# Weak-strong uniqueness for Maxwell–Stefan systems

Xiaokai Huo  Institute for Analysis and Scientific Computing, Vienna
University of Technology, Wiedner Hauptstraße 8–10, 1040 Wien, Austria
<EMAIL_ADDRESS>,  Ansgar Jüngel  Institute for Analysis and
Scientific Computing, Vienna University of Technology, Wiedner Hauptstraße
8–10, 1040 Wien, Austria<EMAIL_ADDRESS>and  Athanasios E. Tzavaras
Computer, Electrical and Mathematical Science and Engineering Division, King
Abdullah University of Science and Technology (KAUST), Thuwal 23955-6900,
Saudi Arabia<EMAIL_ADDRESS>
###### Abstract.

The weak-strong uniqueness for Maxwell–Stefan systems and some generalized
systems is proved. The corresponding parabolic cross-diffusion equations are
considered in a bounded domain with no-flux boundary conditions. The key
points of the proofs are various inequalities for the relative entropy
associated to the systems and the analysis of the spectrum of a quadratic form
capturing the frictional dissipation. The latter task is complicated by the
singular nature of the diffusion matrix. This difficulty is addressed by
proving its positive definiteness on a subspace and using the Bott–Duffin
matrix inverse. The generalized Maxwell–Stefan systems are shown to cover
several known cross-diffusion systems for the description of tumor growth and
physical vapor deposition processes.

###### Key words and phrases:

Cross diffusion, weak-strong uniqueness, relative entropy, Maxwell–Stefan
systems, thin-film solar cell model, tumor-growth model.

###### 2000 Mathematics Subject Classification:

35A02, 35K51, 35K55, 35Q35.

The second author has been partially supported by the Austrian Science Fund
(FWF), grants P30000, P33010, F65, and W1245. This work received funding from
the European Research Council (ERC) under the European Union’s Horizon 2020
research and innovation programme, ERC Advanced Grant NEUROMORPH, no.
101018153.

##  1\. Introduction

The Maxwell–Stefan equations describe the diffusive transport of the
components of gaseous mixtures. Applications arise in, e.g., sedimentation,
dialysis, electrolysis, and ion exchange [34]. They were suggested in 1866 by
James Maxwell [31] for dilute gases and in 1871 by Josef Stefan [33] for
fluids. While there are several works on the existence of local-in-time smooth
solutions [4, 20, 21] and global-in-time weak solutions [27] in the case of
vanishing barycentric velocity, the problem of the uniqueness of solutions is
basically unsolved. The uniqueness of strong solutions has been shown in [21,
24], and uniqueness results for weak solutions in a very special case can be
found in [10]. In this paper, we make a step forward in the uniqueness problem
by showing that strong solutions are unique in the class of weak solutions to
Maxwell–Stefan systems.

###  1.1. Setting

We consider an ideal gaseous mixture consisting of $n$ components with volume
fractions or concentrations $c_{i}(x,t)$, $i=1,\ldots,n$. The dynamics of the
mixture is given by the mass balance equations and the relations between the
driving forces and the fluxes,

(1) $\partial_{t}c_{i}+\operatorname{div}(c_{i}u_{i})=0,\quad\nabla
c_{i}=-\sum_{j=1}^{n}\frac{c_{i}c_{j}}{D_{ij}}(u_{i}-u_{j}),\quad
i=1,\ldots,n,$

where $u_{i}(x,t)$ are the partial velocities and $D_{ij}=D_{ji}>0$ are
diffusion coefficients. The equations are solved in a bounded domain
$\Omega\subset{\mathbb{R}}^{d}$ ($d\geq 1$), supplemented by the initial and
no-flux boundary conditions

(2) $c_{i}(0)=c_{i}^{0}\quad\mbox{in }\Omega,\quad\nabla
c_{i}\cdot\nu=0\quad\mbox{on }\partial\Omega,\ t>0,\ i=1,\ldots,n,$

where $\nu$ is the exterior unit normal vector to $\partial\Omega$.

We assume that the barycentric velocity vanishes, which implies that the sum
of all fluxes vanishes, $\sum_{i=1}^{n}c_{i}u_{i}=0$. Then, supposing that
$c_{i}^{0}\geq 0$ and $\sum_{i=1}^{n}c_{i}^{0}=1$ in $\Omega$, we deduce from
mass conservation that

$\sum_{i=1}^{n}c_{i}=1\quad\mbox{in $\Omega$ for all $t>0$.}$

This constraint is necessary to invert the force-flux relations in (1), i.e.
to express the flux $c_{i}u_{i}$ as a linear combination of the driving forces
$\nabla c_{j}$.

The global existence analysis for (1)–(2) is based on the property that the
system is endowed with the entropy functional

(3) $H(\bm{c})=\sum_{i=1}^{n}\int_{\Omega}c_{i}(\log c_{i}-1)dx,$

where $\bm{c}=(c_{1},\ldots,c_{n})$ solves (1)–(2) and satisfies the entropy
dissipation inequality [27, (1.14)]

(4)
$\frac{dH}{dt}(\bm{c})+C\sum_{i=1}^{n}\int_{\Omega}|\nabla\sqrt{c_{i}}|^{2}dx\leq
0,$

with $C>0$ depending only on $(D_{ij})$. The aim of this paper is to prove the
weak-strong uniqueness for (1)–(2) and generalized systems. Weak-strong
uniqueness means that any weak solution coincides with a strong solution
emanating from the same initial data as long as the latter exists. In other
words, the strong solutions must be unique within the class of weak solutions.
To achieve this aim, we use ideas from our previous work [23] and establish a
relative entropy inequality. This leads to a stability estimate for the
difference of a weak and a strong solution and eventually to the weak-strong
uniqueness property. Here, the relative entropy functional is given by

(5)
$H(\bm{c}|\bar{\bm{c}})=\sum_{i=1}^{n}\int_{\Omega}\bigg{(}c_{i}\log\frac{c_{i}}{\bar{c}_{i}}-(c_{i}-\bar{c}_{i})\bigg{)}dx,$

where $\bm{c}$ and $\bar{\bm{c}}$ are suitable solutions to (1)–(2).

In the literature, relative entropies are known to be useful to prove the
weak-strong uniqueness of solutions. First results were achieved for systems
of hyperbolic conservation laws [14] and later for the compressible
Navier–Stokes equations [16, 17] and general hyperbolic-parabolic systems
endowed with an entropy [13]. The relative entropy technique was applied to,
for instance, entropy-dissipating reaction-diffusion equations [18], reaction-
cross-diffusion systems [11], energy-reaction-diffusion systems [22], nonlocal
cross-diffusion systems [26], and quantum Euler systems [8, 19]. Compared to
the results of, e.g. [11, 22], the diffusion matrix in these works is assumed
to be positive definite if $c_{i}>0$ for all $i=1,\ldots,n$, which is not
satisfied for the Maxwell–Stefan system.

###  1.2. Definitions and assumptions

We impose the following assumptions:

  1. (A1)

Domain: $\Omega\subset{\mathbb{R}}^{d}$ with $d\geq 1$ is a bounded domain.

  2. (A2)

Coefficients: $D_{ij}>0$ and $D_{ij}=D_{ji}$ for all $i,j=1,\ldots,n$, $i\neq
j$.

  3. (A3)

Initial data: $0\leq c_{i}^{0}\in L^{1}(\Omega)$ for $i=1,\ldots,n$,
$H(\bm{c}^{0})<\infty$, and $\sum_{i=1}^{n}c_{i}^{0}=1$ in $\Omega$.

Next, we define the concept of weak and strong solutions employed in this
paper.

We call $\bm{c}=(c_{1},\ldots,c_{n})$ a weak solution to (1)–(2) if $\bm{c}$
satisfies the initial condition (2), $c_{i}\geq 0$, $\sum_{i=1}^{n}c_{i}=1$ in
$\Omega\times(0,\infty)$,

$\sqrt{c_{i}}\in L^{2}_{\rm loc}(0,\infty;H^{1}(\Omega)),\quad c_{i}\in C_{\rm
loc}^{0}([0,\infty);{\mathcal{V}}^{\prime}),\quad i=1,\ldots,n,$

where ${\mathcal{V}}^{\prime}$ is the dual space of ${\mathcal{V}}=\\{w\in
H^{2}(\Omega):\nabla w\cdot\nu=0$ on $\partial\Omega\\}$, and $\bm{c}$
satisfies (1)–(2) in the weak sense, i.e., for any $\phi_{i}\in C_{\rm
loc}^{1}([0,\infty);C^{1}(\overline{\Omega}))$ satisfying
$\nabla\phi_{i}\cdot\nu=0$ on $\partial\Omega$ and any $T>0$, $i=1,\ldots,n$,
we have

$\int_{\Omega}c_{i}(T)\phi_{i}(T)dx-\int_{\Omega}c_{i}^{0}\phi_{i}(0)dx-\int_{0}^{T}\int_{\Omega}c_{i}\partial_{t}\phi_{i}dxdt-\int_{0}^{T}\int_{\Omega}c_{i}u_{i}\cdot\nabla\phi_{i}dxdt=0,$

where $u_{i}$ satisfies the force-flux relations in (1). The last integral is
well defined, since the gradient bound for $\sqrt{c_{i}}$ implies that
$\sqrt{c_{i}}u_{i}\in L_{\rm loc}^{2}(0,\infty;L^{2}(\Omega))$ (see Lemma 7
below) and thus, because of the property $0\leq c_{i}\leq 1$, $c_{i}u_{i}\in
L_{\rm loc}^{2}(0,\infty;L^{2}(\Omega))$. Finally, a weak solution is required
to satisfy the entropy inequality

(6)
$H(\bm{c}(t))+\frac{1}{2}\sum_{i,j=1}^{n}\int_{0}^{t}\int_{\Omega}\frac{c_{i}c_{j}}{D_{ij}}|u_{i}-u_{j}|^{2}dxds\leq
H(\bm{c}^{0}).$

For the Maxwell-Stefan system this is not an additional requirement as it is
guaranteed by the existence theory of [27]; see Section 3.1.

We will use the term strong solution to (1)–(2) to mean that
$\bar{\bm{c}}=(\bar{c}_{i},\ldots,\bar{c}_{n})$ with $0<\bar{c}_{i}<1$ is a
weak solution satisfying additional regularity properties. The necessary
regularity is stated precisely in context. In certain cases, $\bar{c}_{i}$
satisfies (1)–(2) pointwise, as is the traditional notion of strong solutions.

###  1.3. Main results and key ideas of the proofs

Our first main result is concerned with the Maxwell–Stefan system (1)–(2).

######  Theorem 1 (Weak-strong uniqueness).

Let Assumptions (A1)–(A2) hold. Let $\bm{c}$ be a weak solution to (1)–(2) and
let $\bar{\bm{c}}$ be a strong solution to (1)–(2) satisfying
$0<\bar{c}_{i}<1$ in $\Omega$, $t>0$, the regularity properties

$\log\bar{c}_{i}\in H_{\rm
loc}^{1}(\Omega\times(0,\infty)),\quad\bar{u}_{i}\in L_{\rm
loc}^{\infty}(\Omega\times(0,\infty)),$

and $\bar{c}_{i}$ does not have anomalous dissipation, i.e., it satisfies the
entropy identity

$H(\bar{\bm{c}}(t))+\frac{1}{2}\int_{0}^{t}\int_{\Omega}\sum_{i,j=1}^{n}\frac{\bar{c}_{i}\bar{c}_{j}}{D_{ij}}|\bar{u}_{i}-\bar{u}_{j}|^{2}dxds=H(\bar{\bm{c}}^{0})\quad\mbox{for
}t>0.$

The initial data for $\bm{c}$ and $\bar{\bm{c}}$ satisfy Assumption (A3). Then
for any $t>0$, there exists a constant $C(t)>0$, depending on $t$, $\Omega$,
$n$, and $(D_{ij})$, such that

(7)
$H(\bm{c}(t)|\bar{\bm{c}}(t))+\sum_{i=1}^{n}\int_{0}^{t}\int_{\Omega}c_{i}|u_{i}-\bar{u}_{i}|^{2}dxds\leq
C(t)H(\bm{c}^{0}|\bar{\bm{c}}^{0}).$

If the initial data coincide, i.e. $\bm{c}^{0}=\bar{\bm{c}}^{0}$ in $\Omega$,
then $\bm{c}(t)=\bar{\bm{c}}(t)$ in $\Omega$ for $t>0$.

We verify in Section 3.1 that solutions with the stated regularity exist. To
prove Theorem 1 we develop a relative entropy identity and use it as a
yardstick to control the distance between two solutions. First, it is shown
that the relative entropy (5) satisfies the inequality

(8) $\displaystyle\frac{dH}{dt}(\bm{c}|\bar{\bm{c}})$
$\displaystyle+\frac{1}{2}\sum_{i,j=1,\,i\neq
j}^{n}\int_{\Omega}\frac{c_{i}c_{j}}{D_{ij}}|(u_{i}-\bar{u}_{i})-(u_{j}-\bar{u}_{j})|^{2}dx$
$\displaystyle\leq-\sum_{i,j=1,\,i\neq
j}^{n}\int_{\Omega}\frac{c_{i}}{D_{ij}}(c_{j}-\bar{c}_{j})(u_{i}-\bar{u}_{i})\cdot(\bar{u}_{i}-\bar{u}_{j})dx.$

(see Section 3.2). Next, we study how the frictional dissipation (the second
term in (8)) controls the $L^{2}$ norm of $u_{i}-\bar{u}_{i}$. The quadratic
form in (8) captures the dissipative effect of friction in the following way:

$\displaystyle\frac{1}{2}\sum_{i,j=1,\,i\neq j}^{n}$
$\displaystyle\frac{c_{i}c_{j}}{D_{ij}}|(u_{i}-\bar{u}_{i})-(u_{j}-\bar{u}_{j})|^{2}=\sum_{i=1}^{n}c_{i}(u_{i}-\bar{u}_{i})\cdot\sum_{j=1}^{n}\frac{1}{D_{ij}}c_{j}\big{(}(u_{i}-\bar{u}_{i})-(u_{j}-\bar{u}_{j})\big{)}$
(9)
$\displaystyle=\sum_{i,j=1}^{n}A_{ij}(c)(\sqrt{c_{i}}(u_{i}-\bar{u}_{i}))\cdot(\sqrt{c_{j}}(u_{j}-\bar{u}_{j}))=\bm{Y}^{T}A(\bm{c})\bm{Y},$

where the matrix $A(\bm{c})=(A_{ij}(\bm{c}))\in{\mathbb{R}}^{n\times n}$ is
defined by

(10) $A_{ij}(\bm{c})=\left\\{\begin{array}[]{ll}\sum_{k=1,\,k\neq
i}^{n}c_{k}/D_{ik}&\quad\mbox{if }i=j,\\\
-\sqrt{c_{i}c_{j}}/D_{ij}&\quad\mbox{if }i\neq j,\end{array}\right.$

and $\bm{Y}=(Y_{1},\ldots,Y_{n})$ with
$Y_{i}=\sqrt{c_{i}}(u_{i}-\bar{u}_{i})$. The matrix $A(\bm{c})$ is singular
and thus not positive definite. However, we can show that it is positive
definite on the subspace
$L:=\\{\bm{z}\in{\mathbb{R}}^{n}:\sqrt{\bm{c}}\cdot\bm{z}=0\\}$ (here,
$\sqrt{\bm{c}}$ is the vector with components $\sqrt{c_{i}}$) and the
quadratic form satisfies

$\bm{Y}^{T}A(\bm{c})\bm{Y}\geq\mu|P_{L}\bm{Y}|^{2},$

where $\mu>0$ is a uniform lower bound for the positive eigenvalues of
$A(\bm{c})$ and $P_{L}$ is the projection on $L$. This inequality and a
careful estimate of the right-hand side of (8) implies (7) and the weak-strong
uniqueness property.

The $L^{\infty}$ bound on the partial velocities $\bar{u}_{i}$ in Theorem 1
can be avoided at the expense of assuming $\nabla\sqrt{c_{i}}\in L^{\infty}$
and $\bar{c}_{i}$ is uniformly bounded from below by a positive constant. The
uniform lower bound is not needed in Theorem 1, where only positivity is
required.

######  Corollary 2.

Let the assumptions of Theorem 1 hold, replacing $\bar{u}_{i}\in
L^{\infty}(\Omega\times(0,\infty))$ by $\nabla\sqrt{\bar{c}_{i}}\in
L^{\infty}(\Omega\times(0,\infty))$, $i=1,\ldots,n$. Suppose additionally that
there exists $m>0$ such that $\bar{c}_{i}(t)\geq m$ in $\Omega$, $t>0$,
$i=1,\ldots,n$. Then there exist constants $C_{1}>0$ and $C_{2}(t)>0$
(depending on $t$, $\Omega$, $n$, and $(D_{ij})$) such that the following
inequality holds for $t>0$:

(11)
$H(\bm{c}(t)|\bar{\bm{c}}(t))+C_{1}\sum_{i=1}^{n}\int_{0}^{t}\int_{\Omega}|\nabla(\sqrt{c_{i}}-\sqrt{\bar{c}_{i}})|^{2}dxds\leq
C_{2}(t)H(\bm{c}^{0}|\bar{\bm{c}}^{0}).$

The relative entropy inequality (11) is the analogue of the entropy estimate
(4). It can be achieved by working with the square roots $\sqrt{c_{i}}$ as the
main variables. More precisely, we write the force-flux relations in (1) as

(12) $2\nabla\sqrt{c_{i}}=-\sum_{j=1}^{n}A_{ij}(\bm{c})\sqrt{c_{j}}u_{j},\quad
i=1,\ldots,n,$

subject to $\sum_{i=1}^{n}c_{i}u_{i}=0$, where $A(\bm{c})$ is defined in (10).
This system cannot be directly inverted, since $\ker
A(\bm{c})=\operatorname{span}\\{\sqrt{\bm{c}}\\}$. However, introducing the
Bott–Duffin inverse $A^{BD}(\bm{c})$ of $A(\bm{c})$ with respect to
$L:=(\operatorname{span}\\{\sqrt{\bm{c}}\\})^{\perp}$ (see Section 2 and
Appendix A), we can invert (12), leading to

(13)
$\sqrt{c_{i}}u_{i}=-2\sum_{j=1}^{n}A_{ij}^{BD}(\bm{c})\nabla\sqrt{c_{j}},\quad
i=1,\ldots,n,$

and system (1) can be formulated in the concise form

(14)
$\partial_{t}c_{i}=2\operatorname{div}\bigg{(}\sum_{j=1}^{n}\sqrt{c_{i}}A_{ij}^{BD}(\bm{c})\nabla\sqrt{c_{j}}\bigg{)},\quad
i=1,\ldots,n.$

The Bott–Duffin inverse $A^{BD}(\bm{c})$ equals the group inverse studied in
[6], since $L=\operatorname{ran}A(\bm{c})$. Compared to [6], we work here with
the square roots $\sqrt{c_{i}}$ instead of the chemical potentials $\log
c_{i}$ (see [6, (4.25)]). The relative entropy inequality (8) is rewritten in
the form (see Lemma 9)

(15) $\displaystyle\frac{dH}{dt}(\bm{c}|\bar{\bm{c}})$
$\displaystyle+4\sum_{i,j=1}^{n}\int_{\Omega}A_{ij}^{BD}(\bm{c})Z_{i}\cdot
Z_{j}dx$ $\displaystyle\leq
4\sum_{i,j=1}^{n}\int_{\Omega}Z_{i}\cdot\nabla\sqrt{\bar{c}_{j}}\bigg{(}\frac{\sqrt{c_{i}}}{\sqrt{\bar{c}_{i}}}A_{ij}^{BD}(\bar{\bm{c}})-A_{ij}^{BD}(\bm{c})\frac{\sqrt{c_{j}}}{\sqrt{\bar{c}_{j}}}\bigg{)}dx,$

where
$Z_{i}=\nabla\sqrt{c_{i}}-\sqrt{c_{i}/\bar{c}_{i}}\nabla\sqrt{\bar{c}_{i}}$,
$i=1,\ldots,n$. We prove in Lemma 4 that the Bott–Duffin inverse is symmetric
and positive definite on $L$,

$\bm{Z}^{T}A^{BD}(\bm{c})\bm{Z}\geq\lambda|P_{L}\bm{Z}|^{2},\quad\bm{Z}=(Z_{1},\ldots,Z_{n}),$

where $\lambda>0$ is a uniform lower bound for the positive eigenvalues of
$A^{BD}(\bm{c})$. Inequality (11) now follows from this property and suitable
estimates for the right-hand side of (15).

The above-mentioned techniques can be extended to a class of generalized
Maxwell–Stefan systems, which includes several examples of cross-diffusion
systems occurring in applications (see Section 5):

(16) $\displaystyle\partial_{t}c_{i}+\operatorname{div}(c_{i}u_{i})$
$\displaystyle=0,\quad\sum_{j=1}^{n}c_{j}u_{j}=0,$ (17)
$\displaystyle-\sum_{j=1}^{n}K_{ij}(\bm{c})c_{j}u_{j}$
$\displaystyle=c_{i}\nabla\frac{\delta H}{\delta
c_{i}}(\bm{c})-c_{i}\sum_{j=1}^{n}c_{j}\nabla\frac{\delta H}{\delta
c_{j}}(\bm{c}),\quad i=1,\ldots,n,$

together with the initial and boundary conditions (2), where $\delta H/\delta
c_{i}$ denotes the variational derivative of $H$. Again
$\sum_{i=1}^{n}c_{i}^{0}=1$ implies that $\sum_{i=1}^{n}c_{i}(t)=1$ in
$\Omega$, $t>0$. We assume that

$H(\bm{c})=\sum_{i=1}^{n}\int_{\Omega}h_{i}(c_{i})dx,$

which gives $\delta H/\delta c_{i}=h_{i}^{\prime}$, and
$(K_{ij})\in{\mathbb{R}}^{n\times n}$ satisfies
$\sum_{i=1}^{n}K_{ij}(\bm{c})=0$ for all $\bm{c}\in{\mathbb{R}}_{+}^{n}$. This
model was proposed in [23] and can be obtained as the high-friction limit of
multicomponent Euler systems. It can also be derived from elementary
thermodynamic considerations; see Appendix C. If the entropy $H(\bm{c})$
equals (3) and $K_{ij}(\bm{c})=\sqrt{c_{i}}A_{ij}(\bm{c})/\sqrt{c_{j}}$, where
$A_{ij}(\bm{c})$ is defined in (10), then system (16)–(17) reduces to (1). We
refer to [5, 12, 30] for multicomponent diffusion models that account for
other factors, such as thermal conduction, viscous stresses, chemical
reactions, etc.

We introduce the matrix $B(\bm{c})=(B_{ij}(\bm{c}))\in{\mathbb{R}}^{n\times
n}$ by

(18) $B_{ij}(\bm{c})=\frac{1}{\sqrt{c_{i}}}K_{ij}(\bm{c})\sqrt{c_{j}},\quad
i,j=1,\ldots,n,$

and we assume that $B(c)$ is symmetric and as before, we set
$L:=\\{\bm{z}\in{\mathbb{R}}^{n}:\sqrt{\bm{c}}\cdot\bm{z}=0\\}$ and
$L^{\perp}=\mbox{span}\\{\sqrt{\bm{c}}\\}$. We write (17) as (see the
beginning of Section 4)

$-\sum_{j=1}^{n}B_{ij}(\bm{c})\sqrt{c_{j}}u_{j}=\sum_{j=1}^{n}(P_{L})_{ij}\sqrt{c_{j}}\nabla
h_{j}^{\prime}(c_{j}),\quad i=1,\ldots,n.$

We show in Lemma 11 that this system can be inverted, leading to

$\sqrt{c_{i}}u_{i}=-\sum_{j=1}^{n}B_{ij}^{BD}(\bm{c})\sqrt{c_{j}}\nabla
h^{\prime}_{j}(c_{j}),$

where $B^{BD}(\bm{c})$ is the Bott–Duffin inverse of $B(\bm{c})$, and system
(16)–(17) can be formulated as

$\partial_{t}c_{i}=\operatorname{div}\bigg{(}\sum_{j=1}^{n}\sqrt{c_{i}}B_{ij}^{BD}(\bm{c})\sqrt{c_{j}}\nabla\frac{\delta
H}{\delta c_{j}}(\bm{c})\bigg{)},\quad i=1,\ldots,n,$

which, by the way, equals (14) if $H(\bm{c})$ is given by (3) and
$B(\bm{c})=A(\bm{c})$.

We suppose for all $\bm{c}\in[0,1]^{n}$ the following conditions on the matrix
$B(\bm{c})$:

  1. (B1)

$B(\bm{c})$ is symmetric and $L=\operatorname{ran}B(\bm{c})$,
$L^{\perp}=\ker(B(\bm{c})P_{L})$.

  2. (B2)

For all $i,j=1,\ldots,n$ and $s>0$, $B_{ij}(\bm{c})$ is bounded and Lipschitz
continuous for all $\bm{c}\in[s,1]^{n}$.

  3. (B3)

There exists a function $\gamma:(0,\infty)\to(0,\infty)$ such that for all
$m>0$ and all $s\geq m$, it holds that $\gamma(s)\leq\gamma(m)$ and
$\|B(\bm{c})\|_{F}\leq\gamma(\min_{i=1,\ldots,n}c_{i})$.

  4. (B4)

All nonzero eigenvalues of $B(\bm{c})$ are not smaller than a positive
constant $\mu>0$.

The partial free energy functions $h_{i}(c_{i})$ are associated to the
pressures $p_{i}(c_{i})$ via the thermodynamic relations

(19) $p_{i}^{\prime}(c_{i})=c_{i}h_{i}^{\prime\prime}(c_{i}),\quad
p_{i}(c_{i})=c_{i}h_{i}^{\prime}(c_{i})-h_{i}(c_{i}).$

For $h_{i}(c_{i})$ and $p_{i}(c_{i})$, we assume that, for some constants
$K_{1}$, $K_{2}>0$, it holds that

(H) $h_{i}\in C^{3}((0,1]),\quad 0<c_{i}h_{i}^{\prime\prime}(c_{i})\leq
K_{1},\ |p_{i}^{\prime\prime}(c_{i})|\leq
K_{2}h_{i}^{\prime\prime}(c_{i})\mbox{ for }c_{i}\in(0,1]$

for $i=1,\ldots,n$. This hypothesis implies that $h_{i}(c_{i})$ is strictly
convex, $p_{i}(c_{i})$ is Lipschitz on $(0,1]$. The functions
$h_{i}(c_{i})=c_{i}\log c_{i}-c_{i}$ and $h_{i}(c_{i})=c_{i}^{\gamma}$,
$\gamma>1$, satisfy (H).

Our final main result is the weak-strong uniqueness property for (16)–(17).

######  Theorem 3 (Weak-strong uniqueness for the generalized system).

Let Assumptions (A1)–(A3) and (B1)–(B4) hold, and let $h_{i}$ satisfy
Hypothesis (H). Let $\bm{c}$ be a weak solution and $\bar{\bm{c}}$ be a strong
solution to (2), (16)–(17). We suppose that $\bar{\bm{c}}$ satisfies
$\bar{c}_{i}(t)\geq m$ in $\Omega$, $t>0$ for some constant $m>0$,

$h_{i}^{\prime}(\bar{c}_{i})\in H_{\rm loc}^{1}(\Omega\times(0,\infty))\cap
L_{\rm loc}^{\infty}(0,\infty;W^{2,\infty}(\Omega)),\quad i=1,\ldots,n,$

and the entropy identity

(20)
$H(\bar{\bm{c}}(t))+\sum_{i,j=1}^{n}\int_{0}^{t}\int_{\Omega}\sqrt{\bar{c}_{i}\bar{c}_{j}}B_{ij}^{BD}(\bar{\bm{c}})\nabla
h^{\prime}_{i}(\bar{c}_{i})\cdot\nabla
h^{\prime}_{j}(\bar{c}_{j})dxds=H(\bar{\bm{c}}^{0})$

for $t>0$. Then there exists a constant $C(t)>0$, depending on $t$, $m$, and
$(D_{ij})$, such that

$H(\bm{c}(t)|\bar{\bm{c}}(t))\leq
C(t)H(\bm{c}^{0}|\bar{\bm{c}}^{0})\quad\mbox{for }t>0.$

If the initial data coincide, then $\bm{c}(t)=\bar{\bm{c}}(t)$ in $\Omega$ for
$t>0$.

We do not explore the existence of solutions with the stated regularity. The
existence of weak solutions to (2), (16)–(17) can be shown by the techniques
detailed in [28, 29] under suitable assumptions on $K_{ij}$ and
$h^{\prime}_{i}$ that guarantee nonlinear gradient estimates. The existence of
(local-in-time) strong solutions can be shown by following the approach of
[21] by formulating (16)–(17) as

$\partial_{t}c_{i}={\mathcal{M}}_{i}(\bm{c}):=\operatorname{div}\bigg{(}\sum_{j=1}^{n}M_{ij}(\bm{c})\nabla
c_{j}\bigg{)},\quad i=1,\ldots,n,$

where $M_{ij}(\bm{c})$ depends on $B^{BD}(\bm{c})$ and
$h^{\prime\prime}_{i}(c_{i})$, and verifying that the principal part of the
operator
${\mathcal{M}}(\bm{c})=({\mathcal{M}}_{1},\ldots,{\mathcal{M}}_{n})(\bm{c})$,
defined on suitable spaces, is normally elliptic and satisfies the
Lopatinski–Shapiro condition. By [15, Theorem 8.2], the operator
${\mathcal{M}}(\bm{c})$ has maximal regularity of type $L^{p}$ and the local
existence result follows from [21, Theorem A1].

The strategy of the proof of Theorem 3 is similar to that one of Theorem 1,
but it is more involved. First, we show a relative entropy inequality. The
terms of this inequality are estimated by splitting the domain into two
regions: $c_{*}(x,t):=\min_{i=1,\ldots,n}c_{i}(x,t)\leq m/2$ and
$c_{*}(x,t)>m/2$, where $m>0$ is the uniform lower bound for $\bar{c}_{i}$.
The final estimate reads

(21) $\displaystyle\frac{dH}{dt}(\bm{c}|\bar{\bm{c}})$
$\displaystyle+\frac{1}{2}\sum_{i,j=1}^{n}\int_{\Omega}(1-\chi(\bm{c}))\sqrt{c_{i}c_{j}}B_{ij}^{BD}(\bm{c})\nabla
h^{\prime}_{i}(c_{i})\cdot\nabla h^{\prime}_{j}(c_{j})dx$
$\displaystyle{}+C(m)\sum_{i=1}^{n}\int_{\Omega}\chi(\bm{c})|\nabla(c_{i}-\bar{c}_{i})|^{2}dx\leq
CH(\bm{c}|\bar{\bm{c}}),$

where $\chi(\bm{c})$ a cutoff function that vanishes if $c_{i}\leq m/2$ for
some $i$ (see (65) for details). An application of Gronwall’s lemma completes
the proof. Notice, however, that we do not obtain a gradient estimate as in
(11).

By specifying the coefficients $K_{ij}(\bm{c})$ and the entropy densities
$h_{i}$, we prove the weak-strong property for cross-diffusion systems
describing physical vapor deposition processes [1] and for the tumor-growth
model suggested in [25] and analyzed in [27] and the Maxwell-Stefan system
considering different molar masses that is derived in [3, 5]; see Section 5.

The main contributions of this work are, first, the derivation of the relative
entropy inequality (8) for the Maxwell-Stefan system and (21) for generalized
Maxwell-Stefan systems. Second, the introduction of the Bott–Duffin inverse
provides an efficient way to reduce the Maxwell–Stefan system to a
(degenerate) parabolic system formulated in the square roots $\sqrt{c_{i}}$.
(Related formulations using the chemical potentials $\delta H/\delta c_{i}$
can be found in [6].) Third, we show that our technique can be extended to
more general Maxwell–Stefan systems which may have degeneracy at zero.

The paper is organized as follows. We study the properties of the matrix
$A(\bm{c})$, defined in (10), and its Bott–Duffin inverse $A^{BD}(\bm{c})$ in
Section 2. In Section 3, we recall the existence results for global weak and
local strong solutions to (1)–(2), prove the relative entropy inequalities (7)
and (8) as well as Theorem 1 and Corollary 2. Section 4 is devoted to the
existence of the Bott–Duffin inverse of $B(\bm{c})$, defined in (18), and the
proof of the relative entropy inequality (21) eventually leading to the weak-
strong uniqueness Theorem 3. In Section 5, we present some examples that fit
into our framework. Finally, we recall the definition and some properties of
the Bott–Duffin inverse in Appendix A, show two simple inequalities for the
Boltzmann entropy density in Appendix B, and derive the generalized model
(16)–(17) from thermodynamic principles in Appendix C.

### Notation

We set ${\mathbb{R}}_{+}=[0,\infty)$. Elements of the matrix
$A\in{\mathbb{R}}^{n\times n}$ are denoted by $A_{ij}$, $i,j=1,\ldots,n$, and
the elements of a vector $\bm{c}\in{\mathbb{R}}^{n}$ are $c_{1},\ldots,c_{n}$.
If $f:{\mathbb{R}}\to{\mathbb{R}}$ is any function, we define
$f(\bm{c})=(f(c_{1}),\ldots,f(c_{n}))$ for $\bm{c}\in{\mathbb{R}}^{n}$. In the
whole paper, $C>0$, $C_{i}>0$ denote generic constants whose values change
from line to line.

##  2\. Properties of the matrix $A(\bm{c})$

The properties of the matrix $A(\bm{c})$, defined in (10), have been studied
in [4, 21, 27] under the assumption $c_{i}>0$ for all $i=1,\ldots,n$. Our
results are valid for nonnegative concentrations $c_{i}\geq 0$, including
vacuum.

Let $\bm{c}\in{\mathbb{R}}^{n}_{+}$. Since $(D_{ij})$ is symmetric, we have
for all $\bm{z}\in{\mathbb{R}}^{n}$,

$0=\sum_{i,j=1}^{n}A_{ij}(\bm{c})z_{j}=\sum_{i,j=1,\,j\neq
i}^{n}\frac{c_{j}}{D_{ij}}z_{i}-\sum_{i,j=1,\,j\neq
i}^{n}\frac{\sqrt{c_{i}c_{j}}}{D_{ij}}z_{j}=\sum_{i,j=1,\,j\neq
i}^{n}\frac{\sqrt{c_{j}}}{D_{ij}}(\sqrt{c_{j}}z_{i}-\sqrt{c_{i}}z_{j}),$

showing that $\operatorname{span}\\{\sqrt{\bm{c}}\\}=\ker A(\bm{c})$. We set

$\displaystyle\operatorname{ran}A(\bm{c})$
$\displaystyle=L:=\\{\bm{x}\in{\mathbb{R}}^{n}:\sqrt{\bm{c}}\cdot\bm{x}=0\\},$
$\displaystyle\ker A(\bm{c})$
$\displaystyle=(\operatorname{ran}A(\bm{c}))^{\perp}=L^{\perp}=\operatorname{span}\\{\sqrt{\bm{c}}\\},$

and note that $\sum_{i=1}^{n}c_{i}=1$ implies that
$|\sqrt{\bm{c}}|^{2}=c_{1}+\cdots+c_{n}=1$. The projection matrices $P_{L}$ on
$L$ and $P_{L^{\perp}}$ on $L^{\perp}$ are given by

(22)
$(P_{L})_{ij}=\delta_{ij}-\sqrt{c_{i}c_{j}},\quad(P_{L^{\perp}})_{ij}=\delta_{ij}-(P_{L})_{ij}=\sqrt{c_{i}c_{j}},\quad
i,j=1,\ldots,n.$

######  Lemma 4.

Let $\bm{c}\in{\mathbb{R}}_{+}^{n}$ be such that $\sum_{i=1}^{n}c_{i}=1$. Then

(23) $\bm{z}^{T}A(\bm{c})\bm{z}\geq\mu|P_{L}\bm{z}|^{2}\quad\mbox{for all
}\bm{z}\in{\mathbb{R}}^{n},$

where $\mu=\min_{i\neq j}(1/D_{ij})$. Moreover, the Bott–Duffin inverse

$A^{BD}(\bm{c})=P_{L}(A(\bm{c})P_{L}+P_{L^{\perp}})^{-1}$

is well defined, symmetric, and satisfies

(24) $\bm{z}^{T}A^{BD}(\bm{c})\bm{z}\geq\lambda|P_{L}\bm{z}|^{2}\quad\mbox{for
all }\bm{z}\in{\mathbb{R}}^{n},$

where $\lambda=(2\sum_{i\neq j}(1/D_{ij}+1))^{-1}$.

###### Proof.

We first prove (23). Let $0<\alpha\leq\mu$ and suppose that $c_{i}\neq 0$ for
$i=1,\ldots,M$ and $c_{i}=0$ for $i=M+1,\ldots,n$. If necessary, we may
rearrange the indices to achieve this ordering. Since $\sum_{i=1}^{n}c_{i}=1$,
it holds that $M>0$. Thus, we can write $-A(\bm{c})-\alpha P_{L^{\perp}}$ in
block diagonal form as

$-A(\bm{c})-\alpha
P_{L^{\perp}}=\left(\begin{array}[]{c|ccc}\widetilde{A}&0&0&0\\\ \hline\cr
0&a_{M+1}&&0\\\ \vdots&0&\ddots&0\\\ 0&0&&a_{n}\end{array}\right),$

where $\widetilde{A}\in{\mathbb{R}}^{M\times M}$ has the coefficients
$\widetilde{A}_{ii}=-A_{ii}(\bm{c})-\alpha c_{i}$ and
$\widetilde{A}_{ij}=(1/D_{ij}-\alpha)\sqrt{c_{i}c_{j}}$ for $i,j=1,\ldots,M$,
$i\neq j$, and $a_{j}=-\sum_{k=1,\,k\neq j}^{n}c_{k}/D_{kj}$ for
$j=M+1,\ldots,n$. Because of $\alpha\leq\mu$, the matrix $\widetilde{A}$ is
quasi-positive and irreducible. Hence, by the Perron–Frobenius theorem [32,
Chapter 8], the spectral radius of $\widetilde{A}$ is less than or equal to
the Perron–Frobenius eigenvalue that is a simple eigenvalue of $\widetilde{A}$
associated with a strictly positive eigenvector, and all other eigenvalues of
$\widetilde{A}$ have no positive eigenvector. In the present case, the
Perron–Frobenius eigenvalue is given by $\lambda_{\rm PF}=-\alpha$ and is
associated with the eigenvector $(\sqrt{c_{1}},\ldots,\sqrt{c_{M}})$,
recalling that $c_{i}>0$ for all $i=1,\ldots,M$ (also see the proof of Lemma
2.1 in [27]). Since all eigenvalues of $\widetilde{A}$ are not larger than
$\lambda_{\rm PF}$, we have

$\widetilde{\bm{z}}^{T}(-\widetilde{A})\widetilde{\bm{z}}\geq\alpha|\widetilde{\bm{z}}|^{2}\quad\mbox{for
}\widetilde{\bm{z}}=(z_{1},\ldots,z_{M})\in{\mathbb{R}}^{M}.$

This leads, for any $\alpha\leq\mu$ and $\bm{z}\in{\mathbb{R}}^{n}$, to the
inequality

$\displaystyle\bm{z}^{T}$ $\displaystyle(A(\bm{c})+\alpha
P_{L^{\perp}})\bm{z}=\widetilde{z}^{T}(-\widetilde{A})\widetilde{\bm{z}}+\sum_{i=M+1}^{n}\sum_{j=1,\,j\neq
i}^{n}\frac{c_{j}}{D_{ij}}z_{i}^{2}$
$\displaystyle\geq\alpha|\widetilde{\bm{z}}|^{2}+\min_{\begin{subarray}{c}k,\ell=1,\ldots,M\\\
k\neq\ell\end{subarray}}\frac{1}{D_{k\ell}}\sum_{j=1,\,j\neq
i}^{n}c_{j}\sum_{i=M+1}^{n}z_{i}^{2}\geq\alpha|\widetilde{\bm{z}}|^{2}+\alpha\sum_{i=M+1}^{n}z_{i}^{2}=\alpha|\bm{z}|^{2},$

where we have used the fact that $\sum_{j=1,\,j\neq i}^{n}c_{j}=1$ for
$i=M+1,\ldots,n$, since $c_{i}=0$ for exactly these indices. This inequality
implies that for all
$\bm{z}=P_{L}\bm{z}+P_{L^{\perp}}\bm{z}\in{\mathbb{R}}^{n}$,

$\bm{z}^{T}A(\bm{c})\bm{z}+\alpha|P_{L^{\perp}}\bm{z}|^{2}=\bm{z}^{T}(A(\bm{c})+\alpha
P_{L^{\perp}})\bm{z}\geq\alpha|P_{L}\bm{z}|^{2}+\alpha|P_{L^{\perp}}\bm{z}|^{2},$

which shows (23).

The invertibility of $A(\bm{c})P_{L}+P_{L^{\perp}}$ is a consequence of Lemma
17 in the appendix. Consequently, the Bott–Duffin inverse
$A^{BD}(\bm{c})=P_{L}(A(\bm{c})P_{L}+P_{L^{\perp}})^{-1}$ exists.

It remains to show (24). The spectral radius $r(A(\bm{c})P_{L}+P_{L^{\perp}})$
is bounded by the Frobenius norm. Thus, because of $A(\bm{c})P_{L}=A(\bm{c})$
(see Lemma 17 in Appendix A) and $0\leq c_{i}\leq 1$,

$\displaystyle r(A(\bm{c})P_{L}+P_{L^{\perp}})$
$\displaystyle\leq\|A(\bm{c})+P_{L^{\perp}}\|_{F}=\bigg{(}\sum_{i,j=1}^{n}(A_{ij}(\bm{c})+\sqrt{c_{i}c_{j}})^{2}\bigg{)}^{1/2}$
$\displaystyle=\bigg{\\{}\sum_{i=1}^{n}\bigg{(}\sum_{j=1,\,j\neq
i}^{n}\frac{c_{j}}{D_{ij}}+c_{i}\bigg{)}^{2}+\sum_{i,j=1,\,i\neq
j}^{n}\bigg{(}1-\frac{1}{D_{ij}}\bigg{)}^{2}c_{i}c_{j}\bigg{\\}}^{1/2}$
$\displaystyle\leq 2\sum_{i,j=1,\,i\neq
j}\bigg{(}\frac{1}{D_{ij}}+1\bigg{)}=\frac{1}{\lambda}.$

We infer that the eigenvalues of $(A(\bm{c})P_{L}+P_{L^{\perp}})^{-1}$ are
larger than or equal to $\lambda$. Thus, in view of (81), we find that for all
$\bm{z}\in{\mathbb{R}}^{n}$,

$\bm{z}^{T}A^{BD}(\bm{c})\bm{z}=(P_{L}\bm{z})^{T}(A(\bm{c})P_{L}+P_{L^{\perp}})^{-1}P_{L}\bm{z}\geq\lambda|P_{L}\bm{z}|^{2},$

finishing the proof. ∎

Since $(\nabla\sqrt{c_{1}},\ldots,\nabla\sqrt{c_{n}})\in L$, the existence of
the Bott–Duffin inverse guarantees that the solution of (12) can be expressed
via the formula (13); see Appendix A.

##  3\. Weak-strong uniqueness for Maxwell–Stefan systems

###  3.1. Existence theory

We discuss the existence of weak and strong solutions to the Maxwell–Stefan
system (1)–(2). First, we recall the existence theorem for weak solutions,
which was proved in [27].

######  Theorem 5 (Global existence for Maxwell–Stefan systems).

Let Assumptions (A1)–(A3) hold. Then there exists a weak solution to (1)–(2)
satisfying the entropy inequality (6) for $t>0$, or equivalently,

$\displaystyle
H(\bm{c}(t))+4\sum_{i,j=1}^{n}\int_{0}^{t}\int_{\Omega}A^{BD}_{ij}(\bm{c})\nabla\sqrt{c_{i}}\cdot\nabla\sqrt{c_{j}}dxds$
$\displaystyle\leq H(\bm{c}_{0}).$

The existence of strong solutions was proved in [4, Theorem 1] and [21,
Theorem 3.2].

######  Theorem 6 (Strong solutions for Maxwell–Stefan systems).

Let $\Omega\subset{\mathbb{R}}^{d}$ $(d\geq 1)$ be a bounded domain with
$\partial\Omega\in C^{2}$ and let $\bm{c}^{0}\in
W^{2-2/p,p}(\Omega;{\mathbb{R}}^{n})$ with $c_{i}^{0}\geq 0$,
$\sum_{i=1}^{n}c_{i}^{0}=1$ in $\Omega$, where $p>d+2$. Then there exists
$T^{*}>0$ and a unique solution $\bm{c}$ to (1)–(2) satisfying

$c_{i}\in C^{1}((0,T^{*});W^{2-2/p,p}(\Omega))\cap
W^{1,p}(0,T;L^{p}(\Omega))\cap L^{p}(0,T;W^{2,p}(\Omega))$

for $i=1,\ldots,n$.

The strong solution of Theorem 6 has the property of immediate positivity: If
$c_{i}^{0}\geq 0$ in $\Omega$ then $c_{i}(t)>0$ in $\Omega$ for
$0<t<T^{\prime}$, where $T^{\prime}\leq T$ depends on $\bm{c}^{0}$. Moreover,
if the initial data is close to a constant vector, the strong solution can be
extended globally: Let $\bm{c}^{*}\in{\mathbb{R}}_{+}^{n}$. Then there exists
$\varepsilon>0$ such that if the initial data satisfies
$\|\bm{c}^{0}-\bm{c}^{*}\|_{W^{2-2/p,p}(\Omega)}\leq\varepsilon$, then the
strong solution exists globally in time.

If $c_{i}^{0}>0$ in $\Omega$ for $i=1,\ldots,n$, the continuity of the strong
solution implies that there exists $0<T^{\prime\prime}\leq T^{*}$ and $m>0$
such that $c_{i}(t)\geq m>0$ in $\Omega$ for $i=1,\ldots,n$. Therefore,
because of the embedding $W^{2-2/p,p}(\Omega)\hookrightarrow
C^{1}(\overline{\Omega})$, we have $\sqrt{c_{i}}\in
L^{\infty}(0,T^{\prime\prime};W^{1,\infty}(\Omega))$, and
$T^{\prime\prime}=\infty$ if $\|\bm{c}^{0}-\bm{c}^{*}\|_{W^{2-2/p,p}(\Omega)}$
is sufficiently small. This shows that the strong solution satisfies the
regularity assumptions of Corollary 2.

The assumption $\sqrt{c_{i}}\in L_{\rm
loc}^{\infty}(0,\infty;W^{1,\infty}(\Omega))$ and the property $c_{i}(t)\geq
m$ in $\Omega$ imply that the regularity condition $u_{i}\in L_{\rm
loc}^{\infty}(\Omega\times(0,\infty))$ of Theorem 1 is satisfied. This is a
consequence of the following lemma and
$m\sum_{i=1}^{n}|u_{i}|^{2}\leq\sum_{i=1}^{n}c_{i}|u_{i}|^{2}\leq
C\sum_{i=1}^{n}|\nabla\sqrt{c_{i}}|^{2}$. Moreover, the assumption $\log
c_{i}\in H_{\rm loc}^{1}(\Omega\times(0,\infty))$ follows from $|\nabla\log
c_{i}|\leq|\nabla c_{i}|/m\in
C^{0}((0,T^{\prime\prime});C^{0}(\overline{\Omega}))$.

######  Lemma 7.

Let $0\leq c_{i}\leq 1$ and let $u_{i}$ be given by the force-flux relations
in (1) satisfying $\sum_{i=1}^{n}c_{i}u_{i}=0$. Then there exists a constant
$C>0$, only depending on $(D_{ij})$ such that

$\sum_{i=1}^{n}c_{i}|u_{i}|^{2}\leq C\sum_{i=1}^{n}|\nabla\sqrt{c_{i}}|^{2}.$

###### Proof.

It follows from (12) and the symmetry of $A(\bm{c})$, defined in (10), that

$4\sum_{i=1}^{n}|\nabla\sqrt{c_{i}}|^{2}=\sum_{i=1}^{n}\bigg{|}\sum_{j=1}^{n}A_{ij}(\bm{c})\sqrt{c_{j}}u_{j}\bigg{|}^{2}=\sum_{i,j=1}^{n}\sqrt{c_{i}}u_{i}(A(\bm{c})^{2})_{ij}\sqrt{c_{j}}u_{j}.$

Since the eigenvalues of $A(\bm{c})^{2}$ are the square of the eigenvalues of
$A(\bm{c})$, we deduce from (23) that
$\bm{z}^{T}A(\bm{c})^{2}\bm{z}\geq\mu^{2}|P_{L}\bm{z}|^{2}$ for all
$\bm{z}\in{\mathbb{R}}^{n}$. This yields

$4\sum_{i=1}^{n}|\nabla\sqrt{c_{i}}|^{2}\geq\mu^{2}|P_{L}(\sqrt{c_{i}}u_{i})_{i=1,\ldots,n}|^{2}.$

Because of $\sum_{i=1}^{n}\sqrt{c_{i}}(\sqrt{c_{i}}u_{i})=0$, we have
$(\sqrt{c_{i}}u_{i})_{i}\in L$ and hence,
$P_{L}(\sqrt{c_{i}}u_{i})_{i}=(\sqrt{c_{i}}u_{i})_{i}$. The statement of the
lemma follows after setting $C=4/\mu^{2}$. ∎

###  3.2. Relative entropy inequality

We first derive a relative entropy inequality via a formal computation. Using
(5) and (1), we obtain

(25) $\displaystyle\frac{d}{dt}H(\bm{c}|\bar{\bm{c}})$
$\displaystyle=\sum_{i=1}^{n}\int_{\Omega}\bigg{(}\log\frac{c_{i}}{\bar{c}_{i}}\partial_{t}c_{i}+\bigg{(}1-\frac{c_{i}}{\bar{c}_{i}}\bigg{)}\partial_{t}\bar{c}_{i}\bigg{)}dx$
$\displaystyle=\sum_{i=1}^{n}\int_{\Omega}\bigg{(}\nabla\log\frac{c_{i}}{\bar{c}_{i}}\cdot(c_{i}u_{i})-\nabla\log\frac{c_{i}}{\bar{c}_{i}}\cdot(c_{i}\bar{u}_{i})\bigg{)}dx$
$\displaystyle=\sum_{i=1}^{n}\int_{\Omega}c_{i}\nabla(\log
c_{i}-\log\bar{c}_{i})\cdot(u_{i}-\bar{u}_{i})dx.$

To reformulate the integrand of the right-hand side, we insert the second
equation of (1), and use the symmetry of $(D_{ij})$:

$\displaystyle\sum_{i=1}^{n}$ $\displaystyle c_{i}\nabla(\log
c_{i}-\log\bar{c}_{i})\cdot(u_{i}-\bar{u}_{i})=-\sum_{i=1}^{n}c_{i}(u_{i}-\bar{u}_{i})\cdot\sum_{j\neq
i}\frac{1}{D_{ij}}\big{(}c_{j}(u_{i}-u_{j})-\bar{c}_{j}(\bar{u}_{i}-\bar{u}_{j})\big{)}$
$\displaystyle=-\sum_{i=1}^{n}c_{i}(u_{i}-\bar{u}_{i})\cdot\sum_{j\neq
i}\frac{c_{j}}{D_{ij}}\big{(}(u_{i}-\bar{u}_{i})-(u_{j}-\bar{u}_{j})\big{)}$
(26) $\displaystyle\phantom{xx}{}-\sum_{i,j=1,\,i\neq
j}^{n}\frac{1}{D_{ij}}c_{i}(u_{i}-\bar{u}_{i})\cdot\big{(}(c_{j}-\bar{c}_{j})(\bar{u}_{i}-\bar{u}_{j})\big{)}$
$\displaystyle=-\sum_{i,j=1,\,i\neq
j}^{n}\frac{c_{i}c_{j}}{2D_{ij}}\big{|}(u_{i}-\bar{u}_{i})-(u_{j}-\bar{u}_{j})\big{|}^{2}-\sum_{i,j=1,\,i\neq
j}^{n}\frac{c_{i}}{D_{ij}}(c_{j}-\bar{c}_{j})(u_{i}-\bar{u}_{i})\cdot(\bar{u}_{i}-\bar{u}_{j}).$

This shows that

(27) $\displaystyle\frac{d}{dt}H$
$\displaystyle(\bm{c}|\bar{\bm{c}})+\frac{1}{2}\sum_{i,j=1,\,i\neq
j}^{n}\int_{\Omega}\frac{c_{i}c_{j}}{D_{ij}}\big{|}(u_{i}-\bar{u}_{i})-(u_{j}-\bar{u}_{j})\big{|}^{2}dx$
$\displaystyle{}=-\sum_{i,j=1,\,i\neq
j}^{n}\int_{\Omega}\frac{c_{i}}{D_{ij}}(c_{j}-\bar{c}_{j})(u_{i}-\bar{u}_{i})\cdot(\bar{u}_{i}-\bar{u}_{j})dx.$

Our aim is to make this computation rigorous. Since the computation in (26) is
purely algebraic, it holds without any regularity restrictions. In principle,
one would expect that (27) holds under the condition that all the terms are
well defined, which would cover the class of weak solutions subject to the
condition $u_{i}\in L^{\infty}$ (to ensure integrability of the right-hand
side). However, we have not been able to establish (25) for such a class of
solutions, and stricter conditions on one of the solutions are required.

######  Lemma 8.

Let $\bm{c}$ be a weak solution to (1)–(2) and let $\bar{\bm{c}}$ be a strong
solution to (1)–(2) satisfying $0<\bar{c}_{i}(t)<1$ in $\Omega$, the
regularity

$\log\bar{c}_{i}\in L_{\rm loc}^{2}(0,\infty;H^{1}(\Omega)),\
\partial_{t}\log\bar{c}_{i}\in L_{\rm loc}^{2}(\Omega\times(0,\infty)),\
\bar{u}_{i}\in L_{\rm loc}^{\infty}(0,\infty;L^{\infty}(\Omega)),$

and the entropy identity

(28)
$H(\bar{\bm{c}}(t))+\sum_{i,j=1}^{n}\frac{1}{2}\int_{0}^{t}\int_{\Omega}\frac{\bar{c}_{i}\bar{c}_{j}}{D_{ij}}|u_{i}-\bar{u}_{i}|^{2}dxds=H(\bar{\bm{c}}^{0})\quad\mbox{for
}t>0.$

Then

$\displaystyle H(\bm{c}(t)$
$\displaystyle|\bar{\bm{c}}(t))+\sum_{i,j=1}^{n}\int_{0}^{t}\int_{\Omega}A_{ij}(\bm{c})(\sqrt{c_{i}}(u_{i}-\bar{u}_{i}))\cdot(\sqrt{c_{j}}(u_{j}-\bar{u}_{j}))dxds$
(29) $\displaystyle{}\leq H(\bm{c}^{0}|\bar{\bm{c}}^{0})-\sum_{i,j=1,\,i\neq
j}^{n}\int_{0}^{t}\int_{\Omega}\frac{c_{i}}{D_{ij}}(c_{j}-\bar{c}_{j})(u_{i}-\bar{u}_{i})\cdot(\bar{u}_{i}-\bar{u}_{j})dxds.$

###### Proof.

Since

$H(\bm{c}|\bar{\bm{c}})=H(\bm{c})-H(\bar{\bm{c}})-\int_{\Omega}\sum_{i=1}^{n}(c_{i}-\bar{c}_{i})\log\bar{c}_{i}dx,$

we need to formulate the time evolution of each of these terms. According to
Theorem 5, the weak solution $\bm{c}$ satisfies $\nabla\sqrt{c_{i}}$,
$\sqrt{c_{i}}u_{i}\in L_{\rm loc}^{2}(0,\infty;L^{2}(\Omega))$ and

$H(\bm{c}(t))+\frac{1}{2}\sum_{i,j=1}^{n}\int_{0}^{t}\int_{\Omega}\frac{c_{i}c_{j}}{D_{ij}}|u_{i}-u_{j}|^{2}dxds\leq
H(\bm{c}^{0})\quad\mbox{for }t>0.$

The symmetry of $(D_{ij})$ and the force-flux relations in (1) give

$\sum_{i=1}^{n}c_{i}u_{i}\cdot\nabla\log
c_{i}=-\sum_{i,j=1}^{n}c_{i}u_{i}\cdot\frac{c_{j}}{D_{ij}}(u_{i}-u_{j})=-\frac{1}{2}\sum_{i,j=1}^{n}\frac{c_{i}c_{j}}{D_{ij}}|u_{i}-u_{j}|^{2},$

and we formulate the entropy inequality as

(30)
$H(\bm{c}(t))-H(\bm{c}^{0})\leq\sum_{i=1}^{n}\int_{0}^{t}\int_{\Omega}c_{i}u_{i}\cdot\nabla\log
c_{i}dxds\quad\mbox{for }t>0.$

The expression $c_{i}u_{i}\cdot\nabla\log c_{i}$ has to be understood as
$2\nabla\sqrt{c_{i}}\cdot(\sqrt{c_{i}}u_{i})$, which is well defined since
$\nabla\sqrt{c_{i}}$, $\sqrt{c_{i}}u_{i}\in L^{2}(\Omega\times(0,T))$ (see
Lemma 7). In a similar way, we express the entropy identity (28) as

(31)
$H(\bar{\bm{c}}(t))-H(\bar{\bm{c}}^{0})=\sum_{i=1}^{n}\int_{0}^{t}\int_{\Omega}\bar{c}_{i}\bar{u}_{i}\cdot\nabla\log\bar{c}_{i}dxds\quad\mbox{for
}t>0.$

Next, the difference of the weak formulations for $\bm{c}$ and $\bar{\bm{c}}$
gives

$\displaystyle\int_{\Omega}$
$\displaystyle(c_{i}-\bar{c}_{i})(t)\phi_{i}(t)dx-\int_{\Omega}(c_{i}^{0}-\bar{c}_{i}^{0})\phi_{i}(0)dx$
$\displaystyle=\int_{0}^{t}\int_{\Omega}(c_{i}-\bar{c}_{i})\partial_{t}\phi_{i}dxds+\int_{0}^{t}\int_{\Omega}(c_{i}u_{i}-\bar{c}_{i}\bar{u}_{i})\cdot\nabla\phi_{i}dxds$

for test functions $\phi_{i}\in C_{\rm
loc}^{1}([0,\infty);C^{1}(\overline{\Omega}))$. Using a density argument we
see that the test function $\phi_{i}$ can be taken in the class
$H^{1}(\Omega\times(0,T))$ for $T>0$, in which case $\phi_{i}(t)$,
$\phi_{i}(0)$ are well defined by the trace theorem. Selecting
$\phi_{i}=\log\bar{c}_{i}$, we obtain

(32) $\displaystyle\int_{\Omega}$
$\displaystyle(c_{i}-\bar{c}_{i})(t)\log\bar{c}_{i}(t)dx-\int_{\Omega}(c_{i}^{0}-\bar{c}_{i}^{0})\log\bar{c}_{i}^{0}dx$
$\displaystyle=\int_{0}^{t}\int_{\Omega}(c_{i}-\bar{c}_{i})\frac{\partial_{t}\bar{c}_{i}}{\bar{c}_{i}}dxds+\int_{0}^{t}\int_{\Omega}(c_{i}u_{i}-\bar{c}_{i}\bar{u}_{i})\cdot\nabla\log\bar{c}_{i}dxds.$

Taking into account the regularity properties of $\bar{c}_{i}$, we insert
$\partial_{t}\bar{c}_{i}=-\operatorname{div}(\bar{c}_{i}\bar{u}_{i})$ in the
third term and integrate by parts:

$\int_{0}^{t}\int_{\Omega}(c_{i}-\bar{c}_{i})\frac{\partial_{t}\bar{c}_{i}}{\bar{c}_{i}}dxds=\int_{0}^{t}\int_{\Omega}\nabla\bigg{(}\frac{c_{i}}{\bar{c}_{i}}\bigg{)}\cdot(\bar{c}_{i}\bar{u}_{i})dxds.$

We wish to write the integrand on the right-hand side as

$\nabla\bigg{(}\frac{c_{i}}{\bar{c}_{i}}\bigg{)}\cdot(\bar{c}_{i}\bar{u}_{i})=\bigg{(}\nabla
c_{i}-\frac{c_{i}}{\bar{c}_{i}}\nabla\bar{c}_{i}\bigg{)}\cdot\bar{u}_{i}=c_{i}\nabla\log\bigg{(}\frac{c_{i}}{\bar{c}_{i}}\bigg{)}\cdot\bar{u}_{i}.$

Since $c_{i}\geq 0$ only, the expression $\log c_{i}$ may be not integrable.
Therefore, we define

$\nabla\log\bigg{(}\frac{c_{i}}{\bar{c}_{i}}\bigg{)}:=\frac{1}{\sqrt{c_{i}}}(2\nabla\sqrt{c_{i}}-\sqrt{c_{i}}\nabla\log\bar{c}_{i})\quad\mbox{if
}c_{i}>0$

as the product of two functions and $\nabla\log(c_{i}/\bar{c}_{i})$ arbitrary
if $c_{i}=0$. Although this product may be not integrable, the expression
$c_{i}\nabla\log(c_{i}/\bar{c}_{i})$ lies in $L^{2}(\Omega\times(0,T))$ and
consequently, $c_{i}\nabla\log(c_{i}/\bar{c}_{i})\cdot\bar{u}_{i}$ lies in the
same space. Therefore, we can formulate (32) as

(33) $\displaystyle\int_{\Omega}$
$\displaystyle(c_{i}-\bar{c}_{i})(t)\log\bar{c}_{i}(t)dx-\int_{\Omega}(c_{i}^{0}-\bar{c}_{i}^{0})\log\bar{c}_{i}^{0}dx$
$\displaystyle=\int_{0}^{t}\int_{\Omega}c_{i}\nabla\log\bigg{(}\frac{c_{i}}{\bar{c}_{i}}\bigg{)}\cdot\bar{u}_{i}dxds+\int_{0}^{t}\int_{\Omega}(c_{i}u_{i}-\bar{c}_{i}\bar{u}_{i})\cdot\nabla\log\bar{c}_{i}dxds.$

Subtracting (31) and (33) from (30) leads to

(34)
$H(\bm{c}(t)|\bar{\bm{c}}(t))-H(\bm{c}^{0}|\bar{\bm{c}}^{0})\leq\sum_{i=1}^{n}\int_{0}^{t}\int_{\Omega}c_{i}\nabla\log\bigg{(}\frac{c_{i}}{\bar{c}_{i}}\bigg{)}\cdot(u_{i}-\bar{u}_{i})dxds.$

Finally, using (26) and the form (9) of the friction, we obtain (29). ∎

###  3.3. Proof of Theorem 1

We proceed to estimate (29). We set $\bm{Y}=(Y_{1},\ldots,Y_{n})$ with
$Y_{i}=\sqrt{c_{i}}(u_{i}-\bar{u}_{i})$, $i=1,\ldots,n$. Then, using (23), we
have

(35)
$\sum_{i,j=1}^{n}A_{ij}(\bm{c})(\sqrt{c_{i}}(u_{i}-\bar{u}_{i}))\cdot(\sqrt{c_{j}}(u_{j}-\bar{u}_{j}))=\bm{Y}^{T}A(\bm{c})\bm{Y}\geq\mu|P_{L}\bm{Y}|^{2}.$

It follows from the constraints
$\sum_{i=1}^{n}c_{i}u_{i}=\sum_{i=1}^{n}\bar{c}_{i}\bar{u}_{i}=0$ that

$\displaystyle(P_{L^{\perp}}\bm{Y})_{i}$
$\displaystyle=\sum_{j=1}^{n}\sqrt{c_{i}}c_{j}(u_{j}-\bar{u}_{j})=\sqrt{c_{i}}\sum_{j=1}^{n}(\bar{c}_{j}-c_{j})\bar{u}_{j},$
$\displaystyle|P_{L}\bm{Y}|^{2}$
$\displaystyle=|\bm{Y}|^{2}-|P_{L^{\perp}}\bm{Y}|^{2}=\sum_{i=1}^{n}c_{i}|u_{i}-\bar{u}_{i}|^{2}-\sum_{i=1}^{n}c_{i}\bigg{|}\sum_{j=1}^{n}(c_{j}-\bar{c}_{j})\bar{u}_{j}\bigg{|}^{2}$
$\displaystyle\geq\sum_{i=1}^{n}c_{i}|u_{i}-\bar{u}_{i}|^{2}-n\|\bar{u}\|_{L^{\infty}}\sum_{j=1}^{n}(c_{j}-\bar{c}_{j})^{2},$

where
$\|\bar{u}\|_{L^{\infty}}:=\max_{j=1,\ldots,n}\|u_{j}\|_{L^{\infty}(\Omega\times(0,T))}$,
and we used $\sum_{i=1}^{n}c_{i}=1$.

We turn to the last term in (29), which is estimated

(36) $\displaystyle\bigg{|}\int_{0}^{t}$
$\displaystyle\int_{\Omega}\sum_{i,j=1,\,i\neq
j}^{n}\frac{c_{i}}{D_{ij}}(c_{j}-\bar{c}_{j})(u_{i}-\bar{u}_{i})\cdot(\bar{u}_{i}-\bar{u}_{j})dxds\bigg{|}$
$\displaystyle\leq\int_{0}^{t}\int_{\Omega}\sum_{i=1}^{n}\big{(}\sqrt{c_{i}}(u_{i}-\bar{u}_{i}\big{)}\bigg{(}\sqrt{c_{i}}\sum_{j=1}^{n}\frac{|c_{j}-\bar{c}_{j}|}{D_{ij}}|\bar{u}_{i}-\bar{u}_{j}|\bigg{)}dxds$
$\displaystyle\leq\frac{2\|\bar{u}\|_{L^{\infty}}}{\min_{i\neq
j}D_{ij}}\int_{0}^{t}\int_{\Omega}\bigg{(}\sum_{i=1}^{n}c_{i}|u_{i}-\bar{u}_{i}|^{2}\bigg{)}^{1/2}\bigg{(}n\sum_{j=1}^{n}|c_{j}-\bar{c}_{j}|^{2}\bigg{)}^{1/2}dxds$
$\displaystyle\leq\frac{\mu}{2}\int_{0}^{t}\int_{\Omega}\sum_{i=1}^{n}c_{i}|u_{i}-\bar{u}_{i}|^{2}dxds+C(\mu)\int_{0}^{t}\int_{\Omega}\sum_{i=1}^{n}(c_{i}-\bar{c}_{i})^{2}dxds,$

where the constant $C(\mu)>0$ also depends on $\min_{i\neq j}D_{ij}$ and
$\|\bar{u}\|_{L^{\infty}}$. Inserting (35)–(36) into (29) and taking into
account Lemma 18 in Appendix B, we find that

$\displaystyle H($
$\displaystyle\bm{c}(t)|\bar{\bm{c}}(t))+\frac{\mu}{2}\sum_{i=1}^{n}\int_{0}^{t}\int_{\Omega}c_{i}|u_{i}-\bar{u}_{i}|^{2}dxds$
$\displaystyle\leq
H(\bm{c}^{0}|\bar{\bm{c}}^{0})+C(\mu)\int_{0}^{t}\int_{\Omega}\sum_{i=1}^{n}(c_{i}-\bar{c}_{i})^{2}dxds\leq
H(\bm{c}^{0}|\bar{\bm{c}}^{0})+2C(\mu)\int_{0}^{t}H(\bm{c}|\bar{\bm{c}})ds,$

and an application of Gronwall’s lemma finishes the proof.

###  3.4. Proof of Corollary 2

In this section, we express the relative entropy via the Bott–Duffin inverse
$A^{BD}(\bm{c})$.

######  Lemma 9.

Let the assumptions of Lemma 8 hold with $\bar{u}_{i}\in L^{\infty}_{\rm
loc}(0,\infty;L^{\infty}(\Omega))$ replaced by $\sqrt{\bar{c}_{i}}\in L_{\rm
loc}^{\infty}(0,\infty;W^{1,\infty}(\Omega))$. Then, setting
$\bm{Z}=(Z_{1},\ldots,Z_{n})$ with
$Z_{i}=\nabla\sqrt{c_{i}}-(\sqrt{c_{i}}/\sqrt{\bar{c}_{i}})\nabla\sqrt{\bar{c}_{i}}$
for $i=1,\ldots,n$,

(37) $\displaystyle H($
$\displaystyle\bm{c}(t)|\bar{\bm{c}}(t))+4\sum_{i,j=1}^{n}\int_{0}^{t}\int_{\Omega}A_{ij}^{BD}(\bm{c})Z_{i}\cdot
Z_{j}dxds$ $\displaystyle\leq
H(\bm{c}^{0}|\bar{\bm{c}}^{0})+4\sum_{i,j=1}^{n}\int_{0}^{t}\int_{\Omega}Z_{i}\cdot\nabla\sqrt{\bar{c}_{j}}\bigg{(}\frac{\sqrt{c_{i}}}{\sqrt{\bar{c}_{i}}}A_{ij}^{BD}(\bar{\bm{c}})-A_{ij}^{BD}(\bm{c})\frac{\sqrt{c_{j}}}{\sqrt{\bar{c}_{j}}}\bigg{)}dxds.$

###### Proof.

Starting with the relative entropy inequality in the form (34), we express its
right-hand side by using (13):

$\displaystyle\sum_{i=1}^{n}$ $\displaystyle c_{i}\nabla(\log
c_{i}-\log\bar{c}_{i})\cdot(u_{i}-\bar{u}_{i})$
$\displaystyle=-\sum_{i=1}^{n}\bigg{(}\nabla
c_{i}-\frac{c_{i}}{\bar{c}_{i}}\nabla\bar{c}_{i}\bigg{)}\cdot\sum_{j=1}^{n}\bigg{(}\frac{1}{\sqrt{c_{i}}}A_{ij}^{BD}(\bm{c})\frac{1}{\sqrt{c_{j}}}\nabla
c_{j}-\frac{1}{\sqrt{\bar{c}_{i}}}A_{ij}^{BD}(\bar{\bm{c}})\frac{1}{\sqrt{\bar{c}_{j}}}\nabla\bar{c}_{j}\bigg{)}$
$\displaystyle=-4\sum_{i,j=1}^{n}Z_{i}\cdot
A_{ij}^{BD}(\bm{c})Z_{j}-4\sum_{i,j=1}^{n}Z_{i}\cdot\bigg{(}A_{ij}^{BD}(\bm{c})\sqrt{\frac{c_{j}}{\bar{c}_{j}}}-A_{ij}^{BD}(\bar{\bm{c}})\sqrt{\frac{c_{i}}{\bar{c}_{i}}}\bigg{)}\nabla\sqrt{\bar{c}_{j}},$

which gives (37). ∎

We continue with the proof of Corollary 2. We estimate the two integrals of
the relative entropy inequality (37). The integrand of the second term is
estimated, because of (24), as

(38) $\sum_{i,j=1}^{n}A_{ij}^{BD}(\bar{c})Z_{i}\cdot
Z_{j}\geq\lambda|P_{L}\bm{Z}|^{2}.$

The definitions of $P_{L}$ and $\bm{Z}$ yield

$\displaystyle(P_{L}\bm{Z})_{i}$
$\displaystyle=\bigg{(}\nabla\sqrt{c_{i}}-\frac{\sqrt{c_{i}}}{\sqrt{\bar{c}_{i}}}\nabla\sqrt{\bar{c}_{i}}\bigg{)}-\sum_{j=1}^{n}\sqrt{c_{i}c_{j}}\bigg{(}\nabla\sqrt{c_{j}}-\frac{\sqrt{c_{j}}}{\sqrt{\bar{c}_{j}}}\nabla\sqrt{\bar{c}_{j}}\bigg{)}$
$\displaystyle=\nabla(\sqrt{c_{i}}-\sqrt{\bar{c}_{i}})+\frac{\sqrt{\bar{c}_{i}}-\sqrt{c_{i}}}{\sqrt{\bar{c}_{i}}}\nabla\sqrt{\bar{c}_{i}}-\sqrt{c_{i}}\sum_{j=1}^{n}\frac{(\sqrt{\bar{c}_{j}})^{2}-(\sqrt{c_{j}})^{2}}{\sqrt{\bar{c}_{j}}}\nabla\sqrt{\bar{c}_{j}}.$

Using Young’s inequality $(A+B+C)^{2}\geq A^{2}/2-4B^{2}-4C^{2}$ and the
bounds $\bar{c}_{i}\geq m$ and $\sqrt{c_{i}}+\sqrt{\bar{c}_{i}}\leq 2$, we
infer that

$\displaystyle|P_{L}\bm{Z}|^{2}$
$\displaystyle\geq\sum_{i=1}^{n}\bigg{(}\frac{1}{2}|\nabla(\sqrt{c_{i}}-\sqrt{\bar{c}_{i}})|^{2}-4\bigg{|}\frac{\sqrt{\bar{c}_{i}}-\sqrt{c_{i}}}{\sqrt{\bar{c}_{i}}}\nabla\sqrt{\bar{c}_{i}}\bigg{|}^{2}$
$\displaystyle\phantom{xx}{}-4\bigg{|}\sqrt{c_{i}}\sum_{j=1}^{n}\frac{(\sqrt{\bar{c}_{j}})^{2}-(\sqrt{c_{j}})^{2}}{\sqrt{\bar{c}_{j}}}\nabla\sqrt{\bar{c}_{j}}\bigg{|}^{2}\bigg{)}$
$\displaystyle\geq\frac{1}{2}\sum_{i=1}^{n}|\nabla(\sqrt{c_{i}}-\sqrt{\bar{c}_{i}})|^{2}-\frac{4(n+1)}{m}\sum_{i=1}^{n}(\sqrt{c_{i}}-\sqrt{\bar{c}_{i}})^{2}|\nabla\sqrt{\bar{c}_{i}}|^{2}.$

With this estimate, (38) becomes, after integration over $\Omega\times(0,t)$,

(39) $\displaystyle 4$
$\displaystyle\sum_{i,j=1}^{n}\int_{0}^{t}\int_{\Omega}A_{ij}^{BD}(\bar{c})Z_{i}\cdot
Z_{j}dxds\geq
2\lambda\sum_{i=1}^{n}\int_{0}^{t}\int_{\Omega}|\nabla(\sqrt{c_{i}}-\sqrt{\bar{c}_{i}})|^{2}dxds$
$\displaystyle\phantom{xx}{}-\frac{16(n+1)}{m}\max_{j=1,\ldots,n}\|\nabla\sqrt{\bar{c}_{j}}\|_{L^{\infty}(\Omega\times(0,T))}^{2}\sum_{i=1}^{n}\int_{0}^{t}\int_{\Omega}(\sqrt{c_{i}}-\sqrt{\bar{c}_{i}})^{2}dxds.$

Next, we consider the last integral in (37). By Young’s inequality,

(40) $\displaystyle 4$
$\displaystyle\sum_{i,j=1}^{n}\int_{0}^{t}\int_{\Omega}Z_{i}\cdot\nabla\sqrt{\bar{c}_{j}}\bigg{(}\frac{\sqrt{c_{i}}}{\sqrt{\bar{c}_{i}}}A_{ij}^{BD}(\bar{\bm{c}})-A_{ij}^{BD}(\bm{c})\frac{\sqrt{c_{j}}}{\sqrt{\bar{c}_{j}}}\bigg{)}dxds\leq\frac{\lambda}{2}\sum_{i=1}^{n}\int_{0}^{t}\int_{\Omega}|Z_{i}|^{2}dxds$
$\displaystyle{}+C\max_{j=1,\ldots,n}\|\nabla\sqrt{\bar{c}_{j}}\|_{L^{\infty}(\Omega\times(0,T))}^{2}\sum_{i,j=1}^{n}\int_{0}^{t}\int_{\Omega}\bigg{(}\frac{\sqrt{c_{i}}}{\sqrt{\bar{c}_{i}}}A_{ij}^{BD}(\bar{\bm{c}})-A_{ij}^{BD}({\bm{c}})\frac{\sqrt{c_{j}}}{\sqrt{\bar{c}_{j}}}\bigg{)}^{2}dxds,$

and the constant $C>0$ depends on $\lambda$ and $n$. The first term on the
right-hand side is estimated according to

(41) $\displaystyle|Z_{i}|^{2}$
$\displaystyle=\bigg{|}\nabla(\sqrt{c_{i}}-\sqrt{\bar{c}_{i}})-\frac{\sqrt{c_{i}}-\sqrt{\bar{c}_{i}}}{\sqrt{\bar{c}_{i}}}\nabla\sqrt{\bar{c}_{i}}\bigg{|}^{2}$
$\displaystyle\leq
2|\nabla(\sqrt{c_{i}}-\sqrt{\bar{c}_{i}})|^{2}+\frac{2}{m}|\sqrt{c_{i}}-\sqrt{\bar{c}_{i}}|^{2}|\nabla\sqrt{\bar{c}_{i}}|^{2}.$

To estimate the second term on the right-hand side of (40), we need some
preparations. We write

$A^{BD}(\bm{c})=P_{L}(A(\bm{c})+P_{L^{\perp}})^{-1}=\frac{P_{L}\operatorname{adj}(A(\bm{c})+P_{L^{\perp}})}{\det(A(\bm{c})+P_{L^{\perp}})}=:\frac{R(\sqrt{\bm{c}})}{S(\sqrt{\bm{c}})},$

where “adj” denotes the adjugate matrix. We know that the elements of
$A(\bm{c})$, $P_{L}$, and $P_{L^{\perp}}$ are polynomials of $\sqrt{\bm{c}}$.
Therefore, $R(\sqrt{\bm{c}})$ and $S(\sqrt{\bm{c}})$ are also polynomials of
$\sqrt{\bm{c}}$. Any eigenvalue of $A(\bm{c})$ is also an eigenvalue of
$A(\bm{c})+P_{L^{\perp}}$ (since $L^{\perp}=\ker A(\bm{c})$). As $A(\bm{c})$
has the eigenvalue 0 with eigenvector $\sqrt{\bm{c}}$,
$A(\bm{c})+P_{L^{\perp}}$ has the eigenvalue 1 with the same eigenvector.
Moreover, all other eigenvalues of $A(\bm{c})+P_{L^{\perp}}$ are larger than
or equal to $\mu$. Since the determinant of a matrix is the product of its
eigenvalues, we conclude that $S(\sqrt{\bm{c}})\geq\mu^{n-1}>0$. This shows
that $S(\sqrt{\bm{c}})$ is uniformly bounded from below. Thus, we can estimate
as follows, denoting the elements of the matrix $R(\sqrt{\bm{c}})$ by
$R_{ij}(\sqrt{\bm{c}})$:

$\displaystyle\bigg{|}\frac{\sqrt{c_{i}}}{\sqrt{\bar{c}_{i}}}$ $\displaystyle
A_{ij}^{BD}(\bar{\bm{c}})-A_{ij}^{BD}(\bm{c})\frac{\sqrt{c_{j}}}{\sqrt{\bar{c}_{j}}}\bigg{|}=\bigg{|}\frac{\sqrt{c_{i}}R_{ij}(\sqrt{\bar{\bm{c}}})}{\sqrt{\bar{c}_{i}}S(\sqrt{\bar{\bm{c}}})}-\frac{R_{ij}(\sqrt{{\bm{c}}})\sqrt{c_{j}}}{S(\sqrt{{\bm{c}}})\sqrt{\bar{c}_{j}}}\bigg{|}$
$\displaystyle=\frac{1}{S(\sqrt{\bm{c}})S(\sqrt{\bar{\bm{c}}})\sqrt{\bar{c}_{i}{\bar{c}_{j}}}}\big{|}\big{(}\sqrt{c_{i}}S(\sqrt{{\bm{c}}})-\sqrt{\bar{c}_{i}}S(\sqrt{\bar{\bm{c}}})\big{)}R_{ij}(\sqrt{\bar{\bm{c}}})\sqrt{\bar{c}_{j}}$
$\displaystyle\phantom{xx}{}-\big{(}R_{ij}(\sqrt{\bm{c}})\sqrt{c_{j}}-R_{ij}(\sqrt{\bar{\bm{c}}})\sqrt{\bar{c}_{j}}\big{)}\sqrt{\bar{c}_{i}}S(\sqrt{\bar{\bm{c}}})\big{|}$
$\displaystyle\leq C(m)\sum_{i=1}^{n}|\sqrt{c_{i}}-\sqrt{\bar{c}_{i}}|,$

where $C(m)>0$ depends on the Lipschitz constants of the polynomials
$\sqrt{c_{i}}R_{ij}(\sqrt{\bm{c}})$ and $\sqrt{c_{i}}S(\sqrt{\bm{c}})$.
Inserting this estimate into (40), we obtain

(42) $\displaystyle 4\sum_{i,j=1}^{n}$
$\displaystyle\int_{0}^{t}\int_{\Omega}Z_{i}\cdot\nabla\sqrt{\bar{c}_{j}}\bigg{(}\frac{\sqrt{c_{i}}}{\sqrt{\bar{c}_{i}}}A_{ij}^{BD}(\bar{\bm{c}})-A_{ij}^{BD}({\bm{c}})\frac{\sqrt{c_{j}}}{\sqrt{\bar{c}_{j}}}\bigg{)}dxds$
$\displaystyle\leq\lambda\sum_{i=1}^{n}\int_{0}^{t}\int_{\Omega}|\nabla(\sqrt{c_{i}}-\sqrt{\bar{c}_{i}})|^{2}dxds+C\sum_{i=1}^{n}\int_{0}^{t}\int_{\Omega}(\sqrt{c_{i}}-\sqrt{\bar{c}_{i}})^{2}dxds,$

where $C>0$ also depends on the $L^{\infty}$ norm of
$\nabla\sqrt{\bar{c}_{i}}$ through (41).

Finally, we use estimates (39) and (42) in the relative entropy inequality
(37), together with Lemma 18 in Appendix B, to find that

$\displaystyle H($
$\displaystyle\bm{c}(t)|\bar{\bm{c}}(t))+\lambda\sum_{i=1}^{n}\int_{0}^{t}\int_{\Omega}|\nabla(\sqrt{c_{i}}-\sqrt{\bar{c}_{i}})|^{2}dxds$
$\displaystyle\leq
C\sum_{i=1}^{n}\int_{0}^{t}\int_{\Omega}(\sqrt{c_{i}}-\sqrt{\bar{c}_{i}})^{2}dxds\leq
C\sum_{i=1}^{n}\int_{0}^{t}H(\bm{c}|\bar{\bm{c}})ds,$

and an application of Gronwall’s lemma finishes the proof.

######  Remark 10 (Nonhomogeneous total mass).

The condition $\sum_{i=1}^{n}c_{i}^{0}(x)=1$ for $x\in\Omega$ on the initial
total mass can be relaxed to $\sum_{i=1}^{n}c_{i}^{0}(x)=M(x)$ for
$x\in\Omega$ and some strictly positive function $M\in L^{\infty}(\Omega)$. In
this situation, the force-flux relations in (1) change to

$\nabla c_{i}-\frac{c_{i}}{\sum_{j=1}^{n}c_{j}}\sum_{j=1}^{n}\nabla
c_{j}=-\sum_{j=1,\,j\neq i}^{n}\frac{c_{i}c_{j}}{D_{ij}}(u_{i}-u_{j}),\quad
i=1,\ldots,n.$

Notice that the total mass $\sum_{j=1}^{n}c_{j}=M$ is preserved in time. The
previous equation can be expressed in terms of the matrix $A(\bm{c})$, defined
in (10), by

$\sum_{j=1}^{n}(P_{L})_{ij}\nabla\sqrt{c_{j}}=-\sum_{j=1}^{n}A_{ij}(\bm{c})\sqrt{c_{i}}u_{j},$

where the projection matrix $P_{L}$ is now given by
$(P_{L})_{ij}=\delta_{ij}-\sqrt{c_{i}c_{j}}/M(x)$, $i,j=1,\ldots,n$. Lemma 4
still holds in this situation with $\alpha\leq\inf_{x\in\Omega}M(x)\min_{i\neq
j}(1/D_{ij})$. The relative entropy inequalities (29) and (37) do not depend
on the assumption $\sum_{i=1}^{n}c_{i}=1$ such that the relative entropy
inequalities in Theorem 1 and Corollary 2 still hold but with constants
depending on $M$. ∎

##  4\. Weak-strong uniqueness for generalized Maxwell–Stefan systems

We consider the generalized Maxwell–Stefan system (16)–(17). First, we rewrite
(17) in terms of the Bott–Duffin inverse of $B(\bm{c})$. To this end, we
recall the definition of $(P_{L})_{ij}=\delta_{ij}-\sqrt{c_{i}c_{j}}$ and
rewrite the right-hand side of (17),

$\displaystyle c_{i}\nabla
h^{\prime}_{i}(c_{i})-c_{i}\sum_{j=1}^{n}c_{j}\nabla h^{\prime}_{j}(c_{j})$
$\displaystyle=\sqrt{c_{i}}\bigg{(}\sqrt{c_{i}}\nabla
h^{\prime}_{i}(c_{i})-\sum_{j=1}^{n}\sqrt{c_{i}c_{j}}\sqrt{c_{j}}\nabla
h^{\prime}_{j}(c_{j})\bigg{)}$
$\displaystyle=\sqrt{c_{i}}\sum_{j=1}^{n}(P_{L})_{ij}\sqrt{c_{j}}\nabla
h^{\prime}_{j}(c_{j}),$

as well as the left-hand side of (17), using definition (18) of $B(\bm{c})$,

$-\sum_{j=1}^{n}K_{ij}(\bm{c})c_{j}u_{j}=-\sqrt{c_{i}}\sum_{j=1}^{n}B_{ij}(\bm{c})\sqrt{c_{j}}u_{j},\quad
i=1,\ldots,n,$

showing that (17) is equivalent to

$-\sum_{j=1}^{n}B_{ij}(\bm{c})\sqrt{c_{j}}u_{j}=\sum_{j=1}^{n}(P_{L})_{ij}\sqrt{c_{j}}\nabla
h^{\prime}_{j}(c_{j}),\quad i=1,\ldots,n.$

We prove in Lemma 11 below that the Bott–Duffin inverse $B^{BD}(\bm{c})$ of
$B(\bm{c})$ exists. Thus, we can invert the previous system:

$\sqrt{c_{i}}u_{i}=-\sum_{j=1}^{n}(B^{BD}(\bm{c})P_{L})_{ij}\sqrt{c_{j}}\nabla
h_{j}^{\prime}(c_{j})=-\sum_{j=1}^{n}B_{ij}^{BD}(\bm{c})\sqrt{c_{j}}\nabla
h_{j}^{\prime}(c_{j}),$

where we have used the relation $B^{BD}(\bm{c})P_{L}=B^{BD}(\bm{c})$ (see (81)
in Appendix A). This equation generalizes (13). We conclude that system
(16)–(17) can be written as

(43)
$\partial_{t}c_{i}=\operatorname{div}\bigg{(}\sum_{j=1}^{n}\sqrt{c_{i}}B_{ij}^{BD}(\bm{c})\sqrt{c_{j}}\nabla
h^{\prime}_{j}(c_{j})\bigg{)},\quad i=1,\ldots,n.$

###  4.1. Properties of the matrix $B(\bm{c})$

We prove the following lemma.

######  Lemma 11.

Let Assumptions (B1)–(B4) hold for $B(\bm{c})$, defined in (18). Then the
Bott–Duffin inverse $B^{BD}(\bm{c})=P_{L}(B(\bm{c})P_{L}+P_{L^{\perp}})^{-1}$
of $B(\bm{c})$ exists, is symmetric and satisfies the following properties:

  * •

Let $s>0$. Then the elements $B_{ij}^{BD}(\bm{c})$ are bounded and Lipschitz
continuous for all $\bm{c}\in[s,1]^{n}$.

  * •

Let $m>0$. Then there exists $\lambda(m)>0$ such that for all
$\bm{z}\in{\mathbb{R}}^{n}$ and $\bm{c}\in[m,1]^{n}$,

(44) $\bm{z}^{T}B^{BD}(\bm{c})\bm{z}\geq\lambda(m)|P_{L}\bm{z}|^{2}.$

  * •

The matrix $B^{BD}(\mathbf{c})$ satisfies for all $\bm{z}\in{\mathbb{R}}^{n}$
and $\bm{c}\in[0,1]^{n}$,

(45) $\bm{z}^{T}B^{BD}(\bm{c})\bm{z}\leq\frac{1}{\mu}|\bm{z}|^{2},$

recalling that $\mu>0$ is a lower bound for the nonzero eigenvalues of
$B(\bm{c})$; see Assumption (B4).

###### Proof.

Assumption (B1) and Lemma 16 (ii) in Appendix A imply that

$\ker(B(\bm{c})P_{L}+P_{L^{\perp}})=\ker(B(\bm{c})P_{L})\cap L=L^{\perp}\cap
L=\\{0\\}.$

Hence, $B(\bm{c})P_{L}+P_{L^{\perp}}$ is invertible and the Bott–Duffin
inverse is well defined and symmetric.

We continue by studying the eigenvalues of $B(\bm{c})P_{L}+P_{L^{\perp}}$. A
computation shows that for $\sqrt{\bm{c}}\in\ker(B(\bm{c})P_{L})=L^{\perp}$ we
have
$(B(\bm{c})P_{L}+P_{L^{\perp}})\sqrt{\bm{c}}=P_{L^{\perp}}\sqrt{\bm{c}}=\sqrt{\bm{c}}$,
i.e., $\sqrt{\bm{c}}$ is an eigenvector of $B(\bm{c})P_{L}+P_{L^{\perp}}$ with
eigenvalue 1. Let $\xi\notin L^{\perp}$, $\xi\neq 0$, be another eigenvector,

$(B(\bm{c})P_{L}+P_{L^{\perp}})\xi=\rho\xi.$

Then $\rho\neq 0$. Applying $P_{L}$ on both sides, we obtain
$P_{L}B(\bm{c})P_{L}\xi=B(\bm{c})(P_{L}\xi)=\rho P_{L}\xi$, i.e.,
$P_{L}\xi\neq 0$ is an eigenvector of $B(\bm{c})$ with eigenvalue $\rho$. Due
to Assumption (B4), we conclude that $\rho\geq\mu>0$.

We claim that the elements $B_{ij}^{BD}(\bm{c})$ are bounded and Lipschitz
continuous for all $\bm{c}\in[s,1]^{n}$. Indeed, observe that

(46)
$B^{BD}(\bm{c})=P_{L}(B(\bm{c})P_{L}+P_{L^{\perp}})^{-1}=\frac{P_{L}\operatorname{adj}(B(\bm{c})P_{L}+P_{L^{\perp}})}{\det(B(\bm{c})P_{L}+P_{L^{\perp}})},$

where “adj” denotes the adjugate. Since the determinant of a matrix is the
product of its eigenvalues,
$\det(B(\bm{c})P_{L}+P_{L^{\perp}})\geq\mu^{n-1}>0$ and the denominator in
(46) is bounded from below. Assumption (B2) implies that the elements of
$B(\bm{c})$ are bounded. Hence, all elements of
$\operatorname{adj}(B(\bm{c})P_{L}+P_{L^{\perp}})$ are bounded too. We
conclude from (46) that the elements of $B^{BD}(\bm{c})$ are bounded. Since
the product of Lipschitz continuous functions is Lipschitz continuous,
Assumption (B2) further implies that the elements of $B^{BD}(\bm{c})$ are
Lipschitz continuous for all $\bm{c}\in[s,1]^{n}$ for any $s>0$.

We wish to verify (44). Since the spectral radius $r$ of a matrix is bounded
by its Frobenius norm $\|\cdot\|_{F}$ and the Frobenius norm is
submultiplicative, we have

$\displaystyle r(B(\bm{c})P_{L}+P_{L^{\perp}})$
$\displaystyle\leq\|B(\bm{c})P_{L}+P_{L^{\perp}}\|_{F}\leq\|B(\bm{c})P_{L}\|_{F}+\|P_{L^{\perp}}\|_{F}$
$\displaystyle\leq\|B(\bm{c})\|_{F}\|P_{L}\|_{F}+\|P_{L^{\perp}}\|_{F}.$

The Frobenius norms of $P_{L}$ and $P_{L^{\perp}}$ are estimated according to

$\displaystyle\|P_{L}\|_{F}^{2}$
$\displaystyle=\sum_{i,j=1}^{n}(\delta_{ij}-\sqrt{c_{i}c_{j}})^{2}\leq\sum_{i,j=1}^{n}1=n^{2},$
$\displaystyle\|P_{L^{\perp}}\|_{F}^{2}$
$\displaystyle=\sum_{i,j=1}^{n}(\sqrt{c_{i}c_{j}})^{2}=\bigg{(}\sum_{i=1}^{n}c_{i}\bigg{)}^{2}=1.$

Assumption (B3) guarantees that $\|B(\bm{c})\|_{F}\leq\gamma(m)$. This shows
that $r(B(\bm{c})P_{L}+P_{L^{\perp}})\leq{\gamma}(m)n+1$. We infer that the
smallest eigenvalue of $(B(\bm{c})P_{L}+P_{L^{\perp}})^{-1}$ is larger than
$\lambda(m):=1/({\gamma}(m)n+1)$ proving (44).

It remains to prove (45). First, we show that the nonzero eigenvalues of
$B(\bm{c})$ and $B^{BD}(\bm{c})$ are reciprocal to each other. Let
$\ell\in{\mathbb{R}}$ be a nonzero eigenvalue of $B^{BD}(\bm{c})$. Then the
corresponding eigenvector $\bm{y}\in L$ satisfies
$B^{BD}(\bm{c})\bm{y}=\ell\bm{y}$, which is
$P_{L}(B(\bm{c})P_{L}+P_{L^{\perp}})^{-1}\bm{y}=\ell\bm{y}$. Hence,
$\bm{z}:=(B(\bm{c})P_{L}+P_{L^{\perp}})^{-1}\bm{y}$ satisfies
$P_{L}\bm{z}=\ell(B(\bm{c})P_{L}+P_{L^{\perp}})\bm{z}$. Applying $P_{L}$ on
both sides yields $P_{L}\bm{z}=\ell P_{L}B(\bm{c})P_{L}\bm{z}=\ell
B(\bm{c})P_{L}\bm{z}$. Thus, $P_{L}\bm{z}$ is an eigenvector of
$B(\mathbf{c})$ with eigenvalue $1/\ell$. Similarly, we can reverse the above
argument and verify that if $\bm{z}$ is a nonzero eigenvector of $B(\bm{c})$
with eigenvalue $\ell$, then $(B(\bm{c})P_{L}+P_{L^{\perp}})\bm{z}$ is an
eigenvector of $B^{BD}(\bm{c})$ with eigenvalue $1/\ell$. We conclude that the
largest eigenvalue of $B^{BD}(\bm{c})$ is the reciprocal of the smallest
eigenvalue of $B(\bm{c})$, and Assumption (B4) implies that
$\bm{z}^{T}B^{BD}(\bm{c})\bm{z}\leq|\bm{z}|^{2}/\mu$ for all
$\bm{z}\in{\mathbb{R}}^{n}$. ∎

###  4.2. Weak and strong solutions

We call $\bm{c}$ a weak solution to (2), (16)–(17) if

$c_{i}\in C_{\rm loc}^{0}([0,\infty);{\mathcal{V}}^{\prime})\cap L_{\rm
loc}^{2}(0,\infty;H^{1}(\Omega)),\quad i=1,\ldots,n,$

where ${\mathcal{V}}^{\prime}$ is the dual space of ${\mathcal{V}}=\\{w\in
H^{2}(\Omega):\nabla w\cdot\nu=0$ on $\partial\Omega\\}$, it holds for any
test function $\phi_{i}\in C_{\rm
loc}^{1}([0,\infty);C^{1}(\overline{\Omega}))$ with $\nabla\phi_{i}\cdot\nu=0$
on $\partial\Omega$ and all $t>0$ that

$\displaystyle\int_{\Omega}$ $\displaystyle
c_{i}(t)\phi_{i}(t)dx-\int_{\Omega}c_{i}^{0}\phi_{i}(0)dx-\int_{0}^{t}\int_{\Omega}c_{i}\partial_{t}\phi_{i}dxds$
$\displaystyle{}+\sum_{j=1}^{n}\int_{0}^{t}\int_{\Omega}\sqrt{c_{i}}B_{ij}^{BD}(\bm{c})\sqrt{c_{j}}\nabla
h^{\prime}_{j}(c_{j})\cdot\nabla\phi_{i}dxds=0,\quad i=1,\ldots,n,$

and the entropy dissipation inequality

(47)
$H(\bm{c}(t))+\sum_{i,j=1}^{n}\int_{0}^{t}\int_{\Omega}\sqrt{c_{i}c_{j}}B_{ij}^{BD}(\bm{c})\nabla
h^{\prime}_{i}(c_{i})\cdot\nabla h_{j}^{\prime}(c_{j})dxds\leq H(\bm{c}^{0})$

is satisfied for $t>0$.

Furthermore, we call $\bar{\bm{c}}$ a strong solution to (2), (16)–(17) if
$c_{i}\in C_{\rm loc}^{1}([0,\infty);C^{1}(\overline{\Omega}))$ for
$i=1,\ldots,n$, if (2), (16)–(17) are satisfied pointwise, and the entropy
identity (20) is fulfilled.

###  4.3. Relative entropy inequality

The partial free energies $h_{i}(c_{i})$ and pressures $p_{i}(c_{i})$ are
associated through (19). Define the associated relative free energy density
and relative pressure via

(48) $\displaystyle h_{i}(c_{i}|\bar{c}_{i})$
$\displaystyle:=h_{i}(c_{i})-h_{i}(\bar{c}_{i})-h_{i}^{\prime}(\bar{c}_{i})(c_{i}-\bar{c}_{i}),$
$\displaystyle p_{i}(c_{i}|\bar{c}_{i})$
$\displaystyle:=p_{i}(c_{i})-p_{i}(\bar{c}_{i})-p_{i}^{\prime}(\bar{c}_{i})(c_{i}-\bar{c}_{i}).$

We prove a relative entropy inequality associated to the generalized
Maxwell–Stefan system.

######  Lemma 12.

Let $\bm{c}$ be a weak solution to (2), (16)–(17) and let $\bar{\bm{c}}$ be a
strong solution to (2), (16)–(17) satisfying

$h^{\prime}_{i}(\bar{c}_{i})\in L_{\rm loc}^{2}(0,\infty;H^{2}(\Omega)),\
h^{\prime\prime}_{i}(\bar{c}_{i})\in L_{\rm
loc}^{\infty}(0,\infty;L^{\infty}(\Omega)),\ \partial_{t}\bar{c}_{i}\in L_{\rm
loc}^{2}(0,\infty;L^{2}(\Omega)).$

Then the following relative entropy inequality holds:

$\displaystyle H($
$\displaystyle\bm{c}(t)|\bar{\bm{c}}(t))-H(\bm{c}^{0}|\bar{\bm{c}}^{0})+\sum_{i,j=1}^{n}\int_{0}^{t}\int_{\Omega}B_{ij}^{BD}(\bm{c})Y_{i}\cdot
Y_{j}dxds$
$\displaystyle\leq-\sum_{i,j=1}^{n}\int_{0}^{t}\int_{\Omega}\bigg{(}B_{ij}^{BD}(\bm{c})\sqrt{c_{j}}-\frac{\sqrt{c_{i}}B_{ij}^{BD}(\bar{\bm{c}})\sqrt{\bar{c}_{j}}}{\sqrt{\bar{c}_{i}}}\bigg{)}Y_{j}\cdot\nabla
h^{\prime}_{j}(\bar{c}_{j})dxds$ (49)
$\displaystyle\phantom{xx}{}+\sum_{i=1}^{n}\int_{0}^{t}\int_{\Omega}p_{i}(c_{i}|\bar{c}_{i})\operatorname{div}\bigg{(}\sum_{j=1}^{n}B_{ij}^{BD}(\bar{\bm{c}})\frac{\sqrt{\bar{c}_{j}}}{\sqrt{\bar{c}_{i}}}\nabla
h^{\prime}_{j}(\bar{c}_{j})\bigg{)}dxds,$

where

(50)
$Y_{i}=\sqrt{c_{i}}\nabla(h^{\prime}_{i}(c_{i})-h^{\prime}_{i}(\bar{c}_{i})),\quad
i=1,\ldots,n.$

Note that the definition for $Y_{i}$ differs from that one used in Section 3.

###### Proof.

We proceed as in the proof of Lemma 9, but re-arrange the terms in a different
fashion. The difference $c_{i}-\bar{c}_{i}$ satisfies the weak formulation

$\displaystyle 0$
$\displaystyle=\int_{\Omega}(c_{i}-\bar{c}_{i})(t)\phi_{i}(t)dx-\int_{\Omega}(c_{i}^{0}-\bar{c}_{i}^{0})\phi_{i}(0)dx-\int_{0}^{t}\int_{\Omega}(c_{i}-\bar{c}_{i})\partial_{t}\phi_{i}dxds$
$\displaystyle\phantom{xx}{}+\sum_{j=1}^{n}\int_{0}^{t}\int_{\Omega}\big{(}\sqrt{c_{i}}B_{ij}^{BD}(\bm{c})\sqrt{c_{j}}\nabla
h^{\prime}_{j}(c_{j})-\sqrt{\bar{c}_{i}}B_{ij}^{BD}(\bar{\bm{c}})\sqrt{\bar{c}_{j}}\nabla
h^{\prime}_{j}(\bar{c}_{j})\big{)}\cdot\nabla\phi_{i}dxds$

for $i=1,\ldots,n$. We wish to use $\phi_{i}=h^{\prime}_{i}(\bar{c}_{i})$ as a
test function. Strictly speaking, this is not possible, but, as in the proof
of Lemma 8, we can use a density argument. Then, using (43) for the third term
and adding over $i=1,\ldots,n$, we obtain

(51) $\displaystyle 0$
$\displaystyle=\sum_{i=1}^{n}\int_{\Omega}(c_{i}-\bar{c}_{i})(t)h^{\prime}_{i}(\bar{c}_{i}(t))dx-\sum_{i=1}^{n}\int_{\Omega}(c_{i}^{0}-\bar{c}_{i}^{0})h^{\prime}_{i}(\bar{c}_{i0})dx$
$\displaystyle\phantom{xx}{}-\sum_{i=1}^{n}\int_{0}^{t}\int_{\Omega}(c_{i}-\bar{c}_{i})h^{\prime\prime}_{i}(c_{i})\operatorname{div}\bigg{(}\sum_{j=1}^{n}\sqrt{\bar{c}_{i}}B_{ij}^{BD}(\bar{\bm{c}})\sqrt{\bar{c}_{j}}\nabla
h^{\prime}_{j}(\bar{c}_{j})\bigg{)}dxds$
$\displaystyle\phantom{xx}{}+\sum_{i,j=1}^{n}\int_{0}^{t}\int_{\Omega}\big{(}\sqrt{c_{i}}B_{ij}^{BD}(\bm{c})\sqrt{c_{j}}\nabla
h^{\prime}_{j}(c_{j})-\sqrt{\bar{c}_{i}}B_{ij}^{BD}(\bar{\bm{c}})\sqrt{\bar{c}_{j}}\nabla
h^{\prime}_{j}(\bar{c}_{j})\big{)}\cdot\nabla
h^{\prime}_{i}(\bar{c}_{i})dxds.$

Subtracting (51) and the entropy identity (20) for $\bar{\bm{c}}$ from the
entropy inequality (47) for $\bm{c}$, we find that

$\displaystyle H(\bm{c}(t)|\bar{\bm{c}}(t))$ $\displaystyle\leq
H(\bm{c}^{0}|\bar{\bm{c}}^{0})-\sum_{i,j=1}^{n}\int_{0}^{t}\int_{\Omega}B_{ij}^{BD}(\bm{c})\sqrt{c_{i}c_{j}}\nabla(h^{\prime}_{i}(c_{i})-h^{\prime}_{i}(\bar{c}_{i}))\cdot\nabla
h^{\prime}_{j}(c_{j})dxds$ (52)
$\displaystyle\phantom{xx}{}-\sum_{i=1}^{n}\int_{0}^{t}\int_{\Omega}(c_{i}-\bar{c}_{i})h_{i}^{\prime\prime}(\bar{c}_{i})\operatorname{div}\bigg{(}\sum_{j=1}^{n}\sqrt{\bar{c}_{i}}B_{ij}^{BD}(\bar{\bm{c}})\sqrt{\bar{c}_{j}}\nabla
h^{\prime}_{j}(\bar{c}_{j})\bigg{)}dxds.$

In turn, using (50), this is rewritten as

$\displaystyle H($
$\displaystyle\bm{c}(t)|\bar{\bm{c}}(t))-H(\bm{c}^{0}|\bar{\bm{c}}^{0})+\sum_{i,j=1}^{n}\int_{0}^{t}\int_{\Omega}B_{ij}^{BD}(\bm{c})Y_{i}\cdot
Y_{j}dxds$
$\displaystyle\leq-\sum_{i,j=1}^{n}\int_{0}^{t}\int_{\Omega}\bigg{(}B_{ij}^{BD}(\bm{c})\sqrt{c_{j}}-\sqrt{c_{i}}B_{ij}^{BD}(\bar{\bm{c}})\frac{\sqrt{\bar{c}_{j}}}{\sqrt{\bar{c}_{i}}}\bigg{)}Y_{i}\cdot\nabla
h^{\prime}_{j}(\bar{c}_{j})dxds$
$\displaystyle\phantom{xx}{}-\sum_{i,j=1}^{n}\int_{0}^{t}\int_{\Omega}\sqrt{c_{i}}B_{ij}^{BD}(\bar{\bm{c}})\frac{\sqrt{\bar{c}_{j}}}{\sqrt{\bar{c}_{i}}}Y_{i}\cdot\nabla
h^{\prime}_{j}(\bar{c}_{j})dxds$
$\displaystyle\phantom{xx}{}-\int_{0}^{t}\int_{\Omega}\sum_{i=1}^{n}(c_{i}-\bar{c}_{i})\bar{c}_{i}h_{i}^{\prime\prime}(\bar{c}_{i})\operatorname{div}\bigg{(}\sum_{j=1}^{n}B_{ij}^{BD}(\bar{\bm{c}})\frac{\sqrt{\bar{c}_{j}}}{\sqrt{\bar{c}_{i}}}\nabla
h^{\prime}_{j}(\bar{c}_{j})\bigg{)}dxds$
$\displaystyle\phantom{xx}{}-\int_{0}^{t}\int_{\Omega}\sum_{i=1}^{n}(c_{i}-\bar{c}_{i})h_{i}^{\prime\prime}(\bar{c}_{i})\nabla\bar{c}_{i}\cdot\sum_{j=1}^{n}B_{ij}^{BD}(\bar{\bm{c}})\frac{\sqrt{\bar{c}_{j}}}{\sqrt{\bar{c}_{i}}}\nabla
h^{\prime}_{j}(\bar{c}_{j})dxds$ $\displaystyle=:J_{1}+J_{2}+J_{3}+J_{4}.$

The sum $J_{2}+J_{4}$ becomes

$\displaystyle J_{2}+J_{4}$
$\displaystyle=-\int_{0}^{t}\int_{\Omega}\sum_{i=1}^{n}\big{(}c_{i}\nabla
h_{i}^{\prime}(c_{i})-\bar{c}_{i}\nabla
h_{i}^{\prime}(\bar{c}_{i})\big{)}\cdot\sum_{j=1}^{n}B_{ij}^{BD}(\bar{\bm{c}})\frac{\sqrt{\bar{c}_{j}}}{\sqrt{\bar{c}_{i}}}\nabla
h^{\prime}_{j}(\bar{c}_{j})dxds$
$\displaystyle=\int_{0}^{t}\int_{\Omega}\sum_{i=1}^{n}(p_{i}(c_{i})-p_{i}(\bar{c}_{i}))\operatorname{div}\bigg{(}\sum_{j=1}^{n}B_{ij}^{BD}(\bar{\bm{c}})\frac{\sqrt{\bar{c}_{j}}}{\sqrt{\bar{c}_{i}}}\nabla
h^{\prime}_{j}(\bar{c}_{j})\bigg{)}dxds.$

Combining this expression with $J_{3}$ and using definition (48) finally leads
to (49). ∎

###  4.4. The entropy dissipation structure

We state an auxiliary lemma that provides some control of the entropy
inequality (47) and the relative entropy inequality (49).

######  Lemma 13.

Let $\bm{c}$ be a weak solution and $\bar{\bm{c}}$ be a strong solution to
(2), (16)–(17), satisfying the hypotheses of Lemma 12 and $\bar{c}_{i}(t)\geq
m$ in $\Omega$, $t>0$ for some constant $m>0$.

  * (i)

Assume that $c_{i}\geq m/2$ for all $i=1\ldots,n$ and let
$Z_{i}=\sqrt{c_{i}}\nabla h_{i}^{\prime}(c_{i})$. Then, for some constant
$\beta(m)>0$, we have

(53) $\sum_{i,j=1}^{n}B_{ij}^{BD}(\bm{c})Z_{i}\cdot Z_{j}\geq
2\beta(m)\sum_{i=1}^{n}|\nabla c_{i}|^{2}.$

  * (ii)

Assume that ${c}_{i}\geq m/2$ for all $i=1,\ldots,n$ and let
$Y_{i}=\sqrt{c_{i}}\nabla(h^{\prime}_{i}(c_{i})-h^{\prime}_{i}(\bar{c}_{i}))$.
Then, for some $\beta(m)>0$ and $C>0$, we have

(54) $\sum_{i,j=1}^{n}B_{ij}^{BD}(\bm{c})Y_{i}\cdot
Y_{j}\geq\beta(m)\sum_{i=1}^{n}|\nabla
c_{i}-\nabla\bar{c}_{i}|^{2}-C\sum_{i=1}^{n}|c_{i}-\bar{c}_{i}|^{2}.$

  * (iii)

Let the weak solution $\bm{c}$ satisfy (47) and set

(55) $c_{*}(x,t):=\min_{i=1,\ldots,n}c_{i}(x,t).$

Then

(56)
$H(\bm{c}(t))+2\beta(m)\int_{0}^{t}\int_{\Omega}\mathrm{1}_{\\{c_{*}>m/2\\}}\sum_{i=1}^{n}|\nabla
c_{i}|^{2}dxds\leq H(\bm{c}^{0}).$

Note that (56) provides a partial control of the gradients, which however
might degenerate as $m$ tends to zero.

###### Proof.

_Proof of (i)._ Inequality (44) in Lemma 11 implies that

(57) $\displaystyle\sum_{i,j=1}^{n}B_{ij}^{BD}(\bm{c})Z_{i}\cdot Z_{j}$
$\displaystyle\geq\lambda(m/2)|P_{L}\bm{Z}|^{2}=\lambda(m/2)\bm{Z}^{T}P_{L}^{T}P_{L}\bm{Z}$
$\displaystyle=\lambda(m/2)\sum_{i,j=1}^{n}(P_{L})_{ij}Z_{i}\cdot Z_{j}.$

Before we can estimate the right-hand side, we need some preparations.

We define the vector $\widetilde{\bm{c}}:=(c_{1},\ldots,c_{n-1})$ without the
last component and define the entropy density in $n-1$ variables according to

$\widetilde{h}(\widetilde{\bm{c}})=\sum_{i=1}^{n-1}h_{i}(c_{i})+h_{n}\bigg{(}1-\sum_{j=1}^{n-1}c_{j}\bigg{)}.$

Its partial derivative is given by

$\widetilde{h}_{i}^{\prime}(\widetilde{\bm{c}}):=\frac{\partial\widetilde{h}(\widetilde{\bm{c}})}{\partial
c_{i}}=h_{i}^{\prime}(c_{i})-h^{\prime}_{n}\bigg{(}1-\sum_{j=1}^{n-1}c_{j}\bigg{)},\quad
i=1,\ldots,n-1.$

Next, introduce the matrix $E(\bm{c})$ with elements

$E_{ij}(\bm{c})=(P_{L})_{ij}\sqrt{c_{i}c_{j}}=c_{i}\delta_{ij}-c_{i}c_{j}=\begin{cases}c_{i}-c_{i}^{2}&\quad\mbox{if
}i=j.\\\ -c_{i}c_{j}&\quad\mbox{if }i\neq j.\\\ \end{cases}$

The sum of its rows and columns vanishes,
$\sum_{j=1}^{n}E_{ij}(\bm{c})=\sum_{i=1}^{n}E_{ij}(\bm{c})=0$. We deduce from
the symmetry of $E(\bm{c})$ that for all $z_{i}\in{\mathbb{R}}^{d}$,

$\displaystyle\sum_{i,j=1}^{n}E_{ij}(\bm{c})z_{i}\cdot z_{j}$
$\displaystyle=\sum_{i=1}^{n}z_{i}\cdot\bigg{(}\sum_{j=1}^{n-1}E_{ij}(\bm{c})z_{j}+E_{in}(\bm{c})z_{n}\bigg{)}=\sum_{i=1}^{n}z_{i}\cdot\sum_{j=1}^{n-1}E_{ij}(\bm{c})(z_{j}-z_{n})$
$\displaystyle=\sum_{i=1}^{n-1}z_{i}\cdot\sum_{j=1}^{n-1}E_{ij}(\bm{c})(z_{j}-z_{n})+z_{n}\cdot\sum_{j=1}^{n-1}E_{nj}(\bm{c})(z_{j}-z_{n})$
$\displaystyle=\sum_{i,j=1}^{n-1}E_{ij}(\bm{c})(z_{i}-z_{n})\cdot(z_{j}-z_{n}).$

Choosing $z_{i}=\nabla h^{\prime}_{i}(c_{i})$ and observing that
$Z_{i}=\sqrt{c_{i}}z_{i}$, we rewrite the right-hand side of (57):

$\displaystyle\sum_{i,j=1}^{n}(P_{L})_{ij}Z_{i}\cdot Z_{j}$
$\displaystyle=\sum_{i,j=1}^{n-1}E_{ij}(\bm{c})(z_{i}-z_{n})\cdot(z_{j}-z_{n})$
$\displaystyle=\sum_{i,j=1}^{n-1}E_{ij}(\bm{c})\nabla\widetilde{h}_{i}^{\prime}(\widetilde{\bm{c}})\cdot\nabla\widetilde{h}_{j}^{\prime}(\widetilde{\bm{c}}).$

Introducing the matrix $Q(\widetilde{\bm{c}})$ with elements
$Q_{ij}(\widetilde{\bm{c}})=\partial^{2}\widetilde{h}(\widetilde{\bm{c}})/\partial
c_{i}\partial c_{j}$ for $i,j=1,\ldots,n-1$, this expression becomes

(58) $\displaystyle\sum_{i,j=1}^{n}$ $\displaystyle(P_{L})_{ij}Z_{i}\cdot
Z_{j}=\sum_{i,j,k,\ell=1}^{n-1}E_{ij}(\bm{c})Q_{ik}(\widetilde{\bm{c}})\nabla
c_{k}\cdot Q_{j\ell}(\widetilde{\bm{c}})\nabla c_{\ell}.$

We claim that there exists $\zeta(m)>0$ such that for all
$\bm{y}\in{\mathbb{R}}^{n-1}$,

(59)
$\bm{y}^{T}\big{(}Q(\widetilde{\bm{c}})^{T}E(\bm{c})Q(\widetilde{\bm{c}})\big{)}\bm{y}\geq\zeta(m)|\bm{y}|^{2}.$

Then, letting $\bm{y}=\nabla\bm{c}$ in (59) and using (57) and (58) leads to
(53) with $\beta(m)=\zeta(m)\lambda(m/2)/2$. The proof of (59) proceeds in
several steps.

Consider first the matrix $Q(\widetilde{\bm{c}})$. Let
$\eta:=\min_{i=1,\ldots,n}\min_{m/2\leq c_{i}\leq
1}h^{\prime\prime}_{i}(c_{i})>0$ and $\xi\in\mathbb{R}^{n-1}$ and compute

$\displaystyle Q_{ij}(\widetilde{\bm{c}})$
$\displaystyle=h_{i}^{\prime\prime}(c_{i})\delta_{ij}+h_{n}^{\prime\prime}\bigg{(}1-\sum_{k=1}^{n-1}c_{k}\bigg{)},$
$\displaystyle\xi^{T}Q\xi$
$\displaystyle=\sum_{j=1}^{n-1}h_{j}^{\prime\prime}(c_{j})\xi_{j}^{2}+h_{n}^{\prime\prime}\bigg{(}1-\sum_{k=1}^{n-1}c_{k}\bigg{)}(\xi_{1}+\cdots+\xi_{n-1})^{2}\geq\frac{\eta}{2}|\xi|^{2}.$

This implies that $Q(\widetilde{\bm{c}})$ is positive definite with
eigenvalues larger than or equal to $\eta/2$. Consider next the
$(n-1)\times(n-1)$ submatrix
$\widetilde{P}_{L}=((P_{L})_{ij})_{i,j=1,\ldots,n-1}$ of $P_{L}$ and note that
for $\xi\in{\mathbb{R}}^{n-1}$, we have

$\xi^{T}\widetilde{P}_{L}\xi=|\xi|^{2}-(\xi\cdot\sqrt{\widetilde{c}})^{2}\geq|\xi|^{2}-|\xi|^{2}(c_{1}+\cdots+c_{n-1})=c_{n}|\xi|^{2}\geq\frac{m}{2}|\xi|^{2}.$

Finally, let $\widetilde{E}(\bm{c})$ be the first $(n-1)\times(n-1)$ submatrix
of $E(\bm{c})$. Then $\widetilde{E}(\bm{c})=S^{T}\widetilde{P}_{L}S$ with
$S=\operatorname{diag}(\sqrt{c_{1}},\ldots,\sqrt{c_{n-1}})$ and, for all
$\bm{y}\in{\mathbb{R}}^{n-1}$,

$\bm{y}^{T}\widetilde{E}(\bm{c})\bm{y}=(S\bm{y})^{T}\widetilde{P}_{L}(S\bm{y})\geq\frac{m}{2}|S\bm{y}|^{2}=\frac{m}{2}\sum_{i=1}^{n-1}c_{i}y_{i}^{2}\geq\frac{m^{2}}{4}\sum_{i=1}^{n}y_{i}^{2},$

i.e., the eigenvalues of $\widetilde{E}(\bm{c})$ are larger than or equal to
$m^{2}/4$. Since $\widetilde{E}(\bm{c})-(m^{2}/8)I_{n-1}$, with $I_{n-1}$ is
the unit matrix on ${\mathbb{R}}^{(n-1)\times(n-1)}$, and
$Q(\widetilde{\bm{c}})$ are symmetric and positive definite, we deduce that

$Q(\widetilde{\bm{c}})^{T}\widetilde{E}(\bm{c})Q(\widetilde{\bm{c}})\geq
Q(\widetilde{\bm{c}})^{T}\frac{m^{2}}{8}I_{n-1}Q(\widetilde{\bm{c}})\geq\frac{m^{2}\eta^{2}}{32}I_{n-1}\,.$

This proves (59) with $\zeta(m)=m^{2}\eta^{2}/32$.

_Proof of (ii)._ Inequality (44) in Lemma 11 implies that

(60) $\sum_{i,j=1}^{n}B_{ij}^{BD}(\bm{c})Y_{i}\cdot
Y_{j}\geq\lambda(m/2)|P_{L}\bm{Y}|^{2}=\lambda(m/2)\sum_{i,j=1}^{n}(P_{L})_{ij}Y_{i}\cdot
Y_{j}.$

Similarly as the derivation of (58), we compute

$\displaystyle\sum_{i,j=1}^{n}$ $\displaystyle(P_{L})_{ij}Y_{i}\cdot
Y_{j}=\sum_{i,j,k,\ell=1}^{n-1}E_{ij}(\bm{c})\big{(}Q_{ik}(\widetilde{\bm{c}})\nabla
c_{k}-Q_{ik}(\widetilde{\bar{\bm{c}}})\nabla\bar{c}_{k}\big{)}\cdot\big{(}Q_{j\ell}(\widetilde{\bm{c}})\nabla
c_{\ell}-Q_{j\ell}(\widetilde{\bar{\bm{c}}})\nabla\bar{c}_{\ell}\big{)}$
$\displaystyle=\sum_{i,j,k,\ell=1}^{n-1}E_{ij}(\bm{c})\big{(}Q_{ik}(\widetilde{\bm{c}})\nabla(c_{k}-\bar{c}_{k})+(Q_{ik}(\widetilde{\bm{c}})-Q_{ik}(\widetilde{\bar{\bm{c}}}))\nabla\bar{c}_{k}\big{)}$
$\displaystyle\phantom{xx}{}\times\big{(}Q_{j\ell}(\widetilde{\bm{c}})\nabla(c_{\ell}-\bar{c}_{\ell})+(Q_{j\ell}(\widetilde{\bm{c}})-Q_{j\ell}(\widetilde{\bar{\bm{c}}}))\nabla\bar{c}_{\ell}\big{)}.$

We remark that if $E$ is any symmetric positive definite matrix, then for any
$\bm{z}_{1}$, $\bm{z}_{2}\in{\mathbb{R}}^{n}$, the Cauchy–Schwarz and Young’s
inequalities show that

$\displaystyle(\bm{z}_{1}$
$\displaystyle+\bm{z}_{2})^{T}E(\bm{z}_{1}+\bm{z}_{2})=\bm{z}_{1}^{T}E\bm{z}_{1}+\bm{z}_{1}^{T}E\bm{z}_{2}+\bm{z}_{2}^{T}E\bm{z}_{1}+\bm{z}_{2}^{T}E\bm{z}_{2}$
$\displaystyle\geq\bm{z}_{1}^{T}E\bm{z}_{1}-\sqrt{\bm{z}_{1}^{T}E\bm{z}_{1}}\cdot\sqrt{\bm{z}_{2}^{T}E\bm{z}_{2}}-\sqrt{\bm{z}_{1}^{T}E\bm{z}_{1}}\cdot\sqrt{\bm{z}_{2}^{T}E\bm{z}_{2}}+\bm{z}_{2}^{T}E\bm{z}_{2}$
$\displaystyle\geq\bm{z}_{1}^{T}E\bm{z}_{1}-\frac{1}{2}\bm{z}_{1}^{T}E\bm{z}_{1}-2\bm{z}_{2}^{T}E\bm{z}_{2}+\bm{z}_{2}^{T}E\bm{z}_{2}=\frac{1}{2}\bm{z}_{1}^{T}E\bm{z}_{1}-\bm{z}_{2}^{T}E\bm{z}_{2}.$

Using this inequality, (58) is estimated as

(61) $\displaystyle\sum_{i,j=1}^{n}(P_{L})_{ij}Y_{i}\cdot
Y_{j}\geq\frac{1}{2}\sum_{i,j,k,\ell=1}^{n-1}E_{ij}(\bm{c})Q_{ik}(\widetilde{\bm{c}})Q_{j\ell}(\widetilde{\bm{c}})\nabla(c_{k}-\bar{c}_{k})\cdot\nabla(c_{\ell}-\bar{c}_{\ell})$
$\displaystyle\phantom{xx}{}-\sum_{i,j,k,\ell=1}^{n-1}E_{ij}(\bm{c})\big{(}(Q_{ik}(\widetilde{\bm{c}})-Q_{ik}(\widetilde{\bar{\bm{c}}}))\nabla\bar{c}_{k}\big{)}\cdot\big{(}(Q_{j\ell}(\widetilde{\bm{c}})-Q_{j\ell}(\widetilde{\bar{\bm{c}}}))\nabla\bar{c}_{\ell}\big{)}=:J_{5}+J_{6}.$

We infer from (59) that

(62)
$J_{5}\geq\frac{1}{2}\zeta(m)\sum_{i=1}^{n-1}|\nabla(c_{i}-\bar{c}_{i})|^{2}.$

It follows from $\nabla c_{n}=-\sum_{k=1}^{n-1}\nabla c_{k}$ that

$\displaystyle|\nabla(c_{n}-\bar{c}_{n})|^{2}=\bigg{|}\sum_{k=1}^{n-1}\nabla(c_{k}-\bar{c}_{k})\bigg{|}^{2}$
$\displaystyle\leq(n-1)\sum_{k=1}^{n-1}|\nabla(c_{k}-\bar{c}_{k})|^{2},$
$\displaystyle\sum_{k=1}^{n}|\nabla(c_{k}-\bar{c}_{k})|^{2}$
$\displaystyle\leq n\sum_{k=1}^{n-1}|\nabla(c_{k}-\bar{c}_{k})|^{2}.$

Inserting these estimates into (62) yields finally

(63)
$J_{5}\geq\frac{\zeta(m)}{2n}\sum_{i=1}^{n}|\nabla(c_{i}-\bar{c}_{i})|^{2}.$

The estimate of the term $J_{2}$ is easier. Since $E_{ij}(\bm{c})$ is bounded
and the Hessian $Q(\widetilde{\bm{c}})=D^{2}\widetilde{h}$ is Lipschitz
continuous,

$J_{6}\leq
C\sum_{i,k=1}^{n-1}(Q_{ik}(\widetilde{\bm{c}})-Q_{ik}(\widetilde{\bar{\bm{c}}}))^{2}|\nabla\bar{c}_{k}|^{2}\leq
C\sum_{i=1}^{n-1}(c_{i}-\bar{c}_{i})^{2}|\nabla\bar{c}_{i}|^{2}\leq
C\sum_{i=1}^{n-1}(c_{i}-\bar{c}_{i})^{2}.$

Combining the above inequality with (63) and (61) gives

$\sum_{i,j=1}^{n}(P_{L})_{ij}Y_{i}\cdot
Y_{j}\geq\frac{\zeta(m)}{2n}\sum_{i=1}^{n}|\nabla(c_{i}-\bar{c}_{i})|^{2}-C\sum_{i=1}^{n-1}(c_{i}-\bar{c}_{i})^{2}.$

We conclude (54) after inserting the previous estimate into (60).

_Proof of (iii)._ Let $c_{*}(x,t)$ be defined by (55) and split the domain of
integration into the two subdomains

$\Omega\times(0,t)=\bigg{\\{}c_{*}>\frac{m}{2}\bigg{\\}}\cup\bigg{\\{}c_{*}\leq\frac{m}{2}\bigg{\\}}.$

By Lemma 11, the matrix $B^{BD}(\bm{c})$ is symmetric and positive semi-
definite. Using (53), the entropy inequality (47) yields (56). ∎

###  4.5. Proof of Theorem 3

Lemma 12 suggests that the relative entropy inequality can be expressed in two
ways, using either (52) or (49):

$\displaystyle H(\bm{c}|\bar{\bm{c}})(t)$ $\displaystyle\leq
H(\bm{c}^{0}|\bar{\bm{c}}^{0})+\int_{0}^{t}\int_{\Omega}(I_{1}+I_{2})dxds$
(64)
$\displaystyle=H(\bm{c}^{0}|\bar{\bm{c}}^{0})+\int_{0}^{t}\int_{\Omega}(I_{3}+I_{4}+I_{5})dxds,$

where

$\displaystyle I_{1}$
$\displaystyle=-\sum_{i,j=1}^{n}B_{ij}^{BD}(\bm{c})\sqrt{c_{i}c_{j}}\nabla(h^{\prime}_{i}(c_{i})-h^{\prime}_{i}(\bar{c}_{i}))\cdot\nabla
h^{\prime}_{j}(c_{j}),$ $\displaystyle I_{2}$
$\displaystyle=-\sum_{i=1}^{n}(c_{i}-\bar{c}_{i})h_{i}^{\prime\prime}(\bar{c}_{i})\operatorname{div}\bigg{(}\sum_{j=1}^{n}\sqrt{\bar{c}_{i}}B_{ij}^{BD}(\bar{\bm{c}})\sqrt{\bar{c}_{j}}\nabla
h^{\prime}_{j}(\bar{c}_{j})\bigg{)},$ $\displaystyle I_{3}$
$\displaystyle=-\sum_{i,j=1}^{n}Y_{i}\cdot B^{BD}_{ij}(\bm{c})Y_{j},$
$\displaystyle I_{4}$
$\displaystyle=-\sum_{i,j=1}^{n}Y_{i}\cdot\bigg{(}B_{ij}^{BD}(\bm{c})\sqrt{c_{j}}-\frac{\sqrt{c_{i}}B_{ij}^{BD}(\bar{\bm{c}})\sqrt{\bar{c}_{j}}}{\sqrt{\bar{c}_{i}}}\bigg{)}\nabla
h^{\prime}_{j}(\bar{c}_{j}),$ $\displaystyle I_{5}$
$\displaystyle=\sum_{i,j=1}^{n}p_{i}(c_{i}|\bar{c}_{i})\operatorname{div}\bigg{(}\frac{B_{ij}^{BD}(\bar{\bm{c}})\sqrt{\bar{c}_{j}}}{\sqrt{\bar{c}_{i}}}\nabla
h^{\prime}_{j}(\bar{c}_{j})\bigg{)},$

and
$Y_{i}=\sqrt{c_{i}}\nabla(h^{\prime}_{i}(c_{i})-h^{\prime}_{i}(\bar{c}_{i}))$,
$i=1,\ldots,n$.

Step 1: Preparations. Recall that we have assumed that $\bar{c}_{i}(x,t)\geq
m$ for $x\in\Omega$, $t>0$ for some $m>0$. Let
$c_{*}(x,t):=\min_{i=1,\ldots,n}c_{i}(x,t)$. We split the estimations of the
above integrals into two subdomains: one where $c_{*}(x,t)\leq m/2$ and
another one where $c_{*}(x,t)>m/2$. To this end, we use a cutoff function. Let
$\varepsilon>0$ be sufficiently small and $\psi:[0,1]\to[0,1]$ be a
$C^{2}$-function, which takes the values $\psi=0$ on $[0,m/2]$, $\psi=1$ on
$[m/2+\varepsilon,1]$, and $\psi\in(0,1)$ on the complementary interval
$(m/2,m/2+\varepsilon)$. Define $\chi(\bm{c}):{\mathbb{R}}^{n}\to[0,1]$ by

(65) $\chi(\bm{c}):=\prod_{i=1}^{n}\psi(c_{i})=\begin{cases}1&\quad\mbox{if
}m/2+\varepsilon\leq c_{i}\leq 1\mbox{ for all }i=1,\ldots,n,\\\
0&\quad\mbox{if }0<c_{i}\leq m/2\mbox{ for some $i$,}\\\
\alpha(\bm{c})\in(0,1)&\quad\mbox{else}.\end{cases}$

We employ $\chi(\bm{c})$ to split the integral (64) into two parts:

(66) $\displaystyle H(\bm{c}|\bar{\bm{c}})(t)-H(\bm{c}^{0}|\bar{\bm{c}}^{0})$
$\displaystyle\leq\int_{0}^{t}\int_{\Omega}(1-\chi(\bm{c}))(I_{3}+I_{4}+I_{5})dxds$
$\displaystyle\phantom{xx}{}+\int_{0}^{t}\int_{\Omega}\chi(\bm{c})(I_{3}+I_{4}+I_{5})dxds=:J_{L}+J_{H}.$

In the sequel, we estimate $J_{L}$ and $J_{H}$ separately.

Step 2: Case $c_{*}(x,t)\leq m/2+\varepsilon$. We estimate the term $J_{L}$ in
(66). By replacing $p(c_{i}|\bar{c}_{i})$ in $I_{5}$ by definitions (19) and
(48) and tracing backwards the derivation from (52) to (49), we can express
the integral over $(1-\chi(\bm{c})(I_{3}+I_{4}+I_{5})$ by
$(1-\chi(\bm{c})(I_{1}+I_{2})$ except for a term accounting for the cutoff
function:

(67) $\displaystyle J_{L}$
$\displaystyle=\int_{0}^{t}\int_{\Omega}(1-\chi(\bm{c}))(I_{1}+I_{2})dxds$
$\displaystyle\phantom{xx}{}+\int_{0}^{t}\int_{\Omega}\nabla\chi(\bm{c})\cdot\sum_{i=1}^{n}(p_{i}(c_{i})-p_{i}(\bar{c}_{i}))\sum_{j=1}^{n}B_{ij}^{BD}(\bar{\bm{c}})\frac{\sqrt{\bar{c}_{j}}}{\sqrt{\bar{c}_{i}}}\nabla
h^{\prime}_{j}(\bar{c}_{j}).$

In the sequel, we estimate the right-hand side of (67) term-by-term. To
estimate $I_{1}$, we set $\bm{Z}=(Z_{1},\ldots,Z_{n})$ with
$Z_{i}:=\sqrt{c}_{i}\nabla h^{\prime}_{i}(c_{i})$, $i=1,\ldots,n$. By Lemma
11, the matrix $B^{BD}(\bm{c})$ is symmetric and positive semi-definite.
Therefore, using Young’s inequality and the boundedness of $B^{BD}(\bm{c})$
(see Assumption (B2)),

$\displaystyle I_{1}$
$\displaystyle=-\bm{Z}^{T}B^{BD}(\bm{c})\bm{Z}+\sum_{i,j=1}^{n}B_{ij}^{BD}(\bm{c})Z_{i}\sqrt{c_{j}}\nabla
h^{\prime}_{j}(\bar{c}_{j})$
$\displaystyle\leq-\frac{1}{2}\bm{Z}^{T}B^{BD}(\bm{c})\bm{Z}+\frac{1}{2}\sum_{i,j=1}^{n}B_{ij}^{BD}(\bm{c})(\sqrt{c_{i}}\nabla
h^{\prime}_{i}(\bar{c}_{i}))\cdot(\sqrt{c_{j}}\nabla
h^{\prime}_{j}(\bar{c}_{j}))$
$\displaystyle\leq-\frac{1}{2}\bm{Z}^{T}B^{BD}(\bm{c})\bm{Z}+C\sum_{i=1}^{n}|\nabla
h^{\prime}_{i}(\bar{c}_{i})|^{2},$

where $C>0$ depends on $m$ and $\mu$ (defined in Assumption (B4)). For the
term $I_{2}$, we use the regularity for $\bar{c}_{i}$ to conclude that

$I_{2}\leq\sum_{i=1}^{n}(c_{i}-\bar{c}_{i})^{2}+C\sum_{i,j=1}^{n}\big{|}h^{\prime\prime}_{i}(\bar{c}_{i})\operatorname{div}\big{(}\sqrt{\bar{c}_{i}}B_{ij}^{BD}(\bar{\bm{c}})\sqrt{\bar{c}_{j}}\nabla
h^{\prime}_{j}(\bar{c}_{j})\big{)}\big{|}^{2}\leq\sum_{i=1}^{n}(c_{i}-\bar{c}_{i})^{2}+C.$

On the set $\chi(\bm{c})<1$, we have $c_{*}(x,t)<m/2+\varepsilon$, and there
exists $\ell\in\\{1,\ldots,n\\}$ such that $c_{\ell}(x,t)<m/2+\varepsilon$.
Thanks to Assumption (H) on page H, we can apply Lemma 19 to find that

$h_{i_{0}}(c_{i_{0}}|\bar{c}_{i_{0}})\geq\kappa_{m}(c_{i_{0}}-\bar{c}_{i_{0}})^{2}\geq\kappa_{m}\bigg{(}\frac{m}{2}-\varepsilon\bigg{)}^{2}\quad\mbox{when
}\chi(\bm{c})<1.$

We infer that

$I_{2}\leq\sum_{i=1}^{n}(c_{i}-\bar{c}_{i})^{2}+C\sum_{i=1}^{n}h_{i}(c_{i}|\bar{c}_{i})\leq
C\sum_{i=1}^{n}h_{i}(c_{i}|\bar{c}_{i}).$

It remains to estimate the last term in (67), using the fact that
$\nabla\chi(\bm{c})$ vanishes outside the set $c_{*}\in[m/2,m/2+\varepsilon]$,
the Lipschitz continuity of $p_{i}(c_{i})$, entropy inequality (56) and Lemma
19:

$\displaystyle\bigg{|}\int_{0}^{t}\int_{\Omega}\nabla\chi(\bm{c})\cdot\sum_{i=1}^{n}(p_{i}(c_{i})-p_{i}(\bar{c}_{i}))\sum_{j=1}^{n}B_{ij}^{BD}(\bar{\bm{c}})\frac{\sqrt{\bar{c}_{j}}}{\sqrt{\bar{c}_{i}}}\nabla
h^{\prime}_{j}(\bar{c}_{j})dxds\bigg{|}$ $\displaystyle\phantom{xx}{}\leq
C\bigg{(}\max_{j=1,\ldots,n}\sup_{m/2\leq c_{j}\leq
m/2+\varepsilon}\bigg{|}\frac{\partial\chi}{\partial
c_{j}}(\bm{c})\bigg{|}\bigg{)}\int_{0}^{t}\int_{\Omega}\mathrm{1}_{\\{m/2<c_{*}<m/2+\varepsilon\\}}\sum_{i=1}^{n}|\nabla
c_{i}|\sum_{j=1}^{n}|c_{j}-\bar{c}_{j}|dxds$ $\displaystyle\phantom{xx}{}\leq
C\int_{0}^{t}\int_{\Omega}\mathrm{1}_{\\{c_{*}>m/2\\}}\sum_{i=1}^{n}|\nabla
c_{i}|^{2}dxds+C\int_{0}^{t}\int_{\Omega}\mathrm{1}_{\\{c_{*}<m/2+\varepsilon\\}}\sum_{j=1}^{n}|c_{j}-\bar{c}_{j}|^{2}dxds$
$\displaystyle\phantom{xx}{}\leq
C+\int_{0}^{t}\int_{\Omega}\mathrm{1}_{\\{c_{*}<m/2+\varepsilon\\}}\sum_{j=1}^{n}|c_{j}-\bar{c}_{j}|^{2}dxds$
$\displaystyle\phantom{xx}{}\leq
C\int_{0}^{t}\int_{\Omega}\mathrm{1}_{\\{c_{*}<m/2+\varepsilon\\}}\sum_{i=1}^{n}h_{i}(c_{i}|\bar{c}_{i})dxds.$

Note that the final constant $C$, depending on
$\max_{j=1,\ldots,n}\sup_{m/2\leq c_{j}\leq
m/2+\varepsilon}|(\partial\chi/\partial c_{j})(\bm{c})|$, will blow up if we
let $\varepsilon\to 0$. Therefore, we fix $\varepsilon>0$. Combining the
previous estimate, we end up with

(68)
$J_{L}\leq-\frac{1}{2}\int_{0}^{t}\int_{\Omega}(1-\chi(\bm{c}))\bm{Z}^{T}B^{BD}(\bm{c})\bm{Z}dxds+C\int_{0}^{t}\int_{\Omega}\mathrm{1}_{\\{c_{*}<m/2+\varepsilon\\}}\sum_{i=1}^{n}h_{i}(c_{i}|\bar{c}_{i})dxds.$

Step 3: Case $c_{*}(x,t)>m/2$. We proceed to estimate the term $J_{H}$ in
(66). The range of integration now consists of the sets
$\\{m/2<c_{*}<m/2+\varepsilon\\}$, where $0<\chi(\bm{c})<1$, and the set
$\\{m/2+\varepsilon<c_{*}\leq 1\\}$, where $\chi(\bm{c})=1$.

For the term $I_{3}$, we use (54) in Lemma 13:

$\displaystyle-\int_{0}^{T}$
$\displaystyle\int_{\Omega}\chi(\bm{c})I_{3}dxds=\int_{0}^{t}\int_{\Omega}\chi(\bm{c})\sum_{i,j=1}^{n}B^{BD}_{ij}(\bm{c})Y_{i}\cdot
Y_{j}dxds$
$\displaystyle\geq\beta(m)\int_{0}^{t}\int_{\Omega}\chi(\bm{c})\sum_{i=1}^{n}|\nabla(c_{i}-\bar{c}_{i})|^{2}dxds-C\int_{0}^{t}\int_{\Omega}\chi(\bm{c})\sum_{i=1}^{n}|c_{i}-\bar{c}_{i}|^{2}dxds.$

By Young’s inequality with $\delta>0$, the term $I_{4}$ can be estimated as

$\displaystyle\int_{0}^{t}$
$\displaystyle\int_{\Omega}\chi(\bm{c})I_{4}dxds\leq\delta\sum_{i=1}^{n}\int_{0}^{t}\int_{\Omega}\chi(\bm{c})|Y_{i}|^{2}dxds$
$\displaystyle{}+\frac{1}{4\delta}\sum_{i,j=1}^{n}\int_{0}^{t}\int_{\Omega}\chi(\bm{c})\bigg{(}B_{ij}^{BD}(\bm{c})\sqrt{c_{j}}-\frac{\sqrt{c_{i}}B_{ij}^{BD}(\bar{\bm{c}})\sqrt{\bar{c}_{j}}}{\sqrt{\bar{c}_{i}}}\bigg{)}^{2}|\nabla
h^{\prime}_{j}(\bar{c}_{j})|^{2}dxds.$

Recall that we work in the range $c_{i}>m/2$ and $\bar{c}_{i}\geq m$. The
boundedness and Lipschitz continuity of $h_{i}^{\prime\prime}$ imply that

$\displaystyle\sum_{i=1}^{n}|Y_{i}|^{2}$
$\displaystyle=\sum_{i=1}^{n}c_{i}|\nabla(h_{i}^{\prime}(c_{i})-h^{\prime}_{i}(\bar{c}_{i}))|^{2}\leq\sum_{i=1}^{n}c_{i}|h^{\prime\prime}_{i}(c_{i})\nabla(c_{i}-\bar{c}_{i})+(h^{\prime\prime}_{i}(c_{i})-h^{\prime\prime}_{i}(\bar{c}_{i}))\nabla\bar{c}_{i}|^{2}$
$\displaystyle\leq
C\sum_{i=1}^{n}\big{(}|\nabla(c_{i}-\bar{c}_{i})|^{2}+(c_{i}-\bar{c}_{i})^{2}|\nabla\bar{c}_{i}|^{2}\big{)}.$

Furthermore, the boundedness and Lipschitz continuity of $B_{ij}^{BD}$ (see
Lemma 11) yield

$\displaystyle\bigg{|}B_{ij}^{BD}$
$\displaystyle(\bm{c})\sqrt{c_{j}}-\frac{\sqrt{c_{i}}B_{ij}^{BD}(\bar{\bm{c}})\sqrt{\bar{c}_{j}}}{\sqrt{\bar{c}_{i}}}\bigg{|}$
$\displaystyle=\bigg{|}B_{ij}(\bm{c})\sqrt{c_{j}}-B_{ij}^{BD}(\bar{\bm{c}})\sqrt{\bar{c}_{j}}+\frac{(\sqrt{\bar{c}_{i}}-\sqrt{{c}_{i}})B_{ij}^{BD}(\bar{\bm{c}})\sqrt{\bar{c}_{j}}}{\sqrt{\bar{c}_{i}}}\bigg{|}$
$\displaystyle\leq
C\sum_{i=1}^{n}|c_{i}-\bar{c}_{i}|+|\sqrt{c_{i}}-\sqrt{\bar{c}_{i}}|\leq
C\sum_{i=1}^{n}|c_{i}-\bar{c}_{i}|.$

Thus, the choice $\delta=\beta(m)/(2C)$ gives

$\displaystyle\int_{0}^{t}\int_{\Omega}\chi(\bm{c})I_{4}dxds$
$\displaystyle\leq\frac{\beta(m)}{2}\int_{0}^{t}\int_{\Omega}\chi(\bm{c})\sum_{i=1}^{n}|\nabla(c_{i}-\bar{c}_{i})|^{2}dxds$
$\displaystyle\phantom{xx}{}+C\int_{0}^{t}\int_{\Omega}\chi(\bm{c})\sum_{i=1}^{n}|c_{i}-\bar{c}_{i}|^{2}dxds.$

Finally, we use definition (48) of $p_{i}(c_{i}|\bar{c}_{i})$ and Hypothesis
(H) to estimate

$\displaystyle|p_{i}(c_{i}|\bar{c}_{i})|$
$\displaystyle=\bigg{|}(c_{i}-\bar{c}_{i})^{2}\int_{0}^{1}\int_{0}^{s}p_{i}^{\prime\prime}(\tau
c_{i}+(1-\tau)\bar{c}_{i})d\tau ds\bigg{|}$ $\displaystyle\leq
K_{2}(c_{i}-\bar{c}_{i})^{2}\int_{0}^{1}\int_{0}^{s}h_{i}^{\prime\prime}(\tau
c_{i}+(1-\tau)\bar{c}_{i})d\tau ds=K_{2}h_{i}(c_{i}|\bar{c}_{i}).$

In turn, this implies that

$\int_{0}^{t}\int_{\Omega}\chi(\bm{c})I_{5}dxds\leq
C\int_{0}^{t}\int_{\Omega}\chi(\bm{c})\sum_{i=1}^{n}h_{i}(c_{i}|\bar{c}_{i})dxds.$

Summarizing the previous computations and using Lemma 19, we conclude that

(69)
$J_{H}\leq-\frac{\beta(m)}{2}\int_{0}^{t}\int_{\Omega}\chi(\bm{c})\sum_{i=1}^{n}|\nabla(c_{i}-\bar{c}_{i})|^{2}dxds+C\int_{0}^{t}\int_{\Omega}\chi(\bm{c})\sum_{i=1}^{n}h_{i}(c_{i}|\bar{c}_{i})dxds.$

Step 4: End of the proof. We combine the differential inequality (66) with the
estimations (68) and (69) to obtain

(70) $\displaystyle H(\bm{c}$
$\displaystyle(t)|\bar{\bm{c}}(t))+\frac{1}{2}\int_{0}^{t}\int_{\Omega}(1-\chi(\bm{c}))\bm{Z}^{T}B^{BD}(\bm{c})\bm{Z}dxds$
$\displaystyle\phantom{xxxxxx}{}+\frac{\beta(m)}{2}\int_{0}^{t}\int_{\Omega}\chi(\bm{c})\sum_{i=1}^{n}|\nabla(c_{i}-\bar{c}_{i})|^{2}dxds$
$\displaystyle\leq
H(\bm{c}^{0}|\bar{\bm{c}}^{0})+C\int_{0}^{t}H(\bm{c}|\bar{\bm{c}})ds,$

The constant $C>0$ depends in particular on $m$ and the
$L^{\infty}(0,T;W^{2,\infty}(\Omega))$ norm of $\bar{c}_{j}$, $j=1,\ldots,n$.
The proof of Theorem 3 finishes after applying Gronwall’s inequality.

######  Remark 14.

Inequality (70) leads to a slightly stronger version of the relative entropy
inequality than stated in Theorem 3. However, we obtain gradient estimates
only on the set $\\{c_{*}>m/2\\}$, while on $\\{c_{*}\leq m/2\\}$, the
quadratic form $\bm{Z}^{T}B^{BD}(\bm{c})\bm{Z}$ generally does not lead to a
control of the $L^{2}$-norm of $\nabla c_{i}$. ∎

##  5\. Examples

We present some examples for the generalized Maxwell–Stefan system (16)–(17)
satisfying Assumptions (B1)–(B4).

###  5.1. A cross-diffusion system for thin-film solar cells

Thin-film crystalline solar cells can be fabricated by the so-called physical
vapor deposition process. This process produces a metal vapor that can be
deposited on electrically conductive materials as a thin coating. It is shown
in [1] that the evolution of the volume fractions of the thin-film components
can be described by the cross-diffusion system

(71)
$\partial_{t}c_{i}=\operatorname{div}\bigg{(}\sum_{j=1}^{n}a_{ij}(u_{j}\nabla
u_{i}-u_{i}\nabla u_{j})\bigg{)},\quad i=1,\ldots,n,$

where $a_{ij}=a_{ji}>0$ for $i,j=1,\ldots,n$, and $\sum_{i=1}^{n}c_{i}=1$.
This model can be formulated as a generalized Maxwell–Stefan system. Indeed,
let $h_{i}(c_{i})=c_{i}(\log c_{i}-1)$ and
$K_{ij}(\bm{c})=\sum_{j=1}^{n}\sqrt{c_{i}}A_{ij}^{BD}(\bm{c})\sqrt{c_{j}}$ for
$i,j=1,\ldots,n$, where $A(\bm{c})$ is given by (10) with $D_{ij}=1/a_{ij}$.
Then $B(\bm{c})=A^{BD}(\bm{c})$ (see (18)) and hence
$B^{BD}(\bm{c})=A(\bm{c})$. Equation (43) becomes

(72)
$\partial_{t}c_{i}=\operatorname{div}\bigg{(}\sum_{j=1}^{n}\sqrt{c_{i}}A_{ij}(\bm{c})\sqrt{c_{j}}\nabla\log
c_{j}\bigg{)},\quad i=1,\ldots,n.$

Because of (10), the mobility matrix
$(\sqrt{c_{i}}A_{ij}(\bm{c})\sqrt{c_{j}})$ reads as

$\sqrt{c_{i}}A_{ij}(\bm{c})\sqrt{c_{j}}=\left\\{\begin{array}[]{ll}\sum_{k\neq
i}a_{ik}c_{i}c_{k}&\quad\mbox{if }i=j,\\\ -a_{ij}c_{i}c_{j}&\quad\mbox{if
}i\neq j,\end{array}\right.$

and an elementary computation shows that (72) can actually be written as (71).

Although it can be checked that the matrix $B(\bm{c})=A^{BD}(\bm{c})$
satisfies Assumptions (B1)–(B4), we can here directly verify the statements of
Lemma 11. Definition (10) of $A(\bm{c})$ immediately implies that
$B_{ij}^{BD}(\bm{c})$ is bounded and Lipschitz continuous on $[0,1]^{n}$.
Property (44) follows from (23) in Lemma 4 with $\lambda(m)=\mu$. Hence, the
weak-strong uniqueness property holds for this model.

###  5.2. A tumor-growth model

The growth of a symmetric avascular tumor can be modeled on the mechanical
level by diffusion fluxes of the tumor cells, the extracellular matrix (ECM),
and the interstitial fluid (water, nutrients, etc.). The model was suggested
in [25] and analyzed in [27]. The evolution of the volume fractions $c_{i}$ of
the tumor cells, ECM, and interstitial fluid is given by (see, e.g., [29,
Section 4.2])

(73) $\displaystyle\partial_{t}c_{i}+\operatorname{div}(c_{i}u_{i})$
$\displaystyle=0,\quad i=1,\ldots,3,$ (74)
$\displaystyle\nabla(c_{i}P_{i})+c_{i}\nabla p$
$\displaystyle=-\sum_{j=1}^{3}k_{ij}c_{i}c_{j}(u_{i}-u_{j}),\quad i=1,2,$ (75)
$\displaystyle c_{3}\nabla p$
$\displaystyle=-\sum_{j=1}^{3}k_{ij}c_{i}c_{j}(u_{i}-u_{j}),$

where $k_{ij}=k_{ji}>0$ for $i,j=1,2,3$, the partial pressures $P_{1}$,
$P_{2}$ and the phase pressure $p$ are given by

$P_{1}=c_{1},\quad P_{2}=\beta c_{2}(1+\theta c_{1}),\quad
p=-c_{1}P_{1}-c_{2}P_{2},$

$\beta>0$, $\theta>0$ are suitable parameters, and it holds that
$\sum_{i=1}^{3}c_{i}=1$.

We claim that (73)–(75) can be formulated as a generalized Maxwell–Stefan
system. We define the entropy densities as in the previous example,
$h_{i}(c_{i})=c_{i}(\log c_{i}-1)$, $i=1,2,3$. With the matrix

$W(\bm{c})=\begin{pmatrix}2c_{1}(1-c_{1})-\beta\theta c_{1}c_{2}^{2}&-2\beta
c_{1}c_{2}(1+\theta c_{1})&0\\\
-2c_{1}c_{2}+\beta\theta(1-c_{2})c_{2}^{2}&2\beta c_{2}(1-c_{2})(1+\theta
c_{1})&0\\\ -2c_{1}c_{3}-\beta\theta c_{3}c_{2}^{2}&-2\beta
c_{3}c_{2}(1+\theta c_{1})&0\end{pmatrix},$

the left-hand side of (74)–(75) can be written in a more concise form:

$\begin{pmatrix}\nabla(c_{1}P_{1})+c_{1}\nabla p\\\
\nabla(c_{2}P_{2})+c_{2}\nabla p\\\ c_{3}\nabla
p\end{pmatrix}=W(\bm{c})\begin{pmatrix}\nabla c_{1}\\\ \nabla c_{2}\\\ \nabla
c_{3}\end{pmatrix}.$

Let the matrix $A(\bm{c})$ be given by (10) with $D_{ij}=1/k_{ij}$. Then the
right-hand side of (74)–(75) equals (also see (12))

$-\sum_{j=1,\,j\neq
i}^{3}\frac{c_{i}c_{j}}{D_{ij}}(u_{i}-u_{j})=-\sum_{j=1}^{3}\sqrt{c_{i}}A_{ij}(\bm{c})\sqrt{c_{j}}u_{j}.$

Thus, inverting

$\sum_{j=1}^{n}W_{ij}(\bm{c})\nabla
c_{j}=-\sum_{j=1}^{3}\sqrt{c_{i}}A_{ij}(\bm{c})\sqrt{c_{j}}u_{j},\quad
i=1,2,3,$

(which is the same as (74)–(75)) yields

$\sqrt{c_{i}}u_{i}=-\sum_{j,k=1}^{3}A_{ij}^{BD}(\bm{c})\frac{1}{\sqrt{c_{j}}}W_{jk}(\bm{c})\nabla
c_{k},\quad i=1,2,3,$

and system (73)–(75) can be written for $i=1,2,3$ as

$\displaystyle\partial_{t}c_{i}$
$\displaystyle=\operatorname{div}\bigg{(}\sum_{j,k=1}^{3}\sqrt{c_{i}}A_{ij}^{BD}(\bm{c})\frac{1}{\sqrt{c_{j}}}W_{jk}(\bm{c})\nabla
c_{k}\bigg{)}$
$\displaystyle=\operatorname{div}\bigg{(}\sum_{j,k,\ell=1}^{3}\sqrt{c_{i}}A_{ij}^{BD}(\bm{c})\frac{1}{\sqrt{c_{j}}}W_{jk}(\bm{c})\sqrt{c_{k}}(P_{L})_{k\ell}\sqrt{c_{\ell}}\nabla\log
c_{\ell}\bigg{)},$

recalling definition (22) of $P_{L}$. Thus,

$\displaystyle\partial_{t}c_{i}=\operatorname{div}\bigg{(}\sum_{\ell=1}^{n}\sqrt{c_{i}}R_{i\ell}(\bm{c})\sqrt{c_{\ell}}\nabla\log
c_{\ell}\bigg{)},\quad\mbox{where}$ $\displaystyle
R_{i\ell}(\bm{c}):=\sum_{j,k=1}^{3}A_{ij}^{BD}(\bm{c})\frac{1}{\sqrt{c_{j}}}W_{jk}(\bm{c})\sqrt{c_{k}}(P_{L})_{k\ell},\quad
i,\ell=1,2,3.$

Also in this case, it is more convenient to check the statements of Lemma 11
instead of Assumptions (B1)–(B4). Notice that $W(\bm{c})$ is not symmetric, so
$R(\bm{c})$ is not symmetric either. The elements $R_{ij}(\bm{c})$ are bounded
and Lipschitz continuous, since $A_{ij}^{BD}(\bm{c})$ and $W_{ij}(\bm{c})$
have these properties. The second statement of Lemma 11, namely property (44),
is verified only for a special example that was considered in [27].

######  Lemma 15.

Let $k_{ij}=1$ for $i,j=1,2,3$ and $0\leq\theta<4/\sqrt{\beta}$. Then, with
$m=\min_{i=1,2,3}c_{i}>0$, there exists $\lambda(m)>0$ such that

$\bm{z}^{T}R(\bm{c})\bm{z}\geq\lambda(m)|P_{L}\bm{z}|^{2}\quad\mbox{for all
}\bm{z}\in{\mathbb{R}}^{3}.$

###### Proof.

The assumption $k_{ij}=1$ implies that $A(\bm{c})=P_{L}$ and hence
$A^{BD}(\bm{c})=P_{L}(A(\bm{c})P_{L}+P_{L^{\perp}})^{-1}=P_{L}$. Suppose that
for any $\bm{y}\in{\mathbb{R}}^{3}$ satisfying $\bm{y}\in L$ (i.e.
$\sqrt{\bm{c}}\cdot\bm{y}=0$), we have

(76)
$\sum_{i,j=1}^{3}\frac{1}{\sqrt{c_{i}}}W_{ij}(\bm{c})\sqrt{c_{j}}y_{i}y_{j}\geq\lambda(m)|\bm{y}|^{2}.$

Then, for $\bm{z}\in{\mathbb{R}}^{3}$ and $\bm{y}=P_{L}\bm{z}\in L$,

$\bm{z}^{T}R(\bm{c})\bm{z}=\sum_{i,j=1}^{3}(P_{L}\bm{z})_{i}\frac{1}{\sqrt{c_{i}}}W_{ij}(\bm{c})\sqrt{c_{j}}(P_{L}\bm{z})_{j}\geq\lambda(m)|P_{L}\bm{z}|^{2},$

which proves the lemma.

It remains to verify (76). Using
$y_{3}=-(\sqrt{c_{1}}y_{1}+\sqrt{c_{2}}y_{2})/\sqrt{c_{3}}$, we calculate

$\sum_{i,j=1}^{3}\frac{1}{\sqrt{c_{i}}}W_{ij}(\bm{c})\sqrt{c_{j}}y_{i}y_{j}=2\beta(1+\theta
c_{1})(\sqrt{c_{2}}y_{2})^{2}+\beta\theta
c_{2}(\sqrt{c_{1}}y_{1})(\sqrt{c_{2}}y_{2})+2(\sqrt{c_{1}}y_{1})^{2}.$

Since $0\leq\theta<4/\sqrt{\beta}$, the discriminant of the quadratic form is
negative and there exists $\kappa>0$ such that

$\displaystyle\sum_{i,j=1}^{3}\frac{1}{\sqrt{c_{i}}}W_{ij}(\bm{c})\sqrt{c_{j}}y_{i}y_{j}$
$\displaystyle\geq\kappa(c_{1}y_{1}^{2}+c_{2}y_{2}^{2})\geq\frac{\kappa}{3}\big{(}c_{1}y_{1}^{2}+c_{2}y_{2}^{2}+(\sqrt{c_{1}}y_{1}+\sqrt{c_{2}}y_{2})^{2}\big{)}$
$\displaystyle=\frac{\kappa}{3}(c_{1}y_{1}^{2}+c_{2}y_{2}^{2}+c_{3}y_{3}^{2})\geq\frac{\kappa}{3}\Big{(}\min_{i=1,2,3}c_{i}\Big{)}(y_{1}^{2}+y_{2}^{2}+y_{3}^{2}),$

proving the claim with $\lambda(m)=\kappa m/3$. ∎

We deduce from the previous lemma that the weak-strong uniqueness property
holds for the tumor-growth model if $k_{ij}=1$ for $i,j=1,2,3$ and
$0\leq\theta<4/\sqrt{\beta}$. The latter condition is necessary to achieve the
global existence of weak solutions, since it guarantees the positive
semidefiniteness of the mobility matrix; see [27] for details.

###  5.3. A multi-species porous-medium-type model

Another model is a generalization of the first example to illustrate that also
non-logarithmic entropies may be considered. We choose
$h_{i}(c_{i})=c_{i}^{\gamma}/(\gamma-1)$ with $\gamma>1$ and $A(\bm{c})$ as in
(10). The partial pressure becomes
$p_{i}=c_{i}h^{\prime}_{i}(c_{i})-h_{i}(c_{i})=c_{i}^{\gamma}$, and equation
(72) reads here as

$\partial_{t}c_{i}=\operatorname{div}\bigg{(}\sum_{j=1}^{n}\sqrt{c_{i}}A_{ij}(\bm{c})\sqrt{c_{j}}\nabla
h_{j}^{\prime}(c_{j})\bigg{)}=\frac{\gamma}{\gamma-1}\operatorname{div}\bigg{(}\sum_{j=1}^{n}\frac{1}{D_{ij}}(c_{j}\nabla
c_{i}^{\gamma-1}-c_{i}\nabla c_{j}^{\gamma-1})\bigg{)}.$

Hence, the weak-strong property holds for this model.

###  5.4. Maxwell-Stefan system with different molar masses

In equations (1), we have implicitly assumed that all molar masses of the
species are the same. We show that the weak-strong uniqueness property also
holds for the model proposed in [3, 5] without this assumption. In the case of
different molar masses $M_{i}$, we need to distinguish between the mass
densities $\rho_{i}$ and the molar concentrations $c_{i}=\rho_{i}/M_{i}$. The
Maxwell–Stefan equations read for $i=1,\ldots,n$ as

(77)
$\partial_{t}\rho_{i}+\operatorname{div}(\rho_{i}u_{i})=0,\quad-\sum_{j=1}^{n}\frac{c_{i}c_{j}}{c^{2}D_{ij}}(u_{i}-u_{j})=\rho_{i}\nabla\frac{\delta
H}{\delta\rho_{i}}(\bm{\rho})-\rho_{i}\sum_{j=1}^{n}\rho_{j}\nabla\frac{\delta
H}{\delta\rho_{j}}(\bm{\rho}),$

where $c=\sum_{i=1}^{n}c_{i}$. As before, the restriction
$\sum_{j=1}^{n}\rho_{j}=1$ inherited from the initial data is imposed. The
second equation can be rewritten as

$-\sum_{j=1}^{n}\frac{\rho_{i}\rho_{j}}{\widetilde{D}_{ij}(\bm{\rho})}(u_{i}-u_{j})=\rho_{i}\nabla\frac{\delta
H}{\delta\rho_{i}}(\bm{\rho})-\rho_{i}\sum_{j=1}^{n}\rho_{j}\nabla\frac{\delta
H}{\delta\rho_{j}}(\bm{\rho}),$

where
$\widetilde{D}_{ij}(\bm{\rho})=c^{2}M_{i}M_{j}=(\sum_{k=1}^{n}\rho_{k}/M_{k})^{2}M_{i}M_{j}$.
Due to

$\sum_{k=1}^{n}\frac{\rho_{k}}{M_{k}}\geq\sum_{k=1}^{n}\frac{\rho_{k}}{\max_{\ell=1,\ldots,n}{M_{\ell}}}=\frac{1}{\max_{\ell=1,\ldots,n}M_{\ell}},$

the coefficients $\widetilde{D}_{ij}(\bm{\rho})$ are uniformly bounded from
below. Since the proof of Lemma 4 only relies on the uniform boundness of
$D_{ij}$, Lemma 4 also holds for the following matrix $\widetilde{A}(\bm{c})$,
defined similarly as in (10):

$\widetilde{A}_{ij}(\bm{\rho})=\left\\{\begin{array}[]{ll}\sum_{k=1,\,k\neq
i}^{n}\rho_{k}/\widetilde{D}_{ik}(\bm{\rho})&\quad\mbox{if }i=j,\\\
-\sqrt{\rho_{i}\rho_{j}}/\widetilde{D}_{ij}(\bm{\rho})&\quad\mbox{if }i\neq
j.\end{array}\right.$

Therefore, the weak-strong uniqueness holds if $H(\bm{c})$ satisfies the
assumptions of Theorem 3.

Recalling that $H(\bm{\rho})=\sum_{i=1}^{n}\int_{\Omega}h_{i}(\rho_{i})dx$, we
may formulate the entropy in terms of the concentrations $\bm{c}$ as
$\eta(\bm{c})=\sum_{i=1}^{n}\int_{\Omega}\eta_{i}(c_{i})dx,$ where
$\eta_{i}(c_{i})=h_{i}(\rho_{i}/M_{i})$. Then we can rewrite the second
equation in (77) as

$-\sum_{j=1}^{n}\frac{c_{i}c_{j}}{c^{2}D_{ij}}(u_{i}-u_{j})=c_{i}\nabla\mu_{i}-c_{i}M_{i}\sum_{j=1}^{n}c_{j}\nabla\mu_{j},$

where $\mu_{i}=\eta_{i}^{\prime}(c_{i})$ is the molar-based chemical
potential. Using the Gibbs–Duhem equation $\nabla
p=\sum_{i=1}^{n}c_{j}\nabla\mu_{j}$, where $p$ is the pressure, the above
equation can be put into the form

$-\sum_{j=1}^{n}\frac{c_{i}c_{j}}{c^{2}D_{ij}}(u_{i}-u_{j})=c_{i}\nabla\mu_{i}-\rho_{i}\nabla
p,$

which is [5, Formula (203)]. Yet another formulation in terms of the molar
fractions $X_{i}=c_{i}/c$ is

$-\sum_{j=1}^{n}\frac{c_{i}c_{j}}{c^{2}D_{ij}}(u_{i}-u_{j})=c_{i}\nabla_{p}\widetilde{\mu}_{i}+(\phi_{i}-\rho_{i})\nabla
p,$

where $\widetilde{\mu}_{i}$ is given by
$\widetilde{\mu}_{i}(p,X_{1},\ldots,X_{n})=\mu_{i}(c_{i})$,
$\nabla_{p}\widetilde{\mu}_{i}:=\sum_{j=1}^{n}(\partial\widetilde{\mu}_{i}/\partial
X_{j})\nabla X_{j}$, and $\phi_{i}:=\partial\widetilde{\mu}_{i}/\partial p$ is
the volume fraction.

A simple choice is the entropy

$\eta_{i}(c_{i})=c_{i}\log c_{i}-c_{i},\quad i=1,\ldots,n,$

corresponding to $h_{i}(\rho_{i})=(\rho_{i}/M_{i})(\log(\rho_{i}/M_{i})-1)$.
It leads to $\mu_{i}=\log c_{i}$, $p=c$, and the model

$-\sum_{j=1}^{n}\frac{c_{i}c_{j}}{c^{2}D_{ij}}(u_{i}-u_{j})=\nabla
c_{i}-\rho_{i}\nabla c.$

The existence of local strong solutions to this model can be proved as in [4],
while the existence of global weak solutions was shown in [9].

##  Appendix A The Bott–Duffin inverse

For the convenience of the reader, we recall the definition and some
properties of the Bott–Duffin inverse. Let $A\in{\mathbb{R}}^{n\times n}$ be
an arbitrary matrix and $L\subset{\mathbb{R}}^{n}$ be a subspace. The
Bott–Duffin inverse is introduced in connection to the solution of the
constrained inversion problem (see [7], [2, Ch 2.10])

(78) $Ax+y=b,\quad x\in L,\quad y\in L^{\perp}.$

Let $P_{L}$ and $P_{L^{\perp}}$ be the projection operators onto $L$ and
$L^{\perp}$, respectively. The set of solutions of (78) is the same as the set
of solutions to $(AP_{L}+P_{L^{\perp}})z=b$, and $(x,y)$ solves (78) if and
only if $x=P_{L}z$ and $y=P_{L^{\perp}}z=b-AP_{L}z$. Then, if the matrix
$AP_{L}+P_{L^{\perp}}$ is invertible, we define the Bott–Duffin inverse of $A$
with respect to $L$ by

(79) $A^{BD}:=P_{L}(AP_{L}+P_{L^{\perp}})^{-1}.$

and the solution to (78) is expressed in the form

(80) $x=A^{BD}b\,,\quad y=b-Ax.$

If $L=\operatorname{ran}(A)$ and $A$ is symmetric, the Bott–Duffin inverse is
the same as the group inverse, which was investigated in the context of
Maxwell–Stefan systems in [6].

Let $A$ be symmetric. We call $A$ $L$-positive definite if
$\bm{z}^{T}A\bm{z}>0$ for all $\bm{z}\in L\setminus\\{\bm{0}\\}$. For this
class of matrices, a generalized Bott–Duffin inverse is defined in [35], which
coincides with the classical Bott–Duffin inverse when $AP_{L}+P_{L^{\perp}}$
is invertible. The following result is proved in [35, Lemma 2c and 1b].

######  Lemma 16.

Let $A$ be symmetric and $L$-positive definite. Then

$\displaystyle{\rm(i)}$ $\displaystyle\ A^{BD}P_{L^{\perp}}=0,$
$\displaystyle{\rm(ii)}$ $\displaystyle\
\operatorname{ran}(AP_{L}+P_{L^{\perp}})=P_{L}\operatorname{ran}(A)\otimes
L^{\perp},\quad\ker(AP_{L}+P_{L^{\perp}})=\ker(AP_{L})\cap L.$

It follows from property (i) that $A^{BD}$ can be formulated as

(81)
$A^{BD}=P_{L}(AP_{L}+P_{L^{\perp}})^{-1}(P_{L}+P_{L^{\perp}})=P_{L}(AP_{L}+P_{L^{\perp}})^{-1}P_{L}=A^{BD}P_{L}.$

######  Lemma 17.

Let $A$ be symmetric and $L=\operatorname{ran}A$, $L^{\perp}=\ker A$. Then
$AP_{L}=A$, $P_{L}A=A$, $A^{BD}$ is well defined and symmetric.

###### Proof.

The identities $AP_{L}=A$ and $P_{L}A=A$ follow immediately from
$L=\operatorname{ran}A$, We infer from property (ii) that

$\ker(AP_{L}+P_{L^{\perp}})=\ker(AP_{L})\cap L=\ker(A)\cap L=L^{\perp}\cap
L=\\{0\\},$

showing that $AP_{L}+P_{L^{\perp}}$ is invertible. The matrix
$AP_{L}=P_{L}AP_{L}$ is symmetric, since $P_{L}$ and $A$ are symmetric. Also
$P_{L^{\perp}}$ is symmetric, so $AP_{L}+P_{L^{\perp}}$ and its inverse are
symmetric too. Taking into account (81), this implies that
$A^{BD}=P_{L}(AP_{L}+P_{L^{\perp}})^{-1}P_{L}$ is also symmetric. ∎

In our context, we are interested in the constrained inversion

$Ax=b,\quad x\in L,$

where $A$ is a symmetric positive semidefinite matrix, with
$L=\operatorname{ran}(A)$ and thus $L^{\perp}=\ker(A)$, and $b\in L$. Lemma 17
implies that $AP_{L}+P_{L^{\perp}}$ is invertible and $A^{BD}$ is well defined
by (79). Because of (80), we can express the inverse as $x=A^{BD}b$ if $b\in
L$.

##  Appendix B Pointwise estimates for entropy functions

For the convenience of the reader, we recall the following lower bounds.

######  Lemma 18.

The following estimates hold for any $c$, $\bar{c}\in[0,1]$:

$c\log\frac{c}{\bar{c}}-(c-\bar{c})\geq\frac{1}{2}(c-\bar{c})^{2},\quad
c\log\frac{c}{\bar{c}}-(c-\bar{c})\geq(\sqrt{c}-\sqrt{\bar{c}})^{2}.$

###### Proof.

Let $f(c)=c\log c$. Then

$f(c)-f(\bar{c})=f(\theta(c-\bar{c})+\bar{c})\big{|}_{\theta=0}^{1}=(c-\bar{c})\int_{0}^{1}f^{\prime}(\theta(c-\bar{c})+\bar{c})d\theta$

and

$\displaystyle c\log$
$\displaystyle\frac{c}{\bar{c}}-(c-\bar{c})=f(c)-f(\bar{c})-f^{\prime}(\bar{c})(c-\bar{c})$
$\displaystyle=(c-\bar{c})\int_{0}^{1}\big{(}f^{\prime}(\theta(c-\bar{c})+\bar{c})-f^{\prime}(\bar{c})\big{)}d\theta$
$\displaystyle=(c-\bar{c})\int_{0}^{1}f^{\prime}(s(c-\bar{c})+\bar{c})\big{|}_{s=0}^{\theta}d\theta=(c-\bar{c})^{2}\int_{0}^{1}\int_{0}^{\theta}f^{\prime\prime}(s(c-\bar{c})+\bar{c})dsd\theta.$

The first inequality follows after observing that
$f^{\prime\prime}(s(c-\bar{c})+\bar{c})=1/(s(c-\bar{c})+\bar{c})\geq 1$.

For the second inequality, we define $g(c)=(c\log c-c+1)/(\sqrt{c}-1)^{2}$ for
$c\neq 1$ and $g(1)=2$. Then $g$ is continuous and increasing, which implies
that $g(c)\geq g(0)=1$ and proves the statement. ∎

######  Lemma 19.

Let $\bm{c},\bar{\bm{c}}\in\mathbb{R}_{+}^{n}$ satisfy $0\leq c_{i}\leq 1$,
$m\leq\bar{c}_{i}\leq 1$, for $i=1,\ldots,n$, and suppose that $h_{i}\in
C([0,1])\cap C^{2}((0,1])$ satisfies

$h_{i}^{\prime\prime}(c_{i})>0\quad\mbox{for }0<c_{i}\leq 1.$

Then, for some $\kappa_{m}>0$,

(82)
$h_{i}(c_{i}|\bar{c}_{i})=h_{i}(c_{i})-h_{i}(\bar{c}_{i})-h_{i}^{\prime}(\bar{c}_{i})(c_{i}-\bar{c}_{i})\geq\kappa_{m}(c_{i}-\bar{c}_{i})^{2}.$

###### Proof.

By Taylor expansion, the relative entropy density satisfies

$\lim_{c_{i}\to\bar{c}_{i}}\frac{h_{i}(c_{i}|\bar{c}_{i})}{(c_{i}-\bar{c}_{i})^{2}}=\lim_{c_{i}\to\bar{c}_{i}}\int_{0}^{1}\int_{0}^{\theta}h_{i}^{\prime\prime}(s(c_{i}-\bar{c}_{i})+\bar{c}_{i})dsd\theta=\frac{1}{2}h_{i}^{\prime\prime}(\bar{c}_{i})>0.$

Therefore, $h_{i}(c_{i}|\bar{c}_{i})/(c_{i}-\bar{c}_{i})^{2}$ is a continuous
function with a positive minimum:

$\kappa_{m}:=\min_{i=1,\ldots,n}\min_{c_{i}\in[0,1],\,\bar{c}_{i}\in[m,1]}\frac{h_{i}(c_{i}|\bar{c}_{i})}{(c_{i}-\bar{c}_{i})^{2}}>0.$

This shows that
$h_{i}(c_{i}|\bar{c}_{i})\geq\kappa_{m}(c_{i}-\bar{c}_{i})^{2}$ for
$c_{i}\in[0,1]$, $\bar{c}_{i}\in[m,1]$ and proves (82). ∎

##  Appendix C Thermodynamic derivation of the generalized Maxwell–Stefan
system

The aim of this section is to derive (16)–(17) from elementary thermodynamic
principles. We assume that the evolution of the gaseous mixture is given by
the conservation of mass and energy (without chemical reactions),

$\displaystyle\partial_{t}(\rho c_{i})+\operatorname{div}(\rho c_{i}v+J_{i})$
$\displaystyle=0,$ (83) $\displaystyle\partial_{t}(\rho
U)+\operatorname{div}(\rho Uv+q)$ $\displaystyle=0,$
$\displaystyle\partial_{t}\rho+\operatorname{div}(\rho v)$
$\displaystyle=0,\quad i=1,\ldots,n,$

where $\rho_{i}$ is the partial density of the $i$th species,
$\rho=\sum_{i=1}^{n}\rho_{i}$ the total density, $c_{i}=\rho_{i}/\rho$ the
concentration of the $i$th species, $v$ the barycentric velocity, $J_{i}$ the
$i$th flux, $q$ the heat flux, and the internal energy $U$ is given by the
first law of thermodynamics in differential form by

(84) $dU=TdS-pdV+\sum_{i=1}^{n}\mu_{i}dc_{i},$

where $S$ is the entropy, $V=1/\rho$ the volume, and $\mu_{i}=\partial
U/\partial c_{i}$ the $i$th chemical potential. By definition, it holds that
$\sum_{i=1}^{n}c_{i}=1$. Adding the first and last equation in (83), we see
that $\operatorname{div}\sum_{i=1}^{n}J_{i}=0$, which motivates us to assume
that $\sum_{i=1}^{n}J_{i}=0$.

The sum of the fluxes should vanish, $\sum_{i=1}^{n}J_{i}=0$, to be consistent
with the conservation laws.

With the material derivative $D_{t}f=\partial_{t}f+v\cdot\nabla f$, the
conservation laws can be simplified to

$\rho D_{t}c_{i}+\operatorname{div}J_{i}=0,\quad\rho
D_{t}U+\operatorname{div}q=0,\quad D_{t}\rho+\rho\operatorname{div}v=0.$

Inserting these equations into equation (84), formulated as
$D_{t}U=TD_{t}S-pD_{t}V+\sum_{i=1}^{n}\mu_{i}D_{t}c_{i}$, yields the entropy
balance

$\displaystyle\rho D_{t}S$
$\displaystyle=\frac{\rho}{T}D_{t}U+\frac{\rho}{T}pD_{t}\bigg{(}\frac{1}{\rho}\bigg{)}-\sum_{i=1}^{n}\frac{\mu_{i}}{T}D_{t}c_{i}$
$\displaystyle=-\frac{1}{T}\operatorname{div}q+\frac{p}{T}\operatorname{div}v+\sum_{i=1}^{n}\frac{\mu_{i}}{T}\operatorname{div}J_{i}=-\operatorname{div}J_{S}+r_{S},$

where

$J_{S}=\frac{q}{T}-\sum_{i=1}^{n}\frac{\mu_{i}}{T}J_{i},\quad
r_{S}=q\cdot\nabla\frac{1}{T}+p\operatorname{div}v+\sum_{i=1}^{n}J_{i}\cdot\nabla\frac{\mu_{i}}{T}$

are the entropy flux and entropy production, respectively.

In our Maxwell–Stefan model, we assume that $v=0$ and $T=1$. Then the entropy
production simplifies to $r_{S}=\sum_{i=1}^{n}J_{i}\cdot\nabla\mu_{i}$. It can
be reformulated by taking into account that $\sum_{i=1}^{n}J_{i}=0$ and hence
$J_{i}/\sqrt{c_{i}}\in
L=\\{\bm{x}\in{\mathbb{R}}^{n}:\sqrt{\bm{c}}\cdot\bm{x}=0\\}$:

$r_{S}=-\sum_{i=1}^{n}\frac{J_{i}}{\sqrt{c_{i}}}\cdot\sqrt{c_{i}}\nabla\mu_{i}=-\sum_{i,j=1}^{n}\frac{J_{i}}{\sqrt{c_{i}}}\cdot(P_{L})_{ij}\sqrt{c_{j}}\nabla\mu_{j}=-\sum_{i=1}^{n}\frac{J_{i}}{\sqrt{c_{i}}}\cdot\sum_{j=1}^{n}(P_{L})_{ij}\sqrt{c_{j}}\nabla\mu_{j},$

where the projection $P_{L}$ on $L$ is defined in (22). By the second law of
thermodynamics, it should hold that $r_{S}\geq 0$. To guarantee this property,
we introduce a positive semidefinite matrix $B(\bm{c})$ such that

(85)
$\sum_{j=1}^{n}(P_{L})_{ij}\sqrt{c_{j}}\nabla\mu_{j}=-\sum_{j=1}^{n}B_{ij}(\bm{c})\frac{J_{j}}{\sqrt{c_{j}}},\quad
i=1,\ldots,n.$

We claim that these equations correspond to the generalized Maxwell–Stefan
equations (17) after setting $J_{i}=c_{i}u_{i}$ and
$K_{ij}(\bm{c})=\sqrt{c_{i}}B_{ij}(\bm{c})/\sqrt{c_{j}}$ (see (18)). Indeed,
the left-hand side of (85), multiplied by $\sqrt{c_{i}}$, becomes

$\sqrt{c_{i}}\sum_{j=1}^{n}(P_{L})_{ij}\sqrt{c_{j}}\nabla\mu_{j}=c_{i}\nabla\mu_{i}-c_{i}\sum_{j=1}^{n}c_{j}\nabla\mu_{j},$

and the right-hand side of (85), multiplied by $\sqrt{c_{i}}$, equals

$-\sqrt{c_{i}}\sum_{j=1}^{n}B_{ij}(\bm{c})\frac{J_{j}}{\sqrt{c_{j}}}=-\sum_{j=1}^{n}K_{ij}(\bm{c})J_{j}=-\sum_{j=1}^{n}K_{ij}(\bm{c})c_{j}u_{j}.$

Hence, observing that $\mu_{j}=\partial U/\partial c_{j}$ corresponds to
$\delta H/\delta c_{j}$, (85) equals (17).

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