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Higher order terms in the equations

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Old   October 30, 2008, 06:05
Default Higher order terms in the equations
  #1
Argyle
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Hello,

Why do we ignore the second order terms when we derive the continuity, momentum etc. equations? More specifically, if we use Taylor series to derive the substantial derivative, then we omit the higher order terms. I'm sure I'm missing something but it seems odd to have the equations based on an assumption that first order accuracy is ok and then use second or higher order discretization schemes. Please enlighten me.
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Old   October 30, 2008, 07:36
Default Re: Higher order terms in the equations
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Paolo Lampitella
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Why should we use taylor series to obtain the substantial derivative? In my knowledge the substatial derivative is in closed form and simply comes out from the composite function derivation rule :

Df(x,t)/Dt = df/dt + df/dx*dx/dt

because f is a function of t and of x(x0,t) where x0 is the position that was occupied at the time t=t0 by the particle that is in x at time t.
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Old   October 30, 2008, 08:19
Default Re: Higher order terms in the equations
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Argyle
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Well there are different ways in obtaining the substantial derivative. One way is by simply solving the total derivative.

It is fully possible to use a Taylor series approach where, say rho is developed.

rho = rho(x,y,z,t)

then assume we have a small fluid element at two different times (t1 and t2). We will have two different densities, rho1 and rho2.

From a Taylor series expansion we could then obtain:

rho2 = rho1 + (d(rho)/dx)*(x2-x1) + (d(rho)/dy)*(y2-y1) + (d(rho)/dz)*(z2-z1) + (d(rho)/dt)*(t2-t1) + HOT

where HOT is higher order terms.

Rearranging and dividing by (t2-t1) we will get

(rho2-rho1)/(t2-t1) = ......

Taking the limit as t2-->t1 we define this as the substantial derivative D(rho)/Dt.

So I'm confused about the neglecting of the HOT.
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Old   October 30, 2008, 12:03
Default Re: Higher order terms in the equations
  #4
Paolo Lampitella
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Ok, this is another way to obtain it but this is not neglecting, in the limit of t2-->t1 they all go exactly to zero because HOT means they're at least of order dt^2 (but in some cases the next term in time has a bigger order) and so, dividing by dt, they becomes of order dt and go exactly to zero for dt going to zero. It's a derivative.
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Old   October 30, 2008, 12:31
Default Re: Higher order terms in the equations
  #5
Argyle
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First some clarification (I was too sloppy to write it the first time, but I assume this is clear).

(rho2-rho1)/(t2-t1) =(d(rho)/dx)*(x2-x1)/(t2-t1) + (d(rho)/dy)*(y2-y1)/(t2-t1) + (d(rho)/dz)*(z2-z1)/(t2-t1) + (d(rho)/dt)*(t2-t1)/(t2-t1) + HOT/(t2-t1)

Or as t2-->t1

(x2-x1)/(t2-t1) = u (y2-y1)/(t2-t1) = v (z2-z1)/(t2-t1) = w

Since the distance traveled by the volume element approaches zero as t2-->t1

If we look at one of the expanded terms of the HOT (there are 10 extra terms):

d^2(rho)/d(xt)*(x2-x1)*(t2-t1)

I don't see how this vanishes as we divide by dt (t2-t1). Could you please tell me what I'm doing wrong.

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Old   October 30, 2008, 16:46
Default Re: Higher order terms in the equations
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Paolo Lampitella
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I think it vanishes because in the limit (t2-t1) --> 0:

(x2-x1) * (t2-t1) / (t2-t1) --> u * (t2-t1) --> 0

or, simply

x2 = x(t2) and x1 = x(t1)

so every terms in (x2-x1) vanishes for t2 --> t1
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Old   October 31, 2008, 04:47
Default Re: Higher order terms in the equations
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Argyle
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Thank you for a nice discussion.

It is more clear now.
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Old   October 31, 2008, 08:41
Default Re: Higher order terms in the equations
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Paolo Lampitella
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You're welcome. Thank you too.
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