the space of Radon measures on $\overline{Q}$, but not in any Lebesgue
function space. We need to use careful arguments to prove that the sequence of
approximations converge in some sense to a Lebesgue function. Then, we need to
prove that the convergence is sufficiently strong so that we can show that the
elastodynamic equation and the constitutive relation both hold in the limit.

To do this, we require additional regularity of the phase-field function $v$.
To get improved spatial regularity estimates on the approximations of
$\mathbf{T}$, we need $\nabla v$ and its approximations to be uniformly
bounded on $Q$. From the construction of the problem, we expect the gradients
to have absolute value to the order of $\epsilon^{-1}$. Indeed, the parameter
$\epsilon$ gives some sense of the ‘thickness’ of the approximation of the
crack set. However, proving a uniform bound in practice does not seem
possible. Hence, in the spirit of [12], we introduce an extra term in the
minimisation problem (2.2) in order to guarantee this $L^{\infty}(Q)^{d}$
regularity of $\nabla v$. The new term imposes a dependence of the problem on
the speed at which the crack grows. It is referred to as a rate-dependent term
because of the presence of the time derivative $v_{t}$.

In the strain-limiting setting, we have an added issue in the case that the
Neumann part of the boundary is non-empty. As in the steady problem without
fracture [3], we experience a penalisation on this part of the boundary in the
weak elastodynamic equation due to a lack of global integrability of
approximations of the stress tensor. Indeed, we prove that the approximations
of the stress tensor converge in
$L^{\infty}(0,T;\mathcal{M}(\overline{\Omega})^{d\times d})$ and pointwise
a.e. on $Q$ to limits $\overline{\mathbf{T}}$ and $\mathbf{T}$, respectively.
Due to the choice of test function, any singularities on the Dirichlet part of
the boundary do not appear, because the test functions vanish on the whole of
$\Gamma_{D}$. However, we cannot control any difference between
$\overline{\mathbf{T}}$ and $\mathbf{T}$ on the Neumann part of the boundary,
because the test functions do not vanish there. This is the reason for the
presence of the penalty term on $\Gamma_{N}$. This is also the reason for the
existence result being stronger in the case of fully Dirichlet boundary
conditions; it suffices to obtain local convergence with respect to the
spatial domain $\Omega$ in order to prove the required results. As a result of
the error term in the mixed boundary condition case, we not only have a
penalisation in the elastodynamic equation, but we are unable to deduce an
energy-dissipation equality. Even if we assume a priori that the error term
vanishes, the convergence results are not sufficiently strong to show that
equality holds.

The problem of interest in this section is the following. For a fixed
$k\in\mathbb{N}$, we look for a triple
$(\boldsymbol{u},\mathbf{T},v):Q\rightarrow\mathbb{R}^{d}\times\mathbb{R}^{d\times
d}\times\mathbb{R}$ that solves

(3.1) $\displaystyle\boldsymbol{u}_{tt}$
$\displaystyle=\mathrm{div}\big{(}b(v)\mathbf{T}\big{)}+\boldsymbol{f}$
$\displaystyle\text{ in }Q,$
$\displaystyle\boldsymbol{\varepsilon}(\boldsymbol{u}_{t}+\alpha\boldsymbol{u})$
$\displaystyle=F(\mathbf{T})=\frac{\mathbf{T}}{(1+|\mathbf{T}|^{a})^{\frac{1}{a}}}$
$\displaystyle\text{ in }Q,$ $\displaystyle\boldsymbol{u}\big{|}_{\Gamma_{D}}$
$\displaystyle=0,\,v\big{|}_{\Gamma_{D}}=1$ $\displaystyle\text{ on
}[0,T]\times\Gamma_{D},$ $\displaystyle b(v)\mathbf{T}\mathbf{n}$
$\displaystyle=\boldsymbol{g}$ $\displaystyle\text{ on
}[0,T]\times\Gamma_{N},$ $\displaystyle\boldsymbol{u}(0,\cdot)$
$\displaystyle=\boldsymbol{u}_{0},\boldsymbol{u}_{t}(0,\cdot)=\boldsymbol{u}_{1},\,v(0,\cdot)=v_{0}$
$\displaystyle\text{ in }\Omega,$

subject to the minimisation problem

(3.2)
$\mathcal{E}(\boldsymbol{u}(t),v(t))+\mathcal{H}(v(t))+\mathcal{G}_{k}(v(t),v_{t}(t))=\inf_{v\in
H^{k}_{D+1}(\Omega),\,v\leq
v(t)}\Big{\\{}\mathcal{E}(\boldsymbol{u}(t),v)+\mathcal{H}(v)+\mathcal{G}_{k}(v,v_{t}(t))\Big{\\}},$

with energy-dissipation equality

(3.3)
$\displaystyle\mathcal{F}(t;\boldsymbol{u}(t),\boldsymbol{u}_{t}(t),v(t))+\int_{0}^{t}\int_{\Omega}b(v)\big{(}\mathbf{T}-F^{-1}(\boldsymbol{\varepsilon}(\alpha\boldsymbol{u}))\big{]}:\boldsymbol{\varepsilon}(\boldsymbol{u}_{t})+\boldsymbol{f}_{t}\cdot\boldsymbol{u}\,\mathrm{d}x\,\mathrm{d}s$
$\displaystyle\quad+\int_{0}^{t}\int_{\Gamma_{N}}\boldsymbol{g}_{t}\cdot\boldsymbol{u}\,\mathrm{d}S\,\mathrm{d}s+\int_{0}^{t}\|v_{t}(s)\|_{k,2}\,\mathrm{d}s$
$\displaystyle=\mathcal{F}(0;\boldsymbol{u}_{0},\boldsymbol{u}_{1},v_{0}),$

for every $t\in[0,T]$ and the crack non-healing property $v_{t}\leq 0$. All
functionals are defined as in Section 2 but with $F$ given by (3.1). The
functional $\mathcal{G}_{k}$ is defined on $H^{k}(\Omega)\times H^{k}(\Omega)$
by

$\displaystyle\mathcal{G}_{k}(v,w)=(v,w)_{k,2}:=\sum_{|\alpha|\leq
k}\int_{\Omega}\partial^{\alpha}v\cdot\partial^{\alpha}w\,\mathrm{d}x.$

Due to the presence of the rate-dependent term, we cannot guarantee that
$v\geq 0$. For a discussion regarding the physical implications of the
inclusion of this functional on the model, we refer to [12].

To ensure that the nonlinearity $b(v)$ is still uniformly bounded from above,
in this section we define $b$ by

$\displaystyle b(v)=\max\\{0,v\\}^{2}+\eta.$

We later see that the phase-field function and its approximations are
uniformly bounded in the function space $L^{\infty}(0,T;H^{k}(\Omega)^{d})$.
Hence, if $k>\frac{d}{2}$, by the Sobolev embedding theorem, we deduce that
the functions are uniformly bounded on $Q$. As a result, in this case, which
is a standing assumption of the result proven here, the proof is actually
valid for the choice of $b$ that is used in Section 2.

We do not require a compatibility condition of type (2.4) because $v_{t}(0)$
is not defined a priori. Indeed, the minimisation problem does not make sense
at the initial time. However, we require a compatibility condition between the
Neumann boundary data and the initial data. We demand that

(3.4)
$\boldsymbol{g}(0)=b(v_{0})F^{-1}(\boldsymbol{\varepsilon}(\boldsymbol{u}_{1}+\alpha\boldsymbol{u}_{0}))\mathbf{n}\quad\text{
on }\Gamma_{N}.$

This is vital in order to bound approximations of $\boldsymbol{u}_{tt}(0)$
which is needed to obtain higher regularity estimates in the temporal variable
of the approximations of the stress tensor. For (3.4) to make sense for the
given initial data, we also demand the safety strain condition, that is, we
have

(3.5)
$\max\big{\\{}\|\boldsymbol{\varepsilon}(\alpha\boldsymbol{u}_{0})\|_{\infty},\,\|\boldsymbol{\varepsilon}(\boldsymbol{u}_{1}+\alpha\boldsymbol{u}_{0})\|_{\infty}\big{\\}}=C_{*}<1.$

Such a condition is discussed in detail in [3]. To ensure that the total
energy functional is finite at the initial time, we require a bound on
$\boldsymbol{\varepsilon}(\alpha\boldsymbol{u}_{0})$, as well as
$\boldsymbol{\varepsilon}(\boldsymbol{u}_{1}+\alpha\boldsymbol{u}_{0})$ in the
safety strain condition. The bound on
$\boldsymbol{\varepsilon}(\alpha\boldsymbol{u}_{0})$ is non-standard in the
literature. However, the function $\varphi^{*}$ is not necessarily finite
outside the open unit ball, which is why we include the extra assumption.

As with the analysis of time-dependent, strain-limiting problems without a
phase field function (see, for example, the analysis in [9] and [8]), we
approximate (3.1)–(3.3) by a regularised problem. The function $F$ is replaced
by a function $F_{n}$ that contains an elliptic term. As a result, $F_{n}$ is
a bijection from $\mathbb{R}^{d\times d}$ to itself and so the problem can be
written in terms of only $\boldsymbol{u}$ and $v$. We define $F_{n}$ by

$\displaystyle
F_{n}(\mathbf{T})=\frac{\mathbf{T}}{(1+|\mathbf{T}|^{a})^{\frac{1}{a}}}+\frac{\mathbf{T}}{n}.$

with $\varphi_{n}$, $\varphi^{*}_{n}$ and $\mathcal{E}_{n}$ defined
analogously. We approximate the body forces $l$ in the regularised problem to
ensure that a suitable compatibility condition analogous to (3.4) holds in the
setting of the regularised problem. We replace the function $\boldsymbol{g}$
by an approximation $\boldsymbol{g}^{n}$ that we define by

(3.6)
$\boldsymbol{g}^{n}=F^{-1}_{n}(\boldsymbol{\varepsilon}(\boldsymbol{u}_{1}+\alpha\boldsymbol{u}_{0}))\mathbf{n}\cdot\psi(nt)+\boldsymbol{g}(t,x)\cdot(1-\psi(nt)),$

where $\psi\in C^{\infty}_{c}([0,\infty))$ is a non-increasing function with
$\psi=1$ on $[0,\frac{1}{2}]$ and $\psi=0$ on $[1,\infty)$. The external force
approximation is defined by $l^{n}:=\boldsymbol{f}+\boldsymbol{g}^{n}$. We
define the total energy $\mathcal{F}_{n}$ by replacing $\mathcal{E}$ with
$\mathcal{E}_{n}$ and $l$ with $l^{n}$ in the definition of the functional
$\mathcal{F}$. Under the assumption that $l\in
W^{2,1}([0,T];W^{-1,2}_{D}(\Omega)^{d})$, the sequence $(l^{n})_{n}$ is
bounded in this space independent of $n$ and converges weakly to $l$ as
$n\rightarrow\infty$.

######  Remark.

As mentioned in Section 1, the results in this section can be generalised to a
more generalised class of functions $F$. Indeed, the required assumptions on
$F$ are as follows. There exist positive constants $c_{0}$, $c_{1}$ and $a$
such that, for every $\mathbf{T}$, $\mathbf{S}\in\mathbb{R}^{d\times d}$,

$\displaystyle(F(\mathbf{T})-F(\mathbf{S})):(\mathbf{T}-\mathbf{S})$
$\displaystyle\geq 0,$ $\displaystyle F(\mathbf{T})\cdot\mathbf{T}$
$\displaystyle\geq c_{0}|\mathbf{T}|-c_{1},$ $\displaystyle|F(\mathbf{T})|$
$\displaystyle\leq c_{1},$

that is, monotonicity, coercivity and boundedness, respectively. Furthermore,
we assume that there exists a strictly convex function $\varphi\in
C^{2}(\mathbb{R}_{+};\mathbb{R}_{+})$ such that
$\varphi(0)=\varphi^{\prime}(0)=0$ and $|\varphi^{\prime\prime}(s)|\leq
c_{1}(1+s)^{-1}$ for every $s\in\mathbb{R}_{+}$, with $\varphi$ being related
to $F$ by the representation

$F(\mathbf{T})=\frac{\varphi^{\prime}(|\mathbf{T}|)}{|\mathbf{T}|}\mathbf{T}.$

Finally, if we define the inner product
$(\cdot,\cdot)_{\mathcal{A}(\mathbf{T})}$, for
$\mathbf{T}\in\mathbb{R}^{d\times d}$, by

$\displaystyle(\mathbf{S},\mathbf{U})_{\mathcal{A}(\mathbf{T})}=\sum_{i,j,k,l=1}^{d}\frac{\partial
F_{ij}(\mathbf{T})}{\partial\mathbf{T}_{kl}}\mathbf{S}_{ij}\mathbf{U}_{kl},$

then we require the lower bound

$\displaystyle(\mathbf{S},\mathbf{S})_{\mathcal{A}(\mathbf{T})}\geq\frac{c_{0}|\mathbf{S}|^{2}}{(1+|\mathbf{T}|)^{1+a}},$

for every $\mathbf{T}$, $\mathbf{S}\in\mathbb{R}^{d\times d}$.

The structure of this section is as follows. First, we focus on proving the
existence of a weak energy solution to the regularised problem, using a
discretisation in time similar to that which is used in Section 2. Then we
look for $n$-independent bounds on the sequence of solutions, particularly
those which involve higher regularity estimates. At best, we have a bound on
$(\mathbf{T}^{n})_{n}$ in $L^{\infty}(0,T;L^{1}(\Omega)^{d\times d})$, a
function space that does not have nice compactness properties. Hence, to
obtain a pointwise convergence result for $(\mathbf{T}^{n})_{n}$, we use
weighted regularity estimates. Using this pointwise convergence, if
$\Gamma_{N}=\emptyset$, we can then improve the convergence to strong
convergence in $L^{1}(0,T;L^{1}_{loc}(\Omega)^{d\times d})$, making use of the
monotonicity of $F$. If $a\in(0,\frac{2}{d})$, this is further improved to
strong convergence in $L^{1+\delta}(0,T;L^{1+\delta}_{loc}(\Omega)^{d\times
d})$ for every $\delta>0$ such that $a+\delta<\frac{2}{d}$.

We remark that we do not use a Galerkin approximation in space to deduce the
existence of solutions to the regularised problem. This is due to the choice
of regularisation term, which ensures that we are in the Hilbert space
setting. This setting ensures that appropriate bounds are available which
allow us to use the Browder–Minty theorem to prove existence of solutions to
the time discrete problem, without also applying a Galerkin approximation.
This is not the case when the constitutive function has arbitrary $p$-growth,
as in Section 2.

Let us now introduce the time-discrete approximation of the regularised
problem. Although the existence of a solution to the regularised problem
follows from the proof in Section 2, we need the structure of the time
discrete problem to obtain higher time regularity estimates. In particular, we
need to overcome the fact that the second time derivative
$\boldsymbol{u}^{n}_{tt}$ is not necessarily continuous in the time variable.
We do this by working with the time discrete problem and ensuring that
$\delta^{2}\boldsymbol{u}^{\gamma}_{0}$ is suitably defined.

######  Theorem 3.1.

Suppose that $\boldsymbol{u}_{0}$, $\boldsymbol{u}_{1}\in
W^{1,2}_{D}(\Omega)^{d}$ and $l\in C^{1}([0,T];W^{-1,2}_{D}(\Omega)^{d})$. Let
$l_{n}$ be defined by (3.6). Fix $M$ and $n\in\mathbb{N}$, let $\gamma=(n,M)$
and define a time step $h=\frac{T}{M}$. We define initialisations of the time
discrete problem by

$\displaystyle\boldsymbol{u}^{\gamma}_{0}=\boldsymbol{u}_{0},\quad\boldsymbol{u}^{\gamma}_{-1}=\boldsymbol{u}_{0}-h\boldsymbol{u}_{1},\quad
v^{\gamma}_{0}=v_{0}.$

Furthermore, we denote $l^{\gamma}_{m}=l(t^{M}_{m})$ where $t^{M}_{m}=mh$ for
$0\leq m\leq M$. Defining the solution sequence recursively, for every $1\leq
m\leq M$ there exists a unique $\boldsymbol{u}^{\gamma}_{m}\in
W^{1,2}_{D}(\Omega)^{d}$ such that

(3.7)
$\int_{\Omega}\delta^{2}\boldsymbol{u}^{\gamma}_{m}\cdot\boldsymbol{w}+b(v^{\gamma}_{m-1})F^{-1}_{n}(\boldsymbol{\varepsilon}(\delta\boldsymbol{u}^{\gamma}_{m}+\alpha\boldsymbol{u}^{\gamma}_{m})):\boldsymbol{\varepsilon}(\boldsymbol{w})\,\mathrm{d}x=\langle
l^{\gamma}_{m},\boldsymbol{w}\rangle,$

for every $\boldsymbol{w}\in W^{1,2}_{D}(\Omega)^{d}$, and a unique
$v^{\gamma}_{m}\in H^{k}_{D+1}(\Omega)$ such that

(3.8)
$\displaystyle\mathcal{E}_{n}(\boldsymbol{u}^{\gamma}_{m},v^{\gamma}_{m})+\mathcal{H}(v^{\gamma}_{m})+\frac{1}{2h}\mathcal{G}_{k}(v^{\gamma}_{m}-v^{\gamma}_{m-1},v^{\gamma}_{m}-v^{\gamma}_{m-1})$
$\displaystyle=\inf\Big{\\{}\mathcal{E}_{n}(\boldsymbol{u}^{\gamma}_{m},v)+\mathcal{H}(v)+\frac{1}{2h}\mathcal{G}_{k}(v-v^{\gamma}_{m-1},v-v^{\gamma}_{m-1})\,:\,v\in
H^{k}_{D+1}(\Omega),\,,v\leq v^{\gamma}_{m-1}\Big{\\}}.$

The proof follows identical reasoning to that of Theorem 2.3 so is not
repeated here. Similarly, reasoning as with Proposition 2.2, Corollary 2.1 and
Proposition 2.3, we deduce the following four results concerning the
minimisation problem and energy-dissipation inequality.

######  Lemma 3.1.

Let the assumptions of Theorem 3.1 hold and let
$(\boldsymbol{u}^{\gamma}_{m},v^{\gamma}_{m})_{m=1}^{M}$ be the solution of
the time discrete problem. For every $1\leq m\leq M$ and every
$\tilde{\chi}\in H^{k}_{D+1}(\Omega)$ with $\tilde{\chi}\leq
v^{\gamma}_{m-1}$, we have

$\displaystyle 0$
$\displaystyle\leq\big{[}\partial_{v}\mathcal{E}_{n}(\boldsymbol{u}^{\gamma}_{m},v^{\gamma}_{m})+\mathcal{H}^{\prime}(v^{\gamma}_{m})+\frac{1}{2h}\partial_{v}\mathcal{G}_{k}(v^{\gamma}_{m}-v^{\gamma}_{m-1},v^{\gamma}_{m}-v^{\gamma}_{m-1})\big{]}(\tilde{\chi}-v^{\gamma}_{m-1})$
$\displaystyle=\int_{\Omega}\Big{\\{}\frac{b^{\prime}(v^{\gamma}_{m})}{\alpha}(\tilde{\chi}-v^{\gamma}_{m-1})\varphi^{*}_{n}(\boldsymbol{\varepsilon}(\alpha\boldsymbol{u}^{\gamma}_{m}))+\frac{1}{2\epsilon}(v^{\gamma}_{m}-1)(\tilde{\chi}-v^{\gamma}_{m-1})$
$\displaystyle\quad+2\epsilon\nabla
v^{\gamma}_{m}\cdot\nabla(\tilde{\chi}-v^{\gamma}_{m-1})\Big{\\}}\,\mathrm{d}x+\frac{1}{h}\big{(}v^{\gamma}_{m}-v^{\gamma}_{m-1},\tilde{\chi}-v^{\gamma}_{m-1}\big{)}_{k,2}.$

######  Corollary 3.1.

For every $1\leq m\leq M$ and every $\chi\in H^{k}_{D}(\Omega)$ with $\chi\leq
0$, we have

$\displaystyle
0\leq\big{[}\partial_{v}\mathcal{E}_{n}(\boldsymbol{u}^{\gamma}_{m},v^{\gamma}_{m})+\mathcal{H}^{\prime}(v^{\gamma}_{m})+\frac{1}{2h}\partial_{v}\mathcal{G}_{k}(v^{\gamma}_{m}-v^{\gamma}_{m-1},v^{\gamma}_{m}-v^{\gamma}_{m-1})\big{]}(\chi).$

######  Corollary 3.2.

For every $1\leq m\leq M$, we have

$\displaystyle 0$
$\displaystyle=\big{[}\partial_{v}\mathcal{E}_{n}(\boldsymbol{u}^{\gamma}_{m},v^{\gamma}_{m})+\mathcal{H}^{\prime}(v^{\gamma}_{m})+\frac{1}{2h}\partial_{v}\mathcal{G}_{k}(v^{\gamma}_{m}-v^{\gamma}_{m-1},v^{\gamma}_{m}-v^{\gamma}_{m-1})\big{]}(\delta
v^{\gamma}_{h})$
$\displaystyle=\int_{\Omega}\Big{\\{}\frac{b^{\prime}(v^{\gamma}_{m})}{\alpha}\delta
v^{\gamma}_{m}\varphi^{*}_{n}(\boldsymbol{\varepsilon}(\alpha\boldsymbol{u}^{\gamma}_{m}))+\frac{1}{2\epsilon}(v^{\gamma}_{m}-1)\delta
v^{\gamma}_{m}+2\epsilon\nabla v^{\gamma}_{m}\cdot\nabla\delta
v^{\gamma}_{m}\Big{\\}}\,\mathrm{d}x$ $\displaystyle\quad+\|\delta
v^{\gamma}_{m}\|_{k,2}^{2}.$

######  Proposition 3.1.

Let the assumptions of Theorem 3.1 hold and let
$(\boldsymbol{u}^{\gamma}_{m},v^{\gamma}_{m})_{m=1}^{M}$ be the solution of
the time discrete problem. For every $1\leq m\leq M$,

$\displaystyle\mathcal{F}_{n}(t^{M}_{m};\boldsymbol{u}^{\gamma}_{m},\delta\boldsymbol{u}^{\gamma}_{m},v^{\gamma}_{m})+h\sum_{j=1}^{m}\|\delta
v^{\gamma}_{j}\|_{k,2}^{2}+h\sum_{j=1}^{m}\langle\delta
l^{\gamma}_{m},\boldsymbol{u}^{\gamma}_{m-1}\rangle$
$\displaystyle\quad+h\sum_{j=1}^{m}\int_{\Omega}b(v^{\gamma}_{j-1})\big{[}F^{-1}_{n}(\boldsymbol{\varepsilon}(\delta\boldsymbol{u}^{\gamma}_{n}+\alpha\boldsymbol{u}^{\gamma}_{j}))-F^{-1}_{n}(\boldsymbol{\varepsilon}(\alpha\boldsymbol{u}^{\gamma}_{j}))\big{]}:\boldsymbol{\varepsilon}(\delta\boldsymbol{u}^{\gamma}_{j})\,\mathrm{d}x$
$\displaystyle\leq\mathcal{F}_{n}(0;\boldsymbol{u}_{0},\boldsymbol{u}_{1},v_{0}).$

Now we concentrate on obtaining various $M$-independent bounds on the discrete
solution sequence. Then we take the limit as $M\rightarrow\infty$ to obtain a
weak energy solution of the regularised problem. The time continuous problem
is more amenable to work with when it comes to higher spatial regularity
estimates, so we cannot derive these estimates until we have taken the limit
as $M\rightarrow\infty$. To this end, using the fact that

$\displaystyle(F^{-1}_{n}(\mathbf{T})-F^{-1}_{n}(\mathbf{S})):(\mathbf{T}-\mathbf{S})\geq
C(n)|\mathbf{T}-\mathbf{S}|^{2},$

for every $\mathbf{T}$, $\mathbf{S}\in\mathbb{R}^{d\times d}$, where $C=C(n)$
is a constant depending only on $n$, we immediately obtain an $M$-independent
bound from the energy-dissipation inequality.

######  Lemma 3.2.

Let the assumptions of Theorem 3.1 hold and let
$(\boldsymbol{u}^{\gamma}_{m},v^{\gamma}_{m})_{m=1}^{M}$ be the solution of
the time discrete problem. There exists a constant $C=C(n)$, independent of
$M$, such that

$\displaystyle\max_{1\leq m\leq
M}\|\boldsymbol{u}^{\gamma}_{m}\|_{1,2}+\max_{1\leq m\leq
M}\|\delta\boldsymbol{u}^{\gamma}_{m}\|_{2}+\max_{1\leq m\leq
M}\|v^{\gamma}_{m}\|_{k,2}+h\sum_{m=1}^{M}\Big{(}\|\delta\boldsymbol{u}^{\gamma}_{m}\|_{1,2}^{2}+\|\delta
v^{\gamma}_{m}\|_{k,2}^{2}\Big{)}\leq C.$

The next bound in Lemma 3.3 gives us information on the second derivative
$\delta^{2}\boldsymbol{u}^{\gamma}_{m}$. As a result, we are able to improve
the bound on $(\delta\boldsymbol{u}^{\gamma}_{m})_{m=1}^{M}$ also. In Lemma
3.4, we improve the bounds on the phase-field function. This gives sufficient
bounds on the corresponding interpolant functions, which we collect in
Corollary 3.3. Then we take the limit as $M\rightarrow\infty$ in Proposition
3.2 and identify the limiting couple as a weak energy solution of the
regularised problem.

######  Lemma 3.3.

Let the assumptions of Theorem 3.1 hold and let
$(\boldsymbol{u}^{\gamma}_{m},v^{\gamma}_{m})_{m=1}^{M}$ be the solution of
the time discrete problem. There exists a constant $C=C(n)$ independent of $M$
such that

$\displaystyle
h\sum_{m=1}^{M}\|\delta^{2}\boldsymbol{u}^{\gamma}_{m}\|_{2}^{2}+\max_{1\leq
m\leq M}\|\delta\boldsymbol{u}^{\gamma}_{m}\|_{1,2}\leq C.$

###### Proof.

Testing in (3.7) against
$h(\delta^{2}\boldsymbol{u}^{\gamma}_{m}+\alpha\delta\boldsymbol{u}^{\gamma}_{m})$,
we rewrite the term involving the nonlinearity in the following way:

(3.9)
$\displaystyle\int_{\Omega}b(v^{\gamma}_{m-1})F^{-1}_{n}(\boldsymbol{\varepsilon}(\delta\boldsymbol{u}^{\gamma}_{m}+\alpha\boldsymbol{u}^{\gamma}_{m})):\boldsymbol{\varepsilon}(h(\delta^{2}\boldsymbol{u}^{\gamma}_{m}+\alpha\delta\boldsymbol{u}^{\gamma}_{m}))\,\mathrm{d}x$
$\displaystyle=\int_{\Omega}b(v^{\gamma}_{m-1})F^{-1}_{n}(\boldsymbol{\varepsilon}(\delta\boldsymbol{u}^{\gamma}_{m}+\alpha\boldsymbol{u}^{\gamma}_{m})):\boldsymbol{\varepsilon}(\delta\boldsymbol{u}^{\gamma}_{m}+\alpha\boldsymbol{u}^{\gamma}_{m})\,\mathrm{d}x$
$\displaystyle\quad-\int_{\Omega}b(v^{\gamma}_{m-1})F^{-1}_{n}(\boldsymbol{\varepsilon}(\delta\boldsymbol{u}^{\gamma}_{m-1}+\alpha\boldsymbol{u}^{\gamma}_{m-1})):\boldsymbol{\varepsilon}(\delta\boldsymbol{u}^{\gamma}_{m-1}+\alpha\boldsymbol{u}^{\gamma}_{m-1})\,\mathrm{d}x$
$\displaystyle\quad-\int_{\Omega}\Big{\\{}b(v^{\gamma}_{m-1})\big{[}F^{-1}_{n}(\boldsymbol{\varepsilon}(\delta\boldsymbol{u}^{\gamma}_{m}+\alpha\boldsymbol{u}^{\gamma}_{m}))-F^{-1}_{n}(\boldsymbol{\varepsilon}(\delta\boldsymbol{u}^{\gamma}_{m-1}+\alpha\boldsymbol{u}^{\gamma}_{m-1}))\big{]}$
$\displaystyle\quad:\boldsymbol{\varepsilon}(\delta\boldsymbol{u}^{\gamma}_{m-1}+\alpha\boldsymbol{u}^{\gamma}_{m-1})\Big{\\}}\,\mathrm{d}x.$

We define the fourth order tensor $\mathcal{B}_{n}(\mathbf{T})$, where
$\mathbf{T}\in\mathbb{R}^{d\times d}_{sym}$, by

$\displaystyle\mathcal{B}_{n}^{ijkl}(\mathbf{T})=\frac{\partial(F^{-1}_{n})_{ij}}{\partial
T_{kl}}(\mathbf{T}).$

This defines an inner product on $\mathbb{R}^{d\times d}_{sym}$ in the
following way:

$\displaystyle(\mathbf{S},\mathbf{U})_{\mathcal{B}_{n}(\mathbf{T})}=\mathcal{B}^{ijkl}_{n}(\mathbf{T})S_{ij}U_{kl}.$

It follows that, for every $\mathbf{T}$, $\mathbf{S}\in\mathbb{R}^{d\times
d}_{sym}$,

(3.10)
$(F^{-1}_{n}(\mathbf{T})-F^{-1}_{n}(\mathbf{S})):\mathbf{T}=\int_{0}^{1}(\mathbf{T},\mathbf{T}-\mathbf{S})_{\mathcal{B}_{n}(s\mathbf{T}+(1-s)\mathbf{S})}\,\mathrm{d}s.$

On the other hand, we also have

$\displaystyle\varphi_{n}(F^{-1}_{n}(\mathbf{T}))-\varphi_{n}(F^{-1}_{n}(\mathbf{S}))$
$\displaystyle=-\int_{0}^{1}(1-s)(\mathbf{T}-\mathbf{S},\mathbf{T}-\mathbf{S})_{\mathcal{B}_{n}(s\mathbf{T}+(1-s)\mathbf{S})}\,\mathrm{d}s+\int_{0}^{1}(\mathbf{T},\mathbf{T}-\mathbf{S})_{\mathcal{B}_{n}(s\mathbf{T}+(1-s)\mathbf{S})}\,\mathrm{d}s.$

Substituting this into (3.10), it follows that

$\displaystyle(F^{-1}_{n}(\mathbf{T})-F^{-1}_{n}(\mathbf{S})):\mathbf{T}$
$\displaystyle=\varphi_{n}(F^{-1}_{n}(\mathbf{T}))-\varphi_{n}(F^{-1}_{n}(\mathbf{S}))+\int_{0}^{1}(1-s)(\mathbf{T}-\mathbf{S},\mathbf{T}-\mathbf{S})_{\mathcal{B}_{n}(s\mathbf{T}+(1-s)\mathbf{S})}\,\mathrm{d}s$
$\displaystyle\geq\varphi_{n}(F^{-1}_{n}(\mathbf{T}))-\varphi_{n}(F^{-1}_{n}(\mathbf{S})).$

Returning to (3.9), we deduce that

(3.11)
$\displaystyle\int_{\Omega}b(v^{\gamma}_{m-1})F^{-1}_{n}(\boldsymbol{\varepsilon}(\delta\boldsymbol{u}^{\gamma}_{m}+\alpha\boldsymbol{u}^{\gamma}_{m})):\boldsymbol{\varepsilon}(h(\delta^{2}\boldsymbol{u}^{\gamma}_{m}+\alpha\delta\boldsymbol{u}^{\gamma}_{m}))\,\mathrm{d}x$
$\displaystyle\geq\int_{\Omega}b(v^{\gamma}_{m-1})F^{-1}_{n}(\boldsymbol{\varepsilon}(\delta\boldsymbol{u}^{\gamma}_{m}+\alpha\boldsymbol{u}^{\gamma}_{m})):\boldsymbol{\varepsilon}(\delta\boldsymbol{u}^{\gamma}_{m}+\alpha\boldsymbol{u}^{\gamma}_{m})\,\mathrm{d}x$
$\displaystyle\quad-\int_{\Omega}b(v^{\gamma}_{m-1})F^{-1}_{n}(\boldsymbol{\varepsilon}(\delta\boldsymbol{u}^{\gamma}_{m-1}+\alpha\boldsymbol{u}^{\gamma}_{m-1})):\boldsymbol{\varepsilon}(\delta\boldsymbol{u}^{\gamma}_{m-1}+\alpha\boldsymbol{u}^{\gamma}_{m-1})\,\mathrm{d}x$
$\displaystyle\quad+\int_{\Omega}b(v^{\gamma}_{m-1})\big{[}\varphi_{n}(F^{-1}_{n}(\boldsymbol{\varepsilon}(\delta\boldsymbol{u}^{\gamma}_{m-1}+\alpha\boldsymbol{u}^{\gamma}_{m-1})))-\varphi_{n}(F^{-1}_{n}(\boldsymbol{\varepsilon}(\delta\boldsymbol{u}^{\gamma}_{m}+\alpha\boldsymbol{u}^{\gamma}_{m})))\,\mathrm{d}x$
$\displaystyle=\int_{\Omega}b(v^{\gamma}_{m-1})\big{[}\varphi^{*}_{n}(\boldsymbol{\varepsilon}(\delta\boldsymbol{u}^{\gamma}_{m}+\alpha\boldsymbol{u}^{\gamma}_{m}))-\varphi^{*}_{n}(\boldsymbol{\varepsilon}(\delta\boldsymbol{u}^{\gamma}_{m-1}+\alpha\boldsymbol{u}^{\gamma}_{m-1}))\big{]}\,\mathrm{d}x$
$\displaystyle\geq\int_{\Omega}b(v^{\gamma}_{m})\varphi^{*}_{n}(\boldsymbol{\varepsilon}(\delta\boldsymbol{u}^{\gamma}_{m}+\alpha\boldsymbol{u}^{\gamma}_{m}))-b(v^{\gamma}_{m-1})\varphi^{*}_{n}(\boldsymbol{\varepsilon}(\delta\boldsymbol{u}^{\gamma}_{m-1}+\alpha\boldsymbol{u}^{\gamma}_{m-1}))\,\mathrm{d}x.$

We insert (3.11) into the elastodynamic equation and rewrite the terms
involving $\delta^{2}\boldsymbol{u}^{\gamma}_{m}$ using (2.16) to deduce that

$\displaystyle
h\|\delta^{2}\boldsymbol{u}^{\gamma}_{m}\|_{2}^{2}+\frac{\alpha}{2}\Big{(}\|\delta\boldsymbol{u}^{\gamma}_{m}\|_{2}^{2}-\|\delta\boldsymbol{u}^{\gamma}_{m-1}\|_{2}^{2}+\|\delta\boldsymbol{u}^{\gamma}_{m}-\delta\boldsymbol{u}^{\gamma}_{m-1}\|_{2}^{2}\Big{)}$
$\displaystyle\quad+\int_{\Omega}b(v^{\gamma}_{m})\varphi^{*}_{n}(\boldsymbol{\varepsilon}(\delta\boldsymbol{u}^{\gamma}_{m}+\alpha\boldsymbol{u}^{\gamma}_{m}))-b(v^{\gamma}_{m-1})\varphi^{*}_{n}(\boldsymbol{\varepsilon}(\delta\boldsymbol{u}^{\gamma}_{m-1}+\alpha\boldsymbol{u}^{\gamma}_{m-1}))\,\mathrm{d}x$
$\displaystyle\leq\langle
l^{\gamma}_{m},\delta\boldsymbol{u}^{\gamma}_{m}+\alpha\boldsymbol{u}^{\gamma}_{m}\rangle-\langle
l^{\gamma}_{m-1},\delta\boldsymbol{u}^{\gamma}_{m-1}+\alpha\boldsymbol{u}^{\gamma}_{m-1}\rangle+h\langle\delta
l^{\gamma}_{m},\delta\boldsymbol{u}^{\gamma}_{m-1}+\alpha\boldsymbol{u}^{\gamma}_{m-1}\rangle.$

Using this inequality recursively and cancelling appropriate terms, we deduce
that

$\displaystyle
h\sum_{j=1}^{m}\|\delta^{2}\boldsymbol{u}^{\gamma}_{m}\|_{2}^{2}+\|\delta\boldsymbol{u}^{\gamma}_{m}\|_{2}^{2}+\int_{\Omega}b(v^{\gamma}_{m})\varphi^{*}_{n}(\boldsymbol{\varepsilon}(\delta\boldsymbol{u}^{\gamma}_{m}+\alpha\boldsymbol{u}^{\gamma}_{m}))\,\mathrm{d}x$
$\displaystyle\leq
C\Big{[}\|l^{\gamma}_{m}\|_{-1,2}\|\boldsymbol{\varepsilon}(\delta\boldsymbol{u}^{\gamma}_{m}+\alpha\boldsymbol{u}^{\gamma}_{m})\|_{2}+\|l^{\gamma}_{0}\|_{-1,2}\|\boldsymbol{\varepsilon}(\boldsymbol{u}_{1}+\alpha\boldsymbol{u}_{0})\|_{2}+h\sum_{j=1}^{m}\|\delta
l^{\gamma}_{j}\|_{-1,2}^{2}$
$\displaystyle\quad+h\sum_{j=0}^{m-1}\|\boldsymbol{\varepsilon}(\delta\boldsymbol{u}^{\gamma}_{j}+\alpha\boldsymbol{u}^{\gamma}_{j})\|_{2}^{2}+\|\boldsymbol{u}_{1}\|_{2}^{2}+\int_{\Omega}b(v_{0})\varphi^{*}_{n}(\boldsymbol{\varepsilon}(\boldsymbol{u}_{1}+\alpha\boldsymbol{u}_{0}))\,\mathrm{d}x\Big{]}.$

Using Young’s inequality and the fact that $\varphi^{*}_{n}(\mathbf{T})\geq
c(n)|\mathbf{T}|^{2}$ to absorb the first term on the right-hand side into the
left, applying the bound in Lemma 3.2 it follows that for a constant $C=C(n)$,
independent of $h$ and $m$, we have

$\displaystyle
h\sum_{j=1}^{m}\|\delta^{2}\boldsymbol{u}^{\gamma}_{m}\|_{2}^{2}+\|\delta\boldsymbol{u}^{\gamma}_{m}\|_{2}^{2}+\int_{\Omega}b(v^{\gamma}_{m})\varphi^{*}_{n}(\boldsymbol{\varepsilon}(\delta\boldsymbol{u}^{\gamma}_{m}+\alpha\boldsymbol{u}^{\gamma}_{m}))\,\mathrm{d}x\leq
C.$

Optimising with respect to $m$ on the left-hand side, we obtain the required
result. ∎

######  Lemma 3.4.

Let the assumptions of Theorem 3.1 hold and let
$(\boldsymbol{u}^{\gamma}_{m},v^{\gamma}_{m})_{m=1}^{M}$ be the solution of
the time discrete problem. Suppose additionally that $k>\frac{d}{2}$. There
exists a constant $C=C(n)$, independent of $M$, such that

$\displaystyle\max_{1\leq m\leq M}\|\delta v^{\gamma}_{m}\|_{k,2}\leq C.$

###### Proof.

Using Corollary 3.2, we have

$\displaystyle\|\delta v^{\gamma}_{m}\|_{k,2}^{2}$
$\displaystyle=-\int_{\Omega}\Big{\\{}\frac{b^{\prime}(v^{\gamma}_{m})}{\alpha}\delta
v^{\gamma}_{m}\varphi^{*}_{n}(\boldsymbol{\varepsilon}(\alpha\boldsymbol{u}^{\gamma}_{m}))+\frac{1}{2\epsilon}(v^{\gamma}_{m}-1)\delta
v^{\gamma}_{m}+2\epsilon\nabla v^{\gamma}_{m}\cdot\nabla\delta
v^{\gamma}_{h}\Big{\\}}\,\mathrm{d}x$
$\displaystyle\leq\frac{b^{\prime}(1)}{\alpha}\|\delta
v^{\gamma}_{m}\|_{\infty}\|\varphi^{*}_{n}(\boldsymbol{\varepsilon}(\alpha\boldsymbol{u}^{\gamma}_{m}))\|_{1}+\frac{1}{2\epsilon}\|v^{\gamma}_{m}-1\|_{2}\|\delta
v^{\gamma}_{m}\|_{2}+2\epsilon\|\nabla v^{\gamma}_{m}\|_{2}\|\nabla\delta
v^{\gamma}_{m}\|_{2}$ $\displaystyle\leq
C\Big{[}\Big{(}\int_{\Omega}\varphi^{*}_{n}(\boldsymbol{\varepsilon}(\alpha\boldsymbol{u}^{\gamma}_{m}))\,\mathrm{d}x\Big{)}^{2}+\|v^{\gamma}_{m}-1\|_{2}^{2}+\|\nabla
v^{\gamma}_{m}\|_{2}^{2}\Big{]}+\frac{1}{2}\|\delta v^{\gamma}\|_{k,2}^{2}.$

We absorb the final term into the left-hand side and use Lemma 3.2 to bound
the remaining terms. ∎

######  Corollary 3.3.

Let the assumptions of Theorem 3.1 hold and let
$(\boldsymbol{u}^{\gamma}_{m},v^{\gamma}_{m})_{m=1}^{M}$ be the solution of
the time discrete problem. Suppose additionally that $k>\frac{d}{2}$. Using
the notation of Section 2, there exists a constant $C=C(n)$, independent of
$M$, such that

$\displaystyle\max_{t\in[0,T]}\|\overline{\boldsymbol{u}}^{\gamma}(t)\|_{1,2}+\max_{t\in[0,T]}\|\boldsymbol{u}^{\gamma,\pm}(t)\|_{1,2}+\max_{t\in[0,T]}\|\overline{\boldsymbol{u}}^{\gamma,\prime}(t)\|_{1,2}+\max_{t\in[0,T]}\|\boldsymbol{u}^{\gamma,\pm,\prime}(t)\|_{1,2}$
$\displaystyle\quad+\int_{0}^{T}\|\boldsymbol{u}^{\gamma,+,\prime\prime}(t)\|_{2}^{2}\,\mathrm{d}t+\max_{t\in[0,T]}\|\overline{v}^{\gamma}(t)\|_{k,2}+\max_{t\in[0,T]}\|v^{\gamma,\pm}(t)\|_{k,2}+\max_{t\in[0,T]}\|v^{\gamma,+,\prime}(t)\|_{k,2}$
$\displaystyle\leq C.$

######  Proposition 3.2.

Let the assumptions of Theorem 3.1 hold. Suppose additionally that
$k>\frac{d}{2}$. There exists a weak energy solution
$(\boldsymbol{u}^{n},\mathbf{T}^{n},v^{n})$ of the regularised problem in the
following sense. The weak elastodynamic equation holds, that is, for a.e.
$t\in(0,T)$, we have

(3.12)
$\int_{\Omega}\boldsymbol{u}^{n}_{tt}(t)\cdot\boldsymbol{w}+b(v^{n}(t))\mathbf{T}^{n}(t):\boldsymbol{\varepsilon}(\boldsymbol{w})\,\mathrm{d}x=\langle
l^{n}(t),\boldsymbol{w}\rangle,$

for every $\boldsymbol{w}\in W^{1,2}_{D}(\Omega)^{d}$, where the stress tensor
$\mathbf{T}^{n}$ is identified by the constitutive relation

$\displaystyle\boldsymbol{\varepsilon}(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n})=F_{n}(\mathbf{T}^{n})\quad\text{
a.e. in }Q.$

For a.e. $t\in[0,T]$, the following minimisation problem is satisfied:

(3.13)
$\displaystyle\mathcal{E}_{n}(\boldsymbol{u}^{n}(t),v^{n}(t))+\mathcal{H}(v^{n}(t))+\mathcal{G}_{k}(v^{n}(t),v^{n}_{t}(t))$
$\displaystyle=\inf\Big{\\{}\mathcal{E}_{n}(\boldsymbol{u}^{n}(t),v)+\mathcal{H}(v)+\mathcal{G}_{k}(v,v^{n}_{t}(t))\,:\,v\in
H^{k}_{D+1}(\Omega),\,v\leq v^{n}(t)\Big{\\}},$

Furthermore, for every $t\in[0,T]$, we have the energy-dissipation balance

(3.14)
$\displaystyle\mathcal{F}_{n}(t;\boldsymbol{u}^{n}(t),\boldsymbol{u}^{n}_{t}(t),v^{n}(t))+\int_{0}^{t}\langle
l^{n}_{t}(s),\boldsymbol{u}^{n}(s)\rangle+\|v^{n}_{t}(s)\|_{k,2}^{2}\,\mathrm{d}s$
$\displaystyle\quad+\int_{0}^{t}\int_{\Omega}b(v^{n})\big{[}F^{-1}_{n}(\boldsymbol{\varepsilon}(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n}))-F^{-1}_{n}(\boldsymbol{\varepsilon}(\alpha\boldsymbol{u}^{n}))\big{]}:\boldsymbol{\varepsilon}(\boldsymbol{u}^{n}_{t})\,\mathrm{d}x\,\mathrm{d}s$
$\displaystyle=\mathcal{F}_{n}(0;\boldsymbol{u}_{0},\boldsymbol{u}_{1},v_{0}),$

and the non-healing property $v^{n}_{t}\leq 0$. The initial conditions hold in
the sense that

$\displaystyle\lim_{t\rightarrow
0+}\Big{[}\|\boldsymbol{u}^{n}(t)-\boldsymbol{u}_{0}\|_{1,2}+\|\boldsymbol{u}^{n}_{t}(t)-\boldsymbol{u}_{1}\|_{2}+\|v^{n}(t)-v_{0}\|_{k,2}\Big{]}=0.$

Furthermore, there exists a subsequence (not relabelled) in $M$, independent
of $n$, such that the following convergence results hold:

  * •

$\overline{\boldsymbol{u}}^{(n,M)}\overset{\ast}{\rightharpoonup}\boldsymbol{u}^{n}$
weakly-* in $W^{1,\infty}(0,T;W^{1,2}_{D}(\Omega)^{d})$;

  * •

$\boldsymbol{u}^{(n,M),\pm}$,
$\boldsymbol{u}^{(n,M),\pm,\prime}\overset{\ast}{\rightharpoonup}\boldsymbol{u}^{n}$,
$\boldsymbol{u}^{n}_{t}$ weakly-* in
$L^{\infty}(0,T;W^{1,2}_{D}(\Omega)^{d})$, respectively;

  * •

$\overline{\boldsymbol{u}}^{(n,M),\prime}\overset{\ast}{\rightharpoonup}\boldsymbol{u}^{n}_{t}$
weakly-* in $L^{\infty}(0,T;W^{1,2}_{D}(\Omega)^{d})$ and weakly in
$W^{1,2}(0,T;L^{2}(\Omega)^{d})$;

  * •

$\overline{v}^{(n,M)}\overset{\ast}{\rightharpoonup}v^{n}$ weakly-* in
$W^{1,\infty}(0,T;H^{k}(\Omega))$;

  * •

$v^{(n,M),\pm}\overset{\ast}{\rightharpoonup}v^{n}$ weakly-* in
$L^{\infty}(0,T;H^{k}_{D+1}(\Omega))$.

###### Proof.

The stated convergence results follow immediately from standard compactness
results for Bochner spaces. By the Aubin–Lions lemma and the fact that
$k>\frac{d}{2}\geq 1$, it follows that $(\overline{v}^{(n,M)})_{M}$ converges
strongly in $C([0,T];H^{1}(\Omega))$ to $v^{n}$. As a result,
$(v^{(n,M),\pm})_{M}$ converges strongly in $L^{\infty}(0,T;H^{1}(\Omega))$ to
the same limit. Satisfaction of the initial condition for $v^{n}$ follows from
the strong convergence in $C([0,T];H^{1}(\Omega))$. We argue similarly for
$\boldsymbol{u}^{n}(0)$ and $\boldsymbol{u}^{n}_{t}(0)$, applying the
Aubin–Lions lemma again.

Defining the approximate stress tensor
$\mathbf{T}^{(n,M)}=F^{-1}_{n}(\boldsymbol{\varepsilon}(\partial_{t}\overline{\boldsymbol{u}}^{(n,M)}+\alpha\boldsymbol{u}^{(n,M),+}))$,
we see from Lemmas 3.2 and 3.3 that $(\mathbf{T}^{(n,M)})_{M}$ is bounded in
$L^{\infty}(0,T;L^{2}(\Omega)^{d\times d})$. Hence it converges in this space
to a limiting function $\mathbf{T}^{n}\in
L^{\infty}(0,T;L^{2}(\Omega)^{d\times d})$.

Rewriting the elastodynamic equation (3.7) in terms of the interpolants,
multiplying by an arbitrary $\chi\in C([0,T])$ and integrating over $[0,T]$,
we deduce that

$\displaystyle\int_{Q}\partial_{t}\overline{\boldsymbol{u}}^{\gamma,\prime}\cdot(\chi\boldsymbol{w})+b(v^{\gamma,-})\mathbf{T}^{\gamma}:\boldsymbol{\varepsilon}(\chi\boldsymbol{w})\,\mathrm{d}x\,\mathrm{d}t=\int_{0}^{T}\langle
l^{\gamma,+},\chi\boldsymbol{w}\rangle\,\mathrm{d}t,$

for every $\boldsymbol{w}\in W^{1,2}_{D}(\Omega)^{d}$. Letting
$M\rightarrow\infty$ and using Lebesgue’s differentiation theorem, it follows
that

$\displaystyle\int_{\Omega}\boldsymbol{u}_{tt}^{n}(t)\cdot\boldsymbol{w}+b(v^{n}(t))\mathbf{T}^{n}(t):\boldsymbol{\varepsilon}(\boldsymbol{w})\,\mathrm{d}x=\langle
l^{n}(t),\boldsymbol{w}\rangle,$

for a.e. $t\in(0,T)$ and every $\boldsymbol{w}\in W^{1,2}_{D}(\Omega)^{d}$.

Next, noticing that
$(\partial_{t}\overline{\boldsymbol{u}}^{(n,M),\prime}+\alpha\boldsymbol{u}^{(n,M),+})_{M}$
converges strongly in $L^{2}(0,T;L^{2}(\Omega)^{d})$, by the Aubin–Lions
lemma, we deduce that

(3.15)
$\displaystyle\lim_{M\rightarrow\infty}\int_{Q}b(v^{(n,M),-})\mathbf{T}^{\gamma}:\boldsymbol{\varepsilon}(\partial_{t}\overline{\boldsymbol{u}}^{(n,M),\prime}+\alpha\boldsymbol{u}^{(n,M),+})\,\mathrm{d}x\,\mathrm{d}t$
$\displaystyle=\lim_{M\rightarrow\infty}\Big{\\{}-\int_{Q}\partial_{t}\overline{\boldsymbol{u}}^{(n,M),\prime}\cdot(\partial_{t}\overline{\boldsymbol{u}}^{(n,M),\prime}+\alpha\boldsymbol{u}^{(n,M),+})\,\mathrm{d}x\,\mathrm{d}t$
$\displaystyle\quad+\int_{0}^{T}\langle
l^{(n,M),+},\partial_{t}\overline{\boldsymbol{u}}^{(n,M),\prime}+\alpha\boldsymbol{u}^{(n,M),+}\rangle\,\mathrm{d}t\Big{\\}}$
$\displaystyle=-\int_{Q}\boldsymbol{u}^{n}_{tt}\cdot(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n})\,\mathrm{d}x\,\mathrm{d}t+\int_{0}^{T}\langle
l^{n},\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n}\rangle\,\mathrm{d}t$
$\displaystyle=\int_{Q}b(v^{n})\mathbf{T}^{n}:\boldsymbol{\varepsilon}(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n})\,\mathrm{d}x\,\mathrm{d}t.$

Applying Minty’s method, as in the proof of Theorem 2.6, it follows from
(3.15) that we have
$\boldsymbol{\varepsilon}(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n})=F_{n}(\mathbf{T}^{n})$
pointwise a.e. in $Q$.

To prove that the minimisation problem (3.13) is satisfied, we need a strong
convergence result for $(\boldsymbol{u}^{(n,M),+})_{M}$ and
$({\boldsymbol{u}}^{(n,M),\prime,+})_{M}$ in $L^{2}(0,T;W^{1,2}(\Omega)^{d})$.
Such convergence results allow us to deal with the presence of nonlinearities
in the minimisation problem and energy-dissipation balance. Adapting the
reasoning in [28] (namely, Lemmas 3.9, 3.10 and 3.11) to the nonlinear
problem, we deduce that this is the case. Indeed, we have that

(3.16)
$\boldsymbol{u}^{(n,M),+}\rightarrow\boldsymbol{u}^{n},\quad\boldsymbol{u}^{(n,M),\prime,+}\rightarrow\boldsymbol{u}^{n}_{t}\quad\text{
strongly in }L^{2}(0,T;W^{1,2}(\Omega)^{d}).$

As a result, we also have $\mathbf{T}^{(n,M)}\rightarrow\mathbf{T}^{n}$
strongly in $L^{2}(0,T;L^{2}(\Omega)^{d\times d})$ as $M\rightarrow\infty$.
Now, we know that the satisfaction of the minimisation problem (3.13) is
equivalent to

(3.17)
$0\leq\big{[}\partial_{v}\mathcal{E}_{n}(\boldsymbol{u}^{n}(t),v^{n}(t))+\mathcal{H}^{\prime}(v^{n}(t))+\partial_{v}\mathcal{G}_{k}(v^{n}(t),v^{n}_{t}(t))\big{]}(\chi),$

for every $\chi\in H^{k}_{D}(\Omega)$ with $\chi\leq 0$, for a.e. $t\in(0,T)$.
However, (3.17) follows from taking $M\rightarrow\infty$ in Corollary 3.1,
applying the weak convergence results for the approximations of the phase-
field function and the strong convergence result (3.16).

It remains to show that the energy-dissipation equality (3.14) holds. Using
the convergence results, weak lower semi-continuity of norms and Fatou’s
lemma, it follows that (3.14) holds with “$=$” replaced by “$\leq$”. For the
opposite inequality, we argue as in the proof of Proposition 2.6, adapting the
reasoning slightly to take care of the extra rate-dependent term. ∎

Now that we have the existence of a solution to the regularised problem, the
focus is to obtain appropriate $n$-independent bounds that allow us to take
the limit as $n\rightarrow\infty$. Before formulating these bounds, we state a
consequence of the energy-dissipation equality that is analogous to the result
in Corollary 2.1. We differentiate the energy-dissipation equality, carefully
apply the chain rule for weak derivatives and test in the elastodynamic
equation against $\boldsymbol{u}_{t}^{n}$. The remaining terms must vanish and
are exactly the left-hand side of (3.18). This result is vital to the proof of
Lemma 3.7. In turn, Lemma 3.7 is needed to obtain a bound on $(v^{n}_{t})_{n}$
in $L^{\infty}(0,T;H^{k}(\Omega))$. Provided that $k>\frac{d}{2}+1$, by the
Sobolev embedding theorem it follows that $(v^{n})_{n}$ is uniformly bounded
in $W^{1,\infty}(0,T;W^{1,\infty}(\Omega))$, which is required for the proof
of higher regularity estimates for the sequence $(\mathbf{T}^{n})_{n}$.

######  Corollary 3.4.

Let the assumptions of Proposition 3.2 hold and let
$(\boldsymbol{u}^{n},\mathbf{T}^{n},v^{n})$ be the solution triple constructed
there. For a.e. $t\in[0,T]$,

(3.18)
$\big{[}\partial_{v}\mathcal{E}_{n}(\boldsymbol{u}^{n}(t),v^{n}(t))+\mathcal{H}^{\prime}(v^{n}(t))+\partial_{v}\mathcal{G}_{k}(v^{n}(t),v_{t}^{n}(t))\big{]}(v^{n}_{t}(t))=0.$

Now we derive $n$-independent bounds on the solution triple
$(\boldsymbol{u}^{n},\mathbf{T}^{n},v^{n})$. Testing in the elastodynamic
equation (3.12) against $\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n}$, we
reason as in [9] to obtain Lemma 3.5. The bound on the gradient of
$\boldsymbol{u}^{n}$ follows from using the bound on $\mathbf{T}^{n}$, the
constitutive relation, the fundamental theorem of calculus and the
Korn–Poincaré inequality. Indeed, we recall the memory kernel property

(3.19) $\boldsymbol{\varepsilon}(\boldsymbol{u}^{n}(t))=\mathrm{e}^{-\alpha
t}\boldsymbol{\varepsilon}(\boldsymbol{u}_{0})+\int_{0}^{t}\mathrm{e}^{-\alpha(t-\tau)}F_{n}(\mathbf{T}^{n}(\tau))\,\mathrm{d}\tau.$

Furthermore, we have $|F_{n}(\mathbf{T}^{n})|\leq 1+n^{-1}|\mathbf{T}^{n}|$.
The representation (3.19) is vital to obtaining improved bounds on the
solution triple, including using bounds on the linear combination
$\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n}$ to yield bounds on the
function $\boldsymbol{u}^{n}$.

######  Lemma 3.5.

Let the assumptions of Proposition 3.2 hold and let
$(\boldsymbol{u}^{n},\mathbf{T}^{n},v^{n})$ be the solution triple constructed
there. There exists a constant $C$, independent of $n$, such that

$\displaystyle\sup_{t\in[0,T]}\|\boldsymbol{u}^{n}(t)\|_{1,2}+\sup_{t\in[0,T]}\|\boldsymbol{u}^{n}_{t}(t)\|_{2}+\int_{Q}|\mathbf{T}^{n}|+\frac{|\mathbf{T}^{n}|^{2}}{n}+|\boldsymbol{\varepsilon}(\boldsymbol{u}^{n}_{t})|^{2}\,\mathrm{d}x\,\mathrm{d}t\leq
C.$

Combining this with the energy-dissipation inequality immediately yields the
improved bound in Lemma 3.6. Indeed, from Lemma 3.5, we use the bound on
$(\boldsymbol{u}^{n})_{n}$ in $L^{\infty}(0,T;W^{1,2}(\Omega)^{d})$ to bound
the terms in the energy-dissipation equality that result from the presence of
an external force.

At this point, we assume the safety strain condition (3.5) so that the
sequence
$(\varphi^{*}_{n}(\boldsymbol{\varepsilon}(\alpha\boldsymbol{u}_{0})))_{n}$ is
uniformly bounded in $L^{1}(\Omega)$. We recall that convex conjugation is
order reversing so, for every $\mathbf{S}\in\mathbb{R}^{d\times d}$, we have
$\varphi^{*}_{n}(\mathbf{S})\leq\varphi^{*}(\mathbf{S})$. It follows that

$\displaystyle\int_{\Omega}\varphi^{*}_{n}(\boldsymbol{\varepsilon}(\alpha\boldsymbol{u}_{0}))\,\mathrm{d}x\leq\int_{\Omega}\varphi^{*}(\boldsymbol{\varepsilon}(\alpha\boldsymbol{u}_{0}))\,\mathrm{d}x=\int_{\Omega}\int_{0}^{|\boldsymbol{\varepsilon}(\alpha\boldsymbol{u}_{0})|}\frac{s}{(1-s^{a})^{\frac{1}{a}}}\,\mathrm{d}s\,\mathrm{d}x\leq\frac{|\Omega|C_{*}^{2}}{(1-C_{*}^{a})^{\frac{1}{a}}},$

where $C_{*}$ is the constant from the safety strain condition (3.5).

######  Lemma 3.6.

Let the assumptions of Proposition 3.2 hold and let
$(\boldsymbol{u}^{n},\mathbf{T}^{n},v^{n})$ be the solution triple constructed
there. Suppose additionally that the safety strain condition (3.5) holds.
There exists a constant $C$, independent of $n$, such that

$\displaystyle\sup_{t\in[0,T]}\Big{[}\int_{\Omega}\varphi^{*}_{n}(\boldsymbol{\varepsilon}(\alpha\boldsymbol{u}^{n}(t)))\,\mathrm{d}x\Big{]}+\sup_{t\in[0,T]}\|v^{n}(t)\|_{k,2}+\int_{0}^{T}\|v^{n}_{t}(t)\|_{k,2}^{2}\,\mathrm{d}t\leq
C.$

Adapting the argument in Lemma 3.4 and making use of Corollary 3.4, we can
improve the bound on $(v^{n}_{t})_{n}$ in $L^{2}(0,T;H^{k}(\Omega))$ to a
bound in $L^{\infty}(0,T;H^{k}(\Omega))$. As a result and recalling that
$k>\frac{d}{2}$, it follows that $(v^{n}_{t})_{n}$ is bounded in
$L^{\infty}(Q)$, independent of $n$. This is used to improve the time
regularity of the approximations $(\mathbf{T}^{n})_{n}$, which is the subject
of Proposition 3.3.

######  Lemma 3.7.

Let the assumptions of Proposition 3.2 hold and let
$(\boldsymbol{u}^{n},\mathbf{T}^{n},v^{n})$ be the solution triple constructed
there. Suppose additionally that the safety strain condition (3.5) holds.
There exists a constant $C$ independent of $n$ such that

$\displaystyle\sup_{t\in[0,T]}\|v^{n}_{t}(t)\|_{k,2}\leq C.$

With Lemmas 3.5, 3.6 and 3.7 in mind, we work on improving the regularity of
$(\mathbf{T}^{n})_{n}$. Indeed, we look for weighted estimates on the
derivatives $(\mathbf{T}^{n}_{t})_{n}$ and $(\nabla\mathbf{T}^{n})_{n}$ which
allow us to prove a pointwise convergence result on $Q$. For the pointwise
result, we mimic some of the ideas from [3]. However, significant adaptations
are made as we work in the time-dependent setting. Also, we have the added
complication of the phase-field function. To deal with this, Lemma 3.7 is
vital.

We begin with studying the regularity in time. The proof of Proposition 3.3 is
one of the highlights of this work. As mentioned previously, we work directly
with the time discrete approximation. If we were to work with the time
continuous solution, we would require $\boldsymbol{u}^{n}_{tt}$ to be
continuous at $t=0$. This is not necessarily true. However, to work with the
time discrete solution, we must define an extra term
$\boldsymbol{u}^{\gamma}_{-2}$ in the solution sequence in order to define
$\delta^{2}\boldsymbol{u}^{\gamma}_{0}$. By our choice of
$\boldsymbol{u}^{\gamma}_{-2}$, we show that
$\delta^{2}\boldsymbol{u}^{\gamma}_{0}$ is uniformly bounded in
$L^{2}(\Omega)^{d}$, independent of $n$ and $M$. We deduce from this that
$(\boldsymbol{u}^{n}_{tt})_{n}$ is bounded in
$L^{\infty}(0,T;L^{2}(\Omega)^{d})$, independent of $n$. Furthermore, we see
that we have $\mathbf{T}^{n}\in W^{1,2}(0,T;L^{2}(\Omega)^{d\times d})$ and
obtain a weighted bound on the time derivative $\mathbf{T}^{n}_{t}$ that is
independent of $n$.

At this stage, we impose the compatibility condition (3.4). Under the
compatibility condition, we know that the sequence of approximations
$(l^{n})_{n}$ is bounded in $W^{2,1}(0,T;W^{-1,2}_{D}(\Omega)^{d})$ and
converges weakly in this space to $l$. Furthermore, it becomes clear why the
approximation choice (3.6) is made. Indeed, suppose that
$\boldsymbol{u}^{n}_{tt}$ is continuous at $t=0$ and sufficiently smooth so
that the following reasoning is justified. Testing in the elastodynamic
equation at $t=0$ against $\boldsymbol{u}^{n}_{tt}(0)$, we get

$\displaystyle\|\boldsymbol{u}^{n}_{tt}(0)\|_{2}^{2}$
$\displaystyle=\int_{\Omega}-b(v^{n}(0))\mathbf{T}^{n}(0)\cdot\boldsymbol{\varepsilon}(\boldsymbol{u}^{n}_{tt}(0))\,\mathrm{d}x+\langle
l^{n}(0),\boldsymbol{u}^{n}_{tt}(0)\rangle$
$\displaystyle=\int_{\Omega}-b(v_{0})F^{-1}_{n}(\boldsymbol{\varepsilon}(\boldsymbol{u}_{1}+\alpha\boldsymbol{u}_{0}))\cdot\boldsymbol{\varepsilon}(\boldsymbol{u}^{n}_{tt}(0)+\boldsymbol{f}(0)\cdot\boldsymbol{u}^{n}_{tt}(0)\,\mathrm{d}x$
$\displaystyle\quad+\int_{\Gamma_{N}}F^{-1}_{n}(\boldsymbol{\varepsilon}(\boldsymbol{u}_{1}+\alpha\boldsymbol{u}_{0}))\mathbf{n}\cdot\boldsymbol{u}^{n}_{tt}(0)\,\mathrm{d}S$
$\displaystyle=\int_{\Omega}\big{(}\mathrm{div}(b(v_{0})F^{-1}_{n}(\boldsymbol{\varepsilon}(\boldsymbol{u}_{1}+\alpha\boldsymbol{u}_{0})))+\boldsymbol{f}(0)\big{)}\cdot\boldsymbol{u}^{n}_{tt}(0)\,\mathrm{d}x,$

where the boundary integral vanishes due to integration by parts and the
choice of $\boldsymbol{g}^{n}(0)$. It remains to show that
$\mathrm{div}(b(v_{0})F^{-1}_{n}(\boldsymbol{\varepsilon}(\boldsymbol{u}_{1}+\alpha\boldsymbol{u}_{0})))$
is bounded in $L^{2}(\Omega)^{d}$, independent of $n$, to deduce that
$\boldsymbol{u}^{n}_{tt}(0)$ is also bounded in $L^{2}(\Omega)^{d}$,
independent of $n$. We use analogous reasoning in the proof of Proposition 3.3
to show that the discrete time derivative
$\delta^{2}\boldsymbol{u}^{\gamma}_{0}$ is uniformly bounded in
$L^{2}(\Omega)^{d}$. Without the compatibility condition for
$\boldsymbol{g}^{n}$, we get a boundary term involving
$\boldsymbol{u}^{n}_{tt}(0)$, so we would need a bound on
$(\boldsymbol{u}^{n}_{tt}(0))_{n}$ in $W^{1,2}(\Omega)^{d}$. However, this
does not appear to be possible, thus justifying our choice of approximation
and the compatibility condition (3.4).

######  Proposition 3.3.

Let the assumptions of Proposition 3.2 hold and let
$(\boldsymbol{u}^{n},\mathbf{T}^{n},v^{n})$ be the solution triple constructed
there. Suppose additionally that $l\in W^{2,1}(0,T;W^{-1,2}_{D}(\Omega)^{d})$,
the safety strain condition (3.5) holds, the compatibility condition (3.4)
holds and that $\boldsymbol{u}_{1}+\alpha\boldsymbol{u}_{0}\in
W^{2,2}_{D}(\Omega)^{d}$. Then $\mathbf{T}^{n}\in
W^{1,2}(0,T;L^{2}(\Omega)^{d\times d})$ and there exists a constant $C$,
independent of $n$, such that

$\displaystyle\sup_{t\in[0,T]}\|\boldsymbol{u}^{n}_{tt}(t)\|_{2}+\int_{Q}\frac{|\mathbf{T}^{n}_{t}|^{2}}{(1+|\mathbf{T}^{n}|)^{1+a}}+\frac{|\mathbf{T}^{n}_{t}|^{2}}{n}\,\mathrm{d}x\,\mathrm{d}t\leq
C.$

###### Proof.

We define
$\mathbf{T}^{\gamma}_{m}=F^{-1}_{n}(\boldsymbol{\varepsilon}(\delta\boldsymbol{u}^{\gamma}_{m}+\alpha\boldsymbol{u}^{\gamma}_{m}))$
for $0\leq m\leq M$, where
$(\boldsymbol{u}^{\gamma}_{m},v^{\gamma}_{m})_{m=0}^{M}$ is the solution
sequence for the time discrete problem in Theorem 3.1, with $\beta=(n,M)$. We
fix $2\leq m\leq M$, test in the elastodynamic equation (3.7) at time levels
$m$ and $m-1$ against
$\delta^{2}\boldsymbol{u}^{\gamma}_{m}+\alpha\delta\boldsymbol{u}^{\gamma}_{m}$
and subtract the resulting equalities. Using (2.16), we rewrite the interial
terms in the usual way. For the body force $l^{n}$, to avoid having terms
involving spatial derivatives of $\delta^{2}\boldsymbol{u}^{\gamma}_{m}$,
which we cannot necessarily bound, we write

$\displaystyle\langle
l^{\gamma}_{m}-l^{\gamma}_{m-1},\delta^{2}\boldsymbol{u}^{\gamma}_{m}+\alpha\delta\boldsymbol{u}^{\gamma}_{m}\rangle$
$\displaystyle=\langle\delta
l^{\gamma}_{m},\delta\boldsymbol{u}^{\gamma}_{m}+\alpha\boldsymbol{u}^{\gamma}_{m}\rangle-\langle\delta
l^{\gamma}_{m-1},\delta\boldsymbol{u}^{\gamma}_{m-1}+\alpha\boldsymbol{u}^{\gamma}_{m-1}\rangle$
$\displaystyle\quad+h\langle\delta^{2}l^{\gamma}_{m},\delta\boldsymbol{u}_{m-1}^{\gamma}+\alpha\boldsymbol{u}_{m-1}^{\gamma}\rangle.$

For the nonlinear terms, we write

(3.20)
$\displaystyle\int_{\Omega}\big{[}b(v^{\gamma}_{m-1})\mathbf{T}^{\gamma}_{m}-b(v^{\gamma}_{m-2})\mathbf{T}^{\gamma}_{m-1}\big{]}:\boldsymbol{\varepsilon}(\delta^{2}\boldsymbol{u}^{\gamma}_{m}+\alpha\delta\boldsymbol{u}^{\gamma}_{m})\,\mathrm{d}x$
$\displaystyle=\int_{\Omega}\Big{\\{}b(v^{\gamma}_{m-1})\big{(}\mathbf{T}^{\gamma}_{m}-\mathbf{T}^{\gamma}_{m-1}\big{)}:\frac{F_{n}(\mathbf{T}^{\gamma}_{m})-F_{n}(\mathbf{T}^{\gamma}_{m-1})}{h}$
$\displaystyle\quad+\big{(}b(v^{\gamma}_{m-1})-b(v^{\gamma}_{m-2})\big{)}\mathbf{T}^{\gamma}_{m-1}:\frac{F_{n}(\mathbf{T}^{\gamma}_{m})-F_{n}(\mathbf{T}^{\gamma}_{m-1})}{h}\Big{\\}}\,\mathrm{d}x$
$\displaystyle=h\int_{\Omega}\int_{0}^{1}b(v^{\gamma}_{m-1})\Big{(}\frac{\mathbf{T}^{\gamma}_{m}-\mathbf{T}^{\gamma}_{m-1}}{h},\frac{\mathbf{T}^{\gamma}_{m}-\mathbf{T}^{\gamma}_{m-1}}{h}\Big{)}_{\mathcal{A}_{n}(s\mathbf{T}^{\gamma}_{m}+(1-s)\mathbf{T}^{\gamma}_{m-1})}\,\mathrm{d}s\,\mathrm{d}x$
$\displaystyle\quad+h\int_{\Omega}\int_{0}^{1}\Big{(}\frac{b(v^{\gamma}_{m-1})-b(v^{\gamma}_{m-2})}{h}\Big{)}\cdot\Big{(}\mathbf{T}^{\gamma}_{m-1},\frac{\mathbf{T}^{\gamma}_{m}-\mathbf{T}^{\gamma}_{m-1}}{h}\Big{)}_{\mathcal{A}_{n}(s\mathbf{T}^{\gamma}_{m}+(1-s)\mathbf{T}^{\gamma}_{m-1})}\,\mathrm{d}s\,\mathrm{d}x$
$\displaystyle\geq\frac{h}{2}\int_{\Omega}\int_{0}^{1}b(v^{\gamma}_{m-1})\Big{(}\frac{\mathbf{T}^{\gamma}_{m}-\mathbf{T}^{\gamma}_{m-1}}{h},\frac{\mathbf{T}^{\gamma}_{m}-\mathbf{T}^{\gamma}_{m-1}}{h}\Big{)}_{\mathcal{A}_{n}(s\mathbf{T}^{\gamma}_{m}+(1-s)\mathbf{T}^{\gamma}_{m-1})}\,\mathrm{d}s\,\mathrm{d}x$
$\displaystyle\quad-\frac{h}{2\eta}\int_{\Omega}\int_{0}^{1}\Big{(}\frac{b(v^{\gamma}_{m-1})-b(v^{\gamma}_{m-2})}{h}\Big{)}^{2}\big{(}\mathbf{T}^{\gamma}_{m-1},\mathbf{T}^{\gamma}_{m-1}\big{)}_{\mathcal{A}_{n}(s\mathbf{T}^{\gamma}_{m}+(1-s)\mathbf{T}^{\gamma}_{m-1})}\,\mathrm{d}s\,\mathrm{d}x,$

where $\mathcal{A}_{n}(\mathbf{U})$ is the fourth-order tensor defined by

$\displaystyle\mathcal{A}_{n}(\mathbf{U})_{ijkl}=\frac{\partial
F_{n}(\mathbf{U})_{ij}}{\partial U_{kl}}$

with corresponding bilinear form

$\displaystyle(\mathbf{T},\mathbf{S})_{\mathcal{A}_{n}(\mathbf{U})}:=\sum_{i,j,k,l=1}^{d}\mathcal{A}_{n}(\mathbf{U})_{ijkl}T_{ij}S_{kl}.$

We can easily check that $(\cdot,\cdot)_{\mathcal{A}_{n}(\mathbf{U})}$ defines
an inner product on $\mathbb{R}^{d\times d}$, justifying the use of the
Cauchy–Schwarz inequality in (3.20). Summing the result over the indices
$2\leq j\leq m$, we deduce that

(3.21)
$\displaystyle\frac{1}{2}\Big{(}\|\delta^{2}\boldsymbol{u}^{\gamma}_{m}\|_{2}^{2}-\|\delta^{2}\boldsymbol{u}^{\gamma}_{1}\|_{2}^{2}\Big{)}+\alpha(\delta^{2}\boldsymbol{u}^{\gamma}_{m},\delta\boldsymbol{u}^{\gamma}_{m})-\alpha(\delta^{2}\boldsymbol{u}^{\gamma}_{1},\delta\boldsymbol{u}^{\gamma}_{1})$
$\displaystyle\quad+\frac{h}{2}\sum_{j=2}^{m}\int_{\Omega}\int_{0}^{1}b(v^{\gamma}_{j-1})\Big{(}\delta\mathbf{T}^{\gamma}_{j},\delta\mathbf{T}^{\gamma}_{j}\Big{)}_{\mathcal{A}_{n}(s\mathbf{T}^{\gamma}_{j}+(1-s)\mathbf{T}^{\gamma}_{j-1})}\,\mathrm{d}s\,\mathrm{d}x$
$\displaystyle\quad-\alpha
h\sum_{j=2}^{m}(\delta^{2}\boldsymbol{u}^{\gamma}_{j},\delta^{2}\boldsymbol{u}^{\gamma}_{j-1})$
$\displaystyle\leq\langle\delta
l^{\gamma}_{m},\delta\boldsymbol{u}^{\gamma}_{m}+\alpha\boldsymbol{u}^{\gamma}_{m}\rangle-\langle\delta
l^{\gamma}_{1},\delta\boldsymbol{u}^{\gamma}_{1}+\alpha\boldsymbol{u}^{\gamma}_{1}\rangle+h\sum_{j=2}^{m}\langle\delta^{2}l^{\gamma}_{j},\delta\boldsymbol{u}^{\gamma}_{j-1}+\alpha\boldsymbol{u}^{\gamma}_{j-1}\rangle$
$\displaystyle\quad+\frac{h}{2\eta}\sum_{j=1}^{m-1}\int_{\Omega}\int_{0}^{1}\Big{(}\frac{b(v^{\gamma}_{j})-b(v^{\gamma}_{j-1})}{h}\Big{)}^{2}(\mathbf{T}^{\gamma}_{j},\mathbf{T}^{\gamma}_{j})_{\mathcal{A}_{n}(s\mathbf{T}^{\gamma}_{j+1}+(1-s)\mathbf{T}^{\gamma}_{j})}\,\mathrm{d}s\,\mathrm{d}x.$

To bound the terms involving the index 1, in particular, the term
$\delta^{2}\boldsymbol{u}^{\gamma}_{1}$, we need to define
$\delta^{2}\boldsymbol{u}^{\gamma}_{0}$. Equivalently, we look for a
$\boldsymbol{u}_{-2}^{\gamma}\in L^{2}(\Omega)^{d}$ such that, for every
$\boldsymbol{w}\in W^{1,2}_{D}(\Omega)^{d}$,

(3.22)
$\displaystyle\int_{\Omega}\frac{\boldsymbol{u}^{\gamma}_{0}-2\boldsymbol{u}^{\gamma}_{-1}+\boldsymbol{u}^{\gamma}_{-2}}{h^{2}}\cdot\boldsymbol{w}\,\mathrm{d}x$
$\displaystyle=-\int_{\Omega}b(v_{0})F^{-1}_{n}(\boldsymbol{\varepsilon}(\delta\boldsymbol{u}^{\gamma}_{0}+\alpha\boldsymbol{u}^{\gamma}_{0})):\boldsymbol{\varepsilon}(\boldsymbol{w})\,\mathrm{d}x+\langle
l^{n}(0),\boldsymbol{w}\rangle$
$\displaystyle=\int_{\Omega}\Big{[}\mathrm{div}\Big{(}b(v_{0})F^{-1}_{n}(\boldsymbol{\varepsilon}(\boldsymbol{u}_{1}+\alpha\boldsymbol{u}_{0}))\Big{)}+\boldsymbol{f}(0)\Big{]}:\boldsymbol{w}\,\mathrm{d}x,$

using integration by parts and the compatibility condition satisfied by the
approximation $(l^{n})_{n}$. Under the regularity assumptions, namely,
$\boldsymbol{u}_{1}+\alpha\boldsymbol{u}_{0}\in W^{2,2}_{D}(\Omega)^{d}$, we
have

$\displaystyle\Big{\|}\mathrm{div}\Big{(}b(v_{0})F^{-1}_{n}(\boldsymbol{\varepsilon}(\boldsymbol{u}_{1}+\alpha\boldsymbol{u}_{0}))\Big{)}+\boldsymbol{f}(0)\Big{\|}_{2}\leq
C,$

where $C$ is independent of $n$. The bound on
$\mathrm{div}(b(v_{0})F^{-1}_{n}(\boldsymbol{\varepsilon}(\boldsymbol{u}_{1}+\alpha\boldsymbol{u}_{0})))$
in $L^{2}(\Omega)^{d}$ can be determined from the following reasoning.
Performing the differentiation, we have

$\displaystyle\mathrm{div}\big{(}b(v_{0})F^{-1}_{n}(\boldsymbol{\varepsilon}(\boldsymbol{u}_{1}+\alpha\boldsymbol{u}_{0}))\big{)}_{i}$
$\displaystyle=\frac{\partial}{\partial
x_{j}}\Big{(}b(v_{0})F^{-1}_{n}(\boldsymbol{\varepsilon}(\boldsymbol{u}_{1}+\alpha\boldsymbol{u}_{0}))_{ij}\Big{)}$
$\displaystyle=b^{\prime}(v_{0})\frac{\partial v_{0}}{\partial
x_{j}}F^{-1}_{n}(\boldsymbol{\varepsilon}(\boldsymbol{u}_{1}+\alpha\boldsymbol{u}_{0}))_{ij}+b(v_{0})\mathcal{A}_{n}(\boldsymbol{\varepsilon}(\boldsymbol{u}_{1}+\alpha\boldsymbol{u}_{0}))_{ijkl}\frac{\partial\boldsymbol{\varepsilon}(\boldsymbol{u}_{1}+\alpha\boldsymbol{u}_{0})_{kl}}{\partial
x_{j}}.$

Next, we notice that

$\displaystyle|F^{-1}_{n}(\boldsymbol{\varepsilon}(\boldsymbol{u}_{1}+\alpha\boldsymbol{u}_{0}))|\leq|F^{-1}(\boldsymbol{\varepsilon}(\boldsymbol{u}_{1}+\alpha\boldsymbol{u}_{0}))|\leq\frac{C_{*}}{(1+C_{*}^{a})^{\frac{1}{a}}},$

and from direct calculation, we have

$\displaystyle\big{|}\mathcal{A}_{n}(\mathbf{U})_{ijkl}\big{|}=\Big{|}\delta_{ik}\delta_{jl}\Big{(}\frac{1}{(1+|\mathbf{U}|^{a})^{\frac{1}{a}}}+\frac{1}{n}\Big{)}-\frac{|\mathbf{U}|^{a-2}\mathbf{U}_{ij}\mathbf{U}_{kl}}{(1+|\mathbf{U}|^{a})^{1+\frac{1}{a}}}\Big{|}\leq
3.$

Thus we have

$\displaystyle\big{|}\mathrm{div}\big{(}b(v_{0})F^{-1}_{n}(\boldsymbol{\varepsilon}(\boldsymbol{u}_{1}+\alpha\boldsymbol{u}_{0}))\big{)}\big{|}\leq\frac{2C_{*}}{(1-C_{*}^{a})^{\frac{1}{a}}}\Big{|}\frac{\partial
v_{0}}{\partial
x_{j}}\Big{|}+6\Big{|}\frac{\partial\boldsymbol{\varepsilon}(\boldsymbol{u}_{1}+\alpha\boldsymbol{u}_{0})_{kl}}{\partial
x_{j}}\Big{|}.$

It follows that
$\mathrm{div}(b(v_{0})F^{-1}_{n}(\boldsymbol{\varepsilon}(\boldsymbol{u}_{1}+\alpha\boldsymbol{u}_{0})))$
is an element of $L^{2}(\Omega)^{d}$ with bound indepndent of $n$, since
$v_{0}\in H^{1}(\Omega)$ and $\boldsymbol{u}_{1}+\alpha\boldsymbol{u}_{0}\in
W^{2,2}_{D}(\Omega)^{d}$. Returning to (3.22), we deduce that there exists a
unique $\boldsymbol{u}^{\gamma}_{-2}\in L^{2}(\Omega)^{d}$ solving (3.22). As
a result, the difference quotient $\delta^{2}\boldsymbol{u}_{0}^{\gamma}$ is
well-defined and is an element of $L^{2}(\Omega)^{d}$. Testing in (3.22)
against $\delta^{2}\boldsymbol{u}^{\gamma}_{0}$ yields

$\displaystyle\|\delta^{2}\boldsymbol{u}^{\gamma}_{0}\|_{2}^{2}$
$\displaystyle\leq\Big{\|}\mathrm{div}\Big{(}b(v_{0})F^{-1}_{n}(\boldsymbol{\varepsilon}(\boldsymbol{u}_{1}+\alpha\boldsymbol{u}_{0}))\Big{)}+\boldsymbol{f}(0)\Big{\|}_{2}\|\delta\boldsymbol{u}^{\gamma}_{0}\|_{2}$
$\displaystyle\leq C\|\delta^{2}\boldsymbol{u}^{\gamma}_{0}\|_{2}.$

Hence $\delta^{2}\boldsymbol{u}^{\gamma}_{0}$ is bounded in
$L^{2}(\Omega)^{d}$, independent of $n$ and $M$. Subtracting the equality

$\displaystyle\int_{\Omega}\delta^{2}\boldsymbol{u}^{\gamma}_{0}\cdot\boldsymbol{w}+b(v^{\gamma}_{0})\mathbf{T}^{\gamma}_{0}:\boldsymbol{\varepsilon}(\boldsymbol{w})\,\mathrm{d}x=\langle
l^{\gamma}_{0},\boldsymbol{w}\rangle,$

from the elastodynamic equation (3.7) at time level $m=1$, and choosing the
test function
$\boldsymbol{w}=\delta^{2}\boldsymbol{u}^{\gamma}_{1}+\alpha\delta\boldsymbol{u}^{\gamma}_{1}$,
we can manipulate the resulting equation as in the case $m\geq 2$. We add the
corresponding result to (3.21) and deduce that

(3.23)
$\displaystyle\frac{1}{2}\Big{(}\|\delta^{2}\boldsymbol{u}^{\gamma}_{m}\|_{2}^{2}-\|\delta^{2}\boldsymbol{u}^{\gamma}_{0}\|_{2}^{2}\Big{)}+\alpha(\delta^{2}\boldsymbol{u}^{\gamma}_{m},\delta\boldsymbol{u}^{\gamma}_{m})-\alpha(\delta^{2}\boldsymbol{u}^{\gamma}_{0},\delta\boldsymbol{u}^{\gamma}_{1})$
$\displaystyle\quad+\frac{h}{2}\sum_{j=1}^{m}\int_{\Omega}\int_{0}^{1}b(v^{\gamma}_{j-1})\Big{(}\delta\mathbf{T}^{\gamma}_{j},\delta\mathbf{T}^{\gamma}_{j}\Big{)}_{\mathcal{A}_{n}(s\mathbf{T}^{\gamma}_{j}+(1-s)\mathbf{T}^{\gamma}_{j-1})}\,\mathrm{d}s\,\mathrm{d}x$
$\displaystyle\quad-\alpha
h\sum_{j=2}^{m}(\delta^{2}\boldsymbol{u}^{\gamma}_{j},\delta^{2}\boldsymbol{u}^{\gamma}_{j-1})$
$\displaystyle\leq\langle\delta
l^{\gamma}_{m},\delta\boldsymbol{u}^{\gamma}_{m}+\alpha\boldsymbol{u}^{\gamma}_{m}\rangle-\langle\delta
l^{\gamma}_{1},\boldsymbol{u}_{1}+\alpha\boldsymbol{u}_{0}\rangle+h\sum_{j=2}^{m}\langle\delta^{2}l^{\gamma}_{j},\delta\boldsymbol{u}^{\gamma}_{j-1}+\alpha\boldsymbol{u}^{\gamma}_{j-1}\rangle$
$\displaystyle\quad+\frac{h}{2\eta}\sum_{j=1}^{m-1}\int_{\Omega}\int_{0}^{1}\Big{(}\frac{b(v^{\gamma}_{j})-b(v^{\gamma}_{j-1})}{h}\Big{)}^{2}(\mathbf{T}^{\gamma}_{j},\mathbf{T}^{\gamma}_{j})_{\mathcal{A}_{n}(s\mathbf{T}^{\gamma}_{j+1}+(1-s)\mathbf{T}^{\gamma}_{j})}\,\mathrm{d}s\,\mathrm{d}x.$

We have

(3.24) $\alpha
h\sum_{j=2}^{m}(\delta^{2}\boldsymbol{u}^{\gamma}_{j},\delta^{2}\boldsymbol{u}^{\gamma}_{j-1})\leq
C(\alpha)h\sum_{j=1}^{m-1}\|\delta^{2}\boldsymbol{u}^{\gamma}_{j}\|_{2}^{2}+\frac{\|\delta^{2}\boldsymbol{u}^{\gamma}_{m}\|_{2}^{2}}{8},$

and

(3.25)
$\alpha(\delta^{2}\boldsymbol{u}^{\gamma}_{m},\delta\boldsymbol{u}^{\gamma}_{m})\leq\frac{\|\delta^{2}\boldsymbol{u}^{\gamma}_{m}\|_{2}^{2}}{8}+2\alpha^{2}\|\delta\boldsymbol{u}^{\gamma}_{m}\|_{2}^{2}.$

Substituting (3.24) and (3.25) into (3.23) and applying the discrete Gronwall
inequality, it follows that, for every $1\leq m\leq M$,

(3.26)
$\displaystyle\|\delta^{2}\boldsymbol{u}^{\gamma}_{m}\|_{2}^{2}+h\sum_{j=1}^{m}\int_{\Omega}\int_{0}^{1}b(v^{\gamma}_{j-1})\Big{(}\delta\mathbf{T}^{\gamma}_{j},\delta\mathbf{T}^{\gamma}_{j}\Big{)}_{\mathcal{A}_{n}(s\mathbf{T}^{\gamma}_{j}+(1-s)\mathbf{T}^{\gamma}_{j-1})}\,\mathrm{d}s\,\mathrm{d}x$
$\displaystyle\leq
C(\delta)\Big{[}\|\delta^{2}\boldsymbol{u}^{\gamma}_{0}\|_{2}^{2}+\max_{0\leq
j\leq m}\|\delta\boldsymbol{u}^{\gamma}_{j}\|_{2}^{2}+\max_{1\leq j\leq
m}\|\delta
l^{\gamma}_{j}\|_{-1,2}^{2}+\Big{(}h\sum_{j=2}^{m}\|\delta^{2}l^{\gamma}_{j}\|_{-1,2}\Big{)}^{2}$
$\displaystyle\quad+h\sum_{j=1}^{m-1}\int_{\Omega}\int_{0}^{1}\Big{(}\frac{b(v^{\gamma}_{j})-b(v^{\gamma}_{j-1})}{h}\Big{)}^{2}(\mathbf{T}^{\gamma}_{j},\mathbf{T}^{\gamma}_{j})_{\mathcal{A}_{n}(s\mathbf{T}^{\gamma}_{j+1}+(1-s)\mathbf{T}^{\gamma}_{j})}\,\mathrm{d}s\,\mathrm{d}x+1\Big{]}$
$\displaystyle\quad+\delta\max_{0\leq j\leq
m}\|\boldsymbol{\varepsilon}(\delta\boldsymbol{u}^{\gamma}_{j}+\alpha\boldsymbol{u}^{\gamma}_{j})\|_{2}^{2},$

where $C=C(\delta)$ is a constant that is independent of $M$ and $n$, but does
depend on $\delta$, which is fixed sufficiently small so that the final term
on the right can be absorbed into the left-hand side. To prove that $\delta$
can be chosen such that this is possible, we first notice that

$\displaystyle\mathbf{T}^{\gamma}_{m}=h\sum_{j=1}^{m}\delta\mathbf{T}^{\gamma}_{j}+\mathbf{T}^{\gamma}_{0}=h\sum_{j=1}^{m}\delta\mathbf{T}^{\gamma}_{j}+F^{-1}_{n}(\boldsymbol{\varepsilon}(\boldsymbol{u}_{1}+\alpha\boldsymbol{u}_{0})).$

It follows that

$\displaystyle\max_{1\leq j\leq
m}\frac{\|\mathbf{T}^{\gamma}_{j}\|_{2}^{2}}{n}$ $\displaystyle\leq
C\Big{(}h\sum_{j=1}^{m}\frac{\|\delta\mathbf{T}^{\gamma}_{j}\|_{2}^{2}}{n}+\frac{\|F^{-1}_{n}(\boldsymbol{\varepsilon}(\boldsymbol{u}_{1}+\alpha\boldsymbol{u}_{0}))\|_{2}^{2}}{n}\Big{)}$
$\displaystyle\leq
C\Big{(}h\sum_{j=1}^{m}\int_{\Omega}\int_{0}^{1}b(v^{\gamma}_{j-1})\Big{(}\delta\mathbf{T}^{\gamma}_{j},\delta\mathbf{T}^{\gamma}_{j}\Big{)}_{\mathcal{A}_{n}(s\mathbf{T}^{\gamma}_{j}+(1-s)\mathbf{T}^{\gamma}_{j-1})}\,\mathrm{d}s\,\mathrm{d}x+1\Big{)}.$

By definition, we have

$\displaystyle|\boldsymbol{\varepsilon}(\delta\boldsymbol{u}^{\gamma}_{m}+\alpha\boldsymbol{u}^{\gamma}_{m})|\leq
1+\frac{|\mathbf{T}^{\gamma}_{m}|}{n}.$

This yields

$\displaystyle\max_{1\leq j\leq
m}\|\boldsymbol{\varepsilon}(\delta\boldsymbol{u}^{\gamma}_{j}+\alpha\boldsymbol{u}^{\gamma}_{j})\|_{2}^{2}$
$\displaystyle\leq
C\Big{(}1+h\sum_{j=1}^{m}\int_{\Omega}\int_{0}^{1}b(v^{\gamma}_{j-1})(\delta\mathbf{T}^{\gamma}_{j},\delta\mathbf{T}^{\gamma}_{j})_{\mathcal{A}_{n}(s\mathbf{T}^{\gamma}_{j}+(1-s)\mathbf{T}^{\gamma}_{j-1})}\,\mathrm{d}s\,\mathrm{d}x\Big{)}.$

Thus we see that $\delta$ can be chosen sufficiently small so that the final
term on the right-hand side of (3.26) can be absorbed into the left.
Optimising over the indices $1\leq m\leq M$, we deduce that

$\displaystyle\max_{1\leq m\leq
M}\|\delta^{2}\boldsymbol{u}^{\gamma}_{m}\|_{2}^{2}+h\sum_{m=1}^{M}\int_{\Omega}\int_{0}^{1}b(v^{\gamma}_{m})\big{(}\delta\mathbf{T}^{\gamma}_{m},\delta\mathbf{T}^{\gamma}_{m}\big{)}_{\mathcal{A}_{n}(s\mathbf{T}^{\gamma}_{m}+(1-s)\mathbf{T}^{\gamma}_{m-1})}\,\mathrm{d}s\,\mathrm{d}x$
$\displaystyle\leq C\Big{[}1+\max_{1\leq m\leq
M}\|\delta\boldsymbol{u}^{\gamma}_{m}\|_{2}^{2}+\max_{1\leq m\leq M}\|\delta
l^{\gamma}_{m}\|_{-1,2}^{2}+\Big{(}h\sum_{m=2}^{M}\|\delta^{2}l^{\gamma}_{m}\|_{-1,2}\Big{)}^{2}$
$\displaystyle\quad+h\sum_{m=1}^{M-1}\int_{\Omega}\int_{0}^{1}\Big{(}\frac{b(v^{\gamma}_{m})-b(v^{\gamma}_{m-1})}{h}\Big{)}^{2}(\mathbf{T}^{\gamma}_{m},\mathbf{T}^{\gamma}_{m})_{\mathcal{A}_{n}(s\mathbf{T}^{\gamma}_{m+1}+(1-s)\mathbf{T}^{\gamma}_{m})}\,\mathrm{d}s\,\mathrm{d}x\Big{]}.$

Rewriting this in terms of interpolants of the discrete solution sequence, we
obtain

(3.27)
$\displaystyle\sup_{t\in[0,T]}\|\overline{\boldsymbol{u}}^{\gamma,\prime}_{t}(t)\|_{2}^{2}+\sup_{t\in[0,T]}\|\boldsymbol{\varepsilon}(\overline{\boldsymbol{u}}^{\gamma}_{t}+\alpha\boldsymbol{u}^{\gamma,+})\|_{2}^{2}$
$\displaystyle\quad+\int_{Q}\int_{0}^{1}b(v^{\gamma,-})\big{(}\overline{\mathbf{T}}^{\gamma}_{t},\overline{\mathbf{T}}^{\gamma}_{t}\big{)}_{\mathcal{A}_{n}(s\mathbf{T}^{\gamma,+}+(1-s)\mathbf{T}^{\gamma,-})}\,\mathrm{d}s\,\mathrm{d}x\,\mathrm{d}t$
$\displaystyle\leq
C\Big{[}1+\max_{t\in[0,T]}\|\overline{\boldsymbol{u}}^{\gamma}(t)\|_{2}^{2}+\max_{t\in[0,T]}\|\overline{l}^{\gamma}_{t}(t)\|_{-1,2}^{2}+\Big{(}\int_{0}^{T}\|l^{\gamma,+,\prime\prime}(t)\|_{-1,2}\,\mathrm{d}t\Big{)}^{2}$
$\displaystyle\quad+h\int_{Q}\int_{0}^{1}\Big{(}\frac{b(v^{\gamma,+})-b(v^{\gamma,-})}{h}\Big{)}^{2}\cdot(\mathbf{T}^{\gamma,+},\mathbf{T}^{\gamma,+})_{\mathcal{A}_{n}(s\tau^{h}_{t}\mathbf{T}^{\gamma,+}+(1-s)\mathbf{T}^{n\gamma,+})}\,\mathrm{d}s\,\mathrm{d}x\,\mathrm{d}t\Big{]},$

where $C$ is independent of $M$ and $n$ and $\tau^{h}_{t}$ denotes the
translation operator in the time variable of length $h$.

Now we show that $\mathbf{T}^{n}\in W^{1,2}(0,T;L^{2}(\Omega)^{d\times d})$.
To bound the right-hand side of (3.27) independent of $M$, we notice that

$\displaystyle\frac{b(v^{\gamma,+})-b(v^{\gamma,-})}{h}\leq\|\overline{v}^{\gamma}_{t}\|_{\infty}\|b^{\prime}(\overline{v}^{\gamma})\|_{\infty}\leq
C(n).$

Furthermore, for every $\mathbf{T}$, $\mathbf{S}\in\mathbb{R}^{d\times d}$, we
have

$\displaystyle\frac{|\mathbf{T}|^{2}}{n}\leq(\mathbf{T},\mathbf{T})_{\mathcal{A}_{n}(\mathbf{S})}\leq
3|\mathbf{T}|^{2}.$

Using these two statements with the previously determined $M$-independent
bounds and the fact that $(l^{n})_{n}$ is bounded in
$W^{2,1}(0,T;W^{-1,2}_{D}(\Omega)^{d})$, it follows from (3.27) that

(3.28)
$\displaystyle\sup_{t\in[0,T]}\big{(}\|\overline{\boldsymbol{u}}^{\gamma,\prime}_{t}(t)\|_{2}^{2}\big{)}+\sup_{t\in[0,T]}\big{(}\|\boldsymbol{\varepsilon}(\overline{\boldsymbol{u}}^{\gamma}_{t}+\alpha\boldsymbol{u}^{\gamma,+})\|_{2}^{2}\big{)}+\int_{Q}\frac{|\overline{\mathbf{T}}^{\gamma}_{t}|^{2}}{n}\,\mathrm{d}x\,\mathrm{d}t$
$\displaystyle\leq
C\Big{\\{}1+\|l^{\gamma}_{t}\|_{L^{\infty}(W^{-1,2})}+\|l^{\gamma}_{tt}\|_{L^{1}(W^{-1,2})}^{2}+\int_{Q}|\mathbf{T}^{\gamma,+}|^{2}\,\mathrm{d}x\,\mathrm{d}s\Big{\\}}$
$\displaystyle\leq C,$

where $C=C(n)$ is independent of $M$. Hence
$(\overline{\mathbf{T}}^{(n,M)})_{M}$ is bounded in
$W^{1,2}(0,T;L^{2}(\Omega)^{d\times d})$ independent of $M$ and so converges
weakly in this space to
$\mathbf{T}^{n}=F_{n}(\boldsymbol{\varepsilon}(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n}))$.
However, since
$(\overline{\boldsymbol{u}}^{(n,M)}_{t}+\alpha\boldsymbol{u}^{(n,M),+})_{M}$
converges strongly in $L^{2}(0,T;W^{1,2}_{D}(\Omega)^{d})$ by (3.16), it
follows from the constitutive relation that $(\mathbf{T}^{(n,M),+})_{M}$
converges strongly in $L^{2}(Q)^{d\times d}$. Using Lebesgue’s dominated
convergence theorem, we get

(3.29)
$\displaystyle\lim_{M\rightarrow\infty}\int_{Q}\int_{0}^{1}(\mathbf{T}^{(n,M),+},\mathbf{T}^{(n,M),+})_{\mathcal{A}_{n}(s\tau^{h}_{t}\mathbf{T}^{(n,M),+}+(1-s)\mathbf{T}^{(n,M),+})}\,\mathrm{d}s\,\mathrm{d}x\,\mathrm{d}t$
$\displaystyle\quad=\int_{Q}(\mathbf{T}^{n},\mathbf{T}^{n})_{\mathcal{A}_{n}(\mathbf{T}^{n})}\,\mathrm{d}x\,\mathrm{d}t.$

Making further use of the previously determined strong convergence results for
the interpolants, we see that

(3.30)
$\displaystyle\lim_{M\rightarrow\infty}\Big{[}\max_{t\in[0,T]}\|\overline{\boldsymbol{u}}^{(n,M)}(t)\|_{2}^{2}+\max_{t\in[0,T]}\|\overline{l}^{(n,M)}_{t}(t)\|_{-1,2}^{2}+\Big{(}\int_{0}^{T}\|l^{(n,M),+,\prime\prime}(t)\|_{-1,2}\,\mathrm{d}t\Big{)}^{2}\Big{]}$
$\displaystyle=\sup_{t\in[0,T]}\|\boldsymbol{u}^{n}_{t}(t)\|_{2}^{2}+\sup_{t\in[0,T]}\|l^{n}_{t}(t)\|_{-1,2}^{2}+\Big{(}\int_{0}^{T}\|l^{n}_{tt}(t)\|_{-1,2}\,\mathrm{d}t\Big{)}^{2}$
$\displaystyle\leq C$

where $C$ is a constant that is independent of $n$. Ignoring the final term on
the left-hand side of (3.27), and using (3.29) and (3.30), it follows that

$\displaystyle\limsup_{M\rightarrow\infty}\Big{[}\sup_{t\in[0,T]}\|\overline{\boldsymbol{u}}^{(n,M),\prime}_{t}(t)\|_{2}^{2}+\sup_{t\in[0,T]}\|\boldsymbol{\varepsilon}(\overline{\boldsymbol{u}}^{(n,M)}_{t}+\alpha\boldsymbol{u}^{(n,M),+})(t)\|_{2}^{2}\Big{]}\leq
C,$

where $C$ is independent of $n$. By weak lower semi-continuity, we have

(3.31)
$\sup_{t\in[0,T]}\|\boldsymbol{u}^{n}_{tt}(t)\|_{2}+\sup_{t\in[0,T]}\|\boldsymbol{\varepsilon}(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n})(t)\|_{2}\leq
C.$

Next we use (3.31) with the extra regularity of $\mathbf{T}^{n}$ in order to
obtain the bound on $\mathbf{T}^{n}_{t}$ in the statement of the proposition.
Using that $\mathbf{T}^{n}\in W^{1,2}(0,T;L^{2}(\Omega)^{d\times d})$, it
follows that we have $F_{n}(\mathbf{T}^{n})\in
W^{1,2}(0,T;L^{2}(\Omega)^{d\times d})$, with weak derivative given by the
chain rule, that is, $\mathcal{A}_{n}(\mathbf{T}^{n})\mathbf{T}^{n}_{t}$. By
standard properties of differentiable Bochner functions, the following
convergence results hold for a.e. $t\in(0,T)$ with respect to strong
convergence in $L^{2}(\Omega)^{d\times d}$:

(3.32) $\lim_{h\rightarrow
0+}\frac{\mathbf{T}^{n}(t+h)-\mathbf{T}^{n}(t)}{h}=\mathbf{T}^{n}_{t},\quad\lim_{h\rightarrow
0+}\frac{F_{n}(\mathbf{T}^{n}(t+h))-F_{n}(\mathbf{T}^{n}(t))}{h}=\mathcal{A}_{n}(\mathbf{T}^{n})\mathbf{T}^{n}_{t}.$

Let $\Delta^{h}_{t}$ denote the undivided difference quotient of length $h$ in
the time variable and let $t\in(0,T)$. Considering the elastodynamic equation
at times $t$ and $t+h$, for a $h>0$ sufficiently small such that $t+h<T$,
choosing $\Delta^{h}_{t}(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n})(t)$
as the test function we see that

(3.33)
$\displaystyle\langle\Delta^{h}_{t}l^{n}(t),\Delta^{h}_{t}(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n})(t)\rangle$
$\displaystyle=\int_{\Omega}\Delta^{h}_{t}\boldsymbol{u}^{n}_{tt}(t)\cdot\Delta^{h}_{t}(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n})(t)+\Delta^{h}_{t}\Big{(}b(v^{n})\mathbf{T}^{n}\Big{)}(t):\boldsymbol{\varepsilon}\big{(}\Delta^{h}_{t}(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n})\big{)}(t)\,\mathrm{d}x$
$\displaystyle=\frac{\mathrm{d}}{\mathrm{d}t}\Big{(}\frac{\|\Delta^{h}_{t}\boldsymbol{u}^{n}_{t}\|_{2}^{2}}{2}+\alpha\int_{\Omega}\Delta^{h}_{t}\boldsymbol{u}^{n}_{t}\cdot\Delta^{h}_{t}\boldsymbol{u}^{n}\,\mathrm{d}x\Big{)}-\alpha\int_{\Omega}|\Delta^{h}_{t}\boldsymbol{u}_{t}^{n}|^{2}\,\mathrm{d}x$
$\displaystyle\quad+\int_{\Omega}\Delta^{h}_{t}b(v^{n})\tau^{h}_{t}\mathbf{T}^{n}:\Delta^{h}_{t}F_{n}(\mathbf{T}^{n})+b(v^{n})\Delta^{h}_{t}\mathbf{T}^{n}:\Delta^{h}_{t}F_{n}(\mathbf{T}^{n})\,\mathrm{d}x.$

From Lemmas 3.6 and 3.7, we have

$\displaystyle|\Delta^{h}_{t}b(v^{n})|=\Big{|}\int_{0}^{h}b^{\prime}(v^{n}(t+s))v^{n}_{t}(t+s)\,\mathrm{d}s\Big{|}\leq
h\|b^{\prime}(v^{n})\|_{L^{\infty}(Q)}\|v^{n}_{t}\|_{L^{\infty}(Q)}\leq Ch,$

where $C$ is independent of $n$. Using this and recalling that
$(\cdot,\cdot)_{\mathcal{A}_{n}(\mathbf{U})}$ is an inner product on
$\mathbb{R}^{d\times d}$ for every $\mathbf{U}\in\mathbb{R}^{d\times d}$, it
follows that

(3.34)
$\displaystyle\Big{|}\int_{\Omega}\Delta^{h}_{t}b(v^{n})\tau^{h}_{t}\mathbf{T}^{n}:\Delta^{h}_{t}F_{n}(\mathbf{T}^{n})\,\mathrm{d}x\Big{|}$
$\displaystyle\leq
Ch\int_{\Omega}\int_{0}^{1}\big{|}\big{(}\tau^{h}_{t}\mathbf{T}^{n},\Delta^{h}_{t}\mathbf{T}^{n}\big{)}_{\mathcal{A}_{n}(s\tau^{h}_{t}\mathbf{T}^{n}+(1-s)\mathbf{T}^{n})}\big{|}\,\mathrm{d}s\,\mathrm{d}x$
$\displaystyle\leq\frac{h^{2}}{2\eta}\int_{\Omega}\int_{0}^{1}(\tau^{h}_{t}\mathbf{T}^{n},\tau^{h}_{t}\mathbf{T}^{n})_{\mathcal{A}_{n}(s\tau^{h}_{t}\mathbf{T}^{n}+(1-s)\mathbf{T}^{n})}\,\mathrm{d}s\,\mathrm{d}x$
$\displaystyle\quad+\frac{\eta}{2}\int_{\Omega}\int_{0}^{1}\big{(}\Delta^{h}_{t}\mathbf{T}^{n},\Delta^{h}_{t}\mathbf{T}^{n}\big{)}_{\mathcal{A}_{n}(s\tau^{h}_{t}\mathbf{T}^{n}+(1-s)\mathbf{T}^{n})}\,\mathrm{d}s\,\mathrm{d}x.$

Now we notice that there exists a constant $C$, independent of $n$, such that

$\displaystyle|F_{n}(\mathbf{T})-F_{n}(\mathbf{S})|^{2}\leq
C(\mathbf{T}-\mathbf{S}):(F_{n}(\mathbf{T})-F_{n}(\mathbf{S})).$

Thus we have $|\Delta^{h}_{t}F_{n}(\mathbf{T}^{n})|^{2}\leq
C\Delta^{h}_{t}\mathbf{T}^{n}:\Delta^{h}_{t}F_{n}(\mathbf{T}^{n})$. As a
result, we can bound the term involving the external force in the following
way:

(3.35)
$\displaystyle|\langle\Delta^{h}_{t}l^{n},\Delta^{h}_{t}(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n})\rangle|$
$\displaystyle\leq
C\|\Delta^{h}_{t}l^{n}\|_{-1,2}\|\Delta^{h}_{t}\boldsymbol{\varepsilon}(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n})\|_{2}$
$\displaystyle\leq
C\|\Delta^{h}_{t}l^{n}\|_{-1,2}^{2}+\frac{\eta}{4}\int_{\Omega}\Delta^{h}_{t}\mathbf{T}^{n}:\Delta^{h}_{t}F_{n}(\mathbf{T}^{n})\,\mathrm{d}x,$

where $C$ is independent of $n$. Substituting (3.34) and (3.35) into (3.33)
and integrating over $(t_{1},t_{2})\subset[0,T]$, we see that

$\displaystyle{\|\Delta^{h}_{t}\boldsymbol{u}^{n}_{t}(t_{2})\|_{2}^{2}}+\int_{t_{1}}^{t_{2}}\int_{\Omega}{b(v^{n})}\Delta^{h}_{t}\mathbf{T}^{n}:\Delta^{h}_{t}F_{n}(\mathbf{T}^{n})\,\mathrm{d}x\,\mathrm{d}t$
$\displaystyle\leq
C\Big{[}\int_{t_{1}}^{t_{2}}\|\Delta^{h}_{t}l^{n}\|_{-1,2}^{2}\,\mathrm{d}t+h^{2}\int_{t_{1}}^{t_{2}}\int_{\Omega}\int_{0}^{1}(\tau^{h}_{t}\mathbf{T}^{n},\tau^{h}_{t}\mathbf{T}^{n})_{\mathcal{A}_{n}(s\tau^{h}_{t}\mathbf{T}^{n}+(1-s)\mathbf{T}^{n})}\,\mathrm{d}s\,\mathrm{d}x\,\mathrm{d}t$
$\displaystyle\quad+\int_{t_{1}}^{t_{2}}\|\Delta^{h}_{t}\boldsymbol{u}^{n}_{t}(t)\|_{2}^{2}\,\mathrm{d}t+\|\Delta^{h}_{t}\boldsymbol{u}^{n}_{t}(t_{1})\|_{2}^{2}+\|\Delta^{h}_{t}\boldsymbol{u}^{n}(t_{1})\|_{2}^{2}\Big{]}.$

Dividing through by $h^{2}$ and using standard properties of difference
quotients, we get

(3.36)
$\displaystyle\frac{\|\Delta^{h}_{t}\boldsymbol{u}^{n}_{t}(t_{2})\|_{2}^{2}}{h^{2}}+\int_{t_{1}}^{t_{2}}\int_{\Omega}b(v^{n})\frac{\Delta^{h}_{t}\mathbf{T}^{n}}{h}:\frac{\Delta^{h}_{t}F_{n}(\mathbf{T}^{n})}{h}\,\mathrm{d}x\,\mathrm{d}t$
$\displaystyle\leq
C\Big{[}\int_{0}^{T}\|l^{n}_{t}\|_{-1,2}^{2}\,\mathrm{d}t+\int_{t_{1}}^{t_{2}}\int_{\Omega}\int_{0}^{1}\big{(}\tau^{h}_{t}\mathbf{T}^{n},\tau^{n}_{t}\mathbf{T}^{n}\big{)}_{\mathcal{A}_{n}(s\tau^{h}_{t}\mathbf{T}^{n}+(1-s)\mathbf{T}^{n})}\,\mathrm{d}s\,\mathrm{d}x\,\mathrm{d}t$
$\displaystyle\quad+\sup_{t\in[0,T]}\|\boldsymbol{u}^{n}_{tt}(t)\|_{2}^{2}+\sup_{t\in[0,T]}\|\boldsymbol{u}^{n}_{t}(t)\|_{2}^{2}\Big{]}$
$\displaystyle\leq
C\Big{[}1+\int_{t_{1}}^{t_{2}}\int_{\Omega}\int_{0}^{1}\big{(}\tau^{h}_{t}\mathbf{T}^{n},\tau^{n}_{t}\mathbf{T}^{n}\big{)}_{\mathcal{A}_{n}(s\tau^{h}_{t}\mathbf{T}^{n}+(1-s)\mathbf{T}^{n})}\,\mathrm{d}s\,\mathrm{d}x\,\mathrm{d}t\Big{]},$

where $C$ is a constant, independent of $n$ and $h$. Next, we take the limit
as $h\rightarrow 0+$. The strong convergence result
$\tau^{h}_{t}\mathbf{T}^{n}\rightarrow\mathbf{T}^{n}$ in
$L^{2}((t_{1},t_{2})\times\Omega)^{d\times d}$ and the dominated convergence
theorem yield

(3.37) $\displaystyle\lim_{h\rightarrow
0+}\int_{t_{1}}^{t_{2}}\int_{\Omega}\int_{0}^{1}\big{(}\tau^{h}_{t}\mathbf{T}^{n},\tau^{n}_{t}\mathbf{T}^{n}\big{)}_{\mathcal{A}_{n}(s\tau^{h}_{t}\mathbf{T}^{n}+(1-s)\mathbf{T}^{n})}\,\mathrm{d}s\,\mathrm{d}x\,\mathrm{d}t$
$\displaystyle=\int_{t_{1}}^{t_{2}}\int_{\Omega}\big{(}\mathbf{T}^{n},\mathbf{T}^{n})_{\mathcal{A}_{n}(\mathbf{T}^{n})}\,\mathrm{d}x\,\mathrm{d}t$
$\displaystyle\leq
C\int_{t_{1}}^{t_{2}}\int_{\Omega}|\mathbf{T}^{n}|+\frac{|\mathbf{T}^{n}|^{2}}{n}\,\mathrm{d}x\,\mathrm{d}t$
$\displaystyle\leq C.$

Using (3.32), Fatou’s lemma and the monotonicity of $F_{n}$, we have

(3.38) $\displaystyle\liminf_{h\rightarrow
0+}\int_{t_{1}}^{t_{2}}\int_{\Omega}b(v^{n})\frac{\Delta^{h}_{t}\mathbf{T}^{n}}{h}:\frac{\Delta^{h}_{t}F_{n}(\mathbf{T}^{n})}{h}\,\mathrm{d}x\,\mathrm{d}t$
$\displaystyle\geq\int_{t_{1}}^{t_{2}}\int_{\Omega}b(v)\mathbf{T}^{n}_{t}:F_{n}(\mathbf{T}^{n})_{t}\,\mathrm{d}x\,\mathrm{d}t$
$\displaystyle\geq\int_{t_{1}}^{t_{2}}\int_{\Omega}\eta\big{(}\mathbf{T}^{n}_{t},\mathbf{T}^{n}_{t})_{\mathcal{A}_{n}(\mathbf{T}^{n})}\,\mathrm{d}x\,\mathrm{d}t$
$\displaystyle=\eta\int_{t_{1}}^{t_{2}}\int_{\Omega}\frac{|\mathbf{T}^{n}_{t}|^{2}}{(1+|\mathbf{T}^{n}|^{a})^{1+\frac{1}{a}}}+\frac{|\mathbf{T}^{n}_{t}|^{2}}{n}\,\mathrm{d}x\,\mathrm{d}t.$

Substituting (3.37) and (3.38) into (3.36) gives the required result. ∎

Noticing that we have
$|\Delta^{h}_{t}\boldsymbol{\varepsilon}(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n})|\leq
C\Delta^{h}_{t}\mathbf{T}^{n}:\Delta^{h}_{t}F_{n}(\mathbf{T}^{n})$, as well as
the fact that $\boldsymbol{\varepsilon}(\boldsymbol{u}^{n}_{t})\in
L^{\infty}(0,T;L^{2}(\Omega)^{d\times d})$, we immediately obtain the bound in
Corollary 3.5. Then in Lemma 3.8, we use the new bounds to improve the
integrability of $(\mathbf{T}^{n})_{n}$ to
$L^{\infty}(0,T;L^{1}(\Omega)^{d\times d})$.

######  Corollary 3.5.

There exists a constant $C$ independent of $n$ such that

$\displaystyle\int_{0}^{T}\|\boldsymbol{u}^{n}_{tt}(t)\|_{1,2}^{2}\,\mathrm{d}t\leq
C.$

######  Lemma 3.8.

Let the assumptions of Proposition 3.2 hold and let
$(\boldsymbol{u}^{n},\mathbf{T}^{n},v^{n})$ be the solution triple constructed
there. Suppose additionally that $l\in W^{2,1}(0,T;W^{-1,2}_{D}(\Omega)^{d})$,
the safety strain condition (3.5) holds, the compatibility condition (3.4)
holds, and that $\boldsymbol{u}_{1}+\alpha\boldsymbol{u}_{0}\in
W^{2,2}_{D}(\Omega)^{d}$. There exists a constant $C$ independent of $n$ such
that

$\displaystyle\sup_{t\in[0,T]}\int_{\Omega}|\mathbf{T}^{n}(t)|\,\mathrm{d}x\leq
C.$

###### Proof.

Let $t\in(0,T)$ and pick $h>0$ sufficiently small such that $t+h<T$. We test
in (3.12) at times $t$ and $t+h$ against
$(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n})(t)$ and subtract the
resulting equations. Dividing through by $h$, integrating over
$(t_{1},t_{2})\subset[0,T]$ and letting $h\rightarrow 0+$, we deduce that

$\displaystyle\int_{t_{1}}^{t_{2}}\langle
l^{n}_{t},\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n}\rangle\,\mathrm{d}t$
$\displaystyle=\int_{\Omega}\boldsymbol{u}^{n}_{tt}(t_{2})\cdot(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n})(t_{2})-\boldsymbol{u}^{n}_{tt}(t_{1})\cdot(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n})(t_{1})\,\mathrm{d}x$
$\displaystyle\quad+\int_{t_{1}}^{t_{2}}\int_{\Omega}b^{\prime}(v^{n})v^{n}_{t}\mathbf{T}^{n}:F_{n}(\mathbf{T}^{n})+b(v^{n})\mathbf{T}^{n}_{t}:F_{n}(\mathbf{T}^{n})\,\mathrm{d}x\,\mathrm{d}t$
$\displaystyle\quad-\int_{t_{1}}^{t_{2}}\int_{\Omega}\boldsymbol{u}^{n}_{tt}\cdot(\boldsymbol{u}^{n}_{tt}+\alpha\boldsymbol{u}^{n}_{t})\,\mathrm{d}x\,\mathrm{d}t.$

From the chain rule and product rule for weak derivatives, we have
$\varphi_{n}(\mathbf{T}^{n})\in W^{1,1}(0,T;L^{1}(\Omega))$ with weak
derivative given by $\mathbf{T}^{n}_{t}:F_{n}(\mathbf{T}^{n})$. Furthermore,
since $\boldsymbol{u}^{n}_{tt}\in L^{2}(0,T;W^{1,2}(\Omega)^{d})$, we must
have $\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n}\in
C([0,T];W^{1,2}_{D}(\Omega)^{d})$ and so $F_{n}(\mathbf{T}^{n})\in
C([0,T];L^{2}(\Omega)^{d\times d})$. By continuity of $F^{-1}_{n}$, it follows
that $\mathbf{T}^{n}\in C([0,T];L^{2}(\Omega)^{d\times d})$ with
$\mathbf{T}^{n}(0)=F^{-1}_{n}(\boldsymbol{\varepsilon}(\boldsymbol{u}_{1}+\alpha\boldsymbol{u}_{0}))$.
Putting together these facts, we get

(3.39) $\displaystyle\lim_{t_{1}\rightarrow
0+}\int_{t_{1}}^{t_{2}}\int_{\Omega}b(v^{n})\mathbf{T}^{n}_{t}:F_{n}(\mathbf{T}^{n})\,\mathrm{d}x\,\mathrm{d}t$
$\displaystyle=\lim_{t_{1}\rightarrow
0+}\Big{\\{}\Big{[}\int_{\Omega}b(v^{n}(t))\varphi_{n}(\mathbf{T}^{n}(t))\,\mathrm{d}x\Big{]}_{t=t_{1}}^{t=t_{2}}-\int_{t_{1}}^{t_{2}}\int_{\Omega}b^{\prime}(v^{n})v^{n}_{t}\varphi_{n}(\mathbf{T}^{n})\,\mathrm{d}x\,\mathrm{d}t\Big{\\}}$
$\displaystyle=\int_{\Omega}\big{(}b(v^{n})\varphi_{n}(\mathbf{T}^{n})\big{)}(t_{2})-b(v_{0})\varphi_{n}(F_{n}^{-1}(\boldsymbol{\varepsilon}(\boldsymbol{u}_{1}+\alpha\boldsymbol{u}_{0})))\,\mathrm{d}x-\int_{0}^{t_{2}}\int_{\Omega}b^{\prime}(v^{n})v^{n}_{t}\varphi_{n}(\mathbf{T}^{n})\,\mathrm{d}x\,\mathrm{d}t.$

Furthermore, we have

(3.40)
$\displaystyle\Big{[}\int_{\Omega}b(v^{n}(t))\varphi_{n}(\mathbf{T}^{n}(t))\,\mathrm{d}x\Big{]}_{t=t_{1}}^{t=t_{2}}$
$\displaystyle\leq\Big{[}\int_{\Omega}b(v^{n}(t))\varphi_{n}(\mathbf{T}^{n}(t))\,\mathrm{d}x\Big{]}_{t=t_{1}}^{t=t_{2}}-\int_{t_{1}}^{t_{2}}\int_{\Omega}b^{\prime}(v^{n})v^{n}_{t}\varphi_{n}(\mathbf{T}^{n})\,\mathrm{d}x\,\mathrm{d}t$
$\displaystyle=\int_{t_{1}}^{t_{2}}\langle
l^{n}_{t},\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n}\rangle\,\mathrm{d}t-\int_{\Omega}\boldsymbol{u}^{n}_{tt}(t_{2})\cdot(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n})(t_{2})-\boldsymbol{u}^{n}_{tt}(t_{1})\cdot(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n})(t_{1})\,\mathrm{d}x$
$\displaystyle\quad-\int_{t_{1}}^{t_{2}}\int_{\Omega}b^{\prime}(v^{n})v^{n}_{t}\mathbf{T}^{n}:F_{n}(\mathbf{T}^{n})\,\mathrm{d}x\,\mathrm{d}t+\int_{t_{1}}^{t_{2}}\int_{\Omega}\boldsymbol{u}^{n}_{tt}\cdot(\boldsymbol{u}^{n}_{tt}+\alpha\boldsymbol{u}^{n}_{t})\,\mathrm{d}x\,\mathrm{d}t$
$\displaystyle\leq
C\Big{(}\int_{0}^{T}\|l^{n}_{t}\|_{-1,2}^{2}\,\mathrm{d}t+\sup_{t\in[0,T]}\|\boldsymbol{u}^{n}_{tt}(t)\|_{2}^{2}+\sup_{t\in[0,T]}\|(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n})(t)\|_{1,2}^{2}+\sup_{t\in[0,T]}\|\boldsymbol{u}^{n}_{t}(t)\|_{2}^{2}$
$\displaystyle\quad+\|v^{n}_{t}\|_{\infty}\int_{Q}\Big{[}\frac{|\mathbf{T}^{n}|^{2}}{n}+|\mathbf{T}^{n}|\Big{]}\,\mathrm{d}x\,\mathrm{d}t\Big{)}$
$\displaystyle\leq C,$

where $C$ is a constant that is independent of $n$, $t_{1}$ and $t_{2}$.
Letting $t_{1}\rightarrow 0+$ in (3.40) and using (3.39) again, it follows
that

(3.41) $\int_{\Omega}\varphi_{n}(\mathbf{T}^{n}(t_{2}))\,\mathrm{d}x\leq
C\Big{(}1+\int_{\Omega}b(v_{0})\varphi_{n}(F^{-1}_{n}(\boldsymbol{\varepsilon}(\boldsymbol{u}_{1}+\alpha\boldsymbol{u}_{0})))\,\mathrm{d}x\Big{)}\leq
C.$

The second inequality follows from the safety strain condition (3.5). Using
the definition of $\varphi_{n}$, there exists positive constants $c_{1}$ and
$c_{2}$, independent of $n$, such that
$c_{1}|\mathbf{T}|-c_{2}\leq\varphi_{n}(\mathbf{T})$, for every
$\mathbf{T}\in\mathbb{R}^{d\times d}$. Substituting this into (3.41), the
asserted bound follows. ∎

Next, we improve the spatial regularity of the stress tensor. We use arguments
mimicking those in [3] and [8]. However, now we have the extra issue of the
phase-field function being present. It is here that we require $\nabla v^{n}$
to be uniformly bounded in $L^{\infty}(Q)$, independent of $n$. Thus we assume
that $k>\frac{d}{2}+1$ to ensure that we have $H^{k}(\Omega)\subset
C^{1}(\overline{\Omega})$.

######  Proposition 3.4.

Let the assumptions of Proposition 3.2 hold and let
$(\boldsymbol{u}^{n},\mathbf{T}^{n},v^{n})$ be the solution triple constructed
there. Suppose also that $l\in W^{2,1}(0,T;W^{-1,2}_{D}(\Omega)^{d})$,
$\boldsymbol{f}\in L^{2}(0,T;W^{1,2}_{loc}(\Omega)^{d})$, the safety strain
condition (3.5) holds, the compatibility condition (3.4) holds and
$\boldsymbol{u}_{1}+\alpha\boldsymbol{u}_{0}\in W^{2,2}_{D}(\Omega)^{d}$ with
$\boldsymbol{u}_{0}\in W^{2,2}_{loc}(\Omega)^{d}$. Furthermore, assume
$k>\frac{d}{2}+1$. For every compact subset $\Omega_{0}\subset\Omega$, there
exists a constant $C=C(\Omega_{0})$, independent of $n$, such that

$\displaystyle\sup_{t\in[0,T]}\|\nabla^{2}\boldsymbol{u}^{n}(t)\|_{L^{2}(\Omega_{0})}+\int_{0}^{T}\int_{\Omega_{0}}\frac{|\nabla\mathbf{T}^{n}|^{2}}{(1+|\mathbf{T}^{n}|)^{1+a}}+\frac{|\nabla\mathbf{T}^{n}|^{2}}{n}+|\nabla^{2}\boldsymbol{u}^{n}_{t}(t)|^{2}\,\mathrm{d}x\,\mathrm{d}t\leq
C.$

###### Proof.

Let $\zeta\in C^{\infty}_{c}(\Omega)$ be a cut-off function between $\Omega$
and $\Omega_{0}$. We denote by $\Omega_{1}$, a compact subset of $\Omega$ that
contains $\mathrm{supp}(\zeta)$. Suppose that $h>0$ is sufficiently small so
that $\zeta\Delta^{h}_{i}\boldsymbol{w}$ is well-defined on $\Omega$ for an
arbitrary $\boldsymbol{w}\in L^{2}(\Omega)^{d}$, where $\Delta^{h}_{i}$ is the
undivided difference quotient of length $h$ in the $i$-th spatial co-ordinate
direction. For a.e. $t\in(0,T)$, noticing that the Neumann boundary term
vanishes due to the compact support of $\zeta$, using Proposition 3.3 we get

$\displaystyle\int_{\Omega}\zeta^{2}\Delta^{h}_{i}\boldsymbol{u}^{n}_{tt}\cdot\Delta^{h}_{i}(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n})+\Delta^{h}_{i}\big{(}b(v^{n})\mathbf{T}^{n}\big{)}:\boldsymbol{\varepsilon}\big{(}\zeta^{2}\Delta^{h}_{i}(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n})\big{)}\,\mathrm{d}x$
$\displaystyle=\int_{\Omega}\zeta^{2}\Delta^{h}_{i}\boldsymbol{f}(t)\cdot\Delta^{h}_{i}(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n})\,\mathrm{d}x$
$\displaystyle\leq
C(\zeta)h^{2}\|\nabla\boldsymbol{f}(t)\|_{L^{2}(\Omega_{1})}\|\nabla(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n})(t)\|_{L^{2}(\Omega_{1})}$
$\displaystyle\leq Ch^{2}g(t),$

where $g\in L^{2}(0,T;\mathbb{R}_{+})$ with bound in this space independent of
$n$ but depending on $\Omega_{1}$. Similarly, we have

$\displaystyle\int_{\Omega}\zeta^{2}\Delta^{h}_{i}\boldsymbol{u}^{n}_{tt}\cdot\Delta^{h}_{i}(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n})\,\mathrm{d}x$
$\displaystyle\leq
C(\zeta)h^{2}\|\nabla\boldsymbol{u}^{n}_{tt}(t)\|_{L^{2}(\Omega_{1})}\|\nabla(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n})(t)\|_{L^{2}(\Omega_{1})}$
$\displaystyle\leq
C(\zeta)h^{2}\|\nabla\boldsymbol{u}^{n}_{tt}(t)\|_{L^{2}(\Omega_{1})}$
$\displaystyle\leq C(\zeta)h^{2}g(t),$

making $g$ larger if necessary. The right-hand side is in $L^{2}(0,T)$ by
Corollary 3.5, with bound independent of $n$. For the nonlinear terms, we
expand as follows:

$\displaystyle\int_{\Omega}\Delta^{h}_{i}\big{(}b(v^{n})\mathbf{T}^{n}\big{)}:\boldsymbol{\varepsilon}\big{(}\zeta^{2}\Delta^{h}_{i}(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n})\big{)}\,\mathrm{d}x$
$\displaystyle=\int_{\Omega}\Big{(}\Delta^{h}_{i}b(v^{n})\tau^{h}_{i}\mathbf{T}^{n}+b(v^{n})\Delta^{h}_{i}\mathbf{T}^{n}\Big{)}:\Big{(}2\zeta\nabla\zeta\otimes\Delta^{h}_{i}(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n})+\zeta^{2}\boldsymbol{\varepsilon}(\Delta^{h}_{i}(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n}))\Big{)}\,\mathrm{d}x,$

where $\tau^{h}_{i}$ is a translation of length $h$ in the $i$-th spatial co-
ordinate direction. Integration over $(0,T)$ yields

$\displaystyle\int_{Q}\eta\zeta^{2}\frac{|\Delta^{h}_{i}\mathbf{T}^{n}|^{2}}{n}\,\mathrm{d}x\,\mathrm{d}t$
$\displaystyle\leq\int_{Q}\zeta^{2}b(v^{n})\Delta^{h}_{i}\mathbf{T}^{n}:\Delta^{h}_{i}F_{n}(\mathbf{T}^{n})\,\mathrm{d}x\,\mathrm{d}t$
$\displaystyle\leq
Ch^{2}\Big{[}1+\int_{0}^{T}g(t)\,\mathrm{d}t\Big{]}+\int_{Q}\Big{|}2\zeta\Delta^{h}_{i}b(v^{n})\tau^{h}_{i}\mathbf{T}^{n}:\big{(}\nabla\zeta\otimes\Delta^{h}_{i}(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n})\big{)}\Big{|}\,\mathrm{d}x\,\mathrm{d}t$
$\displaystyle\quad+\int_{Q}\Big{|}\zeta^{2}\Delta^{h}_{i}b(v^{n})\tau^{h}_{i}\mathbf{T}^{n}:\Delta^{h}_{i}F_{n}(\mathbf{T}^{n})\Big{|}\,\mathrm{d}x\,\mathrm{d}t+\int_{Q}\Big{|}2\zeta
b(v^{n})\Delta^{h}_{i}\mathbf{T}^{n}:\big{(}\nabla\zeta\otimes\Delta^{h}_{i}(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n})\big{)}\Big{|}\,\mathrm{d}x\,\mathrm{d}t.$

Now we use this to show that $\mathbf{T}^{n}\in
L^{2}(0,T;W^{1,2}_{loc}(\Omega)^{d\times d})$ with bound that does depend on
$n$. Using Young’s inequality and the fact that, for every $\mathbf{T}$,
$\mathbf{S}\in\mathbb{R}^{d\times d}$,
$|F_{n}(\mathbf{T})-F_{n}(\mathbf{S})|\leq 3|\mathbf{T}-\mathbf{S}|$, we see
that

$\displaystyle\frac{\eta}{2}\int_{Q}\zeta^{2}\frac{|\Delta^{h}_{i}\mathbf{T}^{n}|^{2}}{n}\,\mathrm{d}x\,\mathrm{d}t$
$\displaystyle\leq
C(n,\zeta)\Big{[}h^{2}+\int_{0}^{T}\Big{\\{}\||\nabla\zeta|\Delta^{h}_{i}(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n})\|_{2}^{2}+h^{2}\|\nabla
v^{n}\|_{L^{\infty}(\Omega_{1})}^{2}\|\zeta\tau^{h}_{i}\mathbf{T}^{n}\|_{2}^{2}$
$\displaystyle\quad+h\|\nabla
v^{n}\|_{L^{\infty}(\Omega_{1})}\|\zeta\tau^{h}_{i}\mathbf{T}^{n}\|_{2}\||\nabla\zeta|\Delta^{h}_{i}(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n})\|_{2}\Big{\\}}\,\mathrm{d}t\Big{]}$
$\displaystyle\leq C(n,\zeta)h^{2},$

using the $L^{\infty}(0,T;W^{1,\infty}(\Omega))$ bound on $v^{n}$, which is a
consequence of the Sobolev embedding $H^{k}(\Omega)\subset
W^{1,\infty}(\Omega)$. We divide through by $h^{2}$ and let $h\rightarrow 0+$
in order to deduce that $\mathbf{T}^{n}$ is an element of
$L^{2}(0,T;W^{1,2}_{loc}(\Omega)^{d\times d})$. As a result, we have
$\boldsymbol{\varepsilon}(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n})\in
L^{2}(0,T;W^{1,2}_{loc}(\Omega)^{d\times d})$, from which we deduce that
$\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n}\in
L^{2}(0,T;W^{2,2}_{loc}(\Omega)^{d})$, using the Korn–Poincaré inequality.

From the elastodynamic equation (3.12) and this extra regularity, it follows
that

(3.42)
$\boldsymbol{u}^{n}_{tt}=\mathrm{div}\big{(}b(v^{n})\mathbf{T}^{n}\big{)}+\boldsymbol{f}\quad\text{
pointwise a.e. in }Q,$

with each term an element of $L^{2}(0,T;L^{2}_{loc}(\Omega)^{d})$. We take the
scalar product of (3.42) with the function
$\zeta^{2}\nabla\cdot\nabla(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n})$
and integrate over $\Omega$ to obtain

(3.43)
$\displaystyle\int_{\Omega}\zeta^{2}\boldsymbol{u}^{n}_{tt}\cdot\big{(}\nabla\cdot\nabla(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n})\big{)}\,\mathrm{d}x$
$\displaystyle=\int_{\Omega}\zeta^{2}\mathrm{div}\big{(}b(v^{n})\mathbf{T}^{n}\big{)}\cdot\big{(}\nabla\cdot\nabla(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n})\big{)}+\zeta^{2}\boldsymbol{f}\cdot\big{(}\nabla\cdot\nabla(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n})\big{)}\,\mathrm{d}x.$

First we focus on the term containing the nonlinearity $\mathbf{T}^{n}$. We
have

$\displaystyle\int_{\Omega}\zeta^{2}\mathrm{div}\big{(}b(v^{n})\mathbf{T}^{n}\big{)}\cdot\big{(}\nabla\cdot\nabla(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n})\big{)}\,\mathrm{d}x$
$\displaystyle=\int_{\Omega}\zeta^{2}b(v^{n})\frac{\partial
T^{n}_{ij}}{\partial x_{j}}\frac{\partial^{2}}{\partial
x_{k}^{2}}\big{(}\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n}\big{)}_{i}+\zeta^{2}b^{\prime}(v^{n})T^{n}_{ij}\frac{\partial
v^{n}}{\partial x_{j}}\frac{\partial^{2}}{\partial
x_{k}^{2}}\big{(}\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n}\big{)}_{i}\,\mathrm{d}x.$

Using an integration by parts argument, we write

$\displaystyle\int_{\Omega}\zeta^{2}b(v^{n})\frac{\partial
T^{n}_{ij}}{\partial x_{j}}\frac{\partial^{2}}{\partial
x_{k}^{2}}\big{(}\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n}\big{)}_{i}\,\mathrm{d}x$
$\displaystyle=\int_{\Omega}\Big{\\{}\zeta^{2}b(v^{n})\frac{\partial
T^{n}_{ij}}{\partial x_{k}}\frac{\partial}{\partial
x_{k}}\boldsymbol{\varepsilon}(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n})_{ij}-T^{n}_{ij}\frac{\partial^{2}}{\partial
x_{k}^{2}}\big{(}\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n}\big{)}_{i}\Big{(}b^{\prime}(v^{n})\frac{\partial
v^{n}}{\partial x_{j}}\zeta^{2}+2\zeta\frac{\partial\zeta}{\partial
x_{j}}b(v^{n})\Big{)}$ $\displaystyle\quad+T^{n}_{ij}\frac{\partial}{\partial
x_{k}}\boldsymbol{\varepsilon}(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n})_{ij}\Big{(}b^{\prime}(v^{n})\frac{\partial
v^{n}}{\partial x_{k}}\zeta^{2}+2\zeta\frac{\partial\zeta}{\partial
x_{k}}b(v^{n})\Big{)}\Big{\\}}\,\mathrm{d}x.$

This yields

(3.44)
$\displaystyle\int_{\Omega}\zeta^{2}\mathrm{div}\big{(}b(v^{n})\mathbf{T}^{n}\big{)}\cdot\big{(}\nabla\cdot(\nabla(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n}))\big{)}\,\mathrm{d}x$
$\displaystyle=\int_{\Omega}\zeta^{2}b(v^{n})\nabla\mathbf{T}^{n}\vdots\nabla
F_{n}(\mathbf{T}^{n})+T^{n}_{ij}\frac{\partial^{2}}{\partial
x_{k}^{2}}(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n})_{i}\Big{(}\zeta^{2}b^{\prime}(v^{n})\frac{\partial
v^{n}}{\partial x_{j}}\Big{)}$
$\displaystyle\quad-T^{n}_{ij}\frac{\partial^{2}}{\partial
x_{k}^{2}}(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n})_{i}\Big{(}b^{\prime}(v^{n})\frac{\partial
v^{n}}{\partial x_{j}}\zeta^{2}+2\zeta\frac{\partial\zeta}{\partial
x_{j}}b(v^{n})\Big{)}$ $\displaystyle\quad+T^{n}_{ij}\frac{\partial}{\partial
x_{k}}\boldsymbol{\varepsilon}(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n})_{ij}\Big{(}b^{\prime}(v^{n})\frac{\partial
v^{n}}{\partial x_{k}}\zeta^{2}+2\zeta\frac{\partial\zeta}{\partial
x_{k}}b(v^{n})\Big{)}\,\mathrm{d}x$
$\displaystyle=\int_{\Omega}\zeta^{2}b(v^{n})\nabla\mathbf{T}^{n}\vdots\nabla
F_{n}(\mathbf{T}^{n})+\zeta\frac{\partial}{\partial
x_{k}}\boldsymbol{\varepsilon}(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n})_{ij}B^{n,k}_{ij}\,\mathrm{d}x,$

where $\mathbf{B}^{n,k}=(B^{n,k}_{ij})_{i,j=1}^{d}$ is defined, for $1\leq
k\leq d$, by

$\displaystyle B^{n,k}_{ij}$
$\displaystyle=T^{n}_{ij}\big{(}2b(v^{n})\partial_{k}\zeta+\zeta
b^{\prime}(v^{n})\partial_{k}v^{n}\Big{)}-2\delta_{jk}T^{n}_{im}\Big{(}2b(v^{n})\partial_{m}\zeta+\tau
b^{\prime}(v^{n})\partial_{m}v^{n}\Big{)}$
$\displaystyle\quad+\delta_{ij}T^{n}_{km}\Big{(}2b(v^{n})\partial_{m}\zeta+\tau
b^{\prime}(v^{n})\partial_{m}v^{n}\Big{)}+2\delta_{jk}T^{n}_{im}\zeta
b^{\prime}(v^{n})\partial_{m}v^{n}-\delta_{ij}T^{n}_{km}\zeta
b^{\prime}(v^{n})\partial_{m}v^{n},$

making use of the identity

$\displaystyle\frac{\partial^{2}}{\partial
x_{k}^{2}}\boldsymbol{w}_{i}=2\frac{\partial}{\partial
x_{k}}\boldsymbol{\varepsilon}(\boldsymbol{w})_{ik}-\frac{\partial}{\partial
x_{i}}\boldsymbol{\varepsilon}(\boldsymbol{w})_{kk}.$

Using the bound on $(v^{n})_{n}$ in $L^{\infty}(0,T;W^{1,\infty}(\Omega))$ and
the definition of $\mathbf{B}^{n,k}$, there exists a constant $C$, independent
of $n$, such that $|\mathbf{B}^{n,k}|\leq C|\mathbf{T}^{n}|$, for every $1\leq
k\leq d$. Furthermore, we have

$\displaystyle\Big{|}\int_{\Omega}\zeta\frac{\partial}{\partial
x_{k}}\boldsymbol{\varepsilon}(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n}):\mathbf{B}^{n,k}\,\mathrm{d}x\Big{|}$
$\displaystyle=\Big{|}\int_{\Omega}\zeta\frac{\partial}{\partial
x_{k}}F_{n}(\mathbf{T}^{n}):\mathbf{B}^{n,k}\,\mathrm{d}x\Big{|}$
$\displaystyle=\Big{|}\int_{\Omega}\zeta\big{(}\partial_{k}\mathbf{T}^{n},\mathbf{B}^{n,k}\big{)}_{\mathcal{A}_{n}(\mathbf{T}^{n})}\,\mathrm{d}x\Big{|}$
$\displaystyle\leq\int_{\Omega}\frac{\zeta^{2}\eta}{2}\big{(}\partial_{k}\mathbf{T}^{n},\partial_{k}\mathbf{T}^{n}\big{)}_{\mathcal{A}_{n}(\mathbf{T}^{n})}+C(\eta)\mathbb{I}_{\mathrm{supp}(\zeta)}\big{(}\mathbf{B}^{n,k},\mathbf{B}^{n,k}\big{)}_{\mathcal{A}_{n}(\mathbf{T}^{n})}\,\mathrm{d}x$
$\displaystyle\leq\int_{\Omega}\frac{\zeta^{2}\eta}{2}\big{(}\partial_{k}\mathbf{T}^{n},\partial_{k}\mathbf{T}^{n}\big{)}_{\mathcal{A}_{n}(\mathbf{T}^{n})}+C\Big{(}\frac{|\mathbf{B}^{n,k}|^{2}}{n}+\frac{|\mathbf{B}^{n,k}|^{2}}{1+|\mathbf{T}^{n}|}\Big{)}\,\mathrm{d}x$
$\displaystyle\leq\int_{\Omega}\frac{\zeta^{2}\eta}{2}\nabla\mathbf{T}^{n}\vdots\nabla
F_{n}(\mathbf{T}^{n})+C\Big{(}\frac{|\mathbf{T}^{n}|^{2}}{n}+|\mathbf{T}^{n}|\Big{)}\,\mathrm{d}x.$

Using this in (3.44) and substituting the result into (3.43), we obtain

(3.45)
$\displaystyle\int_{\Omega}\zeta^{2}\Big{(}\boldsymbol{u}^{n}_{tt}-\boldsymbol{f}\Big{)}\cdot\Big{(}\nabla\cdot\big{(}\nabla(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n})\big{)}\Big{)}\,\mathrm{d}x+C\int_{\Omega}\frac{|\mathbf{T}^{n}|^{2}}{n}+|\mathbf{T}^{n}|\,\mathrm{d}x$
$\displaystyle\geq\int_{\Omega}\frac{\zeta^{2}b(v^{n})}{2}\nabla\mathbf{T}^{n}\vdots\nabla
F_{n}(\mathbf{T}^{n})\,\mathrm{d}x.$

For the first term on the right-hand side, we integrate by parts to deduce
that

(3.46)
$\displaystyle\int_{\Omega}\zeta^{2}\Big{(}\boldsymbol{u}^{n}_{tt}-\boldsymbol{f}\Big{)}\cdot\Big{(}\nabla\cdot\big{(}\nabla(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n})\big{)}\Big{)}\,\mathrm{d}x$
$\displaystyle=\int_{\Omega}\zeta^{2}\nabla\big{(}\boldsymbol{f}-\boldsymbol{u}^{n}_{tt}\big{)}:\nabla\big{(}\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n}\big{)}+2\zeta\big{[}\nabla\zeta\otimes\big{(}\boldsymbol{f}-\boldsymbol{u}^{n}_{tt}\big{)}\big{]}:\nabla\big{(}\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n}\big{)}\,\mathrm{d}x$
$\displaystyle=-\frac{\mathrm{d}}{\mathrm{d}t}\Big{(}\frac{\|\zeta\nabla\boldsymbol{u}^{n}_{t}\|_{2}^{2}}{2}+\alpha\int_{\Omega}\zeta^{2}\nabla\boldsymbol{u}^{n}_{t}:\nabla\boldsymbol{u}^{n}\,\mathrm{d}x\Big{)}+\alpha\int_{\Omega}\zeta^{2}|\nabla\boldsymbol{u}^{n}_{t}|^{2}\,\mathrm{d}x$
$\displaystyle\quad+\|\zeta\nabla\boldsymbol{f}\|_{2}\|\zeta\nabla(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n})\|_{2}+C(\zeta)\|\boldsymbol{f}\|_{2}\|\nabla(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n})\|_{2}$
$\displaystyle\quad+C(\zeta)\|\boldsymbol{u}^{n}_{tt}\|_{2}\|\nabla(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n})\|_{2}.$

Integrating in time, using the continuity of $\nabla\boldsymbol{u}^{n}$ and
$\nabla\boldsymbol{u}^{n}_{t}$ as maps from $[0,T]$ to $L^{2}(\Omega)^{d\times
d}$, with the assumptions on the initial data, and applying Gronwall’s
inequality, it follows from (3.45) and (3.46) that

(3.47)
$\sup_{t\in[0,T]}\|\zeta\nabla\boldsymbol{u}^{n}_{t}(t)\|_{2}^{2}+\int_{Q}\zeta^{2}\Big{(}\frac{|\nabla\mathbf{T}^{n}|^{2}}{(1+|\mathbf{T}^{n}|)^{1+a}}+\frac{|\nabla\mathbf{T}^{n}|^{2}}{n}\Big{)}\,\mathrm{d}x\,\mathrm{d}t\leq
C(\zeta).$

Recalling that for every $\mathbf{T}$, $\mathbf{S}\in\mathbb{R}^{d\times d}$
we have

$\displaystyle|F_{n}(\mathbf{T})-F_{n}(\mathbf{S})|^{2}\leq
C(\mathbf{T}-\mathbf{S}):(F_{n}(\mathbf{T})-F_{n}(\mathbf{S})),$

where $C$ is a constant independent of $n$, it follows from (3.47) that

$\displaystyle\int_{0}^{T}\int_{\Omega_{0}}|\boldsymbol{\varepsilon}(\partial_{i}(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n}))|^{2}\,\mathrm{d}x\,\mathrm{d}t\leq\int_{0}^{T}\int_{\Omega_{0}}|\nabla(\boldsymbol{\varepsilon}(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n}))|^{2}\leq
C(\Omega_{0}),$

for every $1\leq i\leq d$. Applying Korn’s inequality for functions in the
space $W^{1,2}(\Omega_{0})^{2}$, we deduce that

$\displaystyle\int_{0}^{T}\|\partial_{i}(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n})\|_{W^{1,2}(\Omega_{0})}^{2}\,\mathrm{d}t$
$\displaystyle\leq
C\int_{0}^{T}\|\partial_{i}(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n})\|_{L^{2}(\Omega_{0})}^{2}+\|\boldsymbol{\varepsilon}(\partial_{i}(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n}))\|_{L^{2}(\Omega)}^{2}\,\mathrm{d}t$
$\displaystyle\leq
C\int_{0}^{T}\|\nabla\boldsymbol{u}^{n}_{t}\|_{2}^{2}+\|\nabla\boldsymbol{u}^{n}\|_{2}^{2}+\|\nabla(\boldsymbol{\varepsilon}(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n}))\|_{L^{2}(\Omega_{0})}^{2}\,\mathrm{d}t$
$\displaystyle\leq C(\Omega_{0}).$

Hence $\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n}\in
L^{2}(0,T;W^{2,2}_{loc}(\Omega)^{d})$ with bound independent of $n$. It
remains to prove that $\boldsymbol{u}^{n}\in
L^{\infty}(0,T;W^{2,2}_{loc}(\Omega)^{d})$. From the memory kernel property,
we have

$\displaystyle\boldsymbol{u}^{n}(t)=\mathrm{e}^{-\alpha
t}\boldsymbol{u}_{0}+\int_{0}^{t}\mathrm{e}^{\alpha(s-t)}(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n})\,\mathrm{d}s,$

with equality meant in $L^{2}(\Omega)^{d}$. Each term on the right-hand side
is an element of $W^{2,2}_{loc}(\Omega)^{d}$. Hence, for a.e. $t\in(0,T)$, we
must have $\boldsymbol{u}^{n}(t)\in W^{2,2}_{loc}(\Omega)^{d}$. Furthermore,
we have

$\displaystyle\nabla^{2}\boldsymbol{u}^{n}(t)=\mathrm{e}^{-\alpha
t}\nabla^{2}\boldsymbol{u}_{0}+\int_{0}^{t}\mathrm{e}^{\alpha(s-t)}\nabla^{2}(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n})\,\mathrm{d}s,$

with equality in $L^{2}_{loc}(\Omega)^{d\times d}$. Applying the
$L^{2}(\Omega_{0})$ norm to this and using the previously determined result
for $\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n}$, we conclude the stated
result. ∎

Now we have sufficient $n$-independent estimates to take the limit as
$n\rightarrow\infty$. In Theorem 3.2, we focus on proving a pointwise
convergence result for $(\mathbf{T}^{n})_{n}$ and show that the limiting
triple satisfies the weak elastodynamic equation, up to the Neumann boundary
condition. We also prove the satisfaction of the initial conditions. We delay
the proof of the minimisation problem and the energy-dissipation equality
until Propositions 3.5 and 3.6, respectively, for simplicity of the
presentation.

######  Theorem 3.2.

Assume that $k>\frac{d}{2}+1$ with data $v_{0}\in H^{k}_{D+1}(\Omega)$, and
$\boldsymbol{u}_{0}$, $\boldsymbol{u}_{1}\in W^{1,2}_{D}(\Omega)^{d}$ such
that $\boldsymbol{u}_{0}\in W^{2,2}_{loc}(\Omega)^{d}$,
$\boldsymbol{u}_{1}+\alpha\boldsymbol{u}_{0}\in W^{2,2}_{D}(\Omega)^{d}$ and
the safety strain condition (3.5) holds. Furthermore, assume that $l\in
W^{2,1}(0,T;W^{-1,2}_{D}(\Omega)^{d})$ such that $\boldsymbol{f}\in
L^{2}(0,T;W^{1,2}_{loc}(\Omega)^{d})$ and the compatibility condition (3.4)
holds. Let $((\boldsymbol{u}^{n},\mathbf{T}^{n},v^{n}))_{n}$ be the sequence
of solutions to the regularised problem constructed in Proposition 3.2. There
exists a subsequence in $n$, not relabelled, and a limiting triple
$(\boldsymbol{u},\mathbf{T},v)$ such that

  * •

$\boldsymbol{u}^{n}\overset{\ast}{\rightharpoonup}\boldsymbol{u}$ weakly-* in
$W^{2,\infty}(0,T;L^{2}(\Omega)^{d})\cap
W^{1,\infty}(0,T;W^{1,2}_{D}(\Omega)^{d})$,

  * •

$\boldsymbol{u}^{n}\rightharpoonup\boldsymbol{u}$ weakly in
$W^{2,2}(0,T;W^{1,2}_{D}(\Omega)^{d})\cap
W^{1,2}(0,T;W^{2,2}_{loc}(\Omega)^{d})$,

  * •

$\boldsymbol{\varepsilon}(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n})\rightarrow\boldsymbol{\varepsilon}(\boldsymbol{u}_{t}+\alpha\boldsymbol{u})$
pointwise a.e. in $Q$,

  * •

$v^{n}\overset{\ast}{\rightharpoonup}v$ weakly-* in
$W^{1,\infty}(0,T;H^{k}(\Omega))$ with $v(t)\in H^{k}_{D+1}(\Omega)$ for every
$t\in[0,T]$,

  * •

$\mathbf{T}^{n}\rightarrow\mathbf{T}$ pointwise a.e. in $Q$ where
$\mathbf{T}\in L^{\infty}(0,T;L^{1}(\Omega)^{d\times d})$.

Furthermore, there exists an error term $\tilde{\boldsymbol{g}}\in
L^{\infty}_{w^{*}}(0,T;(C^{1}_{0}(\Gamma_{N})^{d})^{*})$ such that

(3.48)
$\int_{\Omega}\boldsymbol{u}_{tt}(t)\cdot\boldsymbol{w}+b(v(t))\mathbf{T}(t):\boldsymbol{\varepsilon}(\boldsymbol{w})\,\mathrm{d}x=\int_{\Omega}\boldsymbol{f}(t)\cdot\boldsymbol{w}\,\mathrm{d}x+\langle\boldsymbol{g}(t)-\tilde{\boldsymbol{g}}(t),\boldsymbol{w}\rangle_{C_{0}^{1}(\Gamma_{N})},$

for a.e. $t\in(0,T)$ and every $\boldsymbol{w}\in
C^{1}_{D}(\overline{\Omega})^{d}$, where the stress tensor is identified by
the relation
$\boldsymbol{\varepsilon}(\boldsymbol{u}_{t}+\alpha\boldsymbol{u})=F(\mathbf{T})$,
pointwise a.e. in $Q$. Furthermore, the initial conditions hold in the sense
that

(3.49) $\lim_{t\rightarrow
0+}\Big{[}\|\boldsymbol{u}(t)-\boldsymbol{u}_{0}\|_{1,2}+\|\boldsymbol{u}_{t}(t)-\boldsymbol{u}_{1}\|_{1,2}+\|v(t)-v_{0}\|_{k,2}\Big{]}=0.$

###### Proof.

The first, second and fourth convergence results follow immediately from the a
priori bounds in Lemma 3.5, Lemma 3.6, Lemma 3.7, Proposition 3.3 and
Corollary 3.5. Using Propositions 3.3 and 3.4, applying the reasoning in [9]
yields the pointwise convergence result $\mathbf{T}^{n}\rightarrow\mathbf{T}$
a.e. in $Q$. In particular, we consider sequences

$\mathbf{S}^{n}=\frac{\mathbf{T}^{n}}{(1+|\mathbf{T}^{n}|)^{1+a}},\quad
s^{n}=\frac{1}{(1+|\mathbf{T}^{n}|)^{1+a}}.$

By a careful application of the chain rule for weak derivatives, it follows
that $(\mathbf{S}^{n})_{n}$ is bounded in $W^{1,2}(0,T;L^{2}(\Omega)^{d\times
d})$ and $L^{2}(0,T;W^{1,2}_{loc}(\Omega)^{d\times d})$, independent of $n$.
An analogous result holds for $(s^{n})_{n}$. By the Aubin–Lions lemma, it
follows that $\mathbf{S}^{n}\rightarrow\mathbf{S}$ and $s^{n}\rightarrow s$
strongly in $L^{2}(0,T;L^{2}_{loc}(\Omega))$ and hence pointwise a.e. in $Q$,
up to a subsequence that we do not relabel. By Fatou’s lemma and the bound on
$(\mathbf{T}^{n})_{n}$ in $L^{1}(Q)^{d\times d}$, we deduce that
$s^{-\frac{1}{1+a}}\in L^{1}(Q)$. Hence $s>0$ a.e. in $Q$ and so
$\mathbf{T}^{n}=(s^{n})^{-1}\mathbf{S}^{n}$ converges pointwise a.e. in $Q$ to
a limit which we denote by $\mathbf{T}$. By Fatou’s lemma and Lemma 3.8,
$\mathbf{T}\in L^{\infty}(0,T;L^{1}(\Omega)^{d\times d})$. The fifth bullet
point now follows.

By Lemma 3.5, we have that $n^{-1}\mathbf{T}^{n}\rightarrow\mathbf{0}$
strongly in $L^{2}(Q)^{d\times d}$. With the constitutive relation for the
regularised problem and the pointwise convergence of $(\mathbf{T}^{n})_{n}$,
this implies the pointwise convergence result for
$(\boldsymbol{\varepsilon}(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n}))_{n}$,
noting that weak limits in $L^{2}(Q)$ and pointwise limits coincide. Hence all
of the stated convergence results hold. Furthermore, the strain-limiting
constitutive relation between $\mathbf{T}$ and $\boldsymbol{u}$ must hold by
uniqueness of pointwise limits.

The attainment of the initial conditions (3.49) follows from standard
arguments, using the aforementioned regularity, application of the Aubin–Lions
lemma and the initial conditions satisfied by the sequence of approximate
solutions.

Adapting the reasoning in [8], noticing that the Neumann boundary term
vanishes when considering compactly supported test functions, and using the
fact that $(b(v^{n}))_{n}$ is uniformly bounded on $Q$ and converges strongly
in $L^{p}(Q)$ to $b(v)$ for every $p\in[1,\infty)$, we see that, for every
$\boldsymbol{w}\in W^{1,2}_{0}(\Omega)^{d}$ and a.e. $t\in(0,T)$,

(3.50)
$\int_{\Omega}\boldsymbol{u}_{tt}(t)\cdot\boldsymbol{w}+b(v(t))\mathbf{T}(t):\boldsymbol{\varepsilon}(\boldsymbol{w})\,\mathrm{d}x=\langle
l(t),\boldsymbol{w}\rangle=\int_{\Omega}\boldsymbol{f}(t)\cdot\boldsymbol{w}\,\mathrm{d}x.$

Next, we want to be able to test against functions that only vanish on the
Dirichlet part of the boundary, and so obtain some information regarding the
satisfaction of the Neumann boundary condition. First we notice that
$(\mathbf{T}^{n})_{n}$ is bounded in $L^{\infty}(0,T;L^{1}(\Omega)^{d\times
d})$ and hence in
$L^{\infty}_{w^{*}}(0,T;\mathcal{M}(\overline{\Omega})^{d\times d})$, the dual
space of $L^{1}(0,T;C(\overline{\Omega})^{d\times d})$. Up to a further
subsequence that we do not relabel, it follows from the Banach–Alaoglu theorem
that we have $\mathbf{T}^{n}\overset{\ast}{\rightharpoonup}\mathbf{T}$
weakly-* in $L^{\infty}_{w^{*}}(0,T;\mathcal{M}(\overline{\Omega})^{d\times
d})$. Furthermore, this limit satisfies

(3.51)
$\int_{\Omega}\boldsymbol{u}_{tt}(t)\cdot\boldsymbol{w}\,\mathrm{d}x+\langle\overline{\mathbf{T}}(t),\boldsymbol{\varepsilon}(\boldsymbol{w})\rangle_{(\mathcal{M}(\overline{\Omega}),C(\overline{\Omega}))}=\langle
l(t),\boldsymbol{w}\rangle,$

for every $\boldsymbol{w}\in C^{1}_{D}(\overline{\Omega})^{d}$ and a.e.
$t\in(0,T)$, where
$\langle\cdot,\cdot\rangle_{(\mathcal{M}(\overline{\Omega}),C(\overline{\Omega}))}$
denotes the duality pairing between $\mathcal{M}(\overline{\Omega})$ and
$C(\overline{\Omega})$. In the spirit of [3], to define the error on
$\Gamma_{N}$, we define the normal component the stresses $\mathbf{T}(t)$ and
$\overline{\mathbf{T}}(t)$ on the Neumann boundary, respectively, by

$\displaystyle\langle\mathbf{T}(t)\mathbf{n},\boldsymbol{w}\rangle|_{C^{1}_{0}(\Gamma_{N})}=\int_{\Omega}\boldsymbol{u}_{tt}(t)\cdot\tilde{\boldsymbol{w}}+\mathbf{T}(t):\boldsymbol{\varepsilon}(\tilde{\boldsymbol{w}})\,\mathrm{d}x,$

and

$\displaystyle\langle\overline{\mathbf{T}}(t)\mathbf{n},\boldsymbol{w}\rangle_{C^{1}_{0}(\Gamma_{N})}=\int_{\Omega}\boldsymbol{u}_{tt}(t)\cdot\tilde{\boldsymbol{w}}\,\mathrm{d}x+\langle\overline{\mathbf{T}}(t),\tilde{\boldsymbol{w}}\rangle_{(\mathcal{M}(\overline{\Omega}),C(\overline{\Omega}))},$

for a.e. $t\in(0,T)$ and every $\boldsymbol{w}\in C^{1}_{0}(\Gamma_{N})^{d}$,
where $\tilde{\boldsymbol{w}}\in C^{1}_{D}(\overline{\Omega})^{d}$ is an
extension of $\boldsymbol{w}$ to $\overline{\Omega}$ such that it vanishes on
$\Gamma_{D}$. Such an extension exists by the definition of
$C^{1}_{0}(\Gamma_{N})^{d}$. Using (3.50) and (3.51), the definitions of
$\mathbf{T}(t)\mathbf{n}$ and $\overline{\mathbf{T}}(t)\mathbf{n}$ as above
are independent of the choice of extension $\tilde{\boldsymbol{w}}$.
Furthermore, by the discussion in Section 1.1, we can choose the extension
such that $\|\tilde{\boldsymbol{w}}\|_{1,\infty}\leq
C\|\boldsymbol{w}\|_{C^{1}_{0}(\Gamma_{N})}$, where $C$ depends only on
$\Omega$. Hence, $\mathbf{T}(t)\mathbf{n}$,
$\overline{\mathbf{T}}(t)\mathbf{n}$ are well-defined elements of the dual
space $(C^{1}_{0}(\Gamma_{N})^{d})^{*}$. Moreover, by the stated regularity
results, we have $\mathbf{T}(t)\mathbf{n}$,
$\overline{\mathbf{T}}(t)\mathbf{n}\in
L^{\infty}_{w^{*}}(0,T;(C^{1}_{0}(\Gamma_{N})^{d})^{*})$. Then we define the
penalisation $\tilde{\boldsymbol{g}}$ on $(0,T)\times\Gamma_{N}$ by

(3.52)
$\langle\tilde{\boldsymbol{g}}(t),\boldsymbol{w}\rangle=\langle\overline{\mathbf{T}}(t)\mathbf{n}-\mathbf{T}(t)\mathbf{n},\boldsymbol{w}\rangle,$

for a.e. $t\in(0,T)$, for every $\boldsymbol{w}\in C^{1}_{0}(\Gamma_{N})^{d}$.
Using (3.50), (3.51) and (3.52), it follows that (3.48) holds. Hence we
conclude the existence of a solution to the elastodynamic equation with
strain-limiting constitutive relation, up to the Neumann boundary condition. ∎

######  Proposition 3.5.

Let the assumptions of Theorem 3.2 hold and let
$(\boldsymbol{u},\mathbf{T},v,\tilde{\boldsymbol{g}})$ be the solution
quadruple constructed there. For a.e. $t\in(0,T)$, $v(t)$ is the unique
solution of the minimisation problem

(3.53)
$\mathcal{E}(\boldsymbol{u}(t),v(t))+\mathcal{H}(v(t))+\mathcal{G}_{k}(v(t),v_{t}(t))=\inf_{v\in
H^{k}_{D+1}(\Omega),\,v\leq
v(t)}\Big{\\{}\mathcal{E}(\boldsymbol{u}(t),v)+\mathcal{H}(v)+\mathcal{G}_{k}(v,v_{t}(t))\Big{\\}}.$

Furthermore, $v_{t}\leq 0$ a.e. in $Q$.

###### Proof.

Since $v^{n}_{t}\leq 0$ in $Q$ and weak convergence preserves ordering, it is
immediate that $v_{t}\leq 0$ a.e. in $Q$. To prove (3.53), it is enough to
show, for a.e. $t\in(0,T)$ and every $\chi\in H^{k}_{D}(\Omega)$ such that
$\chi\leq 0$, that the following holds:

(3.54)
$0\leq\big{[}\partial_{v}\mathcal{E}(\boldsymbol{u}(t),v(t))+\mathcal{H}^{\prime}(v(t))+\partial_{v}\mathcal{G}_{k}(v(t),v_{t}(t))\big{]}(\chi).$

For every $n$ and every test function $\chi$ as above, we know that the
solution of the regularised problem satisfies

$\displaystyle
0\leq\big{[}\partial_{v}\mathcal{E}_{n}(\boldsymbol{u}^{n}(t),v^{n}(t))+\mathcal{H}^{\prime}(v^{n}(t))+\partial_{v}\mathcal{G}_{k}(v^{n}(t),v_{t}^{n}(t))\big{]}(\chi).$

Using the convergence results of Theorem 3.2, we immediately deduce that

(3.55)
$\lim_{n\rightarrow\infty}\int_{0}^{T}\psi\big{[}\mathcal{H}^{\prime}(v^{n}(t))+\partial_{v}\mathcal{G}_{k}(v^{n}(t),v_{t}^{n}(t))\big{]}(\chi)\,\mathrm{d}t=\int_{0}^{T}\psi\big{[}\mathcal{H}^{\prime}(v(t))+\partial_{v}\mathcal{G}_{k}(v(t),v_{t}(t))\big{]}(\chi)\,\mathrm{d}t,$

for a fixed but arbitrary $\psi\in C([0,T])$ such that $\psi\geq 0$.

For the nonlinear term, first we claim that
$\boldsymbol{\varepsilon}(\boldsymbol{u}^{n})\rightarrow\boldsymbol{\varepsilon}(\boldsymbol{u})$
pointwise a.e. in $Q$. Recalling that $(\boldsymbol{u}^{n})_{n}$ converges
weakly in $W^{1,2}(0,T;W^{2,2}_{loc}(\Omega)^{d})$ and applying the
Aubin–Lions lemma, it follows that
$\boldsymbol{u}^{n}\rightarrow\boldsymbol{u}$ strongly in
$C([0,T];W^{1,2}_{loc}(\Omega)^{d})$. Considering an exhaustion of $\Omega$
yields the required pointwise convergence result.

Next, we claim that
$(\varphi^{*}(\boldsymbol{\varepsilon}(\alpha\boldsymbol{u}^{n})))_{n}$
converges pointwise a.e. on $Q$ to
$\varphi^{*}(\boldsymbol{\varepsilon}(\alpha\boldsymbol{u}))$. Using that
$\boldsymbol{\varepsilon}(\boldsymbol{u}_{t}+\alpha\boldsymbol{u})=F(\mathbf{T})$
with the memory kernel property and the safety strain condition (3.5), we have

$\displaystyle|\boldsymbol{\varepsilon}(\boldsymbol{u}(t))|$
$\displaystyle=\Big{|}\mathrm{e}^{-\alpha
t}\boldsymbol{\varepsilon}(\boldsymbol{u}_{0})+\int_{0}^{t}\mathrm{e}^{\alpha(s-t)}F(\mathbf{T}(s))|\,\mathrm{d}s\Big{|}$
$\displaystyle\leq\frac{\mathrm{e}^{-\alpha
t}}{\alpha}C_{*}+\int_{0}^{t}\mathrm{e}^{\alpha(s-t)}\,\mathrm{d}t$
$\displaystyle=\frac{1}{\alpha}\Big{(}1+\mathrm{e}^{-\alpha
T}(C_{*}-1)\Big{)}$ $\displaystyle=\frac{C_{**}}{\alpha},$

where $C_{**}\in(0,1)$. In particular,
$\|\boldsymbol{\varepsilon}(\alpha\boldsymbol{u})\|_{L^{\infty}(Q)}\leq
C_{**}<1$. For a.e. $(t,x)\in Q$, for $n$ sufficiently large (depending on
$(t,x)$) we have that
$|\boldsymbol{\varepsilon}(\alpha\boldsymbol{u}^{n}(t,x))|$ is bounded away
from $1$ as a consequence of the pointwise convergence result. The function
$\varphi^{*}$ is well-defined and finite on the open unit ball. Also
$\varphi^{*}_{n}(\mathbf{T})\rightarrow\varphi^{*}(\mathbf{T})$ as
$n\rightarrow\infty$ for every $|\mathbf{T}|<1$. Thus, we deduce that

$\displaystyle\varphi^{*}_{n}(\boldsymbol{\varepsilon}(\alpha\boldsymbol{u}^{n}))\rightarrow\varphi^{*}(\boldsymbol{\varepsilon}(\alpha\boldsymbol{u}))\quad\text{
pointwise a.e. in }Q.$

Combining this with the pointwise convergence of $(v^{n})_{n}$, we get that

$\psi\frac{b^{\prime}(v^{n})}{\alpha}\chi\varphi^{*}_{n}(\boldsymbol{\varepsilon}(\alpha\boldsymbol{u}^{n}))\rightarrow\psi\frac{b^{\prime}(v)}{\alpha}\chi\varphi^{*}(\boldsymbol{\varepsilon}(\alpha\boldsymbol{u}))\quad\text{
pointwise a.e. in }Q.$

The sequence on the left is non-positive a.e. in $Q$, using the non-positivity
of $\chi$. Hence, by the reverse Fatou lemma, we deduce that

(3.56)
$\displaystyle\int_{0}^{T}\psi\partial_{v}\mathcal{E}(\boldsymbol{u}(t),v(t))(\chi)\,\mathrm{d}t$
$\displaystyle=\int_{Q}\psi\frac{b^{\prime}(v)}{\alpha}\chi\varphi^{*}(\boldsymbol{\varepsilon}(\alpha\boldsymbol{u}))\,\mathrm{d}x\,\mathrm{d}t$
$\displaystyle\geq\lim_{n\rightarrow\infty}\int_{Q}\psi\frac{b^{\prime}(v^{n})}{\alpha}\chi\varphi^{*}_{n}(\boldsymbol{\varepsilon}(\alpha\boldsymbol{u}^{n}))\,\mathrm{d}x\,\mathrm{d}t.$

Combining (3.56) with (3.55) and recalling that $\psi$ is arbitrary, we see
that (3.54) holds and so we conclude the stated result. ∎

Using the pointwise convergence results, Fatou’s lemma and the weak lower
semi-continuity of norms, taking the limit as $n\rightarrow\infty$ in the
energy-dissipation equality for the regularised solution immediately yields
the following result.

######  Proposition 3.6.

Let the assumptions of Theorem 3.2 hold and let
$(\boldsymbol{u},\mathbf{T},v,\tilde{\boldsymbol{g}})$ be the solution
quadruple constructed there. For every $t\in[0,T]$, the following energy-
dissipation inequality holds:

$\displaystyle\mathcal{F}(t;\boldsymbol{u}(t),\boldsymbol{u}_{t}(t),v(t))+\int_{0}^{t}\|v_{t}(s)\|_{k,2}^{2}\,\mathrm{d}s+\int_{0}^{t}\langle
l_{t}(s),\boldsymbol{u}(s)\rangle\,\mathrm{d}s$
$\displaystyle\quad+\int_{0}^{t}\int_{\Omega}b(v)\big{[}F^{-1}(\boldsymbol{\varepsilon}(\boldsymbol{u}_{t}+\alpha\boldsymbol{u}))-F^{-1}(\boldsymbol{\varepsilon}(\alpha\boldsymbol{u}))\big{]}:\boldsymbol{\varepsilon}(\boldsymbol{u}_{t})\,\mathrm{d}x\,\mathrm{d}s$
$\displaystyle\leq\mathcal{F}(0;\boldsymbol{u}_{0},\boldsymbol{u}_{1},v_{0}).$

Before trying to obtain the opposite inequality, we discuss equivalent
formulations of the energy-dissipation equality to motivate our proof in the
case of fully Dirichlet boundary conditions, as well as to see why it fails in
the case of mixed boundary conditions. We note that we cannot adapt the proof
of Proposition 2.6 to this setting due to the lack of satisfaction of the
elastodynamic equation without a penalisation term.

Differentiating and carefully applying the chain rule for weak derivatives, an
equivalent formulation of the energy-dissipation equality is

$\displaystyle 0$
$\displaystyle=\Big{\\{}\int_{\Omega}\boldsymbol{u}_{tt}(t)\cdot\boldsymbol{u}_{t}(t)+b(v(t))\mathbf{T}(t):\boldsymbol{\varepsilon}(\boldsymbol{u}_{t}(t))\,\mathrm{d}x-\langle
l(t),\boldsymbol{u}_{t}(t)\rangle\Big{\\}}$
$\displaystyle\quad+\Big{[}\int_{\Omega}\frac{b^{\prime}(v(t))v_{t}(t)}{\alpha}\varphi^{*}(\boldsymbol{\varepsilon}(\alpha\boldsymbol{u}(t)))+\frac{1}{2\epsilon}(v(t)-1)v_{t}(t)+2\epsilon\nabla
v(t)\cdot\nabla v_{t}(t)\,\mathrm{d}x+\|v_{t}(t)\|_{k,2}^{2}\Big{]}.$

The terms in the curly brackets correspond to the elastodynamic equation, and
the second set of brackets is exactly

$\big{[}\partial_{v}\mathcal{E}(\boldsymbol{u}(t),v(t))+\mathcal{H}^{\prime}(v(t))+\partial_{v}\mathcal{G}_{k}(v(t),v_{t}(t))\big{]}(v_{t}(t))$

The latter collection of terms in bounded below by zero. Hence, to show that
the sum of terms in the square brackets vanishes, we must prove that

(3.57)
$\displaystyle\int_{\Omega}\frac{b^{\prime}(v(t))v_{t}(t)}{\alpha}\varphi^{*}(\boldsymbol{\varepsilon}(\alpha\boldsymbol{u}(t)))+\frac{1}{2\epsilon}(v(t)-1)v_{t}(t)+2\epsilon\nabla
v(t)\cdot\nabla v_{t}(t)\,\mathrm{d}x+\|v_{t}(t)\|_{k,2}^{2}\leq 0.$

If (3.57) holds, the energy-dissipation equality reduces to showing that

(3.58)
$\int_{\Omega}\boldsymbol{u}_{tt}(t)\cdot\boldsymbol{u}_{t}(t)+b(v(t))\mathbf{T}(t):\boldsymbol{\varepsilon}(\boldsymbol{u}_{t}(t))\,\mathrm{d}x-\langle
l(t),\boldsymbol{u}_{t}(t)\rangle=0,$

for a.e. $t\in(0,T)$. Clearly (3.58) holds if $\Gamma_{N}\neq\emptyset$.
Otherwise, it is sufficient to prove that

$\displaystyle\lim_{n\rightarrow\infty}\int_{Q}b(v^{n})\mathbf{T}^{n}:\boldsymbol{\varepsilon}(\boldsymbol{u}^{n}_{t})\,\mathrm{d}x\,\mathrm{d}t=\int_{Q}b(v)\mathbf{T}:\boldsymbol{\varepsilon}(\boldsymbol{u}_{t})\,\mathrm{d}x\,\mathrm{d}t.$

However, due to a lack of convergence results for $(\mathbf{T}^{n})_{n}$ that
are global in space, it is not clear how to prove such a result in the mixed
boundary case. Hence, from now on, we only consider the case
$\Gamma_{D}=\partial\Omega$ and so, to obtain the energy-dissipation equality
it suffices to prove that (3.57) holds.

Using weak lower semi-continuity and the fact that $k>1$, we see that, for
every $\psi\in C([0,T])$ such that $\psi\geq 0$, we have

$\displaystyle 0$
$\displaystyle=\lim_{n\rightarrow\infty}\int_{0}^{T}\psi(t)\Big{\\{}\int_{\Omega}\Big{[}\frac{b^{\prime}(v^{n}(t))v_{t}^{n}(t)}{\alpha}\varphi^{*}_{n}(\boldsymbol{\varepsilon}(\alpha\boldsymbol{u}^{n}(t)))+\frac{1}{2\epsilon}(v^{n}(t)-1)v_{t}^{n}(t)$
$\displaystyle\quad+2\epsilon\nabla v^{n}(t)\cdot\nabla
v_{t}^{n}(t)\,\mathrm{d}x\Big{]}+\|v_{t}^{n}(t)\|_{k,2}^{2}\Big{\\}}\,\mathrm{d}t$
$\displaystyle\geq\lim_{n\rightarrow\infty}\Big{[}\int_{Q}\psi\frac{b^{\prime}(v^{n})v_{t}^{n}}{\alpha}\varphi^{*}_{n}(\boldsymbol{\varepsilon}(\alpha\boldsymbol{u}^{n}))\,\mathrm{d}x\,\mathrm{d}t\Big{]}$
$\displaystyle\quad+\int_{0}^{T}\psi\Big{\\{}\int_{\Omega}\frac{1}{2\epsilon}(v(t)-1)v_{t}(t)+2\epsilon\nabla
v(t)\cdot\nabla
v_{t}(t)\,\mathrm{d}x+\|v_{t}(t)\|_{k,2}^{2}\Big{\\}}\,\mathrm{d}t.$

If we can show that

(3.59)
$\lim_{n\rightarrow\infty}\Big{[}\int_{Q}\psi\frac{b^{\prime}(v^{n})v_{t}^{n}}{\alpha}\varphi^{*}_{n}(\boldsymbol{\varepsilon}(\alpha\boldsymbol{u}^{n}))\,\mathrm{d}x\,\mathrm{d}t\Big{]}\geq\int_{Q}\psi\frac{b^{\prime}(v)v_{t}}{\alpha}\varphi^{*}(\boldsymbol{\varepsilon}(\alpha\boldsymbol{u}))\,\mathrm{d}x\,\mathrm{d}t,$

then (3.57) follows. At best, we have a weak convergence result for
$(v^{n}_{t})_{n}$. Hence we need a strong convergence result for
$(\varphi^{*}_{n}(\boldsymbol{\varepsilon}(\alpha\boldsymbol{u}^{n})))_{n}$ in
order to compute the limit in (3.59). At present, we only have pointwise
convergence. We show that this pointwise result for
$(\varphi^{*}_{n}(\boldsymbol{\varepsilon}(\alpha\boldsymbol{u}^{n})))_{n}$
can be improved to strong convergence in $L^{1}(0,T;L^{1}_{loc}(\Omega))$.

We first show that $\mathbf{T}^{n}\rightarrow\mathbf{T}$ strongly in
$L^{1}(0,T;L^{1}_{loc}(\Omega)^{d\times d})$. If $a\in(0,\frac{2}{d})$, we
later improve the result on $(\mathbf{T}^{n})_{n}$ to strong convergence in
$L^{1+\delta}(0,T;L^{1+\delta}_{loc}(\Omega)^{d\times d})$ for a
$\delta=\delta(a)>0$. Using that $(v^{n}_{t})_{n}$ and $v_{t}$ vanish on the
boundary of $\Omega$, we then prove that

(3.60)
$\lim_{n\rightarrow\infty}\Big{[}\int_{Q}\psi\frac{b^{\prime}(v^{n})v_{t}^{n}}{\alpha}\varphi^{*}_{n}(\boldsymbol{\varepsilon}(\alpha\boldsymbol{u}^{n}))\,\mathrm{d}x\,\mathrm{d}t\Big{]}=\int_{Q}\psi\frac{b^{\prime}(v)v_{t}}{\alpha}\varphi^{*}(\boldsymbol{\varepsilon}(\alpha\boldsymbol{u}))\,\mathrm{d}x\,\mathrm{d}t.$

The energy-dissipation equality follows from (3.60) using the aforementioned
arguments. As a result, we conclude the existence of a weak energy solution to
the strain-limiting problem with Dirichlet boundary conditions.

######  Lemma 3.9.

Suppose that the hypotheses of Theorem 3.2 hold and additionally assume that
$\Gamma_{N}=\emptyset$. Let $(\boldsymbol{u},\mathbf{T},v)$ be the solution
triple from Theorem 3.2 and $((\boldsymbol{u}^{n},\mathbf{T}^{n},v^{n}))_{n}$
the sequence of solutions to the regularised problem constructed in
Proposition 3.2. Then

$\displaystyle\mathbf{T}^{n}\rightarrow\mathbf{T}\quad\text{ strongly in
}L^{1}(0,T;L^{1}_{loc}(\Omega)^{d\times d}).$

###### Proof.

Using the pointwise convergence result, it is enough to show that
$|\mathbf{T}^{n}|\rightarrow|\mathbf{T}|$ strongly in
$L^{1}(0,T;L^{1}_{loc}(\Omega))$. There exists a constant $C$, independent of
$n$, such that

$\displaystyle\sup_{t\in[0,T]}\int_{\Omega}\frac{|\mathbf{T}^{n}(t)|^{2}}{n}\,\mathrm{d}x+\int_{Q}\frac{|\mathbf{T}^{n}_{t}|^{2}}{n}\,\mathrm{d}x\,\mathrm{d}t+\frac{1}{c(\Omega_{0})}\int_{0}^{T}\int_{\Omega_{0}}\frac{|\nabla\mathbf{T}^{n}|^{2}}{n}\,\mathrm{d}x\,\mathrm{d}t\leq
C,$

for every compact subset $\Omega_{0}\subset\Omega$. Applying the Aubin–Lions
lemma to $(n^{-\frac{1}{2}}\mathbf{T}^{n})_{n}$, it follows that
$n^{-\frac{1}{2}}\mathbf{T}^{n}\rightarrow\mathbf{0}$ strongly in
$L^{2}(0,T;L^{2}_{loc}(\Omega)^{d\times d})$.

On the other hand, using the elastodynamic equation and the convergence
results for $(\boldsymbol{u}^{n})_{n}$, we have that

$\displaystyle\lim_{n\rightarrow\infty}\int_{Q}b(v^{n})\mathbf{T}^{n}:F_{n}(\mathbf{T}^{n})\,\mathrm{d}x\,\mathrm{d}t=\int_{Q}b(v)\mathbf{T}:F(\mathbf{T})\,\mathrm{d}x\,\mathrm{d}t.$

By the pointwise convergence result and the non-negativity of the integrands,
it follows that

$b(v^{n})\mathbf{T}^{n}:F_{n}(\mathbf{T}^{n})\rightarrow
b(v)\mathbf{T}:F(\mathbf{T})\quad\text{ strongly in
}L^{1}(0,T;L^{1}(\Omega)).$

It follows that

$\displaystyle\mathbf{T}^{n}:F_{n}(\mathbf{T}^{n})\rightarrow\mathbf{T}:F(\mathbf{T})\quad\text{
strongly in }L^{1}(0,T;L^{1}(\Omega)).$

Using the local convergence result for $(n^{-\frac{1}{2}}\mathbf{T}^{n})_{n}$
in $L^{2}(0,T;L^{2}_{loc}(\Omega)^{d\times d})$, we see that

(3.61)
$\mathbf{T}^{n}:F(\mathbf{T}^{n})\rightarrow\mathbf{T}:F(\mathbf{T})\quad\text{
strongly in }L^{1}(0,T;L^{1}_{loc}(\Omega)).$

We write

(3.62)
$|\mathbf{T}^{n}|=\Big{(}\frac{|\mathbf{T}^{n}|^{2}}{(1+|\mathbf{T}^{n}|^{a})^{\frac{1}{a}}}\Big{)}\cdot\Big{(}\frac{(1+|\mathbf{T}^{n}|^{a})^{\frac{1}{a}}}{|\mathbf{T}^{n}|}\Big{)}\cdot\mathbb{I}_{\\{|\mathbf{T}^{n}|>1\\}}+|\mathbf{T}^{n}|\cdot\mathbb{I}_{\\{|\mathbf{T}^{n}|\leq
1\\}}.$

The indicator functions $\mathbb{I}_{\\{|\mathbf{T}^{n}|>1\\}}$,
$\mathbb{I}_{\\{|\mathbf{T}^{n}|\leq 1\\}}$ converge pointwise a.e. in $Q$ to
$\mathbb{I}_{\\{|\mathbf{T}|>1\\}}$ and $\mathbb{I}_{\\{|\mathbf{T}|\leq
1\\}}$, respectively. Applying the dominated convergence theorem and the
pointwise convergence result, we get

$\displaystyle|\mathbf{T}^{n}|\cdot\mathbb{I}_{\\{|\mathbf{T}^{n}|\leq
1\\}}\rightarrow|\mathbf{T}|\cdot\mathbb{I}_{\\{|\mathbf{T}|\leq
1\\}}\quad\text{ strongly in }L^{1}(0,T;L^{1}(\Omega)).$

For the other term on the right-hand side of (3.62), the second factor is
uniformly bounded by $2^{\frac{1}{a}}$ and converges pointwise a.e. on $Q$.
The first factor converges strongly in $L^{1}(0,T;L^{1}_{loc}(\Omega))$ by
(3.61). Thus we have

$\displaystyle\Big{(}\frac{|\mathbf{T}^{n}|^{2}}{(1+|\mathbf{T}^{n}|^{a})^{\frac{1}{a}}}\Big{)}\cdot\Big{(}\frac{(1+|\mathbf{T}^{n}|^{a})^{\frac{1}{a}}}{|\mathbf{T}^{n}|}\Big{)}\cdot\mathbb{I}_{\\{|\mathbf{T}^{n}|>1\\}}$
$\displaystyle\rightarrow\Big{(}\frac{|\mathbf{T}|^{2}}{(1+|\mathbf{T}|^{a})^{\frac{1}{a}}}\Big{)}\cdot\Big{(}\frac{(1+|\mathbf{T}|^{a})^{\frac{1}{a}}}{|\mathbf{T}|}\Big{)}\cdot\mathbb{I}_{\\{|\mathbf{T}|>1\\}}$
$\displaystyle=|\mathbf{T}|\cdot\mathbb{I}_{\\{|\mathbf{T}|>1\\}},$

with respect to strong convergence in $L^{1}(0,T;L^{1}_{loc}(\Omega))$. Thus
the convergence asserted in the statement of the lemma follows. ∎

######  Lemma 3.10.

Suppose that the hypotheses of Lemma 3.9 hold. Furthermore, we also assume
that $a\in(0,\frac{2}{d})$. Let $(\boldsymbol{u},\mathbf{T},v)$ be the
solution triple from Theorem 3.2 and
$((\boldsymbol{u}^{n},\mathbf{T}^{n},v^{n}))_{n}$ the sequence of solutions to
the regularised problem constructed in Proposition 3.2. Let
$\Omega_{0}\subset\Omega$ be a compact subset. For every $\delta>0$ such that
$a+\delta<\frac{2}{d}$, there exists a constant $C=C(\Omega_{0})$, independent
of $n$, such that

$\displaystyle\int_{0}^{T}\int_{\Omega_{0}}|\mathbf{T}^{n}|^{1+\delta}+|\mathbf{T}|^{1+\delta}\,\mathrm{d}x\,\mathrm{d}t\leq
C.$

Furthermore, $\mathbf{T}^{n}\rightarrow\mathbf{T}$ strongly in
$L^{1+{\delta}}(0,T;L^{1+{\delta}}_{loc}(\Omega)^{d\times d})$ for every
$\delta$ as above.

###### Proof.

First we recall that there exists a constant $C=C(\Omega_{0})$ independent of
$n$ such that

$\displaystyle C$
$\displaystyle\geq\sup_{t\in[0,T]}\int_{\Omega}|\mathbf{T}^{n}(t)|\,\mathrm{d}x+\int_{0}^{T}\int_{\Omega_{0}}\frac{|\nabla\mathbf{T}^{n}|^{2}}{(1+|\mathbf{T}^{n}|^{a})^{\frac{1}{a}}}\,\mathrm{d}x\,\mathrm{d}t$
$\displaystyle\geq
c_{a}\Big{(}c_{p}\int_{0}^{T}\|(1+|\mathbf{T}^{n}(t)|)^{\frac{1-a}{2}}\|_{L^{p}(\Omega_{0})}\,\mathrm{d}t-1\Big{)}$
$\displaystyle\geq
c_{a}c_{p}\Big{(}\int_{0}^{T}\|\mathbf{T}^{n}\|_{L^{\frac{p(1-a)}{2}}(\Omega_{0})}^{1-a}\,\mathrm{d}t\Big{)},$

where $p\in(2,\infty)$ if $d=2$ and $p\in(2,\frac{2d}{d-2}]$ if $d>2$, by the
Sobolev embedding theorem applied to the function
$(1+|\mathbf{T}^{n}(t)|)^{\frac{1-a}{2}}$. We consider the case $d>2$. The
argument for $d=2$ is similar. Letting $p=\frac{2d}{d-2}$, it follows that

(3.63)
$\sup_{t\in[0,T]}\int_{\Omega}|\mathbf{T}^{n}(t)|\,\mathrm{d}x+\int_{0}^{T}\Big{(}\int_{\Omega_{0}}|\mathbf{T}^{n}(t)|^{\frac{d(1-a)}{d-2}}\,\mathrm{d}x\Big{)}^{\frac{d-2}{d}}\,\mathrm{d}t\leq
C,$

where $C$ is independent of $n$. We fix $\delta=\delta(a,d)>0$ such that
$a+\delta\in(0,\frac{2}{d})$, which is possible due to the assumption on $a$.
A standard manipulation shows that

$\displaystyle\frac{d(1+\delta)-2}{d-2}<\frac{d(1-a)}{d-2}.$

Applying Hölder’s inequality with parameters $\frac{d}{2}$ and
$\frac{d}{d-2}$, we get

$\displaystyle\int_{\Omega_{0}}|\mathbf{T}^{n}(t)|^{1+\delta}\,\mathrm{d}x$
$\displaystyle\leq\Big{(}\int_{\Omega_{0}}|\mathbf{T}^{n}(t)|\,\mathrm{d}x\Big{)}^{\frac{2}{d}}\Big{(}\int_{\Omega_{0}}|\mathbf{T}^{n}(t)|^{1+\frac{\delta
d}{d-2}}\,\mathrm{d}x\Big{)}^{\frac{d-2}{d}}$ $\displaystyle\leq
C\Big{(}\int_{\Omega_{0}}|\mathbf{T}^{n}(t)|^{\frac{d(1+\delta)-2}{d-2}}\,\mathrm{d}x\Big{)}^{\frac{d-2}{d}}$
$\displaystyle\leq
C\Big{(}\int_{\Omega_{0}}|\mathbf{T}^{n}(t)|^{\frac{d(1-a)}{d-2}}\,\mathrm{d}x\Big{)}^{\frac{d-2}{d}}$

Integrating over $(0,T)$ and using (3.63), it follows that

$\displaystyle\int_{0}^{T}\int_{\Omega_{0}}|\mathbf{T}^{n}(t)|^{1+\delta}\,\mathrm{d}x\,\mathrm{d}t\leq
C.$

By reflexivity, we deduce that $\mathbf{T}^{n}\rightharpoonup\mathbf{T}$
weakly in $L^{1+\delta}(0,T;L^{1+\delta}_{loc}(\Omega)^{d\times d})$.
Combining the weak and pointwise convergence results yields the strong
convergence result. ∎

With these strong convergence results in mind, we return to focusing on
proving (3.59). We use the strong convergence of $(\mathbf{T}^{n})_{n}$ to
first show that
$(\varphi^{*}_{n}(\boldsymbol{\varepsilon}(\boldsymbol{u}^{n}_{t}+\alpha\boldsymbol{u}^{n})))_{n}$
converges strongly in $L^{1}(0,T;L^{1}_{loc}(\Omega)^{d\times d})$. Then we
make use of the memory kernel property to show that
$(\varphi^{*}_{n}(\boldsymbol{\varepsilon}(\alpha\boldsymbol{u}^{n})))_{n}$
converges strongly in $L^{1}(0,T;L^{1}_{loc}(\Omega))$ to
$\varphi^{*}(\boldsymbol{\varepsilon}(\alpha\boldsymbol{u}))$. We combine this
with the weak-* convergence of $(v^{n}_{t})_{n}$ in
$L^{\infty}(0,T;L^{\infty}(\Omega))$ to deal with the limit in (3.59) away
from the boundary of $\Omega$. We use the auxiliary result, Corollary 3.6, to
control the limit near the boundary. This is another place where we make use
of the fully Dirichlet boundary condition.

######  Lemma 3.11.