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A Functions with "antisymmetric partial"

  1. Apr 6, 2016 #1
    Sorry for the terribly vague title; I just can't think of a better name for the thread.

    I'm interested in functions ##f:[0,1]^2\to\mathbb{R}## which solve the DE, ##\tfrac{\partial}{\partial y} f(y, x) = -\tfrac{\partial}{\partial x} f(x,y) ##.

    I know this is a huge collection of functions, amounting to everything of the form ##\left\{ \int g(x,y) dx: \ g \text{ antisymmetric} \right\}##, but I'm wondering whether there's a nice description of the family of functions ##f## satisfying the equation. Ideally, I'd like a description which doesn't use derivatives or integrals.
     
  2. jcsd
  3. Apr 6, 2016 #2

    epenguin

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    -1 = (∂ƒ/∂x)/(∂ƒ/∂y) = - (dy/dx)ƒ.
    which is not hard to integrate to
    y - x = K at constant ƒ.

    So ƒ(y - x) = C. , any constant C and any differentiable function ƒ is a solution.

    IMHO not a bad starting point for study of solution of pde's (which I must start one of these days. :oldsmile:)
     
  4. Apr 6, 2016 #3

    andrewkirk

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    Are you sure you mean that and not ##\tfrac{\partial}{\partial y} f(x,y) = -\tfrac{\partial}{\partial x} f(x,y) ##?

    On my calculations, every differentiable anti-symmetric function is a solution of the former (the one you wrote) because both partial derivs are taken wrt the first argument of ##f##. So if ##g:[0,1]^2\to\mathbb{R}## is the partial diff of ##f## wrt its first argument then the DE is just ##g(y,x)=-g(x,y)##, and the problem becomes uninteresting.

    However, if the DE is really ##\tfrac{\partial}{\partial y} f(x,y) = -\tfrac{\partial}{\partial x} f(x,y) ##, or to make it even clearer:

    $$D_2 f(x,y) =-D_1f(x,y) $$
    where ##D_n## indicates partial differentiation wrt the ##n##th argument, the problem becomes interesting and penguin's solution comes into play.
    I wasn't quite sure what you meant by this last part. Do you mean that, for any differentiable function ##g:\mathbb{R}\to\mathbb{R}##, the function ##f:[0,1]^2\to\mathbb{R}## defined by ##f(x,y)=g(y-x)## is a solution of the DE?

    That certainly seems to work, and is a nice solution [of what I think the problem was intended to be, per the above].

    Do you think we can conclude that all solutions are of that form?
     
  5. Apr 7, 2016 #4

    epenguin

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    Aaargh I misread.:redface:
     
  6. Apr 7, 2016 #5
    I did mean ##\tfrac{\partial}{\partial y} f(y, x) = -\tfrac{\partial}{\partial x} f(x,y)##, not ##\tfrac{\partial}{\partial y} f(x, y) = -\tfrac{\partial}{\partial x} f(x,y) ##.

    So I'm after the set of functions whose first partial is an antisymmetric function. I'm wondering if there's a way to specify that class of functions without referencing derivatives or integrals.
     
  7. Apr 7, 2016 #6

    stevendaryl

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    It seems to me that every such [itex]f(x,y)[/itex] can be written in the form [itex]f(x,y) = g(x-y)[/itex], where [itex]g[/itex] is an arbitrary differentiable function.
     
  8. Apr 7, 2016 #7

    andrewkirk

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    A large class of solutions will be of the form

    $$f(u,v)=g(u-v)+h(v)$$

    where
    ##g:[-1,1]\to\mathbb{r}## and ##h:[0,1]\to\mathbb{r}## are both once-differentiable and ##g## is an odd function.
    Note that the ##h## term is a function only of the second argument, and hence disappears upon partial differentiation wrt the first argument.
    The extra requirement that the function ##g## be odd appears necessary. A function that is even, such as ##g(x)=x^2##, or neither even nor odd, such as ##g(x)=e^x##, would not work.

    I can't prove that all solutions are of this form. Perhaps there may be other classes of solutions, but I can't think of any examples.
     
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