User:EGM6321.F10.TEAM1.WILKS/Mtg29

=EGM6321 - Principles of Engineering Analysis 1, Fall 2009= Mtg 29: Thurs, 29Oct09

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Eq.(3) P.28-3 : L2_ODE_VC $$ g_i ( \xi_i) \Rightarrow \ $$ linear $$ \phi_{ij}( \xi_i) \Rightarrow \ $$ linear Simplify: $$ \xi_i \Rightarrow x \ $$ $$ x_i ( \xi_i) \Rightarrow y(x) \ $$ $$ g_i ( \xi_i) \Rightarrow g(x) \ $$ $$ \left [ \sum \phi_{ij}(k_j)^2 \right ]  \Rightarrow a_0(x) \ $$ Eq.(3) P.28-3 : $$ \frac{1}{g(x)} \frac{d}{dx} \left [ g(x) \frac{dy}{dx} \right ] +a_0(x)y=0  \Rightarrow y''+ \frac{g'}{g}y' + a_0y=0 \ $$ Where $$ \frac{g'}{g}y' = a_1 \ $$ L2_ODE_VC, cf. [[media: Egm6321.f09.mtg3.djvu | Eq.(1) P.3-1 ]] Ellipsoidal coordinates: $$ a, b, c=0 \ $$

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$$ g_i ( \xi_i)= \left [ \frac{ \xi_i^2-a^2}{ \xi_i^2-b^2} \right ]^{\frac{1}{2}} \ $$ $$ \phi_{i1}( \xi_i) = 1  \ $$ $$ \phi_{i2}( \xi_i) = \frac{1}{ \xi_i^2-a^2 } \ $$ Heat condition on a sphere Steady State Heat Eq(1): $$ ( \underline{k} \ $$ gradient $$  \psi\ )+f=0 \ $$ Where $$ \underline{k} = \ $$ conductivity and $$ f= \ $$ tensor matrix Homogeneous isotropic material where $$ k= \ $$ condition and $$  \underline{I} = \ $$ Identity matrix If no heat source, $$ f=0 \ $$ Laplace equation Laplace differential operation on orthogonal curviture coordinates (e.g. spherical coordinates, ellipsoidal, etc.)

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$$ \Delta\ \psi\ = \frac{1}{h_1h_2h_3} \sum_{i=1}^3 \frac{\partial}{\partial \xi_i} \left [ \frac{h_1h_2h_3}{(h_i)^2} \frac{\partial \psi\ }{\partial \xi_i} \right ] \ $$ $$ \bar \theta\ \ $$ is " $$ \theta\ \ $$" in $$ K \ $$ Here $$ \theta\ \ $$ is defined such that equator is at $$ 0^o \ $$ latitude $$ x=r \cos \theta\ \cos \varphi \ $$ $$ y=r \cos \theta\ \sin \varphi \ $$ $$ x=r \sin \theta\ \ $$ Eq.(4) P.29-2 $$ ds^2=1dr^2+r^2 \cos ^2 \theta\ d \varphi ^2 + r^2d \theta\ ^2 \ $$ Where $$ 1= (h_1)^2 \ $$, $$ dr^2=(d \xi_1)^2 \ $$ , $$ r^2 \cos ^2 \theta\ = (h_2)^2 \ $$ , $$ d \varphi ^2 = (d \xi_2)^2 \ $$ , $$ r^2 = (h_3)^2 \ $$ and $$ d \theta\ ^2 = (d \xi_3)^2 \ $$ Then $$ h_1 ( \xi\ )=1 \ $$ $$ \xi\ = ( \xi_1, \xi_2, \xi_3) \ $$ $$ h_2 ( \xi\ )= r \cos \theta\ = \xi_1 \cos \xi_2 \ $$ $$ h_3 ( \xi\ )= r = \xi_1 \ $$

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HW END HW HW END HW For axisymmetric problem: no dependancy on $$ \varphi \ $$ cf. [[media: http://books.google.com/books?id=9Cg3HWCnCjAC&printsec=frontcover&dq=differential+equations+billingham&ei=pGR4SpPVLojSMpb07Qw#v=onepage&q=&f=false | Kp.45) ]]

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Where $$ \psi\ (1, \theta\ ) = f( \theta\ ) = T_0 \sin ^4 \bar \theta\ \  \ $$ Where $$ T_0= \ $$ constant and $$ \bar \theta\ = K \ $$ Then $$ \bar \theta\ :=\frac{ \pi\ }{2}- \theta\ \ $$