Clarifying ODE/PDE Integration and Conditions

[member 428835]

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)?

 

[HallsofIvy]

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.

 

[member 428835]

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