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Conservative Fields and Path Independence · Tutorial 789 of 999

Potentials for Polynomial Fields

Use mixed-partial tests to decide whether a polynomial field has a potential, and recover that potential systematically.

Intermediate 10 min read

What You'll Learn

  • Test a two-variable polynomial field by comparing the relevant cross partial derivatives.
  • Apply the corresponding mixed-partial conditions to a three-variable field.
  • Recover a potential by integrating one component and accounting for the other variables.
  • Use compatibility conditions to determine the correction terms.
  • Verify a proposed potential by differentiating it.
  • Recognize when a polynomial field cannot have a potential.

Polynomial Fields and Potential Functions

A polynomial vector field has components built from sums of powers and products of the coordinate variables. This makes its derivatives and integrals manageable, but mixed terms still require care: integrating one component leaves an unknown function of the other variables.

The practical task is to do two things: test whether the components are compatible with being derivatives of a single scalar function, and, if they are, reconstruct that function. The mixed-partial tests from earlier material provide the criterion; component integration turns that criterion into a method.

Definition: A vector field \(\mathbf{F}\) on a domain \(D\subseteq\mathbb{R}^n\) is conservative if there is a scalar function \(\phi:D\to\mathbb{R}\) such that \(\mathbf{F}=\nabla\phi\). The function \(\phi\) is called a potential for \(\mathbf{F}\).

In two variables, write \(\mathbf{F}(x,y)=\langle P(x,y),Q(x,y)\rangle\). A potential must satisfy \(\phi_x=P\) and \(\phi_y=Q\). In three variables, write \(\mathbf{F}(x,y,z)=\langle P,Q,R\rangle\); then the required equations are \(\phi_x=P\), \(\phi_y=Q\), and \(\phi_z=R\).

Compatibility Tests for Polynomial Fields

If the components are polynomials, their first and second partial derivatives are continuous everywhere. Thus, on \(\mathbb{R}^2\), the mixed-partial compatibility test is \(P_y=Q_x\). On \(\mathbb{R}^3\), the corresponding conditions are \(P_y=Q_x\), \(P_z=R_x\), and \(Q_z=R_y\). The domain matters: these tests guarantee a potential on the entire space \(\mathbb{R}^2\) or \(\mathbb{R}^3\), which is open and simply connected. On more general domains, a zero-curl test alone may not guarantee a global potential.

Theorem: A polynomial field on all of \(\mathbb{R}^2\) is conservative if and only if \(P_y=Q_x\) everywhere. A polynomial field on all of \(\mathbb{R}^3\) is conservative if and only if all three mixed-partial conditions above hold everywhere.

Necessity follows from equality of the mixed partial derivatives of a potential. Sufficiency follows from the criterion for curl-free fields on a simply connected domain established in the earlier material. For finding the potential, we can use component integration directly, keeping the unknown terms that integration cannot determine.

A Component-Integration Method

In two variables, integrate \(P\) with respect to \(x\). The result contains an unknown function \(g(y)\), not just a constant, because differentiating any function of \(y\) with respect to \(x\) gives zero. Then differentiate the candidate potential with respect to \(y\) and match it with \(Q\) to determine \(g\).

In three variables, integrating \(P\) with respect to \(x\) leaves an unknown function \(g(y,z)\). Matching the \(y\)-derivative with \(Q\) determines its \(y\)-dependence, but may leave a function of \(z\). Matching the \(z\)-derivative with \(R\) determines what remains.

1
Test compatibility.
Compare the required mixed partial derivatives before attempting to recover a potential.
2
Integrate one component.
Integrate with respect to its coordinate and include an unknown function of the remaining coordinates.
3
Match the remaining components.
Differentiate the candidate with respect to each remaining coordinate to determine the unknown function.
4
Verify.
Compute the full gradient and check that it equals the original field component by component.

This method works because integrating one component enforces one equation, while the unknown function records exactly the freedom left by that equation. The other component equations remove that freedom in turn. The compatibility conditions ensure that the resulting requirements agree.

Worked Examples in Two Variables

Worked Example: A Compatible Field with Mixed Terms

Consider \(\mathbf{F}(x,y)=\langle 3x^2y+2y,\;x^3+2x+4y^3\rangle\). Set \(P=3x^2y+2y\) and \(Q=x^3+2x+4y^3\). First test the mixed partials:

$$ P_y=3x^2+2,\qquad Q_x=3x^2+2. $$

They agree everywhere, so the field is conservative on \(\mathbb{R}^2\). Integrate \(P\) with respect to \(x\):

$$ \phi(x,y)=\int(3x^2y+2y)\,dx=x^3y+2xy+g(y). $$

Differentiate this expression with respect to \(y\) and match it with \(Q\):

$$ \phi_y=x^3+2x+g'(y)=x^3+2x+4y^3. $$

Therefore \(g'(y)=4y^3\), so \(g(y)=y^4+C\), where \(C\) is an arbitrary constant. One potential is \(\phi(x,y)=x^3y+2xy+y^4\). Verify by differentiating:

$$ \phi_x=3x^2y+2y=P,\qquad \phi_y=x^3+2x+4y^3=Q. $$

Both components of the gradient reproduce the field, as required.

Worked Example: A Polynomial Field That Fails the Test

Consider \(\mathbf{F}(x,y)=\langle xy,\;x^2+y\rangle\), so \(P=xy\) and \(Q=x^2+y\). The mixed partials are

$$ P_y=x,\qquad Q_x=2x. $$

These are not equal everywhere; for instance, at \(x=1\), they have values \(1\) and \(2\). Therefore no potential exists on \(\mathbb{R}^2\). The failure also appears if we try component integration. Integrating \(P\) with respect to \(x\) gives

$$ \phi(x,y)=\frac{x^2y}{2}+g(y). $$

Matching its \(y\)-derivative with \(Q\) would require

$$ \frac{x^2}{2}+g'(y)=x^2+y, \qquad g'(y)=\frac{x^2}{2}+y. $$

The left side depends only on \(y\), but the right side also depends on \(x\). No function \(g(y)\) can satisfy this identity for every \(x\) and \(y\). This confirms that a potential cannot be recovered.

Worked Example in Three Variables

Worked Example: Recovering a Potential with Mixed Terms

Let \(\mathbf{F}(x,y,z)=\langle 2xy+z^2,\;x^2+2yz,\;2xz+y^2\rangle\). Set \(P=2xy+z^2\), \(Q=x^2+2yz\), and \(R=2xz+y^2\). Check all three compatibility conditions:

$$ P_y=2x=Q_x,\qquad P_z=2z=R_x,\qquad Q_z=2y=R_y. $$

All conditions hold, so the field is conservative on \(\mathbb{R}^3\). Integrate \(P\) with respect to \(x\), retaining an unknown function of \(y,z\):

$$ \phi(x,y,z)=\int(2xy+z^2)\,dx=x^2y+xz^2+g(y,z). $$

Differentiate with respect to \(y\) and match with \(Q\):

$$ \phi_y=x^2+g_y(y,z)=x^2+2yz, \qquad g_y(y,z)=2yz. $$

Integrating this equation with respect to \(y\), while allowing an undetermined function of \(z\), gives \(g(y,z)=y^2z+h(z)\). Thus

$$ \phi(x,y,z)=x^2y+xz^2+y^2z+h(z). $$

Now differentiate with respect to \(z\) and match with \(R\):

$$ \phi_z=2xz+y^2+h'(z)=2xz+y^2, \qquad h'(z)=0. $$

Hence \(h\) is constant, and one potential is \(\phi(x,y,z)=x^2y+xz^2+y^2z\). Check all components directly:

$$ \nabla\phi = \left\langle 2xy+z^2,\; x^2+2yz,\; 2xz+y^2 \right\rangle =\mathbf{F}(x,y,z). $$

The final gradient matches the given field exactly.

Why the Unknown Terms Matter

A common mistake is to integrate one component and then treat the result as the complete potential. In the first example, integrating \(P\) with respect to \(x\) gives \(x^3y+2xy\), but this is not yet the potential. Its \(y\)-derivative is \(x^3+2x\), which omits the \(4y^3\) term in \(Q\). The correction \(g(y)=y^4\) is essential.

Pitfall: After integrating with respect to one variable, do not add only a numerical constant. Add an arbitrary function of the variables held fixed during integration. Then use the remaining component equations to determine that function.

There is also a useful consistency check in the reverse direction. If matching a remaining component forces a function of \(y\) to depend on \(x\), or forces a function of \(z\) to depend on \(x\) or \(y\), the equations cannot be satisfied as written. A failed mixed-partial test detects this incompatibility before the integration work.

On a connected domain, any two potentials for the same field differ by an additive constant: their difference has zero gradient and is therefore constant. Thus finding one potential is enough; the arbitrary constant does not affect the field.

Check Your Understanding

For each question, use the compatibility test before attempting any integration.

  1. For \(\mathbf{F}(x,y)=\langle 2xy+1,\;x^2+3y^2\rangle\), do the mixed partials agree?
  2. When integrating \(P(x,y,z)\) with respect to \(x\), what variables may appear in the undetermined function?
  3. For a field \(\mathbf{F}=\langle P,Q,R\rangle\) on \(\mathbb{R}^3\), state the three mixed-partial conditions for conservativeness.
  4. If integration yields \(g_y(y,z)=2yz\), what is the general form of \(g(y,z)\)?
  5. Why does a failure of \(P_y=Q_x\) at even one point rule out a potential on all of \(\mathbb{R}^2\)?

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