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Collecting the results of this section, the curve x&which minimizes the integral +, Yf must satisfy J= / ye- ye (1) Euler-Lagrange Equation d -4-l G’f --= af (2) Weierstrass-Erdman Corner Condition (3) Weierstrass Condition 02 af a? 0 (2.4.44~) (2.4.44~) r(x,y, Y’) - f(x,y,y)- (Y’y’) g, (x, y, y’l 2, o :2.4.44c) (4) Legendre Condition a2f - kjf,g’) 20 ay/’ (2.4.44~) In addition to these four conditions, a fifth necessary condition, the classical Jacobi condition, can also be developed by means of Dynamic Programming. Since this condition is rather difficult to apply and fre- quently does not hold in optimal control problems, it will not be developed here. The interested reader should consult Reference (2.4.1), page 103. 2.4-5 Lagrange Problem with Variable End Points In the preceeding sections the problem of minimizing the integral was considered where the limits of integration,(&,,$) and (x4,f6) were fi,zd. In this section a minor variation on this problem will be considered in which the upper limit of integration is not fixed precisely, but is required to lie in the curve p (X,f) = jz ‘xf,fp=o (2.4.45A) 71
The situation is pictured to the right. Note that the minimization of the integral involves both the selection of the optimal curvey(tx) and the optimal terminal point ‘5 ’ Vf ) pW+=O* end p int along the curve As in the fixed case, let */.v Yf Sketch (2.4.4) x Q(X ,y) = M/A/ s Y’ x:)y where the terminal point(+,#f ) lies on the curve of Eq. (2.4.45A). Following the procedure of Section (2.4.2), it can again be shown tnatR(x,y) satisfies the partial differential equation Y (2.4.45B) However, the boundary condition on /? is slightly different in this case. Since R(%,y) is the minimum value of the integral starting from the point CZ:,~) and terminating on the curve #x,5/)=0, it follows that R(z,y) is zero for any (x,y) satisfying g(;xr,$)=O ; that is, the value of the integral is zero since both limits of integration are the same. Hence, the boundary condition for Eq. (2.4.45B) is This condition can be put in an alternate form if the equation. can be solved fory/L'.e-, $ # 0 / . In this case and Eq. (2.4.46) becomes or RC%,y) = 0 ON L/ = Rh, 72 (2.4.46) pcs,y, = 0 (2.4.47) (2.4.48) (2.4.49)
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NASA OI 0 0 P; U 4 cd 4 z . . _ -.
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c FOREWORD This report was prepared
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SECTION PAGE 2.5 Dynamic Programmin
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2.0 STATE OF THE ART 2.1 Developmen
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There are several ways to accomplis
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2.2 Fundamental Concepts and Applic
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II. Although the previous problem w
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I,. Citg B. C optimum cost Path for
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The optimum path can be found by st
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The integral in Equation 2.2.1 can
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2.2.2.1 Shortest Distance Between T
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The cost of the allowable transitio
Page 28 and 29: In a,completely analogous manner th
Page 30 and 31: 2.2.2.2 Variational Problem with Mo
Page 32 and 33: Ii _ 6 5 9 3 I 8 I I The cost integ
Page 34: 2.2.2.3 Simple Guidance Problem As
Page 37 and 38: This process continues in the same
Page 39 and 40: The algorithm for solving this prob
Page 41 and 42: The reader, no doubt, has a reasona
Page 43 and 44: are applied to Min (fl), the range
Page 45 and 46: Note that each diagonal corresponds
Page 47 and 48: The value of X can be found by empl
Page 49 and 50: The initial condition is: K.&P : P(
Page 51 and 52: it is seen that (for a two stage pr
Page 53 and 54: The optimal policy can now be found
Page 55 and 56: 2.3 COMPUTATIONAL CONSIDERATIONS So
Page 57 and 58: If classical techniques were to be
Page 59 and 60: 2.3.3.1 The Curse of Dimensionalitv
Page 61 and 62: Next, fl is evaluated for all allow
Page 63 and 64: % Xl + 5 = 2 (A2 = 2) 5+X2=3 (A2=3.
Page 65 and 66: that minimizes f. This interchange
Page 67 and 68: 2.3.3.2 Stability and Sensitivity I
Page 69 and 70: Let the lower limit of integration
Page 71 and 72: But (2.4.10) since the function on
Page 73 and 74: Now,. noting that the second MIN op
Page 75 and 76: where v 'is determined from and wit
Page 77: NOW, combining these two expression
Page 81 and 82: Hence, the boundary condition OAI f
Page 83 and 84: The boundary condition to be satisf
Page 85 and 86: 2.4.7. Discussion of the Probltsi o
Page 87 and 88: characteristics associated with Eqs
Page 89 and 90: 2.4.8 The Problem of Bolza The prec
Page 91 and 92: The initial position, velocity and
Page 93 and 94: where I is the moment of inertia, F
Page 95 and 96: or, as . -... .-. . .._-_--- to ind
Page 97 and 98: Since R(t, x(t) ) is the minimum va
Page 99 and 100: 2.4.10 Ljnear Problem with Quadrati
Page 101 and 102: where S(t) is some n x n symmetric
Page 103 and 104: The governing equations for the att
Page 105 and 106: y 2.4.11 Dynamic Programming and th
Page 107 and 108: M Since 3t does not depend on u exp
Page 109 and 110: with The P vector for the system is
Page 111 and 112: where With the control known as a f
Page 113 and 114: with and with the boundary conditio
Page 115 and 116: 2.5 DYNAMIC PROGRAMMING AND THE OPT
Page 117 and 118: 2.5.2 Problem Statement Let the sys
Page 119 and 120: Now, introducing the variables qs2,
Page 121 and 122: Thus, the performance index takes t
Page 123 and 124: #R where tr denotes the trace of th
Page 125 and 126: The minimum value of the performanc
Page 127 and 128: variance characterizing Ilt) can be
Page 129 and 130:
The solution takes the form with s
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Since V is positive definite for f
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where the first the second with exp
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Taking the limit and using the expr
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Since V is positive definite for t
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2.5.3 The Treatment of Terminal Con
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F while Equation (2.5.87B) requires
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._ __. ..- . . - which will equal t
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is determined. Let p(z,t') be given
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of the p and $ equations (i.e., Eqs
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(2.5.I.21) Thus, the control is to
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Note, as in.Section (2.5.2.3), the
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.6 t -(h.(S~tj~fitr~f-‘~~~~n~) =
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3.0 RECOMMENDED PROCEDURES - ._ .-_
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4 Section 2.5 (2.5.1) (2.5.2) (2.5.
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