Convert differential equation to finite difference equation

  • Thread starter Hypatio
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  • #1
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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..
 

Answers and Replies

  • #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?
 
  • #3
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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?
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).
 
  • #4
arildno
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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).
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:
  • #5
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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.
 
  • #6
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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).
[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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