14 Higher order forms; divergence theorem

Transcription

1 Tel Aviv University, 2013/14 Analysis-III,IV Higher order forms; divergence theorem 14a Forms of order three b Divergence theorem in three dimensions c Order four, and higher Boundary and derivative are generalized to 3-chains and 2-forms, and higher. Stokes theorem and divergence theorem are generalized accordingly. 14a Forms of order three Similarly to the boundary of a singular 2-box, defined in Sect. 11d as AB + BC + CD + DA = AB + BC DC AD, we define the boundary of a singular 3-box as follows: 1 D A C B (14a1) ADCB + EF GH + ABF E + + DHGC + AEHD + BCGF = = ABCD + EF GH AEF B + + DHGC ADHE + BCGF. F B E A G C H D Similarly to (11d1), (14a2) ( ) = 0 for a singular 3-box. 14a3 Exercise. Similarly to Sect. 11d, find 1 lim ε 0+ ε 3 ε ω where ε : [0, 1] 3 R n, ε (u 1, u 2, u 3 ) = x + εu 1 h 1 + εu 2 h 2 + εu 3 h 3, and ω is an arbitrary 2-form (of class C 1 ) on R n. Answer: ( D h1 ω(, h 2, h 3 ) ) x + ( D h2 ω(, h 3, h 1 ) ) x + ( D h3 ω(, h 1, h 2 ) ) x. We proceed similarly to Def. 11d2. 1 Here we rely on our geometric intuition; for a formal approach see Sect. 14c.

2 Tel Aviv University, 2013/14 Analysis-III,IV a4 Definition. The exterior derivative of a 2-form ω of class C 1 3-form dω defined by is a (dω)(, h 1, h 2, h 3 ) = D h1 ω(, h 2, h 3 ) + D h2 ω(, h 3, h 1 ) + D h3 ω(, h 1, h 2 ). it. Wedge product was defined in Sect. 11e for two 1-forms. Now we extend 14a5 Definition. (a) Let L 1, L 2 be linear forms on R n. Their wedge product L 1 L 2 is an antisymmetric bilinear form L (2) on R n defined by L (2) (a, b) = L 1 (a)l 2 (b) L 1 (b)l 2 (a) for all a, b R n. (b) Let L (1) be a linear form on R n, and L (2) an antisymmetric bilinear form on R n. Their wedge product L (1) L (2) = L (2) L (1) is an antisymmetric trilinear form L (3) on R n defined by L (3) (a, b, c) = L (1) (a)l (2) (b, c) + L (1) (b)l (2) (c, a) + L (1) (c)l (2) (a, b) for all a, b, c R n. (Check the antisymmetry.) This definition is suggested by determinants, as follows. A trilinear form L on R n is generally L(a, b, c) = i,j,k c i,j,ka i b j c k. If L is antisymmetric then L = a i b i c i c i,j,k L i,j,k where L i,j,k (a, b, c) = a j b j c j i<j<k a k b k c k (think, why). Introducing also L i and L i,j by L i (a) = a i, L i,j (a, b) = a i a j b i b j we observe that L i L j = L i,j and L i L j,k = L i,j,k (think, why). Thus, (L i L j ) L k = L i (L j L k ) (since L k,i,j = L i,j,k ). Associativity follows by taking linear combinations: (L 1 L 2 ) L 3 = L 1 (L 2 L 3 ) for all linear forms L 1, L 2, L 3 on R n. Wedge product of differential forms is defined pointwise: (ω 1 ω 2 )(x) = ω 1 (x) ω 2 (x).

3 Tel Aviv University, 2013/14 Analysis-III,IV 223 It follows that (fω 1 ) (gω 2 ) = (fg)(ω 1 ω 2 ) for f, g C 0 (R n ). Note that ω 2 ω 1 = ±ω 1 ω 2 ; the sign is minus for two 1-forms, but plus for a 1-form and 2-form. By associativity, ω 1 ω 2 ω 3 is well-defined for three 1-forms. In particular, (dx i dx j dx k )(x, h 1, h 2, h 3 ) = L i,j,k (h 1, h 2, h 3 ) is the 3 3 determinant. A 2-form (of class C 1 ) is called closed, if its derivative is zero. The 2-form dx i dx j is closed, since (dx i dx j )(x, h, k) does not depend on x. The following two exercises are similar to (11e4) and (11e5). 14a6 Exercise. Prove that for all 1-forms ω of class C 2 on R n. d(dω) = 0 Thus, all exact 2-forms of class C 1 are closed. By the way, the 2-form dx i dx j is exact by 13b18, or just because d(x i dx j ) = dx i dx j by (11e6). Moreover, (14a7) df dg is exact, therefore closed, for all f, g C 1 (R n ). 14a8 Exercise. Prove that d(fω) = df ω + f dω for all f C 1 (R n ) and all 2-forms ω of class C 1 on R n. Therefore (14a9) d(fω) = df ω whenever ω is closed. In particular, d(f dx i dx j ) = df dx i dx j for all f C 1 (R n ). Similarly to 11e7 we get the following definition equivalent to 14a4. 14a10 Definition. The exterior derivative of a 2-form ω of class C 1 is a 3-form dω defined by dω = i<j df i,j dx i dx j for ω = i<j f i,j dx i dx j. We turn to change of variables, treated in Sect. 11f for 2-forms (and 1-forms, and 0-forms). Let ϕ C 1 (R l R n ). Recall the pullback ϕ ω defined by 11f1 for all k-forms ω on R n. We generalize 11f5 and 11f6 as follows.

4 Tel Aviv University, 2013/14 Analysis-III,IV a11 Exercise. Prove that ϕ (ω 1 ω 2 ω 3 ) = (ϕ ω 1 ) (ϕ ω 2 ) (ϕ ω 3 ) for all 1-forms ω 1, ω 2, ω 3 on R n. 1 14a12 Lemma. For every 2-form ω of class C 1 on R n and ϕ C 2 (R l R n ), ϕ (dω) = d(ϕ ω). Proof. We have ω = i<j f i,j dx i dx j and dω = i<j df i,j dx i dx j. It is sufficient to prove that ϕ (df i,j dx i dx j ) = d ( ϕ (f i,j dx i dx j ) ). We denote g i,j = ϕ f i,j, y i = ϕ x i, y j = ϕ x j. By 11f4, ϕ (dx i ) = dy i, ϕ (dx j ) = dy j and ϕ (df i,j ) = dg i,j. By 11f5, ϕ (dx i dx j ) = dy i dy j. By 14a11, ϕ (df i,j dx i dx j ) = dg i,j dy i dy j. On the other hand, d ( ϕ (f i,j dx i dx j ) ) = d(g i,j dy i dy j ) = dg i,j dy i dy j by (14a7), (14a9). 14a13 Theorem. (Stokes theorem for k = 3) Let C be a 3-chain in R n, and ω a 2-form of class C 1 on R n. Then dω = ω. C Proof. It is sufficient to prove the equality dω = ω for every singular 3-box. Similarly to 11g, using (11f2) we transform the needed equality into B (dω) = B ω. Similarly to 11g we may assume that is of class C 2. Thus, 14a12 applies, and the needed equality becomes d( ω) = ω. B Similarly to 11g it remains to prove the equality dω = ω for every B B 2-form ω of class C 1 on the cube B = [0, 1] 3 R 3 ; we consider only ω = f(u 1, u 2, u 3 ) du 1 du 2, since the other two cases are similar. We have dω = df du 1 du 2 = ( ) f u 1 du 1 + f u 2 du 2 + f u 3 du 3 du1 du 2 = f u 3 du 1 du 2 du 3, thus B dω = [0,1] 3 f u 3 du 1 du 2 du 3 = which is equal to ω (see (14a1)). B 1 Hint: similar to 11f5; use the 3 3 determinant L i,j,k. C B 1 f du 1 du 2 du 3 = [0,1] 2 0 u 3 ( = du 1 du 2 f(u1, u 2, 1) f(u 1, u 2, 0) ),

5 Tel Aviv University, 2013/14 Analysis-III,IV a14 Corollary. C 1 C 2 implies C 1 C 2 for arbitrary 3-chains C 1, C 2 in R n. (Similar to 11h1.) 14a15 Exercise. 1 Check that (y dx + x dy) (x dx dz + y dy dz) = (y 2 x 2 ) dx dy dz. 14a16 Exercise. 2 Check that d(x dy dz + y dz dx + z dx dy) = 3 dx dy dz. 14a17 Exercise. 3 Prove that for arbitrary 1-forms ω 1, ω 2 on R n. d(ω 1 ω 2 ) = (dω 1 ) ω 2 ω 1 dω 2 Thus, if ω 1 and ω 2 are closed 1-forms then ω 1 ω 2 is a closed 2-form. (Compare it with 13b18.) 14a18 Exercise. 4 Prove a generalization of the formula for integration by parts, f dω = fω df ω C C for arbitrary 2-form ω (of class C 1 ) on R n, function f C 1 (R n ), and 3-chain C in R n. 14b Divergence theorem in three dimensions A 2-form ω on R 3 corresponds to a vector field H (recall Sect. 12a), namely, H(x) = ( f 2,3 (x), f 3,1 (x), f 1,2 (x) ) ω(x, h 1, h 2 ) = det(h(x), h 1, h 2 ), C for ω = f }{{} 1,2 dx 1 dx 2 + f }{{} 2,3 dx 2 dx 3 + f }{{} 3,1 dx 3 dx 1. H 3 H 1 H 2 1 Sjamaar, p Shurman, p Shurman, Th shows that in general the sign depends on the order of ω 1. 4 Shurman, Ex

6 Tel Aviv University, 2013/14 Analysis-III,IV b1 Exercise. Let a vector field E correspond to a 1-form ω 1, and a vector field H correspond to a 2-form ω 2. Prove that ω 1 ω 2 = E, H dx 1 dx 2 dx 3. For every singular 2-box : B R 3, ω = det ( ) H((u)), (D 1 ) u, (D 2 ) u du = B (recall (12a7)) is the flux of H through. This relation extends by linearity to 2-chains; in particular, ω = H is the flux of H through the boundary of a singular 3-box. The derivative dω (assuming that ω is of class C 1 ), being a 3-form on R 3, is dω = f dx 1 dx 2 dx 3 for some f C 0 (R 3 ). Taking into account that d(h 3 dx 1 dx 2 ) = D 3 H 3 dx 1 dx 2 dx 3 we get dω = (div H) dx 1 dx 2 dx 3, div H = D 1 H 1 + D 2 H 2 + D 3 H 3. Now we finalize the diagram (12a3) (see also (12c9)), 0-form id function H (14b2) d 1-form Euclidean duality vector field (E) d curl 2-form determinant duality vector field (H) d 3-form ω = f dx 1 dx 2 dx 3 div function 14b3 Exercise. 1 Prove that div(fh) = f, H + f div H for all vector fields H (of class C 1 ) on R 3 and all functions f C 1 (R 3 ). 2 1 Zorich, (14.18). 2 Hint: 14a8 and 14b1.

7 Tel Aviv University, 2013/14 Analysis-III,IV b4 Exercise. 1 Prove that div(e 1 E 2 ) = curl E 1, E 2 E 1, curl E 2 for all vector fields E 1, E 2 (of class C 1 ) on R 3. 2 Theorem 14a13 gives the three-dimensional divergence theorem (recall (12c8)): (14b5) H = div H for every vector field H (of class C 1 ) on R 3 and every singular 3-box in R 3. Here (as in 12c) by f we mean f dx 1 dx 2 dx 3. If : B R 3 is such that B is a diffeomorphism between B and an open set G = (B ) R 3 then f(x) dx 1 dx 2 dx 3 = ± f (a similar fact in two dimensions was noted in Sect. 12c, before (12c6)). Assuming that det d > 0 we get (div H) dx 1 dx 2 dx 3 = div H, and G so, (14b6) H = div H similarly to (12c6), (12c8). In particular, spherical coordinates suggest a singular 3-box R that represents a ball of radius R, G G (14b7) R : [0, R] [0, π] [0, 2π] R 3, R (r, θ, ϕ) = (r sin θ cos ϕ, r sin θ sin ϕ, r cos θ). 14b8 Exercise. Prove that f(x) dx 1 dx 2 dx 3 = R for every f C 0 (B R ). 3 1 Zorich, (14.19). 2 Hint: 14a17 and 14b1. 3 Hint: the determinant is equal to r 2 sin θ. B R f

8 Tel Aviv University, 2013/14 Analysis-III,IV 228 Rotation invariance follows (recall 6m4): R T R for every linear isometry T : R 3 R 3. 1 By 14a14 it follows that (14b9) R T R since generally T = (T ) (think, why). 14b10 Exercise. (a) Consider a radial vector field F on R 3, (like 13c3). Check that 2 F (x) = f( x )x, f C 0 [0, ) R F = 4πR 3 f(r) = 4πR 2 f(r)r (the area of the sphere times the length of the vector). (b) More generally, consider F (x) = f(x)x, f C 0 (R 3 ); check that R F = R π 0 2π dθ dϕ R 2 sin θ f(r sin θ cos ϕ, R sin θ sin ϕ, R cos θ). 0 Postponing integration on surfaces in general, for now we define the integral of a function over the sphere B R (the boundary of the ball B R = {x : x R} R 3 ) by (14b11) B R f = π 0 2π dθ dϕ R 2 sin θ f(r sin θ cos ϕ, R sin θ sin ϕ, R cos θ) 0 for arbitrary continuous function f on the sphere. Note that 1 = 4πR 2 ; B R f(x)x = R R f. B R Now we may define the mean value of f on the sphere as 1 4πR 2 B R f. This could not be done via Riemann integral (proper or improper), since the sphere is a set of volume zero. 1 In spherical coordinates this is easy to see for rotations about the z axis, but problematic for other axes. 2 Hint: only one (out of six) face of the boundary contributes; calculate the 3 3 determinant and integrate it.

9 Tel Aviv University, 2013/14 Analysis-III,IV b12 Exercise. Prove that for all f C 0 (B R ). 1 B R f = R 0 dr f B r Therefore B R f = d f ; dr B R rotation invariance follows: f = B R T f B R for every linear isometry T : R 3 R 3. 2 (Compare it with (14b9).) Similarly to Sect. 12d (before (12d4), div f = f, = D 1 D 1 + D 2 D 2 + D 3 D 3 = x 2 1 x 2 2 x 2 3 is the Laplacian. Functions f C 2 (R 3 ) such that f = 0 are called harmonic. Similarly to (12d4) we ll prove the mean value property of a harmonic function u on R 3 : (14b13) u(0) = 1 4πR 2 B R u ; u(x) = 1 4πR 2 B R u(x + ). To this end we need Green formulas (again). Applying (14b5) to H = u we get the first Green formula (recall (12d5)) (14b14) u = u for all u C 2 (R 3 ). Exercise 12d6 holds in all dimensions (with the same proof): (a) div(fh) = f div H + f, H for all f C 1 (R 3 ) and H C 1 (R 3 R 3 ); (b) div(f g) = f g + f, g for all f C 1 (R 3 ) and g C 2 (R 3 ); (c) f g g f = div(f g g f) for all f, g C 2 (R 3 ). 1 Hint: first, replace B R with R. 2 Again, in spherical coordinates this is easy to see for rotations about the z axis, but problematic for other axes.

10 Tel Aviv University, 2013/14 Analysis-III,IV 230 Similarly to (12d7), (12d8) we get the second Green formula (14b15) u v = (u v + u, v ) for all u C 1 (R 3 ) and v C 2 (R 3 ), and the third Green formula (14b16) (u v v u) = (u v v u) for all u, v C 2 (R 3 ). 14b17 Exercise. Similarly to R of (14b7) introduce a singular 3-box R1,R 2 that represents the spherical shell {x : R 1 x R 2 } R 3 (given 0 < R 1 < R 2 < ) and check that R1,R 2 R2 R1. Here is a three-dimensional counterpart of 12d9. 14b18 Exercise. (a) Let u and v be harmonic functions on a spherical shell {x R 3 : a < x < b}; prove that R (u v v u) does not depend on R (a, b). (b) In particular, taking v(z) = 1/ z, prove that 1 u v = 1 u ; R R 2 B R v u = 1 u. R R R (c) Assuming in addition that u is harmonic on the ball {x R 3 : x < b} prove that 1 R 2 B R u does not depend on R (0, b) and is equal to 4πu(0), which proves the first equality of (14b13); the second follows by shift. 14b19 Exercise. (Maximum principle for harmonic functions) Let u be a harmonic function on a connected open set G R 3. sup x G u(x) = u(x 0 ) for some x 0 G then u is constant. Prove it. 2 The mean value may be taken on the ball rather than the sphere: (14b20) u(0) = 3 u ; u(x) = 3 u(x + ). 4πR 3 B R 4πR 3 B R Proof: by 14b12 and (14b13), R u = dr u = B R 0 B r R 0 4πR 2 u(0) dr = 4πR3 3 u(0). 1 Hint: v is harmonic by 13c4. 2 Hint: the set {x 0 : u(x 0 ) = sup x G u(x)} is both open and closed in G. If

11 Tel Aviv University, 2013/14 Analysis-III,IV b21 Proposition. (Liouville s theorem for harmonic functions, dimension three) Every harmonic function R 3 [0, ) is constant. Proof. (Nelson s short proof) For arbitrary x, y R 3 and R > 0 we have u(x) = 3 u(x + ) 3 u(y + ) = 4πR 3 B R 4πR 3 B R+ x y ( ) 3 R + x y u(y), since the R-neighborhood of x is contained in the (R + x y )-neighborhood of y. In the limit R we get u(x) u(y); similarly, u(y) u(x). 14c Order four, and higher In dimension four (and higher) we cannot rely on our geometric intuition as much as we did in (14a1); we need a formal approach to orientation. We introduce three types of cubes: 1 a standard k-cube is the set [ 1, 1] k in R k ; a singular k-cube in R n is a C 1 mapping [ 1, 1] k R n ; a geometric k-cube in R n is a set X R n isometric to [ 1, 1] k. The group 2 G k of all isometries 3 of the standard k-cube (to itself) consists ) of 2 k k! signed permutation matrices, like. The determinant of ( such matrix is ±1. Accordingly, for a given geometric k-cube in R n there exist 2 k k! isometric mappings [ 1, 1] k X. If 1 is such mapping then others are 1 T for T G k ; that is, they are 2 such that G k. All such mappings are singular k-cubes in R n, not all mutually equivalent; rather, 1 2 whenever det( ) = 1, 1 2 whenever det( ) = 1. Thus, a geometric k-cube X R n leads to two equivalence classes of singular k-cubes; these two equivalence classes will be called the two orientations of X. A k-form cannot be integrated over X unless an orientation is chosen; for the other orientation the integral is the opposite number. 1 This time, [ 1, 1] is technically more convenient than [0, 1]. 2 The so-called hyperoctahedral group. 3 Called also automorphisms or congruences. R

12 Tel Aviv University, 2013/14 Analysis-III,IV 232 The simplest case is, k = 1. A geometric 1-cube in R n is a straight interval X = {x : A x + x B = 2} for given A, B R n, A B = 2. An isometry γ : [ 1, 1] X defined by γ(t) = 1 ta + 1+t B is a path; denote it just AB. 2 2 Accordingly, BA is the other isometry [ 1, 1] X, t 1 tb + 1+t A. Note 2 2 that (BA)(t) = (AB)( t). Clearly, BA AB, that is, ω = ω BA AB for all 1-forms ω on R n. The next case is, k = 2. Let X R n be a geometric 2-cube. An isometry : [ 1, 1] 2 X is a singular 2-cube; denote it by ABCD where A = ( 1, 1), B = (1, 1), C = (1, 1), D = ( 1, 1); these are the vertices of X. There are 8 isometries: ABCD, ADCB, BCDA, BADC, CDAB, CBAD, DABC, DCBA; they result from ABCD via elements of the group G 2. For ABCD, BCDA, CDAB and DABC the elements of the group are rotations by 0, π/2, π and 3π/2, of Jacobian +1; for others, the elements of the group are reflections, of Jacobian 1. Thus, ABCD BCDA CDAB DABC is one orientation of X, ADCB BADC CBAD DCBA is the other orientation of X. The standard k-cube has 2k hyperfaces {(u 1,..., u k ) [ 1, 1] k : u i = a} for i {1,..., k} and a { 1, 1} ; each hyperface is a geometric (k 1)-cube. We want to define the boundary X of the standard k-cube X as the sum Y Ỹ of its hyperfaces Y treated as singular (k 1)-cubes Ỹ ; to this end we have to choose orientations of these hyperfaces. We did it already for k = 2, 3. In these two cases the chosen orientations are consistent in the following sense. For every hyperface Y and every T G k such that det T = +1 (that is, T ( X) = X), T (Ỹ ) = T (Y ). This consistency is necessary for Stokes theorem to hold, since T ( X) = X must imply T ( X) = X (recall 14a14). Here is a special case of the consistency condition: (14c1) if T ( X) = X and T (Y ) = Y then T (Ỹ ) = Ỹ. It is worth noting that such a condition fails for edges (rather than faces) of a 3-cube; here is a counterexample.

13 Tel Aviv University, 2013/14 Analysis-III,IV c2 Example. Let X = [ 1, 1] 3, Y = { 1} { 1} [ 1, 1] and T (u 1, u 2, u 3 ) = (u 2, u 1, u 3 ). Then T preserves Y and the orientation of X but does not preserve the orientation of Y. Consider the hyperface Y 0 = {1} [ 1, 1] k 1 of [ 1, 1] k. If T G k, T (Y 0 ) = Y 0, then ( ) 1 0 T = 0 T for some T G k 1. Thus, det T = det T, which ensures (14c1) for Y 0. Now we are in position to ensure the consistency condition in general. (This is somewhat similar to the proof of 13a1.) For each hyperface Y of [ 1, 1] k we choose T Y G k such that det T Y = +1 and T Y (Y 0 ) = Y. We choose an orientation of Y 0 and define Ỹ = T Y (Ỹ0) for all Y. Given hyperfaces Y 1, Y 2 and T G k such that det T = +1 and T (Y 1 ) = Y 2, we have (T 1 Y 2 T T Y1 )(Y 0 ) = Y 0 and det(t 1 Y 2 T T Y1 ) = +1. Y 1 T Y 2 T Y1 T Y2 Y 0 Applying (14c1) to T 1 Y 2 T T Y1 and Y 0 we get (T 1 Y 2 T T Y1 )(Ỹ0) = Ỹ0; thus, T ( T Y1 (Ỹ0) ) = T Y2 (Ỹ0), that is, T (Ỹ1) = Ỹ2. (Similarly to 13a1, the choice of T Y does not really matter; think, why.) Consistent orientations Ỹ are thus constructed in principle; but we need an explicit formula. In terms of singular (k 1)-cubes i,a : [ 1, 1] k 1 [ 1, 1] k for i {1,..., k}, a { 1, +1}, i,a (u 1,..., u k 1 ) = (u 1,..., u i 1, a, u i,..., u k 1 ), we have Ỹi,a ± i,a where Y i,a = {(u 1,..., u k ) [ 1, 1] k : u i = a} are the hyperfaces. But what are the signs? The sign for Y 0 = Y 1,+ is rather a matter of convention; let it be +1. That is, Ỹ 0 1,+. 1 The mapping T i,a : (u 1,..., u k ) (u 2,..., u i, au 1, u i+1,..., u k ) satisfies 2 T i,a (Y 0 ) = Y i,a ; T i,a 1,+ = i,a ; det(t i,a ) = ( 1) i 1 a. 1 More formally, Ỹ0 1,+. 2 Indeed, (u 1,..., u k 1 ) 1,+ (1, u 1,..., u k 1 ) Ti,a (u 1,..., u i 1, a, u i,..., u k 1 ).

14 Tel Aviv University, 2013/14 Analysis-III,IV 234 By the consistency condition, T i,a (Ỹ0) det(t i,a )Ỹi,a, that is, Ỹ i,a det(t i,a )T i,a 1,+ ( 1) i 1 a i,a. 14c3 Definition. The boundary of a singular k-cube : [ 1, 1] k R n is a k-chain k = ( 1) i 1 a( i,a ). i=1 a=±1 14c4 Exercise. Check that the definitions used before for k = 1, 2, 3 conform to 14c3. 14c5 Exercise. Prove that ( ) = 0 for all singular k-cubes in R n. 1 14c6 Exercise. Similarly to 14a3, find 1 lim ε 0+ (2ε) k ε ω where ε : [ 1, 1] k R n, ε (u 1,..., u k ) = x + εu 1 h εu k h k, and ω is an arbitrary (k 1)-form (of class C 1 ) on R n. Answer: k i=1 ( 1)i 1( D hi ω(, h 1,..., h i 1, h i+1,..., h k ) ) x. 14c7 Definition. The exterior derivative of a (k 1)-form ω of class C 1 is a k-form dω defined by (dω)(, h 1,..., h k ) = k ( 1) i 1 D hi ω(, h 1,..., h i 1, h i+1,..., h k ). i=1 14c8 Theorem. (Stokes theorem) Let C be a k-chain in R n, and ω a (k 1)-form of class C 1 on R n. Then dω = ω. C I skip the proof. The general case is somewhat more technical than the case k = 3, but no new ideas appear in the proof. The equivalent definition 14a10 of exterior derivative becomes dω = df i1,...,i k dx i1 dx ik for ω = f i1,...,i k dx i1 dx ik ; i 1 < <i k i 1 < <i k C the form dx i1 dx ik is a determinant similar to L i,j,k of Sect. 14a. Still, d(dω) = 0. 1 Hint: you may use the idea of 14c2, if you like.

15 Tel Aviv University, 2013/14 Analysis-III,IV 235 And still, ϕ (dω) = d(ϕ ω). Similarly to Sect. 14b, an (n 1)-form ω on R n corresponds to a vector field H, namely, ω(x, h 1,..., h n 1 ) = det(h(x), h 1,..., h n 1 ), n ω = ( 1) n 1 H i dx 1 dx i 1 dx i+1 dx n. i=1 1 i n H h 1 h n 1 For every singular (n 1)-box : B R n, ω = det ( ) H((u)), (D 1 ) u,..., (D n 1 ) u du = B is the flux of H through ; and for an n-box, ω = H is the flux of H through the boundary of. We have H dω = n ( 1) n 1 dh i dx 1 dx i 1 dx i+1 dx n = i=1 = n dx 1 dx i 1 dh i dx i+1 dx n = i=1 = n i=1 H i x i dx 1 dx n = (div H) dx 1 dx n, div H = D 1 H D n H n. Thus, Th. 14c8 gives the n-dimensional divergence theorem (recall (14b1)): (14c9) H = div H for every vector field H (of class C 1 ) on R n and every singular n-box in R n.

16 Tel Aviv University, 2013/14 Analysis-III,IV 236 Index boundary, 221, 234 closed, 223 divergence theorem, 227, 235 exterior derivative, 222, 223, 234 flux, 235 Green formulas, 229 harmonic, 229 hyperface, 232 integral over sphere, 228 Laplacian, 229 Liouville s theorem, 231 maximum principle, 230 mean value property, 229 orientation, 231 Stokes theorem, 224, 234 wedge product, 222, 229 i,a, 233 div, 226, 235 dω, 222, 223, 234 R, 227