M344: Calculus III
Section 16.3
Conservative Vector Fields¶
In our study of section 16.1, we concluded the notes with an example that asked the following question.
Given a vector field $\mathbf{F}$ defined on a domain $D$ in $\mathbb{R}^n$, is it possible to find a function $f$ defined on $D$ for which $\mathbf{F} = \nabla f$ at all points in $D$?
In the example in section 16.1, we answered this question for the specific vector field that we were given by brute force integration and inspection. We begin the notes for this section by discussing a theorem which answers our question for vector fields on $\mathbb{R}^2$.
Note: If you have not yet looked at Example 5 in the section 16.1 notes, or if you've forgotten it entirely, please go back and review it now.
Definition¶
A vector field $\mathbf{F}$ defined on a domain $D$ in $\mathbb{R}^n$ is said to be conservative (on $D$) if and only if there exsists a function $f :D \to \mathbb{R}$ satisfying
$$
\mathbf{F}(p) = \nabla f(p)
$$
for all points $p \in D$. A function $f$ that satisfies this property is called a potential function for the vector field $\mathbf{F}$ on $D$.
Note: A conservative vector field $\mathbf{F}$ does not have a unique potential function. Recall that the gradient of a function is a form of derivative. Thus, adding a constant to any potential function for $\mathbf{F}$ will yield another potential function. You're probably thinking, this reminds me a lot of anti-derivatives. You're right! A potential function of a conservative vector field acts just like an anti-derivative.
1. Example¶
Show that the function
$$
f(x,y,z) = \frac{mMG}{r},
$$
where $r = \sqrt{x^2 + y^2 + z^2\ }$ is a potential function for the gravitational vector field.
Video solution
Error Alert: At the end of this video, I lose a cube. Eventually, I will fix this, but for now please be aware and fix my mistake for me!
Theorem¶
Let $\mathbf{F}(x,y) = \langle P(x,y), Q(x,y)\rangle$ be a vector field defined on an open simply-connected region $D$, and suppose that $P$ and $Q$ have continuous first order partial derivatives on $D$. Then $\mathbf{F}$ is conservative on $D$ if and only if
$$
\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 0
$$
at every point in $D$.
We can prove one direction of this theorem right now. In fact, we can even weaken the assumptions on the region $D$ in doing so.
Partial proof
Suppose $\mathbf{F}(x,y) = \langle P(x,y), Q(x,y) \rangle$ is a conservative vector field defined on a domain $D \subseteq \mathbb{R}^2$, such that $P$ and $Q$ have continuous first-order partial derivatives on $D$. Then there exists a potential function $f$ defined on $D$ such that
$$ \frac{\partial f}{\partial x} = P, \quad \text{and}\quad \frac{\partial f}{\partial y} = Q. $$
Taking the mixed partials of each of these, we have
$$ \begin{align}{} \frac{\partial Q}{\partial x} & = \frac{\partial^2 f}{\partial x\,\partial y}, \quad\text{and}\\[2ex] \frac{\partial P}{\partial y} & = \frac{\partial^2 f}{\partial y\,\partial x}. \end{align} $$
Since $P$ and $Q$ are assumed to have continuous first-order partial derivatives on $D$, then Clairaut's theorem applies. This guarantees that the mixed second-order partial derivatives of $f$ are equal, whence
$$ \frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y}. $$
The result now follows immediately.
We'll wait until section 16.4 to prove the other direction of this theorem. By then, we'll have a better understanding of what the extra assumptions on the region $D$ mean, and we'll have some beautiful, high-powered machinery to deal with it.
For a vector field on $\mathbb{R}^3$, the situation is slightly more complicated. Writing the proof of the direction assuming that $\mathbf{F}$ is conservative is a recommended exercise. We'll discuss the other direction of the theorem in a later section.
Theorem¶
Let $\mathbf{F}(x,y,z) = \langle P(x,y,z), Q(x,y,z), R(x,y,z) \rangle$ be a vector field defined on all of $\mathbb{R}^3$, and suppose that $P$, $Q$, and $R$ have continuous first-order partial derivatives. Then $\mathbf{F}$ is conservative if and only if
$$
\begin{align}{}
\frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} & = 0, \\[2ex]
\frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} & = 0, \quad \text{and} \\[2ex]
\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} & = 0.
\end{align}
$$
Recommended Exercise
Assume that $\mathbf{F}$ is conservative on $D$ and its component functions have continuous first-order partial derivatives. Prove that the differences of the indicated partial derivatives must be $0$.
2. Example¶
Determine whether or not the vector field is conservative. If it is, find a potential function.
$$
\mathbf{F}(x,y,z) = \big\langle \sin y, x\cos y + \cos z,-y\sin z \big\rangle
$$
Video solution
The Fundamental Theorem for Path Integrals¶
Let $C$ be a path parametrized by a smooth vector function $\mathbf{r}$, $a \leq t \leq b$. Let $f$ be a differentiable function whose gradient vector field $\nabla f$ is continuous along $C$. Then
$$
\int_C \nabla f\cdot d\mathbf{r} = f\big(\mathbf{r}(b)\big) - f\big(\mathbf{r}(a)\big).
$$
Video proof
In the language of vector fields, the FTPI can be interpreted as follows.
Suppose $\mathbf{F}$ is a conservative vector field on a domain $D$, and $f$ is any potential function for $\mathbf{F}$ on $D$. Let $C$ be any path contained in $D$ with initial point $A$ and terminal point $B$. Then
$$
\int_C \mathbf{F}\cdot d\mathbf{r} = f(B) - f(A).
$$
3. Example¶
Show that the vector field is conservative, find a potential function, then evaluate the path integral
$$
\int_C \mathbf{F}\cdot d\mathbf{r},
$$
where $\mathbf{F}(x,y) = (1 + xy)e^{xy}\,\mathbf{i} + x^2e^{xy}\,\mathbf{j}$ and the path $C$ is parametrized by the vector function $\mathbf{r}(t) = \cos t\,\mathbf{i} + 2\sin t\, \mathbf{j}$ on the interval $0 \leq t \leq \frac{\pi}{2}$.
Video solution
Independence of Path¶
Suppose $\mathbf{F}$ is a continuous vector field on a region $D$, and $C_1$ and $C_2$ are two piecewise smooth paths in $D$ that have the same initial point $A$ and terminal point $B$. In general, $\int_{C_1} \mathbf{F}\cdot d\mathbf{r} \neq \int_{C_2} \mathbf{F}\cdot d\mathbf{r}$. However, if
$$
\int_{C_1} \mathbf{F}\cdot d\mathbf{r} = \int_{C_2} \mathbf{F}\cdot d\mathbf{r}
$$
for any two paths $C_1$ and $C_2$ in $D$ that share the same initial and terminal points, then the path integral $\int_C \mathbf{F}\cdot d\mathbf{r}$ is said to be independent of path in $D$.
Theorem¶
$\int_C\mathbf{F}\cdot d\mathbf{r}$ is independent of path in $D$ if and only if
$$
\int_C \mathbf{F}\cdot d\mathbf{r} = 0
$$
for every closed path in $D$.
Video proof
Theorem¶
Suppose $\mathbf{F}$ is a continuous vector field defined on an open connected region $D$. Then $\int_C \mathbf{F}\cdot d\mathbf{r}$ is independent of path in $D$ if and only if $\mathbf{F}$ is conservative on $D$.
Note: If $\mathbf{F}$ is conservative on $D$, then the Fundamental Theorem of Path Integrals implies that $\int_C \mathbf{F}\cdot d\mathbf{r}$ is independent of path in $D$. Write out the details.
The other direction requires us to construct the potential function for $\mathbf{F}$.
Video proof
Conservation of Energy¶
You've probably been wondering why these mathematical objects are named as they are. Unsurpringly, there is a close link with Physics and the Law of Conservation of Energy. Let's conclude these notes by investigating it!
4. Example¶
Newton's Second Law states that $\mathbf{F} = m\,\mathbf{a}$. Recalling that acceleration is $\mathbf{a} = \ddot{\mathbf{r}}$, where $\mathbf{r}$ represents the position of a particle at time $t$, we obtain a second order differential equation (or SODE).
$$
\mathbf{F}\big(\mathbf{r}(t)\big) = m\, \ddot{\mathbf{r}}(t).
$$
Let $C$ be the path of a particle parametrized by a vector function $\mathbf{r}$ on an interval $a \leq t \leq b$, and suppose $\mathbf{r}(a) = A$ and $\mathbf{r}(b) = B$. The kinetic energy of the particle is defined to be
$$
K(t) =\frac{1}{2} m\, \Vert \mathbf{v}(t) \Vert^2.
$$
Show that the work done by the vector field $\mathbf{F}$ on the particle is $W = K(B) - K(A)$.
Video solution
5. Example¶
Now suppose that $\mathbf{F}$ is a conservative vector field, and suppose $f$ is a potential function. The potential energy of the particle moving in the vector field is defined to be $P(x,y,z) = -f(x,y,z)$, so that $\mathbf{F} = -\nabla P$.
Show that the work done by the vector field $\mathbf{F}$ on the particle is $W = P(A) - P(B)$.
Video solution
Deduce that, for a particle moving in a conservative vector field, $P(A) + K(A) = P(B) + K(B)$. This is the Law of Conservation of Energy!
Discussion¶
Questions? You can ask them here.