2. Distributions and Sobolev spaces

Let $S(\mathbb {R}_{+}^{3})=S(\mathbb {R}^{2}\times [0,\infty ))$Footnote ² denote the set of all $C^{\infty }$ functions $f:R_+^{3}\longrightarrow C$ such that

(2.1)\begin{equation} \left\Vert f\right\Vert_{\alpha,\beta}=\mathop{\sup}\limits_{\mathbb{R}_{+}^{3}}\left\vert w^{\alpha}D^{\beta}f\left(w\right)\right\vert <\infty. \end{equation}

where $\alpha ,\beta$, are (tridimensional) multi-indexes, $w=(x,y,z) \in R_+^{3},\ D=\frac {1}{i}\nabla$ (see [Reference Cordes6, chap. 1, p. 8], [Reference Iorio and Iorio10, chap. 7, p. 323] and [Reference Schecter30, chap. 1, p. 2]). Moreover, the derivatives with respect to $z$ at $z=0,$ are taken from above.

This defines a countable collection of seminorms in $S(\mathbb {R} _{+}^{3}),$ which turns this vector space into a Frèchet space (see [Reference Reed and Simon27]). Let $S^{\prime }(\mathbb {R}_{+}^{3})$ denote the topological dual of $S(\mathbb {R}_{+}^{3})$ that is $f\in S^{\prime }(\mathbb {R}_{+}^{3})$ if and only if $f:S(\mathbb {R}_{+}^{3})\longrightarrow C$ is linear and is continuous in the following sense,Footnote ³ for any convergent net $f_{\lambda }\in \Lambda$ we have

(2.2)\begin{equation} f_{\lambda}\mathop{\longrightarrow}\limits^{\Lambda}f\Longleftrightarrow f_{\lambda}\left(\varphi\right)\mathop{\longrightarrow}\limits^{\mathbb{C}}f\left( \varphi\right)\quad\forall\ \varphi\in\mathfrak{S}\left(\mathbb{R}_{+}^{3}\right).\end{equation}

Now, let $L^{2}(\mathbb {R}_{+}^{3})=L^{2}(\mathbb {R}^{2}\times [0,\infty )).$ It is not difficult to show that

(2.3)\begin{equation} \mathfrak{S}\left(\mathbb{R}_{+}^{3}\right)\hookrightarrow\mathfrak{L}^{2}\left(\mathbb{R}_{+}^{3}\right)\hookrightarrow\mathfrak{S}^{\prime}\left(\mathbb{R}_{+}^{3}\right),\end{equation}

where the symbol $\hookrightarrow$, in the remainder of this article, will always mean that the inclusion is continuous and dense with respect to the relevant topologies involved. Next consider the following operator

(2.4)\begin{equation} \begin{aligned} \mathfrak{D}\left(\widetilde{\Delta}\right) & =\left\{\varphi\in\mathfrak{S}\left(\mathbb{R}_{+}^{3}\right)\left\vert \widetilde {\partial}_{z}\varphi\left(x,y,0\right)=0\right.\right\},\\ -\widetilde{\Delta}\varphi\left(x,y,z\right) & =\left(\partial_{x}^{2}+\partial_{y}^{2}+\widetilde{\partial}_{z}^{2}\right)\varphi\left(x,y,z\right). \end{aligned} \end{equation}

However, it is necessary to explain what the $z$ derivative means. Define $\widetilde {d}_{z}^{2}$ by the equations,

(2.5)\begin{equation} \begin{aligned} \mathfrak{D}\left(\widetilde{d}_{z}^{2}\right) & =\left\{f\in\mathcal{S}(\left[0,\infty\right)\right)\left\vert f^{\prime}\left(0\right)=0\right\}\\ -\widetilde{d}_{z}^{2}f & =\frac{\textrm{d}^{2}f}{\textrm{d}z^{2}},\quad f\in\mathfrak{D}\left(\widetilde{d}_{z}^{2}\right), \end{aligned} \end{equation}

where the derivative at zero is taken from above. Using the Fourier cosine transform and its Inversion formula (see [Reference Churchill4, § 54]),

(2.6)\begin{equation} \begin{aligned} \left(\mathfrak{F}_{c}f\right)\left(\alpha\right) & =\int_{0}^{\infty}f\left(x\right)\cos\left(\alpha x\right)\textrm{d}x,\ x,\ \alpha \in\left[0,\infty\right),\\ \left(\mathfrak{F}_{c}^{{-}1}g\right)\left(x\right) & =\frac{2}{\pi}\int_{0}^{\infty}g\left(\alpha\right)\cos\left(\alpha x\right)\textrm{d}\alpha, \end{aligned} \end{equation}

and the fact that

(2.7)\begin{equation} \left(\mathfrak{F}_{c}f^{\prime\prime}\right) \left(\alpha\right)={-}\alpha^{2}\left(\mathfrak{F}_{c}f\right)\left(\alpha\right) -f^{\prime}\left(0\right), \end{equation}

it is easy to see that $\widetilde {d}_{z}^{2}$ is essentially self-adjoint. Let $d_{z}^{2}$ denote its unique self-adjoint extension. Next, if $(x,y)\in \mathbb {R}^{2}$ is fixed and $\varphi \in S(\mathbb {R}_{+}^{3})$ then $\psi (z) =\varphi (x,y,z)\in S([0,\infty ))$ so we may define

(2.8)\begin{equation} \widetilde{\partial}_{z}^{2}\varphi\left(x,y,z\right) =\widetilde{d}_{z}^{2}\varphi\left(x,y,z\right)=d_{z}^{2}\varphi\left(x,y,z\right). \end{equation}

Once again, it is easy to show that $(-\widetilde {\Delta })$ is essentially self-adjoint. We will denote its unique self-adjoint extension by $(-\Delta )$. Now, it is necessary to introduce a Fourier transform associated with the operator $(-\Delta )$. This can be done noting that

(2.9)\begin{equation} \Theta\left(x,y,z\right)=\exp\left(ix\xi\right) \exp\left(iy\eta\right)\cos\left(\alpha z\right),\end{equation}

satisfies,

(2.10)\begin{equation} \left(-\Delta\right)\Theta\left(x,y,z\right)=\left(\xi^{2}+\eta^{2}+\alpha^{2}\right)\Theta\left(x,y,z\right).\end{equation}

So if $\varphi \in \mathfrak {L}^{1}(\mathbb {R}_{+}^{3})$, we define

(2.11)\begin{align} \widehat{\varphi}\left(\xi,\eta,\alpha\right)&=\left(\mathfrak{F}\varphi\right)\left(\xi,\eta,\alpha\right)\nonumber\\ &=\left(\frac{1}{2\pi}\right)\int_{\mathbb{R}_{+}^{3}}\varphi\left(x,y,z\right)\overline{\Theta\left(x,y,z\right)}\,\textrm{d}x\,\textrm{d}y\,\textrm{d}z. \end{align}

Employing the usual methods [Reference Iorio and Iorio10, Reference Iorio12, Reference Reed and Simon28], we can extend this operator as an unitary map from $\mathfrak {L}^{2}( \mathbb {R}_{+}^{3})$ into itself. Its inverse is given by

(2.12)\begin{align} \mathop{\omega}\limits^{{\vee}}\left(x,y,z\right)& =\left(\mathfrak{F}^{{-}1}\omega\right)\left(x,y,z\right)\nonumber\\ &=\left(\frac{1}{\pi}\right)^{2}\int_{\mathbb{R}_{+}^{3}}\omega\left(\xi,\eta,\alpha\right)\Theta\left(x,y,z\right)\textrm{d}\xi\,\textrm{d}\eta\,\textrm{d}\alpha. \end{align}

The usual methods employed to extend the transform Fourier in $\mathbb {R}^{n}$ can be used in this case to define $\mathfrak {F}$ in $\mathfrak {L}^{2}(\mathbb {R}_{+}^{3})$ and $\mathfrak {S}^{\prime }(\mathbb {R}_{+}^{3})$. Note that

(2.13)\begin{equation} -\Delta f=\mathfrak{F}^{{-}1}\Phi\mathfrak{F}f \end{equation}

where $\Phi$ denotes (with a little abuse of notation) the maximal operator of multiplication by

(2.14)\begin{equation} \Phi\left(\xi,\eta,\alpha\right) =\left(\xi^{2}+\eta^{2}+\alpha^{2}\right)\end{equation}

in $\mathfrak {L}^{2}(\mathbb {R}_{+}^{3})$. It deserves remark that the Fourier transform $\mathfrak {F}$ is a topological isomorphism from $\mathfrak {S}(\mathbb {R}_{+}^{3})$ into itself, so that by the usual duality argument, it has the same property in $\mathfrak {S}^{\prime }(\mathbb {R}_{+}^{3})$. Moreover, it is a unitary operator in $\mathfrak {L}^{2}(\mathbb {R}_{+}^{3})$.Footnote ⁴ We are now in position to introduce the resolvent $z\longrightarrow R(z)$ of $(-\Delta )$ and the Sobolev spaces associated with it. To begin with, it is not difficult to see that

(2.15)\begin{equation} \sum\left(-\Delta\right)=\left[0,\infty\right), \end{equation}

and that the function $z\longrightarrow R(z)$ defined by,

(2.16)\begin{align} R\left(z\right)f & =\left(-\Delta-z\right)^{{-}1}f,\nonumber\\ &=\mathfrak{F}^{{-}1}\left(\xi^{2}+\eta^{2}+\alpha^{2}-z\right)^{{-}1}\mathfrak{F}f,\nonumber\\ z&\in\mathbb{C}\backslash\left[0,\infty\right),\quad f\in\mathfrak{L}\left(\mathbb{R}_{+}^{3}\right), \end{align}

satisfies,

(2.17)\begin{equation} \begin{aligned} R\left(z\right)\left(-\Delta-z\right)f & =f \forall f\in\mathfrak{D}\left(-\Delta\right),\\ \left(-\Delta-z\right)R\left(z\right)g & =g \forall g\in \mathfrak{L}^{2}\left(\mathbb{R}_{+}^{3}\right). \end{aligned} \end{equation}

Next, let $s\in \mathbb {R}$ and denote by $\mathfrak {H}^{s}(\mathbb {R}_{+}^{3})$ the Sobolev space of order $s$, that is,

(2.18)\begin{equation} \mathfrak{H}^{s}\left(\mathbb{R}_{+}^{3}\right)=\left\{f\in\mathfrak{S}^{\prime}\left(\mathbb{R}_{+}^{3}\right)\left\vert \left(1-\Delta\right)^{{s}/{2}}f\in\mathfrak{L}^{2}\left(\mathbb{R}_{+}^{3}\right)\right.\right\}.\end{equation}

These spaces have the same properties as the Sobolev spaces in $\mathbb {R}^{n},$ that is,

1. SB1: $\mathfrak {H}^{s}(\mathbb {R}_{+}^{3})$ are Hilbert spaces when endowed with the inner product

   (2.19)\begin{equation} \left(f\left\vert g\right.\right)_{s}=\left(\left( 1-\Delta\right)^{{s}/{2}}f\left\vert \left(1-\Delta\right)^{{s}/{2}}g\right. \right),\ \forall f,g\in\mathfrak{H}^{s}\left(\mathbb{R}_{+}^{3}\right).\end{equation}

2. SB2: If $s\geqslant \ell$ then $\mathfrak {H}^{s}(\mathbb {R}_{+}^{3})\hookrightarrow \mathfrak {H}^{\ell }(\mathbb {R}_{+} ^{3})$ for all $s,\ell \in \mathbb {R}.$

3. SB3: (Sobolevś lemma.) Let $s>\frac {3}{2}$. Then $\mathfrak {H}^{s}(\mathbb {R}_{+}^{3})\hookrightarrow C_{0}(\mathbb {R}_{+}^{3})$ where $C_{0}(\mathbb {R}_{+}^{3})$ denotes the set of all continuous functions that tend to zero at infinity.

4. SB4: Let $f,g\in \mathfrak {H}^{s}(\mathbb {R}_{+}^{3}),$ $s>\frac {3}{2}.$ Then the pointwise productFootnote ⁵ $fg\in \mathfrak {H}^{s}(\mathbb {R} _+^{3})$ and

   (2.20)\begin{equation} \left\Vert fg\right\Vert_{s}\leq C\left\Vert f\right\Vert _{s}\left\Vert g\right\Vert _{s} \end{equation}
   where $C>0$ is a constant. Note that this turns $\mathfrak {H}^{s}(\mathbb {R}_+^{3})$ into a Banach algebra.

The proofs of these properties are exactly the same as the corresponding ones in the case of $\mathbb {R}^{n},$ and will be omitted. The interested reader can consult [Reference Cordes6, Reference Iorio and Iorio10, Reference Reed and Simon28], for example.

Remark 2.1 Let $\Omega \subseteq \mathbb {R}^{n}$ be an open set. The Dirichlet Laplacian and the Neumann Laplacian, denoted $\Delta _{D}^{\Omega }$ and $\Delta _{N}^{\Omega }$ are the unique self-adjoint operators associated with the quadratic form

(2.21)\begin{equation} q\left(f,g\right)=\int_{\Omega}\nabla f\bullet\overline{\nabla g}\,\textrm{d}x \end{equation}

with domains $C_{0}^{\infty }(\Omega )$ and $H^{1}(\Omega )$ where $\nabla$ denotes the distributional gradient. (See [Reference Reed and Simon29].) This is a very elegant, but rather abstract definition. In many applications one must find the self-adjoint operator in order to deal with actual computations. The Laplacian defined above is the Neumann Laplacian corresponding to $\Omega =\mathbb {R}_{+}^{3}\backslash \{(x,y,z) \left \vert z=0\right .\}$, that is, the interior of $\mathbb {R}_{+}^{3}.$

3. Local well-posedness

We begin reminding the reader of our definition of well-posedness. The Cauchy problem

(3.1)\begin{equation} \begin{aligned} \partial_{t}u & =G\left(t,u\right)\in X,\\ u\left(0\right) & =u_{0}\in Y, \end{aligned} \end{equation}

$Y\hookrightarrow X,$ $t\in [0,T_{0}]$, $G:[ 0,T_{0}] \times Y\longrightarrow X$ is (at least continuousFootnote ⁶ ) is said to be locally well-posed if there exists a $T\in (0,T_{0}]$ and a function $u:[0,T] \longrightarrow Y$ such that $u(0)=u_{0}$ and satisfies the differential equation with respect to the norm of $X$,

(3.2)\begin{equation} \lim_{h\longrightarrow 0}\left\Vert \frac{u\left(t+h\right)-u\left(t\right)}{h}-G\left(t,u\left(t\right)\right)\right\Vert _{X}=0 \end{equation}

Moreover, the solution must depend continuously on the initial data (and on any other relevant parameters occurring in the equation), in appropriate topologies. In what follows, we will consider only the initial data. In that case what we mean is: assume that $u_{0}^{(j)}\in Y,\ j=1,2,3,\ldots ,\infty ,$ let $u^{(j)}$ be the corresponding solutions. Suppose that

(3.3)\begin{equation} \lim_{j\rightarrow\infty}\left\Vert u_{0}^{\left(j\right)}-u_{0}^{\left(\infty\right)}\right\Vert _{Y}=0.\end{equation}

Then, for all $T^{\prime }\in (0,T)$ we have,

(3.4)\begin{equation} \lim_{j\rightarrow\infty}\sup_{t\in\left[0,T^{\prime}\right]}\left\Vert u^{\left(j\right)}\left(t\right)-u^{\left(\infty\right)}\left(t\right)\right\Vert _{Y}=0.\end{equation}

If any of these properties fail, we say that the problem is ill-posed.Footnote ⁷ In case $G$ is defined for all $t\in \mathbb {R}$ and the preceding properties are valid for all $T>0$, we say that the problem is globally well-posed.

Using the definitions and notations of the previous section we can solve for $v$ as indicated in (1.3) and inserting this formula into the first equation of (1.1), we obtain the Cauchy problem

(3.5)\begin{align} \partial_{t}\rho & =\operatorname{div}\left(\rho\left(1-\Delta\right)^{{-}1}\nabla P\left(\rho\right)\right)+F\left(t,\rho\right)\nonumber\\ \rho\left(0\right)&=\rho_{0}. \end{align}

Moreover, the compatibility condition

(3.6)\begin{equation} v_{0}={-}\left(1-\Delta\right) \nabla P\left(\rho_{0}\right)\end{equation}

must be satisfied. Note also that the boundary conditions are inserted in the definition of the operators appearing in (3.5).

Now, there are several ways to solve (3.5). We mention our favourites, namely

• • Katoś theory of quasilinear Equations

• • Parabolic regularization.

3.1 Application of Katoś theory

A very large class of relevant evolution equations can be written in quasilinear form, that is,

(3.7)\begin{equation} \begin{cases} \partial_{t}u+A\left(t,u\right)u=F\left(t,u\right)\in X,\\ u\left(0\right)=\phi\in Y. \end{cases} \end{equation}

Here, $X$ and $Y$ are Banach spaces, as before with $Y$ $\hookrightarrow$ $X$ and $A(t,u)$ is bounded from $Y$ into $X$ (for fixed $t$) and is the (negative) generator of a $C^{0}$ semigroup for each $(t,u) \in \lbrack 0.T]\times W,$ $W$ open in $Y$. In its most general formulation, $X$ and $Y$ may be non-reflexive [Reference Kato18].Footnote ⁸ Since we will deal exclusively with reflexive spaces, we restrict ourselves to a simpler version, which can be found in [Reference Kato19]. (See also [Reference Iorio12].) The essential assumption of the theory is the existence of an isomorphism $S$ from $Y$ onto $X$ such that

(3.8)\begin{equation} SA\left(t,u\right) S^{{-}1}=A\left(t,u\right) +B\left(t,u\right) \end{equation}

where $B(t,u)\in \mathcal {B}(X)$, with the strict domain relation implied by the equation. This is, in fact, a condition on the commutator $[S,A(t,u)]$ because (3.8)Footnote ⁹ can be rewritten as

(3.9)\begin{equation} \left[S,A\left(t,u\right)\right]S^{{-}1}=B\left(t,u\right). \end{equation}

There are also lower requirements, involving Lipschitz conditions on the operators in question. For example, $A(t,u)$ must satisfy

(3.10)\begin{equation} \left\Vert A\left(t,w\right)-A\left(t,\widetilde{w}\right) \right\Vert_{\mathcal{B}\left(Y,X\right)}\leq\theta\left\Vert w-\widetilde {w}\right\Vert_{X},\quad\theta>0,\quad\text{constant} \end{equation}

for all pairs $(t,w),\ (t,\widetilde {w})$ in $[0.T]\times W.$ Both $B(t,u)$ and $F(t,u)$ must satisfy similar conditions. Once these assumptions are satisfied, Kato tells you that (3.7) is locally well-posed.

Now we must write the integrodifferential equation (3.5) in quasilinear form. Consider the linear operator:

(3.11)\begin{equation} f\longmapsto A\left(\rho\right) f={-}\operatorname{div}\left(f\left(1-\Delta\right)^{{-}1}\nabla P\left(\rho\right)\right).\end{equation}

Thus the equation in (3.5) can be written in the form presented in (3.7). Next we choose our function spaces. Due to certain technical estimates needed to control the commutator mentioned above, we take $Y=\mathfrak {H}^{s}(\mathbb {R}_{+}^{3})$, $s>5/2$, $X=\mathfrak {L}^{2}(\mathbb {R}_{+}^{3})$ and $W$ an arbitrary open ball centred at zero in $Y.$

Now assume

• • $P$ maps $\mathfrak {H}^{s}(\mathbb {R})$ into itself, $P(0)=0$ and is Lipschitz in the following sense:

  (3.12)\begin{equation} \left\Vert P\left(\rho\right)-P\left(\widetilde{\rho}\right)\right\Vert _{s}\leq L_{s}\left(\left\Vert \rho\right\Vert _{s},\left\Vert \widetilde{\rho}\right\Vert _{s}\right) \left\Vert \rho-\widetilde{\rho}\right\Vert _{s} \end{equation}

• • $F:[0,T_{0}]\times \mathfrak {H}^{s}(\mathbb {R}) \longrightarrow \mathfrak {H}^{s}(\mathbb {R}),\,F(t,0) =0$ and satisfies the following Lipschitz condition:

  (3.13)\begin{equation} \left\Vert F\left(t,\rho\right)-F\left(t,\widetilde{\rho}\right)\right\Vert _{s}\leq M_{s}\left(\left\Vert \rho\right\Vert _{s},\left\Vert \widetilde{\rho}\right\Vert _{s}\right)\left\Vert \rho-\widetilde{\rho}\right\Vert_{s}. \end{equation}
  where $L_{s},M_{s}:[0,\infty )\times [0,\infty ) \rightarrow [0,\infty )$ are continuous and monotone non-decreasing functions with respect to each of its arguments.

If $T$ is a linear operator and belongs to the class $G(X,1,0)$, that is, if $(-T)$ generates a contraction semigroup, we say that $T$ is maximally accretive (or m-accretive). If $T\in G(X,1,\beta )$, that is, $T$ generates a semigroup $U(t)$ such that $\Vert U(t)\Vert _{B(X)}\leq M\,\textrm {e}^{-t\beta }$, $T$ is said to be quasi-maximally accretive (or quasi m-accretive). Since $X$ is a Hilbert space, it suffices to prove, in our case, that $T=A(\rho )$ is maximally accretive in $X$. (See [Reference Kato15, Reference Pazy26, Reference Reed and Simon27]).

(3.14)\begin{equation} \left\langle A(\rho)f,f\right\rangle\geq{-}\beta\Vert f\Vert^{2},\quad\forall f\in D(A(\rho))=Y;\rho\in W\subset Y \end{equation}

Let

(3.15)\begin{equation} \Theta(\rho)=\left(1-\Delta\right)^{{-}1}\nabla P(\rho), \end{equation}

Integrating by parts and applying Sobolev lemma, we obtain

(3.16)\begin{align} \left\langle A(\rho)f,f\right\rangle& =\left\langle-\text{div}\;(f\,\Theta (\rho)),f\right\rangle={-}\sum_{i=1}^{3}\int f\partial_{x_{i}}(f\,\Theta_{i}(\rho))\,\textrm{d}x\nonumber\\ &=\sum_{i=1}^{3}\int f\partial_{x_{i}}f\,\Theta_{i}(\rho)\,\textrm{d}x=\frac{1}{2}\sum_{i=1}^{3}\int\partial_{x_{i}}(f^{2})\Theta_{i}(\rho)\,\textrm{d}x\nonumber\\ &={-}\frac{1}{2}\sum_{i=1}^{3}\int f^{2}\partial_{x_{i}}\Theta_{i}(\rho)\,\textrm{d}x={-}\frac{1}{2}\int(\text{div}\;\vec{\Theta}(\rho))f^{2} \,\textrm{d}x\nonumber\\ &\geq{-}\underbrace{\frac{\Vert\text{div}\;\Theta(\rho)\Vert_{L^{\infty}}}{2}}_{\beta}\Vert f\Vert^{2} \\ R(A(\rho)+\lambda)&=X=\mathfrak{L}^{2}(\mathbb{R}_{+}^{3}),\quad\forall\lambda>\beta \notag\end{align}

The fact that ${A(\rho )}$ is a closed operator combined with inequality (3.16) shows that $(A(\rho )+\lambda )$ has closed range for all $\lambda >\beta$.

Thus it suffices to show that $(A(\rho )+\lambda )$ has dense range for $\lambda >\beta$. For this, is sufficient to prove that $R(A(\rho )+\lambda )^{\bot }=\{0\}$, because $A(\rho )$ is a linear operator.

Let $g\in \mathfrak {L}^{2}(\mathbb {R}_{+}^{3})$ satisfy,

(3.17)\begin{equation} \left\langle(A(\rho)+\lambda)f,g\right\rangle=0,\quad\forall f\in D(A(\rho))=\mathfrak{H}^{s}(\mathbb{R}_{+}^{3}).\end{equation}

Integrating by parts, yields

(3.18)\begin{align} \left\langle(A(\rho)+\lambda)f,g\right\rangle=0&\Rightarrow\left\langle A(\rho)f,g\right\rangle+\left\langle\lambda f,g\right\rangle=0\nonumber\\ &\Rightarrow\left\langle f,\nabla g\,\Theta(\rho)\right\rangle+\left\langle\lambda f,g\right\rangle=0\nonumber\\ &\Rightarrow\left\langle f,\nabla g\,\Theta(\rho)+\lambda g\right\rangle=0,\quad\forall f\in D(A(\rho))=\mathfrak{H}^{s}(\mathbb{R}^{n})\nonumber\\ &\Rightarrow\nabla g\,\Theta(\rho)+\lambda g=0 \end{align}

Therefore, multiplying by $g$, integrating by parts, and using (3.16) we have:

(3.19)\begin{align} g\nabla g\,\Theta(\rho)+\lambda g^{2}=0 &\Rightarrow\frac{1}{2}\int\nabla(g^{2})\,\Theta(\rho)\,\textrm{d}x+\lambda\Vert g\Vert^{2}=0\nonumber\\ &\Rightarrow\underbrace{-\frac{1}{2}\int g^{2}\text{div}\;\Theta(\rho)\,\textrm{d}x}_{=\left\langle A(\rho)g,g\right\rangle}+\lambda\Vert g\Vert^{2} =0\nonumber\\ &\Rightarrow\left\langle A(\rho)g,g\right\rangle+\lambda\Vert g\Vert^{2}=0\nonumber\\ &\Rightarrow 0\geq{-}\beta\Vert g\Vert^{2}+\lambda\Vert g\Vert^{2}=(\lambda-\beta)\Vert g\Vert^{2}\nonumber\\ &\Rightarrow g=0 \end{align}

Finally, we choose the isomorphism $S:\mathfrak {D}(S) =\mathfrak {h}^{s}(\mathbb {R}_{+}^{3}) \longrightarrow \mathfrak {L}^{2}(\mathbb {R}_{+}^{3})$ to be

(3.20)\begin{equation} S=\left(1-\Delta\right)^{s/2}. \end{equation}

Then the proof of (3.8) is exactly the same of the corresponding fact in $\mathbb {R}^{n}$ (see [Reference Molina Del Sol, Alarcon and Iorio23]).

In view of these remarks, Kato's quasilinear theory implies the following result.

Theorem 3.1 The Cauchy problem (3.5) is locally well-posed in $\mathfrak {H}^{s}(\mathbb {R}_{+}^{3})$ in the sense described at the beginning of this section for all $s>5/2$.

3.2 Parabolic regularization

Considering $F(t,\rho )=0$ in (3.5), for simplicity, it is easy to see that if we integrate with respect to time we obtain

(3.21)\begin{equation} \mathop{\underbrace{\rho\left(t\right)}}\limits_{\mathfrak{H}^{s}\left(\mathbb{R}_{+}^{3}\right)}=\mathop{\underbrace{\phi}}\limits_{\mathfrak{H}^{s}\left(\mathbb{R}_{+} ^{3}\right)}+\mathop{\underbrace{\int_{0}^{t}\operatorname{div}\left(\rho\left(1-\Delta\right)^{{-}1}\nabla P\left(\rho\right)\right)\left(t'\right)\textrm{d}t'}}\limits_{\mathfrak{H}^{s-1}\left( \mathbb{R}_{+}^{3}\right)}.\end{equation}

so we cannot apply Banach's fixed point theorem and Gronwall's inequality to establish local well-posedness. However, we can introduce an artificial viscosity $\mu >0$ to obtain the regularized Cauchy problem

(3.22)\begin{equation} \begin{cases} \partial_{t}\rho_{\mu}=\operatorname{div}\left(\rho_{\mu}\left(1-\Delta\right)^{{-}1}\nabla P\left(\rho_{\mu}\right)\right)+\mu\Delta\rho_{\mu}\\ \rho_{\mu}\left(0\right) =\rho_{0}. \end{cases} \end{equation}

which is equivalent to the integral equation

(3.23)\begin{equation} \rho_{\mu}\left(t\right)=U_{\mu}\left(t\right)\rho_{0}+\int_{0}^{t}U_{\mu}\left(t-t^{\prime}\right)\left[A\left(\rho_{\mu}\left(t^{\prime}\right)\right)\rho_{\mu}\left(t^{\prime}\right)\right]\textrm{d}t^{\prime}, \end{equation}

where $U_{\mu }(t)$ is the infinitely smoothing $C^{0}$ semigroup

(3.24)\begin{equation} U_{\mu}\left(t\right)f=\exp\left(\mu t\Delta\right) f=\mathfrak{F}^{{-}1}\,\textrm{e}^{-\mu t(\xi^{2}+\eta^{2}+\alpha^{2})}\mathfrak{F}f. \end{equation}

Then we can show that (see [Reference Alarcon and Iório1, Reference Molina Del Sol, Alarcon and Iorio23])

Theorem 3.2 Assume that $\mu >0$ and that $P$ satisfy (3.12) for all (fixed) $s>3/2$. Then (3.23) is locally well-posed in $\mathfrak {H} ^{s}(\mathbb {R}_{+}^{3})$. Moreover, if $(0,T_{\mu }]$ is an interval of existence, then $\rho _{\mu }\in C((0,T_{\mu }];\mathfrak {H}^{\infty }(\mathbb {R}_{+}^{3}))$, where $\mathfrak {H}^{\infty }(\mathbb {R}_{+}^{3})=\cap_{s\in \mathbb {R}_{+}^{3}}\mathfrak {H}^{s}(\mathbb {R} _{+}^{3})$ provided with its natural Frechet space topology.

Proof. It should be noted that the proof (even in $\mathbb {R}^{3}$) relies heavily on the inequality

(3.25)\begin{equation} \left\Vert U_{\mu}\left(t\right)\phi\right\Vert_{r+\lambda}\leq K_{\lambda}\left[1+\left(\frac{1}{2\mu t}\right)^{\lambda}\right] ^{1/2}\left\Vert \phi\right\Vert _{r} \end{equation}

where $K_{\lambda }>0$ depends only on $\lambda$ and holds for all $\phi \in H^{r}(\mathbb {R}_{+}^{3})$, $r\in \mathbb {R}$, $\lambda \geq 0$, and $\mu ,\,t>0.$ (See [Reference Alarcon and Iório1, Reference Iorio and Iorio10, Reference Iorio13, Reference Molina Del Sol, Alarcon and Iorio23] e.g.).

The next step, is to employ a bootstrapping argument combining (3.23) and (3.25) (with $\lambda$ fixed in the interval $(1,2)$), to show that $\rho _{\mu }\in C((0,T_{\mu }];\mathfrak {H}^{\infty }(\mathbb {R}_{+}^{3}))$.

Let $t\in (\theta ,\infty ],\,\theta >0$ be fixed (but arbitrary). Then

\begin{align*} \left\Vert \rho_{\mu}\left(t\right)\right\Vert _{s+\lambda}&\leq\left\Vert U_{\mu}\left(t\right)\rho_{0}\right\Vert _{s+\lambda}+\left\Vert\int_{0} ^{t}U_{\mu} \left(t-t^{\prime}\right)\left[A\left(\rho_{\mu}\left(t^{\prime}\right)\right)\rho_{\mu}\left(t^{\prime}\right) \right] \right\Vert_{s+\lambda}\,\textrm{d}t^{\prime},\\ &\leq\left\Vert U_{\mu}(t) \rho_{0}\right\Vert _{s+\lambda}+\int_{0}^{t}\left\Vert U_{\mu}\left(t-t^{\prime}\right)\left[A\left(\rho_{\mu}\left( t^{\prime}\right)\right)\rho_{\mu}\left(t^{\prime}\right) \right]\right\Vert _{s+\lambda}\,\textrm{d}t^{\prime}\\ &\leq K_{\lambda}\left[1+\left(\frac{1}{2\mu t}\right)^{\lambda}\right]^{1/2}\left\Vert \rho_{0}\right\Vert _{s}\\ &\quad +\int_{0}^{t}K_{\lambda }\left[1+\left(\frac{1}{2\mu t}\right)^{\lambda}\right]^{1/2}\left\Vert A\left(\rho_{\mu}\left(t^{\prime}\right)\right)\rho_{\mu }\left(t^{\prime}\right)\right\Vert _{s}\,\textrm{d}t^{\prime}\\ &\leq K_{\lambda}\left[1+\left(\frac{1}{2\mu t}\right)^{\lambda}\right]^{1/2}\left\Vert \rho_{0}\right\Vert_{s}\\ &\quad +K_{\lambda}L_{s} \int_{0}^{t}\left[1+\left(\frac{1}{2\mu t}\right)^{\lambda}\right]^{1/2}\left\Vert \rho_{\mu}\left(t^{\prime}\right)\right\Vert _{s}\,\textrm{d}t^{\prime} \end{align*}

Since$\ t$ is bounded away from $0$ in function

\[ f\left(t\right) =\left[1+\left(\frac{1}{2\mu t}\right)^{\lambda}\right]^{{1}/{2}}, \]

is bounded in any interval $(\theta ,\infty ]$ so that,

\[ \left\Vert \rho_{\mu}\left(t\right)\right\Vert _{s+\lambda}\leq K_{\lambda}\sup f\left(t\right)\left\Vert \rho_{0}\right\Vert +K_{\lambda }L_{s}\int_{0}^{t}\left[1+\left(\frac{1}{2\mu t}\right)^{\lambda}\right]^{1/2}\left\Vert \rho_{\mu}\left(t^{\prime}\right)\right\Vert _{s}\,\textrm{d}t \]

Now, the integral on the left-hand side contributes with a term containing

\[ \frac{1}{t^{1+{\lambda}/{2}}}, \]

which is finite in every interval $(\theta ,T^{\prime }], \theta >0, T^{\prime }>0$. Therefore, Gronwall's inequality implies that

\[ \rho_{\mu}\in C\left((\theta,T_{\mu}];\mathfrak{H}^{\infty}\left(\mathbb{R}_{+}^{3}\right)\right) \]

Next, the, usual limiting process involved in the method of parabolic regularization (see [Reference Iorio and Iorio10, Reference Iorio13]) we are able to show existence and uniqueness of solutions in $AC([0,T];\mathfrak {H} ^{s-1}({\mathbb {R}_{+}^{3}}))\cap L^{\infty }([0,T] ;\mathfrak {H}^{s}({\mathbb {R}_{+}^{3}}))$. Due to technical reasons (lack of invariance under certain changes of variables, see [Reference Iorio12, Reference Iorio13, Reference Kato20]), so far we were unable to prove that the solution we obtained in this way actually belongs to $C([0,T];\mathfrak {H}^{s}( {\mathbb {R}_{+}^{3}})) \cap C^{1}([0,T];\mathfrak {H}^{s-1}({\mathbb {R}_{+}^{3}}))$, $s>3/2$ as we would have liked. However, combining what we already have, with the results in theorem 3.1, proved using Kato's theory when $s>5/2$, we see that the solutions must coincide, due to uniqueness, if $s>5/2$.

4. Comparison principle

To simplify the notation we will write

\[ \mathcal{B}f=R({-}1)f=\mathfrak{F}^{{-}1}\,\left(\xi^{2}+\eta^{2}+\alpha^{2}+1\right)^{{-}1}\,\mathfrak{F}f,\quad f\in\mathfrak{L}^{2}\left( \mathbb{R}_{+}^{3}\right). \]

In order to state our results, we define of the fractional power spaces associated with Neumann Laplacian $-\Delta$. Following the arguments found in [Reference Kato and Fujita16, Reference Weissler31]. For $\alpha >0$ and $f\in L^{2}({\mathbb {R}_{+}^{3}})$, defineFootnote ¹⁰

\[ R^{\alpha}({-}1)\,f=\left(1-\Delta\right)^{-\alpha}\,f=\dfrac{1}{\Gamma(\alpha)}\,\int_{0}^{+\infty}\,t^{\alpha-1}\,\textrm{e}^{{-}t}\,\textrm{e}^{t\,\Delta}f\,\textrm{d}t. \]

Then $(1-\Delta )^{-\alpha }$ is a bounded, one-to-one operator on $\mathfrak {L}^{2}({\mathbb {R}_{+}^{3}})$. We let $\mathfrak {J}^{\alpha }=(1-\Delta )^{\alpha }$ be the inverse of $( 1-\Delta ) ^{-\alpha }$. For $s>0$, the Hilbert space $\mathfrak {H}^{s}({\mathbb {R}_{+}^{3}})$ is the range of $(1-\Delta ) ^{-{s/2}}$ with the inner product

(4.1)\begin{equation} \langle f,\,g\rangle_{\mathfrak{H}^{s}}=\int_{L^{2}({\mathbb{R}_{+}^{3}})}\,\mathfrak{J}^{s/2}f\,\overline{\mathfrak{J}^{s/2}g}\,\textrm{d}x. \end{equation}

Consider the initial value problem (3.5) with $F(t,\rho )=0$,Footnote ¹¹ $P(\rho )=\rho ^{2k},\ k=1,2,3\ldots$.

(4.2)\begin{equation} \left\{\begin{array}{l} \partial_{t}\rho+ {{\mathrm{div}\,}}(\rho \;\mathbf{v}) =0,\ x \in {\mathbb{R}^{3}_+},\ t\in(0,T_{0}]\\ \mathbf{v} ={-} \mathcal{B}\,\nabla \rho^{2k}={-}\vec{\Theta}(\rho^{2 k})\\ (\rho(0),\ \mathbf{v}(0))=(\rho_{0},\,\mathbf{v}_{0}) \end{array}\right. \end{equation}

Theorem 4.1 (Comparison principle) Let $(\rho ,v)$ and $(\eta ,w)$ be solutions of (3.5) with $P(\rho )=\rho ^{2k},P(\eta )=\eta ^{2k},\ k=1,2,3\ldots$;Footnote ¹² and initial values $(\rho _0,v_0)$ and $(\eta _0,w_0)$ respectively. Then

(4.3)\begin{equation} 0\leq\eta_{0}(x)\leq\rho_{0}(x)\ \text{in}\ \Omega\Rightarrow 0\leq \eta(x,t)\leq\rho(x,t)\ \text{in}\ \Omega\times\lbrack 0,T_{0}] \end{equation}

Proof. In this proof, we use the same idea employed by Alarcon, Iorio and Del Sol in the study of Brinkman flow in $\mathbb {R}^{n}$ [Reference Molina Del Sol, Alarcon and Iorio23]. Let

(4.4)\begin{equation} R(t,y)=\rho(\phi(t,y),t);S(t,y)=\eta(\psi(t,y),t) \end{equation}

and

(4.5)\begin{equation} Q(t,y)=R(t,y)-S(t,y) \end{equation}

where $\phi (t,y)$ and $\psi (t,y)$ satisfy the following ordinary differential equations,

(4.6)\begin{align} &\begin{cases} \displaystyle\dfrac{\partial\phi}{\partial t}(t,y)=\mathbf{v}(\phi(t,y),t) & \phi(t,y)=\left(\phi_{1}(t,y),\phi_{2}(t,y), \phi_{3}(t,y)\right)\\ \phi(0,y)=y & v_{i}={-}\partial_{x_{i}}\mathcal{B}(\rho^{2k})\end{cases}, \end{align}

(4.7)\begin{align} &\begin{cases} \displaystyle\dfrac{\partial\psi}{\partial t}(t,y)=\mathbf{w}(\psi(t,y),t) & \psi(t,y)=\left(\psi_{1}(t,y),\psi_{2}(t,y),\psi_{3}(t,y)\right)\\ \psi(0,y)=y & w_{i}={-}\partial_{x_{i}}\mathcal{B}(\eta^{2k})\end{cases}, \end{align}

Now,

(4.8)\begin{equation} \left\{\begin{array}{@{}ll} \displaystyle\dfrac{\textrm{d}R}{\textrm{d}t} ={-}R\,\textrm{div}\;\mathbf{v} & \displaystyle\dfrac{\textrm{d}S}{\textrm{d}t}={-}S\,\textrm{div}\;\mathbf{w}\\ R(0,y) = \rho_{0}(y) & S(0,y) = \eta_{0}(y)\end{array}\right. \end{equation}

Solving (4.8), we obtain:

(4.9)\begin{equation} R(t)=R(0)\exp\left[-\int_{0}^{t}\text{div}\;v(\phi(s,y),s)\,\textrm{d}s\right]\overbrace {\Rightarrow}^{\rho_{0}(y)\geq 0}R(t)\geq 0 \end{equation}

Analogously, we have that:

(4.10)\begin{equation} S(t)=S(0)\exp\left[-\int_{0}^{t}\text{div}\;w(\psi(s,y),s)\,\textrm{d}s\right]\overbrace {\Rightarrow}^{\eta_{0}(y)\geq 0}S(t)\geq 0 \end{equation}

On the other hand, differentiating $Q(t)$:

(4.11)\begin{align} \frac{\textrm{d}Q}{\textrm{d}t}& =\frac{\textrm{d}R}{\textrm{d}t}-\frac{\textrm{d}S}{\textrm{d}t}=(-\text{div}\;v)R(t)+(\text{div}\;\mathbf{w})S(t)\nonumber\\ &={-}\rho\,\text{div}\;v+\eta\,\text{div}\;w\nonumber\\ &={-}(\rho-\eta)\text{div}\;v+\eta(\text{div}\;w-\text{div}\;v)\nonumber\\ &={-}Q(t)(\text{div}\;\mathbf{v})+S(t)(\text{div}\;w-\text{div}\;v) \end{align}

where

(4.12)\begin{equation} \text{div}\;v=\rho^{2k}-\mathcal{B}(\rho^{2k}),\quad\text{div}\;w=\eta^{2k}-\mathcal{B}(\eta^{2k}) \end{equation}

Substituting (4.12) in (4.11), we obtain a new ordinary differential equation for $Q(t)$,

(4.13)\begin{equation} \left\{\begin{array}{@{}l} \displaystyle\dfrac{\textrm{d}Q}{\textrm{d}t} ={-}\left[\textrm{div}\;\mathbf{v}+ S(t)P(R(t),S(t))\right]Q(t) + B(t,Q(t))\\ Q(0)= \rho_{0}(y)-\eta_{0}(y) \end{array}\right. \end{equation}

with

(4.14)\begin{equation} P(R(t),S(t))=P(\rho,\eta)=\sum_{i=0}^{2k-1}\rho^{2k-1-i}\eta^{i} \end{equation}

and

(4.15)\begin{equation} B(t,Q(t))=S(t)(1-\Delta)^{{-}1}\left[Q(t)P(R(t),S(t))\right]. \end{equation}

Thus,

(4.16)\begin{equation} Q(t)=U(t,0)Q(0)+\int_{0}^{t}U(t,s)B(s,Q(s))\,\textrm{d}s \end{equation}

where

(4.17)\begin{equation} U(t,s)=\exp\left[-\int_{s}^{t}[\text{div}\;(v(\phi(\tau,y),\tau))+S(\tau)P(R(\tau),S(\tau))]\,\textrm{d}\tau\right]. \end{equation}

In view of conditions for ${\rho _{0}}$ and $\eta _{0}$, we have that $R(t)\geq 0$ and $S(t)\geq 0$.

Consider the sequence

\[ Q_{n}(t)=\left\{\begin{array}{@{}ll} U(t,0)Q(0)+ \int_{0}^{t}U(t,s)B(s,Q_{n-1}(s))\,\textrm{d}s, & \textrm{se}\ n= 1,2,\ldots;\\ \rho_{0}(y)-\eta_{0}(y) , & \textrm{se}\ n=0. \end{array} \right. \]

If $Q(0)\geq 0$, then $Q_{n}(t)\geq 0$, for all $n$. Therefore,

(4.18)\begin{equation} Q(t)=\rho\left(\phi(t,y),t\right)-\eta\left(\psi(t,y),t\right)=\lim_{n\rightarrow \infty}Q_{n}(t)\geq 0 \end{equation}

To complete the proof we need to show the functions $y\in \Omega \rightarrow \phi (t,y)\in \Omega$ and $y\in \Omega \rightarrow \psi (t,y)\in \Omega$ are onto. To do this, we analyse in detail the map $y\in \Omega \rightarrow \phi (t,y)\in \Omega$.

The Neumann boundary condition $\rho _{z}=0$ for $z=0$ and Brinkman's condition $\ v=-\nabla \,\mathcal {B}(\rho ^{2k})$,Footnote ¹³ implies that

(4.19)\begin{equation} v_{3}((x_{1},\,x_{2},\,0),\,t)=0,\ \forall(x_{1},\,x_{2})\in\mathbb{R} ^{2}\ \text{and}\ t\in\lbrack 0,\,T], \end{equation}

then $\phi _3(t,\,(x_1,\,x_2,\,0)=0$ for all $(x_1,\,x_2)\in \mathbb {R}^{2}$ and $t\in \lbrack 0,\,T]$.

We will show that $\Omega$ it is invariant under the flow $\phi (t,\,y)$, that is,

(4.20)\begin{equation} \phi\left(\Omega\right)\subset\Omega. \end{equation}

By (4.19), the plane $\Pi =\{(x_{1},\,x_{2},\,x_{3})\in \mathbb {R}^{3}\colon x_{3}=0\}$ is invariant under the flow $\phi (t,\,y)$, i.e. $\phi (\Pi )\subset \Pi$. Next we show that (4.20) holds. To this end, it is enough to verify that

(4.21)\begin{equation} \phi(t,\,x_{1},\,x_{2},\,x_{3})\in\Omega,\quad \forall(x_{1},\,x_{2})\in \mathbb{R}^{2},\quad x_{3}>0. \end{equation}

If (4.21) does not hold, there is a $w=(w_{1},\,w_{2},\,w_{3})$ with $w_{3}>0$ and $0< t_{1}< t_{2}\leq T$ such that $\phi (t_{1},\,w)\in \Pi$ and $\phi (t_{2},\,w)\notin \Omega$. But (4.19) implies that

(4.22)\begin{equation} \phi(t_{2},\,w)=\phi(t_{1},\,w)+\int_{t_{1}}^{t_{2}}\,v(\phi(s,\,w),\,s)\,\textrm{d}s, \end{equation}

so that $\phi _{3}(t_{2},\,w)=0$. This contradiction proves (4.20). From (4.6), integrating from 0 to t, we get:

(4.23)\begin{equation} \phi_{i}(t)-y_{i}=\int_{0}^{t}v_{i}(\phi(\tau,y),\tau)\,\textrm{d}\tau;\quad i=1,2,3, \end{equation}

so that

(4.24)\begin{equation} |\phi_{i}(t)-y_{i}|\leq\int_{0}^{t}|v_{i}(\phi(\tau,y),\tau)|\,\textrm{d}\tau\leq a_{i}(\Vert\rho_{0}\Vert_{s},t),\quad i=1,2,3,\,s>\frac{5}{2}, \end{equation}

(4.25)\begin{equation} y_{i}-a_{i}(\Vert\rho_{0}\Vert_{s},t)\leq\phi_{i}(t,y)\leq y_{i}+a_{i}(\Vert\rho_{0}\Vert_{s},t),\ \forall y=(y_{1},\,y_{2},\,y_{3})\in\Omega. \end{equation}

Let $(z_{1},\,z_{2},\,z_{3})\in {\mathbb {R}_{+}^{3}}$. Taking $y_{i}^{(1)}\ll 0$ for $i=1,\,2$; $y_{i}^{(2)}\gg 0$ for $i=1,\,2,\,3$, such that $z_{i}\in (y_{i}^{(1)},y_{i}^{(2)})$ for $i=1,\,2$ and $0< z_{3}< y_{3}^{(2)}$ we have:

(4.26)\begin{equation} y_{i}^{(1)}+a_{i}(\Vert\rho_{0}\Vert_{s},t)< z_{i}< y_{i}^{(2)}-a_{i}(\Vert \rho_{0}\Vert_{s},t) \end{equation}

and

(4.27)\begin{equation} 0< z_{3}<\phi_{i}(t,\,y_{3}^{(2)}). \end{equation}

Therefore

(4.28)\begin{equation} \phi_{i}(t,y_{i}^{(1)})< z_{i}<\phi_{i}(t,y_{i}^{(2)}),\quad \text{for}\ i=1, 2, 3. \end{equation}

Applying the intermediate value theorem to $\phi _{i}$ implies that there exists $y_{i}\in (y_{i}^{(1)},y_{i}^{(2)})$ satisfying $\phi _{i}(t,y_{i})=z_{i}$. For $z_{3}=0$ the proof is analogous, since the plane $x_{3}=0$ is invariant by the flow $\phi (t,\,y)$, a consequence of (4.19).

5. Global results in $\mathfrak {H}^{s}({\mathbb {R}_{+}^{3}}),\ s>5/2$

In this section, we obtain the global $\mathfrak {H}^{s}$-estimate for the solution of the Brinkman flow equation. This will be a consequence of global well-posedness of the regularized problem.

First, we introduce the following estimates

Lemma 5.1 If $s>0$, then

(5.1)\begin{align} &\left\|\sum_{k=1}^{n}\left[\partial_{x_{k}}\mathfrak{J}^{s}(g\partial_{x_{k}}f)-\partial_{x_{k}}f(\partial_{x_{k}}\mathfrak{J}^{s} g)\right]\right\|_{\mathfrak{L}^{2}({\mathbb{R}_{+}^{3}})}\nonumber\\ &\quad\leq c\left(\Vert\mathfrak{J}^{2}f\Vert_{L^{\infty}({\mathbb{R}_{+}^{3}})}\Vert \mathfrak{J}^{s}g\Vert_{\mathfrak{L}^{2}({\mathbb{R}_{+}^{3}})}+\right. \left.\Vert\mathfrak{J}^{s+2}f\Vert_{\mathfrak{L}^{2}({\mathbb{R}_{+}^{3}})}\Vert g\Vert_{L^{\infty}({\mathbb{R}_{+}^{3}})}\right) \end{align}

Lemma 5.2 If $s>0$, then $\mathfrak {H}^{s}({\mathbb {R}^{3}_{+}})\cap L^{\infty }({\mathbb {R}^{3}_{+}})$ is a Banach algebra. Moreover

(5.2)\begin{equation} \|fg\|_{s}\leq c(\|f\|_{L^{\infty}({\mathbb{R}^{3}_{+}})}\|g\|_{\mathfrak{L}^{2}({\mathbb{R}^{3}_{+}})}+\|f\|_{\mathfrak{L} ^{2}({\mathbb{R}^{3}_{+}})}\|g\|_{L^{\infty}({\mathbb{R}^{3}_{+}})}) \end{equation}

Lemma 5.3 Let $f\in X^{s}({\mathbb {R}_{+}^{3}})$, $s>\frac {5}{2},\ k=1,2,\ldots$. Then

\[ \Vert f^{2k}\Vert_{s}\lesssim\Vert f\Vert_{L^{\infty}({\mathbb{R}_{+}^{3}})}^{2k-1}\Vert f\Vert_{s}, \]

where $A\lesssim B$ means that exist a constant $c > 0$ such that $A \leq c\, B$.

Now, we are ready to establish the following result.

Proof. The comparison principle implies that $0\leq \rho (x,t)\leq 1$. Using the regularized initial value problem, with the simplified notations $\rho _{\mu }(t)\equiv \tilde {\rho },\ v\mathbf {_{\mu }}(t)\equiv v$.

(5.3)\begin{equation} \left\{\begin{array}{@{}l} \partial_{t}\tilde{\rho} - \mu\,\Delta_N\, \tilde{\rho} + \textrm{div}\;[\tilde{\rho}\,\mathbf{v}] = 0 \\ \mathbf{v} ={-}\mathcal{B}\,\nabla \,\tilde{\rho}^{2k}\\ (\tilde{\rho}(0),\mathbf{v}(0)) = (\tilde{\rho}_{0},\mathbf{v}_{0}) \end{array}\right.. \end{equation}

Applying $\mathfrak {J}^{s}$ to regularized equation:

(5.4)\begin{equation} \frac{\textrm{d}}{\textrm{d}t}(\mathfrak{J}^{s}\tilde{\rho})-\mu(\mathfrak{J}^{s}\Delta_{N}\,\tilde{\rho})+\mathfrak{J}^{s}\text{div}\;(\tilde{\rho}\,v)=0. \end{equation}

Multiplying (5.4) by $\mathfrak {J}^{s}\tilde {\rho }$ and integrating over ${\mathbb {R}_{+}^{3}}$ we get,

(5.5)\begin{equation} \frac{1}{2}\frac{\textrm{d}}{\textrm{d}t}\int(\mathfrak{J}^{s}\tilde{\rho})^{2}\,\textrm{d}x=\mu \int(\mathfrak{J}^{s}\tilde{\rho})\mathfrak{J}^{s}(\Delta_{N}\,\tilde{\rho })\,\textrm{d}x-\int(\mathfrak{J}^{s}\tilde{\rho})(\mathfrak{J}^{s}\text{div}\;(\tilde{\rho}\,v))\,\textrm{d}x, \end{equation}

so that

(5.6)\begin{equation} \frac{1}{2}\frac{\textrm{d}}{\textrm{d}t}\int(\mathfrak{J}^{s}\tilde{\rho})^{2}\,\textrm{d}x=\underbrace {\mu\int(\mathfrak{J}^{s}\tilde{\rho})\Delta_{N}\,(\mathfrak{J}^{s}\tilde {\rho})\,\textrm{d}x}_{{\leq} 0}-\sum_{i=1}^{3}\int(\mathfrak{J}^{s}\tilde{\rho})\partial_{x_{i}}\mathfrak{J}^{s}(\tilde{\rho}\,v_{i})\,\textrm{d}x. \end{equation}

Using the commutator $[\partial _{x_{i}}\mathfrak {J}^{s},v_{i}]\tilde {\rho }=\partial _{x_{i}}\mathfrak {J}^{s}(\tilde {\rho }\,v_{i})-v_{i}\partial _{x_{i}}J^{s}\tilde {\rho }$, we obtain:

(5.7)\begin{equation} \frac{1}{2}\frac{\textrm{d}}{\textrm{d}t}\int(J^{s}\tilde{\rho})^{2}\,\textrm{d}x\leq{-}\sum_{i=1}^{3} \int(\mathfrak{J}^{s}\tilde{\rho})[\partial_{x_{i}}\mathfrak{J}^{s},v_{i}]\tilde{\rho}\,\textrm{d}x-\sum_{i=1}^{3}\int(\mathfrak{J}^{s}\tilde{\rho} )v_{i}\partial_{x_{i}}\mathfrak{J}^{s}\tilde{\rho}\,\textrm{d}x. \end{equation}

Integration by parts in (5.7) yields,

(5.8)\begin{equation} \frac{1}{2}\frac{\textrm{d}}{\textrm{d}t}\int(\mathfrak{J}^{s}\tilde{\rho})^{2}\,\textrm{d}x\leq{-}\sum_{i=1}^{3}\int(\mathfrak{J}^{s}\tilde{\rho})[\partial_{x_{i}} \mathfrak{J}^{s},v_{i}]\tilde{\rho}\,\textrm{d}x+\frac{1}{2}\int(\mathfrak{J}^{s} \tilde{\rho})^{2}\text{div}\;v\,\textrm{d}x. \end{equation}

Taking (4.12) into (5.8) we obtain,

(5.9)\begin{align} \frac{1}{2}\frac{\textrm{d}}{\textrm{d}t}\int(\mathfrak{J}^{s}\tilde{\rho})^{2}\,\textrm{d}x&\leq{-}\sum\limits_{i=1}^{3}\int(\mathfrak{J}^{s}\tilde{\rho})[\partial_{x_{i}}\mathfrak{J}^{s},v_{i}]\tilde{\rho}\,dx+\frac{1}{2}\int(\mathfrak{J}^{s}\tilde{\rho})^{2}\tilde{\rho}^{2k}\,\textrm{d}x\nonumber\\ &\quad-\frac{1}{2}\int(\mathfrak{J}^{s}\tilde{\rho})^{2}\mathfrak{J}^{{-}1/2}(\tilde{\rho}^{2k})\,\textrm{d}x. \end{align}

From the second equation in (5.3) we have $v_{i}=-\partial _{x_{i}}B_{i}(\tilde {\rho }^{2k})$. Substituting it in (5.9)

(5.10)\begin{align} \frac{\textrm{d}}{\textrm{d}t}\int(\mathfrak{J}^{s}\tilde{\rho})^{2}\,\textrm{d}x&\leq\int(\mathfrak{J}^{s}\tilde{\rho})^{2}\tilde{\rho}^{2k}\,\textrm{d}x-\overbrace{\int(\mathfrak{J} ^{s}\tilde{\rho})^{2}\mathfrak{J}^{{-}1/2}\tilde{\rho}^{2k}\,\textrm{d}x}^{{\geq} 0}\nonumber\\ &\quad+2\int(\mathfrak{J}^{s}\tilde{\rho})\left(\sum\limits_{i=1}^{3}[\partial_{x_{i}}\mathfrak{J}^{s},\partial_{x_{i}}B_{i}\left(\tilde{\rho} ^{2k}\right)]\tilde{\rho}\right)\,\textrm{d}x. \end{align}

Noting that the third term in (5.10) is non-negative, and applying Cauchy Schwartz inequality in the fourth term we get,

(5.11)\begin{align} \frac{\textrm{d}}{\textrm{d}t}\int(\mathfrak{J}^{s}\tilde{\rho})^{2}\,\textrm{d}x&\leq\Vert\tilde{\rho }^{2k}\Vert_{L^{\infty}({\mathbb{R}_{+}^{3}})}\int(\mathfrak{J}^{s}\tilde{\rho})^{2}\,\textrm{d}x\nonumber\\ &\quad+2\Vert\mathfrak{J}^{s}\tilde{\rho}\Vert_{\mathfrak{L}\;^{2}({\mathbb{R}_{+}^{3}})}\left\|\sum\limits_{i=1}^{3}\left[\partial_{x_{i}}\mathfrak{J}^{s},\partial_{x_{i}}\mathfrak{J}^{{-}1/2}\tilde{\rho} ^{2k}\right]\tilde{\rho}\right\|_{\mathfrak{L}^{2}({\mathbb{R}_{+}^{3}})} \end{align}

Using lemma 5.1 in (5.11), with $f=J^{-1/2}\tilde {\rho }^{2k}$ and $g=\tilde {\rho }$, we obtain:

(5.12)\begin{align} \frac{\textrm{d}}{\textrm{d}t}\Vert\tilde{\rho}\Vert_{s}^{2}\leq\Vert\tilde{\rho}^{2k}\Vert_{L^{\infty}({\mathbb{R}_{+}^{3}})}\Vert\tilde{\rho}\Vert_{s}^{2} +2c\Vert\tilde{\rho}\Vert_{s}\left[\Vert\tilde{\rho}^{2k}\Vert_{L^{\infty}({\mathbb{R}_{+}^{3}})}\Vert\tilde{\rho}\Vert_{s}+\Vert\tilde{\rho}^{2k} \Vert_{s}\Vert\tilde{\rho}\Vert_{L^{\infty}({\mathbb{R}_{+}^{3}})}\right] \end{align}

Applying lemma 5.3 in the term $\Vert \tilde {\rho }\Vert _{s}^{2}$, in (5.12):

(5.13)\begin{equation} \frac{\textrm{d}}{\textrm{d}t}\Vert\tilde{\rho}\Vert_{s}^{2}\lesssim\Vert\tilde{\rho} \Vert_{L^{\infty}({\mathbb{R}_{+}^{3}})}^{2k}\Vert\tilde{\rho}\Vert_{s}^{2} \end{equation}

Now, we need to estimate $\Vert \tilde {\rho }\Vert _{L^{\infty }({\mathbb {R} _{+}^{3}})}$. Applying the comparison principle for $\rho$ together with Sobolev's lemma we have

(5.14)\begin{equation} \Vert\tilde{\rho}\Vert_{L^{\infty}({\mathbb{R}_{+}^{3}})}\leq\Vert\tilde{\rho}-\rho\Vert_{L^{\infty}({\mathbb{R}_{+}^{3}})}+\Vert\rho\Vert_{L^{\infty }({\mathbb{R}_{+}^{3}})}\lesssim 1+\Vert\tilde{\rho}-\rho\Vert_{s} \end{equation}

Since $\Vert \tilde {\rho }-\rho \Vert _{s}=\sup _{\Vert \varphi \Vert _{s}=1} |\langle \tilde {\rho }-\rho ,\varphi \rangle _{s}|$ we proceed as follows: in the analysis of the weak convergence of sequence $\rho _{\mu }$ (see the proof of theorem 4.2 in [Reference Molina Del Sol, Alarcon and Iorio23]) we obtained

(5.15)\begin{align} |\langle\rho_{\mu}(t)-\rho_{\eta}(t),\varphi\rangle_{s}|&\leq\Vert\rho_{\mu}(t)-\rho_{\eta}(t)\Vert_{s}\Vert\varphi-\varphi_{\epsilon}\Vert _{s}+\Vert\rho_{\mu}(t)-\rho_{\eta}(t)\Vert_{\mathfrak{L}^{2}({\mathbb{R}_{+}^{3}})}\,\Vert\varphi_{\epsilon}\Vert_{2s}\nonumber\\ &\leq 2M\epsilon+\Vert\rho_{\mu}(t)-\rho_{\eta}(t)\Vert_{\mathfrak{L}^{2}({\mathbb{R}_{+}^{3}})}\,\Vert\varphi_{\epsilon}\Vert_{2s} \end{align}

Taking the limit as $\eta \rightarrow 0$ in (5.15), it follows that,

(5.16)\begin{equation} |\langle\rho_{\mu}(t)-\rho(t),\varphi\rangle_{s}|\leq 2M\epsilon+\Vert\rho _{\mu}(t)-\rho(t)\Vert_{\mathfrak{L}^{2}({\mathbb{R}_{+}^{3}})}\,\Vert \varphi_{\epsilon}\Vert_{2s} \end{equation}

Noting that $\Vert \rho _{\mu }(t)-\rho _{\nu }(t)\Vert _{\mathfrak {L} ^{2}({\mathbb {R}_{+}^{3}})}\leq 2M\sqrt {n\tilde {T}_{s}|\mu -\nu |}\,\textrm {e}^{\tilde {T}_{s}L_{0}(M,M)}$ (see the proof of theorem 4.2 in [Reference Molina Del Sol, Alarcon and Iorio23]) and taking the limit as $\nu \rightarrow 0$, it follows that,

(5.17)\begin{equation} \Vert\rho_{\mu}(t)-\rho(t)\Vert_{\mathfrak{L}^{2}({\mathbb{R}_{+}^{3}})} \leq 2M\sqrt{n\,\tilde{T}_{s}\,\mu}\,\textrm{e}^{\tilde{T}_{s}L_{0}(M,M)}=\widetilde {C}(n,M,\tilde{T}_{s})\sqrt{\mu} \end{equation}

Substituting (5.17) in (5.16) and noting that $\Vert \varphi _{\epsilon }\Vert _{2s}\leq \epsilon ^{-s}\Vert \varphi \Vert _{s}$ with $\varphi _{\epsilon }$ constructed as in [Reference Iorio, Linares and Scialom11, lemma 2.6, p. 900], yields

(5.18)\begin{equation} |\langle\rho_{\mu}(t)-\rho(t),\varphi\rangle_{s}|\leq 2M\epsilon+\widetilde {C}(n,M,\tilde{T}_{s})\sqrt{\mu}\,\epsilon^{{-}s}\Vert\varphi\Vert_{s} \end{equation}

Then

(5.19)\begin{equation} \Vert\tilde{\rho}-\rho\Vert_{s}=\sup_{\Vert\varphi\Vert_{s}=1}|\langle \tilde{\rho}-\rho,\varphi\rangle_{s}|\leq 2M\epsilon+\widetilde{C} (n,M,\tilde{T}_{s})\sqrt{\mu}\,\epsilon^{{-}s} \end{equation}

and

(5.20)\begin{equation} \Vert\tilde{\rho}\Vert_{L^{\infty}({\mathbb{R}_{+}^{3}})}\lesssim 1+2M\epsilon+\widetilde{C}(n,M,\tilde{T}_{s})\sqrt{\mu}\,\epsilon^{{-}s},\quad \forall\epsilon>0 \end{equation}

Since $r(\tau )=\tau ^{2k}$ is a non-decreasing function, it follows that:

(5.21)\begin{equation} \frac{\textrm{d}}{\textrm{d}t}\Vert\tilde{\rho}\Vert_{s}^{2}\lesssim r(1+2M\epsilon +\widetilde{C}(n,M,\tilde{T}_{s})\sqrt{\mu}\,\epsilon^{{-}s})\Vert\tilde{\rho }\Vert_{s}^{2} \end{equation}

Integrating from $0$ to $t$ in (5.21)

(5.22)\begin{equation} \Vert\tilde{\rho}\Vert_{s}^{2}\lesssim\Vert\rho_{0}\Vert_{s}^{2}+r(1+2M\epsilon+\widetilde{C}(n,M,\tilde{T}_{s})\sqrt{\mu}\,\epsilon^{{-}s} )\int_{0}^{t}\Vert\tilde{\rho}(\tau)\Vert_{s}^{2}\,\textrm{d}\tau\end{equation}

Applying Gronwall's inequality to (5.22), we obtain a priori-estimate in $H^{s}(R_{+}^{3});\ s>5/2$

(5.23)\begin{equation} \Vert\tilde{\rho}\Vert_{s}^{2}\lesssim\Vert\rho_{0}\Vert_{s}^{2}\,\textrm{e}^{r\left(1+2M\epsilon+\widetilde{C}(n,M,\tilde{T}_{s})\sqrt{\mu} \,\epsilon^{{-}s}\right)\tilde{T}_{s}},\quad \forall\tilde{T}_{s}>0,\quad\forall \epsilon>0 \end{equation}

Finally, applying [Reference Yosida35, theorem 1, p. 120] in (5.23) we obtain the final estimate

(5.24)\begin{align} \Vert\rho(t)\Vert_{s}^{2}&\leq\liminf_{\mu\rightarrow 0}\Vert\rho_{\mu}(t)\Vert_{s}^{2}\nonumber\\ &\leq\liminf_{\mu\rightarrow 0}\Vert\rho_{0}\Vert_{s}^{2}\,\textrm{e}^{r\left(1+2M\epsilon+\widetilde{C}(n,M,\tilde{T}_{s})\sqrt{\mu} \,\epsilon^{{-}s}\right)\tilde{T}_{s}}\nonumber\\ &=\lim_{\mu\rightarrow 0}\Vert\rho_{0}\Vert_{s}^{2}\,\textrm{e}^{r\left(1+2M\epsilon+\widetilde{C}(n,M,\tilde{T}_{s})\sqrt{\mu}\,\epsilon^{{-}s}\right)\tilde{T}_{s}}\nonumber\\ &=\Vert\rho_{0}\Vert_{s}^{2}\,\textrm{e}^{r(1+2M\epsilon)\tilde{T}_{s}}\ \forall\epsilon>0 \end{align}

Therefore, taking the limit as $\epsilon$ tends to zero, follows the final estimate

(5.25)\begin{equation} \Vert\rho(t)\Vert_{s}^{2}\leq\Vert\rho_{0}\Vert_{s}^{2}\,\textrm{e}^{\tilde{T}_{s}},\quad \forall t\in\lbrack 0,\tilde{T}_{s}], \end{equation}

and the proof is complete.