Green's theorem


In mathematics, Green's theorem gives the relationship between a line integral around a simple closed curve C and a double integral over the plane region D bounded by C. It is named after George Green, but its first proof is due to Bernhard Riemann, and it is the two-dimensional special case of the more general Kelvin–Stokes theorem.

Theorem

Let C be a positively oriented, piecewise smooth, simple closed curve in a plane, and let D be the region bounded by C. If L and M are functions of defined on an open region containing D and having continuous partial derivatives there, then
where the path of integration along C is anticlockwise.
In physics, Green's theorem finds many applications. One is solving two-dimensional flow integrals, stating that the sum of fluid outflowing from a volume is equal to the total outflow summed about an enclosing area. In plane geometry, and in particular, area surveying, Green's theorem can be used to determine the area and centroid of plane figures solely by integrating over the perimeter.

Proof when ''D'' is a simple region

The following is a proof of half of the theorem for the simplified area D, a type I region where C1 and C3 are curves connected by vertical lines. A similar proof exists for the other half of the theorem when D is a type II region where C2 and C4 are curves connected by horizontal lines. Putting these two parts together, the theorem is thus proven for regions of type III. The general case can then be deduced from this special case by decomposing D into a set of type III regions.
If it can be shown that if
and
are true, then Green's theorem follows immediately for the region D. We can prove easily for regions of type I, and for regions of type II. Green's theorem then follows for regions of type III.
Assume region D is a type I region and can thus be characterized, as pictured on the right, by
where g1 and g2 are continuous functions on . Compute the double integral in :
Now compute the line integral in. C can be rewritten as the union of four curves: C1, C2, C3, C4.
With C1, use the parametric equations: x = x, y = g1, a ≤ x ≤ b. Then
With C3, use the parametric equations: x = x, y = g2, a ≤ x ≤ b. Then
The integral over C3 is negated because it goes in the negative direction from b to a, as C is oriented positively. On C2 and C4, x remains constant, meaning
Therefore,
Combining with, we get for regions of type I. A similar treatment yields for regions of type II. Putting the two together, we get the result for regions of type III.

Proof for rectifiable Jordan curves

We are going to prove the following
Theorem. Let be a rectifiable, positively oriented Jordan curve in and let denote its inner region. Suppose that are continuous functions with the property that has second partial derivative at every point of, has first partial derivative at every point of and that the functions, are Riemann-integrable over. Then
We need the following lemmas:
Lemma 1. Assume is a rectifiable, positively oriented Jordan curve in the plane and let be its inner region. For every positive real, let denote the collection of squares in the plane bounded by the lines, where runs through the set of integers. Then, for this, there exists a decomposition of into a finite number of non-overlapping subregions in such a manner that
Each one of the subregions contained in, say, is a square from.
Each one of the remaining subregions, say, has as boundary a rectifiable Jordan curve formed by a finite number of arcs of and parts of the sides of some square from.
Each one of the border regions can be enclosed in a square of edge-length.
If is the positively oriented boundary curve of, then
The number of border regions is no greater than, where is the length of.
Lemma 2. Let be a rectifiable curve in the plane and let be the set of points in the plane whose distance from is at most. The outer Jordan content of this set satisfies.
Lemma 3. Let be a rectifiable curve in and let be a continuous function. Then
Now we are in position to prove the Theorem:
Proof of Theorem. Let be an arbitrary positive real number. By continuity of, and compactness of, given, there exists such that whenever two points of are less than apart, their images under are less than apart. For this, consider the decomposition given by the previous Lemma. We have
Put.
For each, the curve is a positively oriented square, for which Green's formula holds. Hence
Every point of a border region is at a distance no greater than from. Thus, if is the union of all border regions, then ; hence, by Lemma 2. Notice that
We may as well choose so that the RHS of the last inequality is
The remark in the beginning of this proof implies that the oscillations of and on every border region is at most. We have
By Lemma 1,
Combining these, we finally get
for some. Since this is true for every, we are done.

Validity under different hypotheses

The hypothesis of the last theorem are not the only ones under which Green's formula is true. Another common set of conditions is the following:
The functions are still assumed to be continuous. However, we now require them to be Fréchet-differentiable at every point of. This implies the existence of all directional derivatives, in particular, where, as usual, is the canonical ordered basis of. In addition, we require the function to be Riemann-integrable over.
As a corollary of this, we get the Cauchy Integral Theorem for rectifiable Jordan curves:
Theorem. If is a rectifiable Jordan curve in and if is a continuous mapping holomorphic throughout the inner region of, then
the integral being a complex contour integral.
Proof. We regard the complex plane as. Now, define to be such that These functions are clearly continuous. It is well known that and are Fréchet-differentiable and that they satisfy the Cauchy-Riemann equations:.
Now, analysing the sums used to define the complex contour integral in question, it is easy to realize that
the integrals on the RHS being usual line integrals. These remarks allow us to apply Green's Theorem to each one of these line integrals, finishing the proof.

Multiply-connected regions

Theorem. Let be positively oriented rectifiable Jordan curves in satisfying
where is the inner region of. Let
Suppose and are continuous functions whose restriction to is Fréchet-differentiable. If the function
is Riemann-integrable over, then

Relationship to Stokes' theorem

Green's theorem is a special case of the Kelvin–Stokes theorem, when applied to a region in the -plane.
We can augment the two-dimensional field into a three-dimensional field with a z component that is always 0. Write F for the vector-valued function. Start with the left side of Green's theorem:
The Kelvin–Stokes theorem:
The surface is just the region in the plane, with the unit normal defined to have a positive z component in order to match the "positive orientation" definitions for both theorems.
The expression inside the integral becomes
Thus we get the right side of Green's theorem
Green's theorem is also a straightforward result of the general Stokes' theorem using differential forms and exterior derivatives:

Relationship to the divergence theorem

Considering only two-dimensional vector fields, Green's theorem is equivalent to the two-dimensional version of the divergence theorem:
where is the divergence on the two-dimensional vector field, and is the outward-pointing unit normal vector on the boundary.
To see this, consider the unit normal in the right side of the equation. Since in Green's theorem is a vector pointing tangential along the curve, and the curve C is the positively oriented curve along the boundary, an outward normal would be a vector which points 90° to the right of this; one choice would be. The length of this vector is So
Start with the left side of Green's theorem:
Applying the two-dimensional divergence theorem with, we get the right side of Green's theorem:

Area calculation

Green's theorem can be used to compute area by line integral. The area of a planar region is given by
Choose and such that, the area is given by
Possible formulas for the area of include