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Vector fields

  1. Mar 12, 2009 #1
    Can you help me with this? I have a function with domain and range in R^2. What conditions it must have so that a point in the boundary of the domain will have its image in the boundary of the range?

  2. jcsd
  3. Mar 12, 2009 #2


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    Are you saying that every point on the boundary of the domain has image on boundary of the range?
  4. Mar 12, 2009 #3
    Yes. We have a set R in a plane xy and a set R* in a plane uv.

    We have functions f(x, y) and F(u, v) and the variables have this relations:

    u = u(x, y) and v = v(x,y).

    x = x(u, v) and y = y(u,v).

    These functions u, v, x, y are inyective.

    In Apostol' Calculus (vol.2), in the preliminaries to the proof of the Change of Variable theorem for Multiple Integrals, Apostol states this (I am translating from spanish to english):

    "For the proof we suppose that the functions x and y have continuous second partial derivatives and that the jacobian nevers goes null in R*. The J(u, v) is always positive or always negative. The meaning of the sign of J(u,v) is thath when a point (x, y) describes the boundary of R in counterclockwise sense, the image point (u, v) describes the boundary of R* in the same sense if J(u,v) es positive and in contrary sense if J(u,v) is negative.
  5. Mar 12, 2009 #4
    Eh... any idea ??
  6. Mar 12, 2009 #5


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    As stated sufficient conditions is that the functions have continuous second partial derivatives and the Jacobian is not null.

    The necessary conditions may be a bit broader. Hmmmm....

    Recall that by definition a continuous mapping will map open neighborhoods to open neighborhoods. Thus anything defined topologically (open sets, closed sets, boundary, and interior) will be preserved by the mapping. [But note that the inverse mapping may not be defined and so topological properties of the image may not map back to properties of the original set.]

    Certainly a continuous and invertible function will be sufficient (but invertiblity may not be necessary.) I don't remember for certain.

    Now in order to carry out a change of variables you need stronger conditions so that the integrals will be equal and that is the business with the continuous second partial derivatives and non-singular Jacobian.
  7. Mar 12, 2009 #6
    Jambaugh, may be you can give me some hints as why the continuity of the second partial derivatives implies that the image of a boundary point of the domain set is on the boundary of the range set.
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