$\displaystyle 0$ $\displaystyle=$ $\displaystyle
B^{i\;\;b,a}-B^{i\;\;a,b}-c_{\;\;\;jk}^{i}B^{j\;a}B^{k\;b}$ (135)

The third equation is the vanishing of the Yang-Mills field strength on the
$y$-submanifold,

$\mathbf{d}_{\left(y\right)}\mathbf{B}^{i}=-\frac{1}{2}c_{\;\;\;jk}^{i}\mathbf{B}^{j}\wedge\mathbf{B}^{k}$

so that $\mathbf{B}^{k}$ is a pure-gauge connection for any fixed
$x^{\alpha}$. Therefore, for each $x_{0}^{\alpha}$ we may choose an
$SU\left(N\right)$ gauge $\Lambda\left(x_{0}^{\alpha},y_{\beta}\right)$ such
that $\mathbf{B}^{k}=0$. But this makes the value of $\mathbf{B}^{k}$
independent of $x^{\alpha}$ as well, so $\mathbf{B}^{k}=0$ everywhere. As a
result, the fields in terms of the potentials reduce to

$\displaystyle F_{\;\;\;ab}^{i}$ $\displaystyle=$ $\displaystyle
A_{\;\;\;b;a}^{i}-A_{\;\;\;a;b}^{i}-c_{\;\;\;jk}^{i}A_{\;\;\;a}^{j}A_{\;\;\;b}^{k}$
$\displaystyle A_{\;\;\;b}^{i\quad\;\;,a}$ $\displaystyle=$ $\displaystyle 0$

Now, when we write the field equations in terms of the potentials and set
$B^{k\,a}=0$, we have

$\displaystyle 0$ $\displaystyle=$ $\displaystyle F_{k\;ab}^{\;\;\quad,a}$
$\displaystyle 0$ $\displaystyle=$
$\displaystyle\eta^{bc}\left(F_{j\;ac;b}+F_{i\;ac}c_{\;\;\;jk}^{i}A_{\;\;\;a}^{j}\right)$

The first shows that $F_{k\;ab}$ is independent of $y_{\alpha}$ and the second
shows it to be covariantly divergence free.

##  8 Conclusions

Gravitational field theories in doubled dimensions include biconformal gravity
[4, 23, 5, 6], double field theory [13, 14, 15, 12, 33, 35], and gravity on a
Kähler manifold [7, 6]. Each of these cases starts as a fully $2n$-dimensional
theory but ultimately is intended to describe gravity on an $n$-dimensional
submanifold. We have found a satisfactory $2n$-dimensional form of Yang-Mills
matter sources and shown that they also reduce to the expected $n$-dimensional
sources _as a consequence of the field equations_. Our gravitational reduction
and the consequent reduction of the Yang-Mills fields and field equations
_does_ not _require a section condition._

While we discussed the issue in biconformal space, our results hold in the
related forms of double field theory and Kähler manifolds [7, 6].

For matter fields we restrict our attention to gauged $SU\left(N\right)$
sources (Yang-Mills type). While we find that the usual form of
$2n$-dimensional Yang-Mills action gives incorrect coupling to gravity,
including a “twist” matrix in the action corrects the problem.

For the gravitational fields we use the most general action linear in the
biconformal curvatures. The variation is taken with respect to all
$\frac{\left(n+1\right)\left(n+2\right)}{2}$ conformal gauge fields. In the
absence of sources, the use of the gravitational field equations to reduce
fully $2n$-dimensional gravity theory to dependence only on the fields of
$n$-dimensional gravity is well established. The field equations of torsion-
free biconformal space restrict the
$\frac{1}{2}\left(n+1\right)\left(n+2\right)$ curvature components, each
initially dependent on $2n$ independent coordinates, to the usual locally
scale covariant Riemannian curvature tensor in $n$ dimensions. Ultimately, the
$n$-dim solder form determines all fields, up to coordinate and gauge
transformations. Generic, torsion-free, vacuum solutions describe
$n$-dimensional scale-covariant general relativity on the co-tangent bundle.

Here we have shown that the same reduction occurs when gauged
$SU\left(N\right)$ field strengths are included as matter sources. The result
goes well beyond any previous work. With two exceptions [32, 20] studies of
biconformal spaces [23, 5, 20, 8, 9, 7, 10, 6, 22, 11] have considered the
pure gravity biconformal spaces, leading to vacuum general relativity. With
$SU\left(N\right)$ Yang-Mills fields as gravitational sources, the central
issue was to show that a completely general $SU\left(N\right)$ gauge theory
over a $2n$-dimensional biconformal space does not disrupt the gravitational
reduction to general relativity, but rather itself reduces to a suitable
$n$-dim gravitational source and Yang-Mills field equation.

As with the Riemann-Cartan construction of general relativity above, the
development of biconformal spaces from group symmetry made it straightforward
to include the additional symmetry of sources. By extending the quotient to

$\mathcal{M}^{2n}=\left[SO\left(p+1,q+1\right)\times
SU\left(N\right)\right]/\left[SO\left(p,q\right)\times
SO\left(1,1\right)\times SU\left(N\right)\right]$

the local symmetry is enlarged by $SU\left(N\right)$. We considered the
effects of adding an $SU\left(N\right)$ action to the gravitational action
Eq.(8). As central results we successfully showed:

  1. 1.

The number of field components in $2n$ dimensions reduces by a factor of
$\frac{n-2}{2\left(2n-1\right)}$ to the expected number
$\frac{n\left(n-1\right)}{2}\left(N^{2}-1\right)$ on $n$-dimensional
spacetime.

  2. 2.

The functional dependence of the fields reduces from $2n$ to $n$ independent
variables.

  3. 3.

The usual form of Yang-Mills stress-energy tensor provides the source for the
scale-covariant Einstein equation on $n$-dimensional spacetime.

  4. 4.

The usual Yang-Mills field equation holds on the spacetime submanifold.

To accomplish these goals we required two interdependent intermediate steps:

  1. 1.

We considered alternate forms of $2n$-dimensional Yang-Mills action, showing
that the usual action, $S_{YM}^{0}=\int
tr\left(\mathbf{F}\land^{*}\mathbf{F}\right)$ gives nonstandard coupling to
gravity. Instead, including a “twist” matrix in the action $S_{YM}=\int
tr\left(\bar{\mathbf{F}}\land^{*}\mathbf{F}\right)$ with twisted form

$\bar{\mathcal{F}}_{AB}=\frac{1}{2}\left(K_{A}^{\;\;\;C}\mathcal{F}_{CB}+\mathcal{F}_{AC}K_{\;\;\;B}^{C}\right)$

leads to both the usual $n$-dimensional Yang-Mills source to the Einstein
tensor and the usual Yang-Mills equation for the $SU\left(N\right)$ fields. A
similar twist has been found in other double field theory studies [16, 18, 19]
in order to enforce supersymmetry. Here, the twist is required for the bosonic
fields alone. Interestingly, the twist matrix
$K_{\;\;\;B}^{A}=\bar{K}^{AC}g_{CB}$ makes use of both the Kähler and Killing
forms, $g_{AB}$ and $K_{AB}$, respectively.

  2. 2.

We considered two naturally occurring inner products for the orthonormal frame
fields: the restriction to the base manifold of the Killing form, and the
Kähler metric. We showed the Kähler form cannot lead to the usual field
equations while the variation of the Killing form in the twisted action gives
usual Yang-Mills equations and usual coupling to gravity. Previous results in
biconformal gravity did not require the inner product.

We also gave a brief discussion of sources in Weyl geometry and in Weyl
gravity. For Weyl gravity we indicated the possibility of alternative, fourth
order source fields. We give two examples–scalar and electromagnetic fields–of
fourth order generalizations for sources. The fourth order fields satisfy
field equations which differ from the field equations for the corresponding
second order fields only by terms linear in the product of the fields and the
curvature. The difference between solutions to Weyl gravity with fourth order
sources and general relativity with the corresponding second order sources
need not introduce a mass scale.

##  9 Appendices

## Appendix A: Failure of the Kähler inner product

In this Appendix, we show that using the Kähler metric to define
orthonormality of the solder and co-solder forms cannot lead to the usual
Yang-Mills gravitational coupling to gravity, for either the usual Yang-Mills
action or the twisted Yang-Mills action.

We first find the Hodge dual of the Yang-Mills field here using the inverse
Kähler metric $\bar{g}^{AB}$. In general terms the Hodge dual of a 2-form is
given by Eq.(42), where we now substitute the Kähler metric,

${}^{*}\boldsymbol{\mathcal{F}}$ $\displaystyle=$
$\displaystyle\frac{1}{n!\left(n-2\right)!}\left(\frac{1}{2}F_{ab}\bar{g}^{am}\bar{g}^{bn}+G_{\;\;\;b}^{a}\bar{g}_{a}^{\;\;\;m}\bar{g}^{bn}+\frac{1}{2}H^{ab}\bar{g}_{a}^{\;\;\;m}\bar{g}_{b}^{\;\;\;n}\right)\varepsilon_{\;\;\;\qquad
mne\cdots f}^{c\cdots d}\mathbf{f}_{c\cdots d}\mathbf{e}^{e\cdots f}$ (136)
$\displaystyle+\frac{\left(-1\right)^{n-1}}{\left(n-1\right)!\left(n-1\right)!}\left(\frac{1}{2}F_{ab}\bar{g}_{\;\;\;m}^{a}\bar{g}^{bn}+G_{\;\;\;b}^{a}\bar{g}_{am}\bar{g}^{bn}+\frac{1}{2}H^{ab}\bar{g}_{am}\bar{g}_{b}^{\;\;\;n}\right)\varepsilon_{\;\;\;\qquad
ne\cdots f}^{mc\cdots d}\mathbf{f}_{c\cdots d}\mathbf{e}^{e\cdots f}$
$\displaystyle+\frac{\left(-1\right)^{n}}{\left(n-1\right)!\left(n-1\right)!}\left(\frac{1}{2}F_{ab}\bar{g}^{am}\bar{g}_{\;\;\;n}^{b}+G_{\;\;\;b}^{a}\bar{g}_{a}^{\;\;\;m}\bar{g}_{\;\;\;n}^{b}+\frac{1}{2}H^{ab}\bar{g}_{a}^{\;\;\;m}\bar{g}_{bn}\right)\varepsilon_{\;\;\;\qquad
me\cdots f}^{nc\cdots d}\mathbf{f}_{c\cdots d}\mathbf{e}^{e\cdots f}$
$\displaystyle+\frac{1}{n!\left(n-2\right)!}\left(\frac{1}{2}F_{ab}\bar{g}_{\;\;\;m}^{a}\bar{g}_{\;\;\;n}^{b}+G_{\;\;\;b}^{a}\bar{g}_{am}\bar{g}_{\;\;\;n}^{b}+\frac{1}{2}H^{ab}\bar{g}_{am}\bar{g}_{bn}\right)\varepsilon_{\;\;\;\qquad
e\cdots f}^{mnc\cdots d}\mathbf{f}_{c\cdots d}\mathbf{e}^{e\cdots f}$

Forming the usual Yang-Mills Lagrangian density as the wedge product,
$\boldsymbol{\mathcal{F}}\wedge{}^{*}\boldsymbol{\mathcal{F}}$, and
eliminating the basis forms in favor of the volume form $\boldsymbol{\Phi}$
yields

$\displaystyle\boldsymbol{\mathcal{F}}\wedge{}^{*}\boldsymbol{\mathcal{F}}$
$\displaystyle=$
$\displaystyle\left(\frac{1}{2}F_{mn}\bar{g}^{am}\bar{g}^{bn}+G_{\;\;\;n}^{m}\bar{g}_{\;\;\;m}^{a}\bar{g}^{bn}+\frac{1}{2}H^{mn}\bar{g}_{\;\;\;m}^{a}\bar{g}_{\;\;\;n}^{b}\right)F_{ab}\boldsymbol{\Phi}$
(137)
$\displaystyle+\left(F_{mn}\bar{g}_{a}^{\;\;\;m}\bar{g}^{bn}+G_{\;\;\;n}^{m}\left(\bar{g}_{am}\bar{g}^{bn}-\bar{g}_{a}^{\;\;\;n}\bar{g}_{\;\;\;m}^{b}\right)+H^{mn}\bar{g}_{am}\bar{g}_{\;\;\;n}^{b}\right)G_{\;\;\;b}^{a}\boldsymbol{\Phi}$
$\displaystyle+\left(\frac{1}{2}F_{mn}\bar{g}_{a}^{\;\;\;m}\bar{g}_{b}^{\;\;\;n}+G_{\;\;\;n}^{m}\bar{g}_{am}\bar{g}_{b}^{\;\;\;n}+\frac{1}{2}H^{mn}\bar{g}_{am}\bar{g}_{bn}\right)H^{ab}\boldsymbol{\Phi}$

For the diagonal form of the Kähler metric, Eq.(31), this Lagrange density
reduces to

$\displaystyle\boldsymbol{\mathcal{F}}\wedge{}^{*}\boldsymbol{\mathcal{F}}$
$\displaystyle=$
$\displaystyle\left(\frac{1}{2}\eta^{am}\eta^{bn}F_{ab}F_{mn}+\eta_{am}\eta^{bn}G_{\;\;\;n}^{m}G_{\;\;\;b}^{a}+\frac{1}{2}\eta_{am}\eta_{bn}H^{mn}H^{ab}\right)\boldsymbol{\Phi}$
(138)

Varying the potentials in Eq.(138) yields the usual Yang-Mills field equation.
However, metric variation of Eq.(137) gives a nonstandard coupling to gravity.
Varying and reducing the integral of Eq.(137), the resulting gravitational
field equations are:

$\displaystyle\alpha\left(\Omega_{\;\;\;b\quad m}^{n\quad
b}-\Omega_{\;\;\;b\quad a}^{a\quad
b}\delta_{m}^{n}\right)+\beta\left(\Omega_{\;\;\;m}^{n}-\Omega_{\;\;\;a}^{a}\delta_{m}^{n}\right)+\Lambda\delta_{m}^{n}$
$\displaystyle=$
$\displaystyle-\left(2\eta^{be}F_{me}F_{ab}+2G_{\;\;\;a}^{d}G_{\;\;\;m}^{e}\eta_{de}\right)\eta^{an}$
(139) $\displaystyle\left[\alpha\left(\Omega_{\;\;\;n\quad a}^{a\quad
m}-\Omega_{\;\;\;b\quad a}^{a\quad
b}\delta_{n}^{m}\right)+\beta\left(\Omega_{\;\;\;n}^{m}-\Omega_{\;\;\;a}^{a}\delta_{n}^{m}\right)+\Lambda\delta_{n}^{m}\right]$
$\displaystyle=$
$\displaystyle-\left(2G_{\;\;\;c}^{d}G_{\;\;\;b}^{n}\eta_{dm}\eta^{bc}+2\eta_{bc}\eta_{dm}H^{dc}H^{nb}\right)$
(140) $\displaystyle\alpha\Omega_{\;\;\;nam}^{a}+\beta\Omega_{nm}$
$\displaystyle=$
$\displaystyle\left(2\eta^{ca}G_{\;\;\;a}^{b}F_{mc}+2\eta_{ac}G_{\;\;\;m}^{c}H^{ab}\right)\eta_{bn}$
(141) $\displaystyle\alpha\Omega_{\;\;\;b}^{n\quad bm}+\beta\Omega^{nm}$
$\displaystyle=$
$\displaystyle-\left(2\eta^{ca}G_{\;\;\;a}^{m}F_{bc}+2\eta_{ac}G_{\;\;\;b}^{c}H^{am}\right)\eta^{nb}$
(142)

The remaining four field equations involving the torsion and co-torsion are
unchanged.

Notice that the sources on the right sides of Eqs.(141) and (142) differ only
in the overall sign. This means that both the spacetime curvature,
$\Omega_{\;\;\;nam}^{a}$, and the momentum space curvature,
$\Omega_{\;\;\;b}^{n\quad bm}$, are driven with equal strength. Thus, if
spacetime curvature is nonzero, the momentum space must also be
correspondingly curved. In torsion-free solutions the left hand side of
Eq.(142) vanishes independently of the sources, implying a constraint on the
source fields. Thus, for the
$\int\boldsymbol{\mathcal{F}}\wedge^{*}\boldsymbol{\mathcal{F}}$ source and
the Kähler case,

$\displaystyle\eta^{ca}G_{\;\;\;a}^{m}F_{bc}+\eta_{ac}G_{\;\;\;b}^{c}H^{am}$
$\displaystyle=$ $\displaystyle 0$

and this immediately shows that the source for the Einstein equation,
Eq.(141), vanishes.

The necessity for equal spacetime and momentum curvatures suggests the
possibility implementing Born reciprocity. This idea will be explored
elsewhere. However, momentum curvature also requires some part of the torsion
to be nonvanishing, and this in turn requires a different gravitational
solution than that known to reproduce general relativity. Thus, we cannot
maintain vanishing torsion without forcing the spacetime source to vanish.

In addition to the inescapability of torsion and momentum space curvature, the
independence of the sources to Eqs.(139) and (140) also raises issues with the
Kähler variation, because the method of solution employed in [6] makes use of
the near identity of these two equations. At the very least, an entirely
different form of reduction of the equations would be required, with no
guarantee that the Einstein equation would emerge.

A similar calculation shows that the same difficulties arise from the twisted
form of the action using the Kähler metric.

These issues do not arise with the Killing variation, Eqs.(38)–the source for
the spacetime curvature and momentum space curvature are independent while
remaining two variations are identical. The use of the Killing form as metric
also makes good geometric sense, since it arises directly as a symmetric form
in the Lie algebra and thus as metric of the co-tangent space. As such, it
respects the conformal invariance of the full model. The Kähler structure, by
contrast, reflects symmetries and dynamical properties within the conformal
group and depend for their existence on the solution on the biconformal space.
It is not conformally invariant.

## Appendix B: Symmetry of the twist matrix

Writing the twist matrix while keeping factors of $\eta^{ab}$ explicit, we
have

$\displaystyle K_{\;\;\;B}^{A}$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{cc}\bar{K}^{ac}&\bar{K}_{\;\;\;e}^{a}\eta^{ec}\\\
\eta^{ae}\bar{K}_{e}^{\;\;\;c}&\eta^{ae}\eta^{cf}\bar{K}_{ef}\end{array}\right)\left(\begin{array}[]{cc}\eta_{cb}&0\\\
0&\eta_{cm}\eta_{bn}\eta^{mn}\end{array}\right)$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{cc}\bar{K}^{ac}\eta_{cb}&\bar{K}_{\;\;\;b}^{a}\\\
\eta^{ae}\bar{K}_{e}^{\;\;\;c}\eta_{cb}&\eta^{ae}\bar{K}_{eb}\end{array}\right)$
$\displaystyle K_{A}^{\;\;\;B}$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{cc}\eta_{ad}\bar{K}^{db}&\bar{K}_{a}^{\;\;\;b}\\\
\eta_{ad}\bar{K}_{\;\;\;e}^{d}\eta^{eb}&\bar{K}_{ae}\eta^{eb}\end{array}\right)$

Symmetry of $K_{\;\;\;B}^{A}$ follows from the symmetry of the Killing and
Kähler forms, $K_{AB}=K_{BA},g_{AB}=g_{BA}$:

$\displaystyle K_{\;\;\;B}^{A}\equiv\bar{K}^{AC}g_{CB}$ $\displaystyle=$
$\displaystyle g_{BC}\bar{K}^{CA}=K_{B}^{\;\;\;A}$

Comparing the expressions,

$\displaystyle K_{\;\;\;B}^{A}$ $\displaystyle=$ $\displaystyle
K_{B}^{\;\;\;A}$
$\displaystyle\left(\begin{array}[]{cc}\bar{K}^{ac}\eta_{cb}&\bar{K}_{\;\;\;b}^{a}\\\
\eta^{ae}\bar{K}_{e}^{\;\;\;c}\eta_{cb}&\eta^{ae}\bar{K}_{eb}\end{array}\right)$
$\displaystyle=$
$\displaystyle\left(\begin{array}[]{cc}\eta_{bc}\bar{K}^{ca}&\eta_{bc}\bar{K}_{\;\;\;e}^{c}\eta^{ea}\\\
\bar{K}_{b}^{\;\;\;a}&\bar{K}_{bc}\eta^{ca}\end{array}\right)$

Therefore, from the upper right quadrant we must have
$\bar{K}_{\;\;\;b}^{a}=\eta_{bc}\bar{K}_{\;\;\;e}^{c}\eta^{ea}$ and similarly
$\eta^{ae}\bar{K}_{e}^{\;\;\;c}\eta_{cb}=\bar{K}_{b}^{\;\;\;a}$ from the lower
left. However we also have symmetry $\bar{K}^{AB}=\bar{K}^{BA}$ of
$\bar{K}^{AB}$ itself:

$\displaystyle\bar{K}^{AB}$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{cc}\bar{K}^{ab}&\bar{K}_{\;\;\;e}^{a}\eta^{eb}\\\
\eta^{ae}\bar{K}_{e}^{\;\;\;b}&\eta^{ae}\eta^{bf}\bar{K}_{ef}\end{array}\right)$
$\displaystyle\left[\bar{K}^{t}\right]^{BA}$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{cc}\bar{K}^{ba}&\eta^{be}\bar{K}_{e}^{\;\;\;a}\\\
\bar{K}_{\;\;\;e}^{b}\eta^{ea}&\eta^{ae}\eta^{bf}\bar{K}_{ef}\end{array}\right)$

This shows that
$\bar{K}_{\;\;\;e}^{a}\eta^{eb}=\eta^{be}\bar{K}_{e}^{\;\;\;a}$ from which it
follows that

$\displaystyle\bar{K}_{\;\;\;b}^{a}$ $\displaystyle=$
$\displaystyle\bar{K}_{b}^{\;\;\;a}$

Combine this with
$\eta^{ae}\bar{K}_{e}^{\;\;\;c}\eta_{cb}=\bar{K}_{b}^{\;\;\;a}$ and we see
that all forms are equivalent,

$\eta^{ae}\bar{K}_{e}^{\;\;\;c}\eta_{cb}=\bar{K}_{b}^{\;\;\;a}=\bar{K}_{\;\;\;b}^{a}=\eta_{bc}\bar{K}_{\;\;\;e}^{c}\eta^{ea}$

With this, we may write

$\displaystyle K_{\;\;\;B}^{A}$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{cc}\bar{K}^{ac}\eta_{cb}&\bar{K}_{\;\;\;b}^{a}\\\
\bar{K}_{\;\;\;b}^{a}&\eta^{ae}\bar{K}_{eb}\end{array}\right)$ $\displaystyle
K_{B}^{\;\;\;A}$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{cc}\eta_{bd}\bar{K}^{da}&\bar{K}_{b}^{\;\;\;a}\\\
\bar{K}_{b}^{\;\;\;a}&\bar{K}_{be}\eta^{ea}\end{array}\right)=K_{\;\;\;B}^{A}$

## Appendix C: Details of the metric variation of the action

The variation of the twisted action is lengthy and includes some subtleties,
so we include details here.

We have the dual field,

${}^{*}\boldsymbol{\mathcal{F}}$ $\displaystyle=$
$\displaystyle\frac{1}{n!\left(n-2\right)!}\left(\frac{1}{2}F_{ab}\bar{K}^{am}\bar{K}^{bn}+\mathcal{G}_{gb}\eta^{ga}\bar{K}_{a}^{\;\;\;m}\bar{K}^{bn}+\frac{1}{2}\mathcal{H}_{gh}\eta^{ga}\bar{K}_{a}^{\;\;\;m}\eta^{hb}\bar{K}_{b}^{\;\;\;n}\right)\varepsilon_{\;\;\;\qquad
mne\cdots f}^{c\cdots d}\mathbf{f}_{c\cdots d}\mathbf{e}^{e\cdots f}$
$\displaystyle+\frac{\left(-1\right)^{n-1}}{\left(n-1\right)!\left(n-1\right)!}\left(\frac{1}{2}F_{ab}\bar{K}_{\;\;\;m}^{a}\bar{K}^{bn}+\mathcal{G}_{gb}\eta^{ga}\bar{K}_{am}\bar{K}^{bn}+\frac{1}{2}\mathcal{H}_{gh}\eta^{ga}\eta^{hb}\bar{K}_{am}\bar{K}_{b}^{\;\;\;n}\right)\varepsilon_{\;\;\;\qquad
ne\cdots f}^{mc\cdots d}\mathbf{f}_{c\cdots d}\mathbf{e}^{e\cdots f}$
$\displaystyle+\frac{\left(-1\right)^{n}}{\left(n-1\right)!\left(n-1\right)!}\left(\frac{1}{2}F_{ab}\bar{K}^{am}\bar{K}_{\;\;\;n}^{b}+\mathcal{G}_{gb}\eta^{ga}\bar{K}_{a}^{\;\;\;m}\bar{K}_{\;\;\;n}^{b}+\frac{1}{2}\mathcal{H}_{gh}\eta^{ga}\eta^{hb}\bar{K}_{a}^{\;\;\;m}\bar{K}_{bn}\right)\varepsilon_{\;\;\;\qquad
me\cdots f}^{nc\cdots d}\mathbf{f}_{c\cdots d}\mathbf{e}^{e\cdots f}$
$\displaystyle+\frac{1}{n!\left(n-2\right)!}\left(\frac{1}{2}F_{ab}\bar{K}_{\;\;\;m}^{a}\bar{K}_{\;\;\;n}^{b}+\mathcal{G}_{gb}\eta^{ga}\bar{K}_{am}\bar{K}_{\;\;\;n}^{b}+\frac{1}{2}\mathcal{H}_{gh}\eta^{ga}\eta^{hb}\bar{K}_{am}\bar{K}_{bn}\right)\varepsilon_{\;\;\;\qquad
e\cdots f}^{mnc\cdots d}\mathbf{f}_{c\cdots d}\mathbf{e}^{e\cdots f}$

and, changing indices to avoid duplication, the barred field,

$\displaystyle\bar{\boldsymbol{\mathcal{F}}}$ $\displaystyle=$
$\displaystyle\frac{1}{2}\left(F_{rq}\bar{K}^{qt}\eta_{ts}+\bar{K}_{r}^{\;\;\;q}\mathcal{G}_{qs}\right)\mathbf{e}^{r}\wedge\mathbf{e}^{s}$
$\displaystyle+\frac{1}{2}\left(F_{rc}\bar{K}_{\;\;\;s}^{c}+\bar{K}_{r}^{\;\;\;c}\mathcal{H}_{cs}\right)\eta^{sw}\mathbf{e}^{r}\wedge\mathbf{f}_{w}$
$\displaystyle+\frac{1}{2}\left(-\mathcal{G}_{sq}\bar{K}^{qt}\eta_{tr}-\mathcal{G}_{qr}\eta^{qt}\bar{K}_{ts}\right)\eta^{sw}\mathbf{e}^{r}\wedge\mathbf{f}_{w}$
$\displaystyle+\frac{1}{2}\left(\mathcal{G}_{rq}\bar{K}_{\;\;\;s}^{q}+\mathcal{H}_{rq}\eta^{qt}\bar{K}_{ts}\right)\eta^{rw}\eta^{sx}\mathbf{f}_{w}\wedge\mathbf{f}_{x}$

We must wedge these together, then vary the metric.

Wedging,

$\displaystyle\bar{\boldsymbol{\mathcal{F}}}\land{}^{*}\boldsymbol{\mathcal{F}}$
$\displaystyle=$
$\displaystyle\frac{1}{2}\frac{1}{n!\left(n-2\right)!}\left(F_{rq}\bar{K}^{qt}\eta_{ts}+\bar{K}_{r}^{\;\;\;q}\mathcal{G}_{qs}\right)$
$\displaystyle\times\left(\mathcal{G}_{gb}\eta^{ga}\bar{K}_{a}^{\;\;\;m}\bar{K}^{bn}+\frac{1}{2}\mathcal{H}_{gh}\eta^{ga}\bar{K}_{a}^{\;\;\;m}\eta^{hb}\bar{K}_{b}^{\;\;\;n}\right)\varepsilon_{\;\;\;\qquad
mne\cdots f}^{c\cdots d}\mathbf{f}_{c\cdots d}\mathbf{e}^{rse\cdots f}$
$\displaystyle-\frac{1}{2}\frac{1}{\left(n-1\right)!\left(n-1\right)!}\left(F_{rc}\bar{K}_{\;\;\;s}^{c}+\bar{K}_{r}^{\;\;\;c}\mathcal{H}_{cs}\right)\eta^{sw}$
$\displaystyle\times\left(\frac{1}{2}F_{ab}\bar{K}_{\;\;\;m}^{a}\bar{K}^{bn}+\frac{1}{2}\mathcal{H}_{gh}\eta^{ga}\eta^{hb}\bar{K}_{am}\bar{K}_{b}^{\;\;\;n}\right)\varepsilon_{\;\;\;\qquad
ne\cdots f}^{mc\cdots d}\mathbf{f}_{wc\cdots d}\mathbf{e}^{re\cdots f}$
$\displaystyle+\frac{1}{2}\frac{1}{\left(n-1\right)!\left(n-1\right)!}\left(F_{rc}\bar{K}_{\;\;\;s}^{c}+\bar{K}_{r}^{\;\;\;c}\mathcal{H}_{cs}\right)\eta^{sw}$
$\displaystyle\times\left(\frac{1}{2}F_{ab}\bar{K}^{am}\bar{K}_{\;\;\;n}^{b}+\mathcal{G}_{gb}\eta^{ga}\bar{K}_{a}^{\;\;\;m}\bar{K}_{\;\;\;n}^{b}+\frac{1}{2}\mathcal{H}_{gh}\eta^{ga}\eta^{hb}\bar{K}_{a}^{\;\;\;m}\bar{K}_{bn}\right)\varepsilon_{\;\;\;\qquad
me\cdots f}^{nc\cdots d}\mathbf{f}_{wc\cdots d}\mathbf{e}^{re\cdots f}$
$\displaystyle-\frac{1}{2}\frac{1}{\left(n-1\right)!\left(n-1\right)!}\left(-\mathcal{G}_{sq}\bar{K}^{qt}\eta_{tr}-\mathcal{G}_{qr}\eta^{qt}\bar{K}_{ts}\right)\eta^{sw}$
$\displaystyle\times\left(\frac{1}{2}F_{ab}\bar{K}_{\;\;\;m}^{a}\bar{K}^{bn}+\frac{1}{2}\mathcal{H}_{gh}\eta^{ga}\eta^{hb}\bar{K}_{am}\bar{K}_{b}^{\;\;\;n}\right)\varepsilon_{\;\;\;\qquad
ne\cdots f}^{mc\cdots d}\mathbf{f}_{wc\cdots d}\mathbf{e}^{re\cdots f}$
$\displaystyle+\frac{1}{2}\frac{1}{\left(n-1\right)!\left(n-1\right)!}\left(-\mathcal{G}_{sq}\bar{K}^{qt}\eta_{tr}-\mathcal{G}_{qr}\eta^{qt}\bar{K}_{ts}\right)\eta^{sw}$
$\displaystyle\times\left(\frac{1}{2}F_{ab}\bar{K}^{am}\bar{K}_{\;\;\;n}^{b}+\mathcal{G}_{gb}\eta^{ga}\bar{K}_{a}^{\;\;\;m}\bar{K}_{\;\;\;n}^{b}+\frac{1}{2}\mathcal{H}_{gh}\eta^{ga}\eta^{hb}\bar{K}_{a}^{\;\;\;m}\bar{K}_{bn}\right)\varepsilon_{\;\;\;\qquad
me\cdots f}^{nc\cdots d}\mathbf{f}_{wc\cdots d}\mathbf{e}^{re\cdots f}$
$\displaystyle+\frac{1}{2}\frac{1}{n!\left(n-2\right)!}\left(\mathcal{G}_{rq}\bar{K}_{\;\;\;s}^{q}+\mathcal{H}_{rq}\eta^{qt}\bar{K}_{ts}\right)\eta^{rw}\eta^{sx}$
$\displaystyle\times\left(\frac{1}{2}F_{ab}\bar{K}_{\;\;\;m}^{a}\bar{K}_{\;\;\;n}^{b}+\mathcal{G}_{gb}\eta^{ga}\bar{K}_{am}\bar{K}_{\;\;\;n}^{b}\right)\varepsilon_{\;\;\;\qquad
e\cdots f}^{mnc\cdots d}\mathbf{f}_{wxc\cdots d}\mathbf{e}^{e\cdots f}$

Now use the relation between the basis forms and the volume element, and the
Kronecker reduction of pairs of Levi-Civita tensors,

$\displaystyle\mathbf{f}_{c\cdots d}\wedge\mathbf{e}^{e\cdots f}$
$\displaystyle=$ $\displaystyle\bar{e}_{c\cdots d}^{\qquad e\cdots
f}\boldsymbol{\Phi}$ $\displaystyle\varepsilon_{\;\;\;\qquad e\cdots
f}^{mnc\cdots d}\bar{e}_{pqc\cdots d}^{\qquad e\cdots f}$ $\displaystyle=$
$\displaystyle
n!\left(n-2\right)!\left(\delta_{p}^{m}\delta_{q}^{n}-\delta_{p}^{n}\delta_{q}^{m}\right)$
$\displaystyle\varepsilon_{\;\;\;\qquad ne\cdots f}^{mc\cdots
d}\bar{e}_{pc\cdots d}^{\qquad qe\cdots f}$ $\displaystyle=$
$\displaystyle\left(n-1\right)!\left(n-1\right)!\delta_{p}^{m}\delta_{n}^{q}$

to reduce the Lagrange density to

$\displaystyle\bar{\boldsymbol{\mathcal{F}}}\land{}^{*}\boldsymbol{\mathcal{F}}$
$\displaystyle=$
$\displaystyle\frac{1}{2}\left(F_{rq}\bar{K}^{qt}\eta_{ts}+\bar{K}_{r}^{\;\;\;q}\mathcal{G}_{qs}\right)\left(\mathcal{G}_{gb}\eta^{ga}\bar{K}_{a}^{\;\;\;m}\bar{K}^{bn}+\frac{1}{2}\mathcal{H}_{gh}\eta^{ga}\bar{K}_{a}^{\;\;\;m}\eta^{hb}\bar{K}_{b}^{\;\;\;n}\right)\left(\delta_{m}^{r}\delta_{n}^{s}-\delta_{m}^{s}\delta_{n}^{r}\right)\boldsymbol{\Phi}$
$\displaystyle-\frac{1}{2}\left(F_{rc}\bar{K}_{\;\;\;s}^{c}+\bar{K}_{r}^{\;\;\;c}\mathcal{H}_{cs}\right)\eta^{sw}\left(\frac{1}{2}F_{ab}\bar{K}_{\;\;\;m}^{a}\bar{K}^{bn}+\frac{1}{2}\mathcal{H}_{gh}\eta^{ga}\eta^{hb}\bar{K}_{am}\bar{K}_{b}^{\;\;\;n}\right)\delta_{w}^{m}\delta_{n}^{r}\boldsymbol{\Phi}$
$\displaystyle+\frac{1}{2}\left(F_{rc}\bar{K}_{\;\;\;s}^{c}+\bar{K}_{r}^{\;\;\;c}\mathcal{H}_{cs}\right)\eta^{sw}\left(\frac{1}{2}F_{ab}\bar{K}^{am}\bar{K}_{\;\;\;n}^{b}+\mathcal{G}_{gb}\eta^{ga}\bar{K}_{a}^{\;\;\;m}\bar{K}_{\;\;\;n}^{b}+\frac{1}{2}\mathcal{H}_{gh}\eta^{ga}\eta^{hb}\bar{K}_{a}^{\;\;\;m}\bar{K}_{bn}\right)\delta_{w}^{n}\delta_{m}^{r}\boldsymbol{\Phi}$
$\displaystyle-\frac{1}{2}\left(-\mathcal{G}_{sq}\bar{K}^{qt}\eta_{tr}-\mathcal{G}_{qr}\eta^{qt}\bar{K}_{ts}\right)\eta^{sw}\left(\frac{1}{2}F_{ab}\bar{K}_{\;\;\;m}^{a}\bar{K}^{bn}+\frac{1}{2}\mathcal{H}_{gh}\eta^{ga}\eta^{hb}\bar{K}_{am}\bar{K}_{b}^{\;\;\;n}\right)\delta_{w}^{m}\delta_{n}^{r}\boldsymbol{\Phi}$
$\displaystyle+\frac{1}{2}\left(-\mathcal{G}_{sq}\bar{K}^{qt}\eta_{tr}-\mathcal{G}_{qr}\eta^{qt}\bar{K}_{ts}\right)\eta^{sw}\left(\frac{1}{2}F_{ab}\bar{K}^{am}\bar{K}_{\;\;\;n}^{b}+\mathcal{G}_{gb}\eta^{ga}\bar{K}_{a}^{\;\;\;m}\bar{K}_{\;\;\;n}^{b}+\frac{1}{2}\mathcal{H}_{gh}\eta^{ga}\eta^{hb}\bar{K}_{a}^{\;\;\;m}\bar{K}_{bn}\right)\delta_{w}^{n}\delta_{m}^{r}\boldsymbol{\Phi}$
$\displaystyle+\frac{1}{2}\left(\mathcal{G}_{rq}\bar{K}_{\;\;\;s}^{q}+\mathcal{H}_{rq}\eta^{qt}\bar{K}_{ts}\right)\eta^{rw}\eta^{sx}\left(\frac{1}{2}F_{ab}\bar{K}_{\;\;\;m}^{a}\bar{K}_{\;\;\;n}^{b}+\mathcal{G}_{gb}\eta^{ga}\bar{K}_{am}\bar{K}_{\;\;\;n}^{b}\right)\left(\delta_{w}^{m}\delta_{x}^{n}-\delta_{w}^{n}\delta_{x}^{m}\right)\boldsymbol{\Phi}$

Then, absorbing the $\eta_{ab}$ and $\delta_{b}^{a}$ factors,

$\displaystyle\bar{\boldsymbol{\mathcal{F}}}\land{}^{*}\boldsymbol{\mathcal{F}}$
$\displaystyle=$
$\displaystyle\frac{1}{2}\left(F_{mq}\bar{K}^{qt}\eta_{tn}-F_{nq}\bar{K}^{qt}\eta_{tm}+\bar{K}_{m}^{\;\;\;q}\mathcal{G}_{qn}-\bar{K}_{n}^{\;\;\;q}\mathcal{G}_{qm}\right)\left(\mathcal{G}_{gb}\eta^{ga}\bar{K}_{a}^{\;\;\;m}\bar{K}^{bn}+\frac{1}{2}\mathcal{H}_{gh}\eta^{ga}\bar{K}_{a}^{\;\;\;m}\eta^{hb}\bar{K}_{b}^{\;\;\;n}\right)\boldsymbol{\Phi}$
$\displaystyle-\frac{1}{2}\left(F_{nc}\bar{K}_{\;\;\;s}^{c}+\bar{K}_{n}^{\;\;\;c}\mathcal{H}_{cs}\right)\eta^{sm}\left(\frac{1}{2}F_{ab}\bar{K}_{\;\;\;m}^{a}\bar{K}^{bn}+\frac{1}{2}\mathcal{H}_{gh}\eta^{ga}\eta^{hb}\bar{K}_{am}\bar{K}_{b}^{\;\;\;n}\right)\boldsymbol{\Phi}$
$\displaystyle+\frac{1}{2}\left(F_{mc}\bar{K}_{\;\;\;s}^{c}+\bar{K}_{m}^{\;\;\;c}\mathcal{H}_{cs}\right)\eta^{sn}\left(\frac{1}{2}F_{ab}\bar{K}^{am}\bar{K}_{\;\;\;n}^{b}+\mathcal{G}_{gb}\eta^{ga}\bar{K}_{a}^{\;\;\;m}\bar{K}_{\;\;\;n}^{b}+\frac{1}{2}\mathcal{H}_{gh}\eta^{ga}\eta^{hb}\bar{K}_{a}^{\;\;\;m}\bar{K}_{bn}\right)\boldsymbol{\Phi}$
$\displaystyle-\frac{1}{2}\left(-\mathcal{G}_{sq}\bar{K}^{qt}\eta_{tn}-\mathcal{G}_{qn}\eta^{qt}\bar{K}_{ts}\right)\eta^{sm}\left(\frac{1}{2}F_{ab}\bar{K}_{\;\;\;m}^{a}\bar{K}^{bn}+\frac{1}{2}\mathcal{H}_{gh}\eta^{ga}\eta^{hb}\bar{K}_{am}\bar{K}_{b}^{\;\;\;n}\right)\boldsymbol{\Phi}$
$\displaystyle+\frac{1}{2}\left(-\mathcal{G}_{sq}\bar{K}^{qt}\eta_{tm}-\mathcal{G}_{qm}\eta^{qt}\bar{K}_{ts}\right)\eta^{sn}\left(\frac{1}{2}F_{ab}\bar{K}^{am}\bar{K}_{\;\;\;n}^{b}+\mathcal{G}_{gb}\eta^{ga}\bar{K}_{a}^{\;\;\;m}\bar{K}_{\;\;\;n}^{b}+\frac{1}{2}\mathcal{H}_{gh}\eta^{ga}\eta^{hb}\bar{K}_{a}^{\;\;\;m}\bar{K}_{bn}\right)\boldsymbol{\Phi}$
$\displaystyle+\frac{1}{2}\left(\mathcal{G}_{rq}\bar{K}_{\;\;\;s}^{q}\eta^{rm}\eta^{sn}-\mathcal{G}_{rq}\bar{K}_{\;\;\;s}^{q}\eta^{rn}\eta^{sm}+\mathcal{H}_{rq}\eta^{qt}\bar{K}_{ts}\eta^{rm}\eta^{sn}-\mathcal{H}_{rq}\eta^{qt}\bar{K}_{ts}\eta^{rn}\eta^{sm}\right)$
$\displaystyle\times\left(\frac{1}{2}F_{ab}\bar{K}_{\;\;\;m}^{a}\bar{K}_{\;\;\;n}^{b}+\mathcal{G}_{gb}\eta^{ga}\bar{K}_{am}\bar{K}_{\;\;\;n}^{b}\right)\boldsymbol{\Phi}$

After varying the metric, the null-orthonormal form of the metric is restored,
so we may anticipate this and drop terms in the product such as
$\mathcal{G}_{gb}\eta^{ga}\bar{K}_{a}^{\;\;\;m}\bar{K}^{bn}\mathcal{G}_{sq}\bar{K}^{qt}\eta_{tr}$
which will ultimately vanish. Terms with two or more factors of $\bar{K}^{ab}$
and/or $\bar{K}_{ab}$ will vanish, so we have only terms with one of these and
two off diagonal factors such as $\bar{K}_{a}^{\;\;\;b}$, or terms with three
off diagonal factors. In all cases where there is one factor of $\bar{K}^{ab}$
or $\bar{K}_{ab}$, it is this factor that must be varied so the off diagonal
factors may be replaced. For example, once the null orthonormal basis is
restored, the only surviving term of the variation of

$F_{mq}\eta_{tn}\frac{1}{2}\mathcal{H}_{gh}\eta^{ga}\eta^{hb}\bar{K}^{qt}\bar{K}_{a}^{\;\;\;m}\bar{K}_{b}^{\;\;\;n}$

will be

$F_{mq}\eta_{tn}\frac{1}{2}\mathcal{H}_{gh}\eta^{ga}\eta^{hb}\delta\bar{K}^{qt}\bar{K}_{a}^{\;\;\;m}\bar{K}_{b}^{\;\;\;n}$

which we may immediately write as

$F_{mq}\eta_{tn}\frac{1}{2}\mathcal{H}_{gh}\eta^{ga}\eta^{hb}\delta\bar{K}^{qt}\delta_{a}^{\;\;\;m}\delta_{b}^{\;\;\;n}=\frac{1}{2}\left(F_{aq}\eta_{tb}\mathcal{H}_{gh}\eta^{ga}\eta^{hb}\right)\delta\bar{K}^{qt}$

Terms with three off-diagonal components of the metric must be retained until
after variation.

Distributing fully, then making these reductions where possible, we collect
terms

$\displaystyle\bar{\boldsymbol{\mathcal{F}}}\land{}^{*}\boldsymbol{\mathcal{F}}$
$\displaystyle=$
$\displaystyle\frac{1}{2}\left(\frac{1}{2}F_{ec}\eta^{cb}F_{db}-\frac{1}{2}F_{ec}\eta^{ca}F_{ad}+\frac{1}{2}F_{ad}\eta_{eb}H^{ab}+\frac{1}{2}\mathcal{H}_{ec}\eta^{cb}F_{db}-\frac{1}{2}F_{bd}\eta_{ea}H^{ab}-\frac{1}{2}\mathcal{H}_{ec}\eta^{ca}F_{ad}\right)\bar{K}^{de}\boldsymbol{\Phi}$
$\displaystyle+\frac{1}{2}\left(-\mathcal{G}_{ea}\mathcal{G}_{cd}\eta^{ca}+\mathcal{G}_{ae}\mathcal{G}_{cd}\eta^{ca}-\mathcal{G}_{cd}\eta_{ea}\eta^{cb}\mathcal{G}_{gb}\eta^{ga}\right)\bar{K}^{de}\boldsymbol{\Phi}$
$\displaystyle+\frac{1}{2}\left(-\frac{1}{2}F_{bc}\eta^{ce}H^{db}-\mathcal{H}_{bc}\eta^{ce}\frac{1}{2}H^{db}+\frac{1}{2}F_{ac}\eta^{ce}H^{ad}+\frac{1}{2}\mathcal{H}_{ac}\eta^{ce}H^{ad}+\frac{1}{2}H^{ad}\eta^{eb}F_{ab}-\frac{1}{2}H^{bd}\eta^{ea}F_{ab}\right)\bar{K}_{de}\boldsymbol{\Phi}$
$\displaystyle+\frac{1}{2}\left(\mathcal{G}_{rs}\eta^{re}\eta^{sb}\mathcal{G}_{gb}\eta^{gd}-\mathcal{G}_{rs}\eta^{rb}\eta^{se}\mathcal{G}_{gb}\eta^{gd}-\mathcal{G}_{qa}\eta^{qd}\eta^{eb}\mathcal{G}_{gb}\eta^{ga}\right)\bar{K}_{de}\boldsymbol{\Phi}$
$\displaystyle+\frac{1}{2}\left(\frac{1}{2}\mathcal{G}_{qn}H^{ab}\delta_{m}^{c}-\frac{1}{2}\mathcal{G}_{qm}H^{ab}\delta_{n}^{c}+F_{mq}\eta^{cb}\mathcal{G}_{gn}\eta^{ga}\right)\bar{K}_{\;\;\;c}^{q}\bar{K}_{a}^{\;\;\;m}\bar{K}_{b}^{\;\;\;n}\boldsymbol{\Phi}$
$\displaystyle+\frac{1}{2}\left(\mathcal{H}_{qs}\eta^{sb}\mathcal{G}_{gn}\eta^{ga}\delta_{m}^{c}+\frac{1}{2}\mathcal{G}_{rq}\eta^{ra}\eta^{cb}F_{mn}-\frac{1}{2}\mathcal{G}_{rq}\eta^{rb}\eta^{ca}F_{mn}\right)\bar{K}_{\;\;\;a}^{m}\bar{K}_{\;\;\;b}^{n}\bar{K}_{\;\;\;c}^{q}\boldsymbol{\Phi}$

and consolidate using symmetries,

$\displaystyle\bar{\boldsymbol{\mathcal{F}}}\land{}^{*}\boldsymbol{\mathcal{F}}$
$\displaystyle=$
$\displaystyle\frac{1}{2}\left(F_{dc}\eta^{cb}F_{eb}+2F_{ad}H^{ab}\eta_{be}\right)\bar{K}^{de}\boldsymbol{\Phi}$
$\displaystyle+\frac{1}{2}\left(\mathcal{G}_{ae}\mathcal{G}_{cd}\eta^{ca}-2\mathcal{G}_{ea}\mathcal{G}_{cd}\eta^{ca}\right)\bar{K}^{de}\boldsymbol{\Phi}$
$\displaystyle+\frac{1}{2}\left(\mathcal{H}_{ab}H^{ad}\eta^{be}+2F_{ac}H^{ad}\eta^{ce}\right)\bar{K}_{de}\boldsymbol{\Phi}$
$\displaystyle+\frac{1}{2}\left(\mathcal{G}_{ca}\mathcal{G}_{db}\eta^{ab}-2\mathcal{G}_{da}\mathcal{G}_{bc}\eta^{ab}\right)\eta^{de}\eta^{cf}\bar{K}_{ef}\boldsymbol{\Phi}$
$\displaystyle+\frac{1}{2}\left(\frac{1}{2}\mathcal{G}_{dn}H^{ab}\delta_{m}^{c}-\frac{1}{2}\mathcal{G}_{dm}H^{ab}\delta_{n}^{c}+F_{md}\eta^{cb}\mathcal{G}_{gn}\eta^{ga}\right)\bar{K}_{a}^{\;\;\;m}\bar{K}_{b}^{\;\;\;n}\bar{K}_{\;\;\;c}^{d}\boldsymbol{\Phi}$
$\displaystyle+\frac{1}{2}\left(\mathcal{H}_{ds}\eta^{sb}\mathcal{G}_{gn}\eta^{ga}\delta_{m}^{c}+\frac{1}{2}\mathcal{G}_{rd}\eta^{ra}\eta^{cb}F_{mn}-\frac{1}{2}\mathcal{G}_{rd}\eta^{rb}\eta^{ca}F_{mn}\right)\bar{K}_{\;\;\;a}^{m}\bar{K}_{\;\;\;b}^{n}\bar{K}_{\;\;\;c}^{d}\boldsymbol{\Phi}$

Checking the limit in the orthonormal basis, we find the correct form,
$\bar{\boldsymbol{\mathcal{F}}}\land{}^{*}\boldsymbol{\mathcal{F}}=\left(\mathcal{G}_{ab}H^{ab}+F_{ab}\eta^{ac}\eta^{bd}\mathcal{G}_{cd}\right)\boldsymbol{\Phi}$.
Proceeding, we vary the metric

$\displaystyle\delta\left(\bar{\boldsymbol{\mathcal{F}}}\land{}^{*}\boldsymbol{\mathcal{F}}\right)$
$\displaystyle=$
$\displaystyle\frac{1}{2}\left(F_{dc}\eta^{cb}F_{eb}+2F_{ad}H^{ab}\eta_{be}+\left(\mathcal{G}_{ae}-2\mathcal{G}_{ea}\right)\mathcal{G}_{cd}\eta^{ca}\right)\delta\bar{K}^{de}\boldsymbol{\Phi}$
$\displaystyle+\frac{1}{2}\bar{K}_{b}^{\;\;\;n}\bar{K}_{\;\;\;c}^{d}\left(\frac{1}{2}\mathcal{G}_{db}H^{ab}\delta_{m}^{c}-\frac{1}{2}\mathcal{G}_{dm}H^{ab}\delta_{b}^{c}+F_{md}\eta^{cb}\mathcal{G}_{gb}\eta^{ga}\right)\delta\bar{K}_{a}^{\;\;\;m}\boldsymbol{\Phi}$
$\displaystyle+\frac{1}{2}\left(\frac{1}{2}\mathcal{G}_{dn}H^{ab}\delta_{m}^{c}-\frac{1}{2}\mathcal{G}_{dm}H^{ab}\delta_{n}^{c}+F_{md}\eta^{cb}\mathcal{G}_{gn}\eta^{ga}\right)\bar{K}_{a}^{\;\;\;m}\delta\bar{K}_{b}^{\;\;\;n}\bar{K}_{\;\;\;c}^{d}\boldsymbol{\Phi}$
$\displaystyle+\frac{1}{2}\left(\frac{1}{2}\mathcal{G}_{dn}H^{ab}\delta_{m}^{c}-\frac{1}{2}\mathcal{G}_{dm}H^{ab}\delta_{n}^{c}+F_{md}\eta^{cb}\mathcal{G}_{gn}\eta^{ga}\right)\bar{K}_{a}^{\;\;\;m}\bar{K}_{b}^{\;\;\;n}\delta\bar{K}_{\;\;\;c}^{d}\boldsymbol{\Phi}$
$\displaystyle+\frac{1}{2}\left(\mathcal{H}_{ds}\eta^{sb}\mathcal{G}_{gn}\eta^{ga}\delta_{m}^{c}+\frac{1}{2}\mathcal{G}_{rd}\eta^{ra}\eta^{cb}F_{mn}-\frac{1}{2}\mathcal{G}_{rd}\eta^{rb}\eta^{ca}F_{mn}\right)\delta\bar{K}_{\;\;\;a}^{m}\bar{K}_{\;\;\;b}^{n}\bar{K}_{\;\;\;c}^{d}\boldsymbol{\Phi}$
$\displaystyle+\frac{1}{2}\left(\mathcal{H}_{ds}\eta^{sb}\mathcal{G}_{gn}\eta^{ga}\delta_{m}^{c}+\frac{1}{2}\mathcal{G}_{rd}\eta^{ra}\eta^{cb}F_{mn}-\frac{1}{2}\mathcal{G}_{rd}\eta^{rb}\eta^{ca}F_{mn}\right)\bar{K}_{\;\;\;a}^{m}\delta\bar{K}_{\;\;\;b}^{n}\bar{K}_{\;\;\;c}^{d}\boldsymbol{\Phi}$
$\displaystyle+\frac{1}{2}\left(\mathcal{H}_{ds}\eta^{sb}\mathcal{G}_{gn}\eta^{ga}\delta_{m}^{c}+\frac{1}{2}\mathcal{G}_{rd}\eta^{ra}\eta^{cb}F_{mn}-\frac{1}{2}\mathcal{G}_{rd}\eta^{rb}\eta^{ca}F_{mn}\right)\bar{K}_{\;\;\;a}^{m}\bar{K}_{\;\;\;b}^{n}\delta\bar{K}_{\;\;\;c}^{d}\boldsymbol{\Phi}$
$\displaystyle+\frac{1}{2}\left(\mathcal{H}_{ab}H^{ad}\eta^{be}+2F_{ac}H^{ad}\eta^{ce}\right)\bar{K}_{de}\boldsymbol{\Phi}+\frac{1}{2}\mathcal{G}_{ca}\eta^{ab}\left(\mathcal{G}_{db}-2\mathcal{G}_{bd}\right)\eta^{de}\eta^{cf}\bar{K}_{ef}\boldsymbol{\Phi}$
$\displaystyle+\left(\mathcal{G}_{ab}H^{ab}+F_{ab}\eta^{ac}\eta^{bd}\mathcal{G}_{cd}\right)\delta\boldsymbol{\Phi}$

At this point we may replace remaining unvaried metric components, leaving

$\displaystyle\delta\left(\bar{\boldsymbol{\mathcal{F}}}\land{}^{*}\boldsymbol{\mathcal{F}}\right)$
$\displaystyle=$
$\displaystyle\frac{1}{2}\left(F_{dc}\eta^{cb}F_{eb}+2F_{ad}H^{ab}\eta_{be}+\left(\mathcal{G}_{ae}-2\mathcal{G}_{ea}\right)\mathcal{G}_{cd}\eta^{ca}\right)\delta\bar{K}^{de}\boldsymbol{\Phi}$
$\displaystyle+\frac{1}{2}\left(\frac{1}{2}\mathcal{G}_{cb}H^{ab}\delta_{m}^{c}-\frac{1}{2}\mathcal{G}_{cm}H^{ab}\delta_{b}^{c}+F_{mc}\eta^{cb}\mathcal{G}_{gb}\eta^{ga}\right)\delta\bar{K}_{a}^{\;\;\;m}\boldsymbol{\Phi}$
$\displaystyle+\frac{1}{2}\left(\frac{1}{2}\mathcal{G}_{an}H^{ab}-\frac{1}{2}\mathcal{G}_{na}H^{ab}+F_{ac}\eta^{cn}\mathcal{G}_{gb}\eta^{ga}\right)\delta\bar{K}_{b}^{\;\;\;n}\boldsymbol{\Phi}$
$\displaystyle+\frac{1}{2}\left(\frac{1}{2}\mathcal{G}_{db}H^{cb}-\frac{1}{2}\mathcal{G}_{da}H^{ac}+F_{ad}\eta^{cb}\mathcal{G}_{gb}\eta^{ga}\right)\bar{K}_{b}^{\;\;\;n}\delta\bar{K}_{\;\;\;c}^{d}\boldsymbol{\Phi}$
$\displaystyle+\frac{1}{2}\left(\mathcal{H}_{cs}\eta^{sb}\mathcal{G}_{gb}\eta^{ga}\delta_{m}^{c}+\frac{1}{2}\mathcal{G}_{rc}\eta^{ra}\eta^{cb}F_{mb}-\frac{1}{2}\mathcal{G}_{rc}\eta^{rb}\eta^{ca}F_{mb}\right)\delta\bar{K}_{\;\;\;a}^{m}\boldsymbol{\Phi}$
$\displaystyle+\frac{1}{2}\left(\mathcal{H}_{cs}\eta^{sb}\mathcal{G}_{gn}\eta^{gc}+\frac{1}{2}\mathcal{G}_{rc}\eta^{ra}\eta^{cb}F_{an}-\frac{1}{2}\mathcal{G}_{rc}\eta^{rb}\eta^{ca}F_{an}\right)\delta\bar{K}_{\;\;\;b}^{n}\boldsymbol{\Phi}$
$\displaystyle+\frac{1}{2}\left(\mathcal{H}_{ds}\eta^{sb}\mathcal{G}_{gb}\eta^{gc}+\frac{1}{2}\mathcal{G}_{rd}\eta^{ra}\eta^{cb}F_{ab}-\frac{1}{2}\mathcal{G}_{rd}\eta^{rb}\eta^{ca}F_{ab}\right)\delta\bar{K}_{\;\;\;c}^{d}\boldsymbol{\Phi}$
$\displaystyle+\frac{1}{2}\left(\left(\mathcal{H}_{ab}H^{ae}\eta^{bf}+2F_{ac}H^{ae}\eta^{cf}\right)+\mathcal{G}_{ca}\eta^{ab}\left(\mathcal{G}_{db}-2\mathcal{G}_{bd}\right)\eta^{de}\eta^{cf}\right)\bar{K}_{ef}\boldsymbol{\Phi}$
$\displaystyle+\left(\mathcal{G}_{ab}H^{ab}+F_{ab}\eta^{ac}\eta^{bd}\mathcal{G}_{cd}\right)\delta\boldsymbol{\Phi}$

Finally, we collect all terms by type of variation,

$\displaystyle\delta\left(\bar{\boldsymbol{\mathcal{F}}}\land{}^{*}\boldsymbol{\mathcal{F}}\right)$
$\displaystyle=$
$\displaystyle\frac{1}{2}\left(F_{dc}\eta^{cb}F_{eb}+2F_{ad}H^{ab}\eta_{be}+\left(\mathcal{G}_{ae}-2\mathcal{G}_{ea}\right)\mathcal{G}_{cd}\eta^{ca}\right)\delta\bar{K}^{de}\boldsymbol{\Phi}$
$\displaystyle+\frac{1}{2}\left(\frac{1}{2}\mathcal{G}_{cb}H^{nb}\delta_{m}^{c}-\frac{1}{2}\mathcal{G}_{cm}H^{nb}\delta_{b}^{c}+F_{mc}\eta^{cb}\mathcal{G}_{gb}\eta^{gn}\right)\delta\bar{K}_{n}^{\;\;\;m}\boldsymbol{\Phi}$
$\displaystyle+\frac{1}{2}\left(\frac{1}{2}\mathcal{G}_{am}H^{an}-\frac{1}{2}\mathcal{G}_{ma}H^{an}+F_{ac}\eta^{cn}\mathcal{G}_{gm}\eta^{ga}\right)\delta\bar{K}_{n}^{\;\;\;m}\boldsymbol{\Phi}$
$\displaystyle+\frac{1}{2}\left(\frac{1}{2}\mathcal{G}_{mb}H^{nb}-\frac{1}{2}\mathcal{G}_{ma}H^{an}+F_{am}\eta^{nb}\mathcal{G}_{gb}\eta^{ga}\right)\delta\bar{K}_{\;\;\;n}^{m}\boldsymbol{\Phi}$
$\displaystyle+\frac{1}{2}\left(\mathcal{H}_{cs}\eta^{sb}\mathcal{G}_{gb}\eta^{gn}\delta_{m}^{c}+\frac{1}{2}\mathcal{G}_{rc}\eta^{rn}\eta^{cb}F_{mb}-\frac{1}{2}\mathcal{G}_{rc}\eta^{rb}\eta^{cn}F_{mb}\right)\delta\bar{K}_{\;\;\;n}^{m}\boldsymbol{\Phi}$
$\displaystyle+\frac{1}{2}\left(\mathcal{H}_{cs}\eta^{sn}\mathcal{G}_{gm}\eta^{gc}+\frac{1}{2}\mathcal{G}_{rc}\eta^{ra}\eta^{cn}F_{am}-\frac{1}{2}\mathcal{G}_{rc}\eta^{rn}\eta^{ca}F_{am}\right)\delta\bar{K}_{\;\;\;n}^{m}\boldsymbol{\Phi}$
$\displaystyle+\frac{1}{2}\left(\mathcal{H}_{ms}\eta^{sb}\mathcal{G}_{gb}\eta^{gn}+\frac{1}{2}\mathcal{G}_{rm}\eta^{ra}\eta^{nb}F_{ab}-\frac{1}{2}\mathcal{G}_{rm}\eta^{rb}\eta^{na}F_{ab}\right)\delta\bar{K}_{\;\;\;n}^{m}\boldsymbol{\Phi}$
$\displaystyle+\frac{1}{2}\left(\left(\mathcal{H}_{ab}H^{ae}\eta^{bf}+2F_{ac}H^{ae}\eta^{cf}\right)+\mathcal{G}_{ca}\eta^{ab}\left(\mathcal{G}_{db}-2\mathcal{G}_{bd}\right)\eta^{de}\eta^{cf}\right)\bar{K}_{ef}\boldsymbol{\Phi}$
$\displaystyle+\left(\mathcal{G}_{ab}H^{ab}+F_{ab}\eta^{ac}\eta^{bd}\mathcal{G}_{cd}\right)\delta\boldsymbol{\Phi}$

This allows further simplifications, then including the variation of the
volume form we have the final result,

$\displaystyle\delta\left(\bar{\boldsymbol{\mathcal{F}}}\land{}^{*}\boldsymbol{\mathcal{F}}\right)$
$\displaystyle=$
$\displaystyle\frac{1}{2}\left(F_{dc}\eta^{cb}F_{eb}+2F_{ad}H^{ab}\eta_{be}+\left(\mathcal{G}_{ae}-2\mathcal{G}_{ea}\right)\mathcal{G}_{cd}\eta^{ca}\right)\delta\bar{K}^{de}\boldsymbol{\Phi}$
$\displaystyle+H^{na}\left(2\mathcal{G}_{ma}-\mathcal{G}_{am}\right)\delta\bar{K}_{n}^{\;\;\;m}\boldsymbol{\Phi}$
$\displaystyle+F_{ma}\left(\mathcal{G}^{na}-2\mathcal{G}^{an}\right)\delta\bar{K}_{\;\;\;n}^{m}\boldsymbol{\Phi}$
$\displaystyle+\frac{1}{2}\left(\left(\mathcal{H}_{ab}H^{ae}\eta^{bf}+2F_{ac}H^{ae}\eta^{cf}\right)+\mathcal{G}_{ca}\eta^{ab}\left(\mathcal{G}_{db}-2\mathcal{G}_{bd}\right)\eta^{de}\eta^{cf}\right)\bar{K}_{ef}\boldsymbol{\Phi}$
$\displaystyle+\left(\mathcal{G}_{ab}H^{ab}+F_{ab}\eta^{ac}\eta^{bd}\mathcal{G}_{cd}\right)\delta\boldsymbol{\Phi}$

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# On the stability of a double porous elastic system with visco-porous
dampings

Ahmed KEDDI(1), Aicha NEMSI2 & Abdelfeteh FAREH(2)
1 Laboratory of Mathematics, Modelisation and Application (LAMMA)
Department of Mathematics and computer sciences, University of Adrar, Adrar,
Algeria
2 Laboratory of operators theory and PDEs: Foundations and applications
Faculty of Exact sciences, University of El Oued, P.B. 789, El Oued 39000,
Algeria.

###### Abstract

In this paper we consider a one dimensional elastic system with double
porosity structure and with frictional damping in both porous equations. We
introduce two stability numbers $\chi_{0}$ and $\chi_{1}$ and prove that the
solution of the system decays exponentially provided that $\chi_{0}=0$ and
$\chi_{1}\neq 0.$ Otherwise, we prove the lack of exponential decay. Our
results improve the results of [5] and [14].

_2020 Mathematics Subject Classification_ : 35B35; 35B40; 35L15; 35Q74; 74F10;
93D05

_Key words and phrases_ : double porosity, well-posedness, exponential decay,
lack of exponential decay.

††emails<EMAIL_ADDRESS><EMAIL_ADDRESS>
Corresponding<EMAIL_ADDRESS>
##  1 Introduction

In this paper, we are concerned with the following system

$\left\\{\begin{array}[c]{lc}\rho u_{tt}=\mu
u_{xx}+b\varphi_{x}+d\psi_{x}&\mbox{in}~{}(0,\pi)\times\mathbb{R}_{+},\\\
\kappa_{1}\varphi_{tt}=\alpha\varphi_{xx}+\beta\psi_{xx}-bu_{x}-\alpha_{1}\varphi-\alpha_{3}\psi-\tau_{1}\varphi_{t}-\tau_{2}\psi_{t}&\mbox{in}~{}(0,\pi)\times\mathbb{R}_{+},\\\
\kappa_{2}\psi_{tt}=\beta\varphi_{xx}+\gamma\psi_{xx}-du_{x}-\alpha_{3}\varphi-\alpha_{2}\psi-\tau_{3}\varphi_{t}-\tau_{4}\psi_{t}&\mbox{in}~{}(0,\pi)\times\mathbb{R}_{+},\end{array}\right.$
(1.1)

where $u$ is the transversal displacement of a one-dimensional porous elastic
solid of length $\pi$, $\varphi$ and $\psi$ are the porous unknown functions
one associated to the pores in the skeleton and the other associated with the
fissures in the material body. The parameter $\rho,\kappa_{1}$ and
$\kappa_{2},$ which assumed to be strictly positive, are the mass density, and
the products of the mass density by the equilibrated inertia, respectively.
The coefficients
$\mu,\alpha,\beta,\gamma,\alpha_{1},\alpha_{2},\alpha_{3},b,d,\tau_{1}$,
$\tau_{2},\tau_{3}$ and $\tau_{4}$ are parameters related on the properties of
the material. We assume that they satisfy some restrictions that will be
specified later.

The system considered here, represented an elastic solid with double porosity
structure in the framework of the theory of elastic materials with voids
developed by Nunziato-Cowin [9]. This approach has been used by Ieşan and
Quintanilla [12] to derive a new theory of thermoelastic solids which have a
double porosity structure. In contrast to the classical theory the new one is
not based on Darcy’s law, and the porosity structure in the case of
equilibrium is influenced by the displacement field.

The origin of the classical theory of elastic materials with double porosity
goes back to the works of Barenblatt _et al._ [3, 4]. The authors introduced
two liquid pressures at each point of the material which allows the body to
have a double porosity structure: a macro porosity connected to pores in the
body and a micro porosity connected to fissures in the skeleton.

In the last few years, a great interest has been given to the analysis of the
longtime behavior of solutions of porous thermoelastic problems. A part of
this interest stems from the need to have general results that explain the
experimental observations of engineers. The earliest contribution in this
direction was achieved by Quintanilla [17]. He considered the porous elastic
system

$\left\\{\begin{array}[c]{lc}\rho_{0}u_{tt}=\mu
u_{xx}+\beta\varphi_{x}&\mbox{in}~{}~{}(0,\pi)\times(0,+\infty),\\\
\rho_{0}\kappa\varphi_{tt}=\alpha\varphi_{xx}-\beta
u_{x}-\xi\varphi-\tau\varphi_{t}&\mbox{in}~{}~{}(0,\pi)\times(0,+\infty),\end{array}\right.$
(1.2)

where $u$ is the transversal displacement and $\varphi$ is the volume
fraction. He used Hurwitz theorem and showed that the porous dissipation
$\tau\varphi_{t}$ is not powerful enough to produce an exponential stability.

Several dissipative mechanisms have been examined to stabilize system (1.2)
exponentially. Casas and Quintanilla [8] coupled system (1.1) (for $\tau=0$)
with the heat equation, and proved the non-exponential stability. However, if
thermal and porous dissipations or micro-thermal and viscoelastic dissipations
are combined then the solution decays exponentially [7, 13].

Apalara [1] considered the porous thermoelastic system

$\left\\{\begin{array}[c]{ll}\rho u_{tt}-\mu
u_{xx}-b\phi_{x}=0,&in\;\left(0,1\right)\times\left(0,+\infty\right)\\\
J\phi_{tt}-\delta\phi_{xx}+bu_{x}+\xi\phi+\tau\phi_{t}=0,&in\;\left(0,1\right)\times\left(0,+\infty\right),\end{array}\right.$

with different boundary conditions. He investigated the case of equal wave
speeds $\dfrac{\mu}{\rho}=\dfrac{\delta}{J}$ and proved that the unique
dissipation in the porous equation leads to an exponential stability. He also
replaced the frictional damping $\tau\phi_{t}$ by the memory term
$\displaystyle\int_{0}^{t}g(t-s)\phi_{xx}(s)ds$ and obtained a general rate of
decay [2]. We notice that the results of [1, 2] disprove Magaña and
Quintanilla’s claim that a porous-elastic system with a single dissipation
mechanism can not be exponentially stable [13].

In the context of double porous thermoelasticity, Bazarra _et al._ [5]
considered the system

$\left\\{\begin{array}[c]{l}\rho u_{tt}=\mu
u_{xx}+b\varphi_{x}+d\psi_{x}-\beta\theta_{x},\\\
\kappa_{1}\varphi_{tt}=\alpha\varphi_{xx}+b_{1}\psi_{xx}-bu_{x}-\alpha_{1}\varphi-\alpha_{3}\psi+\gamma_{1}\theta-\varepsilon_{1}\varphi_{t}-\varepsilon_{2}\psi_{t},\\\
\kappa_{2}\psi_{tt}=b_{1}\varphi_{xx}+\gamma\psi_{xx}-du_{x}-\alpha_{3}\varphi-\alpha_{2}\psi+\gamma_{2}\theta-\varepsilon_{3}\varphi_{t}-\varepsilon_{4}\psi_{t},\\\
c\theta_{t}=\kappa\theta_{xx}-\beta
u_{tx}-\gamma_{1}\varphi_{t}-\gamma_{2}\psi_{t},\end{array}\right.$ (1.3)

with the boundary conditions

$u\left(x,t\right)=\varphi_{x}\left(x,t\right)=\psi_{x}\left(x,t\right)=\theta_{x}\left(x,t\right)=0,\;x=0,\;x=\pi,\;\;\forall
t\geq 0.$

They proved that the solution decays exponentially when porous dissipation is
assumed for each porous equations. If the dissipation is considered only on
one porous structure, the solution cannot be asymptotically stable in general.
However, they give a sufficient conditions for which the solutions decay
exponentially. See also [6].

Recently, Nemsi and Fareh [14] proved that the solution of the system

$\left\\{\begin{array}[c]{lc}\rho u_{tt}=\mu
u_{xx}+b\varphi_{x}+d\psi_{x}+\lambda
u_{txx},&\mbox{in}~{}~{}(0,L)\times(0,\infty),\\\
\kappa_{1}\varphi_{tt}=\alpha\varphi_{xx}+b_{1}\psi_{xx}-bu_{x}-\alpha_{1}\varphi-\alpha_{3}\psi-\tau_{1}\varphi_{t}&\mbox{in}~{}~{}(0,L)\times(0,\infty),\\\
\kappa_{2}\psi_{tt}=b_{1}\varphi_{xx}+\gamma\psi_{xx}-du_{x}-\alpha_{3}\varphi-\alpha_{2}\psi-\tau_{2}\psi_{t}&\mbox{in}~{}~{}(0,L)\times(0,\infty),\end{array}\right.$
(1.4)

with the boundary conditions

$\begin{array}[c]{ll}u(t,0)=u(t,L)=\varphi_{x}(t,0)=\varphi_{x}(t,L)=\psi_{x}(t,0)=\psi_{x}(t,L)=0&\mbox{in}~{}(0,\infty),\end{array}$

decays exponentially without any assumption on the wave speeds.

In this paper we consider system (1.1) subjected to the initial data

$\displaystyle u\left(x,0\right)$
$\displaystyle=u_{0}\left(x\right),~{}u_{t}\left(x,0\right)=u_{1}\left(x\right),$
$\displaystyle\varphi\left(x,0\right)$
$\displaystyle=\varphi_{0}\left(x\right),~{}\varphi_{t}\left(x,0\right)=\varphi_{1}\left(x\right),$
(1.5) $\displaystyle\psi\left(x,0\right)$
$\displaystyle=\psi_{0}\left(x\right),~{}\psi_{t}\left(x,0\right)=\psi_{1}\left(x\right)$

for all $x\in\left(0,\pi\right)$ and the boundary conditions

$u_{x}\left(0,t\right)=u_{x}\left(\pi,t\right)=\varphi\left(0,t\right)=\varphi\left(\pi,t\right)=\psi\left(0,t\right)=\psi\left(\pi,t\right)=0,\;t\geq
0,$ (1.6)

or

$u\left(0,t\right)=u\left(\pi,t\right)=\varphi_{x}\left(0,t\right)=\varphi_{x}\left(\pi,t\right)=\psi_{x}\left(0,t\right)=\psi_{x}\left(\pi,t\right)=0,\;t\geq
0.$ (1.7)

Note that system (1.1) coincides with (1.3) in the isothermal case
($\beta=\gamma_{1}=\gamma_{2}=0$) and with (1.4) for $\lambda=0.$ Therefore,
system (1.1) lacks thermal and viscoelastic dissipations. Moreover, in somehow
the two porous functions can be viewed as a multivalued function
$(\varphi,\psi)$, consequently, system (1.1) can be viewed as a porous elastic
system with only one dissipation. Thus, our exponential stability extends the
result of [1, 2] and our stability number generalizes the wave speeds
equality.

We assume that the constitutive coefficients
$\mu,\alpha,\beta,\gamma,\alpha_{1}$ and $\alpha_{2}$ are positive and as
coupling is considered the coefficients $b,d$ must not be zero simultaneously.
Next, we define the energy associated with the solution
$\left(u,\varphi,\psi\right)$ of system (1.1) by

$\displaystyle E\left(t\right)$
$\displaystyle:=\frac{1}{2}\int_{0}^{\pi}\left[\rho
u_{t}^{2}+\kappa_{1}\varphi_{t}^{2}+\kappa_{2}\psi_{t}^{2}+\mu
u_{x}^{2}+\alpha\varphi_{x}^{2}+\gamma\psi_{x}^{2}+\alpha_{1}\varphi^{2}+\alpha_{2}\psi^{2}\right.$
(1.8)
$\displaystyle\left.+2\beta\varphi_{x}\psi_{x}+2bu_{x}\varphi+2du_{x}\psi+2\alpha_{3}\varphi\psi\right]dx.$

######  Remark 1.

To guarantee that the energy $E\left(t\right)$ is a positive definite form, we
assume that the matrix

$A=\left(\begin{array}[c]{ccccc}\mu&b&d&0&0\\\ b&\alpha_{1}&\alpha_{3}&0&0\\\
d&\alpha_{3}&\alpha_{2}&0&0\\\ 0&0&0&\alpha&\beta\\\
0&0&0&\beta&\gamma\end{array}\right)$

is positive definite.

Indeed, since any principal submatrix of a positive definite matrix is also
positive definite, then

$\left(\alpha_{1}-\frac{b^{2}}{\mu}\right)\left(\alpha_{2}-\frac{d^{2}}{\mu}\right)-\left(\alpha_{3}-\frac{bd}{\mu}\right)^{2}>0,$
(1.9)
$\alpha_{1}\mu-b^{2}>0,\;\;\alpha_{2}\mu-d^{2}>0,\,\alpha_{1}\alpha_{2}-\alpha_{3}^{2}>0\;\mbox{and}\;\alpha\gamma-\beta^{2}>0.$
(1.10)

Therefore,

$\alpha\varphi_{x}^{2}+\gamma\psi_{x}^{2}+2\beta\varphi_{x}\psi_{x}=\frac{1}{2}\left(\alpha-\frac{\beta^{2}}{\gamma}\right)\varphi_{x}^{2}+\frac{1}{2}\left(\gamma-\frac{\beta^{2}}{\alpha}\right)\psi_{x}^{2}$
$+\frac{\alpha}{2}\left(\varphi_{x}+\frac{\beta}{\alpha}\psi_{x}\right)^{2}+\frac{\gamma}{2}\left(\psi_{x}+\frac{\beta}{\gamma}\varphi_{x}\right)^{2}\geq
0.$

Moreover, there exists a $\varepsilon>0$ such that the matrix

$B=\left(\begin{array}[c]{ccc}\mu-\varepsilon&b&d\\\
b&\alpha_{1}-\varepsilon&\alpha_{3}\\\
d&\alpha_{3}&\alpha_{2}-\varepsilon\end{array}\right)$

still positive definite. Thus,

$E\left(t\right)=\frac{1}{2}\int_{0}^{\pi}\left(\rho
u_{t}^{2}+\kappa_{1}\varphi_{t}^{2}+\kappa_{2}\psi_{t}^{2}+\varepsilon
u_{x}^{2}+\varepsilon\varphi^{2}+\varepsilon\psi^{2}\right)dx$
$+\frac{1}{2}\int_{0}^{\pi}\left[\left(\mu-\varepsilon\right)u_{x}^{2}+2bu_{x}\varphi+2du_{x}\psi+\left(\alpha_{1}-\varepsilon\right)\varphi^{2}+2\alpha_{3}\varphi\psi+\left(\alpha_{2}-\varepsilon\right)\psi^{2}\right]dx$
$+\frac{1}{2}\int_{0}^{\pi}\left[\alpha\left(\varphi_{x}+\frac{\beta}{\alpha}\psi_{x}\right)^{2}+\gamma\left(\psi_{x}+\frac{\beta}{\gamma}\varphi_{x}\right)^{2}\right]dx$
$+\frac{1}{2}\left(\alpha-\frac{\beta^{2}}{\gamma}\right)\int_{0}^{\pi}\left|\varphi_{x}\right|^{2}dx+\frac{1}{2}\left(\gamma-\frac{\beta^{2}}{\alpha}\right)\int_{0}^{\pi}\left|\psi_{x}\right|^{2}dx\geq
0.$

The rest of the paper is organized as follows: in Section 2, we prove the well
posedness of the problem determined by (1.1), (1) and (1.7). In Section 3 we
define two stability numbers $\chi_{0}$ and $\chi_{1}$ and prove, by the use
of the multiplier method, that the solution decays exponentially, provided
that $\chi_{0}=0$ and $\chi_{1}\neq 0$. Section 4 is devoted to the proof of
the lack of the exponential decay when $\chi_{1}=0$ or $\chi_{0}\neq 0$.

##  2 Existence and uniqueness

In this section we prove the existence and the uniqueness of a solution to the
problem determined by system (1.1) and conditions (1) and (1.7), the case of
boundary conditions (1.6) is similar.

As Neumann boundary conditions are considered for $\varphi$ and $\psi,$
Poincaré’s inequality cannot be applied. From the second and the third
equations of (1.1) and boundary conditions (1.7), we have

$\begin{array}[c]{l}\displaystyle\frac{d^{2}}{dt^{2}}\int_{0}^{\pi}\varphi
dx=-\frac{\alpha_{1}}{\kappa_{1}}\int_{0}^{\pi}\varphi
dx-\frac{\alpha_{3}}{\kappa_{1}}\int_{0}^{\pi}\psi
dx-\frac{\tau_{1}}{\kappa_{1}}\frac{d}{dt}\int_{0}^{\pi}\varphi
dx-\frac{\tau_{2}}{\kappa_{1}}\frac{d}{dt}\int_{0}^{\pi}\psi dx,\\\ \\\
\displaystyle\frac{d^{2}}{dt^{2}}\int_{0}^{\pi}\psi
dx=-\frac{\alpha_{3}}{\kappa_{2}}\int_{0}^{\pi}\varphi
dx-\frac{\alpha_{2}}{\kappa_{2}}\int_{0}^{\pi}\psi
dx-\frac{\tau_{3}}{\kappa_{2}}\frac{d}{dt}\int_{0}^{\pi}\varphi_{t}dx-\frac{\tau_{4}}{\kappa_{2}}\frac{d}{dt}\int_{0}^{\pi}\psi_{t}dx.\end{array}$
(2.11)

So if we set $X=\left(\int_{0}^{\pi}\varphi
dx,\int_{0}^{\pi}\varphi_{t}dx,\int_{0}^{\pi}\psi
dx,\int_{0}^{\pi}\psi_{t}dx\right)^{T}$ then (2.11) can be written

$X_{t}\left(t\right)=MX\left(t\right),\ X\left(0\right)=X_{0},$ (2.12)

where

$M=\left(\begin{array}[c]{cccc}0&1&0&0\\\
-\frac{\alpha_{1}}{\kappa_{1}}&-\frac{\tau_{1}}{\kappa_{1}}&-\frac{\alpha_{3}}{\kappa_{1}}&-\frac{\tau_{2}}{\kappa_{1}}\\\
0&0&0&1\\\
-\frac{\alpha_{3}}{\kappa_{2}}&-\frac{\tau_{3}}{\kappa_{2}}&-\frac{\alpha_{2}}{\kappa_{2}}&-\frac{\tau_{4}}{\kappa_{2}}\end{array}\right)$

and

$X_{0}=\left(\int_{0}^{\pi}\varphi_{0}dx,\int_{0}^{\pi}\varphi_{1}dx,\int_{0}^{\pi}\psi_{0}dx,\int_{0}^{\pi}\psi_{1}dx\right)^{T}.$

Solving (2.12) we get

$X\left(t\right)=\exp\left(tM\right)X_{0},$

in particular,

$\begin{array}[c]{cc}\displaystyle\int_{0}^{\pi}\varphi
dx=\sum_{k=1}^{4}\left(\exp\left(tM\right)\right)_{1k}X_{0k},&\displaystyle\int_{0}^{\pi}\psi
dx=\sum_{j=1}^{4}\left(\exp\left(tM\right)\right)_{3k}X_{0k}.\end{array}$

Therefore, if we set

$\begin{array}[c]{cc}\displaystyle\bar{\varphi}=\varphi-\sum_{k=1}^{4}\left(\exp\left(tM\right)\right)_{1k}X_{0k},&\displaystyle\bar{\psi}=\psi-\sum_{j=1}^{4}\left(\exp\left(tM\right)\right)_{3k}X_{0k},\end{array}$

then $\left(u,\overline{\varphi},\overline{\psi}\right)$ solves (1.1) with
boundary conditions (1.7), and we have

$\int_{0}^{\pi}\overline{\varphi}dx=\int_{0}^{\pi}\overline{\psi}dx=0,$

which allows to apply Poincaré’s inequality. In the sequel we will work with
$\bar{\varphi}$ and $\bar{\psi}$ but for convenience, we write $\varphi,\psi$
instead of $\bar{\varphi},\bar{\psi}$ respectively.

Furthermore, as porous dissipations are considered, the weights of porous
dampings $\tau_{1},\tau_{2},\tau_{3}$ and $\tau_{4}$ are assumed to satisfy

$\tau_{1}>0,\;4\tau_{1}\tau_{4}>\left(\tau_{2}+\tau_{3}\right)^{2}.$ (2.13)

######  Lemma 1.

The energy $E\left(t\right)$ satisfies along the solution
$\left(u,\varphi,\psi\right)$ of (1.1)-(1.6) the estimate

$E^{\prime}\left(t\right)=-\tau_{1}\int_{0}^{\pi}\varphi_{t}^{2}dx-\tau_{4}\int_{0}^{\pi}\psi_{t}^{2}dx-\left(\tau_{2}+\tau_{3}\right)\int_{0}^{\pi}\varphi_{t}\psi
dx$ (2.14)

and we have

$E^{\prime}\left(t\right)=-\frac{1}{2}\left(\tau_{1}-\frac{\left(\tau_{2}+\tau_{3}\right)^{2}}{4\tau_{2}}\right)\int_{0}^{\pi}\varphi_{t}^{2}dx-\frac{1}{2}\left(\tau_{2}-\frac{\left(\tau_{2}+\tau_{3}\right)^{2}}{4\tau_{1}}\right)\int_{0}^{\pi}\psi_{t}^{2}dx$
$-\frac{\tau_{1}}{2}\int_{0}^{\pi}\left(\varphi_{t}+\frac{\left(\tau_{2}+\tau_{3}\right)}{2\tau_{1}}\psi_{t}\right)^{2}dx-\frac{\tau_{2}}{2}\int_{0}^{\pi}\left(\psi_{t}+\frac{\left(\tau_{2}+\tau_{3}\right)}{2\tau_{2}}\varphi_{t}\right)^{2}dx\leq
0.$ (2.15)

###### Proof.

Multiplying the equations of (1.1) by $u_{t},\varphi_{t}$ and $\psi_{t}$
respectively, then integrating with respect to $x$ over $\left(0,\pi\right)$
and using integration by parts and boundary conditions (1.7), the estimate
(2.14) follows immediately. ∎

To prove the well-posedness, we use a semigroup approach. First, we introduce
the energy space

$\mathcal{H}=H_{0}^{1}\left(0,\pi\right)\times L^{2}\left(0,\pi\right)\times
H_{\ast}^{1}\left(0,\pi\right)\times L_{\ast}^{2}\left(0,\pi\right)\times
H_{\ast}^{1}\left(0,\pi\right)\times L_{\ast}^{2}\left(0,\pi\right),$

where,

$\displaystyle H_{\ast}^{1}\left(0,\pi\right)$ $\displaystyle=\left\\{\phi\in
H^{1}\left(0,\pi\right):\int_{0}^{\pi}\phi\left(x\right)dx=0\right\\},\,$
$\displaystyle L_{\ast}^{2}\left(0,\pi\right)$ $\displaystyle=\left\\{\phi\in
L^{2}\left(0,\pi\right):\int_{0}^{\pi}\phi\left(x\right)dx=0\right\\}.$

We note that $L_{\ast}^{2}\left(0,\pi\right)$ and
$H_{\ast}^{1}\left(0,\pi\right)$ are closed subspaces of
$L^{2}\left(0,\pi\right)$ and $H^{1}\left(0,\pi\right)$ respectively. Thus,
they are Hilbert spaces and so $\mathcal{H}$ is.

Next, we rewrite the system (1.1) in the setting of Lumer-Phillips theorem, to
do so we introduce the new variables $v=u_{t},\phi=\varphi_{t}$ and
$w=\psi_{t}$ the system (1.1) becomes

$\left\\{\begin{array}[c]{l}u_{t}=v\\\ v_{t}=\dfrac{1}{\rho}\left(\mu
u_{xx}+b\varphi_{x}+d\psi_{x}\right),\\\ \varphi_{t}=\phi\\\
\phi_{t}=\dfrac{1}{\kappa_{1}}\left(\alpha\varphi_{xx}+\beta\psi_{xx}-bu_{x}-\alpha_{1}\varphi-\alpha_{3}\psi-\tau_{1}\phi-\tau_{2}w\right),\\\
\psi_{t}=w\\\
w_{t}=\dfrac{1}{\kappa_{2}}\left(\beta\varphi_{xx}+\gamma\psi_{xx}-du_{x}-\alpha_{3}\varphi-\alpha_{2}\psi-\tau_{3}\phi-\tau_{4}w\right),\end{array}\right.$

which can be written

$\left\\{\begin{array}[c]{l}U_{t}=\mathcal{A}U,\\\
U\left(0\right)=U_{0},\end{array}\right.$ (2.16)

where
$\mathcal{A}:D\left(\mathcal{A}\right)\subset\mathcal{H}\longrightarrow\mathcal{H}$
is the operator defined by

$\mathcal{A=}\left(\begin{array}[c]{cccccc}0&I&0&0&0&0\\\
\frac{\mu}{\rho}\partial_{xx}&0&\frac{b}{\rho}\partial_{x}&0&\frac{d}{\rho}\partial_{x}&0\\\
0&0&0&I&0&0\\\
-\frac{b}{\kappa_{1}}\partial_{x}&0&\frac{\alpha}{\kappa_{1}}\partial_{xx}-\frac{\alpha}{\kappa_{1}}&-\frac{\tau_{1}}{\kappa_{1}}&\frac{\beta}{\kappa_{1}}\partial_{xx}-\frac{\alpha_{3}}{\kappa_{1}}&-\frac{\tau_{2}}{\kappa_{1}}\\\
0&0&0&0&0&I\\\
-\frac{d}{\kappa_{2}}\partial_{x}&0&\frac{\beta}{\kappa_{2}}\partial_{xx}-\alpha_{3}&-\frac{\tau_{3}}{\kappa_{2}}&\frac{\gamma}{\kappa_{2}}\partial_{xx}-\frac{\alpha_{2}}{\kappa_{2}}&-\frac{\tau_{4}}{\kappa_{2}}\end{array}\right)$

with domain

$D\left(\mathcal{A}\right)=\left(H^{2}\left(0,\pi\right)\cap
H_{0}^{1}\left(0,\pi\right)\right)\times H_{0}^{1}\left(0,\pi\right)\times
H_{\ast}^{2}\left(0,\pi\right)\times H_{\ast}^{1}\left(0,\pi\right)\times
H_{\ast}^{2}\left(0,\pi\right)\times H_{\ast}^{1}\left(0,\pi\right).$

Here $I$ is the identity operator, $\partial$ denotes the derivative with
respect to $x$ and

$H_{\ast}^{2}\left(0,\pi\right)=\left\\{\phi\in
H^{2}\left(0,\pi\right):\phi\left(0\right)=\phi\left(\pi\right)=0\right\\}.$

The following two theorems are useful to proof our well posedness result.

######  Theorem 1.

(Lumer-Phillips) [15, 18] Let $\mathcal{A}:D(\mathcal{A})\subset
H\longrightarrow H$ be a densely defined operator. Then $\mathcal{A}$
generates a C0-semigroup of contractions on $H$ if and only if

  * (i)

$\mathcal{A}$ is dissipative;

  * (ii)

there exists $\lambda>0$ such that $\lambda I-\mathcal{A}$ is surjective.

######  Theorem 2.

[18] Let $\mathcal{A}:D(\mathcal{A})\subset H\longrightarrow H$ be the
infinitesimal generator of a C0-semigroup $\\{S(t);t\geq 0\\}$. Then, for each
$\xi\in D(\mathcal{A})$ and each $t\geq 0$, we have $S(t)\xi\in
D(\mathcal{A})$, and the mapping

$\displaystyle t\longrightarrow S(t)\xi$

is of class $C^{1}$ on $\left[0,+\infty\right)$ and satisfies

$\frac{d}{dt}(S(t)\xi)=\mathcal{A}S(t)\xi=S(t)\mathcal{A}\xi.$

Now, we state and prove the well-posedness theorem of the problem (1.1), (1)
and (1.7).

######  Theorem 3.

For any
$U_{0}=\left(u_{0},u_{1},\varphi_{0},\varphi_{1},\psi_{0},\psi_{1}\right)\in\mathcal{H}$,
the problem (1.1),(1) and (1.7) has a unique weak solution
$\left(u,\varphi,\psi\right)$ satisfies the property:

$\displaystyle u$ $\displaystyle\in
C\left(\left[0,+\infty\right[;H_{0}^{1}\left(0,\pi\right)\right)\cap
C^{1}\left(\left[0,+\infty\right[;L^{2}\left(0,\pi\right)\right),$
$\displaystyle\varphi,\psi$ $\displaystyle\in
C\left(\left[0,+\infty\right[;H_{\ast}^{1}\left(0,\pi\right)\right)\cap
C^{1}\left(\left[0,+\infty\right[;L_{\ast}^{2}\left(0,\pi\right)\right).$

Moreover, if $U_{0}\in D\left(\mathcal{A}\right),$ the solution
$\left(u,\varphi,\psi\right)$ satisfies

$\displaystyle u$ $\displaystyle\in C\left(\left[0,+\infty\right[;H^{2}\cap
H_{0}^{1}\left(0,\pi\right)\right)\cap
C^{1}\left(\left[0,+\infty\right[;H_{0}^{1}\left(0,\pi\right)\right)\cap
C^{2}\left(\left[0,+\infty\right[;L^{2}\left(0,\pi\right)\right),$
$\displaystyle\varphi,\psi$ $\displaystyle\in
C\left(\left[0,+\infty\right[;H_{\ast}^{2}\left(0,\pi\right)\right)\cap
C^{1}\left(\left[0,+\infty\right[;H_{\ast}^{1}\left(0,\pi\right)\right)\cap
C^{2}\left(\left[0,+\infty\right[;L_{\ast}^{2}\left(0,\pi\right)\right).$

###### Proof.

According to the Lumer-Phillips theorem, it suffices to prove that the
operator $\mathcal{A}$ is dissipative and maximal.

First, we have for any $U\in D\left(\mathcal{A}\right)$,

$\operatorname{Re}\left\langle\mathcal{A}U,U\right\rangle_{\mathcal{H}}=-\tau_{1}\int_{0}^{\pi}\varphi_{t}^{2}dx-\left(\tau_{2}+\tau_{3}\right)\int_{0}^{\pi}\varphi_{t}\psi_{t}dx-\tau_{4}\int_{0}^{\pi}\psi_{t}^{2}dx\leq
0.$

Therefore, $\mathcal{A}$ is dissipative.

Secondly,

let $F=\left(f_{1},f_{2},f_{3},f_{4},f_{5},f_{6}\right)\in\mathcal{H},$ and
find $U\in D\left(\mathcal{A}\right)$ such that $\mathcal{A}U=\mathcal{F}$,
that is,

$\left\\{\begin{array}[c]{l}v=f_{1}\in H_{0}^{1}\\\ \mu
u_{xx}+b\varphi_{x}+d\psi_{x}=\rho f_{2}\in L^{2},\\\ \phi=f_{3}\in
H_{\ast}^{1}\\\
\alpha\varphi_{xx}+\beta\psi_{xx}-bu_{x}-\alpha_{1}\varphi-\alpha_{3}\psi-\tau_{1}\phi-\tau_{2}w=\kappa_{1}f_{4}\in
L_{\ast}^{2},\\\ w=f_{5}\in H_{\ast}^{1}\\\
\beta\varphi_{xx}+\gamma\psi_{xx}-du_{x}-\alpha_{3}\varphi-\alpha_{2}\psi-\tau_{3}\phi-\tau_{4}w=\kappa_{2}f_{6}\in
L_{\ast}^{2}.\end{array}\right.$

From the first, the third and the fifth equations we have $v\in
H_{0}^{1}\left(0,\pi\right)$ and $\phi,w\in H_{\ast}^{1}\left(0,\pi\right).$
Substituting $v,\phi$ and $w$ by $f_{1},f_{3}$ and $f_{5}$ respectively we
obtain

$\left\\{\begin{array}[c]{l}\displaystyle\mu
u_{xx}+b\varphi_{x}+d\psi_{x}=\rho f_{2}\in L^{2},\\\
\displaystyle\alpha\varphi_{xx}+\beta\psi_{xx}-bu_{x}-\alpha_{1}\varphi-\alpha_{3}\psi=\kappa_{1}f_{4}+\tau_{1}f_{3}+\tau_{2}f_{5}=g_{1}\in
L_{\ast}^{2},\\\
\displaystyle\beta\varphi_{xx}+\gamma\psi_{xx}-du_{x}-\alpha_{3}\varphi-\alpha_{2}\psi=\kappa_{2}f_{6}+\tau_{3}f_{3}+\tau_{4}f_{5}=g_{2}\in
L_{\ast}^{2},\end{array}\right.$ (2.17)

Taking the $L^{2}$-product of (2.17)1 ,(2.17)2 and (2.17)3 by
$u^{\ast},\varphi^{\ast}$ and $\psi^{\ast}$ respectively, using integration by
parts and adding the obtained equations, we arrive at

$a\left(V,V^{\ast}\right)=L\left(V^{\ast}\right),$ (2.18)

where, $a$ is the bilinear form defined on
$\mathcal{W}=\left(H_{0}^{1}\left(0,\pi\right)\times
H_{\ast}^{1}\left(0,\pi\right)\times H_{\ast}^{1}\left(0,\pi\right)\right)$
and for
$V=\left(u,\varphi,\psi\right),~{}\left(u^{\ast},\varphi^{\ast},\psi^{\ast}\right)\in\mathcal{W}$,
by

$\displaystyle a\left(V,V^{\ast}\right)=$
$\displaystyle\mu\int_{0}^{\pi}u_{x}u_{x}^{\ast}dx+b\int_{0}^{\pi}\varphi
u_{x}^{\ast}dx+d\int_{0}^{\pi}\psi
u_{x}^{\ast}dx+\alpha\int_{0}^{\pi}\varphi_{x}\varphi_{x}^{\ast}dx$
$\displaystyle+\beta\int_{0}^{\pi}\psi_{x}\varphi_{x}^{\ast}dx+b\int_{0}^{\pi}u_{x}\varphi^{\ast}dx+\alpha_{1}\int_{0}^{\pi}\varphi\varphi^{\ast}dx+\alpha_{3}\int_{0}^{\pi}\psi\varphi^{\ast}dx$
$\displaystyle+\beta\int_{0}^{\pi}\varphi_{x}\psi_{x}^{\ast}dx+\gamma\int_{0}^{\pi}\psi_{x}\psi_{x}^{\ast}dx+d\int_{0}^{\pi}u_{x}\psi^{\ast}dx$
$\displaystyle+\alpha_{3}\int_{0}^{\pi}\varphi\psi^{\ast}dx+\alpha_{2}\int_{0}^{\pi}\psi\psi^{\ast}dx$

and $L$ is the linear form defined by

$L\left(V^{\ast}\right)=-\rho\int_{0}^{\pi}f_{2}u^{\ast}dx-\int_{0}^{\pi}g_{1}\varphi^{\ast}dx-\int_{0}^{\pi}g_{2}\psi^{\ast}dx.$

Clearly, $a$ and $L$ are continuous. Furthermore, from Remark 1, there exists
$\varepsilon>0,$ such that

$\displaystyle a\left(V,V\right)=$
$\displaystyle\mu\int_{0}^{\pi}u_{x}^{2}dx+\alpha\int_{0}^{\pi}\varphi_{x}^{2}dx+\gamma\int_{0}^{\pi}\psi_{x}^{2}dx+\alpha_{1}\int_{0}^{\pi}\varphi^{2}dx+\alpha_{2}\int_{0}^{\pi}\psi^{2}dx$
$\displaystyle+2b\int_{0}^{\pi}u_{x}\varphi dx+2d\int_{0}^{\pi}\psi
u_{x}dx+2\beta\int_{0}^{\pi}\varphi_{x}\psi_{x}dx+2\alpha_{3}\int_{0}^{\pi}\psi\varphi
dx,$ $\displaystyle\geq$
$\displaystyle\frac{1}{2}\left(\alpha-\frac{\beta^{2}}{\gamma}\right)\int_{0}^{\pi}\varphi_{x}^{2}dx+\frac{1}{2}\left(\gamma-\frac{\beta^{2}}{\alpha}\right)\int_{0}^{\pi}\psi_{x}^{2}dx+\varepsilon\int_{0}^{\pi}\left(u_{x}^{2}+\varphi^{2}+\psi^{2}\right)dx.$

Thus,

$a\left(V,V\right)\geq c\left\|V\right\|_{\mathcal{W}}^{2},$

for
$c=\dfrac{1}{2}\min\\{\displaystyle\alpha-\frac{\beta^{2}}{\gamma},\displaystyle\gamma-\frac{\beta^{2}}{\alpha},2\varepsilon\\},$
which shows that $a$ is coercive. Therefore, Lax-Milgram theorem ensures the
existence of a unique $V=\left(u,\varphi,\psi\right)$

$\in\mathcal{W}$ satisfying

$\begin{array}[c]{cc}a\left(V,V^{\ast}\right)=L\left(V^{\ast}\right),&\forall
V^{\ast}\in\mathcal{W}.\end{array}$

Now, taking $V^{\ast}=(u^{\ast},0,0)$ in (2.18) we get

$\mu\int_{0}^{\pi}u_{x}u_{x}^{\ast}=-\int_{0}^{\pi}\left(\rho
f_{2}-b\varphi_{x}-d\psi_{x}\right)u^{\ast}dx,\,\forall u^{\ast}\in
H_{0}^{1}.$ (2.19)

The elliptic regularity theory shows that

$u\in H^{2}\left(0,\pi\right),$

with

$u_{xx}=\frac{1}{\mu}\left(\rho f_{2}-b\varphi_{x}-d\psi_{x}\right),$

which solves the first equation of (2.17).
Next, let $\varphi^{\ast}\in H_{0}^{1}\left(0,\pi\right)$ and define

$\widetilde{\varphi}\left(x\right)=\varphi^{\ast}\left(x\right)-\int_{0}^{\pi}\varphi^{\ast}\left(x\right)dx,$

clearly, $\widetilde{\varphi}\in H_{\ast}^{1}\left(0,\pi\right).$ Taking
$V^{\ast}=\left(0,\widetilde{\varphi},0\right)$ in (2.18) we get

$\int_{0}^{\pi}\left(\alpha\varphi_{x}+\beta\psi_{x}\right)\widetilde{\varphi}_{x}dx=-\int_{0}^{\pi}\left(g_{1}+bu_{x}+\alpha_{1}\varphi+\alpha_{3}\psi\right)\widetilde{\varphi}dx,\,\forall\widetilde{\varphi}\in
H_{\ast}^{1},$ (2.20)

which means that

$\alpha\varphi+\beta\psi\in H^{2}\left(0,\pi\right),$ (2.21)

with

$\alpha\varphi_{xx}+\beta\psi_{xx}=g_{1}+bu_{x}+\alpha_{1}\varphi+\alpha_{3}\psi.$

Similarly, we obtain

$\beta\varphi+\gamma\psi\in H^{2}\left(0,\pi\right)$ (2.22)
$\beta\varphi_{xx}+\gamma\psi_{xx}=g_{2}+du_{x}+\alpha_{3}\varphi+\alpha_{2}\psi.$

From (2.21) and (2.22) we get

$\varphi,\psi\in H^{2}\left(0,\pi\right).$

To show that $\varphi$ belongs to $H_{\ast}^{2}\left(0,\pi\right)$ we take
$\varphi^{\ast}\in C^{1}\left(0,\pi\right)$ in (2.20) and define
$\widetilde{\varphi}$ as above, then using integration by parts, we obtain,

$\left[\left(\alpha\varphi_{x}+\beta\psi_{x}\right)\widetilde{\varphi}\right]_{0}^{\pi}-\int_{0}^{\pi}\left(\alpha\varphi_{xx}+\beta\psi_{xx}-g_{1}-bu_{x}-\alpha_{1}\varphi-\alpha_{3}\psi\right)\widetilde{\varphi}dx=0,\,\forall\widetilde{\varphi}\in
H_{\ast}^{1}.$ (2.23)

First, we take $\widetilde{\varphi}\in C_{0}^{1}\left(0,\pi\right),$ we get

$\alpha\varphi_{xx}+\beta\psi_{xx}=g_{1}+bu_{x}+\alpha_{1}\varphi+\alpha_{3}\psi,\quad
a.e.\;\mbox{in}\;\left(0,\pi\right).$

Back to (2.23), we get

$\left(\alpha\varphi_{x}\left(\pi\right)+\beta\psi_{x}\left(\pi\right)\right)\widetilde{\varphi}\left(\pi\right)-\left(\alpha\varphi_{x}\left(0\right)+\beta\psi_{x}\left(0\right)\right)\widetilde{\varphi}\left(0\right)=0,\,\forall\widetilde{\varphi}\in
H_{\ast}^{1}.$

As $\widetilde{\varphi}$ is arbitrary in $H_{\ast}^{1}\left(0,\pi\right)$, we
obtain

$\alpha\varphi_{x}\left(\pi\right)+\beta\psi_{x}\left(\pi\right)=0\;\mbox{and}\;\alpha\varphi_{x}\left(0\right)+\beta\psi_{x}\left(0\right)=0.$

Similarly, we obtain

$\beta\varphi_{x}\left(\pi\right)+\gamma\psi_{x}\left(\pi\right)=0\;\mbox{and}\;\beta\varphi_{x}\left(0\right)+\gamma\psi_{x}\left(0\right)=0.$

Therefore, $\varphi,\psi\in H_{\ast}^{2}\left(0,\pi\right),$ consequently
$U\in D\left(\mathcal{A}\right),$ and $0\in\rho\left(\mathcal{A}\right).$
Moreover, using a geometric series argument we prove that $\lambda
I-\mathcal{A}=\mathcal{A}(\lambda\mathcal{A}^{-1}-I)$ is invertible for
$|\lambda|<\|\mathcal{A}^{-1}\|$, then $\lambda\in\rho(\mathcal{A})$, which
completes the proof that $\mathcal{A}$ is the infinitesimal generator of a
C${}_{0}-$semigroup, then the Lumer-Phillips theorem ensures the existence of
unique solution to the problem (1.1),(1) and (1.7) satisfying the statements
of Theorem 3. ∎

######  Remark 2.

We note that if $U_{0}\in D(\mathcal{A})$ then the solution
$U(t)=e^{t\mathcal{A}}U_{0}\in C((0,\infty);D(\mathcal{A}))\cap
C^{1}((0,\infty);\mathcal{H})$ and (2.16) is satisfied in $\mathcal{H}$ for
every $t>0$. It turns out that $u,\varphi,\psi$ satisfy (1.1) in the strong
sense.

If $U_{0}\in\mathcal{H}$ there exists a sequence $U_{0n}\in D(\mathcal{A})$
converging to $U_{0}$ in $\mathcal{H}$. Accordingly, there exists a sequence
of solutions $U_{n}(t)=e^{t\mathcal{A}}U_{0n}$ such that
$u_{n},\varphi_{n},\psi_{n}$ satisfy (1.1) in $L^{2}$ for every $t>0$, and for
any $T>0$, $u_{n}\rightarrow u$ in $C((0,T),H_{0}^{1})\cap
C^{1}((0,T);L^{2})$, $\varphi_{n}\rightarrow\varphi$ and
$\psi_{n}\rightarrow\psi$ in $C((0,T),H_{\ast}^{1})\cap C^{1}((0,T);L^{2})$.
Therefore, if we multiply the equations of (1.1) for
$u_{n},\varphi_{n},\psi_{n}$ by $u^{\ast}\in H_{0}^{1}$ and
$\varphi^{\ast},\psi^{\ast}\in H_{\ast}^{1}$, respectively, then integrate by
parts with respect to $x$ and integrate with respect to $t$, finally passing
to the limit, we find that $u,\varphi$ and $\psi$ are weak solutions to the
variational form of system (1.1).

###  2.1 Exponential stability

In the present section we tackle the main objective of this paper, that is the
prove of the exponential decay of the solution of (1.1). First we introduce
the two following constants

$\chi_{0}=\left(\frac{\mu\kappa_{1}}{\rho}-\alpha\right)\left(\frac{\mu\kappa_{2}}{\rho}-\gamma\right)-\beta^{2},$

and

$\chi_{1}=d^{2}\left(\frac{\mu\kappa_{1}}{\rho}-\alpha\right)+b^{2}\left(\frac{\mu\kappa_{2}}{\rho}-\gamma\right)+2bd\beta.$

Our stability result reads as follow:

######  Theorem 4.

Let __ $\left(u,\varphi,\psi\right)$ be a solution of problem _(_ 1.1) with
boundary conditions (1.7). Assuming that

$\chi_{0}=0\hskip 3.61371pt\mbox{and }\hskip 3.61371pt\chi_{1}\neq 0.$ (2.24)

Then the energy functional __ $E\left(t\right)$ defined by (1.8) satisfies

$E\left(t\right)\leq\lambda e^{-\xi t},\mbox{ }\forall t\geq 0,$ (2.25)

where $\lambda$ and $\xi$ are two positive constants.

######  Remark 3.

The hypothesis (2.24) is equivalent to the following:

There exist two constants $\sigma,\omega\in\mathbb{R}^{\ast},$ such that

$\begin{array}[c]{cc}\dfrac{\mu}{\rho}=\dfrac{\sigma\alpha+\omega\beta}{\sigma\kappa_{1}}=\dfrac{\sigma\beta+\omega\gamma}{\omega\kappa_{2}},&\hskip
2.8903pt\mbox{if }\hskip 2.8903pt\beta\neq 0,\\\
\left(\dfrac{\mu}{\rho}=\dfrac{\alpha}{\kappa_{1}}\hskip 2.8903pt\mbox{and
}\hskip 2.8903ptb\neq 0\right)\hskip 2.8903pt\mbox{or }\hskip
2.8903pt\left(\dfrac{\mu}{\rho}=\dfrac{\gamma}{\kappa_{2}}\hskip
2.8903pt\mbox{and }\hskip 2.8903ptd\neq 0\right),&\hskip 2.8903pt\mbox{if
}\hskip 2.8903pt\beta=0.\end{array}$ (2.26)

It is clear that in the case where $\beta\neq 0$ and $\sigma=b,\omega=d$ solve
(2.26), then $\alpha,\beta,\gamma,\mu,\rho,\kappa_{1}$ and $\kappa_{2}$ solve
(2.24).

The proof of Theorem 4, will be established through several lemmas.

######  Lemma 2.

For $\left(u,\varphi,\psi\right)$ solution of (1.1), there exist positive
constants $\hat{\alpha},\hat{\gamma},\hat{\alpha}_{1}$and $\hat{\alpha}_{2},$
such that the functional

$\displaystyle F_{1}\left(t\right)$
$\displaystyle=\kappa_{1}\int_{0}^{1}\varphi_{t}\varphi
dx+\kappa_{2}\int_{0}^{1}\psi_{t}\psi
dx+\frac{\tau_{1}}{2}\int_{0}^{1}\varphi^{2}dx+\frac{\tau_{4}}{2}\int_{0}^{1}\psi^{2}dx$
$\displaystyle-\frac{\rho}{\mu}\int_{0}^{1}u_{t}\left(\int_{0}^{x}\left(b\varphi+d\psi\right)\left(y\right)dy\right)dx$

satisfies for any $\delta>0$, the estimate

$\displaystyle F_{1}^{\prime}\left(t\right)$
$\displaystyle\leq-\hat{\alpha}\int_{0}^{1}\varphi_{x}^{2}dx-\hat{\gamma}\int_{0}^{1}\psi_{x}^{2}dx-\frac{\hat{\alpha}_{1}}{2}\int_{0}^{1}\varphi^{2}dx-\frac{\hat{\alpha}_{2}}{2}\int_{0}^{1}\psi^{2}dx$
$\displaystyle+\delta\int_{0}^{1}u_{t}^{2}dx+m_{\delta}\int_{0}^{1}\varphi_{t}^{2}dx+m_{\delta}\int_{0}^{1}\psi_{t}^{2}dx.$
(2.27)

###### Proof.

The differentiation of $F_{1}\left(t\right)$ gives

$\displaystyle F_{1}^{\prime}\left(t\right)$
$\displaystyle=\kappa_{1}\int_{0}^{1}\varphi_{tt}\varphi
dx+\kappa_{1}\int_{0}^{1}\varphi_{t}^{2}dx+\kappa_{2}\int_{0}^{1}\psi_{tt}\psi
dx+\kappa_{2}\int_{0}^{1}\psi_{t}^{2}dx$
$\displaystyle+\tau_{1}\int_{0}^{1}\varphi\varphi_{t}dx+\tau_{4}\int_{0}^{1}\psi_{t}\psi
dx-\frac{\rho}{\mu}\int_{0}^{1}u_{tt}\left(\int_{0}^{x}\left(b\varphi+d\psi\right)\left(y\right)dy\right)dx$
$\displaystyle-\frac{\rho}{\mu}\int_{0}^{1}u_{t}\left(\int_{0}^{x}\left(b\varphi+d\psi\right)_{t}\left(y\right)dy\right)dx.$

By exploiting the equations of (1.1) and using integration by parts, we get

$\displaystyle F_{1}^{\prime}\left(t\right)$
$\displaystyle=-\alpha\int_{0}^{1}\varphi_{x}^{2}dx-2\beta\int_{0}^{1}\psi_{x}\varphi_{x}dx-\gamma\int_{0}^{1}\psi_{x}^{2}dx$
$\displaystyle-\left(\alpha_{1}-\frac{b^{2}}{\mu}\right)\int_{0}^{1}\varphi^{2}dx-2\left(\alpha_{3}-\frac{bd}{\mu}\right)\int_{0}^{1}\psi\varphi
dx-\left(\alpha_{2}-\frac{d^{2}}{\mu}\right)\int_{0}^{1}\psi^{2}dx$
$\displaystyle+\kappa_{1}\int_{0}^{1}\varphi_{t}^{2}dx+\kappa_{2}\int_{0}^{1}\psi_{t}^{2}dx-\tau_{2}\int_{0}^{1}\psi_{t}\varphi
dx-\tau_{3}\int_{0}^{1}\varphi_{t}\psi dx$
$\displaystyle-\frac{\rho}{\mu}\int_{0}^{1}u_{t}\left(\int_{0}^{x}\left(b\varphi_{t}+d\psi_{t}\right)\left(y\right)dy\right)dx.$

Then, using Young’s and Cauchy Schwarz inequalities, we obtain

$\displaystyle F_{1}^{\prime}\left(t\right)$
$\displaystyle\leq-\left(\alpha-\beta\varepsilon\right)\int_{0}^{1}\varphi_{x}^{2}dx-\left(\gamma-\frac{b}{\varepsilon}\right)\int_{0}^{1}\psi_{x}^{2}dx$
$\displaystyle-\left[\left(\alpha_{1}-\frac{b^{2}}{\mu}\right)-\left(\alpha_{3}-\frac{bd}{\mu}\right)\eta-\epsilon\right]\int_{0}^{1}\varphi^{2}dx$
$\displaystyle-\left[\left(\alpha_{2}-\frac{d^{2}}{\mu}\right)-\frac{1}{\eta}\left(\alpha_{3}-\frac{bd}{\mu}\right)-\epsilon\right]\int_{0}^{1}\psi^{2}dx$
$\displaystyle+m\left(1+\frac{1}{\epsilon}+\frac{1}{\delta}\right)\int_{0}^{1}\varphi_{t}^{2}dx+m\left(1+\frac{1}{\epsilon}+\frac{1}{\delta}\right)\int_{0}^{1}\psi_{t}^{2}dx+\delta\int_{0}^{1}u_{t}^{2}dx,$

for any $\varepsilon,\eta,\epsilon,\delta>0.$

First, by virtue of (1.10), we can choose $\varepsilon>0$ such that

$\hat{\alpha}=\alpha-\beta\varepsilon>0,\mbox{and
}\,\,\,\hat{\gamma}=\gamma-\dfrac{b}{\varepsilon}>0.$

Similarly, (1.9) allows us to choose $\eta>0$ such that

$\hat{\alpha}_{1}=\left(\alpha_{1}-\frac{b^{2}}{\mu}\right)-\left(\alpha_{3}-\frac{bd}{\mu}\right)\eta>0,$

and

$\hat{\alpha}_{2}=\left(\alpha_{2}-\frac{d^{2}}{\mu}\right)-\frac{1}{\eta}\left(\alpha_{3}-\frac{bd}{\mu}\right)>0.$

Finally, we choose $\epsilon>0$ so that

$\hat{\alpha}_{1}-\epsilon\geq\frac{\hat{\alpha}_{1}}{2},\,\mbox{and
}\,\,\hat{\alpha}_{2}-\epsilon\geq\frac{\hat{\alpha}_{2}}{2}.$

Consequently, the estimate (2.27) follows. ∎

######  Lemma 3.

Let $\sigma$ and $\omega$ be two constants that satisfy (2.26), then the
functional

$F_{2}\left(t\right):=\rho\int_{0}^{\pi}\left(\left(\sigma\alpha+\omega\beta\right)\varphi_{x}+\left(\sigma\beta+\omega\gamma\right)\psi_{x}\right)u_{t}dx+\mu\int_{0}^{\pi}\left(\sigma\kappa_{1}\varphi_{t}+\omega\kappa_{2}\psi_{t}\right)u_{x}dx$

satisfies along the solution $\left(u,\varphi,\psi\right),$ the estimate

$\displaystyle\left(b\sigma+d\omega\right)F_{2}^{\prime}\left(t\right)$
$\displaystyle\leq-\frac{\mu}{2}\left(b\sigma+d\omega\right)^{2}\int_{0}^{\pi}u_{x}^{2}dx$
$\displaystyle+m\left(\int_{0}^{1}\varphi_{x}^{2}dx+\int_{0}^{1}\psi_{x}^{2}dx+\int_{0}^{1}\varphi_{t}^{2}dx+\int_{0}^{1}\psi_{t}^{2}dx\right).$
(2.28)

###### Proof.

Differentiating $F_{2}\left(t\right),$ using integration by parts and boundary
conditions (1.7) we get

$F_{2}^{\prime}\left(t\right)=\rho\left(\frac{\mu\sigma\kappa_{1}}{\rho}-\left(\sigma\alpha+\omega\beta\right)\right)\int_{0}^{1}u_{xt}\varphi_{t}dx+\rho\left(\frac{\mu\omega\kappa_{2}}{\rho}-\left(\sigma\beta+\omega\gamma\right)\right)\int_{0}^{1}u_{xt}\psi_{t}dx$
$-\mu\left(b\sigma+d\omega\right)\int_{0}^{\pi}u_{x}^{2}dx+b\left(\sigma\alpha+\omega\beta\right)\int_{0}^{\pi}\varphi_{x}^{2}dx+d\left(\sigma\beta+\omega\gamma\right)\int_{0}^{\pi}\psi_{x}^{2}dx$
$+\left(\sigma\left(d\alpha+b\beta\right)+\omega\left(d\beta+b\gamma\right)\right)\int_{0}^{\pi}\varphi_{x}\psi_{x}dx-\mu\left(\sigma\alpha_{1}+\omega\alpha_{3}\right)\int_{0}^{\pi}\varphi
u_{x}dx$ $-\mu\left(\sigma\alpha_{3}+\omega\alpha_{2}\right)\int_{0}^{\pi}\psi
u_{x}-\mu\left(\sigma\tau_{1}+\omega\tau_{3}\right)\int_{0}^{\pi}\varphi_{t}u_{x}dx-\mu\left(\sigma\tau_{2}+\omega\tau_{4}\right)\int_{0}^{\pi}\psi_{t}u_{x}dx.$

Thus, estimate (2.28) follows immediately by taking into account (2.26), using
Young’s and Poincaré’s inequalities. ∎

######  Lemma 4.

Along the solution $\left(u,\varphi,\psi\right)$ of (1.1), the functional

$F_{3}\left(t\right)=-\rho\int_{0}^{1}u_{t}udx$

satisfies

$F_{3}^{\prime}\left(t\right)\leq-\rho\int_{0}^{1}u_{t}^{2}dx+2\mu\int_{0}^{1}u_{x}^{2}dx+\frac{b^{2}}{2\mu}\int_{0}^{1}\varphi^{2}dx+\frac{d^{2}}{2\mu}\int_{0}^{1}\psi^{2}dx.$
(2.29)

###### Proof.

Differentiating $F_{3}\left(t\right)$, using integration by parts and Young’s
inequality, estimate (2.29) follows immediately. ∎

End of the proof of Theorem 4

At this point we define the Lyapunov functional $\mathcal{L}\left(t\right)$ as
follows

$\mathcal{L}\left(t\right)=NE\left(t\right)+N_{1}F_{1}\left(t\right)+N_{2}\left(b\sigma+d\omega\right)F_{2}\left(t\right)+F_{3}\left(t\right),$

where $N,N_{1}$ and $N_{2}$ are positive constants to be properly chosen
later.

First, we have

$\left|\mathcal{L}\left(t\right)-NE\left(t\right)\right|\leq
N_{1}\int_{0}^{\pi}\left(\kappa_{1}\left|\varphi_{t}\varphi\right|+\kappa_{2}\left|\psi_{t}\psi\right|+\frac{\tau_{1}}{2}\left|\varphi\right|^{2}+\frac{\tau_{4}}{2}\left|\psi\right|^{2}\right)dx$
$-\frac{\rho}{\mu}\int_{0}^{\pi}\left|u_{t}\left(\int_{0}^{x}\left(b\varphi+d\psi\right)\left(y\right)dy\right)\right|dx+\rho\int_{0}^{1}\left|u_{t}u\right|dx$
$+N_{2}\left|b\sigma+d\omega\right|\int_{0}^{\pi}\left(\rho\left|\sigma\alpha+\omega\beta\right|\left|u_{t}\varphi_{x}\right|+\rho\left|\sigma\beta+\omega\gamma\right|\left|u_{t}\psi_{x}\right|\right)dx$
$+N_{2}\int_{0}^{\pi}\left(\mu\kappa_{1}\left|b\varphi_{t}u_{x}\right|+\mu\kappa_{2}\left|d\psi_{t}u_{x}\right|\right)dx.$

Using Young’s, Cauchy Schwarz and Poincaré’s inequalities, we obtain

$\displaystyle\left|\mathcal{L}\left(t\right)-NE\left(t\right)\right|$
$\displaystyle\leq
c_{0}\int_{0}^{1}\left(u_{t}^{2}+\varphi_{t}^{2}+\psi_{t}^{2}+\left(\varphi_{x}+\psi_{x}\right)^{2}+\psi_{x}^{2}+\left(u_{x}+\varphi+\psi\right)^{2}\right)dx$
$\displaystyle\leq cE\left(t\right).$

Thus,

$\left(N-c\right)E\left(t\right)\leq\mathcal{L}\left(t\right)\leq\left(N+c\right)E\left(t\right).$

Secondly, substituting (2.15),(2.27),(2.28) and (2.29) in the expression of
$\mathcal{L}^{\prime}\left(t\right)$ we get

$\displaystyle\mathcal{L}^{\prime}\left(t\right)$
$\displaystyle\leq-\left[\frac{1}{2}\left(\tau_{1}-\frac{\left(\tau_{2}+\tau_{3}\right)^{2}}{4\tau_{4}}\right)N-m_{\delta}N_{1}-mN_{2}\right]\int_{0}^{1}\varphi_{t}^{2}dx$
$\displaystyle-\left[\frac{1}{2}\left(\tau_{4}-\frac{\left(\tau_{2}+\tau_{3}\right)^{2}}{4\tau_{1}}\right)N-m_{\delta}N_{1}-mN_{2}\right]\int_{0}^{1}\psi_{t}^{2}dx$
$\displaystyle-\mu\left(\frac{\left(\sigma b+\omega
d\right)^{2}}{2}N_{2}-2\right)\int_{0}^{1}u_{x}^{2}dx-\left(\rho-\delta
N_{1}\right)\int_{0}^{1}u_{t}^{2}dx$
$\displaystyle-\left(\hat{\alpha}N_{1}-mN_{2}\right)\int_{0}^{1}\varphi_{x}^{2}dx-\left(\hat{\gamma}N_{1}-mN_{2}\right)\int_{0}^{1}\psi_{x}^{2}dx$
$\displaystyle-\frac{1}{2}\left(\hat{\alpha}_{1}N_{1}-\frac{b^{2}}{\mu}\right)\int_{0}^{1}\varphi^{2}dx-\frac{1}{2}\left(\hat{\alpha}_{2}N_{1}-\frac{d^{2}}{\mu}\right)\int_{0}^{1}\psi^{2}dx.$

Now, we have to choose the coefficients carefully. First, we take

$\delta=\frac{\rho}{2N_{1}}.$

Secondly, We choose $N_{2}$ large enough such that

$\frac{\left(\sigma b+\omega d\right)^{2}}{2}N_{2}-2>0.$

Next, we pick $N_{1}$ large enough such that

$\hat{\alpha}N_{1}-m_{\delta}N_{2}>0,\text{
}\hat{\gamma}N_{1}-m_{\delta}N_{2}>0,$
$\hat{\alpha}_{1}N_{1}-\frac{b^{2}}{\mu}>0,\,\mbox{and }\,\text{
}\hat{\alpha}_{2}N_{1}-\frac{d^{2}}{\mu}>0.$

Finally, we take $N$ large enough such that $\mathcal{L}\left(t\right)\sim$
$E\left(t\right)$ (i.e. $N-c>0$) and

$\displaystyle\frac{1}{2}\left(\tau_{1}-\frac{\left(\tau_{2}+\tau_{3}\right)^{2}}{4\tau_{4}}\right)N-mN_{1}-m_{\delta}N_{2}$
$\displaystyle>0,$
$\displaystyle\frac{1}{2}\left(\tau_{4}-\frac{\left(\tau_{2}+\tau_{3}\right)^{2}}{4\tau_{1}}\right)N-mN_{1}-m_{\delta}N_{2}$
$\displaystyle>0.$

Therefore, there exist $\sigma$ and $\widetilde{\sigma}$ positives constants
such that

$\displaystyle\mathcal{L}^{\prime}\left(t\right)$
$\displaystyle\leq-\sigma\int_{0}^{\pi}\left(\varphi_{t}^{2}+\psi_{t}^{2}+u_{t}^{2}+u_{x}^{2}+\varphi_{x}^{2}+\psi_{x}^{2}+\psi^{2}+\varphi^{2}\right)dx,$
$\displaystyle\leq-\widetilde{\sigma}E\left(t\right),\text{ \ \ \ \ }\ \forall
t\geq 0.$

Since $E\left(t\right)$ is equivalent to $\mathcal{L}\left(t\right),$ we infer
that

$\mathcal{L}^{\prime}\left(t\right)\leq-\omega\mathcal{L}\left(t\right),~{}~{}\forall
t\geq 0,$

for some positive constant $\omega.$ Thus

$\mathcal{L}\left(t\right)\leq\lambda_{1}\mathcal{L}\left(0\right)e^{-\omega
t},~{}~{}\forall t\geq 0.$

Using again the equivalence between $\mathcal{L}\left(t\right)$ and
$E\left(t\right)$ we conclude that

$E\left(t\right)\leq\lambda e^{-\omega t},~{}~{}\forall t\geq 0,$

which completes the proof of Theorem 4.

######  Remark 4.

The same proof is valid for the following boundary conditions

$\begin{array}[c]{cc}\begin{array}[c]{c}u_{x}\left(t,\pi\right)=\varphi\left(t,\pi\right)=\psi\left(t,\pi\right)=0,\\\
u\left(t,0\right)=\varphi_{x}\left(t,0\right)=\psi_{x}\left(t,0\right)=0,\end{array}&t\geq
0.\end{array}$

##  3 Lack of exponential decay

In this section we suppose that (2.24) does not hold, and prove that the
solution $\left(u,\varphi,\psi\right)$ of the system (1.1) lacks exponentially
stability. The proof is based on the following theorem due to Gerhart-Prüss-
Huang [10, 16, 11].

######  Theorem 5.

Let $S\left(t\right)=e^{\mathcal{A}t}$ be a $C_{0}-$semigroup of contractions
on a Hilbert space $\mathcal{H}$, with infinitesimal generator $\mathcal{A}$.
Then $S\left(t\right)$ is exponentially stable if and only if:

  1. $\cdot$

$i\mathbb{R}\subset\rho\left(\mathcal{A}\right),$

  2. $\cdot$

$\underset{\left|\lambda\right|\longrightarrow\infty}{\overline{\lim}}\left\|\left(\lambda
I-\mathcal{A}\right)^{-1}\right\|_{\mathcal{L}\left(\mathcal{H}\right)}<\infty.$

Our result of non exponential stability reads as follow.

######  Theorem 6.

Suppose that (2.24) does not hold, then the energy associated with the
solution $\left(u,\varphi,\psi\right)$ of the system (1.1) is not
exponentially stable.

###### Proof.

It suffices to prove that there exists a sequence
$\left(F_{n}\right)\subset\mathcal{H}$ with bounded norm
$\left\|F_{n}\right\|<1,$ such that

$\underset{\left|\lambda\right|\longrightarrow\infty}{\overline{\lim}}\left\|\left(\lambda
I-\mathcal{A}\right)^{-1}F_{n}\right\|_{\mathcal{H}}=\underset{\left|\lambda\right|\longrightarrow\infty}{\overline{\lim}}\left\|U_{n}\right\|_{\mathcal{H}}=\infty.$
(3.30)

Let $\left(U_{n}\right)_{n\in\mathbb{N}}\subset D\left(\mathcal{A}\right)$ be
the solution of $\left(\lambda I-\mathcal{A}\right)U_{n}=F_{n}$, then,
omitting $n$ we have

$\begin{array}[c]{rl}i\lambda u+v&=f_{1}\\\ i\lambda\rho v+\mu
u_{xx}+b\varphi_{x}+d\psi_{x}&=\rho f_{2}\\\ i\lambda\varphi+\phi&=f_{3}\\\
i\lambda\kappa_{1}\phi+\alpha\varphi_{xx}+\beta\psi_{xx}-bu_{x}-\alpha_{1}\varphi-\alpha_{3}\psi-\tau_{1}\phi-\tau_{2}\chi&=\kappa_{1}f_{4}\\\
i\lambda\psi+\chi&=f_{5}\\\
i\lambda\kappa_{2}\chi+\beta\varphi_{xx}+\gamma\psi_{xx}-du_{x}-\alpha_{3}\varphi-\alpha_{2}\psi-\tau_{3}\phi-\tau_{4}\chi&=\kappa_{2}f_{6}.\end{array}$

Taking $f_{1}=f_{3}=f_{4}=f_{5}=f_{6}=0$ and
$f_{2}=\dfrac{1}{\rho}\sin\left(n\pi x\right),$ then eliminating $v,\phi$ and
$\chi$ we obtain

$\begin{array}[c]{cc}\lambda^{2}\rho u+\mu
u_{xx}+b\varphi_{x}+d\psi_{x}&=\sin\left(n\pi x\right)\\\
\lambda^{2}\kappa_{1}\varphi+\alpha\varphi_{xx}+\beta\psi_{xx}-bu_{x}-\left(\alpha_{1}-i\lambda\tau_{1}\right)\varphi-\left(\alpha_{3}-i\lambda\tau_{2}\right)\psi&=0\\\
\lambda^{2}\kappa_{2}\psi+\beta\varphi_{xx}+\gamma\psi_{xx}-du_{x}-\left(\alpha_{3}-i\lambda\tau_{3}\right)\varphi-\left(\alpha_{2}-i\lambda\tau_{4}\right)\psi&=0.\end{array}$

Taking into account the boundary conditions (1.7), we are looking for
$\left(u,\varphi,\psi\right)$ of the form

$u=A\sin\left(n\pi x\right),\,\varphi=B\cos\left(n\pi
x\right),\,\psi=C\cos\left(n\pi x\right).$

That is

$\left\\{\begin{array}[c]{c}\left(\rho\lambda^{2}-\mu\pi^{2}n^{2}\right)A-bn\pi
B-dn\pi C=1\\\
-b\left(n\pi\right)A+\left[\kappa_{1}\lambda^{2}-\left(n\pi\right)^{2}\alpha-\left(\alpha_{1}-i\lambda\tau_{1}\right)\right]B-\left[\beta\left(n\pi\right)^{2}+\left(\alpha_{3}-i\lambda\tau_{2}\right)\right]C=0\\\
-d\left(n\pi\right)A-\left[\beta\left(n\pi\right)^{2}+\left(\alpha_{3}-i\lambda\tau_{3}\right)\right]B+\left[\kappa_{2}\lambda^{2}-\left(n\pi\right)^{2}\gamma-\left(\alpha_{2}-i\lambda\tau_{4}\right)\right]C=0,\end{array}\right.$

which can be written

$\left(\begin{array}[c]{ccc}p_{1}\left(\lambda\right)&-bn\pi&-dn\pi\\\
-bn\pi&p_{2}\left(\lambda\right)&p_{4}\left(\lambda\right)\\\
-dn\pi&p_{5}\left(\lambda\right)&p_{3}\left(\lambda\right)\end{array}\right)\left(\begin{array}[c]{c}A\\\
B\\\ C\end{array}\right)=\left(\begin{array}[c]{c}1\\\ 0\\\
0\end{array}\right)$ (3.31)

where

$p_{1}\left(\lambda\right):=\rho\lambda^{2}-\mu\left(\pi
n\right)^{2},\,\,p_{2}\left(\lambda\right):=\kappa_{1}\lambda^{2}-\left(n\pi\right)^{2}\alpha-\left(\alpha_{1}-i\lambda\tau_{1}\right),$
$p_{3}\left(\lambda\right):=\kappa_{2}\lambda^{2}-\left(n\pi\right)^{2}\gamma-\left(\alpha_{2}-i\lambda\tau_{4}\right),\;p_{4}\left(\lambda\right):=-\beta\left(n\pi\right)^{2}-\left(\alpha_{3}-i\lambda\tau_{2}\right),$
$p_{5}\left(\lambda\right):=-\beta\left(n\pi\right)^{2}-\left(\alpha_{3}-i\lambda\tau_{3}\right).$

Solving (3.31) we obtain

$A=\frac{K_{1}}{p_{1}K_{1}+K_{2}},$

where,

$K_{1}:=p_{2}p_{3}-p_{4}p_{5},\;K_{2}:=b(n\pi)^{2}\left(dp_{4}-bp_{3}\right)-d\left(n\pi\right)^{2}\left(dp_{2}-bp_{5}\right).$

Let $\lambda$ be such that $p_{1}\left(\lambda\right)=0,$ then
$\left(n\pi\right)^{2}=\dfrac{\rho\lambda^{2}}{\mu}$ and

$K_{1}=\frac{\rho}{\mu}\left[\left(\frac{\mu\kappa_{1}}{\rho}-\alpha\right)\left(\frac{\mu\kappa_{2}}{\rho}-\gamma\right)-\beta^{2}\right]\lambda^{4}$
$+i\frac{\rho}{\mu}\left[\left(\frac{\mu\kappa_{1}}{\rho}-\alpha\right)\tau_{4}+\left(\frac{\mu\kappa_{2}}{\rho}-\gamma\right)\tau_{1}+\beta\left(\tau_{2}+\tau_{3}\right)\right]\lambda^{3}+K_{3},$

that is

$K_{1}=\frac{\rho}{\mu}\chi_{0}\lambda^{4}++i\frac{\rho}{\mu}\left[\left(\frac{\mu\kappa_{1}}{\rho}-\alpha\right)\tau_{4}+\left(\frac{\mu\kappa_{2}}{\rho}-\gamma\right)\tau_{1}+\beta\left(\tau_{2}+\tau_{3}\right)\right]\lambda^{3}+K_{3}$

and

$K_{2}=-\frac{\rho^{2}}{\mu^{2}}\left[b^{2}\left(\frac{\mu\kappa_{2}}{\rho}-\gamma\right)+d^{2}\left(\frac{\mu\kappa_{1}}{\rho}-\alpha\right)+2bd\beta\right]\lambda^{4}$
$+i\dfrac{\rho}{\mu}\left[bd\left(\tau_{2}+\tau_{3}\right)-b^{2}\tau_{4}-d^{2}\tau_{1}\right]\lambda^{3}+K_{4}$

that is

$K_{2}=-\frac{\rho^{2}}{\mu^{2}}\chi_{1}\lambda^{4}+i\dfrac{\rho}{\mu}\left[bd\left(\tau_{2}+\tau_{3}\right)-b^{2}\tau_{4}-d^{2}\tau_{1}\right]\lambda^{3}+K_{4}$

where $K_{3},K_{4}$ are polynomials of degree $2$ in $\lambda.$

At this point we discuss three cases:

1)

Suppose that $\chi_{0}\neq 0$ and $\chi_{1}\neq 0,$then

$A=\frac{K_{1}}{K_{2}}\approx\frac{\mu\chi_{0}}{-\rho\chi_{1}}\equiv c,$

for some constant $c\neq 0.$

2)

Suppose that $\chi_{0}=\chi_{1}=0,$ then

$\frac{\mu\kappa_{1}}{\rho}-\alpha=-\frac{b\beta}{d},\hskip
3.61371pt\frac{\mu\kappa_{2}}{\rho}-\gamma=-\frac{d\beta}{d}.$

Consequently

$\left(\frac{\mu\kappa_{1}}{\rho}-\alpha\right)\tau_{4}+\left(\frac{\mu\kappa_{2}}{\rho}-\gamma\right)\tau_{1}+\beta\left(\tau_{2}+\tau_{3}\right)\neq
0,$ $bd\left(\tau_{2}+\tau_{3}\right)-b^{2}\tau_{4}-d^{2}\tau_{1}\neq 0,$

by virtue of (2.13) and

$A=\frac{K_{1}}{K_{2}}\approx\frac{\left[\left(\frac{\mu\kappa_{1}}{\rho}-\alpha\right)\tau_{4}+\left(\frac{\mu\kappa_{2}}{\rho}-\gamma\right)\tau_{1}+\beta\left(\tau_{2}+\tau_{3}\right)\right]}{\left[bd\left(\tau_{2}+\tau_{3}\right)-b^{2}\tau_{4}-d^{2}\tau_{1}\right]}\equiv
c.$

Therefore,

$\left\|U\right\|^{2}\geq\rho\left\|v\right\|^{2}=\rho
c^{2}\left|\lambda\right|^{2}\int_{0}^{1}\sin^{2}\left(n\pi
x\right)dx=\frac{\rho c^{2}\left|\lambda\right|^{2}}{2},$

and consequently,

$\underset{\left|\lambda\right|\longrightarrow\infty}{\lim}\left\|U\right\|^{2}=\infty.$

3)

Suppose that $\chi_{0}\neq 0$ and $\chi_{1}=0,$ then

$A=\frac{K_{1}}{K_{2}}\approx\frac{\chi_{0}\lambda}{-i\left[bd\left(\tau_{2}+\tau_{3}\right)-b^{2}\tau_{4}-d^{2}\tau_{1}\right]}\approx
c\lambda,$
$\left\|U\right\|^{2}\geq\left\|u_{x}\right\|^{2}=A^{2}\left(n\pi\right)^{2}\int_{0}^{1}\cos^{2}\left(n\pi
x\right)dx=\frac{c^{2}\left|\lambda\right|^{4}\mu}{2\rho}$

and

$\underset{\left|\lambda\right|\longrightarrow\infty}{\lim}\left\|U\right\|^{2}=\infty.$

Therefore, in all cases (3.30) holds and consequently, the proof of Theorem 6
is completed. ∎

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