Proving a Fact About Exact Differentials

[e(ho0n3]

[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

$$\int_C (P \, dx + Q \, dy) = 0 \leftrightarrow Q_x = P_y$$

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

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

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 $g_x = P$ and $g_y = Q$. 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![/itex]

 

[HallsofIvy]

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

$$\int_C (P \, dx + Q \, dy) = 0 \leftrightarrow Q_x = P_y$$

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

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

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

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 $g_x = P$ and $g_y = Q$. 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![/itex]

$If that "b instead of an a" is in the book and not your mistake, then it is a misprint in the book!$

 

[e(ho0n3]

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

Right. Sorry for the typo. That is what screwed me up. It should be:

$$g(x, y) = \int_a^x P(t, b) \, dt + \int_b^y Q(x, s) \, ds$$

And so

$$g_x(x, y) = P_x(x, b) + \int_b^y Q_x(x, s) \, ds$$

I apply the fact that $Q_x = P_y$ and simplify to get that $g_x(x,y) = P(x,y)$. Similarly, $g_y(x,y) = Q(x,y)$ and thus P dx + Q dy is an exact differential.

 

[e(ho0n3]

One more problem: If $Q_x = P_y$, 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.

 

[e(ho0n3]

From Green's theorem,

$$\int_C (P \, dx + Q \, dy) = \iint_R (Q_x - P_y) \, dx dy$$

Given that the above is 0, then

$$\iint_R Q_x \, dx dy = \iint_R P_y \, dx dy$$

This is as much as I can deduce. How is it possible to further deduce this into $Q_x = P_y$?

 

[e(ho0n3]

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 $Q_x - P_y$ 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,

$$\int_a^b f(x) \, dx = 0$$

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,

$$\lim_{x \to c} f(x) = f(c)$$

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,

$$2d(f(c) - e) = \int_{c-d}^{c+d} f(c) - e \, dx < \int_{c-d}^{c+d} f(x) \, dx < \int_{c-d}^{c+d} f(c) + e \, dx = 2d(f(c) + e)$$

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,

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

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 $Q_x - P_y = 0$.

I think this explanation is sound. I will mark this thread as solved.