4 Divergence theorem and its consequences

Transcription

1 Tel Aviv University, 205/6 Analysis-IV 65 4 Divergence theorem and its consequences 4a Divergence and flux b Piecewise smooth case c Divergence of gradient: Laplacian d Laplacian at a singular point e Differential forms of order N The divergence theorem sheds light on harmonic functions and differential forms. 4a Divergence and flux We return to the case treated before, in the end of Sect. 3b: R N is a smooth set. Recall the outward unit normal vector n x for x. 4a Definition. For a continuous F : R N, the (outward) flux of (the vector field) F through is F, n. (The integral is interpreted according to (2d8).) If a vector field F on R 3 is the velocity field of a fluid, then the flux of F through a surface is the amount of fluid flowing through the surface (per unit time). 2 If the fluid is flowing parallel to the surface then, evidently, the flux is zero. We continue similarly to Sect. 3b. Let F C ( R N ), with DF bounded (on ). Recall that, by 3b6, boundedness of DF on ensures that F extends to by continuity (and therefore is bounded). In such cases we always use this extension. The mapping F : R N \ R N defined by { F (x) for x, F (x) = 0 for x / The volume is meant, not the mass. However, these are proportional if the density (kg/m 3 ) of the matter is constant (which often holds for fluids). 2 See also mathinsight.

2 Tel Aviv University, 205/6 Analysis-IV 66 is continuous up to, and F (x 0n x ) = F (x), F (x + 0nx ) = 0 ; div sng F (x) = F (x), nx. By Theorem 3e3 (applied to F and K = ), (4a2) div F = F, n, just the flux. The divergence theorem, formulated below, is thus proved. 4a3 Theorem (Divergence theorem). Let R N be a smooth set, F C ( R N ), with DF bounded on. Then the integral of div F over is equal to the (outward) flux of F through. In particular, if div F = 0, then F, n = 0. 4a4 Exercise. div(ff ) = f div F + f, F whenever f C () and F C ( R N ) Prove it. Thus, the divergence theorem, applied to ff when f C () with bounded f, and F C ( R N ) with bounded DF, gives a kind of integration by parts, similar to (3b2): (4a5) f, F = f F, n f div F. In particular, if div F = 0, then f, F = f F, n Here is a useful special case. We mean by a radial function a function of the form f : x g( x ) where g C (0, ), and by a radial vector field F : x g( x )x. Clearly, f C (R N \ {0}) and F C (R N \ {0} R N ). 4a6 Exercise. (a) If f(x) = g( x ), then f(x) = g ( x ) x; x (b) if F (x) = g( x )x, then div F (x) = x g ( x ) + Ng( x ); (c) if F (x) = g( x )x, then the (outward) flux of F through the boundary of the ball {x : x < r} is cr N g(r), where c = 2πN/2 is the area of the unit Γ(N/2) sphere. Prove it. 2 Divergence is often explained in terms of sources and sinks (of a moving matter). But be careful; the flux of a velocity field is the amount (per unit time) as long as amount means volume. If by amount you mean mass, then you need the vector field of momentum, not velocity; multiply the velocity by the density of the matter. However, the problem disappears if the density is constant (which often holds for fluids). 2 Hint: (b) use (a) and 4a4.

3 Tel Aviv University, 205/6 Analysis-IV 67 Taking = {x : a < x < b} and F (x) = g( x )x, we see that div F = b crn ( rg (r)+ng(r) ) dr by 4a6(b) and (generalized) 3c8; and on the other a hand, F, n = crn g(r) b d r=a by 4a6(c). Well, dr( r N g(r) ) = r N ( rg (r) + Ng(r) ), as it should be according to (4a2). Zero gradient is trivial, but zero divergence is not. For a radial vector field, zero divergence implies that r N g(r) does not depend on r, that is, g(r) = const (and indeed, in this case rg (r) + Ng(r) = 0); r N (4a7) F (x) = const x ; div F (x) = 0 for x 0 ; x N F, n = 0 when / 0 ; note that the latter equality fails for a ball. The flux through a sphere is 2π N/2 (4a8) F, n = const = const Γ(N/2) x =r x = where const is as in (4a7). The same holds for arbitrary smooth set 0: 2π N/2 (4a9) F, n = const Γ(N/2). Proof: we take ε > 0 such that {x : x ε} ; the set ε = {x : x > ε} is smooth; by (4a7), ε F, n = 0; and ε = {x : x = ε}. 4b Piecewise smooth case We want to apply the divergence theorem 4a3 to the open cube = (0, ) N, but for now we cannot, since the boundary is not a manifold. Rather, consists of 2N disjoint cubes of dimension n = N ( hyperfaces ) and a finite number of cubes of dimensions 0,,..., n. For example, {} (0, ) n is a hyperface. Each hyperface is an n-manifold, and has exactly two orientations. Also, the outward unit normal vector n x is well-defined at every point x of a hyperface. For example, n x = e for every x {} (0, ) n. For a function f on we define f as the sum of integrals over the 2N hyperfaces; that is, N (4b) f = f(x,..., x N ) dx j, i= x i =0, (0,) n j:j i n=n In fact, 3 N 2N.

4 Tel Aviv University, 205/6 Analysis-IV 68 provided that these integrals are well-defined, of course. For a vector field F C( R N ) we define the flux of F through as F, n. Note that (4b2) F, n = N i= x i =0, (2x i ) F i (x,..., x N ) (0,) n dx j. j:j i It is surprisingly easy to prove the divergence theorem for the cube. (Just from scratch; no need to use 4a3, nor 3e3.) 4b3 Proposition (divergence theorem for cube). Let F C ( (0, ) N R N), with DF bounded. Then the integral of div F over (0, ) N is equal to the (outward) flux of F through the boundary. (As before, boundedness of DF ensures that F extends to [0, ] N continuity; recall 3b6.) by Proof. 0 D F (x,..., x N ) dx = F (, x 2,..., x N ) F (0, x 2,..., x N ) = = (2x )F (x,..., x N ) ; D F = (0,) N x =0, (2x ) similarly, for each i =,..., N, D i F i = (0,) N it remains to sum over i. x i =0, x =0, (0,) n F (x,..., x N ) dx 2... dx N ; (2x i ) F i (0,) n dx j ; The same holds for every box, of course. A box is only one example of a bounded regular open set R N such that is not an n-manifold and still, the divergence theorem holds as div F = F, n for some closed set Z such that \ Z is an \Z n-manifold of finite n-dimensional volume. For the cube (or box), \ Z is the union of the 2N hyperfaces, and Z is the union of cubes (or boxes) of smaller (than N ) dimensions. j:j i

5 Tel Aviv University, 205/6 Analysis-IV 69 4b4 Definition. We say that the divergence theorem holds for and \ Z, if R N is a bounded regular open set, Z is a closed set, \ Z is an n-manifold of finite n-dimensional volume, and div F = F, n for all F C( \Z RN ) such that F C ( R N ) and DF is bounded on. 4b5 Exercise (product). Let R N, Z, and 2 R N 2, Z 2 2. If the divergence theorem holds for, \ Z and for 2, 2 \ Z 2, then it holds for, \ Z where = 2 R N +N 2 and \ Z = ( ) ( ( \ Z ) 2 ( 2 \ Z 2 ) ). Prove it. 2 An N-box is the product of N intervals, of course. Also, a cylinder {(x, y, z) : x 2 + y 2 < r 2, 0 < z < a} is the product of a disk and an interval. 4c Divergence of gradient: Laplacian Some (but not all) vector fields are gradients of scalar fields. 4c Definition. (a) The Laplacian f of a function f C 2 () on an open set R n is f = div f. (b) f is harmonic, if f = 0. We have f = (D f,..., D n f), thus, div f = D (D f)+ +D n (D n f); in this sense, = D Dn 2 = , x 2 x 2 n the so-called Laplace operator, or Laplacian. Any n-dimensional Euclidean space may be used instead of R n. Indeed, the gradient is well-defined in such space, and the divergence is well-defined even without Euclidean metric. The divergence theorem 4a3 gives, for a smooth, the so-called first reen formula (4c2) f = f, n = D n f, Not a standard terminology. 2 Hint: div F = (D F + + D N F N ) + (D N+F N+ + + D N+N 2 F N+N 2 ).

6 Tel Aviv University, 205/6 Analysis-IV 70 where ( D n f ) (x) = ( D nx f ) x is the directional derivative of f at x in the normal direction n x. Here f C 2 (), with bounded second derivatives. Here is another instance of integration by parts. Let u C (), with bounded gradient, and v C 2 (), with bounded second derivatives. Applying (4a5) to f = u and F = v we get u, v = u v, n u v, that is, (4c3) (u v + u, v ) = u v, n = ud n v, the second reen formula. It follows that (4c4) (u v v u) = (ud n v vd n u), the third reen formula; here u, v C 2 (), with bounded second derivatives. In particular, ud n v = vd n u for harmonic u, v. Rewriting (4c4) as (4c5) u v = v u vd n u + (D n v)u we may say that really (ul ) v = v (ul ) where (ul ) consists of the usual Laplacian ( u)l sitting on and the singular Laplacian sitting on, of two terms, so-called single layer ( D n u) and double layer ud n. Why two layers? Because the Laplacian (unlike gradient and divergence) involves second derivatives. 4c6 Exercise. Consider homogeneous polynomials on R 2 : f(x, y) = m c k x k y m k. k=0 For m =, 2 and 3 find all harmonic functions among these polynomials. 4c7 Exercise. On R 2, (a) a function of the form f(x, y) = m c k e a kx+b k y k= (a k, b k, c k R) In fact, they are Re (x + iy) m, Im (x + iy) m and their linear combinations.

7 Tel Aviv University, 205/6 Analysis-IV 7 is harmonic only if it is constant; (b) a function of the form is harmonic if and only if a = b. Prove it. f(x, y) = e ax cos by Now, what about a radial harmonic function? We seek a radial f such that f is of zero divergence, that is, f(x) = const x (recall (4a7)). By x N 4a6(a), f(x) = g( x ) where g (r) = const ; thus, g(r) = const r r N + const r N 2 2 for N 2. We choose (4c8) f(x) = ; f(x) = 0 for x 0. x N 2 (This works also for N = : f(x) = x is harmonic on R \ {0}.) But for N = 2 we get g (r) = const ; g(r) = const r log r + const 2 ; we choose (4c9) f(x) = log x = log ; f(x) = 0 for x 0. x The flux of f through a sphere is 2 x =r D n f = { (N 2) 2π N/2 Γ(N/2) for N 2, 2π for N = 2; and, similarly to (4a9), the same holds for every smooth set 0. 4d Laplacian at a singular point The function g(x) = / x N 2 is harmonic on R N \{0}, thus, for every f C 2 compactly supported within R N \ {0}, g f = f g = 0. It appears that for f C 2 (R N ) with a compact support, g f = const f(0) ; in this sense g has a kind of singular Laplacian at the origin. That is, f(x, y) = Re ( e x+iy). 2 const = (N 2)const = (N 2) for N 2, and const = const = for N = 2.

8 Tel Aviv University, 205/6 Analysis-IV 72 4d Lemma. R N f(x) 2πN/2 dx = (N 2) x N 2 Γ(N/2) f(0) for every N > 2 and f C 2 (R N ) with a compact support. This improper integral converges, since / x N 2 is improperly integrable near 0. The coefficient 2πN/2 is the (N )-dimensional volume of the unit Γ(N/2) sphere (recall (3c9)). Proof. For arbitrary ε > 0 we consider the function g ε (x) = / ( max( x, ε) ) N 2, and g(x) = / x N 2. Clearly, g ε g 0 (as ε 0), and g ε g f 0, thus, g ε f g f. We take R (0, ) such that f(x) = 0 for x R, introduce smooth sets = {x : x < ε}, 2 = {x : ε < x < R}, and apply (4c4), taking into account that g ε = 0 on and 2 : ( ) ( ) (gε ) g ε f = g ε f = D n f fd n g ε ; however, these D n must be interpreted differently under and 2 : g ε D n f = g ε D n2 f = 2 x =ε x =ε 2 ε D nf, N 2 ε D nf N 2 where n is the outward normal of and inward normal of 2 ; these two summands cancel each other. Further, fd n g ε = x =ε f 0 = 0 since g ε is constant on ; and fd n2 g ε = 2 x =ε f N 2 ε N, since g ε (x) = / x N 2 on 2, and f(x) = 0 when x = R. Finally, g ε f = (N 2) ε N x =ε f = (N 2) 2πN/2 Γ(N/2) f ε, where f ε is the mean value of f on the ε-sphere. By continuity, f ε f(0) as ε 0; and, as we know, g ε f g f.

9 Tel Aviv University, 205/6 Analysis-IV 73 4d2 Remark. For N = 2 the situation is similar: f(x) log dx = 2πf(0) R x 2 for every compactly supported f C 2 (R 2 ). When the boundary consists of a hypersurface and an isolated point, we get a combination of (4c5) and 4d: a singular point and two layers. 4d3 Remark. Let R N be a smooth set, f C 2 () with bounded second derivatives, and 0. Then f(x) 2πN/2 dx = (N 2) x N 2 Γ(N/2) f(0) ( ) ( ) x f(x)d n + x (D x N 2 n f(x)). x N 2 The proof is very close to that of 4d. The case N = 2 is similar to 4d2, of course. The case = {x : x < R} is especially interesting. Here = {x : x = R}; on, thus, x <R x = 2 and D N 2 R N 2 nx = N x N 2 R ; N f(x) 2πN/2 dx = (N 2) x N 2 Γ(N/2) f(0)+ N 2 f + D R N =R R N 2 n f. =R Taking into account that D =R nf = f by (4c2) we get <R ( (N 2) 2πN/2 Γ(N/2) f(0) = x <R x ) f(x) dx + N 2 N 2 R N 2 R N for N > 2; and similarly, 2πf(0) = x <R for N = 2. In particular, for a harmonic f, f(0) = Γ(N/2) 2π N/2 R N ( ) log R log x f(x) dx + f R =R =R =R f = f =R =R for N 2; the following result is thus proved (and holds also for N =, trivially). f

10 Tel Aviv University, 205/6 Analysis-IV 74 4d4 Proposition (ean value property). For every harmonic function on a ball, with bounded second derivatives, its value at the center of the ball is equal to its mean value on the boundary of the ball. 4d5 Remark. Now it is easy to understand why harmonic functions occur in physics ( the stationary heat equation ). Consider a homogeneous material solid body (in three dimensions). Fix the temperature on its boundary, and let the heat flow until a stationary state is reached. Then the temperature in the interior is a harmonic function (with the given boundary conditions). 4d6 Remark. Can the mean value property be generalized to a non-spherical boundary? We leave this question to more special courses (PDE, potential theory). But here is the idea. In 4d3 we may replace f(x) dx x N 2 with ( ) + g(x) f(x) dx where g is a harmonic function satisfying x N 2 + g(x) = 0 for all x (if we are lucky to have such g). Then x N 2 the double layer (D nv)u in (4c5), and the corresponding term in 4d3, disappears, and we get ( (N 2) (x 2πN/2 Γ(N/2) f(0) = )) f(x)d n x + g(x). N 2 4d7 Exercise (aximum principle for harmonic functions). Let u be a harmonic function on a connected open set R N. If sup x u(x) = u(x 0 ) for some x 0 then u is constant. Prove it. 2 (4d8) It appears that ( (mean ) ) f(x) = 2N lim of f on {y : y x = ε} f(x). ε 0 ε 2 4d9 Exercise. (a) Prove that, for N > 2, ( R 2 x ) dx does not depend on R; N 2 R N 2 x <R and for N = 2, R 2 x <R( log R log x ) dx does not depend on R. (No need to calculate these integrals.) 3 In fact, the mean value property is also sufficient for harmonicity, even if differentiability is not assumed. 2 Hint: the set {x 0 : u(x 0 ) = sup x u(x)} is both open and closed in. 3 Hint: change of variable.

11 Tel Aviv University, 205/6 Analysis-IV 75 (b) For f of class C 2 near the origin, prove that the mean value of f on {x : x = ε} is f(0) + c N ε 2 f(0) + o(ε 2 ) as ε 0, for some c 2, c 3, R (not dependent on f). (c) Applying (b) to f(x) = x 2, find c 2, c 3,... and prove (4d8). 4d0 Exercise. (a) For every f integrable (properly) on {x : x < R}, f R <R <R = =r f drn 0 R. =r N (b) For every bounded harmonic function on a ball, its value at the center of the ball is equal to its mean value on the ball. Prove it. 4d Proposition. (Liouville s theorem for harmonic functions) Every harmonic function R N [0, ) is constant. Proof. For arbitrary x, y R N and R > 0 we have f(x) = f(z) dz z x <R z x <R dz ( ) N R + x y = R f(z) dz z x <R dz = z y <R+ x y ( ) f(z) dz N R + x y z y <R+ x y dz = f(y), R z y <R+ x y since the R-neighborhood of x is contained in the (R + x y )-neighborhood of y. In the limit R we get f(x) f(y); similarly, f(y) f(x). 4e Differential forms of order N It is easy to generalize the flux, defined by 4a, as follow. 4e Definition. Let R N be an n-manifold, 2 F : R N a mapping continuous almost everywhere, and n : R N a continuous mapping such that n x is a unit normal vector to at x, for each x. The flux of (the vector field) F through (the hypersurface) in the direction n is F, n. (The integral is treated as improper, and may converge or diverge.) n=n Hint: (a) recall 3c8. 2 Necessarily orientable; see 4e9.

12 Tel Aviv University, 205/6 Analysis-IV 76 It is not easy to calculate this integral, even if is single-chart; the formula is complicated, F, n = F (ψ(u)), n ψ(u) det( (D i ψ) u, (D j ψ) u ) i,j du, and still, n x should be calculated somehow. Fortunately, there is a better formula: (4e2) F, n = ± det ( ) F (ψ(u)), (D ψ) u,..., (D n ψ) u du n=n (and the sign ± will be clarified soon). That is, F, n = ± ω, where ω is the n-form defined by ω(x, h,..., h n ) = det ( ) F (x), h,..., h n. We have to understand better this relation between vector fields and differential forms. Recall two types of integral over an n-manifold: of an n-form ω, ω, defined by (2c2) and (2d4); (,O) of a function f, f, defined by (2d8) and (2d9); they are related by f = fµ (,O), (,O) where µ (,O) is the volume form; that is, f = ω where ω = fµ (,O) (,O). Interestingly, every n-form ω on an orientable n-manifold R N is fµ (,O) for some f C(). This is a consequence of the one-dimensionality 2 of the space of all antisymmetric multilinear n-forms on the tangent space T x. We have f(x) = ω(x, e,..., e n ) for some (therefore, every) orthonormal basis (e,..., e n ) of T x that conforms to O x. But if ω is defined on the whole R N (not just on ), it does not lead to a function f on the whole R N ; indeed, in order to find f(x) we need not just x but also T x (and its orientation). The case n = N is simple: every N-form ω on R N (or on an open subset of R N ) is f det (for some continuous f); here det denotes the volume form on R N ; that is, (4e3) ω(x, h,..., h N ) = f(x) det(h,..., h N ) ; f(x) = ω(x, e,..., e N ). A wonder: the volume form of is not needed; the volume form of R N (the determinant) is used instead. Why so? Since the flux is the volume of fluid flowing through the surface (per unit time), as was noted in 4a. 2 Recall Sect. e and 2c.

13 Tel Aviv University, 205/6 Analysis-IV 77 Note that for every open U R N, (4e4) f det = f(x) dx ; U U U det = v(u). We turn to the case n = N. The space of all antisymmetric multilinear n-forms L on R N is of dimension ( N n) = N. Here is a useful linear one-to-one correspondence between such L and vectors h R N : n=n h,..., h n L(h,..., h n ) = det(h, h,..., h n ). Introducing the cross-product h h n by (4e5) h h, h h n = det(h, h,..., h n ) (it is a vector orthogonal to h,..., h n ), we get L(h,..., h n ) = h, h h n. Doing so at every point, we get a linear one-to-one correspondence between n-forms ω on R N and (continuous) vector fields F on R N : (4e6) ω(x, h,..., h n ) = F (x), h h n = det(f (x), h,..., h n ). Similarly, (n )-forms ω on an oriented n-dimensional manifold (, O) in R N (not just N n = ) are in a linear one-to-one correspondence with tangent vector fields F on, that is, F C( R N ) such that x F (x) T x. Let R N be an orientable n-manifold, ω and F as in (4e6). We know that ω = fµ (,O) for some f. How is f related to F? iven x, we take an orthonormal basis (e,..., e n ) of T x, note that e e n = n x is a unit normal vector to at x, and n=n F (x), n x = F (x), e e n = ω(x, e,..., e n ) = = f(x)µ (,O) (x, e,..., e n ) = ±f(x). In order to get + rather than ± we need a coordination between the orientation O and the normal vector n x. Let the basis (e,..., e n ) of T x For N = 3 the cross-product is a binary operation, but for N > 3 it is not. In fact, it is possible to define the corresponding associative binary operation (the so-called exterior product, or wedge product), not on vectors but on the so-called multivectors, see ultivector and Exterior algebra in Wikipedia.

14 Tel Aviv University, 205/6 Analysis-IV 78 conform to the orientation O x (of at x, or equivalently, of T x, recall Sect. 2b), then µ (,O) (x, e,..., e n ) = +. The two unit normal vectors being ±e e n, we say that n x = e e n conforms to the given orientation, and get F (x), n x = f(x) ; ω = F, n µ (,O). Integrating this over, we get nothing but the flux! Recall 4e: the flux of F through is F, n, that is, F, n µ (,O) (,O) = ω (,O) = ω. (,O) We get (4e2), and moreover, (4e7) F, n = ω (,O) for ω of (4e6) and O conforming to n. In particular, when is single-chart, we have (4e8) F, n = det ( ) F (ψ(u)), (D ψ) u,..., (D n ψ) u du provided that det(n, D ψ,..., D n ψ) > 0. Necessarily, D ψ D n ψ = cn for some c 0 (since both vectors are orthogonal to the tangent space); the sign of c is the sign in (4e2). We summarize the situation with the sign. 4e9 Remark. For an n-dimensional manifold R N, the two orientations O x at a given point x correspond naturally 2 to the two unit normal vectors n x to at x. Namely, for some (therefore, every) orthonormal basis e,..., e n of T x that conforms to O x, (a) det(n x, e,..., e n ) = +; or, equivalently, (b) e e n = n x. Alternatively (and equivalently), for arbitrary (not just orthonormal) basis, (a ) det(n x, e,..., e n ) > 0; (b ) e e n = cn x for some c > 0. iven a chart (, ψ) of around x that conforms to O x, we may take e i = (D i ψ) ψ (x). Orientations (O x ) x of correspond naturally to continuous mappings x n x R N such that for every x, n x is a unit normal vector to at x. Thus, such mappings exist if and only if is orientable (and in this case, there are exactly two of them, provided that is connected). n=n Not unexpectedly, in order to find f(x) we need not just x but also n x. 2 Using the orientation of R N given by the determinant; the other orientation of R N leads to the other correspondence.

15 Tel Aviv University, 205/6 Analysis-IV 79 We turn to a smooth set U R N. Its boundary U is a hypersurface; the outward normal vector leads, according to 4e9, to an orientation of U. In such cases we always use this orientation. iven F C (U R N ) with DF bounded, we may rewrite the divergence theorem 4a3, div F = F, n, U U as (div F ) det = ω U where ω corresponds to F according to (4e6). Taking into account that every n-form of class C corresponds to some vector field, we conclude. 4e0 Proposition. For every n-form ω of class C on R N there exists an N-form ω on R N such that for every smooth set U R N, ω = ω. U 4e Remark. The same holds in the piecewise smooth case: ω = U\Z U ω provided that the divergence theorem holds for U and U \ Z. 4e2 Example. On R 2 consider a vector field F : ( ) ( x y F (x,y) ) F 2 (x,y) and a curve (-manifold) covered by a single chart ψ : (a, b) R 2, ψ(t) = ( ψ (t) ) ψ 2 (t). Using the complicated formula, n ψ(t) = ( ) ψ 2 (t) ψ 2 (t) + ψ 2 2 (t) ψ (t) U U ; J ψ (t) = F (ψ(t)), n ψ(t) =... ( F ψ 2 F 2 ψ ) ; ψ 2 (t) + ψ 2 2 (t) ; n=n flux = b a F (ψ(t)), n ψ(t) J ψ (t) dt = b Alternatively, using (4e8), det ( F (ψ(t)), ψ (t) ) = F ψ F 2 ψ 2 = F ψ 2 F 2 ψ ; flux = 4e3 Exercise. Fill in the details in 4e2. a ( ) F ψ 2 F 2 ψ dt. b a ( ) F ψ 2 F 2 ψ dt. 4e4 Example. Continuing 4e2, consider the -form ω, ω (( x y ), ( dx dy )) = f (x, y) dx + f 2 (x, y) dy; it corresponds to F according to (4e6) when f (x, y) dx + f 2 (x, y) dy = F (x, y) F 2 (x, y) dx dy, that is, f = F 2, f 2 = F.

16 Tel Aviv University, 205/6 Analysis-IV 80 In this case, ω = b a ω ( ψ(t), ψ (t) ) dt = b 4e5 Exercise. Fill in the details in 4e4. a ( f (ψ(t))ψ (t) + f 2 (ψ(t))ψ 2(t) ) dt = = b a ( ) F2 ψ + F ψ 2 dt = flux. 4e6 Remark. Less formally, denoting ψ (t) and ψ 2 (t) by just x(t) and y(t) we have b ( ω = f (x(t), y(t))x (t) + f 2 (x(t), y(t))y (t) ) dt ; a naturally, this is called (f dx + f 2 dy). 4e7 Example. Continuing 4e2 and 4e4, we calculate the divergence: thus, div F = D F + D 2 F 2 = D f 2 D 2 f ; ω = (div F ) det = (D f 2 D 2 f ) det ; ω = (D f 2 D 2 f ) U U for a smooth U R 2. If U is covered (except for a single point) with a chart ψ : (a, b) R 2, ψ(a+) = ψ(b ), then 4e0 gives (f dx + f 2 dy) = (D f 2 D 2 f ). U This is the well-known reen s theorem; in traditional notation, ( (L dx + dy) = x L ) dxdy. y U 4e8 Example. The -form ω = on R 2 (mentioned in Sect. d) 2 corresponds to the vector field F ( ) ( x y = xy ) 2, that is, F (x) = x for x 2 R2. Clearly, div F =, thus, ω = det; by 4e0, ω = v(u) for every smooth U R 2. U U U y dx+x dy

17 Tel Aviv University, 205/6 Analysis-IV 8 4e9 Example. y dx+x dy The -form ω = on R 2 \{0} (treated in x 2 +y 2 Sect. d) corresponds to the vector field F ( ) x ( y = xy ) x 2 +y, that is, F (x) = x for x R 2 \ {0}. 2 x 2 By (4a7), div F = 0 on R 2 \ {0}, thus ω = 0 on R 2 \{0}; by 4e0, ω = 0 for every smooth U U such that U / 0. On the other hand, for every smooth U 0 we have ω = 2π by (4a9); U compare this fact with Sect. d. Similarly, in R 3 the 2-form ω that corresponds to the vector field F (x) = x satisfies ω = 0 whenever U / 0, and ω = 4π whenever U 0. x 3 U U Index cross-product, 77 divergence theorem, 66 divergence theorem for cube, 68 flux, 65, 68, 75 reen formula first, 69 second, 70 third, 70 harmonic, 69 heat, 74 hyperface, 67 Laplacian, 69 layer, 70 Liouville s theorem, 75 maximum principle, 74 mean value property, 74 normal vector conforms to orientation, 78 tangent vector field, 77, 69 h h n, 77