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Convert differential equation to finite difference equation

  1. Sep 15, 2013 #1
    I have the differential equation

    [itex]\frac{dM}{dt}=4\pi \rho(r,t)r(t)^2\frac{dr}{dt}
    [/itex]

    which is the first term from

    [itex]M(t)=4\pi\int_0^{r(t)}C(r,t)r(t)^2dr[/itex]

    This describes the change in mass (M) of a sphere from a change in radius (r) given a density (rho) that depends on radius and time (t).

    My problem is somewhat simple. I tried to convert this equation into a finite difference formula as follows:

    [itex]M_1-M_0=4\pi \rho(r,t)r^2(r_1-r_0)[/itex]

    where subscript 1 indicates the value at a new timestep.

    I must be doing something wrong because the volume of a sphere requires a 1/3 to come from somewhere on the right hand side..
     
  2. jcsd
  3. Sep 15, 2013 #2

    arildno

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    Hmm..
    1. Haven't you forgotten the integral term in dM/dt that sums up the rate of mass change due to the local rate of change of the density function?
     
  4. Sep 15, 2013 #3
    Yes the full form is

    [itex]\frac{dM}{dt}=4\pi \rho(r,t)r(t)^2\frac{dr}{dt}+4\pi\int_0^{r(t)}\frac{\partial \rho}{\partial t}x^2 dx[/itex]

    but I don't see how the 1/3 for the first term can come from the additional term. My problem with the finite-difference conversion would remain if I assumed [itex]\rho[/itex] did not depend on time (thus removing the last term).
     
  5. Sep 15, 2013 #4

    arildno

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    As it should be.
    Let's review the case in which the density is constant.

    We then have:
    [tex]M(t+\bigtriangleup{t})=\frac{4\pi\rho}{3}r(t+\bigtriangleup{t})^{3}\approx\frac{4\pi\rho}{3}(r(t)+\frac{dr}{dt}\bigtriangleup{t})^{3}\approx\frac{4\pi\rho}{3}r(t)^{3}+{4\pi}r(t)^{2}\frac{dr}{dt}\bigtriangleup{t}=M(t)+{4\pi}r(t)^{2}\frac{dr}{dt}\bigtriangleup{t}[/tex]
    when we discard non-linear terms in the time interval.
     
    Last edited: Sep 15, 2013
  6. Sep 15, 2013 #5
    I still don't see where 1/3 and the r^3 are coming from. It looks like it comes out of an integration over time but the dr/dt in such an integral seems to make it go to r^4.
     
  7. Sep 17, 2013 #6
    [tex]\frac{dM}{dt}=\frac{4\pi}{3} \rho(r^3,t)\frac{dr^3}{dt}+\frac{4\pi}{3}\int_0^{r^3(t)}\frac{\partial \rho}{\partial t}dx^3 [/tex]
     
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