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3.6 Green’s theorem for linear fields
3.6.1 Circulation line integrals of linear fields
Only very few vector fields are gradient fields. For most pairs of vector fields and closed curves one may expect the line integral to be different from zero. This section begins a closer investigation of such vector fields. This back-door approach will, quite surprisingly lead to the development of a notion of derivative for vector fields, and associated analogues of the fundamental theorem of calculus. Rather than considering the most general vector fields and general curves, it makes sense to start analyzing the structurally most simple vector fields and curves. Note that every constant vector field is a gradient field. Hence the line integrals of any constant field over any closed curve is zero.
Exercise 3.6.1 For a constant vector field F~ (x, y) = a~ı + b~ find a potential function ϕ(x, y).
The structurally next most simple case is that of linear vector fields
L^ ~(x, y) = (ax + by)~ı + (cx + dy)~ (3.1)
Exercise 3.6.2 (Class-exercise:) Consider the linear vector field ~L(x, y) = (3x − 2 y)~ı + (5x + 8y)~. (i) For each of the closed curves depicted below evaluate the line integral
∮ C ~L^ ·^ dr~.
μ¥
∂≥
μ¥
∂≥ μ¥
∂≥
i
i
Add your own triangles, squares, rectangles, semi-and quarter circles,...
and tabulate the results
Student name Contour Line integral Circle C 1 21. 9911 Square C 2 28 Triangle C 3 3. 5
(ii) Look for patterns. Use the last two columns for additional data as needed. Make conjectures. Test the conjectures by evaluating additional line integrals over other curves, or by modifying the vector field.
Your conjecture should be powerful enough to predict – with only minimal calculation, without paper or calculator! – the value of any line integral of a linear vector field over any closed curve similar to those considered above.
DO NOT READ ON BEFORE COMPLETING THE EXERCISE ABOVE!
Work together. YOU CAN discover some deep math here! Don’t miss the opportunity.
(ii) Argue pictorially, using Riemann sums. Hint: Subdivide the line segment into a number of smaller segments of equal lengths. Group the vectors together analogous to the famous solution supposedly provided by Gauss as a school child when asked to sum 1 + 2 + 3 + ...99 + 100 = 100 + (1 + 99) + (2 +
- +... (49 + 51) + 50 = 50 ∗ 100 + 50 = 5050.
(iii) Argue geometrically using only the abstract characterization of linearity (i.e without referring to any coordinates) and the vector-form of the parameterization of a line segment ~r(t) = ~r 1 + t(~r 2 − ~r 1 ), t = 0.. 1.
Lemma 3.6.2 The line integral of the linear vector field ~L(x, y) = (ax + by)~ı + (cx + dy)~ over the edges of the rectangle with corners (x 0 ± ∆x, y 0 ± ∆y) equals ∮ C
L^ ~ · dr~ = (c − b) · 4∆x∆y (3.4)
Proof. (outline). Use the observation about the exactness of the midpoint rule, and calculate: ∮
C
L^ ~ · dr~ =
∮
C 1
L^ ~ · dr~ +
∮
C 2
~L · dr~ +
∮
C 3
L^ ~ · dr~ +
∮
C 4
~L · dr~
= L~(x 0 , y 0 − ∆y) · (2∆x~ı) + L~(x 0 + ∆x, y 0 ) · (2∆y~) + L^ ~(x 0 , y 0 + ∆y) · (−2∆x~ı) + ~L(x 0 − ∆x, y 0 ) · (−2∆y~)
= (ax 0 + b(y 0 − ∆y)~ı + (.. .)~) · (2∆x~ı) + ((.. .)~ı + c(x 0 + ∆x) + dy 0 ~) · (2∆y~) + (ax 0 + b(y 0 + ∆y)~ı + (.. .)~) · (−2∆x~ı) + ((.. .)~ı + c(x 0 − ∆x) + dy 0 ~) · (−2∆y~) =...
= (c − b) · 4∆x∆y.
Take a closer look how the scaling by area comes to be: It really consists of two components. On one side, there is the contribution of the increments ∆x and ∆y to the change in the vector field, e.g. comparing the vector field along the lower edge and along the upper edge. Dividing by the distance between these edges gives a measure of a rate of change of the vector field. The other increments ∆x and ∆y are contributed by the lengths of the curve segments. On a very small scale the vector field is almost constant, and the value of the line integral along each edge is almost directly proportional to the length of the line segment. Taken together, the line integral over the closed contour is scaled by the square of the linear dimension, or by the area of the region enclosed by the curve.
Exercise 3.6.4 Fill in the omitted details in the calculation above.
Exercise 3.6.5 Replace the rectangle above by a triangle with two sides aligned with the coordinate axes. E.g. choose the corners to be (x 0 , y 0 ), (x 0 + ∆x, y 0 ), and (x 0 , y 0 + ∆y). Carry out the analogous calculation.
Exercise 3.6.6 Replace the rectangle above by a triangle with two sides aligned with the coordinate axes. E.g. choose the corners to be (x 0 , y 0 ), (x 0 + ∆x, y 0 ), and (x 0 , y 0 + ∆y). Carry out the analogous calculation.
Lemma 3.6.3 Suppose C is the curve consisting of the edges of the triangle with corners at (x 1 , y 1 ), (x 2 , y 2 ), and (x 3 , y 3 ) (oriented counter clockwise). If ~L is the linear vector field ~L(x, y) = (ax + by)~ı + (cx + dy)~, then
∮
C
~L · dr~ = (c − b) · ( area of the triangle ) (3.5)
Exercise 3.6.7 Prove the lemma (3.6.3) via direct calculation analogous to that for lemma (3.6.2)
Exercise 3.6.8 Using linearity rewrite the vector field L~ as L~ = L~ 0 + ∆~L where ~L 0 (x, y) = ~L(x 0 , y 0 ) for all (x, y) and ∆~L(x, y) = L~(x − x 0 , y − y 0 ). (i) Explain why
∮ C L~^0 ·^ dr~^ = 0^ and hence^
∮ C L~^ ·^ dr~^ =^
∮ C ∆~L^ ·^ dr~. (ii) Evaluate
−→ ∮^ ∆L^ at the midpoints of the four edges of the rectangle. Use the midpoint rule to evaluate C ∆~L^ ·^ dr~. (iii) Alternatively, observe that
−→ ∆L (x 0 +∆x, y 0 +∆y) = L~(∆x, ∆y), and hence the line integral
∮ C ∆~L·^ dr~ equals
∮ C 0 ~L^ ·^ dr~^ where^ C^0 is the curve^ C^ translated back to the origin (i.e.^ each point^ (x, y)^ is moved to the point (x − x 0 , y − y 0 )). Thus it suffices to consider rectangles centered at the origin! Repeat the calculations for the case of (x 0 , y 0 ) = (0, 0). (iv) Compare these approaches with the calculations above. Which one illustrates best what is going on? (v) Go even a step further. Using linearity to combine some of the terms occurring in the calculation. E.g., simplify L~(∆x, 0) − ~L(−∆x, 0) = 2∆xL~(1, 0), and eventually arrive at
∮
C
~L · dr~ =
( ~L(1, 0) · ~ − ~L(0, 1) ·~ı
) · ( area of the region R). (3.6)
Note that this formulation does not make any explicit reference to the coefficients of the vector field when written out in rectangular coordinates. This very geometric point of view is most useful when working with polar or other curvilinear coordinates.
So far we have proven that the conjecture holds true for (linear vector fields integrated over) rectangles that are aligned with the coordinate axes, and, in the exercises, for any triangles. The next step is to establish that the conjecture is true for (linear vector fields integrated over) arbitrary polygonally bounded regions.
Exercise 3.6.10 Consider a region R that lies between two polygonal curves C 1 and C 2 , oriented as shown (i.e. so that the region always lies to the left as one moves along the curve). By carefully going through the steps of the previous arguments, show that it is still true that
∮ C L~· dr^ ~ = (c−b) ( area of R) if C now denotes the generalized curve consisting of the two connected pieces C 1 and C 2.
Exercise 3.6.11 Develop a formula for the area of a region that is bounded by a polygonal curve. The formula should take as input the list (x 1 , y 1 ),... , (xn, yn) of coordinates of the vertices (corners). Advice: Pick a suitable linear vector field ∮ ~L – there are many possibilities – so that
C L~^ ·^ dr~^ = 1^ ·^ (^ area of^ R). Utilize the midpoint or trapezoidal rule, each of which is gives the exact value for linear vector fields.
Finally consider regions that are bounded by piecewise smooth curves C. Any such curve may be arbitrarily closely approximated by a polygonal curve CP. For the sake of clarity we may assume that the polygonal curve CP lies entirely inside the smooth curve C. On one side we want the areas of the regions inside the curves C and CP to be arbitrarily close together. On the other hand we also want also the line integrals ∮ C ~L^ ·^ dr~^ and^
∮ CP^ ~L^ ·^ dr~^ to be arbitrarily close together. This requires two arguments.
The easy part is that the vector field ~L has almost identical values on corresponding points on the respective curves (by hypothesis it is uniformly continuous). It takes a little bit more work to justify
that the polygonal curve can also be chosen such that also the velocity vectors dr~ dt are arbitrarily close at corresponding points. [[this deserves an elegant argument – idea is to zoom in very much, so that the smooth curve looks practically straight. The rest is just book-keeping, using uniformity, bounds on ~r′′
... ]]. This completes the outline of the proof of the conjecture, formulated now as a proposition.
Proposition 3.6.5 Suppose ~L is the linear vector field ~L(x, y) = (ax + by)~ı + (cx + dy)~, and R is a region in the plane (possibly with “holes”. Let ∂R = C denote the oriented, piecewise smooth boundary of R. Then (^) ∮
∂R
~L · dr~ = (c − b) · ( area of R ) (3.8)
In words: For line integrals of linear vector fields over any region that is bounded by piecewise smooth curves: The ratio of the line integral divided by the area of the enclosed region is a constant. This constant is independent of the location, the shape, and the size of the curve. The constant may be considered as an analogue of the slope of a straight line. As such it will be the precursor for a geometric definition of the scalar curl, one derivative of vector fields.
This proposition also reaffirms the curl test of the previous section in the special case of linear vector fields.
Corollary 3.6.6 A linear vector field L~(x, y) = (ax + by)~ı + (cx + dy)~ is conservative if and only if b = c.
Exercise 3.6.16 Prove the lemma (3.6.9) via direct calculation analogous to that for lemma (3.6.8)
Exercise 3.6.17 In analogy to exercise (3.6.8) use linearity rewrite the vector field ~L as ~L = L~ 0 + ∆~L where ~L 0 (x, y) = ~L(x 0 , y 0 ) for all (x, y) and ∆~L(x, y) = L~(x − x 0 , y − y 0 ). (i) Explain why
∮ C L~^0 ·^ N ds~^ = 0^ and hence^
∮ C ~L^ ·^ N ds~^ =^
∮ C ∆~L^ ·^ N ds~. (ii) Evaluate
−→ ∮^ ∆L^ at the midpoints of the four edges of the rectangle. Use the midpoint rule to evaluate C ∆~L^ ·^ N ds~. (iii) Alternatively, observe that
−→ ∮^ ∆L^ (x^0 + ∆x, y^0 + ∆y) =^ L~(∆x,^ ∆y), and hence the flux line integral C ∆~L^ ·^ N ds~^ equals^
∮ C 0 L^ ~^ ·^ N ds~^ where^ C^0 is the curve^ C^ translated back to the origin (i.e.^ each point (x, y) is moved to the point (x − x 0 , y − y 0 )). Thus it suffices to consider rectangles centered at the origin! Repeat the calculations for the case of (x 0 , y 0 ) = (0, 0). (iv) Compare these approaches with the calculations in the two preceding exercises. Which one illustrates best what is going on? (v) Go even a step further. Using linearity to combine some of the terms occurring in the calculation. E.g., simplify L~(∆x, 0) − ~L(−∆x, 0) = 2∆xL~(1, 0), and eventually arrive at
∮
C
L^ ~ · N ds~ =
( ~ı · L~(1, 0) + ~ · L~(0, 1)
) · ( area of R). (3.12)
Exercise 3.6.18 Similar to exercise (3.6.18) consider a region R that lies between two polygonal curves C 1 and C 2 , oriented as shown in exercise (3.6.18). The region always lies to the left as one moves along the curve, and thus the normal vector always points outward from the region. Note that for the inner curve this means that the outward normal points into the hole. By carefully going through the steps of prior arguments, show that it is still true that
∮ C ~L^ ·^ dr~^ = (c^ −^ b) (^ area of^ R)^ if^ C^ now denotes the generalized curve consisting of the two connected pieces C 1 and C 2.
Exercise 3.6.19 Develop a formula for the area of a region that is bounded by a polygonal curve. The formula should take as input the list (x 1 , y 1 ),... , (xn, yn) of coordinates of the vertices (corners). Advice: Pick a suitable linear vector field ~L – there are many possibilities – so that
∮ C L~^ ·^ dr~^ = 1^ ·^ (^ area of^ R). Utilize the midpoint or trapezoidal rule, each of which is gives the exact value for linear vector fields.
Proposition 3.6.10 Suppose ~L is the linear vector field ~L(x, y) = (ax + by)~ı + (cx + dy)~, and R is a region in the plane (possibly with “holes”. Let ∂R = C denote the oriented, piecewise smooth boundary of R. Then ∮ ∂R
L^ ~ · N ds~ = (a + d) · ( area of R ) (3.13)
In words: For flux line integrals of linear vector fields over any region that is bounded by piecewise smooth curves: The ratio of the line integral divided by the area of the enclosed region is a constant. This constant is independent of the location, the shape, and the size of the curve. The constant may be considered as an analogue of the slope of a straight line. As such it will be the precursor for a geometric definition of the divergence, another derivative of vector fields.
