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Given partials of a function, show the function does not exist

  1. Feb 28, 2009 #1
    1. The problem statement, all variables and given/known data
    You are given the first degree partials of a function. You must prove the function does not exist

    2. Relevant equations

    General question

    3. The attempt at a solution
    I tried to integrate with respect to the variable for the partial derivaties, then equating them together as 3f(x) = g(x) + 3C but that gets me nowhere. Showing the partials are not continuous doesnt really help.

    Nevermind I got it. I integrated the partials with respect to each of their variables, the derived with respect to a fixed variable. Then I showed the contradiction by comparing the constant terms and showing extra variables exist.
    Last edited: Feb 28, 2009
  2. jcsd
  3. Mar 1, 2009 #2


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    This is badly worded! How can you "have first degree partials of a function" that does not exist? It should be "show that two (or three) given functions of x, y (or of x, y, z) cannot be the partial derivatives of any function".

    It would have helped if you had told us the specific functions that purport to be partials. Also what do you mean "showing the partials are not continuous doesn't really help"? Were the partials continuous or not? If, in fact, the derivatives of the partials were continuous then you could look at the mixed second derivatives.
  4. Mar 13, 2009 #3
    Ah sorry, I guess that was a terrible writeup, i think caused by lack of sleep.

    The problem was that I was given the definition of a gradient field and a few functions like F(x,y,z)=(xz,yz,xy) and I had to show that it was not a gradient field. Thus I had to do partial integration. Though my prof did not cover it, and I think he wanted us to figure out what partial integration was. I looked online and saw that the constant term in a partial integral was a function of the variables you did not integrate over.

    Im not sure what I meant by showing the partials are not continuous.

    On the assignment all functions are well behaved unless at obvious points like |x| at 0 or stated.
  5. Mar 13, 2009 #4


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    Why bother integrating at all? What is the curl of a gradient of any function?:wink:
  6. Mar 13, 2009 #5
    Its an analysis class and we are going through the theorems in R^n of calculus, no vector calc besides some random hw question stuff.

    Ive been hearing a lot about these curls, div and grads though.
  7. Mar 14, 2009 #6


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    In that case, perhaps an example will do you some good:

    The vector function [itex]\vec{F}(x,y,z)=(\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2},0)[/tex] has a non-zero curl (along the z-axis anyways), so it can't be a gradient field.

    To show this through integration, you assume that it is a gradient field of some scalar [itex]U[/itex], then you know [tex]\vec{F}(x,y,z)=\vec{\nabla}(U)=(\partial_x U,\partial_y U, \partial_z U)[/tex]

    [tex]\implies U=\int F_x(x,y,z) dx= -\arctan\left(\frac{x}{y}\right)+f(y,z)[/tex]

    (Here, the constant of integration [itex]f(y,z)[/itex] need only be constant in x, it can still have y and z dependence)

    [tex]U=\int F_y(x,y,z) dy= \arctan\left(\frac{y}{x}\right)+g(x,z)[/tex]


    [tex]U=\int F_z(x,y,z) dz=h(x,y)[/tex]

    must all be true simultaneously. And looking at the first two conditions, it should be clear that since [itex]\arctan(u)\neq-\arctan(1/u)[/itex], no single-valued function [itex]U(x,y,z)[/itex] can accomplish this.

    (On a side note, if you stick to simply-connected regions that do not completely encircle the z-axis, or include it, a scalar potential [itex]U(x,y,z)=\tan^{-1}\left(\frac{y}{x}\right)[/itex] will work, but as soon as you include any portion of the z-axis or encircle it completely, the scalar potential becomes multivalued)
    Last edited: Mar 14, 2009
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