Unformatted text preview: Optimal Control Winter 2009
Solutions to Homework 1 Shaunak D. Bopardikar
bshaunakeumail.ucsb.edu Problem 2 Let G : X —r X TAX , H denote the trace function and Y :2 XTAX. Using the property that the
derivative of the composition is the composition of the derivatives. we have [3(§;GJ (1,0)] WM (we )aTDii ()G(mo) —]V.
Now 6(tmce(Y)) _
A1 [ BY (3:0)] — I. shpuicﬂ Muj be wriiiw
SO %[X'FAX](V) = )Z’TAV + VAX. [(?x (x Ax“ (@100 Thus. using the chain rule, 6(tmce(XTAX)) _ ”T "" u ..
[Ti’i/L X AV+VAX. h M 4a.. incurs a”;
Problem 3 We have
6 _c./
am xiii/i:
a _(
—[ A (V): VA: 66X“ —m G. gamma By the linearity property of the derivative operator, we have (A +29") vw (MEN
cumu}. R k. x {ﬁnmi’nt—
Problem 4 Let XTl denote the solution to the Ricatti equation at the n—th iteration. Then the Newton iteration
can be written as %[R(X)](V) = AW + VA + X'RV + VRX. ﬁlmxonxw) = ﬁlm/emu)  7%th A‘Xnn + XnHA + XnRXﬂH + XnHRX = A‘Xn + XnA + 2XnRXn A (Am + XnA + XﬁRXn + Q),
=> A‘XW + XMIA + XHRXHH + Xvi“ RXH = XnRXn — Q Let Qn := XnRXn — Q. Note that QR is symmetric. Thus the above equation becomes
(A + RXn)‘Xn+l + Xn+1(A + RXn) : Qua V which is a Lyapunov equation in Xﬂ+1 since at each iteration, X“ is known. ...
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 Winter '09
 Bamieh

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