## Section10.6Green's theorem

Note: 1 lecture, requires Chapter 9

One of the most important theorems of analysis in several variables is the so-called generalized Stokes' theorem, a generalization of the fundamental theorem of calculus. The two-dimensional version is called Green's theorem 1 . We will state the theorem in general, but we will only prove a special, but important, case.

### Definition10.6.1.

Let $$U \subset \R^2$$ be a bounded connected open set. Suppose the boundary $$\partial U$$ is a disjoint union of (the images of) finitely many simple closed piecewise smooth paths such that every $$p \in \partial U$$ is in the closure of $$\R^2 \setminus \widebar{U}\text{.}$$ Then $$U$$ is called a bounded domain with piecewise smooth boundary in $$\R^2\text{.}$$

The condition about points outside the closure says that locally $$\partial U$$ separates $$\R^2$$ into an “inside” and an “outside.” The condition prevents $$\partial U$$ from being just a “cut” inside $$U\text{.}$$ As we travel along the path in a certain orientation, there is a well-defined left and a right, and either $$U$$ is on the left and the complement of $$U$$ is on the right, or vice versa. The orientation on $$U$$ is the direction in which we travel along the paths. We can switch orientation if needed by reparametrizing the path.

### Definition10.6.2.

Let $$U \subset \R^2$$ be a bounded domain with piecewise smooth boundary, let $$\partial U$$ be oriented , and let $$\gamma \colon [a,b] \to \R^2$$ be a parametrization of $$\partial U$$ giving the orientation. Write $$\gamma(t) = \big(x(t),y(t)\bigr)\text{.}$$ If the vector $$n(t) := \bigl(-y'(t),x'(t)\bigr)$$ points into the domain, that is, $$\epsilon n(t) + \gamma(t)$$ is in $$U$$ for all small enough $$\epsilon > 0\text{,}$$ then $$\partial U$$ is positively oriented. See Figure 10.7. Otherwise it is negatively oriented. Figure 10.7. Positively oriented domain (left), and a positively oriented domain with a hole (right).

The vector $$n(t)$$ turns $$\gamma^{\:\prime}(t)$$ counterclockwise by $$90^\circ\text{,}$$ that is to the left. When we travel along a positively oriented boundary in the direction of its orientation, the domain is “on our left.” For example, if $$U$$ is a bounded domain with “no holes,” that is $$\partial U$$ is connected, then the positive orientation means we are traveling counterclockwise around $$\partial U\text{.}$$ If we do have “holes,” then we travel around them clockwise.

### Proof.

We must show that $$\partial U$$ is a null set. As $$\partial U$$ is a finite union of piecewise smooth paths, which are finite unions of smooth paths, we need only show that a smooth path in $$\R^2$$ is a null set. Let $$\gamma \colon [a,b] \to \R^2$$ be a smooth path. It is enough to show that $$\gamma\bigl((a,b)\bigr)$$ is a null set, as adding the points $$\gamma(a)$$ and $$\gamma(b)\text{,}$$ to a null set still results in a null set. Define

\begin{equation*} f \colon (a,b) \times (-1,1) \to \R^2, \qquad \text{as} \qquad f(x,y) := \gamma(x) . \end{equation*}

The set $$(a,b) \times \{ 0 \}$$ is a null set in $$\R^2$$ and $$\gamma\bigl((a,b)\bigr) = f\bigl( (a,b) \times \{ 0 \} \bigr)\text{.}$$ By Proposition 10.3.10, $$\gamma\bigl((a,b)\bigr)$$ is a null set in $$\R^2$$ and so $$\gamma\bigl([a,b]\bigr)$$ is a null set, and so finally $$\partial U$$ is a null set.

We stated Green's theorem in general, although we will only prove a special version of it. That is, we will only prove it for a special kind of domain. The general version follows from the special case by application of further geometry, and cutting up the general domain into smaller domains on which to apply the special case. We will not prove the general case.

Let $$U \subset \R^2$$ be a domain with piecewise smooth boundary. We say $$U$$ is of type I if there exist numbers $$a < b\text{,}$$ and continuous functions $$f \colon [a,b] \to \R$$ and $$g \colon [a,b] \to \R\text{,}$$ such that

\begin{equation*} U := \{ (x,y) \in \R^2 : a < x < b \text{ and } f(x) < y < g(x) \} . \end{equation*}

Similarly, $$U$$ is of type II if there exist numbers $$c < d\text{,}$$ and continuous functions $$h \colon [c,d] \to \R$$ and $$k \colon [c,d] \to \R\text{,}$$ such that

\begin{equation*} U := \{ (x,y) \in \R^2 : c < y < d \text{ and } h(y) < x < k(y) \} . \end{equation*}

Finally, $$U \subset \R^2$$ is of type III if it is both of type I and type II. See Figure 10.8. Figure 10.8. Domain types for Green's theorem.

Common domains to apply Green's theorem to are rectangles and discs, and these are type III domains. We will only prove Green's theorem for type III domains.

### Proof.

(Proof of Green's theorem for $$U$$ of type III) Let $$f,g,h,k$$ be the functions defined above. Using Exercise 10.5.3, $$U$$ is Jordan measurable and as $$U$$ is of type I, then

\begin{equation*} \begin{split} \int_U \left(- \frac{\partial P}{\partial y} \right) & = \int_a^b \int_{g(x)}^{f(x)} \left(- \frac{\partial P}{\partial y} (x,y) \right) \, dy \, dx \\ & = \int_a^b \Bigl( - P\bigl(x,f(x)\bigr) + P\bigl(x,g(x)\bigr) \Bigr) \, dx \\ & = \int_a^b P\bigl(x,g(x)\bigr) \, dx - \int_a^b P\bigl(x,f(x)\bigr) \, dx . \end{split} \end{equation*}

We integrate $$P\,dx$$ along the boundary. The one-form $$P\,dx$$ integrates to zero along the straight vertical lines in the boundary. Therefore it is only integrated along the top and along the bottom. As a parameter, $$x$$ runs from left to right. If we use the parametrizations that take $$x$$ to $$\bigl(x,f(x)\bigr)$$ and to $$\bigl(x,g(x)\bigr)$$ we recognize path integrals above. However the second path integral is in the wrong direction; the top should be going right to left, and so we must switch orientation.

\begin{equation*} \int_{\partial U} P \, dx = \int_a^b P\bigl(x,g(x)\bigr) \, dx + \int_b^a P\bigl(x,f(x)\bigr) \, dx = \int_U \left(- \frac{\partial P}{\partial y} \right) . \end{equation*}

Similarly, $$U$$ is also of type II. The form $$Q\,dy$$ integrates to zero along horizontal lines. So

\begin{equation*} \int_U \frac{\partial Q}{\partial x} = \int_c^d \int_{k(y)}^{h(y)} \frac{\partial Q}{\partial x}(x,y) \, dx \, dy = \int_a^b \Bigl( Q\bigl(y,h(y)\bigr) - Q\bigl(y,k(y)\bigr) \Bigr) \, dx = \int_{\partial U} Q \, dy . \end{equation*}

Putting the two together we obtain

\begin{equation*} \int_{\partial U} P\, dx + Q \, dy = \int_{\partial U} P\, dx + \int_{\partial U} Q \, dy = \int_U \Bigl(-\frac{\partial P}{\partial y}\Bigr) + \int_U \frac{\partial Q}{\partial x} = \int_U \Bigl( \frac{\partial Q}{\partial x} -\frac{\partial P}{\partial y} \Bigr) . \qedhere \end{equation*}

We illustrate the usefulness of Green's theorem on a fundamental result about harmonic functions.

### Example10.6.5.

Suppose $$U \subset \R^2$$ is open and $$f \colon U \to \R$$ is harmonic, that is, $$f$$ is twice continuously differentiable and $$\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = 0\text{.}$$ We will prove one of the most fundamental properties of harmonic functions.

Let $$D_r := B(p,r)$$ be closed disc such that its closure $$C(p,r) \subset U\text{.}$$ Write $$p = (x_0,y_0)\text{.}$$ We orient $$\partial D_r$$ positively. See Exercise 10.6.1. Then

\begin{equation*} \begin{split} 0 & = \frac{1}{2\pi r} \int_{D_r} \left( \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} \right) \\ & = \frac{1}{2\pi r} \int_{\partial D_r} - \frac{\partial f}{\partial y} \, dx + \frac{\partial f}{\partial x} \, dy \\ & = \frac{1}{2\pi r} \int_0^{2\pi} \biggl( - \frac{\partial f}{\partial y} \bigl(x_0+r\cos(t),y_0+r\sin(t)\bigr) \bigl(-r\sin(t)\bigr) \\ & \hspace{1.2in} + \frac{\partial f}{\partial x} \bigl(x_0+r\cos(t),y_0+r\sin(t)\bigr) r\cos(t) \biggr) \, dt \\ & = \frac{d}{dr} \left[ \frac{1}{2\pi} \int_0^{2\pi} f\bigl(x_0+r\cos(t),y_0+r\sin(t)\bigr) \, dt \right] . \end{split} \end{equation*}

Let $$g(r) := \frac{1}{2\pi} \int_0^{2\pi} f\bigl(x_0+r\cos(t),y_0+r\sin(t)\bigr) \, dt\text{.}$$ Then $$g'(r) = 0$$ for all $$r > 0\text{.}$$ The function is constant for $$r >0$$ and continuous at $$r=0$$ (exercise). Therefore, $$g(0) = g(r)$$ for all $$r > 0\text{,}$$ and

\begin{equation*} g(r) = g(0) = \frac{1}{2\pi} \int_0^{2\pi} f\bigl(x_0+0\cos(t),y_0+0\sin(t)\bigr) \, dt = f(x_0,y_0). \end{equation*}

We proved the mean value property of harmonic functions:

\begin{equation*} f(x_0,y_0) = \frac{1}{2\pi} \int_0^{2\pi} f\bigl(x_0+r\cos(t),y_0+r\sin(t)\bigr) \, dt = \frac{1}{2\pi r} \int_{\partial D_r} f \, ds . \end{equation*}

That is, the value at $$p = (x_0,y_0)$$ is the average over a circle of any radius $$r$$ centered at $$(x_0,y_0)\text{.}$$

### Subsection10.6.1Exercises

#### Exercise10.6.1.

Prove that a disc $$B(p,r) \subset \R^2$$ is a type III domain, and prove that the orientation given by the parametrization $$\gamma(t) = \bigl(x_0+r\cos(t),y_0+r\sin(t)\bigr)$$ where $$p = (x_0,y_0)$$ is the positive orientation of the boundary $$\partial B(p,r)\text{.}$$
Note: Feel free to use what you know about sine and cosine from calculus.

#### Exercise10.6.2.

Prove that a convex bounded domain with piecewise smooth boundary is a type III domain.

#### Exercise10.6.3.

Suppose $$V \subset \R^2$$ is a domain with piecewise smooth boundary that is a type III domain and suppose that $$U \subset \R^2$$ is a domain such that $$\widebar{V} \subset U\text{.}$$ Suppose $$f \colon U \to \R$$ is a twice continuously differentiable function. Prove that $$\int_{\partial V} \frac{\partial f}{\partial x} \, dx + \frac{\partial f}{\partial y} \, dy = 0\text{.}$$

#### Exercise10.6.4.

For a disc $$B(p,r) \subset \R^2\text{,}$$ orient the boundary $$\partial B(p,r)$$ positively.

1. Compute $$\displaystyle \int_{\partial B(p,r)} -y \, dx\text{.}$$

2. Compute $$\displaystyle \int_{\partial B(p,r)} x \, dy\text{.}$$

3. Compute $$\displaystyle \int_{\partial B(p,r)} \frac{-y}{2} \, dx + \frac{x}{2} \, dy\text{.}$$

#### Exercise10.6.5.

Using Green's theorem show that the area of a triangle with vertices $$(x_1,y_1)\text{,}$$ $$(x_2,y_2)\text{,}$$ $$(x_3,y_3)$$ is $$\frac{1}{2}\sabs{x_1y_2 + x_2 y_3 + x_3 y_1 - y_1x_2 - y_2x_3 - y_3x_1}\text{.}$$ Hint: See previous exercise.

#### Exercise10.6.6.

Using the mean value property prove the maximum principle for harmonic functions: Suppose $$U \subset \R^2$$ is a connected open set and $$f \colon U \to \R$$ is harmonic. Prove that if $$f$$ attains a maximum at $$p \in U\text{,}$$ then $$f$$ is constant.

#### Exercise10.6.7.

Let $$f(x,y) := \ln \sqrt{x^2+y^2}\text{.}$$

1. Show $$f$$ is harmonic where defined.

2. Show $$\lim_{(x,y) \to 0} f(x,y) = -\infty\text{.}$$

3. Using a circle $$C_r$$ of radius $$r$$ around the origin, compute $$\frac{1}{2\pi r} \int_{\partial C_r} f \, ds\text{.}$$ What happens as $$r \to 0\text{?}$$

4. Why can't you use Green's theorem?

Named after the British mathematical physicist George Green 2  (1793–1841).
https://en.wikipedia.org/wiki/George_Green_(mathematician)
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