Math2111: Chapter 3: Additional Material: For which curves and regions does Green’s theorem apply?

In this blog entry you can find lecture notes for Math2111, several variable calculus. See also the table of contents for this course. This blog entry printed to pdf is available here.

Here we discuss the conditions under which Green’s theorem (see also here) applies, more precisely for which curves and regions enclosed by the curve.

The formula in Green’s theorem states that

\displaystyle \oint_{\mathcal{C}} \boldsymbol{F} \cdot \mathrm{d} \boldsymbol{s} = \iint_R \left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\,\mathrm{d} A, \qquad\qquad\qquad (1)

where \boldsymbol{F}:\mathbb{R}^2\to\mathbb{R}^2, \boldsymbol{F}=P\widehat{\boldsymbol{i}}+Q\widehat{\boldsymbol{j}}, is a continuously differentiable vector field, \mathcal{C} is a curve which is simple, closed, smooth and oriented positively, and {}R is the region enclosed by the curve \mathcal{C}.

First, the conditions on the curve \mathcal{C} and the region {}R must be such that the line integral and double integral in (1) are well defined.

The curve \mathcal{C}

Let us discuss the line integral and curve \mathcal{C} first. If \mathcal{C} has a parameterisation \boldsymbol{c} which is piecewise continuously differentiable, then the line integral is well defined. On the other hand, the line integral is also defined under weaker conditions: it is enough to assume that \boldsymbol{c}:[a,b]\to\mathbb{R}^2 is continuous and has bounded variation. Let \boldsymbol{c}(t) = x(t) \widehat{\boldsymbol{i}} + y(t) \widehat{\boldsymbol{j}}. Then \boldsymbol{c} has bounded variation if {}x and {}y each have bounded variation, which means that V(x), V(y) \textless \infty, where {}V stands for variation. The variation is defined by

\displaystyle V(x)= \sup_{P \in \mathcal{P}} \sum_{i=1}^{N_P} |x(t_i)-x(t_{i-1})|,

where the supremum is taken over all partitions {}P=\{t_0,\ldots, t_{N_P}\} of the interval [a,b], where a=t_0 \textless t_1 \textless \cdots \textless t_{N_P-1}\textless t_{N_P}=b. Note that if {}x is continuously differentiable, then V(x) = \int_a^b |x^\prime(t)|\,\mathrm{d} t.

It can be shown that if \boldsymbol{c}:[a,b]\to\mathbb{R}^2 is continuous and such that each of the components {}x, {}y have bounded variation, then the parameterised curve \boldsymbol{c} has finite length. We call such curves rectifiable.

Note that assuming that the parameterisation \boldsymbol{c} is only continuous is not sufficient, since there are curves which are continuous everywhere but nowhere differentiable and which have infinite length (see the Koch curve).

Let now \boldsymbol{F}(x,y)=P(x,y) \widehat{\boldsymbol{i}} + Q(x,y) \widehat{\boldsymbol{j}} be a continuous vector field and let \boldsymbol{c}(t)= x(t) \widehat{\boldsymbol{i}} + y(t) \widehat{\boldsymbol{j}} be rectifiable. Then we define

\displaystyle \int_{\boldsymbol{c}} \boldsymbol{F}\cdot\mathrm{d} \boldsymbol{s} = \int_a^b P(x(t),y(t)) \,\mathrm{d} x(t) + \int_a^b Q(x(t)),y(t)) \,\mathrm{d} y(t),

where the last two integrals are Riemann-Stieltjes integrals.

Hence, the line integral is well-defined if \boldsymbol{c} is continuous and each of its components have bounded variation.

The region {}R

We now discuss the region {}R which is enclosed by the curve \mathcal{C} (which is parameterised by \boldsymbol{c}). The double integral is certainly defined (since we assume that Q_x - P_y are continuous) if {}R is Jordan-measurable.

A non-empty bounded set R \subseteq \mathbb{R}^2 is called Jordan-measurable if the function

\displaystyle 1_R(\boldsymbol{z})=\left\{\begin{array}{rl} 1 & \mbox{if } x \in R, \\ &\\ 0&\mbox{otherwise,} \end{array} \right.

is Riemann-integrable.

There is a theorem which states that a set {}R is Jordan-measurable if and only if its boundary \partial R has Lebesgue measure zero. The boundary of {}R is just the curve \mathcal{C}, since {}R is defined as the region enclosed by \mathcal{C}. (The last statement is well-defined by the Jordan-curve theorem.) Since \boldsymbol{c} is rectifiable it has Lebesgue measure zero (the curve is Lebesgue measurable since it is the image of a continuous map defined on a compact set and hence the curve is compact). Therefore, the region {}R enclosed by the simple closed curve which is parameterised by a continuous and rectifiable function \boldsymbol{c} is Jordan-measurable (such curves are also called rectifiable Jordan curves).

Thus the assumptions on \boldsymbol{c} guarantee that the region {}R is Jordan-measurable and, in turn, the double integral is well defined.

Curves and regions for which Green’s theorem applies

So far we investigated curves and regions which ensure that the line integral and double integral in (1) are well defined. We now turn to Green’s theorem itself and investigate in more detail for which regions Green’s theorem holds.

Let [a,b] be an interval with a\textless b and let \phi_1,\phi_2:[a,b]\to\mathbb{R} be two continuous functions such that \phi_1(x) \textless \phi_2(x) for all x \in [a,b]. The set

\displaystyle R_x=\{(x,y)\in\mathbb{R}^2: \phi_1(x) \le y \le \phi_2(x)\}

is called {}x-simple. Analogously, let [c,d] be an interval with c \textless d and let \psi_1,\psi_2: [c,d]\to\mathbb{R} be two continuous functions such that \psi_1(y) \textless \psi_2(y) for all y\in [c,d]. We call the set

\displaystyle R_y=\{(x,y) \in \mathbb{R}^2: \psi_1(y)\le x \le \psi_2(y)\}


For such regions the following two lemmas apply.

Let D be an {}x-simple region and let P:D\to\mathbb{R} be continuously differentiable. Then

\displaystyle \int_{\mathcal{C}^+} P \,\mathrm{d} x= -\int_D \frac{\partial P}{\partial y} \,\mathrm{d} x \,\mathrm{d}y,

where \mathcal{C}^+ is the boundary curve of {}D oriented counterclockwise.

Let D be an {}y-simple region and let Q:D\to\mathbb{R} be continuously differentiable. Then

\displaystyle \int_{\mathcal{C}^+} Q \,\mathrm{d} y= \int_D \frac{\partial Q}{\partial x} \,\mathrm{d} x \,\mathrm{d}y,

where \mathcal{C}^+ is the boundary curve of {}D oriented counterclockwise.

A proof of these results can be found here. Notice that the proof only requires that the functions \phi_1,\phi_2,\psi_1,\psi_2 are continuous – no further smoothness assumption is needed.

Green’s theorem can now be obtained by adding up the two equations from the above lemmas and by assuming that the region {}R is simultaneously {}x-simple and {}y-simple. Regions which are both {}x-simple and {}y-simple are in the following called simple.

What is not yet clear however is the following. In the definition of {}x-simple and {}y-simple we assumed that the boundary is only continuous. However, it is known that there are simple closed curves (the Koch curve for instance) which are not rectifiable and hence for which line integrals over such curves are not well defined. The following result states that this cannot happen. That is, the boundary curve of a region which is simple (i.e. {}x-simple and {}y-simple) has bounded variation.

Let R\subset\mathbb{R}^2 be a region which is {}x-simple and {}y-simple. Then the boundary curve \partial R has bounded variation and is therefore rectifiable.

The proof is left as an exercise.

The last result now implies that the line integral is well-defined, and since the boundary curve is a simple closed curve which is continuous and rectifiable, the region {}R is Jordan-measurable. Hence also the double integral is well defined.

More general regions

Green’s theorem also applies to regions {}R enclosed by a simple closed curve, where {}R can be divided into a finite number of simple regions.

Green’s theorem
Let R\subset \mathbb{R}^2 be a region whose boundary is made up of a finite number of simple closed curves. Suppose that by means of a finite number of line segments parallel to the coordinate axes, {}R can be decomposed into a finite number of simple regions R_1,\ldots, R_N with the boundary of each R_k oriented counterclockwise. Then if {}P and {}Q are continuously differentiable on {}R, then

\displaystyle \int_{\partial R} P \,\mathrm{d} x + Q \,\mathrm{d} y = \iint_R \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)\,\mathrm{d} x\,\mathrm{d}y,

where the boundary curve is oriented counterclockwise.

To prove this result one applied Green’s theorem to each R_k, k=1,\ldots, N. The line integral over the curves inside the region cancel which then implies the result. The details are left as an exercise.


Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s