MATH 255 Line integrals III Fall 216 In general 1. The fundamental theorem of line integrals v T ds depends on the curve between the starting point and the ending point. onsider two was to get from (1, ) to (, 1): (1) 1 : start at (1, ) and first go along the -ais to the origin; then go up the -ais to (, 1). (2) 2 : start at (1, ) and first go straight up to (1, 1); then go parallel to the -ais to (, 1). Suppose v =, 2. To parametrize things easil, let us do four line integrals along lines: L 1 : (, ) to (1, ); L 2 : (, ) to (, 1); L 3 : (, 1) to (1, 1) and L 4 : (1, ) to (1, 1). All t intervals are [, 1]. L 1 : r(t) = t, ; r (t) = 1, dt; v ( r(t) ) 1 =, 2t so v T ds = dt =. L 2 : r(t) =, t; r (t) =, 1 dt; v ( r(t) ) 1 = t, so v T ds = dt =. L 3 : r(t) = t, 1; r (t) = 1, dt; v ( r(t) ) 1 = 1, 2t so v T ds = 1 dt = 1. L 4 : r(t) = 1, t; r (t) =, 1 dt; v ( r(t) ) 1 = t, 2 so v T ds = 2 dt = 2. Then v T ds = v T ds + v T ds = 1 L 1 L 2 v T ds = v T ds v T ds = 2 1 = 1 2 L 4 L 3 Be sure ou understand these equations. The figure below should eplain the various signs. Figure 5: Some oriented lines The four lines L i are labeled and arrows have been added to show the orientations induced b the parametrizations given above. Then 1 = ( L 1 ) (L 2 ) and 2 = (L 4 ) ( L 3 ) as oriented curves. Net v T ds = v T ds and v T ds = v T ds. Finall we have used the L 1 L 1 L 3 L 3 additivit of paths in line integrals (propert P4) to get v T ds from the v T ds. k L i There is a class of vector fields for which the line integral onl depends on the starting point and the ending point. These are precisel the conservative fields. To see this, let v = p so v is conservative and p is a potential function for it. Let g(t) = p ( r(t) ). Then g (t) = p ( r(t) ) r (t) b the hain Rule and so 1
2 so v T ds = b a v ( r(t) ) r (t) dt = v T ds = p ( r(b) ) p ( r(a) ) b a p ( r(t) ) r (t) dt = b a g (t) dt = g(b) g(a) This result is called the Fundamental Theorem of Line Integrals and has man uses, both practical and theoretical. One theoretical use is to show that some fields are not conservative. For eample, we just showed that, 2 is not conservative since 1 and 2 have the same starting points and the same ending points but the two line integrals are not equal. The converse is also true. An vector field v such that the value of v T ds onl depends on the starting point and the ending point of and not on the actual curve is conservative. Here we are require this for all curves which lie in the domain of v. This seems a largel theoretical result since ou have no practical wa to check the hpothesis unless ou actuall see a potential function in which case ou alread know the field is conservative. 2. onservative vector fields lairaut s Theorem supplies another wa to check if a field is conservative which is much more practical. If M, N = p then M = p and N = p. Note M = 2 M and N = 2 N. If these are both continuous, lairaut s Theorem sas M = N The continuit hpothesis is rarel a problem and so we onl need to compute a couple of partials. For the field in the last section,, 2, M = = 1 whereas N = 2 = 2 so here is another reason wh, 2 is not conservative. You can verif the result for fields in space. M = N provided the displaed functions are continuous. M z = P N z = P It would be reall nice if this necessar condition were sufficient but alas it is not. Eample. Let v = 2 +,. 2 2 + 2 M = 2 +, M 2 = + 2 ) (2) 1(2 = 2 2 ( 2 + 2 ) 2 ( 2 + 2 ) 2 N = 2 +, N 2 = 1(2 + 2 ) (2 = 2 2 ( 2 + 2 ) 2 ( 2 + 2 ) 2 so M = N
Let us do v T ds for this v around the unit circle centered at the origin. One parametrization for is r(t) = cos t, sin t t 2π dr = sin t, cos t dt and v ( r(t) ) = sin t, cos t so v dr = 2π sin t, cos t sin t, cos t dt = 2π Hence 2 +, is not conservative. 2 2 + 2 Here is a picture of the associated direction field (the vector field rescaled so that all vectors having length 1). 3 an ou see from the figure wh v = 2 +, 2 2 + 2 v dr? lick here for the answer. Fields which satisf M = N or the 3 dimensional version above are called closed and lairaut s Theorem sas that conservative fields are closed provided the mied partials of a potential function are continuous. The problem of which closed fields are conservative started modern algebraic topolog. One sure wa to show that a field is conservative is to ehibit a potential function for it. Let us start with the 2 dimensional case. Fi a field M, N. We can start with either component so let us find a function p 1 (, ) such that p 1(, ) = M. This is a first ear calculus problem so have at it. Once ou have found one solution, p 1 (, ) all solutions can be found b adding a constant
4 (in or equivalentl, a function of, so the potential function p(, ) = p 1 (, ) + g() for some function g(). Then p = N or p 1(, ) + g () = N so g () = N p 1(, ). This equation has a solution provided h(, ) = N p 1(, ) is a continuous function of onl and this is guaranteed if h(, ) =. But with our mied partial assumptions h(, ) = N 2 p 1 = N 2 p 1 = N M = If M and N are differentiable functions and if the field M, N is closed, then there is a potential function and ou can find it if ou can do some first ear calculus integrals. There is a similar result in three variables. The calculation takes one more step. Eample. Let,. heck (easil) that, is closed. To find p, first solve p = or p = + g() + g(). Then solve = or g () = so p = is one potential function and an other looks like + c. Actuall, there is no reason to check if the field is closed. Here s an eample of what happens if it is not. Eample. Let, 2. To find p, first solve p + g() = or p = +g(). Then solve = 2 or + g () = 2 so g () = and this is impossible. This sort of issue is our clue that the original field is not closed. Here is another eample Eample. Let 2 + 2, 2. This time let s start with p = 2 so p = 2 + h( and then 2 + h( = 2 + h ( = 2 + 2 so h ( = 2 and p = 2 + 3 3 + c. Here is a 3 dimensional eample 2 Eample. Let z 2 + h(, z) solve z = 2 z 2 z + z + g(z) z 2 + z, z, 2 z + + 2 ez = 2 z 2 + h = 2 z + + g (z) = 2 z 2. Start with p = 2 z so p = 2 z + z so h + h(, z). Then = z and h(, z) = z + g(z) and finall + + ez and g (z) = e z so p = 2 z + z + ez + c. Let us tr to appl this technique to the closed form v = 2 +,. 2 2 + 2 p = ( ) 2 + so p = arctan + g( and 2 ( p arctan = + g( = 2 1 + ( ) 2 + g ( = 2 + 2
so ( p(, ) = arctan is a potential function. But v has no potential function. What is wrong? lick here for the answer. 5
6 Recall the question. Appendi A. Line integral question. v = 2 +, 2 2 + 2 is a direction-field and we asked wh is v dr where is the unit circle centered at the origin. The answer is that is a flow line for the direction field and hence a flow line for the field. learl v T > along the curve and hence v dr > if we traverse the circle counterclockwise. From the figure one suspects is a flow line and clearl it is close one. With no calculation it is prett clear that v T > along the curve and hence v dr >. To see wh is a flow line, note ( 2 + 2 ) = 2, 2 is a normal vector to and so, is a tangent vector. Along, 2 + 2 = 1 so, = 2 +, which shows is a flow 2 2 + 2 line for the field.
7 Appendi B. Wh is our potential function not a potential function. Recall the issue. v = 2 +, is a closed vector field which is not conservative. 2 2 + 2 ( p = arctan 1 p = 2 1 + ( ) 2, 1 + ( ) 2 = 2 +, = v 2 2 + 2 which sure looks like the definition of conservative. The solution is to notice that p is not defined on the same domain as v. The domain of v is R 2 { (, ) } but the domain of p is R = R 2 { (, ) } < <, i.e. R 2 minus the -ais. Equivalentl R = { (, ) (, ) R 2 } If we restrict v to R, then v is conservative. Note R is simpl-connected and we have a theorem which sas v restricted to R is conservative. We have actuall ehibited a potential function for it. ( ) 1 Recall arctan(θ) + arctan = π ( ) θ 2 for θ >. Therefore arctan = π ( ) 2 arctan if ( ) >. Hence p 1 = arctan is also a potential function for v: i.e. p 1 = v. Hence v is also conservative on the region R 1 = R 2 { (, ) } { } < < = (, ) R 2. Let S = R 2 { (, ) < < }. Then S = R > R > where R > = { } (, ) R 2 > and R > = { (, ) R 2 } >. Note R > R > = { (, ) R 2 } > and >. Define ( arctan (, ) R > q(, ) = ( ) π 2 arctan (, ) R 1 > Then q = v on S so v is conservative on S which is simpl-connected. The colored part of the plane below is S. R > R > R > is the union of the ellow and blue regions, while R > is the union of the green and blue regions. S
8 FInall, let T = { (, ) R 2 < < }. R > R < The union of the red and green regions is R < = { (, ) R 2 < } ( ). If θ <, arctan(θ) + 1 arctan = π θ 2 so π ( ) ( 2 arctan = π + arctan in R 1 >. Hence ( arctan (, ) > and < < T q 1 (, ) = π 2 arctan ( ) ( π + arctan (, ) (, ) > and < < < and < < is a potential function for v on T. An equivalent wa to define q 1 is ( arctan (, ) > and < < π q 1 (, ) = (, ) > 2 ( π + arctan (, ) < and < < It is harder to see that q 1 is a differentiable function using the second definition. For an <, lim q 1(, t) = π and lim q 1(, t) = so there is no wa to further etend q 1 to t t + include part of <.