{VERSION 3 0 "APPLE_PPC_MAC" "3.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 }{CSTYLE "2D Comment" 2 18 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 256 "" 1 14 0 0 0 0 0 1 0 0 0 0 0 0 0 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 256 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 257 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }} {SECT 0 {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT 256 21 " A Problem in Control" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 257 "" 0 "" {TEXT -1 34 "Presented on the occassion of th e " }{XPPEDIT 18 0 "100^th;" "6#)\"$+\"%#thG" }{TEXT -1 33 " birthday \+ of Joseph W. Herod, Sr." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart; " }}}{PARA 0 "" 0 "" {TEXT -1 58 "This is the differential equation fo r the control problem." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 13 "Find v(s) in " }{XPPEDIT 18 0 "L^2*[0, 1];" "6#*&%\" LG\"\"#7$\"\"!\"\"\"\"\"\"" }{TEXT -1 25 " so that the solution for" } }{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 11 " \+ " }{XPPEDIT 18 0 "diff(y(t),t)+y(t) = v(t);" "6#/,&-%%diffG6$-%\"yG 6#%\"tGF+\"\"\"-F)6#F+F,-%\"vG6#F+" }{TEXT -1 2 ", " }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 52 "with y(0) = 0, y '(0) = 1, y(¹) = 1, and y '(¹) = 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 27 "We re-write the problem as " }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 21 " Z ' = A Z + " }{XPPEDIT 18 0 "matrix([[0], [1]]);" "6#-%'matrixG6#7$7#\"\"!7#\" \"\"" }{TEXT -1 21 " v(s) = A Z + b v(s)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 16 " Z(0) = " }{XPPEDIT 18 0 " matrix([[0], [1]]);" "6#-%'matrixG6#7$7#\"\"!7#\"\"\"" }{TEXT -1 14 " \+ and Z(¹) = " }{XPPEDIT 18 0 "matrix([[1], [0]]);" "6#-%'matrixG6#7$7 #\"\"\"7#\"\"!" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 10 "Here, A = " }{XPPEDIT 18 0 "matrix([[0, 1 ], [-1, 0]]);" "6#-%'matrixG6#7$7$\"\"!\"\"\"7$,$\"\"\"!\"\"F(" } {TEXT -1 24 " and b is as suggested." }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 190 "In order to set up this equation fo r solution we need to compute exponential(t A). We think of writing th e paradigm representation for A. In the context of Maple, the computat ion is easier." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "with(linal g):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "matrix(2,1,[0,1]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "A:=matrix([[0,1],[-1,0]]) ;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "exponential(A,t);" }}} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 45 "The solut ion for the initial value problem is" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 12 " Z(t) = " }{XPPEDIT 18 0 "exp(t*A); " "6#-%$expG6#*&%\"tG\"\"\"%\"AGF(" }{TEXT -1 1 " " }{XPPEDIT 18 0 "ma trix([[0], [1]]);" "6#-%'matrixG6#7$7#\"\"!7#\"\"\"" }{TEXT -1 4 " + \+ " }{XPPEDIT 18 0 "int(exp((t-s)*A)*b*v(s),s = 0 .. t);" "6#-%$intG6$*( -%$expG6#*&,&%\"tG\"\"\"%\"sG!\"\"F-%\"AGF-F-%\"bGF--%\"vG6#F.F-/F.;\" \"!F," }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 71 "The value of Z(¹) is to be as specified above. This gi ves two equations" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 102 "1=cos( Pi)*0+sin(Pi)*1+Int(sin(Pi-s)*v(s),s=0..Pi);\n0=-sin(Pi)*0+cos(Pi)*1+I nt(cos(Pi-s)*v(s),s=0..Pi);" }}}{PARA 0 "" 0 "" {TEXT -1 3 "Or," }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "1=Int(sin(s)*v(s),s=0..Pi); \n1=Int(cos(s)*v(s),s=0..Pi);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 36 "We now define two linear functions: " } {XPPEDIT 18 0 "L[1];" "6#&%\"LG6#\"\"\"" }{TEXT -1 6 "(v) = " } {XPPEDIT 18 0 "int(sin(s)*v(s),s = 0 .. Pi);" "6#-%$intG6$*&-%$sinG6#% \"sG\"\"\"-%\"vG6#F*F+/F*;\"\"!%#PiG" }{TEXT -1 6 " and " }{XPPEDIT 18 0 "L[2];" "6#&%\"LG6#\"\"#" }{TEXT -1 6 "(v) = " }{XPPEDIT 18 0 "in t(cos(s)*v(s),s = 0 .. Pi);" "6#-%$intG6$*&-%$cosG6#%\"sG\"\"\"-%\"vG6 #F*F+/F*;\"\"!%#PiG" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 29 "Th e set M is all x such that " }{XPPEDIT 18 0 "L[1];" "6#&%\"LG6#\"\"\" " }{TEXT -1 12 "(x) = 1 and " }{XPPEDIT 18 0 "L[2];" "6#&%\"LG6#\"\"# " }{TEXT -1 294 "(x) = -1. M is a closed convex set. Let Pm be the clo sest point projection onto M. Pm is not necessarily linear. We want to find a formula for Pm. Then we compute Pm(0). That will be the functi on with minimum norm that makes the boundary condition work. Here's ho w to get the formula for Pm(0)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 24 "(Recall the notes here.)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 17 " 0 - Pm(0) = " } {XPPEDIT 18 0 "alpha;" "6#%&alphaG" }{TEXT -1 10 " sin(s) + " } {XPPEDIT 18 0 "beta;" "6#%%betaG" }{TEXT -1 7 " cos(s)" }}{PARA 0 "" 0 "" {TEXT -1 13 " 0 - 1 = " }{XPPEDIT 18 0 "alpha;" "6#%&alphaG" }{TEXT -1 22 " < sin(s), sin(s) > + " }{XPPEDIT 18 0 "beta;" "6#%%beta G" }{TEXT -1 19 " < cos(s), sin(s) >" }}{PARA 0 "" 0 "" {TEXT -1 33 " \+ 0 -1 = < sin(s), cos(s) > + " }{XPPEDIT 18 0 "beta;" "6#%%betaG" } {TEXT -1 19 " < cos(s), cos(s)>." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 45 " We solve these equations and define \+ v(s)" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 137 "eq1:=-1=alpha*int(s in(s)^2,s=0..Pi)+beta*int(cos(s)*sin(s),s=0..Pi);\neq2:=-1=alpha*int(s in(s)*cos(s),s=0..Pi)+beta*int(cos(s)^2,s=0..Pi);" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 30 "solve(\{eq1,eq2\},\{alpha,beta\});" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "assign(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "v:=s->-alpha*sin(s)+beta*cos(s);" }}} {PARA 0 "" 0 "" {TEXT -1 59 " We solve the differential equation a nd draw the graph." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 57 "dsolve (\{diff(y(t),t,t)+y(t)=v(t),y(0)=0,D(y)(0)=1\},y(t));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "simplify(%,trig);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "ymin:=unapply(rhs(%),t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "plot(ymin(t),t=0..Pi);" }}}{PARA 0 "" 0 "" {TEXT -1 93 "We now establish another \"driver\" by finding a \+ function n(t) in the nullspace of the two L's." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "n:=s->sin(3*s);\nint(sin(s)*n(s),s=0..Pi);\nint( cos(s)*n(s),s=0..Pi);" }}}{PARA 0 "" 0 "" {TEXT -1 44 "Compare the gra phs of v(s) and of v(s)+n(s)." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "plot([v(s),v(s)+n(s)],s=0..Pi);" }}}{PARA 0 "" 0 "" {TEXT -1 98 "We now solve the differential equation with this \"driver\" that does not have the minimum property." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 62 "dsolve(\{diff(y(t),t,t)+y(t)=v(t)+n(t),y(0)=0,D(y)(0)=1\},y(t) );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "simplify(%,trig);" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "yalt:=unapply(rhs(%),t);" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "plot([yalt(t),ymin(t)],t=0 ..Pi);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "9 2" 25 }{VIEWOPTS 1 1 0 1 1 1803 }