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I am talking now about (non-)exact equations (or in some languages: equations of the total differential form).

Usually, when there is a problem stated as: "Find a solution $\omega$ for: $(1)\space M(x,y)dx+N(x,y)dy=d\omega$" and it can be quickly found the given equation is not exact, we know we are about to use an integrating factor $\mu(x)$, $\mu(y)$ or something even better.

But this means we are not going to literally "find a solution" for $(1)$, but only for the modified equation: $(2)\space \mu(x)M(x,y)dx+\mu(x)N(x,y)dy=\mu(x)d\omega=d\sigma$. In another words: the original equation $(1)$, in fact, has no solution. (Solution for $(2)$ usually does not work for $(1)$ if you try to just plug it in.) The actual thing we find in the process is a factor with some desired type of dependence on some of the variables ($x$, $y$, $xy$, $\frac{1}{xy^2}$, etc.) and a potential corresponding to $(2)$ and the chosen factor.

My question(s) is (are): Is my understanding of the subject correct? If yes, is there some meaning behind the fact that almost all textbooks and websites say "find a solution for (a non-exact/integrable) equation $(1)$" when there isn't any? Is it true there is this habit of calling the solution for $(2)$ to be a solution for $(1)$ as well?

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  • $\begingroup$ why do you say: the original equation (1), in fact, has no solution. $\endgroup$
    – felasfa
    Oct 13 '16 at 22:34
  • $\begingroup$ What you're writing in (1) makes absolutely no sense. You're trying to solve $\omega = M\,dx+N\,dy = 0$. You find a nowhere zero function $\mu$ so that $\mu\omega = df$ for some function $f$, and then you say that the level curves $f=C$ give you solutions of $\omega=0$. $\endgroup$ Oct 14 '16 at 1:07
  • $\begingroup$ @felasfa When $(1)$ is not exact there doesn't exist any $\omega$ for which its differential is the left-hand side of the equation. $\endgroup$
    – Degauss
    Oct 14 '16 at 7:01
  • $\begingroup$ @TedShifrin I have edited my question so it is clear that I want to find $\omega$. $C$ was indeed superfluous. Most of problems are not stated that way. $\endgroup$
    – Degauss
    Oct 14 '16 at 7:03
  • $\begingroup$ @TedShifrin After a while, now I see my confusion. Your comment was indeed important. I have answered myself. You could check my answer if you were interested. :) $\endgroup$
    – Degauss
    Sep 12 '17 at 23:22
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After some time I - myself - am going to answer your - my - old question. TedShifrin was right.

Your (my) confusion stemmed from the fact that there are two classes of problems solved by a similar method and by the time of asking, you were studying thermodynamics. So your understanding was truly not quite correct.

In thermodynamics, every thermodynamic potential is a Pfaffian form that can be written as $(1)$. But this follows from the necessary condition that these potentials are exact. Solution $\omega$ can be constructed by a straightforward integration of the left-hand side,

However, when we solve exact and non-exact equations they are always given as

$$ M(x,y)dx+N(x,y)dy=0 \tag{3} $$

Your $d\omega$ on the right-hand side of $(1)$ has now no clear meaning. The problems of the form $\dots=0$ are important. Their use is nicely illustrated on this page.

One of the things is that these equations prescribe some curves in space (like all differential equations after all). In the case of exact equations, we can find a function $F(x,y)$ which has the partial derivatives of $M$ and $N$ and naturally is a constant since $dF$ was set to $0$. And, as we already know from high-school, $F(x,y)=C$ gives us a curve. The slightly confusing thing is that we construct the function $F$ (the potential of the equation) by the same process which was used in the integration of forms.

Aaand, in the case of inexact equations... The use of $\mu$ is reasonable because it doesn't touch the solutions of the original problem $(3)$. The expression $\mu(x,y)(M(x,y)dx+N(x,y)dy)$ is still zero for - at best - the same curves $(x,y)$.

However, here I should be more rigorous: multiplication by an integrating factor is not an equivalent operation and can give the same solutions as for the original problem on a smaller set.

It is possible that I miss a general picture here. If there won't be some general revision or specification by someone I will accept this answer.

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