Comoving mesh method for certain classes of moving boundary problems

Abstract

A Lagrangian-type numerical scheme called the “comoving mesh method” or CMM is developed for numerically solving certain classes of moving boundary problems which include, for example, the classical Hele-Shaw flow problem and the well-known mean curvature flow problem. This finite element scheme exploits the idea that the normal velocity field of the moving boundary can be extended smoothly throughout the entire domain of definition of the problem using, for instance, the Laplace operator. Then, the boundary as well as the finite element mesh of the domain are easily updated at every time step by moving the nodal points along this velocity field. The feasibility of the method, highlighting its practicality, is illustrated through various numerical experiments. Furthermore, in order to examine the accuracy of the proposed scheme, the experimental order of convergences between the numerical and exact solutions for these examples are also calculated.

A Proof of Lemma 1

Proof

Let us now prove Lemma 1. Consider system (5.5) whose solution is given by ${\mathbf {{v}}}^i:={\mathbf {{v}}}({\mathbf {{g}}}^i)$. Also, consider the Dirichlet-to-Neumann map $\Lambda : H^{1/2}(\varGamma ;{\mathbb {R}}^d) \rightarrow H^{-1/2}(\varGamma ;{\mathbb {R}}^d)$. Then, for ${\mathbf {{g}}}^1, {\mathbf {{g}}}^2 \in H^{1/2}(\varGamma ;{\mathbb {R}}^d)$, the binary operation $({\mathbf {{g}}}^1, {\mathbf {{g}}}^2)_{\Lambda } := (\Lambda {\mathbf {{g}}}^1, {\mathbf {{g}}}^2)_{L^2(\varGamma ;{\mathbb {R}}^d)}$ is an inner product on $H^{1/2}(\varGamma ;{\mathbb {R}}^d)$. Indeed, we have the following arguments

• (i) since, for any ${\mathbf {{g}}}^3 \in H^{1/2}(\varGamma ;{\mathbb {R}}^d)$ and $c \in {\mathbb {R}}$, we have $(\Lambda (c {\mathbf {{g}}}^1 + {\mathbf {{g}}}^2), {\mathbf {{g}}}^3)_{L^2(\varGamma ;{\mathbb {R}}^d)} = \int _{\varGamma } (c \nabla {\mathbf {{v}}}^1 + \nabla {\mathbf {{v}}}^2)\cdot \nu \, {\mathbf {{v}}}^3\, \mathrm{d}s = c (\Lambda {\mathbf {{g}}}^1, {\mathbf {{g}}}^3)_{L^2(\varGamma ;{\mathbb {R}}^d)} + (\Lambda {\mathbf {{g}}}^2, {\mathbf {{g}}}^3)_{L^2(\varGamma ;{\mathbb {R}}^d)}$, then $(\, \cdot \,, \, \cdot \,)_{\Lambda }$ is linear with respect to its first argument;

• (ii) the binary operation $(\, \cdot \,, \, \cdot \,)_{\Lambda }$ is positive definite because, for any ${\mathbf {{g}}}^2 \in H^{1/2}(\varGamma ;{\mathbb {R}}^d)$, we have $(\Lambda {\mathbf {{g}}}, {\mathbf {{g}}})_{L^2(\varGamma ;{\mathbb {R}}^d)} = \int _{\varGamma } (\nabla {\mathbf {{v}}} \cdot \nu ) {\mathbf {{v}}}\ \mathrm{d}s = \int _{{\overline{\Omega }}{\setminus } B} |\nabla {\mathbf {{v}}}|^2 \ \mathrm{d}x \geqslant 0;$

• (iii) also, it is point-separating, that is $(\Lambda {\mathbf {{g}}}, {\mathbf {{g}}})_{L^2(\varGamma ;{\mathbb {R}}^d)} = 0$ if and only if ${\mathbf {{g}}} \equiv {\mathbf {{0}}}$; and,

• (iv) lastly, the operation is symmetric because $(\Lambda {\mathbf {{g}}}^1, {\mathbf {{g}}}^2)_{L^2(\varGamma ;{\mathbb {R}}^d)} = \int _{\varGamma } (\nabla {\mathbf {{v}}}^1 \cdot \nu ) {\mathbf {{v}}}^2\ \mathrm{d}s = \int _{{\overline{\Omega }}{\setminus } B} \nabla {\mathbf {{v}}}^1 : \nabla {\mathbf {{v}}}^2 \ \mathrm{d}x = \int _{\varGamma } {\mathbf {{v}}}^1 (\nabla {\mathbf {{v}}}^2 \cdot \nu ) \ \mathrm{d}s = ( {\mathbf {{g}}}^1, \Lambda {\mathbf {{g}}}^2)_{L^2(\varGamma ;{\mathbb {R}}^d)}$.

Additionally, for Lipschitz $\varGamma$, the inner product $(\, \cdot \,, \, \cdot \,)_{\Lambda }$ is equivalent to the natural one in $H^{1/2}(\varGamma ;{\mathbb {R}}^d)$. Here, $H^{1/2}(\varGamma ;{\mathbb {R}}^d)$ is viewed as the image of the trace operator $\gamma _{\varGamma }$ on $\varGamma$ (i.e., ${\text {Im}}(\gamma _{\varGamma }) = \gamma _{\varGamma }(H^1(\Omega ;{\mathbb {R}}^d))$). Consequently, by Riesz representation theorem, together with the embedding $H^{-1/2}(\varGamma ;{\mathbb {R}}^d) \supset L^2(\varGamma ;{\mathbb {R}}^d) \supset H^{1/2}(\varGamma ;{\mathbb {R}}^d)$, we conclude that $\Lambda \in {\text {Isom}}(H^{1/2}(\varGamma ;{\mathbb {R}}^d),H^{-1/2}(\varGamma ;{\mathbb {R}}^d))$. This proves the lemma. $\square$