Cut Singularity of Compressible Stokes Flow

Abstract

In this paper we study the cut singularity governed by a compressible Stokes system. The cut is a non-Lipshitz boundary. The divergence of the leading corner singularity vector, which has the singular exponent 1/2, has different trace values on either side of cut. In the consequence the pressure solution of the continuity equation must have a jump across the streamline emanating from the cut tip. We establish a piecewise regularity of the solution by the corner singularity and the contact singular function.

Appendix

Lemma 6.1

If $\textbf{f}\in \textbf{H}^{-1}$ and $g\in {\textrm{L}}^2$ then there are unique weak solutions $\textbf{u}\in \textbf{H}^1_0$ and $p\in {\textrm{L}}^2$ of (1.1), satisfying the inequality

$$\begin{aligned} \Vert \textbf{u}\Vert _1+\Vert p\Vert _0+\Vert p\Vert _{0,\Gamma _\textrm{out}}\le C(\Vert \textbf{f}\Vert _{-1}+\Vert g\Vert _0), \end{aligned}$$

(6.1)

where C is a generic constant depending on $\Omega$.

Proof

The proof easily follows by a weak formulation on the pair space $\textbf{H}^1_0\times {\textrm{L}}^2$. Letting $(,\,)$ denoting the ${\textrm{L}}^2$ inner product we consider the bilinear forms

$$\begin{aligned} \begin{aligned} a(\textbf{u},\textbf{v})&=\mu (\nabla \textbf{u},\nabla \textbf{v})+\nu (\text{ div }\,\textbf{u},\text{ div }\,\textbf{v}),\\ b(p,\textbf{v})&=-(p,\text{ div }\,\textbf{v}),\\ c(p,\eta )&=(\textbf{U}\cdot \nabla p,\eta ). \end{aligned} \end{aligned}$$

(6.2)

The weak form of problem (1.1) is to find the solutions $\textbf{u}\in \textbf{H}^1_0$ and $p\in {\textrm{L}}^2$ satisfying

$$\begin{aligned} \begin{aligned} a(\textbf{u},\textbf{v})+b(p,\textbf{v})&=(\textbf{f},\textbf{v}),\quad \forall \textbf{v}\in \textbf{H}^1_0,\\ c(p,\eta )-b(\eta ,\textbf{u})&=(g,\eta ),\quad \forall \eta \in {\textrm{L}}^2. \end{aligned} \end{aligned}$$

(6.3)

More detailed proof can be found in [9, Lemma 2.8]. $\square$

We next show Theorem 4.1 by constructing the dual functions used in the stress intensity coefficients.

Lemma 6.2

Let $q'=q/(q-1)$ for $q\ge 2$. For $s>s_1$ there are nontrivial vector functions $\textbf{v}_j\in \textbf{H}^{2-s,q'}$, $j=1,2$, such that $\textbf{v}_j$ satisfies the boundary value problem

$$\begin{aligned} \begin{aligned} {\mathbb {L}}\textbf{v}_j&=0{} & {} \text{ in } \Omega ,\\ \textbf{v}_j&=0{} & {} \text{ on } \Gamma , \end{aligned} \end{aligned}$$

and is orthogonal to the image of $\textbf{H}^{s,q}\cap \textbf{H}^{1,q}_0$ by the Lamé operator ${\mathbb {L}}$ in the ${\textrm{L}}^2$ inner product.

Proof

We define the function $\textbf{v}_j$, $j=1,2$, by $\textbf{v}_j={\mathbf \Phi }_j^*+\textbf{z}_j$, where ${\mathbf \Phi }_j^*=\chi r^{-1/2}\varvec{\Theta }_j^*(\theta )$ with

$$\begin{aligned} \begin{aligned} \varvec{\Theta }_1^*(\theta )&=\begin{pmatrix} \nu _1\sin {\theta /2}-\nu _1\sin {5\theta /2}\\ -(8+5\nu _1)\cos {\theta /2}+\nu _1\cos {5\theta /2} \end{pmatrix},\\ \varvec{\Theta }_2^*(\theta )&=\begin{pmatrix} -(8+3\nu _1)\cos {\theta /2}-\nu _1\cos {5\theta /2}\\ \nu _1\sin {\theta /2}-\nu _1\sin {5\theta /2} \end{pmatrix}, \end{aligned} \end{aligned}$$

and $\textbf{z}_j$ is the solution of the problem

$$\begin{aligned} \begin{aligned} -{\mathbb {L}}\textbf{z}_j&={\mathbb {L}}{\mathbf \Phi }_j^*{} & {} \text{ in } \Omega ,\\ \textbf{z}_j&=0{} & {} \text{ on } \Gamma . \end{aligned} \end{aligned}$$

(6.4)

Since ${\mathbb {L}}(r^{-1/2}\varvec{\Theta }_j^*(\theta ))=0$, ${\mathbb {L}}{\mathbf \Phi }_j^*\in \textbf{L}^q$ and the solution $\textbf{z}_j$ of (6.4) becomes $\textbf{z}_j=\textbf{C}{\mathbf \Phi }+\textbf{z}_{j,R}$ where $\textbf{C}$ is a constant vector and $\textbf{z}_{j,R}\in \textbf{H}^{2,q}$. Since ${\mathbf \Phi }_j\in \textbf{H}^{t,q}$ for $t<s_1$, $\textbf{z}_j\in \textbf{H}^{t,q}$. Also, since ${\mathbf \Phi }_j^*\in \textbf{H}^{2-s,q'}$ we have $\textbf{v}_j\in \textbf{H}^{2-s,q'}$. The vector function $\textbf{v}_j$ satisfies ${\mathbb {L}}\textbf{v}_j=0$ in $\Omega$ and $\textbf{v}_j|_\Gamma =0$. Therefore, for any $\textbf{w}\in \textbf{H}^{s,q}\cap \textbf{H}^{1,q}_0$ for $s>s_1$,

$$\begin{aligned} \begin{aligned} \int \limits _\Omega {\mathbb {L}}\textbf{w}\cdot \textbf{v}_j{\textrm{d}}\textbf{x}&=\int \limits _\Omega \textbf{w}\cdot {\mathbb {L}}\textbf{v}_j{\textrm{d}}\textbf{x}+\int \limits _\Gamma \frac{\partial \textbf{w}}{\partial \textbf{n}}\cdot \textbf{v}_j-\textbf{w}\cdot \frac{\partial \textbf{v}_j}{\partial \textbf{n}} ds\\&\hspace{0.5cm}+\nu _1\int \limits _\Gamma (\text{ div }\,\textbf{w})\textbf{n}\cdot \textbf{v}_j-(\text{ div }\,\textbf{v}_j)\textbf{n}\cdot \textbf{w}ds\\&=0. \end{aligned} \end{aligned}$$

Hence the required result follows. $\square$

To show (4.5), if we write the solution $\textbf{u}$ of (4.2) by $\textbf{u}=\textbf{C}{\mathbf \Phi }+\textbf{u}_R$ where $\textbf{C}=({{\mathcal {C}}}_1,{{\mathcal {C}}}_2)\in \mathbb {R}^2$ and $\textbf{u}_R\in \textbf{H}^{s,q}$, then $\textbf{u}_R=\textbf{g}$ on $\Gamma$ and

$$\begin{aligned} \begin{aligned} \int \limits _\Omega {\mathbb {L}}\textbf{u}_R\cdot \textbf{v}_j{\textrm{d}}\textbf{x}&=-\int \limits _\Gamma \textbf{g}\cdot \Big (\frac{\partial \textbf{v}_j}{\partial \textbf{n}}+\nu _1(\text{ div }\,\textbf{v}_j)\textbf{n}\Big )ds, \end{aligned} \end{aligned}$$

so

$$\begin{aligned} \begin{aligned} \int \limits _\Omega {\mathbb {L}}(\textbf{C}{\mathbf \Phi })\cdot \textbf{v}_j{\textrm{d}}\textbf{x}&=-\int \limits _\Omega ({\textbf{h}}+{\mathbb {L}}\textbf{u}_R)\cdot \textbf{v}_j{\textrm{d}}\textbf{x}\\&=-\int \limits _\Omega {\textbf{h}}\cdot \textbf{v}_j{\textrm{d}}\textbf{x}+\int \limits _\Gamma \textbf{g}\cdot \Big (\frac{\partial \textbf{v}_j}{\partial \textbf{n}}+\nu _1(\text{ div }\,\textbf{v}_j)\textbf{n}\Big )ds. \end{aligned} \end{aligned}$$

Then we have a linear system for ${{\mathcal {C}}}_1$ and ${{\mathcal {C}}}_2$:

$$\begin{aligned} \begin{aligned} a_{11}{{\mathcal {C}}}_1+a_{12}{{\mathcal {C}}}_2&=-\int \limits _\Omega {\textbf{h}}\cdot \textbf{v}_1{\textrm{d}}\textbf{x}+\int \limits _\Gamma \textbf{g}\cdot \Big (\frac{\partial \textbf{v}_1}{\partial \textbf{n}}+\nu _1(\text{ div }\,\textbf{v}_1)\textbf{n}\Big )ds,\\ a_{21}{{\mathcal {C}}}_1+a_{22}{{\mathcal {C}}}_2&=-\int \limits _\Omega {\textbf{h}}\cdot \textbf{v}_2{\textrm{d}}\textbf{x}+\int \limits _\Gamma \textbf{g}\cdot \Big (\frac{\partial \textbf{v}_2}{\partial \textbf{n}}+\nu _1(\text{ div }\,\textbf{v}_2)\textbf{n}\Big )ds, \end{aligned} \end{aligned}$$

(6.5)

where $a_{ij}=\int \limits _\Omega {\mathbb {L}}{\mathbf \Phi }_j\cdot \textbf{v}_i{\textrm{d}}\textbf{x}$. On the other hand, since $\textbf{v}_i={\mathbf \Phi }_i^*+\textbf{z}_i$,

$$\begin{aligned} \begin{aligned} a_{ij}&=\int \limits _\Omega {\mathbb {L}}{\mathbf \Phi }_j\cdot {\mathbf \Phi }_i^*+{\mathbf \Phi }_j\cdot {\mathbb {L}}\textbf{z}_i{\textrm{d}}\textbf{x}\\&=\int \limits _\Omega {\mathbb {L}}{\mathbf \Phi }_j\cdot {\mathbf \Phi }_i^*-{\mathbf \Phi }_j\cdot {\mathbb {L}}{\mathbf \Phi }_i^*{\textrm{d}}\textbf{x}. \end{aligned} \end{aligned}$$

Set $\Omega _\delta =\Omega \cap \{r>\delta \}$ for $\delta <r_0\ll 1$. By integration by parts,

$$\begin{aligned} \int \limits _{\Omega _\delta }{\mathbb {L}}{\mathbf \Phi }_j\cdot {\mathbf \Phi }_i^*-{\mathbf \Phi }_j\cdot {\mathbb {L}}{\mathbf \Phi }_i^*{\textrm{d}}\textbf{x}=\int \limits _{\partial \Omega _\delta }E_1(r,\theta )\text{ d }s+\nu _1\int \limits _{\partial \Omega _\delta }E_2(r,\theta )\text{ d }s, \end{aligned}$$

where

$$\begin{aligned} \begin{aligned} E_1(r,\theta )&=\frac{\partial {\mathbf \Phi }_j}{\partial \textbf{n}}\cdot {\mathbf \Phi }_i^*-{\mathbf \Phi }_j\cdot \frac{\partial {\mathbf \Phi }_i^*}{\partial \textbf{n}},\\ E_2(r,\theta )&=(\text{ div }\,{\mathbf \Phi }_j)\textbf{n}\cdot {\mathbf \Phi }_i^*-(\text{ div }\,{\mathbf \Phi }_i^*)\textbf{n}\cdot {\mathbf \Phi }_j, \end{aligned} \end{aligned}$$

and $\textbf{n}$ is the outward normal unit vector to the boundary $\partial \Omega _\delta$. Since $\chi =0$ for $r>2r_0$, we have

$$\begin{aligned} \begin{aligned} \int \limits _{\partial \Omega _\delta }E_1(r,\theta )\text{ d }s&=\int \limits _{2r_0}^\delta E_1(r,-\pi )dr+\int \limits _{-\pi }^\pi E_1(\delta ,\theta )\delta d\theta +\int \limits _\delta ^{2r_0} E_1(r,\pi )dr\\&=-\int \limits _{-\pi }^\pi \varvec{\Theta }_j\cdot \varvec{\Theta }_i^*d\theta . \end{aligned} \end{aligned}$$

Likewise,

$$\begin{aligned} \int \limits _{\partial \Omega _\delta }E_2(r,\theta )\text{ d }s=-\int \limits _{-\pi }^\pi E_{21}(\theta )+E_{22}(\theta )-E_{23}(\theta )d\theta , \end{aligned}$$

where for $\textbf{e}_1=(\cos \theta ,\sin \theta )^t$ and $\textbf{e}_2=(-\sin \theta ,\cos \theta )^t$,

$$\begin{aligned} \begin{aligned} E_{21}(\theta )&=(\textbf{e}_1\cdot \varvec{\Theta }_j)(\textbf{e}_1\cdot \varvec{\Theta }_i^*),\\ E_{22}(\theta )&=(\textbf{e}_2\cdot \varvec{\Theta }_j')(\textbf{e}_1\cdot \varvec{\Theta }_i^*),\\ E_{23}(\theta )&=(\textbf{e}_1\cdot \varvec{\Theta }_j)(\textbf{e}_2\cdot (\varvec{\Theta }_i^*)'). \end{aligned} \end{aligned}$$

Since $\Omega =\lim _{\delta \rightarrow 0}\Omega _\delta$,

$$\begin{aligned} \begin{aligned} a_{ij}&=\lim _{\delta \rightarrow 0}\int \limits _{\Omega _\delta }{\mathbb {L}}{\mathbf \Phi }_j\cdot {\mathbf \Phi }_i^*-{\mathbf \Phi }_j\cdot {\mathbb {L}}{\mathbf \Phi }_i^*{\textrm{d}}\textbf{x}\\&=-\int \limits _{-\pi }^\pi \varvec{\Theta }_j\cdot \varvec{\Theta }_i^*+\nu _1(E_{21}+E_{22}-E_{23})d\theta . \end{aligned} \end{aligned}$$

Therefore, $a_{11}=a_{22}=-32(\nu _1+2)(\nu _1+1)\pi$ and $a_{12}=a_{21}=0$. Hence, by (6.5), (4.5) follows. $\square$