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 generalized Stokes’ theorem, a generalization of the fundamental theorem of calculus. The two-dimensional version is called Green’s theorem

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

. We will state the theorem in general, but we will only prove an important special 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 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 \(\gamma'(t)\) 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 \(\gamma(b)\) 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 hence \(\partial U\) 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 applying further geometry and cutting up the general domain into smaller domains on which to apply the special case.

🔗

Let \(U \subset \R^2\) be a bounded 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.

🔗

Diagram of three domains. On the left is a type I domain between two graphs where every vertical line intersects the domain in a connected interval and several vertical lines are drawn in the domain. In the middle is a type II domain, which is the same idea but with horizontal lines. On the right is a type III domain, where both horizontal and vertical lines intersect in a connected interval and several horizontal and vertical lines are drawn. — 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,

\begin{equation} \begin{split} \int_U \left(- \frac{\partial P}{\partial y} \right) & = \int_a^b \int_{f(x)}^{g(x)} \left(- \frac{\partial P}{\partial y} (x,y) \right) \, dy \, dx \\ & = \int_a^b \Bigl( - P\bigl(x,g(x)\bigr) + P\bigl(x,f(x)\bigr) \Bigr) \, dx \\ & = \int_a^b P\bigl(x,f(x)\bigr) \, dx - \int_a^b P\bigl(x,g(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 (if there are any). 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 \(\bigl(x,g(x)\bigr)\text{,}\) we recognize path integrals above. However, the second path integral is in the wrong direction; the top should be going right to left. After switching this orientation, we find

\begin{equation} \int_{\partial U} P \, dx = \int_a^b P\bigl(x,f(x)\bigr) \, dx + \int_b^a P\bigl(x,g(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_{h(y)}^{k(y)} \frac{\partial Q}{\partial x}(x,y) \, dx \, dy = \int_c^d \Bigl( Q\bigl(k(y),y\bigr) - Q\bigl(h(y),y\bigr) \Bigr) \, dy = \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 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 are allowed to 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 for each \(j\text{,}\) we find

\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}

and so \(\int_{\partial U} P\, dx + Q \, dy = 0\text{.}\) As \(\partial U\) is \(C\) together with \(\gamma\text{,}\) we find

\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}

making sure to note that \(C\) is oriented clockwise not counterclockwise. The claim follows.

🔗

A diagram of a domain U that is a rectangle with a circular hole around the origin. The boundary of the rectangle goes around counterclockwise and is marked as gamma. The boundary of the inside circle is marked as C and goes clockwise. The domain U is cut in four pieces by four cuts along the axes shown in dashed lines. The four pieces are labeled as U sub 1 through U sub 4. The boundary of each of these four pieces is marked as going counterclockwise, and we note that along the cuts we find that the paths giving the boundaries go in both directions and so will cancel each other out. — Figure 10.16. Changing the box integral to an integral around a small circle around the origin. The domain \(U\) is the entire shaded area between the circle and the box.
🔗

We remark that if \(\gamma\) did 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 \(p = (x_0,y_0)\text{.}\) Orient \(\partial D_r\) positively. See Exercise 10.6.1. Via Green’s and differentiation 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 \((x_0,y_0)\text{,}\) provided \(r\) is small enough so that the entire closed disc fits within \(U\text{.}\)

🔗

Exercises Exercises

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.

🔗

10.6.2.

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

🔗

10.6.3.

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

🔗

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

🔗

🔗

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.

🔗

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.

🔗

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 \(\frac{1}{2\pi r} \int_{C_r} f \, ds\text{.}\) What happens as \(r \to 0\text{?}\)
🔗

🔗
Why can’t you use Green’s theorem in part c)?
🔗

🔗

🔗

Prev Top Next