5
$\begingroup$

As the question says, why are exact differential equations called so?

From Wikipedia, I got "The nomenclature of "exact differential equation" refers to the exact differential of a function". That leads me to ask why is an exact differential called so? Usually, the term 'exact' in the context of English refers to some quantity that is not approximated in any way. How does that definition fit here?

$\endgroup$
1
  • 1
    $\begingroup$ One of the major problems in mathematics is inventing the terminology we need for important notions. Unfortunately, that means that you can't expect to trace the mathematical meanings of terms back to their informal English meanings. E.g., normal subgroups are actually quite special. This applies to the term "exact", although I don't think the terminology is too bad: it is reasonable to try to approximate a $k$-form by the differential of a $k-1$ form and then the exact $k$-forms are the ones that can be approximated exactly. $\endgroup$
    – Rob Arthan
    Commented Sep 9, 2022 at 0:31

3 Answers 3

4
$\begingroup$

A differential $k$ - form $\omega$ is exact if there exists a $(k-1)$ - form $\alpha$ such that $\omega=d \alpha$. This (applied to your context) motivates the name 'exact'.

EDIT: to give an example, consider the equation: $$P(x,y) dx + Q(x,y) dy =0 $$ This equation is exact if there is some $U$ such that: $$\frac{\partial U}{\partial x} = P$$ and $$\frac{\partial U}{\partial y} = Q$$ Notice that: $$ dU = \frac{\partial U}{\partial x} dx +\frac{\partial U}{\partial y} dy = P dx +Qdy$$ That is, $U$ is an exact differential form.

$\endgroup$
4
  • $\begingroup$ I think OP's question is basically why is this called exact. $\endgroup$
    – Randall
    Commented Jun 11, 2020 at 16:37
  • 3
    $\begingroup$ I am an engineering grad student, so your answer just went over my head (I don't even know what a form is). Is there a more intuitive way of understanding it? $\endgroup$
    – Paddy
    Commented Jun 11, 2020 at 16:44
  • $\begingroup$ @Randall thanks for noting this. I have edited the answer. $\endgroup$ Commented Jun 11, 2020 at 16:45
  • $\begingroup$ Maybe, a more intuitive way is to note that a vector field which is conservative can be equivalently seen as an exact $1$-form. $\endgroup$ Commented Jun 11, 2020 at 16:46
2
+50
$\begingroup$

An exact differential has the property that its integral over a path is path-independent: it does not depend upon the taken path, but only upon the origin and end of the path.

In mechanics, the work of a force can be path-dependent (e.g. friction), or path-independent (e.g. gravity), in which case it is also called conservative. When it is path-independent, the force can be associated with a potential energy, and the work is only the (opposite of the) potential energy variation; i.e. the difference between potential energies at origin and end of the path. Work and potential energy (Wikipedia)

Then why the path-independent case was called "exact"? Probably because calculating the work involves no numerical approximation: it is the difference between two values of a potential energy, which has a known expression for usual forces such as gravity.

On the other hand, when there is a non-conservative force such as friction, its contribution must be calculated by an integral and may not have a closed expression, hence the computing should be numerical and give an inexact result.

See also https://physics.stackexchange.com/questions/603147/why-does-an-exact-differential-mean-a-force-is-conservative.

$\endgroup$
0
$\begingroup$

An exact differential equation is defined as an equation which has a solution of the form: $$du(x,y)=P(x,y)dx+Q(x,y)dy$$ if the DE is defined as: $$P(x,y)dx+Q(x,y)dy=0$$ leading to the general solution of: $$u(x,y)=C$$ It may be called an exact equation because it is based on the requirements of continuous pds or that the value of the constant can be worked out easily so values given are "exact" rather than how PDEs are often solved using numerical methods which are effectively good approximations

$\endgroup$
3
  • 1
    $\begingroup$ I think this terminology is very old and may pre-date numerical methods. $\endgroup$
    – Randall
    Commented Jun 11, 2020 at 16:34
  • $\begingroup$ That is a good point I didn't think of that $\endgroup$
    – Henry Lee
    Commented Jun 11, 2020 at 16:35
  • $\begingroup$ Numerical methods to solve differential equations are as old as differential equations themselves, i.e. Newton and Leibniz. That's way older than the idea of an "exact" differential equation or differential form. $\endgroup$ Commented Sep 8, 2022 at 23:32

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .