Optimal bounds for numerical approximations of finite horizon problems based on dynamic programming approach

Optimal b ounds for n umerical appro ximations of finite horizon problems based on dynamic programming approac h Ja vier de F rutos ∗ Julia No v o † F ebruary 19, 2026 Abstract In this paper w e provide optimal b ounds for fully discrete approximations to finite horizon problems via dynamic programming. W e adapt the error analysis in [9] for the infinite horizon case to the finite horizon case. W e pro ve an a priori b ound of size O ( h + k ) for the method, h b eing the time discretization step and k the spatial mesh size. Arguing with piecewise constants con trols we are able to obtain first order of conv ergence in time and space under standard regularit y assumptions, a voiding the more restrictiv e regularity assumptions on the controls required in [9]. W e show that the loss in the rate of con v ergence in time of the infinite case (obtained arguing with piece-wise controls) can be a v oided in the finite horizon case Key words Finite horizon problems, Dynamic programming, Hamilton-Jacobi- Bellman equation, Optimal control. 1 In tro duction The numerical approximation of optimal con trol problems is of imp ortance for man y applications. In this pap er w e consider the dynamic programming approac h for solving finite horizon problems. The dynamic programming principle gives a char- acterization of the v alue function as the unique viscosity solution of a nonlinear partial differential equation, the Hamilton-Jacobi-Bellman (HJB) equation. The v alue function is then used to get a syn thesis of a feedback con trol la w. In the present pap er our concern is to giv e optimal error b ounds for a fully discrete semi-Lagrangian metho d approac hing the v alue function. F or a metho d with a p ositive time step size h and spatial elements of size k we prov e an optimal error b ound of size O ( h + k ) which gives first order of con vergence in time and ∗ Instituto de Inv estigaci´ on en Matem´ aticas (IMUV A), Univ ersidad de V alladolid, Spain. Researc h supp orted by Spanish MINECO under gran t PID2022-136550ND-I00 (frutos@mac.uv a.es) † Departamen to de Matem´ aticas, Universidad Aut´ onoma de Madrid, Spain. Researc h supp orted by Spanish MINECO under gran t PID2022-136550ND-I00. (julia.no vo@uam.es) 1 space for the metho d. W e introduce a characterization of the fully discrete metho d inherited from [9], see also [10]. The temp oral comp onent of the error comes from the approximation of the dynamics b y a discrete one based on the Euler metho d plus the approximation of the time in tegral b y the comp osite rectangle rule. The spatial comp onen t of the error comes from the substitution of the functions, both in the dynamics and in the cost, b y piece-wise linear interpolants in space. Adapting the tec hnique in [9, Section 3.2], based on piece-wise constant con trols in time, w e av oid making regularity assumptions on the controls. W e only need a kind of discrete regularit y assumption on the computed discrete controls to ac hiev e the full order of con vergence, see Remark 1. W e think that the error analysis techniques shown in this pap er are of in terest to analyze similar metho ds applied to the same or analogous problems. T o our kno wledge, there are not many papers getting error estimates for metho ds solving finite horizon con trol problems. In [11], a dynamic programming algorithm based on a tree structure (whic h do es not require a spatial discretization of the problem) to mitigate the curse of dimensionality , see [1], is analyzed. First order b ounds in time are obtained in the first part of this paper. In the second part, it is assumed that the con tinuous set of con trols is replaced by a discrete set. The tree structure considers only spatial no des that result of the discrete dynamics. T o reduce the increase in the total num b er of no des a pruning criterion is applied in [1] that replace a new no de by and old one whenever the distance b etw een them is small enough. In [1] the pruning condition is to o demanding since the difference b etw een no des is tak en O ( h 2 ), h b eing the time step. This fact comes from a factor h that app ears dividing in the error b ounds. This problem can b e solved with the error analysis presen ted in the presen t pap er. The same problem w as fixed with the error analysis of [9] in the infinite horizon case. Reference [9] is the first one in which a rate of con vergence O ( h + k ) is pro ved, improving the rate of conv ergence of size O ( k /h ) sho wn in the literature, see [6, Corollary 2.4], [7, Theorem 1.3]). Although the metho d analyzed in this pap er do es not a v oid the curse of dimen- sionalit y one can apply reduced order tec hniques to this metho d follo wing [8]. In [8] a reduced order metho d based on prop er orthogonal decomp osition (POD) is applied for the numerical appro ximation of infinite horizon optimal control prob- lems. The same ideas extend to finite horizon problems and will b e sub ject of future researc h. F or the error analysis of a reduced order metho d the error analysis of the presen t pap er is essential. Moreov er, as stated b efore, the error analysis shown in this pap er can b e used to analyze or impro ve the analysis of nu merical metho ds for the same problem. The outline of the pap er is as follows. In Section 2 we introduce some notation. The fully discrete approximation is described in Section 3. In Section 4 we carry out the error analysis of the metho d. Finally , some interpolation arguments needed for the pro of of the main theorem are included in the appendix. 2 2 Mo del problem and preliminary results Throughout this section we follow the notation in [1]. Let us consider the system ˙ y ( s ) = f ( y ( s ) , u ( s ) , s ) , s ∈ ( t, T ] , (1) y ( t ) = x ∈ R d . W e will denote b y y : [ t, T ] → R d the solution, by u the con trol u : [ t, T ] → R m , b y f : R d × R m → R d the dynamics, and by U = { u : [ t, T ] → U, measurable } the set of admissible controls where U ⊂ R m is a compact set. W e assume that there exists a unique solution for (1) for each u ∈ U . The cost functional for the finite horizon optimal con trol problem will be given b y J x,t ( u ) := Z T t L ( y ( s, u ) , u ( s ) , s ) e − λ ( s − t ) ds + g ( y ( T )) e − λ ( T − t ) , (2) where L : R d × R m × [ t, T ] → R is the running cost, g : R d → R is the final cost, and λ ≥ 0 is the discount factor. The goal is to find a state-feedback control law u ( t ) = Φ( y ( t ) , t ), in terms of the state v ariable y ( t ), where Φ is the feedback map. T o derive optimalit y conditions, dynamic programming principle (DPP) is used. The v alue function for an initial condition is defined by v ( x, t ) := inf u ∈U J x,t ( u ) . (3) The v alue function (3) satisfies the HHB equation for every x ∈ R d , s ∈ [ t, T ): − ∂ v ∂ s ( x, s ) + λv ( x, s ) + max u ∈ U {− L ( x, u, s ) − ∇ v ( x, s ) · f ( x, u, s ) } = 0 , (4) v ( x, T ) = g ( x ) . If the v alue function is known, then it is p ossible to compute the optimal feedbac k con trol as u ∗ ( t ) := arg max u ∈ U {− L ( x, u, t ) − ∇ v ( x, t ) · f ( x, u, t ) } . (5) Equation (4) is hard to solve analytically . In next section we introduce a semi- Lagrangian metho d to approach the v alue function. As in [11], we assume that the functions f , L and g are con tinuous in all the v ariables and b ounded: ∥ f ( x, u, s ) ∥ ∞ = max 1 ≤ i ≤ n | f i ( x, u, s ) | ≤ M f (6) | L ( x, u, s ) | ≤ M L , | g ( x ) | ≤ M g , ∀ x ∈ R d , u ∈ U, s ∈ [ t, T ] . (7) W e also assume that f and L are Lipschitz-con tin uous with respect to all the argu- men ts: ∥ f ( x, u, s ) − f ( y , u, s ) ∥ 2 ≤ L f ∥ x − y ∥ 2 , ∀ x, y ∈ R d , u ∈ U, s ∈ [ t, T ] (8) ∥ f ( x, u, s 1 ) − f ( x, u, s 2 ) ∥ 2 ≤ L f | s 1 − s 2 | , ∀ x ∈ R d , u ∈ U, s 1 , s 2 ∈ [ t, T ] , (9) ∥ f ( x, u 1 , s ) − f ( x, u 2 , s ) ∥ 2 ≤ L f ∥ u 1 − u 2 ∥ 2 , ∀ x ∈ R d , u 1 , u 2 ∈ U, s ∈ [ t, T ] . (10) 3 | L ( x, u, s ) − L ( y , u, s ) ∥ 2 ≤ L L ∥ x − y ∥ 2 , ∀ x, y ∈ R d , u ∈ U, s ∈ [ t, T ] , (11) | L ( x, u, s 1 ) − L ( x, u, s 2 ) ∥ 2 ≤ L L | s 1 − s 2 | , ∀ x ∈ R d , u ∈ U, s 1 , s 2 ∈ [ t, T ] , (12) | L ( x, u 1 , s ) − L ( x, u 2 , s ) ∥ 2 ≤ L L ∥ u 1 − u 2 ∥ 2 , ∀ x ∈ R d , u 1 , u 2 ∈ U, s ∈ [ t, T ] . (13) Finally , we assume that the cost g is also Lipsc hitz-con tin uous: | g ( x ) − g ( y ) | ≤ L g ∥ x − y ∥ 2 , ∀ x, y ∈ R d . (14) 3 F ully discrete appro ximation Let us define h = ( T − t ) / N , where N is the n umber of temp oral steps. Let Ω a b ounded p olyhedron in R d satisfying the following in ward pointing condition on the dynamics for sufficien tly small h > 0 y + hf ( y , u ) ∈ Ω , ∀ y ∈ Ω , u ∈ U. (15) Let { S j } m s j =1 b e a family of simplices which defines a regular triangulation of Ω Ω = m s [ j =1 S j , k = max 1 ≤ j ≤ m s (diam S j ) . W e assume w e ha ve n s v ertices/no des x 1 , . . . , x n s in the triangulation. Let V k b e the space of piecewise affine functions from Ω to R whic h are con tin uous in Ω having constan t gradien ts in the in terior of an y simplex S j of the triangulation. Then, a fully discrete sc heme for the HJB equations is giv en by: v n h,k ( x i ) = min u ∈ U n hL ( x i , u, t n ) + δ h v n +1 h,k ( x i + hf ( x i , u, t n )) o , n = N − 1 , . . . , 0 v N h,k ( x i ) = g ( x i ) , i = 1 , . . . , n s , (16) where δ h = 1 − λh , t n = t + nh . The functions v n h,k are in V k and are defined b y their v alues at the no des x i , i = 1 , . . . , n s giv en by (16). As in [9], see also [10], we will giv e a c haracterization of the fully discrete metho d that allows to bound the error. F or any x ∈ R d , n = 0 , . . . N − 1 and u l n = n u l,n n , . . . , u l,N − 1 n o with all its comp onen ts u l,k n ∈ U , n ≤ k ≤ N − 1, for 1 ≤ l ≤ n s , let us define the fully discrete functional J n h,k ( x, u 1 n , . . . , u n s n ) := h N − 1 X j = n δ j − n h I k L ( ˆ y j , u 1 ,j n , . . . , u n s ,j n , t j ) + I k g ( ˆ y N ) e − λ ( T − t n ) , (17) ˆ y j +1 = ˆ y j + hI k f ( ˆ y j , u 1 ,j n , . . . , u n s ,j n , t j ) , ˆ y n = x, j = n, . . . N − 1 , (18) where, using the barycentric co ordinates, x = n s X j =1 µ j ( x ) x j , 0 ≤ µ j ( x ) ≤ 1 , n s X j =1 µ j ( x ) = 1 , 4 for any x , any t and any u 1 , . . . , u n s ∈ U the in terp olants I k L ( x, u 1 , . . . , u n s , t ) and I k f ( x, u 1 , . . . , u n s , t ) are defined by I k L ( x, u 1 , . . . , u n s , t ) = n s X j =1 µ j ( x ) L ( x j , u j , t ) , (19) I k f ( x, u 1 , . . . , u n s , t ) = n s X j =1 µ j ( x ) f ( x j , u j , t ) , and I k g is the piecewise linear in terp olant of g in V k . No w, as in [10], for any n = 0 , . . . , N − 1, w e define w n h,k ∈ V k b y w n h,k ( x ) = inf u 1 n ,..., u n s n J n h,k ( x, u 1 n , . . . , u n s n ) . (20) F ollo wing [10, Theorem 1] it follows that Theorem 1 F or any x ∈ Ω , n = 0 , . . . , N , the function w n h,k ( x ) satifies the e quation w n h,k ( x ) = inf u 1 n ,...,u n s n { hI k L ( x, u 1 n , . . . , u n s n , t n ) + δ h w n +1 h,k ( x + hI k f ( x, u 1 n , . . . , u n s n , t n )) } and, as a c onse quenc e, for e ach no de x i , i = 1 , . . . , n s w n h,k ( x i ) = inf u i n n hL ( x i , u i n , t n ) + δ h w n +1 h,k ( x i + hf ( x i , u i n , t n )) o . F ollo wing [10, Theorem 2] w e also hav e the following c haracterization Theorem 2 F or e ach no de x i , i = 1 , . . . , n s , let us denote by u i n the ar gument giving the minimum in w n h,k ( x i ) = inf u i n n hL ( x i , u i n , t n ) + δ h w n +1 h,k ( x i + hf ( x i , u i n , t n )) o . Then, for any x ∈ Ω the subsets of c ontr ols u i n , i = 1 , . . . , n s giving the minimum in w n h,k ( x ) = inf u 1 n ,..., u n s n J n h,k ( x, u 1 n , . . . , u n s n ) ar e determine d by the values of the c ontr ols at the no des, u i n , 1 ≤ i ≤ n s . Mor e pr e cisely, for any j , the values n u 1 ,j n , . . . , u n s ,j n o in (17) - (18) , n ≤ j ≤ N − 1 , ar e u i,j n = u i n if µ i ( ˆ y j )  = 0 and u i,j n = 0 if µ i ( ˆ y j ) = 0 . F rom the ab o ve theorem is clear that the in terp olan ts defined in (19) are alw ays piecewise functions in V k . Then, it is immediate to pro v e w n h,k ∈ V k whic h implies w n h,k is the unique solution defined by (16) i.e., w n h,k = v n h,k and, as a consequence, giv es a characterization of the fully discrete functional. This characterization is used in the presen t pap er to prov e the error b ounds of the metho d. 5 4 Error analysis of the metho d W e follow the error analysis of [9, Section 3.2], based on the analysis in [5], to prov e con vergence of the metho d arguing with piecewise constant controls. Let us denote U pc = { u ∈ U | u ( t ) = u l , t ∈ [ t l , t l +1 ) , 0 ≤ l ≤ N − 1 } , with u l constan t. Let us observ e that we can consider the con tinuous problem for con trols in U pc . The follo wing lemma follows the error analysis in [9, Lemma 1], see also [3, Lemma 1.2, Chapter VI]. Along the error analysis C will represent a generic constant that is not alwa ys necessarily the same and that do es not depend neither on h nor in k . Lemma 1 L et 0 ≤ n ≤ N − 1 , x ∈ R d ⊂ Ω and J x,t n ( u ) , J n h,k ( x, u 1 n , . . . , u n s n ) the functionals define d in (2) and (17) - (18) , r esp e ctively. Assume c onditions (6) , (7) , (8) , (9) , (11) , (12) and (14) hold. Then | J x,t n ( u ) − J n h,k ( x, u 1 n , . . . , u n s n ) | ≤ C ( h + k ) , (21) wher e u ∈ U pc and for i = 1 , . . . , n s , u i n = u = { u n , u n +1 , . . . , u N − 1 } with u l = u ( t ) , t ∈ [ t l , t l +1 ) , l = n, . . . , N − 1 . Pro of Let y ( s ), s ∈ [ t n , T ], be the solution of (1) and let us denote b y ˜ y ( s ) = ˆ y l , l = [ s/h ] where ˆ y l is the solution of (18). Then, ˜ y can b e expressed as ˜ y ( s ) = x + Z [ s/h ] h t n I k f ( ˜ y ( τ ) , u ( τ ) , [ τ /h ] h ) dτ . And, y ( s ) − ˜ y ( s ) = Z [ s/h ] h t n ( f ( y ( τ ) , u ( τ ) , τ ) − I k f ( ˜ y ( τ ) , u ( τ ) , [ τ /h ] h )) dτ + Z s [ s/h ] h f ( y ( τ ) , u ( τ ) , τ ) dτ . F rom the ab o v e equation, applying (6), we get ∥ y ( s ) − ˜ y ( s ) ∥ ∞ ≤ Z [ s/h ] h t n ∥ f ( y ( τ ) , u ( τ ) , τ ) − I k f ( ˜ y ( τ ) , u ( τ ) , [ τ /h ] h ) ∥ ∞ dτ + M f h. (22) Let us b ound now the term inside the in tegral. Adding and subtracting terms we get ∥ f ( y ( τ ) , u ( τ ) , τ ) − I k f ( ˜ y ( τ ) , u ( τ ) , [ τ /h ] h ) ∥ ∞ ≤ (23) ∥ f ( y ( τ ) , u ( τ ) , τ ) − f ( y ( τ ) , u ( τ ) , [ τ /h ] h ) ∥ ∞ + ∥ f ( y ( τ ) , u ( τ ) , [ τ /h ] h ) − I k f ( y ( τ ) , u ( τ ) , [ τ /h ] h ) ∥ ∞ + ∥ I k f ( y ( τ ) , u ( τ ) , [ τ /h ] h ) − I k f ( ˜ y ( τ ) , u ( τ ) , [ τ /h ] h ) ∥ ∞ . 6 F or the first term on the righ t-hand side, applying (9), it is easy to prov e that Z [ s/h ] h t n ∥ f ( y ( τ ) , u ( τ ) , τ ) − f ( y ( τ ) , u ( τ ) , [ τ /h ] h ) ∥ ∞ dτ ≤ C h, C = ( T − t n ) L f . (24) T o bound the other terms on the righ t-hand side of (23) arguing, as in [2], w e observ e that for an y y ∈ Ω there exists an index l with y ∈ S l ⊂ Ω. Let us denote b y J l the index subset such that x i ∈ S l for i ∈ J l . W riting y = n S X i =1 µ i x i , 0 ≤ µ i ≤ 1 , n S X i =1 µ i = 1 , it is clear that µ i = 0 holds for an y i ∈ J l . Now, w e observe that for an y u ∈ U and an y time s , applying (8) we get for 1 ≤ j ≤ d | f j ( y , u, s ) − I k f j ( y , u, s ) | = | n S X i =1 µ i f j ( y , u, s ) − n S X i =1 µ i I k f j ( x i , u, s ) | = | X i ∈ J l µ i ( f j ( y , u, s ) − f j ( x i , u, s ) | ≤ X i ∈ J l µ i L f ∥ y − x i ∥ 2 ≤ L f k , (25) where in the last inequalit y w e ha ve applied ∥ y − x i ∥ 2 ≤ k , for i ∈ J l . F rom the ab o v e inequalit y we get for the second term on the righ t-hand side of (23) Z [ s/h ] h t n ∥ f ( y ( τ ) , u ( τ ) , [ τ /h ] h ) − I k f ( y ( τ ) , u ( τ ) , [ τ /h ] h ) ∥ ∞ ≤ C k . (26) F or the third term on the right-hand side of (23) we observ e that the difference of the in terp olation op erator ev aluated at tw o different p oints can b e b ounded in terms of the constan t gradient of the interpolant in the elemen t to which those p oin ts b elong times the difference of them, i.e., I k f j ( y , u, s ) − I k f j ( ˜ y , u, s ) = ∇ I k f j ( ˜ y , u, s ) · ( y − ˜ y ) ≤ ∥∇ I k f j ( ˜ y , u, s ) ∥ 2 ∥ y − ˜ y ∥ 2 . Moreo ver, ∇ I k f j can b e b ounded in terms of the lipsc hitz constant of f , L f , more precisely , ∥∇ I k f j ( ˜ y , u, s ) ∥ 2 ≤ C √ dL f . Then, | I k f j ( y , u, s ) − I k f j ( ˜ y , u, s ) | ≤ C L f √ d ∥ y − ˜ y ∥ 2 . As a consequence, for the third term on the righ t-hand side of (23) w e get Z [ s/h ] h t n ∥ I k f ( y ( τ ) , u ( τ ) , [ s/h ] h ) − I k f ( ˜ y ( τ ) , u ( τ ) , [ s/h ] h ) ∥ ∞ ≤ (27) C √ dL f Z s t n ∥ y ( τ ) − ˜ y ( τ ) ∥ 2 ≤ C dL f Z s t n ∥ y ( τ ) − ˜ y ( τ ) ∥ ∞ . F rom (24), (26) and (27) we get for C = C dL f ∥ y ( s ) − ˜ y ( s ) ∥ ∞ ≤ C Z s t n ∥ y ( τ ) − ˜ y ( τ ) ∥ ∞ dτ + C ( h + k ) . 7 Applying Gronw all’s lemma we obtain ∥ y ( s ) − ˜ y ( s ) ∥ ∞ ≤ e C s C C ( h + k ) ≤ C ( h + k ) , s ∈ [ t n , T ] . (28) F or simplicity , in the sequel w e will denote b y J n h,k ( x, u ) = J n h,k ( x, u 1 n , . . . , u n s n ) . W e decomp ose | J x,t n ( u ) − J n h,k ( x, u ) | ≤ X 1 + X 2 , (29) with X 1 =       h N − 1 X j = n δ j − n h I k L ( ˆ y j , u j , t j ) − Z T 0 L ( y ( s ) , u ( s ) , s ) e − λ ( s − t n ) ds       , X 2 =    ( g ( y ( T )) − I k g ( ˜ y ( T )) e − λ ( T − t n )    . W e start b ounding the second term. Adding and subtracting terms and using (14) | g ( y ( T )) − I k g ( ˜ y ( T ) | ≤ | g ( y ( T )) − g ( ˜ y ( T )) | + | g ( ˜ y ( T )) − I k g ( ˜ y ( T )) | ≤ L g ∥ y ( T ) − ˜ y ( T ) ∥ 2 + | g ( ˜ y ( T )) − I k g ( ˜ y ( T )) | . T o b ound the first term on the righ t-hand side ab ov e w e apply (28) to get L g ∥ y ( T ) − ˜ y ( T ) ∥ 2 ≤ C ( k + h ) . T o b ound the second term w e apply [4] where it is prov ed that the error in the piece-wise linear interpolant of a Lipsc hitz function can b e b ounded in terms of the lipsc hitz constant of that function so that | g ( ˜ y ( T )) − I k g ( ˜ y ( T )) | ≤ C ( L g ) k . Then, X 2 ≤ e − λ ( T − t n ) C ( h + k ) . (30) T o b ound the first in tegral term w e observe that X 1 =     Z T t n  I k L ( ˜ y ( s ) , u ( s ) , [ s/h ] h ) δ [ s/h ] h − L ( y ( s ) , u ( s ) , s ) e − λs  ds     . Then, we can write X 1 ≤ X 1 , 1 + X 1 , 2 + X 1 , 3 + X 1 , 4 (31) :=     Z T t n I k L ( ˜ y ( s ) , u ( s ) , [ s/h ] h )( δ [ s/h ] − n h − e − λ ( s − t n ) ) ds     +     Z T t n ( I k L ( ˜ y ( s ) , u ( s ) , [ s/h ]) − I k L ( y ( s ) , u ( s ) , [ s/h ])) e − λ ( s − t n ) ds     +     Z T t n ( I k L ( y ( s ) , u ( s ) , [ s/h ] h ) − L ( y ( s ) , u ( s ) , [ s/h ] h )) e − λ ( s − t n ) ds     +     Z T t n ( L ( y ( s ) , u ( s ) , [ s/h ] h ) − L ( y ( s ) , u ( s ) , s )) e − λ ( s − t n ) . ds     8 No w w e b ound the four terms on the righ t-hand side of (31). T o b ound the first term we will apply | I k L ( y , u, s ) | ≤ | L ( · , u, s ) | ∞ ≤ M L (see (7)) to obtain X 1 , 1 = M L Z T t n | δ [ s/h ] − n h − e − λ ( s − t n ) | ds. No w we write δ [ s/h ] − n h = e − λθ ([ s/h ] − n ) h , for θ = − log( δ h ) / ( λh ). Applying the mean v alue theorem to the function e − λ ( s − t n ) and taking into account that since ([ s/h ] − n ) h ≤ s − t n ≤ ([ s/h ] − n ) h + h then | s − t n θ ([ s/h ] − n ) h | ≤ ( θ − 1)( T − t n ) + θ h and that θ − 1 = O ( h ) then we get X 1 , 1 ≤ M L ( T − t n ) λ (( θ − 1)( T − t n ) + θ h ) ≤ C h. (32) T o b ound the next term we argue as in (27) and use (28) to get X 1 , 2 ≤ C L L Z T t n ∥ y ( s ) − ˜ y ( s ) ∥ ∞ e − λ ( s − t n ) ds ≤ C ( h + k ) . F or the third term, arguing as in (25) we get X 1 , 3 ≤ L L k Z T t n e − λ ( s − t n ) ds ≤ C k . (33) Finally the last term X 1 , 4 = O ( h ) applying (12) and arguing as in (24). □ Let us observ e that for any x ∈ Ω | v N h,k ( x ) − v ( x, T ) | = | I k g ( x ) − g ( x ) | ≤ C k . In next theorem we b ound the difference | v n h,k ( x ) − v ( x, t n ) | for n = 0 , . . . , N − 1. As in [9, Theorem 7] for the pro of w e need to assume an additional conv exit y assumption, see [5, (A4)], • (CA) F or every y ∈ R d , and an y s ∈ [ t, T ] { f ( y , u, s ) , L ( y, u, s ) , u ∈ U } is a con v ex subset of R d +1 . W e also need to assume (10) and (13) for the pro of that can b e found in the ap- p endix. Theorem 3 Assume c onditions (6) , (7) , (8) , (9) , (10) , (11) , (12) , (13) , (14) and (CA) hold. F or 0 ≤ n ≤ N − 1 and x ∈ Ω ther e exist c onstants C 1 and C 2 , indep endent of h and k , such that the fol lowing b ound holds | v n h,k ( x ) − v ( x, t n ) | ≤ C 1 h + C 2 k . (34) The c onstant C 2 dep ends line arly on L u = max ∥ u i n − u j n ∥ 2 | x i − x j | , 1 ≤ i, j ≤ n S , (35) wher e u i n ar e the ar guments giving the minimum in v n h,k ( x i ) in (16) . 9 Pro of In view of (20) let us denote by  u 1 n , . . . , u n s n  a con trol giving the minimum v n h,k ( x ) = J n h,k ( x, u 1 n , . . . , u n s n ) , with  u 1 n , . . . , u n s n  , as stated in the assumptions of the present theorem, the controls giving the minim um at the no des. Let us define u = { u n , . . . , u N − 1 } where for l = n, . . . , N − 1, and ˆ y l (defined b y (18)) written as ˆ y l = P n s j =1 µ j ( ˆ y l ) x j then u l = P n s j =1 µ j ( ˆ y l ) u j n . Let us denote by J n h,k ( x, u ) = J n h,k ( x, u , . . . , u ) . Then v ( x, t n ) − v n h,k ( x ) ≤  J x,t n ( u ) − J n h,k ( x, u )  +  J n h,k ( x, u ) − J n h,k ( x, u 1 n , . . . , u n s n )  , where u ∈ U pc suc h that u ( t ) = u l , t ∈ [ t l , t l +1 ), n ≤ l ≤ N − 1. Applying Lemma 1 to bound the first term and applying standard in terp olation arguments for the second one (see (49) in the app endix), there exists a p ositive constan t such that v ( x, t n ) − v n h,k ( x ) ≤ C 1 h + C 2 k . (36) No w, w e need to b ound v n h,k ( x ) − v ( x, t n ). Let us denote by u ∈ U the con trol giving the minimum in (2) for t = t n . so that v ( x, t n ) = J x,t n ( u ) = Z T t n L ( y ( s, u ) , u ( s ) , s ) e − λ ( s − t n ) ds + g ( y ( T )) e − λ ( T − t n ) (37) The following argument uses the ideas from [5, App endix B] and [9, Theorem 7]. F or any t l w e can write y ( t ) = y ( t l ) + Z t t l f ( y ( s ) , u ( s ) , s ) ds. (38) Applying (6) ∥ y ( t ) − y ( t l ) ∥ ∞ ≤ M f h, t ∈ [ t l , t l +1 ] . (39) Then, for an y t ∈ [ t l , t l +1 ], using the ab ov e inequalit y and (8) we obtain     Z t t l f ( y ( s ) , u ( s ) , s ) − f ( y ( t l ) , u ( s ) , s ) ds     ∞ ≤ √ dL f M f h 2 . As a consequence, we get y ( t ) = y ( t l ) + Z t t l f ( y ( t l ) , u ( s ) , s ) ds + ϵ k , ∥ ϵ l ∥ ∞ ≤ √ dL f M f h 2 . (40) On the other hand, as in [5, (B.6a), (B.6b)], [9, Theorem 7], thanks to (CA), for an y l , there exists u l suc h that Z t l +1 t l f ( y ( t l ) , u ( s ) , t l ) ds = hf ( y ( t l ) , u l , t l ) (41) Z t l +1 t l L ( y ( t l ) , u ( s ) , t l ) ds = hL ( y ( t l ) , u l , t l ) . (42) 10 Let us also observe that, applying (41) and (9) Z t l +1 t l ( f ( y ( t l ) , u ( s ) , s ) − f ( y ( t l ) , u l , s )) ds = Z t l +1 t l ( f ( y ( t l ) , u ( s ) , s ) − f ( y ( t l ) , u ( s ) , t l )) ds + Z t l +1 t l ( f ( y ( t l ) , u ( s ) , t l ) − f ( y ( t l ) , u l , t l )) ds + Z t l +1 t l ( f ( y ( t l ) , u l , t l ) − f ( y ( t l ) , u l , s )) ds ≤ 2 L f h 2 . (43) F rom (42) and (7) w e get Z t l +1 t l ( L ( y ( t l ) , u l , s ) − L ( y ( t l ) , u ( s ) , s )) e − λ ( s − t n ) ds ≤ Z t l +1 t l L ( y ( t l ) , u l , s )( e − λ ( s − t n ) − e − λ ( t l − t n ) ) ds + Z t l +1 t l ( L ( y ( t l ) , u l , s ) − L ( y ( t l ) , u ( s ) , t l ) e − λ ( t l − t n ) ds + Z t l +1 t l L ( y ( t l ) , u ( s ) , t l )( e − λ ( t l − t n ) − e − λ ( s − t n ) ) ds ≤ 2 λM L h 2 = C h 2 , C = 2 λM L . (44) F rom (40) and (43) y ( t l +1 ) = y ( t l ) + Z t l +1 t l f ( y ( t l ) , u ( s ) , s ) ds + ϵ l = y ( t l ) + Z t l +1 t l f ( y ( t l ) , u l , s ) ds + Z t l +1 t l ( f ( y ( t l ) , u ( s ) , s ) ds − f ( y ( t l ) , u l , s )) ds + ϵ l ≤ y ( t l ) + Z t l +1 t l f ( y ( t l ) , u l , s ) ds + ϵ l + 2 L f h 2 . (45) Let us denote by y pc the time-contin uous tray ectory solution with the same initial condition as y asso ciated to the control u pc ( t ) = u l , ∀ t ∈ [ t l , t l +1 ) , l = n, . . . , N − 1. Arguing as in (40) w e get y pc ( t ) = y pc ( t l ) + Z t t l f ( y pc ( t l ) , u l , s ) ds + ϵ l , ∥ ϵ l ∥ ∞ ≤ √ dL f M f h 2 . (46) Subtracting (46) from (45) (applied with l − 1) and using (8) w e obtain ∥ y ( t l ) − y pc ( t l ) ∥ ∞ ≤ ∥ y ( t l − 1 ) − y pc ( t l − 1 ) ∥ ∞ + √ dhL f ∥ y ( t l − 1 ) − y pc ( t l − 1 ) ∥ ∞ +2 ∥ ϵ l − 1 ∥ ∞ + 2 L f h 2 ≤ (1 + h √ dL f ) ∥ y ( t l − 1 ) − y pc ( t l − 1 ) ∥ ∞ + C h 2 , where C = 2 √ dL f M f + 2 L f . Since y ( y n ) = y pc ( t n ) = x by standard recursion w e get ∥ y ( t l ) − y pc ( t l ) ∥ ∞ ≤ (1 + h √ dL f ) N − n C ( N − n ) h 2 ≤ e √ dh ( N − n ) L f C ( N − n ) h 2 ≤ C h, (47) 11 for C = C T e √ dT L f . No w, for the control u ∈ U giving the minim um in (2) for t = t n and for u =  u n , . . . u N − 1  , denoting b y J n h,k ( x, u ) = J n h,k ( x, u , . . . , u ) . w e obtain v n h,k ( x ) − v ( x, t n ) ≤ J n h,k ( y , u ) − J x,t n ( u ) = J h,k ( y , u ) − J x,t n ( u pc ) + J x,t n ( u pc ) − J x,t n ( u ) . The first term on the right-hand side abov e is b ounded in Lemma 1 so that v n h,k ( x ) − v ( x, t n ) ≤ C ( h + k ) + J x,t n ( u pc ) − J x,t n ( u ) . T o conclude w e need to b ound the second term. W e write J x,t n ( u pc ) − J x,t n ( u ) = Z T t n ( L ( y pc ( s ) , u pc ( s ) , s ) − L ( y ( s ) , u ( s ) , s )) e − λ ( s − t n ) ds + e − λ ( T − t n ) ( g ( y pc ( T )) − g ( y ( T )) . (48) F or the second term on the righ t-hand side ab o v e , applying (14) w e get | e − λ ( T − t n ) ( g ( y pc ( T )) − g ( y ( T )) | ≤ C L g ∥ y pc ( T ) − y ( T ) ∥ 2 ≤ C h, where in the last inequality we hav e applied (47) with l = N . T o conclude we will b ound the first term on the right-hand side of (48). W e observ e that Z T t n ( L ( y pc ( s ) , u pc ( s ) , s ) − L ( y ( s ) , u ( s ) , s )) e − λ ( s − t n ) ds = N − 1 X l = n Z t l +1 t l ( L ( y pc ( s ) , u l , s ) − L ( y ( s ) , u ( s ) , s )) e − λ ( s − t n ) ds. Adding and subtracting terms w e get Z t l +1 t l ( L ( y pc ( s ) , u l , s ) − L ( y ( s ) , u ( s ) , s )) e − λ ( s − t n ) ds = Z t l +1 t l ( L ( y pc ( s ) , u l , s ) − L ( y ( s ) , u l , s )) e − λ ( s − t n ) ds + Z t l +1 t l ( L ( y ( s ) , u l , s ) − L ( y ( t l ) , u l , s )) e − λ ( s − t n ) ds + Z t l +1 t l ( L ( y ( t l ) , u l , s ) − L ( y ( t l ) , u ( s ) , s )) e − λ ( s − t n ) ds + Z t l +1 t l ( L ( y ( t l ) , u ( s ) , s ) − L ( y ( s ) , u ( s ) , s )) e − λ ( s − t n ) ds Applying (11) and (44) we get Z t l +1 t l ( L ( y pc ( s ) , u l , s ) − L ( y ( s ) , u ( s ) , s )) e − λ ( s − t n ) ds ≤ L L Z t l +1 t l ( ∥ y pc ( s ) − y ( s ) ∥ 2 + 2 ∥ y ( s ) − y ( t l ) ∥ 2 ) e − λ ( s − t n ) ds + C h 2 . 12 T aking into account the following decomp osition ∥ y pc ( s ) − y ( s ) ∥ 2 ≤ ∥ y pc ( s ) − y pc ( t l ) ∥ 2 + ∥ y pc ( t l ) − y ( t l ) ∥ 2 + ∥ y ( t l ) − y ( s ) ∥ 2 , and applying (39) (than can also b e applied to y pc ) and (47) we reach Z t l +1 t l ( L ( y pc ( s ) , u l , s ) − L ( y ( s ) , u ( s ) , s )) e − λ ( s − t n ) ds ≤ C h 2 . And then Z T t n ( L ( y pc ( s ) , u pc ( s ) , s ) − L ( y ( s ) , u ( s ) , s )) e − λ ( s − t n ) ds ≤ C h. As a consequence, we finally obtain v n h,k ( x ) − v ( x, t n ) ≤ C ( h + k ) , whic h finish the pro of. □ Remark 1 W e observ e that the term h 1 / (1+ β ) in [9, Theorem 7] (and, as a conse- quence, the reduction of the error in time in the rate of conv ergence of the metho d) comes from having an infinite horizon problem. In the finite horizon case we handle in this pap er there is no reduction in the rate of con vergence under assumption (CA). The linear dep endence of the constant C 2 in (34) on L u comes from the inter- p olation argumen ts in the app endix. It is clear that since we hav e a finite num b er of terms in (35) L u is alw ays a finite constant for an y h , k . How ever, we cannot pro ve that is indep endent on b oth h and k . On the other hand, in practice one can alw ays compute the v alue of L u in a n umerical exp eriment. In case, L u b eha v es for example as k − α with 0 < α < 1 w e still achiev e con vergence of order k 1 − α in space. It is also clear that to be able to obtain the full order (first order of conv ergence in time and space) we need a constant L u indep enden t of h and k . This assumption is like a discrete Lip c hitz-contin uity condition on the no dal v alues of the computed discrete controls. As can b e seen in the pro of of the app endix, Lipsc hit-con tinuit y of the functions is the minimal requirement to achiev e the full order of con vergence (first order) in the piece-wise linear in terp olation, see [4]. A In terp olation b ounds Lemma 2 F or any x ∈ Ω let us denote by  u 1 n , . . . , u n s n  , u ∈ U the c ontr ol giving the minimum v n h,k ( x ) = J n h,k ( x, u 1 n , . . . , u n s n ) , with  u 1 n , . . . , u n s n  the c ontr ols giving the minimum at the no des. L et u j n = n u j,n n , . . . , u j,N − 1 n o , 1 ≤ j ≤ n S . L et us define u = { u n , . . . , u N − 1 } as fol lows. F or ˆ y l (define d by (18) ) written as ˆ y l = P n s j =1 µ j ( ˆ y l ) x j then u l = P n s j =1 µ j ( ˆ y l ) u j n . L et us denote by J n h,k ( x, u ) = ˆ J n h,k ( x, u , . . . , u ) , 13 then    J n h,k ( x, u ) − ˆ J n h,k ( x, u 1 n , . . . , u n s n )    ≤ C k , (49) wher e the c onstant C dep ends line arly on L u in (35) . Pro of F ollowing (17)-(18) w e ha ve J 1 ,n h,k ( x ) := J n h,k ( x, u 1 n , . . . , u n s n ) = h N − 1 X j = n δ j − n h I k L ( ˆ y j , u 1 ,j n , . . . , u n s ,j n , t j ) + I k g ( ˆ y N ) e − λ ( T − t n ) , ˆ y j +1 = ˆ y j + hI k f ( ˆ y j , u 1 ,j n , . . . , u n s ,j n , t j ) , ˆ y n = x, j = n, . . . N − 1 . And, ˆ J 2 h,k := ˆ J n h,k ( x, u ) := h N − 1 X j = n δ j − n h I k L ( ˆ z j , I k u j , t j ) + I k g ( ˆ z N ) e − λ ( T − t n ) , ˆ z j +1 = ˆ z j + hI k f ( ˆ z j , I k u j , t j ) , ˆ z n = x, j = n, . . . N − 1 , where, for ˆ z j = P n s l =1 µ l ( ˆ z j ) x l then I k f ( ˆ z j , I k u j , t j ) = n s X l =1 µ l ( ˆ z j ) f ( x l , n s X m =1 µ m ( ˆ y j ) u m n , t j ) , I k L ( ˆ z j , I k u j , t j ) = n s X l =1 µ l ( ˆ z j ) L ( x l , n s X m =1 µ m ( ˆ y j ) u m n , t j ) . Let ˜ y ( s ) = ˆ y l and let ˜ z ( s ) = ˆ z l , l = [ s/h ]. Let ˜ u j ( s ) = u j,l n , ˜ u ( s ) = u l , s ∈ [ lh, ( l + 1) h ). Then, ˜ y ( s ) = x + Z [ s/h ] h t n I k f ( ˜ y ( τ ) , ˜ u 1 ( τ ) , . . . , ˜ u n s ( τ ) , [ τ /h ] h ) dτ , ˜ z ( s ) = x + Z [ s/h ] h t n I k f ( ˜ z ( τ ) , I k ˜ u ( τ ) , [ τ /h ] h ) dτ . W e then obtain ∥ ˜ y ( t ) − ˜ z ( t ) ∥ ∞ ≤ Z [ s/h ] h t n ∥ I k f ( ˜ y ( τ ) , ˜ u 1 ( τ ) , . . . , ˜ u n s ( τ ) , [ τ /h ] h ) − I k f ( ˜ z ( τ ) , I k ˜ u ( τ ) , [ τ /h ] h ) ∥ ∞ dτ . W e ha ve to b ound the difference of the t wo p olynomials. F or any s ∈ [ l h, ( l + 1) h ) w e will bound n s X j =1 µ j ( ˆ y l ) f ( x j , u j n , t l ) − n s X j =1 µ j ( ˆ z l ) f ( x j , n s X m =1 µ m ( ˆ y k ) u m n , t l ) . (50) 14 W e decomp ose (50) in t w o terms I = n s X j =1 µ j ( ˆ y k ) f ( x j , u j n , t l ) − n s X j =1 µ j ( ˆ y k ) f ( x j , n s X m =1 µ l ( ˆ y k ) u m n , t l ) , I I = n s X j =1 µ j ( ˆ y k ) f ( x j , n s X m =1 µ l ( ˆ y k ) u m n , t l ) − n s X j =1 µ j ( ˆ z k ) f ( x j , n s X m =1 µ l ( ˆ y k ) u m n , t l ) . Let u ( y ) b e the piecewise linear function satisfying u ( x i ) = u i n , i = 1 , . . . , n s . W e observ e that I is the difference of tw o interpolants. The term on the left-hand side is the interpolant I k g ( y ) that interpolates the function g ( y ) = f ( y , u ( y ) , t l ) at x i for i = 1 , . . . , n s . The term on the righ t-hand side considers the funtion f ( y , u ( y ) , t l ) and interpolates only in the first argument. W e will denote this interpolant by I k f ( y , u ( y ) , t l ) such that for y = P n s j =1 µ j ( y ) x j I k f ( y , u ( y ) , t l ) = n s X j =1 µ j ( y ) f ( x j , u ( y ) , t l ) . W e decomp ose I in tw o terms | I | ≤ | I 1 | + | I 2 | := | I k g ( ˆ y k ) − g ( ˆ y k ) | + | f ( ˆ y k , u ( ˆ y k ) , t ) − I k f ( ˆ y k , u ( ˆ y k ) , t ) | . The first term can b e bounded as the interpolation error in the function g . F ollowing [4] the error is O ( k ) with a constant that dep ends on the Lipschitz constant of g . Let us observ e that applying (8), (10) ∥ g ( y 1 ) − g ( y 2 ) ∥ 2 = ∥ f ( y 1 , u ( y 1 ) , t l ) − f ( y 2 , u ( y 2 ) , t l ) ∥ 2 ≤ ∥ f ( y 1 , u ( y 1 ) , t l ) − f ( y 1 , u ( y 2 ) , t l ) ∥ 2 + ∥ f ( y 1 , u ( y 2 ) , t l ) − f ( y 2 , u ( y 2 ) , t ) |∥ 2 ≤ L f ( ∥ u ( y 1 ) − u ( y 2 ) ∥ 2 + ∥ y 1 − y 2 ∥ 2 ) . Since ∥ u ( y 1 ) − u ( y 2 ) ∥ 2 ≤ L u ∥ y 1 − y 2 ∥ 2 , (51) for L u defined in (35) then g has Lipschitz constan t L f ( L u + 1) and then | I 1 | ≤ C ( L f , L f ( L u + 1)) k . T o b ound the second term we also apply [4] and since the second argument is fixed the b ound dep ends only on the Lipschitz constan t of f so that | I 2 | ≤ C ( L f ) k . T o b ound I I w e observe that w e hav e the difference of t wo in terp olants in the y v ariable, the second argument is fixed, i.e., w e interpolate the function g ( y ) = f ( y , u ). The difference of t w o in terp olan ts can b e b ounded in terms of the v alues at whic h we in terp olate and the Lipsc hitz constant of the function g . Then | I I | ≤ C ( L f ) ∥ ˆ y k − ˆ z k ∥ ∞ = C ( L f ) ∥ ˜ y ( s ) − ˜ z ( s ) ∥ ∞ . 15 Then, we reach ∥ ˜ y ( s ) − ˜ z ( s ) ∥ ∞ ≤ C ( L f ) Z [ s/h ] h t n ∥ ˜ y ( τ ) − ˜ z ( τ ) ∥ ∞ dτ + C ( L f , L u )( T − t n ) k , and arguing as in (28) we conclude ∥ ˜ y ( t ) − ˜ z ( t ) ∥ ∞ ≤ C k . 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