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):
- $\pmb{f}$ is the gradient of some potential function in $S$
- The line integral of $\pmb{f}$ is independent of path in $S$
- 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.
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 $\Bbb{R}^2$).