RA Green’s theorem

Section 10.6 Green’s theorem

Note: 1 lecture, requires Chapter 9

One of the most important theorems in the calculus of 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 theorem¹. We will state the theorem in general, but we will only prove a special, but important, case.

Definition 10.6.1.

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.

Definition 10.6.2.

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 \(n(t) \coloneqq \bigl(-y'(t),x'(t)\bigr)\) 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 10.14. Otherwise it is negatively oriented.

Figure 10.14. 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.

Proposition 10.6.3.

Let \(U \subset \R^2\) be a bounded domain with piecewise smooth boundary. Then \(U\) is Jordan measurable.

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) \coloneqq \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.

Theorem 10.6.4. Green.

Suppose \(U \subset \R^2\) is a bounded domain with piecewise smooth boundary with the boundary positively oriented. Suppose \(P\) and \(Q\) are continuously differentiable functions defined on some open set that contains the closure \(\widebar{U}\text{.}\) Then

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

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 \coloneqq \bigl\{ (x,y) \in \R^2 : a < x < b \text{ and } f(x) < y < g(x) \bigr\} . \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 \coloneqq \bigl\{ (x,y) \in \R^2 : c < y < d \text{ and } h(y) < x < k(y) \bigr\} . \end{equation*}

Finally, \(U \subset \R^2\) is of type III if it is both of type I and type II. See Figure 10.15.

Figure 10.15. 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 Proposition 10.5.8, \(U\) 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 computations 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*}

Let us see how one can use the simple version of Green’s (type III domains only) for a more complex path.

Example 10.6.5.

Suppose \(P(x,y) = \frac{-y}{x^2+y^2}\text{,}\) \(Q(x,y) = \frac{x}{x^2+y^2}\text{.}\) If we think of \((P,Q)\) as a vector, so that we have a so-called vector field, \((P,Q)\) is called the vortex vector field, as it gives the velocity of particles traveling in a vortex around the origin. Variations on this vector field come up often in applications. Suppose that \(\gamma\) is a path that goes counterclockwise around a rectangle whose interior contains the origin. We claim

\begin{equation*} \int_{\gamma} \frac{-y}{x^2+y^2} \, dx + \frac{x}{x^2+y^2} \, dy = 2 \pi . \end{equation*}

First we draw a circle \(C\) of radius \(r > 0\) centered at the origin such that the entire circle is within \(\gamma\) and oriented clockwise. Consider \(U\) to be the domain between \(\gamma\) and \(C\text{.}\) See Figure 10.16. The integral around \(\partial U\) is the integral around \(\gamma\) plus the integral around \(C\text{.}\) Now \(U\) is not a domain of type III, so we cannot just apply the version of Green’s theorem we actually proved. However, if we cut the box along the axis as shown in the figure with dashed lines, the four resulting domains, let us call them \(U_1,U_2,U_3,U_4\text{,}\) are of type III. The dashed lines are oriented in opposite directions for the two \(U_j\) that share them, and so when we integrate along both, the integrals cancel. That is,

\begin{multline*} \int_{\partial U} P\, dx + Q \, dy = \\ \int_{\partial U_1} P\, dx + Q \, dy + \int_{\partial U_2} P\, dx + Q \, dy + \int_{\partial U_3} P\, dx + Q \, dy + \int_{\partial U_4} P\, dx + Q \, dy . \end{multline*}

Now we can apply Green’s theorem to every \(U_j\text{.}\) We leave it to the reader to verify that outside of the origin, \(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 0\text{.}\) So

\begin{equation*} \int_{\partial U_j} P\, dx + Q \, dy = \int_{U_j} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) = \int_{U_j} 0 = 0 . \end{equation*}

Next we notice that

\begin{equation*} \int_C P \, dx + Q \, dy + \int_{\gamma} P \, dx + Q \, dy = \int_{\partial U} P \, dx + Q \, dy = 0 . \end{equation*}

So the integral around \(C\) is minus the integral around \(\gamma\text{.}\) The integral around \(C\) is easy to compute as on \(C\) we have \(x^2+y^2 = r^2\text{,}\) so \(P(x,y) = \frac{-y}{r^2}\) and \(Q(x,y) = \frac{x}{r^2}\text{.}\) We leave it to the reader to compute

\begin{equation*} \int_C P\, dx + Q \, dy = \int_C \frac{-y}{r^2}\, dx + \frac{x}{r^2} \, dy = - 2 \pi . \end{equation*}

The claim follows.

Figure 10.16. Changing the box integral to an integral around a small circle around the origin. The domain \(U\) is the shaded area between the circle and the box.

We remark that if \(\gamma\) would not contain the origin, \(\int_\gamma P\,dx+Q\,dy = 0\text{,}\) as we could just apply Green’s to \(\gamma\text{.}\) So this integral can detect whether the origin is inside \(\gamma\) or not.

As a second example, we illustrate the usefulness of Green’s theorem on a fundamental result about harmonic functions.

Example 10.6.6.

Suppose \(U \subset \R^2\) is open and \(f \colon U \to \R\) is harmonic, that is, \(f\) is twice continuously differentiable and satisfies the Laplace equation, \(\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = 0\text{.}\) Harmonic functions are, for instance, the steady state heat distribution, or the electric potential between charges. We will prove one of the most fundamental properties of these functions.

Let \(D_r \coloneqq B(p,r)\) be a disc such that its closure \(\overline{D_r} = C(p,r) \subset U\text{.}\) Write \(p = (x_0,y_0)\text{.}\) We orient \(\partial D_r\) positively. See Exercise 10.6.1. Then via Green’s and integration under the integral,

\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 \(g(r) \coloneqq \frac{1}{2\pi} \int_0^{2\pi} f\bigl(x_0+r\cos(t),y_0+r\sin(t)\bigr) \, dt\) for \(r \geq 0\) (small enough). The function is continuous at \(r=0\) (exercise), and we have just proved that \(g'(r) = 0\) for all \(r > 0\text{.}\) 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, for a harmonic function, the value at \(p = (x_0,y_0)\) equals the average of its values over a circle of any radius \(r\) centered at \((x_0,y_0)\text{.}\)

Subsection 10.6.1 Exercises

Exercise 10.6.1.

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.

Exercise 10.6.2.

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

Exercise 10.6.3.

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 \(\int_{\partial V} \frac{\partial f}{\partial x} \, dx + \frac{\partial f}{\partial y} \, dy = 0\text{.}\)

Exercise 10.6.4.

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

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

Exercise 10.6.5.

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.

Exercise 10.6.6.

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.

Exercise 10.6.7.

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

Show \(f\) is harmonic where defined.
Show \(\lim\limits_{(x,y) \to 0} f(x,y) = -\infty\text{.}\)
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 \(r \to 0\text{?}\)
Why can’t you use Green’s theorem?

Named after the British mathematical physicist George Green (1793–1841).