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Conversion to closed-form expression with DE's

  1. Oct 27, 2012 #1
    I was wondering, if it is possible to obtain somehow a solution expressed with elementary functions from an equation, which is in non closed-form at first sight.

    For istance let us consider an implicit function F(x,y)=0 with Fx(0,0)≠0, which can't be explicitly solved for y. Nevertheless, according to the implicit function theorem, a function y(x) exists in the neighbourhood of (0,0). We also know from the same theorem that yx(x)=-Fx(x,y)/Fy(x,y).
    One could now claim that the differential equation dy/dx=g(x,y) (whereby g(x,y)=-Fx(x,y)/Fy(x,y)) could have a closed-form solution for y!

    The odds are that this method would never yield any result. If that is the case, can sb, more familiar with the Galois theory or the theory of functions, provide us with a justification/proof?
  2. jcsd
  3. Oct 28, 2012 #2


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    It depends what is meant by closed form and at first sight.
    For some purposes an integral, infinite series, or differential equations is a useful answer to a question and can be considered a closed form. Many textbook problems are silly when seen this way. One might be asked to write an integral as an infinite series, but it is not clear that it is an improvement. Often some clever use of an identity can allow a closed form that is not obvious at first. You are right Galois theory allows one to prove that a closed form does not exist. This holds only for that particular closed form, add some more functions and everything changes.


    http://www.math.huji.ac.il/~kamensky/lectures/diffgalois.pdf [Broken]
    Last edited by a moderator: May 6, 2017
  4. Oct 28, 2012 #3
    For instance the equation F(x,y)=lny+y-x is solvable for y near (1,1) but this solution, the so-called the Lambert function, cannot be expressed in terms of elementary functions.
    According to the method I described, the respective DE we get is: (y+1)y'=-y which has exactly the same solution. So there is nothing gained of course from this attempt.

    Generally, integrals and series are not considered closed forms.

    I guess "at first sight" would mean that our F is not explicitely solvable for a variable, but after some kind of transformation, we could "unravel" an elementary solution. I'm pretty much a beginner with the algebraic Galois theory let alone the differential extension, but the more I read the more obvious it becomes that sth like that is impossible.
  5. Oct 28, 2012 #4


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    Galois theory just allows one to be sure something is impossible. There is nothing special about elementary functions. If you need to solve log y+y-x=0, you need the Lambert function (or similar). Later, maybe you need to solve log y+sin y-x=0 and the Lambert function is not enough. There is no way to get around these things.
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