Similarity between exact equations and potential functions

[leehufford]

Hello,

I'm currently in a differential equations course, and we are learning how to solve exact equations in the form of Mdx + Ndy = 0.

I immediately recognized this from my vector calculus class as it was used to find a potential function of a vector field (assuming the vector field was conservative). We integrated both M dx and N dy, combined them on a term by term basis and came up with a potential function.

In differential equations we start to do the same thing but then we set the resulting function from integration equal to N (or M, whichever one we didn't integrate).

I was just wondering if both of these processes have the same end goal and/or are interchangeable. If they aren't interchangeable why are they different? Seems like we get a function as an answer to each problem. The problems seem to slightly deviate from each other but I was wondering if this is just my professor's style. Thanks in advance,

Lee

 

[UltrafastPED]

Potentials exist if there is an "exact differential"; that is, the expression must be integrable.

For a concise statement see http://mathworld.wolfram.com/ExactDifferential.html

Here is a treatment of the theory: http://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/3.-double-integrals-and-line-integrals-in-the-plane/part-b-vector-fields-and-line-integrals/session-63-potential-functions/MIT18_02SC_MNotes_v2.2to3.pdf

 

[leehufford]

Potentials exist if there is an "exact differential"; that is, the expression must be integrable.

For a concise statement see http://mathworld.wolfram.com/ExactDifferential.html

Here is a treatment of the theory: http://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/3.-double-integrals-and-line-integrals-in-the-plane/part-b-vector-fields-and-line-integrals/session-63-potential-functions/MIT18_02SC_MNotes_v2.2to3.pdf

Thanks for the reply. After reading the MIT document it seems like a potential function is the integral of a solution to an exact differential equation. Is that a correct assessment? Does that make the gradient of the function the original differential equation? Sorry it's still a little hazy.

Lee

 

[UltrafastPED]

You have it just about right.

The potential is the integral of the exact differential; when you take the gradient you obtain the integrated function.

In physics the potential is path independent; the gradient provides the negative of the force.

 

[HallsofIvy]

I get annoyed at physics notation for mathematics concepts! (A pet peeve of mine.)

In mathematics notation, a differential in two (or three) variables, u(x,y)dx+ v(x,y)dy (or df= u(x,y,z)dx+ v(x,y,z)dy+ w(x,y,z)dz) is an exact differential (if there is a function, f such that \dfrac{\partial f}{\partial x}= u(x,y) and \dfrac{\partial f}{\partial y}= v(x,y)
(and that \dfrac{\partial f}{\partial z}= w(x,y,z)). That is the concept introduced in "multi-variable Calculus", usually in connection with path integrals, and is exactly the idea behind "exact differential equations".