Exact subdomain and embedded interface polynomial integration in finite elements with planar cuts

Abstract

The implementation of discontinuous functions occurs in many of today’s state-of-the-art partial differential equation solvers. However, in finite element methods, this poses an inherent difficulty: efficient quadrature rules available when integrating functions whose discontinuity falls in the element’s interior are for low order degree polynomials, not easily extended to higher order degree polynomials, and cover a restricted set of geometries. Many approaches to this issue have been developed in recent years. Among them, one of the most elegant and versatile is the equivalent polynomial technique. This method replaces the discontinuous function with a polynomial, allowing integration to occur over the entire domain rather than integrating over complex subdomains. Although eliminating the issues involved with discontinuous function integration, the equivalent polynomial tactic introduces its problems. The exact subdomain integration requires a machinery that quickly grows in complexity when increasing the polynomial degree and the geometry dimension, restricting its applicability to lower order degree finite element families. The current work eliminates this issue. We provide algebraic expressions to exactly evaluate the subdomain integral of any degree polynomial on parent finite element shapes cut by a planar interface. These formulas also apply to the exact evaluation of the embedded interface integral. We provide recursive algorithms that avoid overflow in computer arithmetic for standard finite element geometries: triangle, square, cube, tetrahedron, and prism, along with a hypercube of arbitrary dimensions.

Appendix

Proposition 6

Let D be a bounded connected domain with smooth boundary ∂D. Let G(x) be a smooth level set function. Let ${\Gamma } = \left \{ \boldsymbol {x}\in D : G(\boldsymbol {x}) = 0 \right \}$ be a continuous smooth embedded interface, that separates D in the two subregions D₁ and D₂, such that G(x) > 0 for all x ∈ D₁ and G(x) < 0 for all x ∈ D₂. Assume the measure μ(Γ ∩ ∂D) = 0. Then, for any differentiable function f(x)

$$ \begin{array}{@{}rcl@{}} -\lim_{t\rightarrow\infty}t& {\int}_{D} f { \text{Li}}_{-1}(-\exp(G t)) \| \nabla G \| d\boldsymbol{x} = {\int}_{D} f \delta(G) \|\nabla(G)\| d\boldsymbol{x}. \end{array} $$

The ∥∇G∥ term in both sides is needed since the level set G(x) only approximates the required condition, ∥∇d∥ = 1, for a true distance d(x), see Appendix in [12].

Proof

Let n on ∂D be the outer unit normal vector to D. Let $\widehat {\boldsymbol {n}} =-\frac {\nabla G}{\| \nabla G \|}$ be defined everywhere on D. $\widehat {\boldsymbol {n}}$ is the unit vector orthogonal to the the level curves G(x) = const, pointing in the direction of maximum decrease. On the interface Γ, $\widehat {\boldsymbol {n}}$ is the unit outer normal to D₁. Let ∂D₁ = ∂D ∩ D₁. Then, the boundary of D₁ is piece-wise-defined by ∂D₁ ∪Γ, with outer unit normal vectors n and $\widehat {\boldsymbol {n}}$, respectively.

Observe that by using the chain rule and the derivative property of the polylogarithm, we have

$$ \begin{array}{@{}rcl@{}} &\nabla&{ \text{Li}}_{0}(-\exp(G t)) \cdot \widehat{\boldsymbol{n}} = \nabla{ \text{Li}}_{0}(-\exp(G t)) \cdot \left( -\frac{\nabla G}{\| \nabla G \|} \right) \\ &=& -t \frac{d}{d G} { \text{Li}}_{0}(-\exp(G t)) \nabla G \cdot\frac{\nabla G}{\| \nabla G \|} = -t { \text{Li}}_{-1}(-\exp(G t)) \| \nabla G \|. \end{array} $$

(A.1)

Then,

$$ \begin{array}{@{}rcl@{}} &-&\lim_{t\rightarrow\infty}t {\int}_{D} f { \text{Li}}_{-1}(-\exp(G t)) \| \nabla G \| d\boldsymbol{x} = \lim_{t\rightarrow\infty}{\int}_{D} \nabla{ \text{Li}}_{0}(-\exp(G t)) \!\cdot\! \left( f \widehat{\boldsymbol{n}}\right) d\boldsymbol{x} \\&=& -\lim_{t\rightarrow\infty}\left( {\int}_{D} { \text{Li}}_{0}(-\exp(G t)) \nabla \cdot \left( f \widehat{\boldsymbol{n}} \right) d\boldsymbol{x} + {\int}_{\partial D} { \text{Li}}_{0}(-\exp(G t)) f \widehat{\boldsymbol{n}} \cdot \boldsymbol{n} dS \right) \\& =& {\int}_{D} {\mathrm{U}}(G) \nabla \cdot \left( f \widehat{\boldsymbol{n}}\right) d\boldsymbol{x} - {\int}_{\partial D} {\mathrm{U}}(G) f \widehat{\boldsymbol{n}}\cdot \boldsymbol{n} dS \\& =& {\int}_{D_{1}} \nabla \cdot \left( f \widehat{\boldsymbol{n}}\right) d\boldsymbol{x} - {\int}_{\partial D_{1}} f \widehat{\boldsymbol{n}} \cdot \boldsymbol{n} dS \\ &=& \left( {\int}_{\partial D_{1}} f \widehat{\boldsymbol{n}}\cdot \boldsymbol{n} d S + {\int}_{\Gamma} f \widehat{\boldsymbol{n}}\cdot \widehat{\boldsymbol{n}} d S \right) - {\int}_{\partial D_{1}} f \widehat{\boldsymbol{n}} \cdot \boldsymbol{n} dS \\&=& {\int}_{\Gamma} f dS = {\int}_{D} f \delta(G) \|\nabla G\| d\boldsymbol{x}, \end{array} $$

Where we used (A.1) for first line equality, divergence Theorem for the second line equality, (2) for the third equality, (3) for the fourth equality, divergence Theorem for the fifth line equality, and the definition of the Dirac delta distribution for a level set in the last line equality. Note that the proof holds only if the measure μ(Γ ∩ ∂D) = 0, for an appropriate product measure μ, since from the third line to the fourth line, the integral equality on the boundary

$${\int}_{\partial D} {\mathrm{U}}(G) f \widehat{\boldsymbol{n}}\cdot \boldsymbol{n} dS = {\int}_{\partial D_{1}} f \widehat{\boldsymbol{n}} \cdot \boldsymbol{n} dS $$

is true only if the Heaviside function U is almost everywhere 1 on ∂D₁ and almost everywhere 0 on its complement. For μ(Γ ∩ ∂D)≠ 0, we would have measurable parts of the boundary ∂D with U = 0.5, and the equality would not hold. □