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Old   November 10, 2005, 20:15
Default Explicit/implicit dependence
Frank Wedburn
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Hello Folks, This little problem is beginning to bug me. If someone could shed some light on the subject I'd really appreciate it.

An isentropic fluid (barotropic) obeys Eulers eqns. Where does the following equation come from mathematically?

Del = del operator (nabla) rho = density P = pressure Int = integration

*** (Del P)/rho = Del (Int(dP/rho))

I've read that it's from Leibnitz's rule and that because rho is explicitly dependent upon rho (and implicitly upon x,y,z) the partial derivative of rho with respect to x is zero. If this is the case then surely the partial derivative of P, which can be written as explicitly dependent upon rho and implicitly with x,y,z, should also be zero?

If this is the case then why is it not removed from Eulers equations when dealing with Barotropic fluids?

I'm genuinely confused here and feel I'm missing something important. If someone could point me in the right direction it'd really help me with what I'm currently working on.

Cheers F
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Old   November 25, 2005, 23:15
Default Re: Explicit/implicit dependence
Ananda Himansu
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Please see my response to your previous post for an abstract answer. Sometimes, a concrete example is the best way to clear up fuzzy ideas. So, here goes. You will need pencil and paper for this.

Case (A): Let p = 2x, and r(p) = p^2, where ^ is the exponentiation operator. Then, dp/dx = 2, and dr/dx = 8x. Now define a function f(p) = Int {from 1 to p} [1/r(q)]dq, where q is the dummy variable of integration. In this case, by substituting for r, and performing the integration, you can directly obtain an explicit expression for f, i.e., f(p) = -1/p + 1. Thus, f(p(x)) = -1/(2x) + 1. Directly differentiating this last equation with respect to x, df/dx = 1/(2x^2). Now arrive at the same result by an alternative method, by applying the Leibniz Equation to the definition of f as an integral. In doing so, you will see how the term involving dr(q)/dx on the RHS of the Leibniz Eqn drops out. This case is the one that you are dealing with in the Euler equations for isentropic flow, where density r is written explicitly as a function of p, with no explicit dependence on x.

Case (B): Let p = 2x, and r(x) = 4x^2, so that as before, dp/dx = 2, and dr/dx = 8x. Now define a function g(p,x) = Int {from 1 to p} [1/r(x)]dq, where q is the dummy variable of integration. Note that this function g(p(x),x) is different from the previous f(p(x)). The latter is related to the "entropy", whereas g is a function of no significance, even though the integrals formally resemble each other. By direct integration, g(p,x) = (p-1)/(4x^2). Although the math is now a little more involved, by directly differentiating this last equation, dg/dx = 1/(2x^2) - (2x-1)/(2x^3), which of course differs from df/dx by an extra term. Arriving at this same result indirectly by applying Leibniz Eqn to the integral definition of g, you will see that in this case both terms on the RHS of the Leibniz Eqn are nonzero, and that the term involving the nonvanishing dr(x)/dx gives rise to the extra term in the final result as compared to df/dx. This case is of course NOT what you are dealing with in the isentropic Euler equations which lead to the potential flow equation. I presented it just to illustrate a case where r is defined explicitly in terms of x, and where the other term in the Leibniz Eqn is nonzero.

Case (C): If you wish to test your understanding of Leibniz Eqn even more precisely, you can define p = 2x, r = (p^2)/2 + 2x^2, and h(p,x) = Int {from 1 to p} [1/r(q,x)]dq. Then find dh/dx both by direct substitution/integration, and by Leibniz Eqn. Good luck!
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Old   November 25, 2005, 23:49
Default Re: Explicit/implicit dependence
Ananda Himansu
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Alas, in my effort to be precise, I lost sight of the forest for the trees. I meant to wind up with a summary directly relating to your question. So, here it is.

In Case(A) of my previous post, you wind up with d/dx[Int[1/r(q)]dq] = 1/r(p)dp/dx, which is what you wished to verify in your formulation. In Case (B) of my previous post, you wind up with d/dx[Int[1/r(x)]dq] = 1/r(x)dp/dx + Int[d/dx(1/r(x))]dq, which is NOT the case you are dealing with, but is a contrasting case to show how dr(x)/dx does not vanish when r is an explicit function of the spatial coordinates and dr/dx then makes the other term in the Leibniz Eqn nonzero. Note that the integrals in the two cases differ from each other, and will take on different values from each other for the same value of x.
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