14 Divergence, flux, Laplacian

Transcription

1 Tel Aviv University, 204/5 Analysis-III,IV Divergence, flux, Laplacian 4a What is the problem b Integral of derivative (again) c Divergence and flux d Divergence of gradient: Laplacian e Laplacian at a singular point Divergence and flux are widely used in order to relate volume integrals and surface integrals in a geometrically natural way. 4a What is the problem The integral of derivative (3b3) deserves a generalization. The most straightforward generalization is (4a) Df = 0 if f C (R n R m ) has a bounded support; R n but this is boring. Indeed, (Df) x may be thought of as a matrix whose rows are gradients of the coordinate functions f,..., f m C (R n ) of f (recall Sect. 2e), and (4a) is just (3b3) applied rowwise. Restricting ourselves to the case m = n, we may think about det(df); definitely an interesting function of Df. We cannot expect det(df) to vanish, since the determinant is a nonlinear function of a matrix. But we know (recall 2e9) that (4a2) det(i + H) = + tr(h) + o(h) for small H. The trace being a linear function of a matrix, we have (4a3) tr(df) = 0 if f C (R n R n ) has a bounded support. R n Now the question is, what is tr(df) good for? Consider a one-parameter family of diffeomorphisms ϕ t : R n R n given for t R; we assume that the mapping (x, t) ϕ t (x) belongs to C 2 (R n+ R n ), and ϕ 0 (x) = x for all x R n. Then (Dϕ 0 ) 0 = I and (Dϕ t ) 0 = I + ta + o(t) where A = d dt t=0 (Dϕ t ) 0 ; thus, det(dϕ t ) 0 = + t tr A + o(t) for small t. If tr A > 0, then det(dϕ t ) 0 > for small t > 0, which means that

2 Tel Aviv University, 204/5 Analysis-III,IV 24 v ( ϕ t (U) ) > v(u) for a small enough neighborhood U of 0 in R n. Moreover, v ( ϕ t (U) ) ( + t tr A)v(U). In mechanics, a flowing matter may be described this way; every point x flows to another point ϕ t (x) during the time interval (0, t). A small drop of the flowing matter inflates if tr A > 0 and deflates if tr A < 0. The rate of this inflation/deflation is tr A. The vector F (x) = d dt t=0 ϕ t (x) is the velocity of the flow at a point x and the instant 0. This mapping F : R n R n is called the velocity field of the flow. We have A = d (Dϕ t ) 0 = dt t=0 ( D ( d )) ϕ t dt t=0 0 = (DF ) 0, thus, the inflation/deflation rate at the origin is tr A = tr(df ) 0, and similarly, at a point x it is tr(df ) x. The velocity field is a vector field. The word field in vector field is not related to the algebraic notion of a field. Rather, it is related to the physical notion of a force field (gravitational, for example), or the velocity field of a moving matter (usually liquid or gas). Mathematically, a vector field formally is just a mapping R n R n ; less formally, a vector is attached to each point. A vector field on an affine space is a mapping from this space to its difference space. Note that the determinant is well-defined in a (finite-dimensional) vector space; metric is irrelevant. The same holds for the trace. 4a4 Definition. The divergence of a mapping ( vector field ) F C (R n R n ) is the function ( scalar field ) div F C(R n ), div F = tr(df ). That is, for F (x) = ( F (x),..., F n (x) ) we have div F = D F + + D n F n = ( F ) + + ( F n ) n ; div F (x,..., x n ) = F (x,..., x n ) + + F n (x,..., x n ). x x n Once again: if F is a velocity field, then div F is the inflation/deflation rate. For a vector field F C (V V ) on an n-dimensional vector space V, still, div F = tr(df ); here (DF ) x : V V. For a vector field F C (S S) on an n-dimensional affine space S, also, div F = tr(df ); here (DF ) x : S S. Clearly, (4a5) div F = 0 if F has a bounded support. R n

3 Tel Aviv University, 204/5 Analysis-III,IV 242 Similarly to the singular gradient (treated in Sect. 3b), we want to introduce singular divergence; and then, similarly to Theorem 3b9, we want to generalize (4a5) to a vector field continuous up to a surface. 4b Integral of derivative (again) Similarly to Sect. 3b we consider a hypersurface, that is, an n-dimensional manifold M in R N, N = n +. Similarly to 3b5, for a vector field F : R N \ M R N we define the notion continuous up to M. Clearly, F = (F,..., F N ) is continuous up to M if and only if F,..., F N are continuous up to M (as defined by 3b5). The one-sided limits F, F + are now vectorvalued, and the jump F + (x 0 ) F (x 0 ) is a vector; its sign depends on the side indicator. Recall the unit normal vector n x R N ; its sign also depends on the side indicator. Here is a definition similar to 3b7. As before, we denote F (x 0n x ) = F (x) and F (x + 0n x ) = F + (x). 4b Definition. The singular divergence div sng F (x) at x M of a mapping F : R N \ M R N continuous up to M is the number div sng F (x) = F (x + 0n x ) F (x 0n x ), n x. As before, the singular divergence does not depend on the side indicator (and n x ). It is a continuous function div sng F : M R. Less formally, the singular divergence is the jump of the normal component of the vector field. Here is the singular counterpart of the formula div F = k ( F k ) k. 4b2 Lemma. div sng F = N ( sng F k ) k. Proof. ( sng F k (x)) k = (( Fk (x + 0n x ) F k (x 0n x ) ) ) n x k = k k = ( F (x + 0nx ) F (x 0n x ) ) k(n x ) k = k k= = F (x + 0n x ) F (x 0n x ), n x = div sng F (x). Not a standard terminology.

4 Tel Aviv University, 204/5 Analysis-III,IV 243 A theorem, similar to 3b9, follows easily. 4b3 Theorem. Let M R n+ be an n-manifold, K M a compact subset, and F : R n+ \ K R n+ a mapping such that (a) F is continuously differentiable (on R n+ \ K); (b) F R n+ \M is continuous up to M; (c) F has a bounded support, and DF is bounded (on R n+ \ K). Then div F + div sng f = 0. R n+ \K Proof. We have F (x) = ( F (x),..., F N (x) ), and Theorem 3b9 applies to each F k, giving F k + sng F k = 0. R n+ \K It remains to take the k-th coordinate, and sum up over k. 4c Divergence and flux We return to the case treated before, in the end of Sect. 3b: R N is a bounded regular open set, and R N a (necessarily compact) hypersurface (that is, n-manifold for n = N ). Recall the outward unit normal vector n x for x. 4c 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 (3a8).) 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 : R N be continuous, F C ( R N ), with DF bounded (on ). Then 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. M M

5 Tel Aviv University, 204/5 Analysis-III,IV 244 is continuous up to, and F (x 0n x ) = F (x), F (x + 0nx ) = 0 ; div sng F (x) = F (x), nx. By Theorem 4b3 (applied to F and K = ), (4c2) div F = F, n, just the flux. The divergence theorem, formulated below, is thus proved. 4c3 Theorem (Divergence theorem). Let R n+ be a bounded regular open set, an n-manifold, F : R n+ continuous, 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. 4c4 Exercise. (a) For every f C (), boundedness of f on ensures that f extends to by continuity (and therefore is bounded). (b) For every F C ( R n+ ), boundedness of DF on ensures that F extends to by continuity (and therefore is bounded). Prove it. 2 In such cases we ll always mean this extension. 4c5 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 (3b3): (4c6) f, F = f F, n f div F. In particular, if div F = 0, then f, F = f F, n 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: recall the proof of 3b4.

6 Tel Aviv University, 204/5 Analysis-III,IV 245 4d Divergence of gradient: Laplacian Some (but not all) vector fields are gradients of scalar fields. 4d 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 affine space may be used instead of R n. Indeed, the gradient is well-defined in such space, and the divergence is welldefined even without Euclidean metric. The divergence theorem 4c3 gives the so-called first reen formula (4d2) f = f, n = D n f, 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 (4c6) to f = u and F = v we get u, v = u v, n u v, that is, (4d3) (u v + u, v ) = u v, n = ud n v, the second reen formula. It follows that (4d4) (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.

7 Tel Aviv University, 204/5 Analysis-III,IV 246 Rewriting (4d4) as (4d5) 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. 4d6 Exercise. Consider homogeneous polynomials on R 2 : m f(x, y) = c k x k y m k. k=0 For m =, 2 and 3 find all harmonic functions among these polynomials. 4d7 Exercise. On R 2, (a) a function of the form m f(x, y) = c k e a kx+b k y k= is harmonic only if it is constant; (b) a function of the form is harmonic if and only if a = b. 2 Prove it. f(x, y) = e ax cos by (a k, b k, c k R) 4d8 Exercise. Consider f : R N \ {0} R of the form f(x) = g( x ) for a given g C 2 (0, ). Prove that 3 (a) f C 2 (R N \ {0}); (b) f(r + ε, δ, 0,..., 0) = g(r) + g (r)ε + 2( g (r)ε 2 + r g (r)δ 2) + o(ε 2 + δ 2 ); (c) f(x) = g ( x ) + N x g ( x ). ( Thus, f is harmonic if and only if g (r) + N g (r) = 0 for all r; that is: r log g (r) ) = N = (N )(log r) ; log g (r) = (N ) log r + const; r g (r) = const r (N ) ; g(r) = const r () + const 2 ; { c + c x (4d9) f(x) = 2 if N 2; c log x + c 2 if N = 2. In fact, they are Re (x + iy) m, Im (x + iy) m and their linear combinations. 2 That is, f(x, y) = Re ( e x+iy). 3 Hint: (a,b) x = x 2 ; (c) rotation invariance.

8 Tel Aviv University, 204/5 Analysis-III,IV 247 4e Laplacian at a singular point The function g(x) = / x 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. 4e Lemma. R N f(x) 2πN/2 dx = (N 2) x 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 is improperly integrable near 0 (recall 0b7(c)). The coefficient 2πN/2 is the (N )-dimensional Γ(N/2) volume of the unit sphere (recall (3c9)). Proof. For arbitrary ε > 0 we consider the function g ε (x) = / ( max( x, ε) ), and g(x) = / x. 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 regular open sets = {x : x < ε}, 2 = {x : ε < x < R}, and apply (4d4), 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 = x =ε ε D nf, g ε D n2 f = 2 x =ε ε D nf where n is the outward normal of and inward normal of 2 ; these two summands cancel each other. Further, fd n g ε = f 0 = 0 since g x =ε ε is constant on ; and fd n2 g ε = f N 2 2 x =ε ε, N 2

9 Tel Aviv University, 204/5 Analysis-III,IV 248 since g ε (x) = / x 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. 4e2 Remark. For N = 2 the situation is similar: f(x) log x dx = 2πf(0) R 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 (4d5) and 4e: a singular point and two layers. 4e3 Remark. Let R N be a bounded regular open set, an n-manifold, f C 2 () with bounded second derivatives, and 0. Then f(x) 2πN/2 dx = (N 2) x Γ(N/2) f(0) ( ) x f(x)d n + x ( ) x (D n f(x)). x The proof is very close to that of 4e. The case N = 2 is similar to 4e2, of course. The case = {x : x < R} is especially interesting. Here = {x : x = R}; on, thus, x <R x = 2 and D R nx = N x R ; N f(x) 2πN/2 dx = (N 2) x Γ(N/2) f(0)+ N 2 f + D R N =R R n f. =R Taking into account that D =R nf = f by (4d2) we get <R ( (N 2) 2πN/2 Γ(N/2) f(0) = x <R x ) f(x) dx + N 2 R R N =R f

10 Tel Aviv University, 204/5 Analysis-III,IV 249 for N > 2; and similarly, 2πf(0) = x <R for N = 2. In particular, for a harmonic f, ( ) log R log x f(x) dx + f R =R f(0) = Γ(N/2) f = 2π N/2 R N =R f =R =R for N 2; the following result is thus proved (and holds also for N =, trivially). 4e4 Proposition (Mean 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. 4e5 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). 4e6 Remark. Can the mean value property be generalized to a nonspherical boundary? We leave this question to more special courses (PDE, potential theory). But here is the idea. In 4e3 we may replace f(x) dx x with ( ) + g(x) f(x) dx where g is a harmonic function satisfying x + g(x) = 0 for all x (if we are lucky to have such g). Then x the double layer (D nv)u in (4d5), and the corresponding term in 4e3, disappears, and we get ( (N 2) (x 2πN/2 Γ(N/2) f(0) = )) f(x)d n x + g(x). 4e7 Exercise (Maximum 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 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.

11 Tel Aviv University, 204/5 Analysis-III,IV 250 (4e8) It appears that ( (mean ) ) f(x) = 2N lim of f on {y : y x = ε} f(x). ε 0 ε 2 4e9 Exercise. (a) Prove that, for N > 2, ( R 2 x ) dx does not depend on R; R 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.) (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 (4e8). 4e0 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. 2 4e Proposition. (Liouville s theorem for harmonic functions) Every harmonic function R N [0, ) is constant. Proof. (Nelson s short 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). Hint: change of variable. 2 Hint: (a) recall 3c8.

12 Tel Aviv University, 204/5 Analysis-III,IV 25 Index divergence, 24 divergence theorem, 244 flux, 243 reen formula first, 245 second, 245 third, 245 harmonic, 245 heat, 249 Laplacian, 245 layer, 246 Liouville s theorem, 250 maximum principle, 249 mean value property, 249 singular divergence, 242 trace, 240 vector field, 24 velocity field, 24, 245 div, 24 div sng, 242 tr, 240