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There are a series of problems in Apostol's Calculus where we are given a specific vector field $\pmb{f}$ and are asked to determine if the vector field is a gradient of a scalar field, ie potential function (and if it is, to find a potential function for the vector field).

My question is about the underlying justification for the procedures involved in solving such problems.

For example, let $\pmb{f}(x,y)=X(x,y)\hat{i}+Y(x,y)\hat{j}$.

Here are the steps and justifications that I think are in play, and my question is if the reasoning below is correct.

Since the vector field $\pmb{f}$ is continuous on all of $\mathbb{R}^2$ (which is an open connected set) then following three statements that are equivalent (and constitute necessary and sufficient conditions for a vector field to be a gradient):

  1. $\pmb{f}$ is the gradient of some potential function in $S$
  2. The line integral of $\pmb{f}$ is independent of path in $S$
  3. The line integral of $\pmb{f}$ is zero around every piecewise smooth closed path in $S$.

However, we don't know if $\pmb{f}$ satisfies any of these conditions.

It seems the strategy used to find a potential involves the following reasoning:

  • If $\pmb{f}$ were a continuous gradient in $S$ (and we can see that it is definitely continuous in $S$) then any line integral on a piecewise smooth path in $S$ would be path independent (ie, the 2nd fundamental theorem for line integrals would apply).
  • This means that if we simply assume $\pmb{f}$ is a gradient, choose two points $\pmb{a}$ and $\pmb{w}$ in $S$ and compute the line integral of $\pmb{f}$ along any path between these points we will obtain an expression that we then equate to $\varphi(\pmb{w})-\varphi(\pmb{a})$ and we can solve for $\varphi(\pmb{w})$ which is our candidate for potential function.
  • Equating the result of the line integral to $\varphi(\pmb{w})-\varphi(\pmb{a})$ is using our assumption of path independence (which is equivalent to assuming that we can apply the 2nd fundamental theorem of calculus).
  • Once we have $\varphi(\pmb{w})$ we can check if it is indeed a potential function for $\pmb{f}$ by computing $\nabla{\varphi}$, and if it coincides with $\pmb{f}$ then we were right with our initial assumption that $\pmb{f}$ was a gradient and we've reached the objective of finding the potential.
  • If we fail, then it means $\pmb{f}$ is definitely not a gradient, because if it were, then our computation should have worked.
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    $\begingroup$ You need some assumptions usually to make this line of argument work, usually to make sure the potential you construct is well-defined. $\endgroup$
    – Randall
    Commented Apr 6, 2023 at 23:18
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    $\begingroup$ @evianpring not true. there is a partial converse, which is applicable here since the various smoothness assumptions on $\mathbf{f}$ are satisfied, and the topological restrictions on the domain are also satisfied. $\endgroup$
    – peek-a-boo
    Commented Apr 6, 2023 at 23:23
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    $\begingroup$ If the domain is open and simply connected, equality of cross derivatives does in fact imply that the field is a gradient. $\endgroup$
    – Randall
    Commented Apr 6, 2023 at 23:26
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    $\begingroup$ how to justify the procedure of finding a potential by integrating? I think you need to reread the proof of the equivalence of the three statements. The proof explicitly tells you why in the case of a continuous vector field whose line integral is path-independent, the function so-defined is $C^1$ and has that vector field as its gradient. $\endgroup$
    – peek-a-boo
    Commented Apr 7, 2023 at 0:02
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    $\begingroup$ but anyway, in practice, if things are smooth enough, then the first thing to check is the condition on partials. If they are equal, then the next question to ask yourself is whether the domain is topologically nice. If it is not, then that's when you start to worry. You'll either have to be clever and spot a potential by-hand, or find some closed loop for which the line-integral is non-zero. In general, the question can get quite hard (see also de Rham cohomology, though that may be a little advanced for now). $\endgroup$
    – peek-a-boo
    Commented Apr 7, 2023 at 0:10

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Why do you not know (1) in this case? Here, 'by observation', we see that $\mathbf{f}$ is the gradient of $\phi(x,y)=x^3y$.

Also, since your $\mathbf{f}$ is smooth and defined everywhere, a necessary and sufficient condition is for the 'cross partials' to be equal (they are both equal to $3x^2$ in this case) (i.e closed implies exact since we're working on all of $\Bbb{R}^2$).

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  • $\begingroup$ Ok, I'm going to remove the example. It's only serving to distract from the main questions I have. Obviously the example I gave is very simple. My question is general. I will remove the example. $\endgroup$
    – xoux
    Commented Apr 6, 2023 at 23:22

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