$\Phi\colon\,\;w\in\mathscr{C}([0,T],H^{1}(\mathbb{R}^{d}))\;\;\mapsto\;\;\Phi(w)\in\mathscr{C}([0,T],H^{1}(\mathbb{R}^{d}))$
the unique solution of the Cauchy problem:

$\displaystyle\begin{cases}v_{t}-\widetilde{\mu}\Delta
u-(\widetilde{\mu}+\widetilde{\lambda})\nabla\operatorname{div}v=f+\operatorname{div}(b_{0}\,\colon
w),\\\ v_{|t=0}=v_{0}\end{cases}$ (C.6)

whose Helmholtz decomposition verifies (C.3). In the following we will set
some tools in order to apply the Banach fixed point theorem. Let us begin by
the following:

There exists a constant $C$ that depends on $b_{0}$, such that:

$\forall\;\varphi\in H^{1}(\mathbb{R}^{d}),\quad\,\forall
t>0\quad\lVert\mathcal{T}(kt)\operatorname{div}(b_{0}\,\colon\varphi)\rVert_{H^{1}(\mathbb{R}^{d})}\leqslant
C(kt)^{-\tfrac{1}{2}}\lVert\varphi\rVert_{H^{1}(\mathbb{R}^{d})}.$

###### Proof.

First, let us begin by writing:

$\mathcal{T}(kt)\operatorname{div}(b_{0}\,\colon\varphi)=G_{kt}*\operatorname{div}(b_{0}\,\colon\varphi)=\nabla
G_{kt}*(b_{0}\,\colon\varphi)$

where $G_{t}$ is the heat kernel. On the first hand, Young’s inequality
yields:

$\lVert\mathcal{T}(kt)\operatorname{div}(b_{0}\,\colon\varphi)\rVert_{L^{2}(\mathbb{R}^{d})}\leqslant\lVert\nabla
G_{kt}\rVert_{L^{1}(\mathbb{R}^{d})}\lVert
b_{0}\rVert_{L^{\infty}(\mathbb{R}^{d})}\lVert\varphi\rVert_{L^{2}(\mathbb{R}^{d})}$

where,

$\lVert\nabla G_{t}\rVert_{L^{1}(\mathbb{R}^{d})}=\dfrac{(4\pi
t)^{-\tfrac{d}{2}}}{2t}(2\sqrt{t})^{d+1}\lvert\partial
B(0,1)\rvert\int_{0}^{\infty}r^{d}e^{-r^{2}}dr.$

On the other hand, one has:

$\displaystyle\lVert\nabla\mathcal{T}(kt)\operatorname{div}(b_{0}\,\colon\varphi)\rVert_{L^{2}(\mathbb{R}^{d})}$
$\displaystyle\leqslant\lVert\nabla
G_{kt}\rVert_{L^{1}(\mathbb{R}^{d})}\lVert\nabla(b_{0}\,\colon\varphi)\rVert_{L^{2}(\mathbb{R}^{d})}$
$\displaystyle\leqslant\lVert\nabla G_{kt}\rVert_{L^{1}(\mathbb{R}^{d})}\lVert
b_{0}\rVert_{W^{1,\infty}(\mathbb{R}^{d})}\lVert\varphi\rVert_{H^{1}(\mathbb{R}^{d})}.$

This concludes the proof of appendix C. ∎

We turn to the stability of the (C.3). So let us set

$R_{T}=\sup_{[0,T]}\left\|\mathcal{T}(kt)\mathcal{A}v_{0}+\int_{0}^{t}\mathcal{T}(k(t-s))\mathcal{A}f(s)ds\right\|_{H^{1}(\mathbb{R}^{d})}.$

There exists an increasing function $g\colon[0,\infty)\;\to\;[0,\infty)$ and a
positive constant $K$, such that the condition

$\displaystyle\forall t>0,\;\lVert w(t)\rVert_{H^{1}(\mathbb{R}^{d})}\leqslant
R_{T}Kg(t)$ (C.7)

implies the same for $\Phi(w)$ for all
$w\in\mathscr{C}([0,T],H^{1}(\mathbb{R}^{d}))$.

###### Proof.

One deduces easily from (C.3) the following:

$\lVert\Phi(w)(t)\rVert_{H^{1}(\mathbb{R}^{d})}\leqslant
R_{T}+\int_{0}^{t}\lVert\mathcal{T}(k(t-s))\mathcal{A}\operatorname{div}(b_{0}\,\colon
w)(s)\rVert_{H^{1}(\mathbb{R}^{d})}ds.$

Next, gathering appendix C and the assumption (C.7), one has:

$\lVert\Phi(w)(t)\rVert_{H^{1}(\mathbb{R}^{d})}\leqslant
R_{T}\left(1+Ck^{-\tfrac{1}{2}}K\int_{0}^{t}\dfrac{g(s)}{\sqrt{t-s}}ds\right).$

For the condition (C.7) to be satisfied for $\Phi(w)$, it is sufficient that

$\displaystyle
1+Ck^{-\tfrac{1}{2}}K\int_{0}^{t}\dfrac{g(s)}{\sqrt{t-s}}ds\leqslant Kg(t).$
(C.8)

Let us look for some $\alpha>0$ and $K>0$ such that $g$ takes the form
$g(t)=e^{\alpha t}$. By change of variable, one has:

$e^{-\alpha t}\int_{0}^{t}\dfrac{e^{\alpha
s}}{\sqrt{t-s}}ds=\dfrac{1}{\sqrt{\alpha}}\int_{0}^{\alpha
t}\dfrac{e^{-s}}{\sqrt{s}}ds\leqslant\dfrac{1}{\sqrt{\alpha}}\int_{0}^{\infty}\dfrac{e^{-s}}{\sqrt{s}}ds=\sqrt{\dfrac{2}{\alpha}}\int_{0}^{\infty}e^{-\tfrac{r^{2}}{2}}ds=\sqrt{\dfrac{\pi}{\alpha}}$

hence,

$1+Ck^{-\tfrac{1}{2}}K\int_{0}^{t}\dfrac{g(s)}{\sqrt{t-s}}ds\leqslant
1+Ck^{-\tfrac{1}{2}}K\sqrt{\dfrac{\pi}{\alpha}}g(t).$

Thus, one can choose $K$ to be

$K=\dfrac{1}{4C}\sqrt{\dfrac{\alpha k}{\pi}}$

and $\alpha$ large such that $K\geqslant 2$. In consequence, the condition
(C.8) becomes, since $g(t)>1$,

$1+Ck^{-\tfrac{1}{2}}K\int_{0}^{t}\dfrac{g(s)}{\sqrt{t-s}}ds\leqslant
1+\dfrac{g(t)}{4}\leqslant Kg(t),$

and this concludes the proof of appendix C. ∎

There exist $\alpha>0$ and a positive constant $0<\kappa<1$ such that for all
$v,\,w\in\mathscr{C}([0,T],H^{1}(\mathbb{R}^{d}))$, one has:

$\sup_{[0,T]}e^{-\alpha
t}\lVert\Phi(v)(t)-\Phi(w)(t)\rVert_{H^{1}(\mathbb{R}^{d})}\leqslant\kappa\sup_{[0,T]}e^{-\alpha
t}\lVert v(t)-w(t)\rVert_{H^{1}(\mathbb{R}^{d})}.$

###### Proof.

Starting from (C.3) and combining appendix C, one has:

$\displaystyle\lVert\Phi(v)(t)-\Phi(w)(t)\rVert_{H^{1}(\mathbb{R}^{d})}$
$\displaystyle=\left\|\int_{0}^{t}\mathcal{T}(k(t-s))\mathcal{A}\operatorname{div}(b_{0}\,\colon(v-w))(s)ds\right\|_{H^{1}(\mathbb{R}^{d})}$
$\displaystyle\leqslant
Ck^{-\tfrac{1}{2}}\int_{0}^{t}\dfrac{1}{\sqrt{t-s}}\lVert
v(s)-w(s)\rVert_{H^{1}(\mathbb{R}^{d})}ds$ $\displaystyle\leqslant
Ck^{-\tfrac{1}{2}}\sup_{s\in[0,T]}e^{-\alpha s}\lVert
v(s)-w(s)\rVert_{H^{1}(\mathbb{R}^{d})}\int_{0}^{t}\dfrac{e^{\alpha
s}}{\sqrt{t-s}}ds.$

It then turns out that,

$\sup_{[0,T]}e^{-\alpha
t}\lVert\Phi(v)(t)-\Phi(w)(t)\rVert_{H^{1}(\mathbb{R}^{d})}\leqslant
Ck^{-\tfrac{1}{2}}\sqrt{\dfrac{\pi}{\alpha}}\sup_{t\in[0,T]}e^{-\alpha
t}\lVert v(t)-w(t)\rVert_{H^{1}(\mathbb{R}^{d})}$

and appendix C follows by choosing $\alpha$ large enough. ∎

We turn to the application of the Banach fixed point theorem in the following
vector space

$E:=\left\\{v\in\mathscr{C}([0,T],H^{1}(\mathbb{R}^{d}))\colon\sup_{t\in[0,T]}e^{-\alpha
t}\lVert v(t)\rVert_{H^{1}(\mathbb{R}^{d})}<\infty\right\\}$

endowed with the norm:

$\forall v\in E,\quad\lVert v\rVert_{E}=\sup_{t\in[0,T]}e^{-\alpha t}\lVert
v(t)\rVert_{H^{1}(\mathbb{R}^{d})}.$

It is obvious that $(E,\lVert\,\cdot\,\rVert_{E})$ is a Banach space and, in
particular the closed ball $E_{T}=B(0,KR_{T})$ is complete with respect to the
norm $\lVert\,\cdot\,\rVert_{E}$. On the first hand, appendix C ensures that
$\Phi$ maps $E_{T}$ into itself and the appendix C ensures that $\Phi$ is a
contracting application. Hence, by Banach fixed point theorem, $\Phi$ admits
an unique fixed point $v$ in $E_{T}$ that solves the equation (C.1). Moreover,
since $v$ verifies

$\mathcal{A}v=\mathcal{T}(kt)\mathcal{A}v_{0}+\int_{0}^{t}\mathcal{T}(k(t-s))\mathcal{A}f(s)ds+\int_{0}^{t}\mathcal{T}(k(t-s))\mathcal{A}\operatorname{div}(b_{0}\,\colon
v)(s)ds$

one deduces using [8, Lemma 2.2] that:

$\int_{0}^{t}\lVert
v_{t},\nabla^{2}v\rVert_{L^{2}(\mathbb{R}^{d})}\leqslant\lVert
v_{0}\rVert_{H^{1}(\mathbb{R}^{d})}^{2}+\int_{0}^{t}\lVert
f,\operatorname{div}(b_{0}\,\colon v)\rVert_{L^{2}(\mathbb{R}^{d})}^{2}.$

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