{VERSION 2 3 "IBM INTEL NT" "2.3" } {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 Output" 2 20 "" 0 1 0 0 255 1 0 0 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 "Maple Output" 0 11 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 3 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 11 12 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }1 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Maple Plot" 0 13 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 "" 0 259 1 {CSTYLE "" -1 -1 "Times" 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 }} {SECT 0 {EXCHG {PARA 259 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 19 "Homework Problem #2" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 95 "In the above, we solved the wave equation u_tt = u_xx with Dirichlet boundary conditions. This" }}{PARA 0 "" 0 "" {TEXT -1 87 "meant that we needed an odd reflection and used a si ne series. The wave equation with " }}{PARA 0 "" 0 "" {TEXT -1 85 "Ne umann boundary condition will need an even reflection and will use a c osine series." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 85 "a) Modify the above program so that it solves the wave eq uation with Neumann boundary" }}{PARA 0 "" 0 "" {TEXT -1 10 "condition ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 65 "b) V erify that your code is correct by considering special cases " }} {PARA 0 "" 0 "" {TEXT -1 39 " 1) phi(x) = cos(2x), psi(x) = 0" }}{PARA 0 "" 0 "" {TEXT -1 39 " 2) phi(x) = sin(2x), psi(x) = 0 " }}{PARA 0 "" 0 "" {TEXT -1 39 " 3) phi(x) = 0, psi(x) = cos(3 x)" }}{PARA 0 "" 0 "" {TEXT -1 39 " 4) phi(x) = 0, psi(x) = sin (4x)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 109 "NOTE: You have to be careful with the \+ zero Fourier component terms. If b_0(t) is the coefficient of the 0t h" }}{PARA 0 "" 0 "" {TEXT -1 108 "cosine term then d^2/dt^2 b_0 = 0. \+ Thus b_0(t) = b_psi[0] t + b_phi[0] where b_phi[0] is the coefficient of" }}{PARA 0 "" 0 "" {TEXT -1 89 "the 0th cosine term of phi and b_p si[0] is the coefficient of the 0th cosine term of psi." }}{PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "N := 5: b_phi := array(0..N) : b_psi := array(0..N): " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "psi := \+ x -> 0: phi := x -> cos(2*x):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 127 "f or j from 1 to N do b_phi[j]:= evalf(Int(phi(x)*cos(j*x),x=-Pi..Pi))/( Pi) od: b_phi[0] := evalf(Int(phi(x),x=-Pi..Pi))/(2*Pi):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 128 "for j from 1 to N do b_psi[j]:= evalf(Int(ps i(x)*cos(j*x),x=-Pi..Pi))/(Pi) od: b_psi[0] := evalf(Int(psi(x),x=-Pi. .Pi))/(2*Pi): " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 117 "u := b_phi[0] + \+ t*b_psi[0]:\nfor j from 1 to N do u := u + (b_phi[j]*cos(j*t) + b_psi[ j]*sin(j*t)/(j))*cos(j*x) od:\nu;" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#, .*$%#PiG!\"\"$\"+^\"=[1#!#A*(F%F&-%$cosG6#%\"tG\"\"\"-F,6#%\"xGF/$!+F2 (*RTF)*(F%F&-F,6#,$F.\"\"#F/-F,6#,$F2F9F/$\"+aEfTJ!\"**(F%F&-F,6#,$F. \"\"$F/-F,6#,$F2FDF/$!+7w.UTF)*(F%F&-F,6#,$F.\"\"%F/-F,6#,$F2FNF/$\"+w =I>TF)*(F%F&-F,6#,$F.\"\"&F/-F,6#,$F2FXF/$!+#Qrh9%F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "N := 5: b_phi := array(0..N): b_psi := ar ray(0..N): " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "psi := x -> 0: phi : = x -> sin(2*x):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 127 "for j from 1 t o N do b_phi[j]:= evalf(Int(phi(x)*cos(j*x),x=-Pi..Pi))/(Pi) od: b_phi [0] := evalf(Int(phi(x),x=-Pi..Pi))/(2*Pi):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 128 "for j from 1 to N do b_psi[j]:= evalf(Int(psi(x)*cos (j*x),x=-Pi..Pi))/(Pi) od: b_psi[0] := evalf(Int(psi(x),x=-Pi..Pi))/(2 *Pi): " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 117 "u := b_phi[0] + t*b_psi[ 0]:\nfor j from 1 to N do u := u + (b_phi[j]*cos(j*t) + b_psi[j]*sin(j *t)/(j))*cos(j*x) od:\nu;" }}{PARA 0 "" 0 "" {TEXT -1 121 "The initial data phi(x) = sin(2x), psi(x) = 0 has the exact solution u(x,t) = 0 s ince all the cosine transforms are zero." }{MPLTEXT 1 0 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "N := 5: b_phi := array(0..N): b_psi := array(0..N): " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "psi := x -> cos(3*x): phi := x -> 0:" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 127 "for j from 1 to N do b_phi[j]:= ev alf(Int(phi(x)*cos(j*x),x=-Pi..Pi))/(Pi) od: b_phi[0] := evalf(Int(phi (x),x=-Pi..Pi))/(2*Pi):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 128 "for j f rom 1 to N do b_psi[j]:= evalf(Int(psi(x)*cos(j*x),x=-Pi..Pi))/(Pi) od : b_psi[0] := evalf(Int(psi(x),x=-Pi..Pi))/(2*Pi): " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 117 "u := b_phi[0] + t*b_psi[0]:\nfor j from 1 to N do u := u + (b_phi[j]*cos(j*t) + b_psi[j]*sin(j*t)/(j))*cos(j*x) od:\nu; " }}{PARA 0 "" 0 "" {TEXT -1 115 "The initial data phi(x) = 0, psi(x) \+ = cos(3x) has exact solution u(x,t) = sin(3t) cos(3x)/3, which is what we have," }}{PARA 0 "" 0 "" {TEXT -1 24 "up to machine precision." } {MPLTEXT 1 0 0 "" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#,.*&%\"tG\"\"\"%#P iG!\"\"$!+@>&*o?!#A*(F'F(-%$sinG6#F%F&-%$cosG6#%\"xGF&$\"+s+xLTF+*(F'F (-F.6#,$F%\"\"#F&-F16#,$F3F:F&$!+1)=52#F+*(F'F(-F.6#,$F%\"\"$F&-F16#,$ F3FDF&$\"+^v>Z5!\"**(F'F(-F.6#,$F%\"\"%F&-F16#,$F3FOF&$!+n&fq.\"F+*(F' F(-F.6#,$F%\"\"&F&-F16#,$F3FYF&$\"+a9%*z#)!#B" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "N := 5: b_phi := array(0..N): b_psi := array(0.. N): " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "psi := x -> sin(4*x): phi : = x -> 0:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 127 "for j from 1 to N do \+ b_phi[j]:= evalf(Int(phi(x)*cos(j*x),x=-Pi..Pi))/(Pi) od: b_phi[0] := \+ evalf(Int(phi(x),x=-Pi..Pi))/(2*Pi):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 128 "for j from 1 to N do b_psi[j]:= evalf(Int(psi(x)*cos(j*x),x=-Pi ..Pi))/(Pi) od: b_psi[0] := evalf(Int(psi(x),x=-Pi..Pi))/(2*Pi): " }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 117 "u := b_phi[0] + t*b_psi[0]:\nfor j from 1 to N do u := u + (b_phi[j]*cos(j*t) + b_psi[j]*sin(j*t)/(j))*c os(j*x) od:\nu;" }}{PARA 11 "" 0 "" {TEXT -1 117 "The initial data phi (x) = 0, psi(x) = sin(4x) has exact solution u(x,t) = 0 since all the \+ cosine transforms are zero." }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 151 "Amp:=2: eps := .5: x_L : = Pi-2*eps: x_R := Pi - eps: phi := x -> Amp*((Heaviside(x-x_L) + Hea viside(x_R-x)-1)+(Heaviside(-x-x_L)+Heaviside(x_R+x)-1)):" }}{PARA 0 " > " 0 "" {MPLTEXT 1 0 14 "psi := x -> 0:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "N := 20: b_phi := array(0..N): b_psi := array(0..N): \+ " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 127 "for j from 1 to N do b_phi[j]: = evalf(Int(phi(x)*cos(j*x),x=-Pi..Pi))/(Pi) od: b_phi[0] := evalf(Int (phi(x),x=-Pi..Pi))/(2*Pi):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 128 "for j from 1 to N do b_psi[j]:= evalf(Int(psi(x)*cos(j*x),x=-Pi..Pi))/(Pi ) od: b_psi[0] := evalf(Int(psi(x),x=-Pi..Pi))/(2*Pi): " }}{PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 114 "u := b_phi[0] + t*b_psi[0]:\nfor j from 1 to \+ N do u := u + (b_phi[j]*cos(j*t) + b_psi[j]*sin(j*t)/(j))*cos(j*x) od: " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "plot(\{phi(x),subs(t=0,u)\},x=0..Pi);" }}{PARA 13 "" 1 "" {INLPLOT "6&-%'CURVESG6$7co7$\"\"!F(7$$\"1]0\"3%zxZo!#F0F(7$$\"1byw%3o^i#F0F(7$$\"1*\\]U[nkH$F0 F(7$$\"1![v)fz%)=RF0F(7$$\"1t5q#*oGjXF0F(7$$\"1A(3*elwH_F0F(7$$\"1#Guj 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