{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 "" -1 259 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 260 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 261 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 262 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 263 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 264 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 265 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 267 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 268 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 269 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 270 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 271 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 275 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 276 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 277 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 278 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 } {CSTYLE "" -1 279 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 280 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 281 "" 0 1 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 "Headi ng 1" 0 3 1 {CSTYLE "" -1 -1 "" 1 18 0 0 0 0 0 1 0 0 0 0 0 0 0 }1 0 0 0 6 6 0 0 0 0 0 0 -1 0 }{PSTYLE "Heading 2" 3 4 1 {CSTYLE "" -1 -1 "" 1 14 0 0 0 0 0 0 0 0 0 0 0 0 0 }0 0 0 -1 4 4 0 0 0 0 0 0 -1 0 } {PSTYLE "Title" 0 18 1 {CSTYLE "" -1 -1 "" 1 18 0 0 0 0 0 1 1 0 0 0 0 0 0 }3 0 0 -1 12 12 0 0 0 0 0 0 19 0 }{PSTYLE "" 4 256 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 }} {SECT 0 {PARA 18 "" 0 "" {TEXT -1 45 "Orthogonal Series and Boundary V alue Problems" }}{PARA 256 "" 0 "" {TEXT -1 15 "Traveling Waves" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 17 "d 'Alembert's idea" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 400 " The solution we obtained for the wave equation in the previous module is a natural extension of the development for \+ solutions of the heat diffusion model. The mathematics is accurate, an d the method is satisfactory in that the solutions provided for the mo del meet with our expectations. There was no new mathematical insight, except that we can do the process again for this different equation. " }}{PARA 0 "" 0 "" {TEXT -1 133 " In the 18th Century, Jean Le Ro nd d'Alembert had formulated a different way to come to this solution. Morris Kline in his book, " }{TEXT 262 49 "Mathematical Thought from \+ Ancient to Modern Times" }{TEXT -1 531 ", reminds us that several math ematical concepts had to evolve with time. Kline says that d'Alembert \+ accepted only analytic curves and solutions. He wanted functions that \+ could be given with equations of relations, for example, between the e lementary algebraic and trigonometric functions. A part of the undevel oped understanding was that an initial function need not be expressibl e by a single analytic expression. There was also the struggle for the concept of an infinite sum -- a series. It is easy to imagine the dif ficulty in " }}{PARA 0 "" 0 "" {TEXT -1 52 "(1) having the idea of an \+ infinite sum of functions," }}{PARA 0 "" 0 "" {TEXT -1 91 "(2) having \+ a series of cosine functions to converge to, for example, the sine fun ction, and" }}{PARA 0 "" 0 "" {TEXT -1 122 "(3) having the series to c onverge to an analytic function on a finite interval, but not to the f unction off that interval." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 325 " When a creative person is stuck on an idea, interesting things can happen. We present d'Alembert's solution here. If he could see what follows, d\"Alembert might not admit that the pr esentation is the way he had thought of the solution. But, we should t ake advantage of the fact that we stand on the shoulders of giants! " }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 32 "Solutions for The Wave Equation." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 274 " Take note that nothing has been s aid about initial condtions or boundary conditons for this wave equati on. One might conceive these function as solutions of the wave equatio n for all t > 0 and all x > 0. We found that functions satisfying the \+ wave equation had the form" }}{PARA 0 "" 0 "" {TEXT -1 22 " \+ u(t, x) = " }{XPPEDIT 18 0 "psi;" "6#%$psiG" }{TEXT -1 12 "( x + t ) \+ + " }{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 10 "( x - t )." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 69 "Furthermo re, all functions defined in that form satisfy the equation." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 58 "We conceive of \+ finding all solutions for the wave equation" }}{PARA 0 "" 0 "" {TEXT -1 28 " " }{XPPEDIT 18 0 "diff(u(t,x),`$`(t ,2)) = diff(u(t,x),`$`(x,2));" "6#/-%%diffG6$-%\"uG6$%\"tG%\"xG-%\"$G6 $F*\"\"#-F%6$-F(6$F*F+-F-6$F+\"\"#" }}{PARA 0 "" 0 "" {TEXT -1 46 "as \+ the problem for finding the nullspace of a " }{TEXT 263 13 "wave opera tor" }{TEXT -1 1 ":" }}{PARA 0 "" 0 "" {TEXT -1 27 " \+ " }{XPPEDIT 18 0 "diff(u(t,x),`$`(t,2))-diff(u(t,x),`$`(x,2)) = 0;" "6#/,&-%%diffG6$-%\"uG6$%\"tG%\"xG-%\"$G6$F+\"\"#\"\"\"-F&6$-F) 6$F+F,-F.6$F,\"\"#!\"\"\"\"!" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 156 "We define a wave operator in Maple. We test that functions of \+ the special form satisfy the wave equation -- they are in the null spa ce of the wave operator." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 " WaveOp:=f->D[1,1](f)-D[2,2](f);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "u:=(t,x)->sin(exp(cos(x-t)));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "WaveOp(u)(t,x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 26 " We have seen th at if " }{TEXT 264 1 "f" }{TEXT -1 66 " is any function depending on t he combination x + t or x - t, then" }{TEXT 265 2 " f" }{TEXT -1 48 " \+ will lie in the nullspace of the wave operator." }}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 71 "f:=(t,x)->psi(x+t);\nWaveOp(f)(t,x);\nf:=(t,x) ->phi(x-t);\nWaveOp(f)(t,x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 145 " Since the wave equation i s linear, it is not a surprise that given any two twice differentiable functions of a single variable, the function" }}{PARA 0 "" 0 "" {TEXT -1 23 " " }{XPPEDIT 18 0 "psi(x+t)+phi(x-t );" "6#,&-%$psiG6#,&%\"xG\"\"\"%\"tGF)F)-%$phiG6#,&F(F)F*!\"\"F)" }} {PARA 0 "" 0 "" {TEXT -1 167 "is also a solution. D'Alembert had the i nsight that not only is this combination a solution, it is the general solution. To understand this we need to think about how " }{XPPEDIT 18 0 "phi" "6#%$phiG" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "psi" "6#%$ps iG" }{TEXT -1 40 " could be determine for a specific case." }}{PARA 0 "" 0 "" {TEXT -1 46 " This is the subject for the next section." } }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 41 "Initial conditions for the wave equation." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 95 " Suppose that we have an infinitely long string, with initi al conditions of the usual form:" }}{PARA 0 "" 0 "" {TEXT -1 29 " \+ " }{XPPEDIT 18 0 "u(0,x) = f(x)" "6#/-%\"uG6$ \"\"!%\"xG-%\"fG6#F(" }{TEXT -1 4 ", " }{XPPEDIT 18 0 "u[t](0,x)=g(x )" "6#/-&%\"uG6#%\"tG6$\"\"!%\"xG-%\"gG6#F+" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 8 "Here, - " }{XPPEDIT 18 0 "infinity" "6#%)infinit yG" }{TEXT -1 7 " < x < " }{XPPEDIT 18 0 "infinity" "6#%)infinityG" } {TEXT -1 106 ", so that we have no boundary conditions to impose. The \+ question is: How do we find a solution of the form" }}{PARA 0 "" 0 "" {TEXT -1 37 " u(t,x)= " }{XPPEDIT 18 0 "ph i(x+c*t)+psi(x-c*t)" "6#,&-%$phiG6#,&%\"xG\"\"\"*&%\"cGF)%\"tGF)F)F)-% $psiG6#,&F(F)*&F+F)F,F)!\"\"F)" }}{PARA 0 "" 0 "" {TEXT -1 26 "which s olves this problem?" }}{PARA 0 "" 0 "" {TEXT -1 132 " We break t he problem into two parts. Suppose first that g(x) = 0 for all x. Then , as in the text, we substitute and find that" }}{PARA 0 "" 0 "" {TEXT -1 29 " " }{XPPEDIT 18 0 "phi(x) = f (x)/2" "6#/-%$phiG6#%\"xG*&-%\"fG6#F'\"\"\"\"\"#!\"\"" }{TEXT -1 6 " \+ and " }{XPPEDIT 18 0 "psi(x) = f(x)/2" "6#/-%$psiG6#%\"xG*&-%\"fG6#F' \"\"\"\"\"#!\"\"" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 299 "For \+ example, if f is a function with one bump near x = 0 and there is no i nitial velocity, we can anticipate how this initial condition will evo lve with increasing time: we should see the bump break into the superp osition of two bumps of half the height, one moving to the right and o ne to the left." }}{PARA 0 "" 0 "" {TEXT -1 24 " Take, for example , " }{XPPEDIT 18 0 "f(x)=exp(-x^2)" "6#/-%\"fG6#%\"xG-%$expG6#,$*$F'\" \"#!\"\"" }{TEXT -1 125 " as the initial function. We construct a solu tion to the wave equation with this initial value and watch it evolve \+ with time." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "f:=x->exp(-x^2 );\nplot(f(x),x=-4..4);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 " with(plots):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "animate((f( x+t)+f(x-t))/2,x=-4..4,t=0..10);" }}}{PARA 0 "" 0 "" {TEXT -1 90 " \+ Sometimes it is instructive to view u as a graph in t and x, but not \+ as an animation." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 87 "plot3d(( f(x+t)+f(x-t))/2,x=-10..10,t=0..6,axes=NORMAL,\n orientation=[ -120,65]);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 101 " We now investigate the second case -- that f(x) = 0 and g (x) ­ 0. As in the text, we learn that" }}{PARA 0 "" 0 "" {TEXT -1 37 " " }{XPPEDIT 18 0 "psi(x) = -p hi(x)" "6#/-%$psiG6#%\"xG,$-%$phiG6#F'!\"\"" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 3 "and" }}{PARA 0 "" 0 "" {TEXT -1 38 " \+ " }{XPPEDIT 18 0 "phi(x)=int(g(x),x)" "6#/-% $phiG6#%\"xG-%$intG6$-%\"gG6#F'F'" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 81 "Consequently, the solutio n for the wave equation with these initial conditions is" }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 39 " \+ " }{XPPEDIT 18 0 "u(t,x)= int(g(s),s=x-c*t..x+c*t )/(2*c)" "6#/-%\"uG6$%\"tG%\"xG*&-%$intG6$-%\"gG6#%\"sG/F0;,&F(\"\"\"* &%\"cGF4F'F4!\"\",&F(F4*&F6F4F'F4F4F4*&\"\"#F4F6F4F7" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 33 " He re is an example. We take " }{XPPEDIT 18 0 "f(x) = 0 " "6#/-%\"fG6#%\" xG\"\"!" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "g(x) = x/(1+x+x^2)" "6#/- %\"gG6#%\"xG*&F'\"\"\",(\"\"\"F)F'F)*$F'\"\"#F)!\"\"" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "g:=x->x/(1+x+x^2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "u:=(t,x)->int(g(s),s=x-t..x+t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "u(t,x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 " animate(u(t,x),x=-10..10,t=0..6);" }}}{PARA 0 "" 0 "" {TEXT -1 90 " \+ Sometimes it is instructive to view u as a graph in t and x, but not as an animation." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 77 "plot3d( (u(t,x),x=-10..10,t=0..6,axes=NORMAL,\n orientation=[-120,65]) ;" }}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 8 "Exercise" }}{EXCHG {PARA 0 " " 0 "" {TEXT -1 141 "Draw the three dimensional graphs u(t, x) for sol utions of the wave equation if f(x) = sin(x) and g(x) = 0 and if f(x) \+ = 0 and g(x) = cos(x)." }}}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 40 "A general solution for the wave equation " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 64 "Herea fter we acknowledge that d'Alembert's solution has the form" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 22 " u(t , x) = " }{XPPEDIT 18 0 "psi(x+c*t);" "6#-%$psiG6#,&%\"xG\"\"\"*&%\"cG F(%\"tGF(F(" }{TEXT -1 3 " + " }{XPPEDIT 18 0 "phi(x-c*t);" "6#-%$phiG 6#,&%\"xG\"\"\"*&%\"cGF(%\"tGF(!\"\"" }{TEXT -1 1 "." }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 120 "Note that we have not \+ used any information about boundary conditions to arrive at this resul t. to get information about " }{XPPEDIT 18 0 "psi;" "6#%$psiG" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 119 " we used information about the initial conditions. We review the situation wit h the assumption that u(0,x) = f(x) and " }{XPPEDIT 18 0 "u[t];" "6#&% \"uG6#%\"tG" }{TEXT -1 15 " (0, x) = g(x)." }}{PARA 0 "" 0 "" {TEXT -1 100 " We use the ideas of the previous section to construct a g eneral solution for the wave equation." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "f:='f': g:='g': u:='u':" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 87 "u:=(f,g,t,x,c)->(f(x+c*t)+f(x-c*t))/2\n \+ +int(g(s),s=x-c*t..x+c*t)/(2*c);" }}}{PARA 0 "" 0 "" {TEXT -1 71 " \+ To illustrate this general solution, we take a particular example. " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "f:=x->4*cos(4*x);\ng:=x- >x^2-1/3;" }}}{PARA 0 "" 0 "" {TEXT -1 52 " Here is an evaluation \+ of this general solution." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "u(f,g,t,x,1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "animate(u (f,g,t,x,1),x=-2..2,t=0..2);" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {SECT 1 {PARA 3 "" 0 "" {TEXT -1 21 "Half Infinite Strings" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 137 "This solution is for all number x. It is as though we had a string that was infinite i n both directons. That is, this solution holds for " }{XPPEDIT 18 0 "- infinity < x;" "6#2,$%)infinityG!\"\"%\"xG" }{TEXT -1 5 " and " } {XPPEDIT 18 0 "x < infinity;" "6#2%\"xG%)infinityG" }{TEXT -1 1 "." }} {PARA 0 "" 0 "" {TEXT -1 80 " Our next idea is to suppose that we \+ solve the equation on the interval [0, " }{XPPEDIT 18 0 "infinity;" "6 #%)infinityG" }{TEXT -1 144 " ). We begin by imposing only one boundar y condition. We suppose that x > 0. With this assumption, we have a bo undary condition at the left end." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 32 "Boundary Condition: u(t, 0) = 0." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 39 "Initial C onditions: u(0, x) = f(x) and " }{XPPEDIT 18 0 "u[t];" "6#&%\"uG6#%\"t G" }{TEXT -1 24 "(0, x) = g(x) for x > 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 79 " With no initial conditions or boundary conditions, we have concluded that " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 32 " u(t , x) = " }{XPPEDIT 18 0 "psi(x+c*t);" "6#-%$psiG6#,&%\"xG\"\"\"*&%\"cG F(%\"tGF(F(" }{TEXT -1 3 " + " }{XPPEDIT 18 0 "phi(x-c*t);" "6#-%$phiG 6#,&%\"xG\"\"\"*&%\"cGF(%\"tGF(!\"\"" }{TEXT -1 1 " " }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 52 "as t increases. With in itial conditions, we got that" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 31 " u(t, x) = " } {XPPEDIT 18 0 "(f(x+c*t)+f(x-c*t))/2;" "6#*&,&-%\"fG6#,&%\"xG\"\"\"*&% \"cGF*%\"tGF*F*F*-F&6#,&F)F**&F,F*F-F*!\"\"F*F*\"\"#F2" }{TEXT -1 3 " \+ + " }{XPPEDIT 18 0 "(G(x+c*t)-G(x-c*t))/2;" "6#*&,&-%\"GG6#,&%\"xG\"\" \"*&%\"cGF*%\"tGF*F*F*-F&6#,&F)F**&F,F*F-F*!\"\"F2F*\"\"#F2" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 34 "where G is an antiderivative of g." }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 69 " Now, we are supposing only that \+ x > 0. We must do extensions of " }{XPPEDIT 18 0 "psi;" "6#%$psiG" } {TEXT -1 5 " and " }{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 77 " for as t increases for c t will exceed x and we will need the definition \+ of " }{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 96 " for negative valu es. Happily, there is information available through the boundary cond itions. " }}{PARA 0 "" 0 "" {TEXT -1 35 " The first boundary condi ton is" }}{PARA 0 "" 0 "" {TEXT -1 24 " 0 = u(t, 0) = " } {XPPEDIT 18 0 "psi;" "6#%$psiG" }{TEXT -1 11 " ( c t ) + " }{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 12 " (- c t ) = " }{XPPEDIT 18 0 "(f( c*t)+G(c*t)+C)/2;" "6#*&,(-%\"fG6#*&%\"cG\"\"\"%\"tGF*F*-%\"GG6#*&F)F* F+F*F*%\"CGF*F*\"\"#!\"\"" }{TEXT -1 6 " + " }{XPPEDIT 18 0 "(f(-c* t)-G(-c*t)-C)/2;" "6#*&,(-%\"fG6#,$*&%\"cG\"\"\"%\"tGF+!\"\"F+-%\"GG6# ,$*&F*F+F,F+F-F-%\"CGF-F+\"\"#F-" }{TEXT -1 3 " ." }}{PARA 0 "" 0 "" {TEXT -1 6 " Thus," }}{PARA 0 "" 0 "" {TEXT -1 65 " 0 = \+ f( c t ) + f( - c t ) + G( c t ) - G( - c t )." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 62 "Since this holds for all \+ f and for all g, then it must be that" }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 77 " f( c t ) = - f( - c t ) a nd G( c t ) = G( - c t ) for all t > 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 73 "Thus, f should have an odd extensi on and G should have an even extension." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 30 "Making even and odd extensio ns" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 245 "It 's clear that in order to illustrate these ideas, we are going to need to be able to make even and odd extensions for functions. Here is a p rocedure. You must execute these two parts inorder to be able to work \+ what the rest of this work sheet." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 80 "ef:=proc(f,x) \n if x>0 then f(x);\n el se f(-x);\n fi;\n end;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 77 "of:=proc(f,x) \n if x>0 then f(x);\n else -f(-x);\n \+ fi;\n end;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {PARA 0 "" 0 "" {TEXT -1 57 "Having defined this part we proceed with \+ the development." }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 " " 0 "" {TEXT -1 46 "Comparing Infinite and Half-Infinite Problems." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 281 "It is wi se to constrast this \"half\" infinite string with the infinite string of the previous module. Let's do one example twice: first we let x ra nge over all numbers and use just the original d'Alembert solution, an d then we let x only be positive and use the odd extension of f. " }} {PARA 0 "" 0 "" {TEXT -1 20 " Here's f and g." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "f:=x->-(Heaviside(2*Pi-x)-Heaviside(Pi-x))*si n(x);\ng:=x->0;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "plot(f(x ),x=0..10);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 121 "Now, we solve the PDE with initial conditions and no boundary \+ conditions. we use the d'Alembert solution, and take c = 1." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "u:=(t,x)->(f(x+t)+f(x-t))/2;" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "plot3d(u(t,x),x=-10..10,t=0. .5,orientation=[-70,25]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "with(plots):\nanimate(u(t,x),x=-10..10,t=0..5);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 81 "Next, we solve the PDE with initial conditions and boundary conditon u(t, 0) = 0." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "u:=(t,x)->(of(f,x+t)+of(f,x-t))/2;" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "plot3d('u(t,x)',x=0..4*Pi,t= 0..10,orientation=[-90,10]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 74 "with(plots):\nanimate('u(t,x)',x=0..4*Pi,t=0..3*Pi,numpoints=75, color=RED);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 44 "The wave equation on an interval: an ex ample" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 196 " We review how to draw an odd, periodic extension of a function d efined on the interval [0,1]. This has been described as making a refl ection about both end points. For this illustration, take" }}{PARA 0 " " 0 "" {TEXT -1 26 " " }{XPPEDIT 18 0 "f(x) = x*(1-x)" "6#/-%\"fG6#%\"xG*&F'\"\"\",&\"\"\"F)F'!\"\"F)" }{TEXT -1 22 " on the inteval [0,1]." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 213 "f:=piecewise(-2<=x and x<-1, (2+x) *(1-(2+x)),\n -1<=x and x<0, x*(1+x),\n 0<=x \+ and x<=1, x*(1-x),\n 1 " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "plot(f( x),x=-2..3);" }}}{PARA 0 "" 0 "" {TEXT -1 196 " We could continue \+ to extend f in a piecewise fashion, or we could use the modular arithm etic. We assume that the function is defined on the interval [0,1]. T he extension is made by using the" }{TEXT 261 14 " floor functio" } {TEXT -1 33 "n in a manner we describe next.. " }}{PARA 0 "" 0 "" {TEXT -1 25 " The floor function, " }{TEXT 259 8 "floor(x)" } {TEXT -1 93 ", is the greatest integer less than or equal to x. This \+ function is related to the function " }{TEXT 260 7 "frac(x)" }{TEXT -1 61 " which is the fractional part of x as follows: The function, " }{XPPEDIT 18 0 "x-floor(x)" "6#,&%\"xG\"\"\"-%&floorG6#F$!\"\"" } {TEXT -1 72 " is frac(x) if x > 0 and is 1 + frac(x) if x < 0. This n ew function -- " }{XPPEDIT 18 0 "x-floor(x)" "6#,&%\"xG\"\"\"-%&floorG 6#F$!\"\"" }{TEXT -1 54 " -- is a periodic extension of frac(x). See t he graph." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "plot(x-floor(x),x=-2..2,discont=true);" }}}{PARA 0 " " 0 "" {TEXT -1 85 " To save typing, we define this periodic exten sion of frac(x) for this worksheet." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "Pfrac:=x->x-floor(x);" }}}{PARA 0 "" 0 "" {TEXT -1 119 " Here, now, is the definition of the even, periodic extension of a function f and the odd, periodic extension of f." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 105 "OPf:=x->if (0<=Pfrac(x/2) and Pfra c(x/2)<1/2) then f(Pfrac(x))\n else -f(1-Pfrac(x)) fi; \+ " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 93 "EPf:=x->if (0<=Pfrac( x/2) and Pfrac(x/2)<1/2) then f(Pfrac(x))\n else f(1-Pfrac(x)) \+ fi;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "f:=x->x*(x-1);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "plot('OPf(x)',x=-2..2);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "plot('EPf(x)',x=-2..2);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "g:=x->0;" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 54 "plot3d('u(OPf,g,t,x,1)',x=-1..1,'t'=0..1,axe s=NORMAL);" }}}{PARA 0 "" 0 "" {TEXT -1 53 " It would be interesti ng to animate the solution." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 52 "Even pe riodic extensions and odd periodic extensions" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 295 "In order to be able to g raph solutions for the finite string, we need to be able to make a dif ferent extensions of functions from simply even or odd. We need to mak e odd and even, 2 L periodic extensions of functions of functions de fined on the interval [0, L]. Here is a procedure to do this." }} {PARA 0 "" 0 "" {TEXT -1 53 " First, we make an even, 2 L periodi c extension." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart:" }}} {PARA 0 "" 0 "" {TEXT -1 109 "First we make the extension on the inter val [0, L] aware that the goal is to get an even, periodic extension. " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 113 "pfe:= proc(f,x)\n g lobal L;\n\011\011if 0 <= x and x < L then f(x)\n\011\011elif L <= x a nd x <= 2*L then f(2*L-x)\n\011\011fi end;" }}}{PARA 0 "" 0 "" {TEXT -1 57 "Here is where we make extend the function to all numbers." }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "epf:= (f,x)->pfe(f,frac(abs( x)/(2*L))*2*L);" }}}{PARA 0 "" 0 "" {TEXT -1 270 "We test this setup. \+ We define an interval L and a function f and draw the graph. Be aware \+ that we want an even, 2 L extension. Look at the following to see that on the interval [0, L], we have the graph of L, that the graph is eve n, and that the function has period 2 L." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "L:=2;\nf:=x->x;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "plot('epf(f,x)',x=-3*L..3*L);" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 15 "L:='L':\nf:='f':" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 47 " Next, we make an odd, 2 L periodic extension." }}{PARA 0 "" 0 "" {TEXT -1 108 "First we make the extension on the interval [0 , L] aware that the goal is to get an odd, periodic extension." }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 114 "pfo:= proc(f,x)\n globa l L;\n\011\011if 0 <= x and x < L then f(x)\n\011\011elif L <= x and x <= 2*L then -f(2*L-x)\n\011\011fi end;" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 57 "Here is where we make extend the f unction to all numbers." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "o pf:= (f,x)->sign(x)*pfo(f,frac(abs(x)/(2*L))*2*L);" }}}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 291 "As before, we test thi s setup. We define an interval L and a function f and draw the graph. \+ This time, be aware that we want an even, 2 L extension. Look at the f ollowing to see that on the interval [0, L], we have the graph of L, t hat the graph is odd, and that the function has period 2 L." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "L:=2;\nf:=x->x;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "plot('opf(f,x)',x=-3*L..3*L);" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 45 "An alternative method for periodic extens ions" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 137 " Get periodic extensions of functions is an interesting problem in the \+ context of these notes. Here is another method from the literature." } }{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 3 "" 0 "" {TEXT -1 44 "Generati ng periodic extensions of functions." }}{PARA 0 "" 0 "" {TEXT -1 201 " Suppose that f is defined on the interval [a,b], in MapleTech Vol. 3, \+ NO.3, Monagan and Lopez provided the following method to output a func tion which extends the definition of f to the real line using" }} {PARA 0 "" 0 "" {TEXT -1 23 " f(a + k (b-a) + " }{XPPEDIT 18 0 " delta" "6#%&deltaG" }{TEXT -1 10 ") = f(a + " }{XPPEDIT 18 0 "delta" " 6#%&deltaG" }{TEXT -1 1 ")" }}{PARA 0 "" 0 "" {TEXT -1 14 "for integer k." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 165 "PeriodicExtender:=pr oc(f,d::range)\nsubs( \{'F' = f, 'L'=lhs(d),\n 'D'=rhs(d)-lhs(d )\},\nproc(x::algebraic) local y;\n y:=floor((x-L)/D);\n F(x-y*D );\nend)\nend;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "sw:=Perio dicExtender(signum,-1..1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "plot('sw(x)','x'=-4..4);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "g:=x->piecewise(x<0,-x,x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "st:=PeriodicExtender(g,-1..1);" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 24 "plot('st(x)','x'=-3..3);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 8 "Examples" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 64 "Take c = 1 and L, f, and g are specified:\n1. L = 2, f(x ) = sin(" }{XPPEDIT 18 0 "pi;" "6#%#piG" }{TEXT -1 47 "x), g(x) = 0.\n 2. L = 2, f(x) = 0, g(x) = sin(" }{XPPEDIT 18 0 "pi" "6#%#piG" } {TEXT -1 11 "x).\n3. L =" }{XPPEDIT 18 0 "pi" "6#%#piG" }{TEXT -1 9 " , f(x) = " }{XPPEDIT 18 0 "pi" "6#%#piG" }{TEXT -1 8 "/2 - |x-" } {XPPEDIT 18 0 "pi" "6#%#piG" }{TEXT -1 188 "/2|, g(x) = 0.\n4. L = 4, f(x) = (x-1)*(Heaviside(x-1)-Heaviside(x-2)) + \n\011\011\011(3-x)*(H eaviside(x-2)-Heaviside(x-3))\n\011\011g(x) = 0.\n5. a = 4, f(x) = 0, and g(x) = Heaviside(x-3)-Heaviside(x-4)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 92 "In order to work example, we need \+ to have the functions epf and opf from the section above. " }}{PARA 0 "" 0 "" {TEXT -1 18 "Here is example 1." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "L:=2;\nf:=x->sin(Pi*x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "plot('opf(f,x)',x=-2*L..2*L);" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 36 "u:=(t,x)->(opf(f,x+t)+opf(f,x-t))/2;" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "plot3d('u(t,x)',x=0..L,t=0.. 2*L,axes=NORMAL,orientation=[-135,45]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "with(plots):\nanimate('u(t,x)',x=0..L,t=0..L);" }}} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 18 "Here is example 2." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "L:=2;\ng:=x->sin(Pi*x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "int(g(x),x);\nG:=unapply(%,x);" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 29 "plot('epf(G,x)',x=-2*L..2*L);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "u:=(t,x)->(epf(G,x+t)-epf(G,x-t))/2 ;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "plot3d('u(t,x)',x=0..L ,t=0..2*L,axes=NORMAL,orientation=[-135,45]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "with(plots):\nanimate('u(t,x)',x=0..L,t=0..L);" }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 29 "The pdesolve command in Maple " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 246 " \+ This solution by d'Alembert has such a simple structure, it is no sur prise that it has been programmed in Maple. (Indeed, it is programmed \+ into every really good computer algebra system!) Here is a way to use \+ Maple to identify this solution." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "wave:=diff(u(t,x),t,t)-c^2*diff(u(t,x),x,x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "sol:=pdesolve(wave,u(t,x));" }}}{PARA 0 " " 0 "" {TEXT -1 109 " The solution is in terms of two arbitrary fu nctions just as we had before. Maple chose not to call them " } {XPPEDIT 18 0 "phi" "6#%$phiG" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "psi " "6#%$psiG" }{TEXT -1 141 " but this is of no consequence. To see how to plot a solution, as before, we must identify _F1 and _F2. Here is \+ a way to think of doing this." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 57 "c:=1;\nf1:=s->0;\nf2:=s->piecewise(-1/2 " 0 "" {MPLTEXT 1 0 24 "subs(_F1=f1,_F2=f2,sol);" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "subs(%,u(t,x));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "U:=unapply(%,(t,x));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "plot3d('U(t,x)','x'=-4..4,t=0..5,ax es=NORMAL);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 26 "The structure of solutions" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 132 "In this section we want to contrast the \+ three situations that we have seen. We take three examples which are d ifferent, but similar." }}{PARA 0 "" 0 "" {TEXT -1 31 " We compare the situations:" }}{PARA 0 "" 0 "" {TEXT -1 54 "(1) x is any number o n the real line and call this an " }{TEXT 267 15 "infinite string" } {TEXT -1 4 ", or" }}{PARA 0 "" 0 "" {TEXT -1 34 "(2) x is positive and call this a " }{TEXT 268 20 "half infinite string" }{TEXT -1 4 ", or " }}{PARA 0 "" 0 "" {TEXT -1 40 "(3) x is in an interval and call this a " }{TEXT 269 13 "finite string" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 56 " We begin with a function f which has only one bump. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "f:=x->piecewise(x<=0,0, x<=3,x/3,x<=4,4-x,0);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "pl ot('f(x)',x=0..4);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {PARA 0 "" 0 "" {TEXT -1 243 "If we are going to assume we have three \+ strings, one fitting each of the situations described above, that we g ive the string an intiial displacement determined by f, and give the s tring an initial velocity of zero, then we need to recall that " }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 20 " \+ U(t, x) = " }{XPPEDIT 18 0 "(F(x+t)+F(x-t))/2;" "6#*&,&-%\"FG6#,&%\"x G\"\"\"%\"tGF*F*-F&6#,&F)F*F+!\"\"F*F*\"\"#F/" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 111 "and that this F is obtained from the original \+ f through some extension. Thus, we will need the extension tools." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 16 "E xtensions Tools" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 30 "Making even and odd extensions" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 241 "It's clear that in order to illustrate these ideas, we are going to need to be able to make ev en and odd extensions for functions. Here is a procedure. You must exe cute these two parts in order to be able to work the rest of this work sheet." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 80 "ef:=proc(f,x) \+ \n if x>0 then f(x);\n else f(-x);\n fi;\n end;" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 77 "of:=proc(f,x) \n if x >0 then f(x);\n else -f(-x);\n fi;\n end;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 52 "Even periodic extensions and od d periodic extensions" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 283 "In order to be able to graph solutions for the finite string, we need to be able to make a different extensions of function s than simply even or odd. We need to make odd and even, 2 L periodi c extensions of functions defined on the interval [0, L]. Here is a p rocedure to do this." }}{PARA 0 "" 0 "" {TEXT -1 53 " First, we ma ke an even, 2 L periodic extension." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 116 "To do this we make \+ the extension to the interval [0, 2 L] aware that the goal is to get a n even, periodic extension." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 113 "pfe:= proc(f,x)\n global L;\n\011\011if 0 <= x and x < L then f(x)\n\011\011elif L <= x and x <= 2*L then f(2*L-x)\n\011\011fi end; " }}}{PARA 0 "" 0 "" {TEXT -1 52 "Here is where we extend the function to all numbers." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "epf:= (f ,x)->pfe(f,frac(abs(x)/(2*L))*2*L);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 47 " Next, we make an odd, 2 L periodic extension." }}{PARA 0 "" 0 "" {TEXT -1 115 "To do this we make the extension to the interv al [0, 2 L] aware that the goal is to get an odd, periodic extension. " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 114 "pfo:= proc(f,x)\n g lobal L;\n\011\011if 0 <= x and x < L then f(x)\n\011\011elif L <= x a nd x <= 2*L then -f(2*L-x)\n\011\011fi end;" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 57 "Here is where we make extend th e function to all numbers." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "opf:= (f,x)->sign(x)*pfo(f,frac(abs(x)/(2*L))*2*L);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 45 "We use both these extensions in what follows." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 85 " For Situation 1, we define F1 to agree with f and to be defined for all numbers." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "F1:=x->piecewise(x<=0,0,x<=3,x/3,x<=4,4-x,0);" }}}{PARA 0 "" 0 "" {TEXT -1 116 "For Situation 2, we define F2 to agree with f(x) fo r positive x's and to be the odd extension of f for negative x' s" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "F2:=x->of(F1,x);" }}}{PARA 0 "" 0 "" {TEXT -1 104 "For Situation 3, we make the odd, periodic ext ension of f. For this example, the period will be 2 L = 8." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "L:=4;\nF3:=x->opf(F1,x);" }}}{PARA 0 "" 0 "" {TEXT -1 237 "We draw a graph of each of these. For the purp oses of drawing these graphs, which will have considerable overlap, we not only color them differently, but also artificially off-set them. \+ This off-setting is for illustration purposes only." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 74 "plot(['F1(x)','F2(x)'+.015,'F3(x)'-.015],x= -8..8,color=[black,red,green]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 42 "We define the solution u for each example." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 95 "u1:=(t,x)->(F1(x+t)+F1(x-t))/2:\nu2:=(t,x)->(F2( x+t)+F2(x-t))/2;\nu3:=(t,x)->(F3(x+t)+F3(x-t))/2;" }}}{PARA 0 "" 0 "" {TEXT -1 62 "We will draw the graph of each u, and make animations of \+ each." }}{PARA 0 "" 0 "" {TEXT -1 40 "Here is u1 as a solution in Situ ation 1." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "plot3d('u1(t,x)' ,x=-8..8,t=0..8,axes=NORMAL,orientation=[-115,25]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "with(plots):\nanimate('u1(t,x)',x=-8..8,t =0..20);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "plot([u1(0,x),u 1(3,x),u1(5,x)],x=-6..10,color=[BLACK,RED,GREEN]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 40 "Here is u2 as a solution in Situation 2." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 66 "plot3d('u2(t,x)',x=0..8,t=0..8,axes=NORMAL,orientation=[-115,25]);" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "with(plots):\nanimate('u2( t,x)',x=0..10,t=0..20);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 86 " plot(['u2(0,x)','u2(3,x)'+.015,'u2(5,x)'+.025],x=0..10,\n color=[B LACK,RED,GREEN]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 40 "Here is u3 as a solution in Situation 3." }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 66 "plot3d('u3(t,x)',x=0..4,t=0. .8,axes=NORMAL,orientation=[-115,25]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "with(plots):\nanimate('u3(t,x)',x=0..4,t=0..20);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 85 "plot(['u3(0,x)','u3(1,x)'+.0 15,'u3(2,x)'+.025],x=0..4,\n color=[BLACK,RED,GREEN]);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 0 "" 0 "" {TEXT -1 137 "Before we leave these examples too far behind, they call our attention to at least four properties of the wave equation that are central." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 162 "(1) Bump s in the initial distribution are split into two parts each having hal f the height of the original. One bump moves to the right and one move s to the left." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 290 "(2) We are willing to talk about functions being solutio ns to the wave equation that are not even differentible once, much les s twice. This luxury is a result of having d'Alembert's formulation of solutions. Such an extension of the idea of solution can be made prec ise with the concept of " }{TEXT 270 14 "weak solutions" }{TEXT -1 44 ". We will not pursue this idea further here." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 18 "(3) Choose points " } {XPPEDIT 18 0 "[t[0], x[0]];" "6#7$&%\"tG6#\"\"!&%\"xG6#F'" }{TEXT -1 137 " out in the plane and ask what points of the extended f and g inf luence the behavior of the solution at this chosen point. We see that \+ u(" }{XPPEDIT 18 0 "t[0],x[0];" "6$&%\"tG6#\"\"!&%\"xG6#F&" }{TEXT -1 32 ") involves the values of f at " }{XPPEDIT 18 0 "x[0]+c*t[0];" "6 #,&&%\"xG6#\"\"!\"\"\"*&%\"cGF(&%\"tG6#F'F(F(" }{TEXT -1 8 " and " }{XPPEDIT 18 0 "x[0]-c*t[0];" "6#,&&%\"xG6#\"\"!\"\"\"*&%\"cGF(&%\"tG6 #F'F(!\"\"" }{TEXT -1 50 " and involves the values of g over the inte rval [" }{XPPEDIT 18 0 "x[0]-c*t[0];" "6#,&&%\"xG6#\"\"!\"\"\"*&%\"cGF (&%\"tG6#F'F(!\"\"" }{TEXT -1 3 " , " }{XPPEDIT 18 0 "x[0]+c*t[0]" "6# ,&&%\"xG6#\"\"!\"\"\"*&%\"cGF(&%\"tG6#F'F(F(" }{TEXT -1 32 " ]. This i nterval, then, is the " }{TEXT 271 20 "domain of dependence" }{TEXT -1 5 " for " }{XPPEDIT 18 0 "[t[0], x[0]];" "6#7$&%\"tG6#\"\"!&%\"xG6# F'" }{TEXT -1 150 " . If we change the initial conditions ourside this interval and leave it the same inside, the change will not influence \+ the value of u at this point." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 79 "(4) If f and g are zero outside some inte rval [ a, b], then u will be zero for " }{XPPEDIT 18 0 "b+c*t <= x;" " 6#1,&%\"bG\"\"\"*&%\"cGF&%\"tGF&F&%\"xG" }{TEXT -1 9 " and for " } {XPPEDIT 18 0 "x <= a-c*t;" "6#1%\"xG,&%\"aG\"\"\"*&%\"cGF'%\"tGF'!\" \"" }{TEXT -1 80 " . Thus, information travels no faster than speed c \+ to the left or to the right." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 0 {PARA 4 "" 0 " " {TEXT -1 8 "Exercise" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 86 "Give the solution u for all three situations if g = 0 and f = sin on the inter val [0, " }{XPPEDIT 18 0 "2*Pi;" "6#*&\"\"#\"\"\"%#PiGF%" }{TEXT -1 28 " ] and zero everywhere else." }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 3 "" 0 "" {TEXT -1 22 "Structure of Solutions" }}}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 30 "A string in a v iscous medium " }}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 23 "Setting up the equation" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}{PARA 0 "" 0 "" {TEXT -1 35 "The problem i s to find w to satisfy" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 29 " " }{XPPEDIT 18 0 "diff( w,`$`(x,2))-k*diff(w,t)-g = diff(w,`$`(t,2));" "6#/,(-%%diffG6$%\"wG-% \"$G6$%\"xG\"\"#\"\"\"*&%\"kGF.-F&6$F(%\"tGF.!\"\"%\"gGF4-F&6$F(-F*6$F 3\"\"#" }}{PARA 0 "" 0 "" {TEXT -1 3 "(*)" }}{PARA 0 "" 0 "" {TEXT -1 65 " w(t, 0) = 0 = w(t, L) " }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 59 " \+ w( 0, x) = F(x)" }}{PARA 0 "" 0 "" {TEXT -1 44 " " } {XPPEDIT 18 0 "w[t]" "6#&%\"wG6#%\"tG" }{TEXT -1 17 " ( 0 , x) = G(x). " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 37 "To wo rk this problem, we define v by " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 58 " v(t, x) = w( t , x) + g x (L-x)/2" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 29 "v:=(t,x)->w(t,x)+g/2*x*(L-x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "diff(v(t,x),x,x)-k*diff(v(t,x),t)=d iff(v(t,x),t,t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "v(t,0); v(t,L);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "v(0,x); D[1](v) (0,x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 105 "Thus, we see that w s atisfies (*) if and only if v satisfies the following partial differen tial equation." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 28 " " }{XPPEDIT 18 0 "diff(v,`$`( x,2))-k*diff(v,t) = diff(v,`$`(t,2));" "6#/,&-%%diffG6$%\"vG-%\"$G6$% \"xG\"\"#\"\"\"*&%\"kGF.-F&6$F(%\"tGF.!\"\"-F&6$F(-F*6$F3\"\"#" }} {PARA 0 "" 0 "" {TEXT -1 4 "(**)" }}{PARA 0 "" 0 "" {TEXT -1 65 " \+ v(t, 0) = 0 = v(t, L)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 77 " \+ v( 0 , x) = F(x) + g x ( L - x)/2" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 44 " \+ " }{XPPEDIT 18 0 "v[t]" "6#&%\"vG6# %\"tG" }{TEXT -1 17 " ( 0 , x) = G(x)." }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 59 "To solve this equation (**), we defin e a new function u by " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 59 " u( t , x) = exp(t k/2 ) v (t , x)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 83 "We check u in the following partial differential equation with bou ndary conditions:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "v:='v';\nu:=(t,x)->exp(k/2*t)*v(t,x);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "diff(u(t,x),x,x)+k^2/4*u(t,x )-diff(u(t,x),t,t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "coll ect(%,exp(k/2*t));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "u(t,0 ); u(t,L);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "u(0,x); D[1]( u)(0,x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 105 "Thus, we see that v satisfies (**) if and only if u sati sfies the following partial differential equation" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 39 " \+ " }{XPPEDIT 18 0 "diff(u,`$`(x,2))+k^2/4*u = diff(u,`$`( t,2));" "6#/,&-%%diffG6$%\"uG-%\"$G6$%\"xG\"\"#\"\"\"*(%\"kG\"\"#\"\"% !\"\"F(F.F.-F&6$F(-F*6$%\"tG\"\"#" }}{PARA 0 "" 0 "" {TEXT -1 5 "(***) " }}{PARA 0 "" 0 "" {TEXT -1 65 " \+ u(t, 0) = 0 = u(t, L)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 65 " u( \+ 0 , x) = v( 0 , x)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 44 " " }{XPPEDIT 18 0 "u[t]" "6#&%\"uG6#%\"tG" }{TEXT -1 12 " ( 0 , x) = " }{XPPEDIT 18 0 "k/2*v(0,k)+v[t](0,x))" "6#,&*(%\"kG\"\"\"\"\"#!\"\"-%\"vG6$\"\"! F%F&F&-&F*6#%\"tG6$F,%\"xGF&" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 34 "So here is how the procedure goes:" }}{PARA 0 " " 0 "" {TEXT -1 53 "To solve (*) solve (***) with u(t, 0) = 0 = u(t, L ), " }}{PARA 0 "" 0 "" {TEXT -1 46 " u(0, x) = F(x)+ g x (L - \+ x)/2 , and " }{XPPEDIT 18 0 "u[t]" "6#&%\"uG6#%\"tG" }{TEXT -1 13 " ( 0 , x ) = " }{XPPEDIT 18 0 "k*(F(x)+g*x*(L-x)/2)/2+G(x);" "6#,&*(%\"k G\"\"\",&-%\"FG6#%\"xGF&**%\"gGF&F+F&,&%\"LGF&F+!\"\"F&\"\"#F0F&F&\"\" #F0F&-%\"GG6#F+F&" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 12 "Create v \+ by " }}{PARA 0 "" 0 "" {TEXT -1 66 " \+ v( t , x) = exp(-k/2 t) u(t, x)" }}{PARA 0 "" 0 "" {TEXT -1 12 "Creat e w by " }}{PARA 0 "" 0 "" {TEXT -1 107 " \+ w( t , x ) = v(t, x) - g x (L-x)/2 = exp(-k/2 t) u(t, x) - g x (L-x)/2" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 32 "Here is a check that this works." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "u:='u';" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 " w:=(t,x)->exp(-k/2*t)*u(t,x)-g*x*(L-x)/2;" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 53 "diff(w(t,x),x,x)-k*diff(w(t,x),t)-g-diff(w(t,x),t,t );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "simplify(%);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "collect(%,exp(-k/2*t));" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "w(t,0); w(t,L);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "w(0,x); D[1](w)(0,x);" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 20 "Solving the equation" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}{PARA 0 "" 0 "" {TEXT -1 54 " We o ften set c = 1 for convenience, so we define the " }{TEXT 276 23 "modi fied wave operator." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "wave op:=diff(u(t,x),x,x)+k^2/4*u(t,x)-diff(u(t,x),t,t);" }}}{PARA 0 "" 0 " " {TEXT -1 151 " We have used method of separation of variables: One assumes that solutions can be written as products of separate functio ns of t and x. We make the " }{TEXT 275 7 "ansatz " }{TEXT -1 83 "that u(t, x) is of the special form T(t) X(x), which we know as a product \+ solution." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "eq:=subs(u(t,x) =T(t)*X(x),waveop)=0;" }}}{PARA 0 "" 0 "" {TEXT -1 55 " This simpl ifies if we divide through by T(t) X(x)." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "eq/X(x)/T(t):\nexpand(%);" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 51 "sep:=(%)+(diff(T(t),t,t)/T(t)=diff(T(t),t,t)/T(t)); " }}}{PARA 0 "" 0 "" {TEXT -1 34 " The left side of the equation" }{TEXT 277 4 " sep" }{TEXT -1 195 " depends only on x and the right si de depends only on t. Thus, each side must be constant. We do not know the value of this constant, yet. For small k, the constant will be ne gative; we call it -" }{XPPEDIT 18 0 "mu^2" "6#*$%#muG\"\"#" }{TEXT -1 12 " and define " }{XPPEDIT 18 0 "alpha" "6#%&alphaG" }{TEXT -1 3 " by" }}{PARA 0 "" 0 "" {TEXT -1 73 " \+ - " }{XPPEDIT 18 0 "alpha^2" "6#* $%&alphaG\"\"#" }{TEXT -1 5 " = - " }{XPPEDIT 18 0 "mu^2-k^2/4;" "6#,& *$%#muG\"\"#\"\"\"*&%\"kG\"\"#\"\"%!\"\"F," }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 99 "We get the general solution to the resulting diff erential equation and use the boundary conditions." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "dsolve(\{dif f(X(x),x,x)=-alpha^2*X(x)\},X(x));" }}}{PARA 0 "" 0 "" {TEXT -1 65 "In order to satisfy the boundary condition at x = 0, we must have" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "X:=x->sin(alpha*x);" }}} {PARA 0 "" 0 "" {TEXT -1 68 "The second boundary condition makes a dif ferent kind of requirement." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "solve(X(L)=0,L);" }}}{PARA 0 "" 0 "" {TEXT -1 84 "There are an inf inite number of solutions for this equation, they change with L and " }{XPPEDIT 18 0 "alpha" "6#%&alphaG" }{TEXT -1 113 ". Maple did not pic k out any solution except L = 0. We know, however where the sine funct ion is zero. Thus, take " }{XPPEDIT 18 0 "alpha*L=n*Pi" "6#/*&%&alphaG \"\"\"%\"LGF&*&%\"nGF&%#PiGF&" }{TEXT -1 15 " and solve for " } {XPPEDIT 18 0 "alpha" "6#%&alphaG" }{TEXT -1 1 "." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "alpha:=solve(alpha*L=n*Pi,alpha);" }}}{PARA 0 "" 0 "" {TEXT -1 35 " We now have the function X(x)." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "X(x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "diff(X(x),x,x)/X(x)=-mu^2-k^2/4;" }}}{PARA 0 "" 0 "" {TEXT -1 71 " We now look at the other part of the PDE which we \" separated off\"." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 39 "dsolve(diff(T(t),t,t)=-mu^2*T(t),T(t));" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "mu:=sqrt(n^2*Pi^2/L^2-k^2/4) ;" }}}{PARA 0 "" 0 "" {TEXT -1 217 " The general solution we have \+ found is a linear combination of the particular solutions we get by se parating the equation. It is not yet obvious that this is a completely general solution. We will see that it is. " }}{PARA 0 "" 0 "" {TEXT -1 56 " The most general linear combination is of the form." }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 95 "u:=(t,x)->sum((A[n]*sin(mu*t )+B[n]*cos(mu*t))*sin(n*Pi*x/L),\n n=1..infinity);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 5 " " }}{PARA 0 "" 0 "" {TEXT -1 27 " To get a soluti on for " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 17 " " }{XPPEDIT 18 0 "diff(w(t,x),x,x)-k*diff(w(t,x), t) = diff(w(t,x),`$`(t,2));" "6#/,&-%%diffG6%-%\"wG6$%\"tG%\"xGF,F,\" \"\"*&%\"kGF--F&6$-F)6$F+F,F+F-!\"\"-F&6$-F)6$F+F,-%\"$G6$F+\"\"#" } {TEXT -1 1 "," }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 27 "recall that we make w from " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 85 " w( t , x ) = v(t,x ) - g x (L-x)/2 = exp(-k/2 t) u(t,x) - g x (L-x)/2" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 114 "The coefficients A and B have to be determined from the \+ intial conditions in a manner that are, by now, familiar. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 8 "Example;" }}{PARA 0 "" 0 "" {TEXT -1 20 "Find a solution for " }}{PARA 0 "" 0 "" {TEXT -1 17 " " }{XPPEDIT 18 0 "diff(w(t,x),`$`(x,2))-k*diff(w(t,x),t) = diff(w(t,x),`$`(t,2));" "6 #/,&-%%diffG6$-%\"wG6$%\"tG%\"xG-%\"$G6$F,\"\"#\"\"\"*&%\"kGF1-F&6$-F) 6$F+F,F+F1!\"\"-F&6$-F)6$F+F,-F.6$F+\"\"#" }{TEXT -1 1 "." }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 26 "with the requireme nt that " }}{PARA 0 "" 0 "" {TEXT -1 59 " \+ w( t, 0) = 0 = w( t, ¹) " }}{PARA 0 "" 0 "" {TEXT -1 3 "and" }} {PARA 0 "" 0 "" {TEXT -1 25 " " }{XPPEDIT 18 0 "w(0,x)=sin(2*x)" "6#/-%\"wG6$\"\"!%\"xG-%$sinG6#*&\"\"#\"\"\"F(F." }{TEXT -1 5 " and " }{XPPEDIT 18 0 "w[t](0,x) = 0" "6#/-&%\"wG6#%\"tG6 $\"\"!%\"xGF*" }}{PARA 0 "" 0 "" {TEXT -1 3 "and" }}{PARA 0 "" 0 "" {TEXT -1 45 " k = 1/4." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 72 "From the above an d with g = 0, we know the solution is given as follows:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 43 " w( t , x ) = exp(-k/2 t) u(t,x) " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 3 "and" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 7 " " }{XPPEDIT 18 0 "u(t,x) = sum((A[n]*sin(mu[n]*t) +B[n]*cos(mu[n]*t))*sin(n*Pi*x/L),n=1..infinity)" "6#/-%\"uG6$%\"tG%\" xG-%$sumG6$*&,&*&&%\"AG6#%\"nG\"\"\"-%$sinG6#*&&%#muG6#F2F3F'F3F3F3*&& %\"BG6#F2F3-%$cosG6#*&&F96#F2F3F'F3F3F3F3-F56#**F2F3%#PiGF3F(F3%\"LG! \"\"F3/F2;\"\"\"%)infinityG" }}{PARA 0 "" 0 "" {TEXT -1 5 "where" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 10 " \+ " }{XPPEDIT 18 0 "mu[n] =sqrt(n^2*Pi^2/L^2-k^2/4)" "6#/&%#muG6#%\"nG- %%sqrtG6#,&*(F'\"\"#%#PiG\"\"#*$%\"LG\"\"#!\"\"\"\"\"*&%\"kG\"\"#\"\"% F3F3" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 33 "The initial conditions imply that" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 47 " sin (2 x) = u( 0, x) = = " }{XPPEDIT 18 0 "sum(B[n]*sin(n*Pi*x/L)" "6#-%$ sumG6#*&&%\"BG6#%\"nG\"\"\"-%$sinG6#**F*F+%#PiGF+%\"xGF+%\"LG!\"\"F+" }}{PARA 0 "" 0 "" {TEXT -1 3 "and" }}{PARA 0 "" 0 "" {TEXT -1 18 " \+ " }{XPPEDIT 18 0 "k/2" "6#*&%\"kG\"\"\"\"\"#!\"\"" } {TEXT -1 12 " sin(2 x) = " }{XPPEDIT 18 0 "u[t](0,x)" "6#-&%\"uG6#%\"t G6$\"\"!%\"xG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "sum(A[n]*mu[n]*sin(n* Pi*x/L)" "6#-%$sumG6#*(&%\"AG6#%\"nG\"\"\"&%#muG6#F*F+-%$sinG6#**F*F+% #PiGF+%\"xGF+%\"LG!\"\"F+" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 36 "where L = ¹. This implies that al l " }{XPPEDIT 18 0 "B[n]" "6#&%\"BG6#%\"nG" }{TEXT -1 20 "'s are zero, except " }{XPPEDIT 18 0 "B[2]" "6#&%\"BG6#\"\"#" }{TEXT -1 12 "=1, an d all " }{XPPEDIT 18 0 "A[n]" "6#&%\"AG6#%\"nG" }{TEXT -1 19 "'s are z ero except " }{XPPEDIT 18 0 "A[2]" "6#&%\"AG6#\"\"#" }{TEXT -1 3 " = \+ " }{XPPEDIT 18 0 "k/(2*mu[n])" "6#*&%\"kG\"\"\"*&\"\"#F%&%#muG6#%\"nGF %!\"\"" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 35 " Finally, here is the solution:" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 109 "k:=1/4;\nu:=(t,x)->sin(2*x)*cos(sqrt(4-k ^2/4)*t)\n + k/(2*sqrt(4-k^2/4))*sin(2*x)*sin(sqrt((4-k^2/4))*t) ;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "w:=(t,x)->exp(-k/2*t)* u(t,x);" }}}{PARA 0 "" 0 "" {TEXT -1 34 "We verify the boundary condit ions." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "w(t,0); w(t,Pi);" } }}{PARA 0 "" 0 "" {TEXT -1 33 "We verify the initial conditions." }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "w(0,x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "diff(w(t,x),t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "simplify(subs(t=0,%));" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 28 "We verify the PDE is solved. " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 79 "simplify(diff(diff(w(t,x ),x),x)-k*diff(w(t,x),t)-\n diff(diff(w(t,x),t),t));" }}}{PARA 0 " " 0 "" {TEXT -1 38 "We plot solutions at different times." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 79 "plot([w(0,x),w(1,x),w(2,x),w(3,x)], x=0..Pi,\n color=[black,red,blue,green]);" }}}{PARA 0 "" 0 "" {TEXT -1 24 "We animate the solution." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(plots):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "animate(w(t,x),x=0..Pi,t=0..20);" }}}{PARA 0 "" 0 "" {TEXT -1 29 "We draw the solution surface." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "plot3d(w(t,x),x=0..Pi,t=0..6,orientation=[-135,45],ax es=NORMAL);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "resta rt;" }}{PARA 0 "" 0 "" {TEXT -1 35 "The problem is to find w to satisf y" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 29 " \+ " }{XPPEDIT 18 0 "diff(w,`$`(x,2))-k*diff(w,t )-g = diff(w,`$`(t,2));" "6#/,(-%%diffG6$%\"wG-%\"$G6$%\"xG\"\"#\"\"\" *&%\"kGF.-F&6$F(%\"tGF.!\"\"%\"gGF4-F&6$F(-F*6$F3\"\"#" }}{PARA 0 "" 0 "" {TEXT -1 3 "(*)" }}{PARA 0 "" 0 "" {TEXT -1 65 " \+ w(t, 0) = 0 = w(t, L)" }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 59 " \+ w( 0, x) = F(x)" }}{PARA 0 "" 0 "" {TEXT -1 44 " \+ " }{XPPEDIT 18 0 "w[t]" "6#& %\"wG6#%\"tG" }{TEXT -1 17 " ( 0 , x) = G(x)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 37 "To work this problem, we \+ define v by " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 58 " v(t, x) = w( t , x) + g x (L-x)/2" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "v:=(t,x)->w(t,x)+g/2*x*(L-x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "diff(v(t,x),x,x)-k*diff(v(t,x),t)=diff(v(t,x),t,t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "v(t,0); v(t,L);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "v(0,x); D[1](v)(0,x);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 105 "Thus, we see that w satisfies \+ (*) if and only if v satisfies the following partial differential equa tion." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 28 " " }{XPPEDIT 18 0 "diff(v,`$`(x,2))-k*diff( v,t) = diff(v,`$`(t,2));" "6#/,&-%%diffG6$%\"vG-%\"$G6$%\"xG\"\"#\"\" \"*&%\"kGF.-F&6$F(%\"tGF.!\"\"-F&6$F(-F*6$F3\"\"#" }}{PARA 0 "" 0 "" {TEXT -1 4 "(**)" }}{PARA 0 "" 0 "" {TEXT -1 65 " \+ v(t, 0) = 0 = v(t, L)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 77 " \+ v( 0 , x) = F(x) + g x ( L - x)/2" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 44 " \+ " }{XPPEDIT 18 0 "v[t]" "6#&%\"vG6#%\"tG" }{TEXT -1 17 " ( 0 , x) = G(x)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 59 "To solve this equation (**), we define a new funct ion u by " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 59 " u( t , x) = exp(t k/2 ) v(t , x)" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 83 "We check \+ u in the following partial differential equation with boundary conditi ons:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "v:='v';\nu:=(t,x)->exp(k/2*t)*v(t,x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "diff(u(t,x),x,x)+k^2/4*u(t,x)-diff( u(t,x),t,t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "collect(%,e xp(k/2*t));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "u(t,0); u(t, L);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "u(0,x); D[1](u)(0,x) ;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 105 "Thus, we see that v satisfies (**) if and only if u satisfies \+ the following partial differential equation" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 39 " \+ " }{XPPEDIT 18 0 "diff(u,`$`(x,2))+k^2/4*u = diff(u,`$`(t,2)); " "6#/,&-%%diffG6$%\"uG-%\"$G6$%\"xG\"\"#\"\"\"*(%\"kG\"\"#\"\"%!\"\"F (F.F.-F&6$F(-F*6$%\"tG\"\"#" }}{PARA 0 "" 0 "" {TEXT -1 5 "(***)" }} {PARA 0 "" 0 "" {TEXT -1 65 " \+ u(t, 0) = 0 = u(t, L)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 65 " u( 0 , x) = v( 0 , x)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 44 " " }{XPPEDIT 18 0 "u[t]" "6#&%\"uG6#%\"tG" }{TEXT -1 12 " ( 0 , x) = " }{XPPEDIT 18 0 "k/2*v(0,k)+v[t](0,x))" "6#,&*(%\"kG\"\"\"\"\"#!\"\"-%\"vG6$\"\"! F%F&F&-&F*6#%\"tG6$F,%\"xGF&" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 34 "So here is how the procedure goes:" }}{PARA 0 " " 0 "" {TEXT -1 53 "To solve (*) solve (***) with u(t, 0) = 0 = u(t, L ), " }}{PARA 0 "" 0 "" {TEXT -1 46 " u(0, x) = F(x)+ g x (L - \+ x)/2 , and " }{XPPEDIT 18 0 "u[t]" "6#&%\"uG6#%\"tG" }{TEXT -1 13 " ( 0 , x ) = " }{XPPEDIT 18 0 "k*(F(x)+g*x*(L-x)/2)/2+G(x);" "6#,&*(%\"k G\"\"\",&-%\"FG6#%\"xGF&**%\"gGF&F+F&,&%\"LGF&F+!\"\"F&\"\"#F0F&F&\"\" #F0F&-%\"GG6#F+F&" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 12 "Create v \+ by " }}{PARA 0 "" 0 "" {TEXT -1 66 " \+ v( t , x) = exp(-k/2 t) u(t, x)" }}{PARA 0 "" 0 "" {TEXT -1 12 "Creat e w by " }}{PARA 0 "" 0 "" {TEXT -1 107 " \+ w( t , x ) = v(t, x) - g x (L-x)/2 = exp(-k/2 t) u(t, x) - g x (L-x)/2" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 32 "Here is a check that this works." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "u:='u';" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 " w:=(t,x)->exp(-k/2*t)*u(t,x)-g*x*(L-x)/2;" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 53 "diff(w(t,x),x,x)-k*diff(w(t,x),t)-g-diff(w(t,x),t,t );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "simplify(%);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "collect(%,exp(-k/2*t));" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "w(t,0); w(t,L);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "w(0,x); D[1](w)(0,x);" }}}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 29 "Differe nt boundary conditions" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 196 "We examine a variety of boundary conditi ons. We will provide illustrations for how these boundary conditions w ould look in a vibrating string. We will need the plots package to pro vide animations." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(pl ots):" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 278 20 "(1) Fixed endpoints." }{TEXT -1 176 " This is the situation we hav e considered to this point in these notes. Here is a graph of a funct ion that has this boundary condition. Mathematically, the conditions w ould be" }}{PARA 0 "" 0 "" {TEXT -1 38 " u(t, 0) = 0 and u(t, L) = 0." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "u:=(t,x)->2*sin( x)*cos(t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "diff(u(t,x),t ,t)-diff(u(t,x),x,x);\nu(t,0);\nu(t,L);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "plot(u(0,x),x=0..Pi);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "animate(u(t,x),x=0..Pi,t=0..8);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "plot3d(u(t,x),x=0..Pi,t=0..2*Pi,axes=NORMAL,o rientation=[-160,60]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 279 23 "(2) E lastic attachment." }{TEXT -1 199 " This situation assumes we have a s pring, or other elastic device attached to the string. Such an arrange ment would tend to bring the string back to the displacement. Such con ditions could be written" }}{PARA 0 "" 0 "" {TEXT -1 10 " " } {XPPEDIT 18 0 "diff(u,x);" "6#-%%diffG6$%\"uG%\"xG" }{TEXT -1 27 "(t, \+ 0) = k u(t, 0) and " }{XPPEDIT 18 0 "diff(u,x);" "6#-%%diffG6$%\"u G%\"xG" }{TEXT -1 21 "(t, L) = - k u(t, L)." }}{PARA 0 "" 0 "" {TEXT -1 81 "Such conditions are reminiscent of radiation cooling for heat d iffusion problems." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "u:=(t, x)->cos(t)*sin(x+Pi/4);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 86 " diff(u(t,x),t,t)-diff(u(t,x),x,x);\nD[2](u)(t,0)-k*u(t,0);\nD[2](u)(t, Pi/2)-k*u(t,Pi/2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "plot( u(0,x),x=0..Pi/2,y=0..1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "animate(u(t,x),x=0..Pi/2,t=0..5);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "plot3d(u(t,x),x=0..Pi/2,t=0..Pi,axes=NORMAL,orientati on=[25,60]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 280 24 "(3) Frictionle ss sleeve." }{TEXT -1 233 " This models having the string attached to \+ carts which hold the string with a horizontal tangent at the ends, but allows the string to move up and down as the cart runs along a fricti onless track. This situation can be described with" }}{PARA 0 "" 0 "" {TEXT -1 10 " " }{XPPEDIT 18 0 "diff(u,x);" "6#-%%diffG6$%\"u G%\"xG" }{TEXT -1 15 "(t, 0) = 0 and " }{XPPEDIT 18 0 "diff(u,x);" "6# -%%diffG6$%\"uG%\"xG" }{TEXT -1 12 "(t, L) = 0. " }}{PARA 0 "" 0 "" {TEXT -1 78 "In this illustration, one end is fixed, the other is on a frictionless sleeve." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "u:= (t,x)->sin(t+Pi/2)*cos(x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "diff(u(t,x),t,t)-diff(u(t,x),x,x);\nD[2](u)(t,0);\nu(t,Pi/2);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "plot(u(0,x),x=0..Pi/2);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "animate(u(t,x),x=0..Pi/2,t=0 ..8);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "plot3d(u(t,x),x=0. .Pi/2,t=0..2*Pi,axes=NORMAL,orientation=[-70,55]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 281 31 "(4) Changing Boundary Condition" }{TEXT -1 362 ". We suppose we have a half infinite string in this model and we \+ change the boundary condition with time. Take the initial conditions t o be zero, and take U(t, 0) = b(t). We expect to see a signal move dow n the string. Here is an analysis of the problem. With no appeal to in itial conditions or boundary conditions, we found that solutions shoul d have the form " }{XPPEDIT 18 0 "psi;" "6#%$psiG" }{TEXT -1 12 "(x + c t) + " }{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 98 "(x - c t). Us ing that the initial conditions are zero for x > 0, recall that we can conclude that " }{XPPEDIT 18 0 "psi;" "6#%$psiG" }{TEXT -1 12 "(x) = \+ 0 and " }{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 30 "(x) = 0 for x > 0. The term " }{XPPEDIT 18 0 "psi;" "6#%$psiG" }{TEXT -1 70 "(x + c t) will be zero since c > 0 and t > 0. Thus, what ever happens," }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 15 " u(t, x) = " }{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 10 "(x - c t)," }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 59 "and this \+ is zero as long as x > c t. We ask how to extend " }{XPPEDIT 18 0 "ph i;" "6#%$phiG" }{TEXT -1 145 " to the negative numbers. The answer mus t lie in the boundary condition. Recall that it did earlier, also. Wel l, we know that u(t, 0) = b(t), so " }}{PARA 0 "" 0 "" {TEXT -1 10 " \+ " }{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 27 "(0 - c t) = u (t, 0) = b(t)." }}{PARA 0 "" 0 "" {TEXT -1 36 "So, for negative number s n, we have " }{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 34 "(n) = b( -n/c). Here is an example:" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "b:=x->sin(x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "u:=(t, x)->piecewise(x " 0 "" {MPLTEXT 1 0 33 "animate(u(t,x),x=0..10, t=0..10);" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 65 "plot3d(u(t,x),x=0..10,t=0..10,axes=NORMAL,orientati on=[-115,55]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 8 "Ex ercise" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 96 "Send a signal down a hal f infinite string that consists of only one period of the sine functio n." }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 94 "We can not leave the subject of vibrati ng strings without saying a word or two about music and" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 14 "Standing wav es" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 67 "Her e we say a bit about standing waves. Look first at these graphs." }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "plot([sin(x),sin(2*x),sin(3* x)],x=0..Pi,color=[BLACK,RED,BLUE]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 164 "With these as initi al functions, we know what are the solutions for the wave equation.\n \+ sin( x ) cos( t ), or sin( 2 x ) cos( 2 t), or sin( 3 x ) cos( \+ 3 t )." }}{PARA 0 "" 0 "" {TEXT -1 81 "To see what happens to any othe r initial distribution, we make the Fourier series" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 12 " f(x) = " }{XPPEDIT 18 0 "sum(a[n]*sin(n*x),n);" "6#-%$sumG6$*&&%\"aG6#%\"nG\"\"\"-%$sinG6 #*&F*F+%\"xGF+F+F*" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 6 "to ge t" }}{PARA 0 "" 0 "" {TEXT -1 23 " u( t, x ) = " }{XPPEDIT 18 0 "sum(a[n]*sin(n*x)*cos(n*t),n);" "6#-%$sumG6$*(&%\"aG6#%\"nG\"\" \"-%$sinG6#*&F*F+%\"xGF+F+-%$cosG6#*&F*F+%\"tGF+F+F*" }{TEXT -1 2 ". \+ " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 18 "Here' s an example." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "f:=x->piece wise(x " 0 "" {MPLTEXT 1 0 19 "pl ot(f(x),x=0..Pi);" }}}{PARA 0 "" 0 "" {TEXT -1 36 "Fourier coefficient s can be found by" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 76 "for n f rom 1 to 5 do\n a[n]:=2/Pi*int(f(x)*sin(n*x),x=0..Pi);\nod;\nn:='n ';" }}}{PARA 0 "" 0 "" {TEXT -1 51 "Here is a check that this is a goo d representation." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "plot([f (x),sum(a[n]*sin(n*x),n=1..5)],x=0..Pi,color=[BLACK,RED]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 42 "He re are the standing waves that are used." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "plot([sin(x),sin(3*x),sin(5*x)],x=0..Pi,color=[BLACK, RED,BLUE]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 70 "Here are the standing waves multiplied by the app ropriate coefficient." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 85 "plo t([a[1]*sin(x),a[3]*sin(3*x),a[5]*sin(5*x)],x=0..Pi,\n color=[BLAC K,RED,BLUE]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 398 "What sounds go od? Even if you are not a musician, you probably know that the C-major chord is C - E -G. This sounds good. Why it sounds good is a combinat ion of mathematics and biology. What to note here is the good approxim ation for the plucked string could resemble C, together with faint ove rtones of E and G. Let me remind you that if C = 256 vibrations per se cond, then E = ~323 and G = ~ 384." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "256,256*3/2,256*5/4;" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 283 "This plucked string would be a bout C if the vibrations were 256 vibrations per second. There are tho se other terms. I have become deaf to high pitched signals, especially as faint as the fourth, fifth, and higher terms would be. Thus, I wou ld hear only the first three terms anyway. " }}{PARA 0 "" 0 "" {TEXT -1 209 " Other initial distributions than this string plucked in t he middle might not have this same distribution. there might be a prom inent C# overtone. C and C# played together sounds BAD. Try it on your piano." }}{PARA 0 "" 0 "" {TEXT -1 90 " The mathematics of music \+ would be an interesting direction to go from here. We won't." }}}}} {MARK "20" 0 }{VIEWOPTS 1 1 0 1 1 1803 }