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First order partial diff eqn

  1. Apr 17, 2010 #1
    Hi there, how can i solve this first order partial differential equation:

    grad p= (0,0,-ρg)

    where p=p(x,y,z,t) is pressure
    where ρ=ρ(x,y,z,t) is density
    where grad p is in three dimensional Cartesian coordinates

    can i just separately solve the three differential equations?
    ie dpx/dx = 0
    dpy/dy = 0
    dpz/dz=-ρg

    But im confused as p is not a vector function itself. so the final answer cannot be in the form p=(px,py,pz)
     
  2. jcsd
  3. Apr 17, 2010 #2

    Redbelly98

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    Since p is a scalar, instead you would have
    ∂p/∂x = 0
    ∂p/∂y = 0
    ∂p/∂z=-ρg​
     
  4. Apr 17, 2010 #3
    I see, so how do i combine these solutions for p?
    i get three solutions p=A (constant), p=B (constant) and p=-ρgz + C.
    so what does my solution look like for p?
     
  5. Apr 17, 2010 #4

    Redbelly98

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  6. Apr 17, 2010 #5
  7. Apr 18, 2010 #6

    HallsofIvy

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    [tex]\frac{\partial p}{\partial x}= 0[/tex]
    and
    [tex]\frac{\partial p}{\partial y}= 0[/tex]
    say that p is not a function of x or y but a function of z only.

    To solve
    [tex]\frac{dp}{dz}= -\rho g[/tex]
    just intgrate to get [itex]p(x, y, z)= -\rho gz+ C[/itex].

    Partial derivatives do NOT give three different solutions. If you had, for example,
    [tex]\frac{\partial f}{\partial x}= x+ y[/tex]
    and
    [tex]\frac{\partial f}{\partial y}= e^y+ x[/tex]
    From the first equation, you would get [itex]f(x,y)= (1/2)x^2+ xy+ C(y)[/tex]. Because, in taking the partial derivative with respect to x, we treat y as a constant, the "constant of integration" may be a function of y.

    Now differentiate that f with respect to y:
    [tex]\frac{\partial f}{\partial y}= x+ C'(y)= e^y+ x[/tex]
    The "x" terms cancel (that had to happen for this system to have a solution) giving just [itex]C'(y)= e^y[/itex] so [itex]C(y)= e^y+ D[/itex] where D now really is a constant.

    That gives [itex]f(x,y)= (1/2)x^2+ xy+ e^y+ D[/itex].

    Your first two equations,
    [tex]\frac{\partial p}{\partial x}= 0[/tex]
    [tex]\frac{\partial p}{\partial y}= 0[/tex]
    do NOT give "p= A" or "p= B". Since differentiation with respect to x treats both y and z as constant, the "constant of integration" may be a function of y and z: p= A(y,z). Then
    [tex]\frac{\partial p}{\partial y}= \frac{0\partial A}{\partial y}= 0[/tex]
    which says that A is a "constant" with respect to y- but may be a function of z.
     
    Last edited: Apr 18, 2010
  8. Apr 20, 2010 #7
    Thanks this is very well explained, i understand it now:cool:
     
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