Handbook of mathematics for engineers and scienteists part 148 pptx

19.1. CALCULUS OF VARIATIONS AND OPTIMAL CONTROL 997 Example 3 (Hilbert’s example). In this example, a unique admissible extremal exists that provides a global extremum but is not a differentiable function. Let J[x]=  1 0 t 2/3 (x  t ) 2 dt → min; x = x(t), x(0)=0, x(1)=1. The Euler equation becomes d dt  2t 2/3 x  t  = 0. The extremals are given by the equation x(t)=C 1 t 1/3 + C 2 . The unique extremal satisfying the endpoint conditions is the function ˆx(t)=t 1/3 , which does not belong to the class C 1 [0, 1]. Nevertheless, it provides the global minimum in this problem on the set of all absolutely continuous functions x(t) that satisfy the boundary conditions and for which the integral J is finite. Indeed, J[x]=J[ˆx + h]=  1 0 t 2/3  1 3t 2/3 + h  t  2 dt = J[ˆx]+ 2 3  1 0 h  t dt +  1 0 t 2/3 (h  t ) 2 dt ≥ J[ˆx]. Example 4 (Weierstrass’s example). In this example, the problem has no solutions and no admissible extremals even in the class of absolutely continuous functions. Let J[x]=  1 0 t 2 (x  t ) 2 dt → min; x = x(t), x(0)=0, x(1)=1. The Euler equation becomes d dt (2t 2 x  t )=0. The extremals are given by the equation x(t)=C 1 t –1 + C 2 ; none of them satisfies the boundary condition x(0)=0.Furthermore,J[x] ≥ 0,andJ[x]>0 for any absolutely continuous function x(t). Obviously, the lower bound in this problem is zero. To prove this, it suffices to consider the sequence of admissible functions x n (t) = arctan(nt)/ arctan n. One has J[x]=  1 0 t 2 n 2 (1 + n 2 t 2 ) 2 arctan 2 n dt ≤  1/n 0 dt arctan 2 n +  1 1/n dt n 2 t 2 arctan 2 n → 0. Example 5. In this example, there exists a unique admissible extremal that does not provide an extremum. Let J[x]=  3π/2 0  (x  t ) 2 – x 2  dt → min; x = x(t), x(0)=x(3π/2)=0. The Euler equation becomes x  tt + x = 0. The extremals are given by the equation x(t)=C 1 sin t + C 2 cos t. The only extremal satisfying the endpoint conditions is ˆx(t)=0. The admissible extremal ˆx(t)=0 does not provide a local minimum. Consider the sequence of admissible functions x n (t)=n –1 sin(2t/3). Obviously, x n (t) → 0 in C 1 [0, 3π/2], but J[x n ]=– 5π 12n 2 < 0 = J[ˆx]. This example shows that the Euler equation is a necessary but not sufficient condition for extremum. 19.1.2-4. Broken extremals. Weierstrass–Erdmann conditions. If the Euler equation has a piecewise smooth solution, i.e., if x(t) has corner points (the function x  t (t) has a discontinuity), then the Weierstrass–Erdmann conditions L x  t [c, x(c), x  t (c – 0)] = L x  t [c, x(c), x  t (c + 0)], (19.1.2.16) L[c, x(c), x  t (c – 0)] – x  t (c – 0)L x  t [c, x(c), x  t (c – 0)] = L[c, x(c), x  t (c + 0)] – x  t (c + 0)L x  t [c, x(c), x  t (c + 0)] (19.1.2.17) hold at each point c that is the abscissa of a corner point. If x  t L x  t ≠ 0 (t 0 ≤ t ≤ t 1 ), then the Euler equation has only smooth solutions. Curves consisting of pieces of extremals and satisfying the Weierstrass–Erdmann con- ditions at the corner points are called broken extremals. 998 CALCULUS OF VARIATIONS AND OPTIMIZATION 19.1.2-5. Second-order necessary conditions. Legendre condition: If an extremal provides a minimum (resp., maximum) of the functional, then the following inequality holds: L x  t x  t ≥ 0 (resp., L x  t x  t ≤ 0)(t 0 ≤ t ≤ t 1 ). (19.1.2.18) Weierstrass condition: If a curve x(t) provides a strong minimum (resp., maximum) of the classical functional, then the Weierstrass function E(t, x, x  t , k) ≡ L(t, x, k)–L(t, x, x  t )–(k – x  t ) L x  t (t, x, x  t )(19.1.2.19) is nonnegative (resp., nonpositive) for arbitrary finite values k at all points (t, x)ofthe extremal. The equation L xx h + L xx  t h  t – d dt  L xx  t h + L x  t x  t h  t  = 0 (19.1.2.20) is called the Jacobi equation.Hereh(t) is an arbitrary smooth function satisfying the conditions h(t 0 )=h(t 1 )=0. If the Legendre condition L x  t x  t ≠ 0 (t 0 ≤ t ≤ t 1 ) is satisfied, then the Jacobi equation is a second-order linear equation that can be solved for the second derivative. It follows from the conditions h(t 0 )=h(t 1 )=0 that h(t) ≡ 0. A point τ is said to be conjugate to the point t 0 if there exists a nontrivial solution of the Jacobi equation such that h(t 0 )=h(τ)=0. Jacobi condition: If the extremal x(t)(t 0 ≤ t ≤ t 1 ) provides an extremum of the functional (19.1.2.1), then it does not contain points conjugate to t 0 . 19.1.2-6. Multidimensional case. The vector problem is posed and necessary conditions for an extremum are stated by analogy with the one-dimensional simplest problem of calculus of variations. Let x(t)= (x 1 (t), , x n (t)) be an n-dimensional vector function, and let the Lagrangian L be a function of 2n + 1 variables: L = L(t, x, x  t ). The vector problem has the form J[x] ≡  t 1 t 0 L(t, x, x  t ) dt → extremum; x = x(t), x i (t j )=x ij (i = 1, 2, , n, j = 0, 1). (19.1.2.21) A necessary condition for admissible vector function x(t) to provide weak extremum in problem (19.1.2.21) is that the function x(t) satisfies the system of n Euler differential equations L x i – d dt L (x i )  t = 0 (i = 1, 2, , n), (19.1.2.22) whereweassumethatthefunctionsL, L x , L x  t are continuous asfunctions of 2n+1 variables (t, x i ,and(x i )  t , i = 1, 2, , n), and L (x i )  t C 1 [t 0 , t 1 ]. The solutions of the system of Euler differential equations (19.1.2.22) are called extremals. Admissible functions satisfying the system of Euler differential equations are called admissible extremals. The derivation of conditions (19.1.2.5) and (19.1.2.22) from the relation δJ = 0 is based on the fundamental lemmas of calculus of variations. 19.1. CALCULUS OF VARIATIONS AND OPTIMAL CONTROL 999 L EMMA 1(FUNDAMENTAL LEMMA OR LAGRANGE’S LEMMA). Let f(t) C[t 0 , t 1 ] and assume that  t 1 t 0 f(t)g(t) dt = 0 for each function g(t) C 1 [t 0 , t 1 ] such that g(t 0 )=g(t 1 )=0 .Then f(t) ≡ 0 on [t 0 , t 1 ] . LEMMA 2. Let f(t) C[t 0 , t 1 ] and assume that  t 1 t 0 f(t)g(t) dt = 0 for each function g(t) C 1 [t 0 , t 1 ] such that  t 1 t 0 g(t) dt = 0. Then f(t)=const on [t 0 , t 1 ] . 19.1.2-7. Weierstrass–Erdmann conditions in multidimensional case. If the system of Euler differential equations has a piecewise smooth solution, i.e., if x(t) has corner points (the function x  t (t) has discontinuities), then the Weierstrass–Erdmann conditions L (x i )  t [c, x(c), x  t (c – 0)] = L (x i )  t [c, x(c), x  t (c + 0)] (i = 1, 2, , n), L[c, x(c), x  t (c – 0)] – n  i=1 (x i )  t    t=c–0 L (x i )  t [c, x(c), x  t (c – 0)] = L[c, x(c), x  t (c + 0)] – n  i=1 (x i )  t    t=c+0 L (x i )  t [c, x(c), x  t (c + 0)] hold at each point c that is the abscissa of a corner point. If (x i )  t L (x i )  t ≠ 0 (t 0 ≤ t ≤ t 1 , i = 1, 2, , n), then the system of Euler differential equations has only smooth solutions. Curves consisting of pieces of extremals and satisfying the Weierstrass–Erdmann con- ditions at the corner points are called broken extremals. 19.1.2-8. Higher-order necessary conditions in multidimensional case. We consider the simplest problem J[x] ≡  t 1 t 0 L(t, x, x  t ) dt → min (or max), x = x(t), x(t 0 )=x 0 , x(t 1 )=x 1 . (19.1.2.23) Legendre condition: If an extremal provides a minimum (resp., maximum) of the functional, then the following inequality holds: L x  t x  t ≡  L (x i )  t (x j )  t  ≥ 0 (resp., L x  t x  t ≤ 0)(t 0 ≤ t ≤ t 1 ; i, j = 1, 2, , n). (19.1.2.24) Strengthened Legendre condition: If an extremal provides a minimum (resp., maximum) of the functional, then the following inequality holds: L x  t x  t > 0 (resp., L x  t x  t < 0)(t 0 ≤ t ≤ t 1 ; i, j = 1, 2, , n). (19.1.2.25) 1000 CALCULUS OF VARIATIONS AND OPTIMIZATION Remark. In the vector case, L x  t x  t is an n × n matrix. The condition L x  t x  t ≥ 0 means that the matrix is positive semidefinite, and the condition L x  t x  t > 0 means that the matrix is positive definite. The Jacobi equation for the vector problem has a form similar to that in the one- dimensional case, i.e., L xx h + L xx  t h  t –(L xx  t h + L x  t x  t h  t )  t = 0,(19.1.2.26) where L xx ≡  L (x i )(x j )  , L xx  t ≡  L (x i )(x j )  t  ,andh=h(t) is an column-vector with components h i (t), which are arbitrary smooth functions satisfying the conditions h i (t 0 )=h i (t 1 )=0 (i = 1, 2, , n). Suppose that the strengthened Legendre condition is satisfied on an extremal x(t). A point τ is said to be conjugate to the point t 0 if there exists a nontrivial solution h(t)ofthe Jacobi equation such that h(t 0 )=h(τ)=0.TheJacobi condition (resp., the strengthened Jacobi condition)issaidtobesatisfied on an extremal x(t)iftheinterval(t 0 , t 1 ) (resp., the half-open interval (t 0 , t 1 ]) does not contain points conjugate to t 0 . We find the fundamental system of solutions of the Jacobi equation for the functions x(t) in the matrix form H(t, t 0 ) ≡ (h 1 (t), , h n (t)), h i (t) ≡ ⎛ ⎝ h i 1 (t) . . . h i n (t) ⎞ ⎠ , where H(t 0 , t 0 )=0 and H  t (t 0 , t 0 ) ≠ 0 or H  t (t 0 , t 0 )=I. Column-vectors h i (t)arethe solutions of the Jacobi system of equations. A point τ is conjugate to the point t 0 if and only if the matrix H(τ, t 0 ) is degenerate. Necessary conditions for weak minimum (resp., maximum): 1. If x(t) provides a weak minimum (resp., maximum), then the function x(t)isanadmis- sible extremal on which the Legendre and Jacobi conditions are satisfied. 2. If x(t) provides a strong local minimum (resp., maximum), then the Weierstrass function E(t, x, x  t , k)=L(t, x, k)–L(t, x, x  t )– n  i=1 [k i –(x i )  t ]L (x i )  t (t, x, x  t )(19.1.2.27) is nonnegative (resp., nonpositive) for arbitrary finite values k =(k 1 , , k n ) at all points (t, x) of the extremal. Sufficient condition for the weak minimum or maximum: If the strengthened Legendre and Jacobi conditions are satisfied on an admissible extremal, then this extremal provides a weak local minimum (or maximum). Example 6. J[x]=  1 0  [(x 1 )  t ] 2 +[(x 2 )  t ] 2 + 2x 1 x 2  dt → min; x 1 (0)=x 2 (0)=0, x 1 (1)=sin1, x 2 (1)=–sin1 (i = 1, 2). The Lagrangian is L =[(x 1 )  t ] 2 +[(x 2 )  t ] 2 + 2x 1 x 2 . A necessary condition is given by the system (19.1.2.22) of Euler equations –(x 1 )  tt + x 2 = 0,–(x 2 )  tt + x 1 = 0 =⇒ (x 1 )  tttt = x 1 ,(x 2 )  tttt = x 2 . The general solution of the Euler equations is x 1 (t)=C 1 sinh t + C 2 cosh t + C 3 sin t + C 4 cos t, x 2 (t)=C 1 sinh t + C 2 cosh t – C 3 sin t – C 4 cos t. 19.1. CALCULUS OF VARIATIONS AND OPTIMAL CONTROL 1001 It follows from the initial conditions that C 1 = C 2 = C 4 = 0 and C 3 = 1. The only admissible extremal is ˆx 1 (t)=sint,ˆx 2 (t)=–sint. Let us apply sufficient conditions. The strengthened Legendre condition (19.1.2.25) is satisfied. The system of Jacobi equations (19.1.2.26) coincides with the system of Euler equations, i.e., –(h 1 )  tt + h 2 = 0,–(h 2 )  tt + h 1 = 0. For H(t, 0), such that H(0, 0)=0 and H  t (0, 0)=I,wetakethematrix  1 2 (sinh t +sint) 1 2 (sinh t –sint) 1 2 (sinh t –sint) 1 2 (sinh t +sint)  . Then det H(t, 0)=sinht sin t. The conjugate points are τ = kπ, k = 1, 2, 3, The half-open interval (0, 1] does not contain conjugate points, and so the strengthened Jacobi condition is satisfied. Thus the vector ˆ x(t)=(ˆx 1 (t), ˆx 2 (t)) provides a local minimum of the functional J. 19.1.2-9. Bolza problem. 1 ◦ .TheBolza problem is the following extremal problem without constraints in the space C 1 [t 0 , t 1 ]: B[x] ≡  t 1 t 0 L(t, x, x  t ) dt + l(x(t 0 ), x(t 1 )) → extremum. (19.1.2.28) The function L(t, x, x  t ) is called the Lagrangian, the function l = l(x(t 0 ), x(t 1 )) is called the terminal cost function, and the functional B is called the Bolza functional.TheBolza problem is an elementary problem of classical calculus of variations. Any functions of class C 1 [t 0 , t 1 ] are admissible. An admissible function ˆx(t) C 1 [t 0 , t 1 ] provides a weak local minimum (maximum) in problem (19.1.2.28) if there exists a δ > 0 such that B[x] ≥ B[ˆx](B[x] ≤ B[ˆx]) (19.1.2.29) for any admissible function x(t) satisfying x –ˆx 1 < δ (see Paragraph 19.1.2-1). A necessary condition for a weak extremum of the Bolza functional has the same character as the necessary condition in the classical problem; i.e., the function providing an extremum of the Bolza functional must be a solution of a boundary value problem for the Euler equation (19.1.2.5). The boundary conditions are given by the relations L x  t (t 0 , x(t 0 ), x  t (t 0 )) = ∂l ∂[x(t 0 )] , L x  t (t 1 , x(t 1 ), x  t (t 1 )) = – ∂l ∂[x(t 1 )] (19.1.2.30) and are called the transversality conditions. If the function part of the Bolza functional is lacking, i.e., l ≡ 0, then conditions (19.1.2.30) acquire the form L x  t (t 0 , x(t 0 ), x  t (t 0 )) = L x  t (t 1 , x(t 1 ), x  t (t 1 )) = 0, which means in many applied problems that the extremal trajec- tories are orthogonal to the boundary vertical lines t = t 0 and t = t 1 .Ifafixing condition is given at one endpoint of the interval, then the fixing condition at this endpoint and the transversality condition at the other endpoint serve as the boundary conditions for the Euler equation. 1002 CALCULUS OF VARIATIONS AND OPTIMIZATION 2 ◦ .Then-dimensional problem is posed and necessary conditions for extremum are stated by analogy with the one-dimensional Bolza problem. Let x(t)=(x 1 (t), , x n (t)) be an n-dimensional vector function, and let the Lagrangian L be a function of 2n + 1 variables: L = L(t, x, x  t ). Then the vector problem reads  t 1 t 0 L(t, x, x  t ) dt + l(x(t 0 ), x(t 1 )) → extremum. (19.1.2.31) Example 7. Consider the problem B[x]=  1 0  (x  t ) 2 – x  dt + x 2 (1) → min . The Euler equation (19.1.2.5) has the form –1 – 2x  tt = 0, and the transversality conditions (19.1.2.30) are as follows: 2x  t (0)=0, 2x  t (1)=–2x(1). The extremals are given by the equation x(t)=– 1 4 t 2 +C 1 t+C 2 . The unique extremal satisfying the transversality conditions is ˆx(t)= 1 4 (3 –t 2 ). This admissible extremal provides the absolute minimum in the problem. Indeed, B[ˆx + h]–B[ˆx]=  1 0 2 ˆx  t h  t dt +  1 0 (h  t ) 2 dt –  1 0 hdt+ 2 ˆx(1)h(1)+[h  t (1)] 2 =  1 0 (h  t ) 2 dt +[h  t (1)] 2 ≥ 0 for an arbitrary function h(t) C 1 [t 0 , t 1 ]. Thus ˆx(t)= 1 4 (3 – t 2 ) provides the absolute minimum in the problem. Furthermore, B[ˆx]=– 1 3 . 19.1.3. Isoperimetric Problem 19.1.3-1. Statement of problem. Necessary condition for extremum. The isoperimetric problem (with fixed endpoints) in calculus of variations is the following extremal problem in the space C 1 [t 0 , t 1 ](orPC 1 [t 0 , t 1 ]): J 0 [x] ≡  t 1 t 0 f 0 (t, x, x  t ) dt → extremum; (19.1.3.1) J i [x] ≡  t 1 t 0 f i (t, x, x  t ) dt = α i (i = 1, 2, , m); (19.1.3.2) x(t 0 )=x 0 , x(t 1 )=x 1 ,(19.1.3.3) where α 1 , , α m R are given numbers. Constraints of the form (19.1.3.2) are said to be isoperimetric. The functions f i (t, x, x  t )(i = 1, 2, , m) are called the Lagrangians of the problem. Functions x(t) C 1 [t 0 , t 1 ] satisfying the isoperimetric conditions (19.1.3.2) and conditions (19.1.3.3) at the endpoints are said to be admissible. An admissible function ˆx(t) provides a weak local minimum (resp., maximum) in prob- lem (19.1.3.1) if there exists a δ > 0 such that the inequality J 0 [x] ≥ J 0 [ˆx] (resp., J 0 [x] ≤ J 0 [ˆx]) holds for any admissible function x(t) C 1 [t 0 , t 1 ] satisfying x –ˆx 1 < δ. 19.1. CALCULUS OF VARIATIONS AND OPTIMAL CONTROL 1003 Necessary condition for extremum (the Lagrange multiplier rule): Let f i (t, x, x  t )(i = 1, 2, , m) be functions continuous together with their partial derivatives (f i ) x and (f i ) x  t ,andletx(t) C 1 [t 0 , t 1 ]. If x(t) provides a weak local extremum in the isometric problem (19.1.3.1), then there exist Lagrange multipliers λ 0 , λ 1 , , λ m , not all zero simultaneously, such that for the Lagrangian L(t, x, x  t )= m  i=0 λ i f i (t, x, x  t )(19.1.3.4) the Euler equation L x – d dt L x  t = 0 (19.1.3.5) is satisfied. If the regularity condition is satisfied [the functions – d dt (f i ) x  t +(f i ) x (i = 1, 2, , m), are linearly independent], then λ 0 ≠ 0. One of the best-known isoperimetric problems, after which the entire class of problems was named, is the Dido problem. Example 1. Find a smooth curve of given length fixed at two points of a straight line and, together with the segment of the straight line, bounding the largest area. The formalized problem is  T 0 –T 0 xdt → max;  T 0 –T 0  1 +(x  t ) 2 dt = l, x = x(t), x(–T 0 )=x(T 0 )=0, where T 0 is given. The Lagrangian (19.1.3.4) has the form L = λ 0 x + λ  1 +(x  t ) 2 . A necessary condition is given by the Euler equation (19.1.3.5) λ 0 – d dt λx  t  1 +(x  t ) 2 = 0. Elementary calculations result in the first-order equation (C + x)  1 +(x  t ) 2 =–λ. We integrate this equation and obtain (C + x) 2 +(C 1 + t) 2 = λ 2 , which is the equation of a circle. It follows from the endpoint conditions x(–T 0 )=x(T 0 )thatC 1 = 0, i.e., (C + x) 2 + t 2 = λ 2 . The unknown constants C and λ are uniquely determined by the condition x(T 0 )=0 and the isoperimetric condition  T 0 –T 0  1 +(x  t ) 2 dt = l. For 2T 0 < l < πT 0 , there is a unique (up to the sign) extremal that is an arc of length l of the circle passing through the points ( T 0 , 0) and centered on the OX-axis. Since this is a maximization problem, we must take the extremal lying in the upper half-plane. For l < 2T 0 , there are no admissible functions in the problem, and for l > πT 0 , there are no extremals. 19.1.3-2. Higher-order necessary conditions. Sufficient conditions. Consider the isoperimetric problem J 0 [x] → min (or max); J i [x]=α i (i = 1, 2, , m); x(t 0 )=x 0 , x(t 1 )=x 1 . (19.1.3.6) . calculus of variations. 19.1. CALCULUS OF VARIATIONS AND OPTIMAL CONTROL 999 L EMMA 1(FUNDAMENTAL LEMMA OR LAGRANGE’S LEMMA). Let f(t) C[t 0 , t 1 ] and assume that  t 1 t 0 f(t)g(t) dt = 0 for each. system of solutions of the Jacobi equation for the functions x(t) in the matrix form H(t, t 0 ) ≡ (h 1 (t), , h n (t)), h i (t) ≡ ⎛ ⎝ h i 1 (t) . . . h i n (t) ⎞ ⎠ , where H(t 0 , t 0 )=0 and H  t (t 0 ,. serve as the boundary conditions for the Euler equation. 1002 CALCULUS OF VARIATIONS AND OPTIMIZATION 2 ◦ .Then-dimensional problem is posed and necessary conditions for extremum are stated by analogy