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"" -1 362 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 363 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 364 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 365 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 366 "" 1 12 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }{PSTYLE "Normal " -1 0 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Heading 1" -1 3 1 {CSTYLE "" -1 -1 "Times" 1 18 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 8 4 1 0 1 0 2 2 0 1 }{PSTYLE "Title" 0 18 1 {CSTYLE "" -1 -1 "" 1 18 0 0 0 0 0 1 1 0 0 0 0 0 0 0 }3 0 0 -1 12 12 0 0 0 0 0 0 19 0 }{PSTYLE "Author" 0 19 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 0 0 -1 8 8 0 0 0 0 0 0 -1 0 }{PSTYLE "Normal" -1 256 1 {CSTYLE "" -1 -1 "Times" 1 18 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }3 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Nor mal" -1 257 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }3 1 0 0 0 0 1 0 1 0 2 2 0 1 }} {SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT -1 40 "Partial Differential Equa tions PowerTool" }}{PARA 19 "" 0 "" {TEXT -1 16 "by Dr. Jim Herod" }}} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 259 57 "Section \+ 4.5: Structure for Solutions of the Heat Equation" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 3 "" 0 "" {TEXT 346 30 "Maple Packages f or Section 4.5" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart:" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 55 "We review structure for s olutions of the heat equation:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 44 "The Partial Differential Equation: \+ " }{XPPEDIT 18 0 "diff(u,t);" "6#-%%diffG6$%\"uG%\"tG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "diff(u,`$`(x,2));" "6#-%%diffG6$%\"uG-%\"$G6$%\" xG\"\"#" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 41 "The Boundary Conditions: u(t, 0) = " } {XPPEDIT 18 0 "T[0];" "6#&%\"TG6#\"\"!" }{TEXT -1 11 " and u(t, " } {TEXT 358 1 "a" }{TEXT -1 4 ") = " }{XPPEDIT 18 0 "T[1];" "6#&%\"TG6# \"\"\"" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 24 "The Initial Conditions: " }{TEXT 260 1 "u" }{TEXT -1 4 "(0, " }{TEXT 261 1 "x" }{TEXT -1 4 ") = " }{TEXT 262 1 "f" } {TEXT -1 1 "(" }{TEXT 263 1 "x" }{TEXT -1 9 "), where " }{TEXT 264 1 " f" }{TEXT -1 1 "(" }{TEXT 265 1 "x" }{TEXT -1 23 ") is continuous on [ 0, " }{TEXT 266 1 "a" }{TEXT -1 6 "] and " }{TEXT 267 1 "f" }{TEXT -1 6 "(0) = " }{XPPEDIT 18 0 "T[0];" "6#&%\"TG6#\"\"!" }{TEXT -1 3 " , " }{TEXT 268 1 "f" }{TEXT -1 1 "(" }{TEXT 269 1 "a" }{TEXT -1 4 ") = " } {XPPEDIT 18 0 "T[1];" "6#&%\"TG6#\"\"\"" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 256 23 "Existence of soluti ons:" }{TEXT -1 77 " There is one and only one solution for the above problem and it exists for " }{TEXT 270 1 "t" }{TEXT -1 8 " in [0, " } {XPPEDIT 18 0 "infinity;" "6#%)infinityG" }{TEXT -1 11 " ) and for " } {TEXT 271 1 "x" }{TEXT -1 35 " in [0, a]. Moreover, the solution " } {TEXT 272 1 "u" }{TEXT -1 35 " is infinitely differentiable for " } {TEXT 273 1 "t" }{TEXT -1 8 " in [0, " }{XPPEDIT 18 0 "infinity;" "6#% )infinityG" }{TEXT -1 11 " ) and for " }{TEXT 274 1 "x" }{TEXT -1 8 " \+ in [0, " }{TEXT 359 1 "a" }{TEXT -1 2 "]." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 318 "Note that this implies that the s olutions have an intolerance for sharp corners. The initial conditions may have some points where the derivative fails to exist, but the sol utions become infinitely differentiable instantaneously. We illustrate with an initial value having a sharp corner and examine the solution \+ near " }{TEXT 275 1 "t" }{TEXT -1 69 " = 0. We expect to see that the \+ sharp corner is immediately smoothed." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart:" }}}{PARA 0 "" 0 "" {TEXT -1 59 "Here is the \+ definition of the function with a sharp corner." }}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 52 "f:=x->Heaviside(1/2-x)*x^2+Heaviside(x-1/2)*(x -1)^2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "plot(f(x),x=0..1) ;" }}}{PARA 0 "" 0 "" {TEXT -1 33 "We make the Fourier coefficients." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 98 "for n from 1 to 40 do\n \+ c[n]:=int(f(x)*sin(n*Pi*x),x=0..1)/int(sin(n*Pi*x)^2,x=0..1):\nod:\nn: ='n';" }{TEXT -1 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 64 "With the coeffic ients, we make a solution for the heat equation." }}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 57 "u:=(t,x)->sum(c[n]*exp(-n^2*Pi^2*t)*sin(n*Pi*x ),n=1..40);" }}}{PARA 0 "" 0 "" {TEXT -1 52 "Check to see that the gra ph is smooth for any t > 0." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "plot3d(u(t,x),x=0..1,t=0..1/10,axes=NORMAL,orientation=[-60,75]); " }}}{PARA 0 "" 0 "" {TEXT -1 66 "It is interesting to view that the s moothing happens immediately. " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "plot([f(x),u(1/10000,x)],x=0..1,color=[red,blue,green]);" }}} {PARA 0 "" 0 "" {TEXT -1 90 "This example was created to have another \+ property. The initial value is concave up at any " }{TEXT 347 1 "x" } {TEXT -1 47 " except 1/2. It should be that if you pick any " }{TEXT 348 1 "x" }{TEXT -1 68 " in the interval [0, 1] except 1/2, then there is a positive number " }{TEXT 357 1 "t" }{TEXT -1 11 " for which " } {TEXT 353 1 "u" }{TEXT -1 1 "(" }{TEXT 349 1 "t" }{TEXT -1 2 ", " } {TEXT 350 1 "x" }{TEXT -1 4 ") > " }{TEXT 351 1 "f" }{TEXT -1 1 "(" } {TEXT 352 1 "x" }{TEXT -1 62 "). This is true because the second deriv ative with respect to " }{TEXT 354 1 "x" }{TEXT -1 17 " is positive, s o " }{TEXT 355 1 "u" }{TEXT -1 28 " must be increasing at that " } {TEXT 356 1 "x" }{TEXT -1 1 "." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 257 36 "Asymptotic Properties for Solutions:" }{TEXT -1 4 " As " }{TEXT 276 1 "t" }{TEXT -1 12 " increases, " }{TEXT 277 1 "u" }{TEXT -1 1 "( " }{TEXT 278 1 "t" }{TEXT -1 2 ", " }{TEXT 279 1 "x" }{TEXT -1 48 ") a pproaches a solution which is independent of " }{TEXT 280 1 "t" } {TEXT -1 66 ". For the above equation, this time independent solution \+ would be " }}{PARA 0 "" 0 "" {TEXT -1 17 " ( " } {XPPEDIT 18 0 "T[1]-T[0];" "6#,&&%\"TG6#\"\"\"F'&F%6#\"\"!!\"\"" } {TEXT -1 5 " ) ( " }{TEXT 360 1 "x" }{TEXT -1 3 " - " }{TEXT 361 1 "a " }{TEXT -1 4 ") / " }{TEXT 362 1 "a" }{TEXT -1 3 " + " }{XPPEDIT 18 0 "T[1];" "6#&%\"TG6#\"\"\"" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 39 "That is, no matter what initial value, " }{TEXT 281 1 "f" }{TEXT -1 1 "(" }{TEXT 282 1 "x" }{TEXT -1 152 "), we begin with, all solutions for the above equation would have limit this steady state solution. We compute the solution for the abo ve equation with " }}{PARA 0 "" 0 "" {TEXT -1 23 " \+ " }{TEXT 283 1 "f" }{TEXT -1 1 "(" }{TEXT 284 1 "x" }{TEXT -1 12 ") = ( 1+sin(" }{XPPEDIT 18 0 "pi;" "6#%#piG" }{TEXT -1 1 " " }{TEXT 285 1 "x" }{TEXT -1 10 " / 2) ) ( " }{TEXT 286 1 "x" }{TEXT -1 6 " + 1 ) " }}{PARA 0 "" 0 "" {TEXT -1 10 "and with " }{XPPEDIT 18 0 "T[0];" "6#&%\"TG6#\"\"!" }{TEXT -1 6 " = 1, " }{XPPEDIT 18 0 "T[2];" "6#&%\"T G6#\"\"#" }{TEXT -1 53 " = 2. We compute the solution and check the r esults." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "f:=x->(sin(Pi*x)+1)*(x+1);\ng:=x->f(x)-x-1;" }}}{PARA 0 "" 0 "" {TEXT -1 33 "We make the Fourier coefficients." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 98 "for n from 1 to 20 do\n c[n]:=int(g(x)*sin(n*P i*x),x=0..1)/int(sin(n*Pi*x)^2,x=0..1):\nod:\nn:='n':" }}}{PARA 0 "" 0 "" {TEXT -1 21 "We make the solution." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 61 "u:=(t,x)->1+x+sum(c[n]*exp(-n^2*Pi^2*t)*sin(n*Pi*x),n =1..20);" }}}{PARA 0 "" 0 "" {TEXT -1 99 "We check that this really is a solution, checking that it satisfies the PDE and boundary condition ." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "diff(u(t,x),t)-diff(u(t ,x),x,x);\nu(t,0);\nu(t,1);" }}}{PARA 0 "" 0 "" {TEXT -1 59 "We check \+ to see that we are close to the initial condition." }}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 50 "plot([u(1/1000,x),f(x)],x=0..1,color=[red,bl ack]);" }}}{PARA 0 "" 0 "" {TEXT -1 30 "Finally, we draw the graph of \+ " }{TEXT 287 1 "f" }{TEXT -1 122 ". What do we expect? We expect that \+ the solution begins looking like the initial value and ends looking li ke the graph of " }{TEXT 288 1 "x" }{TEXT -1 5 " + 1." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "plot3d(u(t,x),x=0..1,t=0..1/2,axes=NORMAL ,orientation=[-125,65]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 " " }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 378 "The next subject is the maximal principl e. We ask where can the maximum value of solutions lie. One might try \+ the following experiment. Take an initial distribution of heat with tw o or more peaks and see if, as the heat diffuses, the heat from one pe ak might combine with the heat from another peak to produce a temperat ure greater than that for either of the two initial peaks." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 29 "Too bad. That wil l not work. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 258 40 "Maximal Principle for the Heat Equation:" }{TEXT -1 46 " The s olution for the above heat equation for " }{TEXT 289 1 "t" }{TEXT -1 15 " in [0, T] and " }{TEXT 290 1 "x" }{TEXT -1 138 " in [0, a] attain s it maximum and minimum on one of four sides of the boundary of the r egion in the plane described as all points given by" }}{PARA 0 "" 0 " " {TEXT -1 5 " " }{TEXT 291 1 "t" }{TEXT -1 9 " = 0 and " }{TEXT 292 1 "x" }{TEXT -1 8 " in [0, " }{TEXT 293 1 "a" }{TEXT -1 5 "], or" }}{PARA 0 "" 0 "" {TEXT -1 5 " " }{TEXT 294 1 "t" }{TEXT -1 15 " i n [0, T] and " }{TEXT 295 1 "x" }{TEXT -1 8 " = 0, or" }}{PARA 0 "" 0 "" {TEXT -1 5 " " }{TEXT 296 1 "t" }{TEXT -1 15 " in [0, T] and " }{TEXT 297 1 "x" }{TEXT -1 3 " = " }{TEXT 298 1 "a" }{TEXT -1 4 ", or " }}{PARA 0 "" 0 "" {TEXT -1 5 " " }{TEXT 299 1 "t" }{TEXT -1 9 " \+ = T and " }{TEXT 300 1 "x" }{TEXT -1 8 " in [0, " }{TEXT 301 1 "a" } {TEXT -1 2 "]." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 45 "That is, the temperature in the interval [0, " }{TEXT 302 1 "a" }{TEXT -1 131 "] can never exceed the larger of the maximum \+ of the initial temperature or the maximum temperature at the ends of t he interval [0, " }{TEXT 303 1 "a" }{TEXT -1 78 "] as time increases. \+ In a similar manner, the temperature in the interval [0, " }{TEXT 304 1 "a" }{TEXT -1 120 "] is not smaller than the minimum of the initial \+ temperature or the minimum temperature at the ends of the interval [0, " }{TEXT 305 1 "a" }{TEXT -1 20 "] as time increases." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 33 "Here's an example to try out for " }{TEXT 306 1 "t" }{TEXT -1 29 " in the interval [0, 10] and " }{TEXT 307 1 "x" }{TEXT -1 21 " in the interval [0, " } {XPPEDIT 18 0 "pi;" "6#%#piG" }{TEXT -1 41 "/2]. We ask what is the ma ximum value of " }{TEXT 308 1 "u" }{TEXT -1 1 "(" }{TEXT 309 1 "t" } {TEXT -1 2 ", " }{TEXT 310 1 "x" }{TEXT -1 26 ") and where does it occ ur?" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "u:=(t,x)->exp(-t)*sin (x)+x+1;" }}}{PARA 0 "" 0 "" {TEXT -1 16 "Check that this " }{TEXT 311 1 "u" }{TEXT -1 59 " satisfies the heat equation with some boundar y conditions." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "diff(u(t,x) ,t)-diff(u(t,x),x,x);\nu(t,0);\nu(t,Pi/2);\nu(0,x);" }}}{PARA 0 "" 0 " " {TEXT -1 52 "Now, ask what is the maximum on the four boundaries." } }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 132 "maximize(u(0,x),x=0..evalf (Pi/2));\nmaximize(u(t,0),t=0..10);\nmaximize(u(t,Pi/2),t=0..10);\nmax imize(u(t,x),t=0..10,x=0..evalf(Pi/2));" }}}{PARA 0 "" 0 "" {TEXT -1 44 "See if it is clear where the maximum occurs." }}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 67 "plot3d(u(t,x),x=0..Pi/2,t=0..10,axes=NORMAL,or ientation=[-125,60]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "ev alf(u(0,Pi/2));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 194 "A conseq uence of the Maximum Principle for the Heat Equation is that order is \+ preserved. Consider this: Suppose that u satisfies the differential eq uation above and that v satisfies the equation" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 44 "The Partial Differential Equation: " }{XPPEDIT 18 0 "d iff(v,t);" "6#-%%diffG6$%\"vG%\"tG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 " diff(v,`$`(x,2));" "6#-%%diffG6$%\"vG-%\"$G6$%\"xG\"\"#" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 31 "The Boundary Conditions: " }{TEXT 312 1 "v" }{TEXT -1 1 "(" }{TEXT 313 1 "t" }{TEXT -1 7 ", 0) = " }{XPPEDIT 18 0 "T[0];" "6#&%\"TG6#\"\" !" }{TEXT -1 6 " and " }{TEXT 314 1 "v" }{TEXT -1 1 "(" }{TEXT 315 1 "t" }{TEXT -1 2 ", " }{TEXT 363 1 "a" }{TEXT -1 4 ") = " }{XPPEDIT 18 0 "T[1];" "6#&%\"TG6#\"\"\"" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 24 "The Initial Conditions: " } {TEXT 316 1 "v" }{TEXT -1 4 "(0, " }{TEXT 317 1 "x" }{TEXT -1 4 ") = \+ " }{TEXT 318 1 "g" }{TEXT -1 1 "(" }{TEXT 319 1 "x" }{TEXT -1 11 "), w here g(" }{TEXT 320 1 "x" }{TEXT -1 23 ") is continuous on [0, " } {TEXT 321 1 "a" }{TEXT -1 6 "] and " }{TEXT 322 1 "g" }{TEXT -1 6 "(0) = " }{XPPEDIT 18 0 "T[0];" "6#&%\"TG6#\"\"!" }{TEXT -1 10 " , g(a) = \+ " }{XPPEDIT 18 0 "T[1];" "6#&%\"TG6#\"\"\"" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 18 "Suppose also that " }{XPPEDIT 18 0 "f <= g;" "6#1%\"fG%\"gG" }{TEXT -1 24 ". We argue that for all \+ " }{TEXT 324 1 "t" }{TEXT -1 5 " and " }{TEXT 323 1 "x" }{TEXT -1 31 " in the appropriate intervals, " }{XPPEDIT 18 0 "u(t,x) <= v(t,x);" "6 #1-%\"uG6$%\"tG%\"xG-%\"vG6$F'F(" }{TEXT -1 69 " . The suggestion for \+ doing this problem is to consider the function " }{TEXT 325 1 "w" } {TEXT -1 1 "(" }{TEXT 326 1 "t" }{TEXT -1 2 ", " }{TEXT 327 1 "x" } {TEXT -1 4 ") = " }{TEXT 328 1 "u" }{TEXT -1 1 "(" }{TEXT 329 1 "t" } {TEXT -1 2 ", " }{TEXT 330 1 "x" }{TEXT -1 4 ") - " }{TEXT 331 1 "v" } {TEXT -1 1 "(" }{TEXT 332 1 "t" }{TEXT -1 2 ", " }{TEXT 333 1 "x" } {TEXT -1 24 ") and ask what equation " }{TEXT 334 1 "w" }{TEXT -1 30 " satisfies. You will see that " }{TEXT 335 1 "w" }{TEXT -1 24 " satisf ies the equation." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 44 "The Partial Differential \+ Equation: " }{XPPEDIT 18 0 "diff(w,t);" "6#-%%diffG6$%\"wG%\" tG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "diff(w,`$`(x,2));" "6#-%%diffG6$ %\"wG-%\"$G6$%\"xG\"\"#" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 31 "The Boundary Conditions: " } {TEXT 336 1 "w" }{TEXT -1 1 "(" }{TEXT 337 1 "t" }{TEXT -1 7 ", 0) = \+ " }{XPPEDIT 18 0 "0;" "6#\"\"!" }{TEXT -1 6 " and " }{TEXT 338 1 "w" }{TEXT -1 1 "(" }{TEXT 339 1 "t" }{TEXT -1 7 ", 1) = " }{XPPEDIT 18 0 "0;" "6#\"\"!" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 24 "The Initial Conditions: " }{TEXT 345 1 "w " }{TEXT -1 4 "(0, " }{TEXT 344 1 "x" }{TEXT -1 4 ") = " }{TEXT 343 1 "f" }{TEXT -1 1 "(" }{TEXT 342 1 "x" }{TEXT -1 4 ") - " }{TEXT 341 1 " g" }{TEXT -1 1 "(" }{TEXT 340 1 "x" }{TEXT -1 2 ")." }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 21 "We can conclude that " }{XPPEDIT 18 0 "w(t,x) <= 0;" "6#1-%\"wG6$%\"tG%\"xG\"\"!" }{TEXT -1 65 " because the maximum must occur on the boundary and that is zero. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 339 "In this section, we have discussed the t heoretical structure for solutions of the heat equation. We will see t hat not everything that happens here is true for the wave equation. Th ere, sharp edges can propagate in time. It will not longer be true tha t solutions must be differentiable. This is an important change which \+ we will illustrate." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{SECT 0 {PARA 3 "" 0 "" {TEXT 366 16 "Unassisted Map le" }}{PARA 0 "" 0 "" {TEXT -1 277 "In the example to illustrate that \+ the maximum (and the minimum) value of a function satisfying the diffu sion equation had to occur on the boundary, each boundary was examined separately and then the graph was drawn to conclude that the maximum \+ occurred at the boundary point [ " }{TEXT 364 1 "t" }{TEXT -1 2 ", " } {TEXT 365 1 "x" }{TEXT -1 9 " ] = [0, " }{XPPEDIT 18 0 "Pi;" "6#%#PiG " }{TEXT -1 76 "/2 ]. Never mind. Maple could have found the maximum a nd minimum unassisted." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "ma ximize(exp(-t)*sin(x)+x+1, x=0..Pi/2, t=0..10, location);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "minimize(exp(-t)*sin(x)+x+1, x=0..P i/2, t=0..10, location);" }}}{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 50 "EMAIL: herod@math.gatech. edu or jherod@tds.net" }}{PARA 0 "" 0 "" {TEXT -1 38 "URL: http:// www.math.gatech.edu/~herod" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 257 "" 0 "" {TEXT -1 36 "Copyright \251 2003 by James V. Herod" }} {PARA 257 "" 0 "" {TEXT -1 19 "All rights reserved" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{MARK "0 0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 } {PAGENUMBERS 1 1 2 33 1 1 }