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4 Divergence Theorem and Its Consequences Tel Aviv University, 2015/16 Analysis-IV 65 4 Divergence theorem and its consequences 4a Divergence and flux . 65 4b Piecewise smooth case . 67 4c Divergence of gradient: Laplacian . 69 4d Laplacian at a singular point . 71 4e Differential forms of order N − 1 . 75 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: G ⊂ RN is a smooth set. Recall the outward unit normal vector nx for x 2 @G. 4a1 Definition. For a continuous F : @G ! RN , the (outward) flux of (the vector field) F through @G is Z hF; ni : @G (The integral is interpreted according to (2d8).) If a vector field F on R3 is the velocity field of a fluid, then the flux of F through a surface is the amount1 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 2 C1(G ! RN ), with DF bounded (on G). Recall that, by 3b6, boundedness of DF on G ensures that F extends to G by continuity (and therefore is bounded). In such cases we always use this extension. The mapping F~ : RN n @G ! RN defined by ( F (x) for x 2 G; F~(x) = 0 for x2 = G 1The volume is meant, not the mass. However, these are proportional if the density (kg=m3) of the matter is constant (which often holds for fluids). 2See also mathinsight. Tel Aviv University, 2015/16 Analysis-IV 66 is continuous up to @G, and ~ ~ F (x − 0nx) = F (x) ; F (x + 0nx) = 0 ; ~ divsng F (x) = −hF (x); nxi : By Theorem 3e3 (applied to F~ and K = @G), Z Z (4a2) div F = hF; ni ; G @G just the flux. The divergence theorem, formulated below, is thus proved.1 4a3 Theorem (Divergence theorem). Let G ⊂ RN be a smooth set, F 2 C1(G ! RN ), with DF bounded on G. Then the integral of div F over G is equal to the (outward) flux of F through @G. R In particular, if div F = 0, then @GhF; ni = 0. 4a4 Exercise. div(fF ) = f div F + hrf; F i whenever f 2 C1(G) and F 2 C1(G ! RN ) Prove it. Thus, the divergence theorem, applied to fF when f 2 C1(G) with bounded rf, and F 2 C1(G ! RN ) with bounded DF , gives a kind of integration by parts, similar to (3b12): Z Z Z (4a5) hrf; F i = fhF; ni − f div F: G @G G R R In particular, if div F = 0, then Ghrf; F i = @G fhF; ni Here is a useful special case. We mean by a radial function a function of the form f : x 7! g(jxj) where g 2 C1(0; 1), and by a radial vector field F : x 7! g(jxj)x. Clearly, f 2 C1(RN n f0g) and F 2 C1(RN n f0g ! RN ). g0(jxj) 4a6 Exercise. (a) If f(x) = g(jxj), then rf(x) = jxj x; (b) if F (x) = g(jxj)x, then div F (x) = jxjg0(jxj) + Ng(jxj); (c) if F (x) = g(jxj)x, then the (outward) flux of F through the boundary N 2πN=2 of the ball fx : jxj < rg is cr g(r), where c = Γ(N=2) is the area of the unit sphere. Prove it.2 1Divergence 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). 2Hint: (b) use (a) and 4a4. Tel Aviv University, 2015/16 Analysis-IV 67 R Taking G = fx : a < jxj < bg and F (x) = g(jxj)x, we see that G div F = R b N−1 0 a cr rg (r)+Ng(r) dr by 4a6(b) and (generalized) 3c8; and on the other R N b d N N−1 0 hand, @GhF; ni = cr g(r)jr=a by 4a6(c). Well, dr r g(r) = r 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 rN g(r) does not depend on r, that is, const 0 g(r) = rN (and indeed, in this case rg (r) + Ng(r) = 0); const F (x) = x ; div F (x) = 0 for x 6= 0 ; jxjN (4a7) Z hF; ni = 0 when G 3= 0 ; @G note that the latter equality fails for a ball. The flux through a sphere is Z Z 2πN=2 (4a8) hF; ni = const · 1 = const · jxj=r jxj=1 Γ(N=2) where 'const' is as in (4a7). The same holds for arbitrary smooth set G 3 0: Z 2πN=2 (4a9) hF; ni = const · : @G Γ(N=2) Proof: we take " > 0 such that fx : jxj ≤ "g ⊂ G; the set G" = fx 2 G : jxj > "g is smooth; by (4a7), R hF; ni = 0; and @G = @G ] fx : jxj = "g. @G" " 4b Piecewise smooth case We want to apply the divergence theorem 4a3 to the open cube G = (0; 1)N , but for now we cannot, since the boundary @G is not a manifold. Rather, @G consists of 2N disjoint cubes of dimension n = N − 1 (\hyperfaces") and a finite number1 of cubes of dimensions 0; 1; : : : ; n − 1. n=N −1 For example, f1g × (0; 1)n is a hyperface. Each hyperface is an n-manifold, and has exactly two orientations. Also, the outward unit normal vector nx is well-defined at every point x of a hyperface. n For example, nx = e1 for every x 2 f1g × (0; 1) . R For a function f on @G we define @G f as the sum of integrals over the 2N hyperfaces; that is, N Z X X Z Z Y (4b1) f = ··· f(x1; : : : ; xN ) dxj ; @G i=1 x =0;1 j:j6=i i (0;1)n 1In fact, 3N − 1 − 2N. Tel Aviv University, 2015/16 Analysis-IV 68 provided that these integrals are well-defined, of course. For a vector field F 2 C(@G ! RN ) we define the flux of F through @G R as @GhF; ni. Note that N Z X X Z Z Y (4b2) hF; ni = (2xi − 1) ··· Fi(x1; : : : ; xN ) dxj : @G i=1 x =0;1 j:j6=i i (0;1)n 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 2 C1(0; 1)N ! N N R , with DF bounded. Then the integral of div F over (0; 1) is equal to the (outward) flux of F through the boundary. (As before, boundedness of DF ensures that F extends to [0; 1]N by continuity; recall 3b6.) Proof. Z 1 D1F1(x1; : : : ; xN ) dx1 = F1(1; x2; : : : ; xN ) − F1(0; x2; : : : ; xN ) = 0 X = (2x1 − 1)F1(x1; : : : ; xN ); x1=0;1 Z Z X Z Z ··· D1F1 = (2x1 − 1) ··· F1(x1; : : : ; xN ) dx2 ::: dxN ; x =0;1 (0;1)N 1 (0;1)n similarly, for each i = 1;:::;N, Z Z X Z Z Y ··· DiFi = (2xi − 1) ··· Fi dxj ; x =0;1 j:j6=i (0;1)N i (0;1)n it remains to sum over i. The same holds for every box, of course. A box is only one example of a bounded regular open set G ⊂ RN such that @G is not an n-manifold and still, the divergence theorem holds as R R G div F = @GnZ hF; ni for some closed set Z ⊂ @G such that @G n Z is an n-manifold of finite n-dimensional volume. For the cube (or box), @G n Z is the union of the 2N hyperfaces, and Z is the union of cubes (or boxes) of smaller (than N − 1) dimensions. Tel Aviv University, 2015/16 Analysis-IV 69 4b4 Definition. We say1 that the divergence theorem holds for G and @G n Z, if G ⊂ RN is a bounded regular open set, Z ⊂ @G is a closed set, @G n Z is an n-manifold of finite n-dimensional volume, and R R N G div F = @GnZ hF; ni for all F 2 C(G ! R ) such that F jG 2 C1(G ! RN ) and DF is bounded on G. N1 N2 4b5 Exercise (product). Let G1 ⊂ R , Z1 ⊂ @G1, and G2 ⊂ R , Z2 ⊂ @G2. If the divergence theorem holds for G1, @G1 n Z1 and for G2, N1+N2 @G2 n Z2, then it holds for G, @G n Z where G = G1 × G2 ⊂ R and @G n Z = (@G1 n Z1) × G2 ] G1 × (@G2 n Z2) . Prove it.2 An N-box is the product of N intervals, of course. Also, a cylinder f(x; y; z): x2 + y2 < r2; 0 < z < ag 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. 4c1 Definition. (a) The Laplacian ∆f of a function f 2 C2(G) on an open set G ⊂ Rn is ∆f = div rf : (b) f is harmonic, if ∆f = 0.
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