3D2D FLUIDS.pdf

8/14/2019 3D2D FLUIDS.pdf

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Twistor theory and fluids

Complex techniques for 3D applied mathematics andapplications to viscous flow

William T. Shaw

[email protected]

O.C.I.A.M., Mathematical Institute, Oxford

Twistor theory and fluids p. 1/41


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Paper and Acknowledgements

Complex Variable Methods for 3D AppliedMathematics: 3D Twistors and an application toStokes Flow

Paper at (for now - KCL from Jan 2006):www.maths.ox.ac.uk/shaww/fluids.pdf



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Bateman, Whittaker, Penrose, Hitchin, Mason




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Bateman, Whittaker, Penrose, Hitchin, Mason

Technology translation for applied mathematicsof NH work.




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Plan of Talk

Topics to look at: Remedial Fluid Dynamics



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Plan of Talk


Characterization of Navier-Stokes



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Plan of Talk


Characterization of Navier-Stokes Summary of 3D Twistor Methods



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Plan of Talk



Laplaces Equation



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Plan of Talk



Laplaces Equation

Biharmonic Equation



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Plan of Talk



Laplaces Equation

Biharmonic Equation

(Reduction to 2D - skip here, in paper)



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Plan of Talk



Laplaces Equation

Biharmonic Equation

(Reduction to 2D - skip here, in paper)

Contour Integrals for Axis-symmetric StokesStream Function



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Remedial Fluids

A large class of fluids can be characterized by theirdensity,, a scalar field not presumed to be constant,and their dynamic viscosity. The flow is

characterized by a velocity vector fieldv, and anassocicated scalar pressure fieldp. Conservation ofmass is expressed by the continuity equation

t+ .(v) = 0 (1)

and the conservation of momentum is expressed bythe Navier-Stokes equations

(v

t +v. v) = p+2v (2)



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Incompressibility

If the fluid is incompressible in the sense thatis aconstant in both time and space, we have thecondition:

.v = 0 (3)To analyse matters further, we introduce the vorticityvector

= v (4)

We demand incompressibility but allow for non-zerovorticity. We let

H=p+1

2v2 (5)



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Recasting of Navier-Stokes

If .v= 0then 2v= . Use the identity

(1

2v.v) =v. v+v (6)

Recast Navier-Stokes equations as

(vt

v ) + H= . (7)

Taking curl, we get vorticity equation

t +v. . v=2 (8)

where kinematic viscosity=/.Twistor theory and fluids p. 6/41


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Vector potential

Since the velocity field is divergence-free, we mayintroduce a vector potentialsuch that

v= (9)and furthermore we may choose it so that it isdivergence free (a gauge condition):

. = 0 (10)

It isperhapsa tacet assumption of fluid dynamics thatthis object is of no use except when it can be reducedto a single function. For example, planar 2D flow isobtained by setting (gauge condition guaranteed)

= (x, y)ez (11)Twistor theory and fluids p. 7/41


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Good idea in 3D too

We note that under the assumption thatsatisfies . = 0

= 2 (12)

and the vorticity equation becomes, denoting t

by:

4 =1

(( ) 2) + 2

(13)

Well-known in the 2D planar case where it reduces to:

4=1

(, 2)

(x, y) + 2

(14)



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Difficult (impossible?) Goal

Can we: Understand



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Can we: Understand

Solve in 2D (e.g. more insight into K. Rangerswork)



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Can we: Understand


In 3D, generalize and solve the 3D analogue of

complex 2D (steady state) form:



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Can we: Understand




i 4

w2w2 = 1

2

w

3

ww2

w

3

w2w(18)



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Can we: Understand




i 4

w2w2 = 1

2

w

3

ww2

w

3

w2w(19)

w=x+iy - bad idea to use z here!Twistor theory and fluids p. 9/41


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Potential Flow

Hereis zero and 2 = 0. Vorticity equationsatisfied as identity.The potentialis

(r) =

r( ).dr (20)

and is harmonic conjugate ofin 2D case.

2= 0, 2 = 0 (21)



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The biharmonic limit

When viscosity , ignore non-linearities.Time-independent Navier-Stokes equations reduce to

4

= 0 (22)which is the biharmonic limit, also known as Stokesflow. We want to understand the holomorphic

structure for the 3D vector version of:

4

w2w2= 0 (23)

Key issues: what do we do aboutw =x + iywhen wealso have az? How can we make it all holomorphic

when we have all these wlittered around?Twistor theory and fluids p. 11/41


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Holomorphic methods for 3D

We need a new picture to proceed. It is very wellknown that the Laplace equation can be solved interms of holomorphic functions in two dimensions.

Among devotees of twistor methods, and students ofBateman, Whittaker, it is known that this can becarried out in three dimensions. Want to extend to thebiharmonic case. This can be done. This is at least anopportunity to explain how to use complex methods in3 dimensions. In 2D we letw=x+iy. What if wealso have az? (Never putz=x+iy!) But whatcomplex structure do we use? The key is twistorspace for 3D, a la N. Hitchin.



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Twistor space for 3D

The twistor space associated with R3 is first, as a realspace, the set of oriented straight lines in R3. Relativeto some originO, letrdenote the position vector of

the point on a given line nearest toO. Thenrisorthogonal to the the direction of the line, which wedenote byuwithu.u = 1. So the set of oriented

straight lines is the set

T S2 = (r, u) R3 S2 | r.u= 0 (24)

This tangent bundle for the unit sphere is also the

tangent bundle to a complex manifold,S2 is theRiemann sphere CP1.TCP1 is our twistor space.


D fi i P i I


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Defining Points I

How do we define apointin ordinary space in termsof structure onTCP1. A point may be regarded as theintersection of all straight lines through it. So a point

is necessarily some vector field inTCP1 that isdefinedglobally.


D fi i P i II


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Defining Points II

To see the implications of this we introduce two opensets that cover CP1. We can take coordinates for thesphere ason one patch (covering everything except

infinity), and = 1/on another patch, coveringeverything except= 0. Over each of theserespective patches we can define coordinates for the

tangent bundle as(, )and(,), where the relevantvector fields are, respectively

,

(25)


D fi i P i t III


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Defining Points III

Consider now a holomorphic vector field. On thepatch it can be written as

f0()

(26)

for somef0

, and on the patch, it can be written as

f1()

(27)

for somef1.


D fi i P i t IV


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Defining Points IV

On the intersection of the two patches we have:

f1(1)(2)

=f0()

(28)

Taylor series expansion of both functions,

fi() = n=0

ai

n

n, we deduce that the coefficients

ainvanish ifn >2. Global vector fields must be of theform, for example on thepatch:

() =a+b+c2 (29)

Quadratics are only global holomorphic vector fields,

correspond to points ofC3.Twistor theory and fluids p. 17/41

S f lit d t i


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Summary of reality and metric

Further analysis allows the identification of real pointsin R3, and the construction of a natural metric. Thepoints are real if and only if

c= a AND b=b (30)

Metric discriminant of the quadratic. Normalize as

ds2 =dx2 +dy2 +dz2 =1

4db2 dadc (31)

The metric for real points:

ds2

=dx2

+dy2

+dz2

=

1

4db2

+dada (32)Twistor theory and fluids p. 18/41

Th fi l i t d


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The final point correspondence

If we pick our coordinate system such that the realpart ofaisx, we see that we can take the imaginarypart ofato be yand setb= 2z. The convention is

to set:

r() = (x+iy) + 2z (x iy)2 (33)

This gives us the correspondence between real pointsin 3D and global holomorphic vector fields


The scalar Laplace eq ation


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The scalar Laplace equation

We consider a functionf(, )defined on twistorspace. This can then be thought of as restricted to thespecial global sections of twistor space represented by

r(), and the-dependence integrated out byintegration over a contourC. We set:

(r) =C

f(r(), )d (34)

Note that what really matters isfmodulo otherfunctions that are holomorphic inside or outsideC.Cauchys Theorem ensures that these do notcontribute.


Laplace II


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Laplace II

It is easy to check thatsatisfies the scalar Laplacesequation. Observe that

k

f(r(), )xk = (1 2)k

k

fk |=r (35)

k

f(r(), )yk

=ik(1 +2)k k

fk

|=r (36)

k

f(r(), )zk = (2)k

k

fk |

=r (37)

and that adding these three expressions withk= 2

gives zero identically for any choice off.Twistor theory and fluids p. 21/41

Not just a modal trick


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Not just a modal trick

It would be wrong to think that this is just a trick forbuilding modes that solve Laplaces equation andthen adding them up by integration. There is much

more to it than that. See NH Monopoles paper -consider curves in twistor space and field ofosculating quadratics. The real part of this solves theminimal surface problem. This is another story - butthe point is that the geometric structure is veryimportant.


The scalar biharmonic equation


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The scalar biharmonic equation

How do we modify the integrandf(r(), ), say tosome holomorphic functiong, to arrange that4g= 0but 2g= 0? We try to buildh fromfby

multiplying by some prefactorh(r, ), so that

g=h(r, )f(r(), ) (38)

Now

2g=f2h+h2f+ 2 h. f (39)

That is, asfsatisfies Laplace equation:

2g=f2h+ 2 h. f (40)


Bih i II


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Biharmonics II

If we furthermore choosehto be linear inrmatterssimplify further and we have

2

g= 2 h. f (41)Let us set, w.l.o.g.,h=u().r, so that h=u().We also note that

f=f

=

f

(1 2, i(1 +2), 2). (42)


Bih i III


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Biharmonics III

Putting this all together, we arrive at

2g= 2u().(1 2, i(1 + 2), 2)f

= 2u()()f

(43)We can now see that 2 of this last expressionvanishes identically, while this expression does notitself vanish unless

u()() 0. (44)


Biharmonics IV


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Biharmonics IV

To see what is happening, we can now make mattersmore explicit. We letu() = (u1(), u2(), u3()),then

u.r=u1()x+u3()y+u3()z (45)and

u()() = (u1()+iu2())+2u3()(u1()iu2())2

(46)


Biharmonics V


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Biharmonics V

In terms of these variables the proposed integralrepresentation for solutions of the 3Dscalarbiharmonic equation is just

=

C

d

xu1() +yu2() +zu3()

f(r(), )

(47)or indeed, withw =x+iy:

=

1

2Cd

wg() + wg+() + 2zu3()

f(r(), )

(48)whereg() =u1() iu2())


Relationship to 2D


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Relationship to 2D

Suppose we want noz-dependence. Setu3= 0andw=x+iy:

=C

d

wg() + wg+()

f(r(), ) (49)

=wC

df1(r(), ) + wC

df2(r(), ) (50)

Consider second term. This is w(x,y,z), where

(x,y,z) =

C

df2(r(), ) (51)

is a solution of Laplaces equation.Twistor theory and fluids p. 28/41

Relationship to 2D II


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Relationship to 2D II

(x,y,z) =C

df2(r(), ) (52)

We want this not to depend onzeither. But this looksawkward givenr() = (x+iy) + 2z (x iy)2.

Actually it is not!

(x,y,z+h/2) =C

df2(r() +h, ) (53)

The equation we need is

(x,y,z+h/2) =(x,y,z) (54)


Relationship to 2D III


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Relationship to 2D III

This doesnotrequire that

f2(r() +h, ) =f2(r(), ) (55)

Instead we need

f2(r()+h,) =f2(r(), )+g0( , , h)g1( , , h)

(56)whereg0is holomorphic on and inside C andg1islikewise outside. (Cauchy!) Lets takeCto be unit

circle. Now differentiate w.r.th then seth= 0.

f2

=G0(, ) G1(, ) (57)


Relationship to 2D IV


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Relationship to 2D IV

Integrated w.r.t.and divide by.

f2=H0(, )

H1(, )

(58)

Now evaluate the integral using calculus of residues.First term is easy, get

2iH0(r(0), 0) =K(w) (59)

and a contribution to the field ofwK(w). Calculationof second piece gives another function ofw, to bemultiplied by w. Other terms; we get

= wK(w) + wH( w) +wK( w) +wH(w) (60)Twistor theory and fluids p. 31/41

Relationship to 2D V


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Relationship to 2D V

A fully holomorphic picture in 3D projects to 2Dpicture and generates the familiar yet superficiallynon-holomorphic 2D representation of solutions to

biharmonic and Laplace equations. In 3D ourfunctions are contour integrals.


Axis-symmetric Stokes Flow


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Axis symmetric Stokes Flow

This is traditionally modelled in terms of the Stokesstream functionS(r, ). The components of thevelocity field are given by

ur = 1

r2 sin()

S

, u = 1

r sin()

Sr

(61)

What this representation isreallytelling us, as ismade clear in modern fluid theory, is that thevectorpotentialfor the flow is given by

= Sr sin()

e (62)


The PDE


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The PDE

For an axis-symmetric functionf(r, )

( f

r sin()

e) = 1

r sin()

(D2f)e (63)

where the operatorD2 is given by

D2f= 2f

r2 +sin()

r2

1

sin()f

(64)

The biharmonic condition may be expressed as thescalar PDE

D4S= 0 (65)


Going to Cartesian Basis


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g

e=yex+xey

x2 +y2

=yex+xey

r sin() =

1

r sin(){w[iex+ey]

(66)wherew=x+iy as before.

= Sr sin()e (67)

and ifSis real we can write the vector potential as

=

Sw

r2 sin2()[iex+ey]

(68)


Cartesian Analysis


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y

Cartesian components must satisfy the scalarbiharmonic equation

S=r2 sin2()g(r, ) =ww 1

2i

d1

f

(69)for some complex functionf.fcan be expanded as aLaurent series:

S=ww

1

2i

d

n=

an

n+1 (w+ 2z w2

)n

(70)The terms in the series can be evaluated in terms ofLegendre functions.


Adding Harmonic Components


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g p

A corresponding analysis of the purely harmonicpieces gives their contribution as

SH= w

2i

d

+ w

2i

d1

2

(71)

for some choice of complex functionsand. Canbe worked out in spatial terms.


The Stokes stream function


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The Stokes stream function

The contour integral solution for the Stokes streamfunction for axis-symmetric biharmonic flow:

S=ww 1

2i

d1

f

+ w2i

d

+ w2i

d12

(72)

whereis written in terms ofx, y, z and wheref, ,

have Laurent series expansions that generateexpansions in terms of powers ofrand regular (f)and modified (, ) functions.


Simple Twistor Functions


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p

fconstant gives a contribution toSproportional to

ww=r2 sin2() (73)

f(z) = 1/zgives a Coulomb field and a contributiontoSproportional to

wwr

=r sin2() (74)

(z) = 1/z2, a contribution toSof the form

ww

r3 =

1

rsin2() (75)


Does it all add up?


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The particular combination

S=U

2

sin2()r2

3ar

2

+a3

2r (76)

gives the well-known stream function for very viscousflow around a sphere of radiusaand uniform flow atat rateUat infinity. In general we have a noveltechnique for solving the PDE

D4

S= 0 (77)


Where next?


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More serious explicit solution representations


Where next?


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More serious explicit solution representations Can we add non-ininfite viscosity somehow?


Where next?


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Clearer ideas about 2D NS? (work by Ranger)


Where next?


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3D NS? (work by Roulstone)


Where next?


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Where next?


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3D2D FLUIDS.pdf