ES.182A Topic 41 Notes Jerem Orloff 41 Etensions and applications of Green s theorem 41.1 eview of Green s theorem: Tangential (work) form: F T ds = curlf d d M d + N d = N M d d. Normal (flu) form: F n ds = divf d d M d N d = M + N d d. T n Both forms require (for now): A curve that is a positivel oriented (interior on left), simple closed curve and is its interior. T = unit tangent vector to. n = unit normal to = clockwise rotation of T. (It points out from. F is defined and differentiable on all of. 41.2 Simpl connected regions We will need the notion of a simpl connected region. We will stick with an informal definition of simpl connected that will be sufficient for our purposes. (For those who are interested: We will assume that a simple closed curve has an inside and an outside. This is intuitive and is eas to show if is a smooth curve, but turns out to surprisingl hard if we allow to be strange, e.g. a Koch snowflake.) Definition: A region D in the plane is called simpl connected if, for ever simple closed curve that lies entirel in D the interior of also lies entirel in D. 1
41 EXTENSIONS AND APPLIATIONS OF GEEN S THEOEM 2 Eamples: D 1 D 2 D 3 D 4 D 5 = whole plane D1-D5 are simpl connected, since for an simple closed curve inside them its interior is entirel inside the region. This is sometimes phrased as each region has no holes. Note. An alternative definition, which works in higher dimensions is that the region is simpl connected if an curve in the region can be continuousl shrunk to a point without leaving the region. The regions at below are not simpl connected. That is, the interior of the curve is not entirel in the region. Annulus Puntured plane 41.3 Potential theorem and conservative fields As an application of Green s theorem we can now give a more complete answer to our question of how to tell if a field is conservative. The theorem does not have a standard name, so we choose to call it the Potential Theorem. You should check that it is largel a restatement for simpl connected regions of Theorem 1 in Topic 38. Theorem. (Potential theorem) Take F = M, N defined and differentiable on a region D. (a) If F = f then curlf = N M = 0. (b) If D is simpl connected and curlf = 0 on D, then F = f for some f. Notes. 1. We know that on a connected region, being a gradient field is equivalent to being conservative. So we can restate the Potential theorem as: On a simpl connected region, F is conservative is equivalent to curlf = 0. 2. ecall that once we know work integral is path independent, we can compute the potential function f b picking a base point P 0 in D and letting f(q) = F dr, where is an path in D from P 0 to Q.
41 EXTENSIONS AND APPLIATIONS OF GEEN S THEOEM 3 Proof of (a): This was proved in Theorem 1 of Topic 38. Proof of (b): Suppose is a simple closed curve in D. Since D is simpl connected the interior of is also in D. Therefore, using Green s theorem we have, curlf da = 0. D This shows that F is conservative in D. Therefore b Theorem 1 in Topic 38, F is a gradient field. 41.4 Summar Suppose F = Mi + Nj = M, N on a simpl connected region D. The following statements are equivalent. 1. 2. Q P Q P M d + N d is path independent. M d + N d = 0 for an closed path. 3. F = f for some f in D. (Equivalentl M d + N d = df.) If F is continuousl differentiable then all of these impl 4. curlf = N M = 0 in D, i.e. M = N in D. Note. The most important eample of a simpl connected D is the entire -plane. 41.5 Wh we need D simpl connected in the Potential theorem The basic idea is that if there is a hole in D, then F might not be defined on the interior of. This is illustrated in the net eample. Eample 41.1. (What can go wrong if D is not simpl connected.) Here we will repeat the super-duper reall important eample from Topic 36. i + j Let F = ( tangential field ). r 2 F is defined on D = plane - (0,0) = punctured plane
41 EXTENSIONS AND APPLIATIONS OF GEEN S THEOEM 4 Puntured plane Several times now we have shown that curlf = 0. (If ou ve forgotten this, ou should recompute it now.) We also know that on an circle of radius a centered at the origin F T = 1/a, so F T ds = 2π. So, the conclusion of the above theorem that curlf = 0 implies F is conservative does not hold. The problem is that D is not simpl connected and, in fact, F is not defined on the entire region inside, so we can t appl Green s theorem to conclude that the line integral is 0. 41.6 Etended Green s theorem We can etend Green s theorem to a region which has multiple boundar curves. The figures below show regions bounded b 2 or more curves. You will see that this gives us awa to work around singularities in the field F. 41.6.1 egions with multiple boundar curves onsider the following three regions. B 3 A 7 1 2 8 4 5 6 The region on the left, A is bounded b 1 and 2. We sa that the boundar is 1 + 2. Note that the wa it is drawn, the region is alwas to the left as ou traverse either boundar curve. The region on the right, is bounded b 7 and 8. We sa that the boundar is 7 8. The reason for the minus sign is that the boundar curves should be oriented so that the region is to our left as ou traverse the curve. As shown, the region is to the right of 8, but to the left of 8. Likewise, in the middle figure, B has boundar 3 + 4 + 5 + 6. You should check that our signs are consistent with the orientation of the curves.
41 EXTENSIONS AND APPLIATIONS OF GEEN S THEOEM 5 41.7 Etended Green s theorem Theorem. Etended Green s theorem. Suppose A is the region in the left-hand figure above then, for an vector field F differentiable in all of A we have curlf d d. 1 + 2 A Likewise for more than two curves: If B has boundar 3 + 4 + 5 + 6 and F is differentiable on all of B then curlf d d. 3 + 4 + 5 + 6 B Proof. We will prove the formula for A. The case of more than two curves is essentiall the same. The ke is to make the cut shown in the figure below, so that the resulting curve is simple. A 3 3 2 1 In the figure the curve 1 + 3 + 2 3 surrounds the region A. (You have to imagine that the cut is infinitesimall wide so 3 and 3 are right on top of each other.) Now the original Green s theorem applies: 1 + 3 + 2 3 curlf d d Since the contributions of 3 and 3 will cancel, we have proved the etended form of Green s theorem. curlf d d. QED 1 + 2 i + j Eample 41.2. Again, let F be the tangential field F = r 2. What values can F dr take for a simple closed positivel oriented curve that doesn t go through the origin? answer: We have two cases (i) 1 does not go around 0; (ii) 2 goes around 0 1 1 2 2
41 EXTENSIONS AND APPLIATIONS OF GEEN S THEOEM 6 ase (i) We know F is defined and curlf = 0 in the entire region inside 1, so Green s theorem implies curlf d d = 0. 1 ase (ii) We can t appl Green s theorem directl because F is not defined everwhere inside 2. Instead, we use the following trick. Let 3 be a circle centered on the origin and small enough that is entirel inside 2. 2 3 2 The region 2 has boundar 2 3 and F is defined and differentiable in 2. We know that curlf = 0 in 2, so etended Green s theorem implies curlf d d = 0. 2 3 2 So F dr. 2 3 Since 3 is a circle centered on the origin we can compute the line integral directl we ve done this man times alread. On 3, F T = 1/a, so 1 2πa F T ds = ds = 3 3 a a = 2π. Therefore, for a simple closed curve and F as given, the line integral F dr is either 2π or 0, depending on whether surrounds the origin or not. Answer to the question: The onl possible values are 0 and 2π. Eample 41.3. Use the same F as in the previous eample. What values can the line integral take if is not simple. answer: If is not simple we can break it into a sum of simple curves. 1 2
41 EXTENSIONS AND APPLIATIONS OF GEEN S THEOEM 7 In the figure, we can think of the entire curve as 1 + 2. Since each of these curves surrounds the origin we have 1 + 2 F dr + 1 2π + 2π = 4π. 2 In general, 2πn, where n is the number of times goes around (0,0) in a counterclockwise direction. Aside for those who are interested: The integer n is called the winding number of around 0. The number n also equals the number of times crosses the positive -ais, counting +1 if it crosses from below to above and 1 if it crosses from above to below. 41.8 More eamples Eample 41.4. Is d d 2 eact? If so find a potential function. answer: The functions M = 1, N = 2 are continuosl differentiable whenever 0 i.e, in the two half-planes 1 and 2. Both 1 and 2 are simpl connected 1 2 It is eas to compute that M = 1/ 2 = N. Thus, in each half-plane M d + N d is eact. We use method 2 to find a potential function f. f = M = 1/ f = / + g(); f = / 2 + g () = N = / 2. Thus, g () = 0, so g() = c. We get f(, ) = / + g() = / + c. Eample 41.5. Let F = r n (i + j). For n 0, F is defined on the entire plane. For n < 0, F is defined on the -plane minus the origin (the punctured plane). Use etended Green s theorem to show that F is conservative on the punctured plane for all integers n. Then, find a potential function. answer: We start b computing the curl: M = r n M = nr n 2 N = r n N = nr n 2
41 EXTENSIONS AND APPLIATIONS OF GEEN S THEOEM 8 So, curlf = N M = 0. To show that F is conservative in the punctured plane, we will show that all simple closed curves that don t go through the origin. 0 for If is a simple closed curve not around 0 then F is differentiable on the entire region inside and Green s theorem implies curlf d d = 0. If is a simple closed curve that surrounds 0, then we can use the etended form of Green s theorem as in Eample 41.2. 2 We put a small circle 2 centered at the origin and inside. Since F is radial, it is orthogonal to 2. So, on 2, F T = 0, which implies 2 F T ds = 0. Now, on the region with boundar 2 we can appl the etended 2 Green s theorem curlf d d = 0. 2 Thus, F dr, which, as we saw, equals 0. 2 Thus 0 for all closed loops, which implies F is conservative. QED To find the potential function we use method 1 over the curve = 1 + 2 shown. 2 ( 1, 1 ) 1 (1, 1) The following calculation works for n 2. For n = 2 everthing is the same ecept we get natural logs instead of powers. Parametrize 1 using : = 0, = ; from 1 to 1. So,
41 EXTENSIONS AND APPLIATIONS OF GEEN S THEOEM 9 d = 0, d = d, skip M, since d = 0, N = r n = (1 + 2 ) n/2. So, 1 M d + N d = (1 + 2 ) n/2 d 1 1 1 = (1 + 2 ) (n+2)/2 1 (u-substitution: u = 1 + 2 ) n + 2 Parametrize 2 using : =, = (1 + 2 1 )(n+2)/2 n + 2 1 2(n+2)/2 n + 2. = 1 ; from 1 to 1. So, d = d, d = 0, M = r n = ( 2 + 1) 2 n/2 skip, N since d = 0. So, 1 M d + N d = ( 2 + 1) 2 n/2 d 2 2 1 = (2 + 1 2 1 )(n+2)/2 (u-substitution: u = 2 + 2 n + 2 1) = (2 1 + 2 1 )(n+2)/2 n + 2 1 (1 + 2 1 )(n+2)/2. n + 2 Adding these we get f( 1, 1 ) f(1, 1) = (2 1 + 2 1 )(n+2)/2 2 (n+2)/2. So, n + 2 f(, ) = rn+2 n + 2 + c. (If n = 2 we get f(, ) = ln r +.) (Note, we ignored the fact that if ( 1, 1 ) is on the negative -ais we should have used a different path that doesn t go through the origin. This isn t reall an issue because we know there is a potential function. Because our function f is known to be a potential function everwhere ecept the negative -ais, b continuit it also works on the negative -ais.)