DAE - 3D pendulum � p e 3 � Model is a semi-explicit DAE with x = v O e 1 F ( x , z , u ) � ˙ � �� � e 2 u � � � p v = ˙ x = m − g e 3 − z u ˙ v m p 0 = p ⊤ p − L 2 p � �� � G ( x ) 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 7 / 25
DAE - 3D pendulum � p e 3 � Model is a semi-explicit DAE with x = v O e 1 F ( x , z , u ) � ˙ � �� � e 2 u � � � p v = ˙ x = m − g e 3 − z u ˙ v m p 0 = p ⊤ p − L 2 p � �� � G ( x ) Consider the root-finding problem to be solved in ˙ x , z : � ˙ � x − F ( x , z , u ) r ( ˙ x , x , z , u ) = = 0 G ( x ) 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 7 / 25
DAE - 3D pendulum � p e 3 � Model is a semi-explicit DAE with x = v O e 1 F ( x , z , u ) � ˙ � �� � e 2 u � � � p v = ˙ x = m − g e 3 − z u ˙ v m p 0 = p ⊤ p − L 2 p � �� � G ( x ) Consider the root-finding problem to be solved in ˙ x , z : � ˙ � x − F ( x , z , u ) r ( ˙ x , x , z , u ) = = 0 G ( x ) Then: I 0 0 x , z r ⊤ = is rank-deficient. The Newton step does not exist !! ∇ ˙ 0 I p 0 0 0 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 7 / 25
DAE - 3D pendulum � p e 3 � Model is a semi-explicit DAE with x = v O e 1 F ( x , z , u ) � ˙ � �� � e 2 u � � � p v = ˙ x = m − g e 3 − z u ˙ v m p 0 = p ⊤ p − L 2 p � �� � G ( x ) Note that ∂ G ( x ) = 0 !! ∂ z Consider the root-finding problem to be solved in ˙ x , z : � ˙ � x − F ( x , z , u ) r ( ˙ x , x , z , u ) = = 0 G ( x ) Then: I 0 0 x , z r ⊤ = is rank-deficient. The Newton step does not exist !! ∇ ˙ 0 I p 0 0 0 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 7 / 25
DAE - Delta Robot Lagrange model yields a semi-explicit DAE with: � p − p 1 � 2 − L 2 � p − p 2 � 2 − L 2 G ( x ) = � p − p 3 � 2 − L 2 where cos γ k sin γ k 0 L cos α k p k = R z k = − sin γ k cos γ k 0 0 0 0 1 L sin α k � � 0 , 2 π 3 , 4 π using γ 1 , 2 , 3 = . 3 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 8 / 25
DAE - Delta Robot Lagrange model yields a semi-explicit DAE with: � p − p 1 � 2 − L 2 � p − p 2 � 2 − L 2 G ( x ) = � p − p 3 � 2 − L 2 where cos γ k sin γ k 0 L cos α k p k = R z k = − sin γ k cos γ k 0 0 0 0 1 L sin α k � � 0 , 2 π 3 , 4 π using γ 1 , 2 , 3 = . 3 Algebraic variables z for the forces in the arms: ∂ G ( x ) = 0 ∂ z 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 8 / 25
DAE - Delta Robot Lagrange model yields a semi-explicit DAE with: � p − p 1 � 2 − L 2 � p − p 2 � 2 − L 2 G ( x ) = � p − p 3 � 2 − L 2 where cos γ k sin γ k 0 L cos α k p k = R z k = − sin γ k cos γ k 0 0 0 0 1 L sin α k � � 0 , 2 π 3 , 4 π using γ 1 , 2 , 3 = . 3 Algebraic variables z for the forces in the arms: ∂ G ( x ) = 0 ∂ z Such that the DAE: x = F ( x , z , u ) ˙ 0 = G ( x ) ... cannot be solved for z , because ∂ G ( x ) = 0 !! ∂ z 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 8 / 25
DAE from Lagrange Mechanics Is that a general problem in Lagrange mechanics ? Pretty much ... 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 9 / 25
DAE from Lagrange Mechanics Is that a general problem in Lagrange mechanics ? Pretty much ... The difficulty comes from having holonomic (aka purely position-dependent) constraints: G ( q ) = 0 which ”hold the system together”. 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 9 / 25
DAE from Lagrange Mechanics Is that a general problem in Lagrange mechanics ? Pretty much ... The difficulty comes from having holonomic (aka purely position-dependent) constraints: G ( q ) = 0 which ”hold the system together”. Then the Euler-Lagrange equations: ∂ L q − ∂ L d ∂ q = 0 ∂ ˙ d deliver an explicit ODE for the accelerations ¨ q , involving the algebraic variables z . 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 9 / 25
DAE from Lagrange Mechanics Is that a general problem in Lagrange mechanics ? Pretty much ... The difficulty comes from having holonomic (aka purely position-dependent) constraints: G ( q ) = 0 which ”hold the system together”. Then the Euler-Lagrange equations: ∂ L q − ∂ L d ∂ q = 0 ∂ ˙ d deliver an explicit ODE for the accelerations ¨ q , involving the algebraic variables z . But the forces generated by the algebraic variables z are not defined by the algebraic equations because: ∂ G ( q ) ∂ G ( x ) = 0 and therefore = 0 ∂ z ∂ z 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 9 / 25
DAE from Lagrange Mechanics Is that a general problem in Lagrange mechanics ? Pretty much ... The difficulty comes from having holonomic (aka purely position-dependent) constraints: G ( q ) = 0 which ”hold the system together”. Then the Euler-Lagrange equations: ∂ L q − ∂ L d ∂ q = 0 ∂ ˙ d deliver an explicit ODE for the accelerations ¨ q , involving the algebraic variables z . But the forces generated by the algebraic variables z are not defined by the algebraic equations because: ∂ G ( q ) ∂ G ( x ) = 0 and therefore = 0 ∂ z ∂ z What is going on ?!? 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 9 / 25
Outline ”Easy” & ”Hard” DAEs 1 Differential Index 2 Index Reduction 3 Constraints drift 4 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 10 / 25
DAE - Differential Index 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 11 / 25
DAE - Differential Index Definition: The DAE differential index is the minimum i such that: d i d t i F ( ˙ x , x , z , u ) = 0 is a pure ODE 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 11 / 25
DAE - Differential Index Example: � x 1 − ˙ � Definition: x 1 + 1 F ( ˙ x , x ) = = 0 x 1 x 2 + 2 ˙ The DAE differential index is the minimum i such that: d i d t i F ( ˙ x , x , z , u ) = 0 is a pure ODE 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 11 / 25
DAE - Differential Index Example: � x 1 − ˙ � Definition: x 1 + 1 F ( ˙ x , x ) = = 0 x 1 x 2 + 2 ˙ The DAE differential index is the minimum i Note that: such that: � − 1 � ∂ F 0 x = → this is a DAE d i 1 0 ∂ ˙ d t i F ( ˙ x , x , z , u ) = 0 is a pure ODE 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 11 / 25
DAE - Differential Index Example: � x 1 − ˙ � Definition: x 1 + 1 F ( ˙ x , x ) = = 0 x 1 x 2 + 2 ˙ The DAE differential index is the minimum i Note that: such that: � − 1 � ∂ F 0 x = → this is a DAE d i 1 0 ∂ ˙ d t i F ( ˙ x , x , z , u ) = 0 For i = 1 reads as: is a pure ODE � � x 1 − ¨ ˙ x 1 ˙ F (¨ x , ˙ x , x ) = = 0 x 1 x 2 + ˙ ¨ x 1 ˙ x 2 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 11 / 25
DAE - Differential Index Example: � x 1 − ˙ � Definition: x 1 + 1 F ( ˙ x , x ) = = 0 x 1 x 2 + 2 ˙ The DAE differential index is the minimum i Note that: such that: � − 1 � ∂ F 0 x = → this is a DAE d i 1 0 ∂ ˙ d t i F ( ˙ x , x , z , u ) = 0 For i = 1 reads as: is a pure ODE � � x 1 − ¨ ˙ x 1 ˙ F (¨ x , ˙ x , x ) = = 0 x 1 x 2 + ˙ ¨ x 1 ˙ x 2 Using (to write a 1 st -order ODE) x 1 s 1 − s 3 ˙ ˙ s ≡ x 2 we have F (˙ s , s ) = s 3 − ˙ s 3 x 1 ˙ s 3 s 2 + s 3 ˙ ˙ s 2 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 11 / 25
DAE - Differential Index Example: � x 1 − ˙ � Definition: x 1 + 1 F ( ˙ x , x ) = = 0 x 1 x 2 + 2 ˙ The DAE differential index is the minimum i Note that: such that: � − 1 � ∂ F 0 x = → this is a DAE d i 1 0 ∂ ˙ d t i F ( ˙ x , x , z , u ) = 0 For i = 1 reads as: is a pure ODE � � x 1 − ¨ ˙ x 1 ˙ F (¨ x , ˙ x , x ) = = 0 x 1 x 2 + ˙ ¨ x 1 ˙ x 2 Using (to write a 1 st -order ODE) x 1 s 1 − s 3 ˙ ˙ s ≡ x 2 we have F (˙ s , s ) = s 3 − ˙ s 3 x 1 ˙ s 3 s 2 + s 3 ˙ ˙ s 2 And 1 0 0 � ∂ F � ∂ F , s = 0 0 − 1 with det = s 3 ⇒ now we have an ODE ∂ ˙ ∂ ˙ s 0 s 3 s 2 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 11 / 25
DAE - Differential Index Example: � x 1 − ˙ � Definition: x 1 + 1 F ( ˙ x , x ) = = 0 x 1 x 2 + 2 ˙ The DAE differential index is the minimum i Note that: such that: � − 1 � ∂ F 0 x = → this is a DAE d i 1 0 ∂ ˙ d t i F ( ˙ x , x , z , u ) = 0 For i = 1 reads as: is a pure ODE � � x 1 − ¨ ˙ x 1 ˙ F (¨ x , ˙ x , x ) = = 0 x 1 x 2 + ˙ ¨ x 1 ˙ x 2 Using (to write a 1 st -order ODE) x 1 s 1 − s 3 ˙ F is an ˙ s ≡ x 2 we have F (˙ s , s ) = s 3 − ˙ s 3 index-1 DAE x 1 ˙ s 3 s 2 + s 3 ˙ ˙ s 2 And 1 0 0 � ∂ F � ∂ F , s = 0 0 − 1 with det = s 3 ⇒ now we have an ODE ∂ ˙ ∂ ˙ s 0 s 3 s 2 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 11 / 25
DAE - Differential Index Definition The DAE differential index is the minimum i such that: How does the differential index relate to the DAE being ”easy” to solve ?? d i d t i F ( ˙ x , x , z , u ) = 0 is a pure ODE 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 12 / 25
DAE - Differential Index Definition The DAE differential index is the minimum i such that: How does the differential index relate to the DAE being ”easy” to solve ?? d i d t i F ( ˙ x , x , z , u ) = 0 is a pure ODE For an index-1 DAE : d x , x , z , u ) = ˙ d t F ( ˙ F = 0 yields a pure ODE. 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 12 / 25
DAE - Differential Index Definition The DAE differential index is the minimum i such that: How does the differential index relate to the DAE being ”easy” to solve ?? d i d t i F ( ˙ x , x , z , u ) = 0 is a pure ODE For an index-1 DAE : d x , x , z , u ) = ˙ d t F ( ˙ F = 0 yields a pure ODE. Observe that: d F = ∂ F x + ∂ F x + ∂ F z + ∂ F d t F = ˙ x ¨ ∂ x ˙ ∂ z ˙ ∂ u ˙ u = 0 ∂ ˙ 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 12 / 25
DAE - Differential Index Definition The DAE differential index is the minimum i such that: How does the differential index relate to the DAE being ”easy” to solve ?? d i d t i F ( ˙ x , x , z , u ) = 0 is a pure ODE For an index-1 DAE : d x , x , z , u ) = ˙ d t F ( ˙ F = 0 yields a pure ODE. Observe that: d F = ∂ F x + ∂ F x + ∂ F z + ∂ F d t F = ˙ x ¨ ∂ x ˙ ∂ z ˙ ∂ u ˙ u = 0 ∂ ˙ Then the ODE reads as (use v = ˙ x ): x = v ˙ � ˙ � � ∂ F � v x + ∂ F � � − 1 ∂ F ∂ F = − ∂ x ˙ ∂ u ˙ u ∂ ˙ z ˙ x ∂ z 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 12 / 25
DAE - Differential Index Definition The DAE differential index is the minimum i such that: How does the differential index relate to the DAE being ”easy” to solve ?? d i d t i F ( ˙ x , x , z , u ) = 0 is a pure ODE For an index-1 DAE : d x , x , z , u ) = ˙ d t F ( ˙ F = 0 yields a pure ODE. Observe that: An index-1 DAE has d F = ∂ F x + ∂ F x + ∂ F z + ∂ F d t F = ˙ x ¨ ∂ x ˙ ∂ z ˙ ∂ u ˙ u = 0 � � ∂ F ∂ F ∂ ˙ ∂ ˙ x ∂ z Then the ODE reads as (use v = ˙ x ): full rank and is therefore x = v ˙ ”easy” to solve !! � ˙ � � ∂ F � v x + ∂ F � � − 1 ∂ F ∂ F = − ∂ x ˙ ∂ u ˙ u ∂ ˙ z ˙ x ∂ z 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 12 / 25
Semi-explicit DAEs - Differential Index For a semi-explicit DAE the differential index is the minimum i such that: x = F ( x , z , u ) ˙ 0 = d i d t i G ( x , z , u ) is an ODE 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 13 / 25
Semi-explicit DAEs - Differential Index Remark: for an index-1 semi-explicit DAE: For a semi-explicit DAE the d t G ( x , z , u ) = ∂ G d ∂ x F + ∂ G z + ∂ G differential index is the ∂ z ˙ ∂ u ˙ u = 0 minimum i such that: yields a pure ODE. x = F ( x , z , u ) ˙ 0 = d i d t i G ( x , z , u ) is an ODE 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 13 / 25
Semi-explicit DAEs - Differential Index Remark: for an index-1 semi-explicit DAE: For a semi-explicit DAE the d t G ( x , z , u ) = ∂ G d ∂ x F + ∂ G z + ∂ G differential index is the ∂ z ˙ ∂ u ˙ u = 0 minimum i such that: yields a pure ODE. We have: x = F ( x , z , u ) ˙ − 1 � ∂ G � 0 = d i z = − ∂ G ∂ x F + ∂ G ˙ ∂ u ˙ u d t i G ( x , z , u ) ∂ z is an ODE such that ∂ G ∂ z is full rank !! 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 13 / 25
Semi-explicit DAEs - Differential Index Remark: for an index-1 semi-explicit DAE: For a semi-explicit DAE the d t G ( x , z , u ) = ∂ G d ∂ x F + ∂ G z + ∂ G differential index is the ∂ z ˙ ∂ u ˙ u = 0 minimum i such that: yields a pure ODE. We have: x = F ( x , z , u ) ˙ − 1 � ∂ G � 0 = d i z = − ∂ G ∂ x F + ∂ G ˙ ∂ u ˙ u d t i G ( x , z , u ) ∂ z is an ODE such that ∂ G ∂ z is full rank !! Example: � ˙ � 0 � � x 1 � 0 � � � x 1 1 = + z x 2 ˙ 0 0 x 2 1 0 = 1 � � x 2 1 + x 2 2 − 1 2 � �� � G ( x ) 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 13 / 25
Semi-explicit DAEs - Differential Index Remark: for an index-1 semi-explicit DAE: For a semi-explicit DAE the d t G ( x , z , u ) = ∂ G d ∂ x F + ∂ G z + ∂ G differential index is the ∂ z ˙ ∂ u ˙ u = 0 minimum i such that: yields a pure ODE. We have: x = F ( x , z , u ) ˙ − 1 � ∂ G � 0 = d i z = − ∂ G ∂ x F + ∂ G ˙ ∂ u ˙ u d t i G ( x , z , u ) ∂ z is an ODE such that ∂ G ∂ z is full rank !! Example: � ˙ � 0 � � x 1 � 0 � � � x 1 1 = + z x 2 ˙ 0 0 x 2 1 0 = 1 � � x 2 1 + x 2 2 − 1 2 � �� � G ( x ) Then d d t G = x 1 x 2 + x 2 z = 0 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 13 / 25
Semi-explicit DAEs - Differential Index Remark: for an index-1 semi-explicit DAE: For a semi-explicit DAE the d t G ( x , z , u ) = ∂ G d ∂ x F + ∂ G z + ∂ G differential index is the ∂ z ˙ ∂ u ˙ u = 0 minimum i such that: yields a pure ODE. We have: x = F ( x , z , u ) ˙ − 1 � ∂ G � 0 = d i z = − ∂ G ∂ x F + ∂ G ˙ ∂ u ˙ u d t i G ( x , z , u ) ∂ z is an ODE such that ∂ G ∂ z is full rank !! Example: � ˙ � 0 � � x 1 � 0 � � � x 1 1 = + z x 2 ˙ 0 0 x 2 1 0 = 1 � � x 2 1 + x 2 2 − 1 2 � �� � G ( x ) Then d d t G = x 1 x 2 + x 2 z = 0 d 2 d t 2 G = ˙ x 1 x 2 + x 1 ˙ x 2 + ˙ x 2 z + x 2 ˙ z = 0 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 13 / 25
Semi-explicit DAEs - Differential Index Remark: for an index-1 semi-explicit DAE: For a semi-explicit DAE the d t G ( x , z , u ) = ∂ G d ∂ x F + ∂ G z + ∂ G differential index is the ∂ z ˙ ∂ u ˙ u = 0 minimum i such that: yields a pure ODE. We have: x = F ( x , z , u ) ˙ − 1 � ∂ G � 0 = d i z = − ∂ G ∂ x F + ∂ G ˙ ∂ u ˙ u d t i G ( x , z , u ) ∂ z is an ODE such that ∂ G ∂ z is full rank !! Example: � ˙ � 0 � � x 1 � 0 � � � x 1 1 = + z x 2 ˙ 0 0 x 2 1 0 = 1 � � x 2 1 + x 2 2 − 1 2 � �� � G ( x ) Example is an index-2 DAE Then d d t G = x 1 x 2 + x 2 z = 0 d 2 d t 2 G = ˙ x 1 x 2 + x 1 ˙ x 2 + ˙ x 2 z + x 2 ˙ z = 0 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 13 / 25
For a semi-explicit DAE Differential Index - 3D pendulum the differential index is Example: 3D pendulum the minimum i such that: m ¨ p = u − mg e 3 − z p x = F ( x , z , u ) ˙ 0 = 1 � p ⊤ p − L 2 � 0 = d i 2 d t i G ( x , z , u ) � �� � G ( x ) is an ODE e 3 O e 1 e 2 u p 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 14 / 25
For a semi-explicit DAE Differential Index - 3D pendulum the differential index is Example: 3D pendulum the minimum i such that: m ¨ p = u − mg e 3 − z p x = F ( x , z , u ) ˙ 0 = 1 � p ⊤ p − L 2 � 0 = d i 2 d t i G ( x , z , u ) � �� � G ( x ) is an ODE Perform two time differentiations on G yields: e 3 � � G = 1 p ⊤ ˙ p ⊤ ˙ ¨ p ⊤ p + p ⊤ ¨ = p ⊤ ¨ ¨ p + 2 ˙ p + ˙ p = 0 p 2 O e 1 e 2 u p 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 14 / 25
For a semi-explicit DAE Differential Index - 3D pendulum the differential index is Example: 3D pendulum the minimum i such that: m ¨ p = u − mg e 3 − z p x = F ( x , z , u ) ˙ 0 = 1 � p ⊤ p − L 2 � 0 = d i 2 d t i G ( x , z , u ) � �� � G ( x ) is an ODE Perform two time differentiations on G yields: e 3 � � G = 1 p ⊤ ˙ p ⊤ ˙ ¨ p ⊤ p + p ⊤ ¨ = p ⊤ ¨ ¨ p + 2 ˙ p + ˙ p = 0 p 2 O e 1 Substitute ¨ p from m ¨ p = u − mg e 3 − z p yields: e 2 u � 1 � m u − g e 3 − 1 p ⊤ ˙ p ⊤ m z p + ˙ p = 0 p 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 14 / 25
For a semi-explicit DAE Differential Index - 3D pendulum the differential index is Example: 3D pendulum the minimum i such that: m ¨ p = u − mg e 3 − z p x = F ( x , z , u ) ˙ 0 = 1 � p ⊤ p − L 2 � 0 = d i 2 d t i G ( x , z , u ) � �� � G ( x ) is an ODE Perform two time differentiations on G yields: e 3 � � G = 1 p ⊤ ˙ p ⊤ ˙ ¨ p ⊤ p + p ⊤ ¨ = p ⊤ ¨ ¨ p + 2 ˙ p + ˙ p = 0 p 2 O e 1 Substitute ¨ p from m ¨ p = u − mg e 3 − z p yields: e 2 u � 1 � m u − g e 3 − 1 p ⊤ ˙ p ⊤ m z p + ˙ p = 0 p i.e. 1 � � p ⊤ ˙ p ⊤ u − mg p ⊤ e 3 + m ˙ z = p p ⊤ p 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 14 / 25
For a semi-explicit DAE Differential Index - 3D pendulum the differential index is Example: 3D pendulum the minimum i such that: m ¨ p = u − mg e 3 − z p x = F ( x , z , u ) ˙ 0 = 1 � p ⊤ p − L 2 � 0 = d i 2 d t i G ( x , z , u ) � �� � G ( x ) is an ODE Perform two time differentiations on G yields: e 3 � � G = 1 p ⊤ ˙ p ⊤ ˙ ¨ p ⊤ p + p ⊤ ¨ = p ⊤ ¨ ¨ p + 2 ˙ p + ˙ p = 0 p 2 O e 1 Substitute ¨ p from m ¨ p = u − mg e 3 − z p yields: e 2 u � 1 � m u − g e 3 − 1 p ⊤ ˙ p ⊤ m z p + ˙ p = 0 p i.e. 1 � � p ⊤ ˙ p ⊤ u − mg p ⊤ e 3 + m ˙ z = p p ⊤ p A third time differentiation yields an ODE for z : � �� � z = d 1 p ⊤ ˙ p ⊤ u − mg p ⊤ e 3 + m ˙ ˙ p d t p ⊤ p 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 14 / 25
For a semi-explicit DAE Differential Index - 3D pendulum the differential index is Example: 3D pendulum the minimum i such that: m ¨ p = u − mg e 3 − z p x = F ( x , z , u ) ˙ 0 = 1 � p ⊤ p − L 2 � 0 = d i 2 d t i G ( x , z , u ) � �� � G ( x ) is an ODE Perform two time differentiations on G yields: e 3 � � G = 1 p ⊤ ˙ p ⊤ ˙ ¨ p ⊤ p + p ⊤ ¨ = p ⊤ ¨ ¨ p + 2 ˙ p + ˙ p = 0 p 2 O e 1 Substitute ¨ p from m ¨ p = u − mg e 3 − z p yields: e 2 u � 1 � m u − g e 3 − 1 p ⊤ ˙ p ⊤ m z p + ˙ p = 0 p i.e. 1 � � p ⊤ ˙ p ⊤ u − mg p ⊤ e 3 + m ˙ z = p The 3D pendulum in p ⊤ p Lagrange is an index-3 A third time differentiation yields an ODE for z : DAE !! � �� � z = d 1 p ⊤ ˙ p ⊤ u − mg p ⊤ e 3 + m ˙ ˙ p d t p ⊤ p 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 14 / 25
For a semi-explicit DAE Differential Index - 3D pendulum the differential index is Example: 3D pendulum the minimum i such that: m ¨ p = u − mg e 3 − z p x = F ( x , z , u ) ˙ 0 = 1 � p ⊤ p − L 2 � 0 = d i 2 d t i G ( x , z , u ) � �� � G ( x ) is an ODE Perform two time differentiations on G yields: e 3 � � G = 1 p ⊤ ˙ p ⊤ ˙ ¨ p ⊤ p + p ⊤ ¨ = p ⊤ ¨ ¨ p + 2 ˙ p + ˙ p = 0 p 2 O e 1 Assemble: e 2 u m ¨ p + z p = u − mg e 3 p ⊤ ˙ p ⊤ ¨ p = − ˙ p p 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 14 / 25
For a semi-explicit DAE Differential Index - 3D pendulum the differential index is Example: 3D pendulum the minimum i such that: m ¨ p = u − mg e 3 − z p x = F ( x , z , u ) ˙ 0 = 1 � p ⊤ p − L 2 � 0 = d i 2 d t i G ( x , z , u ) � �� � G ( x ) is an ODE Perform two time differentiations on G yields: e 3 � � G = 1 p ⊤ ˙ p ⊤ ˙ ¨ p ⊤ p + p ⊤ ¨ = p ⊤ ¨ ¨ p + 2 ˙ p + ˙ p = 0 p 2 O e 1 Assemble: e 2 u m ¨ p + z p = u − mg e 3 p ⊤ ˙ p ⊤ ¨ p = − ˙ p p in matrix form yields: � mI � � ¨ � u − mg e 3 � � p p = p ⊤ ˙ p ⊤ 0 z − ˙ p 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 14 / 25
For a semi-explicit DAE Differential Index - 3D pendulum the differential index is Example: 3D pendulum the minimum i such that: m ¨ p = u − mg e 3 − z p x = F ( x , z , u ) ˙ 0 = 1 � p ⊤ p − L 2 � 0 = d i 2 d t i G ( x , z , u ) � �� � G ( x ) is an ODE Perform two time differentiations on G yields: e 3 � � G = 1 p ⊤ ˙ p ⊤ ˙ ¨ p ⊤ p + p ⊤ ¨ = p ⊤ ¨ ¨ p + 2 ˙ p + ˙ p = 0 p 2 O e 1 Assemble: e 2 u m ¨ p + z p = u − mg e 3 p ⊤ ˙ p ⊤ ¨ p = − ˙ p p in matrix form yields: � mI � � ¨ � u − mg e 3 � � p p = p ⊤ ˙ p ⊤ 0 z − ˙ p This is an index-1 (i.e. ”easy”) DAE !! 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 14 / 25
For a semi-explicit DAE Differential Index - 3D pendulum the differential index is Example: 3D pendulum the minimum i such that: m ¨ p = u − mg e 3 − z p x = F ( x , z , u ) ˙ 0 = 1 � p ⊤ p − L 2 � 0 = d i 2 d t i G ( x , z , u ) � �� � G ( x ) is an ODE Perform two time differentiations on G yields: e 3 � � G = 1 p ⊤ ˙ p ⊤ ˙ ¨ p ⊤ p + p ⊤ ¨ = p ⊤ ¨ ¨ p + 2 ˙ p + ˙ p = 0 p 2 O e 1 Assemble: e 2 u m ¨ p + z p = u − mg e 3 p ⊤ ˙ p ⊤ ¨ p = − ˙ p p in matrix form yields: � mI � � ¨ � u − mg e 3 � � p p We have converted the = p ⊤ ˙ p ⊤ 0 z − ˙ index-3 DAE into an p index-1 DAE !! This is an index-1 (i.e. ”easy”) DAE !! 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 14 / 25
For a semi-explicit DAE Differential Index - 3D pendulum the differential index is Example: 3D pendulum the minimum i such that: m ¨ p = u − mg e 3 − z p x = F ( x , z , u ) ˙ 0 = 1 � p ⊤ p − L 2 � 0 = d i 2 d t i G ( x , z , u ) � �� � G ( x ) is an ODE Perform two time differentiations on G yields: e 3 � � G = 1 p ⊤ ˙ p ⊤ ˙ ¨ p ⊤ p + p ⊤ ¨ = p ⊤ ¨ ¨ p + 2 ˙ p + ˙ p = 0 p 2 O e 1 e 2 u Transforming a high-index DAE into an equivalent lower-index one is labelled index reduction p We have converted the index-3 DAE into an index-1 DAE !! 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 14 / 25
Outline ”Easy” & ”Hard” DAEs 1 Differential Index 2 Index Reduction 3 Constraints drift 4 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 15 / 25
DAEs from Lagrange Mechanics Index-3 DAE from Lagrange: d ∂ L q − ∂ L ∂ q = F g d ∂ ˙ c ( q ) = 0 q ) − V ( q ) − z ⊤ c ( q ) with L ( q , ˙ q , z ) = T ( q , ˙ 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 16 / 25
DAEs from Lagrange Mechanics For most mechanical applications: Index-3 DAE from Lagrange: q ) = 1 d ∂ L q − ∂ L q ⊤ M ( q ) ˙ ∂ q = F g T ( q , ˙ 2 ˙ q d ∂ ˙ c ( q ) = 0 such that: ⊤ d ∂ L q ) − V ( q ) − z ⊤ c ( q ) with L ( q , ˙ q , z ) = T ( q , ˙ q + ˙ = M ( q )¨ M ( q , ˙ q ) ˙ q d ∂ ˙ q 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 16 / 25
DAEs from Lagrange Mechanics For most mechanical applications: Index-3 DAE from Lagrange: q ) = 1 d ∂ L q − ∂ L q ⊤ M ( q ) ˙ ∂ q = F g T ( q , ˙ 2 ˙ q d ∂ ˙ c ( q ) = 0 such that: ⊤ d ∂ L q ) − V ( q ) − z ⊤ c ( q ) with L ( q , ˙ q , z ) = T ( q , ˙ q + ˙ = M ( q )¨ M ( q , ˙ q ) ˙ q d ∂ ˙ q Then the differential part of the DAE model reads as: q + ˙ M ( q )¨ M ( q , ˙ q ) ˙ q − ∇ q ( T ( q , ˙ q ) − V ( q )) + ∇ c ( q ) z = F g 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 16 / 25
DAEs from Lagrange Mechanics For most mechanical applications: Index-3 DAE from Lagrange: q ) = 1 d ∂ L q − ∂ L q ⊤ M ( q ) ˙ ∂ q = F g T ( q , ˙ 2 ˙ q d ∂ ˙ c ( q ) = 0 such that: ⊤ d ∂ L q ) − V ( q ) − z ⊤ c ( q ) with L ( q , ˙ q , z ) = T ( q , ˙ q + ˙ = M ( q )¨ M ( q , ˙ q ) ˙ q d ∂ ˙ q Then the differential part of the DAE model reads as: q + ˙ M ( q )¨ M ( q , ˙ q ) ˙ q − ∇ q ( T ( q , ˙ q ) − V ( q )) + ∇ c ( q ) z = F g The 1 st and 2 nd -order time derivatives of c ( q ) read as: d d t c ( q ) = ∇ c ( q ) ⊤ ˙ q , 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 16 / 25
DAEs from Lagrange Mechanics For most mechanical applications: Index-3 DAE from Lagrange: q ) = 1 d ∂ L q − ∂ L q ⊤ M ( q ) ˙ ∂ q = F g T ( q , ˙ 2 ˙ q d ∂ ˙ c ( q ) = 0 such that: ⊤ d ∂ L q ) − V ( q ) − z ⊤ c ( q ) with L ( q , ˙ q , z ) = T ( q , ˙ q + ˙ = M ( q )¨ M ( q , ˙ q ) ˙ q d ∂ ˙ q Then the differential part of the DAE model reads as: q + ˙ M ( q )¨ M ( q , ˙ q ) ˙ q − ∇ q ( T ( q , ˙ q ) − V ( q )) + ∇ c ( q ) z = F g The 1 st and 2 nd -order time derivatives of c ( q ) read as: d 2 � � ⊤ d d t c ( q ) = ∇ c ( q ) ⊤ ˙ d t 2 c ( q ) = ∇ c ( q ) ⊤ ¨ ∇ c ( q ) ⊤ ˙ q , q + ∇ q q q ˙ 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 16 / 25
DAEs from Lagrange Mechanics For most mechanical applications: Index-3 DAE from Lagrange: q ) = 1 d ∂ L q − ∂ L q ⊤ M ( q ) ˙ ∂ q = F g T ( q , ˙ 2 ˙ q d ∂ ˙ c ( q ) = 0 such that: ⊤ d ∂ L q ) − V ( q ) − z ⊤ c ( q ) with L ( q , ˙ q , z ) = T ( q , ˙ q + ˙ = M ( q )¨ M ( q , ˙ q ) ˙ q d ∂ ˙ q Then the differential part of the DAE model reads as: q + ˙ M ( q )¨ M ( q , ˙ q ) ˙ q − ∇ q ( T ( q , ˙ q ) − V ( q )) + ∇ c ( q ) z = F g The 1 st and 2 nd -order time derivatives of c ( q ) read as: d 2 � � ⊤ d d t c ( q ) = ∇ c ( q ) ⊤ ˙ d t 2 c ( q ) = ∇ c ( q ) ⊤ ¨ ∇ c ( q ) ⊤ ˙ q , q + ∇ q q q ˙ Index-1 DAE model: � F g − ˙ � � ¨ � � � M ( q ) ∇ q c ( q ) M ( q , ˙ q ) ˙ q + ∇ q ( T ( q , ˙ q ) − V ( q )) q = � � ⊤ ∇ q c ( q ) ⊤ ˙ z ∇ q c ( q ) ⊤ −∇ q ˙ q q 0 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 16 / 25
DAEs from Lagrange Mechanics For most mechanical applications: Index-3 DAE from Lagrange: q ) = 1 d ∂ L q − ∂ L q ⊤ M ( q ) ˙ ∂ q = F g T ( q , ˙ 2 ˙ q d ∂ ˙ c ( q ) = 0 such that: ⊤ d ∂ L q ) − V ( q ) − z ⊤ c ( q ) with L ( q , ˙ q , z ) = T ( q , ˙ q + ˙ = M ( q )¨ M ( q , ˙ q ) ˙ q d ∂ ˙ q Index-1 DAE model: � F g − ˙ � � ¨ � � � M ( q ) ∇ c ( q ) M ( q , ˙ q ) ˙ q + ∇ q ( T ( q , ˙ q ) − V ( q )) q = � � ⊤ ∇ c ( q ) ⊤ ˙ z ∇ c ( q ) ⊤ −∇ q ˙ q q 0 Models based on Lagrange mechanics typically are index-3 DAEs , making them intrinsically difficult to use. The best approach to treat them is usually to proceed with an index reduction down to index 1 for which very classical integration tools work well. 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 16 / 25
Index reduction for semi-explicit DAEs - A general view High-index semi-explicit DAE x = F ( x , z , u ) ˙ 0 = G ( x , z , u ) Algorithm ( see ”Nonlinear Programming”, L.T. Biegler ) 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 17 / 25
Index reduction for semi-explicit DAEs - A general view High-index semi-explicit DAE x = F ( x , z , u ) ˙ 0 = G ( x , z , u ) Algorithm ( see ”Nonlinear Programming”, L.T. Biegler ) ∂ G Check if the DAE system is index 1 (i.e. ∂ z full rank). 1 If yes, stop. Identify a subset of algebraic equations that can be 2 solved for a subset of algebraic variables. d Apply d t on the remaining algebraic equations that 3 contain the differential variables x j . Terms ˙ x j will appear in these differentiated equations. 4 Substitute the ˙ x j with F j ( x , z , u ). This leads to new 5 algebraic equations. With this new DAE system, go to step 1. 6 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 17 / 25
Index reduction for semi-explicit DAEs - A general view High-index semi-explicit DAE x = F ( x , z , u ) ˙ 0 = G ( x , z , u ) Algorithm ( see ”Nonlinear Programming”, L.T. Biegler ) ∂ G Check if the DAE system is index 1 (i.e. ∂ z full rank). 1 If yes, stop. Identify a subset of algebraic equations that can be 2 solved for a subset of algebraic variables. d Apply d t on the remaining algebraic equations that 3 contain the differential variables x j . Terms ˙ x j will appear in these differentiated equations. 4 Substitute the ˙ x j with F j ( x , z , u ). This leads to new 5 algebraic equations. With this new DAE system, go to step 1. 6 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 17 / 25
Index reduction for semi-explicit DAEs - A general view High-index semi-explicit DAE x = F ( x , z , u ) ˙ 0 = G ( x , z , u ) Algorithm ( see ”Nonlinear Programming”, L.T. Biegler ) ∂ G Check if the DAE system is index 1 (i.e. ∂ z full rank). 1 If yes, stop. Writing a general-purpose Identify a subset of algebraic equations that can be ”Index-reduction 2 solved for a subset of algebraic variables. algorithm” can be very tricky, as one of the steps d Apply d t on the remaining algebraic equations that 3 is not easily automated contain the differential variables x j . Terms ˙ x j will appear in these differentiated equations. 4 Substitute the ˙ x j with F j ( x , z , u ). This leads to new 5 algebraic equations. With this new DAE system, go to step 1. 6 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 17 / 25
e 3 DAE Consistency - 3D pendulum O e 1 Does the index reduction really yield equivalent models ? e 2 u p Index-3 DAE Index-1 DAE � mI � u − mg e 3 � � � � ¨ p p m ¨ p = u − mg e 3 − z p = p ⊤ ˙ p ⊤ 0 z − ˙ p � p ⊤ p − L 2 � 0 = 1 2 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 18 / 25
e 3 DAE Consistency - 3D pendulum O e 1 Does the index reduction really yield equivalent models ? e 2 u p Index-3 DAE Index-1 DAE � mI � u − mg e 3 � � � � ¨ p p m ¨ p = u − mg e 3 − z p = p ⊤ ˙ p ⊤ 0 z − ˙ p � p ⊤ p − L 2 � 0 = 1 2 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 18 / 25
e 3 DAE Consistency - 3D pendulum O e 1 Does the index reduction really yield equivalent models ? e 2 u p Index-3 DAE Index-1 DAE � mI � u − mg e 3 � � � � ¨ p p m ¨ p = u − mg e 3 − z p = p ⊤ ˙ p ⊤ 0 z − ˙ p � p ⊤ p − L 2 � 0 = 1 2 What is going on ?? 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 18 / 25
DAE Consistency - 3D pendulum Index-3 DAE m ¨ p = u − mg e 3 − z p 0 = 1 � p ⊤ p − L 2 � 2 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 19 / 25
DAE Consistency - 3D pendulum Index-3 DAE m ¨ p = u − mg e 3 − z p 0 = 1 � p ⊤ p − L 2 � 2 Index reduction c = 1 � p ⊤ p − L 2 � 2 c = p ⊤ ˙ ˙ p p ⊤ ˙ c = p ⊤ ¨ ¨ p + ˙ p 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 19 / 25
DAE Consistency - 3D pendulum Index-3 DAE m ¨ p = u − mg e 3 − z p 0 = 1 � p ⊤ p − L 2 � 2 Index reduction c = 1 � p ⊤ p − L 2 � 2 c = p ⊤ ˙ ˙ p p ⊤ ˙ c = p ⊤ ¨ ¨ p + ˙ p Index-1 DAE � � ¨ � mI � u − mg e 3 � � p p = p ⊤ ˙ p ⊤ 0 − ˙ z p ... is built to impose ¨ c = 0 at all time. 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 19 / 25
DAE Consistency - 3D pendulum Index-3 DAE m ¨ p = u − mg e 3 − z p 0 = 1 � p ⊤ p − L 2 � 2 Index reduction c = 1 � p ⊤ p − L 2 � 2 c = p ⊤ ˙ ˙ p p ⊤ ˙ c = p ⊤ ¨ ¨ p + ˙ p Index-1 DAE � � ¨ � mI � u − mg e 3 � � p p = p ⊤ ˙ p ⊤ 0 − ˙ z p ... is built to impose ¨ c = 0 at all time. But it does not ensure c = 0 ˙ and c = 0 !! 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 19 / 25
DAE Consistency - 3D pendulum Index-3 DAE m ¨ p = u − mg e 3 − z p 0 = 1 � p ⊤ p − L 2 � 2 Index reduction c = 1 � p ⊤ p − L 2 � 2 4 c = p ⊤ ˙ ˙ p 3 p ⊤ ˙ 2 c = p ⊤ ¨ c ¨ p + ˙ p 1 0 Index-1 DAE � � ¨ 0 5 10 � mI � u − mg e 3 � � p p t = p ⊤ ˙ p ⊤ 0 − ˙ z p 0.5 ... is built to impose ¨ c = 0 at all time. But it does 0 c ˙ not ensure c = 0 ˙ and c = 0 !! -0.5 0 5 10 t 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 19 / 25
DAE Consistency - 3D pendulum Index-3 DAE m ¨ p = u − mg e 3 − z p 0 = 1 � p ⊤ p − L 2 � 2 Index reduction c = 1 � p ⊤ p − L 2 � 2 4 c = p ⊤ ˙ ˙ p 3 p ⊤ ˙ 2 c = p ⊤ ¨ c ¨ p + ˙ p 1 0 Index-1 DAE � � ¨ 0 5 10 � mI � u − mg e 3 � � p p t = p ⊤ ˙ p ⊤ 0 − ˙ z p 0.5 ... is built to impose ¨ c = 0 at all time. But it does 0 c ˙ not ensure c = 0 ˙ and c = 0 !! -0.5 0 5 10 t How can we address that ?? 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 19 / 25
DAE Consistency - 3D pendulum Index-1 DAE � � ¨ � mI � u − mg e 3 � � p p = p ⊤ ˙ p ⊤ 0 − ˙ z p ... is built to impose ¨ c = 0 at all time. 4 3 2 c 1 0 0 5 10 t 0.5 0 c ˙ -0.5 0 5 10 t 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 20 / 25
DAE Consistency - 3D pendulum Index-1 DAE � � ¨ � mI � u − mg e 3 � � p p = p ⊤ ˙ p ⊤ 0 − ˙ z p ... is built to impose ¨ c = 0 at all time. Then if c = 0 and ˙ c = 0 are satisfied at any time on 4 the trajectory, then they are satisfied at all time. 3 2 c 1 0 0 5 10 t 0.5 0 c ˙ -0.5 0 5 10 t 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 20 / 25
DAE Consistency - 3D pendulum Index-1 DAE � � ¨ � mI � u − mg e 3 � � p p = p ⊤ ˙ p ⊤ 0 − ˙ z p ... is built to impose ¨ c = 0 at all time. Then if c = 0 and ˙ c = 0 are satisfied at any time on 4 the trajectory, then they are satisfied at all time. 3 2 c An index-reduced DAE must come with consistency 1 conditions . E.g. for the 3D pendulum, the index-1 0 DAE should be given as: 0 5 10 t � � ¨ � mI � u − mg e 3 � � p p 0.5 = p ⊤ ˙ p ⊤ 0 z − ˙ p 0 c ˙ -0.5 0 5 10 t 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 20 / 25
DAE Consistency - 3D pendulum Index-1 DAE � � ¨ � mI � u − mg e 3 � � p p = p ⊤ ˙ p ⊤ 0 − ˙ z p ... is built to impose ¨ c = 0 at all time. Then if c = 0 and ˙ c = 0 are satisfied at any time on 4 the trajectory, then they are satisfied at all time. 3 2 c An index-reduced DAE must come with consistency 1 conditions . E.g. for the 3D pendulum, the index-1 0 DAE should be given as: 0 5 10 t � � ¨ � mI � u − mg e 3 � � p p 0.5 = p ⊤ ˙ p ⊤ 0 z − ˙ p 0 c ˙ with the consistency conditions: � p ⊤ p − L 2 � c = 1 c = p ⊤ ˙ -0.5 = 0 , ˙ p = 0 0 5 10 2 t ... to be satisfied e.g. at t 0 . 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 20 / 25
DAE Consistency - 3D pendulum Index-1 DAE � � ¨ � mI � u − mg e 3 � � p p = p ⊤ ˙ p ⊤ 0 − ˙ z p ... is built to impose ¨ c = 0 at all time. Then if c = 0 and ˙ c = 0 are satisfied at any time on 4 the trajectory, then they are satisfied at all time. 3 2 c An index-reduced DAE must come with consistency 1 conditions . E.g. for the 3D pendulum, the index-1 0 DAE should be given as: 0 5 10 t � � ¨ � mI � u − mg e 3 � � p p 0.5 = p ⊤ ˙ p ⊤ 0 z − ˙ p 0 c ˙ with the consistency conditions: � p ⊤ p − L 2 � c = 1 c = p ⊤ ˙ -0.5 = 0 , ˙ p = 0 0 5 10 2 t ... to be satisfied e.g. at t 0 . 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 20 / 25
Consistency of DAEs from Lagrange Mechanics Index-3 DAE from Lagrange: ∂ L q − ∂ L d ∂ q = F g For most mechanical applications: ∂ ˙ d q ) = 1 c ( q ) = 0 q ⊤ M ( q ) ˙ T ( q , ˙ 2 ˙ q q ) − V ( q ) − z ⊤ c ( q ) with L ( q , ˙ q , z ) = T ( q , ˙ Index reduction based on: d 2 � � ⊤ d d t c ( q ) = ∇ c ( q ) ⊤ ˙ d t 2 c ( q ) = ∇ c ( q ) ⊤ ¨ ∇ c ( q ) ⊤ ˙ and q + ∇ q ˙ q q q Index-1 DAE model: � F g − ˙ � � ¨ � � M ( q ) ∇ q c ( q ) � M ( q , ˙ q ) ˙ q + ∇ q ( T ( q , ˙ q ) − V ( q )) q = � � ⊤ ∇ c ( q ) ⊤ ˙ z ∇ q c ( q ) ⊤ −∇ q q q ˙ 0 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 21 / 25
Consistency of DAEs from Lagrange Mechanics Index-3 DAE from Lagrange: ∂ L q − ∂ L d ∂ q = F g For most mechanical applications: ∂ ˙ d q ) = 1 c ( q ) = 0 q ⊤ M ( q ) ˙ T ( q , ˙ 2 ˙ q q ) − V ( q ) − z ⊤ c ( q ) with L ( q , ˙ q , z ) = T ( q , ˙ Index reduction based on: d 2 � � ⊤ d d t c ( q ) = ∇ c ( q ) ⊤ ˙ d t 2 c ( q ) = ∇ c ( q ) ⊤ ¨ ∇ c ( q ) ⊤ ˙ and q + ∇ q ˙ q q q Index-1 DAE model: � F g − ˙ � � ¨ � � M ( q ) ∇ q c ( q ) � M ( q , ˙ q ) ˙ q + ∇ q ( T ( q , ˙ q ) − V ( q )) q = � � ⊤ ∇ c ( q ) ⊤ ˙ z ∇ q c ( q ) ⊤ −∇ q q q ˙ 0 with the consistency conditions: d d t c ( q ) = ∇ c ( q ) ⊤ ˙ c ( q ) = 0 and q 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 21 / 25
Outline ”Easy” & ”Hard” DAEs 1 Differential Index 2 Index Reduction 3 Constraints drift 4 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 22 / 25
Constraints drift - 3D pendulum Index-1 DAE � mI � � ¨ � u − mg e 3 � � p p = p ⊤ ˙ p ⊤ 0 z − ˙ p ... is built to impose ¨ c = 0 at all time. Then if c = 0 and ˙ c = 0 are satisfied at any time on the trajectory, then they are satisfied at all time. 4 3 2 c 1 0 0 5 10 t 0.5 0 c ˙ -0.5 0 5 10 t 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 23 / 25
Constraints drift - 3D pendulum Index-1 DAE � mI � � ¨ � u − mg e 3 � � p p = p ⊤ ˙ p ⊤ 0 z − ˙ p ... is built to impose ¨ c = 0 at all time. Then if c = 0 and ˙ c = 0 are satisfied at any time on the trajectory, then they are satisfied at all time. 4 3 Index-1 DAE : 2 c � mI � � ¨ � u − mg e 3 1 � � p p 0 = p ⊤ ˙ p ⊤ 0 z − ˙ p 0 5 10 t 0.5 0 c ˙ -0.5 0 5 10 t 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 23 / 25
Constraints drift - 3D pendulum Index-1 DAE � mI � � ¨ � u − mg e 3 � � p p = p ⊤ ˙ p ⊤ 0 z − ˙ p ... is built to impose ¨ c = 0 at all time. Then if c = 0 and ˙ c = 0 are satisfied at any time on the trajectory, then they are satisfied at all time. 4 3 Index-1 DAE : 2 c � mI � � ¨ � u − mg e 3 1 � � p p 0 = p ⊤ ˙ p ⊤ 0 z − ˙ p 0 5 10 t with the consistency conditions : 0.5 c = 1 � p ⊤ p − L 2 � c = p ⊤ ˙ = 0 , ˙ p = 0 2 0 c ˙ ... imposed at e.g. t 0 . -0.5 0 5 10 t 22 nd of February, 2016 S. Gros Optimal Control with DAEs, lecture 11 23 / 25
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