Method of Moving Frames to Solve (An)isotropic Diffusion Equations on Curved Surfaces

Abstract

As a continuing effort to develop the method of moving frames (MMF) ensuing (Chun in J Sci Comput 53(2):268–294, 2012), a novel MMF scheme is proposed to solve (an)isotropic diffusion equations on arbitrary curved surfaces. First we show that if the divergence of a vector is computed exactly on the surface, the mixed formulations expanded in the moving frames are equivalent to the Laplace–Beltrami operator. Otherwise, the divergence error dominates, but it can be made negligible by either way; the use of a higher order differentiation scheme more than the first order or the alignment of the moving frames. Moreover, the propagational property of the media along a specific direction, known as anisotropy, is represented by the rescaling of the moving frames, not by repetitive multiplications of the diffusivity tensor, without adding any schematic complexity nor deterioration of the accuracy and stability to the isotropic diffusion scheme. Convergence results for a spherical shell, an irregular surface, and a non-convex surface are displayed with several examples of modeling anisotropy on various curved surfaces. A computational simulation of atrial reentry is illustrated as an exemplary use of the MMF scheme for practical applications.

Appendices

Appendix 1: Proof of Proposition 1

With the same $y_m$ and $\xi _m$ as defined previously, consider the following equalities:

$$\begin{aligned} \nabla u = \sum _{m=1}^2 \frac{1}{\Vert \varvec{\nu }^m \Vert ^2} \left( \frac{\partial u}{ \partial x^m} \right) \mathbf{e}^m = \sum _{m=1}^2 \frac{1}{\Vert \varvec{\nu }^m \Vert ^2} \frac{\partial u}{\partial y_m} \left( \frac{\partial y_m}{\partial x_m} \right) \varvec{\nu }^m \left( \frac{\partial x_m}{\partial \xi _m} \right) . \end{aligned}$$

(46)

By using the proposition (3) by the exponential extension, the above equation becomes

$$\begin{aligned} \nabla u = \sum _{m=1}^2 \frac{1}{\Vert \varvec{\nu }^m \Vert ^2} \left( \frac{\partial u}{ \partial \xi _m} \right) \varvec{\nu }^m. \end{aligned}$$

(47)

By using the above equality and the well-known equality $\Vert \partial / \partial \xi _m \Vert ^{-2} = g^{mm}$ in the orthogonal curved axis [29], the Eq. (7) becomes

$$\begin{aligned} \mathcal{L }_c (u) = \nabla \cdot \left( \sum _{m=1}^2 \frac{1}{ (\varvec{\nu }^m)^2 } \frac{\partial u}{\partial x^m} \mathbf{e}^m \right) = \nabla \cdot \left( \sum _{m=1}^2 g^{mm} \frac{\partial u}{\partial \xi _m} \varvec{\nu }^m \right) . \end{aligned}$$

Applying the divergence operator to each term inside the bracket and using Eq. (3) again leads to

$$\begin{aligned} \mathcal{L }_c (u) = \sum _{m=1}^2 \left[ \frac{\partial u}{\partial \xi _m} \left( \nabla g^{mm} \cdot \varvec{\nu }^m \right) + g^{mm} \left( \nabla \frac{\partial u}{\partial \xi _m} \cdot \varvec{\nu }^m \right) + g^{mm} \frac{\partial u}{\partial \xi _m} (\nabla \cdot \varvec{\nu }^m ) \right] . \end{aligned}$$

Using Eq. (4) for each gradient in the first and the second terms, we obtain

$$\begin{aligned} \mathcal{L }_c (u) = \sum _{m=1}^2 \left[ \left( \frac{\partial u}{\partial \xi _m} \frac{\partial g^{mm}}{\partial \xi _m} + g^{mm} \frac{\partial ^2 u}{\partial \xi _m^2} \right) + g^{mm} \frac{\partial u}{\partial \xi _m} (\nabla \cdot \varvec{\nu }^m ) \right] . \end{aligned}$$

(48)

The first term can be expressed with the second-order Christoffel symbol $\Gamma ^i_{jk}$, but it is zero as

$$\begin{aligned} \frac{\partial g^{mm} }{\partial \xi _m} = - 2 g^{mk} \Gamma ^k_{km} = 0, \quad \text {for}\, m = 1,~2 ~~\text {and}\, ~k \ne m , \end{aligned}$$

(49)

because $g^{mk}=0$ when $m \ne k$ and $\Gamma ^m_{mm}=0$ due to the orthogonality of the curved axes. For the third term in Eq. (48), we use the fact that the divergence of the tangent vector can be expressed as

$$\begin{aligned} \nabla \cdot \varvec{\nu }^m = \frac{1}{\sqrt{g}} \frac{\partial \sqrt{g}}{\partial \xi _m}, \quad \text {for } m = 1,~2. \end{aligned}$$

(50)

Substituting Eqs. (49) and (50) into Eq. (48) yields

$$\begin{aligned} \mathcal{L }_c (u) = \sum _{m=1}^2 \left[ g^{mm} \frac{\partial ^2 u}{\partial \xi _m^2 } + \frac{ g^{mm}}{\sqrt{g}} \frac{\partial u}{\partial \xi _m} \frac{\partial \sqrt{g}}{\partial \xi _m} \right] . \end{aligned}$$

Finally, if we use Eq. (49) again for the second term, we obtain the Laplace–Beltrami operator

$$\begin{aligned} \mathcal{L }_c (u) = \sum _{m=1}^2 \left[ g^{mm} \frac{\partial ^2 u}{\partial \xi _m^2 } + \frac{1}{\sqrt{g}} \frac{\partial u}{\partial \xi _m} \frac{\partial (\sqrt{g} g^{mm} )}{\partial \xi _m} \right] = \frac{1}{\sqrt{g}} \sum _{m=1}^2 \left[ \frac{\partial }{\partial \xi _m} \left( \sqrt{g} g^{mm} \frac{\partial u}{\partial \xi _m} \right) \right] \, \square . \end{aligned}$$

Appendix 2: Proof of Proposition 2

Substituting the MMF error estimate for $\partial u / \partial y_m$ (Eq. 6) into the gradient of $u$ (Eq. 46) yields

$$\begin{aligned} \nabla u = \frac{ \varvec{\nu }^m}{\Vert \varvec{\nu }^m \Vert ^2} \left( \frac{\partial u}{ \partial \xi _m} \right) + \frac{\varvec{\nu }^m}{\Vert \varvec{\nu }^m \Vert ^2} \mathcal{O } \left( \frac{h^2_m}{R^2} \right) , \end{aligned}$$

and subsequently the operator ${\mathcal{L }}_c^h$ changes to

$$\begin{aligned} {\mathcal{L }}_c^h (u) = \frac{1}{R^2} \left[ \left( \frac{\partial ^2 u}{\partial \xi _m^2 } \right) + \mathcal{O } \left( \frac{h^2_m}{R^2} \right) + \frac{\partial u}{\partial \xi _m} ( \nabla \cdot \varvec{\nu }^m ) + ( \nabla \cdot \varvec{\nu }^m ) \mathcal O \left( \frac{h^2_m}{R^2} \right) \right] . \end{aligned}$$

(51)

Let us pay attention to $\nabla \cdot \varvec{\nu }^m$ of the third component of which order of convergence with respect to $h_m$ is only unknown. Note that the curved axes $\xi _1$ and $\xi _2$ are dependent on the direction of the moving frames $y_1$ and $y_2$ and these moving frames are aligned along the polar angle $\theta$ and the azimuthal angle $\phi$, respectively. Therefore, $\sqrt{g}$ is derived as $\sqrt{g} = R^2 \cos ( h_m / R )$ and consequently, we obtain the divergence of the tangent vector $\varvec{\nu }^m$ as

$$\begin{aligned} \nabla \cdot \varvec{\nu }^m = \frac{\partial (\log \sqrt{g} )}{\partial \xi _m} = \frac{\partial (\log ( R^2 \cos (h_m / R)) )}{\partial \xi _m} = - \tan \left( \frac{h_m}{ R} \right) . \end{aligned}$$

By the Taylor expansion of the tangent of $h_m/R$, the above equality can be identified as the first order with respect to $h_m/R$ as

$$\begin{aligned} \nabla \cdot \varvec{\nu }^m = \mathcal O \left( \frac{h_m }{R} \right) . \end{aligned}$$

(52)

Substituting Eq. (52) in Eq. (51) yields the error estimation such as

$$\begin{aligned} \left\| \mathcal{L }_c (u) - {\mathcal{L }}_c^h (u) \right\| \le \mathcal{O } \left( \frac{h^2_m}{R^4} \right) + \frac{\partial u}{\partial \xi _m} \mathcal O \left( \frac{h_m}{R^3} \right) + \mathcal{O } \left( \frac{h^3_m}{R^5} \right) . \end{aligned}$$

(53)

Substituting the condition on $\partial u / \partial \xi _m$ (Eq. 14) into Eq. (53) and by the rule of the predominance of the lowest order for a sufficiently small $h_m$, we obtain

$$\begin{aligned} \left\| \mathcal{L }_c (u) - {\mathcal{L }}_c^h (u) \right\| \le \Vert u' \Vert \mathcal{O } \left( \frac{h_m^2}{R^3} \right) \, \square . \end{aligned}$$

Appendix 3: Galerkin Scheme for Isotropic Diffusion Equations

Consider a curved element $\mathcal{M }^e$ that is a two-dimensional manifold and locally Euclidean. The weak form of the mixed formulations of Eqs. (7) and (8) can be written in the context of discontinuous Galerkin methods such as

$$\begin{aligned} \int \limits _{\mathcal{M }^e} f \varphi dx&= \int \limits _{\mathcal{M }^e} \nabla \cdot \mathbf q \varphi dx = \int \limits _{\mathcal{M }^e} \mathbf q \cdot \nabla \varphi dx + \int \limits _{\partial \mathcal{M }^e} (\tilde{\mathbf{q }} \cdot \mathbf{n}) \varphi ds , \end{aligned}$$

(54)

$$\begin{aligned} \int \limits _{\mathcal{M }^e} \mathbf{q} \cdot \varvec{\omega } dx&= \int \limits _{\mathcal{M }^e} \nabla u \cdot \varvec{\omega } dx = - \int \limits _{\mathcal{M }^e} (\nabla \cdot \varvec{\omega } ) u dx + \int \limits _{\partial \mathcal{M }^e} ( \varvec{\omega } \cdot \mathbf{n}) \tilde{u} ds , \end{aligned}$$

(55)

with the boundary conditions from Eqs. (9) and (10)

$$\begin{aligned} \int \limits _{ \partial \mathcal{M }_D } u \varphi d s&= \int \limits _{ \partial \mathcal{M }_D } g_D \varphi d s, \quad \text {in}~~ \partial \mathcal{M }_D , \end{aligned}$$

(56)

$$\begin{aligned} \int \limits _{ \partial \mathcal{M }_N } \mathbf{q} \cdot \mathbf{n} \varphi d s&= \int \limits _{ \partial \mathcal{M }_N } g_N \varphi d s,\quad \text {in}~~ \partial \mathcal{M }_N , \end{aligned}$$

(57)

where $\varphi$ and $\varvec{\omega }$ are the scalar test function and the vector test function, respectively. The second equality in Eqs. (54) and (55) is obtained by multidimensional integration by parts [20]. The vector n is the edge-normal vector and It has been shown in Ref. [14] that $\partial \xi / \partial x_m$ is bounded by, for any index $i$,

$$\begin{aligned} \left\| \frac{\partial u}{\partial \xi _m} - \frac{\partial u}{\partial y_m} \right\| \le M' \left\| \frac{\partial u}{\partial s^i_m} \right\| \mathcal{O } \left( K \ell _{max}^3 p^{-3} \right) ,\quad M' > 0 , \end{aligned}$$

where $K$ is the Gaussian curvature of the axis $\xi _m, \ell _{max}$ is the maximum length of the edges of the elements and $p$ is the polynomial order of approximation. According to the previous error estimate for the finite difference scheme, we suppose that the order of convergence of $\nabla \varphi \cdot \mathbf{e}^m$ is higher than that of the divergence term in Eqs. (38) and (39). But, by aligning the moving frames according to proposition 3, with an $m$th-order locally Euclidean element, the divergence component is bounded by

$$\begin{aligned} \int \limits _{\mathcal{M }^e} (\nabla \cdot \mathbf{e}^m ) u dx \le C \left( \frac{\Vert \mathbf{e}^m \Vert }{\Vert \varvec{\nu ^1} \Vert } \right) _{\min } \mathcal{O } \left( \ell _{max}^{m-1} {p}^{1-m} \right) \Vert u \Vert _{\infty } ,\quad C > 0 , \end{aligned}$$

(58)

The subscript $min$ indicates that the corresponding quantity is the minimum among the quantities for the index $m$ and $\Vert ~\Vert$ is $L_{\infty }$-norm. The uniform grid interval $h$ is approximated as $h = \ell _{max}/p$. Note that if the moving frames are aligned along the $4$th-order locally Euclidean element, then the above divergence term has the same order as $\partial u / \partial y_m$. But, if they are aligned along the second-order locally Euclidean element, then we also need to consider the order of the overall discretization error for Eqs. (37–39). For this purpose, we use the error estimate of the following one-dimensional Helmholtz problem [4] along the axis $\xi$:

Let $\ell$ be the size of the one-dimensional domain. Let $\xi$ be the axis of the domain and let $f$ and $u$ be a scalar variable. For the interval $0 \le h \le \ell$,

$$\begin{aligned} \mathcal{L }_H = \frac{\partial ^2 u}{\partial \xi ^2} - \lambda u + f = 0,\quad \lambda \in \mathbb{R }^+ , \end{aligned}$$

(59)

with the following boundary conditions:

$$\begin{aligned} u(0) = g_{D},\quad \frac{\partial u}{\partial \xi } (\ell ) = g_N ,\quad g_D,~g_N \in \mathbb{R } \end{aligned}$$

For the test function $v$, the weak form of Eq. (59) is obtained as follows:

$$\begin{aligned} \int \limits _0^{\ell } \left[ \frac{\partial v}{\partial \xi } \frac{\partial u }{\partial \xi } + \lambda v u \right] dx = \int \limits _0^{\ell } v f dx, \end{aligned}$$

(60)

where the second derivative of $u$ is bounded and continuous such that $|u''| \le C,~ C>0$. According to Babuška and Suri [4], the error estimate for the Helmholtz problem for a discretization on a uniform mesh of equispaced elements of size $h$ is expressed in the energy norm $\Vert \cdot \Vert _E$ such as

$$\begin{aligned} \Vert \varepsilon \Vert _E \equiv \int \limits _0^{\ell } \left[ \left( \frac{\partial \varepsilon }{\partial \xi } \right) ^2 + \lambda u^2 \right] \le C \ell _{\max }^{\min (k, P+1) - 1} P^{-k+1} \Vert u \Vert _k, \end{aligned}$$

(61)

where $\varepsilon$ is the interpolation error, $P$ is the polynomial order, and $C$ is only dependent on $k$. If we assume that the discretization error for the elliptic problem is the same as that for the Helmholtz problem in Eq. (60), then the direct comparison between Eqs. (58) and (61) yields the following proposition:

Proposition D

Suppose that the curved element $\mathcal{M }^e$ is a two-dimensional manifold and the moving frames are aligned along a $m$th-order locally Euclidean element. If the order $m$ is larger than $1 + \min (k, P+1)$, i.e.

$$\begin{aligned} m > 1 + \min ( k, P+1 ) . \end{aligned}$$

(62)

Then, by reducing $h$ for $m$, the MMF error for Eqs. (37) and (38) can be made negligible compared to the energy norm of the approximation error for a one-dimensional Helmholtz problem. Similarly, if

$$\begin{aligned} m>k, \end{aligned}$$

(63)

then the same conclusion can be drawn by increasing the polynomial order $p$.

Proposition D implies that if the element is sufficiently of high-order locally Euclidean, then the divergence error of Eq. (58) decreases much faster than the energy norm of the error estimate of Eq. (61). The conditions for this boundedness are displayed by inequalities (62) and (63) which are relatively strict, but the computational tests for the diffusion problems indicate that these conditions seem to be much more lenient.

Appendix 4: Test Problem for Isotropic Diffusion Equation

Let $\mathbb{S }^2$ be the spherical shell of radius $r$. On $\mathbb{S }^2$, the following system of linear reaction-diffusion equations is considered for the scalar variable $u$ and $v$: For any $\mathbf{x} \in \mathbb{S }^2$,

$$\begin{aligned} \frac{\partial u (\mathbf{x}) }{\partial t}&= \mu \nabla ^2 u (\mathbf{x}) + a u (\mathbf{x}) + b v (\mathbf{x}),\quad \mu \in \mathbb{R }^+,~a,b \in \mathbb{R }, \end{aligned}$$

(64)

$$\begin{aligned} \frac{\partial v (\mathbf{x}) }{\partial t}&= \nu \nabla ^2 v (\mathbf{x}) + c u (\mathbf{x}) + d v (\mathbf{x}),\quad \nu \in \mathbb{R }^+,~c,d \in \mathbb{R }, \end{aligned}$$

(65)

with the boundary conditions $u_D$ and $v_D$ for the variable $u$ and $v$ respectively at the boundaries $\partial \mathbb{S }^2$ as

$$\begin{aligned} u (\mathbf{x}) = u_D,~~ v (\mathbf{x}) = v_D,\quad \mathbf{x} \in \partial \mathbb{S }^2, \end{aligned}$$

(66)

where $\mu$ and $\nu$ are the diffusivity coefficients, while the parameters $a,~b,~c$, and $d$ are for the linear reaction functions. Since all the points $\mathbf{x}$ lie on the spherical shell, the variables can be expanded in terms of spherical harmonics $Y^m_n(\theta , \varphi )$ such as

$$\begin{aligned} u (\theta , \varphi , t) = \sum _{n=0}^{\infty } \sum _{m=-n}^{m=n} A^m_n (t) Y^m_n (\theta , \varphi ),~~~v (\theta , \varphi , t) = \sum _{n=0}^{\infty } \sum _{m=-n}^{m=n} B^m_n (t) Y^m_n (\theta , \varphi ) , \end{aligned}$$

where $\theta$ is the polar angle from the $xy$ plane in the range of  $0 \le \theta \le \pi$, while $\varphi$ is the azimuthal angle in the $xy$ plane from the positive $x$ axis in the range of $-\pi \le \varphi \le \pi$ (Fig. 2). Note that the polynomial coefficients $A^m_n(t)$ and $B^m_n(t)$ only depend on the time variable $t$. Using the following properties of the spherical harmonics $Y^m_n(\theta , \varphi ) , \nabla ^2 Y^m_n = - {n(n+1)} / {r^2} Y^m_n$, we can express the analytic solution of Eqs. (64) and (65) as

$$\begin{aligned} u (\theta , \varphi , t)&= \sum _{n=0}^{\infty } \sum _{m=-n}^{m=n} e^{\gamma _n t} ( \tilde{A}^m_n \cosh (\delta _n t) + \tilde{B}^m_n \sinh (\delta _n t) ) Y^m_n (\theta , \varphi ) , \end{aligned}$$

(67)

$$\begin{aligned} v (\theta , \varphi , t)&= \sum _{n=0}^{\infty } \sum _{m=-n}^{m=n} e^{\gamma _n t} ( \tilde{C}^m_n \cosh (\delta _n t) + \tilde{D}^m_n \sinh (\delta _n t) ) Y^m_n (\theta , \varphi ) , \end{aligned}$$

(68)

where each parameter has the following value [37]:

$$\begin{aligned} a&= -B -1,~~ b = A^2, ~~ c = B,~~ d = - A^2 , \\ \tilde{A}^m_n&= A^m_n (0),~~\tilde{B}^m_n = \frac{1}{ \delta _n } \left[ (a_n - {\gamma }_n ) A^m_n(0) + b B^m_n (0) \right] , \\ \tilde{C}^m_n&= B^m_n (0),~~ \tilde{D}^m_n = \frac{1}{ \delta _n } \left[ c A^m_n (0) + (d_n - {\gamma }_n ) B^m_n(0) \right] , \\ a_n&= a - \mu \frac{n (n + 1)}{r^2} , ~~~~d_n = d - \nu \frac{ n (n + 1) }{r^2} , \\ \gamma _n&= \frac{1}{2} ( a_n + d_n ), ~~~~~~ \delta _n = \frac{1}{2} \sqrt{(a_n + d_n)^2 - 4 (a_n d_n -bc ) } . \end{aligned}$$

For the implementation of the test problem, the initial conditions are chosen with $n=5$ and $A^m_5 = [ -0.5839, -0.8436, -0.4764, -0.6475, 0.1886, 0.8709, -0.8338, 0.1795, -0.7873, 0.8842, 0.2943]$ and $B^m_5 = [-0.6263, 0.9803, 0.7222, 0.5945, 0.6026, -0.2076, 0.4556, 0.6024, 0.9695, -0.4936, 0.1098]$ for $m \in [-5,5]$. Also, we used $A=2, B = 5, l = 1.0e-3$, and $m = 2.0e-3$. In every curved element, the moving frames are continuous and differentiable within each element, but may not be continuous across the interfaces of the elements as aforementioned. The diffusion operator is generated as a global matrix and solved implicitly by the Helmholtz solver at every time step of the Runge–Kutta implicit–explicit 3rd-order scheme [48]. The time step ($\Delta t$) is $1.0e-4$ and the final time is 1.0.