16}}({\mathbb {T}}))$ and we can re-apply the same argument with $u(0+) \in H^{s+\frac{1}{16}}({\mathbb {T}})$ as initial datum to get $u\in C(]0,T_s]; H^{s+\frac{1}{8}}({\mathbb {T}}))$. Actually, the argument can be repeated as long as $s+\frac{n}{16} \le s'+1+\frac{1}{16}$. This shows that $u\in C(]0,T^*(0)]; H^{s'+1}({\mathbb {T}}))$ since, according to the above considerations, $T^*(s'+1)=T^*(0)$. $\square$

3.2 The case $-1/2< s< 0$.

Taking $k=0$, a classical dilation argument (see [8, 2]) ensures that $H^{-\frac{1}{2}}({\mathbb {T}})$ is critical for the well-posedness of (1.1)–(1.2). To prove the well-posedness of (1.1)–(1.2) in $H^s({\mathbb {T}})$ for $-1/2< s< 0$ we will use a fixed point argument in the Kato-Fujita’s space

$$\begin{aligned} X_{M,T} =\{ u\in C(]0,T];H^{s_0})\,:\, \Vert u\Vert _{X_{T}} =\sup _{t\in ]0,T]} t^\frac{1}{4} \Vert u(t)\Vert _{H^{s+\frac{1}{2}}({\mathbb {T}})} \le M \} \;. \end{aligned}$$

Proposition 3.4

Let $\mu >0$, $\beta \in {\mathbb {R}}^*$, and let $-1/2<s<0$. For any $u_0\in H^s_0({\mathbb {T}})$ and any $k\in L^\infty ({\mathbb {R}}_+; L^2_0({\mathbb {T}}))$ there exists $T=T(\Vert u_0\Vert _{H^s({\mathbb {T}})}, \Vert k\Vert _{L^\infty ({\mathbb {R}}^*_+; L^2({\mathbb {T}}))} )>0$ and a solution $u\in C([0,T];H^s_0({\mathbb {T}}))$ of (1.1)–(1.2). This solution is the unique solution to (1.1)–(1.2) that belongs to $X_T$. Finally, for any $M>0$, the map $u_0\mapsto u$ is continuous from the ball of $H^s_0({\mathbb {T}})$ with radius M into $C([0,T(M)]; H^s_0({\mathbb {T}}))$.

Proof

Again, in view of the hypothesis, $\Lambda _{u_0}$ maps zero-mean functions to zero-mean functions. Applying (3.3) with $\theta =1/2$ it follows that

$$\begin{aligned} t^{1/4} \Vert e^{\mu t \partial _x^2} u_0\Vert _{H^{s+\frac{1}{2}}({\mathbb {T}})} \le C_\mu \, \Vert u_0 \Vert _{H^s({\mathbb {T}})} \;. \end{aligned}$$

(3.12)

Now, we apply (3.3) with $\theta =\frac{5}{4}-\frac{s}{2} >0$ and (3.2) with $(r_1,r_2,t)=(s+\frac{1}{2},s+\frac{1}{2},\frac{3\,s}{2}+\frac{1}{4})$. Note that the hypotheses of Lemma 3.1 are fulfilled since for $0>s>-1/2$ we have

$$\begin{aligned}&r_1=r_2=s+\frac{1}{2}>0\,, \\&r_1=r_2=s+\frac{1}{2}> t=(s+\frac{1}{2}) +\frac{s}{2}-\frac{1}{4}\,,\\&r_1+r_2=2(s+\frac{1}{2})>t+1/2=\frac{3}{2}(s+\frac{1}{2})\,. \end{aligned}$$

We then obtain in the same way as above

$$\begin{aligned}&t^{1\over 4} \Bigl \Vert \int _0^t e^{\mu (t-t')\partial _x^2} \partial _x (u v + \mathcal {H}u \mathcal {H}v )(t') \, dt' \Bigr \Vert _{H^{s+\frac{1}{2}}({\mathbb {T}})}\nonumber \\&\quad \le C_\mu \, t^{1\over 4 } \int _0^t (t-t')^{\frac{s}{4}-\frac{5}{8}} \Vert u(t')v(t')\Vert _{H^{\frac{3s}{2}+\frac{1}{4}}({\mathbb {T}})} \, dt' \nonumber \\&\quad \le C_\mu \, t^{1\over 4} \int _0^t (t-t')^{\frac{s}{4}-\frac{5}{8}} \Vert u(t')\Vert _{H^{s+\frac{1}{2}}({\mathbb {T}})} \Vert v(t')\Vert _{H^{s+\frac{1}{2}}({\mathbb {T}})}\, dt' \nonumber \\&\quad \le C_\mu \, \sup _{\tau \in ]0,t[} \Bigl (\tau ^{{1\over 4}}\Vert u(\tau )\Vert _{H^{s+\frac{1}{2}}({\mathbb {T}})} \Bigr ) \sup _{\tau \in ]0,t[} \Bigl (\tau ^{1\over 4}\Vert v(\tau )\Vert _{H^{s+\frac{1}{2}}({\mathbb {T}})} \Bigr ) \; t^{1\over 4} \int _0^t (t-t')^{\frac{s}{4}-\frac{5}{8}} \, t'^{-\frac{1}{2}} \, dt' \nonumber \\&\quad \le C_\mu \, \Vert u\Vert _{X_T} \Vert v\Vert _{X_T} \; t^{1\over 4}\, t^{{s\over 4} -{1\over 8}} \int _0^1 (1-\tau )^{\frac{s}{4}-\frac{5}{8}} \, \tau ^{-\frac{1}{2}} \, d\tau \nonumber \\&\quad \le C_\mu \, T^{s+1/2\over 4} \Vert u\Vert _{X_T} \Vert v\Vert _{X_T} \; \end{aligned}$$

(3.13)

where in the last step we used that $s/4-5/8>-1/8-5/8>-1$ and in the next-to-last step we performed the change of variables $t'=t\tau$. Moreover, it is easy to check that

$$\begin{aligned}&t^{1\over 4} \Bigl \Vert \int _0^t e^{\mu (t-\tau )\partial _x^2} (\beta \mathcal {H}u(\tau ) -k(\tau )) \, d\tau \Bigr \Vert _{H^{s+\frac{1}{2}}({\mathbb {T}})} \\&\quad \le C_\mu \, t^{1\over 4} \int _0^t \Bigl ( |\beta | \Vert u(\tau ) \Vert _{H^{s+\frac{1}{2}}({\mathbb {T}})} + (t-\tau )^{-\frac{1}{4}} \Vert k(\tau )\Vert _{H^s({\mathbb {T}})} \Bigr ) \; d\tau \\&\quad \le C_\mu \,\Bigr ( t^{\frac{1}{4}} |\beta | \Vert u\Vert _{X_T} \int _0^t \tau ^{-\frac{1}{4}}\, d\tau +T \Vert k\Vert _{L^\infty _T L^2}\Bigr ) \\&\quad \le C_\mu \, T \Bigl ( |\beta | \Vert u\Vert _{X_T}+ \Vert k\Vert _{L^\infty _T L^2} \Bigr ) \; . \end{aligned}$$

In view of (3.13) we easily get for $0<T<1$ and $v_i\in X_T$, $i=1,2$, that

$$\begin{aligned} \Vert \Lambda _{u_0}(v_i) \Vert _{X_T} \le C_\mu \, \Bigl ( \Vert u_0\Vert _{H^s({\mathbb {T}})} + T^{s+1/2\over 4} \Vert v_i\Vert _{X_T}^2+ T\Bigl ( |\beta | \Vert v_i\Vert _{X_T}+ \Vert k\Vert _{L^\infty _TL^2} \Bigr )\Bigr ) \;\nonumber \\ \end{aligned}$$

(3.14)

and

$$\begin{aligned} \Vert \Lambda _{u_0}(v_1-v_2) \Vert _{X_T} \le C_\mu \, \Bigl ( T^{s+1/2\over 4} (\Vert v_1\Vert _{X_T} +\Vert v_2\Vert _{X_T}) + T |\beta |\Bigr ) \Vert v_1-v_2\Vert _{X_T} \;.\nonumber \\ \end{aligned}$$

(3.15)

Combining these estimates with (3.12) we infer that, for $s>-1/2$, $\Lambda _{u_0}$ is 1/2-contractive on $X_{M,T}$ with $M\sim \Vert u_0\Vert _{H^s({\mathbb {T}})}$ and

$$\begin{aligned} T_s\sim \min \Bigl \{ \frac{1}{|\beta |+1}, \Vert u_0\Vert _{H^s({\mathbb {T}})}^{-{4\over s+\frac{1}{2}}}, \Vert u_0\Vert _{H^s}\Vert k\Vert _{L^\infty _T L^2}^{-1}\Bigr \}\;. \end{aligned}$$

This leads to the existence and uniqueness in $X_{T_s}$ of solutions to (1.1)–(1.2) with $u_0\in H^s_0 ({\mathbb {T}})$. Now to prove that the solution u belongs to $C([0,T_s]; H^s_0({\mathbb {T}}))$ we start by noticing that (3.4) ensures

$$\begin{aligned} e^{\mu t \partial _x^2} u_0 \in C({\mathbb {R}}_+; H^s_0({\mathbb {T}})) . \end{aligned}$$

Then, we make use of (3.3) with $\theta = \frac{3}{4}-\frac{s}{2}>0$ and of (3.2) with $(r_1,r_2,t)=(s+\frac{1}{2},s+\frac{1}{2}, \frac{3\,s}{2}+ \frac{1}{4})$, that is, the same triplet as above. Proceeding as above, we get

$$\begin{aligned} \qquad&\sup _{t\in ]0,T[} \Bigl \Vert \int _0^t e^{\mu (t-t')\partial _x^2} \partial _x (uv+\mathcal {H}u \mathcal {H}v )(t') \, dt' \Bigr \Vert _{H^{s}({\mathbb {T}})} \end{aligned}$$

(3.16)

$$\begin{aligned}&\quad \le C_\mu \, \sup _{t\in ]0,T[} \int _0^t (t-t')^{s/4-3/8} \Vert (u v+\mathcal {H}u \mathcal {H}v )(t')\Vert _{H^{\frac{1}{4}+\frac{3s}{2}}({\mathbb {T}})} \, dt' \nonumber \\&\qquad \le C_\mu \, \sup _{t\in ]0,T[} \int _0^t (t-t')^{s/4-3/8} \Vert u(t')\Vert _{H^{s+\frac{1}{2}} ({\mathbb {T}})}\Vert v(t')\Vert _{H^{s+\frac{1}{2}}({\mathbb {T}})} \, dt' \nonumber \\&\qquad \le C_\mu \, \sup _{\tau \in ]0,t[} \Bigl (\tau ^{{1\over 4}}\Vert u(\tau )\Vert _{H^{s+\frac{1}{2}}({\mathbb {T}})} \Bigr ) \sup _{\tau \in ]0,t[} \Bigl (\tau ^{{1\over 4}}\Vert v(\tau )\Vert _{H^{s+\frac{1}{2}}({\mathbb {T}})} \Bigr ) \; \nonumber \\&\qquad \int _0^t (t-t')^{s/4-3/8} \, t'^{-1/2} \, dt' \nonumber \\&\qquad \le C_\mu \, \Vert u\Vert _{X_T}\Vert v\Vert _{X_T}\; T^{s+1/2\over 4} \int _0^1 (1-\theta )^{s/4-3/8} \theta ^{-1/2} \, d\theta \nonumber \\&\qquad \le C_\mu \, T^{s+1/2\over 4} \Vert u\Vert _{X_T}\Vert v\Vert _{X_T} \,, \end{aligned}$$

(3.17)

where in the last step we used that $s/4-3/8>-1/8-3/8>-1$. On the other hand,

$$\begin{aligned}&\sup _{t\in ]0,T[} \Bigl \Vert \int _0^t e^{\mu (t-t')\partial _x^2} (\beta \mathcal {H}u +k)(t')\, dt' \Bigr \Vert _{H^{s} ({\mathbb {T}})} \nonumber \\&\quad \le C_\mu \, \sup _{t\in ]0,T[} \int _0^t ( |\beta | \Vert \mathcal {H}u(t')\Vert _{H^{s} ({\mathbb {T}})}+ \Vert k(t')\Vert _{H^s({\mathbb {T}})}) \, dt'\nonumber \\&\quad \le C_\mu \, \Bigl ( |\beta | \Vert u\Vert _{X_T} \sup _{t\in ]0,T[} \int _0^t \tau ^{-1/4} +T \Vert k\Vert _{L^\infty _T L^2} \Bigr ) d\tau \nonumber \\&\quad \le C_\mu \, \Bigl (T^{3/4} |\beta | \Vert u\Vert _{X_T} +T \Vert k\Vert _{L^\infty _T L^2}\Bigr ) \; . \end{aligned}$$

(3.18)

This ensures that, starting with the continuous function $v=e^{\mu t \partial _x^2} u_0 \in C\big ([0,T_s];$$H^s_0 ({\mathbb {T}})\big ) \cap X_{M,T_s}$, the function sequence provided by the Picard iterative scheme that, according to the contraction theorem, converges to the solution in $u\in X_T$, is a Cauchy sequence in $C\big ([0,T_s]; H^s_0 ({\mathbb {T}})\big )$ and thus $u\in C\big ([0,T_s]; H^s_0 ({\mathbb {T}})\big )$. Finally, the continuous dependence with respect to initial data in $H^s ({\mathbb {T}})$ follows also easily from (3.17)–(3.18).

$\square$

4 Global Existence for Small Data

To fully justify the following calculations we may assume that $u\in C([0,T];H^4_0({\mathbb {T}}))$ and then extend the obtained $L^2({\mathbb {T}})$-bound to $u\in C([0,T];L^2_0({\mathbb {T}}))$ by invoking the continuity with respect to initial data, established in Proposition 3.3.

Taking the $L^2({\mathbb {T}})$-scalar product of (1.1) with u and integrating by parts we get

$$\begin{aligned} \frac{1}{2} \frac{d}{dt} \Vert u\Vert ^2_{L^2({\mathbb {T}})} = - \mu \Vert u_x\Vert _{L^2({\mathbb {T}})}^2 - \int _{{\mathbb {T}}} (\mathcal {H}u) (\mathcal {H}u_x) u + \int _{{\mathbb {T}}} k \, u . \end{aligned}$$

(4.1)

Since u has zero mean-value, we have

$$\begin{aligned} \Vert u\Vert _{L^2({\mathbb {T}})} \le \Vert u_x\Vert _{L^2({\mathbb {T}})}\quad \text { and } \quad \Vert u\Vert _{L^\infty ({\mathbb {T}})}^2 \le \Vert u_x\Vert _{L^2({\mathbb {T}})} \Vert u\Vert _{L^2({\mathbb {T}})} \;. \end{aligned}$$

Therefore

$$\begin{aligned} \Bigl |\int _{{\mathbb {T}}} k u \Bigr | \le \frac{2}{ \mu } \Vert k\Vert _{L^2({\mathbb {T}})}^{2}+ \frac{\mu }{8} \Vert u \Vert _{L^2({\mathbb {T}})}^2\le \frac{2}{ \mu } \Vert k\Vert _{L^2({\mathbb {T}})}^{2}+ \frac{\mu }{8} \Vert u_x \Vert _{L^2({\mathbb {T}})}^2, \end{aligned}$$

(4.2)

and

$$\begin{aligned} \Bigl | \int _{{\mathbb {T}}} \mathcal {H}u \mathcal {H}u_x u \Big |&\le \frac{\mu }{8} \Vert u_{x}\Vert _{L^2({\mathbb {T}})}^2+ \frac{2}{\mu } \int ( u \mathcal {H}u )^2 \nonumber \\&\le \frac{\mu }{8} \Vert u_{x}\Vert _{L^2({\mathbb {T}})}^2+ \frac{2}{\mu } \Vert u\Vert _{L^\infty ({\mathbb {T}})}^2 \Vert u \Vert _{L^2({\mathbb {T}})}^2 \nonumber \\&\le \frac{\mu }{8} \Vert u_{x}\Vert _{L^2({\mathbb {T}})}^2+ \frac{2}{\mu } \Vert u_x \Vert _{L^2({\mathbb {T}})} \Vert u \Vert _{L^2({\mathbb {T}})}^3 \nonumber \\&\le \frac{\mu }{4} \Vert u_{x}\Vert _{L^2({\mathbb {T}})}^2+ \frac{8}{\mu ^3} \Vert u \Vert _{L^2({\mathbb {T}})}^6 \; , \end{aligned}$$

(4.3)

using the inequality $ab \le a^2+b^2/4$ with $a=\frac{\mu }{4} \Vert u_{x}\Vert _{L^2({\mathbb {T}})}$ and $b=\frac{4}{\mu } \Vert u \Vert _{L^2({\mathbb {T}})}^3$ in the last step of (4.3). Combining (4.1), (4.2) and (4.3), we obtain

$$\begin{aligned} \frac{1}{2} \frac{d}{dt} \Vert u\Vert _{L^2({\mathbb {T}}))}^2\le & {} -\frac{5\mu }{8} \Vert u_x\Vert _{L^2({\mathbb {T}})}^2 + \frac{2}{\mu } \Vert k(t)\Vert _{L^2({\mathbb {T}})}^2 +\frac{8}{\mu ^3} \Vert u \Vert _{L^2({\mathbb {T}})}^6\nonumber \\\le & {} -\frac{5\mu }{8} \Vert u\Vert _{L^2({\mathbb {T}})}^2 + \frac{2}{\mu } \Vert k(t)\Vert _{L^2({\mathbb {T}})}^2 + \frac{8}{\mu ^3} \Vert u \Vert _{L^2({\mathbb {T}})}^6 \;. \end{aligned}$$

(4.4)

Assuming that

$$\begin{aligned} \quad \Vert u_0\Vert _{L^2({\mathbb {T}})}^2 \le \frac{\mu ^2}{2^{2}} \quad \text {and} \quad \Vert k\Vert _{L^\infty ({\mathbb {R}}_+;L^2({\mathbb {T}}))}^2 \le \frac{\mu ^4}{2^{7}} \;, \end{aligned}$$

(4.5)

yields therefore $\Vert u(t)\Vert _{L^2({\mathbb {T}})}^2 \le \frac{\mu ^2}{2^4}$ for all positive existence times since

$$\begin{aligned} -\frac{5\mu }{8} \frac{\mu ^2}{2^{2}} + \frac{\mu ^3}{2^6} + \frac{2^3}{\mu ^3} \Bigl (\frac{\mu ^2}{2^{2}}\Bigr )^3= - \frac{\mu ^3}{2^6}<0 \;. \end{aligned}$$

In view of Proposition 3.3, we thus obtain the following global existence result for small data:

Proposition 4.1

Let $k\in L^\infty ({\mathbb {R}}_+;L^2_0({\mathbb {T}}))$, $\mu >0$ and $u_0\in H^s_0({\mathbb {T}})$, with $s\ge 0$, satisfying (4.5). Then the solution u to (1.1)–(1.2) belongs to $C({\mathbb {R}}_+;H^s_0({\mathbb {T}}))$ and is bounded in $L^2({\mathbb {T}})$ uniformly in time.

5 Global Existence for Large Data

In this section we prove a global existence result for large data in the case $k \equiv 0$, corresponding to an isobaric level set $z=z_0$.

For spatially $2\pi$-periodic square-integrable real functions of mean zero we have the Fourier modes decomposition

$$\begin{aligned} {{\mathcal {H}}}u= -\textrm{i} \sum _{j \in {{\mathbb {Z}}}} \text {sgn}(j)\, \eta _j\,\textrm{e}^{\textrm{i}j x} \quad \text {for}\quad u=\sum _{j \in {{\mathbb {Z}}} \setminus \{0\}} \eta _j\,\textrm{e}^{\textrm{i}j x}, \end{aligned}$$

and, since u is real-valued, we also have

$$\begin{aligned} \eta _{-j}=\overline{\eta _j},\qquad j \in {{\mathbb {Z}}}\setminus \{0\}. \end{aligned}$$

(5.1)

The evolution equation (1.1) decomposes into the infinite system of ordinary differential equations

$$\begin{aligned}{} & {} {\dot{\eta }}_m + \sum _{l,\,j \in {{\mathbb {Z}}}:\ l+j=m} \textrm{i} l\, \eta _l \eta _j \Big [ 1 - \text {sgn}(l)\,\text {sgn}(j)\Big ] + \Big [ \mu \, m^2 - \textrm{i} \beta \, \text {sgn}(m) \Big ] \,\eta _m=0,\\{} & {} m \in {{\mathbb {Z}}} \setminus \{0\}, \end{aligned}$$

or, equivalently,

$$\begin{aligned}{} & {} {\dot{\eta }}_m + \textrm{i} m\sum _{l \in {{\mathbb {Z}}}} \eta _l \eta _{m-l} \Big [ 1 - \text {sgn}(l)\,\text {sgn}(m-l)\Big ] + \Big [ \mu \, m^2 - \textrm{i} \beta \, \text {sgn}(m) \Big ] \,\eta _m =0,\\{} & {} m \in {{\mathbb {Z}}}\setminus \{0\} . \end{aligned}$$

Thus

$$\begin{aligned} {\dot{\eta }}_n + \textrm{i} n\sum _{j \in {{\mathbb {Z}}}} \eta _j \eta _{n-j} \Big [ 1 - \text {sgn}(j)\,\text {sgn}(n-j)\Big ] + \Big [ \mu n^2 - \textrm{i} \beta \Big ] \,\eta _n =0, \qquad n \ge 1, \end{aligned}$$

the case of $\eta _m$ with $m<0$ being covered by (5.1). Consequently we have

$$\begin{aligned} {\dot{\eta }}_n + 4\textrm{i} n \sum _{j >n} \eta _j \eta _{n-j} + \Big [ \mu n^2 - \textrm{i} \beta \Big ] \,\eta _n =0,\qquad n \ge 1. \end{aligned}$$

(5.2)

This ensures that an initial data $u_0$ supported by a finite number of Fourier modes (with $\eta _j(0)=0$ for $|j| \ge N$) gives rise to a solution supported by a finite of Fourier modes with vanishing Fourier modes of index j with $|j| >N$. In particular, if

$$\begin{aligned} u_0(x)= 2{\Re } \Big \{\eta _1(0)\,\textrm{e}^{\textrm{i} x} + \eta _N(0)\,\textrm{e}^{\textrm{i} Nx} \Big \} \end{aligned}$$

for some integer $N \ge 2$, where $\Re \{\zeta \}$ and $\Im \{\zeta \}$ stand for the real and imaginary part of the complex number $\zeta$, respectively, then (5.2) particularizes for $N \ge 3$ to the linear system

$$\begin{aligned} {\left\{ \begin{array}{ll} &{}{\dot{\eta }}_1 + \Big [ \mu - \textrm{i} \beta \Big ] \,\eta _1 =0 ,\\ &{}{\dot{\eta }}_N + \Big [ \mu N^2 - \textrm{i} \beta \Big ] \,\eta _N =0\,, \end{array}\right. } \end{aligned}$$

(5.3)

and for $N=2$ to the nonlinear system

$$\begin{aligned} {\left\{ \begin{array}{ll} &{}{\dot{\eta }}_1 + 4\textrm{i} \,\eta _2 \overline{\eta _1} + \Big [ \mu - \textrm{i} \beta \Big ] \,\eta _1 =0 ,\\ &{}{\dot{\eta }}_2 + \Big [ 4\mu - \textrm{i} \beta \Big ] \,\eta _2 =0. \end{array}\right. } \end{aligned}$$

(5.4)

Both (5.3) and (5.4) can be solved explicitly and their solutions are defined for all $t \ge 0$. We have therefore proved the following result:

Proposition 5.1

For $k \equiv 0$ and any integer $N \ge 2$, an initial data of the form

$$\begin{aligned} u_0(x)= 2{\Re } \Big \{\eta _1(0)\,\textrm{e}^{\textrm{i} x} + \eta _j(0)\,\textrm{e}^{\textrm{i} Nx} \Big \} \end{aligned}$$

develops into a solution of (1.1) that is defined for all $t \ge 0$.

6 A Blow-Up Result

We show that for $k \equiv 0$ and $\beta >0$ the solution of (1.1) that emerges from a suitable initial data $u_0$ with mean-value zero and only supported by the first three Fourier modes blows-up in finite time:

Proposition 6.1

For $k \equiv 0$ and $\beta >0$ an initial data of the form

$$\begin{aligned} u_0(x)= 2{\Re } \Big \{ \eta _1(0)\,\textrm{e}^{\textrm{i} x} + \eta _2(0)\,\textrm{e}^{2\textrm{i} x} + \eta _3(0)\,\textrm{e}^{3\textrm{i} x} \Big \} \end{aligned}$$

with

$$\begin{aligned}&\quad&0 < \eta _3(0) \le \frac{\beta }{10 [ 1- \cos (\frac{\pi }{12})]}\,\textrm{e}^{\frac{\mu \pi }{2\beta }},\quad \eta _2(0)=\frac{\beta (1-\textrm{i})}{2(1- \cos (\frac{\pi }{12}))}\, \textrm{e}^{\frac{\mu \pi }{2\beta }}, \\{} & {} \quad \eta _1(0)=\frac{\beta ^2(1-2\textrm{i})}{16[ 1- \cos (\frac{\pi }{12}]\,\eta _3(0)}\,\textrm{e}^{\frac{\mu \pi }{\beta }}, \end{aligned}$$

develops into a solution

$$\begin{aligned} u(t,x)= 2{\Re } \Big \{ \eta _1(t)\,\textrm{e}^{\textrm{i} x} + \eta _2(t)\,\textrm{e}^{2\textrm{i} x} + \eta _3(t)\,\textrm{e}^{3\textrm{i} x} \Big \} \end{aligned}$$

to (1.1) with a finite maximal existence time $0<t_0 < \frac{\pi }{12\beta }$ such that

$$\begin{aligned} \lim _{t \uparrow t_0} \Im \{\eta _1(t)\}=-\infty . \end{aligned}$$

Proof

The equations (5.2) show that the solution emanating from $u_0$ will also have mean-value zero and will be supported by the three first Fourier modes for as long as it is defined. In particular, (5.2) simplifies to the following differential system

$$\begin{aligned} \left\{ \begin{array}{l} {\dot{\eta }}_1 +4 \textrm{i} (\eta _3 \overline{\eta _{2}} +\eta _2 \overline{\eta _1} ) + (\mu -\textrm{i}\beta ) \eta _1=0,\\ {\dot{\eta }}_2+ 8 \textrm{i} \eta _3 \overline{\eta _1} +(4\mu -\textrm{i}\beta )\eta _2=0 ,\\ {\dot{\eta }}_3+(9\mu - \textrm{i} \beta ) \eta _3=0 , \end{array} \right. \end{aligned}$$

(6.1)

The third equation can be solved explicitly, giving

$$\begin{aligned} \eta _3(t)=\eta _3(0)\,\textrm{e}^{-(9\mu - \textrm{i}\beta )t}, \end{aligned}$$

(6.2)

so that, by setting

$$\begin{aligned} {\left\{ \begin{array}{ll} {\tilde{\eta }}_1(t) &{}= \eta _1(t)\,\textrm{e}^{(\mu - \textrm{i}\beta )t},\\ {\tilde{\eta }}_2(t) &{}= \eta _2(t)\,\textrm{e}^{(4\mu - \textrm{i}\beta )t} , \end{array}\right. } \end{aligned}$$

(6.3)

we transform (6.1) to the equivalent system

$$\begin{aligned} \left\{ \begin{array}{l} \dot{{\tilde{\eta }}}_1 +4 \textrm{i} \,\eta _3(0) \overline{{\tilde{\eta }}_2(t)} \,\textrm{e}^{-(12\mu + \textrm{i}\beta )t} + 4 \textrm{i} \,{\tilde{\eta }}_2(t) \overline{{\tilde{\eta }}_1(t)} \,\textrm{e}^{-(4\mu + \textrm{i}\beta )t}=0,\\ \dot{{\tilde{\eta }}}_2+ 8 \textrm{i} \,\eta _3(0) \overline{{\tilde{\eta }}_1(t)} \,\textrm{e}^{-(6\mu + \textrm{i}\beta )t}=0 . \end{array} \right. \end{aligned}$$

(6.4)

Denoting now

$$\begin{aligned} {\tilde{\eta }}_1(t)=a_1(t) + \textrm{i}\,b_1(t),\qquad {\tilde{\eta }}_2(t)=a_2(t) + \textrm{i}\,b_2(t),\qquad \eta _3(0)=A + \textrm{i} \,B, \end{aligned}$$

(6.5)

with real coefficients $a_1$, $b_1$, $a_2$, $b_2$, A, and B, we can write (6.1) as the real $4 \times 4$ system

$$\begin{aligned}&{\dot{a}}_1 + \,4 \,\textrm{e}^{-12\mu t} \{ (Ab_2 -Ba_2) \cos (\beta t) + (Aa_2 +Bb_2)\sin (\beta t) \} \nonumber \\&\quad + \,4 \,\textrm{e}^{- 4\mu t} \{(a_2b_1-a_1b_2)\cos (\beta t) + (a_1a_2 + b_1b_2) \sin (\beta t) \}=0\,, \end{aligned}$$

(6.6)

$$\begin{aligned}&\quad {\dot{b}}_1 + \,4 \,\textrm{e}^{- 12\mu t} \{ (Aa_2 + Bb_2) \cos (\beta t) + (Ba_2 -Ab_2)\sin (\beta t) \} \nonumber \\&\quad + \,4 \,\textrm{e}^{- 4\mu t} \{(a_1a_2 + b_1b_2) \cos (\beta t) + (b_2a_1-a_2b_1)\sin (\beta t) \}=0\,, \end{aligned}$$

(6.7)

$$\begin{aligned} {\dot{a}}_2&\quad + \,8 \,\textrm{e}^{- 6\mu t} \{ (Ab_1 -Ba_1) \cos (\beta t) + (Aa_1 +Bb_1)\sin (\beta t) \} =0\,, \end{aligned}$$

(6.8)

$$\begin{aligned} {\dot{b}}_2&\quad + \,8 \,\textrm{e}^{- 6\mu t} \{ (Aa_1 +Bb_1) \cos (\beta t) + (Ba_1 -Ab_1)\sin (\beta t) \} =0 \,. \end{aligned}$$

(6.9)

For $\beta >0$ we have

$$\begin{aligned} \cos (\beta t)> 3 \sin (\beta t) >0 ,\qquad t \in \big (0, \tfrac{\pi }{12\beta }\big ),\nonumber \\ \end{aligned}$$

(6.10)

Denoting by $t_0 >0$ the maximal existence time of the solution, we introduce the open subset ${{\mathcal {A}}}$ of the linear space of continuously differentiable functions $(a_1,b_1,a_2,b_2)$ on $(0,t_0)$ by requiring:

$$\begin{aligned} b_1(t)< - a_1(t)<0 \quad \text {and}\quad -3a_2(t)< b_2(t) <0 \quad \text {for all}\quad t \in \big (0,\,\min \{t_0,\,\tfrac{\pi }{12\beta }\}\big ).\nonumber \\ \end{aligned}$$

(6.11)

We now prove that if the initial data satisfies

$$\begin{aligned} A> B=0,\quad a_1(0) > 3A,\quad b_1(0)< -a_1(0), \quad - 3a_2(0)< b_2(0) <0,\nonumber \\ \end{aligned}$$

(6.12)

then the unique solution to the system (6.6)–(6.9) belongs to ${{\mathcal {A}}}$. By continuity this solution verifies (6.11) on some interval [0, t) with $t>0$ small enough. If the claim does not hold, there exists some minimal value $t^*\in \big (0,\,\min \{t_0,\,\tfrac{\pi }{12\beta }\}\big )$ with $(a_1(t),\,b_1(t),\,a_2(t),$$b_2(t)) \in {{\mathcal {A}}}$ for all $t \in (0,t^*)$ and with $(a_1(t^*),\,b_1(t^*),\,a_2(t^*),\,b_2(t^*)) \not \in {{\mathcal {A}}}$. However, from (6.10) and (6.8) we get

$$\begin{aligned} {\dot{a}}_2&= -8A \,\textrm{e}^{- 6\mu t} \{ b_1 \cos (\beta t) + a_1 \sin (\beta t) \} \\&> -8A \,\textrm{e}^{- 6\mu t} \{ b_1 \cos (\beta t) + a_1 \cos (\beta t) \} \\&= -8A \,\textrm{e}^{- 6\mu t} [a_1 +b_1] \cos (\beta t) >0 \,,\qquad \qquad \qquad t \in (0,t^*)\,, \end{aligned}$$

while from (6.10) and (6.9) we obtain

$$\begin{aligned} {\dot{b}}_2&= -8A \,\textrm{e}^{- 6\mu t} \{ a_1 \cos (\beta t) - b_1 \sin (\beta t) \} \\&< -8A \,\textrm{e}^{- 6\mu t} \{ a_1 \sin (\beta t) - b_1 \sin (\beta t) \} \\&= -8A \,\textrm{e}^{- 6\mu t} [a_1 - b_1] \sin (\beta t) < 0 \,,\qquad \qquad \qquad t \in (0,t^*)\,, \end{aligned}$$

and therefore (6.12) ensures

$$\begin{aligned} a_2(t^*)> 0 > b_2(t^*) . \end{aligned}$$

(6.13)

Similarly, (6.10), (6.7) and (6.11) yield

$$\begin{aligned} {\dot{b}}_1&= - 4 \,\textrm{e}^{- 12\mu t} \Big \{ Aa_2 + \textrm{e}^{8\mu t} [a_1a_2 + b_1b_2] \Big \} \cos (\beta t) \\&\qquad - 4 \,\textrm{e}^{- 12\mu t} \Big \{ -Ab_2 + \textrm{e}^{8\mu t} [a_1b_2 - a_2b_1] \Big \} \sin (\beta t) \\&\le - 4 \,\textrm{e}^{- 12\mu t} \Big \{ A a_2 + \textrm{e}^{8\mu t} [a_1a_2 + b_1b_2] \Big \} \sin (\beta t) \\&\qquad - 4 \,\textrm{e}^{- 12\mu t} \Big \{ -Ab_2 + \textrm{e}^{8\mu t} [a_1b_2 - a_2b_1] \Big \} \sin (\beta t) \\&= - 4A \,\textrm{e}^{- 12\mu t} \Big \{ a_2 -b_2 \Big \} \sin (\beta t) \\&\qquad - 4 \,\textrm{e}^{- 4\mu t} \Big \{ b_2 [a_1 + b_1] - a_2 (b_1-a_1) \Big \} \sin (\beta t) < 0 \,,\qquad \qquad t \in (0,t^*)\,, \end{aligned}$$

so that (6.12) ensures

$$\begin{aligned} b_1(t^*)<-3A < 0 . \end{aligned}$$

(6.14)

On the other hand, from (6.6) and (6.7) we get

$$\begin{aligned} {\dot{a}}_1 + {\dot{b}}_1&= - 4 \,\textrm{e}^{- 12\mu t} \Big \{ Aa_2 + Ab_2 + \textrm{e}^{8\mu t} [a_1a_2 + b_1b_2 + a_2b_1 -a_1b_2] \Big \} \cos (\beta t) \\&\qquad - 4 \,\textrm{e}^{- 12\mu t} \Big \{ Aa_2 - Ab_2 + \textrm{e}^{8\mu t} [a_1a_2 + b_1b_2 - a_2b_1 + a_1b_2] \Big \} \sin (\beta t) \\&= - 4 \,\textrm{e}^{- 12\mu t} \Big \{ Aa_2 + (A+\textrm{e}^{8\mu t}b_1)b_2 + \textrm{e}^{8\mu t} [(a_1+b_1)a_2 -a_1b_2] \Big \} \cos (\beta t) \\&\qquad - 4 \,\textrm{e}^{- 12\mu t} \Big \{ Aa_2 - Ab_2 + \textrm{e}^{8\mu t} [a_1a_2 + (a_1+b_1)b_2 - a_2b_1 ] \Big \} \sin (\beta t) \,, \end{aligned}$$

which, since (6.12) and the fact that $b_1$ is negative and strictly decreasing on $(0,t^*)$ ensure

$$\begin{aligned} A+\textrm{e}^{8\mu t}b_1(t) \le A+b_1(0) < 0 ,\qquad \qquad t \in (0,t^*), \end{aligned}$$

leads to

$$\begin{aligned} {\dot{a}}_1 + {\dot{b}}_1&\le - 4 \,\textrm{e}^{- 12\mu t} \Big \{ Aa_2 + \textrm{e}^{8\mu t} [(a_1+b_1)a_2 -a_1b_2] \Big \} \cos (\beta t) \\&\quad - 4 \,\textrm{e}^{- 12\mu t} \Big \{ A(a_2 - b_2) + \textrm{e}^{8\mu t} [a_1a_2 + (a_1+b_1)b_2 - a_2b_1 ] \Big \} \sin (\beta t) \\&\quad t \in (0,t^*)\,, \end{aligned}$$

Thus

$$\begin{aligned}&({\dot{a}}_1 + {\dot{b}}_1) + 4 \,\textrm{e}^{- 4\mu t} (a_1+b_1)\Big (a_2 \cos (\beta t)+b_2 \sin (\beta t)\Big ) \\&\qquad \le - 4 \textrm{e}^{- 12\mu t} \Big \{ [Aa_2 -\textrm{e}^{8\mu t} a_1 b_2] \cos (\beta t) + \Bigl (A(a_2 - b_2) \\&\qquad + \textrm{e}^{8\mu t} [a_1a_2 - a_2b_1 ] \Bigr ) \sin (\beta t)\Big \} <0\,, \quad t \in (0,t^*)\,, \end{aligned}$$

so that the function

$$\begin{aligned} t \mapsto [a_1(t)+b_1(t)] \,\exp \Big \{ \int _0^t 4 \,\textrm{e}^{- 4\mu s}[a_2(s)\cos (\beta s)+b_2(s)\sin (\beta s)]\,\textrm{d}s\Big \} \end{aligned}$$

is strictly decreasing on $(0,t^*)$. Since $a_1(0)+b_1(0)<0$ by (6.12), we obtain that

$$\begin{aligned} a_1(t) + b_1(t) < 0 ,\qquad t \in [0,t^*]. \end{aligned}$$

(6.15)

We use now (6.15) in combination with (6.8), (6.9) and (6.10) to get

$$\begin{aligned} 3{\dot{a}}_2 + {\dot{b}}_2&= - 8A \,\textrm{e}^{- 6\mu t} \Big \{ ( 3b_1 + a_1) \cos (\beta t) +( 3a_1 - b_1) \sin (\beta t) \Big \} \\&\ge - 8A \,\textrm{e}^{- 6\mu t} \Big \{ 3( 3b_1 + a_1) \sin (\beta t) +( 3a_1 - b_1) \sin (\beta t) \Big \}\\&= - 8A \,\textrm{e}^{- 6\mu t} \Big \{ 8b_1 + 6a_1\Big \} \sin (\beta t) >0\,,\qquad \qquad t \in (0,t^*)\,. \end{aligned}$$

In particular, since $3a_2(0)+b_2(0)>0$ due to (6.12), we deduce that

$$\begin{aligned} 3a_2(t) + b_2(t) > 0 ,\qquad t \in [0,t^*]. \end{aligned}$$

(6.16)

These considerations show that the only option left is that $a_1(t^*)=0$. But then, taking into account (6.13), (6.14), (6.9) and (6.15), we would obtain from (6.6) and (6.10) that

$$\begin{aligned} {\dot{a}}_1(t^*)&= - 4 \,\textrm{e}^{- 12\mu t^*} \Big \{ Ab_2(t^*) + \textrm{e}^{8\mu t^*} a_2(t^*) b_1(t^*)\Big \} \cos (\beta t^*) \\&\qquad - 4 \,\textrm{e}^{- 12\mu t^*} \Big \{ Aa_2(t^*) + \textrm{e}^{8\mu t^*} b_1(t^*)b_2(t^*) \Big \} \sin (\beta t^*) \\&\ge - 12 \,\textrm{e}^{- 12\mu t^*} \Big \{ Ab_2(t^*) + \textrm{e}^{8\mu t^*} a_2(t^*) b_1(t^*)\Big \} \sin (\beta t^*) \\&\qquad - 4 \,\textrm{e}^{- 12\mu t^*} \Big \{ Aa_2(t^*) + \textrm{e}^{8\mu t^*} b_1(t^*)b_2(t^*) \Big \} \sin (\beta t^*) \\&= - 4 \,\textrm{e}^{- 12\mu t^*} \Big \{ 3Ab_2(t^*) + 3\textrm{e}^{8\mu t^*} a_2(t^*) b_1(t^*) \\&+ Aa_2(t^*) + \textrm{e}^{8\mu t} b_1(t^*)b_2(t^*) \Big \} \sin (\beta t^*) \\&> - 4 \,\textrm{e}^{-12\mu t^*} \Big ( 3A + \textrm{e}^{8\mu t^*} b_1(t^*)\Big ) \Big ( 3 a_2(t^*) + b_2(t^*)\Big ) \sin (\beta t^*) > 0 \,, \end{aligned}$$

which is a contradiction with the assumption $a_1(t)>0=a_1(t^*)$ for $t \in (0,t^*)$. This proves the claim about the solution developing from a data subject to the constraints (6.12) remaining in the set ${{\mathcal {A}}}$ for all $t \in \big [0,\,\min \{t_0,\,\tfrac{\pi }{12\beta }\}\big )$.

We now claim that for a suitable choice of initial data subject to (6.12), the resulting solution will blow-up in the finite time $t_0 \in \big (0,\, \tfrac{\pi }{12\beta }\big ]$, with

$$\begin{aligned} \lim _{t \uparrow t_0} b_1(t) = -\infty . \end{aligned}$$

(6.17)

Indeed, if the negative function $b_1(t)$ stays bounded from below on $[0,t_0)$, then the relation $0< a_1(t) < -b_1(t)$ yields that $a_1(t)$ stays bounded on $[0,t_0)$ and the linear equations (6.8)–(6.9) now ensure this also for the remaining two Fourier modes $a_2(t)$ and $b_2(t)$.

To prove blow-up, setting

$$\begin{aligned} \gamma = 4A \textrm{e}^{- \frac{\mu \pi }{2\beta }} >0, \end{aligned}$$

the validity of (6.10)–(6.11) in combination with (6.7) and (6.9) with $B=0$ yields

$$\begin{aligned} {\dot{b}}_1&=- \,4A \,\textrm{e}^{- 12\mu t} \{ a_2 \cos (\beta t) - b_2\sin (\beta t) \} - \,4 \,\textrm{e}^{- 4\mu t} \{(a_1a_2 + b_1b_2) \cos (\beta t) \\&\quad + (b_2a_1-a_2b_1)\sin (\beta t) \}\\&\le - \,4 \,\textrm{e}^{- 4\mu t} \{(a_1a_2 + b_1b_2) \cos (\beta t) + b_2a_1 \sin (\beta t) \}\\&= - \,4 \,\textrm{e}^{- 4\mu t} b_1b_2 \cos (\beta t) - \,4 \,\textrm{e}^{- 4\mu t} \{a_1a_2 \cos (\beta t) + b_2a_1 \sin (\beta t) \}\\&\le - \,4 \,\textrm{e}^{-4\mu t} b_1b_2 \cos (\beta t) - \,4 \,\textrm{e}^{- 4\mu t} \{3a_1a_2 \sin (\beta t) + b_2a_1 \sin (\beta t) \}\\&= - \,4 \,\textrm{e}^{- 4\mu t} b_1b_2 \cos (\beta t) - \,4 \,\textrm{e}^{- 4\mu t} a_1(3a_2 +b_2) \sin (\beta t) \\&\le - \,4 \,\textrm{e}^{-4\mu t} b_1b_2 \cos (\beta t) \le - \frac{\gamma }{A} \,b_1b_2 \,\cos (\beta t) \\&\le - \frac{\gamma }{A} \,b_1b_2 \,\sin (\beta t)\,,\quad t \in \big (0,\,\min \{t_0,\,\tfrac{\pi }{12\beta }\}\big )\,, \end{aligned}$$

and

$$\begin{aligned} {\dot{b}}_2&= - \,8A \,\textrm{e}^{- 6\mu t} \{ a_1 \cos (\beta t) - b_1\sin (\beta t) \} \\&\le 8A \,\textrm{e}^{- 6\mu t} b_1\sin (\beta t) \le 2\gamma \,b_1 \,\sin (\beta t)\,,\qquad \qquad t \in \big (0,\,\min \{t_0,\,\tfrac{\pi }{12\beta }\}\big )\,, \end{aligned}$$

Therefore

$$\begin{aligned} {\left\{ \begin{array}{ll} {\dot{b}}_1 \le - \displaystyle \frac{\gamma }{A} \,b_1b_2 \,\sin (\beta t),\\ {\dot{b}}_2 \le 2\gamma \,b_1 \,\sin (\beta t), \end{array}\right. }\qquad t \in \big (0,\,\min \{t_0,\,\tfrac{\pi }{12\beta }\}\big ). \end{aligned}$$

(6.18)

with

$$\begin{aligned} b_1(t)<0 \quad \text {and} \quad b_2(t) <0 \quad \text {for all }\quad t \in \big [0,\,\min \{t_0,\,\tfrac{\pi }{12\beta }\}\big ). \end{aligned}$$

(6.19)

Since for every fixed $t \in \big [0,\,\min \{t_0,\,\tfrac{\pi }{12\beta }\}\big )$ the vector function

$$\begin{aligned} F(t,b_1,b_2) =\Big (- \frac{\gamma }{A}\, b_1b_2 \sin (\beta t),\,2\gamma \,b_1 \sin (\beta t)\Big ) \end{aligned}$$

is of Kamke type in the region $\{(b_1,b_2) \in {{\mathbb {R}}}^2:\ b_1<0,\, b_2 <0\}$ (that is, its first component is a nondecreasing function of $b_2$ and its second component is a nondecreasing function of $b_1$), the comparison method applies (see [7, 9]): if the continuously differentiable functions $B_1,\,B_2: [0,T) \rightarrow (-\infty ,0)$ solve the differential system

$$\begin{aligned} {\left\{ \begin{array}{ll} {\dot{B}}_1 = - \frac{\gamma }{A} \,B_1B_2 \,\sin (\beta t),\\ {\dot{B}}_2 = 2\gamma B_1\,\sin (\beta t), \end{array}\right. }\qquad t \in \big (0,\,\min \{t_0,\,\tfrac{\pi }{12\beta }\}\big ). \end{aligned}$$

(6.20)

on some interval $(0,T) \subset \big (0,\,\min \{t_0,\,\tfrac{\pi }{12\beta }\}\big )$ and if

$$\begin{aligned} b_1(0) \le B_1(0)<0 ,\qquad b_2(0) \le B_2(0) <0, \end{aligned}$$

(6.21)

then

$$\begin{aligned} b_1(t) \le B_1(t) \quad \text {and}\quad b_2(t) \le B_2(t) \quad \text {for all} \quad t \in [0,T). \end{aligned}$$

(6.22)

Setting

$$\begin{aligned} d= \cos \big (\tfrac{\pi }{12}\big ) \in (0,1), \end{aligned}$$

we verify by direct computation that

$$\begin{aligned} B_1(t)= - \frac{\beta ^2 A}{\gamma ^2 [\cos (\beta t) -d]^2},\qquad B_2(t)=- \frac{2\beta A}{\gamma [\cos (\beta t) -d]},\qquad t \in \Big [0,\frac{\pi }{12\beta }\Big ), \end{aligned}$$

are strictly decreasing negative functions that solve (6.20) and satisfy

$$\begin{aligned} \lim _{t \uparrow \frac{\pi }{12\beta }} B_1(t)=\lim _{t \uparrow \frac{\pi }{12\beta }} B_2(t)=-\infty . \end{aligned}$$

Now we observe that for $0<A\le \frac{\beta }{10(1-d)} \,\textrm{e}^{\frac{\mu \pi }{2\beta }}$ we have

$$\begin{aligned} \frac{\beta ^2 A}{2 \gamma ^2(1-d)^2}=\frac{\beta ^2 A}{2\Big (4A e^{-\frac{\mu \pi }{2\beta }} \Big )^2(1-d)^2}= \frac{\beta ^2}{32 A(1-d)^2} \textrm{e}^{\frac{\mu \pi }{\beta }}>3A . \end{aligned}$$

Therefore, due to (6.11) and (6.22), the solution to (6.6)–(6.9) with initial data

$$\begin{aligned} a_1(0)= \frac{\beta ^2 A}{2 \gamma ^2(1-d)^2}> 0,\quad b_1(0)= - \frac{\beta ^2 A}{\gamma ^2(1-d)^2}<0,a_2(0)= - b_2(0)= \frac{2\beta A}{\gamma (1-d)} >0, \end{aligned}$$

blows-up at some finite time $t_0 \in (0,\frac{\pi }{12\beta }]$, since (6.12) is verified. $\square$