7 OPTIMAL CONTROL 7.1 EXERCISE 1. Solve the following optimal control problem. max. (x u 2 )dt x = u x(0) = Solution with the first variation

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1 7 OPTIMAL CONTROL 7. EXERCISE Solve the following optimal control problem max 7.. Solution with the first variation The Lagrangian is L(x, u, λ, μ) = (x u )dt x = u x() =. [(x u ) λ(x u)]dt μ(x() ). Now perform the first variation δl(x, u, λ, μ) = d L(x + αδ dα x,u+ αδ u,λ+ αδ λ,μ+ αδ μ ) α= Making the substitution yields d δl = lim [x + αδ x (u + αδ u ) (λ + αδ λ )(x + αδ x u αδ u )]dt α dα (μ + αδ μ )(x() + αδ x() ) After differentiation one has δl = lim α And collecting terms [δ x (u + αδ u )δ u δ λ (x + αδ x u αδ u ) (λ + αδ λ )(δ x δ u )]dt δμ(x() + αδ x() ) (μ + αδ μ )δ x() = [δ x uδ u δ λ (x u) λ(δ x δ u )]dt δ μ (x() ) μδ x() 97

2 98 OPTIMAL CONTROL One can derive directly this last passage skipping the above calculations if performs the differentiation of each variable. Now there is the problem to express the variation δ x in terms of the other variations. To see that consider the term involving δ x, that is λδ x.if one differentiates λδ x with respect to t has d dt [λδ x]=λ δ x + λδ x = λδ x = d dt [λδ x] λ δ x. Using this relation in the above expression gives δl = [δ x uδ u δ λ (x u) d dt (λδ x)+λ δ x + λδ u ]dt δ μ (x() ) μδ x(). Now collecting the variations in order to use the du Bois-Reymond theorem leads to δl = [δ x ( + λ )+δ u ( u + λ)+δ λ (u x )]dt λ()δ x() +(λ() μ)δ x() δμ(x() ) This leads to this system of ordinary differential equations, +λ = λ() = u + λ = λ() μ = u x = x() = From the first differential equation one has λ = t + c, then using the initial condition λ() = yields λ() = +c = = c =. So the multiplier is λ(t) = t +. From the multiplier one can resolve the optimal control u(t), in facts from the second differential equation u + λ = = u(t) = λ(t) = t+. Finally one can reconstruct the state variable x(t). From the first differential equation x = t+ = x(t) = 4 t + t + c. Now from the initial condition on x(t) one has x() = c =. In general, to prove the fact that this control maximizes the integral is a difficult task, therefore it is better to rely on special theorems. 7. EXERCISE Solve the following optimal control problem (OCP). J = u(t) x(t) dt with x (t) = u(t) x() =, x() =.

3 7. EXERCISE Solution with variations calculus To solve the problem, one can use calculus of variation in order to find a maximum of the functional J, so J =max J =max The Lagrangian function is given by L(x, u, λ, μ,μ )= Now perforg the first variation of L yields δl = u(t) +x(t) dt. u +x λ(x +u) dt μ (x() ) μ (x() ) uδ u +δ x λ(δ x +δ u ) δ λ (x +u) dt μ δ x() δ μ (x() ) μ δ x() δ μ (x() ). To simplify the variation δ x it is enough to derive λδ x, (λδ x ) = λ δ x + λδ x = λδ x =(λδ x ) λ δ x, so the previous expression becomes δl = uδ u +δ x + λ δ x λδ u δ λ (x +u) dt λ()δ x() + λ()δ x() μ δ x() δ μ (x() ) μ δ x() δ μ (x() ). Collecting the expression of each variation gives the associated boundary value problem. δ u : u λ δ x : +λ δ λ : x +u δ x() : λ() μ δ x() : λ() μ δ μ : x() δ μ : x() From the variation δ x one can solve the multiplier λ, in facts λ = = λ(t) = t + c c R. From variation δ u one can solve the optimal control u, u λ = u ( t + c) = = u =t c. From the differential equation given by the variation of the multiplier λ x = u = x = t + c = x(t) = t +(c +)t + d d R.

4 OPTIMAL CONTROL Now the constants c, d can be evaluated using the boundary condition x() = and x() =. x() = d = = d = x() = 4+c ++= = c =. In conclusion the optimal control is u(t) =t and the associated trajectory or status is x(t) = t + 3 t +. The extremal value of the functional is therefore ( max u(t) +x(t) dt = t + ( t ) + 3 ) t + dt = 6t +5t +7/4 dt = t t t = 5. If one tries another function that satisfies the required constraints, for example x(t) = (t )(t ) and the associated control u(t) =t 5, that the functional J gives < It can be showed, for example using the Hamiltonian function or with the second variation, that it is a maximum point, so for the original problem it will be a imum. 7.3 EXERCISE 3 Solve the following OCP, x u dt x =x + u x() = Solution with the Hamiltonian The OCP has a quadratic control, so we can apply the Hamiltonian to solve it, H = x u + λ(x + u). The equation for the control is H/ u = u + λ =which implies u = λ/. The adjoint equation is λ =(x λ), thus we have to solve a system of two differential equations. The boundary conditions for this system are x() = and λ() =. To H u = <

5 7.4 EXERCISE 4 solve the system there are various techniques as the Laplace transform or differential equations methods, we choose the latter. Differentiating the adjoint equation yields, λ =(x λ )=(x + u x +λ) =(λ + u) =4λ +u =4λ + λ =5λ. This is a second order differential equation which has boundary conditions given by λ() = and λ () = 5λ() =. Its general solution is λ(t) =c e 5t + c e 5t, substituting the boundary conditions gives c = c =therefore λ(t) =. This implies u(t) =and the differential equation for the state x becomes x =x. The last equation solves in x(t) =ke t with x() = k =, that is x(t) =e t. Let us check if the candidate solution u(t) =is a imum, looking at the H/ u = <, shows that u(t) =can not be a imum, so this problem does not have solution. 7.4 EXERCISE 4 Solve the following OCP, /3 u +x dt x =u + x x() = Solution with the Hamiltonian The OCP has a quadratic control, so we can apply the Hamiltonian to solve it, H = u +x + λ(u + x). The equation for the control is H/ u =u +λ =which implies u = λ. The adjoint equation is λ = 4x λ, thus we have to solve a system of two differential equations. The boundary conditions for this system are x() = and λ(/3) =. Differentiating the adjoint equation yields, λ = 4x λ = 4(u + x) λ = 8u 4x +4x + λ = 8u + λ =9λ. This is a second order differential equation which has boundary conditions given by λ(/3) = and λ (/3) = 9λ(/3) =. Its general solution is λ(t) =c + c e 5t, substituting the boundary conditions gives c = c =therefore λ(t) =. This implies u(t) =and the differential equation for the state x becomes x =x. The last equation solves in x(t) =ke t with x() = k =, that is x(t) =e t. Let us check if the candidate solution u(t) =is a imum, looking at the H/ u = <, shows that u(t) =can not be a imum, so this problem does not have solution.

6 OPTIMAL CONTROL 7.5 EXERCISE 5 x dt x = u x() =, x() =, u Solution with the Hamiltonian t t 6 x(t) =, λ(t) =t 6, u(t) =. t + 6<t H = x + λu. The adjoint equation is λ = H/ x =, which implies that λ(t) =t + c for an unknown constant c. From the Pontryagin maximum principle u = arg λu = signλ. Hence t + c< u =? t + c = t + c>. The problem can have one or zero switching points, depending on the value of c. If c then λ(t) < on the whole [, ] and there are not switching points, thus x(t) =t + k for a constant k to be detered from the boundary conditions. It is clear that a single line x(t) =t + k can not match x() = and x() =, therefore this case does not occur. The same conclusion holds for the case c because x(t) = t + k can not meet the boundary conditions. If there is a switching point, it occurs when λ(t )=t + c =, that is if t = c. This is possible if and only if c (, ), so in the first segment λ <, u = and x(t) = t + k for t =[, c]. Substituting the left boundary condition x(t) =t. In the second segment for t [ c, ] we have u = because λ>, thus x(t) = t + k and the right boundary condition yields k =. From the fact that x(t) is a continuous function, the two segments have to join for t = t = c, i.e. c = c +which gives c = EXERCISE 6 4 u +u x dt x = x + u x() =, u 3.

7 7.7 EXERCISE Solution with the Hamiltonian ( e 4 x(t) = 4 ) e t e4 4 e t + 3, λ(t) = e4 e t, u(t) = 3+e4 e t. The adjoint equation is λ H = u +u x + λ(x + u) = H/ x = λ with λ(4) =, which implies that λ(t) =+c e t for an unknown constant c. Imposing the boundary condition gives c = e 4, thus λ(t) = e 4 e t. The control is quadratic in the Hamiltonian and it can be solved by H/ u =u ++λ =, that implies u = λ = +e4 e t = 3+e4 e t > 3. We have to solve now the differential equation for the state x, x = x + u. Wedoit applying the Laplace transform. sx x() = X + e4 s + 3 s (s )X =+e4 s + 3 s. Dividing for the LHS and perforg partial fraction decomposition we have X = s + e4 s 3 = s + e4 = s 3 s(s ) ( s s s + e4 s + 3 s. The solution is obtained by the inverse Laplace transform and is x(t) = et + e4 sinh t + 3 ( e 4 = 4 ) e t e4 4 e t + 3. ) 7.7 EXERCISE 7 x() + x u dt x =+u x() =.

8 4 OPTIMAL CONTROL 7.7. Solution with the Hamiltonian x(t) =t + 4(3 t) +, λ(t) = t +3, u(t) = (3 t). H = x u + λ( + u ) The adjoint equation is λ = H/ x = with λ() =, which implies that λ(t) = t + c for an unknown constant c. From the final condition we have that c =3and λ(t) = t +3. The control is quadratic in the Hamiltonian and it can be solved by H/ u = + λu =, that implies u(t) = λ(t) = (3 t). The differential equation for the state is x =+u which yields to x (t) =+ 4(3 t) = x(t) =t t + k for a constant k that can be retrieved by the initial condition x() =, i.e. x() = + + k = = k =. 7.8 EXERCISE 8 3x +u dt x = e u + t 3 x() = e, u Solution with the Hamiltonian x(t) = 4 t4 + e t, 4 λ(t) = 3t +6, u(t) =. H =3x +u + λ(e u + t 3 ). The adjoint equation is λ = H/ x = 3 with λ() =, which implies that λ(t) = 3t + c for an unknown constant c. From the final condition we have that c =6and λ(t) = 3t +6. The control is not linear in the Hamiltonian and it can be solved by H/ u = λe u =, that implies ( ) 3t +6 λ = e u u =ln ln λ = u(t) = ln( 3t +6) ln = ln,

9 7.9 EXERCISE 9 5 but this control violates the constraint u 3, in facts it results u<, therefore u(t) =. The differential equation for the state is x = e + t 3 which yields to x(t) = 4 t4 + e t + k = x() = e = 4 + e + k = k = EXERCISE 9 xe t dt x = et u + x x() =, u Solution with the Hamiltonian (t )et x(t) =, λ(t) =e t e 4 e t, u(t) =. ( ) e H = xe t t + λ u(t) + x. The adjoint equation is λ = H/ x =e t λ with λ() =, which implies that λ(t) =c e t + c e t for an unknown constant c,c. From the condition λ() = we have that c =and c = e 4 (in facts λ + λ =c e t =e t ), hence λ(t) =e t e 4 e t. The control can be solved by Pontryagin theorem and gives λe t u = et e 4 u for t [, ] and u>, therefore it violates the constraint and thus u =. The differential equation for the state yields to x(t) = (t )et.