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A solution of Euler's equation

  1. Jul 6, 2010 #1
    1. The problem statement, all variables and given/known data
    The velocity vector for a flow is u = (xt, yt, -2zt). Given that the density is constant and that the body force is F = (0,0,-g) find the pressure, P(x,t) in the fluid which satisfies [tex]P = P_0(t) [/tex] at x = 0

    2. Relevant equations
    Euler's equation: [tex] \frac{Du}{Dt}=-\frac{1}{\rho}\nabla P + F [/tex]

    3. The attempt at a solution

    [tex] \nabla P = \rho(-x-xt^2, -y - yt^2, 2z - 4zt^2 - g) [/tex]
    How do you get P from this. Back of the book gives
    [tex] P = - \frac {1}{2} \rho (x^2+y^2)(1+t^2) + \rho z^2(1-2t^2)-\rho gz + P_0(t) [/tex]
    How did they get that?
  2. jcsd
  3. Jul 6, 2010 #2
    What do you understand by the del operator? You simply have to solve a set of three partial differential equations to reconstruct P.
  4. Jul 6, 2010 #3


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    Science Advisor

    More specifically, you have
    [tex]\frac{\partial P}{\partial x}= \rho(x- xt^2)= \rho(1- t^2)x[/tex]
    [tex]\frac{\partial P}{\partial y}= \rho(y- yt^2)= \rho(1- t^2)y[/tex]
    [tex]\frac{\partial P}{\partial z}= \rho(z- zt^2)= \rho(1- t^2)z[/tex]

    Which can be treated as three ordinary differential equations with 't' as a fixed parameter.
  5. Jul 6, 2010 #4


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    Homework Helper

    The arbitrary function of t comes into play as you're doing a partial integration, so when you take the gradient again, all functions of t get knocked out as you're doing partial differentiation.
  6. Jul 6, 2010 #5
    Thanks I was getting delta mixed up with nabla!
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