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{SECT 0 {EXCHG {PARA 256 "" 0 "" {TEXT 256 0 "" }{TEXT -1 54 "PHYS 312
:\nSOLUTIONS TO PROBLEMS\nSect 5.1\nProblem 5.1:1" }}{PARA 0 "" 0 "" 
{TEXT -1 3877 "\\n \{\\bf Problem 1:\}\\\\\n\{\\bf a:\}\\\\\nIf $x=L\\
theta$ represents distances along the perimeter of the ring the\ndiffe
rential equation is\n\\[\\frac\{\\pa^2 T\}\{\\pa x^2\}=\\frac\{1\}\{k
\}\\frac\{\\pa T\}\{\\pa t\}\\]\nIn terms of the angle $\\theta$ the d
ifferential equation becomes\n\\[\\frac\{\\pa^2 T\}\{\\pa \\theta^2\}=
\\frac\{L^2\}\{k\}\\frac\{\\pa T\}\{\\pa t\}\\]\nSeparating the variab
les\n\\[T(\\theta,t)=\\phi(\\theta)\\tau(t)\\]\nwe find \n\\[\\tau^\\p
rime=-\\frac\{k^2\\lambda^2\}\{L^2\}\\tau\\]\n\\[\\phi^\{\\prime\\prim
e\}+\\lambda^2\\phi=0\\]\nThe general solution to the equation for $\\
phi$ is \n\\[\\phi=A\\cos(\\lambda\\theta)+B\\sin(\\lambda\\theta)\\]
\nIf the angle is incremented by $2\\pi$ we are back were we started,
\nhence $\\lambda=n=$ integer or zero. The solution for $\\tau$ is\n\\
[\\tau=const\\exp(-\\frac\{kn^2t\}\{L^2\})\\]\nThe solution to the pro
blem can thus be expressed on the\nFourier series form\n\\[T(\\theta,t
)=A_0+\\sum_\{n=0\}^\\infty\n\\exp(-\\frac\{kn^2t\}\{L^2\})(A_n\\cos(n
\\theta)+B_n\\sin(n\\theta))\\]\nwhere\n\\[A_0=\\frac\{1\}\{2\\pi\}\\i
nt_0^\{2\\pi\}d\\theta f(\\theta)\\]\n\\[A_n=\\frac\{1\}\{\\pi\}\\int_
0^\{2\\pi\}d\\theta f(\\theta)\\cos(n\\theta)\\]\n\\[B_n=\\frac\{1\}\{
\\pi\}\\int_0^\{2\\pi\}d\\theta f(\\theta)\\sin(n\\theta)\\]\n\{\\bf b
\}\\\\\nIn this case $\\cos\\theta$ is an eigenfunction with eigenvalu
e $\\lambda=n=1$\nHence we put $A_n=0$ for $n\\neq 1$ and $B_n=0$ for \+
all $n$\n\\[T(\\theta,t)=T_0\\exp(-\\frac\{kt\}\{L^2\})\\cos\\theta\\]
\nThe time it takes for the temperature difference between the hottest
 \nand coldest spot to halve is thus\n\\[t_\{1/2\}=\\frac\{L^2\}\{k\}
\\ln 2\\]\n\\n \{\\bf Problem 2:\}\\\\\nThe general solution to  the 2
-dimensional Laplace equation \n\\[\\nabla^2u(r,\\theta)=0\\]\nin pola
r coordinates is\n\\[u=a+b\\ln r +\\sum_\{n=1\}^\\infty(a_n\\cos(n\\th
eta)+b_n\\sin(n\\theta))\n(\\frac\{\\alpha_n\}\{r^n\}+\\beta r^n)\\]\n
Because of the \n boundary conditions \n\\[u(a,\\theta)=\\cos\\theta;
\\;u(2a,\\theta)=\\cos\\theta\\]\nonly the cosine term with $n=1$ will
 contribute to the solution so we can write\n\\[u=(\\frac\{\\alpha\}\{
r\}+\\beta r)\\cos\\theta\\]\nSubstituting for the conditions at $r=a$
 and $r=2a$ we \nfind\n\\[\\frac\{\\alpha\}\{a\}+\\beta a=1\\]\n\\[\\f
rac\{\\alpha\}\{2a\}+\\beta 2a=1\\]\nwe find \n\\[\\alpha=2a/3, \\beta
=1/(3a)\\]\n and\n\\[u(r)=\\frac\{2a\}\{3r\}+\\frac\{r\}\{3a\}\\]\n\\v
space\{0.5cm\}\n\n\\n \{\\bf Problem 3:\}\\\\\nThe differential equati
on\n\\[\\nabla^2u=1\\]\nbecomes in spherical coordinates\n\\[\\frac\{1
\}\{r^2\}\\frac\{\\pa\}\{\\pa r\}(r^2\\frac\{\\pa u\}\{\\pa r\})+\n\\f
rac\{1\}\{r^2\\sin\\theta\}\\frac\{\\pa\}\{\\pa\\theta\}(\\sin\\theta
\\frac\{\\pa u\}\{\\pa\\theta\})+\\frac\{1\}\{r^2\\sin^2\\theta\}\\fra
c\{\\pa^2 u\}\{\\pa\\phi^2\}=\n1\\]\nThe boundary conditions are such \+
that $u$ does not depend on either \n$\\phi$ nor $\\theta$. The differ
ential equation then simplifies to\n\\[\\frac\{1\}\{r^2\}\\frac\{d\}\{
d r\}(r^2\\frac\{d u\}\{d r\})=1\\]\nThe general solution to the homog
eneous equation\n\\[\\frac\{d\}\{d r\}(r^2\\frac\{d u\}\{d r\})=0\\]\n
is\n\\[u_h=\\frac\{c\}\{r\}+b\\]\nwhere $c$ and $b$ are constants. We \+
require that $u$ is bounded \nat $r=0$ hence $c=0$. We must add to the
 homogeneous solution a particular solution to the inhomogeneous \nequ
ation. It was suggested to try a solution on the form\n$u=ar^2$. Subst
ituting into the differential equation\ngives\n\\[\\frac\{1\}\{r^2\}\\
frac\{d\}\{\\d r\}(r^22br)=6a=1\\]\nGiving $a=1/6)$\nThe boundary cond
ition \n\\[u(1,\\theta,\\phi)=0\\]\ngives $b=1/6$ Hence\n\\[u(r)=(r^2-
1)/6\\]\n\n\n\\n \{\\bf Problem 4:\}\\\\\n\{\\bf a:\} \n\\[\\frac\{1\}
\{r\}\\frac\{\\pa\}\{\\pa r\}(r\\frac\{\\pa u\}\{\\pa r\})=\\frac\{r\}
\{r\}\\frac\{\\pa^2u\}\{\\pa r^2\}+\\frac\{1\}\{r\}\\frac\{\\pa r\}\{
\\pa r\}\\frac\{\\pa u\}\{\\pa r\}=\\frac\{\\pa^2u\}\{\\pa r^2\}+\\fra
c\{1\}\{r\}\\frac\{\\pa u\}\{\\pa r\}\\]\nq.e.d.\\\\\n\{\\bf b:\} If\n
\\[\\nabla^2u=\\frac\{1\}\{r\}\\frac\{d\}\{d r\}(r\\frac\{du\}\{d r\})
= f(r)\\]\nwe have a particular solution\n\\[r\\frac\{du\}\{dr\}=\\int
_b^\{r\}r_2dr_2f(r_2)\\]\n\\[u(r)=\\int_a^r\\frac\{dr_1\}\{r_1\}\\int_
b^\{r_1\}dr_2\\;r_2f(r_2)\\]\nfor any value of the lower limits of int
egration $a$ and $b$.\\\\\n\{\\bf c:\} Since the boundary condition is
 given for $r=0$ we choose $a=b=0$\n\\[u(r)=\\int_0^r\\frac\{dr_1\}\{r
_1\}\\int_0^\{r_1\}dr_2\\;r_2f(r_2)+u(0)\\]\nWith\n$f(r)=1;\\;u(0)=0$ \+
we find\n\\[u(r)=\\frac\{r^2\}\{4\}\\]\n\n\\vspace\{0.5cm\}\n\n\\vspac
e\{0.5cm\}\n\n" }}}}{MARK "0 1 0" 3160 }{VIEWOPTS 1 1 0 1 1 1803 1 1 
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