vector analysis
I'm trying to do some practice for an electromagnetism course, and am trying to calculate the grad, curl and div of a vector $A = (2xy, 3zx, yx^2)$
I know:
Div = $\nabla . A$
Curl = $\nabla \times A$
Grad = $\nabla A$
I have worked out the first two, but I seem to be having a mind blank and I'm getting myself all confused with the Grad component if someone could clarify this?
Also if someone could explain when I might need to use the Laplace Operator? What is it used for?
I would suggest Vector Calculus by Marsden and Tromba. The preface says,
"We require only the bare rudiments of matrix algebra, and the necessary concepts are developed in the text. If this course is preceded by a course in linear algebra, the instructor will have no difficulty enhancing the material. However, we do assume a knowledge of the fundamentals of one-variable calculus $-$ the process of differentiation and integration and their geometric and physical meaning as well as a knowledge of the standard functions, such as the trigonometric and exponential functions."
Let's formulate the definition of curl slightly more precisely in the form of a definition/theorem. I'll also not use boldface objects, simply for ease of typing
Definition/Theorem.
Let $A\subset \Bbb{R}^3$ be open, $F: A \to \Bbb{R}^3$ be $C^1$. Then, there is a unique continuous function $H:A \to \Bbb{R}^3$, such that for every $p\in A$, for every $\epsilon > 0$, for every "nice" surface $S\subset A$, there is an open neighbourhood $U$ of $p$ in $S$ such that for every "nice" oriented surface $M$ with $p\in M\subset U$, with outward normal vector field $n(\cdot)$ (which is simply the restriction of the outward normal of $S$ to $M$), boundary $\partial M$ and tangent vector field $\tau(\cdot)$ on $\partial M$, we have: \begin{align} \left|\dfrac{1}{|M|}\int_{\partial M}\langle F, \tau\rangle\, dl - \langle H(p), n(p)\rangle\right| < \epsilon\tag{1} \end{align} (this is the more precise meaning of the limit you're talking about) In this case, because $H$ is unique, we can give it the name $\text{curl}(F)$. In fact, we can show that \begin{align} H = \left(\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}\right) \boldsymbol{\hat\imath} + \left(\frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} \right) \boldsymbol{\hat\jmath} + \left(\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \right) \boldsymbol{\hat k} \tag{2} \end{align}
In the above paragraph, "nice oriented surface" means nice enough so that Stoke's theorem can be applied; for example a smooth two-dimensional, oriented manifold-with-boundary (or however much you wanna weaken the hypothesis... because each book presents it with varying levels of generality... just as long as Stoke's theorem can be applied).
Note that for the above definition of curl to make sense, we have to first show the existence and uniqueness of such a vector field $H$. We shall start by showing the uniqueness of $H$. So, we assume such a $H$ exists, and then prove its components are determined according to the formula $(2)$; this will complete the proof of uniqueness.
Proof of Uniqueness
I'll carry out the computation in detail for proving $H_x = \dfrac{\partial F_z}{\partial y} - \dfrac{\partial F_y}{\partial z}$, and leave the other two to you (it's simply a matter of renaming $x,y,z$). We prove this equality pointwise of course. So, fix a point $p \in A$; then for any $\delta > 0$ such that the closed cube $C_{p,\delta} = p + [-\delta,\delta]^3$ (which is the closed cube centered at $p$ of sidelength $2\delta$) lies entirely inside $A$ (note that since $A$ is open, there are infinitely many such $\delta>0$), we define $M^{\delta} := \{p_1\}\times [p_2-\delta, p_2 + \delta]\times [p_3-\delta, p_3 + \delta]$. This is a piece of a plane which we shall orient so that it has a constant outward normal vector field $n = e_1 \equiv \boldsymbol{i}$. Now, we calculate: note that $\partial M^{\delta}$ has $4$-pieces, and the unit tangent vector along these boundary paths is constant, so (if you're very careful with signs... which I hope I didn't make any sign mistakes), we get \begin{align} \dfrac{1}{|M^{\delta}|} \int_{\partial M^{\delta}}\langle F, \tau \rangle\, dl &= \dfrac{1}{4\delta^2} \bigg[\int_{-\delta}^{\delta} F_2(p_1, p_2 + y, p_3-\delta) - F_2(p_1, p_2 + y, p_3+\delta) \, dy\bigg]\\ &+\dfrac{1}{4\delta^2}\bigg[ \int_{-\delta}^{\delta} F_3(p_1, p_2+\delta, p_3+z) - F_3(p_1, p_2-\delta, p_3+z)\, dz\bigg] \end{align} For each term, we apply the mean-value theorem for integrals (which we can use since everything is continuous), to obtain some $\eta \in [p_2-\delta, p_2+\delta]$ and some $\zeta\in [p_3-\delta, p_3+\delta]$ such that
\begin{align} \dfrac{1}{|M^{\delta}|} \int_{\partial M^{\delta}}\langle F, \tau \rangle\, dl &= \dfrac{1}{4\delta^2}\bigg[2\delta \cdot F_2(p_1, \eta, p_3-\delta) - 2\delta \cdot F_2(p_1, \eta, p_3+\delta)\bigg] \\ &+ \dfrac{1}{4\delta^2}\bigg[2\delta \cdot F_3(p_1, p_2+\delta, \zeta) - 2\delta \cdot F_3(p_1, p_2-\delta, \zeta)\bigg] \\\\ &=\dfrac{F_3(p_1, p_2+\delta, \zeta) - F_3(p_1, p_2-\delta, \zeta)}{2\delta} -\dfrac{F_2(p_1, \eta, p_3+\delta) - F_2(p_1, \eta, p_3-\delta)}{2\delta} \\ &= \dfrac{\partial F_3}{\partial y}(p_1, \alpha, \zeta) - \dfrac{\partial F_2}{\partial z}(p_1, \eta, \beta), \end{align} for some $\alpha\in [p_2-\delta, p_2+\delta], \beta\in [p_3-\delta, p_3+\delta]$, using the mean-value theorem for derivatives (which can certainly be applied since we assumed $F$ is $C^1$).
Quick summary: what we showed so far is that for every $p\in A$ and every $\delta>0$ such that the cube $C_{p,\delta}$ lies inside $A$, if we define $M^{\delta}$ as above to be the plane centered at $p$ with normal pointing in $e_1$ direction, then, there exist points $a_{p,\delta},b_{p,\delta} \in M^{\delta} \subset C_{p,\delta}$ inside the surface (in particular inside the cube) such that
\begin{align} \dfrac{1}{|M^{\delta}|}\int_{\partial M^{\delta}}\langle F, \tau\rangle \, dl &= \dfrac{\partial F_3}{\partial y}(a_{p,\delta}) - \dfrac{\partial F_2}{\partial z}(b_{p,\delta}) \end{align} From here, it is a simple matter of using the continuity of partial derivatives. Here's the full $\epsilon,\delta$ argument to finish it off: let $p\in A$ and $\epsilon> 0$ be arbitrary. By our hypothesis of $(1)$, there is an open $U$ such that blablabla. Now, for this given $\epsilon > 0$, let's choose $\delta > 0$ small enough so that
(so really we have to take a minimum of several $\delta$'s). Then, we choose the oriented plane $M^{\delta}$ as defined above (this plane lies inside $U$ by construction, because of how small $\delta$ is). \begin{align} \left|\left(\dfrac{\partial F_3}{\partial y}(p) - \dfrac{\partial F_2}{\partial z}(p)\right) - H_1(p)\right| &= \left|\left(\dfrac{\partial F_3}{\partial y}(p) - \dfrac{\partial F_2}{\partial z}(p)\right) - \langle H(p), n(p)\rangle\right| \\\\ &\leq \left|\dfrac{\partial F_3}{\partial y}(p) - \dfrac{\partial F_3}{\partial y}(a_{p,\delta})\right| + \left|-\dfrac{\partial F_2}{\partial z}(p) + \dfrac{\partial F_2}{\partial z}(b_{p,\delta})\right|\\ &+ \left|\left(\dfrac{\partial F_3}{\partial y}(a_{p,\delta}) - \dfrac{\partial F_2}{\partial z}(b_{p,\delta})\right) - \dfrac{1}{|M^{\delta}|}\int_{\partial M^{\delta}}\langle F, \tau\rangle\, dl \right| \\ &+ \left|\dfrac{1}{|M^{\delta}|}\int_{\partial M^{\delta}}\langle F, \tau\rangle\, dl - \langle H(p), n(p) \rangle \right| \\\\ &\leq 4\epsilon \end{align} (each absolute value is $\leq \epsilon$ based on everything I've said above, and by the choice of $\delta$). Since the point $p$ and $\epsilon > 0$ are arbitrary, the inequality above shows that \begin{align} H_1 &= \dfrac{\partial F_3}{\partial y} - \dfrac{\partial F_2}{\partial z} \end{align}
As a recap of the proof idea: choose a small plane $M^{\delta}$ with outward normal pointing along $e_1$; it is the flatness of the plane (which is inherently adapted to cartesian coordinates), along with the ease of boundary parametrization which makes the resulting line integral easy to calculate. Then, simply calculate everything, and use the mean-value theorems for derivatives and integrals (this is one way to fill in the gaps for the arugments you typically see in physics texts, which say "let's keep things up to first order only" and where they use $\approx$ everywhere); finally we complete it off with a standard $\epsilon,\delta$ continuity argument.
Some remarks is that for this argument to work I've had to assume $F$ is $C^1$, so that I can apply the mean-value theorems twice, and finally finish it off with a continuity argument. I'm not sure if this proof can be strengthened so that we only have to assume $F$ is differentiable (rather than $C^1$).
Proof of Converse
Now we show the existence of such a vector field $H$; for this we'll show that $\text{curl}F$, defined by equation $(2)$ satisfies the conditions of $(1)$. Like I mentioned in the comments, I'm not sure how to do this without already appealing to Stokes theorem. With Stokes' theorem, this becomes quite simple.
Let $p\in A$, $\epsilon > 0$ and let $S\subset A$ be any "nice surface". Since $\langle\text{curl}(F), n\rangle$ is a continuous function on $S$, there is an open neighbourhood $U$ around $p$ in $S$ such that for all $q\in U$, \begin{align} \left|\langle \text{curl}F(q), n(q)\rangle - \langle \text{curl}F(p), n(p)\rangle\right| & \leq \epsilon \end{align} Now, for any "nice surface" $M\subset U$ (with unit normal being the restriction of the one already on $S$), we have by Stokes theorem: \begin{align} \left|\dfrac{1}{|M|}\int_{\partial M}\langle F,\tau\rangle \, dl - \langle \text{curl} F(p), n(p) \rangle\right| &= \dfrac{1}{|M|}\left|\int_M \langle \text{curl}F, n\rangle \, dA - \int_M \langle \text{curl} F(p), n(p) \rangle\, dA\right| \\ &\leq \dfrac{1}{|M|}\int_M \left|\langle\text{curl }F, n\rangle - \langle\text{curl }F(p), n(p)\rangle \right| \, dA \\ & \leq \dfrac{1}{|M|} \epsilon |M| \\ &= \epsilon. \end{align} This completes the proof of existence.
Best Answer
This might help.
You don't take the gradient of a vector field, so $\nabla A$ makes no sense. Instead, the gradient is an operator that acts on scalar functions in $\mathbb{R}^n$ and produces a vector field on $\mathbb R^n$.
Now if $f(x,y)$ is a scalar function, then $\nabla f:=\langle f_x,f_y\rangle$.
As for the Laplacian operator, it is given by $\Delta=\nabla\cdot \nabla$, i.e. the divergence of the gradient. So for example, with the $f(x,y)$ above we would have $$\Delta f:=\nabla \cdot \nabla f=\nabla \cdot \langle f_x,f_y\rangle=f_{xx}+f_{yy},$$ and similarly for higher dimensions.
The Laplacian operator is an extremely important and ubiquitous operator throughout pure and applied mathematics, certainly in areas of PDE and physics.
A simple application of why the Laplacian operator is so central is because many important problems in applied mathematics and physics can be distilled down to this:
The answer is quickly revealed: $$\mathbf F=\nabla f \implies \nabla \cdot\mathbf F=\nabla\cdot \nabla f\implies 0=\Delta f.$$ The potential $f$ is a solution of Laplace's equation! (In fact, this is why some books refer to it as the potential equation.