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Trouble understanding differential k form

  1. May 2, 2015 #1
    1. The problem statement, all variables and given/known data
    data

    Solving differential k forms.

    2. Relevant equations
    I don't want to give any exact problems from my problem set.

    3. The attempt at a solution
    solution.


    The text I'm using, CH Edwards, is very abstract in this section and the explanation over a sped up, last class and a half of the semester was only more vague.

    I think I understand the definition of a k form and how they go through the grad, curl, and div as k increases but I'm losing it on the proof which jumps over what it considers "routine but brute force computations".

    Is there a bit more concrete way to explain the direction I'm trying to go?
     
  2. jcsd
  3. May 3, 2015 #2

    robphy

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    Can you give more information? It's not clear to me what you are looking for.
     
  4. May 3, 2015 #3
    So, this first example, i have,

    w=(3x^2y^2+8xy^3)dx + (2x^3y+12x^2y^2+4y)dy

    I'm asked to find da such that da=w

    Which reminds me of a differential equation but I'm not sure what to do with it. What does the dx and dy represent?
     
  5. May 4, 2015 #4

    HallsofIvy

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    For any function, f(x,y), of two variables, [itex]df= f_xdx+ f_ydy[/itex]. In particular, if you are given that [itex]da= (3x^2y^2+ 8xy^3)dx+ (3x^3y+ 12x^2y^2+ 4y)dy[/itex] then [itex]a_x= 3x^2y^2+ 8xy^3[/itex] and [itex]a_y= 2x^3y+ 12x^2y^2+ 4y[/itex].

    I presume you know that, for any function, a(x,y), with continuous second derivatives, the two mixed second derivatives must be equal: [itex]a_{xy}= a_{yx}[/itex]. We can use that to check if there is, in fact, a solution to this problem- in general, just putting to simple functions, say, f(x,y) and g(x,y) together as f(x,y)dx+ g(x,y) does NOT result in an "exact differential" because [itex]f_y\ne g_x[/itex]. But here, [itex]a_{xy}= (3x^2y^2+ 8xy^3)_y= 6x^2y+ 24xy^2= (2x^3y+ 12x^2y^2+ 4y)_x[/itex].

    So, knowing that [itex]a_x= 3x^2y^2+ 8xy^3[/itex] and that the partial derivative with respect to x treats y like a constant, we can find a by integrating with respect to x, treating y as a constant. HOWEVER, that means that the "constant" of integration might actually be a function of y- call that, say F(y). So now differentiate the "a" you just got with respect to y, leaving the derivative of F as F' and set that equal to [itex]2x^3y+ 12x^2y^2+ 4y[/itex]. Because of the check above, we know that the "x" terms will cancel leaving an expression, involving only y, for F'. Integrate that to find F.

    (I am puzzled by your reference to "differential k forms". They usually arise in an advanced differential geometry course. But the problem you give is a standard "third semester Calculus" or "Calculus of functions of several variables" problem.)
     
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