MATH 251 Lecture 31

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Next written homework will cover Div, Grad, Curl (and all that)

Laplace Operator

• $\mathrm{\nabla }\times (\mathrm{\nabla }f)=0$
• $\mathrm{\nabla }\cdot (\mathrm{\nabla }\times f)=0$
• $\mathrm{\nabla }\cdot (\mathrm{\nabla }f)={\mathrm{\nabla }}^{2}f=\mathrm{\Delta }f$ ← this is the Laplace Operator

${\mathrm{\nabla }}^{2}f=f_{xx}+f_{yy}+f_{zz}$

Partial differential equation: ${\mathrm{\nabla }}^{2}f=\mathrm{\Delta }f=0$, where $f$ is an unknown function.

For example:

• $\mathrm{\Delta }(x^2-y^2)=2-2=0$ (this satisfies the diff eq)
• $\mathrm{\Delta }(x^2+y^2)=2+2=4$

Version of Laplace's Equation for Electrostatics

Let $u(x)$ represent the electric potential at point $x$. In a domain, take a point $p$ and a vector $\overrightarrow{v}$.

We want to find the voltage drop at $p$ in the direction of $\overrightarrow{v}$:

$V=u(p+\overrightarrow{v})-u(p)={\mathrm{D}}_{\overrightarrow{v}}u=\mathrm{\nabla }u\cdot \overrightarrow{v}$

Therefore $I=\frac{\mathrm{\nabla }u\cdot \overrightarrow{v}}{\rho }=\sigma \mathrm{\nabla }u\cdot \overrightarrow{v}$

Think of current $\sigma \mathrm{\nabla }u$ as a vector field. By conservation of charge, its divergence should be 0:

$\mathrm{\nabla }\cdot (\sigma \mathrm{\nabla }u)=0$

If $\sigma$ is a constant, we can factor it out of the divergence operator, leaving

$\begin{array}{rl}\sigma \mathrm{\nabla }\cdot \mathrm{\nabla }u& =0\\ \sigma \mathrm{\Delta }u& =0\\ \mathrm{\Delta }u& =0\end{array}$

Product Rule for Div, Grad, Curl

$\begin{array}{rl}\overrightarrow{G}& =P\hat{ı}+Q\hat{ȷ}\\ f(x,y)\\ \mathrm{d}\mathrm{i}\mathrm{v}(f\overrightarrow{G})& =\mathrm{\nabla }f\cdot \overrightarrow{G}+f\mathrm{\nabla }\cdot \overrightarrow{G}\\ \mathrm{c}\mathrm{u}\mathrm{r}\mathrm{l}(f\overrightarrow{G})& =\mathrm{\nabla }f\times \overrightarrow{G}+f\mathrm{\nabla }\times \overrightarrow{G}\end{array}$

Green's Theorem

(See Vector Analysis Theorems#Green's Theorem→)

For a vector field $\overrightarrow{F}(x,y)=P(x,y)\hat{ı}+Q(x,y)\hat{ȷ}$

$\mathrm{\nabla }\times \overrightarrow{F}=(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})\hat{k}$

Recall that curl measures the circulation of $\overrightarrow{F}$ around a point $(x,y)$.

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For a domain $\mathrm{\Omega }$ and a boundary curve $C=\partial \mathrm{\Omega }$. Let $\overrightarrow{F}(x,y)=P(x,y)\hat{ı}+Q(x,y)\hat{ȷ}$ be differentiable throughout domain (even at boundary).

We want to find $\underset{C}{\int }\overrightarrow{F}\cdot \mathrm{d}\overrightarrow{r}$. This could be interpreted as the circulation of $\overrightarrow{F}$ around $\mathrm{\Omega }$

This could also be the sum of $\mathrm{\nabla }\times \overrightarrow{F}$ around a bunch of little pieces inside $\mathrm{\Omega }$.

$\sum (\mathrm{\nabla }\times \overrightarrow{F})\cdot \hat{k}(\mathrm{\Delta }{A}_i)⟶\underset{\mathrm{\Omega }}{\iint }(\mathrm{\nabla }\times \overrightarrow{F}\cdot \hat{k}\mathrm{d}A$

Therefore,

$\underset{C}{\int }\overrightarrow{F}\cdot \mathrm{d}\overrightarrow{r}=\underset{\mathrm{\Omega }}{\iint }\mathrm{\nabla }\times \overrightarrow{F}\mathrm{d}A$

Example

Let $C$ be the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$ in the CCW direction, and let $\overrightarrow{F}=-y\hat{ı}+x\hat{ȷ}$