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C. Results from Analytic Function Theory


C.3 Simply Connected Domains

Roughly speaking, a domain $D$ is simply connected if it has no holes. More precisely, $D$ is simply connected if, for every simple closed curve $C$ in $D$, the region $R$ enclosed by $C$ lies wholly in $D$. For simply connected domains we have the following:

Theorem C.4 (Green's theorem)   Let $D$ be a simply connected domain, and let $C$ be a piecewise-smooth simple closed curve in $D$. Let $P(x,y)$ and $Q(x,y)$ be functions that are continuous and that have continuous first partial derivatives in $D$. Then


\begin{displaymath}\oint (P dx +Q dy) =\int \int_R \left (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right )dx dy \end{displaymath} (C.3.1)

where $R$ is the region bounded by $C$.

Proof

We first consider a simple case in which $R$ is representable in both of the forms:


$\displaystyle f_1(x)\leq f_2(x) \hspace{10mm}$ $\textstyle \mbox{for}$ $\displaystyle a\leq x\leq b \hspace{3mm}$ (C.3.2)
$\displaystyle g_1(y)\leq g_2(y) \hspace{10mm}$ $\textstyle \mbox{for}$ $\displaystyle c\leq y\leq d \hspace{3mm}$ (C.3.3)

Then


\begin{displaymath}\int\int_R \frac{\partial P}{\partial y}dx dy = \int_a^b\int_{f_1(x)}^{f_2(x)} \frac{\partial P}{\partial y}dx dy \end{displaymath} (C.3.4)

One can now integrate to achieve


$\displaystyle \int\int_R \frac{\partial P}{\partial y}dx dy$ $\textstyle =$ $\displaystyle \int_a^b[P(x,f_2(x))-P(x,f_1(x))]dx$ (C.3.5)
  $\textstyle =$ $\displaystyle \int_a^b P(x,f_2(x))dx-\int_a^bP(x,f_1(x))dx$ (C.3.6)
  $\textstyle =$ $\displaystyle \oint_CP(x,y)dx$ (C.3.7)

By a similar argument,


\begin{displaymath}\int\int_R \frac{\partial Q}{\partial x}dx dy =\oint_CQ(x,y)dy \end{displaymath} (C.3.8)

For more complex regions, we decompose into simple regions as above. The result then follows.

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We then have the following converse to Theorem C.3.

Theorem C.5  Let $P(x,y)$ and $Q(x,y)$ have continuous derivatives in $D$ and let $D$ be simply connected. If $\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x} $, then $\oint P dx +Q dy$ is independent of path in $D$.

Proof

Suppose that


\begin{displaymath} \frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x} \hspace{10mm}\mbox{in $D$ } \end{displaymath} (C.3.9)

Then, by Green's Theorem (Theorem C.4),


\begin{displaymath}\oint_c Pdx + Q dy=\int\int_R \left( \frac{\partial Q}{\partial x}- \frac{\partial P}{\partial y}\right)dx dy =0 \end{displaymath} (C.3.10)

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