BELTRAMI FLOWS

Athesissubmittedtothe Kent State University Honors College in partial fulfillment of the requirements for Departmental Honors

by

Alexander Margetis

May, 2018 Thesis written by

Alexander Margetis

Approved by

, Advisor

, Chair, Department of Mathematics

Accepted by

, Dean, Honors College

ii Table of Contents

Acknowledgments...... iv

1 Introduction ...... 1

2 Sobolev-Spaces & the Lax-Milgram theorem ...... 2

2.0.1 HilbertspaceandtheLax-Milgramtheorem ...... 3

2.0.2 Dirichlet Problem ...... 5

2.0.3 NeumannProblem ...... 8

3 Constructing a solution of (1.1) ...... 11

3.1 The ...... 12

3.2 A solution of (3.3) results in a solution of (3.1) ...... 16

References ...... 19

iii Acknowledgments

I would like to thank my research advisor, Dr. Benjamin Jaye for everything he has done. His support and advice in this thesis and everything else has been invaluable. I would also like to thank everyone on my defense committee. I know you did not have to take that time out of your busy schedule.

iv Chapter 1

Introduction

ABeltramiflowinthree-dimensionalspaceisanincompressible(divergencefree) vector field that is everywhere parallel to its curl. That is, B = curl(B)= r⇥ B for some function .Theseflowsarisenaturallyinmanyphysicalproblems.

In astrophysics and in plasma fusion Beltrami fields are known as force-free fields.

They describe the equilibrium of perfectly conducting pressure-less plasma in the presence of a strong magnetic field. In fluid mechanics, Beltrami flows arise as steady states of the 3D Euler equations. Numerical evidence suggests that in certain regimes turbulent flows organize into a coherent hierarchy of weakly interacting superimposed approximate Beltrami flows [PYOSL]. Given the importance of Beltrami fields, there are several approaches to proving existence of solutions, for instance use the calculus of variations [LA], and use fixed point arguments [BA]. In this thesis we instead use a approach through the Lax-Milgram lemma.

Our goal will be to find a weak solution to the Beltrami flow, for a constant in

⌦andaboundaryfunctiong on @⌦satisfying @⌦ gdS =0, R

B = B in ⌦ 8r⇥ > (1.1) > B =0in⌦ >r· <> B n = g on @⌦ > · > > :> 1 Chapter 2

Sobolev-Spaces & the Lax-Milgram theorem

Sobolev Spaces give rise to what are called weak solutions to di↵erential equations

- and do so because they are based on the notion of the weak derivative. We recall the classical integration by parts formula for an open set ⌦with a smooth boundary:

If f and g are continuously di↵erentiable functions in ⌦(we shall write C1(⌦) for the collection of such functions), then

@f @g (2.1) gdx = f dx + fgn dS, @x @x i Z⌦ j Z⌦ j Z@⌦ where n =(n1,n2,n3)istheoutwardunitnormalto⌦. Wesaythatanintegrable function f has a weak derivative Djf if Djf is an integrable function, and for any

C1(⌦) function g with g =0on@⌦,

@g (D f)gdx = f dx. j @x Z⌦ Z⌦ j

We will see that by moving away from the classical, or strong, derivative we will be able to prove some quite powerful theorems about the existence of solutions to uniformly elliptic equations. The Lax–Milgram theorem, gives conditions under which a bilinear function can be ”inverted” to show the existence and uniqueness of a weak solution to a given boundary value problem.

2 3

2.0.1 Hilbert space and the Lax-Milgram theorem

Definition 2.0.1. An inner product space is a V over R equipped with a function , : V V R satisfying: h· ·i ⇥ !

1. (Bilinearity) af + bg, h = a f,h + b g, h , and f, ag + bh = a f,g + b f,h , h i h i h i h i h i h i for any f,g,h V and a, b R 2 2

2. (Symmetry) f,g = g, f for any x, y V . h i h i 2

3. (Non-degeneracy) For f V , f,f > 0 for f =0. 2 h i 6

Given an inner product space (V, , ), we can associate a norm to V through the h i formula

f = f,f . k k h i p The function satisfies the following three properties k·k

1. (Non-degeneracy) f =0ifandonlyiff =0. k k

2. (Homoegeneity) if f V and F,then f = f . 2 2 k k | |k k

3. (The triangle inequality) if f,h V ,then f + g f + g . 2 k kk k k k

Therefore, we may define the distance between f and g as f g .Thisgivesus k k notions of convergence and continuity, etc. We say that V is a Hilbert space if it is a complete inner product space, meaning that every sequence f that is Cauchy, i.e. { n}

for every ">0thereexistsN N (2.2) 2 such that f f <"for every n, m N, k n mk 4

has a limit in V ,meaningthatthereisf V with 2

for every ">0thereexistsN N such that fn f <"for every n N. 2 k k

The most common example of a Hilbert space that we shall see is a . We first define L2(⌦) as the collection of (Lebesgue) function f :⌦ R with ! f 2dx < .WeputaninnerproductonL2(⌦) with the formula ⌦ | | 1 R f,g = fgdx. h i2 Z⌦

The Cauchy-Schwarz inequality

1/2 1/2 (2.3) f g dx f 2dx g 2dx for any f,g L2(⌦) ⌦ | || |  ⌦ | | ⌦ | | 2 Z ⇣Z ⌘ ⇣Z ⌘ ensures that f,g is finite if f,g L2(⌦). We next define h i2 2

W 1,2(⌦) = f such that f L2(⌦) and, for every j =1, 2, 3 (2.4) 2 n the weak derivative D f L2(⌦) . j 2 o We can endow this space with an inner product via

3

f,g 1,2 = f,g + D f,D g , h iW h i2 h j j i2 j=1 X where each term makes sense because of (2.3). With an abuse of notation, we define the weak gradient

f =(D ,f,D f,D f), r 1 2 3 5

and set 3 f, g = D f,D g . hr r i2 h j j i2 j=1 X We next state the Lax-Migram lemma, which may be found in Chapter 6 of the book by Evans [E].

Theorem 2.0.2 (Lax–Milgram Lema). Suppose that V is a Hilbert space, and : A V V R satisfies the properties: ⇥ !

1. is continuous: there is >0 such that (v, ) v for every v, A |A | || |||| | 2 V .

2. is elliptic: there is ↵>0 such that (u, u) ↵ u 2 for every u V . A A || || 2

Then, for every L : V R that is continuous in the sense that there is >0 such ! that L() , there is a unique solution u V of (u, )=L() for every | | || ||v 2 A V . 2

We first show how the Lax-Milgram lemma can be used to solve some classical problems in PDE.

2.0.2 Dirichlet Problem

Define the Laplacian u =div( u). For a function f L2(⌦), our goal is to find r 2 asolutionu of the following problem

u = f in ⌦ (2.5) 8 > <>u =0 on@⌦ > :> 6

For the Dirichlet Problem we pick the function space

V = W 1,2(⌦) = u W 1,2,u=0 in @⌦ , 0 2

1 which is formally the closure of the space C (⌦) with u =0on@! in the norm 1,2 k·kW associated to the Sobolev inner product , 1,2 introduced in the last chapter. We h· ·iW choose and L as A

(v, )= u dxand L()= fdx. A r ·r Z Z

Assuming for the moment that we can verify conditions (1) and (2) of the Lax-

Milgram theorem, we would arrive at a function v such that

u dx = fdx for all W 1,2(⌦). r ·r 2 0 Z⌦ Z⌦

This is considered a weak solution of the Dirichlet problem (2.5) because if we were permitted to integrate by parts on the left hand side, we would arrive at (due to the zero boundary condition)

( u)dx = fdx for all W 1,2(⌦), 2 0 Z⌦ Z⌦ that is, u = f in ⌦. 7

We now return to check (1) and (2). Applying the Cauchy-Schwarz inequality

(v, ) = u dx |A | | r ·r Z 1/2 1/2 (2.6) v 2 dx 2 dx  |r | |r | ⇣Z ⌘ ⇣Z ⌘ v 1,2 1,2 , || ||W || ||W we get that a is continuous so (1) holds.

For property (2), we observe that

(u, u)= u udx= u 2 dx. A r ·r |r | Z⌦ Z⌦

Therefore, in order to establish (2), we need to compare u 2 dx with the ⌦ |r | 2 2 2 Sobolev norm u 1,2 = u dx + u dx. Noticing thatR the Sobolev norm k kW (⌦) ⌦ ⌦ |r | has an extra term in it, we needR an inequalityR of the following form:

Theorem 2.0.3 (Sobolev). There exists C0 > 0 (depending on ⌦) such that

(2.7) u2 dx C u 2 dx for u W 1,2(⌦).  0 |r | 2 0 Z⌦ Z⌦

Inequalities such as (2.7) are called either Friedrich, Sobolev, or Poincar´einequal- ities – we shall state a number of them in this thesis and refer to Chapter 1 of [Maz] for the proofs. Notice that the obstacle to achieving a statement such as (2.7) are constant functions, since if u is constant then u = 0. However, the only constant r 1,2 function in W0 (⌦) is the 0 function, so we do not face this obstacle. Sobolev and Poincar´einequalities essentially state that the constant functions are the only obsta- cle to obtaining an inequalities which says that the integral of a function is dominated 8

by the integral of its (weak) gradient.

Returning to verifying property (2) of the Lax-Milgram lemma, we apply the inequaly (2.7) to get

2 2 2 v 1,2 = v dx + v dx k kW (⌦) |r | (2.8) Z⌦ Z⌦ u 2 dx + C u 2 dx =(C +1) u 2 dx.  |r | 0 |r | 0 |r | Z⌦ Z⌦ Z⌦

Therefore

2 1 2 u u 1,2 , |r | C +1|| ||W (⌦) Z⌦ 0 which gives the ellipticity property (2) with ↵ = 1 . C0+1

With this, we therefore may apply the Lax–Milgram theorem to get a solution u of (u, )=F ()forevery in V ,whichistheweakformulationoftheDirichlet A problem.

2.0.3 Neumann Problem

Our second classical problem we consider will be the Neumann problem:

u = f in ⌦ (2.9) 8 > @u < @n =0 on@⌦ > :> where @u = u n denotes the normal derivative. We first suppose we have a classical @n r · solution of (2.9), and see what this tells us about f.Thedivergencetheorem,which 9

follows from (2.1) ensures that, if u solves (2.9), then

@u fdx = udx = dS =0, @n Z⌦ Z⌦ Z@⌦ we therefore arrive at a computability condition

(2.10) fdx =0. Z⌦

Our weak formulation of (2.9) is that, for f L2(⌦) with fdx =0, 2 ⌦ R v dx = fdx r ·r Z⌦ Z⌦ for every v, that belongs to the function space

V = u W 1,2(⌦) with udx=0 , 2 ⇢ Z⌦ with the same inner product as W 1,2(⌦). Consequently, our bilinear form a and function L are the same as in the Dirichlet problem:

(v, )= v dx,L()= f dx, A r ·r Z⌦ Z⌦ where f satisfies (2.10).

We again need to check properties (1) and (2) of the Lax-Milgram lemma for a.

The verification of (1) is identical to the Dirichlet problem. For property (2), we need 10

another Sobolev type inequality. Recall that

(u, u)= u 2 dx. A |r | Z⌦

2 Again we need to compare with u 1,2 ,asinthecaseoftheDirichletproblem, k kW (⌦) but now the set of functions V is di↵erent. This time, we have ⌦ vdx=0,soweuse the Poincare Inequality,whichisagainprovedinChapter1of[Maz]:R

Theorem 2.0.4 (Poincare). There exists C>0, depending on ⌦, such that

1 2 g(x) g(y) dy dx C g 2 dx, ⌦ vol(⌦) ⌦  ⌦ |r | Z ⇣ Z ⌘ Z for every g W 1,2(⌦). Here vol(⌦) is the volume of ⌦ 2 As we discussed before, the enemy for the validity of such inequalities are the constant functions. In this case, this case is handled by subtracting the integral average on the left hand side, so both sides of the inequality are 0 for a constant function.

For u V , we have u =0,sothePoincar´einequalitywithg = u becomes 2 ⌦ R

2 2 1 2 u dx = u u dx C1 u dx. ⌦ ⌦ vol(⌦) ⌦  ⌦ |r | Z Z ⇣ Z ⌘ Z So, in the same way as in the Dirichlet problem case, we have the ellipticity property (2) with ↵ = 1 .Therefore,theLax–Milgramthenallowsustofindu in C1+1 V with

(u, )=L()forevery V, A 2 which is the weak solution of the Neumann problem. Chapter 3

Constructing a solution of (1.1)

Since here we will be constructing a gradient field, it will be useful to introduce the notation

X3 = B =(B ,B ,B ):B ,B ,B X { 1 2 3 1 2 3 2 } where X is any given space of functions.

For our smooth boundary function g on @⌦, we first solve the classical Neumann problem

u0 =0in⌦ 8 > <> @u = u n = g on @⌦, @n r 0 · > which can be done, for instance,:> by a modification of the Neumann problem in the previous chapter.

We then consider the problem, for a vector field G (W 1,2(⌦))3 with G =0 2 r· in ⌦.

B = (B + G)in⌦ 8r⇥ > (3.1) > B =0in⌦ >r· <> B n =0on@⌦ > · > > :> 11 12

Lemma 3.0.1. If we can solve (3.1) with G = u , then we can find a solution B r 0 of (1.1), that is, as solution of

B = B in ⌦ 8r⇥ > (3.2) > B =0in ⌦ >r· <> B n = g on @⌦ > · > > . :>

Proof. Set B to be the solution of (3.1) with G = u ,andset r 0 e B = B + G.

e Then

B = B + G =0+u =0, and r· r· r· 0 e B n = B n + u n =0+g = g, · · r 0 · e and since, by vector calculus, u =0,so r⇥r 0

B = B + u = B = (B + G)=B. r⇥ r⇥ r⇥r 0 r⇥ e e e So B solves (1.1).

3.1 The bilinear form

We consider the following Sobolev spaces

1,2 3 3 1 (Wn (⌦)) = B :⌦ R : B fdx = ( B)fdx for every f H (⌦) { ! ·r r· 2 } Z⌦ Z⌦ 13

From the integration by parts formula, we observe that this is a weak way of saying

“B n =0on@⌦00. ·

1,2 3 3 1 (Wtan(⌦)) = B :⌦ R : B Fdx = ( B) Fdx for every F H (⌦) , { ! ·r⇥ r⇥ · 2 } Z⌦ Z⌦ which is a weak formulation of “B n =0on@⌦00. ⇥ We introduce the following bilinear form:

(B, F)= ( B) ( F)+( B)( F)dx B ( F)dx. A r⇥ · r⇥ r· r· · r⇥ Z⌦ Z⌦

With a view to solving (3.1), we seek B (W 1,2(⌦))3 with 2 n

(3.3) (B, F)= G ( F)dx A · r⇥ Z⌦ for every F (W 1,2(⌦))3.WeshallfirstfindsuchaB using the Lax-Milgram, then 2 n show why it indeed solves (3.1).

Theorem 3.1.1. Provided that is small enough depending on ⌦, there exists a unique solution B (W 1,2(⌦))3 of (3.3). 2 n

Proof. We shall verify the assumptions of the Lax-Milgram lemma. The verification of (1) follows from the Cauchy-Schwarz inequality in the same way as for the Dirichlet and Neumann problems in the last chapter.

To verify ellipticity condition (2) from the Lax-Milgram lemma, we will need some 14

more Sobolev type inequalities. Consider

(B, B)= B 2dx + ( B)2dx B ( B)dx. A |r ⇥ | r· · r⇥ Z⌦ Z⌦ Z

a2 b2 Notice that, since ab < 2 + 2 ,

B ( B)dx B B dx · r⇥  | ||r ⇥ | (3.4) Z Z⌦ B 2dx + B 2dx.  2 | | 2 | | Z⌦ Z⌦

Consequently,

(B, B) 1 B 2dx + B 2dx B 2dx. A 2 ⌦ |r ⇥ | ⌦ |r· | 2 ⌦ | | ⇣ ⌘hZ Z i Z

Using one type of Sobolev inequality (see [Maz]),

1,2 3 Theorem 3.1.2. There is a constant C3 > 0 such that if B lies in either (Wn (⌦))

1,2 3 or (Wtan(⌦)) ,

(3.5) B 2dx C ( B)3 + B 2dx. |r |  3 r· |r ⇥ | Z⌦ Z⌦

we get that

1 B 2dx + ( B)2dx DB 2dx, |r ⇥ | r· C | | Z⌦ Z⌦ 3 Z⌦ and using another type of Sobolev inequality (see [Maz]), 15

Theorem 3.1.3. There is a constant C > 0 such that if B (W 1,2(⌦))3, then 1 2 n

(3.6) B 2dx C DB 2dx | |  1 | | Z⌦ Z⌦

we get that 1 B 2dx DB 2dx. | | C | | Z⌦ 1 Z⌦ Inserting these inequalities into the expression for (B, B)resultsin A

1 (B, B) 1 DB 2dx DB 2dx A 2 C | | C | | 3 Z⌦ 1 Z⌦ ⇣ 1 ⌘ = 1 DB 2dx. C3 2 C1 ⌦ | | h ⇣ ⌘ iZ But now using (3.6) once more, we get

1 1 DB 2dx = DB 2dx + DB 2dx | | 2 | | 2 | | Z⌦ Z⌦ Z⌦ 1 1 DB 2dx + B 2dx 2 | | 2C | | Z⌦ 1 Z⌦ 1 B H1(⌦). 2C1 k k

After all this, we arrive at

1 1 (B, B) 1 B H1(⌦), A C3 2 C1 2C1 k k h ⇣ ⌘ i which verifies the ellipticity condition if is small enough. Consequently, we may use the Lax-Milgram lemma to find a weak solution of (3.3) 16

3.2 A solution of (3.3) results in a solution of (3.1)

Suppose B (W 1,2(⌦))3 solves (3.3). 2 n

Lemma 3.2.1. B =0in ⌦. r·

Proof. Consider the solution u of the Neumann problem

u = B in ⌦ 8 r· > <> u n =0on@⌦, r · > :> which is solvable since B n =0on@⌦so Bdx =0.Since B L2(⌦), · ⌦ r· r· 2 we actually have u H2(⌦) by elliptic regularityR (see Chapter 6 of [E]). Then 2 F = u (W 1,2(⌦))3.ThereforewecanplugthisF as a test function in (3.3). r 2 n Since u =0,(3.3)collapsesto r⇥r

( B)( u)dx =0, r· r·r Z⌦ but

( B)( u)dx = ( B)(u)dx = ( B)2dx r· r·r r· r· Z⌦ Z⌦ Z⌦ Therefore B =0in⌦. r·

With this lemma, one of the terms in (3.3) disappears, and we are left with

(3.7) ( B) ( F)dx B ( F)dx = G ( F)dx. r⇥ · r⇥ · r⇥ · r⇥ Z⌦ Z Z⌦ 17

Recall that we would like to conclude that

B B = G in ⌦, r⇥ which is (3.1). Observe that the vector field X = B B G lies in (W 1,2(⌦))3 r⇥ because of (3.7) and has X =0in⌦.Therefore,itwillsucetoprovethefollowing r· lemma:

Lemma 3.2.2. Suppose that X L2(⌦) satisfies X =0in ⌦ and 2 r·

(3.8) X ( F) dx =0 · r⇥ Z⌦ for every F (W 1,2(⌦))3. Then X =0in ⌦. 2 n

Proof. First notice that if F has F =0on@⌦, then F (W 1,2(⌦))3,soforallsuch 2 n F ,

( X) Fdx = X ( F)dx =0. r⇥ · · r⇥ Z⌦ Z⌦ But this means that X =0in⌦. r⇥ If F (W 1,2(⌦))3 then by definition 2 tan

X ( F)dx = ( X) Fdx, · r⇥ r⇥ · Z⌦ Z⌦ and the right-hand side equals zero since X =0in⌦. r⇥ Since any F C1(⌦) can be decomposed as a sum of F = F + F where F 2 1 2 1 2 18

(W 1,2(⌦))3 and F (W 1,2(⌦))3 (see [Maz]) we get that n 2 2 tan

X ( F) dx =0foreveryF (W 1,2(⌦))3. · r⇥ 2 Z⌦

Therefore X (W 1,2(⌦))3.Butnow,since X =0and X =0in⌦and 2 tan r⇥ r· X (W 1,2(⌦))3, we see from the following Sobolev inequality: 2 tan Theorem 3.2.3. There is a constant C > 0 such that if B (W 1,2(⌦))3, then 2 2 tan

(3.9) B 2dx C DB 2dx | |  2 | | Z⌦ Z⌦

that DX 0in⌦,soX is a constant vector field. But then X is the zero field, ⌘ because of (3.8).

With this we have completed our task of finding a solution of (1.1) References

[BA] T. Z. Boulmezaoud, T. Amari, On the existence of non-linear force-free fields in three-dimensional domains,ZeitschriftfrangewandteMathe- matik und Physik ZAMP, November 2000, Volume 51, Issue 6, pp 942–967 [CDG] Jason Cantarella, Dennis DeTurck and Herman Gluck, Vector Calculus and the Topology of Domains in 3-Space, The American Mathematical Monthly, Vol. 109, No. 5 (May, 2002), pp. 409–442 [CC] C. Chae and P. Constantin Remarks on a Liouville-type theorem for Bel- trami flow Int. Math. Res. Not. 2015 10012-6 [EP] A.EncisoandD.Peralta-Salas Knots and links in steady solutions of the Euler equation Ann. Math. 175 345-67 [E] L. C. Evans, Partial Di↵erential Equations, 2nd Edition. AMS Graduate Studies in Mathematics. [LA] Laurence and Avellanda, On Woltjer’s variational principle for force-free fields. J. Math. Phys. 32 (5), 1991 [Maz] V. Maz’ya, Sobolev spaces with applications to elliptic partial di↵eren- tial equations. Second, revised and augmented edition. Grundlehren der Mathematischen Wissenschaften, 342. Springer, Heidelberg, 2011. [N] Nadirashvili, N. Liouville theorem for Beltrami flow,toappearinGeo- metric and Functional Analysis. [PYOSL] Pelz, R., Yakhot, V., Orszag, S.A., Shtilman, L., Levich, E. Velocity– vorticity patterns in turbulent flow. Phys. Rev. Lett. 54, 2505–2509

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