Conservative Vector Fields

Recall from the Vector Fields page that if $z = f(x, y)$ is a two variable real-valued function, then the gradient of $f$, $\nabla f(x, y) = \frac{\partial f}{\partial x} \vec{i} + \frac{\partial f}{\partial y} \vec{j}$ defines a vector field on $\mathbb{R}^2$. Similarly, for a three variable real-valued function $w = f(x, y, z)$, the gradient of $f$, $\nabla f(x, y, z) = \frac{\partial f}{\partial x} \vec{i} + \frac{\partial f}{\partial y} \vec{j} + \frac{\partial f}{\partial z} \vec{k}$ defines a vector field on $\mathbb{R}^3$.

Now we have already seen that the gradient of a scalar field produces a vector field, however, there exists vector fields in which the gradient of any scalar field does not equal this vector field.

Definition: Let $\mathbf{F} = \mathbf{F}(x, y)$ be a vector field on $\mathbb{R}^2$. $\mathbf{F}$ is said to be a Conservative vector field on the domain $D$ if there exists a function $\phi$ such that the gradient of $\phi$ equals $\mathbf{F}$, that is $\mathbf{F} (x, y) = \nabla \phi (x, y)$. The function $\phi$ is called a Potential for $\mathbf{F}$ on $D$.

Conservative vector fields can be defined on higher dimensions as well in an analogous manner. Furthermore, we note that the potential cannot have any singular points in $D$ since the first partial derivatives of $\phi$ must exist for every point in $D$ to ensure that $\mathbf{F}(x, y) = \nabla \phi (x, y)$ for all points in $D$.

Now the following theorem will give us necessary conditions to ensure that a vector field on $\mathbb{R}^2$ is conservative.

Theorem 1: If the vector field $\mathbf{F}(x, y) = P(x, y) \vec{i} + Q(x, y) \vec{j}$ on $\mathbb{R}^2$ is conservative on the domain $D$, then $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$ for all points $(x, y) \in D$.

• Proof: Suppose that $\mathbf{F}(x, y) = P(x, y) \vec{i} + Q(x, y) \vec{j}$ is a conservative vector field on $\mathbb{R}^2$. Then there exist a potential $\phi (x, y)$ such that:

(1)

\begin{align} \quad \mathbf{F} (x, y) = \nabla \phi (x, y) \\ \quad P(x, y) \vec{i} + Q(x, y) \vec{j} = \frac{\partial \phi}{\partial x} \vec{i} + \frac{\partial \phi}{\partial u} \vec{j} \end{align}

• The equation above implies that:

(2)

\begin{align} \quad P(x, y) = \frac{\partial \phi}{\partial x} \quad \quad Q(x, y) = \frac{\partial \phi}{\partial y} \end{align}

• For the lefthand equation, take the partial derivatives of both sides with respect to $y$. For the righthand equation, take the partial derivatives of both sides with respect to $x$. We thus get the following equations:

(3)

\begin{align} \quad \frac{\partial P}{\partial y}= \frac{\partial^2 \phi}{\partial y \partial x} \quad \quad \frac{\partial Q}{\partial x} = \frac{\partial^2 \phi}{\partial x \partial y} \end{align}

• The mixed partial derivatives $\frac{\partial^2 \phi}{\partial y \partial x} = \frac{\partial^2 \phi}{\partial x \partial y}$ by Clairaut's Theorem on Higher Order Partial Derivatives and thus:

(4)

\begin{align} \quad \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \quad \blacksquare \end{align}

Theorem 2: If the vector field $\mathbf{F}(x, y, z) = P(x, y, z) \vec{i} + Q(x, y, z) \vec{j} + R(x, y, z) \vec{k}$ on $\mathbb{R}^3$ is conservative on the domain $D$, then $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$, $\frac{\partial P}{\partial z} = \frac{\partial R}{\partial x}$, and $\frac{\partial Q}{\partial z} = \frac{\partial R}{\partial y}$ for all points $(x, y, z) \in D$.

• Proof: Suppose that $\mathbf{F}(x, y, z) = P(x, y, z) \vec{i} + Q(x, y, z) \vec{j} + R(x, y, z) \vec{k}$ is a conservative vector field on $\mathbb{R}^3$. Then there exists a potential $\phi(x, y, z)$ such that:

(5)

\begin{align} \quad \mathbf{F}(x, y, z) = \nabla \phi (x, y, z) \\ \quad P(x, y, z) \vec{i} + Q(x, y, z) \vec{j} + R(x, y, z) \vec{k} = \frac{\partial \phi}{\partial x} \vec{i} + \frac{\partial \phi}{\partial y} \vec{j} + \frac{\partial \phi}{\partial z} \vec{j} \end{align}

• The equation above implies that:

(6)

\begin{align} \quad P(x, y, z) = \frac{\partial \phi}{\partial x} \quad \quad Q(x, y, z) = \frac{\partial \phi}{\partial y} \quad \quad R(x, y, z) = \frac{\partial \phi}{\partial z} \end{align}

• If we partial differentiate the first equation with respect to $y$ and the second equation with respect to $x$, then we have that $\frac{\partial P}{\partial y} = \frac{\partial^2 \phi}{\partial y \partial x} = \frac{\partial ^2 \phi}{\partial x \partial y} = \frac{\partial Q}{\partial x}$.

• If we partial differentiate the first equation with respect to $z$ and the third equation with respect to $z$, then we have that $\frac{\partial P}{\partial z} = \frac{\partial^2 \phi}{\partial z \partial x} = \frac{\partial^2 \phi}{\partial x \partial z} = \frac{\partial R}{\partial x}$.

• If we partial differentiate the second equation with respect to $z$ and the third equation with respect to $y$, then we have that $\frac{\partial Q}{\partial z} = \frac{\partial^2 \phi}{\partial z \partial y} = \frac{\partial^2 \phi}{\partial y \partial z} = \frac{\partial R}{\partial y}$.

• Thus Theorem 2 holds. $\blacksquare$

Remark 1: Theorem 1 and Theorem 2 above are NOT sufficient to guarantee that a vector field $\mathbf{F}$ on $\mathbb{R}^2$ or $\mathbb{R}^3$ is conservative, that is, the converses of these theorems are not necessarily true.

Let's look at an example of showing that a vector field is conservative.

Example 1

Determine whether the vector field $\mathbf{F}(x, y, z) = x \vec{i} - 2y \vec{j} + 3z \vec{k}$ is conservative or not.

We will first see if we can immediately determine if $\mathbf{F}$ is not conservative by seeing whether or not Theorem 2 holds. We calculate the necessary partial derivatives as follows:

(7)

\begin{align} \quad \frac{\partial P}{\partial y} = 0 = \frac{\partial Q}{\partial x} \\ \quad \frac{\partial P}{\partial z} = 0 = \frac{\partial R}{\partial x} \\ \quad \frac{\partial Q}{\partial z} = 0 = \frac{\partial R}{\partial y} \end{align}

So we suspect that $\mathbf{F}$ may be a conservative vector field. If so, then we have that for some functions $a, b, m, n, p,$ and $q$:

(8)

\begin{align} \quad P(x, y, z) = \frac{\partial \phi}{\partial x} \Leftrightarrow \phi (x, y, z) = \int x \: dx = \frac{x^2}{2} + a(y) + b(z) \\ \quad Q(x, y, z) = \frac{\partial \phi}{\partial y} \Leftrightarrow \phi (x, y, z) = \int -2y \: dy = -y^2 + m(x) + n(z) \\ \quad R(x, y, z) = \frac{\partial \phi}{\partial z} \Leftrightarrow \phi (x, y, z) = \int 3z \: dz = \frac{3z^2}{2} + p(x) + q(y) \end{align}

Thus we have that:

(9)

\begin{align} \quad \frac{x^2}{2} + a(y) + b(z) = m(x) -y^2 + n(z) = p(x) + q(y) + \frac{3z^2}{2} \end{align}

The equation above implies that $\frac{x^2}{2} = m(x) = p(x)$, $-y^2 = a(y) = q(y)$ and $\frac{3z^2}{2} = b(z) = n(z)$. For any constant $C$, a potential for $\mathbf{F}$ is:

(10)

\begin{align} \quad \phi(t) = \frac{x^2}{2} - y^2 + \frac{3z^2}{2} + C \end{align}