Proving a Fact About Exact Differentials

In summary: Since f is continuous on D, this implies that for some x,f(x) > 0This seems like a pretty easy proof, but I can't seem to get it to work. Any help would be much appreciated.In summary, the problem was discussing how to prove that P dx + Q dy is an exact differential. The hint given was to use a function g(x, y) with g_x = P and g_y = Q. However, there was a problem with the hint which was a typo. After fixing the typo, the problem was that there was a b instead of a y. The anti-derivatives of P and
  • #1
e(ho0n3
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0
[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!
 
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  • #2
e(ho0n3 said:
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]
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]


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!
If that "b instead of an a" is in the book and not your mistake, then it is a misprint in the book!
 
  • #3
HallsofIvy said:
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]

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.
 
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  • #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.
 
  • #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]?
 
  • #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.
 
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Related to Proving a Fact About Exact Differentials

What is an exact differential?

An exact differential is a type of differential that can be integrated directly without the need for an integrating factor. It is a mathematical concept used in calculus and physics to describe changes in a system.

How can I prove that a differential is exact?

To prove that a differential is exact, you need to show that it satisfies the condition of being conservative. This means that the differential must be the total derivative of a function, also known as a potential function.

What is the importance of proving a fact about exact differentials?

Proving a fact about exact differentials is important because it allows us to solve problems related to changes in a system. It also helps us understand the underlying mathematical concepts and principles involved in these changes.

What are some common techniques used to prove a fact about exact differentials?

Some common techniques used to prove a fact about exact differentials include the use of partial derivatives, the use of exact differential equations, and the use of line integrals.

Are there any real-world applications of proving facts about exact differentials?

Yes, there are many real-world applications of proving facts about exact differentials. One example is in thermodynamics, where exact differentials are used to describe changes in temperature and energy in a system. They are also used in economics, fluid mechanics, and other fields to model and analyze changes in various systems.

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