�� 28. Manifolds with boundary Definition 28.1. Upper half space is the set Hm = { (x1,x2,...,xm) | xm ≥ 0 }⊂ Rm . The boundary of Hm, is ∂Hm = { (x1,x2,...,xm) | xm =0 }⊂ Mm . Definition 28.2. A subset M ⊂ Rk is a smooth m-manifold with boundary if for every �a ∈ M there is an open subset W ⊂ Rk and an open subset U ⊂ Rm, and a diffeomorphism �g : Hm ∩ U −→ M ∩ W. The boundary of M is the set of points �a which map to a point of the boundary of Hm . Example 28.3. The solid ellipse, ��2 ��2 M = { (x,y) ∈ R2 | xa + yb ≤ 1 }, is a 2-manifold with boundary. Let U1 = { (u,v) | 0 0 for every (u,v) ∈ H2 ∩ U. 3 � � � � � � � � � � � � � � � � � � � � � In this case, parametrise ∂M ∩ W as follows; define �s:(−a,a) −→ ∂M ∩ W, by the rule �s(u)= �g(�x(u)) = �g(u, 0). Note that this is compatible with the orientation, as we are assuming that the Jacobian of g is positive. a F�d�s = F�(�s(u)) �s�(u)du·· ∂M �−a a = F�(�g(�x(u)))D�g(�x(u)) �x�(u)du· �−a = G�d�s, ∂H2 · where G�: H2 ,−→ R2 is defined by the rule G�(u,v)= F�(�g(u,v))D�g(u,v) if(u,v) ∈ U �0 otherwise. Now we compute, ∂G2 ∂G1 ∂ ∂x ∂y ∂∂x ∂y∂u − ∂v = ∂u F1 ∂v + F2 ∂v − ∂v F1 ∂u + F2 ∂u∂F1 ∂x ∂F1 ∂y ∂x ∂2x =+ + F1∂x ∂u ∂y ∂u ∂v ∂u∂v ∂F2 ∂x ∂F2 ∂y ∂y ∂2y++ + F2∂x ∂u ∂y ∂u ∂v ∂u∂v ∂F1 ∂x ∂F1 ∂y ∂x ∂2x − ∂x ∂v + ∂y ∂v ∂u − F1 ∂v∂u ∂F2 ∂x ∂F2 ∂y ∂y ∂2x − ∂x ∂v + ∂y ∂v ∂u − F2 ∂v∂u ∂F2 ∂x ∂y ∂x ∂y ∂F1 ∂x ∂y ∂x ∂y = ∂x ∂u ∂v − ∂x ∂u − ∂y ∂u ∂v − ∂x ∂u ∂F2 ∂F1 ∂(x,y) = ∂x − ∂y ∂(u,v) . 4 �� � � � � � � � Using this, we get �� � ��� � � ∂G2 ∂G1 ∂G2 ∂G1 H2 ∂u − ∂v du dv = H2∂u − ∂v du dv �� ∩U �� ∂F2 ∂F1 ∂(x,y) =du dv H2∩U ∂x − ∂y ∂(u,v) ∂F2 ∂F1 = ∂x − ∂y dx dy �� M ∩�W � ∂F2 ∂F1 = ∂x − ∂y dx dy. M Putting all of this together, we have �� � ��� � � ∂F2 ∂F1 ∂G2 ∂G1 ∂x − ∂y dx dy = ∂u − ∂v du dv M H2 = G�d�s· �∂H2 = F�d�s.· ∂M This completes step 2. We now turn to the third and final step. To complete the proof, we need to invoke the existence of partitions of unity. Starting with F�,I claim that there are vector finitely many fields F1,F2,...,Fk, each of which satisfy the hypotheses of step 2, such that kF�= F�1 + F�2 ++ F�k = F�i.··· i=1 Indeed, start with a partition of unity, m1= ρi, i=1 multiply both sides by F�, to get m F� = i = 1mρiF� = F�i. i=1 5 � � � Granted this, Green’s Theorem follows very easily, �� �� �� � � M ∂F2 ∂x − ∂F1 ∂y kdx dy = � i=1 M ∂Fi,2 ∂x − ∂Fi,1 ∂y dx dy k� � = �Fi · d�s. i=1� ∂M = �F · d�s. � ∂M Lemma 28.5. Let K ⊂ Rn . Suppose that K is containedin the union of closed balls B1,B2,...,Bm, such that any point of K belongs to the interior of at least one of B1,B2,...,Bm. Then we may find smooth functions ρ1,ρ2,...,ρm such that ρi is zero outside Bi and m1= ρi. i=1 Proof. We prove the case n = 2. The general case is similar, only notationally more involved. First observe that it is enough to find smooth functions σ1,σ2,...,σm, such that σi is zero outside Bi and such that mσ = σi, i=1 does not vanish at any point of K. Indeed, if we let σiρi = ,σ then ρi is smooth, it vanishes outside Bi and dividing both sides of the equation above by σ, we have m1= ρi. i=1 In fact it suffices to find functions σ1,σ2,...,σm, such that σi vanishes outside Bi and which is non-zero on the interior of Bi (replacing σi by σi 2, so that σi 2 is positive on the interior of Bi, we get rid of the annoying possibility that the sum is zero because of cancelling). It is enough to do this for one solid circle Bi and we might as well assume that B = Bm = B1 is the solid unit circle. Using polar coordinates, we want a function of one variable r which is zero outside [0, 1] and which is non-zero on (0, 1), so we are now down to a one variable question. 6 At this point we realise we want a smooth function, f : R −→ R, all of whose derivatives are zero at 0 and yet the function f is not the zero function. Such a function is given by f(x)= e−1/x2 . � 7 MIT OpenCourseWarehttp://ocw.mit.edu 18.022 Calculus of Several Variables Fall 2010 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.