The solution for stress free boundaries has been discussed in [1, 3, 9]. Here we only emphasize the solution aspects which have not been discussed before. We solve the eigenvalue problem with two stress free boundaries to study the onset of instability in the ferromagnetic fluid. We consider a small variation in time scale *τ* = *ϵ*^{2}*t* such that stationary convection occurs at lower orders of *ϵ* and introduce the following asymptotic expansions

$$R{a}^{\frac{1}{2}}={\left(R{a}^{\alpha}\right)}^{\frac{1}{2}}+{\in}^{2}R{a}_{2}^{{}^{\frac{1}{2}}}+{\in}^{4}R{a}_{4}^{{}^{\frac{1}{2}}}+\dots ,$$(27)$$w=\in {w}_{1}+{\in}^{2}{w}_{2}+{\in}^{3}{w}_{3}+\dots \text{\hspace{0.17em}},$$(28)$$\zeta =\in {\zeta}_{1}+{\in}^{2}{\zeta}_{2}+{\in}^{3}{\zeta}_{3}+\dots \text{\hspace{0.17em}},$$(29)$$\theta =\in {\theta}_{1}+{\in}^{2}{\theta}_{2}+{\in}^{3}{\theta}_{3}+\dots ,$$(30)$$\varphi =\in {\varphi}_{1}+{\in}^{2}{\varphi}_{2}+{\in}^{3}{\varphi}_{3}+\cdots .$$(31)Substituting Eqs. (27)–(31) into Eqs. (21)–(24), at the lowest order of *ϵ* we obtain

$$B{Z}_{1}={R}_{1},$$(32)where

$$B=\left(\begin{array}{cccc}-{\nabla}^{4}& -\sqrt{Ta\frac{\partial}{\partial z}}& -\sqrt{R{a}^{\alpha}\left(1-{M}_{1}\right){\nabla}_{l}^{2}}& \sqrt{R{a}^{\alpha}\left(1-{M}_{1}\right){\nabla}_{l}^{2}}\\ -\sqrt{Ta\frac{\partial}{\partial z}}& -{\nabla}^{2}& 0& 0\\ -\sqrt{R{a}^{\alpha}\left(1-{M}_{2}\right)}& 0& -{\nabla}^{2}& 0\\ 0& 0& -\frac{\partial}{\partial z}& \frac{{\partial}^{2}}{\partial {z}^{2}}+{M}_{3}{\nabla}_{l}^{2}\end{array}\right),$$${z}_{1}=\left(\begin{array}{c}{w}_{1}\\ {\zeta}_{1}\\ {\theta}_{1}\\ {\varphi}_{1}\end{array}\right)\text{and}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{R}_{1}\text{\hspace{0.17em}}=\left(\begin{array}{c}0\\ 0\\ 0\\ 0\end{array}\right)\text{\hspace{0.17em}}.$ This equation corresponds

to the linear equations in [1, 9]. Solving Eq. (32) we obtain the solution

$${w}_{1}\phantom{\rule{thinmathspace}{0ex}}=A\left(\tau \right)\mathrm{sin}ax\phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}\phantom{\rule{thinmathspace}{0ex}}\pi z,$$(33)$${\zeta}_{1}=-\frac{T{a}^{\frac{1}{2}}\pi}{{\pi}^{2}+{a}^{2}}A\left(\tau \right)\mathrm{sin}\text{\hspace{0.17em}}ax\text{\hspace{0.17em}}\mathrm{sin}\pi z,$$(34)$${\theta}_{1}=\frac{{\left(R{a}^{\alpha}\right)}^{\frac{1}{2}}\left(1-{M}_{2}\right)}{{\pi}^{2}+{a}^{2}}A\left(\tau \right)\mathrm{sin}ax\mathrm{cos}\text{\hspace{0.17em}}\pi z,$$(35)$$\begin{array}{rl}{\varphi}_{1}& =-\left(\frac{(R{a}^{\alpha}{)}^{\frac{1}{2}}(1-{M}_{2})\pi}{({\pi}^{2}+{a}^{2})({\pi}^{2}+{a}^{2}{M}_{3})}\right)\end{array}$$(36)*A*(*τ*) sin *ax* sin *πz*,

where *a* is a dimensionless wave number. Thus, the stationary Rayleigh number is given as

$$R{a}^{\alpha}=\frac{\left({\left({\pi}^{2}+{a}^{2}\right)}^{3}+{\pi}^{2}Ta\right)\left({\pi}^{2}+{a}^{2}{M}_{3}\right)}{\left(1-M2\right)\left({a}^{2}{\pi}^{2}{M}_{1}+{a}^{2}\left({M}_{1}+1\right)\left({\pi}^{2}+{a}^{2}{M}_{3}\right)\right)}.$$(37)To find the critical wave number and the corresponding critical Rayleigh number we set *a*^{2} = *π*^{2}*x*. Then the stationary Rayleigh number can be written as

$$R{a}^{\alpha}\text{\hspace{0.17em}}=\frac{\left({\pi}^{4}{\left(1+x\right)}^{3}\right)+Ta\left(1+x{M}_{3}\right)}{\left(1-{M}_{2}\right)+\left(1+x{M}_{3}\left({M}_{1}+1\right)\right)}.$$(38)This result agrees with [1, 3, 9]. Since M_{2} is very small as indicated in [1, 9] it can be neglected in the subsequent analysis.

The critical wave number and the corresponding critical Rayleigh number are obtained from

$$\frac{\partial R{a}^{\alpha}}{\partial x}\text{\hspace{0.17em}}=0$$then we have

$$\begin{array}{l}-Ta-2Ta\left(x{M}_{3}+x{M}_{1}{M}_{3}\right)\hfill \\ -Ta\left({x}^{2}{M}_{3}^{2}+{x}^{2}{M}_{1}{M}_{{}_{3}}^{2}\right)\hfill \\ +{\pi}^{4}{\left(1+x\right)}^{2}(-1+2x-2x{M}_{3}-2x{M}_{1}{M}_{3}-4{x}^{2}{M}_{3}\hfill \\ +{x}^{2}{M}_{3}^{2}{M}_{1}+2{x}^{3}{M}_{3}^{2}+2{x}^{3}{M}_{1}{M}_{3})=0\hfill \end{array}$$(39)For the case *Ta* = 0, *M*_{1} = 0 and *M*_{3} = 0, the classical critical wave number is

$${a}_{c}=\frac{\pi}{\sqrt{2}}$$with corresponding classical critical Rayleigh number

$$R{a}_{c}^{\alpha}\text{\hspace{0.17em}}=\frac{27}{4}{\pi}^{4}.$$The magnetic Rayleigh number is also of an interest and can be expressed as

$${N}^{\alpha}\text{\hspace{0.17em}}=R{a}^{\alpha}{M}_{1}\text{\hspace{0.17em}}=\frac{\left({\pi}^{4}{\left(1+x\right)}^{3}+Ta\right){M}_{1}\left(1+x{M}_{3}\right)}{x\left(1+x{M}_{3}\left({M}_{1}+1\right)\right)}.$$(40)For large values of *M*_{1} the magnetic Rayleigh number in the absence of buoyancy effects is obtained as

$${N}^{\alpha}=\frac{\left({\pi}^{4}{\left(1+x\right)}^{3}+Ta\right)\left(1+x{M}_{3}\right)}{{x}^{2}{M}_{3}}.$$(41)The critical wave number and corresponding critical magnetic Rayleigh number are obtained from solving the equation

$$-2Ta+M3x+{\pi}^{4}{\left(1+x\right)}^{2}\left(-2+x-M3x+2M3{x}^{2}\right)=0.$$(42)The critical wave number and corresponding Rayleigh numbers are given for different values of the Taylor number *Ta* in and .

At second order *O*(*ϵ*^{2}) we obtain the following equations:

$$B{Z}_{2}={R}_{2}$$(43)where ${Z}_{2}=\left(\begin{array}{c}{w}_{2}\\ \zeta \\ {\theta}_{2}\\ \varphi \end{array}\right)\text{and}\text{\hspace{0.17em}}{\text{R}}_{2}=\left(\begin{array}{c}{R}_{21}\\ {R}_{22}\\ {R}_{23}\\ {R}_{24}\end{array}\right)$with

$${R}_{21}=0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}{R}_{22\text{\hspace{0.17em}}}=0,\text{\hspace{0.17em}}{R}_{24\text{\hspace{0.17em}}}=0,$$(44)Table 1 Comparison of the critical wave number and corresponding Rayleigh number for *M*_{1} = 1 and *M*_{3} = 5.

Table 2 Comparison of the critical wave number and corresponding Magnetic Rayleigh number for *M*_{1} *→* ∞and *M*_{3} = 3.

$${R}_{23}=-\frac{\mathrm{Pr}\pi {\left(R{a}^{\alpha}\right)}^{\frac{1}{2}}}{2\left({\pi}^{2}+{a}^{2}\right)}{A}^{2}\left(\tau \right){\mathrm{sin}}^{2}ax\text{\hspace{0.17em}}\mathrm{sin}2\pi z.$$The solution at the second order is

$${w}_{2}=0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\zeta}_{2}\text{\hspace{0.17em}}=0,$$(45)$$\begin{array}{l}{\theta}_{2}=-\left(\frac{\mathrm{Pr}\pi {\left(R{a}^{\alpha}\right)}^{\frac{1}{2}}}{4\left({\pi}^{2}+{a}^{2}\right)\left(2{\pi}^{2}+{a}^{2}\right)}\right)\\ {A}^{2}\left(\tau \right){\mathrm{sin}}^{2}ax\text{\hspace{0.17em}}\mathrm{sin}2\pi z,\end{array}$$(46)$${\varphi}_{2}=-\left(\frac{\mathrm{Pr}{\pi}^{2}{\left(R{a}^{\alpha}\right)}^{\frac{1}{2}}}{4\left(2{\pi}^{2}+{a}^{2}\right)\left({\pi}^{2}+{a}^{2}\right)\left(2{\pi}^{2}+{a}^{2}{M}_{3}\right)}\right)$$(47)*A*^{2}(*τ*) sin^{2} *ax* cos 2*πz*.

At the third order, we obtain

$$B{Z}_{3}\text{\hspace{0.17em}}={R}_{3},$$(48)where ${Z}_{3}=\left(\begin{array}{c}{w}_{3}\\ {\zeta}_{3}\\ {\theta}_{3}\\ {\varphi}_{3}\end{array}\right)\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}{\text{R}}_{2}=\left(\begin{array}{c}{R}_{31}\\ {R}_{32}\\ {R}_{33}\\ {R}_{34}\end{array}\right)$with

$$\begin{array}{l}{R}_{31}=\{-{\gamma}^{2}\frac{dA}{d\tau}+\frac{{a}^{2}{\left(R{a}^{\alpha}\right)}^{\frac{1}{2}}{R}_{{a}^{2}}^{\frac{1}{2}}\left(1+{M}_{1}\right)}{{\gamma}^{2}\left({\pi}^{2}+{a}^{2}{M}_{3}\right)}\\ \left(\left(1+{M}_{1}\right){\pi}^{2}+{a}^{2}{M}_{3}\right)A\left(\tau \right)\}\mathrm{sin}ax\mathrm{cos}\pi z,\end{array}$$(49)$${R}_{32}=-\frac{T{a}^{\frac{1}{2}}{\pi}^{2}}{{\gamma}^{2}}\frac{dA}{d\tau}\mathrm{sin}ax\mathrm{sin}\text{\hspace{0.17em}}\pi z,$$(50)$$\begin{array}{l}{R}_{33}=-(\left({\left(R{a}^{\alpha}\right)}^{\frac{1}{2}}\frac{\partial F}{\partial z}-R{a}_{2}^{\frac{1}{2}}\right)A\left(\tau \right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+\frac{\mathrm{Pr}{\left(R{a}^{\alpha}\right)}^{\frac{1}{2}}}{{\gamma}^{2}}\frac{dA}{d\tau})\mathrm{sin}ax\mathrm{cos}\pi z+\frac{{\mathrm{Pr}}^{2}{\pi}^{2}{\left(R{a}^{\alpha}\right)}^{\frac{1}{2}}}{2{\gamma}^{2}\left(2{\pi}^{2}+{a}^{2}\right)}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{A}^{3}\left(\tau \right){\mathrm{sin}}^{3}ax\text{\hspace{0.17em}}\mathrm{cos}\pi z\text{\hspace{0.17em}}\mathrm{cos}2\pi z,\end{array}$$(51)$${R}_{34}=0.$$(52)Here, *γ*^{2} = *a*^{2} + *π*^{2}. To obtain the Ginzburg Landau equation we applied the Fredholm solvability condition [11, 18]

$$\underset{0}{\overset{\frac{1}{2}}{\int}}{\displaystyle \underset{0}{\overset{\frac{2\pi}{a}}{\int}}\left[{\widehat{w}}_{1}{R}_{31}+{\widehat{\zeta}}_{1}{R}_{32}+\widehat{\theta}{R}_{33}\right]}\text{\hspace{0.17em}}dxdz=0$$(53)where ŵ_{1}, *ζ̂*_{1} and *θ̂*_{1} are the solutions of the adjoint system of the first order. This gives

$$\begin{array}{l}\left[\frac{\pi {\gamma}^{2}}{4a}+\frac{\mathrm{Pr}R{a}^{\alpha}\pi}{4{\gamma}^{2}}-\frac{Ta{\pi}^{4}}{4a{\gamma}^{4}}\right]\frac{dA}{d\tau}=\{\frac{\pi R{a}^{\alpha \frac{1}{2}}R{a}_{2}^{\frac{1}{2}}}{4a{\gamma}^{2}}\\ -\frac{\pi R{a}^{\alpha}}{a{\gamma}^{2}}I+\frac{a\pi R{a}^{\alpha \frac{1}{2}}{R}_{2}^{\frac{1}{2}}\left(1+{M}_{1}\right)}{4{\gamma}^{2}\left({\pi}^{2}+{a}^{2}{M}_{3}\right)}\\ \left(\left(1+{M}_{1}\right){\pi}^{2}+{a}^{2}{M}_{3}\right)\}A-\frac{3{\mathrm{Pr}}^{2}{\pi}^{2}R{a}^{\alpha}}{64a{\gamma}^{4}\left(2{\pi}^{2}+{a}^{2}\right)}{A}^{3}.\end{array}$$(54)The above equation reduces to

$$\frac{dA}{d\tau}={\Delta}_{1}A-{\Delta}_{2}{A}^{3\text{\hspace{0.17em}}},$$(55)where ${\Delta}_{1}={\gamma}^{2}/\left({\gamma}^{6}+\mathrm{Pr}R{a}^{\alpha}-Ta{\pi}^{3}\right)$

$$\begin{array}{l}\left(R{a}^{*}-4R{a}^{\alpha}I+\frac{{a}^{2}R{a}^{*}\left(1+{M}_{1}\right)\left(\left(1+{M}_{1}\right){\pi}^{2}+{a}^{2}{M}_{3}\right)}{{\pi}^{2}+{a}^{2}{M}_{3}}\right),\\ R{a}^{*}=R{a}^{\alpha \frac{1}{2}}\text{\hspace{0.17em}}R{a}_{2}^{\frac{1}{2}}\text{\hspace{0.17em}},{\Delta}_{2}=3{\mathrm{Pr}}^{2}{\pi}^{3}\text{\hspace{0.17em}}R{a}^{\alpha}/\left(\left(2{\pi}^{2}+{a}^{2}\right)\right),\end{array}$$and

$$I={\displaystyle \underset{0}{\overset{\frac{1}{2}}{\int}}\frac{dF}{dz}{\mathrm{cos}}^{2}\left(\pi z\right)}dz.$$In this study we are also interested in heat transfer in ferromagnetic fluids. The Nusselt number for ferromagnetic fluids is defined as 2 1

$$\begin{array}{l}Nu\left(\tau \right)=\frac{\text{Heat}\text{\hspace{0.17em}}\text{transfer}\text{\hspace{0.17em}}\text{by}\text{\hspace{0.17em}}\text{conduction}\text{\hspace{0.17em}}\text{+}\text{\hspace{0.17em}}\text{convection}\text{\hspace{0.17em}}}{\text{Heat}\text{\hspace{0.17em}}\text{transfer}\text{\hspace{0.17em}}\text{by}\text{\hspace{0.17em}}\text{conduction}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=1+\frac{\mathrm{Pr}{\pi}^{2}{\left(R{a}^{\alpha}\right)}^{\frac{1}{2}}}{4{\gamma}^{2}\left(2{\pi}^{2}+{a}^{2}\right)}{A}^{2}\left(\tau \right).\end{array}$$(56)
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