Small time heat kernel expansion for a class of model hypoelliptic H - PowerPoint PPT Presentation

An explicit form of the heat kernel Theorem Let A ∈ R n × n and B ∈ R n × k with the rank condition � � B , AB , A 2 B , · · · , A m − 1 B rank = n for some m ∈ N . Then: The heat operator L − ∂ ∂ t is hypoelliptic and admits a smooth fundamental solution p ( t ; x , y ) ∈ C ∞ � R + × R n × R n � ( ∗ ) . W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 8 / 33

An explicit form of the heat kernel Theorem Let A ∈ R n × n and B ∈ R n × k with the rank condition � � B , AB , A 2 B , · · · , A m − 1 B rank = n for some m ∈ N . Then: The heat operator L − ∂ ∂ t is hypoelliptic and admits a smooth fundamental solution p ( t ; x , y ) ∈ C ∞ � R + × R n × R n � ( ∗ ) . The kernel ( ∗ ) is explicitly known: 1 � − 1 � � y − e tA x ) ∗ D − 1 ( y − e tA x ) p ( t ; x , y ) = 2 √ det D t exp , n t 2 (2 π ) W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 8 / 33

An explicit form of the heat kernel Theorem (continued) Here D t ∈ R n × n is the matrix: �� t � D t = e tA e − sA BB ∗ e − sA ∗ ds e tA ∗ . 0 W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 9 / 33

An explicit form of the heat kernel Theorem (continued) Here D t ∈ R n × n is the matrix: �� t � D t = e tA e − sA BB ∗ e − sA ∗ ds e tA ∗ . 0 In particular, this matrix is invertible for all t > 0 (we will prove this later). Next goal: From what is known in the elliptic set-up (next slide), one expect that, that the small time expansion of the heat kernel includes some geometric data and data on the drift term X 0 . W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 9 / 33

An explicit form of the heat kernel Theorem (continued) Here D t ∈ R n × n is the matrix: �� t � D t = e tA e − sA BB ∗ e − sA ∗ ds e tA ∗ . 0 In particular, this matrix is invertible for all t > 0 (we will prove this later). Next goal: From what is known in the elliptic set-up (next slide), one expect that, that the small time expansion of the heat kernel includes some geometric data and data on the drift term X 0 . Q: What happens in the sub-elliptic case? W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 9 / 33

Laplace operator with drift term Theorem Let ( M , g ) be a Riemannian manifold with Laplacian ∆ g and L = ∆ g + X 0 . W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 10 / 33

Laplace operator with drift term Theorem Let ( M , g ) be a Riemannian manifold with Laplacian ∆ g and L = ∆ g + X 0 . Then the heat kernel of L has the following on-diagonal asymptotic expansion for small times: � � div( X 0 ) � � + � X 0 ( x 0 ) � 2 1 − S ( x 0 ) t + O ( t 2 ) p ( t ; x 0 , x 0 ) = 1 − , n 2 2 6 (4 π t ) 2 where S denotes the scalar curvature of the Riemannian metric g. W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 10 / 33

Basics on optimal control problems Let Ω ⊂ R k be a set and f : R n × Ω → R n . With a given control function α : [0 , ∞ ) → Ω and x 0 ∈ R n consider the system of ODE: � � � x ( t ) = ˙ f x ( t ) , α ( t ) , t > 0 ( ∗ ) x (0) = x 1 . We will call the solution the response of the system. W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 11 / 33

Basics on optimal control problems Let Ω ⊂ R k be a set and f : R n × Ω → R n . With a given control function α : [0 , ∞ ) → Ω and x 0 ∈ R n consider the system of ODE: � � � x ( t ) = ˙ f x ( t ) , α ( t ) , t > 0 ( ∗ ) x (0) = x 1 . We will call the solution the response of the system. Definition The set of admissible controls is � � A := α : [0 , ∞ ) → Ω : α measurable . W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 11 / 33

Basics on optimal control There is the question of controllability: W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 12 / 33

Basics on optimal control There is the question of controllability: Controllability problem (special case) Given an initial point x 1 ∈ R n and an end point x 2 ∈ R n . Does there exist a control α ( t ) and a time t 0 > 0 with x ( t 0 ) = x 2 ∈ R n , where x ( t ) is a solution of the system ( ∗ ) of ODE’s? W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 12 / 33

Basics on optimal control There is the question of controllability: Controllability problem (special case) Given an initial point x 1 ∈ R n and an end point x 2 ∈ R n . Does there exist a control α ( t ) and a time t 0 > 0 with x ( t 0 ) = x 2 ∈ R n , where x ( t ) is a solution of the system ( ∗ ) of ODE’s? For our later purpose it is sufficient to consider linear systems where we can answer the question: Let A = ( a jh ) ∈ R n × n and B = ( b il ) ∈ R n × k . With T > 0 consider: W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 12 / 33

Basics on optimal control There is the question of controllability: Controllability problem (special case) Given an initial point x 1 ∈ R n and an end point x 2 ∈ R n . Does there exist a control α ( t ) and a time t 0 > 0 with x ( t 0 ) = x 2 ∈ R n , where x ( t ) is a solution of the system ( ∗ ) of ODE’s? For our later purpose it is sufficient to consider linear systems where we can answer the question: Let A = ( a jh ) ∈ R n × n and B = ( b il ) ∈ R n × k . With T > 0 consider: � ˙ x = A x + Bu u ∈ L ∞ � [0 , T ] , R k � where ( ∗∗ ) . = x 1 ∈ R n , x (0) W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 12 / 33

A linear control problem For a given control u we write x u : [0 , T ] → R n for the solution of the initial value problem ( ∗∗ ). These are the admissible curves. W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 13 / 33

A linear control problem For a given control u we write x u : [0 , T ] → R n for the solution of the initial value problem ( ∗∗ ). These are the admissible curves. Solution: � t x u ( t ) = e tA x 1 + e tA e − sA Bu ( s ) ds . 0 Lemma The following are equivalent: W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 13 / 33

A linear control problem For a given control u we write x u : [0 , T ] → R n for the solution of the initial value problem ( ∗∗ ). These are the admissible curves. Solution: � t x u ( t ) = e tA x 1 + e tA e − sA Bu ( s ) ds . 0 Lemma The following are equivalent: (a) A solution to the controllability problem for ( ∗∗ ) with end point x 2 ∈ R n and time T > 0 exists. W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 13 / 33

A linear control problem For a given control u we write x u : [0 , T ] → R n for the solution of the initial value problem ( ∗∗ ). These are the admissible curves. Solution: � t x u ( t ) = e tA x 1 + e tA e − sA Bu ( s ) ds . 0 Lemma The following are equivalent: (a) A solution to the controllability problem for ( ∗∗ ) with end point x 2 ∈ R n and time T > 0 exists. (b) There is a control u ∈ L ∞ ([0 , T ] , R k ) such that � T x 2 = e TA x 1 + e TA e − sA Bu ( s ) ds . 0 W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 13 / 33

Linear control problem Example Consider the following special case: Let ( x 0 , y 0 ) t = 0 and � � � � � � � � � � x ˙ x u 1 0 0 0 = A + B = where A = 0 B = . y ˙ y u 2 u 2 0 1 Since the x -component of a solution is constant end points ( x 1 , y 1 ) with x 1 � = 0 cannot be reached for any control u ” . W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 14 / 33

Linear control problem Example Consider the following special case: Let ( x 0 , y 0 ) t = 0 and � � � � � � � � � � x ˙ x u 1 0 0 0 = A + B = where A = 0 B = . y ˙ y u 2 u 2 0 1 Since the x -component of a solution is constant end points ( x 1 , y 1 ) with x 1 � = 0 cannot be reached for any control u ” . Let t > 0 and consider the following two reachable sets: C ( t ) : = initial points x 0 for which there is a control u such that x u ( t ) = 0 . � C : = C ( t ) = overall reachable set . t > 0 W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 14 / 33

Kalman condition There is an algebraic condition which guarantees that C is a zero-neighbourhood. W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 15 / 33

Kalman condition There is an algebraic condition which guarantees that C is a zero-neighbourhood. Definition The controllability matrix for the system ( ∗∗ ) is defined by: � � B , AB , A 2 B , · · · , A n − 1 B G ( A , B ) = . � �� � = n × ( n · k ) − matrix W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 15 / 33

Kalman condition There is an algebraic condition which guarantees that C is a zero-neighbourhood. Definition The controllability matrix for the system ( ∗∗ ) is defined by: � � B , AB , A 2 B , · · · , A n − 1 B G ( A , B ) = . � �� � = n × ( n · k ) − matrix Theorem (rank condition) The following statements are equivalent: a (i) rank G ( A , B ) = n, ◦ (ii) 0 ∈ C (interior of C ). a J. Macki, A. Strauss, introduction to optimal control , Springer, 1982 W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 15 / 33

Optimal control problem Now we add a cost functional to the controlled ODE. With T > 0 consider � u = ( u 1 , · · · , u k ) ∈ L ∞ ([0 , T ] , R k ) = A x + B u , where x ˙ � T � k = 1 i =1 | u i ( s ) | 2 ds . J T ( u ) 2 0 W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 16 / 33

Optimal control problem Now we add a cost functional to the controlled ODE. With T > 0 consider � u = ( u 1 , · · · , u k ) ∈ L ∞ ([0 , T ] , R k ) = A x + B u , where x ˙ � T � k = 1 i =1 | u i ( s ) | 2 ds . J T ( u ) 2 0 Problem : Among all solutions x u := [0 , T ] → R n corresponding to the control u we want to minimize the cost J T ( u ) . W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 16 / 33

Optimal control problem Now we add a cost functional to the controlled ODE. With T > 0 consider � u = ( u 1 , · · · , u k ) ∈ L ∞ ([0 , T ] , R k ) = A x + B u , where x ˙ � T � k = 1 i =1 | u i ( s ) | 2 ds . J T ( u ) 2 0 Problem : Among all solutions x u := [0 , T ] → R n corresponding to the control u we want to minimize the cost J T ( u ) . Consider the value function: � � J T ( u ) : u ∈ L ∞ ([0 , T ] , R k ) , x u (0) = x 1 , x u ( T ) = x 2 S T ( x 1 , x 2 ) = inf . This function is finite for all T > 0 and x 1 , x 2 ∈ R n by the rank condition. W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 16 / 33

Optimal control problem Now we add a cost functional to the controlled ODE. With T > 0 consider � u = ( u 1 , · · · , u k ) ∈ L ∞ ([0 , T ] , R k ) = A x + B u , where x ˙ � T � k = 1 i =1 | u i ( s ) | 2 ds . J T ( u ) 2 0 Problem : Among all solutions x u := [0 , T ] → R n corresponding to the control u we want to minimize the cost J T ( u ) . Consider the value function: � � J T ( u ) : u ∈ L ∞ ([0 , T ] , R k ) , x u (0) = x 1 , x u ( T ) = x 2 S T ( x 1 , x 2 ) = inf . This function is finite for all T > 0 and x 1 , x 2 ∈ R n by the rank condition. Definition A control u that realizes the minimum is called an optimal control. The corresponding trajectory x u : [0 , T ] → R n is an optimal trajectory. W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 16 / 33

Optimal control problem Q: How to find an optimal control? We assign to the optimal control problem an Hamiltonian, i.e. a function on the cotangent bundle: H ( x , p ) := p ∗ Ax + 1 ( x , p ) ∈ T ∗ R n ∼ = R 2 n . 2 p ∗ BB ∗ p , where W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 17 / 33

Optimal control problem Q: How to find an optimal control? We assign to the optimal control problem an Hamiltonian, i.e. a function on the cotangent bundle: H ( x , p ) := p ∗ Ax + 1 ( x , p ) ∈ T ∗ R n ∼ = R 2 n . 2 p ∗ BB ∗ p , where This induces a Hamilton system: � = − H x = − A ∗ p p ˙ (HS) = H p = Ax + BB ∗ p . x ˙ W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 17 / 33

Optimal control problem Q: How to find an optimal control? We assign to the optimal control problem an Hamiltonian, i.e. a function on the cotangent bundle: H ( x , p ) := p ∗ Ax + 1 ( x , p ) ∈ T ∗ R n ∼ = R 2 n . 2 p ∗ BB ∗ p , where This induces a Hamilton system: � = − H x = − A ∗ p p ˙ (HS) = H p = Ax + BB ∗ p . x ˙ Proposition Optimal trajectories are projections x ( t ) of the solution ( x ( t ) , p ( t )) of (HS). The control realizing the optimal trajectory is uniquely given by: u op ( t ) = B ∗ p ( t ) . W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 17 / 33

Optimal control problem Here is the explicit solution of the Hamilton system (HS) with initial condition ( x 1 , p 1 ) ∈ T x 1 R n :  = e − tA ∗ p 1 p ( t )   = e tA � � � t 0 e sA B B ∗ e − sA ∗ p 1 (SHS) x ( t ) x 1 + ds .  � �� �  = u op ( s ) W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 18 / 33

Optimal control problem Here is the explicit solution of the Hamilton system (HS) with initial condition ( x 1 , p 1 ) ∈ T x 1 R n :  = e − tA ∗ p 1 p ( t )   = e tA � � � t 0 e sA B B ∗ e − sA ∗ p 1 (SHS) x ( t ) x 1 + ds .  � �� �  = u op ( s ) For each t > 0 we define � t e sA BB ∗ e − sA ∗ ds ∈ R n × n . Γ t = 0 W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 18 / 33

Optimal control problem Here is the explicit solution of the Hamilton system (HS) with initial condition ( x 1 , p 1 ) ∈ T x 1 R n :  = e − tA ∗ p 1 p ( t )   = e tA � � � t 0 e sA B B ∗ e − sA ∗ p 1 (SHS) x ( t ) x 1 + ds .  � �� �  = u op ( s ) For each t > 0 we define � t e sA BB ∗ e − sA ∗ ds ∈ R n × n . Γ t = 0 Next step: We want to show that the matrix Γ t is invertible. Then we solve the last equation in ( SHS ) for p 1 . W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 18 / 33

Optimal control problem Here is the explicit solution of the Hamilton system (HS) with initial condition ( x 1 , p 1 ) ∈ T x 1 R n :  = e − tA ∗ p 1 p ( t )   = e tA � � � t 0 e sA B B ∗ e − sA ∗ p 1 (SHS) x ( t ) x 1 + ds .  � �� �  = u op ( s ) For each t > 0 we define � t e sA BB ∗ e − sA ∗ ds ∈ R n × n . Γ t = 0 Next step: We want to show that the matrix Γ t is invertible. Then we solve the last equation in ( SHS ) for p 1 . Recall the controllability matrix: � � B , AB , A 2 B , · · · , A m − 1 B G ( A , B ) = . � �� � = n × ( m · k ) − matrix W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 18 / 33

An invertibility condition Another consequence of the rank condition is the following: W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 19 / 33

An invertibility condition Another consequence of the rank condition is the following: Lemma Assume that rank G ( A , B ) = n, then for all t > 0 the matrix-valued integral � t e − sA BB ∗ e − sA ∗ ds ∈ R n × n Γ t = 0 is invertible. W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 19 / 33

An invertibility condition Another consequence of the rank condition is the following: Lemma Assume that rank G ( A , B ) = n, then for all t > 0 the matrix-valued integral � t e − sA BB ∗ e − sA ∗ ds ∈ R n × n Γ t = 0 is invertible. Proof: Let x ∈ R n such that Γ t x = 0. Then � � t � t � � � 2 e − sA BB ∗ e − sA ∗ ds · x , x � � B ∗ e − sA ∗ x � 0 = = ds . � 0 0 Therefore, we have 0 = B ∗ e − sA ∗ x for all s ∈ [0 , t ]. W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 19 / 33

Proof (continued) Taking the transpose of the last equation, we find for all s ∈ [0 , t ]: x ∗ e − sA B = 0 . W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 20 / 33

Proof (continued) Taking the transpose of the last equation, we find for all s ∈ [0 , t ]: x ∗ e − sA B = 0 . Taking derivatives of order ℓ ∈ N with respect to the parameter s gives: 0 = d ℓ � � x ∗ e − sA B = ( − 1) ℓ x ∗ A ℓ e − sA B . ds ℓ W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 20 / 33

Proof (continued) Taking the transpose of the last equation, we find for all s ∈ [0 , t ]: x ∗ e − sA B = 0 . Taking derivatives of order ℓ ∈ N with respect to the parameter s gives: 0 = d ℓ � � x ∗ e − sA B = ( − 1) ℓ x ∗ A ℓ e − sA B . ds ℓ In particular, we may choose s = 0. Then we find: 0 = x ∗ B = x ∗ AB = · · · = x ∗ A m − 1 B . Since the controllability matrix G ( A , B ) = [ B , AB , A 2 B , · · · A m − 1 B ] has linear independent rows (maximal rank n ) we conclude that x = 0. Hence Γ t is injective and therefore invertible. W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 20 / 33

The value of the value function Next goal: Calculate the value function. Let us go back to the solution of the Hamilton system , which stands behind the optimal control problem: � = e − tA ∗ p 1 p ( t ) = e tA � � (SHS) x ( t ) x 1 + Γ t · p 1 . 2 ( x 1 , p 1 ) was the initial value in (HS) W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 21 / 33

The value of the value function Next goal: Calculate the value function. Let us go back to the solution of the Hamilton system , which stands behind the optimal control problem: � = e − tA ∗ p 1 p ( t ) = e tA � � (SHS) x ( t ) x 1 + Γ t · p 1 . Since Γ t is invertible for any t > 0 we can solve the 2nd equation for p 1 with x 2 = x ( T ) 2 : � � p 1 = Γ − 1 e − TA x 2 − x 1 ( T > 0) . T 2 ( x 1 , p 1 ) was the initial value in (HS) W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 21 / 33

The value of the value function Next goal: Calculate the value function. Let us go back to the solution of the Hamilton system , which stands behind the optimal control problem: � = e − tA ∗ p 1 p ( t ) = e tA � � (SHS) x ( t ) x 1 + Γ t · p 1 . Since Γ t is invertible for any t > 0 we can solve the 2nd equation for p 1 with x 2 = x ( T ) 2 : � � p 1 = Γ − 1 e − TA x 2 − x 1 ( T > 0) . T The optimal control is given by u op ( t ) = B ∗ p ( t ) and therefore one can calculate: � � J T ( u ) : u ∈ L ∞ ([0 , T ] , R k ) , x u (0) = x 1 , x u ( T ) = x 2 S T ( x 1 , x 2 ) = inf . 2 ( x 1 , p 1 ) was the initial value in (HS) W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 21 / 33

The value of the value function Using u op ( t ) = B ∗ p ( t ) and p ( t ) = e − tA ∗ p 1 gives: W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 22 / 33

The value of the value function Using u op ( t ) = B ∗ p ( t ) and p ( t ) = e − tA ∗ p 1 gives: S T ( x 1 , x 2 ) = J T ( u op ) � T = 1 � � � 2 ds � B ∗ p ( s ) 2 0 � T = 1 � � B ∗ e − sA ∗ p 1 , B ∗ e − sA ∗ p 1 ds 2 0 = 1 � � = 1 2 p ∗ Γ T p 1 , p 1 1 Γ T p 1 . 2 W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 22 / 33

The value of the value function Using u op ( t ) = B ∗ p ( t ) and p ( t ) = e − tA ∗ p 1 gives: S T ( x 1 , x 2 ) = J T ( u op ) � T = 1 � � � 2 ds � B ∗ p ( s ) 2 0 � T = 1 � � B ∗ e − sA ∗ p 1 , B ∗ e − sA ∗ p 1 ds 2 0 = 1 � � = 1 2 p ∗ Γ T p 1 , p 1 1 Γ T p 1 . 2 Since � � p 1 = Γ − 1 e − TA x 2 − x 1 ( T > 0) T we have: W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 22 / 33

The value of the value function Using u op ( t ) = B ∗ p ( t ) and p ( t ) = e − tA ∗ p 1 gives: S T ( x 1 , x 2 ) = J T ( u op ) � T = 1 � � � 2 ds � B ∗ p ( s ) 2 0 � T = 1 � � B ∗ e − sA ∗ p 1 , B ∗ e − sA ∗ p 1 ds 2 0 = 1 � � = 1 2 p ∗ Γ T p 1 , p 1 1 Γ T p 1 . 2 Since � � p 1 = Γ − 1 e − TA x 2 − x 1 ( T > 0) T we have: Corollary The value function S T ( x 1 , x 2 ) is smooth in ( T , x 1 , x 2 ) ∈ R + × R n × R n . W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 22 / 33

The geodesic cost Let x 1 ∈ R n be fixed and let x u ( t ) be an optimal trajectory of the problem: � u = ( u 1 , · · · , u k ) ∈ L ∞ ([0 , T ] , R k ) x ˙ = A x + B u , where � T � k = 1 i =1 | u i ( s ) | 2 ds J T ( u ) 2 0 (i.e. u realizes the minimum of the cost functional J T ( u )). W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 23 / 33

The geodesic cost Let x 1 ∈ R n be fixed and let x u ( t ) be an optimal trajectory of the problem: � u = ( u 1 , · · · , u k ) ∈ L ∞ ([0 , T ] , R k ) x ˙ = A x + B u , where � T � k = 1 i =1 | u i ( s ) | 2 ds J T ( u ) 2 0 (i.e. u realizes the minimum of the cost functional J T ( u )). Definition The geodesic cost corresponding to x u is the family { c t } t of functions: � � x ∈ R n . c t ( x ) = − S t x , x u ( t ) , where W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 23 / 33

The geodesic cost Let x 1 ∈ R n be fixed and let x u ( t ) be an optimal trajectory of the problem: � u = ( u 1 , · · · , u k ) ∈ L ∞ ([0 , T ] , R k ) x ˙ = A x + B u , where � T � k = 1 i =1 | u i ( s ) | 2 ds J T ( u ) 2 0 (i.e. u realizes the minimum of the cost functional J T ( u )). Definition The geodesic cost corresponding to x u is the family { c t } t of functions: � � x ∈ R n . c t ( x ) = − S t x , x u ( t ) , where There is a unique minimizer of the cost functional for all trajectories connecting x and x u ( t ). W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 23 / 33

The geodesic cost We can calculate the geodesic cost explicitly from our previous formulas: W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 24 / 33

The geodesic cost We can calculate the geodesic cost explicitly from our previous formulas: Recall that x u ( t ) is the solution of the Hamilton system � = − A ∗ p p ˙ (HS) = Ax + BB ∗ p x ˙ with some initial data ( x 1 , p 1 ). The optimal trajectory was obtained by W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 24 / 33

The geodesic cost We can calculate the geodesic cost explicitly from our previous formulas: Recall that x u ( t ) is the solution of the Hamilton system � = − A ∗ p p ˙ (HS) = Ax + BB ∗ p x ˙ with some initial data ( x 1 , p 1 ). The optimal trajectory was obtained by x u ( t ) = e tA � � x 1 + Γ t p 1 . W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 24 / 33

The geodesic cost We can calculate the geodesic cost explicitly from our previous formulas: Recall that x u ( t ) is the solution of the Hamilton system � = − A ∗ p p ˙ (HS) = Ax + BB ∗ p x ˙ with some initial data ( x 1 , p 1 ). The optimal trajectory was obtained by x u ( t ) = e tA � � x 1 + Γ t p 1 . Lemma The geodesic cost is obtained by = − 1 1 ( x − x 1 ) − 1 � � 2 p ∗ 1 Γ t p 1 + p ∗ 2( x − x 1 ) ∗ Γ − 1 c t ( x ) = − S t x , x u ( t ) t ( x − x 1 ) . W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 24 / 33

Proof of the Lemma Proof : We use our explicit formula for S t ( x , x u ( t )): Let v ( s ) be an optimal trajectory which connects x and x u ( t ). Then v ( s ) = e sA � � p 1 ∈ R n . x + Γ s ˜ p 1 with some ˜ We use the condition v ( t ) = x u ( t ) = e tA ( x 1 + Γ t p 1 ) to determine ˜ p 1 : p 1 = Γ − 1 ˜ t ( x 1 − x ) + p 1 . W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 25 / 33

Proof of the Lemma Proof : We use our explicit formula for S t ( x , x u ( t )): Let v ( s ) be an optimal trajectory which connects x and x u ( t ). Then v ( s ) = e sA � � p 1 ∈ R n . x + Γ s ˜ p 1 with some ˜ We use the condition v ( t ) = x u ( t ) = e tA ( x 1 + Γ t p 1 ) to determine ˜ p 1 : p 1 = Γ − 1 ˜ t ( x 1 − x ) + p 1 . Insert this expression into our previous formula for the value function c t ( x ) = − S t ( x , x u ( t )) = − 1 p ∗ 2˜ 1 Γ t ˜ p 1 � � ∗ � � = 1 Γ − 1 Γ − 1 t ( x 1 − x ) + p 1 Γ t t ( x 1 − x ) + p 1 . 2 W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 25 / 33

Proof of the Lemma Proof : We use our explicit formula for S t ( x , x u ( t )): Let v ( s ) be an optimal trajectory which connects x and x u ( t ). Then v ( s ) = e sA � � p 1 ∈ R n . x + Γ s ˜ p 1 with some ˜ We use the condition v ( t ) = x u ( t ) = e tA ( x 1 + Γ t p 1 ) to determine ˜ p 1 : p 1 = Γ − 1 ˜ t ( x 1 − x ) + p 1 . Insert this expression into our previous formula for the value function c t ( x ) = − S t ( x , x u ( t )) = − 1 p ∗ 2˜ 1 Γ t ˜ p 1 � � ∗ � � = 1 Γ − 1 Γ − 1 t ( x 1 − x ) + p 1 Γ t t ( x 1 − x ) + p 1 . 2 Combining terms give the result. � W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 25 / 33

A family of quadratic forms Define for t > 0 a family of quadratic forms on R k corresponding to c t ( x ): W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 26 / 33

A family of quadratic forms Define for t > 0 a family of quadratic forms on R k corresponding to c t ( x ): Definition With the previous notation put: Q ( t ) = B ∗ � � B = − d d 2 dt B ∗ Γ − 1 x 0 ˙ c t t B . Note: Q ( t ) does not depend on the initial data ( x 1 , p 1 ) and is the same for any geodesic (intrinsic object of the control system and cost). W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 26 / 33

A family of quadratic forms Define for t > 0 a family of quadratic forms on R k corresponding to c t ( x ): Definition With the previous notation put: Q ( t ) = B ∗ � � B = − d d 2 dt B ∗ Γ − 1 x 0 ˙ c t t B . Note: Q ( t ) does not depend on the initial data ( x 1 , p 1 ) and is the same for any geodesic (intrinsic object of the control system and cost). Remark : W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 26 / 33

A family of quadratic forms Define for t > 0 a family of quadratic forms on R k corresponding to c t ( x ): Definition With the previous notation put: Q ( t ) = B ∗ � � B = − d d 2 dt B ∗ Γ − 1 x 0 ˙ c t t B . Note: Q ( t ) does not depend on the initial data ( x 1 , p 1 ) and is the same for any geodesic (intrinsic object of the control system and cost). Remark : The family Q ( t ) is associated to each optimal trajectory. W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 26 / 33

A family of quadratic forms Define for t > 0 a family of quadratic forms on R k corresponding to c t ( x ): Definition With the previous notation put: Q ( t ) = B ∗ � � B = − d d 2 dt B ∗ Γ − 1 x 0 ˙ c t t B . Note: Q ( t ) does not depend on the initial data ( x 1 , p 1 ) and is the same for any geodesic (intrinsic object of the control system and cost). Remark : The family Q ( t ) is associated to each optimal trajectory. The coefficient matrices in the t -expansion of Q ( t ) which intrinsically is induced by the underlying optimal control problem play a role in the W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 26 / 33

A family of quadratic forms Define for t > 0 a family of quadratic forms on R k corresponding to c t ( x ): Definition With the previous notation put: Q ( t ) = B ∗ � � B = − d d 2 dt B ∗ Γ − 1 x 0 ˙ c t t B . Note: Q ( t ) does not depend on the initial data ( x 1 , p 1 ) and is the same for any geodesic (intrinsic object of the control system and cost). Remark : The family Q ( t ) is associated to each optimal trajectory. The coefficient matrices in the t -expansion of Q ( t ) which intrinsically is induced by the underlying optimal control problem play a role in the ”small time heat kernel expansion” . W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 26 / 33

Towards the coefficients in the heat kernel expansion Theorem (A. Agrachev, D. Barilari, L Rizzi) Let x u : [0 , T ] → R n be an optimal trajectory of the optimal control problem and Q ( t ) the corresponding family of quadratic forms: W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 27 / 33

Towards the coefficients in the heat kernel expansion Theorem (A. Agrachev, D. Barilari, L Rizzi) Let x u : [0 , T ] → R n be an optimal trajectory of the optimal control problem and Q ( t ) the corresponding family of quadratic forms: (a) t �→ t 2 Q ( t ) extends to a smooth family of symmetric operators. W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 27 / 33

Towards the coefficients in the heat kernel expansion Theorem (A. Agrachev, D. Barilari, L Rizzi) Let x u : [0 , T ] → R n be an optimal trajectory of the optimal control problem and Q ( t ) the corresponding family of quadratic forms: (a) t �→ t 2 Q ( t ) extends to a smooth family of symmetric operators. (b) The small time expansion ℓ Q ( t ) = 1 � O ( i ) t i + O ( t i +1 ) t 2 I + as t → 0 i =0 defines symmetric matrices O ( i ) , i ≥ 0 and I . W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 27 / 33

Towards the coefficients in the heat kernel expansion Theorem (A. Agrachev, D. Barilari, L Rizzi) Let x u : [0 , T ] → R n be an optimal trajectory of the optimal control problem and Q ( t ) the corresponding family of quadratic forms: (a) t �→ t 2 Q ( t ) extends to a smooth family of symmetric operators. (b) The small time expansion ℓ Q ( t ) = 1 � O ( i ) t i + O ( t i +1 ) t 2 I + as t → 0 i =0 defines symmetric matrices O ( i ) , i ≥ 0 and I . (c) There is a trace formula: with k i = dim span { B , AB , · · · , A i − 1 B } : m � � � trace I = (2 i − 1) k i − k i − 1 . i =1 W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 27 / 33

Main results Theorem (D. Barilari, E. Paoli, 2017) Let A ∈ R n × n , B ∈ R n × k and x 0 ∈ R n . Consider the hypo-elliptic operator: L = Ax · ∇ + 1 2div( BB ∗ ∇ ) (with rank condition) with heat kernel p ( t ; x , y ) ∈ C ∞ ( R + × R n × R n ) . Assume that Ax 0 = 0 . � ℓ � t − 1 2 tr I � a i t i + O ( t ℓ +1 ) p ( t , x 0 , x 0 ) = ( t → 0) , 2 √ c 0 n (2 π ) i =0 where � tr A , tr O (0) , · · · , tr Q ( i − 2) � a i = P i and P i = polynomials . , W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 28 / 33

Main results Theorem (D. Barilari, E. Paoli, 2017) Let A ∈ R n × n , B ∈ R n × k and x 0 ∈ R n . Consider the hypo-elliptic operator: L = Ax · ∇ + 1 2div( BB ∗ ∇ ) (with rank condition) with heat kernel p ( t ; x , y ) ∈ C ∞ ( R + × R n × R n ) . Assume that Ax 0 = 0 . � ℓ � t − 1 2 tr I � a i t i + O ( t ℓ +1 ) p ( t , x 0 , x 0 ) = ( t → 0) , 2 √ c 0 n (2 π ) i =0 where � tr A , tr O (0) , · · · , tr Q ( i − 2) � a i = P i and P i = polynomials . , and a 2 = (tr A ) 2 + tr O (0) In particular: a 1 = − tr A . 2 8 4 W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 28 / 33

Main results Theorem (D. Barilari, E. Paoli, 2017) With x 1 , x 2 ∈ R n consider the minimal cost function again: � � J T ( u ) : u ∈ L ∞ ([0 , T ] , R k ) , x u (0) = x 1 , x u ( T ) = x 2 S T ( x 1 , x 2 ) = inf . W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 29 / 33

Main results Theorem (D. Barilari, E. Paoli, 2017) With x 1 , x 2 ∈ R n consider the minimal cost function again: � � J T ( u ) : u ∈ L ∞ ([0 , T ] , R k ) , x u (0) = x 1 , x u ( T ) = x 2 S T ( x 1 , x 2 ) = inf . Then there is the following off-diagonal small time heat kernel asymptotic: � ℓ � t − 1 2 tr I � a i t i + O ( t ℓ +1 ) e − S t ( x 1 , x 2 ) p ( t ; x 1 , x 2 ) ( t → 0) . 2 √ c 0 n (2 π ) i =0 W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 29 / 33

Main results Theorem (D. Barilari, E. Paoli, 2017) With x 1 , x 2 ∈ R n consider the minimal cost function again: � � J T ( u ) : u ∈ L ∞ ([0 , T ] , R k ) , x u (0) = x 1 , x u ( T ) = x 2 S T ( x 1 , x 2 ) = inf . Then there is the following off-diagonal small time heat kernel asymptotic: � ℓ � t − 1 2 tr I � a i t i + O ( t ℓ +1 ) e − S t ( x 1 , x 2 ) p ( t ; x 1 , x 2 ) ( t → 0) . 2 √ c 0 n (2 π ) i =0 The coefficients a i are the ones from the last theorem. W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 29 / 33

Main results In the final result we consider the case Ax 0 � = 0. With i = 1 , · · · , m put: � � A i Bx : x ∈ R k , 0 ≤ j ≤ i − 1 ⊂ R n . E i = span W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 30 / 33

Main results In the final result we consider the case Ax 0 � = 0. With i = 1 , · · · , m put: � � A i Bx : x ∈ R k , 0 ≤ j ≤ i − 1 ⊂ R n . E i = span From the rank condition it is clear that we obtain a filtration of R n : E 1 = { span of columns of B } ⊂ E 2 ⊂ · · · ⊂ E m = R n . W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 30 / 33

Main results In the final result we consider the case Ax 0 � = 0. With i = 1 , · · · , m put: � � A i Bx : x ∈ R k , 0 ≤ j ≤ i − 1 ⊂ R n . E i = span From the rank condition it is clear that we obtain a filtration of R n : E 1 = { span of columns of B } ⊂ E 2 ⊂ · · · ⊂ E m = R n . Observation Now, the small time heat kernel expansion depends on the level E j in which we find Ax 0 : W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 30 / 33

Main results Theorem (D. Barilari, E. Paoli, 2017) Let Ax 0 � = 0 . The following two cases show different behaviour: W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 31 / 33

Main results Theorem (D. Barilari, E. Paoli, 2017) Let Ax 0 � = 0 . The following two cases show different behaviour: (i) If Ax 0 ∈ E 1 , then we have polynomial decay as t → 0 : t − 1 � � tr A � � 2 tr I + | Ax 0 | 2 t + O ( t 2 ) p ( t ; x 0 , x 0 ) = 1 − . 2 √ c 0 n 2 2 (2 π ) W. Bauer (Leibniz U. Hannover ) Small time heat kernel expansion March 4-10. 2018 31 / 33