# Motion and Flow by Jacques Vanneste 2010/2011 Notes of Paul Boeck

## Full text

(1)

### Motion and Flow by Jacques Vanneste

2010/2011 Notes of Paul Boeck

Last changes: November 26, 2010

### Contents

I Scalar and vector fields 2

I.1 Notation . . . 2

I.2 Gradient . . . 2

I.3 Divergence. . . 3

I.4 Curl . . . 4

II Line, surface and volume integrals 5 II.1 Line integrals. . . 5

II.2 Surface integrals . . . 6

II.3 Gauß’ and Stokes’ theorem . . . 7

IV Solutions of Laplace’s equation (dipole) 10 V PDEs for continuous media 11 V.1 Introduction: heat equation. . . 11

V.2 Motion of an inviscid fluid . . . 12

d) Vorticity eqn for barotropic fluids . . . 14

Kelvin’s circulation thm (barotropic fluids) . . . 14

e) Bernoulli’s eqn and applications . . . 15

V.3 Wave eqn . . . 16

b) String under tensionT . . . 16

d) Standing waves . . . 17

V.4 Viscous fluids . . . 18

(2)

I SCALAR AND VECTOR FIELDS

Motivation

Mathematical description ofcontinous models, physical models in which properties vary contiously in time and space.

Example 1 a river is best modelled as a continuum with

• density̺(x, t) = lim∆V→0∆M

∆V,

• a velocityu(x, t) = lim∆V→0 P

imiui

∆V

### I.1 Notation

pointP ∈R3(Euclidean space) withr=OP= position vectors =xi+yj+zk.

The natural objects are:

• scalar fields: f =f(r, t) defines a scalar property at ptrand timet

• vector fields: F=F(r, t) defines a vector at each ptrand timet

The gradient of ascalar field f(r) is the vector field

f(r) =g(r) where r=|r|=p

x2+y2+z2

∇f =g(r)

∂r

∂xi+∂y∂rj+∂r∂zk

=g(r)

r

xi+yrj+rzk

=g(r)r

r =g(r)ˆr (unit vetor in direction ofr) Meaning of gradient: CurveCand a scalar field f along it.

f(r+δr) =f(r) +∂f∂xδx+∂f∂yδy+∂f∂zδz+O(|δr|2)

=f(r) +∇f·δr+O(|δr|2) If the curve is parametricised,C:x=x(s), y=y(s), z=z(s), s∈R

df ds = lim

δs→0∇f· δr

δs =∇f· dr

ds=∇f· t

tangent|{z}

toCatr

Thus for anyt,∇f·tmeasures the rate of change off in the direction oft.

Ifsis arclength,|t|= 1.

Remark

(i) ∇f is perpendicular to the surfacesf(r) =c. Taketto be tangent to f(r) =c, then df

ds = 0 =∇f·t =⇒ ∇f⊥t

(3)

I.3 Divergence I SCALAR AND VECTOR FIELDS

Example 3 surfaces of constant pressurep(r) =p(r) are (almost) sperical.

∇p=p(r)ˆr

(ii) dfds =∇f·t=|∇f||t|cosθ is maxf orθ= 0. So∇f gives the direction of maximum growth off.

Example 4 topology: heightz=f(x, y), the steepest ascent curves are||to∇f.

Example 5 (Newton’s law of gravitation) observation indicate that the force exerted by a massM on a mass mis:

F=−GM m

r2 ˆr=−GM m r3 r

whereGis the gravitational constant. If there are several massesM1, M2, . . . , Mn

F=−G· Xn

i=1

Mim

|r−ri|3(r−ri) The gravitational fieldE=mF =−GMr2 ˆr

Observe that:

E=−∇φ with φ=−GM

r the gravitational potential

∇φ=φ(r)ˆr= GM r2 ˆr The potential ofN masses is φ=−GX

i

Mi

|r−ri| a continous distribution of mass with densityρ(r).

φ=−G ZZZ

V

ρ(r)dr

|r−r|

Example 6 (Electrostatics) Coulomb’s law for the force between 2 chargesqandQis F= 1

4πε0

qQ

r2ˆr repulsive force field E= 1

4πε0

Q

r2ˆr=−∇φ φ= Q 4πε0|r|

### I.3 Divergence

The divergence of a vector fieldF= (F1, F2, F3) is

∇ ·F= ∂F1

∂x +∂F2

∂y +∂F3

∂z

∇ ·Fmeasures the rate of expansion (¿0) (compression (¡0)) described by a vector field.

Example 7 • v=ar. a radial vector field. ∇ ·v= 3a

• v=f(r)r.

∇ ·v= ∂

∂x(f(r)x) + ∂

∂y(f(r)y) + ∂

∂z(f(r)z)

= 3f(r) +f(r)

x∂r

∂x+y∂r

∂y +z∂r

∂z

= 3f(r) +f(r) In particular,f(r) =rn,∇ ·v= 3rn+nrn= (3 +n)rn. n=−3,∇ ·v=∇(rr3) = 0

(4)

I.4 Curl I SCALAR AND VECTOR FIELDS

Interpretation as a rate of compression: take a small volume (area) ABCD. ρis a constant.

Flux of mass through AB is ρv2(x, y)dx DC is −ρv2(x, y+dy)dx AD is ρv1(x, y)dy BC is −ρv1(x+dx, y)dy

Net flux is ρ(v2(x, y)−v2(x, y+dy))dx+ρ(v1(x, y)−v1(x+dx, y))dy

=−ρ∂v2

∂y dxdy−ρ∂v1

∂xdxdy=−ρ ∂v1

∂x +∂v2

∂y

dxdy Remark ∇ ·(fF) =∇f·F+f∇F.

### I.4 Curl

The curl of a vector fieldF= (F1, F2, F3) is

∇ ×F= curlF= ∂F3

∂y −∂F2

∂z

i− ∂F1

∂z −∂F3

∂x

j+ ∂F2

∂x −∂F1

∂y

k

=

i j k

∂x ∂y ∂z F1 F2 F3

εijk∂jFk

Here we use

• Einstein’s notation: εijkjFk meansP3 j=1

P3

k=1εijkjFk ”Repeated indices are summed”

• εijk is the permutation symbol

= 1 if i= 1, j= 2, k= 3 or 312 or 231

=−1 if i= 1, j= 3, k= 2 or 213 or 321

= 0 otherwise

• εijk =−εikj etc

• εijkεilmjlδkm−δjmδkl

The curl measures the amount of rotation inF.

Remark

∇ ×(fF) =∇f ×F+f∇ ×F check: (∇ ×(fF))1= ∂

∂y(f F3)− ∂

∂z(f F2)

= ∂

∂yf F3− ∂

∂zf F2+f( ∂

∂yF3− ∂

∂zF2)

= (∇f×F)1+f(∇ ×F)1

Example 8

∇ ×(f(r)r) =∇f×r+f∇ ×r

=f(r) ˆr×r

| {z }

0

+f ×0 = 0 rigid body rotating with angular velocityw=wk.

|v|=wr v=

−wy wx

0

∇ ×v=

i j k

xyz

−wy wx 0

= 0i+ 0j+ 2wk

Remark • ∇ ×E= 0 (gravitational/electric field).

(5)

II LINE, SURFACE AND VOLUME INTEGRALS

• More generally,∇ × ∇φ= 0 since

εijkjkφ=εijkjk2 φ= 1

2(εijkjk2 φ+εikjjk2 φ) =1

2(εijkikj)

| {z }

=0

2jkφ

and∇ ·(∇ ×F) = 0 since∂(εijkjF k) = 0.

Successive applications of∇

• ∇ · ∇f =∇2f = ∆f =∂xx2 f +∂2yyf+∂zz2 f (Laplacian)

• ∇ × ∇f = 0

• ∇(∇ ·F

• ∇ ·(∇ ×F) = 0

• ∇ ×(∇ ×F) =∇(∇ ·F)− ∇2F

| {z }

2F1i+∇2F2j+∇2F3k

Example 9 since∇ ·E= 0, E=−∇φ =⇒ ∇2φ= 0. (check directly∇2(1r) = 0 atr6= 0).

Terminology

• a vector field s.t. F=∇φis said to beconservative

• a conservative cector field satisfies ∇ ×F= 0 curlfree

• in R3, if∇ ×F= 0,F=∇φprovidesFdecays asr→ ∞.

• if∇F= 0, solenoidal (divergence free)

### II.1 Line integrals

LetCbe a curve in R3 andFa vector field. Parametrise C:r(t) =

 x(t) y(t) z(t)

, t∈[a, b]

The line integral ofFalongC is Z

C

F·dr= Z b

a

F·dr dtdt=

Z b a

(F1

dx dt +F2

dy dt +F3

dz dt)dt Example 10 F=

 z x y

andC:r(t) =

 cost sint 3t

, 0≤t≤2π(helix)

Z

C

F·dr= Z

0

 z x y

·

−sint cost

3

dt

= Z 2

0

π

 3t cost sint

·

−sint cost

3

dt= Z

0

(−3tsint+ cos2t+ 3 sintdt

= [3tcost]0 +π= 7π Remark if the curve is closed, we use the special notation

I

C

F·dr In fluid mechanics,H

Cu·dris the circulation.

(6)

II.2 Surface integrals II LINE, SURFACE AND VOLUME INTEGRALS

Example 11 ConservativeF, i.e. F=∇φ, (t∈[a, b]) then Z

C

Fdr= Z b

a

∇φdr dtdt=

Z b a

∂φ

∂x dx dt +∂φ

∂y dy dt +∂φ

∂z dz dt

= Z b

a

dtdt=φ(b)−φ(a) independant ofC (endpoints only) in particular

I

C

Fdr= I

C

∇φdr= 0

Example 12 in physics, ifFis a force acting on a particle with positionr(t), the work done by the force W =

Z

C

Fdr= Z b

a

Fdr dtdt=

Z b a

Fvdt Newton’s law: mdvdt =F

W =m Z t=b

t=a

dv

dtvdt= m 2

Z b a

d

dtkvk2dt=m

2 kv(b)k2− kv(a)k2 Work = change in kinetic energy

In particular, ifFis conservative, there is no change in kinetic energy along closed paths.

### II.2 Surface integrals

A surface is defined:

• explicitly: z=f(x, y)

• implicitly: g(x, y, z) = 0

• parametrically: r=r(u, v) =

 x(u, v) y(u, v) z(u, v)

Example 13 A sphere of radiusa:

z=±p

a2−(x2+y2) a2=x2+y2+z2

 x y z

=

acosφsinθ asinφsinθ

acosθ

 0≤θ < π 0≤φ≤2π Normal to a surface

• forg(x, y, z) = 0, the normal is||to∇gn= k∇gk∇g

• forz=f(x, y) the normal is||to

−fx

−fy

1

• forr=r(u, v), 2 tangent vectors are ∂u∂r,∂r∂v and the normal isn= k∂r∂u∂r×∂r∂v

∂u×∂r∂vk

Example 14 sphere (i) g=z−p

a2−x2−y2= 0n||

 x/z y/z 1

 =⇒ n||r.

(ii) in polar spherical coordinates:

∂r

∂φ =

−asinφsinθ acosφsinθ

acosθ

, ∂r

∂θ =

acosφcosθ asinφcosθ

−asinθ

N= ∂r

∂φ×∂r

∂θ =· · ·=−asinθ(xi+yj+zk) =−a2sinθˆr

(7)

II.3 Gauß’ and Stokes’ theorem II LINE, SURFACE AND VOLUME INTEGRALS

Definition 1

The surface integral of a vector field F ZZ

S

F·ndS, wheredS is the area element is called the fluxof Fthrough S.

Interpretationifq(=ρu) is the amount of a quantity flowing inR3 per unit area and unit time (q·n)δS= amount of mass crossing the small surfaceδS

Computation:

n=

∂r

∂u×∂r∂v

k k dS=

∂r

∂u×∂r

∂v dudv Hence

ZZ

S

F·ndS= ZZ

F·n

∂r

∂u×∂r

∂v dudv

= ZZ

F·Ndudv with N= ∂r

∂u ×∂r

∂v

### II.3 Gauß’ and Stokes’ theorem

Theorem 2 (Gauß’ Theorem (divergence thm))

Let V be a volume in R3,∂V be its boundary and Fa vector field. Then ZZZ

V

∇ ·FdV = ZZ

∂V

F·ndS

Remark TakingF=

 ψ

0 0

gives ZZZ

V

xψV = ZZ

∂V

ψn1dS. SimilarlyF=

 0 ψ 0

gives ZZZ

V

yψdV = ZZ

∂V

ψn2dS.

Theorem 3 (Green’s identities) Consider two scalar fields φandψ.

Since ∇ ·(ψ∇φ) =∇ψ∇φ˙ +ψ∇2φ ZZZ

V

∇ ·(ψ∇φ)dV = ZZZ

V

(∇ψ·) Swapping the roles ofψ andφand subtracting gives

ZZZ

V

(ψ∇2φ−φ∇2ψ)dV = ZZ

∂V

(ψ∂φ

∂n−φ∂ψ

∂n)dS Theorem 4 (Stokes)

Let S be a surface and∂S be its boundary. Letnbe the normal to S anddlbe the line element on the curve∂S.

ZZ

S

(∇ ×F)·ndS= Z

∂S

F·dl The orientation ofnanddlare related by the ”right hand rule”.

Remark • RR

S(∇ ×F)ndS depends in∂S only.

• RR

∂V(∇ ×F·ndS= 0 (surface without boundary) Theoretical applications

(i) Given the scalar fieldφ= 1r, the flux of∇φis the same through any sphere centred at the origin.

n= ˆr ∇φ=φ(r)ˆr=−1

r2ˆr ∇φ·n=−1 r2 Hence

ZZ

∂S

∇φ·ndS=−1 r2

ZZ

∂S

dS=−1 r2

4

πr2=−4π

(8)

II.3 Gauß’ and Stokes’ theorem II LINE, SURFACE AND VOLUME INTEGRALS

(ii) The flux of∇φis in fact the same through any surface enclosing the origin.

• geometrically:

Show that ∇φ·n∂S=∇φ·ˆr∂S Sphere of Radiusr

∇φ·n∂S=−ˆr

r2 ·n∂S=−cosθ

r2 ∂S=−∂S

=−∂S Hence RR

∂V(∇φ·n)dS=−4π.

• using Gauß’ thm: compare the flux through∂V with the flux through the unit sphere ZZ

∂V

(∇φ·n)dS− ZZ

Sphere

(∇φ·n)dS

= ZZ

∂shell

(∇φ·n)dS= ZZZ

shell

= ZZZ

shell

2φdV = 0

Index notation & Einstein’s notation Coordinates:

 x y z

=

 x1

x2

x3

and Basis vectorsi=e1,j=e2,k=e3

Position: r=xi+yj+zk=P3

i=1xiei=xiei (repeated indices are summed over) Gradient: ∇φ=∂xφi+∂yφj+∂zφk=P3

i=1

∂φ

∂xiei=P3

i=1iφei=∂iφei

Dot Product: a·b=P3 i=1aiei

·P3 j=1bjej

=P3 i=1

P3

j=1aibj(ei·ej) =P3 i=1

P3

j=1aibjδij =P3 i=1ai

P3 j=1bjδij

= P3

i=1aibi=aibi

Short: a·b=aiei·bjej =aibjei·ej=aibjδij =aibi

Cross product: (a×b)iijkajbk (sum over j&k). a×b=εijkajbkei(Sum over i,j,k)

Example: a×(b×c) =a×(εijkbjckei) =εlmiamεijkbjckelijkεlmiambjckelijkεilmamb−jckel= (δjlδkm− δjmδlk)ambjckel=akbjckej−akbjckek = (a·c)b−(a·b)c

SummaryFor the gravitational fieldE=−GMr2 ˆrandE=−∇φwithφ=−GMr ZZ

∂V

E·ndS=

(−4πGM ifM ∈V

0 ifM /∈V

This generalises to a number of masses:

ZZ

∂V

E·ndS=−4πG X

i mi∈V

mi

For a continuos distribution of mass:

ZZ

∂V

E·ndS=−4πG ZZZ

V

ρ(r)dv Using Gauß’s thm

ZZZ

V

|{z}∇E

−∇2φ

dv=−4πG ZZZ

V

ρ(r)dv

Hence,−∇2φ=−4πGρ ⇐⇒ ∇2φ= 4πGρPoisson equation.

Remark • this is a generalisation of ∇2φ= 0 (Laplace eqn) which holds whereρ= 0.

• the electrostatic equivalent replaces Gby−4πε10 Then∇ ·E=ερ0,∇2φ=−ερ0 (ρcharge density.) and ZZ

∂V

E·ndS= 1 ε0

ZZZ

V

ρdv= 1 ε0

total charge inV

• The solution φ=−GMr can be interpreted as the solution of ∇2φ= 4πGM δ(r). RRR

V δ(rdv= 1.

(9)

II.3 Gauß’ and Stokes’ theorem II LINE, SURFACE AND VOLUME INTEGRALS

Practical applications

Gauß’ thm + symmetry arguments can be used to computeEfor simple mass distributions.

1) Gravitational field of a spherical planet: Assume the densityρ(r) =ρ(|r|) for 0≤r≤aandρ= 0 forr≥a (i) Show thatφ(r) =−GMr whereM = 4πRa

0 ρ(r)r2dris the total mass of the planet, forr≥a.

(ii) Relateφ(r) toρ(r) for 0≤r≤a (iii) Findφ(r) ifρ(r) =ρ0= const.

(i)

ZZ

∂V

E·ndS=−4πG ZZZ

V

φ(r)dV

∂V: sphere of sizer > a. Bexause of spherical symmetry, we will assume thatφ=φ(r). E=−∇φ=−φ(r)·ˆr.

∂V =Sr2,dS=r2sinθdθdϕ=r2dΩ2 andn= ˆr.

(1) = ZZ

S2r

(−φ2(r)) ˆrˆr

|{z}1

r2dΩ2

=−r2φ(r)4π=−4πG ZZZ

Vr

f(r)dV =−4πG Z a

0

dr Z π

0

dθ Z

0

dϕρ(r)r2sinθ≡Mp

4πr2φ(R) = 4πGMp φ(r) = GMp

r2 φ(r) =−GMp

r +C (φ(r)→0 =⇒ C= 0)

Conclusion: For any (spherical) distribution of matter the gravitational field outside the object is the same as a point particle ofM =Mtotal located at the origin.

(ii) ∂V: sphere of sizer < a. Same symmetry: E=−φ(r)ˆr,n= ˆr,dS=r2sinθdθdϕ.

(1) =−r2φ(r)4π=−4πGM(r) where M(r) = 4π Z r

0

r′2ρ(r)dr φ(r) = GM(r)

r2 φ(r) =

Z rGM(r) r′2 dr+C (iii) ρ(r) =ρ0 const. M(r) =ρ0r33. φ+(r) =23πGρ0r2+ ˜C.

Requiring continuity of φ(r)|r=a φ+(a) =−GMap(a) =⇒ fixes ˜C=−32GMap. Conclusion:

φ(r) = (GM

p

2a (ar22 −3) for 0≤r≤a

GMrp forr≥a 2) Electric fields of a sharped sphere:

E=−∇φ= 1 4πε0

Z ρ(r)

|r−r|3(¯r−r¯)d3r φ= 1

4πε0

Z ρ(r)

|¯r−r¯|d3r φ(r)≡ charge density Gauß’ thm, when applied to electric fields:

ZZ

∂V

E·ndS= 1 ε0

Z

V

ρ(r)d3r≡Q

Question: Consider a spherical metallicconductor of radiusawith total chargeQ. Its charge is distributeduniformly over its surface. (density of charge = 4πaQ2). FindE andφoutside and inside of the conductor.

Same symmetry, same considerations as before. Let’s have a look atr > a.

(1) = ZZ

∂V

EndS=|E|4πr2= 1 ε0

Q =⇒ same as beforeφ∼Q r +C Inside: r < a.

(1) =|E|4πr2= 0 =⇒ |Ein|= 0 =⇒ φin=const.

(10)

IV SOLUTIONS OF LAPLACE’S EQUATION (DIPOLE)

Remark • E is⊥conductor surface (⊥to surface of constantφ).

• φconst. on the surface

• φconst. in the interior of conductor (∇2φ= 0)

• |Ein= 0.

### IV Solutions of Laplace’s equation (dipole)

The potentialsφ=−GMr andφ=4πε1

0

Q

r are solutions of Laplace’s equation

2φ= 0 for r6= 0 with

ZZ

sphere

∇φndS=

(4πGM

εQ

0

This can be written as ∇2φ= (GM

εQ

0

×δ(r)

Dirac–distribution RRR

volume enclosing 0

δ(r)dv= 1 ZZZ

V

2φdv= ZZ

∂V

∇φ·ndS=

(4πGM

εQ

0

There are other useful soultions.

Dipole:

Consider two chargesQand−Qat positionsr=±ak. The total electrostatic potential is given by:

φ(r) = Q 4πε0

1

|r−ak|− 1 r+ak|

Note: |r∓ak|=s

r2+a2∓2ar·k

|{z}ζ

=p

r2+a2∓2ζ

φ(r) = Q 4πε0

1

pr2+a2−2aζ − 1 r2+a2+ 2aζ

!

For a

r ≪1 , 1

r2+a2∓2aζ =1 r

1 q

1∓2r2 +ar22

=1 r

1±aζ

r2 +O a2

r2

Hence φ(r) = Q 4πε0

2aζ

r3 +O(ar32) Take the limita→0 with 2Q=pfixed. Then

φ(r) = pζ

4πε0r3 dipole solution withpdipole strengh In general: φ(r) = p(r−r0)

4πε0|r−r0|3. the dipole strengthpis vector,r0 is the location of the dipole.

The dipole potential clearly solves Laplace’s eqn (r 6=r0). Its form follows from the fact that if φis a solution of

2φ= 0, so is−p· ∇φfor anyp. Withφ= 1r

−p· ∇φ= p·r

r3 dipole Carrying on, we can findmultipole solutions.

Qijij2φ=Qij

3rirj−r2δij

r5

Note that the trace of 3rirjr−r5 3δii =3r2r−r523 = 0. tr∂ij2φ=∂2iiφ=∇2φ= 0.

(11)

V PDES FOR CONTINUOUS MEDIA

The multipole solutions arise when computing approximation to the potential caused by a distribution of charges.

φ(r) = 1 4πε0

ZZZ

V

ρ(r)

|r−r|dv To find and approximation for larger, we use Taylor expansion nearr= 0.

Solutions of Laplace’s eqn∇2φ= 0: multipoles φm= 1

4πε0

1

r monopole

φd= 1 4πε0

p·r

r3 dipole

φq = 1 4πε0

Qij

(3rirj−r2δij)

Response to a distribution of charges:

φ(r) = 1 4πε0

ZZZ

V

ρ(r)

|r−r|dv

At a large distancer, we find using a Taylor expansion of|r−r|−1 nearr= 0:

4piε0φ(r) = Z

V

ρ(r)

|r| dv+ Z

V

φ(r)ri

∂ri r=0

1

|r−r

dv+1 2

Z

V

ρ(r)·rirj2

∂ri∂rj r=0

(r−r|)dv+· · ·

= 1

|r|

Z

V

ρ(r)dv− ∂

∂ri

1

|r|

Z

V

riρ(r)dv+1 2

2

∂ri∂rj

1

|r|

Z

V

rirjρ(r)dv+· · ·

= 1 r

Z

V

ρ(r)dv+ Z

V

riρ(r)dv· ri

r3 + Z

V

rirjρ(r)dv

3rirj−r2δij

r5

= Q r +p·r

r3 +Qij

3rirjr2δij

r5 +· · · where Q=

Z

ρ(r)dv= total charge and p= Z

rρ(r)dv= dipole strength and Qij =

Z

V

Remark Since Laplaces eqn is linear, various solns (e.g. monopoles and dipole) can be superposed.

Ifφ1 andφ2 are solns, then so isφ12. (∇2φ1= 0,∇2φ2= 0 =⇒ ∇212) = 0).

Example 15 earthed spherical conductor in an electric field.

E→E0k as r→ ∞

What isφandE? We need to imposeφ= 0 on the spherer= 0 (=earthed). We try superposing (∇φ=−E).

φ=−E0z +Az

r3 (dipole) whereAis to be determinded φ−(r= 0) = 0

z=acosθ =⇒ −E0acosθ+Aacosθ

a3 = 0 =⇒ A=Ea3 Hence φ=−E0z(1−ar33) solves ∇2φ= 0

E=−∇φ=−E0(1−ar33)k−3ar35E0zr

### V.1 Introduction: heat equation

Many media can be described using a continuous description. The properties are described by (average) densities:

f(r, t) =





mass density at ptr,f(r, t) =ρ(r, t) momentum density in x–direction =ρ(r, t)u(r, t)

energy density =ε(r, t)

(12)

V.2 Motion of an inviscid fluid V PDES FOR CONTINUOUS MEDIA

Together with the densityf(r, t), we are given a flux: q(r, t).

The evolution of the density is governed by aconservation law: balance between rate of change of the property in a volumeV and its flux across∂V.

d dt

ZZZ

V

f(r, t)dv=− ZZ

∂V

q(r, t)·ndS

⇐⇒

ZZZ

V

∂f

∂tdv=− ZZZ

V

∇ ·qdv

This holds for anyV, hence

∂f

∂t +∇ ·q= 0 source - sink conservation off This is a PDE which can be solved forf given initial and boundary conditions.

Remark Ifq·n= 0 on∂V thenRRR

V f(r, t) =const.

Example 16 (heat conduction) The energy of the material is conserved. The energy densityε(r, t) can be related to the temperature: ε(r, t) =cT(r, t). c specific heat, taken as a constant. The flux of heat is modeled by Fourier’s law:

q=−k∇T, k >0 The conservation law reads:

d dt

ZZZ

V

ε(r, t)dv=− ZZ

∂V

q·ndS d

dt ZZZ

V

(cT)dv=− ZZZ

V

∇ ·qdv =⇒ c∂T

∂t +∇ ·q= 0 c∂T

∂t =∇ ·(k∇T) ifkis independent ofr.

∂T

∂t = k

c∇2T heat eqn

### V.2 Motion of an inviscid fluid

a) Mass conservation

Consider a fixed volumeV of a fluid. The mass isM =RRR

V ρ(r, t)dvand the flux of mass across∂V isRRR

q·ndS. q is given by

q(r, t) =ρ(r, t)(r, t) Then

d dt

ZZZ

V

ρ(r, t)dv= ZZ

q·ndS=− ZZZ

V

∇ ·qdv=− ZZZ

V

∇ ·(ρu)dv Hence

∂ρ

∂t +∇ ·(ρu) = 0 b) Pressure

In a fluid there is a force acting on any surface in the direction of the normal. The pressure p(r, t) s this (negative) force per unit area.

force =−pndS Integral of−pnover some ∂V gives the total force acting onV.

B=− ZZ

∂V

pndS=− ZZZ

V

(∇p)·dv

(13)

V.2 Motion of an inviscid fluid V PDES FOR CONTINUOUS MEDIA

In the presence of gravity,Bbalances the gravitational force. −M gk, where M =RRR

V ρdvis the mass ofV. Hence B−M gk= 0

− ZZZ

V

∇pdv− ZZZ

V

gkρdv= 0

Since this holds for anyV: ∇p=−ρgk. Hencep=p(z) and dpdz =−ρg.

−g Z

ρdz+c=p hydrostatic relation

Example 17 (i) Pressure in a container for a constant density fluid. ρ=ρ0=const. dpdz =−ρ0g =⇒ p=−ρ·gz+c at z= 0, p0=c. Hence p=p0−ρ0gz. At the bottomz=−h,p=p00gh.

(ii) A (perfect) gas at constant temperature satisfiesp=αρ,α=const. What is the hydrostatic pressure?

dp

dz =−ρg=−pg α − g

αz p=Cegαz=p0eαgz (iii) Force on an immersed body. The fluid exerts a force:

F=− ZZ

∂V

pndS=g ZZZ

V

ρdv whereρis the fluid density

= weight of the displaced fluid Archimedes’ principle, Heureka!!

c) Momentum conservation

Newtons law dtd(mu) = Force, states what the rate of change of the momentum (mass× velocity) is equal to the force acting. It can be applied to continuous media, for inviscid fluids, the only force acting is pressure.

Consider a volumeV Let’s write the balance ofx–momentum

• x–momentum density: ρu

• x–momentum flux: ρu·uwith =(u, v, w).

Then

d dt

ZZZ

V

ρudv

| {z }

rate of change of momentum

=− ZZ

∂V

ρuu·ndS

| {z }

flux of momentum

− ZZ

∂V

pndS

·i

| {z }

Force in x–direction

ZZZ

V

∂t(ρu)dv=− ZZZ

V

∇ ·(ρuu)dv− ZZZ

V

∇pdV ·i

Hence





∂t(ρu) +∇(ρuu) =−∂xp (1)

∂t(ρv) +∇(ρvu) =−∂yp (2)

∂t(ρw) +∇(ρwu) =−∂zp (3) Note: ∂

∂t(ρu) +∇ ·(u·ρu) =ρ∂u

∂t +u∂ρ

∂t +ρu· ∇u+u∇ ·(ρu)

∇ ·(aF) =F· ∇a+a∇ ·F=ρ(∂u

∂t +u∇u) +u(∂ρ

∂t +∇ ·(ρu))

| {z }

=0 mass conservation

(1),(2),(3) becomes





∂u

∂t+u· ∇u=−1ρxp

∂v

∂t +u· ∇v=−1ρyp

∂w

∂t +u· ∇w=−1ρzp

∂u

∂t +u· ∇u=−1 ρ∇p

(14)

V.2 Motion of an inviscid fluid V PDES FOR CONTINUOUS MEDIA

In summary: compressibleEuler equations

∂ρ

∂t +∇ ·(ρu) = 0 ∂u

∂t +u· ∇u=−1 ρ∇p

This is complemented by anequation of state relatingρto pand other quantities (e.g. temperature) Remark ∂u

∂t +u· ∇u= du

dt, where d dt = ∂

∂t+u· ∇denotes thematerial derivative, i.e. time derivative following particles moving with velocityu. Particles moving atu have position position r(t), s.t.

dr

dt =u(r, t) dx

dt =u dy

dt =v dz dt =w Material derivative off(r, t):

d

dtf(r(t), t) =∂f

∂t +dr

dt · ∇f =∂f

∂t +u· ∇f = ∂

∂t +u· ∇

f

The momentum eqn can be written as ρd

dtu=−∇p (Newton’s law applied to a fluid particle) Simple equations of states:

• p=p(ρ): pressure is a function of density alone,barotropic fluid.

Note: defining w= Z dp

ρ, the eqn of momentum becomes du

dt =−∇w=−1 ρ∇p.

• ρ=const=ρ0: incompressible fluid. The Euler eqn become (d

dtu =−∇(p/ρ0)

∇ ·u = 0 Euler equations

d) Vorticity eqn for barotropic fluids

Define thevorticity ω=∇ ×u. Note that ω×u=u· ∇u−12∇|u|2. We can rewrite the momentum eqn as

∂u

∂t +ω×u=−∇(w+1 2|u|2) Take curl:

∂ω

∂t +∇ ×(w×u) = 0 ω=∇ ×u

Remark ifω(r,0) = 0, thenω(r, t) = 0, for all t >0. So ifω is zero initially, it remains zero: irrotational flow = potential flow. Irrotational flows are described using a potentialφsinceω=∇ ×u= 0, thenu=∇φ.

If in addition, the fluid is incompressible,∇ ·u= 0 =⇒ ∇2φ= 0: Laplace’s eqn.

φ6= 0 if boundary conditions imposeu6= 0 on boundaries.

Kelvin’s circulation thm (barotropic fluids)

Consider a material (moves with the fluid) closed curve. We’re interested in the circulation I

C(t)

u·dl Now

d dt

I

C(t)

u·dl= I

C(t)

d

dt(u·dl) = I

C(t)

du

dt ·dl+ud dt(dl) =

I

−∇w·dl+n·du= I

−∇w·dl+d(|u|2/2)

= 0 Hence the circulationH

C(t)u·dl= const.

(15)

V.2 Motion of an inviscid fluid V PDES FOR CONTINUOUS MEDIA

e) Bernoulli’s eqn and applications

Consider the steady flow∂t·= 0) of a barotropic fluid (p=p(ρ)). It satisfies u· ∇u=−∇w w=

Z dp ρ ω×u=−∇(w+12|u|2) (∗) Taking·(∗) gives 0 =u· ∇(w+12|u|2) Recall: n· ∇f =∂n f is the derivative in the direction ofn.

w+12|u|2= constant along particle trajectories i.e. streamlines. For an incompressible flow:

p ρ0

+12|u|2= const. along streamlines Applications:

• flow over an obsticale

• Pitot tubep+12u2=ptube+02

For potential flows (ω= curlu= 0,u=∇φ), Bernoulli’s eqn has a simpler form:

∇(w+12|u|2) = 0 sinceω= 0 =⇒ w+12|u|2= const everywhere For incompressible fluids replace ρp

0 +12|u|2=const.

Example 18 consider a two–dimensional flow of an irrotational, incompressible fluid around a cylinder. Velocity of the flow is u = (u, v). Incompressible means, that ∂xu+∂yv = 0. We can introduce a streamfunction ψ s.t.

u=−∂yψ andv=∂xψ. R∇ψwithR= rotation by π2.

Since the flow is irrotational, ω= (∂xv−∂yu)·k= 0 and therefore∇2ψ= 0. ψ also satisfies: ψ =const forr=a (the radius of the cylinder) andψ∼ −U y forr→ ∞.

A possibleψ isψ=−U y(1−ar22)−γlogr(check that∇2ψ= 0). Hence u=U(2−a2

r2) +2U y2a2 r4 +γy

r2 v=2U xya2

r4 −γx r2 Atr=a,x=acosθ, y=asinθ

u= 2Usin2θ+γ asinθ v=−2Ucosθsinθ−γ

acosθ Use Bernoulli: p

ρ0

+12U2= p ρ0

+12h

2Usinθ+γ a

i2

Hence p(cylinder) = ρ 2

U2

2Usinθ+γ a

2 +p

The force (per unit length) resulting from this pressure is F=−

Z

Z 0

(U2−(2Usinθ+γ

a) +p)·(cosθi+ sinθj)dθ

= 2πρU γj is the left on the cylinder

In the presence of a force per unit mass f, −∇p is replaced by −∇p+ρf. If the force derives from a potential:

f =−∇φ, then−∇pis replaced by−(∇p+ρ∇φ). Bernoulli’s thm for an compressible fluid becomes p

ρ0

+12|u|2+φ= const Example 19

f =−gk φ=gz p

ρ0

+12|u|2+gz= const

(16)

V.3 Wave eqn V PDES FOR CONTINUOUS MEDIA

### V.3 Wave eqn

a) Acoustic waves

tρ+∇ ·(ρ(u) = 0

tu+u· ∇u=−1 ρ∇p

p=p(ρ) (barotropic) Consider small–amplitude perturbations to a constant–density state of rest:

ρ=ρ0 ρ≪ρ0

u=u

p=p(ρ0) =p(ρ0) +dp

dρ(ρ0+· · ·=p(ρ0) +c2ρ+· · ·, where c= s

dp dρ(ρ0)

| {z }

sound speed

Introducing into (∗) and ignoring quadratic terms

(∂tρ0∇ ·u= 0

tu=−ρc2

0∇ρ withu=∇φ ∂tρ02φ= 0,∂t∇φ=−ρc2

0∇ρ =⇒ ∂tφ=−cρ2

0ρ.

Finally, ∂tt2 =−c22φ the wave equation In spatial dimensionφ(x, t) satisfies∂ttφ=c2xx2 φ(eg. gas in a tube).

b) String under tension T

y=η(x, t) is the displacement. y–component of Newton’s law:

| {z }

mass

tt2η

|{z}

acceleration

=ATsin(θ(x+dx))−ATsin(θ(x)) (†)

Small displacements: sinθ(x) =θ(x) andθ(x) = tanθ(x) =∂x∂η(x).

Hence sin(θ(x+dx))−sinθ(x) = ∂η∂x(x+dx)−∂η∂x(x) = ∂x2η(x)dxas dx→0.

(†) becomes ρA∂2ttη=AT ∂xx2 η

2ttη=c2xx2 η with c= s

T

ρ wave speed

Properties of the one–dimensional wave eqn

i) d’Alemtant’s solution: The general soln in an∞te domain is

φ(x, t) =f(x−ct) +g(x+ct) = superposition of right and left traveling waves ii) Seperable solns Solutions can be sought in the form

φ(x, t) =X(x)T(t) In∂tt2φ=c22xxφ

XT′′=c2X′′T =⇒ T′′

T =c2X′′

X =β = const

(17)

V.3 Wave eqn V PDES FOR CONTINUOUS MEDIA

d) Standing waves

Waves in bounded domainsx∈[0, L]. Boundary conditions:

• string, attached at both ends: η(0, t) =η(L, t) = 0

• string, attached atx= 0 only: η(0, t) = 0∂xη(L, t) = 0.

• gas in a tube, both ends are closedφ(x, t),u=∂xφ. u(0, t) =u(L, t) = 0, i.e. ∂xφ(0, t) =∂xφ(L, t) = 0.

• open end atx=L,p(L,0) = 0

tt2η=c22xxη,η(ρ, t) =η(L, t) = 0. Solve by separation: η(x, t) =X(x)T(t) XT′′=c2X′′T T′′

T =c2X′′

X =β = const if β >0 : X =Acosh(

qβ

c2x) +Bsinh(

qβ c2x) X(0) = 0 =X(L) =⇒ A=B= 0

β = 0 X =Ax+B, X(0) =X(L) = 0 =⇒ A=B= 0 β <0 X =Acos(

q−β

c2x) +Bsin(

q−β c2 x)

X(0) = 0 =⇒ A= 0 X(L) = 0 =⇒ Bsin(

q−β c2L) = 0 Hence

q−β

c2 L=nπ,n= 1,2, . . .. ie. β =−c2(L)2 andX(x) = sin(nπxL ) =Xn(x).

T′′−βT = 0,T′′+c2(L)2T = 0 =⇒ Tn(t) =αncos(nπcL t+φn) =αncos(ωt+φn).

whereωnπcL is the ’angular frequency’. αn is the amplitude andφn is the phase and both are arbitrary constants.

We have the solution

ηn(x, t) =αncos(ωnt+φn) sin(nπx

L standing wave standing waves:

• periodic solutions with period T =ω

n = 2πLnπc

• ηn(x, t) = 0∀tat the nodesx=kLn ,k= 0,1,2, . . . , n General soln is a superposition of standing waves:

η(x, t) = X

n=1

ancos(ωnt) +bnsin(ωnt) sin(nπx L

With initial conditionsη(x,0) =f(x) and∂t(x,0) =g(x) an= 2

L Z L

0

f(x) sin(nπx L dx bn= 2

L Z L

0

g(x) sin(nπx L )dx

Remark relation between standing waves and d’Alemtat’s soln ηn(x, t) =αncos(ωnt+φn) sin(nπx

L )

= αn

2

sin(nπx

L +ωnn) + sin(nπx

L −ωnt−φn)

= αn

2

sin(nπ

L (x+ct) +φn) + sin(nπ

L (x−ct)−φn)

= superposition of 2 traveling waves

(18)

V.4 Viscous fluids V PDES FOR CONTINUOUS MEDIA

### V.4 Viscous fluids

In ideal fluids, the only force exerted by the fluid particles is the pressure force−pn, normal to any surface.

In a viscous fluid, particles also exert a frictional force which has a tangential component. The frictional force is specified by a thestress tensor σij,i, j= 1,2,3 st. σ·n=σijnjei is the frictional force per unit surface in a surface with normaln.

Some physical input is necessary to model the stress tensorσ. The simplest model is as follows:

since σ is caused by differential motion, we expect σ to depend on velocity gradients ∂u∂xji. The simplest relation betweenσand ∂u∂xi

j is linear. σij∂u∂xi

j∂u∂xj

i∂u∂xk

kδij. Sinceσ= 0 for pure rotationu=Ω×r.

This leads to the model σij

∂ui

∂xj

+∂uj

∂xi

+γ∂uk

∂xk

δij =η ∂ui

∂xj

+∂uj

∂xi

−2 3

∂uk

∂xk

δij

+ζ∂uk

∂xk

δij

whereη is the shear viscosity andζ is the bulk viscosity.

Momentum eqn.

d dt

ZZZ

V

ρuidv=− ZZ

S

ρuiujnjds− ZZ

S

pnidS+ ZZ

S

σijnjdS ZZZ

V

∂t(ρui)dv=− ZZZ

V

∂xj

(ρuiuj)− ∂

∂xi

p+ ∂

∂xj

ij)dv Hence ∂

∂t(ρui) + ∂

∂xj

(ρuiuj) =− ∂

∂xi

p+ ∂

∂xj

ij) ρ(∂

∂tui+u· ∇ui) =− ∂

∂xi

p+µ(∇2ui+ ∂

∂xi

(∇ ·u)−2 3

∂xi

(∇ ·u) +ζ ∂

∂xi

(∇ ·u)

We thus the compressible Navier–Stokes:

∂u∂t +u· ∇u

=−∇p+µ∇2u+ (ζ+13µ)∇(∇ ·u)

∂ρ

∂t +∇ ·(ρu) = 0 For an incompressible fluid: ρ=const,∇ ·u= 0

∂u

∂t +u· ∇u=−∇p+ν∇2u and ∇ ·u= 0 with ν =µ

ρ and p→ p ρ Boundary conditions: u= 0 at the boundaries.

Example 20 (of viscous flows) Steady flows, so∂t= 0, incompressible (i) Flow in a pipe: Assumeu=u(z)i. Then 0 =−∂p∂x∂z2u2

• Plane Corette: p=const,u(H) =U (moving upper boundary) andu(0) = 0. Then u(z) =U zH.

• Plane Poiseuille: ∂x∂p = dpdx =const. Then ∂z2u2 = 1νdxdp, u(0) =u(H) = 0 and we getu(z) = ν1dpdxz(z−H).

The pressure gradient dpdx can be related to the mass fluxQ=ρRH

0 u(z)dz=−ρ12νH3dxdp.

(ii) Axisymetric Poiseuille flow. u=u(r)i, ∇2u = 1r∂r (r∂u∂r)i ∂θ, ∂z = 0 (cylindrical coordinates). The NS eqns reduce to:

0 =−∂p

∂x+ν1 r

∂r(r∂u

∂r)

(19)

V.4 Viscous fluids V PDES FOR CONTINUOUS MEDIA

Assuming ∂p∂x == dpdx =const.

d dr(rdu

dr) = 1 ν

dp dxr rdu

dr = 1 ν

dp dx(r2

2 +C) du

dr = 1 ν

dp dx(r

2 +C

r) C= 0 foru(0) finite u(r) = 1

ν dp dx(r2

4 +D)

Imposeu(R) = 0

u(r) = 1 4ν

dp

dx r2−R2 Mass flux Q=ρ

Z R 0

Z 0

u(r)rdrdθ=2πρ 4ν

dp dx

Z R 0

(r2−R2)rdr

=−πρR4

dp dx

Updating...

## References

Related subjects :