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Shape derivative of geometric constraints without integration along rays Florian Feppon Gr egoire Allaire, Charles Dapogny Julien Cortial, Felipe Bordeu ENGOPT September 18th, 2018 Thickness control in structural optimization Some


  1. Shape derivative of geometric constraints without integration along rays Florian Feppon Gr´ egoire Allaire, Charles Dapogny Julien Cortial, Felipe Bordeu ENGOPT – September 18th, 2018

  2. Thickness control in structural optimization Some recent advances in level-set based shape optimization: geometric constraints. [1][2] : Figure: Michailidis (2014) [1] Michailidis2014Manufacturing . [2] Allaire2016Thickness .

  3. Outline 1. Shape derivatives of geometric constraints based on the signed distance function 2. A variational method for avoiding integration along rays 3. Numerical comparisons and applications to shape and topology optimization

  4. Outline 1. Shape derivatives of geometric constraints based on the signed distance function 2. A variational method for avoiding integration along rays 3. Numerical comparisons and applications to shape and topology optimization

  5. 1. Shape derivatives of geometric constraints The signed distance function d Ω to the domain Ω ⊂ D is defined by:  − min y ∈ ∂ Ω || y − x || if x ∈ Ω ,   ∀ x ∈ D , d Ω ( x ) = y ∈ ∂ Ω || y − x || min if x ∈ D \ Ω .  

  6. 1. Shape derivatives of geometric constraints The signed distance function allows to formulate geometric constraints. ◮ Maximum thickness constraint : ∀ x ∈ Ω , | d Ω ( x ) | ≤ d max / 2 ◮ Minimum thickness constraint: ∀ y ∈ ∂ Ω , | ζ − ( y ) | ≥ d min / 2 .

  7. 1. Shape derivatives of geometric constraints For shape optimization, one formulates geometric constraints using penalty functionals P (Ω) as follows: � min Ω J (Ω) , s.t. P (Ω) ≤ 0 , where P (Ω) := j ( d Ω ( x )) d x . D We rely on the method of Hadamard (figure from [3] ): [3] dapogny2017geometrical .

  8. 1. Shape derivatives of geometric constraints For shape optimization, one formulates geometric constraints using penalty functionals P (Ω) as follows: � min Ω J (Ω) , s.t. P (Ω) ≤ 0 , where P (Ω) := j ( d Ω ( x )) d x . D The shape derivative of P (Ω) reads � � P ′ (Ω)( θ ) = j ′ ( d Ω ( x )) d ′ Ω ( θ )( x ) d x = u θ · n d y D \ Σ ∂ Ω with � � j ′ ( d Ω ( x )) ∀ y ∈ ∂ Ω , u ( y ) = − (1 + κ i ( y ) d Ω ( x )) d x . x ∈ ray ( y ) 1 ≤ i ≤ n − 1

  9. 1. Shape derivatives of geometric constraints � j ′ ( d Ω ( x )) � ∀ y ∈ ∂ Ω , u ( y ) = − (1 + κ i ( y ) d Ω ( x )) d x . x ∈ ray ( y ) 1 ≤ i ≤ n − 1 Computing u requires:

  10. 1. Shape derivatives of geometric constraints � j ′ ( d Ω ( x )) � ∀ y ∈ ∂ Ω , u ( y ) = − (1 + κ i ( y ) d Ω ( x )) d x . x ∈ ray ( y ) 1 ≤ i ≤ n − 1 Computing u requires: 1. Integrating along rays on the discretization mesh:

  11. 1. Shape derivatives of geometric constraints � j ′ ( d Ω ( x )) � ∀ y ∈ ∂ Ω , u ( y ) = − (1 + κ i ( y ) d Ω ( x )) d x . x ∈ ray ( y ) 1 ≤ i ≤ n − 1 Computing u requires: 1. Integrating along rays on the discretization mesh: 2. Estimating the principal curvatures κ i ( y ).

  12. Outline 1. Shape derivatives of geometric constraints based on the signed distance function 2. A variational method for avoiding integration along rays 3. Numerical comparisons and applications to shape and topology optimization

  13. 2. A variational method for avoiding integration along rays More precisely, the shape derivative of P (Ω) reads � � P ′ (Ω)( θ ) = j ′ ( d Ω ( x )) d ′ Ω ( θ )( x ) d x = u θ · n d y D \ Σ ∂ Ω with d ′ Ω ( θ ) satisfying � ∇ d ′ Ω ( θ ) · ∇ d Ω = 0 in D \ Σ d ′ Ω ( θ ) = − θ · n on ∂ Ω . θ | θ · n | Ω

  14. 2. A variational method for avoiding integration along rays More precisely, the shape derivative of P (Ω) reads � � P ′ (Ω)( θ ) = j ′ ( d Ω ( x )) d ′ Ω ( θ )( x ) d x = u θ · n d y D \ Σ ∂ Ω with d ′ Ω ( θ ) satisfying � ∇ d ′ Ω ( θ ) · ∇ d Ω = 0 in D \ Σ d ′ Ω ( θ ) = − θ · n on ∂ Ω . Our method : u solves the following variational problem (with ω > 0 rather arbitrary): Find u ∈ V ω such that ∀ v ∈ V ω , � � � j ′ ( d Ω ( x )) v ( x ) d x , uv d s + ω ( ∇ d Ω · ∇ u )( ∇ d Ω · ∇ v ) d x = − ∂ Ω D \ Σ D \ Σ

  15. 2. A variational method for avoiding integration along rays More precisely, the shape derivative of P (Ω) reads � � P ′ (Ω)( θ ) = j ′ ( d Ω ( x )) d ′ Ω ( θ )( x ) d x = u θ · n d y D \ Σ ∂ Ω with d ′ Ω ( θ ) satisfying � ∇ d ′ Ω ( θ ) · ∇ d Ω = 0 in D \ Σ d ′ Ω ( θ ) = − θ · n on ∂ Ω . Take v = d ′ Our method : Ω ( θ ): Find u ∈ V ω such that ∀ v ∈ V ω , � � � j ′ ( d Ω ( x )) v ( x ) d x , uv d s + ω ( ∇ d Ω · ∇ u )( ∇ d Ω · ∇ v ) d x = − ∂ Ω D \ Σ D \ Σ

  16. 2. A variational method for avoiding integration along rays More precisely, the shape derivative of P (Ω) reads � � P ′ (Ω)( θ ) = j ′ ( d Ω ( x )) d ′ Ω ( θ )( x ) d x = u θ · n d y D \ Σ ∂ Ω with d ′ Ω ( θ ) satisfying � ∇ d ′ Ω ( θ ) · ∇ d Ω = 0 in D \ Σ d ′ Ω ( θ ) = − θ · n on ∂ Ω . Take v = d ′ Our method : Ω ( θ ): Find u ∈ V ω such that ∀ v ∈ V ω , � � � j ′ ( d Ω ( x )) v ( x ) d x , uv d s + ω ( ∇ d Ω · ∇ u )( ∇ d Ω · ∇ v ) d x = − ∂ Ω D \ Σ D \ Σ � � � ud ′ ω ( ∇ d Ω ·∇ u )( ∇ d Ω · ∇ d ′ j ′ ( d Ω ( x )) d ′ Ω ( θ ) d s + Ω ( θ )) d x = − Ω ( θ )( x ) d x , ∂ Ω D \ Σ D \ Σ

  17. 2. A variational method for avoiding integration along rays More precisely, the shape derivative of P (Ω) reads � � P ′ (Ω)( θ ) = j ′ ( d Ω ( x )) d ′ Ω ( θ )( x ) d x = u θ · n d y D \ Σ ∂ Ω with d ′ Ω ( θ ) satisfying � ∇ d ′ Ω ( θ ) · ∇ d Ω = 0 in D \ Σ d ′ Ω ( θ ) = − θ · n on ∂ Ω . Take v = d ′ Our method : Ω ( θ ): Find u ∈ V ω such that ∀ v ∈ V ω , � � � j ′ ( d Ω ( x )) v ( x ) d x , uv d s + ω ( ∇ d Ω · ∇ u )( ∇ d Ω · ∇ v ) d x = − ∂ Ω D \ Σ D \ Σ � � � ud ′ ω ( ∇ d Ω ·∇ u )( ∇ d Ω · ∇ d ′ j ′ ( d Ω ( x )) d ′ Ω ( θ ) d s + Ω ( θ )) d x = − Ω ( θ )( x ) d x , ∂ Ω D \ Σ D \ Σ

  18. 2. A variational method for avoiding integration along rays More precisely, the shape derivative of P (Ω) reads � � P ′ (Ω)( θ ) = j ′ ( d Ω ( x )) d ′ Ω ( θ )( x ) d x = u θ · n d y D \ Σ ∂ Ω with d ′ Ω ( θ ) satisfying � ∇ d ′ Ω ( θ ) · ∇ d Ω = 0 in D \ Σ d ′ Ω ( θ ) = − θ · n on ∂ Ω . Take v = d ′ Our method : Ω ( θ ): Find u ∈ V ω such that ∀ v ∈ V ω , � � � j ′ ( d Ω ( x )) v ( x ) d x , uv d s + ω ( ∇ d Ω · ∇ u )( ∇ d Ω · ∇ v ) d x = − ∂ Ω D \ Σ D \ Σ � � � ud ′ ω ( ∇ d Ω ·∇ u )( ∇ d Ω · ∇ d ′ j ′ ( d Ω ( x )) d ′ Ω ( θ ) d s + Ω ( θ )) d x = − Ω ( θ )( x ) d x , ∂ Ω D \ Σ D \ Σ � � j ′ ( d Ω ( x )) d ′ u ( − θ · n ) d s + 0 = − Ω ( θ )( x ) d x . ∂ Ω D \ Σ

  19. 2. A variational method for avoiding integration along rays Our theoretical results for the variational problem: Find u ∈ V ω such that ∀ v ∈ V ω , � � � j ′ ( d Ω ( x )) v ( x ) d x uv d s + ω ( ∇ d Ω · ∇ u )( ∇ d Ω · ∇ v ) d x = − (1) ∂ Ω D \ Σ D \ Σ

  20. 2. A variational method for avoiding integration along rays Our theoretical results for the variational problem: Find u ∈ V ω such that ∀ v ∈ V ω , � � � j ′ ( d Ω ( x )) v ( x ) d x uv d s + ω ( ∇ d Ω · ∇ u )( ∇ d Ω · ∇ v ) d x = − (1) ∂ Ω D \ Σ D \ Σ 1. Under rather unrestrictive assumptions, the trace of the solution u is independent on the weight ω and is given by � j ′ ( d Ω ( x )) � ∀ y ∈ ∂ Ω , u ( y ) = − (1 + κ i ( y ) d Ω ( x )) d x . (2) x ∈ ray ( y ) 1 ≤ i ≤ n − 1

  21. 2. A variational method for avoiding integration along rays Our theoretical results for the variational problem: Find u ∈ V ω such that ∀ v ∈ V ω , � � � j ′ ( d Ω ( x )) v ( x ) d x uv d s + ω ( ∇ d Ω · ∇ u )( ∇ d Ω · ∇ v ) d x = − (1) ∂ Ω D \ Σ D \ Σ 1. Under rather unrestrictive assumptions, the trace of the solution u is independent on the weight ω and is given by � j ′ ( d Ω ( x )) � ∀ y ∈ ∂ Ω , u ( y ) = − (1 + κ i ( y ) d Ω ( x )) d x . (2) x ∈ ray ( y ) 1 ≤ i ≤ n − 1 (1) can be solved with FEM while (2) requires computing rays and curvatures!

  22. 2. A variational method for avoiding integration along rays Our theoretical results for the variational problem: Find u ∈ V ω such that ∀ v ∈ V ω , � � � j ′ ( d Ω ( x )) v ( x ) d x uv d s + ω ( ∇ d Ω · ∇ u )( ∇ d Ω · ∇ v ) d x = − (1) ∂ Ω D \ Σ D \ Σ 1. Under rather unrestrictive assumptions, the trace of the solution u is independent on the weight ω and is given by � j ′ ( d Ω ( x )) � ∀ y ∈ ∂ Ω , u ( y ) = − (1 + κ i ( y ) d Ω ( x )) d x . (2) x ∈ ray ( y ) 1 ≤ i ≤ n − 1 (1) can be solved with FEM while (2) requires computing rays and curvatures! 2. It is possible to show the well-posedeness of ( ?? ) for a large class of weights ω in a suitable space V ω .

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