{VERSION 4 0 "IBM INTEL NT" "4.0" } {USTYLETAB {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 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Normal" -1 256 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }} {SECT 0 {EXCHG {PARA 256 "" 0 "" {TEXT -1 54 "PHYS 312:\nSOLUTIONS TO \+ PROBLEMS\nSect 4.6\nProblem 4.6:1" }}{PARA 0 "" 0 "" {TEXT -1 1548 "\n \\n \{\\bf Problem 2:\}\\\\\n\{\\bf a:\} Substituting\n\\[T(x,t)=const . e^\{-px\}e^\{i(\\omega t-px)\}\\]\ninto the one-dimensional heat equ ation.\n\\[\\frac\{\\pa^2 T\}\{\\pa x^2\}=\\frac\{1\}\{k\}\\frac\{\\pa T\}\{\\pa t\}\\]\ngives\n\\[(-p-ip)^2=i\\omega\\]\nor\n\\[p=\\pm\\sqr t\{\\frac\{\\omega\}\{2k\}\}\\]\n\n\\n\{\\bf b:\}\n\\[T(x,t)=e^\{i\\om ega t\}F(x)\\]\n\\[F(x)=(B\\exp(-x(1+i)\\sqrt\{\\frac\{\\omega\}\{2k\} \})+C\\exp(x(1+i)\\sqrt\{\\frac\{\\omega\}\{2k\}\}))\\]\nDefine\n\\[Z= \\exp(a(1+i)\\sqrt\{\\frac\{\\omega\}\{2k\}\})\\]\nWe have\n\\[B+C=A\\ ]\n\\[\\frac\{B\}\{Z\}+ZC=A\\]\nSo that \n\\[B=\\frac\{AZ\}\{1+Z\},\\; C=\\frac\{A\}\{1+Z\}\\]\nWe thus have\n\\[T(x,t)=Ae^\{i\\omega t\}(\\f rac\{Z\}\{1+Z\}\\exp(-x(1+i)\\sqrt\{\\frac\{\\omega\}\{2k\}\})+\\frac \{1\}\{1+Z\}\\exp(x(1+i)\\sqrt\{\\frac\{\\omega\}\{2k\}\}))\\]\n\{\\bf c:\} Since the heat equation is linear, the real and imaginary part o f the solution\nsolve the equation separately. The particular solution which satisfies the boundary condition\n\\[T(0,t)=T(a,t)=A\\cos(\\ome ga t)\\]\nwill therefore be\n\\[T(x,t)=\\cos(\\omega t)\{\\cal R\}(F(x ))-\\sin(\\omega t)\{\\cal I\}(F(x))\\]\nwhere $\\cal R$ and $\\cal I$ stand for real and imaginary part, respectively.\\\\\n\{\\bf d:\} We \+ can add to the solution found above any solution to \n\\[\\frac\{\\pa^ 2 u(x,t)\}\{\\pa x^2\}=\\frac\{1\}\{k\}\\frac\{\\pa u(x,t)\}\{\\pa t\} \\]\nwith boundary conditions\n\\[u(0,t)=u(a,t)=0\\]\n\\[u(x,0)=f(x)\\ ]\nThis solution that can be written\n\\[u(x,t)=\\sum_\{n=1\}^\\infty \\alpha_n\\sin\\frac\{n\\pi x\}\{a\}\\exp(-\\frac\{tkn^2\\pi^2\}\{a^2 \})\\]\nwhere the coefficients $\\alpha_n$ can be obtained by making a Fourier sine expansion of $f(x)$.\nSuch a solution will always be \{ \\bf damped\} and approach zero for large times.\n\\vspace\{0.5cm\}\n \n" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}}{MARK "0 1 0" 1548 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }