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Thread: Method of charasteristics

  1. #1
    Feb 2013

    Method of charasteristics

    How can I solve this problem?
    \frac{\mathrm{d}r}{0}=\frac{\mathrm{d}z}{C}=\frac{  \mathrm{d}u}{0}=\frac{\mathrm{d}v}{0}=\frac{d \Theta}{0}, where C is a constant.
    I gained it via the solution of a linear partial differential equation and \Theta=\Theta(r,z),\ u=u(r,z),\ v=v(r,z).
    I have a thought, could anybody check, whether it is good or not?
    From the first equality 0 \mathrm{d} z=C \mathrm{d} r \implies  r=k_1. Similarly, 0 \mathrm{d} z=C \mathrm{d} u \implies  u=k_2, etc. So the general solution is u=u(r),\ v=v(r),\ \Theta=\Theta(r).

    Thanks, Zoli
    Last edited by Zoli; Jan 22nd 2014 at 02:37 AM.
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