MATH 251 Lecture 31

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

Laplace Operator

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

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

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

For example:

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

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 ${\vec {v}}$.

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

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

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

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

$\nabla \cdot (\sigma \nabla u)=0$

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

${\begin{aligned}\sigma \nabla \cdot \nabla u&=0\\\sigma \Delta u&=0\\\Delta u&=0\end{aligned}}$

Product Rule for Div, Grad, Curl

${\begin{aligned}{\vec {G}}&=P{\hat {\imath }}+Q{\hat {\jmath }}\\f(x,y)\\\mathrm {div} (f{\vec {G}})&=\nabla f\cdot {\vec {G}}+f\nabla \cdot {\vec {G}}\\\mathrm {curl} (f{\vec {G}})&=\nabla f\times {\vec {G}}+f\nabla \times {\vec {G}}\end{aligned}}$

Green's Theorem

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

For a vector field ${\vec {F}}(x,y)=P(x,y){\hat {\imath }}+Q(x,y){\hat {\jmath }}$

$\nabla \times {\vec {F}}=\left({\frac {\partial Q}{\partial x}}-{\frac {\partial P}{\partial y}}\right){\hat {k}}$

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

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

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

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

$\sum (\nabla \times {\vec {F}})\cdot {\hat {k}}(\Delta A_{i})\quad \longrightarrow \quad \iint _{\Omega }(\nabla \times {\vec {F}}\cdot {\hat {k}}\,\mathrm {d} A$

Therefore,

$\int _{C}{\vec {F}}\cdot \mathrm {d} {\vec {r}}=\iint _{\Omega }\nabla \times {\vec {F}}\,\mathrm {d} A$

Example

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