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Proving a Fact About Exact Differentials

  1. Feb 17, 2008 #1
    [SOLVED] Proving a Fact About Exact Differentials

    Problem. Let D be a disc and let P and Q be functions on D with continuous partial derivatives with respect to x and y. Let C be any closed curve in D.

    Given the fact that

    [tex]\int_C (P \, dx + Q \, dy) = 0 \leftrightarrow Q_x = P_y[/tex]

    show that P dx + Q dy is an exact differential. (Hint: Let (a, b) be any point in D and set

    [tex]g(x, y) = \int_a^b P(t, b) \, dt + \int_b^y Q(a, s) \, ds[/tex]

    Show that g is the desired function.

    Attempt. P dx + Q dy is an exact differential if there is a function g(x, y) with [itex]g_x = P[/itex] and [itex]g_y = Q[/itex]. So if I take the partial derivative with respect to x of the function g defined in the hint, I should get P right?

    Let p be the anti-derivative of P with respect to the first variable and let q be the anti-derivative of Q with respect to the second variable. Then by the fundamental theorem of calculus

    g(x, y) = p(x, b) - p(a, b) + q(a, y) - q(a, b)

    and so [itex]g_x(x, y) = p_x(x, b) = P(x, b)[/tex] right? The problem here is that there's a b instead of a y. I'm mostly likely doing something terrible wrong here because I haven't used the fact stated in the problem. Help!
  2. jcsd
  3. Feb 17, 2008 #2


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    I assume this was supposed to be
    [tex]g(x, y) = \int_a^x P(t, b) \, dt + \int_b^y Q(a, s) \, ds[/tex]

    If that "b instead of an a" is in the book and not your mistake, then it is a misprint in the book!
  4. Feb 17, 2008 #3
    Right. Sorry for the typo. That is what screwed me up. It should be:

    [tex]g(x, y) = \int_a^x P(t, b) \, dt + \int_b^y Q(x, s) \, ds[/tex]

    And so

    [tex]g_x(x, y) = P_x(x, b) + \int_b^y Q_x(x, s) \, ds[/tex]

    I apply the fact that [itex]Q_x = P_y[/itex] and simplify to get that [itex]g_x(x,y) = P(x,y)[/itex]. Similarly, [itex]g_y(x,y) = Q(x,y)[/itex] and thus P dx + Q dy is an exact differential.
    Last edited: Feb 18, 2008
  5. Feb 18, 2008 #4
    One more problem: If [itex]Q_x = P_y[/itex], it is easy to show that the line integral of P dx + Q dy around any closed curve is 0 by applying Green's theorem. However, the converse does not seem to be necessarily true.
  6. Feb 20, 2008 #5
    From Green's theorem,

    [tex]\int_C (P \, dx + Q \, dy) = \iint_R (Q_x - P_y) \, dx dy[/tex]

    Given that the above is 0, then

    [tex]\iint_R Q_x \, dx dy = \iint_R P_y \, dx dy[/tex]

    This is as much as I can deduce. How is it possible to further deduce this into [itex]Q_x = P_y[/itex]?
  7. Feb 22, 2008 #6
    The problem of my previous two posts were discussed in class today. The prof. said that because the double integral is 0 for any region R, the function [itex]Q_x - P_y[/itex] must be 0.

    This sounds reasonable. I would like to examine this idea for the one-dimensional case. Let f(x) be a function continuous on a domain D. Now suppose for any interval (a,b) in D,

    [tex]\int_a^b f(x) \, dx = 0[/tex]

    To prove that f(x) = 0, I have to show that for any x in D, f(x) = 0. Arguing by contradiction, suppose for some c, f(c) is not 0. Because f is continuous,

    [tex]\lim_{x \to c} f(x) = f(c)[/tex]

    Thus, for any e > 0, there is a d > 0 such that for all x with c - d < x < c + d, f(c) - e < f(x) < f(c) + e. Therefore,

    [tex]2d(f(c) - e) = \int_{c-d}^{c+d} f(c) - e \, dx < \int_{c-d}^{c+d} f(x) \, dx < \int_{c-d}^{c+d} f(c) + e \, dx = 2d(f(c) + e)[/tex]

    For e << f(c), if f(c) is positive, f(c) - e and f(c) + d are both positive and so the integral in the middle must be positive. If f(c) is negative, then the middle integral is negative. This is a contradiction as that integral should be 0. Hence, for all values of x in D, f(x) = 0.

    That's cool. Concerning the problem in the previous two posts, the region R enclosed by C may be described as the points (x, y) such that x is in the interval (a, b) and y is in the interval (c, d). In this manner,

    [tex]\iint_R Q_x - P_y \, dxdy = \int_c^d \int_a^b Q_x(x, y) - P_y(x, y) \, dx dy = 0[/tex]

    This is true for every region R enclosed by a curve C in the domain D. Thus, the bounds of the outer integral range over the domain of x-values in D and the bounds of the inner integral range over the domain in y-values of D.

    This implies that the inner integral must equal 0 and consequently, that [itex]Q_x - P_y = 0[/itex].

    I think this explanation is sound. I will mark this thread as solved.
    Last edited: Feb 23, 2008
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