Unfitted Finite Element Methods for Axisymmetric Two-Phase Flow

Abstract

We propose and analyze unfitted finite element approximations for the two-phase incompressible Navier–Stokes flow in an axisymmetric setting. The discretized schemes are based on an Eulerian weak formulation for the Navier–Stokes equation in the 2d-meridian halfplane, together with a parametric formulation for the generating curve of the evolving interface. We use lowest order Taylor–Hood and piecewise linear elements for discretizing the Navier–Stokes formulation in the bulk and the moving interface, respectively. We discuss a variety of schemes, amongst which is a linear scheme that enjoys an equidistribution property on the discrete interface and good volume conservation. An alternative scheme can be shown to be unconditionally stable and to conserve the volume of the two phases exactly. Numerical results are presented to show the robustness and accuracy of the introduced methods for simulating both rising bubble and oscillating droplet experiments.

Appendix: Derivation of (3.11)

Let $\vec x= (x,y,z)$ be the Cartesian coordinates, and $(r,z,\varphi )$ be the cylindrical coordinates such that

$$\begin{aligned} x = r\cos \varphi ,\qquad y = r\sin \varphi , \end{aligned}$$

where $r=\sqrt{x^2+y^2}$ is the radial distance and $\varphi$ is the azimuthal angle. This gives rise to the basis vectors $\big \{\vec e_r,\vec e_\varphi ,\vec e_z\big \}$

$$\begin{aligned} \vec e_r = (\cos \varphi ,\sin \varphi , 0)^T,\qquad \vec e_\varphi = (-\sin \varphi ,\cos \varphi , 0)^T,\qquad \vec e_z=(0,0,1)^T, \end{aligned}$$

(A.1)

which satisfy

$$\begin{aligned} \frac{\partial \vec e_r}{\partial \varphi } = \vec e_\varphi ,\qquad \frac{\partial \vec e_\varphi }{\partial \varphi } = -\vec e_r,\qquad \frac{\partial \vec e_z}{\partial \varphi } = \vec 0,\qquad \frac{\partial \vec e_r}{\partial r} = \frac{\partial \vec e_\varphi }{\partial r}=\frac{\partial \vec e_z}{\partial r}=\vec {0}. \end{aligned}$$

(A.2)

For any scalar function $\breve{b}(\vec x)$, we write in cylindrical coordinates $b(r,\varphi ,z) = \breve{b}(\vec x)$. Then we have

$$\begin{aligned} \nabla \breve{b}(\vec x) = \frac{\partial b}{\partial r}\,\vec e_r + \frac{1}{r}\frac{\partial b}{\partial \varphi }\,\vec e_\varphi + \frac{\partial b}{\partial z}\,\vec e_z. \end{aligned}$$

(A.3)

For any vector-valued function $\breve{\vec b}(\vec x, t)$, similarly we write $\breve{\vec b}(\vec x)=\vec b(r,\varphi , z) = b^r\,\vec e_r + b^\varphi \,\vec e_\varphi + b^z\,\vec e_z$. Then using the relations in (A.2) gives

$$\begin{aligned} \nabla \breve{\vec b} =[\vec e_r\quad \vec e_\varphi \quad \vec e_z]\left[ \begin{array}{lll} \frac{\partial b^r}{\partial r} &{}\frac{1}{r}\frac{\partial b^r}{\partial \varphi }-\frac{b^\varphi }{r} &{}\frac{\partial b^r}{\partial z}\\ \frac{\partial b^\varphi }{\partial r}&{}\frac{b^r}{r}+ \frac{1}{r}\frac{\partial b^\varphi }{\partial \varphi } &{}\frac{\partial b^\varphi }{\partial z}\\ \frac{\partial b^z}{\partial r}&{}\frac{1}{r}\frac{\partial b^z}{\partial \varphi } &{}\frac{\partial b^z}{\partial z} \end{array}\right] \left[ \begin{array}{l} (\vec e_r)^T\\ (\vec e_\varphi )^T\\ (\vec e_z)^T \end{array} \right] . \end{aligned}$$

(A.4)

Moreover, for the divergence of $\breve{\vec b}$ it holds that

$$\begin{aligned} \nabla \cdot \breve{\vec b} = \frac{\partial \vec b}{\partial r}\cdot \vec e_r + \frac{1}{r}\frac{\partial \vec b}{\partial \varphi }\cdot \vec e_\varphi + \frac{\partial \vec b}{\partial z}\cdot \vec e_z = \frac{\partial b^r}{\partial r} + \frac{b^r}{r}+\frac{1}{r}\frac{\partial b^\varphi }{\partial \varphi } +\frac{\partial b^z}{\partial z}. \end{aligned}$$

(A.5)

We now follow the axisymmetric setting in Sect. 3.1 and denote

$$\begin{aligned} \breve{\vec u}(\vec x, t)=u^r(r,z,t)\,\vec e_r + u^z(r,z,t)\,\vec e_z,\qquad \breve{{\vec {\chi }}}(\vec x, t) =\chi ^r(r,z,t)\,\vec e_r + \chi ^z(r,z,t)\,\vec e_z. \end{aligned}$$

Then it follows from (3.12a) that

$$\begin{aligned} \Bigl (\rho \breve{\vec u},~\breve{{\vec {\chi }}}\Bigr )_\varOmega&= 2\pi \int _{\mathcal {R}_-(t)}\rho _-(u^r\,\chi ^r + u^z\,\chi ^z)r\,\textrm{d}r\textrm{d}z+ 2\pi \int _{\mathcal {R}_+(t)}\rho _+(u^r\,\chi ^r + u^z\,\chi ^z)r\,\textrm{d}r\textrm{d}z\nonumber \\&=2\pi \int _{\mathcal {R}}\rho _{_c}(u^r\,\chi ^r + u^z\,\chi ^z)r\,\textrm{d}r\textrm{d}z=2\pi \,\Bigl (\rho _{_c}\vec u,~{\vec {\chi }}\,r\Bigr ), \end{aligned}$$

(A.6)

where $\vec u = (u^r, u^z)^T$ and ${\vec {\chi }} = (\chi ^r,\chi ^z)^T$ are vectors in the 2d-meridian halfplane $\mathcal {R}$. Similarly we have

$$\begin{aligned} \Bigl (\rho \partial _t\breve{\vec u},~\breve{{\vec {\chi }}}\Bigr )_\varOmega = 2\pi \,\Bigl (\rho _{_c}\partial _t\vec u,~{\vec {\chi }}\,r\bigr ),\qquad \Bigl (\rho \breve{\vec u},~\partial _t\breve{{\vec {\chi }}}\Bigr )_\varOmega = 2\pi \,\Bigl (\rho _{_c}\vec u,~\partial _t{\vec {\chi }}\,r\Bigr ). \end{aligned}$$

(A.7)

On recalling (A.5) and (3.8), it holds that

$$\begin{aligned} \Bigl (\breve{q},\,\nabla \cdot \breve{\vec u}\Bigr ) = 2\pi \int _\mathcal {R}\left( \frac{\partial u^r}{\partial r} + \frac{u^r}{r} + \frac{\partial u^z}{\partial z}\right) q\,r\textrm{d}r\textrm{d}z= 2\pi \,\Bigl (\nabla _c\cdot ([r\vec u],~q\Bigr ). \end{aligned}$$

(A.8)

Using (A.4) for $\breve{\vec u}$ and $\breve{{\vec {\chi }}}$, it is not difficult to show that

$$\begin{aligned} \int _{\varOmega _\pm (t)}\left( [\breve{\vec u}\cdot \nabla ]\breve{\vec u}\cdot \breve{{\vec {\chi }}} -[\breve{\vec u}\cdot \nabla ]\breve{{\vec {\chi }}}\cdot \breve{\vec u}\right) \,\textrm{d}V= 2\pi \int _{\mathcal {R}_\pm (t)}\,\bigl ([\vec u\cdot \nabla _c]\vec u\cdot {\vec {\chi }} - [\vec u\cdot \nabla _c]{\vec {\chi }}\cdot \vec u\bigr )\,r\,\textrm{d}r\textrm{d}z. \end{aligned}$$

(A.9)

Multiplying (A.9) with $\rho _\pm$ and combining these two equations gives rise to

$$\begin{aligned} \Bigl (\rho ,~[\breve{\vec u}\cdot \nabla ]\breve{\vec u}\cdot \breve{{\vec {\chi }}} -[\breve{\vec u}\cdot \nabla ]\breve{{\vec {\chi }}}\cdot \breve{\vec u}\Bigr )_\varOmega =2\pi \, \Bigl (\rho _{_c}r,~[\vec u\cdot \nabla _c]\vec u\cdot {\vec {\chi }} - [\vec u\cdot \nabla _c]{\vec {\chi }}\cdot \vec u\Bigr ). \end{aligned}$$

(A.10)

Similarly we can compute

$$\begin{aligned} \underline{\underline{\mathbb {D}}}(\breve{\vec u}) = \frac{1}{2}\left( \nabla \breve{\vec u}+ (\nabla \breve{\vec u})^T\right) =[\vec e_r\quad \vec e_\varphi \quad \vec e_z]\left[ \begin{array}{lll} \frac{\partial u^r}{\partial r} &{}0 &{}\frac{1}{2}\left( \frac{\partial u^r}{\partial z}+\frac{\partial u^z}{\partial r}\right) \\ 0&{}\frac{u^r}{r} &{}0\\ \frac{1}{2}\left( \frac{\partial u^r}{\partial z}+\frac{\partial u^z}{\partial r}\right) &{}0 &{}\frac{\partial u^z}{\partial z} \end{array}\right] \left[ \begin{array}{l} (\vec e_r)^T\\ (\vec e_\varphi )^T\\ (\vec e_z)^T \end{array} \right] . \end{aligned}$$

This implies that

$$\begin{aligned}&\int _{\varOmega _\pm (t)}\underline{\underline{\mathbb {D}}}(\breve{\vec u}):\underline{\underline{\mathbb {D}}}(\breve{{\vec {\chi }}})\,\textrm{d}V\nonumber \\&\quad =2\pi \int _{\mathcal {R}_\pm (t)}\left\{ \frac{\partial u^r}{\partial r}\frac{\partial \chi ^r}{\partial r} + \frac{1}{2}\left[ \frac{\partial u^r}{\partial z} + \frac{\partial u^z}{\partial r}\right] \left[ \frac{\partial \chi ^r}{\partial z} + \frac{\partial \chi ^z}{\partial r}\right] + \,\frac{\partial u^z}{\partial z}\frac{\partial \chi ^z}{\partial z} + \frac{u^r\,\chi ^r}{r^2}\right\} r\,\textrm{d}r\textrm{d}z\nonumber \\&\quad =2\pi \int _{\mathcal {R}_\pm (t)}\underline{\underline{\mathbb {D}_c}}(\vec u):\underline{\underline{\mathbb {D}_c}}({\vec {\chi }})\,r\textrm{d}r\textrm{d}z+2\pi \int _{\mathcal {R}_\pm (t)}(\vec u\cdot \vec e_1)({\vec {\chi }}\cdot \vec e_1)\,r^{-1}\,\textrm{d}r\textrm{d}z. \end{aligned}$$

(A.11)

Multiplying (A.11) with $\mu _\pm$ and combining these two equations yields that

$$\begin{aligned} \Bigl (\mu \,\underline{\underline{\mathbb {D}}}(\breve{\vec u}),~\underline{\underline{\mathbb {D}}}(\breve{{\vec {\chi }}})\Bigr )_\varOmega = 2\pi \,\Bigl (\mu _{_c}r\,\underline{\underline{\mathbb {D}_c}}(\vec u),~\underline{\underline{\mathbb {D}_c}}({\vec {\chi }})\Bigr ) +2\pi \, \Bigl (\mu _{_c}r^{-1}(\vec u\cdot \vec e_1),~({\vec {\chi }}\cdot \vec e_1)\Bigr ).\nonumber \\ \end{aligned}$$

(A.12)

On the axisymmetric surface $S(t)$, by (3.1) we can set

$$\begin{aligned} \mathcal {\vec {\hspace{0.0pt}V}}(\vec x, t) = {\mathfrak {X}}^r_t\,\vec e_r + {\mathfrak {X}}^z_t\,\vec e_z,\qquad \vec n_{_S}(\vec x, t) = -{\mathfrak {X}}^z_s\,\vec e_r + {\mathfrak {X}}^r_s\,\vec e_z,\nonumber \\ \breve{{\vec {\eta }}}(\vec x) = \eta ^r\vec e_r + \eta ^z\vec e_z\quad \text{ for }\quad \vec x\in S(t). \end{aligned}$$

(A.13)

Denote $\zeta (\alpha , t) = \breve{\zeta }(\vec x, t)$ for $\vec x\in S(t)$, $\alpha \in \mathbb {I}$. Then it is easy to get

$$\begin{aligned} \int _{S(t)}(\mathcal {\vec {\hspace{0.0pt}V}}-\breve{\vec u})\cdot \vec n_{_S}\,\breve{\zeta }\,\;\textrm{d}S= & {} 2\pi \int _{\varGamma (t)}({\vec {{\mathfrak {X}}}}_t - \vec u)\cdot {\vec {\nu }}\,\zeta \,({\vec {{\mathfrak {X}}}}\cdot \vec e_1)\;\textrm{d}s \nonumber \\= & {} 2\pi \,\Big \langle ({\vec {{\mathfrak {X}}}}\cdot \vec e_1)({\vec {{\mathfrak {X}}}}_t-\vec u)\cdot {\vec {\nu }},\,\zeta \,|{\vec {{\mathfrak {X}}}}_\alpha |\Big \rangle . \end{aligned}$$

(A.14)

Similarly, it is straightforward to obtain

$$\begin{aligned} \int _{S(t)}\mathcal {H}\,\vec n_{_S}\cdot \breve{{\vec {\eta }}}\,\;\textrm{d}S =2\pi \int _{\varGamma (t)}\varkappa \,{\vec {\nu }}\cdot {\vec {\eta }}\;({\vec {{\mathfrak {X}}}}\cdot \vec e_1)\;\textrm{d}s = 2\pi \,\Big \langle ({\vec {{\mathfrak {X}}}}\cdot \vec e_1)\,\varkappa \,{\vec {\nu }},~{\vec {\eta }} \,|{\vec {{\mathfrak {X}}}}_\alpha |\Big \rangle ,\nonumber \\ \end{aligned}$$

(A.15)

where ${\vec {\eta }}=(\eta ^r,~\eta ^z)^T$ is the vector in the 2d-meridian plane. Finally, we have

$$\begin{aligned} \int _{S(t)}\nabla _s{\vec {\textrm{id}}}:\nabla _s\breve{{\vec {\eta }}}\,\;\textrm{d}S&= \int _{S(t)}\nabla _s\cdot \breve{{\vec {\eta }}}\,\;\textrm{d}S = 2\pi \int _{\varGamma (t)}({\vec {{\mathfrak {X}}}}\cdot \vec e_1)\left( {\vec {{\mathfrak {X}}}}_s\cdot {\vec {\eta }}_s + \frac{{\vec {\eta }}\cdot \vec e_1}{{\vec {{\mathfrak {X}}}}\cdot \vec e_1}\right) \;\textrm{d}s\nonumber \\&= 2\pi \int _{\mathbb {I}}({\vec {{\mathfrak {X}}}}\cdot \vec e_1){\vec {{\mathfrak {X}}}}_s\cdot {\vec {\eta }}_s\,|{\vec {{\mathfrak {X}}}}_\alpha | + ({\vec {\eta }}\cdot \vec e_1)\,|{\vec {{\mathfrak {X}}}}_\alpha |\,\;\textrm{d}\alpha , \end{aligned}$$

(A.16)

where we used the fact

$$\begin{aligned} \nabla _s\cdot \breve{{\vec {\eta }}} = {\mathfrak {X}}^r_s\eta ^r_s + {\mathfrak {X}}^z_s\eta ^z_s + \frac{\eta ^r}{{\vec {{\mathfrak {X}}}}\cdot \vec e_1}. \end{aligned}$$

(A.17)

Collecting the above results, we obtain (3.11) from (2.5).