Lecture 3. Vector fields with given vorticity, divergence and the normal trace, and the divergence and curl estimates

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1 Lecture 3. Vector fields with given vorticity, divergence and the normal trace, and the divergence and curl estimates Ching-hsiao (Arthur) Cheng Department of Mathematics National Central University Taiwan, ROC The National Center for Theoretic Sciences, Summer 2012

2 The divergence and the vorticity Definition (Divergence) Let u : Ω R n be a vector field. The divergence of u is defined by divu = ui x i. Definition (Vorticity) Suppose that n = 2 or 3, and let u : Ω R n be a vector field. The vorticity of u, denoted by ω, is defined by ω = u2 u1 if n = 2 x 1 x 2 or ω i u k = ɛ ijk if n = 3, x j where ɛ ijk is the permutation symbol which is defined as 1 ( 1) if (i, j, k) is even (odd) permutation of (1, 2, 3).

3 Notations and some useful identities Remarks: 1 With the permutation symbol, the cross product of vectors u and v becomes (u v) i = ɛ ijk u j v k. Therefore, (curlu) i = ( u) i = ɛ ijk xj u k. 2 In the following, we use the notation, j to denote the (weak) partial derivative with respect to x j for the sake of simplicity. For example, divu = u i, i and (curlu) i = ɛ ijk u j,k. 3 One of the most useful identity in the vector-related computation is ɛ ijk ɛ irs = δ jr δ ks δ js δ kr.

4 Some useful identities 4 Some identities that can be easily computed via the identity of permutation symbol: 1 divcurlu = 0 if n = 3. 2 curl ϕ = 0 if n = 2 or 3. 3 curlcurlu = u + divu. 4 N (u N) = u (u N)N. 5 (u v) w = u (v w) = v (w u). 6 By the divergence theorem, curlv ϕdx = v curlϕdx ϕ (v N)dS. Ω Ω Ω

5 The normal and tangential trace estimates Lemma (The normal trace estimate) Let u : Ω R n be a vector field, and u L 2 (Ω; R n ) with divu L 2 (Ω). Then the normal trace u N of u belongs to H 0.5 ( Ω), and u N H 0.5 ( Ω) C [ ] u L 2 (Ω;R n ) + divu L 2 (Ω). Lemma (The tangential trace estimate) Let u : Ω R n be a vector field, and u L 2 (Ω; R n ) with curlu L 2 (Ω). Then the tangential trace N (u N) of u belongs to H 0.5 ( Ω), and [ ] N (u N) H 0.5 ( Ω) C u L 2 (Ω;R n ) + curlu L 2 (Ω).

6 The Hodge decomposition The origin: Let Ω = R 3 or T 3. Since v = curlcurlv divv, Each vector field u H m (Ω; R 3 ) can be decomposed as u = curlw + p, where w = curl( ) 1 u and p = div( ) 1 u. Moreover, 1 The decomposition is unique since if u = curlw 1 + p 1 = curlw 2 + p 2, then curl(w 1 w 2 ) = (p 2 p 1 ) which implies that (p 2 p 1 ) L 2 (R 3 ) = 0. 2 u H m (Ω) w H m+1 (Ω;R 3 ) + p H m+1 (Ω) [ ] C curlu H m 1 (Ω;R 3 ) + divu H m 1 (Ω).

7 The Hodge decomposition Theorem For Ω = R 3 (or T 3 ) and m N, each vector field u H m (Ω; R 3 ) can be uniquely decomposed as u = curlw + p for some w H m+1 (Ω; R 3 ) with divw = 0, and p H m+1 (Ω). Theorem For Ω = R 3 (or T 3 ) and m N, each vector field u H m (Ω) satisfies u H m (Ω) C [ ] curlu H m 1 (Ω;R 3 ) + divu H m 1 (Ω). Question: What if Ω has boundary?

8 The Hodge decomposition Define Hdiv k (Ω) := { u H k (Ω; R n ) divu = 0 in Ω, u N = 0 on Ω }. Lemma There exists a linear operator P : H k (Ω; R n ) Hdiv k (Ω; Rn ) such that P(Pu) = Pu (that is, P is a projection) and Pu H k (Ω;R n ) C u H k (Ω;R n ). This operator P is called the Leray projector. Proof: Given u H k (Ω; R n ), let p solves p = divu in Ω, p = u N N on Ω. Then Pu = v u p is what we want.

9 The Hodge decomposition Let u H k (Ω; R 3 ). Question: Is it possible to write u = curlw + p for some divergence-free vector w H k+1 div (Ω; R3 ) and p H k+1 (Ω)? Answer: Yes. Let V solve V = u in Ω, V = 0 on Ω. and let w = PcurlV. Then 1 curlw = curlcurlv = V + divv = u + divv. Letting p = divv leads to u = curlw + p. 2 Estimates: w H k+1 (Ω;R 3 ) + p H k+1 (Ω) C V H k+2 (Ω;R 3 ) C u H k (Ω;R 3 ).

10 The Hodge decomposition Question: Is the decomposition unique? Analysis: Suppose u = curlw 1 + p 1 = curlw 2 + p 2. Then curl(w 1 w 2 ) = (p 2 p 1 ), and (p 2 p 1 ) 2 L 2 (Ω) = curl(w 1 w 2 ) (p 2 p 1 )dx Ω = curl(w 1 w 2 ) N(p 2 p 1 )ds. Ω The decomposition does not seem to be unique unless curlw 1 N = curlw 2 N or p 1 = p 2 on Ω. Remark: In fact, a choice of different boundary condition of V might produce different decomposition!

11 Vector fields with given vorticity, divergence and the normal trace We are interested in finding a vector field u such that curlu = f in Ω, divu = g in Ω, u N = h on Ω. for some given f, g and h. Solvability conditions: 1 Since divcurlw = 0 for all vector w, we must have divf = 0. 2 By the divergence theorem, gdx = divudx = Ω Ω Ω (u N)dS = Ω hds.

12 Vector fields with given vorticity, divergence and the normal trace Case I: If Ω = R 3, since curlcurlu = u + divu, we find that u solves u = curlf g in R n. To solve for u, no boundary condition is needed. Therefore, if f H m 1 (R n ) and g H m 1 (R n ), there exists a unique solution u H m (R n ) satisfying u H m (R n ) C [ ] f H m 1 (R n ) + g H m 1 (R n ). We note that since curl ϕ = 0 for all scalar ϕ, u satisfies curlu = curl( ) 1 curlf = ( ) 1( f + divf ) = f + ( ) 1 divf ; thus u is not a solution unless divf = 0.

13 Vector fields with given vorticity, divergence and the normal trace Case II: If Ω = R 3 + { x R 3 x 3 > 0 }, then N = (0, 0, 1). So the problem becomes curlu = f in Ω, divu = g in Ω, u 3 = h on Ω. By the solvability condition, there exists a solution p to p = g in Ω, p = h N on Ω. Let v = u p, then v satisfies curlv = f in Ω, divv = 0 in Ω, v 3 = 0 on Ω.

14 Vector fields with given vorticity, divergence and the normal trace The same as the previous case, v satisfies v = curlf in R In order to solve for v, boundary conditions are needed. 2 However, there seems to be just one boundary condition: v 3 = 0 on R 2 {x 3 = 0} which can be used to solve for v 3. 3 The boundary condition for v 1 and v 2 : the first two components of the identity curlv = f implies that v,3 1 = f 2 v,1 3 on R 2 {x 3 = 0}, v,3 2 = f 2 + v,2 3 on R 2 {x 3 = 0}. Note that they are Neumann boundary conditions for v 1 and v 2, and the right-hand sides are known as long as v 3 is solved.

15 Vector fields with given vorticity, divergence and the normal trace Question: Does v = (v 1, v 2, v 3 ) satisfy the equation? Answer: Yes, because of the following: 1 Note that the identity (v 1,1 + v 2,2 ), 3 = (v 3,11 + v 3,22 ) + (curlf )3 = v 3,33 implies that divv N = 0 on R2 {x 3 = 0}. 2 Since v satisfies v = curlf, (divv) = 0 in R If v H 3 (R 3 + ; R3 ), we must have divv = 0 in R Since divv = 0, curl(curlv f ) = 0 in R 3 + ; thus curlv f is conservative. Since R 3 + is simply connected, there exists a scalar potential ϕ such that curlv f = ϕ in R 3 +.

16 Vector fields with given vorticity, divergence and the normal trace 5 By the Neumann boundary condition that we use to solve for v 1 and v 2, ϕ = 0 on R 2 {x 3 = 0}. Therefore, ϕ is constant on R 2 {x 3 = 0}. 6 Since ϕ = curlv f and divf = 0, ϕ is harmonic in R 3 +. By assuming f L 2 (R 3 + ) and v H1 (R 3 + ; R3 ), ϕ must be constant in R 3 +. Therefore, curlv = f in R3 +. Estimates: Since v 3 solves a Dirichlet problem, and v 1, v 2 solve two Neumann problems, elliptic regularity suggests that v H m (R 3 + ;R3 ) C f H m 1 (R 3 + ;R3 ) which in turn implies ] u Hm (R 3 + ;R3 ) [ f C H m 1 (R 3 + ;R3 ) + g H m 1 (R 3 + ) + h H m 0.5 (R 2 {x 3 =0}).

17 Vector fields with given vorticity, divergence and the normal trace Case III: Ω R 3 is a convex domain. Since u = curlw + p for some divergence-free w and scalar p, it amounts to find (one) w and p satisfying w = curlcurlw divw = curlu = f in Ω. and p = divu = g in Ω. In order to solve for w and p, boundary conditions are needed! Question: Can we assign the boundary conditions for w and p arbitrarily (but in the form of Dirichlet, Neumann or Robin BCs)?

18 Vector fields with given vorticity, divergence and the normal trace Since g and h satisfies the solvability condition, we may choose p to satisfy By elliptic regularity, Then w solves p H m+1 (Ω) C p = g in Ω, p = h N on Ω. [ ] g H m 1 (Ω) + h H m 0.5 ( Ω). w = f in Ω, divw = 0 on Ω, curlw N = 0 on Ω. We note that since divf = 0, we treat divw = 0 as a boundary condition of w instead of an interior equation.

19 A quick review on differential geometry Definition A (bounded) domain Ω R n is said to be of class C k if Ω is an n 1 dimensional C k sub-manifold of R n, or equivalently, 1 there are open sets O l R n such that Ω N O l ; l=1 2 there are maps ϕ l : U l := Ω O l R n 1 so that ψ l := ϕ 1 l : ϕ l (U l ) R n is a C k injective immersion; that is, for each l, ψ l is C k, one-to-one, and for a given coordinate (y 1,, y n 1 ) in ϕ l (U l ), the set of vectors {ψ l,1 (y),, ψ l,n 1 (y)} are linearly independent for all y ϕ l (U l ) ; 3 the transition map ϕ l1 ϕ 1 l 2 : U l1 U l2 R n 1 is C r for some r k.

20 A quick review on differential geometry Given (U, ϕ) some chart at a point y Ω, of associated coordinates y α, let T y Ω denote the tangent space at y ; that is, if N is the unit normal on Ω, T y Ω = { v R n v N(y) = 0 }. It is a n 1 dimensional linear subspace of R n, therefore has n 1 linearly independent vectors spanning the space. Note that vectors ( y α ) y := ψ l,α(y) are tangent to Ω. By 2 in the Definition of C k -domain, these vectors span T y Ω.

21 A quick review on differential geometry For y n R, define the map Ψ : ϕ(u) R R n by Ψ(y 1,, y n 1, y n ) = ψ(y 1,, y n 1 ) + y n N(ψ(y 1,, y n 1 )), where N denotes the outward-pointing unit normal on Ω. We assume that O is small enough such that Ψ : V = Φ(O) O is bijective; thus for every y O with Cartesian coordinate x, there is a unique (y 1,, y n ) Φ(O) so that x = Ψ(y) ; hence (y 1,, y n ) can be viewed as a curvilinear coordinate in O. Φ(O + ) (y 1,,y n 1 ) R n 1 Ψ Ω ψ = ϕ 1 O + ϕ y Ω Φ=Ψ 1 Ω

22 A quick review on differential geometry By adding an open set O 0 Ω, we assume that Ω N O l. Let {χ l } N l=0 be a partition of unity (of Ω) subordinate to O l; i.e., 1 χ l 0 for all 0 l N, and N χ l (x) = 1 for all x Ω. 2 spt(χ l ) O l. 3 χ l : O l R is smooth. For any f defined on Ω, let f l = χ l f. Then f = N (χ l f ) = N f l. For 1 l n, f l Ψ l is defined on Φ(O l ), or in other words, f l Ψ l is a function of y. Let A l = ( y Ψ l ) 1, by the chain rule, f l x i Ψ l = n (A l ) j i j=1 l=0 (f l Ψ l ) n 1 = y j α=1 l=0 l=0 l=0 (A l ) α i f l,α + (A l ) n i f l,n.

23 A quick review on differential geometry Example Assume that n = 2. In each chart (U l, ϕ l ), let g l = ψ l,1 ψ l,1 and b l = n l,1 ψ l,1 be the metric and curvature tensor. Then n l = ( ψ2 l,1 gl, ψ1 l,1 gl ), and n l,1 = g 1 l b l ψ l,1. By the definition of Ψ l, ψl,1 1 + y 2 nl,1 1 ψ2 l,1 gl y Ψ l = ψ 1 ψl,1 2 + y 2 nl,1 2 l,1 gl = ψ 1 l,1(1 + y 2b l g l ) ψ2 l,1 gl ψ 2 l,1(1 + y 2b l g l ) ψ 1 l,1 gl ; hence (next page)

24 Example (Continued...) gl A l = g l + y 2 b l As a consequence τ 1 l := (A l ) 1 = ψ 1 l,1 gl ψ 2 l,1 gl ψ 2 l,1(1 + y 2b l g l ) ψ 1 l,1(1 + y 2b l g l ). 1 [ ψ 1 g l + y 2 b l,1, ψ 2 ] l,1 // Ω, and (Al ) 2 = n l ; l thus the gradient can be decomposed into the tangential and normal derivatives: f l x i Ψ l = τ 1 li f l,1 + n li f l,n.

25 A quick review on differential geometry Fix one l and omit the index l in all the following discussion (and resume the Einstein summation convection). Then f x i = A α i f, α +A n i f, n. Let e α = (0,, 0, 1, 0,, 0) T be the α-th tangent vector on {y n = 0}, and e n = (0,, 0, 1) T is the outward-pointing normal to {y n 0}. Define τ α = A T e α and ν = A T e n (where A = ( y Ψ) 1 ). We then have τ α = A T e α = G αβ Ψ,β and n = A T e n, where G αβ is the inverse of G αβ Ψ,α Ψ,β. Note that now τ α and n are also defined away from y n = 0, and τ α n = 0 for all α = 1,, n 1.

26 A quick review on differential geometry Definition Given (U, ϕ) some chart at a point y Ω with associated coordinates (y, y n ). 1 The tangential derivative of a function f in the α-th (f Ψ) direction, denoted by f,α or α f, is the function. y α 2 The normal derivative of f, denoted by f,n or N f, is the (f Ψ) function. y n For any function f, we have or simply in notation, x f Ψ = τ α f, α +nf, n = G αβ Ψ, β f, α +nf, n i = τ α i α + n i N = G αβ Ψ i,β α + n i N.

27 A quick review on differential geometry Let g αβ = ψ, α ψ, β be the metric tensor on Ω with inverse g αβ, and Γ γ αβ = 1 2 gγδ[ g αδ,β + g βδ,α g αβ,δ ] be the Christoffel symbol. Definition (Covariant derivatives) Let X = X α ψ, α and Y = Y β ψ, β be two vector fields on Ω. The covariant derivative of Y along the direction X, denoted by X Y, is the vector fields on Ω given by Definition (Divergence) X Y = X α[ Y β,αψ, β +Y β Γ γ αβ ψ, γ The divergence of a vector field X = X α ψ, α = X β g αβ ψ, β on Ω is defined by div Ω X = X α,α + X α Γ β αβ = gαβ[ X α,β + X γ Γ γ αβ]. ].

28 A quick review on differential geometry Some useful formulas: 1 Any vector v can be written as v i = (v Ψ, β )G αβ Ψ i,α + (v n)n i, and on Ω, v i = v β g αβ ψ, α +v n n i = v β ψ,β i + v nn i. 2 The Christoffel symbol satisfies the relation Γ γ αβ = gγδ ψ, δ ψ, αβ = τ γ ψ, αβ. 3 The mean curvature H of Ω is computed by H = 1 2 gαβ ψ, αβ n = 1 2 gαβ ψ, α n, β.

29 curlw N on the boundary By all the terminologies above, near the boundary we have curlw N = N i ɛ ijk j w k = N i [ ɛ ijk G αβ Ψ j,β α + n j ][ N wδ G γδ Ψ k,γ + w n n k], where w δ = w Ψ, δ and w n = w n are the tangential and normal component of w, respectively. Further computations eventually leads to the following identity: curlw N = 1 ] [ψ, 2 ψ,1 (P tan w) ψ, 1 ψ,2 (P tan w) g on Ω, where 1 with g αβ ψ, α ψ, β denoting the metric tensor on Ω, g = det(g αβ ), and 2 P tan is the projection of w onto the tangent plane defined by P tan w = N (w N) = w (w N)N.

30 divw on the boundary Similarly, where divw = i w i = [ G αβ Ψ i,β α + n i N ][ wδ G γδ Ψ i,γ + w n n i] = div Ω (P tan w) + 2Hw n + w n N on Ω, 1 div Ω is the divergence operator on the boundary. 2 H 1 2 gαβ ψ,αβ n is the mean curvature of Ω. Therefore, if we assume that P tan w = 0; that is, the tangential components of w vanishes on Ω, then 1 curlw N = 0. 2 divw = w n N + 2Hw n.

31 Vector fields with given vorticity, divergence and the normal trace Back to our problem w = f in Ω, divw = 0 on Ω, curlw N = 0 on Ω. Suppose that P tan w = 0. The equation above then becomes the following elliptic problem w = f in Ω, P tan w = 0 on Ω, (W) w n N + 2Hw n = 0 on Ω. Question: Is there a solution to this problem?

32 Vector fields with given vorticity, divergence and the normal trace Suppose now that Ω is a bounded, convex domain. Then H 0 and H 0. Define the solution space Hτ 1 (Ω) as { } Hτ 1 (Ω; R 3 ) w H 1 (Ω; R 3 ) P tan w = 0. Definition A vector-valued function w Hτ 1 (Ω; R 3 ) is said to be a weak solution to (W) if w ϕdx + 2 Hw n ϕ n ds = f ϕdx ϕ Hτ 1 (Ω; R 3 ). Ω Ω Remark: To compute the weak formulation, one needs to use the identity Ω w,j i ϕi N j ds = Ω Ω w n N ϕ nds ϕ H 1 τ (Ω; R 3 ).

33 Vector fields with given vorticity, divergence and the normal trace Similar to the existence of the weak solution to the elliptic equations, if the bilinear form B : H 1 τ (Ω; R 3 ) H 1 τ (Ω; R 3 ) R given by B(u, ϕ) w ϕdx + 2 Hw n ϕ n ds Ω Ω is bounded and coercive. Nevertheless, the coercivity of B is guaranteed by the Poincaré inequalities, and the boundedness is trivial. Regularity: If f H m 1 (Ω; R 3 ) with divf = 0, then the same proof of elliptic regularity suggests that w H m+1 (Ω; R 3 ) and w H m+1 (Ω;R n ) C f H m 1 (Ω;R 3 ).

34 Vector fields with given vorticity, divergence and the normal trace Theorem Let Ω R 3 be a convex, bounded and smooth domain. For any given f H m 1 (Ω; R 3 ), g H m 1 (Ω) and h H m 0.5 ( Ω) satisfying the solvability condition divf = 0 and there exists u H m (Ω; R 3 ) satisfying and Ω gdx = curlu = f in Ω, divu = g in Ω, u N = h on Ω, Ω hds, [ ] u H m (Ω;R 3 ) C f H m 1 (Ω;R 3 ) + g H m 1 (Ω) + h H m 0.5 ( Ω).

35 Vector fields with given vorticity, divergence and the normal trace Question: What if Ω is NOT convex? Analysis: If Ω is not convex, then the mean curvature H will be negative somewhere. As a result, there may be no solution to equation (W). Nevertheless, we may switch the course by finding w first and solve for p accordingly, where w and p give the decomposition u = curlw + p. In other words, we try to find w satisfying only w = f in Ω, (W2) divw = 0 in Ω, and leave the boundary value of curlw N undetermined.

36 Vector fields with given vorticity, divergence and the normal trace Analysis: (continued...) As long as one such w is found, we look for p solving p = g in Ω, p = h curlw N on Ω. N Note that the solvability condition guarantees the existence of such p. Then u = curlw + p solves the problem of given vorticity, divergence and the normal trace. Question: How does one find one w satisfying (W2)? Idea: We have already converted the problem again all into the interior, and no boundary condition has to be satisfied. So, we can enlarge the domain of interest to a convex domain.

37 Vector fields with given vorticity, divergence and the normal trace An extension of equation (W2): Let Ω B(0, R) for some R > 0. 1 We need to extend f to a divergence-free vector field defined on B(0, R). 2 It is easy to extend f if f L 2 (Ω; R 3 ) or H 1 (Ω; R 3 ): let f be defined by a solution to div f = 0 in B(0, R)\Ω, f N = 0 ( f = 0) on B(0, R), f N = f N ( f = f ) on Ω. 3 What if f H m (Ω; R 3 ) for some m 2? We do NOT have any tool to extend an divergence-free f H m (Ω; R 3 ) so that it preserves the regularity and the div-free constraint.

38 Vector fields with given vorticity, divergence and the normal trace Therefore, there exists an extension f of f defined as f outside Ω, although this particular extension has limited regularity even if f H m (Ω; R 3 ) with divf = 0 for m 2. Nevertheless, we can still solve for w from the equation w = f in B(0, R), P tan w = 0 on B(0, R), w n N + 2H w n = 0 on B(0, R), and the restriction of w to Ω, denoted by w, solves (W2). Regularity: w H 2+l (B(0, R); R 3 ) if f H l (B(0, R); R 3 ) for l = 0, 1, and w H 2+l (Ω;R 3 ) w H 2+l (B(0,R);R 3 ) C f H l (B(0,R);R 3 ) C f H l (Ω;R 3 ).

39 Vector fields with given vorticity, divergence and the normal trace As a consequence, u = curlw + p has the following properties: 1 u H 1 (Ω; R 3 ) if f L 2 (Ω; R 3 ), g L 2 (Ω), and h H 0.5 ( Ω). 2 u H 2 (Ω; R 3 ) if f H 1 (Ω; R 3 ), g H 1 (Ω), and h H 1.5 ( Ω). 3 u is a vector field with given vorticity f, divergence g, and the normal trace h: curlu = f in Ω, divu = g in Ω, u N = h on Ω. However, even if f, g, h are all smooth, so far we at best expect that u H 2 (Ω; R 3 ).

40 The divergence and curl estimates Question: Can the regularity of u be improved? Answer: The divergence and the curl estimates: Theorem Let Ω R n, n = 2 or 3, be a bounded and smooth domain. Suppose that u L 2 (Ω; R n ) has the properties that 1 curlu H m 1 (Ω; R n ), 2 divu H m 1 (Ω), and 3 u N H m 0.5 ( Ω). Then u H m (Ω; R n ), and for some constant C = C(Ω), [ u H m (Ω;R n ) C u L2 (Ω;R n ) + curlu H m 1 (Ω;R n ) ] + divu H m 1 (Ω) + u N H m 0.5 ( Ω).

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