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{SECT 0 {EXCHG {PARA 256 "" 0 "" {TEXT 256 0 "" }{TEXT -1 54 "PHYS 312
:\nSOLUTIONS TO PROBLEMS\nSect 5.7\nProblem 5.7:1" }}{PARA 0 "" 0 "" 
{TEXT -1 1099 "\n\\n\{\\bf Problem 2:\}\\\\\n\{\\bf a:\}\\\\\nWe subst
itute\n\\[u(r,t)=t^\{b\}\\exp(-ar^2/t)\\]\ninto the two dimensional he
at equation in polar coordinates.\n\\[\\nabla^2u-\\frac\{\\pa u\}\{k\\
pa t\}=(\\frac\{1\}\{r\}\\frac\{\\pa \}\{\\pa r\}r\\frac\{\\pa\}\{\\pa
 r\}-\\frac\{\\pa\}\{k\\pa t\})\nt^b\\exp(\\frac\{-ar^2\}\{t\})\\]\\[=
\\frac\{1\}\{r\}\\frac\{\\pa \}\{\\pa r\}(-2ar^2t^\{b-1\}\\exp(-ar^2/t
))-(\\frac\{ar^2t^\{b-2\}\}\{k\}\n+\\frac\{bt^\{b-1\}\}\{k\})\\exp(-ar
^2/t)\\]\n\\[=(-4at^\{b-1\}+4a^2r^2t^\{b-2\}-\\frac\{ar^2t^\{b-2\}\}\{
k\}\n-\\frac\{bt^\{b-1\}\}\{k\})\\exp(-ar^2/t)\\]\nFor this expression
 to be zero we must have\n\\[a=\\frac\{1\}\{4k\},\\;\\;b=-1\\]\n\{\\bf
 b:\}\\\\\nWe again substitute\n\\[u(r,t)=t^\{b\}\\exp(-ar^2/t)\\]\nno
w into the three dimensional heat equation in polar coordinates.\n\\[
\\nabla^2u-\\frac\{\\pa u\}\{k\\pa t\}=(\\frac\{1\}\{r^2\}\\frac\{\\pa
 \}\{\\pa r\}r^2\\frac\{\\pa\}\{\\pa r\}-\\frac\{\\pa\}\{k\\pa t\})\nt
^b\\exp(\\frac\{-ar^2\}\{t\})\\]\\[=\\frac\{1\}\{r^2\}\\frac\{\\pa \}
\{\\pa r\}(-2ar^3t^\{b-1\}\\exp(-ar^2/t))-(\\frac\{ar^2t^\{b-2\}\}\{k
\}\n+\\frac\{bt^\{b-1\}\}\{k\})\\exp(-ar^2/t)\\]\n\\[=(-6at^\{b-1\}+4a
^2r^2t^\{b-2\}-\\frac\{ar^2t^\{b-2\}\}\{k\}\n-\\frac\{bt^\{b-1\}\}\{k
\})\\exp(-ar^2/t)\\]\nFor this expression to be zero we must have\n\\[
a=\\frac\{1\}\{4k\},\\;\\;b=-\\frac\{3\}\{2\}\\]\n\n" }{TEXT 257 13 "P
roblem 5.7:2" }}{PARA 0 "" 0 "" {TEXT -1 1367 "\\n\{\\bf Problem 3\}\\
\\\n\{\\bf a:\}\\\\\n\\[\\nabla^2u(r,t)-\\frac\{1\}\{c^2\}\\frac\{\\pa
^2u(r,t)\}\{\\pa t^2\}\n=\\frac\{1\}\{r^2\}\\frac\{\\pa\}\{\\pa r\}r^2
\\frac\{\\pa u\}\{\\pa r\}-\n\\frac\{1\}\{c^2\}\\frac\{\\pa^2u(r,t)\}
\{\\pa t^2\}=0\\]\n\\[u=\\rho(r)\\tau(t)\\]\n\\[\\frac\{1\}\{\\rho r^2
\}\\frac\{d\}\{dr\}(r^2\\frac\{d\\rho\}\{dr\})=\\frac\{1\}\{c^2\\tau\}
\\frac\{d^2\\tau\}\{dt^2\}=-\\lambda^2\\]\n\\[\\tau=A_\\lambda\\sin(\\
lambda ct)+B_\\lambda\\cos(\\lambda ct)\\]\nFrom \\htmladdnormallink\{
lecture 27\}\{../p312l23/index.html\}  we know that the  solution to t
he radial equation\nwhich remains finite as $r\\rightarrow 0$ are prop
ortional to\n\\[\\frac\{\\sin(\\lambda r)\}\{r\}\\]\nFor\n\\[u(R,t)=0
\\]\nwe must have \n\\[\\lambda=\\frac\{n\\pi\}\{R\}\\]\nwhere $n$ is \+
an integer. So we have\n\\[u(r,t)=\\sum_\{n=1\}^\\infty\\frac\{\\sin\\
frac\{n\\pi r\}\{R\}\}\{r\}(A_n\\cos\\frac\{n\\pi ct\}\{R\}+B_n\\sin\\
frac\{n\\pi ct\}\{R\})\\]\n\{\\bf b:\}\\\\\nSubstituting \n\\[u(r,t)=
\\frac\{1\}\{r\}(\\phi(r-ct)+\\psi(r+ct))\\]\ninto $\\nabla^2u$\n\\[\\
frac\{1\}\{r^2\}\\frac\{\\pa\}\{\\pa r\}r^2\\frac\{\\pa u\}\{\\pa r\}=
\\frac\{1\}\{r^2\}\\frac\{d\}\{dr\}r^2(-\\frac\{\\phi+\\psi\}\{r^2\}+
\\frac\{\\phi^\\prime+\\psi^\\prime\}\{r\}\n)=\\frac\{1\}\{r\}(\\phi^
\{\\prime\\prime\}-\n\\psi^\{\\prime\\prime\})\\]\nwhile \n\\[\\frac\{
\\pa^2 u\}\{c^2\\pa t^2\}=\\frac\{1\}\{r\}(\\phi^\{\\prime\\prime\}-\n
\\psi^\{\\prime\\prime\})\\]\nSince the two expressions agree the equa
tion is satisfied.\\\\\n\{\\bf c:\}\\\\\nWe must have\n\\[\\phi(ct)=-
\\psi(-ct)\\]\nso the solution must be on the form\n\\[u(r,t)=\\frac\{
1\}\{r\}(\\psi(ct+r)-\\psi(ct-r))\\]\n\\vspace\{0.5cm\}\n\n" }}}}
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