If xt C2[t0, t1] provides a weak minimum resp., maximum in problem 19.1.3.6 and the regularity condition is satisfied i.e., the functions g i t are linearly independent on any of the int
Trang 1Let x(t)C2[t
0, t1] be an extremal of problem (19.1.3.6) with λ0=1; i.e., the Euler equation (19.1.3.4) is satisfied on this extremal for the Lagrangian
L (t, x, x t ) = f0(t, x, x t) +
m
i=1
λ i f i (t, x, x t
with some Lagrange multipliers λ i
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) (t0 ≤t≤t1). (19.1.3.7)
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) (t0 ≤t≤t1) (19.1.3.8) The equation
xL xx +x t L x
t x–dt d xL x
t x +x t L x
t x t
+
m
i=1
μ i g i=0, g i= –dt d (f i)x
t +(f i)x (19.1.3.9)
is called the (inhomogeneous) Jacobi equation for isoperimetric problem (19.1.3.6) on the extremal x(t); μ i are Lagrange multipliers (i =1,2, , n).
Suppose that the strengthened Legendre condition (19.1.3.8) is satisfied on the
ex-tremal x(t) A point τ is said to be conjugate to the point t0 if there exists a nontrivial solution of the Jacobi equation such that
0 g i (t)h(t) dt =0 (i =1,2, , m), where h(t) is an arbitrary smooth function satisfying the conditions h(0) = h(τ ) =0
We say that the Jacobi condition (resp., strengthened Jacobi condition) is satisfied on the extremal x(t) if the interval (t0, t1) (resp., the half-interval (t0, t1]) does not contain
points conjugate to t0
A point τ is conjugate to t0if and only if the matrix
H (τ ) =
⎛
⎜
⎜
⎝
h0(τ ) · · · h m (τ )
7τ
t0g1(t)h0(t) dt · · · 7t τ0 g1(t)h m (t) dt
7τ
t0g m (t)h0(t) dt · · · 7t τ0g m (t)h m (t) dt
⎞
⎟
⎟
⎠
is degenerate
Trang 2Necessary conditions for weak minimum (maximum):
Suppose that the Lagrangians f i (t, x, x t ) (i = 0,1, , m) in problem (19.1.3.6) are sufficiently smooth If x(t) C2[t0, t1] provides a weak minimum (resp., maximum) in
problem (19.1.3.6) and the regularity condition is satisfied (i.e., the functions g i (t) are linearly independent on any of the intervals [t0, τ ] and [τ , t1] for any τ ), then x(t) is an extremal of problem (19.1.3.6) and the Legendre and Jacobi conditions are satisfied on x(t) Sufficient conditions for a strong minimum (resp., maximum):
Suppose that the Lagrangian
L = f0+
m
i=1
μ i f i
is sufficiently smooth and the strengthened Legendre and Jacobi conditions, as well as the
regularity condition, are satisfied on an admissible extremal x(t) Then x(t) provides a
strong minimum (resp., maximum)
THEOREM Suppose that the functionalJ0in problem (19.1.3.6) is quadratic, i.e.,
J0 [x] =
t1
t0
A0(x t 2+ B0x2
dt,
and the functionalsJ iare linear,
J i [x] =
t1
t0
a i (x t2+ b i x2
dt (i =1,2, , m).
Moreover, assume that the functions A0, a1, , a m are continuously differentiable, the
functions B0, b1, , b m are continuous, and the strengthened Legendre condition and the regularity condition are satisfied If the Jacobi condition does not hold, then the lower bound in the problem is –∞ (resp., the upper bound is +∞) If the Jacobi condition holds,
then there exists a unique admissible extremal that provides the absolute minimum (resp., maximum)
Example 2 Consider the problem
J =
2π
0
(x t)2– x2
dt → min;
2π
0
x dt, x(0) = x(2π) =0
A necessary condition is given by the Lagrange multiplier rule (19.1.3.5): x tt + x – λ =0 The general
solution of the resulting equation with the condition x(0 ) = 0taken into account is x(t) = A sin t + B(cos t –1 ).
The set of admissible extremals always contains the admissible extremal ˆx(t)≡ 0
The Legendre condition (19.1.3.8) is satisfied: L x
t x
t (t, ˆx, ˆx t) = 2 > 0 The Jacobi equation (19.1.3.9)
coincides with the Euler equation (19.1.3.5) The solution h0(t) of the homogeneous equation with the conditions h0 ( 0 ) = 0and (h0 ) t( 0 ) = 1is the function sin t The solution h1(t) of the homogeneous equation
x tt + x +1 = 0with the conditions h1 ( 0 ) = 0and (h1 ) t( 0 ) = 0is the function cos t –1 The matrix H(τ ) acquires
the form
H (τ ) =
h
0(τ ) h0(τ )
7τ
0 h0g1dt
7τ
0 h m g1dt
=
sin τ cos τ –1
1– cos τ sin τ – τ
Thus the conjugate points are the solutions of the equation
det H(τ ) =2 – 2cos τ – τ sin τ =0 ⇔ sin τ2 = 0 , τ
2 = tan
τ
2.
The conjugate point nearest to zero is τ =2π.
Thus the admissible extremals have the form ˆx(t) = C sin t and provide the absolute minimum J [ˆx] =0
Trang 319.1.4 Problems with Higher Derivatives
19.1.4-1 Statement of problem Necessary condition for extremum
A problem with higher derivatives (with fixed endpoints) in classical calculus of variations
is the following extremal problem in the space C n [t0, t1]:
J [x] =
t1
t0
f0(t, x, x t , , x(t n) ) dt → extremum; (19.1.4.1)
x(k)
t (t j ) = x k j (k =0,1, , n –1, j =0,1) (19.1.4.2)
Here L is a function of n +2variables, which is called the Lagrangian Functions x(t)
C n [t0, t1] satisfying conditions (19.1.4.2) at the endpoints are said to be admissible.
An admissible function ˆx(t) is said to provide a weak local minimum (or maximum) in problem (19.1.4.1) if there exists a δ >0such that the inequality
J [x]≥J [ˆx] (resp., J [x]≤J [ˆx])
holds for any admissible function x(t)C n [t0, t1] satisfyingx – ˆx n < δ.
An admissible function ˆx(t)P C n [t0, t1] is said to provide a strong minimum (resp., maximum) in problem (19.1.4.1) if there exists an ε >0such that the inequality
J [x]≥J [ˆx] (resp., J [x]≤J [ˆx])
holds for any admissible function x(t)P C n [t0, t1] satisfyingx(t) – ˆx(t) n–1 < ε Necessary condition for extremum:
Suppose that the Lagrangian L is continuous together with its derivatives with respect
to x, x t , , x(t n) (the smoothness condition) for all t[t0, t1] If the function x(t) provides
a local extremum in problem (19.1.4.1), then L x(k)
t
C k [t0, t1] (k = 1,2, , n) and the Euler–Poisson equation holds:
n
k=0
(–1)k d k
dt k L x(k)
t =0 (t0 ≤t≤t1). (19.1.4.3)
For n =1, the Euler–Poisson equation coincides with the Euler equation (19.1.2.5) For
n=2, the Euler–Poisson equation has the form
d2
dt2L x
tt– d
dt L x
t + L x =0
The general solution of equation (19.1.4.3) contains 2n arbitrary constants These constants can be determined from the boundary conditions (19.1.4.2)
19.1.4-2 Higher-order necessary and sufficient conditions
Consider the problem
J [x] =
t1
t0
f0(t, x, x t , , x(t n) ) dt → min (or max);
x(k)
t (t j ) = x k j (k =0,1, , n –1, j =0,1)
(19.1.4.4)
Trang 4with higher derivatives, where L is the function of n +1 variables Suppose that x(t)
C2n [t0, t1] is an extremal of problem (19.1.4.4), i.e., the Euler-Poisson equation is satisfied
on this extremal
Legendre condition: If an extremal provides a minimum (resp., maximum) of the
functional, then the following inequality holds:
L x(n)
t x(n)
t ≥ 0 (resp., L x(n)
t x(n)
t ≤ 0) (t0 ≤t≤t1). (19.1.4.5)
Strengthened Legendre condition: If an extremal provides a minimum (resp., maximum)
of the functional, then the following inequality holds:
L x(n)
t x(n)
t >0 (resp., L x(n)
t x(n)
t <0) (t0≤t≤t1) (19.1.4.6) The functionalJ has the second derivative at the point x(t): J
tt [x, x] = K[x], where
K[x] =
t1
t0
n
i,j=0
L x(i)
t x(j)
t (t, x, x t , , x(t n) )x(t i) x(j)
t dt. (19.1.4.7)
The Euler–Poisson equation (19.1.4.3) for the functionalK is called the Jacobi equation
for problem (19.1.4.4) on the extremal ˆx(t).
For a quadratic functionalK of the form
K[x] =
t1
t0
n
i=0
L x(i)
t x(i)
t (t, x, x t , , x(t n)) x(i)
t 2
dt, (19.1.4.8)
the Jacobi equation reads
n
i=0
(–1)i d
i
dt i
L x(i)
t x(i)
t x(i) t
=0
Suppose that the strengthened Legendre condition (19.1.4.6) is satisfied on an extremal
x (t) A point τ is said to be conjugate to the point t0if there exists a nontrivial solution
h (t) of the Jacobi equation such that h(t i) (t0) = h(t i) (τ ) =0(i =0,1, , n –1) One says that
the Jacobi condition (resp., the strengthened Jacobi condition) is satisfied on the extremal
x (t) if the interval (t0, t1) (resp., the half-interval (t0, t1]) does not contain points conjugate
to t0
The Jacobi equation is a 2nth-order linear equation that can be solved for the higher
derivative Suppose that h1(t), , h n (t) are solutions of the Jacobi equation such that
H (t0) =0and H t(n) (t0) is a nondegenerate matrix, where
H (τ ) =
⎛
⎝ h1(τ ) . · · · h n .(τ )
[h1(τ )](t n–1) · · · [h n (τ )](t n–1)
⎞
⎠, H(n)
t (τ ) =
⎛
⎝ [h1(τ )]
(n)
t · · · [h n (τ )](t n)
[h1(τ )](t2n–1) · · · [h n (τ )](t2n–1)
⎞
⎠
A point τ is conjugate to t0if and only if the matrix H(τ ) is degenerate.
Trang 5Necessary conditions for weak minimum (resp., maximum):
Suppose that the Lagrangian L of problem (19.1.4.4) satisfies the smoothness condition.
If a function x(t)C2n [t0, t1] provides a weak minimum (resp., maximum), then x(t) is an extremal and the Legendre and Jacobi conditions hold on x(t).
Sufficient conditions for strong minimum (resp., maximum):
Suppose that the Lagrangian L is sufficiently smooth If x(t) C2n [t0, t1] is an
admissible extremal and the strengthened Legendre condition and the strengthened Jacobi
condition are satisfied on x(t), then x(t) provides a strong minimum (resp., maximum) in
problem (19.1.4.4)
For quadratic functionals of the form (19.1.4.8), the problem can be examined com-pletely
THEOREM Suppose that the functional has the form (19.1.4.8), L x(i)
t x(i) t
C i [t0, t1], and the strengthened Legendre condition is satisfied If the Jacobi condition does not hold, then the lower bound in the problem is –∞ (the upper bound is +∞) If the Jacobi condition is
satisfied, then there exists a unique admissible extremal that provides the absolute minimum (maximum)
Example 3 Consider the problem
J [x] =
2π
0
(x tt)2– (x t)2
dt → extremum; x = x(t), x(0) = x(2π) = x t( 0) = x t( 2π) =0
A necessary condition is given by the Euler–Poisson equation (19.1.4.3): x tttt + x tt= 0 The general
solution of this equation is x(t) = C1sin t + C2cos t + C3t+ C4 The set of admissible extremals always contains
the admissible extremal ˆx(t)≡ 0
The Legendre condition L x
t x
t (t, ˆx, ˆx t , ˆx tt) = 2 > 0 is satisfied The Jacobi equation coincides with the
Euler-Poisson equation If we set h1(t) =1– cos t and h2(t) = sin t – t, then the matrix H(t) acquires the form
H(t) =
h
1(t) h2(t) [h1(t)] t [h2(t)] t
=
1– cos t sin t – t
sin t cos t –1
.
Then H(0 ) = 0 and
det H tt ( 0 ) = [h
1(t)] tt [h2(t)] tt [h1(t)] ttt [h2(t)] ttt
= 1 0
≠ 0 Thus the conjugate points are the solutions of the equation
det H(t) =2(cos t –1) – t sin t =0 ⇔ sin2t = 0 , 2t = tan
t
2.
The conjugate point nearest to zero is t1 = 2π.
Thus the admissible extremals have the form ˆx(t) = C(1–cos t) and provide the absolute minimum J [ˆx] =0
19.1.5 Lagrange Problem
19.1.5-1 Lagrange principle
The Lagrange problem is the following problem:
B0 (γ) → min;
B i (γ)≤ 0 (i =1,2, , m ),
B i (γ) =0 (i = m +1, m +2, , m),
(19.1.5.1)
(xα) t – ϕ(t, x) =0 for all t T, (19.1.5.2)
Trang 6where x ≡ x(t) ≡ (xα, xβ) ≡ (x1(t), , x n (t)) P C1(Γ, Rn), xα ≡ (x1(t), , x k (t))
P C1(Γ, Rk), x
β ≡ (x k+1(t), , x n (t)) P C1(Γ, Rn–k ), γ = (x, t0, t1), ϕ
P C(Γ, Rn),
t0, t1 Γ, t0< t1,Γ is a given finite interval, and
B i (x, t0, t1) =
t1
t0
f i (t, x, (x β) t ) dt + ψ i (t0, x(t0), t1, x(t1)) (i =0,1, , m) Here P C(Γ, Rn) is the space of piecewise continuous vector functions on the closed interval
derivative onΓ
The constraint (19.1.5.2) is the differential equation that is called the differential con-straint The differential constraint can be imposed on all coordinates x (i.e., k = n
in (19.1.5.2)) or be lacking altogether (k = 0) The element γ is called an admissible
element.
An admissible element ˆγ = (ˆx, ˆt0, ˆt1) provides a weak local minimum in the Lagrange problem if there exists a δ > 0 such that the inequality B0 (γ) ≥ B0 ( ˆγ) holds for any admissible element γ satisfying the condition γ – ˆγ C1 < δ, |t– ˆt0|< δ, and|t– ˆt1|< δ,
wherex C1 = max
t T |x|+ max
t T |x
t|
19.1.5-2 Necessary conditions for extremum Euler–Lagrange theorem
Suppose that ˆγ provides a weak local minimum in the Lagrange problem (19.1.5.1), and, moreover, the functions ϕ = (ϕ1, , ϕ n ) and f i (i =0,1, , m) and their partial derivatives are continuous in x in a neighborhood of{(t, ˆx|tΓ}and the functions ψ i (i =0,1, , m) are continuously differentiable in a neighborhood of the point (ˆt0, ˆx(ˆt0), ˆt1, ˆx(ˆt1)) (the smooth-ness condition)
Then there exist Lagrange multipliers λ i (i = 0, , m) and p j ≡ p j (t) P C1(T )
(j =1, , k) that are not zero simultaneously, such that the Lagrange function
Λ =
t1
t0
i=0
λ i f i (t, x, (x β) t) +
k
i=1
p i
*
(x i) t – ϕ i (t, x)+4
dt+
m
i=0
λ i ψ i (t0, x(t0), t1, x(t1))
satisfies the following conditions:
1 The conditions of stationarity with respect to x, i.e., the Euler equations
dp i
dt +
k
j=1
p j ∂ϕ ∂x j
i =
m
j=0
λ j ∂f ∂x j
i (i =1,2, , k) for all tT,
where all derivatives with respect to x k are evaluated at (t, ˆx).
2 The conditions of transversality with respect to x,
p i (ˆt j) = (–1)
k
j=0
λ j ∂x ∂ψ j
i (t j (j =0,1; i =1,2, , k),
where all derivatives with respect to x i (t k ) (k =0,1) are evaluated at (ˆt0, ˆx(ˆt0), ˆt1, ˆx(ˆt1))
3 The conditions of stationarity with respect to t k (only for movable endpoints of the integration interval),
Λt k (ˆt k) =0 (k =0,1)
Trang 74 The complementary slackness conditions
λ iBi ( ˆγ) =0 (i =1,2, , m )
5 The nonnegativity conditions
λ i≥ 0 (i =1,2, , m )
19.1.6 Pontryagin Maximum Principle
19.1.6-1 Statement of problem
The optimal control problem (in Pontryagin’s form) is the problem
B0 (ω) → min;
B i (ω)≤ 0 (i =1,2, , m ),
B i (ω) =0 (i = m +1, m +2, , m),
(19.1.6.1)
x t – ϕ(t, x, u) =0 for all tT, (19.1.6.2)
where x≡ x(t) P C1(Γ, Rn), u ≡ u(t) P C(Γ, Rr ), ω = (x, u, t0, t1), ϕ
P C(Γ, Rn),
t0, t1 Γ, t0< t1,Γ is a given finite interval, U ⊂ R r is an arbitrary set, T ⊂ Γ is the set of
continuity points of u, and
B i (x, u, t0, t1) =
t1
t0
f i (t, x, u) dt + ψ i (t0, x(t0), t1, x(t1)) (i =0,1, , m).
Here P C(Γ, Rn) is the space of piecewise continuous vector functions on the closed interval
derivative onΓ
The vector function x = (x1(t), , x n (t)) is called the phase variable, and the vector
function u = (u1(t), , u r (t)) is called the control The constraint (19.1.6.2) is a differential equation that is called a differential constraint In contrast with the Lagrange problem, this
problem contains the inclusion-type constraint (19.1.6.3), which should be satisfied at all
points tΓ, and, moreover, the phase variable x can be less smooth.
An element ω = (x, u, t0, t1) for which all conditions and constraints of the problem
are satisfied is called an admissible controlled process An admissible controlled process
ˆ
ω = (ˆx, ˆu, ˆt0, ˆt1) is called a (locally) optimal process (or a process optimal in the strong sense) if there exists a δ >0such thatB0 (ω)≥B0( ˆω) for any admissible controlled process
ω = (x, u, t0, t1) such that
ω – ˆω C < δ, |t– ˆt0|< δ, |t– ˆt1|< δ,
wherex C = max
t Γ |x|