UC DAVIS Eric B. Chin Jean B. Lasserre LAAS-CNRS N. Sukumar UC DAVIS Atlanta, GA // October 28, 2015 Research support of the NSF is gratefully acknowledged NUMERICAL INTEGRATION of HOMOGENEOUS FUNCTIONS and POLYNOMIALS on POLYTOPES POEMs 2015
2/33 Determine Problem statement ∫ R d f ( x ) d x P ⊂ I ▶ f ( x ) is a homogeneous function ▶ P is a convex or nonconvex polytope
3/33 Three methods to integrate functions on polytopes Triangulation Divergence theorem Moment fjtting Existing methods ∫ ∫ V ∇ · F dV = S F · n dS x x x V x x x x S
4/33 Background Euler’s homgeneous function theorem and Generalized Stokes’s theorem Extension of Lasserre’s method Integration over convex and nonconvex polytopes Main results Application: Elastic fracture with the X-FEM Conclusions and outlook Contents
4/33 Background Euler’s homgeneous function theorem and Generalized Stokes’s theorem Extension of Lasserre’s method Integration over convex and nonconvex polytopes Main results Application: Elastic fracture with the X-FEM Conclusions and outlook Contents
4/33 Background Euler’s homgeneous function theorem and Generalized Stokes’s theorem Extension of Lasserre’s method Integration over convex and nonconvex polytopes Main results Application: Elastic fracture with the X-FEM Conclusions and outlook Contents
4/33 Background Euler’s homgeneous function theorem and Generalized Stokes’s theorem Extension of Lasserre’s method Integration over convex and nonconvex polytopes Main results Application: Elastic fracture with the X-FEM Conclusions and outlook Contents
4/33 Background Euler’s homgeneous function theorem and Generalized Stokes’s theorem Extension of Lasserre’s method Integration over convex and nonconvex polytopes Main results Application: Elastic fracture with the X-FEM Conclusions and outlook Contents
Background 5/33 Background
6/33 and then it also satisfjes: Background Euler’s homogeneous func. thm. A continuously difgerentiable function f ( x ) is said to be positive homogeneous of degree q if: f ( λ x ) = λ q f ( x ) ∀ λ > 0 , { R d I if q > 0 qf ( x ) = ⟨∇ f ( x ) , x ⟩ ∀ x ∈ R d \{ 0 } I if q < 0 ⟨· , ·⟩ : inner product d : dimension
Background 7/33 Proof By defjnition, a homogeneous function of degree q has the property λ q f ( x ) = f ( λ x ) Defjne x ′ := λ x and calculate ∂ ∂λ : ∂ x ′ · ∂ x ′ qλ q − 1 f ( x ) = ∂f ∂λ = ∂f ∂ x ′ · x Let λ = 1 : qf ( x ) = ∂f ∂ x · x = ⟨∇ f ( x ) , x ⟩ ▶ Converse is also readily established
Background 8/33 : , where : , where : Examples of homogeneous fns. q = 0 : f ( x ) = 1 q = 1 : f ( x ) = x + y q = 2 : f ( x ) = 3 x 2 +2 xy
8/33 where where Background Examples of homogeneous fns. q = − 1 2 : 1 f ( x ) = √ r , q = 0 : x 2 + y 2 √ f ( x ) = 1 r = q = 1 : q = 0 : f ( x ) = x + y f ( x ) = cos θ , q = 2 : θ = tan − 1 y x f ( x ) = 3 x 2 +2 xy q = 1 f ( x ) = √ r cos θ 2 :
Background 9/33 (See: Taylor, PDEs: Basic Theory , 2011 ) Generalized Stokes’s theorem ∫ ∫ dω = ω M ∂M ⇓ ∫ ∫ ∫ ( ∇ · X ) f ( x ) d x + X · ∇ f ( x ) d x = ( X · n ) f ( x ) dσ M M ∂M ▶ X : vector fjeld ▶ M : region of integration ▶ dσ : Lebesgue measure on ∂M
Extension of Lasserre’s method 10/33 Extensionof Lasserre’smethod
Extension of Lasserre’s method 11/33 Soc., 1998, 1999 ) for convex regions Mech., 2011 ) Mech., 2015, doi 10.1007/s00466-015-1213-7 ) [PDF] generalized Stokes’s theorem History ▶ Method developed by Lasserre ( Proc. Am. Math. ▶ First applied to X-FEM by Mousavi and S ( Comp. ▶ Extended to nonconvex regions by Chin et al. ( Comp. ▶ Method uses properties of homogeneous functions and
Main results 12/33 Mainresults
13/33 Main results Reducing integration to bdry. Apply Stokes’s theorem with X := x and f ( x ) is q -homogeneous: m ∫ ∫ ∫ ∑ d f ( x ) d x + ⟨∇ f ( x ) , x ⟩ d x = ( x · n i ) f ( x ) dσ P P F i i =1 P : polygon F i : boundary facets Apply Euler’s homogeneous fn. theorem, qf ( x ) = ⟨∇ f ( x ) , x ⟩ : m ∫ ∫ ∫ ∑ d f ( x ) d x + q f ( x ) d x = ( x · n i ) f ( x ) dσ P P F i i =1 m 1 ∫ ∫ ∑ f ( x ) d x = ( x · n i ) f ( x ) dσ d + q P F i i =1
the polytope Main results 14/33 (*) Integral of f ( x ) m ∫ 1 ∫ ∑ f ( x ) d x = ( x · n i ) f ( x ) dσ d + q P F i i =1 ▶ F i ⊂ a i · x = b i : equation of a hyperplane a i ▶ n i = ∥ a i ∥ : unit normal to hyperplane ∥ a i ∥ = x · a i a i b i ▶ x · n i = x · ∥ a i ∥ = ∥ a i ∥ m ∫ 1 ∫ b i ∑ f ( x ) d x = ∥ a i ∥ f ( x ) dσ ∴ d + q P F i i =1 ▶ Using (*), one can reduce integration to the boundary of
Main results 15/33 one gives the correct answer in (*). Which is correct? Answer: Given the vertices of the polygon, travel around the Sign of a i and b i Question: x = 1 and − x = − 1 produce the same line, yet only polygon in a clockwise direction. 2 F 1 P start 1
Main results 15/33 one gives the correct answer in (*). Which is correct? Answer: Given the vertices of the polygon, travel around the Sign of a i and b i Question: x = 1 and − x = − 1 produce the same line, yet only polygon in a clockwise direction. F 2 2 3 F 1 P 1
Main results 15/33 one gives the correct answer in (*). Which is correct? Answer: Given the vertices of the polygon, travel around the Sign of a i and b i Question: x = 1 and − x = − 1 produce the same line, yet only polygon in a clockwise direction. F 2 2 F 3 4 3 F 1 P 1
15/33 one gives the correct answer in (*). Which is correct? Answer: Given the vertices of the polygon, travel around the Main results Sign of a i and b i Question: x = 1 and − x = − 1 produce the same line, yet only polygon in a clockwise direction. F 2 2 F 3 4 3 F 1 F 4 5 P 1
Main results 15/33 one gives the correct answer in (*). Which is correct? Answer: Given the vertices of the polygon, travel around the Sign of a i and b i Question: x = 1 and − x = − 1 produce the same line, yet only polygon in a clockwise direction. F 2 2 F 3 4 3 F 1 F 4 5 F 5 P 1 6
15/33 one gives the correct answer in (*). Which is correct? Answer: Given the vertices of the polygon, travel around the Main results Sign of a i and b i Question: x = 1 and − x = − 1 produce the same line, yet only polygon in a clockwise direction. F 2 2 F 3 4 3 F 1 F 4 5 F 5 P 1 6 F 6 7
15/33 one gives the correct answer in (*). Which is correct? Answer: Given the vertices of the polygon, travel around the Main results Sign of a i and b i Question: x = 1 and − x = − 1 produce the same line, yet only polygon in a clockwise direction. 2 F 2 4 F 3 3 F 1 F 4 5 F 5 P finish 1 6 F 7 F 6 7
16/33 Main results Sign of a i and b i F i ( x 1 , y 1 ) ( x 2 , y 2 ) F 1 � � x y 1 � � � � � � det = ( y 1 − y 2 ) x + ( x 2 − x 1 ) y + ( x 1 y 2 − y 1 x 2 ) x 1 y 1 1 � � � � � x 2 y 2 1 � � � ⇓ a i = { y 1 − y 2 , x 2 − x 1 } T b i = − ( x 1 y 2 − y 1 x 2 )
Main results 17/33 Reapplying Stokes’s theorem Select: Further reducing the integration ∫ ∫ ∫ ( ∇ · X ) f ( x ) d x + X · ∇ f ( x ) d x = ( X · n ) f ( x ) dσ M M ∂M ▶ M as F i ▶ ∂M as v ij (vertices in I R 2 – intersection of F i and F j ) ▶ X := x − x 0 ( x 0 : any point on hyperplane containing F i ) ▶ f ( x ) is a homogeneous function of degree q
18/33 Reapplying Stokes’s theorem (**) Main results Integral of f ( x ) on F i ∫ ∫ ∫ ( ∇ · X ) f ( x ) d x + X · ∇ f ( x ) d x = ( X · n ) f ( x ) dσ M M ∂M ⇓ [ 2 ] 1 ∫ ∫ ∑ f ( x ) dσ = d ij f ( v ij ) + ⟨∇ f ( x ) , x 0 ⟩ dσ d + q − 1 F i F i j =1 ▶ d ij := ⟨ x − x 0 , n ij ⟩ – algebraic distance from v ij to x 0 ▶ ∫ F i ⟨∇ f ( x ) , x 0 ⟩ dσ can be applied recursively
cubature rule is exact (*) 19/33 vertices of the polytope (**) Main results Results m 1 b i ∫ ∫ ∑ f ( x ) d x = ∥ a i ∥ f ( x ) dσ d + q P F i i =1 [ 2 ] ∫ 1 ∫ ∑ f ( x ) dσ = d ij f ( v ij ) + ⟨∇ f ( x ) , x 0 ⟩ dσ d + q − 1 F i F i j =1 ▶ These formulas can be used to reduce integration to the ▶ Further, a closed-form cubature rule can be developed ▶ If the partial derivatives of f ( x ) eventually vanish, this
20/33 Combine (*) with (**): Main results where and 2D closed-form cubature rule ∑ m b i ∑ j ̸ = i d ij I ( v ij ) ∫ i =1 ∥ a i ∥ f ( x ) d x = ( q + 2)( q + 1) P q 2 D | α | f ( v ij ) Q k ( v ij ) ∑ ∑ ∏ ( x 0 ℓ ) α ℓ I ( v ij ) := Q k ( v ij ) := ( q ) α ! k k =0 | α | = q − k ℓ =1 ▶ α is an n -tuple of nonnegative integers ▶ D is the difgerential operator in multi-index notation
Main results Apply Stokes’s theorem: 21/33 Polar transformation: Curved regions Consider a region V bounded by homogeneous functions h i ( x ) = b i ( i = 1 , . . . , m ) of degree q i m ∫ 1 ∫ ∑ ∥∇ h i ∥ − 1 f ( x ) dσ f ( x ) d x = q i b i d + q V A i i =1 ∫ β i m ∫ 1 ∑ H 2 f ( x ) d x = i ( θ ) f ( x ( θ )) dθ 2 + q V α i i =1 ▶ Region bounded by equations of the form r = H i ( θ )
Numerical examples 22/33 Numericalexamples
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