Finding Potential Functions

Two fields. One has a hidden hill behind it.

Here are two plane vector fields. They look roughly the same - swirly arrows, no obvious peak. But one of them is secretly the gradient of some scalar function $f$, and the other isn't. If we could recover that $f$, we'd know the work along any curve just from the endpoints. So: which is which, and how do we tell?

Field A

$\mathbf{u} = 2xy\,\mathbf{i} + (x^2 + 1)\,\mathbf{j}$

Field B

$\mathbf{u} = -y\,\mathbf{i} + x\,\mathbf{j}$

Given a vector field, can we always recover a scalar function it came from? And if we can, how do we actually find it?

Step 0 - the two-step diagnostic

Before we integrate anything, we should ask whether a potential is even possible. There's a quick local test, and a global one. Both have to pass.

Step 0a - check the curl

A potential exists only if $\nabla \times \mathbf{u} = \mathbf{0}$. In two dimensions that's a single equation; in three dimensions it's three.

$$\text{2D:}\quad \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 0.$$ $$\text{3D for } \mathbf{u} = L\,\mathbf{i} + M\,\mathbf{j} + N\,\mathbf{k}: \quad \frac{\partial M}{\partial z} = \frac{\partial N}{\partial y}, \quad \frac{\partial N}{\partial x} = \frac{\partial L}{\partial z}, \quad \frac{\partial L}{\partial y} = \frac{\partial M}{\partial x}.$$

If any cross-partial fails, stop. There is no $f$ - the integration we're about to set up is doomed, and forging ahead just produces inconsistent equations.

Step 0b - check the domain

Curl-zero is necessary but not quite sufficient. The four theorems of §5.6 tell us what's actually needed: a curl-zero field on a simply connected domain has a global potential. On a domain with holes, we may only get a local one.

If $D$ is simply connected (no holes) and curl is zero, you'll find a single $f$ that works everywhere on $D$.
If $D$ has a hole - the punctured plane, the $z$-axis removed from $\mathbb{R}^3$, the slit annulus - curl-zero gives only a local potential. The global integral may be multi-valued. See §5.7 for how to handle that systematically.

For the rest of this page we'll assume Step 0 has passed: curl is zero, domain is simply connected. Now we go hunting for $f$.

Step 1 Integrate $P$ in $x$

We want $\nabla f = \mathbf{u}$, which means $f_x = P$, $f_y = Q$, $f_z = R$ all simultaneously. Pick the $x$-equation first and integrate it, treating $y$ and $z$ as parameters.

$$f(x, y, z) = \int P(x, y, z)\,dx + g(y, z).$$

The "$+\,g$" is the constant of integration with respect to $x$ - but here's the thing.

The integration "constant" is a function of the uneliminated variables, not just a single number $C$. In 2D it's $g(y)$; in 3D it's $g(y, z)$. Forgetting to keep $g$ as a function of the remaining variables is the most common mistake at this step - and it silently kills the recipe, because we'll need $g_y$ and $g_z$ to be nonzero in the next steps to absorb $Q$ and $R$.

So at the end of Step 1 we have a candidate $f$ that's correct in $x$ by construction, but with an unknown function $g(y, z)$ riding along. The remaining equations $f_y = Q$ and $f_z = R$ will pin $g$ down.

Step 2 Match $Q$ to determine $g_y$

Differentiate the candidate $f$ from Step 1 with respect to $y$:

$$f_y \;=\; \frac{\partial}{\partial y} \int P(x, y, z)\,dx \;+\; g_y(y, z).$$

We need $f_y = Q$. Set the two expressions equal and solve for $g_y$:

$$g_y(y, z) \;=\; Q(x, y, z) \;-\; \frac{\partial}{\partial y} \int P(x, y, z)\,dx.$$

Sanity check. Look at the right-hand side. It seems to depend on $x$. But $g_y$ should be a function of $y$ and $z$ only - $x$ has no business being there. So every $x$ on the right must cancel.

Why does it cancel? Because the cross-partial condition $\partial L/\partial y = \partial M/\partial x$ from Step 0 guarantees it. If you do the algebra and $x$ doesn't cancel, two things are possible: an arithmetic slip somewhere, or the field wasn't conservative to begin with (Step 0 was botched). Either way, stop and recheck.

Once $g_y$ is in hand, integrate in $y$:

$$g(y, z) \;=\; \int g_y(y, z)\,dy \;+\; h(z).$$

Same trap as before: the new integration "constant" $h$ is a function of the one variable we still haven't pinned down, namely $z$. In 2D the recipe ends here ($h$ collapses to a literal constant $C$). In 3D we have one more equation to use.

Step 3 Match $R$ to determine $h'(z)$ (3D only)

We now have a candidate $f(x, y, z)$ that's correct in $x$ and $y$, with one loose end: an unknown $h(z)$. Differentiate in $z$ and demand $f_z = R$:

$$h'(z) \;=\; R(x, y, z) \;-\; \frac{\partial}{\partial z}\!\left[\int P\,dx + \int g_y\,dy\right].$$

Same sanity check as Step 2. The right-hand side looks like it could depend on $x$ and $y$, but it can't - $h$ is a function of $z$ alone. The other two curl-zero conditions, $\partial N/\partial x = \partial L/\partial z$ and $\partial M/\partial z = \partial N/\partial y$, are exactly what guarantees the $x$ and $y$ dependence cancels. If they don't cancel, the field isn't conservative or the algebra is off.

Integrate to get $h(z) + C$, where now $C$ is a genuine constant. Plug everything back in to assemble the final $f$.

Step 4 - always recompute $\nabla f$

The recipe has three integrations and three differentiations woven together. There are a lot of places to drop a sign or miscarry a $2$. Every time we finish, we go back and verify - not by re-reading the work, but by computing $\nabla f$ from scratch and comparing with $\mathbf{u}$.

$$\text{Take final } f. \quad \text{Compute } f_x, f_y, f_z. \quad \text{Confirm } (f_x, f_y, f_z) = (P, Q, R).$$

This is cheap (three differentiations, no integration) and catches just about every mistake the recipe is prone to: dropped terms, sign errors, forgotten coefficients. Don't skip it. We'll do it on the worked example next.

Recipe at a glance

Step 0. Check curl is zero and the domain is simply connected.
Step 1. $f = \int P\,dx + g(y, z)$.
Step 2. Set $f_y = Q$, solve for $g_y$, integrate to get $g(y, z) = \int g_y\,dy + h(z)$.
Step 3 (3D). Set $f_z = R$, solve for $h'(z)$, integrate to get $h(z) + C$.
Step 4. Verify $\nabla f = \mathbf{u}$ component-by-component.

Worked example - a 3D field

Let's run the whole recipe on a clean 3D example. Take

$$\mathbf{u} \;=\; (2xy - z^2)\,\mathbf{i} \;+\; (x^2 + 2z)\,\mathbf{j} \;+\; (2y - 2xz)\,\mathbf{k}.$$

So $L = 2xy - z^2$, $M = x^2 + 2z$, $N = 2y - 2xz$. We're hunting for $f$ with $\nabla f = \mathbf{u}$.

Step 0 Curl check

Compute the three cross-partials.

$$\frac{\partial M}{\partial z} = 2, \quad \frac{\partial N}{\partial y} = 2 \;\;\checkmark$$ $$\frac{\partial N}{\partial x} = -2z, \quad \frac{\partial L}{\partial z} = -2z \;\;\checkmark$$ $$\frac{\partial L}{\partial y} = 2x, \quad \frac{\partial M}{\partial x} = 2x \;\;\checkmark$$

All three match. Curl is zero on $\mathbb{R}^3$, which is simply connected, so a global potential exists.

Step 1 Integrate $L$ in $x$

$$f(x, y, z) \;=\; \int (2xy - z^2)\,dx + g(y, z) \;=\; x^2 y - x z^2 + g(y, z).$$

Step 2 Match $M$, find $g_y$

Differentiate in $y$:

$$f_y \;=\; x^2 + g_y(y, z).$$

We need $f_y = M = x^2 + 2z$. So

$$g_y(y, z) \;=\; (x^2 + 2z) - x^2 \;=\; 2z.$$

Sanity check: $g_y$ has no $x$ in it. Good. Integrate in $y$:

$$g(y, z) \;=\; \int 2z\,dy + h(z) \;=\; 2yz + h(z).$$

So our updated candidate is

$$f(x, y, z) \;=\; x^2 y - x z^2 + 2yz + h(z).$$

Step 3 Match $N$, find $h'(z)$

Differentiate in $z$:

$$f_z \;=\; -2xz + 2y + h'(z).$$

We need $f_z = N = 2y - 2xz$. So

$$h'(z) \;=\; (2y - 2xz) - (-2xz + 2y) \;=\; 0.$$

Sanity check: $h'(z)$ has no $x$ or $y$ - in fact it's identically zero, which is allowed. Integrating gives $h(z) = C$, a genuine constant.

Final answer

$$\boxed{\,f(x, y, z) \;=\; x^2 y - x z^2 + 2yz + C.\,}$$

Step 4 Verify $\nabla f = \mathbf{u}$

Compute each partial from scratch and compare to the original components $L, M, N$.

$f_x = 2xy - z^2$vs.$L = 2xy - z^2$✓

$f_y = x^2 + 2z$vs.$M = x^2 + 2z$✓

$f_z = -2xz + 2y$vs.$N = 2y - 2xz$✓

All three components match. The potential is correct.

Edge case - curl zero, but no global potential

The recipe assumes Step 0b passes: the domain is simply connected. What if it isn't? The classic example is the planar vortex

$$\mathbf{u} \;=\; \frac{-y}{x^2 + y^2}\,\mathbf{i} + \frac{x}{x^2 + y^2}\,\mathbf{j}, \qquad D = \mathbb{R}^2 \setminus \{(0,0)\}.$$

A direct computation gives $Q_x - P_y = 0$ everywhere $\mathbf{u}$ is defined - so locally the recipe should work. But $D$ has a hole at the origin; it isn't simply connected. And indeed: the line integral of $\mathbf{u}$ around the unit circle gives $2\pi$, not zero. So no single-valued $f$ exists on all of $D$.

What still works: pick any simply connected sub-region of $D$ (a slit plane $\{x > 0\}$, or any open disc not containing the origin), and run the recipe there. You'll recover $f = \arctan(y/x) + C$ on the slit plane, $f = \theta$ in polar terms - exactly the angle function. Globally on $D$, $\theta$ is multi-valued: it jumps by $2\pi$ each time you loop the origin.

Multi-valued potentials are a real, useful object - not a failure mode. See §5.7 for how to use them: pick a system of "fundamental periods" (one per hole), and any line integral on $D$ becomes (single-valued part of $f$) + (integer combination of periods).

Practice Problems - §3.6 / §5.6

5(a) Find all $f$ with $\nabla f = \mathbf{v}$ for $\mathbf{v} = 2xyz\,\mathbf{i} + x^2 z\,\mathbf{j} + x^2 y\,\mathbf{k}$. (Kaplan §3.6)

For $\mathbf{v} = 2xyz\,\mathbf{i} + x^2 z\,\mathbf{j} + x^2 y\,\mathbf{k}$, verify that $\operatorname{curl} \mathbf{v} = \mathbf{0}$ and find every scalar function $f$ such that $\nabla f = \mathbf{v}$.

Step 0 - curl check. $\partial M/\partial z = x^2$, $\partial N/\partial y = x^2$ ✓. $\partial N/\partial x = 2xy$, $\partial L/\partial z = 2xy$ ✓. $\partial L/\partial y = 2xz$, $\partial M/\partial x = 2xz$ ✓. Curl is zero on $\mathbb{R}^3$, which is simply connected.

Step 1. $f = \int 2xyz\,dx + g(y, z) = x^2 yz + g(y, z)$.

Step 2. $f_y = x^2 z + g_y$. Set $= x^2 z$. So $g_y = 0$, hence $g(y, z) = h(z)$.

Step 3. $f_z = x^2 y + h'(z)$. Set $= x^2 y$. So $h'(z) = 0$, hence $h(z) = C$.

Final. $f = x^2 yz + C$. Verify: $f_x = 2xyz$ ✓, $f_y = x^2 z$ ✓, $f_z = x^2 y$ ✓.

5(b) Find all $f$ with $\nabla f = \mathbf{v}$ for $\mathbf{v} = e^{xy}\big[(2y^2 + yz^2)\mathbf{i} + (2xy + xz^2 + 2)\mathbf{j} + 2z\,\mathbf{k}\big]$. (Kaplan §3.6)

For $\mathbf{v} = e^{xy}\big[(2y^2 + yz^2)\mathbf{i} + (2xy + xz^2 + 2)\mathbf{j} + 2z\,\mathbf{k}\big]$ (Kaplan writes the $\mathbf{k}$ component without the $e^{xy}$ factor; we keep his convention $\mathbf{v} = e^{xy}[(\cdots)\mathbf{i} + (\cdots)\mathbf{j} + 2z\,\mathbf{k}]$ as printed), verify $\operatorname{curl} \mathbf{v} = \mathbf{0}$ and find all $f$ with $\nabla f = \mathbf{v}$.

Step 0. The cross-partials all match (long but mechanical - distribute $e^{xy}$ and use $\partial e^{xy}/\partial x = y e^{xy}$, $\partial e^{xy}/\partial y = x e^{xy}$). Domain is $\mathbb{R}^3$, simply connected.

Step 1. Try the ansatz $f = e^{xy} \cdot \phi(x, y, z) + g(y, z)$. Computing $f_x$ should give $e^{xy}(2y^2 + yz^2)$, which suggests $\phi = y \cdot \text{something}$. A direct integration: $\int e^{xy}(2y^2 + yz^2)\,dx = e^{xy}(2y + z^2)$ (since $y$ is constant in $x$, $\int e^{xy}\,dx = e^{xy}/y$, multiplied by $y(2y + z^2)$). So $f = e^{xy}(2y + z^2) + g(y, z)$.

Step 2. $f_y = x e^{xy}(2y + z^2) + e^{xy} \cdot 2 + g_y = e^{xy}(2xy + xz^2 + 2) + g_y$. Set $= e^{xy}(2xy + xz^2 + 2)$, so $g_y = 0$, hence $g(y, z) = h(z)$.

Step 3. $f_z = e^{xy} \cdot 2z + h'(z) = 2z e^{xy} + h'(z)$. Set $= 2z e^{xy}$ (the $\mathbf{k}$ component, with the $e^{xy}$ factor restored). So $h'(z) = 0$, hence $h(z) = C$.

Final. $f = e^{xy}(2y + z^2) + C$. Verify by computing $f_x, f_y, f_z$ and checking against the components of $\mathbf{v}$.

1(a) Determine $F(x, y)$ with $dF = 2xy\,dx + x^2\,dy$, then evaluate $\int_C dF$ along $C: y = x^3$ from $(0,0)$ to $(2, 8)$. (Kaplan after §5.6)

Determine by inspection a function $F(x, y)$ whose differential is $2xy\,dx + x^2\,dy$, and integrate $\displaystyle \int_C 2xy\,dx + x^2\,dy$ along $C: y = x^3$ from $(0, 0)$ to $(2, 8)$.

Step 0. $P = 2xy$, $Q = x^2$. $\partial Q/\partial x = 2x$, $\partial P/\partial y = 2x$ ✓. The plane is simply connected.

Step 1. $F = \int 2xy\,dx + g(y) = x^2 y + g(y)$.

Step 2. $F_y = x^2 + g'(y)$. Set $= x^2$, so $g'(y) = 0$ and $g(y) = C$. So $F(x, y) = x^2 y + C$. (This is what "by inspection" means - both terms come from differentiating $x^2 y$.)

Evaluate the line integral. Since the integrand is $dF$ and the field is conservative, $\int_C dF = F(2, 8) - F(0, 0) = 4 \cdot 8 - 0 = 32$. The path $y = x^3$ never enters the calculation - that's the payoff of finding the potential.

6(a) Show $\displaystyle \int_{(0,0)}^{(1,1)} 2xy\,dx + (x^2 - y^2)\,dy$ is path independent and evaluate. (Kaplan after §5.6)

Show that $\displaystyle \int_{(0,0)}^{(1,1)} 2xy\,dx + (x^2 - y^2)\,dy$ is independent of path in the $xy$-plane and evaluate it.

Step 0. $P = 2xy$, $Q = x^2 - y^2$. $\partial Q/\partial x = 2x$, $\partial P/\partial y = 2x$ ✓. The plane is simply connected, so a global potential exists and the integral is path independent.

Step 1. $F = \int 2xy\,dx + g(y) = x^2 y + g(y)$.

Step 2. $F_y = x^2 + g'(y)$. Set $= x^2 - y^2$, so $g'(y) = -y^2$ and $g(y) = -y^3/3 + C$. Hence $F(x, y) = x^2 y - y^3/3 + C$.

Evaluate. $\int_{(0,0)}^{(1,1)} dF = F(1, 1) - F(0, 0) = (1 - 1/3) - 0 = 2/3$.

Verify. $F_x = 2xy$ ✓, $F_y = x^2 - y^2$ ✓.