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Optimal Control, Third Edition Solutions Manual Optimal Control, Third Edition Solutions Manual By Frank L. Lewis, Draguna Vrabie, Vassilis L. Syrmos Prepared by Ravi K.A. V. Aripirala BL $b = tem t350 a sae a }este-a, afe-t. 5 u i ad + iS a 1 2 Sof a [B]. moe [ i -(2 | [ 0 | eal =f °) uae | GE B09 | aly Suuluco = . HL a 13y/ 0 0 Gyar Luu} = 0-= origin is a singular point. c). f(a,y) = a? + yt. £(0,0) = 0, 2? > 0 for alla £0. y* > 0 for ally #0. f(a,y) = 2? + > 0 for all xf 0 and y ¢ 0 = (0,0) is & minimum point. Tsland L214 La,y) = Fle — 20)" + ly — 80)" we Hla) = V¥e~ y= 0, Alaa) = LNT = 5a 20)? + Ey = 0)! + ABs —v), Hye Viz y = 0 A= 1.16 Ae W+AV38=0 > > 2 = 17.99 Hy=y-30-4=0 y= 31.16 .y) = (17.99, 31.16), distance = JOE = 2.82. ~36 hours. ¥ Closest point! Time = MEE 1.2-2 L= minz,(d}), a 2 fe dad. Hos H(yd)=(e-m)+-ny eA [(e— mi) + yo wn)? = (2 — 22)? — (v= 92) oH dz an oy Uy) + Ay — m1) — Aly ~ yo) = 0 a (2 — an)? + (y— 1)? = (2 = 22)? — (y~ 2)? = 0 2(w — 94) + 2M 2 ~ 24) ~ 2A(e a2) = 0 tH L+ Ajai - Are y= (1+ d)u Ave ay Ba vit ye SS ye SS 2 1,2-3 feast NN elxyu,)e conek. -— 1 i > 2 a? Hos Me, — a) + Hum) +a(S-8-1) Bra» a Hey = ~(a1~ 22) +92 =0, . ees Hy, = (uy 92) + ai = oo = 2_# = Hy = go ple he a?(my — 24) Pn = 92) ‘ Qe2 Qy. From the above equation we obtain 3 ow a” oF ' 1 aay, th yey = (a? + b*)zey “Al 1.2-6 ipfi of 1_f2 2 _ {i 0 L= 3° lt fete 1 4i% Q= a al: R= r _ ap 0 Hos 37 |4 a)e+ ge | . “2 2] - Hy = 2+ (7 o|%" [- |= 1 0 - Hy, = [o afeta=o, — fal -2 _ Hy = [} tee (4 r=0, — fp pt -l pt =|. -.| ° 4 wos (E+ STORY'S ac= | sgis|> w=[ 4 i}: at oe LP RIOCe hl +2 op = [718 oo = tt arye= |g ras a= {Tit}, « 2 lerye 9] ik = 5c r= [te 1.2-7 a). Ko = (n+ BTQB) ATO ue “KC al -U- BRIC a“ QU - BKC Shel 5c7all — BKIC b). So = Q-QB(B"QH + R-)B7Q = QU ~ BK) Ko = (R+B7QB) BTQ o = BTQ-{R+ATQB\)K Furthermore, So Lt c). From {1)-(3), we have a? BQ — BK) ~ RK ~KT BT QU ~ BK) +KTRK ttf 9 Q i] QU - BR)- KT BTQU~BKJ+KTRK = QU- BR) where K7 BT QU — BK) + KTRK = 0. So, QU - BK) =U — K7B7)QU — BK) + KTRK (2- BRY'QU- BK) + K7RE 5078 Ox 5oTU ~ BRY JO" MONE - BIC tO VRKG 5 Vou ~ BK) ol" [Yeu - Bxyc] + 3 (VaKc)" Vie So = Q-QB(R+BTQB) BTQ = (Q°4BR BT) H Ay = Hy, = Qeytz? boy =0 (} Hy = Qx?y2? +2hy = 0 (2) H, = (3) 2 Hy = $t-rPadae =F re 3 yo — us (=) 2\2 1.2-9 Let H Ha, Ha, Hs Ny Ht, Ai(ay2) = 32+2y+2-1=0 A{eyz) = ety 32- 1 1 Hay2) = ge? + gu? + guTwe iy mit ; Se--2y-bz-1=0 e+ 2y—324+4=0 yt Bry taos ytDAy + 2d2 = 0 rbhAp~ 34. =0 From (7)-(9) we obtain and result in Ay = gy, Ap = ge. Hence, 2 = 1.2-10 ~3Ay ~ Ag Dh — D3 z = A+ Bhp co y= TMt2. = Dr + Ds a } ge & H ox a). Here we have, =0 lia 2 1 3 ~1 ~4 2 | y z 1 a o Lfzy) = fy) = r 0 Hay) = 7 He = (4) Hy = (22) Ay = (13) Equations (11)-(13) result in Nos 2a? +e? a OS Vaak ye y= tb? fy iy Ly = Ay a Hey + poles A Bhy#a? 2 ay? = Dog + 2| Fal <2 iff 4<0 b). Lay) = H(e,y) = 2h He, = Ayt yt =f (14) 2d ; Hy = 44+557=0 (15) Hy = (18) Equations (14)-(16) result in 
