Suppose we have a differential equation of the form $M(x,\ y)\ dx + N(x,\ y)\ dy = 0$, where $\frac{\partial N}{\partial x} – \frac{\partial M}{\partial y} \neq 0$. It is thus inexact, and absent multiplication by an integrating factor, has no potential function.
We can write this equation as $\vec f(x,\ y)\cdot d\vec r = \vec 0\cdot d\vec r$, where $\vec f = \begin{bmatrix}M(x,\ y) \\ N(x,\ y)\end{bmatrix}$ and $\vec d\vec r = \begin{bmatrix}dx \\ dy\end{bmatrix}$, without altering the differential form. Now let's take the closed line integral of both sides along an arbitrary curve, yielding $\oint\vec f\cdot d\vec r = \oint\vec 0\cdot d\vec r$.
Regardless of the curve, the RHS simplifies to $0$. But then $\oint\vec f\cdot d\vec r = 0$, implying that $\vec f$ is conservative, and therefore should have a potential function, allowing us to solve the differential equation. On the other hand, we assumed earlier that the scalar curl of $\vec f$ was nonzero, and therefore should have no potential function. What is causing the apparent contradiction here?
Best Answer
$\vec f(x,\ y)\cdot d\vec r = 0$ is not an identity valid for all $x$, $y$, $dx$ and $dy$, it is only true on certain curves (the integral curves of the differential equation). So you can't integrate this around an arbitrary curve and expect to get $0$.