{VERSION 5 0 "IBM INTEL NT" "5.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 1 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 1 }{CSTYLE "2D Comment" 2 18 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 256 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 257 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 258 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 259 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 260 "" 1 14 0 0 0 0 0 1 1 0 0 0 0 0 0 1 } {PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Heading 1" 0 3 1 {CSTYLE "" -1 -1 "" 1 18 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }1 0 0 0 8 4 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 256 1 {CSTYLE "" -1 -1 "" 1 14 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }0 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 -1 72 "Module 27 Laplace's Equation (or the Potential Equation) on a rectang le." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 122 " \+ From what has come before, we can expect that a study of the heat \+ equation on a rectangle to have a predictable form:" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 36 " \+ " }{XPPEDIT 18 0 "diff(u,t) = diff(u,`$`(x,2))+diff(u,`$` (y,2));" "6#/-%%diffG6$%\"uG%\"tG,&-F%6$F'-%\"$G6$%\"xG\"\"#\"\"\"-F%6 $F'-F-6$%\"yGF0F1" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 74 "Boundary conditions: u(t, x, 0) = f(x), \+ u(t, x, L) = g(x)" }}{PARA 0 "" 0 "" {TEXT -1 88 " \+ u(t, 0, y) = h(y), u(t, M, y) = J(y)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 64 "Initial condition: u(0, x, y) = K(x, y)." }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 83 "It is not a surpri se, either, that one should solve the steady-state problem first." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 34 " \+ " }{XPPEDIT 18 0 "0 = diff(u,`$`(x,2))+diff(u ,`$`(y,2));" "6#/\"\"!,&-%%diffG6$%\"uG-%\"$G6$%\"xG\"\"#\"\"\"-F'6$F) -F+6$%\"yGF.F/" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 68 "Boundary conditions: u(x, 0) = f(x), u(x , L) = g(x)" }}{PARA 0 "" 0 "" {TEXT -1 82 " \+ u(0, y) = h(y), u(M, y) = J(y)" }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 31 "and then the trans ition problem" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 34 " " }{XPPEDIT 18 0 "diff( u,t) = diff(u,`$`(x,2))+diff(u,`$`(y,2));" "6#/-%%diffG6$%\"uG%\"tG,&- F%6$F'-%\"$G6$%\"xG\"\"#\"\"\"-F%6$F'-F-6$%\"yGF0F1" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 68 "Boundary conditions: \+ u(t, x, 0) = 0, u(t, x, L) = 0" }}{PARA 0 "" 0 "" {TEXT -1 82 " u(t, 0, y) = 0, u(t, M, y) = 0" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 64 "Initial condition: u(0, x, y) = k(x, y)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 111 "The wave equation leads to this same steady state proble m. We consider this problem in this part of the notes. " }}{PARA 0 "" 0 "" {TEXT -1 98 " We begin consideration of this steady-state pro blem. It is what is called Laplace's equation." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 83 "This steady state problem is commonly broken into two problems. Here is one of them" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 47 " " } {XPPEDIT 18 0 "0 = diff(u,`$`(x,2))+diff(u,`$`(y,2));" "6#/\"\"!,&-%%d iffG6$%\"uG-%\"$G6$%\"xG\"\"#\"\"\"-F'6$F)-F+6$%\"yGF.F/" }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 68 "Boundary condition s: u(x, 0) = f(x), u(x, L) = g(x)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 76 " \+ u(0, y) = 0, u(M, y) = 0" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 54 "Here are the steps: (1) Make a separation of variables" }}{PARA 0 "" 0 "" {TEXT -1 111 " \+ (2) Identify the resulting ordinary different ial equation and boundary conditions" }}{PARA 0 "" 0 "" {TEXT -1 55 " \+ (3) Solve these equations" }}{PARA 0 "" 0 "" {TEXT -1 80 " (4) Construct the gene ral solution for the problem" }}{PARA 0 "" 0 "" {TEXT -1 111 " \+ (5) Use the boundary conditions to get the solut ion for this particular equation." }}{PARA 0 "" 0 "" {TEXT -1 27 "Step 1: Separate variables." }}{PARA 0 "" 0 "" {TEXT -1 96 " The partial \+ differential equation leads to X '' Y + X Y'' = 0, with X(0) Y(y) = 0 \+ = X(M) Y(y)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 44 "Step 2: Identify the differential equations." }}{PARA 0 "" 0 " " {TEXT -1 39 " The differential equations are X '' + " }{XPPEDIT 18 0 "lambda^2;" "6#*$%'lambdaG\"\"#" }{TEXT -1 36 " X = 0, X(0) = 0 = X( M), and Y '' = " }{XPPEDIT 18 0 "lambda^2;" "6#*$%'lambdaG\"\"#" } {TEXT -1 3 " Y." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 30 "Step 3: Solve these equations." }}{PARA 0 "" 0 "" {TEXT -1 5 " " }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT -1 5 " = n \+ " }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 18 "/M , X(x) = sin(n" } {XPPEDIT 18 0 "pi;" "6#%#piG" }{TEXT -1 23 "x/M), and Y(y) = sinh(n" } {XPPEDIT 18 0 "pi;" "6#%#piG" }{TEXT -1 15 "y/M) and sinh(n" } {XPPEDIT 18 0 "pi;" "6#%#piG" }{TEXT -1 9 "(L-y)/M)." }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 58 "Step 4: Construct the g eneral solution. If L & M were 1, " }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{PARA 0 "" 0 "" {TEXT -1 15 " u(x, y) = " }{XPPEDIT 18 0 "sum(a[n ]*sinh(n*Pi*y)+b[n]*sinh(n*Pi*(1-y))*sin(n*Pi*x),n);" "6#-%$sumG6$,&*& &%\"aG6#%\"nG\"\"\"-%%sinhG6#*(F+F,%#PiGF,%\"yGF,F,F,*(&%\"bG6#F+F,-F. 6#*(F+F,F1F,,&F,F,F2!\"\"F,F,-%$sinG6#*(F+F,F1F,%\"xGF,F,F,F+" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 53 "Step 5: Use the boundary conditions to determine the " }{TEXT 256 1 " a" }{TEXT -1 8 " 's and " }{TEXT 257 1 "b" }{TEXT -1 4 " 's." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 25 "Check t he Behold solution" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 247 "Here, I suppose that n is an integer and check that the \+ terms u(x,y) = X(x) Y(y) form a solution to Laplaces equation with zer o boundary conditions at x = 0 and at x = M. I also check that at y = \+ 0 and at y = L, we are set to use a Fourier Series." }}{PARA 0 "" 0 " " {TEXT -1 2 " " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "assume(n ,integer);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 77 "u:=(x,y)->(a* sinh(n*Pi*y/M)+b*sinh(n*Pi*(L-y)/M))*\n sin(n*Pi*x/M);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "simplify(diff(u(x,y),x,x)+di ff(u(x,y),y,y));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "u(0,y); u(M,y);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "u(x,0); u(x,L); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{PARA 0 "" 0 "" {TEXT -1 42 "The point is to check and check and check." }}} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 24 "Details for this problem" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 155 "We are now ready to apply this to a situation with and both f and g specified. We consid er the problem with f = g and as given below. We choose L = M = 1." }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 102 "u:=(x,y)->sum((a[n]*sinh(n*Pi*y)+b[n]*sinh( n*Pi*(1-y)))*sin(n*Pi*x),\n n=1..infinity);" }}} {PARA 0 "" 0 "" {TEXT -1 64 "We wish to determine the coefficients. Ho w will they be defined?" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "f (x)=u(x,0);\ng(x)=u(x,1);" }}}{PARA 0 "" 0 "" {TEXT -1 83 "This remind s us to use the Fourier coefficients. ( Don't forget to divide by sinh (n" }{XPPEDIT 18 0 "pi;" "6#%#piG" }{TEXT -1 4 "). )" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "f:=x->piecewise(x<1/2,2*x,2*(1-x));" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "plot(f(x),x=0..1);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 127 "for n from 1 to 3 do\n b[ n]:=int('f(x)'*sin(n*Pi*x),x=0..1)/\n int(sin(n*Pi*x)^2,x=0..1 )/sinh(n*Pi);\n a[n]:=b[n]:\nod;" }}}{PARA 0 "" 0 "" {TEXT -1 35 "We incorporate these values into u." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 118 "u:=(x,y)->8/Pi^2*sum(sin(n*Pi/2)/n^2*\n (sin h(n*Pi*y)+sinh(n*Pi*(1-y)))/sinh(n*Pi)*sin(n*Pi*x),n=1..3);\nn:='n':" }}}{PARA 0 "" 0 "" {TEXT -1 59 "As a safety check, we plot the ends ag ainst the graph of f." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "plo t([u(x,0),'f(x)'],x=0..1);" }}}{PARA 0 "" 0 "" {TEXT -1 56 "We are now ready to see a plot of u. What do you expect?" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "plot3d(u(x,y),x=0..1,y=0..1,axes=NORMAL);" }}} {PARA 0 "" 0 "" {TEXT -1 46 "It is of value sometimes to see contour p lots." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(plots):" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "contourplot(u(x,y),x=0..1,y= 0..1);" }}}{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 85 "We now do a problem with two sides insula ted. Here are the equations for the problem." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 47 " \+ " }{XPPEDIT 18 0 "0 = diff(u,`$`(x,2))+diff(u,`$`(y,2) );" "6#/\"\"!,&-%%diffG6$%\"uG-%\"$G6$%\"xG\"\"#\"\"\"-F'6$F)-F+6$%\"y GF.F/" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 36 " Boundary conditions: " }{XPPEDIT 18 0 "u;" "6#%\"uG" } {TEXT -1 17 "(x, 0) = f(x), " }{XPPEDIT 18 0 "u;" "6#%\"uG" }{TEXT -1 13 "(x, L) = g(x)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 50 " " } {XPPEDIT 18 0 "u[x];" "6#&%\"uG6#%\"xG" }{TEXT -1 14 "(0, y) = 0, " }{XPPEDIT 18 0 "u[x];" "6#&%\"uG6#%\"xG" }{TEXT -1 11 "(M, y) = 0." }} {PARA 0 "" 0 "" {TEXT -1 168 "This says there is no passage across the left, or right, boundary of the rectangle, but the top and bottom of \+ the distribution are held as g(x) and f(x), respectively. " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 111 "spacecurve(\{[x,0,f(x)],[x,1,f(x)] ,[1,x,0],[x,1,0]\},x=0..1,color=BLACK,\n axes=NORMAL,orientation=[ -25.,50]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 31 "We perform the sam e five steps." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 27 "Step 1: Separate variables." }}{PARA 0 "" 0 "" {TEXT -1 100 " The partial differential equation leads to X '' Y + X Y'' = 0, \+ with X '(0) Y(y) = 0 = X '(M) Y(y)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{PARA 0 "" 0 "" {TEXT -1 44 "Step 2: Identify the differential equati ons." }}{PARA 0 "" 0 "" {TEXT -1 39 " The differential equations are X '' + " }{XPPEDIT 18 0 "lambda^2;" "6#*$%'lambdaG\"\"#" }{TEXT -1 40 " X = 0, X '(0) = 0 = X '(M), and Y '' = " }{XPPEDIT 18 0 "lambda^2;" " 6#*$%'lambdaG\"\"#" }{TEXT -1 3 " Y." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 30 "Step 3: Solve these equations." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 7 " if " } {XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT -1 38 " = 0, then X(x) = \+ 1 and Y(y) = 1 or y." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 6 " if " }{XPPEDIT 18 0 "lambda <> 0;" "6#0%'lambdaG\"\"! " }{TEXT -1 8 ", then " }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" } {TEXT -1 5 " = n " }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 18 "/M , \+ X(x) = cos(n" }{XPPEDIT 18 0 "pi;" "6#%#piG" }{TEXT -1 23 "x/M), and Y (y) = sinh(n" }{XPPEDIT 18 0 "pi;" "6#%#piG" }{TEXT -1 15 "y/M) and si nh(n" }{XPPEDIT 18 0 "pi;" "6#%#piG" }{TEXT -1 9 "(L-y)/M)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 39 "Step 4: Constru ct the general solution." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 14 " u(x, y) =" }{XPPEDIT 18 0 "a[0]+(b[0]-a[0])*y /L;" "6#,&&%\"aG6#\"\"!\"\"\"*(,&&%\"bG6#F'F(&F%6#F'!\"\"F(%\"yGF(%\"L GF0F(" }{TEXT -1 2 " +" }{TEXT -1 1 " " }{XPPEDIT 18 0 "sum((a[n]*sinh (n*Pi*y/M)+b[n]*sinh(n*Pi*(L-y)/M))*cos(n*Pi*x/M),n);" "6#-%$sumG6$*&, &*&&%\"aG6#%\"nG\"\"\"-%%sinhG6#**F,F-%#PiGF-%\"yGF-%\"MG!\"\"F-F-*&&% \"bG6#F,F--F/6#**F,F-F2F-,&%\"LGF-F3F5F-F4F5F-F-F--%$cosG6#**F,F-F2F-% \"xGF-F4F5F-F," }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 53 "Step 5: Use the boundary conditions to de termine the " }{TEXT 258 1 "a" }{TEXT -1 8 " 's and " }{TEXT 259 1 "b " }{TEXT -1 4 " 's." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 24 "Details for this problem" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 155 "We are now ready to appl y this to a situation with and both f and g specified. We consider th e problem with f = g and as given below. We choose L = M = 1." }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 120 "u:=(x,y)->a[0]+(b[0]-a[0])*y +\n sum((a[n ]*sinh(n*Pi*y)+b[n]*sinh(n*Pi*(1-y)))*cos(n*Pi*x),\n n=1..i nfinity);" }}}{PARA 0 "" 0 "" {TEXT -1 64 "We wish to determine the co efficients. How will they be defined?" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "f(x)=u(x,0);\ng(x)=u(x,1);" }}}{PARA 0 "" 0 "" {TEXT -1 83 "This reminds us to use the Fourier coefficients. ( Don't forget to divide by sinh(n" }{XPPEDIT 18 0 "pi;" "6#%#piG" }{TEXT -1 4 "). ) " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "f:=x->piecewise(x<1/2,2* x,2*(1-x));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "plot(f(x),x= 0..1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 180 "b[0]:=int(f(x),x =0..1)/int(1^2,x=0..1);\na[0]:=b[0];\nfor n from 1 to 3 do\n b[n]:=i nt('f(x)'*cos(n*Pi*x),x=0..1)/\n int(cos(n*Pi*x)^2,x=0..1)/si nh(n*Pi);\n a[n]:=b[n]:\nod;" }}}{PARA 0 "" 0 "" {TEXT -1 35 "We inc orporate these values into u." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 123 "u:=(x,y)->a[0]+(b[0]-a[0])*y +sum((a[n]*sinh(n*Pi*y)+b[n]*sinh( n*Pi*(1-y)))*cos(n*Pi*x),\n n=1..3);\nn:='n':" }}} {PARA 0 "" 0 "" {TEXT -1 59 "As a safety check, we plot the ends again st the graph of f." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "plot([ u(x,0),'f(x)'],x=0..1);" }}}{PARA 0 "" 0 "" {TEXT -1 56 "We are now re ady to see a plot of u. What do you expect?" }}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 41 "plot3d(u(x,y),x=0..1,y=0..1,axes=NORMAL);" }}} {PARA 0 "" 0 "" {TEXT -1 46 "It is of value sometimes to see contour p lots." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(plots):" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "contourplot(u(x,y),x=0..1,y= 0..1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT -1 2 " " }} {PARA 0 "" 0 "" {TEXT 260 11 "Assignment:" }{TEXT -1 33 " Graph a solu tion for the problem" }}{PARA 0 "" 0 "" {TEXT -1 32 " \+ " }{XPPEDIT 18 0 "0 = diff(u,`$`(x,2))+diff(u,`$`(y,2)) ;" "6#/\"\"!,&-%%diffG6$%\"uG-%\"$G6$%\"xG\"\"#\"\"\"-F'6$F)-F+6$%\"yG F.F/" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 50 "B oundary conditions: u(x, 0) = sin(" }{XPPEDIT 18 0 "Pi; " "6#%#PiG" }{TEXT -1 18 " x), u(x, 1) = 0" }}{PARA 0 "" 0 "" {TEXT -1 64 " u(0, y) = sin (" }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 18 " y), u(1, y) = 0" }} }{MARK "1 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }