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Mixed derivative in DE

  1. Apr 11, 2005 #1
    Hi, has anyone here ever seen anything like this?

    f = f(x,y,t)
    g = g(x,y,t)

    [tex]\frac{ \partial^{2}{f} }{ \partial {y}\partial {t} } } +
    \frac{ \partial^{2}{f} }{ \partial {x}\partial {t} } }+
    \frac{\partial{f}}{\partial{t}}+
    \frac{\partial{f}}{\partial{y}}+
    \frac{\partial{f}}{\partial{x}}+f = g[/tex]

    Personnally, my blood pressure has begun to drop dangerously. You can drop in a real constant by each term. Any info or sites on DE's with mixed derivatives would be helpful. Can any DE with mixed derivatives at all be solved analytically? A good direction for numerical solving would be helpful too.

    If it's as satanic as it first seems, I'll probably do approximations to simplify my model.

    Thanks.
     
  2. jcsd
  3. Apr 12, 2005 #2
    Actually, the units don't work without the constants, so it should be :

    [tex]C_1\frac{ \partial^{2}{f} }{ \partial {y}\partial {t} } } + C_2\frac{ \partial^{2}{f} }{ \partial {x}\partial {t} } }+C_3\frac{\partial{f}}{\partial{t}}+C_4\frac{\partial{f}}{\partial{y}}+C_5\frac{\partial{f}}{\partial{x}}+C_6f = g[/tex]

    g(x,y,t) is known
    Finding f(x,y,t) is the problem.
     
  4. Apr 12, 2005 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    It is always possible, by rotating the coordinate system, to convert mixed derivatives to 'non-mixed' derivatives. One way is to set up the second derivative coefficients as a matrix and find the eigen-vectors. Those should be the new axes.
     
  5. Apr 13, 2005 #4

    saltydog

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    Yea, Gonzolo, I'd like to see that one solved too. However, I think a good approach is to first analyze it in just 2-D. You know, look first at f(x,y) and g(x,y):

    [tex]\frac{\partial^2f}{\partial x\partial y}+\frac{\partial f}{\partial x}+f=g[/tex]

    I mean, just any solution, any initial condition, any boundary conditions. Can anyone here propose a method for solving this one?
     
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