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

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Motion and Flow by Jacques Vanneste

2010/2011 Notes of Paul Boeck

Last changes: November 26, 2010

I Scalar and vector fields

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

I.2 Gradient

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

gradf =∇f =∂xfi+∂yfj+∂zfk Example 2

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

x²+y²+z²

∇f =g^′(r)

∂r

∂xi+_∂y^∂rj+^∂r_∂zk

=g^′(r)

r

xi+_y^rj+^r_zk

=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

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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) ^df_ds =∇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=_m^F =−^GM_r2 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=∇(_r^r3) = 0

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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: εijk∂jFk meansP3 j=1

P3

k=1εijk∂jFk ”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

• εijkεilm=δjlδkm−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

∂x ∂y ∂z

−wy wx 0

= 0i+ 0j+ 2wk

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

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II LINE, SURFACE AND VOLUME INTEGRALS

• More generally,∇ × ∇φ= 0 since

εijk∂j∂kφ=εijk∂_jk² φ= 1

2(εijk∂_jk² φ+εikj∂_jk² φ) =1

2(εijk+εikj)

| {z }

=0

∂²_jkφ

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

Successive applications of∇

• ∇ · ∇f =∇²f = ∆f =∂_xx² f +∂²_yyf+∂_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(¹_r) = 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 Line, surface and volume integrals

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 2π

0



 z x y



·





−sint cost

3



dt

= Z 2

0

π



 3t cost sint



·





−sint cost

3



dt= Z 2π

0

(−3tsint+ cos²t+ 3 sintdt

= [3tcost]^2π₀ +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.

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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

dφ

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: m^dv_dt =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

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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= ¹_r, 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

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(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

r² sphere of Radiusr

=−∂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=1∂iφei=∂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)i=εijkajbk (sum over j&k). a×b=εijkajbkei(Sum over i,j,k)

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

SummaryFor the gravitational fieldE=−^GM_r2 ˆrandE=−∇φwithφ=−^GM_r 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=_ε^ρ₀,∇²φ=−_ε^ρ₀ (ρcharge density.) and ZZ

∂V

E·ndS= 1 ε0

ZZZ

V

ρdv= 1 ε0

total charge inV

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

V δ(rdv= 1.

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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) =−^GM_r 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 =S_r²,dS=r²sinθdθdϕ=r²dΩ2 andn= ˆr.

(1) = ZZ

S²_r

(−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 2π

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) =ρ04π^r₃³. φ+(r) =²₃πGρ0r²+ ˜C.

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

φ(r) = (_GM

p

2a (_a^r²2 −3) for 0≤r≤a

−^GM_r^p 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 = _4πa^Q2). 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.

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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=−^GM_r andφ=_4πε¹

0

Q

r are solutions of Laplace’s equation

∇²φ= 0 for r6= 0 with

ZZ

sphere

∇φndS=

(4πGM

−_ε^Q

0

This can be written as ∇²φ= (GM

−_ε^Q

0

×δ(r)

Dirac–distribution RRR

volume enclosing 0

δ(r)dv= 1 ZZZ

V

∇²φdv= ZZ

∂V

∇φ·ndS=

(4πGM

−_ε^Q

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∓2^aζ_r2 +^a_r²2

=1 r

1±aζ

r² +O a²

r²

Hence φ(r) = Q 4πε0

2aζ

r³ +O(^a_r3²) 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= ¹_r

−p· ∇φ= p·r

r³ dipole Carrying on, we can findmultipole solutions.

Qij∂_ij²φ=Qij

3rirj−r²δij

r⁵

Note that the trace of ^3rⁱ^r^j_r^−r5 ³^δⁱⁱ =^3r²_r^−r5²³ = 0. tr∂_ij²φ=∂²_iiφ=∇²φ= 0.

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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 φ^m= 1

4πε0

1

r monopole

φ^d= 1 4πε0

p·r

r³ dipole

φ^q = 1 4πε0

Qij

(3rirj−r²δij)

r⁵ quadrupole

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^′)r_i^′ ∂

∂r_i^′ _r′=0

1

|r−r^′

dv^′+1 2

Z

V

ρ(r^′)·r^′_ir^′_j ∂²

∂r_i^′∂r^′_j _r′=0

(r−r^′|)dv^′+· · ·

= 1

|r|

Z

V

ρ(r^′)dv^′− ∂

∂ri

1

|r|

Z

V

r_i^′ρ(r^′)dv^′+1 2

∂²

∂ri∂rj

1

|r|

Z

V

r^′_ir_j^′ρ(r^′)dv^′+· · ·

= 1 r

Z

V

ρ(r)dv^′+ Z

V

r_i^′ρ(r)dv^′· ri

r³ + Z

V

r^′_ir_j^′ρ(r)dv^′

3rirj−r²δij

r⁵

= Q r +p·r

r³ +Qij

3rirjr²δij

r⁵ +· · · where Q=

Z

ρ(r^′)dv^′= total charge and p= Z

r^′ρ(r^′)dv^′= dipole strength and Qij =

Z

V

r_i^′r^′_jρ(r^′)dv^′ = quadrupole (moment)

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

Ifφ1 andφ2 are solns, then so isφ1+φ2. (∇²φ1= 0,∇²φ2= 0 =⇒ ∇²(φ1+φ2) = 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−^a_r3³) solves ∇²φ= 0

E=−∇φ=−E0(1−^a_r³3)k−3^a_r³5E0zr

V PDEs for continuous media

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)

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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

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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 ^dp_dz =−ρg.

−g Z

ρdz+c=p hydrostatic relation

Example 17 (i) Pressure in a container for a constant density fluid. ρ=ρ0=const. ^dp_dz =−ρ0g =⇒ p=−ρ·gz+c at z= 0, p0=c. Hence p=p0−ρ0gz. At the bottomz=−h,p=p0+ρ0gh.

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

dp

dz =−ρg=−pg α − g

αz p=Ce⁻^g^α^z=p0e⁻^α^g^z (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 _dt^d(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=−¹_ρ∂xp

∂v

∂t +u· ∇v=−¹_ρ∂yp

∂w

∂t +u· ∇w=−¹_ρ∂zp

∂u

∂t +u· ∇u=−1 ρ∇p

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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.

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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 _ρ^p

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−^a_r²2)−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

pnadθ=ρ 2

Z 2π 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

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V.3 Wave eqn V PDES FOR CONTINUOUS MEDIA

V.3 Wave eqn

a) Acoustic waves

Start with the compressible Euler eqns

∂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^′ ρ^′≪ρ0

u=u^′

p=p(ρ0+ρ^′) =p(ρ0) +dp

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

dp dρ(ρ0)

| {z }

sound speed

Introducing into (∗) and ignoring quadratic terms

(∂tρ^′+ρ0∇ ·u^′= 0

∂tu^′=−_ρ^c²

0∇ρ^′ withu^′=∇φ ∂tρ^′+ρ0∇²φ= 0,∂t∇φ=−_ρ^c²

0∇ρ^′ =⇒ ∂tφ=−^c_ρ²

0ρ.

Finally, ∂_tt² =−c²∇²φ the wave equation In spatial dimensionφ(x, t) satisfies∂ttφ=c²∂_xx² φ(eg. gas in a tube).

b) String under tension T

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

ρAdx

| {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) = ^∂_∂x²^η(x)dxas dx→0.

(†) becomes ρA∂²_ttη=AT ∂_xx² η

∂²_ttη=c²∂_xx² 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=c²∂²_xxφ

XT^′′=c²X^′′T =⇒ T^′′

T =c²X^′′

X =β = const

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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=c²∂²_xxη,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(^nπ_L)² andX(x) = sin(^nπx_L ) =Xn(x).

T^′′−βT = 0,T^′′+c²(^nπ_L)²T = 0 =⇒ Tn(t) =αncos(^nπc_L t+φn) =αncos(ωt+φn).

whereω^nπc_L 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^2π

n = ^2πL_nπc

• ηn(x, t) = 0∀tat the nodesx=^kL_n ,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 +ωn+φn) + sin(nπx

L −ωnt−φn)

= αn

2

sin(nπ

L (x+ct) +φn) + sin(nπ

L (x−ct)−φn)

= superposition of 2 traveling waves

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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_∂x_jⁱ. The simplest relation betweenσand ^∂u_∂xⁱ

j is linear. σij=α^∂u_∂xⁱ

j +β^∂u_∂x^j

i +γ^∂u_∂x^k

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

This leads to the model σij =α

∂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

(σij)dv Hence ∂

∂t(ρui) + ∂

∂xj

(ρuiuj) =− ∂

∂xi

p+ ∂

∂xj

(σij) ρ(∂

∂tui+u· ∇ui) =− ∂

∂xi

p+µ(∇²ui+ ∂

∂xi

(∇ ·u)−2 3

∂

∂xi

(∇ ·u) +ζ ∂

∂xi

(∇ ·u)

We thus the compressible Navier–Stokes:

(ρ ^∂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+ν^∂_∂z²^u2

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

• Plane Poiseuille: _∂x^∂p = ^dp_dx =const. Then ^∂_∂z²^u2 = ¹_ν_dx^dp, u(0) =u(H) = 0 and we getu(z) = _ν¹^dp_dxz(z−H).

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

0 u(z)dz=−ρ_12ν^H³_dx^dp.

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

0 =−∂p

∂x+ν1 r

∂

∂r(r∂u

∂r)

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V.4 Viscous fluids V PDES FOR CONTINUOUS MEDIA

Assuming ^∂p_∂x == ^dp_dx =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 2π 0

u(r)rdrdθ=2πρ 4ν

dp dx

Z R 0

(r²−R²)rdr

=−πρR⁴ 8ν

dp dx

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

Full text