# Holographic Anisotropic Background in 5D Einstien-Gauss-Bonnet Gravity

S. N. Sajadi
School of Physics, Institute for Research in Fundamental Sciences (IPM),
P. O. Box 19395-5531, Tehran, Iran
Electronic address<EMAIL_ADDRESS>
###### Abstract

In this paper, we extend the work on the AdS/QCD model to quadratic gravity to
gain insight into the influence of gravity. We obtain an anisotropic black
brane solution to a $5D$ Einstein-Gauss-Bonnet-two Maxwell-dilaton system. The
background is specified by an arbitrary exponent, a dilaton field, a time
component of the first Maxwell field, and a magnetic component of the second
Maxwell field. The system in three cases has been investigated and in each
case the effect of the parameter of theory, the anisotropic parameter has been
considered. The blackening function supports the thermodynamical phase
transition between small/large and AdS/large black brane for a suitable
chemical potential and other parameters.

##  1 Introduction

Quantum chromodynamics (QCD) is a non-abelian gauge theory that describes the
strong interaction between quarks and gluons. QCD at low temperatures exhibits
confinement whereas at high temperature undergoes a phase transition to a
chiral symmetry. The investigation and understanding of the phase diagram of
QCD and the search for new phases of matter are of attracting attention in the
theoretical and experimental communities. The gauge/gravity duality provided
another way to further understand the dynamics of the strong-couple system,
where standard methods do not work [1], [2]. The quark-gluon plasma (QGP) is
one such system created in a short time in heavy ion collisions, it is
believed to be anisotropic during this time [3], [4]. Therefore, various
properties of QCD have been investigated in an anisotropic background [5]. In
[6], [7] the confinement-deconfinement phase transition in the framework of
the Einstein-dilaton-Maxwell theory for the isotropic case has been studied.
In [8], the confinement-deconfinement phase transition in the framework of
$5D$ Einstein-dilaton two-Maxwell theory with an anisotropic background has
been studied. In [9], the authors extended the work of [8], by introducing a
background magnetic field to gain insight into the influence of such field on
QCD observables.
Higher-order gravitational models have recently received attention [10]-[14],
in part because string theory predicts that at low energies Einstein’s
equations are subject to first-order corrections [15]. In AdS/CFT context,
higher-order gravities have been used as tools to characterize numerous
properties of strongly coupled conformal field theories [16]-[18]. From
quantum gravity viewpoint, in order to unify quantum mechanics and
gravitational interactions, going beyond the Einstein gravity is necessary
[19]. The first correction of Lovelock gravity to the Einstein-Hilbert action
appears in five and higher dimensions and is given by a precise combination of
quadratic curvature terms yields the second-order field equations known as the
Gauss-Bonnet density [20], [21], [22]. Cosmological models, including in the
inflation, and in the framework of Brane cosmology have been well studied in
this theory [23]. Black hole solutions of the theory have been studied in
[24]-[27]. The thermodynamics of black holes has also been studied in the
framework of this theory [28]. The Gauss-Bonnet term in 4D gives a non-zero
contribution to the field equations in the presence of the dilatonic scalar
field $\phi$ [29], [30], [31]. In this paper, we extend the work of [8] to the
Einstein Quadratic Gravity, which is general relativity extended by quadratic
curvature invariants in the action to find the effect of higher derivative
terms on QCD.
The paper is organized as follows. In section 2 we construct the anisotropic
$5$-dimensional solution with an arbitrary dynamical exponent, an exponential
quadratic warp function, a non-zero time component of the first Maxwell field
and a non-zero magnetic component of the second Maxwell field in the framework
of EGB gravity. In section 2.1 first we consider zero warp function and obtain
the exact solution for blackening function and other unknown quantities. We
have shown the behavior of the quantities with plots and we discuss the
thermodynamics of the constructed background. In section 2.2, we consider
exponential quadratic warp function and zero chemical potential and solved the
differential equations approximatly and show that for negative exponential
warp function the dilaton field is real. Then, we discuss the thermodynamics
of the constructed background and find out the small/large phase transition.
In section 2.3 we consider the non-zero warp function and non-zero chemical
potential and obtained the approximatly solution for the unknown functions. In
this case we study the thermodynamics of the black brane and find out the
small/large and AdS/large phase transitions. We finish the paper with some
concluding remarks in section 3.

##  2 Basic Formalism

We consider a $5$-dimensional Einstein-quadratic-dilaton-two-Maxwell system.
The action of the system in the Einstein frame is specified as [8]

$S=\dfrac{1}{16\pi G_{5}}\int d^{5}x\sqrt{-g}L,$ (1)

where the Lagrangian is

$L=R+\gamma R_{abcd}R^{abcd}+\beta R_{ab}R^{ab}+\alpha
R^{2}-\dfrac{1}{4}f_{1}(\phi)F_{(1)}^{2}-\dfrac{1}{4}f_{2}(\phi)F_{(2)}^{2}-\dfrac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi-V(\phi),$
(2)

and $F_{(i)}^{2}=F_{\mu\nu}F^{\mu\nu}$, $\phi$ is the dilaton field,
$f_{1}(\phi)$ and $f_{2}(\phi)$ are the gauge functions representing the
coupling between the two $U(1)$ gauge fields on one hand and the dilaton on
the other hand. $V(\phi)$ is the potential of the dilaton field, and $G_{5}$
is the Newton constant in five dimensions. ($\alpha,\beta,\gamma$) are
coupling constants of theory. We use the metric ansatz $g_{\mu\nu}$, dilaton
field $\phi$ and field strength tensor $F_{(i)}^{\mu\nu}$ in the following
form:

$ds^{2}=\dfrac{l^{2}b(z)}{z^{2}}\left(-g(z)dt^{2}+\dfrac{dz^{2}}{g(z)}+dx^{2}+P(z)(dy_{1}^{2}+dy_{2}^{2})\right),$
(3)

with

$A^{(1)}_{\mu}=A_{t}(z)\delta^{0}_{\mu},\;\;\;\;\;F_{(2)}=qdy^{1}\wedge
dy^{2},\;\;\;\;\phi=\phi(z),$ (4)

where $b(z)$ is the warp function, $g(z)$ is the metric function and $l$ is
the AdS length scale. $z=0$ corresponds to the boundary of the $5d$ spacetime.
The first gauge field ($F^{(1)}$) is the electric part of the Maxwell tensor
which causes the black hole to become electrically charged. In relation (11),
we relate the charge of the black hole to the chemical potential of the dual
quantum field system. The second gauge field ($F^{(2)}$) is the magnetic part
of the Maxwell tensor on a plane $y_{1}y_{2}$ and causes the anisotropy of the
metric spatial part. The variation of the action (1) over metric $g_{\mu\nu}$,
the scalar field $\phi$ and $A_{t}$ gives the field equations as follows

$\displaystyle
E_{\mu\nu}=G_{\mu\nu}+\alpha\left[2R(R_{\mu\nu}-\frac{1}{4}g_{\mu\nu}R)+2(g_{\mu\nu}\square-\nabla_{\mu}\nabla_{\nu})R\right]+\beta\left[(g_{\mu\nu}\square-\nabla_{\mu}\nabla_{\nu})R+\square
G_{\mu\nu}+2R^{\lambda\rho}(R_{\mu\lambda\nu\rho}\right.$
$\displaystyle\left.-\frac{1}{4}g_{\mu\nu}R_{\lambda\rho})\right]+\gamma\left[-\frac{1}{2}g_{\mu\nu}R_{\alpha\beta\gamma\eta}R^{\alpha\beta\gamma\eta}+2R_{\mu\lambda\rho\sigma}R_{\nu}{}^{\lambda\rho\sigma}+4R_{\mu\lambda\nu\rho}R^{\lambda\rho}-4R_{\mu\sigma}R^{\sigma}_{\nu}+4\square
R_{\mu\nu}-2\nabla_{\mu}\nabla_{\nu}R\right]$
$\displaystyle=\dfrac{1}{2}f_{(i)}\left(F^{(i)}_{\mu\rho}F_{\nu}^{(i)\rho}-\dfrac{1}{4}g_{\mu\nu}F^{2(i)}\right)+\dfrac{1}{2}\left(\partial_{\mu}\phi\partial_{\nu}\phi-\dfrac{1}{2}g_{\mu\nu}(\partial\phi)^{2}-g_{\mu\nu}V\right),\;\;\;\;\;\;\;\nabla^{2}\phi=\dfrac{\partial
V}{\partial\phi}+\dfrac{1}{4}\dfrac{\partial
f_{(i)}}{\partial\phi}(F^{2(i)}),$
$\displaystyle\nabla_{\mu}\left(f_{(i)}F^{\mu\nu(i)}\right)=0,\;\;\;\;\;\;\;\;\;(i=1,2)$
(5)

where $G_{\mu\nu}$ is the Einstien tensor. Using the ansatz of the metric, the
Maxwell fields and the dilaton field (2), it is easy to obtain the equations
of motion for the background fields. The explicit components of the field
equation are large and bulky and we have not included them here. The field
equations for $\phi$ and $A_{t}$ are given by:

$\displaystyle-P^{2}z^{4}f_{1}^{\prime}h^{\prime
2}+q^{2}z^{4}\dfrac{df_{2}}{d\phi}+2b^{2}l^{4}P^{2}\dfrac{dV}{d\phi}-3l^{2}P^{2}gz^{2}\phi^{\prime}b^{\prime}-2l^{2}bP^{2}z^{2}g\phi^{\prime\prime}+6l^{2}zbP^{2}g\phi^{\prime}$
$\displaystyle-2l^{2}z^{2}bP^{2}\phi^{\prime}g^{\prime}=0,$ (6) $\displaystyle
f_{1}A_{t}^{\prime}b^{\prime}zP-2f_{1}A_{t}^{\prime}bP+2f_{1}bPA_{t}^{\prime\prime}z+2bPzf^{\prime}_{1}A_{t}^{\prime}=0,$
(7)

where prime is differential with respect to $z$. One can check that the
equation of motion for the second Maxwell field will not give any additional
equation. To find the solution for the field equations, we assume [8]

$b(z)=e^{-\frac{cz^{2}}{2}},\;\;\;\;\;f_{1}=z^{-2+\frac{2}{\nu}},\;\;\;\;P(z)=z^{2-\frac{2}{\nu}},$
(8)

where $\nu$ is a parameter that specified the anisotropic backgrounds. To
solve the background, we also impose the boundary conditions in the form

${b(0)=1},\;\;\;\;g(0)=1,\;\;\;\;g(z_{h})=0,\;\;\;\;A_{t}(0)=\mu,\;\;\;\;A_{t}(z_{h})=0$
(9)

where $z_{h}$ is the horizon and $\mu$ is the chemical potential of the
boundary theory. The boundary conditions are used to fix the integration
constants. Now, we are going to solve the field equations. First, by solving
the differential equation (7), one can get

$A_{t}(z)=\mathfrak{c}_{1}+\mathfrak{c}_{2}e^{\frac{cz^{2}}{4}},$ (10)

where

$\displaystyle\mathfrak{c}_{1}=\dfrac{\mu
e^{\frac{cz_{h}^{2}}{4}}}{-1+e^{\frac{cz_{h}^{2}}{4}}},\;\;\;\;\mathfrak{c}_{2}=\dfrac{\mu}{-1+e^{\frac{cz_{h}^{2}}{4}}}.$
(11)

By inserting solution (10) into the equation $E_{tt}$, one can obtain
$V(\phi)$. Then, by inserting $V(\phi)$ into the field equation $E_{xx}$, one
can obtain $f_{2}(\phi)$. By inserting $f_{2}(\phi)$ and $V(\phi)$ into
$E_{zz}$ one can obtain $\phi^{\prime}$. Finally, from equation
$E_{y_{1}y_{1}}$, the differential equation for $g(z)$ obtains as follows

$\displaystyle
4\nu^{4}z^{4}l(4\alpha+\beta)e^{-\frac{cz^{2}}{4}}g(z)g^{\prime\prime\prime\prime}-4\nu^{3}lz^{3}e^{-\frac{cz^{2}}{4}}[(4\alpha+\beta)(\nu
cz^{2}-2\nu+4)g-2z\nu(\alpha-\gamma)g^{\prime}]g^{\prime\prime\prime}$
$\displaystyle-6\nu
zle^{-\frac{cz^{2}}{4}}[-4l^{2}z\nu^{3}e^{-\frac{cz^{2}}{2}}-4\nu^{2}z^{2}g^{\prime}(20\gamma-4\alpha+4\beta+(10\alpha+6\beta+14\gamma)\nu+c\nu
z^{2}(5\beta+3\alpha+17\gamma))$ $\displaystyle+\nu
g(c^{2}\nu^{2}z^{4}(5\beta+24\gamma)+4c\nu
z^{2}(16\gamma+2\beta-4\alpha+\nu(4\alpha+\beta))+\nu^{2}(20\beta+32\alpha+48\gamma)+16\nu(\beta+4\gamma)$
$\displaystyle+48\gamma-32\alpha)]g^{\prime\prime}-\nu
zle^{-\frac{cz^{2}}{4}}[l^{2}\nu^{2}e^{-\frac{cz^{2}}{2}}(4\nu+6\nu
cz^{2}+8)-g((3\beta+12\gamma+2\alpha)c^{3}\nu^{3}z^{6}+c^{2}\nu^{2}z^{4}(80\gamma+$
$\displaystyle 16\alpha+20\beta-(19\beta+16\alpha+72\gamma)+4c\nu
z^{2}(3(\beta+2\alpha+2\gamma)\nu^{2}-\nu(12\alpha+4\beta+8\gamma)+12\alpha+10\beta+38\gamma)$
$\displaystyle+4(\nu+2)(\nu^{2}(5\beta+8\alpha+12\gamma)+\nu(16\gamma-8\alpha+2\beta)+4\beta+12\gamma+8\alpha)))]g^{\prime}-4z^{4}\nu^{3}l(2\gamma+\beta+2\alpha)$
$\displaystyle e^{-\frac{cz^{2}}{4}}g^{\prime\prime 2}-\nu
z^{2}le^{-\frac{cz^{2}}{4}}g^{\prime
2}(112\alpha+32\beta+32\gamma+(128\alpha+32\beta+8cz^{2}(18\alpha+5\beta+4\gamma))\nu+(48\alpha+20\beta$
$\displaystyle+32\gamma-16cz^{2}(3\alpha+\beta+\gamma)+c^{2}z^{4}(40\alpha+8\gamma+11\beta))\nu^{2})-c^{2}\mathfrak{c}_{2}^{2}\nu^{3}le^{\frac{cz^{2}}{4}}z^{\frac{4\nu+2}{\nu}}=0.$
(12)

For generic coupling constant $\alpha$, $\beta$ and $\gamma$ this fourth order
differential equation analytically cannot be solved, therefore we consider the
case where $\gamma=\alpha,\beta=-4\alpha$. In this case, the theory reduced to
Einstien-Gauss-Bonnet gravity (EGB), and the field equation (2), reduced to
second order differential equation for metric function $g(z)$ as follows

$\displaystyle-4zl((2+\nu cz^{2})^{2}z\nu g(z)\alpha
e^{-\frac{cz^{2}}{4}}-z\nu^{3}l^{2}e^{-\frac{3cz^{2}}{4}})g^{\prime\prime}-4\nu
z^{2}l\alpha(2+\nu cz^{2})^{2}e^{-\frac{cz^{2}}{4}}g^{\prime 2}-$
$\displaystyle 2zl((2+\nu cz^{2})(z^{4}c^{2}\nu^{2}-6cz^{2}\nu^{2}+6\nu
cz^{2}+4\nu+8)g(z)\alpha e^{-\frac{cz^{2}}{4}}+2l^{2}\nu^{2}(4+2\nu+3\nu
cz^{2})e^{-\frac{3cz^{2}}{4}})g^{\prime}$
$\displaystyle-c^{2}\nu^{3}l\mathfrak{c}_{2}^{2}z^{\frac{4\nu+2}{\nu}}e^{\frac{cz^{2}}{4}}=0.$
(13)

In the following, we solve the above differential equations in special cases:

###  2.1 The case $c=0$

In this case the warp function $b(z)=1$ and the field equation for $g(z)$
becomes:

$z\nu(l^{2}\nu^{2}-4\alpha g(z))g^{\prime\prime}-g^{\prime}(4\nu\alpha
zg^{\prime}+(2+\nu)(l^{2}\nu^{2}-4\alpha g))=0,$ (14)

one can exactly solve it and obtain analytic solution for $g(z)$ as

$\displaystyle g(z)=$
$\displaystyle\dfrac{1}{4\alpha(1+\nu)}[l^{2}\nu^{2}+l^{2}\nu^{3}-$
$\displaystyle\sqrt{l^{4}\nu^{4}+2l^{4}\nu^{5}+l^{4}\nu^{6}-4\nu^{2}\alpha
c_{1}z^{\frac{2(1+\nu)}{\nu}}-16\nu\alpha c_{2}-8\nu^{2}\alpha c_{2}-4\alpha
c_{1}\nu z^{\frac{2(\nu+1)}{\nu}}-8\alpha c_{2}}],$ (15)

and by taking into account the boundary conditions (9), we get

$c_{1}=-\dfrac{2(\nu^{3}l^{2}+\nu^{2}l^{2}-2\alpha\nu-2\alpha)}{\nu
z_{h}^{\frac{2(\nu+1)}{\nu}}},\;\;\;\;\;c_{2}=\nu^{2}l^{2}-2\alpha.$ (16)

For $\alpha\ll 1$, the metric function is given as

$g(z)\approx
1-\left(\dfrac{z}{z_{h}}\right)^{\frac{2(\nu+1)}{\nu}}-\dfrac{2\alpha}{\nu^{2}l^{2}}\left(\dfrac{z}{z_{h}}\right)^{\frac{2(\nu+1)}{\nu}}\left(1-\left(\dfrac{z}{z_{h}}\right)^{\frac{2(\nu+1)}{\nu}}\right)+\mathcal{O}(\alpha^{2}).$
(17)

The second term is the correction from the Gauss-Bonnet gravity and in the
case of $\alpha\to 0$ the metric function goes to [8] for Einstein gravity.

(a) $\nu=4.5,l=1,z_{h}=2$ (b) $\alpha=0.1,l=1,z_{h}=2$ (c)
$\alpha=0.1,l=1,\nu=4.5$

Figure 1: Plot of $g(z)$ in terms of $z$ for
$\alpha={\color[rgb]{1,0,0}0},{\color[rgb]{0,0,1}0.5},1,{\color[rgb]{0,1,0}1.5},{\color[rgb]{1,.5,0}2}$
(left),
$\nu={\color[rgb]{1,0,0}1},{\color[rgb]{0,0,1}2},3,{\color[rgb]{0,1,0}4},{\color[rgb]{1,.5,0}5},6$
(middle) and
$z_{h}={\color[rgb]{1,0,0}1},{\color[rgb]{0,0,1}2},3,{\color[rgb]{0,1,0}4},{\color[rgb]{1,.5,0}5},6$
(right).

The behavior of the metric function is depicted in Fig.(1). The main feature
is that the metric function values decrease faster for larger $\alpha$
(Fig.1a). In the isotropic case ($\nu=1$) the metric function values are
larger than in the anisotropic ones ($\nu\neq 1$) (Fig.1b). In this panel by
increasing $\nu$ the metric function values decrease faster. Changing the
values of $\alpha$ and $\nu$ does not influence the horizon position. In the
following we look at the behavior of Ricci and Kretschmann scalar
$K=R_{abcd}R^{abcd}$ of the black brane. The Ricci scalar is given as follows

$\displaystyle R$
$\displaystyle=\dfrac{1}{2\alpha\nu^{2}l^{2}\left((\nu^{2}l^{2}-4\alpha)^{2}z_{h}^{\frac{2\nu+2}{\nu}}+8\alpha(\nu^{2}l^{2}-2\alpha)z^{\frac{2\nu+2}{\nu}}\right)^{\frac{3}{2}}}[-l^{2}\nu^{2}(4\nu+3(\nu^{2}+1))$
$\displaystyle\left((\nu^{2}l^{2}-4\alpha)^{2}z_{h}^{\frac{2\nu+2}{\nu}}+8\alpha(\nu^{2}l^{2}-2\alpha)z^{\frac{2\nu+2}{\nu}}\right)^{\frac{3}{2}}(3(\nu^{2}+1)+4\nu)(\nu^{2}l^{2}-4\alpha)^{4}z_{h}^{\frac{3\nu+3}{\nu}}$
$\displaystyle+4(\nu^{2}l^{2}-2\alpha)((\nu^{2}l^{2}-4\alpha)^{2}(9\nu^{2}+11\nu+10)(z_{h}z^{2})^{\frac{\nu+1}{\nu}}+8\alpha(\nu^{2}l^{2}-2\alpha)(2\nu^{2}$
$\displaystyle+\nu+3)z_{h}^{\frac{-\nu-1}{\nu}}z^{\frac{4\nu+4}{\nu}})].$ (18)

The scalars are smooth inside the black hole and start to diverge for
$z>z_{h}$. In larger $\alpha$ it happens earlier for $R$, while for $K$ it
happens earlier for smaller $\alpha$.

(a) $\nu=4.5,l=1,z_{h}=2$ (b) $\nu=4.5,l=1,z_{h}=2$

Figure 2: Plot of Ricci scalar and Kretschmann scalar in terms of $z$ for
$\alpha={\color[rgb]{1,0,0}0.1},{\color[rgb]{0,0,1}0.2},0.3,{\color[rgb]{0,1,0}0.4},{\color[rgb]{1,.5,0}0.5}$.

By inserting (2.1) in $E_{tt}$, one can obtain $\phi$ as follows:

$\displaystyle\phi(z)=\dfrac{-2\sqrt{-2\mathcal{K}(\nu-1)}\left(\sqrt{\mathcal{H}\sqrt{\mathcal{D}}}\arctan\left(\sqrt{\frac{\mathcal{G}}{\mathcal{F}}}\right)-\mathcal{L}\sqrt{\mathcal{F}}\arctan\left(\sqrt{\frac{\mathcal{G}}{\mathcal{D}}}\right)-2\sqrt{\mathcal{DFG}(l^{4}\nu^{4}-8\alpha
c_{2})}\right)}{(\nu+1)l\sqrt{\mathcal{GFE}\nu}\sqrt{l^{4}\nu^{4}-8\alpha
c_{2}}\sqrt{l^{2}\nu^{3}+l^{2}\nu^{2}-\sqrt{\mathcal{E}}}}+c_{3},$ (19)

by imposing the condition $\phi(z_{h})=0$, one can obtain $c_{3}=0$. The
constants $\mathcal{K},\mathcal{H},...$ are provided in (Constants). In the
case of $\alpha\ll 1$, one can get

$\displaystyle\phi(z)$
$\displaystyle=\dfrac{2\sqrt{\nu-1}}{\nu}\ln\left(\dfrac{z}{z_{h}}\right)-\dfrac{4\alpha\sqrt{\nu-1}}{l^{2}\nu^{2}(\nu+1)}\left[(\nu+1)\ln\left(\dfrac{z}{z_{h}}\right)+\dfrac{\nu}{2}\left(1-\left(\dfrac{z}{z_{h}}\right)^{\frac{2\nu+2}{\nu}}\right)\right]+\mathcal{O}(\alpha^{2}).$
(20)

The first term is the contribution of the Einstien term and the second term is
from the Gauss-Bonnet term. In figure (3), the real and imaginary parts of the
scalar field in terms of $z$ for different values of parameters have been
shown. As can be seen the imaginary part of scalar field inside and outside
the black brane has a non-zero value and is unstable. By increasing $\nu$ in
$0<z<z_{h}$, the real part and imaginary part of the scalar field increase and
decrease respectively and for $z>z_{h}$ vice versa (Fig.(3)a). In panel b, by
increasing the coupling of theory in $0<z<z_{h}$ the real and imaginary parts
decrease and increase respectively, and for $z>z_{h}$ vice versa.

(a) $\alpha=0.1,l=1,z_{h}=2$ (b) $\alpha=0.1,l=1,\nu=4.5$ (c)
$z_{h}=2,\alpha=0.1,l=1$

Figure 3: Plots of imagenary (solid lines) and real part (dashed lines) of
$\phi$ in terms of $z$ for
$\nu={\color[rgb]{1,0,0}2},{\color[rgb]{0,0,1}3},4,{\color[rgb]{0,1,0}5},{\color[rgb]{1,.5,0}6}$
(left), for
$\alpha={\color[rgb]{1,0,0}0.1},{\color[rgb]{0,0,1}0.2},0.3,{\color[rgb]{0,1,0}0.4},{\color[rgb]{1,.5,0}0.5}$
(middle), for
$z_{h}={\color[rgb]{1,0,0}2},{\color[rgb]{0,0,1}3},4,{\color[rgb]{0,1,0}5},{\color[rgb]{1,.5,0}6}$
(right).

By inserting (20) into $E_{xx}$, one can obtain $f_{2}$ as follows:

$\displaystyle f_{2}$
$\displaystyle=-\dfrac{2}{\nu^{3}q^{2}\alpha(\nu+1)\bar{\mathcal{A}}^{\frac{5}{2}}(\nu^{2}l^{2}(\nu+1)-\sqrt{\bar{\mathcal{A}}})}[-6\nu(\nu-1)(\nu+1)^{5}\alpha
c_{1}\bar{\mathcal{C}}(\nu^{4}l^{4}-8\alpha c_{2})(\nu l^{2}(\nu-\frac{1}{2})$
$\displaystyle\sqrt{\bar{\mathcal{A}}}-\bar{\mathcal{K}})z^{\frac{2\nu-2}{\nu}}-8\alpha^{3}c_{1}^{3}\bar{\mathcal{C}}\nu^{2}(2\nu^{2}-5\nu-2+\nu^{3})(\nu+1)^{3}z^{\frac{2+6\nu}{\nu}}+2z^{4}\nu(\nu+1)^{4}\alpha^{2}c_{1}^{2}\bar{\mathcal{C}}(\nu^{2}l^{2}\sqrt{\bar{\mathcal{A}}}$
$\displaystyle(2\nu^{2}+5-5\nu)-\bar{\mathcal{F}})+z^{-\frac{4}{\nu}}[\frac{1}{2}(\nu^{2}l^{2}(\nu+1)^{2}\bar{\mathcal{E}}\bar{\mathcal{A}}^{\frac{3}{2}})+\frac{1}{2}(\nu^{2}l^{2}(\nu^{2}-1)\bar{\mathcal{B}}\bar{\mathcal{A}}^{\frac{5}{2}})+\nu
l^{2}(\nu-1)(\nu+1)^{6}$
$\displaystyle(\nu-\frac{1}{2})\bar{\mathcal{C}}\sqrt{\bar{\mathcal{A}}}(\nu^{4}l^{4}-8\alpha
c_{2})^{2}-\nu^{15}l^{8}\bar{\mathcal{C}}\bar{\mathcal{G}}-5\nu^{14}l^{8}\bar{\mathcal{C}}\bar{\mathcal{G}}-9\nu^{13}l^{8}\bar{\mathcal{C}}\bar{\mathcal{G}}-5\nu^{12}l^{8}\bar{\mathcal{C}}\bar{\mathcal{G}}+5\nu^{11}l^{4}\bar{\mathcal{C}}\bar{\mathcal{G}}(l^{4}+$
$\displaystyle\frac{16}{5}\alpha
c_{2})+9\nu^{10}l^{4}\bar{\mathcal{C}}\bar{\mathcal{G}}(l^{4}+\frac{80}{9}\alpha
c_{2})+5\nu^{9}l^{4}\bar{\mathcal{C}}\bar{\mathcal{G}}(l^{4}+\frac{144}{5}\alpha
c_{2})+l^{4}\nu^{8}\bar{\mathcal{C}}\bar{\mathcal{G}}(l^{4}+80\alpha
c_{2})-80\alpha\nu^{7}c_{2}\bar{\mathcal{C}}\bar{\mathcal{G}}$
$\displaystyle(l^{4}+\frac{4}{5}\alpha c_{2})-144\alpha
c_{2}\nu^{6}\bar{\mathcal{C}}\bar{\mathcal{G}}(l^{4}+\frac{20}{9}\alpha
c_{2})-80\alpha\nu^{5}c_{2}\bar{\mathcal{C}}\bar{\mathcal{G}}(l^{4}+\frac{36}{5}\alpha
c_{2})-16\alpha\nu^{4}c_{2}\bar{\mathcal{C}}\bar{\mathcal{G}}(l^{4}+20\alpha
c_{2})$
$\displaystyle+320\alpha^{2}\nu^{3}c_{2}^{2}\bar{\mathcal{C}}\bar{\mathcal{G}}+\nu^{2}\bar{\mathcal{L}}+\nu\bar{\mathcal{J}}+\frac{1}{2}\bar{\mathcal{B}}\bar{\mathcal{A}}^{3}+64\alpha^{2}c_{2}^{2}\bar{\mathcal{C}}\bar{\mathcal{G}}]],$
(21)

where the constants $\bar{\mathcal{A}}$, $\bar{\mathcal{B}}$… are provided in
(Constants). In the case of $\alpha\ll 1$, one can get

$\displaystyle f_{2}$
$\displaystyle=\dfrac{4l^{2}(\nu^{2}-1)z^{-\frac{4}{\nu}}}{\nu^{2}q^{2}}-\dfrac{8(\nu^{2}-1)z^{4}\alpha}{q^{2}\nu^{4}\left(z^{\frac{2\nu+2}{\nu}}-z_{h}^{\frac{2\nu+2}{\nu}}\right)^{2}}\times$
$\displaystyle\left[-4\nu\left(\dfrac{z_{h}}{z}\right)^{\frac{2\nu+2}{\nu}}+3\left(\dfrac{z}{z_{h}}\right)^{\frac{4\nu+4}{\nu}}-6\left(\dfrac{z}{z_{h}}\right)^{\frac{2\nu+2}{\nu}}+\nu\left(\dfrac{z_{h}}{z}\right)^{\frac{4\nu+4}{\nu}}+(2\nu+3)\right]+\mathcal{O}(\alpha^{2}).$
(22)

The first terms is the contribution of Einstien gravity and the second term is
related to the Gauss-Bonnet gravity. In figure (4), the behavior of $f_{2}$ in
terms of $z$ are shown. As can be seen, by increasing $\alpha$ and $q$,
$f_{2}$ decreases and by increasing $\nu$, $f_{2}$ increases.

(a) $\nu=4.5,l=1,z_{h}=2,q=0.5$ (b) $\nu=4.5,l=1,z_{h}=2,\alpha=0.1$ (c)
$l=1,\alpha=0.1,q=0.5,z_{h}=2$

Figure 4: Plots of $f_{2}$ in terms of $z$ for
$\alpha={\color[rgb]{1,0,0}0.1},{\color[rgb]{0,0,1}0.2},0.3,{\color[rgb]{0,1,0}0.4},{\color[rgb]{1,.5,0}0.5}$
(left),
$q={\color[rgb]{1,0,0}0.1},{\color[rgb]{0,0,1}0.2},0.3,{\color[rgb]{0,1,0}0.4},{\color[rgb]{1,.5,0}0.5}$
(middle),
$\nu={\color[rgb]{1,0,0}1},{\color[rgb]{0,0,1}2},3,{\color[rgb]{0,1,0}4},{\color[rgb]{1,.5,0}5}$
(right).

Finally from $E_{tt}$, one can get $V(z)$ (We did not bring it here due to its
bulk). In the case of $\alpha\ll 1$, one can get

$\displaystyle V(z)$
$\displaystyle=-\dfrac{2(\nu+1)(2\nu+1)}{\nu^{2}l^{2}}+\dfrac{\alpha}{\left(z^{\frac{2\nu+2}{\nu}}-z_{h}^{\frac{2\nu+2}{\nu}}\right)^{3}\nu^{4}l^{4}}[-4z^{\frac{4\nu+4}{\nu}}z_{h}^{\frac{2\nu+2}{\nu}}(-1-7\nu+6\nu^{3}+20\nu^{2})$
$\displaystyle+8\nu(3\nu^{2}+7\nu-1)z^{\frac{2\nu+2}{\nu}}z_{h}^{\frac{4\nu+4}{\nu}}+(8\nu^{3}+64\nu^{2}-36\nu-12)z^{\frac{6\nu+6}{\nu}}+4(\nu-1)(2\nu+1)z^{\frac{10\nu+10}{\nu}}z_{h}^{\frac{-4\nu-4}{\nu}}$
$\displaystyle-4(\nu-1)(8\nu+3)z^{\frac{8\nu+8}{\nu}}z_{h}^{\frac{-2\nu-2}{\nu}}-8\nu^{2}(\nu+2)z_{h}^{\frac{6\nu+6}{\nu}}]+\mathcal{O}(\alpha^{2}).$
(23)

In Fig. (5), the behavior of scalar potential in terms of $z$ for anisotropic
case has been shown. In the left panel, between $0<z<z_{h}$ by increasing
$\alpha$, the potential decreases. In the right panel, between $0<z<z_{h}$ by
increasing $\nu$ the scalar potential increases.

(a) $\nu=4.5,z_{h}=2,l=1$ (b) $\nu=4.5,l=1,z_{h}=2$

Figure 5: Plots of $V$ in terms of $z$ for
$\alpha={\color[rgb]{1,0,0}0.1},{\color[rgb]{0,0,1}0.2},0.3,{\color[rgb]{0,1,0}0.4},{\color[rgb]{1,.5,0}0.5}$
(left),
$\nu={\color[rgb]{1,0,0}1},{\color[rgb]{0,0,1}2},3,{\color[rgb]{0,1,0}4},{\color[rgb]{1,.5,0}5}$
(right).

####  2.1.1 Thermodynamics of the background

In this subsection, we explore the thermodynamics of the black brane solution
(2.1). In order to investigate the thermodynamic properties of the black brane
we need to obtain some relevant thermodynamic quantities. The temperature of
the black brane is obtained as follows:

$\displaystyle T=\left|\dfrac{g^{\prime}}{4\pi}\right|=\dfrac{\nu+1}{2\pi
z_{h}\nu}\left(1-\dfrac{2\alpha}{\nu^{2}l^{2}}\right).$ (24)

It is noticed that, the temperature monotonically decreases with the increase
of the horizon. By increasing $\alpha$, the temperature is decreased, and for
$\alpha=0$, one can get the result of [8]. The entropy is given as follows
[32],[33]

$\displaystyle S=-\dfrac{1}{8}\int_{\Sigma}d^{n-2}x\sqrt{\eta}\dfrac{\delta
L}{\delta R_{\mu\alpha\beta\nu}}\epsilon_{\mu\alpha}\epsilon_{\beta\nu},$ (25)

where

$\displaystyle\dfrac{\delta L}{\delta R_{\mu\alpha\beta\nu}}=$
$\displaystyle\left(\dfrac{1}{2}+\alpha
R\right)\left(g^{\mu\beta}g^{\alpha\nu}-g^{\mu\nu}g^{\alpha\beta}\right)+\dfrac{1}{2}\beta\left(R^{\mu\beta}g^{\alpha\nu}-R^{\alpha\beta}g^{\mu\nu}-R^{\mu\nu}g^{\alpha\beta}+R^{\alpha\nu}g^{\mu\beta}\right)$
$\displaystyle+2\gamma R^{\mu\alpha\beta\nu},$ (26)

and $\epsilon_{\mu\nu}=-2\sqrt{-h}\delta^{t}_{[\mu}\delta^{z}_{\nu]}$. For
$c=0$ we have

$s=\dfrac{S}{\mathcal{V}}=\dfrac{l^{3}P(z_{h})b(z_{h})^{\frac{3}{2}}}{4z_{h}^{3}}=\dfrac{l^{3}}{4z_{h}^{\frac{\nu+2}{\nu}}},$
(27)

which is independent of parameter of the EGB gravity. In terms of the
temperature, the entropy is given as

$s=\dfrac{l^{3}}{4}\left(\dfrac{2\pi
T\nu^{3}l^{2}}{(\nu+1)(\nu^{2}l^{2}-2\alpha)}\right)^{\frac{\nu+2}{\nu}}.$
(28)

For isotropic case $s\approx T^{3}$ and for anisotropic case $s\approx
T^{\frac{\nu+1}{\nu}}$. The free energy density $F(T)$ can be calculated from
the entropy density $s(T)$ by integrating as follows

$F=\int sdT=\dfrac{\nu}{2(\nu+1)}Ts,$ (29)

which is related to temperature as $F\approx T^{\frac{2\nu+1}{\nu}}$. The
sound velocity $c_{s}^{2}$ which can directly measure the conformality of the
system, can be obtained from the temperature and entropy:

$c_{s}^{2}=\dfrac{d\log T}{d\log s}=\dfrac{\nu}{\nu+2}.$ (30)

For isotropic case ($\nu=1$), $c_{s}^{2}=1/3$, the system is conformal, for
anisotropic ($\nu\neq 1$), $c_{s}^{2}\neq 1/3$ the system is non-conformal.
The heat capacity is given as

$C_{V}=T\dfrac{ds}{dT}=\dfrac{(\nu+2)}{\nu}s=\dfrac{s}{c_{s}^{2}}.$ (31)

In terms of temperature,

$\dfrac{C_{V}}{T^{3}}=\dfrac{l^{3}(\nu+2)}{4\nu}\left(\dfrac{2\pi\nu^{3}l^{2}}{(\nu+1)(\nu^{2}l^{2}-2\alpha)}\right)^{\frac{\nu+2}{\nu}}T^{\frac{2-2\nu}{\nu}}.$
(32)

For isotropic case, the right hand side of (32) has a constant value and for
$\nu>1$ depends to the temperature and at high temperature goes to zero. Since
entropy is positive therefore $C_{V}$ is positive and the black hole is
stable. In figure 6, the behavior of $s$, $F$ and $C_{V}$ in terms of $T$ for
isotropic (dashed lines) and anisotropic (solid lines) and different values of
$\alpha$ have been shown. As can be seen, by increasing $\alpha$, the
thermodynamical quantities $s$, $F$, and $C_{V}$ increase. For
$T<T^{i}_{cross}$, the entropy, free energy, and heat capacity of anisotropic
case is larger than isotropic, and for $T>T^{i}_{cross}$ vice versa. Where
$i=s,F,C_{V}$ and $T^{i}_{cross}$ are given as follows

(a) $l=1$ (b) $l=1$ (c) $l=1$

Figure 6: Plots of $s$, $F$ and $C_{V}$ in terms of $T$ for
$\alpha={\color[rgb]{1,0,0}0.1},{\color[rgb]{0,0,1}0.2},0.3,{\color[rgb]{0,1,0}0.4},{\color[rgb]{1,.5,0}0.5}$
and $\nu=4.5$ (solid line) and $\nu=1$ (dashed lines) .

$\displaystyle T^{s}_{cross}=$
$\displaystyle\left[\dfrac{l^{4}((l^{2}-2\alpha)^{3}\nu^{\frac{3(\nu+2)}{\nu}}\pi^{\frac{2(1-\nu)}{\nu}}2^{\frac{\nu+2}{\nu}})^{\nu}}{l^{4\nu}((\nu+1)(l^{2}\nu^{2}-2\alpha))^{2}(\nu+1)(l^{2}\nu^{2}-2\alpha)^{\nu}}\right]^{\frac{1}{2\nu-2}},$
(33) $\displaystyle T_{cross}^{F}=$
$\displaystyle\dfrac{\left(\dfrac{(l^{2}-2\alpha)^{2}4^{\frac{\nu+1}{\nu}}\pi^{\frac{2-2\nu}{\nu}}}{\nu^{\frac{-3\nu-6}{\nu}}+\nu^{\frac{-4\nu-6}{\nu}}}\right)^{\frac{\nu}{2(\nu-1)}}}{[(\nu+1)(l^{2}\nu^{2}-2\alpha)^{\nu+2}l^{4(\nu-1)}]^{\frac{1}{2\nu-2}}},$
(34) $\displaystyle T^{C_{V}}_{cross}=$
$\displaystyle\left[\dfrac{l^{4(\nu+1)}((l^{2}\nu^{2}-2\alpha)^{3}(\nu+2)\nu^{\frac{2(\nu+1)}{\nu}}\pi^{\frac{2(1-\nu)}{\nu}}2^{\frac{\nu+2}{\nu}})^{\nu}}{3^{\nu}(\nu+1)(l^{2}\nu^{2}-2\alpha)^{\nu+2}}\right]^{\frac{1}{2\nu-2}}.$
(35)

###  2.2 The case $c\neq 0,\mu=0$

In this case $A_{t}=0$ and the differential equation (2), becomes

$\displaystyle 4\nu
lz^{2}e^{-\frac{cz^{2}}{4}}(l^{2}\nu^{2}e^{-\frac{cz^{2}}{2}}-\alpha(2+\nu
cz^{2})^{2}g(z))g^{\prime\prime}-2zle^{-\frac{cz^{2}}{4}}(\nu^{2}l^{2}(4+(2+3z^{2})\nu)-$
$\displaystyle\alpha(2+\nu cz^{2})(8+4\nu+6\nu
cz^{2}-6c\nu^{2}z^{2}+c^{2}\nu^{2}z^{4})g)g^{\prime}-4\alpha\nu
z^{2}le^{-\frac{cz^{2}}{4}}(2+\nu cz^{2})^{2}g^{\prime 2}=0.$ (36)

In order to solve equation (2.2), we assume $g(z)$ as follows

$g(z)=1+\epsilon g_{1}(z)+\mathcal{O}(\epsilon^{2}),$ (37)

by inserting it into the (2.2), one can achieve a homogeneous differential
equation for $g_{1}(z)$ as

$g_{1}^{\prime\prime}+\dfrac{[-l^{2}\nu^{3}(2+3cz^{2})-4\nu^{2}(l^{2}+4\alpha
cz^{2})+4\alpha\nu(2+cz^{2})+16\alpha]g_{1}^{\prime}}{\nu
z(l^{2}\nu^{2}-4\alpha)}=0.$ (38)

Solving (38) give $g_{1}(z)$ as

$\displaystyle
g_{1}(z)=c_{1}+c_{2}\text{erf}\left(\dfrac{1}{2}\sqrt{\dfrac{c(-6l^{2}\nu^{2}-32\nu\alpha+8\alpha)}{(l^{2}\nu^{2}-4\alpha)}}z\right),$
(39)

where $c_{1}$ and $c_{2}$ are constants of integration. Using (39), the metric
(37) becomes

$g(z)=1+c_{1}+c_{2}\text{erf}\left(\dfrac{1}{2}\sqrt{\dfrac{c(-6l^{2}\nu^{2}-32\nu\alpha+8\alpha)}{(l^{2}\nu^{2}-4\alpha)}}z\right).$
(40)

The conditions (9) give us $c_{1}$ and $c_{2}$ as

$c_{1}=0,\;\;\;\;c_{2}=-\dfrac{1}{\text{erf}\left(\dfrac{1}{2}\sqrt{\dfrac{c(-6l^{2}\nu^{2}-32\nu\alpha+8\alpha)}{(l^{2}\nu^{2}-4\alpha)}}z_{h}\right)}.$
(41)

Finally, using (41) the metric function becomes

$g(z)=1-\dfrac{\text{erf}\left(\dfrac{1}{2}\sqrt{\dfrac{c(-6l^{2}\nu^{2}-32\nu\alpha+8\alpha)}{(l^{2}\nu^{2}-4\alpha)}}z\right)}{\text{erf}\left(\dfrac{1}{2}\sqrt{\dfrac{c(-6l^{2}\nu^{2}-32\nu\alpha+8\alpha)}{(l^{2}\nu^{2}-4\alpha)}}z_{h}\right)}.$
(42)

In the case of $\alpha\ll 1$ and $c<0$, the blakening function become

$g(z)\approx
1-\dfrac{\text{erf}\left(\dfrac{\sqrt{-6c}z}{2}\right)}{\text{erf}\left(\dfrac{\sqrt{-6c}z_{h}}{2}\right)}-\dfrac{4(2\nu+1)\sqrt{-6c}\alpha\left(ze^{\frac{3cz^{2}}{2}}\text{erf}\left(\dfrac{\sqrt{-6c}z_{h}}{2}\right)-z_{h}\text{erf}\left(\dfrac{\sqrt{-6c}z}{2}\right)e^{\frac{3cz_{h}^{2}}{2}}\right)}{3\sqrt{\pi}l^{2}\nu^{2}\text{erf}\left(\dfrac{\sqrt{-6c}z_{h}}{2}\right)^{2}}+\mathcal{O}\left(\alpha^{2}\right),$
(43)

and for $\alpha\ll 1$ and $c>0$, we have

$g(z)\approx
1-\dfrac{z}{z_{h}}e^{\frac{3c(z^{2}-z_{h}^{2})}{2}}-\dfrac{2\alpha
c(2\nu+1)}{\nu^{2}l^{2}}\dfrac{z(z^{2}-z_{h}^{2})}{z_{h}}e^{\frac{3c(z^{2}-z_{h}^{2})}{2}}+\mathcal{O}(\alpha^{2}).$
(44)

In figure (7), the behavior of $g(z)$ for positive and negative values of warp
function is depicted. As $c$ increases, the metric slope becomes more
decreasing. Also, changing the value of $c$ has no effect on the value of the
horizon. By substituting the obtained metric (42), we arrive at the
differential equation for the scalar field as:

(a) $\nu=4.5,\alpha=0.1,l=1,z_{h}=2$ (b) $\alpha=0.1,l=1,z_{h}=2$

Figure 7: Plots of $g(z)$ in terms of $z$ for
$c={\color[rgb]{1,0,0}0.5},{\color[rgb]{0,0,1}0.4},0.3,{\color[rgb]{0,1,1}0.2},{\color[rgb]{1,.5,0}0.1}$
(dashed lines) and
${\color[rgb]{1,0,0}-0.5},{\color[rgb]{0,0,1}-0.4},-0.3,{\color[rgb]{0,1,0}-0.2},{\color[rgb]{1,.5,0}-0.1}$
(solid lines) (left), $\nu=1,2,3,4,5,6$, ${\color[rgb]{1,0,0}c=0.5}$ and
${\color[rgb]{0,0,1}c=-0.5}$ (right).

$\displaystyle\phi^{\prime 2}$
$\displaystyle=\dfrac{e^{\frac{cz^{2}}{2}}}{z^{2}l^{2}\nu^{3}{2\pi(l^{2}\nu^{2}-4\alpha)\text{erf}\left(\frac{\sqrt{-2c\mathbb{A}}z_{h}}{2}\right)\left[\text{erf}\left(\frac{\sqrt{-2c\mathbb{A}}z_{h}}{2}\right)-\text{erf}\left(\frac{\sqrt{-2c\mathbb{A}}z}{2}\right)\right]}}[2\alpha\sqrt{-2c\pi\mathbb{A}}z(2+\nu
cz^{2})$
$\displaystyle\left[-\text{erf}\left(\frac{\sqrt{-2c\mathbb{A}}z_{h}}{2}\right)+\text{erf}\left(\frac{\sqrt{-2c\mathbb{A}}z}{2}\right)\right]e^{\frac{\mathbb{A}cz^{2}}{2}}(\nu^{2}c^{2}z^{4}(5l^{2}\nu^{2}+32\alpha\nu-4\alpha)-4(\nu+2)(l^{2}\nu^{2}-4\alpha)$
$\displaystyle+2\nu
cz^{2}(3\nu^{3}l^{2}+20\alpha\nu+3l^{2}\nu^{2}+4\alpha))-8c\alpha\nu
z^{2}(2+\nu
cz^{2})^{2}(3l^{2}\nu^{2}+16\alpha\nu-4\alpha)e^{\mathbb{A}cz^{2}}$
$\displaystyle+2l^{2}\nu^{2}z\sqrt{-2c\pi\mathbb{A}}\text{erf}\left(\frac{\sqrt{-2c\mathbb{A}}z_{h}}{2}\right)e^{\frac{cz^{2}\nu(\nu
l^{2}+8\alpha)}{\nu^{2}l^{2}-4\alpha}}(\nu
cz^{2}(3l^{2}\nu^{2}+32\alpha\nu+4\alpha)-2(\nu+2)(l^{2}\nu^{2}-4\alpha))$
$\displaystyle-\pi(l^{2}\nu^{2}-4\alpha)\left(\text{erf}(\frac{\sqrt{-2c\mathbb{A}}z}{2})-\text{erf}(\frac{\sqrt{-2c\mathbb{A}}z_{h}}{2})\right)(l^{2}\nu
e^{-\frac{cz^{2}}{2}}\text{erf}(\frac{\sqrt{-2c\mathbb{A}z_{h}}}{2})(-8+8\nu+3\nu^{2}c^{2}z^{4}$
$\displaystyle+18\nu^{2}cz^{2})+\alpha(2+\nu
cz^{2})(\text{erf}(\frac{\sqrt{-2c\mathbb{A}}z}{2})-\text{erf}(\frac{\sqrt{-2c\mathbb{A}}z_{h}}{2}))(-16+16\nu+3\nu^{2}z^{6}c^{3}+2\nu
c^{2}z^{4}(1+$ $\displaystyle 11\nu)+4cz^{2}(6\nu^{2}-2+5\nu)))]$ (45)

where

$\mathbb{A}=\dfrac{3l^{2}\nu^{2}+16\nu\alpha-4\alpha}{l^{2}\nu^{2}-4\alpha}.$
(46)

In figure 8, the behavior of imaginary and real part of $\phi(z)$ in terms of
$z$ for different values of parameters has been shown. As can be seen, the
imaginary part of scalar field inside the black hole is zero and outside the
black hole the scalar field is unstable, and by increasing $\alpha$
instability increase.

(a) $\nu=4.5,\alpha=0.1,l=1,z_{h}=2$ (b) $\nu=4.5,l=1,z_{h}=2,c=-0.5$

Figure 8: Plots of real (dashed lines) and imagenary (solid lines) part of
$\phi$ in terms of $z$ for
$c={\color[rgb]{1,0,0}-0.1},{\color[rgb]{0,0,1}-0.2},-0.3,{\color[rgb]{0,1,0}-0.4},{\color[rgb]{1,1,0}-0.5}$
(left),
$\alpha={\color[rgb]{0,0,1}0.1},0.2,{\color[rgb]{0,1,0}0.3},{\color[rgb]{1,1,0}0.4}$,0.5
(right).

In equation (2.2) the exact coupling function $f_{2}$ and approximatly to
first order in $\alpha$ in equation (2.2) has been obtained. In figure (9),
$f_{2}$ for positive/negative $c$ and for different values of parameters has
been plotted. The important feature of the figures is that for the negative
$c$, $f_{2}$ goes to the negative values by increasing $z$, while it does not
become negative anywhere for the positive $c$.

$\displaystyle f_{2}$
$\displaystyle=\dfrac{(1-\nu)z^{-\frac{4}{\nu}}}{\pi^{\frac{3}{2}}\nu^{3}q^{2}(l^{2}\nu^{2}-4\alpha)\text{erf}\left(\frac{\sqrt{-2c\mathbb{A}}z_{h}}{2}\right)^{2}}[\pi
z\alpha\sqrt{\mathbb{A}}e^{\frac{c\mathbb{A}z^{2}}{2}}\left[\text{erf}\left(\frac{\sqrt{-2c\mathbb{A}}z}{2}\right)-\text{erf}\left(\frac{\sqrt{-2c\mathbb{A}}z_{h}}{2}\right)\right](cz^{2}l^{2}\nu^{4}(cz^{2}-6)$
$\displaystyle+\nu^{3}(32c^{2}z^{4}\alpha-2l^{2}(cz^{2}+10))+\nu^{2}(-8l^{2}+\alpha(12c^{2}z^{4}+88cz^{2}))+40\alpha\nu(cz^{2}+2)+32\alpha)-4\sqrt{\pi}\alpha
cz^{2}\nu(2+$ $\displaystyle\nu
cz^{2})(3l^{2}\nu^{2}+16\alpha\nu-4\alpha)e^{c\mathbb{A}z^{2}}+(l^{2}\nu^{2}-4\alpha)(-2l^{2}\nu^{2}\pi
z\sqrt{-2c\mathbb{A}}e^{\frac{cz^{2}\nu(l^{2}\nu+8\alpha)}{l^{2}\nu^{2}-4\alpha}}\text{erf}\left(\frac{\sqrt{-2c\mathbb{A}}z_{h}}{2}\right)+\pi^{\frac{3}{2}}$
$\displaystyle\left[\text{erf}\left(\frac{\sqrt{-2c\mathbb{A}}z}{2}\right)-\text{erf}\left(\frac{\sqrt{-2c\mathbb{A}}z_{h}}{2}\right)\right](l^{2}\nu\text{erf}\left(\frac{\sqrt{-2c\mathbb{A}}z_{h}}{2}\right)e^{-\frac{cz^{2}}{2}}(3\nu
cz^{2}+4\nu+4)+[\text{erf}\left(\frac{\sqrt{-2c\mathbb{A}}z}{2}\right)-$
$\displaystyle\text{erf}\left(\frac{\sqrt{-2c\mathbb{A}}z_{h}}{2}\right)]\alpha(16+8cz^{2}+\nu^{2}(z^{6}c^{3}-2c^{2}z^{4})+\nu(16+6c^{2}z^{4}+12cz^{2}))],$
(47)

In the case of $\alpha\ll 1$ and $c<0$ one can get

$\displaystyle
f_{2}\approx\dfrac{l^{2}(\nu-1)e^{-\frac{cz^{2}}{2}}\left[2\sqrt{-6c}z\nu
e^{\frac{3cz^{2}}{2}}-\pi^{\frac{1}{2}}\left(4+4\nu+3\nu
cz^{2}\right)\left(\text{erf}\left(\frac{\sqrt{-6c}z}{2}\right)-\text{erf}\left(\frac{\sqrt{-6c}z_{h}}{2}\right)\right)\right]}{\nu^{2}q^{2}z^{\frac{4}{\nu}}q^{2}\pi^{\frac{1}{2}}\text{erf}\left(\dfrac{\sqrt{-6c}z_{h}}{2}\right)}$
$\displaystyle-\dfrac{(\nu-1)\alpha}{\nu^{4}q^{2}\pi^{3}z^{\frac{4}{\nu}}\text{erf}\left(\frac{\sqrt{-6c}z_{h}}{2}\right)^{2}}[-12z^{2}\pi^{2}c\nu^{2}(2+\nu
cz^{2})e^{3cz^{2}}-4(2\nu+1)ze^{cz^{2}}(4\pi^{2}c\nu
z_{h}e^{\frac{3cz_{h}}{2}}+\pi^{\frac{5}{2}}$
$\displaystyle\sqrt{-6c}\text{erf}\left(\frac{\sqrt{-6c}z_{h}}{2}\right)\left(\nu
cz^{2}-\frac{2}{3}\nu-\frac{4}{3}\right)+\pi^{\frac{5}{2}}\sqrt{-6c}\nu
z)e^{\frac{3cz^{2}}{2}}(c^{2}z^{4}\nu^{2}-6c\nu^{2}z^{2}-2\nu cz^{2}-20\nu-8)$
$\displaystyle\left(\text{erf}\left(\frac{\sqrt{-6c}z}{2}\right)-\text{erf}\left(\frac{\sqrt{-6c}z_{h}}{2}\right)\right)-4(2\nu+1)\sqrt{-6c}\pi^{\frac{5}{2}}z_{h}(\frac{4}{3}+\frac{4}{3}\nu+\nu
cz^{2})\text{erf}\left(\frac{\sqrt{-6c}z}{2}\right)e^{-\frac{cz^{2}}{2}}e^{\frac{3cz^{2}}{2}}$
$\displaystyle+\nu\pi^{3}\left(\text{erf}\left(\frac{\sqrt{-6c}z}{2}\right)-\text{erf}\left(\frac{\sqrt{-6c}z_{h}}{2}\right)\right)^{2}(16+8cz^{2}+\nu(16+16c^{2}z^{4}+12cz^{2})+\nu^{2}(z^{6}c^{3}-2c^{2}z^{4}))]$
$\displaystyle+\mathcal{O}(\alpha^{2}).$ (48)

(a) $\nu=4.5,\alpha=0.1,l=1,z_{h}=2,q=0.5$ (b) $\alpha=0.1,l=1,z_{h}=2,q=0.5$
(c) $l=1,z_{h}=2,q=0.5,\nu=4.5$

Figure 9: Plots of $f_{2}(z)$ in terms of $z$ for
$c={\color[rgb]{1,0,0}0.1},{\color[rgb]{0,0,1}0.2},0.3,{\color[rgb]{0,1,0}0.4},{\color[rgb]{1,1,0}0.5}$
(solid lines) and
$c={\color[rgb]{1,0,0}-0.1},{\color[rgb]{0,0,1}-0.2},-0.3,{\color[rgb]{0,1,0}-0.4},{\color[rgb]{1,1,0}-0.5}$
(dashed lines) (left),
$\nu={\color[rgb]{0,0,1}2},3,{\color[rgb]{0,1,0}4},{\color[rgb]{1,1,0}5}$,
$c=0.5$ (solid lines) and $c=-0.5$ (dahsed lines) (middle),
$\alpha={\color[rgb]{0,0,1}0.1},0.2,{\color[rgb]{0,1,0}0.3},{\color[rgb]{1,1,0}0.4}$,
0.5, $c=0.5$ (solid lines) and $c=-0.5$ (dashed lines) (right).

From equation $E_{tt}$ we get the expression for the scalar potential $V$ as a
function of $z$ as follows:

$\displaystyle V$
$\displaystyle=\dfrac{e^{cz^{2}}}{4\nu^{3}l^{4}\pi^{\frac{3}{2}}(l^{2}\nu^{2}-4\alpha)\text{erf}\left(\frac{\sqrt{-2c\mathbb{A}}z_{h}}{2}\right)^{2}}[-2\alpha\pi
ze^{\frac{c\mathbb{A}z^{2}}{2}}\sqrt{-2c\mathbb{A}}\left[\text{erf}\left(\frac{\sqrt{-2c\mathbb{A}}z}{2}\right)-\text{erf}\left(\frac{\sqrt{-2c\mathbb{A}}z_{h}}{2}\right)\right]$
$\displaystyle[l^{2}z^{2}c\nu^{5}(5cz^{2}+3c^{2}z^{4}+6)+\nu^{4}(-32c^{2}z^{4}\alpha+l^{2}(20cz^{2}+20+c^{2}z^{4}))+\nu^{3}(4l^{2}+10l^{2}cz^{2}-\alpha(88cz^{2}$
$\displaystyle+4c^{2}z^{4}+12c^{3}z^{6}))-\nu^{2}(8l^{2}+\alpha(80+36c^{2}z^{4}+48cz^{2}))-\nu\alpha(8cz^{2}+48)+32\alpha]+\nu[8c\alpha
z^{2}\sqrt{\pi}(1$ $\displaystyle-\nu)(2+\nu
cz^{2})(3l^{2}\nu^{2}+16\alpha\nu-4\alpha)e^{c\mathbb{A}z^{2}}+(l^{2}\nu^{2}-4\alpha)[-2\pi\nu
zl^{2}\sqrt{-2c\mathbb{A}}(3\nu
cz^{2}+4\nu+2)\text{erf}\left(\frac{\sqrt{-2c\mathbb{A}}z_{h}}{2}\right)$
$\displaystyle e^{\frac{c\nu
z^{2}(l^{2}\nu+8\alpha)}{l^{2}\nu^{2}-4\alpha}}+\left[\text{erf}\left(\frac{\sqrt{-2c\mathbb{A}}z}{2}\right)-\text{erf}\left(\frac{\sqrt{-2c\mathbb{A}}z_{h}}{2}\right)\right][l^{2}\text{erf}\left(\frac{\sqrt{-2c\mathbb{A}}z_{h}}{2}\right)e^{-\frac{cz^{2}}{2}}(8+\nu^{2}(9c^{2}z^{4}+12cz^{2}+16)$
$\displaystyle+\nu(6cz^{2}+8))+(2+cz^{2})\alpha\left[\text{erf}\left(\frac{\sqrt{-2c\mathbb{A}}z}{2}\right)-\text{erf}\left(\frac{\sqrt{-2c\mathbb{A}}z_{h}}{2}\right)\right](\nu^{2}(3z^{6}c^{3}-2c^{2}z^{4})+\nu(20cz^{2}+16+$
$\displaystyle 14c^{2}z^{4}+32+16cz^{2}))]]]].$ (49)

(a) $\nu=4.5,\alpha=0.1,l=1,z_{h}=2$ (b) $\alpha=0.1,l=1,z_{h}=2$

Figure 10: Plots of $V(z)$ in terms of $z$ for
$c={\color[rgb]{1,0,0}0.1},{\color[rgb]{0,0,1}0.2},0.3,{\color[rgb]{0,1,0}0.4},{\color[rgb]{1,1,0}0.5}$
(dashed lines) and
$c={\color[rgb]{1,0,0}-0.1},{\color[rgb]{0,0,1}-0.2},-0.3,{\color[rgb]{0,1,0}-0.4},{\color[rgb]{1,1,0}-0.5}$
(solid lines) (left),
$\alpha={\color[rgb]{1,0,0}0.1},{\color[rgb]{0,0,1}0.2},0.3,{\color[rgb]{0,1,0}0.4},{\color[rgb]{1,1,0}0.5}$,
$c=0.5$ (dashed lines) and $c=-0.5$ (solid lines) (right).

Regardless of the sign of $c$, $V(z)$ is negative under the horizon. In panel
a, with the increase of $|c|$, $|V|$ decreases. In panel b, as $\alpha$
increases, $|V|$ increases.

####  2.2.1 Thermodynamics of background

The Hawking temperature can be obtained by using the surface gravity
interpretation

$T=\dfrac{\kappa}{2\pi}=\left|\dfrac{g^{\prime}}{4\pi}\right|=\dfrac{e^{\frac{c\mathbb{A}z_{h}^{2}}{2}}\sqrt{-2c\mathbb{A}}}{4\pi^{\frac{3}{2}}\text{erf}\left(\dfrac{z_{h}}{2}\sqrt{-2c\mathbb{A}}\right)},$
(50)

where $\mathbb{A}$ provided in (46). In the case of small $\alpha$ and $c<0$,
one can get

$\displaystyle
T\approx\dfrac{\sqrt{-6c}e^{\frac{3cz_{h}^{2}}{2}}}{4\pi^{\frac{3}{2}}\text{erf}\left(\frac{\sqrt{-6c}z_{h}}{2}\right)}+\dfrac{(2\nu+1)e^{\frac{3cz_{h}^{2}}{2}}\alpha[\pi\sqrt{-6c}(cz_{h}^{2}+\frac{1}{3})\text{erf}\left(\frac{\sqrt{-6c}z_{h}}{2}\right)-2\sqrt{\pi}cz_{h}e^{\frac{3cz_{h}^{2}}{2}}]}{\pi^{\frac{5}{2}}l^{2}\nu^{2}\text{erf}\left(\frac{\sqrt{-6c}z_{h}}{2}\right)^{2}}+\mathcal{O}\left(\alpha^{2}\right).$
(51)

The behavior of temperature is shown in Figs. (11). In Fig. (11)a, the
variation of temperature with respect to the horizon radius $z_{h}$ for
different values of the coupling of theory $\alpha$ is shown. As can be seen
there exists a minimum temperature $T_{min}$ below which no black hole
solution exist (thermal gas). However, for $T>T_{min}$, there are two black
hole solutions, a large and a small one (deconfined quark gluon plasma phase).
The small black hole phase for which $T$ increases with $z_{h}$ whereas the
large black hole phase for which $T$ decreases with $z_{h}$. In Fig. (11)b, to
study the stability of the solutions, we have shown the behavior of heat
capacity $C_{V}$ and temperature. As can be seen the large black hole has
positive heat capacity and therefore is stable and small black hole is
unstable and thus not physical.

(a) $\nu=4.5,l=1,c=-0.5$ (b) $\nu=4.5,l=1,c=-0.5$

Figure 11: Plots of $T$ in terms of $z_{h}$ for
$\alpha={\color[rgb]{1,0,0}0.1},{\color[rgb]{0,0,1}0.2},0.3,{\color[rgb]{0,1,0}0.4},{\color[rgb]{1,.5,0}0.5}$
and $\nu=4.5$ (left). Plots of $C_{V}$ and $T$ in terms of $z_{h}$ for
$\alpha={\color[rgb]{1,0,0}0.1},{\color[rgb]{0,0,1}0.2},0.3,{\color[rgb]{0,1,0}0.4},{\color[rgb]{1,.5,0}0.5}$
(right)

Following the standard Bekenstein-Hawking formula (27), one can easily read
the black hole entropy density $s$, which is defined as

$s=\dfrac{l^{3}e^{-\frac{3cz_{h}^{2}}{4}}}{4z_{h}^{\frac{\nu+1}{\nu}}}.$ (52)

The scaled entropy density $s/T^{3}$ as a function of scaled temperature
$T/T_{min}$ is shown in Fig.12a. The red lines correspond to the large stable
solution and blue lines are for the small unstable solution. The numerical
result of the square of the sound velocity is shown in 12b. At $T_{min}$, the
sound velocity square is around $0$ which is in agreement with lattice data
$0.05$. At high temperature, the sound velocity square goes to $0.45$ for
$\nu=1$, which means that the system is approximatly asymptotically conformal,
while for $\nu=4.5$, the sound velocity square goes to $0.8$. The numerical
result of the specific heat is shown in Fig. 12c. It can be clearly seen that
the specific heat $C_{V}$ diverges at $T_{min}$. At $T\to\infty$, the scaled
specific heat $C_{V}/T^{3}$ approaches to the zero for this approximate
solution [34].

(a) $l=1,c=-0.5,\alpha=0.1$ (b) $l=1,c=-0.5,\alpha=0.1$ (c)
$l=1,c=-0.5,\alpha=0.1$

Figure 12: Plots of scaled entropy density
$s/T^{3}\;(\text{left}),c_{s}^{2}\;(\text{middle})$ and
$C_{V}/T^{3}\;(\text{right})$ in terms of scaled temperature $T/T_{min}$ for
$\nu=1,T_{min}=0.19457\;(\text{dashed
lines}),\nu=4.5,T_{min}=0.1357\;(\text{solid lines})$. In each panel red lines
correspond to the large stable solution and blue lines correspond to unstable
small solution.

In figure 13, the behavior of $c_{s}^{2}$ and $C_{V}$ have been shown. As can
be seen from the figure and obtained the result in the previous section, it
can be concluded that the heat capacity has an inverse relationship with the
speed of sound.

(a) $l=1,c=-0.5,\alpha=0.1$ Figure 13: Plots of $c_{s}^{2}$ (red lines) and
$C_{V}$ (blue lines) in terms of $z_{h}$ for $\nu=1\;(\text{dashed
lines}),4.5\;(\text{solid lines})$.

###  2.3 The case $c\neq 0,\mu\neq 0$

In this case the differential equation (2), becomes

$\displaystyle 4\nu
lz^{2}e^{-\frac{cz^{2}}{4}}(l^{2}\nu^{2}e^{-\frac{cz^{2}}{2}}-\alpha(2+\nu
cz^{2})^{2}g(z))g^{\prime\prime}-2zle^{-\frac{cz^{2}}{4}}(\nu^{2}l^{2}(4+(2+3z^{2})\nu)-$
$\displaystyle\alpha(2+\nu cz^{2})(8+4\nu+6\nu
cz^{2}-6c\nu^{2}z^{2}+c^{2}\nu^{2}z^{4})g)g^{\prime}-4\alpha\nu
z^{2}le^{-\frac{cz^{2}}{4}}(2+\nu cz^{2})^{2}g^{\prime 2}$
$\displaystyle-c^{2}\nu^{3}l\mathfrak{c}_{2}^{2}z^{\frac{4\nu+2}{\nu}}e^{\frac{cz^{2}}{4}}=0$
(53)

in order to solve the above differential equation, we assume

$g(z)=1+\epsilon g_{1}(z)+\mathcal{O}(\epsilon^{2}),$ (54)

by inserting it, one can achieve a non-homogeneous differential equation as

$g_{1}^{\prime\prime}+\dfrac{[-l^{2}\nu^{3}(2+3cz^{2})-4\nu^{2}(l^{2}+4\alpha
cz^{2})+4\alpha\nu(2+cz^{2})+16\alpha]g_{1}^{\prime}}{\nu
z(l^{2}\nu^{2}-4\alpha)}+\dfrac{c\nu\mathfrak{c}_{2}^{2}z^{\frac{2}{\nu}}e^{\frac{cz^{2}}{2}}}{4\alpha(4+c\nu
z^{2})}=0.$ (55)

In order to solve the non-homogeneous differential equation (55) in the case
of $c<0,\mathfrak{c}_{2}\neq 0$, we consider the following particular
solution:

$g_{2}(z)=e^{\frac{cz^{2}}{2}}z^{-\frac{2(\nu-1)}{\nu}}\sum_{i}h_{i}z^{i},$
(56)

where $h_{i}$ are coefficients of expansions. By inserting the above solution
into the differential equation (55), and solving order by order for
coefficients, one can get

$\displaystyle\hskip 142.26378pth_{0}=h_{1}=h_{3}=h_{5}=0,$ $\displaystyle
h_{2}$
$\displaystyle=\dfrac{\mathfrak{c}_{2}^{2}\nu^{2}}{8\alpha(\nu-2)},\;\;\;h_{4}=-\dfrac{c\nu^{3}\mathfrak{c}_{2}^{2}(-20\alpha\nu+l^{2}\nu^{3}+l^{2}\nu^{2}-12\alpha)}{16\alpha(\nu-2)(\nu^{4}l^{2}+3l^{2}\nu^{3}-4\alpha\nu^{2}+2l^{2}\nu^{2}-12\alpha\nu-8\alpha)},$
$\displaystyle h_{6}$
$\displaystyle=\dfrac{\nu^{4}c^{2}\mathfrak{c}_{2}^{2}}{32\alpha(2+3\nu)(l^{2}\nu^{2}-4\alpha)^{2}(1+2\nu)(1+\nu)(\nu^{2}-4)}(3l^{4}\nu^{6}+6l^{5}\nu^{5}-40\nu^{4}l^{2}\alpha+3\nu^{4}l^{4}+256\alpha^{2}\nu^{3}$
$\displaystyle-56\nu^{3}l^{2}\alpha+624\nu^{2}\alpha^{2}-24\nu^{2}l^{2}\alpha+448\alpha^{2}\nu+112\alpha^{2}),$
(57)

where $\mathfrak{c}_{2}$ used from (11). Therefore, the metric function by
considering a particular solution becomes

$g(z)=1-\dfrac{\text{erf}\left(\dfrac{1}{2}\sqrt{\dfrac{c(-6l^{2}\nu^{2}-32\nu\alpha+8\alpha)}{(l^{2}\nu^{2}-4\alpha)}}z\right)}{\text{erf}\left(\dfrac{1}{2}\sqrt{\dfrac{c(-6l^{2}\nu^{2}-32\nu\alpha+8\alpha)}{(l^{2}\nu^{2}-4\alpha)}}z_{h}\right)}+e^{\frac{cz^{2}}{2}}z^{-\frac{2(\nu-1)}{\nu}}\sum_{i}h_{i}z^{i},$
(58)

with $h_{i}$ are provided in (2.3). In figure 14, the effect of $\mu$ on the
behavior of $f_{2}$ and $V$ have been shown. As can be seen in left panel by
increasing chemical potential, $f_{2}$ decreases and in right panel $V$
increases.

(a) $\nu=4.5,l=1,\alpha=0.1,c=\pm 0.5$ (b) $\nu=4.5,l=1,\alpha=0.1,c=\pm 0.5$

Figure 14: Plots of $f_{2}$ and $V$ in terms of $z$ for
$\mu={\color[rgb]{1,0,0}0},{\color[rgb]{0,0,1}0.05},{\color[rgb]{0,0,0}0.1},{\color[rgb]{0,1,0}0.15},{\color[rgb]{1,.5,0}0.2},{\color[rgb]{1,1,0}0.25}$.

(a) $\nu=4.5,l=1,c=-0.5,\alpha=0.1$ (b) $\nu=4.5,l=1,c=-0.5,\alpha=0.1$ (c)
$\nu=4.5,l=1,c=-0.5,\alpha=0.1$

Figure 15: Plots of $s,T$ and $C_{V}$ in terms of $z_{h}$ for
$\mu={\color[rgb]{0,1,0}0.1},{\color[rgb]{0,0,1}0.15},{\color[rgb]{0,0,0}0.2},{\color[rgb]{1,0,0}0.22728},{\color[rgb]{1,.5,0}0.3}$
(left) and $\mu={\color[rgb]{0,0,1}0.2},{\color[rgb]{1,0,0}0.22728},0.3$
(middle and right).

In fig. 15a, the behavior of entropy is shown in terms of temperature. The
figure shows the minimum and maximum temperature. As $\mu$ increases, the
minimum temperature decreases. This plot also shows that for $0<\mu<\mu_{c}$,
there are minimal $T_{min}$ and maximal $T_{max}$ temperatures, between which
the entropy is a function of $T$ with three branches. When we decrease the
temperature, the entropy decreases along the first branch
($T_{min}<T<\infty$). Then the entropy decreases along the second branch with
an increase of temperature from $T_{min}$ to $T_{max}$, i.e. here the black
holes are unstable. Finally the entropy increases along the third branch with
an increase of temperature for $0<T<T_{max}$. Such a behavior in terms of
event horizon, one can see in the 15b and c. In each panel, the critical point
has been shown in red color curve. Upon varying the Hawking temperature, a
phase transition from the large black hole phase to the thermal AdS phase
takes place at a critical temperature $T_{c}=0.1242$. This is the famous black
hole-thermal AdS Hawking-Page phase transition which occurs in the presence of
chemical potential $\mu_{c}=0.22728$. In fig. 16 by using the approximate
analysis, the phase diagram of the holographic QCD model for anisotropic
background and for Einstien gravity (long dashed line), Einstien-Gauss-Bonnet
gravity (solid line) has been shown. As one can see at $\mu=0$ the system
undergoes a black hole to thermal gas first order phase transition so that
$T^{(EGB)}(\mu=0)>T^{(E)}(\mu=0)$. For $0<\mu<\mu_{c}$ (in the transition
lines), the system undergoes a large black hole to a small black hole first-
order phase transition. For $0<\mu<\mu_{I}$, the temperature of the black hole
to black hole transition of Einstien gravity ($T^{(E)}(\mu)$) is less than
Einstien-Gauss-Bonnet gravity ($T^{(EGB)}(\mu)$) and for $\mu_{I}<\mu<\mu_{c}$
vice versa. The first order phase transition stops at the critical point
($\mu_{c},T_{c}$), where the phase transition becomes second order, herewith
$T^{(EGB)}_{c}<T^{(E)}_{c}$, $\mu_{c}^{(EGB)}<\mu_{c}^{(E)}$. For
$\mu>\mu_{c}$, the system has a sharp but smooth crossover. These thermal AdS
and black hole phases in the usual language of gauge-gravity duality are dual
to the confinement and deconfinement phases in the dual boundary theory.

(a) $\nu=4.5,l=1,c=-0.5$ Figure 16: The phase diagram in $T$ and $\mu$ plane
for anisotropic background. At small $\mu$, the system undergoes a first order
phase transition at finite $T$. The first order phase transition stops at the
critical point $(\mu_{c},T_{c})\sim(0.22728,0.1242)$, where the phase
transition becomes second order. The solid line is for $\alpha=0.1$ and long
dashed line is for $\alpha=0$.

##  3 Conclusion

In this work, we extended the AdS/QCD model to quadratic gravity to gain
insight into the influence of gravity on QCD. To do so, we considered an
anisotropic black hole metric as a solution to a system of $5D$ Einstein-
quadratic-two Maxwell-dilaton fields. The anisotropic background is specified
by an arbitrary exponent, a non-zero dilaton field, a non-zero time component
of the first Maxwell field, and a non-zero longitudinal magnetic component of
the second Maxwell field. The field equations for the considered theory are
coupled and bulky differential equations for six unknown functions. Therefore,
obtaining the solution to such field equations is too hard, this is why we
considered the special case of field equation, i.e.
$\gamma=\alpha,\beta=-4\alpha$ (Einstien-Gauss-Bonnet gravity). The
differential equation for the metric function in EGB gravity is a nonlinear
second-order equation that has been solved in special cases. At the first, we
obtained the exact solutions for the differential equations with zero warp
functions. In this case, it doesn’t occur any thermodynamical phase
transitions to the black brane. The second case that we have considered is the
case with zero chemical potential. The blackening function in this case
supports the Van der Waals-like phase transition between small and large black
holes for suitable values of parameters. The third case that has been
considered is nonzero warp function and chemical potential. In this case, in
addition to the small/large phase transition, the blackening function
supported the phase transition from the large black hole phase to the thermal
AdS phase at a critical temperature. Holographically, this phase transition
corresponds to the confinement-deconfinement phase transition in QCD. In each
case, we investigated the anisotropy influence and the effect of parameters of
theory on the thermodynamic properties of our background, in particular, on
the small/large black holes phase transition diagram. In fig. (16), the effect
of the Gauss-Bonnet term on the phase transition has been shown. This figure
shows that before/after the intersection point for constant chemical
potential, the temperature of a black hole in EGB gravity is more/less than
that of Einstien gravity. Clearly, before the intersection point, $\alpha$ has
a dominant impact on the temperature (compared to the effect of $\mu$ on the
temperature). In the isotropic case corresponding to $\nu=1$ (zero magnetic
fields) and $,\alpha\to 0$ reproduces previously known results [8].
For future work, one can consider the Weyl-squared term by using the
combination $\gamma=6\alpha$ and $\beta=-4/3\gamma$ for the parameters of the
theory. But since in this case, the field equation for the metric has a 4th-
order derivative, the differential equations should be solved numerically.
Also, following the paper [9], one can study the effect of the magnetic field
on the system in the framework of EGB.

## Acknowledgements

I would like to thank the referee for her/his fruitful comments. I also would
like to thank the School of Physics of the Institute for Research in
Fundamental Sciences (IPM).

## Constants

The constants related to equation(19):

$\displaystyle\mathcal{E}$ $\displaystyle=(\nu+1)((\nu+1)(\nu^{4}l^{4}-8\alpha
c_{2})-4\alpha\nu c_{1}z^{\frac{2\nu+2}{\nu}}),$ $\displaystyle\mathcal{D}$
$\displaystyle=\nu l^{2}-\nu^{2}l^{2}-\sqrt{\nu^{4}l^{4}-8\alpha c_{2}},$
$\displaystyle\mathcal{F}$ $\displaystyle=\nu
l^{2}-\nu^{2}l^{2}+\sqrt{\nu^{4}l^{4}-8\alpha c_{2}},$
$\displaystyle\mathcal{G}$ $\displaystyle=\dfrac{\nu^{3}l^{2}-\nu
l^{2}-\sqrt{\mathcal{E}}}{\nu+1},$ $\displaystyle\mathcal{H}$
$\displaystyle=-l^{2}\nu(\nu-1)\sqrt{\nu^{4}l^{4}-8\alpha
c_{2}}+\nu^{4}l^{4}-8\alpha c_{2},$ $\displaystyle\mathcal{K}$
$\displaystyle=-l^{2}\nu\left(\nu-\frac{1}{2}\right)\sqrt{\mathcal{E}}-2\alpha
c_{1}\nu z^{\frac{2\nu+2}{\nu}}+(\nu+1)\left(\nu^{4}l^{4}-4\alpha
c_{2}-\dfrac{l^{4}\nu^{3}}{2}\right),$ $\displaystyle\mathcal{L}$
$\displaystyle=l^{2}\nu(\nu-1)\sqrt{\nu^{4}l^{4}-8\alpha
c_{2}}+\nu^{4}l^{4}-8\alpha c_{2}.$ (59)

The constants related to equation(2.1):

$\displaystyle\bar{\mathcal{A}}$ $\displaystyle=(\nu+1)(-4\alpha c_{1}\nu
z^{\frac{2\nu+2}{\nu}}+(\nu+1)(\nu^{4}l^{4}-8\alpha c_{2})),$
$\displaystyle\bar{\mathcal{B}}$
$\displaystyle=-4\alpha\nu^{2}c_{1}z^{\frac{2\nu+2}{\nu}}+(\nu+1)(\nu+\frac{3}{2})(\nu^{4}l^{4}-8\alpha
c_{2}),$ $\displaystyle\bar{\mathcal{C}}$ $\displaystyle=-4\alpha\nu
c_{1}z^{\frac{2\nu+2}{\nu}}+(\nu+1)(\nu^{4}l^{4}-8\alpha c_{2}),$
$\displaystyle\bar{\mathcal{E}}$
$\displaystyle=-4c_{1}^{2}\alpha^{2}\nu^{2}(3\nu^{3}-4\nu^{2}+3\nu+2)z^{\frac{2(2\nu+2)}{\nu}}+$
$\displaystyle
2c_{1}\nu\alpha(\nu^{2}-1)(\nu^{4}l^{4}+2l^{4}\nu^{3}-16\alpha\nu
c_{2}-8\alpha c_{2})z^{\frac{2\nu+2}{\nu}}+$
$\displaystyle(\nu-1)(\nu+1)^{2}(\nu^{4}l^{4}-8\alpha c_{2})(4\alpha
c_{2}+\nu^{5}l^{5}-l^{4}\nu^{3}-\frac{\nu^{4}l^{4}}{2}),$
$\displaystyle\bar{\mathcal{F}}$
$\displaystyle=\nu^{7}l^{4}-\frac{7}{2}\nu^{6}l^{4}+2\nu^{5}l^{4}+\frac{5}{2}\nu^{4}l^{4}+4\nu^{3}\alpha
c_{2}+24\alpha\nu^{2}c_{2}-36\alpha\nu c_{2}-8\alpha c_{2},$
$\displaystyle\bar{\mathcal{H}}$
$\displaystyle=4c_{1}^{2}\alpha^{2}(5\nu+2-2\nu^{2}+\nu^{3})z^{\frac{2(2\nu+2)}{\nu}}-4c_{1}\alpha(\nu^{2}-1)(l^{4}\nu^{3}-8\alpha\nu
c_{2}+4\alpha c_{2})z^{\frac{2\nu+2}{\nu}}$
$\displaystyle-(\nu+1)^{2}(\nu-1)\nu^{2}l^{4}(\nu^{2}-\frac{1}{2}\nu-1)(\nu^{4}l^{4}-8\alpha
c_{2}),$ $\displaystyle\bar{\mathcal{G}}$ $\displaystyle=\nu^{5}l^{4}+2\alpha
c_{2}+\frac{1}{2}\nu^{4}l^{4}-\frac{1}{2}\nu^{3}l^{4},\;\;\;\;\;\bar{\mathcal{K}}=\nu^{5}l^{4}+\frac{1}{3}\nu^{4}l^{4}-\frac{1}{2}l^{4}\nu^{3}+4\alpha
c_{2},$ $\displaystyle\bar{\mathcal{L}}$
$\displaystyle=576\alpha^{2}c_{2}^{2}\bar{\mathcal{G}}\bar{\mathcal{C}}+\frac{1}{2}\bar{\mathcal{A}}^{2}\bar{\mathcal{H}},\;\;\;\;\bar{\mathcal{J}}=320\alpha^{2}c_{2}^{2}\bar{\mathcal{G}}\bar{\mathcal{C}}-\frac{1}{2}(\bar{\mathcal{A}}^{2}(\bar{\mathcal{A}}\bar{\mathcal{B}}-\bar{\mathcal{H}})).$
(60)

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