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# lec10 - MIT OpenCourseWare http/ocw.mit.edu 16.323...

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MIT OpenCourseWare http://ocw.mit.edu 16.323 Principles of Optimal Control Spring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .

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16.323 Lecture 10 Singular Arcs Bryson Chapter 8 Kirk Section 5.6
�� Spr 2008 16.323 10–1 Singular Problems There are occasions when the PMP u ( t ) = arg min H ( x , u , p , t ) u ( t ) ∈U fails to define u ( t ) can an extremal control still exist? Typically occurs when the Hamiltonian is linear in the control, and the coeﬃcient of the control term equals zero . Example: on page 9-10 we wrote the control law: u m b < p 2 ( t ) u ( t ) = 0 b < p 2 ( t ) < b u m p 2 ( t ) < b but we do not know what happens if p 2 = b for an interval of time. Called a singular arc . Bottom line is that the straightforward solution approach does not work, and we need to investigate the PMP conditions in more detail. Key point : depending on the system and the cost, singular arcs might exist, and we must determine their existence to fully characterize the set of possible control solutions. Note: control on the singular arc is determined by the requirements that the coeﬃcient of the linear control terms in H u remain zero on the singular arc and so must the time derivatives of H u . Necessary condition for scalar u can be stated as d 2 k ( 1) k ∂u dt 2 k H u 0 k = 0 , 1 , 2 . . . June 18, 2008

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Spr 2008 Singular Arc Example 1 16.323 10–2 With x ˙ = u , x (0) = 1 and 0 u ( t ) 4 , consider objective 2 min ( x ( t ) t 2 ) 2 dt 0 First form standard Hamiltonian H = ( x ( t ) t 2 ) 2 + p ( t ) u ( t ) which gives H u = p ( t ) and p ˙( t ) = H x = 2( x t 2 ) , with p (2) = 0 (10.15) Note that if p ( t ) > 0 , then PMP indicates that we should take the minimum possible value of u ( t ) = 0 . Similarly, if p ( t ) < 0 , we should take u ( t ) = 4 . Question: can we get that H u 0 for some interval of time? Note: H u 0 implies p ( t ) 0 , which means p ˙( t ) 0 , and thus p ˙( t ) 0 x ( t ) = t 2 , u ( t ) = x ˙ = 2 t Thus we get the control law that 0 p ( t ) > 0 u ( t ) = 2 t when p ( t ) = 0 4 p ( t ) < 0 June 18, 2008
Spr 2008 16.323 10–3 Can show by contradiction that optimal solution has x ( t ) t 2 for t [0 , 2] . And thus we know that p ˙( t ) 0 for t [0 , 2] But p (2) = 0 and p ˙( t ) 0 imply that p ( t ) 0 for t [0 , 2] So there must be a point in time k [0 , 2] after which p ( t ) = 0 (some steps skipped here...) Check options: k = 0 ? contradiction Check options: k = 2 ? contradiction So must

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