M344: Calculus III
Section 16.5
Div, Grad, Curl, and all that*¶
*I borrowed this title from this little text.
In this section, we begin a brief study of differential operators. These will provide the framework moving forward for us to reframe all that we've learned about multivariable calculus.
Differential Operators¶
For the sake of this course, we give the following definitions.
An operator is a function whose domain is a function space and codomain is a (possibly different) function space. A differential operator is an operator that uses derivatives to map a function from one function space to another.
1. Example¶
We have already seen one fundamental example of a differential operator. Suppose $f :D \to \mathbb{R}$ is a differentiable function from a domain $D \subseteq \mathbb{R}^3$. The gradient field of $f$ is the vector field
$$
\nabla f = \left\langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right
\rangle.
$$
Noticing that the $f$ shows up systematically on both sides of this equation, we may write
$$
\nabla f = \left\langle \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right
\rangle f.
$$
Allowing $f$ to vary in the space of all differentiable functions on $D$, we may define the gradient operator
$$
\nabla = \left\langle \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right
\rangle.
$$
The symbol $\nabla$ is called "del," and we frequently refer to $\nabla f$ as "del- $f$."
Curl and Divergence¶
The $\nabla$ operator can be used to define two differential operators on vector fields.
Let $\mathbf{F} = \langle P, Q, R \rangle$ be a vector field defined on a domain $D \subseteq \mathbb{R}^3$ for which the partial derivatives of $P$, $Q$, and $R$ all exist.
The curl of $\mathbf{F}$ is the vector field on $D$ defined by
$$
\text{curl}\,\mathbf{F} = \left\langle \left(\dfrac{\partial R}{\partial y} - \dfrac{\partial Q}{\partial z}\right), \left(\dfrac{\partial P}{\partial z} - \dfrac{\partial R}{\partial x}\right), \left(\dfrac{\partial Q}{\partial x} - \dfrac{\partial P}{\partial y}\right) \right\rangle.
$$
2. Exercise¶
The curl of a vector field can be represented formally by a cross product.
$$
\text{curl}\,\mathbf{F} = \nabla \times \mathbf{F} = \left\vert \begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z} \\
P & Q & R \end{array} \right\vert
$$
Work out the details to verify this claim.
The divergence of $\mathbf{F}$ is the function defined by
$$
\text{div}\,\mathbf{F} = \dfrac{\partial P}{\partial x} + \dfrac{\partial Q}{\partial y} + \dfrac{\partial R}{\partial z}.
$$
3. Exercise¶
The divergence of a vector field $\mathbf{F}$ can also be represented by a formal dot product with the gradient operator. Work out the details to show that
$$
\text{div}\,\mathbf{F} = \nabla \cdot \mathbf{F}.
$$
4. Example¶
Compute the curl and divergence of the vector field
$$
\mathbf{F}(x,y,z) = \Big\langle \ln\big(2y+3z\big), \ln\big(x+3z\big), \ln\big(x+2y\big) \Big\rangle.
$$
Video solution
5. Theorem¶
If $f$ is a function of three variables that has continuous second-order partial derivatives, then
$$
\text{curl}\,\nabla f = \mathbf{0}.
$$
Video proof
6. Theorem¶
The contrapositive of the preceding theorem gives us a way to check if a given vector field is conservative or not. Recall that if $\mathbf{F}$ is conservative, then $\mathbf{F}$ has a potential function $f$ satisfying
$$
\mathbf{F}(x,y,z) = \nabla f\,(x,y,z).
$$
Therefore, the preceding theorem says that if $\mathbf{F}$ is conservative on a domain $D$, then $\text{curl}\,\mathbf{F} = \mathbf{0}$. The contrapositive of this statement says
If $\text{curl}\,\mathbf{F} \neq \mathbf{0}$, then $\mathbf{F}$ is not conservative on $D$.
The converse of this theorem is not true in general, but it is true if $D = \mathbb{R}^3$.
We stated this theorem in section 16.3. We restate it now.
If $\mathbf{F}$ is a vector field defined on all of $\mathbb{R}^3$ whose component functions have continuous partial derivatives and $\text{curl}\,\mathbf{F} = \mathbf{0}$, then $\mathbf{F}$ is a conservative vector field on $\mathbb{R}^3$.
7. Example¶
Show that the vector field is conservative and find a potential function.
$$
\mathbf{F}(x,y,z) = \big\langle y^2z^3,2xyz^3,3xy^2z^2 \big\rangle
$$
Video solution
8. Exercise¶
Let $\mathbf{F} = \langle P,Q,R \rangle$ be a vector field on $\mathbb{R}^3$ such that the component functions $P$, $Q$, and $R$ have continuous second-order partial derivatives. Show that
$$
\text{div}\,\big(\text{curl}\,\mathbf{F}\big) = 0.
$$
Video solution
9. Example¶
Is there a vector field $\mathbf{G}$ on $\mathbb{R}^3$ such that
$$
\text{curl}\, \mathbf{G} = \big\langle x\sin y, \cos y, z-xy \big\rangle?
$$
Explain.
Video solution
Physical Interpretations¶
Next, we briefly consider the physcial and geometric interpretations of the curl and divergence of a vector field. For both examples, it is useful to think of the vector field as the velocity field of a fluid flow.
Curl¶
The curl of a vector field $\mathbf{F}$ measures rotations induced by the vector field. If the vector field $\mathbf{F}$ represents the velocity of a fluid flow, then the curl measures how much a small paddle wheel rotates as it moves in the fluid. If the curl of a velocity field $\mathbf{F}$ is $\mathbf{0}$ at a point, then it means that there is no vortex or eddy at the point. A fluid flow that has $\text{curl}\,\mathbf{F} = \mathbf{0}$ at every point is called irrotational.
Divergence¶
We can again understand the divergence of a vector field by regarding the vector field $\mathbf{F}$ as the velocity field of a fluid flow. The divergence of $\mathbf{F}$ represents the net change in the mass of a small parcel of fluid per unit of volume. A vector field that obeys $\text{div}\,\mathbf{F} = 0$ is said to be divergence free. If the vector field represents the velocity field of a fluid flow and $\text{div}\,\mathbf{F} = 0$, then the flow is said to be incompressible.
The Laplace Operator¶
Let $f$ be a function defined on a domian $D \subseteq \mathbb{R}^3$ whose second partial derivatives are all continuous. The Laplacian of $f$ is the function defined by
$$
\Delta f := \nabla \cdot \nabla f = \text{div}\,\big( \text{grad}\, f\big) = \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} + \frac{\partial^2 f}{\partial z^2}.
$$
Some engineering and physics texts denote the Laplacian as $\Delta f = \nabla^2 f$. I personally find this to be unnecessarily misleading, as there is no well-defined way to square a vector operator. But, lo, we must live with it.
Vector forms of Green's theorem¶
Let $\mathbf{F} = \langle P,Q \rangle$ be a vector field on a Jordan region $D \subseteq \mathbb{R}^2$ for which the component functions $P$ and $Q$ satisfy the hypotheses of Green's theorem.
The curl of $\mathbf{F}$ may be computed by regarding the vector field $\mathbf{F}$ in $\mathbb{R}^3$ as $\mathbf{F} = \langle P,Q,0 \rangle$ and then computing it the normal way.
Exercise: Show that $\text{curl}\,\mathbf{F} = \big\langle 0, 0, \partial_xQ - \partial_yP\big\rangle$.
Recalling that $\mathbf{k} = \langle 0,0,1 \rangle$ is the unit vector in the direction of the positive $z$-axis, show that Green's theorem may be written as
$$
\int_{\partial D} \mathbf{F}\cdot d\mathbf{r} = \int\!\!\!\!\int_D \big(\text{curl}\,\mathbf{F}\big)\cdot\mathbf{k}\, dA.
$$
Video lecture
Suppose $\partial D = C$ is parametrized by a smooth vector function $\mathbf{r}(t) = \big\langle x(t),y(t)\big\rangle$, $a\leq t\leq b$. Assume this parametrization traverses $C$ in the positive orientation. Recall that the unit tangent vector field along $C$ is given by
$$
\mathbf{T}(t) = \frac{\dot{\mathbf{r}}(t)}{\Vert\dot{\mathbf{r}}(t)\Vert} = \left\langle \frac{\dot{x}(t)}{\Vert\dot{\mathbf{r}}(t)\Vert},\frac{\dot{y}(t)}{\Vert\dot{\mathbf{r}}(t)\Vert}\right\rangle.
$$
The outward unit normal vector field along $C$ is then given by
$$
\mathbf{n}(t) = \left\langle \frac{\dot{y}(t)}{\Vert\dot{\mathbf{r}}(t)\Vert},\frac{-\dot{x}(t)}{\Vert\dot{\mathbf{r}}(t)\Vert}\right\rangle.
$$
Green's theorem can be formulated with respect to this outward normal vector field as
$$
\int_{\partial D} \mathbf{F}\cdot d\mathbf{n} = \int\!\!\!\!\int_D \text{div}\,\mathbf{F}\,dA,
$$
where $d\mathbf{n} = \mathbf{n}\,ds$.
Video lecture
Green's Identities¶
Suppose the hypotheses of Green's theorem are satisfied for all curves, regions, and vector fields under consideration, and suppose the appropriate partial derivatives of the functions $f$ and $g$ exist and are continuous. Then the following identities hold.
Green's first identity¶
$$
\int\!\!\!\!\int_D f\,\Delta g\, dA = \int_{\partial D} f\, \nabla g \cdot d\mathbf{n} - \int\!\!\!\!\int_D \nabla f \cdot \nabla g\, dA.
$$
Green's Second Identity¶
$$
\int\!\!\!\!\int_D \Big( f\Delta g - g\Delta f\Big)\, dA = \int_{\partial D} \Big(f\,\nabla g - g\,\nabla f\Big)\cdot d\mathbf{n}.
$$
The derivations of these formulas are left as a strongly recommended exercise. The vector forms of Green's theorem can be used together with integration by parts techniques to derive the first idenity. Once proved, the first identity may be used to prove the second identity.
Discussion¶
Questions? You can ask them here.