Conservative Fields, Potentials and Path Independence

Sometimes it does not. Take $\mathbf{F}=(2xy,x^2)$ instead and repeat all three routes: $1$, $1$, $1$. This field is the gradient of $f=x^{2}y$, and the integral is just $f(1,1)-f(0,0)=1$ whichever way you travel, as climbing a hill gains the same height regardless of the path taken.

So one class of fields collapses a route integral into a subtraction, and the practical question is how to recognise them. There is a quick test, compare $\partial P/\partial y$ against $\partial Q/\partial x$, which correctly rejects the first field, where they are $-1$ and $1$.

But the test has a hole in it, and the hole is literal. The field

passes the test at every single point where it is defined. Integrate it around the unit circle and the answer is $2\pi$, not the $0$ that a conservative field must give on a closed loop. The origin is missing from its domain, and that missing point is enough to break the implication.

This unit covers recognising such fields, constructing a potential when one exists, and the domain condition that separates a test that works from one that merely looks like it does.

Why the cross-partial test is only one direction.

The first direction is immediate: if $P=f_x$ and $Q=f_y$ then $P_y=f_{xy}$ and $Q_x=f_{yx}$, equal by Clairaut when the second partials are continuous. The converse is the substantive theorem, and its proof constructs the potential by integrating along paths that can be deformed into one another, which requires that any loop can be shrunk to a point inside the region.

What "simply connected" is doing. It is exactly the condition that permits that shrinking. Remove one point from the plane and a loop encircling it cannot contract without leaving the region, so the construction fails, and the punctured-plane example shows the failure is real rather than an artefact of the proof.

The consequence: conservativeness is a property of a field together with a region. The same formula can be conservative on a half-plane and not on a punctured plane.

A potential is unique up to a constant. Adding a constant changes no partial derivative, so potentials are never unique; but any two differ by a constant on a connected region, and the constant cancels in $f(\text{end})-f(\text{start})$. Endpoint differences are therefore unambiguous, which is what makes the fundamental theorem usable.

Scope. All of this is stated in the plane. In three dimensions the single cross-partial comparison becomes a condition on the curl, and the corresponding theorems are those of Stokes and the divergence theorem, outside this unit.

A height function. If $\mathbf{F}=\mathrm{\nabla }f$ on a region, then ${\int }_C\mathbf{F}\cdot d\mathbf{r}=f(\text{end})-f(\text{start})$ for any path $C$ in that region, so the integral depends only on the endpoints and every closed path gives $0$. Reading $f$ as height makes this the statement that height gained depends on start and finish rather than on route.

These two cases are not exhaustive. A vector field need not be a gradient field, and a field that is not a gradient field need not circulate uniformly; $\mathbf{F}=(x,y)$ is a gradient field of $f=(x^2+y^2)/2$ and does not circulate, while a general field is neither of the worked examples. What is exhaustive is the equivalence: on a region, $\mathbf{F}$ with continuous components is a gradient field if and only if its line integrals are path independent there, if and only if every closed line integral in the region is zero.

The local test. For $\mathbf{F}=(P,Q)$ with continuous first partials, $\partial Q/\partial x-\partial P/\partial y$ measures local circulation. For $(-y,x)$ it equals $1-(-1)=2$. For a gradient field with continuous second partials it is zero everywhere, by equality of mixed partials. A nonzero value therefore rules out a potential.

Why zero is not sufficient. Vanishing local circulation implies a potential when the region is simply connected. On the plane with the origin removed, $\mathbf{F}=(\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2})$ has $\partial P/\partial y=\partial Q/\partial x$ at every point of its domain, and its integral around the unit circle is $2\pi$ rather than $0$, so it has no potential on that domain. It does have one on any simply connected subregion, such as the right half-plane, where $f=\arctan (y/x)$.

$\mathbf{F}=(-y,x)$ — circulating, path dependent.

Cross-partials $-1$ and $1$: unequal, so no potential. From $(0,0)$ to $(1,1)$ the diagonal gives $0$, the route through $(1,0)$ gives $1$, the route through $(0,1)$ gives $-1$. Around the unit square counterclockwise, $2$.

The swirl $\partial Q/\partial x-\partial P/\partial y=2$ is constant and positive, matching a uniform counterclockwise rotation.

$\mathbf{F}=(2xy,x^2)$ — a gradient, path independent.

Cross-partials both $2x$: equal, and the plane is simply connected, so a potential exists. Integrating $P$ in $x$ gives $x^{2}y+g(y)$; matching $f_y$ against $Q=x^2$ forces $g^{\prime }=0$. The potential is $f=x^{2}y$, verified by $f_x=2xy$ and $f_y=x^2$.

All three paths from $(0,0)$ to $(1,1)$ give $1$, equal to $f(1,1)-f(0,0)$. Every closed path gives $0$.

$\mathbf{F}=({\displaystyle \frac{-y}{x^2+y^2}},{\displaystyle \frac{x}{x^2+y^2}})$ — passes the test, not conservative.

Both cross-partials equal ${\displaystyle \frac{y^2-x^2}{(x^2+y^2{)}^2}}$ at every point of the domain. The plane minus the origin. Yet on the unit circle $\mathbf{r}(t)=(\cos t,\sin t)$ the integrand is

so the integral is ${\int }_0^{2\pi }1dt=2\pi$. A conservative field gives $0$ on every closed path, so this one is not conservative, despite passing the test everywhere it is defined.

On a simply connected subregion avoiding the origin, such as the right half-plane, the same formula is conservative and does have a potential there. Conservativeness depends on the region as well as the formula.

$\mathbf{F}=(-y,x)$ used as an area meter.

The swirl is the constant $2$, so half the closed integral gives area. On the triangle $(0,0)$, $(4,0)$, $(0,3)$: the side along the $x$-axis and the side along the $y$-axis each contribute nothing, and the hypotenuse parametrised as $(4-4t,3t)$ gives integrand $(4-4t)(3)-(3t)(-4)=12$, hence ${\int }_0^{1}12dt=12$. Half of that is $6$, matching $\frac{1}{2}\times 4\times 3$.

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The first two are the basic dichotomy: circulation means the route matters, a gradient means it does not, and one pair of derivatives distinguishes them. The third shows the test's converse failing on a region with a hole, which is the only way the implication can fail. The fourth shows the same circulating field from the first example put to work, its uniform swirl is a nuisance for path independence and exactly what makes it an area meter.

To test for a potential.

1. Compute $\partial P/\partial y$ and $\partial Q/\partial x$.
2. If they differ, stop: no potential exists, the integral is path dependent, and there is nothing further to look for.
3. If they agree, check the domain before concluding. On a simply connected region a potential exists. On a region with a point or hole removed, agreement establishes nothing and a closed integral around the hole must be computed to settle it.

To construct a potential.

1. Integrate $P$ with respect to $x$, treating $y$ as constant, and write the constant of integration as an unknown function $g(y)$, not a constant, since anything depending only on $y$ differentiates to zero in $x$.
2. Differentiate the result with respect to $y$ and set it equal to $Q$.
3. Solve for $g^{\prime }(y)$ and integrate to get $g$.
4. Verify by differentiating the finished potential back to both components. This is cheap and catches sign errors that otherwise propagate into every subsequent evaluation.
5. Evaluate by endpoints: $f(\text{end})-f(\text{start})$. The additive constant cancels, so it need not be determined.

Checks. Confirm a computed potential differentiates back to both components before using it. And when a closed integral comes out nonzero, ask whether the field has a singularity inside before concluding an arithmetic error: a nonzero loop integral is the signature of an enclosed hole.

Part 2: a field whose integral does not.

Take $\mathbf{F}=(2xy,x^2)$, same endpoints.

Test. ${\displaystyle \frac{\partial P}{\partial y}}=2x$ and ${\displaystyle \frac{\partial Q}{\partial x}}=2x$. Equal, and the plane is simply connected, so a potential exists.

Find it. Integrating $P$ in $x$: $f=\int 2xydx=x^{2}y+g(y)$. Differentiating in $y$ gives $f_y=x^2+g^{\prime }(y)$, which must equal $Q=x^2$, so $g^{\prime }(y)=0$ and $g$ is constant. Take

Verify by differentiating back. $f_x=2xy=P$ and $f_y=x^2=Q$.

Evaluate by endpoints. $f(1,1)-f(0,0)=1-0=1$.

Check against all three paths.

• Path A, $\mathbf{r}(t)=(t,t)$: ${\int }_0^1[2t^2+t^2]dt={\int }_0^{1}3t^{2}dt=1$
• Path B: horizontal leg has $y=0$ so $P=0$, contributing $0$; vertical leg has $x=1$ so $Q=1$, contributing ${\int }_0^{1}1dy=1$. Total $1$
• Path C: vertical leg has $x=0$ so $Q=0$, contributing $0$; horizontal leg has $y=1$ so $P=2x$, contributing ${\int }_0^{1}2xdx=1$. Total $1$

All three agree with the potential difference.

Two fields, same endpoints, same three paths.

The first row decides everything below it, at the cost of two derivatives. When the cross-partials disagree, no amount of integration will produce a consistent answer; when they agree on a simply connected region, one subtraction replaces every integral in the column.

A test that fails against one that passes and proves nothing.

For $(-y,x)$ the cross-partials are $-1$ and $1$: unequal, and the field is definitively not conservative. For the punctured-plane field they are equal everywhere in the domain, and the field is still not conservative, as its $2\pi$ loop integral shows.

So a failed test is conclusive and a passed test is not. The asymmetry is the practical content of the theorem: use the test to rule out, and use it to rule in only after checking the region.

The same formula on two regions.

$(\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2})$ on the punctured plane: not conservative, loop integral $2\pi$.

The identical formula on the right half-plane : conservative, with a potential, every loop integral $0$.

Nothing about the formula changed. Conservativeness is a property of a field together with a region, and reporting it without naming the region is reporting half a claim.

A closed path with a singularity inside against one without.

Green's theorem requires continuous partials on the enclosed region, not merely on the curve. The punctured-plane field is perfectly well behaved on the unit circle itself and has a singularity at the centre, so the theorem does not apply, which is how a field with zero swirl everywhere it is defined can have a nonzero loop integral. A field with no enclosed singularity cannot do this.

Two orientations of one curve.

Traversing the unit square counterclockwise gives $2$ for $(-y,x)$; clockwise gives $-2$. The curve is the same set of points, and the integral is a property of the oriented curve. A stated value without a stated direction is ambiguous by a sign.

An interval against a route.

${\int }_a^{b}fdx$ has one path from $a$ to $b$, so path dependence cannot arise and the question never comes up. Adding a second dimension creates the choice, and with it a genuine possibility that the answer differs. The fundamental theorem for line integrals is the statement that gradient fields behave like the one-dimensional case, and the point of this unit is that most fields do not.

Concluding conservativeness from equal cross-partials. The implication runs one way without conditions and the other way only on a simply connected region. The punctured-plane field passes the test at every point of its domain and integrates to $2\pi$ around the unit circle, where a conservative field must give $0$. Check the region before concluding, every time.

Stopping the potential construction at the first integration. Integrating $P$ in $x$ leaves an unknown function of $y$, not a constant. Treating it as a constant gives a function that usually fails to differentiate back to $Q$, and the error surfaces only later, in an endpoint evaluation that quietly gives a wrong number.

Skipping the verification step. Differentiating a proposed potential back to both components takes seconds and catches sign errors before they propagate into every subsequent evaluation. A potential is used many times once found, so an unverified one is a compounding mistake.