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PDE 2Ux + 3Uy + U = 0 with change of variables V(x,y)=ln[U(x,y)]

  1. Sep 14, 2009 #1
    [note: Ux=∂U/∂x, Uy=∂U/∂y]

    Example: Solve the partial differential equation 2Ux + 3Uy + U = 0 by using the change of variables V(x,y)=ln[U(x,y)]

    Vx = Ux/U
    Vy = Uy/U

    2Ux + 3Uy + U = 0
    Dividing both sides by U, we have
    2Ux/U + 3Uy/U + 1 = 0
    => 2Vx + 3Vy +1 = 0
    => 2Vx + 3Vy = -1

    The corresponding homogeneous equation has the general solution V = f(3x-2y) where f is arbitrary function.

    2Vx + 3Vy = -1
    Set V(x,y)=f(x)
    => 2f ' + 0 = -1
    => f ' = -1/2
    => f= -x/2 + C
    => f= -x/2 (take C=0)
    Therefore, a particular solution to 2Vx + 3Vy = -1 is V = -x/2

    So the general solution to 2Vx + 3Vy = -1 is V = -x/2 + f(3x-2y)
    => the general solution to the original PDE is U = exp(-x/2) exp[f(3x-2y)] = exp(-x/2) g(3x-2y)

    Now, I don't understand the parts in red:

    1) In the solution, they divided both sides by U. Why is this always allowed? How do we know that U is not 0?

    2) For the part of finding a particular solution to 2Vx + 3Vy = -1, they first set V(x,y)=f(x). What is the logic behind this step? Why would this lead us to a particular solution? I just don't get the idea.

    3) At the end, they claimed that the general solution to the original PDE is
    U = exp(-x/2) exp[f(3x-2y)]
    i.e. U= exp(-x/2) g(3x-2y) (final answer)
    I don't understand why exp[f(3x-2y)] can be replaced by g(3x-2y). Why do we have to do that? Is g here an arbitrary function?

    Could someone please kindly explain? Any help is greatly appreciated!:smile:
  2. jcsd
  3. Sep 14, 2009 #2


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    Regarding dividing by U, you have already assumed U not zero when you made the substitution V(x,y)=ln[U(x,y)]. This means your method may miss a possible solution U = 0. This is, in fact, a solution although likely uninteresting.

    On setting V(x,y) = f(x), you are looking for any particular solution on the nonhomogeneous equation. So you are trying to see if perchance a pure function of x works. Luckily, it does.
    You can check that V = -x/2 works.

    As to why exp(-x/2) exp[f(3x-2y)] = exp(-x/2) g(3x-2y), you are just observing that if f can be anything, so can exp(f). g can't be completely arbitrary; it would have to be positive. But you can get any such g by taking f = ln(g).

    In the end, sometimes although your steps rule out, for example, g being negative, it may turn out that in the equation itself it may be OK for g to be negative. You can check that by seeing if it satisfies the equation with arbitrary g.
  4. Sep 17, 2009 #3
    1) Then that looks even worse to me, it is a very strong restriction. Why can we assume that U is always non-negative in this PDE?

    2) To find a particular solution to the PDE 2Vx + 3Vy = -1, they set V(x,y)=f(x). But why can they be so sure that a solution of a function that only depends on x even EXISTS? What if it fails, then what else can we do? Does this method always work for finding particular solutions to linear non-homogeneous first order PDEs?

    3) I think that typically:
    Every function of the form exp[f(3x-2y)] can be expressed in the form g(3x-2y), and every function of the form g(3x-2y) can be expressed in the form exp[f(3x-2y)] by taking f=ln(g). And I think that's why they can replace exp[f(3x-2y)] by g(3x-2y).
    But exp(.) is always positive, so I think g here is an arbitrary function, but MUST be positive.

    To be precise, we should say that the general solution is U = exp(-x/2) g(3x-2y) where g is an arbitrary non-negative function, right?

    Thank you! :)
  5. Sep 17, 2009 #4


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    Because V(x,y)=ln[U(x,y)], and V, U are real functions?
  6. Sep 17, 2009 #5


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    (2) There may be a test that indicates a pure function of x works; I don't recall for sure.
    But even if you don't know ahead of time it is going to work, nothing is lost by trying. And if that fails try a function of y. Nothing guarantees that either will work.

    As I said before, your method of solution precludes g being negative. But that doesn't preclude the DE itself from allowing g to be negative. And, in fact, if you try substituting

    U = exp(-x/2) g(3x-2y)

    in the equation with no restriction on g except differentiability, you will see that it works.
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