Mathetmatical Methods of Physics Problem Solutions

8/10/2019 Mathetmatical Methods of Physics Problem Solutions

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Physics 384

Problem Set 8: Solutions

1. We are asked to find the potential between two concentric cylinders, andhence we need the form of the solution of Laplaces equation in polar co-ordinates, which is

(r, ) =c0+d0ln(r) +

m=1

(cmrm +dmr

m)cos(m) +

m=1

(fmrm +gmr

m)sin(m)

Now, we need to apply the boundary conditions to this general solution.Firstly we note that at r = a, the potential is constant, and hence we musthave that there is no contribution from sin(m) or cos(m) terms, i.e.

c0+d0ln(a) = V0

cmam +dma

m = 0

fmam +gma

m = 0.

We now consider the boundary conditions at r = b and note that these havean angular dependence, but that (b, ) is even in , hence sin(m) termscannot contribute, and hence

fmbm +gmbm = 0,

which implies fm= gm= 0 for all m. Thus we have that

(b, ) =c0+d0ln(b) +

m=1

(cmbm +dmb

m)cos(m),

and this is a Fourier series in , so we can use this to determine the coefficients(and noting that (b, ) is even simplifies the integrals)

c0+d0ln(b) = 2

2

0

V02

2

V0 = 0,

hencec0 = d0ln(b), and we also have that c0= V0 d0ln(a), so

c0= V0ln(b)

lnab

, d0= V0lnab

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Turning to the remainder of the coefficients:

cmbm +dmb

m = 2

2

0

V0cos(m) 2

2

V0cos(m)

= 2V0

2

0

cos(m)

2

cos(m)

= 2V0

m

[sin(m)]

20 [sin(m)]

2

=

2V0m

sin

m2

sin(m) + sin

m2

=

4V0

m

sin m2

=

0, if m even4V0

(2j+1)(1)j, if m= 2j+ 1 odd

Thus we have

d2j+1 = c2j+1a4j+2

d2j+1 = b2j+1

4V0

(2j+ 1)(1)j c2j+1b

2j+1

which we can solve to get

c2j+1=

4V0

(2j+ 1) (1)

j b2j+1

b4j+2 a4j+2 , d2j+1=

4V0

(2j+ 1) (1)

j a4j+2b2j+1

b4j+2 a4j+2 ,

and hence we get our full solution

(r, ) = V0

lnab

ln rb

+

4V0

j=0

(1)j

2j+ 1

b2j+1

b4j+2 a4j+2

r2j+1 a4j+2r(2j+1)

cos((2j+ 1)

2. a) We consider a rod of length L, with both ends held at temperatureT0. Thus, to get the temperature distibution in the rod, we should solve thediffusion equation

D2T T

t = 0.

WriteT(x, t) =T0+T(x, t), and Tclearly also satisfies the diffusion equation,with T(0, t) = T(L, t) = 0. Using separation of variables, write T(x, t) =X(x)(t), in which case

X

X

D= 0,

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and introducing a separation constant k2, we getX =k2X, =k2D,

so we have the solutions(t) =eDk

2t, X(x) =A cos(kx) +Bsin(kx).

Applying the boundary conditions on T implies A= 0 and sin(kL) = 0, sothe allowed values ofk are k = n/L, and the general solution for T(x, t) is

T(x, t) =T0+n=1

Ansinnx

L

eDk

2t.

We now apply the initial conditions to determine the coefficients in theFourier series:

An = 2L

L0

dx(T(x, 0) T0)sin

nxL

= 2

L

2

LT1

L2

0

dx x sinnx

L

+

LL2

dx (L x)sinnx

L

= 4T1

L2

L

nx cos

nxL

L20

+ L

n

L2

0

dx cosnx

L

+

L

n(L x)cos

nxL

LL2

L

n

LL2

dx cosnx

L

= 4T1L2

L2

2ncos

n2

+ L2

2ncos

n2

+

L

n

2 sin

nxL

L2

0

L

n

2 sin

nxL

LL2

= 8T1(n)2

sinn

2

,

and sinn2

= (1)

n12 ifn odd, and 0 ifn is even. Thus

An= 8T1

(n)2

(1)n12 ,

so ifn = 2j+ 1, then

T(x, t) =T0+8T1

2

j=0

()j

(2j+ 1)2sin

(2j+ 1)x

L

e

2(2j+1)2

L2 Dt.

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b) i) Now, we are given

D d2

T1(x)dx2 =(x),

and to solve this, we introduce a Greens function G(x|) which satisfies theequation

D d2

dx2G(x|) =(x ),

and then

T1(x) =

L0

d G(x|)(),

and

Dd2T1(x)

dx2 =

L

0

dD

d2

dx2G(x|)

() =

L

0

d(x )() =(x).

Now, we need to solve for G(x|); we note that G has Dirichlet boundaryconditions, and that

d2G

dx2 = 0,

for x =. Hence

G(x|) = Ax+B, 0< x < Cx+D, < x < L

Now,G(0|) =G(L|) = 0, henceB = 0 andCL + D= 0, and applying con-tinuity atx = givesA= C(L ), which in combination with integratingthe given equation in a small interval about x = :

dG

dx

x=+

dG

dx

x=

=1

D,

gives (as we have seen previously)

A=

(L )

DL , C=

DL ,

and hence

G(x|) =

(L)x

DL , 0< x <

(Lx)DL

, < x < L

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ii) Having determined the Greens function in part i), we can now find the

temperature distribution T1(x) when(x) =x by using

T1(x) =

L0

dG(x|)().

This gives us

T1(x) =

x0

d(L x)

DL +

Lx

d(L )x

DL

= x

DL

1

3x2(L x) +

1

22L

L

x

1

33

L

x

= x6D

(L2 x2).

3. a) The given boundary conditions imply T(r,,t) =T0 at r =a and theinsulation at r = b impliesn T

r= 0, i.e.

T

r(b,,t) = 0.

b) The diffusion equation is

T

t D2

T = 0,

and to solve the problem, write

T(r,,t) =T0+ T(r,,t),

in which case Talso satisifies the diffusion equation:

T

t D2T = 0,

and the boundary conditions become

T(a,,t) = 0

T

r(b,,t) = 0.

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Use separation of variables to write T(r,,t) = (t)(r, ) and then the

diffusion equation takes the form

D

2

= 0.

Introduce a separation constant k2, then

+k2D = 0,

and we can solve to get(t) =ek

2Dt.

Now, noting that in cylindrical co-ordinates

2 =1

r

rr

r+

1

r22

2,

we are left to solve1

r

rr

r+

1

r22

2

(r, ) +k2(r, ) = 0.

Try writing (r, ) =R(r)A(), then

1R

rd

drrd

dr+k2r2

R+ A

A = 0.

Introduce a second separation constant m2, and then

A +m2A= 0,

and

rd

drr

dR

dr + (k2r2 m2)R= 0,

and the first of these equations can be solved to give

A() = cos(m) +sin(m),

and requiring that A is single-valued, i.e. A(+ 2) =A() implies that mis an integer.

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The radial equation is Bessels equation which has solutions

R(r) =Jm(kr) +Nm(kr).

The boundary conditions discussed in part a) imply that R(a) = 0 andR(b) = 0. The first condition thus implies

Jm(ka) +Nm(ka) = 0,

or

= Jm(ka)

Nm(ka),

and thus we may write ((r) is to be distinguished from (r, ) introducedabove)(r) =Nm(ka)Jm(kr) Jm(ka)Nm(kr).

The second condition gives an eigenvalue equation for kmn which are thesolutions of the equation

Nm(kmna)J

m(kmnb) Jm(kmna)N

m(kmnb) = 0.

Using linear superposition, the general form ofT(r,,t) consistent with theboundary conditions is

T(r,,t) = T0+ T(r,,t)

= T0+n=1

m=0

m(kmnr) [Amncos(m) +Bmnsin(m)] eDk2mnt,(1)

wherem(kr) =Nm(ka)Jm(kr) Jm(ka)Nm(kr),

and the kmn are expressed in the eigenvalue equation given above.

c) We are given that T(r,, 0) = T1. Comparing this to the expression inpart b) we get

T1 T0=n=1

m=0

m(kmnr) [Amncos(m) +Bmnsin(m)] .

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The left hand side is an even function of and hence Bmn = 0 for allm and

n. The left hand side is also independent of and hence only the m = 0term can contribute to the sum, so Amn = 0 is m = 0. Hence we have

n=1

An0(knr) =T1 T0,

whereAn= A0nandkn= k0n. The0 are the solutions of a Sturm-Liouvilleproblem and hence are orthogonal for different kn, so we determine that

An=

ba

dr r (T1 T0)0(knr)

ba dr r [0(knr)]2 ,

so we may write

T(r,,t) =T0+n=1

An0(knr)eDk2nt,

as required, withAn given above and kn= k0n.

4. We are asked to find the allowed modes and their associated frequenciesfor sound waves in a pipe of lengthLand radiusawhich is open at both ends.

In solving the wave equation in a cylindrical geometry we can use separationof variables to write

(r, t) =S(r)T(t),

whereis the velocity potential. Subsitution into the wave equation

2 1

c22u

t2 = 0,

and following the steps outlined in the lecture notes leads to

S(r) =v0a0Jm(qr) [A cos(m) +Bsin(m)] Ceikz +Deikz ,where m is an integer (required for the solution to be single valued). Theradial solution contains the origin, hence there is only a Bessel function andno Neumann function, and the boundary condition on the surface of the pipeis thatn= 0, corresponding to/r= 0 and hencedJm(qr)/dr|r=a= 0,

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which impliesqa = mn, thenth zero ofJm. The frequency of a mode is given

by 2 =c2(q2 +k2z).

The boundary condition at each end of the pipe that follows from the pipebeing open is that the fluctuations in the pressure (p1) vanish at z= 0 andL. This corresponds to

p1

t

z=0

=

t

z=L

= 0,

which givesC+D= 0,

andCeikzL +DeikzL = 0,

so sin(kzL) = 0, which implies

kz =p

L,

wherepis an integer.Thus the normal modes are of the form

S(r) =v0a0Jm(qmnr) cos(m)

sin(m)

sinpz

L

,

with frequencies

= c

2mna2

+p22

L2 .

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Mathetmatical Methods of Physics Problem Solutions