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Simple ODE/PDE clarification

  1. Nov 17, 2013 #1


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    hey pf!

    here's the question: $$ u \frac{ \partial u}{ \partial x} = \rho \frac{ d P}{ d x}$$ may i generally state $$ \rho P+1/2 u^2 = const. $$

    the book does, and it seems the [itex]dx[/itex] cancels the [itex]\partial x[/itex] on both sides and we simply integrate through. this seems to be mathematically untrue. can someone confirm/reject this? also, what conditions would be necessary to have the above true (if it is indeed untrue generally)?
  2. jcsd
  3. Nov 17, 2013 #2


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    I doubt very much that you book says exactly that! I suspect it says instead that
    [itex]\rho P- (1/2)u^2= const[/itex]. (Notice the negative.)

    The partial derivative is simply the ordinary derivative while treating other variables as if they were constants. What ever the other variable(s) in u might be, since they do not appear in the equation, this would be solved exactly as if it were
    [tex]u\dfrac{du}{dx}= \rho\dfrac{dP}{dx}[/tex].

    Now, you can treat [itex]du/dx[/itex] and [itex]dP/dx[/itex] as if they were ratios of differentials as we do in Ordinary Caculus.
  4. Nov 18, 2013 #3


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    ahh yes, my mistake. the negative is definitely there. sorry. but thanks for answering the crux of the question
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