Lecture 26: Conservative Vector Fields

Size: px

Start display at page:

Download "Lecture 26: Conservative Vector Fields"

Maximillian Marvin Andrews
6 years ago
Views:

1 Lecture 26: onservative Vector Fields 26. The line integral of a conservative vector field Suppose f : R n R is differentiable the vector field f : R n R n is continuous. Let F (x) = f(x). Then F is a conservative vector field. If ϕ : [a, b] R n is a smooth parametrization of a curve, then f dr = = b a b a f(ϕ(t)) ϕ (t)dt d dt f(ϕ(t))dt = f(ϕ(t)) b a = f(ϕ(b)) f(ϕ(a)) = f(b) f(a), where a = ϕ(a) is the initial point of b = ϕ(b) is the terminal point of. Example The vector field is a conservative vector field with potential F (x, y, z) = (x, y, z) (x 2 + y 2 + z 2 ) 3 2 f(x, y, z) = x2 + y 2 + z 2. if is a curve with initial point (,, ) terminal point ( 2, 2, 3), then f( 2, 2, ) f(,, ) = 3 = Path independence Definition Suppose F : R n R n is a continuous vector field. If for any two curves 2 in the domain of F with the same initial terminal points we have F dr, 2 then we say F dr is independent of path. 26-

2 Lecture 26: onservative Vector Fields 26-2 Definition curve. A curve whose initial terminal points are the same is called a closed Proposition F dr is independent of path if only if for every closed path in the domain of F. Proof First suppose F dr is independent of path let be a closed curve. Let a b be two points on. Let be the part of from a to b let 2 be the part of from b to a. Then F dr + 2 F dr, 2 where the final equality follows from the fact that 2 have the same initial terminal points. Now suppose for any closed curve. Let 2 be two curves, both with initial point a terminal point b. Let be the closed curve obtained by first traversing then traversing 2. Then = F dr + 2 F dr F dr, 2 so F dr. 2 Definition We say a region D in R n is path connected, or connected, if for any two points a b in D there exists a path from a to b which lies entirely within D. Proposition Suppose F : R n R n is a continuous vector field defined on an open connected region D. If F dr is independent of path in D, then F is a conservative vector field. Proof We will prove the proposition for n = 2. Let (a, b) be a point in D. We define a scalar field f : R n R by f(x, y) = F dr, where is a path with initial point (a, b) terminal point (x, y). Note that, because of path independence, the value of f(x, y) depends only on (x, y), not on the choice for. Let F (x, y) = P (x, y)i + Q(x, y)j. Now f(x, y) = lim x h f(x + h, y) f(x, y). h

3 Lecture 26: onservative Vector Fields 26-3 Let be a path from (a, b) to (x, y) let be a horizontal path from (x, y) to (x+h, y). Then f(x + h, y) f(x, y) = F dr + F dr F dr. = F dr If we parametrize by ϕ(t) = (x + t, y), t h, then ϕ (t) = (, ) h (P (x + t, y), Q(x + t, y)) (, )dt = h f(x, y) = lim x h A similar calculation shows that so f(x, y) = F (x, y). P (x + t, y)dt h h = lim h P (x + h, y) (using l Hôpital s rule) = P (x, y). f(x, y) = Q(x, y), y P (x + t, y)dt. Proposition If F (x, y) = P (x, y)i + Q(x, y)j is a conservative vector field P Q have continuous partial derivatives, then P y = Q x. Proof Let f be a potential function for F. Then P y = 2 f(x, y) = 2 yx Q f(x, y) = xy x. Example The vector field F (x, y) = (x 2 y, xy 3 ) is not conservative since y (x2 y) = x 2 x (xy3 ) = y 3.

4 Example If we let then Let Lecture 26: onservative Vector Fields 26-4 F (x, y) = x 2 ( y, x). + y2 P (x, y) = y x 2 + y 2 Q(x, y) = x x 2 + y 2, P y = (x2 + y 2 ) + 2y 2 (x 2 + y 2 ) 2 = y2 x 2 (x 2 + y 2 ) 2 Q x = (x2 + y 2 ) 2x 2 (x 2 + y 2 ) 2 = y2 x 2 (x 2 + y 2 ) 2. P y = Q x. However, F is not a conservative vector field. For example, if is the unit circle centered at the origin parametrized by ϕ(t) = (cos(t), sin(t)), t 2π, then 2π ( sin(t), cos(t)) ( sin(t), cos(t))dt = F dr, so F cannot be conservative. It follows from the previous example that the condition P y = Q x, 2π dt = 2π. although a necessary condition for a vector field F (x, y) = P (x, y)i + Q(x, y)j to be conservative, is not a sufficient condition. The problem in the previous example turns out to be the nature of the domain of the vector field. Note that the closed curve which yields the nonzero line integral contains the origin, a point at which F is not defined. We will see in the next section that if P Q have continuous partial derivatives on an open region D, P y = Q x for all (x, y) in D, D has the property that for any closed curve in D, all points inside lie in D (that is, D has no holes ), then F is conservative. We say that such a region D is simply connected.

5 Example If we let then Suppose Lecture 26: onservative Vector Fields 26-5 F (x, y) = (2xy + 2x, x 2 6y). P (x, y) = 2xy + 2x Q(x, y) = x 2 6y, P (x, y) = 2x = Q(x, y). y x Since the domain of F is all of R 2, it follows that F is conservative. That is, there exists a scalar field f for which f(x, y) = 2xy + 2x x y f(x, y) = x2 6y. From the first of these two equations, we have f(x, y) = (2xy + 2x)dx = x 2 y + x 2 + g(y) for some function g which depends only on y. Now we must have x 2 6y = y f(x, y) = y (x2 y + x 2 + g(y)) = x 2 + g (y), from which it follows that g (y) = 6y. g(y) = 3y 2 + c for some constant c. Thus for any constant c the function f(x, y) = x 2 y + x 2 3y 2 + c is a potential function for the vector field F.

47. Conservative Vector Fields

47. Conservative Vector Fields 47. onservative Vector Fields Given a function z = φ(x, y), its gradient is φ = φ x, φ y. Thus, φ is a gradient (or conservative) vector field, and the function φ is called a potential function. Suppose