October 02, 2025
We analyze the nonlinear inertial instability of Couette flow under Coriolis forcing in \(\mathbb{R}^{3}\). For the Coriolis coefficient \(f \in (0,1)\), we show that the non-normal operator associated with the linearized system admits only continuous spectrum. Hence, there are no exponentially growing eigenfunctions for the linearized system. Instead, we construct unstable solutions in the form of pseudo-eigenfunctions that exhibit non-ideal spectral properties. Then through a bootstrap argument and resolving the challenges posed by the non-ideal spectral behavior of pseudo-eigenfunctions, we establish the velocity instability of Couette flow in the Hadamard sense for \(f \in \Big(\frac{2}{17} \left(5-2 \sqrt{2}\right), \frac{2}{17} \left(5 + 2 \sqrt{2}\right) \Big)\).
Couette flow; inertial instability; Coriolis force; nonlinear instability.
Inertial instability of shear flows is a crucial concept in fluid dynamics, manifesting in diverse natural and engineered fluid systems. It occurs when the equilibrium between the pressure gradient and the Coriolis force in a rotating fluid is disrupted. The inertial instability of shear flows drives the redistribution of momentum, heat, and mass within the fluid. It can give rise to complex flow patterns and vortices, affecting the overall energy balance and dynamic processes of the system. Understanding this instability is essential for accurately predicting and explaining various phenomena in fields ranging from climate science [1]–[3] to astrophysics [4], [5].
In this article, we consider inertial instability governed by the Navier-Stokes equations with Coriolis force in \(\mathbb{R} ^3\) \[\begin{align} \begin{cases} \frac{\partial{\mathbf{v}}}{\partial{t}}+ \mathbf{v} \cdot \nabla \mathbf{v} + \mathbf{f} \times \mathbf{v} = \nu \Delta \mathbf{v} - \nabla p , \\ \operatorname{div}\mathbf{v}= 0, \end{cases} \end{align}\] where \(\mathbf{v} = (v_1, v_2, v_3)\) is the velocity, \(p\) is the pressure, \(\mathbf{f} = (0, 0, f )\) is the Coriolis vector and \(f>0\) is called the Coriolis coefficient. A steady shear flow takes the form of \[\label{steady95state} \overline{\mathbf{v}}=\big(U(y),0,0\big),\quad \overline{p} =\overline{p}(y),\tag{1}\] where \(U\) and \(\overline{p}\) satisfy the geostrophic balance \[\begin{align} f U(y) = - \frac{\partial{\overline{p}}}{\partial{y}}. \end{align}\] The inertial instability of shear flows is determined by the Rayleigh discriminant [4], also known as the Bradshaw-Richardson number [6], [7], defined as \[\mathrm{R}(y;f,U(y)) \overset{\mathrm{def}}{=}f (f-U'(y)).\] When \(\mathrm{R}(y;f,U(y)) <0\) at some \(y_0\), the flow becomes inertially unstable. In this article, we consider the Couette flow \[\label{cff} U(y) = y.\tag{2}\] Thus, the Rayleigh discriminant becomes a constant \(\mathrm{R}(f) = \mathrm{R}(y;f,y)=f (f- 1).\)
Couette flow is a phenomenon where viscous fluids move between parallel plates or concentric cylinders, driven by viscous shear forces as one surface moves tangentially. The Couette flow along with other monotonic shear flows are common in geophysical fluids and have been comprehensively studied from both the physical [8]–[12] and the mathematical perspectives [13]–[21]. As the Couette flow is widespread within geophysical fluids, the primary factor leading to the instability of this flow is the Coriolis force. This type of instability is geophysically referred to as inertial instability [22].
Introducing the perturbations around the Couette flow 2 \[\mathbf{u} = (u_1, u_2 ,u_3) = \mathbf{v} - \overline{\mathbf{v}}, \quad q = p - \overline{p},\] we derive the perturbed equations \[\label{nonlinear95perturbed95equation} \begin{cases} \frac{\partial{\mathbf{u}}}{\partial{t}} + \mathbf{u} \cdot \nabla \mathbf{u} + \begin{pmatrix} u_2 \\0 \\0 \end{pmatrix} + y \partial_x \mathbf{u} + f \begin{pmatrix} -u_2 \\ u_1 \\0 \end{pmatrix} = \nu \Delta \mathbf{u}- \nabla q , \\ \operatorname{div}\mathbf{u}= 0, \end{cases}\tag{3}\] where the pressure \(q\) is given by \(q \overset{\mathrm{def}}{=}q^{L} + q^{NL}\) and \[\require{physics} \begin{align} q^{L} &\overset{\mathrm{def}}{=}- \Delta ^{-1} \operatorname{div}\qty( \mathbf{u} \cdot \nabla \overline{\mathbf{v}} +\overline{\mathbf{v}}\cdot \nabla \mathbf{u} + \mathbf{f} \times \mathbf{u}) \\&= - \Delta ^{-1} \qty( 2\partial_x u_2 + f ( \partial_y u_1 - \partial_x u_2 )), \\ q^{NL} &\overset{\mathrm{def}}{=}- \Delta ^{-1} \operatorname{div}\qty(\mathbf{u} \cdot \nabla \mathbf{u}). \end{align}\] The system is supplemented with the initial data \[\label{initial95conditions95for95perturbed95equation} \left. \mathbf{u}(t,\mathbf{x}) \right|_{t = 0} = \mathbf{u}_{\mathrm{in}}(\mathbf{x}) \overset{\mathrm{def}}{=}(u_{1,\mathrm{in}}(\mathbf{x}),u_{2,\mathrm{in}}(\mathbf{x}),u_{3,\mathrm{in}}(\mathbf{x}))\tag{4}\] and the far-field condition at infinity \[\label{boundary95conditions95for95perturbed95equation} \lim_{ \abs{\mathbf{x}} \to + \infty} \abs{\mathbf{u}(t,\mathbf{x})} = 0, \quad \forall t\geq 0.\tag{5}\]
Specifically, we aim to establish the nonlinear inertial instability in the following sense:
Definition 1 (Instability in the sense of Hadamard). The steady state solution \(\mathbf{\overline{v}}\) is nonlinearly unstable if there are constants \(\sigma\) and \(C\) such that for every \(\delta\) arbitrarily small there exists a solution \({\boldsymbol{u}}\) of 3 satisfying \[\begin{align} &||{\boldsymbol{u}}(0,\mathbf{x}) ||_{H^{1}} \leq\delta,\quad ||\mathbf{u}(T^{\delta},\mathbf{x}) ||_{L^{2}} \geq\sigma, \end{align}\] where \(T^{\delta}\leq C\ln \delta^{-1} + C\) is an escape time.
It is clear that Definition 1 implies violation of the continuous dependence of the solution on the initial data. We refer to [23]–[26] among others for the analysis of Rayleigh-Taylor instability in the sense of Hadamard.
The linear part of system 3 is identified as \[\require{physics} \label{linearized95perturbed95equation} \begin{cases} \frac{\partial{\mathbf{u}}}{\partial{t}} + \begin{pmatrix} u_2 \\0 \\0 \end{pmatrix} + y \partial_x \mathbf{u} + f \begin{pmatrix} -u_2 \\ u_1 \\0 \end{pmatrix} = \nu \Delta \mathbf{u} + \nabla \Delta ^{-1} \qty( 2\partial_x u_2 + f ( \partial_y u_1 - \partial_x u_2 )) , \\ \operatorname{div}\mathbf{u}= 0, \end{cases}\tag{6}\] with the initial-boundary condition 4 5 . A key step in the analysis is to obtain the spectral properties of the linearized operator \(\mathscr{L}\) \[\label{Linear95eq95abstract95form} \partial_t \mathbf{u} = \mathscr{L} \mathbf{u} ,\tag{7}\] with the domain of definition \[\require{physics} \label{domain95of95L} \begin{align} \mathcal{D} (\mathscr{L}) & \overset{\mathrm{def}}{=}\qty{ \mathbf{u} =(u_1,u_2,u_3) \in[L^2(\mathbb{R}^3)]^3 \;\middle| \; \nu \Delta\mathbf{u} , y \partial_x \mathbf{u} \in[L^2(\mathbb{R}^3)]^3 } . \\ \end{align}\tag{8}\] We split the operator \(\mathscr{L}\) as follows \[\require{physics} \label{definition95of95L} \begin{align} \mathscr{L} &\overset{\mathrm{def}}{=}\mathscr{L}_0 + \mathscr{L}_1 + \mathscr{L}_{2}, \quad \mathscr{L}_{2} \overset{\mathrm{def}}{=}- \mathop{\mathrm{diag}} (y \partial_x, y \partial_x, y \partial_x), \\ \mathscr{L}_0 & \overset{\mathrm{def}}{=} \begin{pmatrix} \nu \Delta & f - 1 & 0 \\ -f + f \Delta^{-1} \partial_{y}^2 & \nu \Delta & 0\\ f \Delta^{-1} \partial_{yz} & 0 &\nu \Delta \\ \end{pmatrix} , \\ \mathscr{L}_1 & \overset{\mathrm{def}}{=}\Delta ^{-1} \begin{pmatrix} f \partial_{xy} & \qty(2 - f )\partial_x^2 & 0 \\ 0 & \qty(2 - f )\partial_{xy} & 0 \\ 0 & \qty(2 - f )\partial_{xz} & 0 \\ \end{pmatrix}. \end{align}\tag{9}\] That is
\(\mathscr{L}_0\) denotes the component whose form remains invariant when \(\mathbf{u}\) is independent of \(x\);
\(\mathscr{L}_1\) is the bounded part of the operator involving partial derivatives with respect to \(x\);
\(\mathscr{L}_2\) is the bad part with \(\partial_x\) that gives non-normal feature to the operator.
Note that \(\mathscr{L}_0\) is a normal operator for which the analysis is standard, cf. [27].
****Remark** 1**. As observed in [28], [29] and references therein, the set of unknowns \(w_{i} = \Delta u_{i}\), \(i = 1,2,3\) are more convenient to use than \(\mathbf{u}\) in the study of stability threshold, see also [30]. In this article, however, the main focus is on nonlinear instability of \(\mathbf{u}\).
For nonlinear instability analysis, one seeks the linearly growing modes and the bound of growth rate of the linear equation, cf. [25], [31]. However, the growth rate of the linear operator does not need be optimal as long as it is less than twice the growth rate of the linearly growing modes. This is sufficient to overcome the effect of the nonlinear terms, thereby proving nonlinear instability in the Hadamard sense. In what follows, we define the exponential growth rates \[\overline{\Lambda} \overset{\mathrm{def}}{=}\frac{2 - f}{2} \qand \underline{\Lambda} \overset{\mathrm{def}}{=}\sqrt{f( 1 - f)},\] and we will prove later that these rates act as the bounds on the growth rates of the linear operator and the linearly growing modes, respectively. Then, one imposes \(\overline{\Lambda} < 2 \underline{\Lambda}\) in order to obtain nonlinear instability in the Hadamard sense .
For the operator \(\mathscr{L}\), we have the following resolvent estimate:
Lemma 1. Suppose \(\nu > 0\), \(0<f < 1\) and \(\lambda \in \mathbb{C}\) satisfying \(\Re\lambda > \overline{\Lambda}\), then we have the resolvent estimate \[\begin{align} &\norm{(\lambda - \mathscr{L})^{-1} }_{L^2} \leq \frac{1}{\Re \lambda - \overline{\Lambda}}. \end{align}\]
The proof of Lemma 1, though lengthy, is standard [32]. The details are provided in the appendix. Then the Hille-Yosida theorem, [33] implies
Lemma 2. Suppose \(0<f < 1\), the semigroup \(e^{t \mathscr{L}}\) generated by \(\mathscr{L}\) on \([L^2(\mathbb{R}^3)]^3\) satisfies \[\norm{e^{t \mathscr{L}}}_{L^2} \leq e ^{t \overline{\Lambda}} \qfor t \geq 0.\]
For the construction of linearly unstable modes, one notes that the linear system 6 neither possesses a natural variational structure nor admits an integrable separation-of-variables solution in \(L^2\), cf. [25], [26], [34]–[36] for applications of the variational method. Nonetheless, we observe that the zonal-independent variant of the equations 6 admits separation-of-variables solutions, though these solutions are not integrable in \(\mathbb{R}^3\) and thus not eigenfunctions of the linear operator. Indeed, as shown in 5, the linear operator does not have any point spectrum–this implies that it is impossible to find an exponentially growing eigenfunction for the linearized system. To tackle this problem, we take a perturbative approach and look for the linear solution near the exponentially growing solution of the zonal-independent version of the equations 6 . We have
Theorem 1. Suppose \(0 < f < 1\), then for any given time \(T > 0\) and \(\epsilon > 0\), equations 6 with initial-boundary condition 4 5 has a real valued solution \(\mathbf{u}^{T,\epsilon}(t) \in C([0,T];[H^k(\mathbb{R}^3)]^3)\) and \(\nabla q^{T,\epsilon} (t) \in C([0,T];[H^k(\mathbb{R}^3)]^3)\), \(k \geq 0\) satisfying
For any \(t\in [0,T]\) \[\begin{align} & (e^{t \underline{\Lambda} } - \epsilon) \norm{ \mathbf{u}^{T,\epsilon}(0)}_{L^2} \leq \norm{\mathbf{u}^{T,\epsilon}(t)}_{L^2}\leq (e^{t \underline{\Lambda} } + \epsilon) \norm{\mathbf{u}^{T,\epsilon}(0)}_{L^2}; \\ \end{align}\]
For any positive integer \(k\), there exists a constant \(C_k > 0\) independent of \(T\) and \(\epsilon\) such that \[\norm{\mathbf{u}^{T,\epsilon}(0)}_{H^k} \leq C_k k \norm{\mathbf{u}^{T,\epsilon}(0)}_{L^2}.\]
The nonlinear instability of the perturbed system 3 is established in the following theorem.
Theorem 2. The Couette flow is unstable in the sense of Hadamard. That is, for any \(\frac{2}{17} \left(5-2 \sqrt{2}\right) < f < \frac{2}{17} \left(5 + 2 \sqrt{2}\right)\), \(\nu > 0\), there exists constants \(\overline{\delta}_0 \leq 1\), \(\varepsilon_0 > 0\) and an initial condition \(\mathbf{u}_{\mathrm{in}} \in [\mathcal{S}(\mathbb{R}^3)]^3\), such that for any \(\delta \in (0,\overline{\delta}_0)\), there exists a unique strong solution \(\mathbf{u}^{\delta} \in C([0,T^{\max});H^1(\mathbb{R}^3))\cap L^2(0,T^{\max};H^2(\mathbb{R}^3))\) of 3 5 emanating from the initial data \(\mathbf{u}_{\mathrm{in}}^{\delta} \overset{\mathrm{def}}{=}\delta \mathbf{u}_{\mathrm{in}}\) with an associated pressure \(q^{\delta} \in C([0, T^{\max});H^{1}(\mathbb{R}^3))\), such that \[\begin{align} \norm{\mathbf{u}^{\delta}(T^{\delta})}_{L^2} > \varepsilon_0 \end{align}\] for some escape time \(T^{\delta} =\frac{1}{\underline{\Lambda}} \ln \frac{2\varepsilon_0}{\delta} < T^{\max}\), where \(T^{\max}\) denotes the maximal time of existence of solution \(\mathbf{u}^{\delta}\).
Note that the linearly unstable solution is given by a pseudo-eigenfunction, that does not behave like an exponential function for large time. Indeed, as time increases, the error between the pseudo-eigenfunction and an exponential function accumulates. Only for a fixed time \(T\), one can find an unstable solution that is uniformly close to an exponential in \([0,T]\). To establish the nonlinear instability result, one needs a precise estimate of the dependence of \(T\) on the size of the initial condition, and a careful construction of the initial condition.
Without the Coriolis effect \(f = 0\), the linearized system around Couette flow is known to be spectrally stable, in the sense that there are no unstable eigenmodes. However, it has been observed that the linear system has large pseudo-spectra and may lead to significant transient growth [37]. In this case, the study of nonlinear stability, namely the transition threshold, of 3D Couette flow is carried out in some recent works [28], [38]–[41]. For nonlinear instability, Li et al. [30] adopted a dynamical approach to study the instability of steady-state profiles near the Couette flow, and showed that the vorticity field is unstable in the \(L^2\) norm. Recently, boundary-driven instability [42] and viscosity-driven instability [43] of shear flows have also been investigated.
The nonlinear stability of Couette flow with Coriolis force is established in [44], [45] when \(\mathrm{R}(f) > 0\). In particular, Guo et al. [46] constructed a class of axisymmetric global solutions near the stationary state to the 3d Euler equations with uniform rigid body rotation \(f = 1\) (\(R(f)= 0\)). To the best of our knowledge, when \(R(f) < 0\), the inertial instability of the Couette flow in \(\mathbb{R}^3\) has not been resolved in the literature, which is the major undertaking of this article. Recent results regarding the effect of the Coriolis force on the stability and instability of Couette flow are summarized in the table below.
| References | domain | range of \(f\) | small perturbation is |
|---|---|---|---|
| [38], [40] | \(\mathbb{T} \times \mathbb{R} \times \mathbb{T}\) | \(f = 0\) | nonlinear stable |
| [45] | \(\mathbb{T} \times \mathbb{R} \times \mathbb{T}\) | \(( - \infty, - 2] \cup [2, + \infty)\) | nonlinear stable |
| [47] | \(\mathbb{T} \times \mathbb{R} \times \mathbb{T}\) | \(f = 1\) | nonlinear stable |
| [44] | \(\mathbb{T} \times \mathbb{R} \times \mathbb{T}\) | \(( - \infty,0) \cup (1, + \infty)\) | nonlinear stable |
| Current paper | \(\mathbb{R}^3\) | \(\require{physics} \qty(\frac{2}{17} \left(5-2 \sqrt{2}\right), \frac{2}{17} \left(5 + 2 \sqrt{2}\right) )\) | nonlinear unstable |
From the linear analysis, it is expected that the Couette flow is nonlinearly unstable when \(0 < f < 1\), while remaining stable outside this interval. Our result shows that when \(f\) is in \[\require{physics} \label{f95range} \qty(\frac{2}{17} \left(5-2 \sqrt{2}\right), \frac{2}{17} \left(5 + 2 \sqrt{2}\right) ) \approx (0.255479,0.920991) \subseteq (0,1),\tag{10}\] the velocity is nonlinearly unstable in the Hadamard sense. This means that deriving results regarding the nonlinear stability threshold is infeasible. It also implies that the stabilizing effect, such as inviscid damping and enhanced dissipation, fails to suppress the linear instability effect generated by the Coriolis force.
However, our ability to establish the system’s instability is currently limited to 10 , i.e., \(2\underline{\Lambda} > \overline{\Lambda}\), which is a constraint rooted in the linear operator’s non-normality: the semigroup may grow more rapidly than even the fastest-growing (pseudo)-eigenfunctions. So, our future work will involve establishing the (in)stability of the system for \[\require{physics} f\in \qty(0,\frac{2}{17} \left(5-2 \sqrt{2}\right)) \cup \qty(\frac{2}{17} \left(5 +2 \sqrt{2}\right),1),\] and exploring the competition between destabilizing effects (lift-up, Coriolis force) and stabilizing effects (inviscid damping, enhanced dissipation, viscosity).
The article is structured as follows. 2 recalls some preliminary concepts. 3 covers semigroup, spectral analysis and instability of linearized system. 4 proves the nonlinear instability 2. The appendix provides some analysis tools.
Throughout, \(\left \langle \cdot, \cdot \right \rangle\) is the \(L^2\) inner product. \(\cdot ^{\intercal}\) denotes the transpose of a matrix or a vector. The space \(\mathcal{H} = [L^2(\mathbb{R}^3)]^2\). Suppose \(\mathscr{T} :\mathcal{H}_1 \to \mathcal{H}_2\) is an operator from the Hilbert space \(\mathcal{H}_1\) to another Hilbert space \(\mathcal{H}_2\) in which the linear subspace \(\mathcal{D}(\mathscr{T})\) is the domain of \(\mathscr{T}\). We denote the range of \(\mathscr{T}\) by \[\require{physics} \mathcal{R}(\mathscr{T}) \overset{\mathrm{def}}{=}\mathscr{T}(\mathcal{D}(\mathscr{T})) = \qty{\mathscr{T}(x) \;| \;x\in \mathcal{D}(\mathscr{T}) }.\] We write the kernel of \(\mathscr{T}\) as \[\require{physics} \mathcal{N}(\mathscr{T}) \overset{\mathrm{def}}{=}\qty{x \in \mathcal{D}(\mathscr{T}) \;| \;\mathscr{T}(x) = 0 }.\] The graph of \(\mathscr{T}\) is the set \[\require{physics} \mathcal{G}(\mathscr{T} ) = \qty{(x, \mathscr{T} x)\;| \;x \in \mathcal{D}(\mathscr{T} )}.\] The graph norm on \(\mathcal{D}(\mathscr{T})\) is defined as \(\norm{x}_{\mathscr{T}} = \norm{x}_{\mathcal{H}_1} + \norm{\mathscr{T}x}_{\mathcal{H}_2}\) is called of the operator \(\mathscr{T}\).
Definition 2. An operator \(\mathscr{T}\) is called closed if its graph \(\mathcal{G}(\mathscr{T} )\) is a closed subset of the Hilbert space \(\mathcal{H}_1 \times \mathcal{H}_2\), and \(\mathscr{T}\) is called closable (or pre-closed) if there exists a closed linear operator \(\mathscr{\overline{T}}\) from \(\mathcal{H}_1\) to \(\mathcal{H}_2\) such that \(\mathscr{T} \subseteq \mathscr{\overline{T}}\). The operator \(\mathscr{\overline{T}}\) is called the closure of the closable operator \(\mathscr{T}\).
Another useful notion is that of a core of an operator which allows us to prove statements of a closed operator on its core rather than the domain.
Definition 3. A linear subspace \(\mathcal{D}\) of \(\mathcal{D}(\mathscr{T})\) is called a core for \(\mathscr{T}\) if \(\mathcal{D}\) is dense in \((\mathcal{D}(\mathscr{T}), \norm{\cdot}_{\mathscr{T}} )\), that is, for each \(x\in \mathcal{D}(\mathscr{T})\), there exists a sequence \((x_{n})_{n\in \mathbb{N}}\) of vectors \(x_n \in \mathcal{D}\) such that \(x = \lim_{n \to \infty} x_n\) in \(\mathcal{H}_1\) and \(\mathscr{T}x = \lim_{n \to \infty} \mathscr{T}x_n\) in \(\mathcal{H}_2\).
Let us define \(\rho(\mathscr{T})\) and \(\sigma(\mathscr{T})\) to be the resolvent set and the spectrum of a closed operator \(\mathscr{T}\) on a Banach space \(\mathcal{X}\), respectively. The spectral bound is defined by \(\alpha(\mathscr{T}) = \sup_{\lambda \in \sigma(\mathscr{T})} \Re \lambda.\)
Theorem 3 ([48] p.150). Let \(\mathscr{T}\) be a closed linear operator on a Banach space \(\mathcal{X}\) generating a \(C_0\)-semigroup. Then \[\norm{e^{t \mathscr{T}}} \geq e^{t \alpha(\mathscr{T})} ,\quad \forall t \geq 0,\] where we use \(e^{t \mathscr{T}}\) to denote the \(C_0\)-semigroup generated by \(\mathscr{T}\).
The following concept of pseudo-spectra is from the monograph by Lloyd N. Trefethen and Mark Embree [48].
Definition 4 ([48] p.31). Let \(\mathscr{T}\) be a closed operator on a Banach space \(\mathcal{X}\) and \(\varepsilon > 0\) be arbitrary. The \(\varepsilon\)-pseudo-spectrum \(\sigma_{\varepsilon}(\mathscr{T})\) of \(\mathscr{T}\) is the set of \(\zeta\in \mathbb{C}\) defined equivalently by any of the conditions
(i) \(\norm{(\zeta - \mathscr{T})^{-1} } > \varepsilon^{-1}\);
(ii) \(\zeta \in \sigma(\mathscr{T} + \mathscr{E})\) for some bounded operator \(\mathscr{E}\) with \(\norm{\mathscr{E}} < \varepsilon\);
(iii) \(\zeta\in \sigma(\mathscr{T})\) or \(\norm{(\zeta - \mathscr{T}) u} < \varepsilon\) for some \(u \in \mathcal{D}(\mathscr{T})\) with \(\norm{u} = 1\). Then \(\zeta\) is an \(\varepsilon\)-pseudo-eigenvalue of \(\mathscr{T}\) and \(u\) is the corresponding \(\varepsilon\)-pseudo-eigenfunction.
The \(\varepsilon\)-spectral bound is defined as \(\alpha_{\varepsilon}(\mathscr{T}) = \sup_{\lambda \in \sigma_{\varepsilon}(\mathscr{T})} \Re \lambda.\)
Theorem 4 ([48] p.31). Given a closed operator \(\mathscr{T}\) on a Banach space \(\mathcal{X}\), the pseudo-spectra \(\{ \sigma_{\varepsilon}(\mathscr{T})\}_{\varepsilon> 0}\) have the following properties. They can be defined equivalently by any of the conditions (i)-(iii) in 4. Each \(\sigma_{\varepsilon}(\mathscr{T})\) is a nonempty open subset of \(\mathbb{C}\), and any bounded connected component of \(\sigma_{\varepsilon}(\mathscr{T})\) has a nonempty intersection with \(\sigma(\mathscr{T})\). The pseudo-spectra are strictly nested supersets of the spectrum: \(\cap_{\varepsilon > 0 }\sigma_{\varepsilon}(\mathscr{T}) =\sigma(\mathscr{T})\), and conversely, for any \(\delta > 0\), \(\sigma_{\varepsilon + \delta}(\mathscr{T}) \supseteq \sigma_{\varepsilon }(\mathscr{T}) + B(\delta)\); where \(B(\delta)\) is the open disk of radius \(\delta\).
In this section, we analyze the linearized operator \(\mathscr{L}\) defined in 7 –9 .
In this subsection, we prove that the operator \(\mathscr{L}\) generates a \(C_0\)-semigroup on \([L^2(\mathbb{R}^3)]^3\). For the sake of convenience, we define the key part of \(\mathscr{L}\) by \[\mathscr{L}_{\star} \overset{\mathrm{def}}{=}\nu \Delta - y \partial_x,\] the domain of which is defined by \[\require{physics} \label{domian95of95L95star} \mathcal{D}(\mathscr{L}_{\star}) \overset{\mathrm{def}}{=}\qty{u\in L^2(\mathbb{R}^3)\;| \;\Delta u, y\partial_x u \in L^2(\mathbb{R}^3)} .\tag{11}\] One may recall from 9 that we can rewrite the definition of \(\mathscr{L}\) by \[\label{decomposition95of95L} \mathscr{L} = \mathscr{L}_{\star }I_3 + \mathscr{L}_{\bullet } ,\tag{12}\] where \(I_3\) is identity matrix of order 3 and \[\require{physics} \mathscr{L}_{\bullet} \overset{\mathrm{def}}{=}\begin{pmatrix} f \Delta^{-1} \partial_{xy} & f - 1 + \qty(2 - f ) \Delta^{-1}\partial_x^2 & 0 \\ - f + f \Delta^{-1} \partial_{y}^2 & \qty(2 - f ) \Delta^{-1}\partial_{xy} & 0 \\ f \Delta^{-1} \partial_{yz} & \qty(2 - f ) \Delta^{-1}\partial_{xz} & 0 \\ \end{pmatrix} ,\] is a bounded operator on \([L^2(\mathbb{R}^3)]^3\).
Lemma 3. The operator \(\mathscr{L}\) is closed with the domain \(\mathcal{D}(\mathscr{L})\) given by 8 .
Proof. First, note that the closure of \(\mathscr{L}_{\star}\) on \(C_0^{ \infty}(\mathbb{R}^3)\) is closable (cf. [49]). For any \(u \in C_0^{ \infty}(\mathbb{R}^3)\) we have, using integration by parts, that \[\require{physics} \label{sec951:eq951} \begin{align} \norm{\mathscr{L}_{\star} u}_{L^2}^2 & = \norm{\nu \Delta u}_{L^2}^2 + \norm{y \partial_x u}_{L^2}^2 - 2 \left \langle \nu \Delta u, y \partial_x u \right \rangle \\ & = \norm{\nu \Delta u}_{L^2}^2 + \norm{y \partial_x u}_{L^2}^2 + 2 \nu \int_{\mathbb{R}^3}^{} y \partial_x \nabla u \cdot\nabla u +\partial_x u \partial_y u \dd{\mathbf{x}} \\ & = \norm{\nu \Delta u}_{L^2}^2 + \norm{y \partial_x u}_{L^2}^2 + 2 \nu \int_{\mathbb{R}^3}^{} \partial_x u \partial_y u \dd{\mathbf{x}} . \\ \end{align}\tag{13}\] Now, in view of Plancherel’s formula, one notices that \[\require{physics} \begin{align} &\abs{\int_{\mathbb{R}^3}^{} \partial_x u \partial_y u \dd{\mathbf{x}}} = \abs{\int_{\mathbb{R}^3}^{} u \partial_x\partial_y u \dd{\mathbf{x}}} \\&\leq \norm{u}_{L^2} \norm{\partial_x\partial_y u}_{L^2} =\norm{u}_{L^2} \norm{\xi_1 \xi_2 \widehat{u}}_{L^2} \\ &\leq \norm{u}_{L^2} \norm{\frac{\xi_1^2 + \xi_2^2 + \xi_3^2}{2} \widehat{u}}_{L^2} = \frac{1}{2} \norm{u}_{L^2} \norm{\Delta u}_{L^2}. \end{align}\] Plugging the above estimate into 13 , we obtain \[\require{physics} \begin{align} \norm{\mathscr{L}_{\star} u}_{L^2}^2 \geq {} &\norm{\nu \Delta u}_{L^2}^2 + \norm{y \partial_x u}_{L^2}^2 -\nu\norm{u}_{L^2} \norm{\Delta u}_{L^2} \\ \geq {} &\norm{\nu \Delta u}_{L^2}^2 + \norm{y \partial_x u}_{L^2}^2 - \frac{1}{2} \qty(\norm{u}_{L^2}^2 + \nu^2\norm{\Delta u}_{L^2}^2) ,\\ \end{align}\] which further leads to \[\require{physics} \begin{align} \norm{\mathscr{L}_{\star} u}_{L^2}^2 + \norm{u}_{L^2}^2\geq C \qty(\norm{\nu \Delta u}_{L^2}^2 + \norm{y \partial_x u}_{L^2}^2 + \norm{u}_{L^2}^2),\\ \end{align}\] for some constant \(C > 0\). And the inverse inequality \[\require{physics} \begin{align} \norm{\mathscr{L}_{\star} u}_{L^2}^2 + \norm{u}_{L^2}^2 \leq C \qty(\norm{\nu \Delta u}_{L^2}^2 + \norm{y \partial_x u}_{L^2}^2 + \norm{u}_{L^2}^2),\\ \end{align}\] is obvious in virtue of the definition of \(\mathscr{L}_{\star}\). That is, we found the equivalent norm for the graph norm of \(\mathscr{L}_{\star}\). Hence \(\mathscr{L}_{\star}\) is closed with domain defined in 11 . Therefore \(\mathscr{L}\) is closed. ◻
Lemma 4. The operator \(\mathscr{L}\) generates a \(C_0\)-semigroup on \([L^2(\mathbb{R}^3)]^3\).
Proof. We prove the lemma following a perturbation argument. Recalling 12 , we only need to show that \(\mathscr{L}_{\star }\) generates a \(C_0\)-semigroup, since \(\mathscr{L}_{\bullet}\) is a bounded operator on \([L^2(\mathbb{R}^3)]^3\). According to the Hille-Yosida Theorem [32], the operator \(\mathscr{L}_{\star}\) generates a \(C_0\)-semigroup on \(L^2(\mathbb{R}^3)\) if
(i) \(\mathscr{L}_{\star}\) is closed and \(\mathcal{D} (\mathscr{L}_{\star})\) is dense in \(L^2(\mathbb{R}^3)\);
(ii) The resolvent set \(\rho(\mathscr{L}_{\star})\) contains \((0, + \infty)\), and for every \(\lambda > 0\) the following resolvent estimate holds \[\label{resolvent95estimate} \norm{(\lambda I -\mathscr{L}_{\star})^{-1} } \leq \frac{1}{\lambda }.\tag{14}\]
The domain is obviously dense in \(L^2(\mathbb{R}^3)\) since \(C_0^{ \infty}(\mathbb{R}^3) \subset \mathcal{D}(\mathscr{L}_{\star})\). And the operator \(\mathscr{L}_{\star}\) is closed by 3. Therefore, \(C_0^{ \infty}(\mathbb{R}^3)\) is a core of \(\mathscr{L}_{\star}\).
Now, we derive the resolvent estimate 14 . Consider for any \(F \in C_0^{ \infty}(\mathbb{R}^3) \subset \mathcal{D}(\mathscr{L}_{\star})\), \(\lambda > 0\), \[\require{physics} \begin{align} \norm{(\lambda I-\mathscr{L}_{\star})F }_{L^2}^2 = {} & \lambda^2 \norm{F}_{L^2}^2 + \norm{\mathscr{L}_{\star} F}_{L^2}^2 - 2\lambda \left \langle F, \mathscr{L}_{\star} F \right \rangle \\ \geq & \lambda^2 \norm{F}_{L^2}^2 - 2\lambda \int_{\mathbb{R}^3}^{} ( \nu \Delta F- y \partial_x F)F \dd{\mathbf{x}} \\ \geq & \lambda^2 \norm{F}_{L^2}^2 + 2\lambda \int_{\mathbb{R}^3}^{} \nu \abs{\nabla F}^2 \dd{\mathbf{x}} \\ \end{align}\] where the last inequality follows from integrating by parts. Hence, owing to the fact that \(C_0^{ \infty}(\mathbb{R}^3)\) is a core of \(\mathscr{L}_{\star}\) we obtain \[\label{resolvent95estimate951} \frac{1}{\lambda} \norm{(\lambda I -\mathscr{L}_{\star})F }_{L^2} \geq \norm{F}_{L^2}, \quad \forall F \in \mathcal{D}(\mathscr{L}_{\star}).\tag{15}\] This shows that \(\lambda I -\mathscr{L}_{\star}\) is injective and \((\lambda I-\mathscr{L}_{\star})^{-1} :\mathcal{R}(\lambda I -\mathscr{L}_{\star}) \to \mathcal{D}(\mathscr{L}_{\star})\) is bounded by \(\frac{1}{\lambda}\). Also, \(\mathcal{R}(\lambda I -\mathscr{L}_{\star})\) is closed in \(L^2(\mathbb{R}^3)\) [49].
Finally, we show that the range \(\mathcal{R}(\lambda -\mathscr{L}_{\star})\) is also dense in \(L^2(\mathbb{R}^3)\). Due to [49], we deduce that \(L^2(\mathbb{R}^3) = \mathcal{N}(\lambda -\mathscr{L}_{\star}^{*}) \oplus \mathcal{R} (\lambda -\mathscr{L}_{\star})\), which suggests that we only have to verify \(\mathcal{N}(\lambda -\mathscr{L}_{\star}^{*}) = \{0\}\). It is clear that, for \(F\in \mathcal{D}(\mathscr{L}_{\star})\) and \(G\in C_0^{ \infty} (\mathbb{R}^3)\) \[\left \langle \mathscr{L}_{\star} F, G \right \rangle = \left \langle F, ( \nu \Delta + y \partial_x)G \right \rangle.\] Hence the restriction of \(\mathscr{L}_{\star}^{*}\) on \(C_0^{ \infty}(\mathbb{R}^3)\) is \(\nu \Delta + y \partial_x\). Since \(\mathscr{L}_{\star}^{*}\) is closed and \(C_0^{ \infty}(\mathbb{R}^3) \subset \mathcal{D}( \mathscr{L}_{\star}^{*} )\) is a core of \(\mathscr{L}_{\star}^{*}\), then \(\mathscr{L}_{\star}^{*}\) is the closure of \(\nu \Delta + y \partial_x\). Now, similar to the derivation of 15 , one obtains that for \(\lambda > 0\) and \(G \in \mathcal{D}(\mathscr{L}_{\star}^{*})\) \[\frac{1}{\lambda} \norm{(\lambda I-\mathscr{L}_{\star}^{*})G }_{L^2} \geq \norm{G}_{L^2}.\] This shows \(\mathcal{N}(\mathscr{L}_{\star}^{*}) = \{0 \}\). The proof is complete. ◻
Lemma 5. All the spectrum of the operator \(\mathscr{L} :\mathcal{D}(\mathscr{L}) \to \mathcal{H}\) are continuous spectrum.
Proof. We prove the lemma by showing that neither the operator \(\mathscr{L}\) nor the adjoint \(\mathscr{L}^{*}\) has point spectrum. Indeed, if \(\lambda_r\) is a residual spectrum of \(\mathscr{L}\), then \(\mathcal{R}(\lambda_r - \mathscr{L})^{\perp} \neq \{ \boldsymbol{0}\}\). Since \(\mathcal{R}(\lambda_r - \mathscr{L})^{\perp} = \mathcal{N}(\overline{\lambda_r} - \mathscr{L}^{*})\) (see [49]), then \(\overline{\lambda_r}\) is a point spectrum for \(\mathscr{L}^{*}\). So if both \(\mathscr{L}\) and \(\mathscr{L}^{*}\) have no point spectrum, we can conclude that all the spectrum of the operator \(\mathscr{L}\) are continuous spectrum.
Suppose the pair \((\lambda, \mathbf{u}) \in \mathbb{C}\times \mathcal{D}(\mathscr{L})\) are the corresponding eigenvalue and eigenfunction such that \[\label{eigen95prob950} \mathscr{L} \mathbf{u} (\mathbf{x}) = \lambda \mathbf{u} (\mathbf{x}).\tag{16}\] Then \[\label{eigen95prob951-1} \mathscr{\widehat{L}} \widehat{ \mathbf{u} }(\boldsymbol{\xi}) = \lambda \widehat{ \mathbf{u} }(\boldsymbol{\xi}),\tag{17}\] holds for a.e. \(\boldsymbol{\xi} \in \mathbb{R}^3\). Here the operator \(\mathscr{\widehat{L}}\) is defined as \[\begin{align} \mathscr{\widehat{L}} \overset{\mathrm{def}}{=} \begin{pmatrix} - \nu \abs{\boldsymbol{\xi}}^2 + \xi_1 \partial_{\xi_2} + f \frac{\xi_1 \xi_2}{ \abs{\boldsymbol{\xi}}^2 }& f - 1 + (2 - f) \frac{\xi_1^2}{ \abs{\boldsymbol{\xi}}^2 } & 0\\ - f + f \frac{\xi_2^2}{ \abs{\boldsymbol{\xi}}^2 }& - \nu \abs{\boldsymbol{\xi}}^2 + \xi_1 \partial_{\xi_2} + (2 - f) \frac{\xi_1 \xi_2}{ \abs{\boldsymbol{\xi}}^2 }& 0\\ f \frac{\xi_2 \xi_3}{ \abs{\boldsymbol{\xi}}^2 }& (2 - f) \frac{\xi_1 \xi_3}{ \abs{\boldsymbol{\xi}}^2 }& - \nu \abs{\boldsymbol{\xi}}^2 + \xi_1 \partial_{\xi_2} \end{pmatrix} , \end{align}\] with the domain \[\require{physics} \mathcal{D}(\mathscr{\widehat{L}}) \overset{\mathrm{def}}{=}\qty{\widehat{ \mathbf{u} } \in [L^2(\mathbb{R}^3)]^3\;\middle|\;\abs{\boldsymbol{\xi}}^2 \widehat{ \mathbf{u} },\; \xi_1 \partial_{\xi_2}\widehat{ \mathbf{u} } \in [L^2(\mathbb{R}^3)]^3 }.\]
Next, we show by contradiction that there exists no non-zero \(\mathbf{\widehat{u}}\) such that 17 holds. We rewrite the eigenvalue problem as the \(\xi_1\) and \(\xi_3\) parameterized ODE system of the independent variable \(\xi_2\): \[\label{linear95analysis954} \xi_1 \partial_{\xi_2} \mathbf{\widehat{u}} = \begin{pmatrix} \lambda + \nu \abs{\boldsymbol{\xi}}^2 - f \frac{\xi_1 \xi_2}{ \abs{\boldsymbol{\xi}}^2 }& -f + 1 - (2- f) \frac{\xi_1^2}{ \abs{\boldsymbol{\xi}}^2 } & 0\\ f - f \frac{\xi_2^2}{ \abs{\boldsymbol{\xi}}^2 }& \lambda + \nu \abs{\boldsymbol{\xi}}^2 - (2 - f) \frac{\xi_1 \xi_2}{ \abs{\boldsymbol{\xi}}^2 }& 0\\ -f \frac{\xi_2 \xi_3}{ \abs{\boldsymbol{\xi}}^2 }& - (2 -f) \frac{\xi_1 \xi_3}{ \abs{\boldsymbol{\xi}}^2 }& \lambda + \nu \abs{\boldsymbol{\xi}}^2 \end{pmatrix} \mathbf{\widehat{u}}.\tag{18}\]
Since \(\mathbf{u}\not\equiv 0\), there exists \(\boldsymbol{\xi}^\ast\) with \(\xi_1^\ast \neq 0\) such that \(\widehat{ \mathbf{u} }(\boldsymbol{\xi}^\ast)\neq 0\). Without loss of generality, we assume \(\xi_1^\ast>0\) throughout the proof. By the uniqueness of solutions to the ODE system 18 , one has \(\widehat{ \mathbf{u} }(\xi_2;\xi_1^\ast, \xi_3^\ast) \neq \boldsymbol{0}\) for all \(\xi_2 \in \mathbb{R}\). Since \(\widehat{ \mathbf{u} } \in \mathcal{D} (\mathscr{\widehat{L}})\), then \(\abs{\boldsymbol{\xi}}^2 \widehat{ \mathbf{u} } \in [L^2(\mathbb{R}^3)]^3\) and \(\xi_1^\ast \partial_{\xi_2}\widehat{ \mathbf{u} } \in [L^2(\mathbb{R}^3)]^3\), hence \(\widehat{ \mathbf{u} }(\xi_2;\xi_1^\ast,\xi_3^\ast) \in[ C(\mathbb{R}) ]^3\) is bounded by Fubini’s theorem and the Sobolev embedding.
On the other hand, one obtains by multiplying 18 by \(\overline{\widehat{\mathbf{u}}}(\xi_2;\xi_1^\ast,\xi_3^\ast)\) that \[\require{physics} \begin{align} \frac{\xi_1^\ast}{2} \partial_{\xi_2} \abs{\widehat{\mathbf{u}}} ^2 & = (\Re \lambda +\nu \abs{\boldsymbol{\xi}}^2) \abs{\widehat{\mathbf{u}}} ^2 + \Re \qty( \mathscr{M} \mathbf{\widehat{u}} \cdot \mathbf{\overline{\widehat{u}}}), \end{align}\] where \(|\boldsymbol{\xi}|^2=|\xi_1^\ast|^2+|\xi_2|^2+|\xi_3^\ast|^2\), and the operator \(\mathscr{M} = \mathscr{M}(\xi_2;\xi_1^\ast,\xi_3^\ast)\) is identified as follows \[\mathscr{M}(\xi_2;\xi_1^\ast,\xi_3^\ast) = \begin{pmatrix} - f \frac{\xi_1^\ast \xi_2}{ \abs{\boldsymbol{\xi}}^2 }& -f + 1 - (2 -f) \frac{{\xi_1^\ast}^2}{ \abs{\boldsymbol{\xi}}^2 } & 0\\ f - f \frac{\xi_2^2}{ \abs{\boldsymbol{\xi}}^2 } & - (2 - f) \frac{\xi_1^\ast \xi_2}{ \abs{\boldsymbol{\xi}}^2 }& 0\\ ^ -f \frac{\xi_2 \xi_3^\ast}{ \abs{\boldsymbol{\xi}}^2 }& - (2 -f) \frac{\xi_1^\ast \xi_3^\ast}{ \abs{\boldsymbol{\xi}}^2 }& 0 \end{pmatrix}.\] It is clear that \(\abs{\mathscr{M} \widehat{\mathbf{u}}} \leq C(f) \abs{\widehat{\mathbf{u}}}\) for a constant \(C(f) > 0\) depending on \(f\). It follows that \[\require{physics} \begin{align} \frac{\xi_1^\ast}{2} \partial_{\xi_2} \abs{\widehat{\mathbf{u}}} ^2 & \geq \qty(\Re \lambda +\nu \abs{\boldsymbol{\xi}}^2) \abs{\widehat{\mathbf{u}}} ^2 - C(f) \abs{\widehat{\mathbf{u}}}^2. \end{align}\] Then, for large enough \(\abs{\xi_2}\) satisfying \[\require{physics} \abs{\xi_2} \geq \xi_2^{*}(f,\nu,\lambda) \overset{\mathrm{def}}{=}\sqrt{\frac{1}{\nu} \qty(\frac{C(f)}{2} -\Re \lambda )}\] we have \[\begin{align} \xi_1^\ast\partial_{\xi_2} \abs{\widehat{\mathbf{u}}} ^2 & \geq (\Re \lambda +\nu \abs{\boldsymbol{\xi}}^2) \abs{\widehat{\mathbf{u}}} ^2. \end{align}\] One concludes that \[\begin{align} |\widehat{\mathbf{u}}( \xi_2;\xi_1^{*},\xi_3^{*})|^2 \geq |\widehat{\mathbf{u}}(\xi_2^{*};\xi_1^{*},\xi_3^{*})|^2 \cdot \exp\left( \frac{1}{\xi_1^{*}} \left[ (\Re\lambda + \nu (\xi_1^{*} + \xi_3^{*})^2)(\xi_2 - \xi_2^*) + \frac{\nu}{3} \left( \xi_2^3 - (\xi_2^*)^3 \right) \right] \right), \end{align}\] which is a contradiction. Hence \(\widehat{\mathbf{u}}\equiv 0\). Therefore, \(\mathscr{L}\) does not possess point spectrum.
The adjoint \(\mathscr{L}^{*}\) is given by \[\require{physics} \begin{align} \mathscr{L}^{*} &\overset{\mathrm{def}}{=}\mathscr{L}_0^{*} + \mathscr{L}_1^{*} - \mathscr{L}_{2}, \\ \mathscr{L}_{2} & \overset{\mathrm{def}}{=}- \mathop{\mathrm{diag}} (y \partial_x, y \partial_x, y \partial_x) \\ \mathscr{L}_0^{*} & \overset{\mathrm{def}}{=} \begin{pmatrix} \nu \Delta &-f + f \Delta^{-1} \partial_{y}^2& f \Delta^{-1} \partial_{yz} \\ f - 1 & \nu \Delta & 0 \\ 0 & 0 &\nu \Delta \\ \end{pmatrix} , \\ \mathscr{L}_1^{*} & \overset{\mathrm{def}}{=}\Delta ^{-1} \begin{pmatrix} f \partial_{xy} & 0 & 0 \\ \qty(2 - f )\partial_x^2 & \qty(2 - f )\partial_{xy} & \qty(2 - f )\partial_{xz} \\ 0 & 0 & 0 \\ \end{pmatrix} , \end{align}\] with the domain \(\mathcal{D}(\mathscr{L}) = \mathcal{D}(\mathscr{L}^{*})\). The same argument shows that \(\mathscr{L}^{*}\) also has no point spectrum. Consequently, the spectrum of both \(\mathscr{L}\) and \(\mathscr{L}^{*}\) consist entirely of continuous spectrum. This completes the proof. ◻
In this subsection we construct a linearly unstable solution of the system 7 under the assumption that \(0<f < 1\), i.e. \(\mathrm{R}(f) < 0\). In this case it is known that the \(x\)-independent system is unstable.
Assume \(\mathbf{u}\) is \(x\)-independent. Recalling 9 , the Fourier transform of 7 gives that \[\label{Linear95system95x95independent95Fourier95form} \begin{align} \partial_t \mathbf{\widehat{u}} = \mathscr{\widehat{L}}_{0} (\xi_2,\xi_3) \mathbf{\widehat{u}} \overset{\mathrm{def}}{=} \begin{pmatrix} - \nu \abs{\boldsymbol{\xi}}^2 & f - 1 & 0 \\ -f + f \frac{\xi_2^2}{ \abs{\boldsymbol{\xi}}^2 } & - \nu \abs{\boldsymbol{\xi}}^2 & 0\\ f \frac{\xi_2 \xi_3}{ \abs{\boldsymbol{\xi}}^2 } & 0 & - \nu \abs{\boldsymbol{\xi}}^2 \\ \end{pmatrix} \mathbf{\widehat{u}}, \end{align}\tag{19}\] where \(\abs{\boldsymbol{\xi}}^2=\xi_2^2+\xi_3^2\) and \(\mathscr{\widehat{L}}_{0}(\xi_2,\xi_3) \equiv \mathscr{\widehat{L}}(\boldsymbol{\xi})|_{\xi_1 = 0}\). The eigenvalues and eigenfunctions are as follows \[\require{physics} \label{streak95unstable95system} \begin{align} \lambda_1(\xi_2,\xi_3) & = - \nu (\xi_2^2 + \xi_3^2) + \sqrt{f(1 - f) \frac{\xi_3^2}{ (\xi_2^2 + \xi_3^2)} } , \\ \mathbf{\widehat{u}}_1(\xi_2,\xi_3) &= \qty(-\frac{ \abs{\boldsymbol{\xi}} \sqrt{f(1 - f) \xi_3^2 }}{f }, - \xi_3^2,\xi_2 \xi_3), \\ \lambda_2(\xi_2,\xi_3) & = - \nu (\xi_2^2 + \xi_3^2) - \sqrt{f(1 - f) \frac{\xi_3^2}{ (\xi_2^2 + \xi_3^2)} } , \\\mathbf{\widehat{u}}_2(\xi_2,\xi_3) &= \qty( \frac{ \abs{\boldsymbol{\xi}} \sqrt{ f(1 - f) \xi_3^2}}{f }, - \xi_3^2, \xi_2 \xi_3), \\ \lambda_3(\xi_2,\xi_3) &= - \nu (\xi_2^2 + \xi_3^2), \quad \mathbf{\widehat{u}}_3(\xi_2,\xi_3) = \qty(0,0,1). \end{align}\tag{20}\] It is clear that if \(f(1 - f) > \nu^2 \frac{(\xi_2^2 + \xi_3^2)^3}{\xi_3^2},\) the eigenvalue \(\lambda_1 > 0\), hence the system admits exponentially growing solution. Since the eigenfunctions are \(x\)-independent, they are not integrable in \(\mathbb{R}^3\). Thus, the unstable eigenfunctions above are not eigenfunctions in \([L^2(\mathbb{R}^3)]^3\), and they do not imply the existence of unstable eigenvalues for \(\mathscr{L}\). However, the above unstable eigenfunction assists us to prove the instability of the 3D system.
For the general case when \(\mathbf{u}\) is \(x\)-dependent, the following statement for \(\varepsilon\)-pseudo-spectra of \(\mathscr{L}\) is true.
Lemma 6. Let \(\lambda_{j}(\xi_2,\xi_3)\), \(j = 1,2,3\) be the eigenvalues of the matrix \(\mathscr{\widehat{L}}_{0}(\xi_2,\xi_3)\) defined in 19 –20 . If \(0 < f < 1\), then for any \(\varepsilon > 0\) and \(\require{physics} (\xi_2,\xi_3) \in \mathbb{R}^2 \setminus \qty{\boldsymbol{0}}\), the eigenvalues \(\lambda_{j}(\xi_2,\xi_3)\), \(j = 1,2,3\) lie in the \(\varepsilon\)-pseudo-spectrum of \(\mathscr{L}\), i.e. \(\lambda_{j}(\xi_2,\xi_3) \in \sigma_{\varepsilon}(\mathscr{L})\) for \(j = 1 ,2,3\). Moreover, \(\lambda_{j}(\xi_2,\xi_3) \in \sigma(\mathscr{L})\) for \(j = 1,2,3\).
Proof. The proof is the same for \(j = 1 ,2,3\). We provide the details for \(\lambda_{1} (\xi_2,\xi_3)\).
By the virtue of Plancherel’s identity, for any \(\mathbf{u} \in \mathcal{D}(\mathscr{L}) \subset [L^2(\mathbb{R}^3)]^3\), \[\require{physics} \norm{ \qty(\lambda_1(\xi_2,\xi_3) - \mathscr{L}) \mathbf{u} }_{L^2} = \norm{(\lambda_1(\xi_2,\xi_3) - \widehat{\mathscr{L}}) \widehat{\mathbf{u}} }_{L^2}.\] Hence, we can show \(\lambda_1 (\xi_2,\xi_3)\) is an \(\varepsilon\)-pseudo-eigenvalue by finding a vector-valued function \(\widehat{\mathbf{u}}_{\varepsilon} (\boldsymbol{\xi}) \in [L^2(\mathbb{R}^3)]^3\) satisfying \(\norm{\widehat{\mathbf{u}}_{\varepsilon}}_{L^2} = 1\) and \(\mathcal{F}^{-1}\widehat{\mathbf{u}}_{\varepsilon} \in \mathcal{D} (\mathscr{L})\) such that \[\label{linear95analysis951} \norm{(\lambda_1(\xi_2,\xi_3) - \widehat{\mathscr{L}}) \widehat{\mathbf{u}}_{\varepsilon} }_{L^2} < \varepsilon .\tag{21}\]
Apply the Fourier transformation to 7 , one obtains \[\begin{align} \partial_t \mathbf{\widehat{u}} = \begin{pmatrix} - \nu \abs{\boldsymbol{\xi}}^2 + \xi_1 \partial_{\xi_2} + f \frac{\xi_1 \xi_2}{ \abs{\boldsymbol{\xi}}^2 }& f - 1 + (2 - f) \frac{\xi_1^2}{ \abs{\boldsymbol{\xi}}^2 } & 0\\ - f + f \frac{\xi_2^2}{ \abs{\boldsymbol{\xi}}^2 }& - \nu \abs{\boldsymbol{\xi}}^2 + \xi_1 \partial_{\xi_2} + (2 - f) \frac{\xi_1 \xi_2}{ \abs{\boldsymbol{\xi}}^2 }& 0\\ f \frac{\xi_2 \xi_3}{ \abs{\boldsymbol{\xi}}^2 }& (2 - f) \frac{\xi_1 \xi_3}{ \abs{\boldsymbol{\xi}}^2 }& - \nu \abs{\boldsymbol{\xi}}^2 + \xi_1 \partial_{\xi_2} \end{pmatrix} \widehat{\mathbf{u}} . \end{align}\] One introduces fixed frequencies \[\label{frequencies951} \boldsymbol{\xi}_1^{*} = (0,\xi_2^{*},\xi_3^{*}) ,\quad \boldsymbol{\xi}_2^{*} = (0, -\xi_2^{*},\xi_3^{*}) ,\quad \boldsymbol{\xi}_3^{*} = (0,\xi_2^{*}, -\xi_3^{*}) ,\quad \boldsymbol{\xi}_4^{*} = (0, -\xi_2^{*}, -\xi_3^{*}) ,\tag{22}\] such that \(\xi_2^{*}, \xi_3^{*} > 0\). For the degenerate case \(\xi_2^{*} \xi_3^{*} = 0\) and \((\xi_2^{*},\xi_3^{*}) \neq (0,0)\), we only need to fix two frequencies. For example, if \(\xi_2^{*} = 0\) and \(\xi_3^{*} \neq 0\), we define \[\label{frequencies952} \boldsymbol{\xi}_1^{*} = (0,0,\xi_3^{*}) ,\quad \boldsymbol{\xi}_2^{*} = (0,0, -\xi_3^{*}) .\tag{23}\] Hereafter, we only prove the non-degenerate case, the proof of the degenerate case is similar.
Let \(\theta (x)\) be the standard mollifier with compact support in the interval \([ - 1,1]\), and \(\theta_{\delta}(x)\) be defined by \[\require{physics} \label{1D95mollifier} \theta_{\delta}(x) \overset{\mathrm{def}}{=}\frac{1}{\delta} \theta \qty(\frac{x}{\delta}), \quad \delta \in (0, 1].\tag{24}\] Then \[\eta(\boldsymbol{\xi}) \overset{\mathrm{def}}{=}\theta (\xi_1)\theta (\xi_2)\theta (\xi_3),\quad \eta_{\delta,\delta'}(\boldsymbol{\xi}) \overset{\mathrm{def}}{=}\theta_{\delta} (\xi_1)\theta_{\delta'} (\xi_2) \theta_{\delta} (\xi_3), \quad \delta' \in 90,1],\] and \[\label{construction95of95eigenfunction} \widehat{\mathbf{u}}_{\delta,\delta'}(\boldsymbol{\xi};\xi_2^{*},\xi_3^{*}) \overset{\mathrm{def}}{=}\sum_{j = 1}^{4} \eta_{\delta,\delta'} (\boldsymbol{\xi} - \boldsymbol{\xi}_{j}^{*})\widehat{\mathbf{u}}_{1}(\xi_2^{*},\xi_3^{*}) ,\tag{25}\] where \(\widehat{\mathbf{u}}_{1}(\xi_2^{*},\xi_3^{*})\) is the eigenvector of the \(\mathscr{\widehat{L}}_{0}\) corresponding to the eigenvalue \(\lambda_1(\xi_2^{*},\xi_3^{*})\) given by 20 . Hereafter without causing confusion, we use the shorthand notations \(\lambda_1 = \lambda_1(\xi_2^{*},\xi_3^{*})\), \(\widehat{\mathbf{u}}_{1} =\widehat{\mathbf{u}}_{1}(\xi_2^{*},\xi_3^{*})\) and \(\widehat{\mathbf{u}}_{\delta,\delta'} (\boldsymbol{\xi})= \widehat{\mathbf{u}}_{\delta,\delta'}(\boldsymbol{\xi};\xi_2^{*},\xi_3^{*})\), with the understanding that the frequency \((\xi_2^{*},\xi_3^{*})\) is fixed.
Notice that by construction \(\widehat{\mathbf{u}}_{\delta,\delta'}(\boldsymbol{\xi})\) is even in variables \(\xi_{j}\), \(j = 1,2,3\). Hence \(\mathbf{u}_{\delta,\delta'} (\mathbf{x}) = \mathcal{F}^{-1} \widehat{\mathbf{u}}_{\delta,\delta'} (\mathbf{x})\) is a real function. Moreover, since \(\widehat{\mathbf{u}}_{\delta,\delta'}\) is compactly supported and smooth, \(\mathbf{u}_{\delta,\delta'}(\mathbf{x})\) lies in the Schwarz space \([\mathcal{S}(\mathbb{R}^3)]^3\), and thus in \(\mathcal{D}(\mathscr{L})\).
Step 3. We verify that \(\mathbf{u}_{\delta,\delta'}(\mathbf{x})\) is indeed an \(\varepsilon\)-eigenvector corresponding to \(\lambda_1\) for suitably small \(\delta\) and \(\delta'\), namely 21 holds for \(\widehat{\mathbf{u}}_{\delta,\delta'}(\boldsymbol{\xi})\).
We assume without loss of generality that for any \(\delta,\delta' \in (0,1]\), the support of \(\eta_{\delta,\delta'} (\boldsymbol{\xi} - \boldsymbol{\xi}_{j}^{*})\) is mutually disjoint. Then, one gets by change of variables \[\require{physics} \norm{ \widehat{\mathbf{u}}_{\delta,\delta'} }_{L^2}^2 = \sum_{j = 1}^{4} \int_{\mathbb{R}^3}^{} \abs{\eta_{\delta,\delta'} (\boldsymbol{\xi} - \boldsymbol{\xi}_{j}^{*}) \widehat{\mathbf{u}}_{1} }^2 \dd{\boldsymbol{\xi}} = \frac{4}{\delta^2 \delta'} \abs{\widehat{\mathbf{u}}_{1} }^2 \norm{\eta(\boldsymbol{\xi})}_{L^2}^2.\]
Furthermore, recalling 9 , a direct calculation shows for \(\lambda_1\) that \[\require{physics} \begin{align} \frac{1}{\norm{\widehat{\mathbf{u}}_{\delta,\delta'}}_{L^2} } \norm{ (\lambda_1 - \widehat{\mathscr{L}}) \widehat{\mathbf{u}}_{\delta,\delta'} }_{L^2} \leq {} & \frac{1}{\norm{\widehat{\mathbf{u}}_{\delta,\delta'}}_{L^2} } \sum_{j = 1}^{4} \bigg( \norm{\eta_{\delta,\delta'}(\boldsymbol{\xi} - \boldsymbol{\xi}_{j}^{*}) \qty[ \qty(\lambda_1 - \mathscr{\widehat{L}}_0(\boldsymbol{\xi}) - \mathscr{\widehat{L}}_1(\boldsymbol{\xi}) )\widehat{\mathbf{u}}_{1}] }_{L^2} \\ & {} + \norm{\xi_1 \partial_{\xi_2} \eta_{\delta,\delta'}(\boldsymbol{\xi} - \boldsymbol{\xi}_{j}^{*}) \widehat{\mathbf{u}}_{1} }_{L^2} \bigg) \\ \eqqcolon & I_1 + I_2. \end{align}\] Denote \(\require{physics} h (\boldsymbol{\xi}) = \qty(\lambda_1 - \mathscr{\widehat{L}}_0 (\boldsymbol{\xi}) - \mathscr{\widehat{L}}_1(\boldsymbol{\xi}) ) \frac{\widehat{\mathbf{u}}_{1}}{ \abs{\widehat{\mathbf{u}}_{1}} }\). Then, taking the symmetry of \(\eta_{\delta,\delta'}(\boldsymbol{\xi} )\) and \(\abs{\boldsymbol{\xi}}^2\) into account, we have \[\require{physics} \begin{align} I_1 \leq {} & \frac{1}{\norm{\widehat{\mathbf{u}}_{\delta,\delta'}}_{L^2} }\sum_{j = 1}^{4} \norm{\eta_{\delta,\delta'} (\boldsymbol{\xi} - \boldsymbol{\xi}_{j}^{*})}_{L^2} \norm{ \qty(\lambda_1 - \mathscr{\widehat{L}}_0 (\boldsymbol{\xi}) - \mathscr{\widehat{L}}_1(\boldsymbol{\xi}) )\widehat{\mathbf{u}}_{1}}_{L^{ \infty}(\mathop{\mathrm{supp}}\eta_{\delta,\delta'} )} \\ = {} & \frac{4}{\norm{\widehat{\mathbf{u}}_{\delta,\delta'}}_{L^2} } \norm{\eta_{\delta,\delta'} (\boldsymbol{\xi} )}_{L^2} \abs{\widehat{\mathbf{u}}_{1}} \norm{ h (\boldsymbol{\xi}) }_{L^{ \infty}( [ -\delta,\delta] \times [ \xi_2^{*} -\delta',\xi_2^{*} +\delta'] \times [ \xi_3^{*}-\delta,\xi_3^{*} +\delta] )} \\ = {} & \frac{4}{\frac{2}{\delta \delta'^{\frac{1}{2}}} \abs{\mathbf{\widehat{u}}_{1} } \norm{\eta(\boldsymbol{\xi})}_{L^2} } \frac{1}{\delta \delta'^{\frac{1}{2}} }\norm{\eta (\boldsymbol{\xi} )}_{L^2} \abs{\widehat{\mathbf{u}}_{1}} \norm{ h (\boldsymbol{\xi}) }_{L^{ \infty}( [ -\delta,\delta] \times [ \xi_2^{*} -\delta',\xi_2^{*} +\delta'] \times [ \xi_3^{*}-\delta,\xi_3^{*} +\delta] )} \\ = {} & 2 \norm{ h (\boldsymbol{\xi}) }_{L^{ \infty}( [ -\delta,\delta] \times [ \xi_2^{*} -\delta',\xi_2^{*} +\delta'] \times [ \xi_3^{*}-\delta,\xi_3^{*} +\delta] )}, \end{align}\] where we utilized the fact that \[\norm{\eta_{\delta,\delta'} (\boldsymbol{\xi} - \boldsymbol{\xi}_{j}^{*})}_{L^2} = \frac{1}{\delta \delta'^{\frac{1}{2}} }\norm{\eta (\boldsymbol{\xi} )}_{L^2}.\] One observes that by the construction of \(\widehat{\mathbf{u}}_{1}\), the function \(h (\boldsymbol{\xi})\) is a smooth function in the neighborhood of \(\boldsymbol{\xi}_{j}^{*}\), \(j = 1,2,3,4\) and satisfies \(h (\boldsymbol{\xi}_{j}^{*}) = \boldsymbol{0}\). Thus, for any \(\varepsilon > 0\), there exists \(\delta' > 0\) and \(\delta_0(\delta',\varepsilon) > 0\) such that for any \(0 < \delta < \delta_0(\delta',\varepsilon)\), \[I_1 < \frac{\varepsilon}{2} .\]
For \(I_2\), we have \[\begin{align} I_2 = {} & \frac{1}{\norm{\widehat{\mathbf{u}}_{\delta,\delta'}}_{L^2} }\sum_{j = 1}^{4} \norm{\xi_1 \partial_{\xi_2} \eta_{\delta,\delta'} (\boldsymbol{\xi} - \boldsymbol{\xi}_{j}^{*}) \widehat{\mathbf{u}}_{1}}_{L^2} \\ \leq {} & \frac{4}{\frac{2}{\delta \delta'^{\frac{1}{2}}} \abs{\widehat{\mathbf{u}}_{1} } \norm{\eta(\boldsymbol{\xi})}_{L^2} }\abs{\widehat{\mathbf{u}}_{1}} \norm{\xi_1 \partial_{\xi_2} \eta_{\delta,\delta'}(\boldsymbol{\xi} ) }_{L^2}, \\ \end{align}\] where \[\require{physics} \begin{align} &\norm{\xi_1 \partial_{\xi_2} \eta_{\delta,\delta'} (\boldsymbol{\xi} ) }_{L^2}^2 = \int_{\mathbb{R}^3}^{} \abs{\xi_1 \partial_{\xi_2} \qty(\frac{1}{\delta^2\delta'} \eta \qty( \frac{\xi_1}{\delta}, \frac{\xi_2}{\delta'}, \frac{\xi_3}{\delta} ))}^2 \dd{\boldsymbol{\xi}} \\ & = \int_{\mathbb{R}^3}^{} \abs{\frac{\xi_1}{\delta'} \frac{1}{\delta^2\delta'} \partial_{\xi_2}\eta \qty( \frac{\xi_1}{\delta}, \frac{\xi_2}{\delta'}, \frac{\xi_3}{\delta} )}^2 \dd{\boldsymbol{\xi}} \leq \frac{1}{\delta^2\delta'} \abs{\frac{\delta}{\delta'} }^2 \int_{\mathbb{R}^3}^{} \abs{ \xi_1\partial_{\xi_2} \eta \qty(\boldsymbol{\xi} )}^2 \dd{\boldsymbol{\xi}}. \\ \end{align}\] Hence \[\require{physics} \begin{align} I_2 \leq {} & \frac{4}{\frac{2}{\delta \delta'^{\frac{1}{2}}} \abs{\widehat{\mathbf{u}}_{1} } \norm{\eta(\boldsymbol{\xi})}_{L^2} } \abs{\widehat{\mathbf{u}}_{1}} \frac{1}{\delta\delta'^{\frac{1}{2}}} \frac{\delta}{\delta'} \norm{\xi_1\partial_{\xi_2} \eta \qty(\boldsymbol{\xi} )}_{L^2} \\ \leq {} & \frac{2 \norm{\xi_1\partial_{\xi_2} \eta \qty(\boldsymbol{\xi} )}_{L^2} }{ \norm{\eta(\boldsymbol{\xi})}_{L^2}} \frac{\delta}{\delta'} \leq C \frac{\delta}{\delta'}, \end{align}\] for a constant \(C > 0\) determined by \(\eta\). Therefore, for any \(\varepsilon > 0\), there exists \(\delta' > 0\) and \(\delta_1(\varepsilon,\delta') > 0\) such that for any \(0 < \delta < \delta_1(\varepsilon,\delta')\) \[I_2 < \frac{\varepsilon}{2}.\]
Combining the estimates for \(I_1\) and \(I_2\), we have for any \(\varepsilon > 0\), there exists small enough \(\delta' > 0\), such that for any \(\require{physics} \delta < \min \qty{\delta_1(\varepsilon,\delta'),\delta_2(\varepsilon,\delta')}\) \[\norm{(\lambda_1 - \widehat{\mathscr{L}}) \frac{ \widehat{\mathbf{u}}_{\delta,\delta'}}{ \norm{ \widehat{\mathbf{u}}_{\delta,\delta'}}_{L^2} } }_{L^2} < \varepsilon.\] This implies that \(\lambda_1(\xi_2^\ast,\xi_3^\ast)\) is an \(\varepsilon\)-pseudo-eigenvalue of \(\mathscr{L}\) with corresponding \(\varepsilon\)-pseudo-eigenfunction \(\mathbf{u}_{\delta,\delta'}\).
Finally, by 4, one has \(\cap_{\varepsilon > 0} \sigma_{\varepsilon}(\mathscr{L}) = \sigma(\mathscr{L})\). The proof is complete. ◻
Corollary 1. Under the conditions in 6, for any \(\require{physics} (\xi_2^{*},\xi_3^{*}) \in \mathbb{R}^2 \setminus \qty{\boldsymbol{0}}\), \(\xi_3^{*}\neq 0\), there exist an \(\varepsilon\)-pseudo-eigenfunction of \(\lambda_1(\xi_2^{*},\xi_3^{*})\), denoted by \(\mathbf{u}_\varepsilon = (u_{1,\varepsilon},u_{2,\varepsilon},u_{3,\varepsilon})\), such that \[\label{ratio95of95u} u_{1,\varepsilon}(\mathbf{x}) = \frac{ \sqrt{ (\xi_2^{*})^2 + (\xi_3^{*})^2} }{ \abs{\xi_3^{*}} } \frac{\sqrt{1 - f }}{\sqrt{f}} u_{2,\varepsilon}(\mathbf{x}), \quad u_{3,\varepsilon}(\mathbf{x}) = - \frac{ \xi_2^{*} }{ \xi_3^{*} } u_{2,\varepsilon}(\mathbf{x}),\qquad{(1)}\] for all \(\mathbf{x} \in \mathbb{R}^3\). Moreover, the support of \(\widehat{\mathbf{u}}_\varepsilon\) satisfies \[\require{physics} \label{supp95cond} \mathop{\mathrm{supp}}\mathbf{\widehat{u}}_\varepsilon \subseteq \bigcup_{j = 1}^{4} B \qty(\boldsymbol{\xi}_j^{*},\frac{1}{2})\qquad{(2)}\] where \(\boldsymbol{\xi}_j^{*} ,j = 1,2,3,4\) are defined in 22 and \(\require{physics} B \qty(\boldsymbol{\xi}_j^{*} ,\frac{1}{2})\) is the ball centered at \(\boldsymbol{\xi}_j^{*}\) of radius \(\frac{1}{2}\).
Proof. The proof of 6 shows that, there exists a small enough \(\delta' > 0\) and \(\require{physics} \delta < \min \qty{\delta_1(\varepsilon,\delta'),\delta_2(\varepsilon,\delta')}\) such that \(\mathcal{F}^{-1} \widehat{\mathbf{u}}_{\delta,\delta'}(\mathbf{x};\xi_2^{*},\xi_3^{*})\) is an \(\varepsilon\)-pseudo-eigenfunction of \(\lambda_1(\xi_2^{*},\xi_3^{*})\). Denote the three components of \(\widehat{\mathbf{u}}_{\delta,\delta'}(\boldsymbol{\xi};\xi_2^{*},\xi_3^{*})\) by \(\widehat{\mathbf{u}}_{\delta,\delta'}(\boldsymbol{\xi};\xi_2^{*},\xi_3^{*}) = (\widehat{u}_{1,\delta,\delta'},\widehat{u}_{2,\delta,\delta'},\widehat{u}_{3,\delta,\delta'})(\boldsymbol{\xi};\xi_2^{*},\xi_3^{*})\). Then by the construction 25 and 20 , we have \[\label{linear_analysis_{12}} \begin{align} &\widehat{u}_{1,\delta,\delta'}(\boldsymbol{\xi}) = \frac{ \sqrt{ (\xi_2^{*})^2 + (\xi_3^{*})^2} }{ \abs{\xi_3^{*}} } \frac{\sqrt{1 - f }}{\sqrt{f}} \widehat{u}_{2,\delta,\delta'}(\boldsymbol{\xi}), \quad \widehat{u}_{3,\delta,\delta'}(\boldsymbol{\xi}) = - \frac{ \xi_2^{*} }{ \xi_3^{*} } \widehat{u}_{2,\delta,\delta'}(\boldsymbol{\xi}), \\ \end{align}\tag{26}\] for all \(\boldsymbol{\xi} \in \mathbb{R}^3\) when \(\xi_3^{*}\neq 0\).
Then, from [linear_analysis_{12}] and the inverse Fourier transform, we have \[\label{linear_analysis_{13}} u_{1,\delta,\delta'} (\mathbf{x}) = \frac{ \sqrt{ (\xi_2^{*})^2 + (\xi_3^{*})^2} }{ \abs{\xi_3^{*}} } \frac{\sqrt{1 - f }}{\sqrt{f}} u_{2,\delta,\delta'}(\mathbf{x}), \quad u_{3,\delta,\delta'}(\mathbf{x}) = - \frac{ \xi_2^{*} }{ \xi_3^{*} } u_{2,\delta,\delta'}(\mathbf{x}),\tag{27}\] for all \(\mathbf{x} \in \mathbb{R}^3\). Moreover, [linear_analysis_{13}] is valid for all \(\delta,\delta'\).
The support of \(\widehat{\mathbf{u}}_{\delta,\delta'}\) satisfies \[\mathop{\mathrm{supp}}\widehat{\mathbf{u}}_{\delta,\delta'}\subseteq [ -\delta,\delta] \times A_{2,\delta'}\times A_{3,\delta}.\] where \(A_{2,\delta'}\) and \(A_{3,\delta}\) are defined by \[\begin{align} A_{2,\delta'} &= [ \xi_2^{*}-\delta',\xi_2^{*} +\delta'] \cup [ -\xi_2^{*}-\delta', -\xi_2^{*} +\delta'] ,\\ A_{3,\delta} &= [ \xi_3^{*}-\delta,\xi_3^{*} +\delta] \cup [ -\xi_3^{*}-\delta, -\xi_3^{*} +\delta] . \end{align}\] And for any \(0 <\delta'_1 < \delta'\) and \(0 < \delta_1 < \delta\), the function \(\mathcal{F}^{-1} \widehat{\mathbf{u}}_{\delta_1,\delta'_1}(\mathbf{x};\xi_2^{*},\xi_3^{*})\) is also the \(\varepsilon\)-pseudo-eigenfunction of \(\lambda_1 (\xi_2^{*},\xi_3^{*})\), hence ?? and ?? hold for \(\mathbf{u}_{\varepsilon}(\mathbf{x}) = \mathcal{F}^{-1} \widehat{\mathbf{u}}_{\delta_1,\delta'_1}(\mathbf{x};\xi_2^{*},\xi_3^{*})\) with small enough \(\delta_1\) and \(\delta_1'\). This completes the proof. ◻
Corollary 2. Suppose \(0 < f < 1\), then for any real number \(\lambda \leq \sqrt{f(1 - f) }\), \(\lambda \in \sigma(\mathscr{L})\).
Proof. The 6 shows that \(\lambda_1\) defined in 20 lies in the spectrum of \(\mathscr{L}\). By the definition of \(\lambda_1\), one has \[\require{physics} \overline{ \qty{ \lambda_1(\xi_2,\xi_3) | \xi_2,\xi_3 \in \mathbb{R}^2 \setminus \{\boldsymbol{0}\} } } = ( - \infty, \sqrt{f(1 - f) } ].\] This completes the proof. ◻
Lemma 7. Suppose \(0 < f < 1\), then for any \(T > 0\) and \(\epsilon > 0\), there exists a \(\mathbf{u}_{T,\epsilon} = (u_{1,T,\epsilon},u_{2,T,\epsilon},u_{3,T,\epsilon}) \in [L^2(\mathbb{R}^3)]^3\) such that
for all \(t\in[0,T]\) and \(\underline{\Lambda} = \sqrt{f(1 - f) } > 0\) \[\require{physics} \label{linear95analysis957} \norm{ \qty(e^{t \underline{\Lambda} } -e^{t \mathscr{L}})\mathbf{u}_{T,\epsilon}}_{L^2} \leq \epsilon\norm{\mathbf{u}_{T,\epsilon}}_{L^2} ;\qquad{(3)}\]
\(\abs{u_{1,T,\epsilon}(\mathbf{x})} = \frac{ \sqrt{1 - f }}{\sqrt{f} } \abs{u_{2,T,\epsilon}(\mathbf{x})}, \quad \abs{u_{3,T,\epsilon}(\mathbf{x})} = 0 \qq{for all} \mathbf{x} \in \mathbb{R}^3;\)
for all \(t\in[0,T]\), \(e^{t \mathscr{L} }\mathbf{u}_{T,\epsilon} \in [H^k(\mathbb{R}^3)]^3\) for any \(t\in[0,T]\), \(k \geq 0\);
\(\mathop{\mathrm{supp}}\widehat{\mathbf{u}_{T,\epsilon}} \subseteq B(0,1)\), where \(B(0,1)\) is the unit ball at origin.
Proof. Proof of (i): First, we notice that, for any \(\mathbf{u} \in [L^2(\mathbb{R}^3)]^3\) and \(\zeta \in \mathbb{C}\), \[\norm{e^{t \underline{\Lambda} }\mathbf{u}- e^{t \mathscr{L}} \mathbf{u}}_{L^2} \leq \norm{e^{t \underline{\Lambda} }\mathbf{u} - e^{t \zeta }\mathbf{u}}_{L^2} + \norm{e^{t \zeta }\mathbf{u}- e^{t \mathscr{L}} \mathbf{u}}_{L^2} .\] Hence, the conclusion ?? is valid if \[\label{linear95analysis958} \norm{e^{t \underline{\Lambda} }\mathbf{u} - e^{t \zeta }\mathbf{u}}_{L^2} \leq \frac{\epsilon}{2} \norm{\mathbf{u}}_{L^2} ,\quad \norm{e^{t \zeta }\mathbf{u}- e^{t \mathscr{L}} \mathbf{u}}_{L^2} \leq \frac{\epsilon}{2} \norm{\mathbf{u}}_{L^2}\tag{28}\] holds for all \(t\in[0,T]\) and for suitable \(\zeta\) and \(\mathbf{u}\).
Second, let us choose \(\lambda_1 = \lambda_1(0,\xi_3^{\epsilon})\), where \(\xi_3^{\epsilon}\) being a small enough positive number satisfying \[\label{linear_analysis_{11}} \abs{\xi_3^{\epsilon}} \leq \begin{cases} \sqrt{ -\frac{1}{\nu T} \ln\left(1 - \frac{\epsilon}{2e^{T \underline{\Lambda}}}\right) }, & \text{when} \quad \epsilon < 2e^{T \underline{\Lambda}}, \\ + \infty, & \text{when} \quad \epsilon \geq 2e^{T \underline{\Lambda}}, \end{cases}\tag{29}\] Then we have \[e^{t \underline{\Lambda} }- e^{t \lambda_{1}} \leq \frac{\epsilon}{2}, \qq{for all t\in[0,T],}\] hence \[\norm{e^{t \underline{\Lambda} }\mathbf{u} - e^{t \lambda_1 }\mathbf{u}}_{L^2} \leq \frac{\epsilon}{2} \norm{\mathbf{u}}_{L^2}.\] Hence, we now fix some \(\xi_3^{\epsilon }\) satisfying [linear_analysis_{11}], then \(\zeta = \lambda_{1}\) meets the first requirement in 28 for all \(\mathbf{u} \in [L^2(\mathbb{R}^3)]^3\).
Third, with the help of 2, we are able to estimate the difference of \(e^{t\mathscr{L}} \mathbf{u}_{\gamma}\) and \(e^{t \lambda_{1} } \mathbf{u}_{\gamma}\) with \(\mathbf{u}_{\gamma}\) being the \(\gamma\)-pseudo-eigenfunction corresponds to \(\lambda_{1}\). We have \[\require{physics} \begin{align} &\norm{(e^{t\mathscr{L}} - e^{t \lambda_{1} }) \mathbf{u}_{\gamma}}_{L^2} = \norm{e^{t \lambda_{1} } (e^{t(\mathscr{L}- \lambda_{1} )} - 1) \mathbf{u}_{\gamma}}_{L^2} \\&=\norm{e^{ t \lambda_{1} } \int_{0}^{t} e^{s(\mathscr{L}- \lambda_{1} )}(\mathscr{L} - \lambda_{1} ) \mathbf{u}_{\gamma} \dd{s}}_{L^2} \\ &= \norm{ \int_{0}^{t} e^{ (t- s) \lambda_{1} }e^{s\mathscr{L}}(\mathscr{L} - \lambda_{1} ) \mathbf{u}_{\gamma} \dd{s}}_{L^2} \\&\leq \int_{0}^{t} e^{ (t- s) \lambda_{1} } \norm{e^{s\mathscr{L}}}_{L^2 \to L^2} \norm{(\mathscr{L} - \lambda_{1} )\mathbf{u}_{\gamma}}_{L^2} \dd{s} \\ & \leq \int_{0}^{t} e^{ t \lambda_{1} } e^{s \qty(\overline{\Lambda} - \lambda_{1} )} \norm{(\mathscr{L} - \lambda_{1} )\mathbf{u}_{\gamma}}_{L^2} \dd{s} \leq \gamma \norm{\mathbf{u}_{\gamma}}_{L^2} e^{ t \lambda_{1} } \frac{1}{\overline{\Lambda} - \lambda_{1} } e^{s \qty(\overline{\Lambda} - \lambda_{1} )} \Big|^{t}_{0} \\ & \leq \gamma \norm{\mathbf{u}_{\gamma}}_{L^2} e^{ t \lambda_{1} } \frac{1}{\overline{\Lambda} - \lambda_{1} } \qty(e^{t \qty(\overline{\Lambda} - \lambda_{1} )} - 1) . \\ \end{align}\] Recalling 20 , we have \(\lambda_{1} =\lambda_{1}(0,\xi_3^{\epsilon}) \leq \underline{\Lambda}\) and \[\require{physics} \gamma e^{ t \lambda_{1} } \frac{1}{\overline{\Lambda} - \lambda_{1} } \qty(e^{t \qty(\overline{\Lambda} - \lambda_{1} )} - 1) < \gamma e^{ t \lambda_{1} } \frac{1}{\overline{\Lambda} - \underline{\Lambda} } \qty(e^{t \qty(\overline{\Lambda} - \lambda_{1} )} - 1) < \gamma \frac{1}{\overline{\Lambda} - \underline{\Lambda} } e^{t\overline{\Lambda} } .\] Then we can choose \(\gamma\) small enough such that \(\gamma \frac{1}{\overline{\Lambda} - \underline{\Lambda} } e^{t\overline{\Lambda} } < \frac{\epsilon}{2}\) for all \(t \in [0,T]\). Therefore, we have \[\norm{(e^{t\mathscr{L}} - e^{t \lambda_{1} }) \mathbf{u}_{\gamma}}_{L^2}< \frac{\epsilon}{2} \norm{\mathbf{u}_{\gamma}}_{L^2}.\] Then 28 holds for \(\zeta = \lambda_{1}(0,\xi_3^{\epsilon})\) and \(\mathbf{u}_\gamma\), which implies ?? with \(\mathbf{u}_{T,\epsilon} = \mathbf{u}_{\gamma}\).
Proof of (ii): We apply 1 on \(\lambda_{1}(0,\xi_3^{\epsilon})\) to obtain \[u_{1,\gamma}(\mathbf{x}) = \frac{\sqrt{1 - f }}{\sqrt{f}} u_{2,\gamma}(\mathbf{x}), \quad u_{3,\gamma}(\mathbf{x}) = 0 .\]
Proof of (iii): we prove \(e^{t \mathscr{L}} \mathbf{u}_{\gamma} \in \bigcap_{k \geq 0} [H^k(\mathbb{R}^3)]^3\) for all \(t \in[0,T]\). For this matter, we recall that \(\mathcal{D}(\mathscr{L}^{\infty})\) is an invariant set of the semigroup \(e^{t \mathscr{L}}\) for all \(t \geq 0\). Then, by the fact that \[\mathbf{u}_{\gamma} \in [\mathcal{S}(\mathbb{R}^3)]^3 \subseteq \mathcal{D}(\mathscr{L}^{\infty}) \subseteq \bigcap_{k \geq 0}[H^k(\mathbb{R}^3)]^3 ,\] we have \(e^{t \mathscr{L}} \mathbf{u}_{\gamma} \in \bigcap_{k \geq 0}[H^k(\mathbb{R}^3)]^3\) for all \(t \geq 0\).
Proof of (iv): (iv) is from ?? with \((\xi_2^{*},\xi_3^{*}) = (0,\xi_3^{\epsilon})\) and \(\xi_3^{\epsilon} < \frac{1}{2}\).
The proof is now complete. ◻
Proof of 1:
Define \(\mathbf{u}^{T,\epsilon} (t) = e^{t \mathscr{L}} \mathbf{u}_{T,\epsilon}\), where \(\mathbf{u}_{T,\epsilon}\) is given in 7. Then, (i) is a corollary of 7. Indeed, we have \[\require{physics}
\begin{align} \norm{\mathbf{u}^{T,\epsilon}(t)} \leq \norm{e^{t \underline{\Lambda} } \mathbf{u}_{T,\epsilon}} + \norm{\mathbf{u}^{T,\epsilon}(t) - e^{t \underline{\Lambda} } \mathbf{u}_{T,\epsilon}}_{L^2} \leq (e^{t \underline{\Lambda} } + \epsilon)
\norm{\mathbf{u}_{T,\epsilon}}_{L^2} , \\ \norm{\mathbf{u}^{T,\epsilon}(t)} \geq \norm{e^{t \underline{\Lambda} } \mathbf{u}_{T,\epsilon}} - \norm{\mathbf{u}^{T,\epsilon}(t) - e^{t \underline{\Lambda} } \mathbf{u}_{T,\epsilon}}_{L^2} \geq \qty(e^{t
\underline{\Lambda} } - \epsilon) \norm{\mathbf{u}_{T,\epsilon}}_{L^2} .
\end{align}\] By 7 (iv), \(\mathop{\mathrm{supp}}\widehat{\mathbf{u}_{T,\epsilon}}\) is uniformly bounded
for \(T\) and \(\epsilon\). Then (ii) follows since the support of \(\widehat{\mathbf{u}_{T,\epsilon}}\) is compact. The proof of 7 is complete.
This section is devoted to proving 2. The constants in this section are denoted by \(\mathsf{C}_{j}\), which represents the generic positive constants depending on \(\nu\) and \(f\) unless otherwise specified. Moreover, we continue to use \(C\) to denote the generic positive constant depending on those parameters, which need not be labeled and may vary from line to line.
By 1 (i), for any given parameter set \(\kappa = ( T, \epsilon)\) with \(T > 0\) and \(\epsilon > 0\) one can construct the unstable solution \((\mathbf{u}^{ \kappa},q^{ \kappa})\) for the linearized system 6 satisfying \[\label{growth95estimate} (e^{t \Lambda } - \epsilon) \norm{\mathbf{u}^{ \kappa}(0)}_{L^2} \leq \norm{\mathbf{u}^{ \kappa}(t)}_{L^2}\leq (e^{t \Lambda } + \epsilon) \norm{\mathbf{u}^{ \kappa}(0)}_{L^2}\tag{30}\] for any \(t\in [0, T ]\). We denote by \(\mathbf{u}_{\mathrm{in}}^{\kappa}\) the initial data of \(\mathbf{u}^{ \kappa}\). In view of (ii) of 1, there exists a constant \(L >0\) independent of \(\kappa\) such that \[\label{def95of95L32} \frac{\norm{\mathbf{u}_{\mathrm{in}}^{ \kappa}}_{L^2}}{\mathcal{E}_{\mathrm{in}}^{\kappa}} \geq L .\tag{31}\]
Now, define the initial data with a rescaling factor \(\delta \in (0, \overline{\delta}_0)\) by \[\mathbf{u} ^{\delta}_{\mathrm{in}} \overset{\mathrm{def}}{=}\frac{\delta}{\mathcal{E}_{\mathrm{in}}^{\kappa}} \mathbf{u} _{\mathrm{in}}^{\kappa}.\] The parameter \(\overline{\delta}_0\) is small enough such that 1 holds. Clearly, the size of the initial value in \(H^1\) satisfies \[\begin{align} \mathcal{E}(\mathbf{u}^{\delta}_{\mathrm{in}}) = \delta < 1 . \end{align}\] In addition, we denote the approximate solution \(\mathbf{u}^{\mathrm{a}}\) with the associated pressure by \[\label{Hadamard95estimate953} \begin{align} \mathbf{u}^{\mathrm{a}}(\mathbf{x},t) \overset{\mathrm{def}}{=}\frac{\delta}{\mathcal{E}_{\mathrm{in}}^{\kappa}} \mathbf{u}^{ \kappa}(\mathbf{x},t) ,\quad q ^{\mathrm{a}}(\mathbf{x},t) \overset{\mathrm{def}}{=}\frac{\delta}{\mathcal{E}_{\mathrm{in}}^{\kappa}} q^{ \kappa}(\mathbf{x},t), \end{align}\tag{32}\] which also obeys estimate 30 .
Now, the parameters we have introduced are \(\kappa = ( T, \epsilon)\) and \(\delta\), where \(\kappa\) determines the generation of the linear approximate solution while \(\delta\) dictates size of the initial data. Hereafter, we fix \(\delta \in (0,\overline{\delta}_0)\) and the parameter set \(\kappa = (T, \epsilon)\) satisfying \[\require{physics} \label{fixed95parameters} \begin{align} \delta^{ - 4} \lesssim T \leq T_0, \quad \epsilon < \frac{1}{2} \min \qty{L \varepsilon,1}, \quad \frac{\delta}{2}e^{\frac{\underline{\Lambda} T }{2} } \geq \mathsf{C}_0 , \end{align}\tag{33}\] where \(\varepsilon = \varepsilon(\overline{\delta}_0, f,\nu)\) is small and will be fixed later in 38 , \(\mathsf{C}_0 > 0\) is a constant that only depends on \(\nu\) and \(f\), \(T_0\) given in ?? is the maximal time that the energy estimate holds. The rationale behind the requirements in 33 is as follows:
The condition \(\delta^{ - 4} \lesssim T\) is to make sure \(T\) is large enough, such that there is enough time for the nonlinear solution \(\mathbf{u}^{\delta}\) (defined below) to exceeds a constant value;
The second condition guarantees the approximate solution is close enough to an exponential;
The condition \(\frac{\delta}{2}e^{\frac{\underline{\Lambda} T }{2} } \geq \mathsf{C}_0\) is so that \(\varepsilon\) does not depend on \(\delta\), otherwise we have to modify the definition of \(\varepsilon\). That is, we need \(\varepsilon\) to satisfy \(\varepsilon \leq \frac{\delta}{2}e^{\frac{\underline{\Lambda} T }{2} }\), which goes to 0, as \(\delta \to 0\). Since \(\varepsilon\) represent the threshold of the instability, small \(\varepsilon\) weakens our conclusion for the instability.
Next, we also would like to elaborate why \(\frac{\delta}{2}e^{\frac{\underline{\Lambda} T }{2} } \geq \mathsf{C}_0\) can be achieved for some constant \(\mathsf{C}_0 > 0\) and how to determine the constant. By ?? , there exists a constant \(K = K(\nu) > 0\) such that \[T_0 \geq K \delta^{ - 4} .\] Therefore, if we set \(T = K \delta^{ - 4}\), then \(\frac{\delta}{2}e^{\frac{\underline{\Lambda} T }{2} } \geq \mathsf{C}_0\) translated to \[\frac{\delta}{2}e^{\frac{1}{2}\underline{\Lambda} K \frac{1}{ \delta^4 } } \geq \mathsf{C}_0, \qq{i.e.} \underline{\Lambda} \geq 2\delta^4 K^{-1} \ln(\frac{2 }{\delta} \mathsf{C}_0) .\] Recalling \(\delta \leq 1\), one may determine \(\mathsf{C}_0 = \mathsf{C}_0(\nu,f)\) from \[\underline{\Lambda} = \sup_{\delta\in[0,1]} 2 \delta^4 K^{-1} \ln(\frac{2 }{\delta}\mathsf{C}_0).\]
Let \(\mathbf{u}^{\delta}\) be a local nonlinear strong solution to 3 , emanating from \(\mathbf{u} ^{\delta}_{\mathrm{in}}\), with the associated pressure \(q ^{\delta}\). Define the time of instability \[\begin{align} T^{\delta} \overset{\mathrm{def}}{=}\frac{1}{\underline{\Lambda}} \ln \frac{2\varepsilon}{\delta}, \qq{i.e.} \delta e^{\underline{\Lambda} T^{\delta}} = 2 \varepsilon, \end{align}\] and the escape times \[\require{physics} \label{escape95times} \begin{align} T^{*} & \overset{\mathrm{def}}{=}\sup \qty{ t > 0\;\middle|\;\mathcal{E}(\mathbf{u}^{\delta}(t)) \leq \overline{\delta}_0 }, \\ T^{* *} & \overset{\mathrm{def}}{=}\sup \qty{ t > 0\;\middle|\;\norm{\mathbf{u}^{\delta}(t)}_{L^2} \leq 2 \delta e ^{\underline{\Lambda} t}}. \end{align}\tag{34}\] Then, \(\mathbf{u}^{\delta}\) satisfying 39 for \(T^{*}\) and \(\overline{\delta}_0\). Thus, recalling the estimate ?? , one obtains that \[\require{physics} \label{Hadamard95estimate951} \begin{align} &\norm{( \mathbf{u}^{\delta},\nabla \mathbf{u}^{\delta})(t)}_{L^2}^2 + \int_{0}^{t} \norm{(\nabla \mathbf{u}^{\delta}, \Delta \mathbf{u}^{\delta})(\tau)}_{L^2}^2 \dd{\tau}\\& \leq C \qty(\mathcal{E}^2(\mathbf{u}^{\delta}_{\mathrm{in}}) + \int_{0}^{t} \|\mathbf{u}^{\delta}(\tau)\|_{L^2}^2 \dd{\tau} ) \\ &\leq C \qty( \delta^2 + \frac{2}{\underline{\Lambda}} \delta^2 e ^{2\underline{\Lambda} t}) \leq \mathsf{C}_1 \delta^2 e^{2\underline{\Lambda} t}, \end{align}\tag{35}\] for any \(\require{physics} t \leq \min \qty{T^{\delta}, T^{*}, T^{* *} }\), where \(\mathsf{C}_1\) does not depend on \(\delta\).
Subsequently, we denote the difference of nonlinear solution \(\mathbf{u}^{\delta}\) and the approximate solution by \[\mathbf{u}^{\mathrm{d}} = \mathbf{u}^{\delta} -\mathbf{u}^{\mathrm{a}} ,\quad q ^{\mathrm{d}}= q ^{\delta} - q ^{\mathrm{a}} .\] The nonlinear solution has the integral expression \[\require{physics} \mathbf{u}^{\delta}(t) = e^{t\mathscr{L}} \mathbf{u}_{\mathrm{in}}^{\delta} + \int_{0}^{t} e^{(t - s)\mathscr{L}} \mathscr{N}(\mathbf{u}^{\delta} (s)) \dd{s} ,\] where the nonlinear term \(\mathscr{N}\) is defined as \[\require{physics} \mathscr{N}(\mathbf{u}) \overset{\mathrm{def}}{=}\nabla \Delta ^{-1} \operatorname{div}\qty(\mathbf{u} \cdot \nabla \mathbf{u}) -\mathbf{u} \cdot \nabla \mathbf{u}.\] Since the approximate solution satisfies \(\mathbf{u}^{\mathrm{a}} = e^{t\mathscr{L}} \mathbf{u}_{\mathrm{in}}^{\delta}\), then the difference admits the integral representation \[\require{physics} \mathbf{u}^{\mathrm{d}}(t) = \int_{0}^{t} e^{(t - s)\mathscr{L}} \mathscr{N}(\mathbf{u}^{\delta} (s)) \dd{s} .\] For the nonlinear term, we have the following estimate using Gagliardo-Nirenberg inequality \[\norm{\mathscr{N}(\mathbf{u}) }_{L^2} \leq \norm{ \mathbf{u} \cdot \nabla \mathbf{u}}_{L^2} \leq \norm{\mathbf{u}^{\delta} }_{L^{3}} \norm{ \nabla \mathbf{u}^{\delta} }_{L^6} \leq \norm{\mathbf{u}^{\delta} }_{L^{2}} ^{\frac{1}{2}} \norm{ \nabla ^2 \mathbf{u}^{\delta} }_{L^2}^{\frac{3}{2}} .\] In light of this estimate and 35 , we have \[\require{physics} \label{Hadamard95estimate954} \begin{align} \norm{ \mathbf{u}^{\mathrm{d}}(t)} & \leq C \int_{0}^{t} e^{(t - s)\overline{\Lambda}} \norm{\mathbf{u}^{\delta}(s) }_{L^{2}} ^{\frac{1}{2}} \norm{ \nabla ^2 \mathbf{u}^{\delta}(s) }_{L^2}^{\frac{3}{2}} \dd{s} \\&\leq C \int_{0}^{t} e^{(t - s)\overline{\Lambda}} \delta^{\frac{1}{2}} e^{ \frac{1}{2} s \underline{\Lambda}} \norm{ \nabla ^2 \mathbf{u}^{\delta} (s)}_{L^2}^{\frac{3}{2}} \dd{s} \\ & \leq C \delta^{\frac{1}{2}} e^{t \overline{\Lambda}} \int_{0}^{t} e^{ s \qty( \frac{1}{2} \underline{\Lambda} - \frac{1}{4} \overline{\Lambda})} e^{ - \frac{3}{2} \cdot \frac{1}{2}\overline{\Lambda} s } \norm{ \nabla ^2 \mathbf{u}^{\delta}(s) }_{L^2}^{\frac{3}{2}} \dd{s} \\ & \leq C \delta^{\frac{1}{2}} e^{t \overline{\Lambda}} \qty(\int_{0}^{t} e^{ s \qty( 2 \underline{\Lambda} - \overline{\Lambda})} \dd{s})^{\frac{1}{4}} \qty( \int_{0}^{t} e^{ - \overline{\Lambda} s }\norm{ \nabla ^2 \mathbf{u}^{\delta} (s) }_{L^2}^{2} \dd{s} )^{\frac{3}{4}}. \end{align}\tag{36}\] By using 8 and 35 , we have \[\require{physics} \int_{0}^{t} e^{ - \overline{\Lambda} s } \norm{ \nabla ^2 \mathbf{u}^{\delta} (s) }_{L^2}^{2} \dd{s} \leq C \delta^2 e^{(2 \underline{\Lambda} - \overline{\Lambda})t},\] 36 then becomes \[\require{physics} \begin{align} \norm{ \mathbf{u}^{\mathrm{d}}(t)} & \leq C \delta^{2} e^{t \overline{\Lambda}} \qty( e^{t \qty( 2 \underline{\Lambda} - \overline{\Lambda})} )^{\frac{1}{4}} \qty( e^{t \qty( 2 \underline{\Lambda} - \overline{\Lambda})} )^{\frac{3}{4}} \leq \mathsf{C}_2 \delta^{2} e^{ 2 \underline{\Lambda} t } \qfor t \leq \min \qty{T^{\delta}, T^{*}, T^{* *} }, \\ \end{align}\] provided that \(\underline{\Lambda} > \frac{1}{2} \overline{\Lambda}\), which is valid if \(\frac{2}{17} \left(5-2 \sqrt{2}\right) < f < \frac{2}{17} \left(5 + 2 \sqrt{2}\right)\).
Next, we show that \[\require{physics} \label{Hadamard95estimate952} \begin{align} T^{\delta} = \min \qty{T^{\delta},T^{*}, T^{* *}, T }, \end{align}\tag{37}\] with \(\varepsilon\) defined as \[\require{physics} \label{def95of95epsilon950} \varepsilon \overset{\mathrm{def}}{=}\min \qty{ \frac{\overline{\delta}_0}{4\sqrt{\mathsf{C}_1}} , \frac{1}{ 4 \mathsf{C}_2}, \mathsf{C}_0, \frac{L}{ 4 \mathsf{C}_2} }.\tag{38}\] If \(T^{*} < T^{\delta}\), then we have \[\overline{\delta}_0 = \mathcal{E}(\mathbf{u}^{\delta}(T^{*})) \leq \sqrt{\mathsf{C}_1} \delta e^{\underline{\Lambda} T^{*}} \leq \sqrt{\mathsf{C}_1} \delta e^{\underline{\Lambda} T^{\delta}} = 2\sqrt{\mathsf{C}_1} \varepsilon \leq \frac{\overline{\delta}_0}{2} < \overline{\delta}_0,\] which is a contradiction. If \(T^{* *} < T^{\delta}\), then for \(\epsilon < \frac{1}{2}\), one gets \[\begin{align} \norm{\mathbf{u}^{\delta}(T^{* *})}_{L^2} \leq {} &\norm{\mathbf{u}^{\mathrm{a}}(T^{* *})}_{L^2} + \norm{\mathbf{u}^{\mathrm{d}}(T^{* *})}_{L^2} \\\leq {} & \delta (e ^{\underline{\Lambda} T^{* *}} + \epsilon) + \mathsf{C}_2 \delta^2 e^{ 2\underline{\Lambda} T^{* *}} \\ \leq {} & \delta e ^{\underline{\Lambda} T^{* *}} (1 + 2\mathsf{C}_2 \varepsilon ) + \delta \epsilon \leq \frac{3}{2}\delta e^{\underline{\Lambda} T^{* *}}+ \delta \epsilon < 2\delta e^{\underline{\Lambda} T^{* *}}, \end{align}\] which contradicts 34 . By 33 and 38 , we have \[T^{\delta} = \frac{1}{\underline{\Lambda}} \ln \frac{2\varepsilon}{\delta} \leq \frac{ T }{2} < T .\] Thus, \(T > T^{\delta}\). 37 is now verified.
Finally, we show that the velocity is unstable in the \(L^2\) norm. Recalling 32 , 33 and 38 , we have \[\require{physics} \begin{align} \norm{\mathbf{u}^{\delta}(T^{\delta})}_{L^2} \geq {} & \norm{\mathbf{u}^{\mathrm{a}}(T^{\delta})}_{L^2} - \norm{\mathbf{u}^{\mathrm{d}}(T^{\delta})}_{L^2} \\ \geq {} &\frac{\delta}{\mathcal{E}_{\mathrm{in}}^{\kappa}} \norm{ \mathbf{u}^{ \kappa}(T^{\delta})}_{L^2} - \mathsf{C}_2 \delta^{2} e^{ 2 \underline{\Lambda} T^{\delta} } \\ \geq {} & \delta (e^{T^{\delta} \underline{\Lambda} } - \epsilon) \frac{\norm{\mathbf{u}_{\mathrm{in}}^{\kappa}}_{L^2}}{\mathcal{E}_{\mathrm{in}}^{\kappa}} -\mathsf{C}_2 \delta^{2} e^{ 2 \underline{\Lambda} T^{\delta} }\\ \geq {} & 2 \varepsilon \qty(L - 2 \mathsf{C}_2 \varepsilon) - \epsilon \delta \geq \frac{3}{2}L \varepsilon - \epsilon \delta > L \varepsilon, \\ \end{align}\] where we used the fact from 33 : \[\begin{align} \epsilon \delta \leq \epsilon < \frac{1}{2}L \varepsilon. \end{align}\] By setting \(\varepsilon_0 = L\varepsilon\), the proof of 2 is now complete.
In this section, we derive some nonlinear energy estimates for the perturbed problem, which are used in the proof of the nonlinear instability, cf. [50]. To this end, let us assume that \((\mathbf{u},q)\) is a sufficiently regular solution in \([0,T]\times \mathbb{R}^3\) for some \(T > 0\) to the perturbed system 3 .
Hereafter, we define \[\mathcal{E} (\mathbf{u}(t)) \overset{\mathrm{def}}{=}\norm{\mathbf{u}(t)}_{H^1} ,\quad \mathcal{E}_{\mathrm{in}} \overset{\mathrm{def}}{=}\norm{\mathbf{u}(0)}_{H^1}= \norm{\mathbf{u}_{\mathrm{in}}}_{H^1}.\] and assume \[\label{small95energy95assumption} \mathcal{E}(\mathbf{u}(t))\leq \overline{\delta}_0 \leq 1,\qfor t \in[0,T] .\tag{39}\]
One has \[\require{physics} \label{energy95estimate95195new} \begin{align} \frac{1}{2} \dv{}{t} \norm{\mathbf{u}}_{L^2}^2 + \nu \norm{\nabla \mathbf{u}}_{L^2}^2 = -\int_{\mathbb{R}^3}^{} u_1 u_2 \dd{\mathbf{x}} \leq \norm{u_1}_{L^2}\norm{u_2}_{L^2} \leq \norm{\mathbf{u}}_{L^2}^2, \end{align}\tag{40}\] and \[\require{physics} \begin{align} &\frac{1}{2} \dv{}{t} \|\nabla \mathbf{u}\|_{L^2}^2 + \nu \|\Delta \mathbf{u}\|_{L^2}^2 \\ &= \int_{\mathbb{R}^3} \nabla q \cdot \Delta \mathbf{u} \dd{\mathbf{x}} + \int_{\mathbb{R}^3} ( \mathbf{u} \cdot \nabla) \mathbf{u} \cdot \Delta \mathbf{u} \dd{\mathbf{x}} \\&\quad+ \int_{\mathbb{R}^3}^{} u_2 \Delta u_1 \dd{\mathbf{x}} + \int_{\mathbb{R}^3}^{} y \partial_x \mathbf{u} \cdot \Delta \mathbf{u} \dd{\mathbf{x}} \\ &= - \int_{\mathbb{R}^3} q \Delta(\nabla \cdot \mathbf{u}) \dd{\mathbf{x}} - \int_{\mathbb{R}^3} \nabla(( \mathbf{u} \cdot \nabla) \mathbf{u} ): \nabla \mathbf{u} \dd{\mathbf{x}} \\&\quad - \int_{\mathbb{R}^3}^{} \nabla u_2 \cdot \nabla u_1 \dd{\mathbf{x}} - \int_{\mathbb{R}^3}^{} \partial_x \mathbf{u}\cdot \partial_y \mathbf{u} \dd{\mathbf{x}} \\ &= - \int_{\mathbb{R}^3} \mathbf{u} \cdot \nabla \frac{|\nabla \mathbf{u}|^2}{2} \dd{\mathbf{x}} - \int_{\mathbb{R}^3} (\nabla \mathbf{u} \cdot (\nabla \mathbf{u})^{\intercal} ): \nabla \mathbf{u}\dd{\mathbf{x}} \\&\quad- \int_{\mathbb{R}^3}^{} \nabla u_2 \cdot \nabla u_1 \dd{\mathbf{x}} - \int_{\mathbb{R}^3}^{} \partial_x \mathbf{u}\cdot \partial_y \mathbf{u} \dd{\mathbf{x}} \\ &\leq \|\nabla \mathbf{u}\|_{L^3}^3 + 2\|\nabla \mathbf{u}\|_{L^2}^2 \leq C \|\nabla \mathbf{u}\|_{L^2}^{3/2} \|\Delta \mathbf{u}\|_{L^2}^{3/2} + 2\|\nabla \mathbf{u}\|_{L^2}^2 \\ &\leq \frac{\nu}{2} \|\Delta \mathbf{u}\|_{L^2}^2 + \frac{C}{\nu^3} \|\nabla \mathbf{u}\|_{L^2}^6 + 2\|\nabla \mathbf{u}\|_{L^2}^2. \end{align}\] Hence \[\label{energy95estimate95395new} \frac{1}{2} \dv{}{t} \|\nabla \mathbf{u}\|_{L^2}^2 + \frac{\nu}{2} \|\Delta \mathbf{u}\|_{L^2}^2 \leq \frac{C}{\nu^3} \|\nabla \mathbf{u}\|_{L^2}^6 + 2\|\nabla \mathbf{u}\|_{L^2}^2 .\tag{41}\] Combining 40 and 41 , we have \[\require{physics} \dv{}{t} \norm{( \mathbf{u},\nabla \mathbf{u})}_{L^2}^2 + \norm{(\nabla \mathbf{u}, \Delta \mathbf{u})}_{L^2}^2 \leq C \qty( \norm{\nabla \mathbf{u}} _{L^2}^6 + \|\mathbf{u}\|_{L^2}^2).\] For small enough \(\overline{\delta}_0\), the term \(\norm{\nabla \mathbf{u}} _{L^2}^6\) on the right-hand side can be absorbed by the term \(\norm{\nabla \mathbf{u}} _{L^2}^2\) of the left-hand side. One has
****Proposition** 1**. There exists a constant \(\overline{\delta}_0 \leq 1\), such that if a strong solution \(\mathbf{u}(t)\) of the system 3 satisfies the assumption 39 for some \(T > 0\), then the solution satisfies \[\require{physics} \label{estimate95with95small95energy95assumption95for95Hadamard95new} \begin{align} & \norm{( \mathbf{u},\nabla \mathbf{u})(t)}_{L^2}^2 + \int_{0}^{t} \norm{(\nabla \mathbf{u}, \Delta \mathbf{u})(\tau)}_{L^2}^2 \dd{\tau} \leq C \qty(\mathcal{E}^2_{\mathrm{in}}+ \int_{0}^{t} \|\mathbf{u}(\tau)\|_{L^2}^2 \dd{\tau} ), \end{align}\qquad{(4)}\] for all \(t\in [0,T]\), where the constant \(C\) depends on \(\nu\).
In view of the energy method in [38], [50], [51] we have the existence of the solution to the system 3 . Let \[L^2_\sigma(\mathbb{R}^3)=\left\{\mathbf{u}\in L^2(\mathbb{R}^3) : \nabla\cdot \mathbf{u} = 0\right\},\] we then have the following theorem.
Theorem 5 (Local existence). Fix \(\nu>0\) and \(\mathbf{u}_{\mathrm{in}}\in H^1_\sigma(\mathbb{R}^3) := L^2_\sigma(\mathbb{R}^3)\cap H^1(\mathbb{R}^3)\). Then there exists a constant \(C_0 > 0\) such that \[\label{maximal95time95of95energy95estimate} T_0 = C_0\frac{\nu^3}{\|\nabla \mathbf{u}_{\mathrm{in}}\|_{L^2}^4}> 0\qquad{(5)}\] and a unique solution \(\mathbf{u}\in C([0,T_0);H^1_\sigma)\cap L^2((0,T_0);H^2_\sigma)\) of the Cauchy problem for the Navier-Stokes equation 3 , with initial value \(\mathbf{u}_{\mathrm{in}}\).
Lemma 8. Let \(a > b > 0\), \(T > 0\) and \(K > 0\) be constants, and let \(F \in L^2([0,T])\) be a function such that for all \(0 \leq t \leq T\), \[\require{physics} \int_0^t |F(s)|^2 \dd{s} \leq K e^{2at}.\] Then there exists a constant \(C = \frac{aK}{a - b}\) such that for all \(0 \leq t \leq T\), \[\require{physics} \int_0^t |F(s)|^2 e^{-2bs} \dd{s} \leq C e^{2(a - b)t}.\]
Proof. Define \(\require{physics} G(t) = \int_0^t |F(s)|^2 \dd{s}\), so \(G(t) \leq K e^{2at}\) for all \(0 \leq t \leq T\), and \(G'(s) = |F(s)|^2\) almost everywhere.
To estimate \(\require{physics} \int_0^t |F(s)|^2 e^{-2bs} \dd{s}\), we use integration by parts, which results in \[\require{physics} \begin{align} \int_0^t |F(s)|^2 e^{-2bs} \dd{s} & = \left. \left[ G(s) e^{-2bs} \right] \right|_0^t + 2b \int_0^t G(s) e^{-2bs} \dd{s} \\ & = G(s) e^{-2bt} + 2b \int_0^t G(s) e^{-2bs} \dd{s} \\ & \leq K e^{2(a -b)t} + 2b \int_0^t K e^{2as} \cdot e^{-2bs} \dd{s} \\ & \leq K e^{2(a -b)t} + \frac{bK}{a - b} \left( e^{2(a - b)t} - 1 \right). \end{align}\] Using the condition \(a > b > 0\), one has \[\require{physics} \begin{align} \int_0^t |F(s)|^2 e^{-2bs} \dd{s} & \leq \frac{aK}{a - b} e^{2(a - b)t} . \end{align}\] ◻
Proof. Let us define the real part of numerical range of the operator \(\mathscr{L}\): \[\require{physics} \mathcal{W}(\mathscr{L}) \overset{\mathrm{def}}{=} \qty{ \Re \left \langle \mathscr{L} \mathbf{u}, \mathbf{u} \right \rangle \;: \; \mathbf{u}\in \mathcal{D}(\mathscr{L}),\;\norm{\mathbf{u}}_{L^2} = 1 },\] and \[\alpha \overset{\mathrm{def}}{=}\sup \mathcal{W}(\mathscr{L}).\] Then, for \(\lambda \in \mathbb{C}\) with \(\Re \lambda > \alpha\) and \(\mathbf{u} \in \mathcal{D} (\mathscr{L})\), we have \[\norm{(\lambda - \mathscr{L}) \mathbf{u}} \norm{\mathbf{u}} \geq \Re \left \langle (\lambda - \mathscr{L}) \mathbf{u}, \mathbf{u} \right \rangle \geq (\Re \lambda - \alpha )\norm{\mathbf{u}} ^2 ,\] from which we deduce that \[\norm{(\lambda - \mathscr{L}) \mathbf{u}} \geq (\Re \lambda - \alpha )\norm{\mathbf{u}} .\] Since \(\Re(\sigma(\mathscr{L})) \subset \overline{\mathcal{W}(\mathscr{L})}\), one has \(\lambda \in \rho(\mathscr{L})\), which means \((\lambda - \mathscr{L})\) is invertible. Thus, for any \(\mathbf{u} \in \mathcal{D}(\mathscr{L})\), we denote \(\mathbf{w} = (\lambda - \mathscr{L}) \mathbf{u}\). Then \[\frac{ \norm{(\lambda - \mathscr{L}) ^{-1} \mathbf{w} } }{ \norm{\mathbf{w} } } = \frac{ \norm{\mathbf{u}} }{ \norm{(\lambda - \mathscr{L}) \mathbf{u}} } \leq \frac{1}{\Re \lambda - \alpha} .\] Therefore, we have \[\norm{(\lambda - \mathscr{L})^{-1} } \leq \frac{1}{\Re\lambda - \alpha} .\]
In what follows, we estimate the value of \(\alpha\). The upper bound of \(\mathcal{W}(\mathscr{L})\) is controlled by the self-adjoint part of \(\mathscr{L}\), since one may notice that \[\require{physics} \begin{align} \Re \left \langle \mathscr{L} \mathbf{u}, \mathbf{u} \right \rangle & = \Re \left \langle \qty(\frac{\mathscr{L} + \mathscr{L}^{*}}{2} + \frac{\mathscr{L} - \mathscr{L}^{*}}{2}) \mathbf{u}, \mathbf{u} \right \rangle \\&= \Re \left \langle \mathbf{u}, \frac{\mathscr{L} + \mathscr{L}^{*}}{2} \mathbf{u} \right \rangle + \Re\left \langle \mathbf{u}, -\frac{\mathscr{L} - \mathscr{L}^{*}}{2} \mathbf{u} \right \rangle \\ & = \left \langle \mathbf{u}, \frac{\mathscr{L} + \mathscr{L}^{*}}{2} \mathbf{u} \right \rangle = \left \langle \mathbf{\widehat{u}}, \frac{\mathscr{\widehat{L}} + \mathscr{\widehat{L}}^{*}}{2} \mathbf{\widehat{u}} \right \rangle = \left \langle \mathbf{\widehat{u}}, \mathscr{\widehat{L}}_{\text{sym}} \mathbf{\widehat{u}} \right \rangle \end{align}\] where \(\mathscr{L}^{*}\) is the adjoint of \(\mathscr{L}\) and \(\left \langle \cdot , \cdot \right \rangle\) is the \(L^2\) inner product, and the self-adjoint part of the original operator under Fourier transform is defined by \[\mathscr{\widehat{L}}_{\text{sym}} \overset{\mathrm{def}}{=}\frac{1}{2}(\mathscr{\widehat{L}} + \mathscr{\widehat{L}}^*).\] The self-adjoint operator takes the following form in the Cartesian coordinates \(\boldsymbol{\xi} = (\xi_1, \xi_2, \xi_3)\): \[\mathscr{\widehat{L}}_{\text{sym}} = \begin{pmatrix} - \nu |\boldsymbol{\xi}|^2 + f \frac{\xi_1 \xi_2}{|\boldsymbol{\xi}|^2} & \frac{1}{2}\left[ - 1 + (2-f)\frac{\xi_1^2}{|\boldsymbol{\xi}|^2} + f \frac{\xi_2^2}{|\boldsymbol{\xi}|^2} \right] & \frac{1}{2}f \frac{\xi_2 \xi_3}{|\boldsymbol{\xi}|^2} \\ \frac{1}{2}\left[ - 1 + (2-f)\frac{\xi_1^2}{|\boldsymbol{\xi}|^2} + f \frac{\xi_2^2}{|\boldsymbol{\xi}|^2} \right] & - \nu |\boldsymbol{\xi}|^2 + (2-f) \frac{\xi_1 \xi_2}{|\boldsymbol{\xi}|^2} & \frac{1}{2}(2-f) \frac{\xi_1 \xi_3}{|\boldsymbol{\xi}|^2} \\ \frac{1}{2}f \frac{\xi_2 \xi_3}{|\boldsymbol{\xi}|^2} & \frac{1}{2}(2-f) \frac{\xi_1 \xi_3}{|\boldsymbol{\xi}|^2} & - \nu |\boldsymbol{\xi}|^2 \end{pmatrix}.\] To convert this to spherical coordinates, we use the transformations \[\xi_1 = r\sin\theta\cos\phi, \quad \xi_2 = r\sin\theta\sin\phi, \quad \xi_3 = r\cos\theta, \quad |\boldsymbol{\xi}| = r,\] where \(r \geq 0\) is the radial coordinate, \(\theta \in [0, \pi]\) is the polar angle, and \(\phi \in [0, 2\pi)\) is the azimuthal angle. Introducing the following shorthands \[\begin{align} A & = \sin^2\theta \cos\phi\sin\phi, \quad & B &= \frac{1}{2}\left[ -1 + (2-f)\sin^2\theta \cos^2\phi + f \sin^2\theta \sin^2\phi \right], \quad \\ C & = \frac{1}{2}f \sin\theta\cos\theta \sin\phi, \quad & D &= \frac{1}{2}(2-f) \sin\theta\cos\theta \cos\phi, \end{align}\] the self-adjoint part in spherical coordinates becomes \[\mathscr{\widehat{L}}_{\text{sym}} = \begin{pmatrix} - \nu r^2 + f A & B & C \\ B & - \nu r^2 + (2-f) A & D \\ C & D & - \nu r^2 \end{pmatrix}.\] Since \(\mathscr{\widehat{L}}_{\mathrm{sym}}\) is real symmetric matrix, the range of \(\left \langle \mathscr{\widehat{L}}_{\mathrm{sym}} \mathbf{\widehat{u}}, \mathbf{\widehat{u}} \right \rangle\) is determined by its eigenvalues. And, aiming for the upper bound of the eigenvalues when \(\boldsymbol{\xi}\) varies in \(\mathbb{R}^3\), one may consider the case \(\nu = 0\). Precisely, we have \[\left \langle \mathscr{\widehat{L}}_{\mathrm{sym}} \mathbf{\widehat{u}}, \mathbf{\widehat{u}} \right \rangle \leq \left \langle \mathscr{\widehat{L}}_{\mathrm{sym}} \mathbf{\widehat{u}}, \mathbf{\widehat{u}} \right \rangle + \nu \norm{\mathbf{u}}_{L^2}^2 .\]
By setting \(\nu = 0\), the characteristic polynomial of \(\mathscr{\widehat{L}}_{\mathrm{sym}}\) is \[g(x) = a_3 x^3 + a_2 x^2 + a_1 x + a_0,\] where \[\begin{align} a_0 & = \frac{1}{32} f(2 - f) \sin ^2 2 \theta \sin 2 \phi , \quad a_2 =-\sin ^2 \theta \sin 2 \phi ,\quad a_3 = 1, \\ a_1 & = \frac{1}{64} \big[ -4 ( 2 f(2 - f) +1) \cos 2 \theta -8 \sin ^4 \theta \cos 4 \phi+\cos 4 \theta + 8 f(2 - f)-13 \big]. \end{align}\] To prove the eigenvalues are not greater than \(\frac{2 - f}{2}\), we substitute \(y = x - \frac{2 - f}{2}\) into the polynomial \[\require{physics} g \qty(y + \frac{2 - f}{2} ) = h(y) = b_3 y^3 + b_2 y^2 + b_1 y + b_0,\] where \[\begin{align} b_0 & = -\frac{1}{128} (2 - f) [8 (8-5 f) \sin ^2 \theta \sin 2 \phi \\&\quad+\cos 2 \theta \left(- 8 f \sin ^2 \theta \sin 2 \phi + 8 f(2 - f) +4\right) \\ &\quad +8 \sin ^4 \theta \cos 4 \phi -\cos 4 \theta + 8 f(6 - f) -51 ], \\ b_1 & = \frac{1}{64} [ - 64 (2 - f) \sin ^2 \theta \sin 2 \phi \\&\quad- 4 (2 f(2 - f) +1) \cos 2 \theta -8 \sin ^4 \theta \cos 4 \phi \\ & \quad +\cos 4 \theta - 8 f (22 - 5f)+179] ,\quad b_3 = 1. \end{align}\] and \[\begin{align} b_2 = 3-\frac{3 f}{2} -\sin ^2 \theta \sin 2 \phi , \quad b_3 = 1. \end{align}\] Now, we use Routh-Hurwitz criterion to prove the real parts of roots of \(h(y)\) are non-positive. Hence, in what follows, we prove \[\label{eq95proof951} b_3 ,b_2 > 0 , \quad b_1 , b_0 \geq 0, \qand b_2 b_1 - b_3 b_0 \geq 0.\tag{42}\] Once the above is proved, we infer the real parts of roots of \(g(x)\) are not greater than \(\frac{2 - f}{2}\), which means \[\left \langle \mathscr{\widehat{L}}_{\mathrm{sym}} \mathbf{\widehat{u}}, \mathbf{\widehat{u}} \right \rangle \leq \left \langle \mathscr{\widehat{L}}_{\mathrm{sym}} \mathbf{\widehat{u}}, \mathbf{\widehat{u}} \right \rangle + \nu \norm{\mathbf{u}}_{L^2}^2 \leq \frac{2 - f}{2} \norm{\mathbf{u}}_{L^2}^2 .\] Then we have the proof of 1.
In the subsequent paragraphs, we prove 42 step by step.
Step 1. \(b_3, b_2 > 0\) is evident for \(f\in(0,1)\).
Step 2. Next, we prove \(b_1 > 0\). By solving \(b_1 = 0\), we have \[f = \frac{1}{b_{10}}(b_{11} \pm \sqrt{b_{12}})\] where \[\begin{align} b_{10} & = 16 (\cos 2 \theta +5) > 0, \quad b_{11} =16 \left(-4 \sin ^2 \theta \sin 2 \phi +\cos 2 \theta +11\right) > 0 , \\ b_{12} & = - 32 ( \cos 2 \theta + 5) \left(-128 \sin ^2 \theta \sin 2 \phi -8 \sin ^4 \theta \cos 4 \phi -4 \cos 2 \theta +\cos 4 \theta +179\right) \\ & \quad + 16^2 \left(-4 \sin ^2\theta \sin 2 \phi +\cos 2 \theta +11\right)^2. \end{align}\] We aim to show that \(\frac{1}{b_{10}}(b_{11} \pm \sqrt{b_{12}}) \not \in (0,1)\), thus \(b_1\) does not change sign when \(f\in (0,1)\). To this end, we first show \(b_{12} > 0\). Utlizing the trigonometric identities \[\begin{align} \sin ^4\theta & = \frac{1}{8} (-4 \cos 2\theta +\cos 4\theta +3),\quad \cos 4\theta = \cos ^2 2\theta -\sin ^2 2\theta ,\quad \\ \cos 4 \phi & = \cos ^2 2 \phi -\sin ^2 2 \phi ,\quad \cos 6\theta = \cos ^3 2\theta -3 \sin ^2 2\theta \cos 2\theta ,\quad \\ \sin ^2 2\theta & = 1-\cos ^2 2\theta ,\quad \cos ^2 2 \phi = 1-\sin ^2 2 \phi \end{align}\] and the substitution \[\label{substitution} \cos 2 \theta = s \in [ - 1,1],\quad \sin 2 \phi = t\in [ - 1,1],\tag{43}\] we have \[\begin{align} \frac{1}{128} b_{12} & = 2 (11 + s^2) - (s-1 )^2 t (8 + (s-3 ) t) \\&\geq 2 (11 + s^2) - (s-1 )^2 (8 + (s-3 ) ) \\ & = 17 + 9 s - s^2 - s^3 > 0 \qfor s\in[ - 1,1]. \end{align}\] Then, let us prove \(b_{11} > b_{10}\). Direct calculation shows that \[b_{11} - b_{10} = 6-4 \sin ^2\theta \sin 2 \phi > 0\] Next, we show that \(\frac{1}{b_{10}}(b_{11} - \sqrt{b_{12}}) \geq 1\). To this end, we prove \[\require{physics} \qty( b_{11} - b_{10}) ^2 - b_{12} > 0.\] Then, using trigonometric identities with preivous substitution, we have \[\require{physics} \begin{align} \qty( b_{11} - b_{10}) ^2 - b_{12} & = 8 (5 + s) (2 (5 - s) + (s-1 ) t (8 + (s - 1) t)) \\ & \geq 8 (5 + s) (2 (5 - s) + (s-1 ) (8 + (s - 1))) \\ & = 8 (1 + s) (3 + s) (5 + s) \geq 0. \end{align}\] Then, one can conclude \[\frac{1}{b_{10}}(b_{11} \pm \sqrt{b_{12}}) \geq 1,\] which means \(b_1\) does not change sign when \(f \in(0,1)\) and we have \(b_1 > 0\) by taking some specific value.
Step 3. Next, we prove \(b_0 \geq 0\). The strategy is very similar to Step 2. By solving \(b_0 = 0\), we have \[f = \frac{1}{b_{00}}(b_{01} \pm \sqrt{b_{02}}) \qor f = 2,\] where \[\begin{align} b_{00} & = 32 \cos ^2\theta \geq 0, \quad b_{01} = 16 (\cos 2 \theta +3)-8 \sin ^2\theta (\cos 2 \theta +5) \sin 2 \phi , \\ b_{02} & = 64 \cos ^2\theta \left(64 \sin ^2\theta \sin 2 \phi +8 \sin ^4\theta \cos 4 \phi +4 \cos 2 \theta -\cos 4 \theta -51\right) \\ & \quad + 64 \left(\sin ^2\theta (\cos 2 \theta +5) \sin 2 \phi -2 (\cos 2 \theta +3)\right)^2. \end{align}\] Since the above fomula make since when \(\cos ^2\theta > 0\), we then make the assumption that \(\cos ^2\theta > 0\) in this step.
First, we notice that \[b_{01} \geq 16 (\cos 2 \theta +3)-8 \sin ^2\theta (\cos 2 \theta +5) = 8 \cos ^2\theta (\cos 2 \theta +7) \geq 0.\] Second, we consider \(b_{02}\). Using trigonometric indentities, one has \[\begin{align} \frac{1}{16} b_{02} & = 16 (3 + s^2) + (s-1 )t ( 8(7 + s^2) + (s -1 ) (17 + s (2 + s)) t) \\ & \geq 16 (3 + s^2) + (s-1 ) ( 8(7 + s^2) + (s -1 ) (17 + s (2 + s)) ) \\ & = (1 + s)^2 (3 + s)^2 \geq 0. \end{align}\] Third, \[b_{01} \geq 8 \cos ^2\theta (\cos 2 \theta +7) \geq 48\cos ^2\theta \geq b_{00} .\] Then, using trigonometric identities, we have \[\require{physics} \begin{align} \qty( \frac{b_{01}}{b_{00}} - 1) ^2 - \qty(\frac{\sqrt{b_{02}}}{b_{00}})^2 & = \tan ^2\theta (\sin 2 \phi -1) \left(\sin ^2\theta \sin 2 \phi -1\right) \geq 0. \end{align}\] Then \(b_0\) does not change sign for \(f\in (0,1)\) when \(\cos \theta \not = 0\), and thus \(b_0 > 0\) in this case.
For the case \(\cos \theta = 0\), we have \[b_0 = \frac{1}{8} (2 - f) ( 1 -\sin 2 \phi ) (3 - 2 f - \sin 2 \phi)\geq 0 \qfor f\in(0,1).\] The equlity \(b_0 = 0\) holds when \(\theta = \frac{\pi}{2} ,\phi = \frac{\pi}{4}\). Thus, \(b_0 \geq 0\).
Step 4. Next, we prove \(b_2 b_1 - b_3 b_0 \geq 0\). Direct calculation gives that \[\begin{align} b(s,t;f) \overset{\mathrm{def}}{=}32(b_2 b_1 - b_3 b_0) & = 4 (2 - f) (30(1 - f) + 7 f^2 - f(2 - f) s) \\ & \quad + (s-1 ) (124 - 122 f + 29 f^2 - f(2 - f) s) t \\ & \quad + 10 (2 - f) (s -1 )^2 t^2 + (s -1 )^3 t^3 . \end{align}\] Taking partial derivative of \(b(s,t;f)\) with respect to the \(t\), we get \[\begin{align} \partial_t b(s,t;f) & = (s- 1) (124 + f^2 (29 + s) - 2 f (61 + s)\\&\quad + 20 (2 - f) (s-1) t + 3 (s-1)^2 t^2) \\ \partial_t^2 b(s,t;f) & = (s - 1) (20 (2 - f) (s-1) + 6 (s-1)^2 t)\\& \geq (s - 1) (20 (2 - f) (s-1) + 6 (s-1)^2 ) \\ & = 2 (17 - 10 f + 3 s) (s- 1 )^2 \geq 0, \qfor s\in[ - 1,1],t\in[ - 1,1], \end{align}\] which means \(b(s,t;f)\) is convex with respect to \(t\) in the set \(s\in[ - 1,1],t\in[ - 1,1]\). Then, aiming to find the minimal value of \(b(s,t;f)\), we solve \(\partial_t b(s,t;f) = 0\) to get \[\require{physics} \begin{align} t & = \frac{1}{t_0} \qty(t_1 \pm \sqrt{t_2}), \qq{where} t_0 =3 (1 - s) \geq 0 , \quad t_1 = 10 (2 - f) \geq 0, \\ t_2 & = f (3 (2 - f) s + 13 f - 34) +28 . \end{align}\] One may notcite that \[t_2 \geq f (3 (2 - f) ( - 1) + 13 f - 34) +28 = 28 + 8 f (-5 + 2 f) > 0 \qfor f\in(0,1).\] And, \[t_1 - t_0 = 17 - 10 f + 3 s >0 \qfor f\in(0,1), s\in[ - 1,1].\] Next, using the substitution \(s = \cos 2\theta\), one get \[\require{physics} \begin{align} \qty( \frac{t_{1}}{t_{0}} - 1) ^2 - \qty(\frac{\sqrt{t_{2}}}{t_{0}})^2 & = \frac{1}{6} \left[ (62 - f (62 - 15 f ) )\csc ^4\theta - (20- f) (2 - f) \csc ^2\theta +6\right] . \end{align}\] Then, denoting \(\csc \theta = u^{\frac{1}{2}}\) with \(u \geq 1\), one can define \[\begin{align} \phi(u) & = \frac{1}{6} \left[ (62 - f (62 - 15 f ) )u^2 - (20- f) (2 - f) u +6\right] . \end{align}\] For \(0 < f < 1\) one may notice that \((62 + f (-62 + 15 f)) > 0\). The axis of symmetry of \(\phi(u)\) is \(u = \frac{(20 - f)(2 - f)}{2(62 - f (62 - 15 f ))} < 1\). Then \(\phi(u)\) is increasing for \(u \geq 1\). Calculating \(\phi(1)\) gives that \[\begin{align} \phi(1) =\frac{1}{3} (14 + f (-20 + 7 f)) > 0 \qfor f\in(0,1). \end{align}\] Then we have \(\require{physics} \frac{1}{t_0} \qty(t_1 \pm \sqrt{t_2}) > 1\). Hence, the extreme value for \(b(s,t;f)\) is at \(t = \pm 1\). Substituting \(t = 0,s = 0\) into \(\partial_t b(s,t;f)\), we have \(\partial_t b(s,t;f) < 0\), which means the minimal value of \(b(s,t;f)\) takes at \(t = 1\).
Subsitituting \(t =1\), into \(b(s,t;f)\), we have \[\begin{align} b(s,1;f) & = (5 - 4 f + s) (f^2 (7 + s) + (3 + s) (9 + s) - 4 f (7 + 2 s)) \\ & > (5 - 4 f + s) ( (7 + s) + (3 + s) (9 + s) - 4 (7 + 2 s)) \\ &=(5 - 4 f + s)(2 + s) (3 + s) > 0 \qfor f\in (0,1),s\in[ - 1,1]. \end{align}\] Then, \(b_2 b_1 - b_3 b_0 > 0\) is proved.
Finally, \(\Re \left \langle \mathscr{L} \mathbf{u}, \mathbf{u} \right \rangle \leq \frac{2 - f}{2}\) for \(\norm{\mathbf{u}}_{L^2} = 1\), and the proof is finished. ◻
The work of Q.Wang is supported by the National Natural Science Foundation of Sichuan Province (No.2025ZNSFSC0072) and the National Science Foundation of China (No.12301131 and No. 11901408).
email:daozhiha@buffalo.edu↩︎
email:xihujunzi@scu.edu.cn↩︎