2 Eigenvalue equation for the energy stability of plane parallel flows

When the fluctuation in the temperature of the fluid is small, the motion of the fluid between parallel plates is governed by the Boussinesq equation (Straughan Reference Straughan1992, p. 56),

(2.1) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{\unicode[STIX]{x2202}\boldsymbol{U}^{\ast }}{\unicode[STIX]{x2202}t^{\ast }}+(\boldsymbol{U}^{\ast }\boldsymbol{\cdot }\unicode[STIX]{x1D735}^{\ast })\boldsymbol{U}^{\ast }=-\frac{1}{\unicode[STIX]{x1D70C}_{r}}\unicode[STIX]{x1D735}^{\ast }P^{\ast }+\unicode[STIX]{x1D708}\unicode[STIX]{x1D6FB}^{\ast 2}\boldsymbol{U}^{\ast }+\frac{1}{\unicode[STIX]{x1D70C}_{r}}F_{x}^{\ast }\boldsymbol{e}_{x}+[1-\unicode[STIX]{x1D6FE}(T^{\ast }-T_{r}^{\ast })]\boldsymbol{g},\qquad & \displaystyle\end{eqnarray}$$

(2.2) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{\unicode[STIX]{x2202}T^{\ast }}{\unicode[STIX]{x2202}t^{\ast }}+(\boldsymbol{U}^{\ast }\boldsymbol{\cdot }\unicode[STIX]{x1D735}^{\ast })T^{\ast }=\unicode[STIX]{x1D705}\unicode[STIX]{x1D6FB}^{\ast 2}T^{\ast }+\frac{1}{\unicode[STIX]{x1D70C}_{r}c_{p}}Q^{\ast }, & \displaystyle\end{eqnarray}$$

(2.3) $$\begin{eqnarray}\displaystyle & \displaystyle \unicode[STIX]{x1D735}^{\ast }\boldsymbol{\cdot }\boldsymbol{U}^{\ast }=0, & \displaystyle\end{eqnarray}$$

(2.4) $$\begin{eqnarray}\displaystyle & \displaystyle \left.\left(U^{\ast }\pm l^{\ast }\frac{\unicode[STIX]{x2202}U^{\ast }}{\unicode[STIX]{x2202}y^{\ast }}\right)\right|_{y^{\ast }=\pm h^{\ast }}-U_{w,\pm h^{\ast }}^{\ast }=V^{\ast }|_{y^{\ast }=\pm h^{\ast }}=\left.\left(W^{\ast }\pm l^{\ast }\frac{\unicode[STIX]{x2202}W^{\ast }}{\unicode[STIX]{x2202}y^{\ast }}\right)\right|_{y^{\ast }=\pm h^{\ast }}=0,\qquad & \displaystyle\end{eqnarray}$$

(2.5) $$\begin{eqnarray}\displaystyle & \displaystyle \left.\left(T^{\ast }\pm l_{T}^{\ast }\frac{\unicode[STIX]{x2202}T^{\ast }}{\unicode[STIX]{x2202}y^{\ast }}\right)\right|_{y^{\ast }=\pm h^{\ast }}-T_{w,\pm h^{\ast }}^{\ast }=0, & \displaystyle\end{eqnarray}$$

where $x^{\ast }$ , $y^{\ast }$ and $z^{\ast }$ are the coordinates in the streamwise, wall-normal and spanwise directions, respectively; $t^{\ast }$ is the time; $\boldsymbol{U}^{\ast }=(U^{\ast },V^{\ast },W^{\ast })$ , $T^{\ast }$ and $P^{\ast }$ are the velocity, temperature and pressure of the fluid, respectively; $\unicode[STIX]{x1D70C}_{r}$ is the reference density of the fluid at the reference temperature $T_{r}^{\ast }$ ; $\unicode[STIX]{x1D708}$ , $\unicode[STIX]{x1D6FE}$ , $\unicode[STIX]{x1D705}$ and $c_{p}$ are the kinematic viscosity, coefficient of thermal expansion, thermal diffusivity and specific heat capacity at constant pressure of the fluid, respectively; $\boldsymbol{g}=-g\boldsymbol{e}_{y}$ is the gravitational acceleration; $\boldsymbol{e}_{x}$ and $\boldsymbol{e}_{y}$ are the unit vectors in the streamwise and wall-normal directions. Both the streamwise component of the body force $F_{x}^{\ast }$ and the volumetric heat source $Q^{\ast }$ are functions of only the wall-normal coordinate $y^{\ast }$ . The distance between the parallel plates is $2h^{\ast }$ ; $U_{w,\pm h^{\ast }}^{\ast }$ and $T_{w,\pm h^{\ast }}^{\ast }$ are the streamwise velocities and the temperatures of the walls at $y^{\ast }=\pm h^{\ast }$ .

The region is assumed to be periodic in the streamwise and spanwise directions. The velocity of the fluid satisfies the slip boundary condition at the walls, and $l^{\ast }$ is the slip length. The temperature of the fluid satisfies the convective boundary condition at the walls, and $l_{T}^{\ast }$ is the ratio between the thermal conductivity of the fluid and the convective heat transfer coefficient of the flow. The no-slip boundary condition and the fixed temperature boundary condition correspond to $l^{\ast }=0$ and $l_{T}^{\ast }=0$ , respectively.

Introducing non-dimensional quantities,

(2.6a-d ) $$\begin{eqnarray}(x,y,z)=\frac{(x^{\ast },y^{\ast },z^{\ast })}{h^{\ast }},\quad \boldsymbol{U}=\frac{\boldsymbol{U}^{\ast }}{U_{c}^{\ast }},\quad t=\frac{U_{c}^{\ast }}{h^{\ast }}t^{\ast },\quad T=\frac{T^{\ast }-T_{r}^{\ast }}{\unicode[STIX]{x0394}T_{c}^{\ast }},\end{eqnarray}$$

(2.7a-d ) $$\begin{eqnarray}P=\frac{P^{\ast }+\unicode[STIX]{x1D70C}_{r}gy^{\ast }}{\unicode[STIX]{x1D70C}_{r}U_{c}^{\ast 2}},\quad F_{x}=\frac{h^{\ast }}{\unicode[STIX]{x1D70C}_{r}U_{c}^{\ast 2}}F_{x}^{\ast },\quad Q=\frac{h^{\ast }}{\unicode[STIX]{x1D70C}_{r}c_{p}U_{c}^{\ast }\unicode[STIX]{x0394}T_{c}^{\ast }}Q^{\ast },\quad (l,l_{T})=\frac{(l^{\ast },l_{T}^{\ast })}{h^{\ast }},\end{eqnarray}$$

where $U_{c}^{\ast }$ and $\unicode[STIX]{x0394}T_{c}^{\ast }$ are the characteristic velocity and temperature difference in the fluid, we have the non-dimensional governing equation,

(2.8) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{\unicode[STIX]{x2202}\boldsymbol{U}}{\unicode[STIX]{x2202}t}+(\boldsymbol{U}\boldsymbol{\cdot }\unicode[STIX]{x1D735})\boldsymbol{U}=-\unicode[STIX]{x1D735}P+\frac{1}{Re}\unicode[STIX]{x1D6FB}^{2}\boldsymbol{U}+F_{x}\boldsymbol{e}_{x}+\frac{Gr}{Re^{2}}T\boldsymbol{e}_{y}, & \displaystyle\end{eqnarray}$$

(2.9) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{\unicode[STIX]{x2202}T}{\unicode[STIX]{x2202}t}+(\boldsymbol{U}\boldsymbol{\cdot }\unicode[STIX]{x1D735})T=\frac{1}{PrRe}\unicode[STIX]{x1D6FB}^{2}T+Q, & \displaystyle\end{eqnarray}$$

(2.10) $$\begin{eqnarray}\displaystyle & \displaystyle \unicode[STIX]{x1D735}\boldsymbol{\cdot }\boldsymbol{U}=0, & \displaystyle\end{eqnarray}$$

(2.11) $$\begin{eqnarray}\displaystyle & \displaystyle \left.\left(U\pm l\frac{\unicode[STIX]{x2202}U}{\unicode[STIX]{x2202}y}\right)\right|_{y=\pm 1}-U_{w,\pm 1}=V|_{y=\pm 1}=\left.\left(W\pm l\frac{\unicode[STIX]{x2202}W}{\unicode[STIX]{x2202}y}\right)\right|_{y=\pm 1}=0, & \displaystyle\end{eqnarray}$$

(2.12) $$\begin{eqnarray}\displaystyle & \displaystyle \left.\left(T\pm l_{T}\frac{\unicode[STIX]{x2202}T}{\unicode[STIX]{x2202}y}\right)\right|_{y=\pm 1}-T_{w,\pm 1}=0, & \displaystyle\end{eqnarray}$$

where the Reynolds number, the Prandtl number and the Grashof number are defined as

(2.13a-c ) $$\begin{eqnarray}Re=\frac{U_{c}^{\ast }h^{\ast }}{\unicode[STIX]{x1D708}},\quad Pr=\frac{\unicode[STIX]{x1D708}}{\unicode[STIX]{x1D705}},\quad Gr=\frac{g\unicode[STIX]{x1D6FE}h^{\ast 3}\unicode[STIX]{x0394}T_{c}^{\ast }}{\unicode[STIX]{x1D708}^{2}}.\end{eqnarray}$$

Here we assume $l\geqslant 0$ and $l_{T}\geqslant 0$ . When $l_{T}>0$ , the Nusselt number can be defined as $l_{T}^{-1}$ .

The base flow $(\boldsymbol{U}_{0},T_{0},P_{0})$ is a parallel flow (with $V_{0}=W_{0}=0$ and only depending on the wall-normal coordinate $y$ ), and satisfies

(2.14) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{1}{Re}\frac{\text{d}^{2}U_{0}}{\text{d}y^{2}}+F_{x}=0, & \displaystyle\end{eqnarray}$$

(2.15) $$\begin{eqnarray}\displaystyle & \displaystyle -\frac{\text{d}P_{0}}{\text{d}y}+\frac{Gr}{Re^{2}}T_{0}=0, & \displaystyle\end{eqnarray}$$

(2.16) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{1}{PrRe}\frac{\text{d}^{2}T_{0}}{\text{d}y^{2}}+Q=0, & \displaystyle\end{eqnarray}$$

(2.17) $$\begin{eqnarray}\displaystyle & \displaystyle \left.\left(U_{0}\pm l\frac{\text{d}U_{0}}{\text{d}y}\right)\right|_{y=\pm 1}-U_{w,\pm 1}=\left.\left(T_{0}\pm l_{T}\frac{\text{d}T_{0}}{\text{d}y}\right)\right|_{y=\pm 1}-T_{w,\pm 1}=0. & \displaystyle\end{eqnarray}$$

The governing equation of the disturbance $(\boldsymbol{u},T^{\prime },p)=(\boldsymbol{U},T,P)-(\boldsymbol{U}_{0},T_{0},P_{0})$ is

(2.18) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{\unicode[STIX]{x2202}\boldsymbol{u}}{\unicode[STIX]{x2202}t}+U_{0}\frac{\unicode[STIX]{x2202}\boldsymbol{u}}{\unicode[STIX]{x2202}x}+v\frac{\text{d}\boldsymbol{U}_{0}}{\text{d}y}+(\boldsymbol{u}\boldsymbol{\cdot }\unicode[STIX]{x1D735})\boldsymbol{u}=-\unicode[STIX]{x1D735}p+\frac{1}{Re}\unicode[STIX]{x1D6FB}^{2}\boldsymbol{u}+\frac{Gr}{Re^{2}}T^{\prime }\boldsymbol{e}_{y}, & \displaystyle\end{eqnarray}$$

(2.19) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{\unicode[STIX]{x2202}T^{\prime }}{\unicode[STIX]{x2202}t}+U_{0}\frac{\unicode[STIX]{x2202}T^{\prime }}{\unicode[STIX]{x2202}x}+v\frac{\text{d}T_{0}}{\text{d}y}+(\boldsymbol{u}\boldsymbol{\cdot }\unicode[STIX]{x1D735})T^{\prime }=\frac{1}{PrRe}\unicode[STIX]{x1D6FB}^{2}T^{\prime }, & \displaystyle\end{eqnarray}$$

(2.20) $$\begin{eqnarray}\displaystyle & \displaystyle \unicode[STIX]{x1D735}\boldsymbol{\cdot }\boldsymbol{u}=0, & \displaystyle\end{eqnarray}$$

(2.21) $$\begin{eqnarray}\displaystyle & \displaystyle \left.\left(u\pm l\frac{\unicode[STIX]{x2202}u}{\unicode[STIX]{x2202}y}\right)\right|_{y=\pm 1}=v|_{y=\pm 1}=\left.\left(w\pm l\frac{\unicode[STIX]{x2202}w}{\unicode[STIX]{x2202}y}\right)\right|_{y=\pm 1}=\left.\left(T^{\prime }\pm l_{T}\frac{\unicode[STIX]{x2202}T^{\prime }}{\unicode[STIX]{x2202}y}\right)\right|_{y=\pm 1}=0. & \displaystyle\end{eqnarray}$$

Define the energy of the disturbance as

(2.22) $$\begin{eqnarray}E(Pr,\unicode[STIX]{x1D706})=\int _{\unicode[STIX]{x1D6FA}}\frac{u^{2}+v^{2}+w^{2}+\unicode[STIX]{x1D706}Pr{T^{\prime }}^{2}}{2}\,\text{d}V,\end{eqnarray}$$

where $\unicode[STIX]{x1D706}>0$ is a coupling parameter (Joseph Reference Joseph1966), and $\text{d}V=\text{d}x\,\text{d}y\,\text{d}z$ . The region is $\unicode[STIX]{x1D6FA}=(0,L_{x})\times (-1,1)\times (0,L_{z})$ , where $L_{x}$ and $L_{z}$ are the wavelengths of the disturbance in the streamwise and spanwise directions, respectively. From (2.18)–(2.21), we have

(2.23) $$\begin{eqnarray}\frac{\text{d}E}{\text{d}t}=\int _{\unicode[STIX]{x1D6FA}}\left(-uv\frac{\text{d}U_{0}}{\text{d}y}-\unicode[STIX]{x1D706}PrvT^{\prime }\frac{\text{d}\tilde{T}_{0}}{\text{d}y}\right)\text{d}V-\frac{1}{Re}(\Vert \unicode[STIX]{x1D735}\boldsymbol{u}\Vert ^{2}+\unicode[STIX]{x1D706}\Vert \unicode[STIX]{x1D735}T^{\prime }\Vert ^{2}),\end{eqnarray}$$

where

(2.24) $$\begin{eqnarray}\tilde{T}_{0}=T_{0}-\frac{Gr}{\unicode[STIX]{x1D706}PrRe^{2}}y,\end{eqnarray}$$

(2.25) $$\begin{eqnarray}\displaystyle \Vert \unicode[STIX]{x1D735}\boldsymbol{u}\Vert ^{2} & = & \displaystyle \int _{\unicode[STIX]{x1D6FA}}\left(\frac{\unicode[STIX]{x2202}u_{i}}{\unicode[STIX]{x2202}x_{j}}\frac{\unicode[STIX]{x2202}u_{i}}{\unicode[STIX]{x2202}x_{j}}\right)\text{d}V+l\int _{0}^{L_{z}}\int _{0}^{L_{x}}\left[\left(\left.\frac{\unicode[STIX]{x2202}u}{\unicode[STIX]{x2202}y}\right|_{y=1}\right)^{2}+\left(\left.\frac{\unicode[STIX]{x2202}u}{\unicode[STIX]{x2202}y}\right|_{y=-1}\right)^{2}\right]\text{d}x\,\text{d}z\nonumber\\ \displaystyle & & \displaystyle +\,l\int _{0}^{L_{z}}\int _{0}^{L_{x}}\left[\left(\left.\frac{\unicode[STIX]{x2202}w}{\unicode[STIX]{x2202}y}\right|_{y=1}\right)^{2}+\left(\left.\frac{\unicode[STIX]{x2202}w}{\unicode[STIX]{x2202}y}\right|_{y=-1}\right)^{2}\right]\text{d}x\,\text{d}z,\end{eqnarray}$$

(2.26) $$\begin{eqnarray}\Vert \unicode[STIX]{x1D735}T^{\prime }\Vert ^{2}=\int _{\unicode[STIX]{x1D6FA}}\left(\frac{\unicode[STIX]{x2202}T^{\prime }}{\unicode[STIX]{x2202}x_{j}}\frac{\unicode[STIX]{x2202}T^{\prime }}{\unicode[STIX]{x2202}x_{j}}\right)\text{d}V+l_{T}\int _{0}^{L_{z}}\int _{0}^{L_{x}}\left[\left(\left.\frac{\unicode[STIX]{x2202}T^{\prime }}{\unicode[STIX]{x2202}y}\right|_{y=1}\right)^{2}+\left(\left.\frac{\unicode[STIX]{x2202}T^{\prime }}{\unicode[STIX]{x2202}y}\right|_{y=-1}\right)^{2}\right]\text{d}x\,\text{d}z,\end{eqnarray}$$

and the Einstein summation convention is used.

In the energy stability analysis, the critical Reynolds number $Re_{E}(Pr,\unicode[STIX]{x1D706})$ is defined by

(2.27) $$\begin{eqnarray}\frac{1}{Re_{E}(Pr,\unicode[STIX]{x1D706})}=\max \frac{\displaystyle \int _{\unicode[STIX]{x1D6FA}}\left(-uv{\displaystyle \frac{\text{d}U_{0}}{\text{d}y}}-\unicode[STIX]{x1D706}PrvT^{\prime }{\displaystyle \frac{\text{d}\tilde{T}_{0}}{\text{d}y}}\right)\text{d}V}{\Vert \unicode[STIX]{x1D735}\boldsymbol{u}\Vert ^{2}+\unicode[STIX]{x1D706}\Vert \unicode[STIX]{x1D735}T^{\prime }\Vert ^{2}},\end{eqnarray}$$

where the maximum is searched among all divergence-free disturbances satisfying the boundary condition (2.21). Therefore, we have

(2.28) $$\begin{eqnarray}\frac{\text{d}E}{\text{d}t}\leqslant \left(\frac{1}{Re_{E}(Pr,\unicode[STIX]{x1D706})}-\frac{1}{Re}\right)(\Vert \unicode[STIX]{x1D735}\boldsymbol{u}\Vert ^{2}+\unicode[STIX]{x1D706}\Vert \unicode[STIX]{x1D735}T^{\prime }\Vert ^{2})\leqslant 0,\end{eqnarray}$$

when $Re\leqslant Re_{E}(Pr,\unicode[STIX]{x1D706})$ . We also have $E\rightarrow 0$ when $Re<Re_{E}(Pr,\unicode[STIX]{x1D706})$ , because $\Vert \boldsymbol{u}\Vert$ and $\Vert T^{\prime }\Vert$ are controlled by $\Vert \unicode[STIX]{x1D735}\boldsymbol{u}\Vert$ and $\Vert \unicode[STIX]{x1D735}T^{\prime }\Vert$ in the bounded domain according to the Poincaré inequality.

The Euler–Lagrange equation corresponding to (2.27) is

(2.29) $$\begin{eqnarray}\displaystyle & \displaystyle \unicode[STIX]{x1D6FB}^{2}u-\frac{R}{2}\frac{\text{d}U_{0}}{\text{d}y}v-\frac{\unicode[STIX]{x2202}q}{\unicode[STIX]{x2202}x}=0, & \displaystyle\end{eqnarray}$$

(2.30) $$\begin{eqnarray}\displaystyle & \displaystyle \unicode[STIX]{x1D6FB}^{2}v-\frac{R}{2}\frac{\text{d}U_{0}}{\text{d}y}u-\frac{\unicode[STIX]{x1D706}PrR}{2}\frac{\text{d}\tilde{T}_{0}}{\text{d}y}T^{\prime }-\frac{\unicode[STIX]{x2202}q}{\unicode[STIX]{x2202}y}=0, & \displaystyle\end{eqnarray}$$

(2.31) $$\begin{eqnarray}\displaystyle & \displaystyle \unicode[STIX]{x1D6FB}^{2}w-\frac{\unicode[STIX]{x2202}q}{\unicode[STIX]{x2202}z}=0, & \displaystyle\end{eqnarray}$$

(2.32) $$\begin{eqnarray}\displaystyle & \displaystyle \unicode[STIX]{x1D6FB}^{2}T^{\prime }-\frac{PrR}{2}\frac{\text{d}\tilde{T}_{0}}{\text{d}y}v=0, & \displaystyle\end{eqnarray}$$

(2.33) $$\begin{eqnarray}\displaystyle & \displaystyle \unicode[STIX]{x1D735}\boldsymbol{\cdot }\boldsymbol{u}=0, & \displaystyle\end{eqnarray}$$

(2.34) $$\begin{eqnarray}\displaystyle & \displaystyle \left.\left(u\pm l\frac{\unicode[STIX]{x2202}u}{\unicode[STIX]{x2202}y}\right)\right|_{y=\pm 1}=v|_{y=\pm 1}=\left.\left(w\pm l\frac{\unicode[STIX]{x2202}w}{\unicode[STIX]{x2202}y}\right)\right|_{y=\pm 1}=\left.\left(T^{\prime }\pm l_{T}\frac{\unicode[STIX]{x2202}T^{\prime }}{\unicode[STIX]{x2202}y}\right)\right|_{y=\pm 1}=0,\qquad & \displaystyle\end{eqnarray}$$

where $R$ and $q(x,y,z)$ are the Lagrange multipliers.

From (2.25), (2.26) and (2.29)–(2.34), we have

(2.35) $$\begin{eqnarray}\displaystyle \Vert \unicode[STIX]{x1D735}\boldsymbol{u}\Vert ^{2}+\unicode[STIX]{x1D706}\Vert \unicode[STIX]{x1D735}T^{\prime }\Vert ^{2} & = & \displaystyle -\int _{\unicode[STIX]{x1D6FA}}(u\unicode[STIX]{x1D6FB}^{2}u+v\unicode[STIX]{x1D6FB}^{2}v+w\unicode[STIX]{x1D6FB}^{2}w+\unicode[STIX]{x1D706}T^{\prime }\unicode[STIX]{x1D6FB}^{2}T^{\prime })\,\text{d}V\nonumber\\ \displaystyle & = & \displaystyle R\int _{\unicode[STIX]{x1D6FA}}\left(-uv\frac{\text{d}U_{0}}{\text{d}y}-\unicode[STIX]{x1D706}PrvT^{\prime }\frac{\text{d}\tilde{T}_{0}}{\text{d}y}\right)\text{d}V.\end{eqnarray}$$

Comparing (2.35) with (2.27), we notice that the critical Reynolds number $Re_{E}(Pr,\unicode[STIX]{x1D706})$ is just the minimum positive eigenvalue $R$ .

Introducing the normal mode $(u,v,w,T^{\prime },q)=(\hat{u} ,\hat{v},{\hat{w}},\hat{T},\hat{q})\exp [\text{i}(\unicode[STIX]{x1D6FC}x+\unicode[STIX]{x1D6FD}z)]+\text{c.c.}$ , where $\unicode[STIX]{x1D6FC}$ and $\unicode[STIX]{x1D6FD}$ are the streamwise and spanwise wavenumbers, and c.c. denotes the complex conjugate, we have

(2.36) $$\begin{eqnarray}\displaystyle & \displaystyle (\text{D}^{2}-k^{2})\hat{u} -\frac{R}{2}(\text{D}U_{0})\hat{v}-\text{i}k(\cos \unicode[STIX]{x1D703})\hat{q}=0, & \displaystyle\end{eqnarray}$$

(2.37) $$\begin{eqnarray}\displaystyle & \displaystyle (\text{D}^{2}-k^{2})\hat{v}-\frac{R}{2}(\text{D}U_{0})\hat{u} -\frac{\unicode[STIX]{x1D706}PrR}{2}(\text{D}\tilde{T}_{0})\hat{T}-\text{D}\hat{q}=0, & \displaystyle\end{eqnarray}$$

(2.38) $$\begin{eqnarray}\displaystyle & \displaystyle (\text{D}^{2}-k^{2}){\hat{w}}-\text{i}k(\sin \unicode[STIX]{x1D703})\hat{q}=0, & \displaystyle\end{eqnarray}$$

(2.39) $$\begin{eqnarray}\displaystyle & \displaystyle (\text{D}^{2}-k^{2})\hat{T}-\frac{PrR}{2}(\text{D}\tilde{T}_{0})\hat{v}=0, & \displaystyle\end{eqnarray}$$

(2.40) $$\begin{eqnarray}\displaystyle & \displaystyle \text{i}k(\cos \unicode[STIX]{x1D703})\hat{u} +\text{D}\hat{v}+\text{i}k(\sin \unicode[STIX]{x1D703}){\hat{w}}=0, & \displaystyle\end{eqnarray}$$

(2.41) $$\begin{eqnarray}\displaystyle & \displaystyle (\hat{u} \pm l\text{D}\hat{u} )(\pm 1)=\hat{v}(\pm 1)=({\hat{w}}\pm l\text{D}{\hat{w}})(\pm 1)=(\hat{T}\pm l_{T}\text{D}\hat{T})(\pm 1)=0, & \displaystyle\end{eqnarray}$$

where $\text{D}\equiv \text{d}/\text{d}y$ , $k=(\unicode[STIX]{x1D6FC}^{2}+\unicode[STIX]{x1D6FD}^{2})^{1/2}$ and $\unicode[STIX]{x1D703}=\arctan (\unicode[STIX]{x1D6FD}/\unicode[STIX]{x1D6FC})$ . If we denote the positive eigenvalues of (2.36)–(2.41) as $R_{1}(k,\unicode[STIX]{x1D703},Pr,\unicode[STIX]{x1D706})\leqslant R_{2}(k,\unicode[STIX]{x1D703},Pr,\unicode[STIX]{x1D706})\leqslant \cdots \,$ , then the critical Reynolds number for the energy stability is

(2.42) $$\begin{eqnarray}Re_{E}(Pr,\unicode[STIX]{x1D706})=\min _{k,\unicode[STIX]{x1D703}}R_{1}(k,\unicode[STIX]{x1D703},Pr,\unicode[STIX]{x1D706}).\end{eqnarray}$$

Note that the $n$ th positive eigenvalue $R_{n}$ and the corresponding eigenvector may be only piecewise continuously differentiable where $R_{n}=R_{n+1}$ or $R_{n}=R_{n-1}$ (figure 1). Due to the symmetry of (2.36)–(2.41), only the cases with $k>0$ and $0\leqslant \unicode[STIX]{x1D703}\leqslant \unicode[STIX]{x03C0}/2$ need to be considered.

Figure 1.

The fifth and sixth positive eigenvalues for the energy stability of the parallel shear flow with $\text{D}U_{0}=\sin (8\unicode[STIX]{x03C0}y)$ , $Pr\,\text{D}\tilde{T}_{0}=0$ , $l=0$ and $k=1$ .

The critical Reynolds number for the energy stability $Re_{E}$ depends on the definition of the energy of the disturbance (2.22), and is therefore a function of the Prandtl number $Pr$ and the coupling parameter $\unicode[STIX]{x1D706}$ . For given Prandtl number, an optimal critical Reynolds number for the energy stability can be further defined as the maximum of $Re_{E}(Pr,\unicode[STIX]{x1D706})$ for all $\unicode[STIX]{x1D706}>0$ (Joseph Reference Joseph1966).

3 The monotonicity of the positive eigenvalues $R_{n}\,(n\geqslant 1)$ in the wavenumber plane

3.1 The expressions for $\unicode[STIX]{x2202}R_{n}/\unicode[STIX]{x2202}k$ and $\unicode[STIX]{x2202}R_{n}/\unicode[STIX]{x2202}\unicode[STIX]{x1D703}$

To study the monotonicity of $R_{n}$ in the $(k,\unicode[STIX]{x1D703})$ plane, we first derive the expressions of $\unicode[STIX]{x2202}R_{n}/\unicode[STIX]{x2202}k$ and $\unicode[STIX]{x2202}R_{n}/\unicode[STIX]{x2202}\unicode[STIX]{x1D703}$ .

Denote (2.36)–(2.40) as $\unicode[STIX]{x1D647}\hat{\boldsymbol{u}}=\mathbf{0}$ , where

(3.1a,b ) $$\begin{eqnarray}\unicode[STIX]{x1D647}=\left[\begin{array}{@{}ccccc@{}}\text{D}^{2}-k^{2} & -{\textstyle \frac{1}{2}}R(\text{D}U_{0}) & 0 & 0 & -\text{i}k\cos \unicode[STIX]{x1D703}\\ -{\textstyle \frac{1}{2}}R(\text{D}U_{0}) & \text{D}^{2}-k^{2} & 0 & -{\textstyle \frac{1}{2}}\unicode[STIX]{x1D706}PrR(\text{D}\tilde{T}_{0}) & -\text{D}\\ 0 & 0 & \text{D}^{2}-k^{2} & 0 & -\text{i}k\sin \unicode[STIX]{x1D703}\\ 0 & -{\textstyle \frac{1}{2}}PrR(\text{D}\tilde{T}_{0}) & 0 & \text{D}^{2}-k^{2} & 0\\ \text{i}k\cos \unicode[STIX]{x1D703} & \text{D} & \text{i}k\sin \unicode[STIX]{x1D703} & 0 & 0\end{array}\right],\quad \hat{\boldsymbol{u}}=\left[\begin{array}{@{}c@{}}\hat{u} \\ \hat{v}\\ {\hat{w}}\\ \hat{T}\\ \hat{q}\end{array}\right].\end{eqnarray}$$

For any $\unicode[STIX]{x1D706}>0$ , assume the geometric multiplicity of the $n$ th eigenvalue $R_{n}(k,\unicode[STIX]{x1D703},Pr)$ to be 1 in an open set $S\subset (0,+\infty )\times [0,\unicode[STIX]{x03C0}/2]\times [0,+\infty )$ . If the $n$ th positive eigenvalue and the corresponding eigenvector of (2.36)–(2.41) are $(R_{n},\hat{\boldsymbol{u}}_{n})$ for parameters $(k,\unicode[STIX]{x1D703},Pr)\in S$ , and are $(R_{n}+\text{d}R_{n},\hat{\boldsymbol{u}}_{n}+\text{d}\hat{\boldsymbol{u}}_{n})$ for parameters $(k+\text{d}k,\unicode[STIX]{x1D703}+\text{d}\unicode[STIX]{x1D703},Pr+\text{d}Pr)\in S$ , we can neglect the higher-order terms and obtain

(3.2) $$\begin{eqnarray}\displaystyle & \displaystyle \unicode[STIX]{x1D647}(\text{d}\hat{\boldsymbol{u}}_{n})+(\text{d}k)\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D647}}{\unicode[STIX]{x2202}k}\hat{\boldsymbol{u}}_{n}+(\text{d}\unicode[STIX]{x1D703})\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D647}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}\hat{\boldsymbol{u}}_{n}+(\text{d}Pr)\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D647}}{\unicode[STIX]{x2202}Pr}\hat{\boldsymbol{u}}_{n}+(\text{d}R_{n})\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D647}}{\unicode[STIX]{x2202}R_{n}}\hat{\boldsymbol{u}}_{n}=\mathbf{0}, & \displaystyle\end{eqnarray}$$

(3.3) $$\begin{eqnarray}\displaystyle & \displaystyle [\text{d}\hat{u} _{n}\pm l\text{D}(\text{d}\hat{u} _{n})](\pm 1)=(\text{d}\hat{v}_{n})(\pm 1)=[\text{d}{\hat{w}}_{n}\pm l\text{D}(\text{d}{\hat{w}}_{n})](\pm 1)=0, & \displaystyle\end{eqnarray}$$

(3.4) $$\begin{eqnarray}\displaystyle & \displaystyle [\text{d}\hat{T}_{n}\pm l_{T}\text{D}(\text{d}\hat{T}_{n})](\pm 1)=0. & \displaystyle\end{eqnarray}$$

Define the inner product between two vectors $\hat{\boldsymbol{u}}^{\prime }$ and $\hat{\boldsymbol{u}}^{\prime \prime }$ as

(3.5) $$\begin{eqnarray}\langle \hat{\boldsymbol{u}}^{\prime },\hat{\boldsymbol{u}}^{\prime \prime }\rangle =\int _{-1}^{1}(\bar{\hat{u} }^{\prime }\hat{u} ^{\prime \prime }+\bar{\hat{v}}^{\prime }\hat{v}^{\prime \prime }+\bar{{\hat{w}}}^{\prime }{\hat{w}}^{\prime \prime }+\unicode[STIX]{x1D706}\bar{\hat{T}}^{\prime }\hat{T}^{\prime \prime }+\bar{\hat{q}}^{\prime }\hat{q}^{\prime \prime })\,\text{d}y,\end{eqnarray}$$

where the overlines denote the complex conjugates. Because $\unicode[STIX]{x1D647}$ is a self-adjoint operator with respect to the inner product (3.5) and the boundary conditions (2.41), (3.3) and (3.4), we have

(3.6) $$\begin{eqnarray}\langle \hat{\boldsymbol{u}}_{n},\unicode[STIX]{x1D647}(\text{d}\hat{\boldsymbol{u}}_{n})\rangle =\langle \unicode[STIX]{x1D647}\hat{\boldsymbol{u}}_{n},\text{d}\hat{\boldsymbol{u}}_{n}\rangle =\langle \mathbf{0},\text{d}\hat{\boldsymbol{u}}_{n}\rangle =0,\end{eqnarray}$$

and then it follows from (3.2) that

(3.7) $$\begin{eqnarray}\left\langle \hat{\boldsymbol{u}}_{n},\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D647}}{\unicode[STIX]{x2202}k}\hat{\boldsymbol{u}}_{n}\right\rangle \text{d}k+\left\langle \hat{\boldsymbol{u}}_{n},\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D647}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}\hat{\boldsymbol{u}}_{n}\right\rangle \text{d}\unicode[STIX]{x1D703}+\left\langle \hat{\boldsymbol{u}}_{n},\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D647}}{\unicode[STIX]{x2202}Pr}\hat{\boldsymbol{u}}_{n}\right\rangle \text{d}Pr+\left\langle \hat{\boldsymbol{u}}_{n},\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D647}}{\unicode[STIX]{x2202}R_{n}}\hat{\boldsymbol{u}}_{n}\right\rangle \text{d}R_{n}=0.\end{eqnarray}$$

Therefore, we have

(3.8) $$\begin{eqnarray}\displaystyle & \displaystyle {\displaystyle \frac{\unicode[STIX]{x2202}R_{n}}{\unicode[STIX]{x2202}k}}=-{\displaystyle \frac{\left\langle \hat{\boldsymbol{u}}_{n},{\displaystyle \frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D647}}{\unicode[STIX]{x2202}k}}\hat{\boldsymbol{u}}_{n}\right\rangle }{\left\langle \hat{\boldsymbol{u}}_{n},{\displaystyle \frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D647}}{\unicode[STIX]{x2202}R_{n}}}\hat{\boldsymbol{u}}_{n}\right\rangle }}, & \displaystyle\end{eqnarray}$$

(3.9) $$\begin{eqnarray}\displaystyle & \displaystyle {\displaystyle \frac{\unicode[STIX]{x2202}R_{n}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}}=-{\displaystyle \frac{\left\langle \hat{\boldsymbol{u}}_{n},{\displaystyle \frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D647}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}}\hat{\boldsymbol{u}}_{n}\right\rangle }{\left\langle \hat{\boldsymbol{u}}_{n},{\displaystyle \frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D647}}{\unicode[STIX]{x2202}R_{n}}}\hat{\boldsymbol{u}}_{n}\right\rangle }}. & \displaystyle\end{eqnarray}$$

The inner products in (3.7)–(3.9) are

(3.10) $$\begin{eqnarray}\displaystyle & \hspace{-6.0pt}\displaystyle \left\langle \hat{\boldsymbol{u}}_{n},\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D647}}{\unicode[STIX]{x2202}k}\hat{\boldsymbol{u}}_{n}\right\rangle =-2k\int _{-1}^{1}(|\hat{u} _{n}|^{2}+|\hat{v}_{n}|^{2}+|{\hat{w}}_{n}|^{2}+\unicode[STIX]{x1D706}|\hat{T}_{n}|^{2})\,\text{d}y-\frac{1}{k}\int _{-1}^{1}(\bar{\hat{q}}_{n}\text{D}\hat{v}_{n}+\hat{q}_{n}\text{D}\bar{\hat{v}}_{n})\,\text{d}y, & \displaystyle \nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

(3.11) $$\begin{eqnarray}\displaystyle & \hspace{-6.0pt}\displaystyle \left\langle \hat{\boldsymbol{u}}_{n},\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D647}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}\hat{\boldsymbol{u}}_{n}\right\rangle =-\int _{-1}^{1}(\bar{\hat{q}}_{n}\hat{\unicode[STIX]{x1D702}}_{n}+\hat{q}_{n}\bar{\hat{\unicode[STIX]{x1D702}}}_{n})\,\text{d}y, & \displaystyle\end{eqnarray}$$

(3.12) $$\begin{eqnarray}\displaystyle & \hspace{-6.0pt}\displaystyle \left\langle \hat{\boldsymbol{u}}_{n},\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D647}}{\unicode[STIX]{x2202}Pr}\hat{\boldsymbol{u}}_{n}\right\rangle =-\frac{\unicode[STIX]{x1D706}R_{n}}{2}\int _{-1}^{1}(\text{D}\tilde{T}_{0})(\bar{\hat{v}}_{n}\hat{T}_{n}+\hat{v}_{n}\bar{\hat{T}}_{n})\,\text{d}y, & \displaystyle\end{eqnarray}$$

(3.13) $$\begin{eqnarray}\displaystyle & \hspace{-6.0pt}\displaystyle \left\langle \hat{\boldsymbol{u}}_{n},\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D647}}{\unicode[STIX]{x2202}R_{n}}\hat{\boldsymbol{u}}_{n}\right\rangle =-\frac{1}{2}\int _{-1}^{1}(\text{D}U_{0})(\bar{\hat{u} }_{n}\hat{v}_{n}+\hat{u} _{n}\bar{\hat{v}}_{n})\,\text{d}y-\frac{\unicode[STIX]{x1D706}Pr}{2}\int _{-1}^{1}(\text{D}\tilde{T}_{0})(\bar{\hat{v}}_{n}\hat{T}_{n}+\hat{v}_{n}\bar{\hat{T}}_{n})\,\text{d}y, & \displaystyle \nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

where $\hat{\unicode[STIX]{x1D702}}_{n}=\text{i}k\hat{u} _{n}\sin \unicode[STIX]{x1D703}-\text{i}k{\hat{w}}_{n}\cos \unicode[STIX]{x1D703}$ is the wall-normal component of the disturbance vorticity.

Using (2.36)–(2.41) in (3.10)–(3.13), we have

(3.14) $$\begin{eqnarray}\displaystyle & \hspace{-6.0pt}\displaystyle \left\langle \hat{\boldsymbol{u}}_{n},\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D647}}{\unicode[STIX]{x2202}k}\hat{\boldsymbol{u}}_{n}\!\right\rangle =-\frac{2}{k}\int _{-1}^{1}[2(|\text{D}\hat{v}_{n}|^{2}+k^{2}|\hat{v}_{n}|^{2})+|\hat{\unicode[STIX]{x1D702}}_{n}|^{2}+\unicode[STIX]{x1D706}k^{2}|\hat{T}_{n}|^{2}]\,\text{d}y+\frac{1}{k^{3}}(I_{v}+I_{\unicode[STIX]{x1D702}}+\unicode[STIX]{x1D706}k^{2}I_{T}), & \displaystyle \nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

(3.15) $$\begin{eqnarray}\displaystyle & \hspace{-12.0pt}\displaystyle \left\langle \hat{\boldsymbol{u}}_{n},\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D647}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}\hat{\boldsymbol{u}}_{n}\right\rangle =\frac{R_{n}}{2}\int _{-1}^{1}(\text{D}U_{0})(\bar{\hat{v}}_{n}{\hat{w}}_{n}+\hat{v}_{n}\bar{{\hat{w}}}_{n})\,\text{d}y, & \displaystyle\end{eqnarray}$$

(3.16) $$\begin{eqnarray}\displaystyle & \hspace{-12.0pt}\displaystyle \left\langle \hat{\boldsymbol{u}}_{n},\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D647}}{\unicode[STIX]{x2202}Pr}\hat{\boldsymbol{u}}_{n}\right\rangle =\frac{2\unicode[STIX]{x1D706}}{Pr}I_{T}, & \displaystyle\end{eqnarray}$$

(3.17) $$\begin{eqnarray}\displaystyle & \hspace{-12.0pt}\displaystyle \left\langle \hat{\boldsymbol{u}}_{n},\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D647}}{\unicode[STIX]{x2202}R_{n}}\hat{\boldsymbol{u}}_{n}\right\rangle =\frac{1}{k^{2}R_{n}}(I_{v}+I_{\unicode[STIX]{x1D702}}+\unicode[STIX]{x1D706}k^{2}I_{T}), & \displaystyle\end{eqnarray}$$

where

(3.18) $$\begin{eqnarray}\displaystyle & \displaystyle I_{v}=\int _{-1}^{1}|(\text{D}^{2}-k^{2})\hat{v}_{n}|^{2}\,\text{d}y+l[|(\text{D}^{2}\hat{v}_{n})(1)|^{2}+|(\text{D}^{2}\hat{v}_{n})(-1)|^{2}], & \displaystyle\end{eqnarray}$$

(3.19) $$\begin{eqnarray}\displaystyle & \displaystyle I_{\unicode[STIX]{x1D702}}=\int _{-1}^{1}(|\text{D}\hat{\unicode[STIX]{x1D702}}_{n}|^{2}+k^{2}|\hat{\unicode[STIX]{x1D702}}_{n}|^{2})\,\text{d}y+l[|(\text{D}\hat{\unicode[STIX]{x1D702}}_{n})(1)|^{2}+|(\text{D}\hat{\unicode[STIX]{x1D702}}_{n})(-1)|^{2}], & \displaystyle\end{eqnarray}$$

(3.20) $$\begin{eqnarray}\displaystyle & \displaystyle I_{T}=\int _{-1}^{1}(|\text{D}\hat{T}_{n}|^{2}+k^{2}|\hat{T}_{n}|^{2})\,\text{d}y+l_{T}[|(\text{D}\hat{T}_{n})(1)|^{2}+|(\text{D}\hat{T}_{n})(-1)|^{2}]. & \displaystyle\end{eqnarray}$$

3.2 The dependence of the positive eigenvalues $R_{n}$ $(n\geqslant 1)$ on $k$

Theorem 1. For any continuously differentiable real functions $\text{D}U_{0}$ and $\text{D}\tilde{T}_{0}$ , and any $\unicode[STIX]{x1D706}>0$ , $Pr\geqslant 0$ , $l\geqslant 0$ , $l_{T}\geqslant 0$ and $\unicode[STIX]{x1D703}\in [0,\unicode[STIX]{x03C0}/2]$ , $kR_{n}$ is an increasing function of $k$ when $k>0$ , and $R_{n}$ is a decreasing function of $k$ when $0<k<k_{0}(l,l_{T})$ , where $R_{n}$ is the $n$ th positive eigenvalue of (2.36)–(2.41) $(n\geqslant 1)$ , and $k_{0}(l,l_{T})=\min \{k_{0}^{\prime }(l),\unicode[STIX]{x1D70E}(\max \{l,l_{T}\})\}$ . Here $k_{0}^{\prime }(l)$ is the solution of the equation

(3.21) $$\begin{eqnarray}(\sqrt{3}k_{0}^{\prime })\tan (\sqrt{3}k_{0}^{\prime })+k_{0}^{\prime }\tanh (k_{0}^{\prime })+4k_{0}^{\prime 2}l=0\end{eqnarray}$$

in the interval $(\unicode[STIX]{x03C0}/\sqrt{12},\unicode[STIX]{x03C0}/\sqrt{3})$ ; $\unicode[STIX]{x1D70E}(l)$ is the solution of the equation $\unicode[STIX]{x1D70E}^{-1}\cot \unicode[STIX]{x1D70E}=l$ in the interval $(0,\unicode[STIX]{x03C0}/2]$ . Specifically, when $l=l_{T}=0$ and $0<k<k_{0}(0,0)=k_{0}^{\prime }(0)\approx 1.534$ , $R_{n}$ is a decreasing function of $k$ .

The following lemmas, which can be proved easily with the variational method, will be used in the proof of theorem 1.

Lemma 1. For any $l\geqslant 0$ and any complex-valued function $\hat{f}$ , if $(\hat{f}\pm l\text{D}\hat{f})(\pm 1)=0$ , then

(3.22) $$\begin{eqnarray}\int _{-1}^{1}|\text{D}\hat{f}|^{2}\,\text{d}y+l[|(\text{D}\hat{f})(1)|^{2}+|(\text{D}\hat{f})(-1)|^{2}]\geqslant [\unicode[STIX]{x1D70E}(l)]^{2}\int _{-1}^{1}|\,\hat{f}|^{2}\,\text{d}y,\end{eqnarray}$$

where $\unicode[STIX]{x1D70E}(l)$ is the solution of the equation $\unicode[STIX]{x1D70E}^{-1}\cot \unicode[STIX]{x1D70E}=l$ in the interval $(0,\unicode[STIX]{x03C0}/2]$ . The equality in (3.22) holds when $\hat{f}=C\cos (\unicode[STIX]{x1D70E}y)$ , where $C$ is any complex constant.

Lemma 2. For any $l\geqslant 0$ and any complex-valued function ${\hat{g}}$ , if ${\hat{g}}(\pm 1)=(\text{D}{\hat{g}}\pm l\text{D}^{2}{\hat{g}})(\pm 1)=0$ , then

(3.23) $$\begin{eqnarray}\displaystyle & & \displaystyle \int _{-1}^{1}(|\text{D}^{2}{\hat{g}}|^{2}+k^{2}|\text{D}{\hat{g}}|^{2})\,\text{d}y+l[|(\text{D}^{2}{\hat{g}})(1)|^{2}+|(\text{D}^{2}{\hat{g}})(-1)|^{2}]\nonumber\\ \displaystyle & & \displaystyle \quad \geqslant [\unicode[STIX]{x1D70E}^{\prime }(k,l)]^{2}\int _{-1}^{1}(|\text{D}{\hat{g}}|^{2}+k^{2}|{\hat{g}}|^{2})\,\text{d}y,\end{eqnarray}$$

where $\unicode[STIX]{x1D70E}^{\prime }(k,l)$ is the solution of the equation $\unicode[STIX]{x1D70E}^{\prime }\tan \unicode[STIX]{x1D70E}^{\prime }+k\tanh k+({\unicode[STIX]{x1D70E}^{\prime }}^{2}+k^{2})l=0$ in the interval $(\unicode[STIX]{x03C0}/2,\unicode[STIX]{x03C0})$ . The equality in (3.23) holds when ${\hat{g}}=C^{\prime }[\cos (\unicode[STIX]{x1D70E}^{\prime })\cosh (ky)-\cosh (k)\cos (\unicode[STIX]{x1D70E}^{\prime }y)]$ , where $C^{\prime }$ is any complex constant.

To prove theorem 1, it is sufficient to prove for any given $\unicode[STIX]{x1D706}>0$ , $l\geqslant 0$ and $l_{T}\geqslant 0$ ,

(3.24) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}k}(kR_{n})\geqslant 0,\quad (k>0), & \displaystyle\end{eqnarray}$$

(3.25) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{\unicode[STIX]{x2202}R_{n}}{\unicode[STIX]{x2202}k}\leqslant 0,\quad (0<k<k_{0}(l,l_{T})), & \displaystyle\end{eqnarray}$$

in each subset $S\subset (0,+\infty )\times [0,\unicode[STIX]{x03C0}/2]\times [0,+\infty )$ where the geometric multiplicity of the $n$ th eigenvalue $R_{n}(k,\unicode[STIX]{x1D703},Pr)$ is 1.

Substituting (3.14) and (3.17) into (3.8), we have $\unicode[STIX]{x2202}R_{n}/\unicode[STIX]{x2202}k\geqslant -R_{n}/k$ , and then (3.24) holds.

Using lemma 1 for $\hat{\unicode[STIX]{x1D702}}_{n}$ and $\hat{T}_{n}$ , and using lemma 2 for $\hat{v}_{n}$ , we have

(3.26) $$\begin{eqnarray}\displaystyle & \displaystyle I_{\unicode[STIX]{x1D702}}\geqslant [k^{2}+(\unicode[STIX]{x1D70E}(l))^{2}]\int _{-1}^{1}|\hat{\unicode[STIX]{x1D702}}_{n}|^{2}\,\text{d}y, & \displaystyle\end{eqnarray}$$

(3.27) $$\begin{eqnarray}\displaystyle & \displaystyle I_{T}\geqslant [k^{2}+(\unicode[STIX]{x1D70E}(l_{T}))^{2}]\int _{-1}^{1}|\hat{T}_{n}|^{2}\,\text{d}y, & \displaystyle\end{eqnarray}$$

(3.28) $$\begin{eqnarray}\displaystyle & \displaystyle I_{v}\geqslant [k^{2}+(\unicode[STIX]{x1D70E}^{\prime }(k,l))^{2}]\int _{-1}^{1}(|\text{D}\hat{v}_{n}|^{2}+k^{2}|\hat{v}_{n}|^{2})\,\text{d}y. & \displaystyle\end{eqnarray}$$

Substituting (3.26)–(3.28) into (3.14), we have

(3.29) $$\begin{eqnarray}\displaystyle \left\langle \hat{\boldsymbol{u}}_{n},\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D647}}{\unicode[STIX]{x2202}k}\hat{\boldsymbol{u}}_{n}\right\rangle & {\geqslant} & \displaystyle \frac{(\unicode[STIX]{x1D70E}^{\prime }(k,l))^{2}-3k^{2}}{k^{3}}\int _{-1}^{1}(|\text{D}\hat{v}_{n}|^{2}+k^{2}|\hat{v}_{n}|^{2})\,\text{d}y+\frac{(\unicode[STIX]{x1D70E}(l))^{2}-k^{2}}{k^{3}}\int _{-1}^{1}|\hat{\unicode[STIX]{x1D702}}_{n}|^{2}\,\text{d}y\nonumber\\ \displaystyle & & \displaystyle +\,\frac{\unicode[STIX]{x1D706}[(\unicode[STIX]{x1D70E}(l_{T}))^{2}-k^{2}]}{k}\int _{-1}^{1}|\hat{T}_{n}|^{2}\,\text{d}y.\end{eqnarray}$$

When $0<k\leqslant k_{0}^{\prime }(l)$ , we have $[\unicode[STIX]{x1D70E}^{\prime }(k,l)]^{2}-3k^{2}\geqslant [\unicode[STIX]{x1D70E}^{\prime }(k_{0}^{\prime }(l),l)]^{2}-3[k_{0}^{\prime }(l)]^{2}=0$ , because $\unicode[STIX]{x1D70E}^{\prime }(k,l)$ is a decreasing function of $k$ for any given $l\geqslant 0$ . Noticing that $\unicode[STIX]{x1D70E}(l)$ is a decreasing function of $l$ when $l\geqslant 0$ , we have $[\unicode[STIX]{x1D70E}(l)]^{2}-k^{2}\geqslant 0$ and $[\unicode[STIX]{x1D70E}(l_{T})]^{2}-k^{2}\geqslant 0$ when $0<k\leqslant \unicode[STIX]{x1D70E}(\max \{l,l_{T}\})$ . Therefore, we have $\langle \hat{\boldsymbol{u}}_{n},(\unicode[STIX]{x2202}\unicode[STIX]{x1D647}/\unicode[STIX]{x2202}k)\hat{\boldsymbol{u}}_{n}\rangle \geqslant 0$ when $0<k<k_{0}(l,l_{T})=\min \{k_{0}^{\prime }(l),\unicode[STIX]{x1D70E}(\max \{l,l_{T}\})\}$ , and then (3.25) holds because of (3.8) and (3.17).

Specifically, when $l=l_{T}=0$ , we have $k_{0}(0,0)=k_{0}^{\prime }(0)$ because $k_{0}^{\prime }(0)\approx 1.534<\unicode[STIX]{x03C0}/2=\unicode[STIX]{x1D70E}(0)$ . Thus we have proved theorem 1.

A direct corollary of theorem 1 is that $R_{n}=O(k^{-1})$ when $k\rightarrow 0$ . Another corollary is that the least stable mode in the energy method must have $k\geqslant k_{0}(l,l_{T})$ . The contours of $k_{0}$ in the $(l,l_{T})$ plane are plotted in figure 2.

Figure 2.

The contours of $k_{0}$ in the $(l,l_{T})$ plane.

3.3 The dependence of the positive eigenvalues $R_{n}$ $(n\geqslant 1)$ on $\unicode[STIX]{x1D703}$

Theorem 2. For any continuously differentiable real functions $\text{D}U_{0}$ and $\text{D}\tilde{T}_{0}$ , and any $\unicode[STIX]{x1D706}>0$ , $Pr\geqslant 0$ , $l\geqslant 0$ , $l_{T}\geqslant 0$ and $k>0$ , $(\cos \unicode[STIX]{x1D703}\pm 1)R_{n}$ are decreasing functions of $\unicode[STIX]{x1D703}$ when $0\leqslant \unicode[STIX]{x1D703}\leqslant \unicode[STIX]{x03C0}/2$ , where $R_{n}$ is the $n$ th positive eigenvalue of (2.36)–(2.41)  $(n\geqslant 1)$ .

To prove theorem 2, it is sufficient to prove for any given $\unicode[STIX]{x1D706}>0$ , $l\geqslant 0$ and $l_{T}\geqslant 0$ ,

(3.30) $$\begin{eqnarray}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}[(\cos \unicode[STIX]{x1D703}\pm 1)R_{n}]\leqslant 0\end{eqnarray}$$

in each subset $S\subset (0,+\infty )\times [0,\unicode[STIX]{x03C0}/2]\times [0,+\infty )$ where the geometric multiplicity of the $n$ th eigenvalue $R_{n}(k,\unicode[STIX]{x1D703},Pr)$ is 1.

Substituting (3.13) and (3.15) into (3.9), we have

(3.31) $$\begin{eqnarray}\frac{1}{R_{n}}\frac{\unicode[STIX]{x2202}R_{n}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}=\frac{\displaystyle \int _{-1}^{1}(\text{D}U_{0})(\bar{\hat{v}}_{n}{\hat{w}}_{n}+\hat{v}_{n}\bar{{\hat{w}}}_{n})\,\text{d}y}{\displaystyle \int _{-1}^{1}(\text{D}U_{0})(\bar{\hat{u} }_{n}\hat{v}_{n}+\hat{u} _{n}\bar{\hat{v}}_{n})\,\text{d}y+\unicode[STIX]{x1D706}Pr\int _{-1}^{1}(\text{D}\tilde{T}_{0})(\bar{\hat{v}}_{n}\hat{T}_{n}+\hat{v}_{n}\bar{\hat{T}}_{n})\,\text{d}y}.\end{eqnarray}$$

(i) When $\unicode[STIX]{x1D703}=0$ , we have ${\hat{w}}_{n}=0$ from (2.38) and (2.41), and then $\unicode[STIX]{x2202}R_{n}/\unicode[STIX]{x2202}\unicode[STIX]{x1D703}=0$ follows from (3.31). As a result,

(3.32) $$\begin{eqnarray}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}[(\cos \unicode[STIX]{x1D703}\pm 1)R_{n}]=0.\end{eqnarray}$$

(ii) When $\unicode[STIX]{x1D703}=\unicode[STIX]{x03C0}/2$ , if $(\hat{u} _{n},\hat{v}_{n},{\hat{w}}_{n},\hat{T}_{n},\hat{q}_{n})$ is an eigenvector corresponding to the eigenvalue $R_{n}$ , then $(\bar{\hat{u} }_{n},\bar{\hat{v}}_{n},-\bar{{\hat{w}}}_{n},\bar{\hat{T}}_{n},\bar{\hat{q}}_{n})$ is also an eigenvector corresponding to the same eigenvalue. In each subset $S$ where the geometric multiplicity of the eigenvalue $R_{n}$ is 1, there exists a complex constant $C$ such that $(\bar{\hat{u} }_{n},\bar{\hat{v}}_{n},-\bar{{\hat{w}}}_{n},\bar{\hat{T}}_{n},\bar{\hat{q}}_{n})=C(\hat{u} _{n},\hat{v}_{n},{\hat{w}}_{n},\hat{T}_{n},\hat{q}_{n})$ , and then $\bar{\hat{v}}_{n}{\hat{w}}_{n}+\hat{v}_{n}\bar{{\hat{w}}}_{n}=0$ . Therefore, we have $\unicode[STIX]{x2202}R_{n}/\unicode[STIX]{x2202}\unicode[STIX]{x1D703}=0$ from (3.31), and then

(3.33) $$\begin{eqnarray}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}[(\cos \unicode[STIX]{x1D703}\pm 1)R_{n}]=-R_{n}\leqslant 0.\end{eqnarray}$$

(iii) When $0<\unicode[STIX]{x1D703}<\unicode[STIX]{x03C0}/2$ , using (2.36)–(2.41) in (3.15), we have

(3.34) $$\begin{eqnarray}\left\langle \hat{\boldsymbol{u}}_{n},\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D647}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}\hat{\boldsymbol{u}}_{n}\right\rangle =-\frac{(\sin ^{2}\unicode[STIX]{x1D703})I_{v}-(1+\cos ^{2}\unicode[STIX]{x1D703})I_{\unicode[STIX]{x1D702}}-\unicode[STIX]{x1D706}k^{2}(\sin ^{2}\unicode[STIX]{x1D703})I_{T}}{k^{2}\sin \unicode[STIX]{x1D703}\cos \unicode[STIX]{x1D703}}.\end{eqnarray}$$

Define

(3.35) $$\begin{eqnarray}I(\unicode[STIX]{x1D707})=\int _{-1}^{1}|{\hat{h}}_{\unicode[STIX]{x1D707}}|^{2}\,\text{d}y+l[|{\hat{h}}_{\unicode[STIX]{x1D707}}(1)|^{2}+|{\hat{h}}_{\unicode[STIX]{x1D707}}(-1)|^{2}],\end{eqnarray}$$

for $\unicode[STIX]{x1D707}\in \mathbb{R}$ , where

(3.36) $$\begin{eqnarray}{\hat{h}}_{\unicode[STIX]{x1D707}}=(\text{D}^{2}-k^{2})\hat{v}_{n}+\unicode[STIX]{x1D707}\text{D}\hat{\unicode[STIX]{x1D702}}_{n}.\end{eqnarray}$$

Using (2.36)–(2.41) in (3.35), we have

(3.37) $$\begin{eqnarray}I(\unicode[STIX]{x1D707})=I_{v}-\unicode[STIX]{x1D707}(\tan \unicode[STIX]{x1D703})(I_{v}-I_{\unicode[STIX]{x1D702}}-\unicode[STIX]{x1D706}k^{2}I_{T})+\unicode[STIX]{x1D707}^{2}\left(I_{\unicode[STIX]{x1D702}}-k^{2}\int _{-1}^{1}|\hat{\unicode[STIX]{x1D702}}_{n}|^{2}\,\text{d}y\right).\end{eqnarray}$$

Substituting (3.17) and (3.34) into (3.9), we have

(3.38) $$\begin{eqnarray}\frac{1}{R_{n}}\frac{\unicode[STIX]{x2202}R_{n}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}=\frac{(\sin ^{2}\unicode[STIX]{x1D703})I_{v}-(1+\cos ^{2}\unicode[STIX]{x1D703})I_{\unicode[STIX]{x1D702}}-\unicode[STIX]{x1D706}k^{2}(\sin ^{2}\unicode[STIX]{x1D703})I_{T}}{(\sin \unicode[STIX]{x1D703}\cos \unicode[STIX]{x1D703})(I_{v}+I_{\unicode[STIX]{x1D702}}+\unicode[STIX]{x1D706}k^{2}I_{T})},\end{eqnarray}$$

and then

(3.39) $$\begin{eqnarray}\displaystyle & & \displaystyle \frac{1}{R_{n}}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}[(\cos \unicode[STIX]{x1D703}\pm 1)R_{n}]\nonumber\\ \displaystyle & & \displaystyle \quad =\pm \frac{(\sin ^{2}\unicode[STIX]{x1D703})I_{v}-(1\pm \cos \unicode[STIX]{x1D703})^{2}I_{\unicode[STIX]{x1D702}}-\unicode[STIX]{x1D706}k^{2}(\sin ^{2}\unicode[STIX]{x1D703})(1\pm 2\cos \unicode[STIX]{x1D703})I_{T}}{(\sin \unicode[STIX]{x1D703}\cos \unicode[STIX]{x1D703})(I_{v}+I_{\unicode[STIX]{x1D702}}+\unicode[STIX]{x1D706}k^{2}I_{T})}\nonumber\\ \displaystyle & & \displaystyle \quad =-\frac{(\sin ^{2}\unicode[STIX]{x1D703})I\left({\displaystyle \frac{\cos \unicode[STIX]{x1D703}\pm 1}{\sin \unicode[STIX]{x1D703}}}\right)+\unicode[STIX]{x1D706}k^{2}(\sin ^{2}\unicode[STIX]{x1D703})I_{T}+k^{2}(1\pm \cos \unicode[STIX]{x1D703})^{2}\displaystyle \int _{-1}^{1}|\hat{\unicode[STIX]{x1D702}}_{n}|^{2}\,\text{d}y}{(\sin \unicode[STIX]{x1D703})(I_{v}+I_{\unicode[STIX]{x1D702}}+\unicode[STIX]{x1D706}k^{2}I_{T})}\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant 0,\end{eqnarray}$$

where we have used (3.37) for $\unicode[STIX]{x1D707}=(\cos \unicode[STIX]{x1D703}\pm 1)/\sin \unicode[STIX]{x1D703}$ . Noticing that $R_{n}>0$ , we have proved theorem 2.

According to theorem 2, we have

(3.40) $$\begin{eqnarray}\frac{R_{n}\left(k,{\displaystyle \frac{\unicode[STIX]{x03C0}}{2}},Pr,\unicode[STIX]{x1D706}\right)}{1+\cos \unicode[STIX]{x1D703}}\leqslant R_{n}(k,\unicode[STIX]{x1D703},Pr,\unicode[STIX]{x1D706})\leqslant \min \left\{\frac{R_{n}\left(k,{\displaystyle \frac{\unicode[STIX]{x03C0}}{2}},Pr,\unicode[STIX]{x1D706}\right)}{1-\cos \unicode[STIX]{x1D703}},\frac{2R_{n}(k,0,Pr,\unicode[STIX]{x1D706})}{1+\cos \unicode[STIX]{x1D703}}\right\},\end{eqnarray}$$

for any $k>0$ , $0<\unicode[STIX]{x1D703}\leqslant \unicode[STIX]{x03C0}/2$ , $Pr\geqslant 0$ , $\unicode[STIX]{x1D706}>0$ , $l\geqslant 0$ , $l_{T}\geqslant 0$ and $n\geqslant 1$ . From (2.42) and (3.40), we have

(3.41) $$\begin{eqnarray}\frac{1}{2}\min _{k>0}R_{1}\left(k,\frac{\unicode[STIX]{x03C0}}{2},Pr,\unicode[STIX]{x1D706}\right)\leqslant Re_{E}(Pr,\unicode[STIX]{x1D706})\leqslant \min _{k>0}R_{1}\left(k,\frac{\unicode[STIX]{x03C0}}{2},Pr,\unicode[STIX]{x1D706}\right).\end{eqnarray}$$

Calculating the critical Reynolds number for the energy stability $Re_{E}$ according to (2.42) requires solving the eigenvalue equation (2.36)–(2.41) for all wavenumber pairs $(\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD})$ or $(k,\unicode[STIX]{x1D703})$ . However, the critical Reynolds number can be estimated by only solving the eigenvalue equation for the streamwise vortices with $\unicode[STIX]{x1D6FC}=k\cos \unicode[STIX]{x1D703}=0$ according to (3.41).

As a direct result of (3.41), the minimum $R_{1}$ for all two-dimensional waves is also bounded from below by a half of the minimum $R_{1}$ for all streamwise vortices, i.e.

(3.42) $$\begin{eqnarray}\min _{k>0}R_{1}(k,0,Pr,\unicode[STIX]{x1D706})\geqslant \frac{1}{2}\min _{k>0}R_{1}\left(k,\frac{\unicode[STIX]{x03C0}}{2},Pr,\unicode[STIX]{x1D706}\right),\end{eqnarray}$$

for any $Pr\geqslant 0$ , $\unicode[STIX]{x1D706}>0$ , $l\geqslant 0$ and $l_{T}\geqslant 0$ . Kaiser & Schmitt (Reference Kaiser and Schmitt2001) have proved an inequality stronger than (3.42),

(3.43) $$\begin{eqnarray}\min _{k>0}R_{1}(k,0)\geqslant \frac{16}{27}\min _{k>0}R_{1}\left(k,\frac{\unicode[STIX]{x03C0}}{2}\right),\end{eqnarray}$$

for $Pr\,\text{D}\tilde{T}_{0}=0$ and $l=l_{T}=0$ .

4 Numerical results

In order to calculate the first positive eigenvalue $R_{1}$ of the eigenvalue equation (2.36)–(2.41) for various base flows, we first discretise the equation using 129 Chebyshev–Gauss–Lobatto collocation points, and then solve it with the QZ function in MATLAB (Dongarra, Straughan & Walker Reference Dongarra, Straughan and Walker1996). The first positive eigenvalue $R_{1}$ is found to decrease with increasing $\unicode[STIX]{x1D703}$ for any given $k$ in the energy stability analysis of the plane Couette flow, the plane Poiseuille flow, the shear flow with a cubic velocity profile and the inclined buoyancy layer when $\unicode[STIX]{x1D706}=1$ (figure 3). In figure 3, we plot the contours of $R_{1}$ in the $(k,\unicode[STIX]{x1D703})$ plane, instead of plotting them in the usual wavenumber plane $(\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD})$ as in the previous works (Reddy & Henningson Reference Reddy and Henningson1993; Xiong & Tao Reference Xiong and Tao2017). Although Sagalakov & Shtern (Reference Sagalakov and Shtern1971) also plotted their figure 2 in the $(k,\unicode[STIX]{x1D703})$ plane, the contour $R_{1}=35$ in their figure is slightly different from that in our figure 3(c), implying that $R_{1}$ increases with increasing $\unicode[STIX]{x1D703}$ at $(k,\unicode[STIX]{x1D703})\approx (1.9,\unicode[STIX]{x03C0}/2)$ . Noticing that their contour $R_{1}=35$ and the boundary $\unicode[STIX]{x1D703}=\unicode[STIX]{x03C0}/2$ do not intersect at a right angle at $(k,\unicode[STIX]{x1D703})\approx (1.9,\unicode[STIX]{x03C0}/2)$ , we believe our result is more accurate. Actually, we have shown $\unicode[STIX]{x2202}R_{1}/\unicode[STIX]{x2202}\unicode[STIX]{x1D703}=0$ at the boundaries $\unicode[STIX]{x1D703}=0$ and $\unicode[STIX]{x1D703}=\unicode[STIX]{x03C0}/2$ in the proof of theorem 2.

Figure 3.

The contours of the first positive eigenvalue $R_{1}$ for the energy stability of various base flows. (a) The plane Couette flow with the no-slip boundary condition ( $\text{D}U_{0}=1$ , $Pr\,\text{D}\tilde{T}_{0}=0$ and $l=0$ ) (Reddy & Henningson Reference Reddy and Henningson1993). (b) The plane Poiseuille flow with the no-slip boundary condition ( $\text{D}U_{0}=-2y$ , $Pr\,\text{D}\tilde{T}_{0}=0$ and $l=0$ ) (Reddy & Henningson Reference Reddy and Henningson1993). (c) The plane Couette flow with the slip boundary condition ( $\text{D}U_{0}=(1+l)^{-1}$ , $Pr\,\text{D}\tilde{T}_{0}=0$ and $l=0.1$ ). (d) The plane Poiseuille flow with the slip boundary condition ( $\text{D}U_{0}=-2(1+3l)^{-1}y$ , $Pr\,\text{D}\tilde{T}_{0}=0$ and $l=0.1$ ). (e) The shear flow with a cubic velocity profile ( $\text{D}U_{0}=1-3y^{2}$ , $Pr\,\text{D}\tilde{T}_{0}=0$ and $l=0$ ) (Sagalakov & Shtern Reference Sagalakov and Shtern1971). (f) The inclined buoyancy layer ( $\text{D}U_{0}=-0.5L^{2}\text{e}^{-y^{\prime }}(\sin y^{\prime }-\cos y^{\prime })$ , $Pr=0.72$ , $\text{D}\tilde{T}_{0}=-0.5L^{2}\text{e}^{-y^{\prime }}(\sin y^{\prime }+\cos y^{\prime })$ , $\unicode[STIX]{x1D706}=1$ and $l=l_{T}=0$ , where $y^{\prime }=L(y+1)$ and $L=20$ ) (Xiong & Tao Reference Xiong and Tao2017).

Under the no-slip boundary condition and the slip boundary condition with the slip length $l=0.1$ , the least stable modes in the energy stability analysis of the plane Couette flow have $(k,\unicode[STIX]{x1D703})=(1.558,\unicode[STIX]{x03C0}/2)$ (figure 3 a) and $(k,\unicode[STIX]{x1D703})=(1.426,\unicode[STIX]{x03C0}/2)$ (figure 3 c), respectively. These wavenumbers $k$ are close to the corresponding lower bounds $k_{0}(0,0)=1.534$ and $k_{0}(0.1,0)=1.411$ given in theorem 1 (figure 2).

When there is temperature variation in the base flow ( $Pr\,\text{D}\tilde{T}_{0}\neq 0$ ), the first positive eigenvalue $R_{1}$ may increase with $\unicode[STIX]{x1D703}$ , e.g. when $\text{D}U_{0}=\cos (\unicode[STIX]{x03C0}y)+\cos (3\unicode[STIX]{x03C0}y)$ , $Pr\,\text{D}\tilde{T}_{0}=10\sin (\unicode[STIX]{x03C0}y)$ , $\unicode[STIX]{x1D706}=1$ and $l=l_{T}=0$ (figure 4 a). Figure 4(b) shows that theorem 2 still holds for this base flow.

Figure 4.

The contours of $R_{1}$ (a) and $(1+\cos \unicode[STIX]{x1D703})R_{1}$ (b) for the energy stability of the mixed convection with $\text{D}U_{0}=\cos (\unicode[STIX]{x03C0}y)+\cos (3\unicode[STIX]{x03C0}y)$ , $Pr\,\text{D}\tilde{T}_{0}=10\sin (\unicode[STIX]{x03C0}y)$ , $\unicode[STIX]{x1D706}=1$ and $l=l_{T}=0$ .

For shear flows under the no-slip boundary condition ( $l=0$ ) and without variations in temperature ( $Pr\,\text{D}\tilde{T}_{0}=0$ ), we define

(4.1) $$\begin{eqnarray}J(k,\unicode[STIX]{x1D703},\boldsymbol{a})=-\frac{1}{R_{1}}\frac{\text{d}R_{1}}{\text{d}\unicode[STIX]{x1D703}}=-\frac{\displaystyle \int _{-1}^{1}(\text{D}U_{0})(\bar{\hat{v}}_{1}{\hat{w}}_{1}+\hat{v}_{1}\bar{{\hat{w}}}_{1})\,\text{d}y}{\displaystyle \int _{-1}^{1}(\text{D}U_{0})(\bar{\hat{u} }_{1}\hat{v}_{1}+\hat{u} _{1}\bar{\hat{v}}_{1})\,\text{d}y},\end{eqnarray}$$

where we have used (3.31). Here $\text{D}U_{0}=\sum _{m\geqslant 0}a_{m}T_{m}(y)$ , $T_{m}$ is the Chebyshev polynomial of degree $m$ , and $\boldsymbol{a}=(a_{0},a_{1},\ldots )$ is the coefficient. We also define

(4.2) $$\begin{eqnarray}J_{min}(k,\unicode[STIX]{x1D703})=\min _{\boldsymbol{a}}J(k,\unicode[STIX]{x1D703},\boldsymbol{a}).\end{eqnarray}$$

Note that $R_{1}$ and $\text{D}U_{0}$ only appear in the form of $R_{1}\text{D}U_{0}$ in (2.36)–(2.41), provided that $Pr\,\text{D}\tilde{T}_{0}=0$ . The eigenvector $\hat{\boldsymbol{u}}_{1}$ therefore does not change when $\text{D}U_{0}$ is multiplied by a non-zero real constant, and then $J(k,\unicode[STIX]{x1D703},C\boldsymbol{a})=J(k,\unicode[STIX]{x1D703},\boldsymbol{a})$ for any real constant $C\neq 0$ .

We use two methods to approximate $J_{min}$ at $(k_{j},\unicode[STIX]{x1D703}_{s})$ , where $k_{j}=10^{-1+(j-1)/5}$   $(1\leqslant j\leqslant 11)$ and $\unicode[STIX]{x1D703}_{s}=(s/20)\unicode[STIX]{x03C0}$ $(1\leqslant s\leqslant 9)$ . In the following calculations, 65 Chebyshev–Gauss–Lobatto collocation points are used. In the first calculation, for given wavenumber pair $(k_{j},\unicode[STIX]{x1D703}_{s})$ , $J_{min}$ is approximated by the minimum of $J(k_{j},\unicode[STIX]{x1D703}_{s},\boldsymbol{a})$ among all $\boldsymbol{a}$ such that $|a_{m}|\leqslant 5$ $(0\leqslant m\leqslant 5)$ are integers and $a_{m}=0$ for $m\geqslant 6$ (figure 5 a). In the second calculation, $J_{min}$ is approximated with a gradient descent method among all $\boldsymbol{a}$ such that $a_{m}$ $(0\leqslant m\leqslant 15)$ are real numbers and $a_{m}=0$ for $m\geqslant 16$ . The minimum found with the gradient descent method is generally a local minimum, and depends on the initial value of the coefficient $\boldsymbol{a}$ . The result shown in figure 5(b) is the minimum of $J$ calculated from 100 sets of initial $\boldsymbol{a}$ that are randomly chosen. The detail of the gradient descent algorithm will be introduced in appendix B. In both calculations, we did not find $J<0$ for any base flow at any $(k_{j},\unicode[STIX]{x1D703}_{s})$ ( $1\leqslant j\leqslant 11$ and $1\leqslant s\leqslant 9$ ), which means $R_{1}$ is probably a decreasing function of $\unicode[STIX]{x1D703}$ for the base flows that we have examined. Note that $J=0$ when $\unicode[STIX]{x1D703}=0$ or $\unicode[STIX]{x1D703}=\unicode[STIX]{x03C0}/2$ , according to the proof of theorem 2.

Figure 5.

The contours of $J_{min}$ in the wavenumber plane approximated with two methods: (a) $J_{min}$ is approximated by the minimum of $J$ for all $\text{D}U_{0}=\sum _{0\leqslant m\leqslant 5}a_{m}T_{m}(y)$ , where $T_{m}$ is the Chebyshev polynomial of degree $m$ , and $|a_{m}|\leqslant 5$ $(0\leqslant m\leqslant 5)$ are integers; (b) $J_{min}$ is approximated by the local minimum of $J$ among all $\text{D}U_{0}=\sum _{0\leqslant m\leqslant 15}a_{m}T_{m}(y)$ with a gradient descent method.

5 Conjecture

In the parallel shear flows without variations in temperature, we have $\text{D}T_{0}=0$ and $Gr=0$ , and then $\text{D}\tilde{T}_{0}=0$ due to (2.24). Under this circumstance, $Pr$ , $\unicode[STIX]{x1D706}$ and $l_{T}$ have no influence on the eigenvalue equation (2.36)–(2.41). According to the numerical results in § 4, we propose the following conjecture:

If conjecture 1 is correct, the least stable mode for the energy stability of any parallel shear flow between no-slip walls without variations in temperature will be a streamwise vortex, because

(5.1) $$\begin{eqnarray}Re_{E}=\min _{k>0}R_{1}\left(k,\frac{\unicode[STIX]{x03C0}}{2}\right)\end{eqnarray}$$

follows from (2.42), and $\unicode[STIX]{x1D703}=\unicode[STIX]{x03C0}/2$ implies the streamwise wavenumber $\unicode[STIX]{x1D6FC}=k\cos \unicode[STIX]{x1D703}=0$ .

Conjecture 1 implies that

(5.2) $$\begin{eqnarray}\frac{\unicode[STIX]{x2202}R_{1}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}\leqslant 0\end{eqnarray}$$

in each subset $S\subset (0,+\infty )\times [0,\unicode[STIX]{x03C0}/2]$ where the geometric multiplicity of the first eigenvalue $R_{1}(k,\unicode[STIX]{x1D703})$ is 1.

According to (3.9), (3.11) and (3.17), conjecture 1 is equivalent to the following conjecture:

Conjecture 2 (A conjecture equivalent to conjecture 1).

For any continuously differentiable real function $\text{D}U_{0}$ , if $\text{D}\tilde{T}_{0}=0$ , $l=0$ , $k>0$ and $0\leqslant \unicode[STIX]{x1D703}\leqslant \unicode[STIX]{x03C0}/2$ , then the eigenvector corresponding to the first positive eigenvalue of (2.36)–(2.41) satisfies

(5.3) $$\begin{eqnarray}-\int _{-1}^{1}(\bar{\hat{q}}_{1}\hat{\unicode[STIX]{x1D702}}_{1}+\hat{q}_{1}\bar{\hat{\unicode[STIX]{x1D702}}}_{1})\,\text{d}y\geqslant 0.\end{eqnarray}$$

Now the monotonicity of the eigenvalue in conjecture 1 has been transformed to a more tractable inequality of the eigenvector in conjecture 2, where more mathematical tools can be used in the proof, such as the lemmas used in the proof of theorem 1.

Using (3.15) instead of (3.11), we have another condition which is equivalent to (5.3),

(5.4) $$\begin{eqnarray}\frac{R_{1}}{2}\int _{-1}^{1}(\text{D}U_{0})(\bar{\hat{v}}_{1}{\hat{w}}_{1}+\hat{v}_{1}\bar{{\hat{w}}}_{1})\,\text{d}y\geqslant 0.\end{eqnarray}$$

When $\text{D}\tilde{T}_{0}=0$ , we have $\hat{T}_{1}=0$ from (2.39) and (2.41), and then $I_{T}=0$ . The condition (5.3) is therefore also equivalent to

(5.5) $$\begin{eqnarray}(1+\cos ^{2}\unicode[STIX]{x1D703})I_{\unicode[STIX]{x1D702}}-(\sin ^{2}\unicode[STIX]{x1D703})I_{v}\geqslant 0\end{eqnarray}$$

after using (3.34) instead of (3.11).