Mathematical Methods of Physics-Collection of Problems and Solutions

8/12/2019 Mathematical Methods of Physics-Collection of Problems and Solutions

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Mathematical Methods of PhysicsCollection of problems and solutions

Maxim Zabzine, Joel Ekstrand

Department of Physics and Astronomy, Uppsala University

2010

1 Vector fields, curved coordinates, operators, integrals1.1 (Yngve, vektoranalys, 1.8.2) Show that uvis divergence-free ifu and vare rotation-free.

1.2 (Yngve, vektoranalys, 1.8.4) Show that v=r is divergence-free if is a constantvector field (ris the position vector).

1.3 (Yngve, vektoranalys, 2.4.7) Use cylindrical coordinates and determine

1. v=r, where v is the velocity, is a constant angle velocity, and ris the positionvector.

2. the rotation ofv,i.e.

v.

1.4 (Yngve, vektoranalys,1.8.16) The potential from an electric dipolpis given by

(r) = p r40r3

.

DetermineE= at the pointr0.

1.5 (Exam FFM NV 060322, 1) Show that the vector fieldF= 20x9yz x + (2x10z + 10y9) y + (2x10y + 25z24)z

is conserved and there exists a scalar potential. Calculate the line integralC

F drfor the curve

starting at the point(0, 0, 0)and finishing at(1, 1, 1).

1.6 (Yngve, vektoranalys, 2.4.13) The vector fieldFhas the cartesian components

Fx= yx2 +y2

, Fy= x

x2 +y2, Fy=0.

1. Determine the components ofFin cylindrical coordinates.

2. What is the rotation ofF,i.e. F?3. Calculate W=Fdr, where the integration is along the unit circle.

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4. Is the results of2and3in agreement with Stokes theorem?

1.7 (Yngve, vektoranalys, 1.10.1) Calculate the contour integral

(4,4)

(1,1)

F

drwhere the co-

ordinates are cartesian(x,y)-coordinates.Fhas the cartesian components

Fx= kx, Fy= ky, Fy= 0,

wherekis a constant. Calculate the integral along different contours:

1. Along(1, 1) (4, 1) (4, 4)in thexy-plane.2. Alongx=y.

1.8 (Homework 1 FFM NV 2006, 2) Calculate

mrr3

,wheremis constant vector.

1.9 Use Gauss law to show that

Su dS=0 if Laplace equation is fulfilled.

1.10 (Homework 1 FFM QX 2008, 1) Calculate

mrr7

+ 3r

,

wheremis constant vector and r=x x +y y +zz.

1.11 (Homework 1 FFM QX 2008, 2) The vector fieldFis defined byF= (yz + 2xz2) x +xz y + (xy + 2x2z)z.

Calculate the line integralC

FdrwhereCis any curve you wish starting at point (0, 0, 0)andfinishes at point(1, 2, 1).

1.12 Calculate

B dS, whereB= A.

1.13 (Homework 2 FFM NV 2006, 1) The vector field is defined byF= 2xz x + 2yz2 y + (x2 + 2y2z1)z.

Show that for any curve Cthe integral C

F

dr=0

1.14 (Homework 2 FFM NV 2006, 2) Evaluate the surface integral S

r dSover the spherex2 +y2 +z2 =R2.

1.15 (Exam 011012, 1) Calculate the contour integral

Cr dr, whereC is the contour de-scribed byr() =2a(1cos )( x cos +y sin ), ais a constant and [0, 2]

1.16 Show that

Crdr= 2Az, whereA is the area bounded by the curveC, andCis in thexy-plane.

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1.17 (Exam FMM QX 061023, 1) Consider the vector field

F=r

r3+ 5r,

where

r=x x +y y +zz, r= |r|.Show thatFis conserved and there exists a scalar potential. Calculate the line integral

C

Fdr

for the curve starting at the point with the cartesian coordinate(1, 0, 0)and finishing at the pointwith the cartesian coordinate(1, 1, 1).

2 Heat and wave equation. One dimensional problems.

2.1 (Homework 3 FFM NV 2006, 1) Determine the steady-state concentration distributionoutsidethe sphere of radiusR, if the concentration on the sphere isc0.

Hint: use the diffusion equation and think about the correct boundary condition at infinity.

2.2 (Exam FFM NV 060322, 2) Assume that the temperature is spherically symmetric,u(r, t).Consider the heat flow (without sources) between two concentric spheres of radii R1 and R2,

R2>R1. Determine the steady-state temperature distribution, if the temperature of outer sphere

is u2 and the inner sphere is u1. Assuming the constant specific heatc and the constant massdensity , calculate the total heat energy for this state-state solution.

2.3 (Homework 2 FFM QX 2008) Consider the diffusion of a chemical pollutant describedby the diffusion equation without sources:

1. Determine the steady-state concentration distribution between two concentric spheres

with radii R1 and R2, R1< R2. The concentration on the outer sphere is c2 and on theinner sphere isc1.

2. Calculate the total mass of the pollutant between two spheres for this steady-state con-

centration distribution.

3. Calculate the total flow of pollutant through the outer sphere.

2.4 (Homework 3 FFM NV 2006, 2) Consider a steady-state temperature distribution in avolumeVwith the boundary given by surface S. Please calculate the total heat flow through

surfaceSand give possible physical explanation for the answer.

2.5 (Exam FFM QX 061023, 2) Assume that the temperature is spherically symmetric,u(r, t).Consider the heat flow (without sources) between two concentric spheres of radii R1 and R2,

R2>R1. Determine the steady-state temperature distribution, if the temperature of outer sphereisu2 and the inner sphere is perfectly insulated. Assuming the constant specific heat c and the

constant mass density , calculate the total heat energy for this steady-state solution.

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3 Two and three dimensional problems. Separation of variables.

3.1 (Homework 3 FFM QX 2007) Solve the following PDE

u

t =

2u

x2+ 2

u

x+ u,

with the boundary conditionsu(0, t) =0, u(L, t) =0.The initial condition isu(x, 0) = f(x).

3.2 (Exam FFM NV 060322, 3) Consider the vibrationu(r, , t)of a circular membrane ofradiusa which is described by the two-dimensional wave equation

2u

t2 =c2u

Find a solution which satisfiesu(a, , t) =0, u(r, , 0) =0,

u

t(r, , 0) = f(r) sin3.

3.3 (Exam FFM NV 060322, 4) Solve the Laplace equation insider the sphere of the radiusawith the following boundary condition

u(a, , ) = f() cos

Hint: use the following conventions for the Laplace operator in the spherical coordinates:

u= 1

r2

r

r2

u

r

+

1

r2 sin

sin

u

+

1

r2 sin2

2u

2.

3.4 (Exam FFM NV 060322, 5) On the interval [0, ] solve the heat equation with the source

u

t=

2u

x2+ 10sin(6x)et

with the boundary conditions

u(0, t) =1, u(, t) =3

and the initial condition

u(x, 0) =1 +2

x + 6sin6x + 3sinx

3.5 (Exam FFM QX 070109, 4) Consider the two-dimensional heat equation on the disk ofradiusb

u

t=2u

Find a solutionu(, , t)which satisfies

u(b, , t) =0, u(, , 0) = f() sin3.

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3.6 (Exam FFM QX 070109, 5) Solve the Laplace equation outside of sphere of the radiusawith the following boundary condition

u(a, , ) =P13(cos ) sin

3.7 (Exam FFM QX 070109, 3) Using the separation of variables write the general solutionfor the PDE

u

t =

2u

x2+ 5u

whereu(x, t)satisfies the following boundary conditions

u(0, t) =0, u(1, t) =0

3.8 (Homework 4 FFM NV 2006) Solve the PDE ut

=k2ux2u,wherekand are con-

stants. The boundary conditions are u(0, t) = 0, u(L, t) = 0 and the initial condition is u(x, 0) =f(x).What happens with the solution whent ? [Hint: the limit depends fromkand .]

3.9 (Homework 5 FFM NV 2006) Solve the heat equationu

t=ku

inside a cylinder (of radiusa and heightH) subject to the initial condition,

u(r, ,z, 0) = f(r,z)

independent of, if the boundary conditions are

u(r, , 0, t) =0

u(r, ,H, t) =0

u

r(a, ,z, t) =0

[Hint: try to use the independence of f fromas soon as possible, this will simplify some of

the calculations]

3.10 (Homework 4 FFM QX 2007) Consider the two-dimensional heat equation on the diskof radiusb

u

t=4u.

Find a solutionu(, , t)which satisfies

u(b, , t) =0,

u(, , 0) = f() sin(7) + g() sin(3).

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3.11 (Homework 5 FFM QX 2007) Solve the Laplace equationu=0

between two concentric spheres of the radiusa and b,a


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4 Inhomogeneous problems.

4.1 (Homework 7 FFM NV 2008,1) Solve the heat equation with the source

u

t =

2u

x2+ sin7xe3t

on the interval[0, ]subject to

u(0, t) =1, u(, t) =0, u(x, 0) =0

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Solutions

S 1.1 Use (ab) =b (a)a (b)

S 1.2 We are free to choose a coordinate system so =z z

(z zr) =z [z (x x +y y +zz)] =z (y x +x y) =z(y

xx +

x

yy) =0.

S 1.5 The vector field has a potential f if the following is satisfied

F= f.

In components it can be written as the following set of PDEs

f

x=20x9yz

f

y=2x10z + 10y9

f

z =2x10

y + 25z24

These equations have the solution

f(x,y,z) =2x10yz +y10 +z25 + constant

The line integral is calculates asC

Fdr=

C

d f= f(1, 1, 1) f(0, 0, 0) =4

S 1.8 The answer is ( mrr3

) =3 mrr5

r mr3

.

S 1.10 The answer is

mrr7

+ 3r

=9.

S 1.11 Let us show thatFis conserved vector field, namely

F= f=f

xx +

f

yy +

f

zz.

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This implies the following set of PDEs

f

x=yz + 2xz2

f

y=xzy

f

z=xy + 2x2z

which have the following solution

f(x,y,z) =xyz +x2z2 + constant

Since the vector field is conserved the line integral is independent from the concrete curve C

and it is given by the following expressionC

Fdr=C

f dr=C

d f= f(1, 2, 1) f(0, 0, 0) =3.

S 1.13 Let us show thatFis conserved vector field, namely

F= f=f

xx +

f

yy +

f

zz.

This implies the following set of PDEs

f

x=2xz

f

y =2yz2

f

z =x2 + 2y2z1

which have the following solution

f(x,y,z) =x2

z +y2

z2

z + constantUsing the Stockes theorem we have

C

f dr=S

(f) ds,where we have used that(f) =0.

S 1.14 Using the Gauss theorem we getS

rds=V

( r)dV=3V

dV

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Next we have to calculate the volume of the ball of radius R (x2 +y2 +z2 R2). It is easy to doin the spherical coordinates

V

dV=R

0

0

20

drddr2 sin =4

3R3,

where we use the following formula

dV= h1h2h3du1du2du3=r2 sin drdd.

The final answer is S

rds=4R3

S 1.17 The vector field has a potential fif the following is satisfied

F= f.

In components it can be written as the following set of PDEs

f

x=

x

(x2 +y2 +z2)3/2+ 5x

f

y =

y

(x2

+y2

+z2

)

3/2+ 5y

f

z =

z

(x2 +y2 +z2)3/2+ 5z

These equations have the solution

f(x,y,z) = 1x2 +y2 +z2

+5

2(x2 +y2 +z2) + constant

Alternatively one can find the function f in spherical coordinates. In spherical coordninatesFis written as

F= r2 + 5r rand the gradient of f in spherical coordinates has a form

f=f

rr+

1

r

f

+

1

rsin

f

.

Thus we arrive to the following equation

f

r=r2 + 5r

and thus f is

f(r) =

1

r

+5

2

r2,

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which is the same as above when we rewrite it in the cartesian coordinates.

The line integral is calculates as

C

F dr=C

d f= f(1, 1, 1) f(1, 0, 0) =6 13

S 2.1 The diffusion equation isc

t=kc,

wherec(r, t) is concentration and kis the diffusion coefficient. We have to look for a steady-state solution, c

t =0 and thus c satisfies the Laplace equation,

c=0.

Moreover the problem has spherical symmetry and therefore cdepends only on the radial coor-

dinate . Using the form of in spherical coordinates

c= 1

2

2

c

+

1

2 sin

sin

c

+

1

2 sin2

2c

2

we arrive at the following equation

1

2d

d 2dc

d=0.This equation has the following solution

c() =B A

,

whereB and A are constants which can be fixed by imposing the correct boundary conditions.

We look for the solution outside the sphere,R . On the sphere we have

c(R) =c0.

At infinite distance from the sphere we would expect that there will be no any chemical and

thus , c() 0.

We can conclude that the solution of the problem is

c() =c0R

.

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S 2.2 To find a spherically symmetric steady-state we have to solve

1

r2

d

dr(r2

du

dr) =0,

which is the Laplace equation for u(r)written in the spherical coordinates. The equation has asolution

u(r) = Ar

+B,

whereA and B are constants. The constants can be fixed by imposing the boundary conditions

u(R1) =u1, u(R2) =u2.

Thus we get

A=(u2 u1)R1R2

R2R1, B=

u2R2u1R1R2R1

and

u(r) =(u1u2)R1R2

(R2 R1)r +u2R2 u1R1

R2 R1The total heat energy is

c

V

udV= c

R2R1

dr

0

d

20

dsin r2(Ar

+B) =2cA(R21 R22) +4

3 cB(R32R31)

S 2.3 1. The diffusion equation isc

t =kc,

wherec(r, t) is concentration and kis the diffusion coefficient. We have to look for a steady-state solution, c

t =0 and thus c satisfies the Laplace equation,

c=0.

Moreover the problem has spherical symmetry and therefore cdepends only on the radial coor-

dinater. Using the form of in spherical coordinates

c= 1

r2

r

r2

c

r

+ 1

r2 sin

sin

c

+ 1

r2 sin2 2c 2

we arrive at the following equation

1

r2d

dr

r2

dc

dr

=0.

This equation has the following solution

c(r) =B Ar

,

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whereB and A are constants which can be fixed by imposing the correct boundary conditions.

We look for the solution between two concentric spheres, R1 rR2. On the sphere we have

c(R1) =c1, c(R2) =c2.

We can conclude that the solution of the problem is

c(r) =(c2 c1)R1R2

(R1R2)r +c2R2 c1R1

R2R1 .

A=(c1 c2)R1R2

R1R2 , B=c2R2 c1R1

R2 R1 .2. The mass of pollutant is given by the volume integral over the space between two concentric

spheres

V

c dV=

R2R1

dr

0

d

20

d r2 sin

B Ar

=4B

3(R32R31)4A2 (R22 R21).

3. The flux vector is proportional to the gradient of the concentrationc

J= K0c.In the spherical coordinates we have

J= K0 Ar2

r.

The total heat flow through the outer sphereSis given by surface integral

S

Jds= K0

0

d2

0

d A

R22R22 sin = 4AK0.

S 2.4 The steady-state temperature distribution satisfies the Laplace equation

u=0.

The heat flow vector is proportional to the gradient of the temperature

=

K0u.

The total heat flow through the surfaceSis given by surface integralS

ds= K0S

u ds= K0V

2u dV= 0,

where we have applied the Gauss theorem and used the fact thatusatisfies the Laplace equation

(remember that =2 = ).Thus for the steady state temperature distribution the total flow through any closed surface is

zero. This is natural since the total heat energy in Vdoes not change in time (there are no

sources).

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S 2.5 To find a spherically symmetric steady-state we have to solve

1

r2

d

dr

(r2du

dr

) =0,

which is the Laplace equation for u(r)written in the spherical coordinates. The equation has asolution

u(r) = Ar

+B,

whereA and B are constants. The constants can be fixed by imposing the boundary conditions

du

dr(R1) =0, u(R2) =u2,

where the first condition corresponds to the perfectly insulated boundary. Thus we get

B=u2

and

u(r) =u2

The total heat energy is

c

V

udV=c

R2R1

dr

0

d

20

dsin r2u2=4

3 cu2(R

32 R31)

S 3.1 The equation should be solved by the method of separation of variables. Let us look forthe solutions of the form

u(x, t) =(x)G(t).

Plugging this into the equation we get

dG

dt=G

d2

dx2+ 2

d

dx+

,

and1

G

dG

dt =

1

d2

dx2 + 2

d

dx+

= .

Thus we get two ODEsdG

dt+ G=0,

d2

dx2 + 2

d

dx+ = ,

which should be supplemented by the boundary conditions

(0) =0, (L) =0.

The equation for G is solved very easily

G(t) =Aet .

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Forwe can look for the solutions of the form = ex, where

(+ 1)2 =

and we get

= i

1.If> 0 then the solution can be written as

(x) =ex

A cos(

x) +B sin(

x)

.

Imposing the boundary conditions we get

(x) =Aex sin

nx

L , =

n

L 2

, n=1, 2, 3,...

There will be no non-trivial solutions for 0 (see the lectures notes for the derivation).The general solution of the problem will be

u(x, t) =

n=1

Ane( nL)

2tex sin

nxL

.

Using the orthogonality ofsins we get

An=2

L

L0

dx f(x)ex sinnx

L

.

S 3.2 We have to solve the following equation

2u

t2 =c2

1

r

r

r

u

r

+

1

r22u

2

Assuming that

u(t, r, ) =G(t)F(r)g()

we get the following set of ODEs

d2gd2

= g, g(+ 2) =g()

r2d2F

dr2 + r

dF

dr + (r2 )F= 0, F(a) =0, |F(0)|


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and the second

F(r) = CJm(nr), n= zmna2 , Jm(zmn) =0, n=1, 2,...

The solution forG is

G(t) =D cos(c

nt) +Esin(c

nt),

where we can drop cos in order to satisfy the first initial condition. Thus we get

u(t, r, ) =

m=0

n=1

[Bmn sin(m) +Amn cos(m)] sin(c

nt)Jm(

nr)

Imposing the remaining initial condition we arrive at the solution of the problem

u(t, r, ) =

n=1Bn sin(c

nt) sin(3)J3(

nr)

with

Bn=

a0

dr f(r)rJ3(

n)

c

na

0

drrJ23 (

nr)

S 3.3 We have to solve

1r2

r

r2 u

r

+ 1

r2 sin

sin u

+ 1

r2 sin2

2

u 2

=0

The problem is solved by the separation of the variables

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

We get the following set of ODEs

d2g

d2= g, g(+ 2) =g()

dd

sin d h

d

+

sin m2

sin

h=0

r2d2f

dr2+ 2r

d f

drf=0

The first equation has a solution

g() =A sin(m) +B cos(m), = m2, m=0, 1, 2,...

For the second equation in order to have a finite solution everywhere we have to require =l(l + 1)withl m

h() = CPml (cos )

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For fwe have

f(r) =Drl +Erl1

and we keep only the solution finite at the origin. Thus the solution of Laplace equation insidea sphere is

u(r, , ) =

m=0

l=m

rl[Aml cos(m) +Bml sin(m)]Pml (cos )

Imposing the boundary conditions we arrive at the final solution of the problem

u(r, , ) =

l=1

alAl cos P1l (cos )

with

Al=

0

f()P1l (cos ) sin d

al0

(P1l (cos ))2 sin d

S 3.4 First we would like to switch to the homogeneous BC

u(x, t) =1 +2

x + v(x, t)

wherev(0, t) = 0, v(, t) = 0 and v satisfies the same equation as u. We can look for v in theform

v(x, t) =

n=1

an(t) sin(nx)

Substituting this into PDE we get

dan

dt= n2an, n =6

da6

dt= 36a6+ 10et

with the initial conditions

a6(0) =6, a1(0) =3, an(0) =0, n =1, 6So the final solution is

u(x, t) =1 +2

x + 3etsinx +

(6 10

35)e36t +

10

35et

sin6x

S 3.5 We have to solve the following equation

u

t =2

1

u

+ 1

22u

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Assuming that

u(, , t) =G(t)F()g()


d2g

d2= g, g(+ 2) =g()

2d2F

d2+

dF

d+ ( 2)F= 0, F(b) =0, |F(0)|


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d2g

d2 = g

, g

(+ 2) =g

()

d

d

sin

d h

d

+

sin m

2

sin

h=0

r2d2f

dr2+ 2r

d f

drf=0


g() =A sin(m) +B cos(m), = m2, m=0, 1, 2,...

For the second equation in order to have a finite solution everywhere we have to require =

l(l + 1)withl mh() = CPml (cos )

For fwe have

f(r) =Drl +Erl1

and we keep only the solution finite at the infinity. Thus the solution of Laplace equation inside

a sphere is

u(r, , ) =

m=0

l=m

rl1[Aml cos(m) +Bml sin(m)]Pml (cos )

Imposing the boundary conditions and using the orthogonality of associated Legendre functions(together with the orthogonality of sin and cos) we arrive at the final solution of the problem

u(r, , ) =a

r

4P13(cos ) sin

S 3.7 We have to solveu

t=

2u

x2+ 5u.

Assuming thatu(x, t) =G(t)(x)we get

dG

dt=G

d2

x2+ 5

which implies the following ODEsdG

t= G

d2

dx2+ (+ 5)=0

with the last equation supplemented by the boundary conditions

(0) =0, (1) =0.

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The equation foris solved by

(x) =A sin(nx), = (n)2

5, n=1, 2,...

It is important to notice that >0 for all n = 1, 2, .. since for 2 >5. The equation forG issolved by

G(t) =Aet

Combing all together we get the following solution

u(x, t) =

n=1

Ane(5(n)2)tsin(nx)

S 3.8 We have a linear homogeneous PDE with the linear homogeneous boundary conditions

and thus we can solve the problem by the separation of variables.

Let us solve PDEu

t=k

2u

x2u

assuming the following form of the solution

u(x, t) =h(t)(x).

PDE reduces to the following equation

1

h

dh

dt=k

1

d2

dx2

Using the standard argument for the separation of variables we get

d2

dx2 =

withbeing a constant anddh

dt= (k+ )h.

The last equation can be easily solved

h(t) =Ae(k+)t.

Forwe have to specify the boundary conditions

(0) =0, (L) =0.

The system forhas been considered many times during the lectures and it has the following

solution

(x) =A sinnx

L

, =

nL

2, n=1, 2, 3,...

The solution of the original PDE can be written as

u(x, t) =

n

=1

Ane(( nL )2k+)tsin

nx

L 20


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Finally we have to impose the initial conditions

u(x

,0

) =

n=1A

nsinnx

L

= f

(x

)

Using the orthogonality of sines we get the following expressions for As

An=2

L

L0

f(x) sinnx

L

dx

Now about the limitt . Ifk+ > 0 then

u(x, t) 0

Otherwiseu(x, t)will go to infinity.

S 3.9 We have a linear homogeneous PDE with the linear homogeneous boundary conditions.

Therefore we can apply the method of separation of variables. Assume our solution of the form

u(r, ,z, t) =G(t)(r, ,z)

then we arrive to the followin problems

dG

dt = kG

+ = 0, (r, , 0) =0, (r, ,H) =0,

r(a, ,z) =0.

First equation can be easily solved by

G(t) =Aekt

and second problem should be solved by the separation of variables. Assuming that

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

we arrive to the following systems of equations

d2h

dz2 = h, h(0) =0, h(H) =0

d2g

d2= g, g(+ 2) =g(), , dg

d(+ 2) =

dg

d()

r2d2f

dr2+ r

d f

dr+

()r2 f=0, |f(0)| < , d fdr

(a) =0

First two equations can be solved easily as follows

h(z) =A sinnzH

, = nH

2

, n=1, 2, 3,...

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g() =A sin(m) +B cos(m), = m2, m=0, 1, 2,....

However we can observe at this stage that the initial conditions do not have any dependence on

and therefore the solution is independent from . Thus we can conclude that= 0. Thus theequation for f becomes

r2d2f

dr2+ r

d f

dr+ r2f=0, |f(0)| 0 the solution (which is finite at r=0) is

f(r) = CJ0(

r).

Next we have to impose the boundary conditions at r= a. We use the following property ofBessel functions (see the lecture notes)

J1(x) = dJ0dx

.

Thus we get that

J1(

a) =0, l =z1l

a

2, l=1, 2, 3,...

wherez1l are zeros ofJ1. For 0 no solution exists due to the properties of modified Bessel

functions (no zeros).

Thus the solution of the PDE with the boundary conditions is

u(r, ,z, t) =

n=1

l=

1

Anl enl ktsin

nz

H J0(l r),

where

nl= l+

nH

2Finally we have to impose the initial conditions

u(r, ,z, 0) =

n=1

l=1

Anl sinnz

H

J0(

l r) = f(r,z)

Using the orthogonality of sines andJ0 we get

Anl=

2H

0

dza

0

dr r f (r,z) sinnzHJ0(

l

r)

HH0

dr rJ20 (

l r)

S 3.10 (Some details of the solution which have been presented on the lectures is not written

out). We have to solve the heat equation in cylindrical coordinates ignoring the dependence

fromz. We solve the following equation

u

t =4

1

u

+ 1

22u

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Assuming that

u(, , t) =G(t)F()g()


d2g

d2= g, g(+ 2) =g()

2d2F

d2+

dF

d+ ( 2)F= 0, F(b) =0, |F(0)|


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S 3.11 We have to solve

1

r2

rr2 u

r+ 1

r2 sin

sin u

+ 1

r2 sin2

2u

2=0

The problem is solved by the separation of the variables

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


d2g

d2= g, g(+ 2) =g()

d

dsin d h

d+sin

m2

sin h=0

r2d2f

dr2+ 2r

d f

drf=0


g() =A sin(m) +B cos(m), = m2, m=0, 1, 2,...

For the second equation in order to have a finite solution everywhere we have to require =l(l + 1)withl being a positive integer such thatl m

h() = CPml (cos )

For fwe have

f(r) =Drl +Erl1

and we keep both the solutions, since we solve the problem between two spheres. Thus the

solution of Laplace equation between two spheres is

u(r, , ) =

m=0

l=m

rl1[Aml cos(m) +Bml sin(m)]Pml (cos )+

+

m=0

l=m

rl[Dml cos(m) +Cml sin(m)]Pml (cos )

Imposing the boundary conditions we arrive at the final solution of the problem

u(r, , ) =

l=1

rl1Al+ rlBl

P1l (cos ) cos

with

al1Al+ alBl =

0

f()P1l (cos ) sin d

0

(P1l (cos ))2 sin d

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bl1Al+ blBl=

0

g()P1l (cos ) sin d

0 (P

1l (cos ))

2

sin d

.

Finally the coefficients are

Al=

0

blf()alg()P1l (cos ) sin d

(bl al1al bl1)0

(P1l (cos ))2 sin d

Bl=

0

bl1f()al1g()P1l (cos ) sin d

(bl1al al1bl )0

(P1l (cos ))2 sin d

S 3.12 We have a linear homogeneous PDE with linear homogeneous BC and thus we can

apply the method of separation of variables. Our equation is

1

c22u

t2 =

1

r2

r

r2

u

r

+

1

r2 sin

sin

u

+

1

r2 sin2

2u

2

We could observe that initial conditions require that uis independent of. Thus we can simplify

the equation 1

c22u

t2 =

1

r2

r

r2

u

r

+

1

r2 sin2

2u

2.

However although we assumed the independence of the solution from the right hand side

depends on. Thus we can proceed only if

2u

2=0

and this will contradict the IC unlessF(r) =0.The problem has a trivial solution only ifF(r) =0 otherwise there is no solution.Alternatively you could separate the variables

u(r, , , t) =G(t)f(r)h()g()

(see the lecture notes) and find among other things

h() =Pml (cos ), g() =A cos(m) +B sin(m)

with m l. The initial conditions require thatm= 2 and at the same time there is no de-pendence. So we will arrive at the same conclusion, that there exists only trivial solution when

F(r) =0.

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S 3.13 We have a linear homogeneous PDE with the linear homogeneous boundary conditions

and thus we can solve the problem by the separation of variables.

Let us solve PDE

ut

=k2u

x2u,

assuming the following form of the solution

u(x, t) =h(t)(x).

PDE reduces to the following equation

1

h

dh

dt =k

1

d2

dx2.

Using the standard argument for the separation of variables we get

d2

dx2 =

withbeing a constant anddh

dt= (k+ )h.

The last equation can be easily solved

h(t) =Ae(k+)t .

Forwe have to specify the boundary conditions

d

dx(0) =0, (L) =0.

Lets consider the different cases. For the case >0 the general solution is

(x) =A sin(

x) +B cos(

x).

Imposing the boundary conditions we arrive at the solution

(x) =A cos(n + 1

2)x

L , = (n + 1

2)

L 2

, n=0, 1, 2, 3,...

For the case 0 there are no solutions satisfying the boundary conditions. Thus the solutionof the original PDE can be written as

u(x, t) =

n=0

Ane(( (n+1/2)L )2k+)tcos

(n + 1

2)x

L

.

Finally we have to impose the initial conditions

u(x, 0) =

n=0

An cos

(n + 1

2)x

L

= f(x)

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Using the orthogonality of coss we get the following expressions for As

An=2

L

L0

f(x) cos(n + 1

2

)x

L

dx

Now about the limitt . Ifk

(n + 1

2)

L

2+ > 0

(for allns) then

u(x, t) 0Otherwiseu(x, t)will go to infinity.

S 3.14 We have to solve the heat equation in cylindrical coordinates ignoring the dependence

fromz. We solve the following equation

u

t=

1

u

+

1

22u

2

Assuming that

u(, , t) =G(t)F()g()


d2g

d2 = g, g(+ 2) =g()

2d2F

d2+

dF

d+ ( 2)F= 0, F(1) =0, |F(0)|


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Imposing the remaining initial condition we arrive at the solution of the problem

u(

t,

r,

) =

n=1B

1ne

(z1n)2tsin

(

)J

1(z

1n

) +B

2ne

(z2n)2tsin

(2

)J

2(z

2n

)

with

B1n=

a0

d g()J1(z1n)

a0

d (J1(z1n))2

and

B2n=

a0

d f()J2(z2n)

a0

d (J2

(z2n

))2

S 3.15 Solve the problem inside of a sphere first.

We have a linear homogeneous PDE with linear homogeneous BC and thus we can apply the

method of separation of variables. Our equation is

1

c22u

t2 =

1

r2

r

r2

u

r

+

1

r2 sin

sin

u

+

1

r2 sin2

2u

2

We could observe that initial conditions require that uis independent of. Thus we can simplify

the equation1

c22u

t2 =

1

r2

r

r2

u

r

+

1

r2 sin

sin

u

.

We separate the variables

u(r, , , t) =G(t)f(r)h()

and arrive at the following equations and BC

d

d

sin

d h

d

+ sin h=0, |h(0)| < , |h()| < ,

d

dr

r2 d f

dr

+ (r2

)f(r) =0, f(a) =0, f< d2G

dt2 + c2G=0,

The solutions of the first equation are

h() =Pl(cos ), l= 0, 1, 2,...

and the second equation is solved by

f(r) =r1/2Jl+ 12

(lkr), lk= z(l+1/2)k

a 2

, k=1, 2, 3,...

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wherez(l+1/2)kare the zeros ofJl+ 12

. The last equation is solved as follows

G(t) =A cos(

ct) +B sin(

ct).

Imposing the one of ICu

t(r, , , t) =0

we get

u(r, , t) =

l=0

k=1

Alkr1/2Jl+ 1

2(

lkr)Pl(cos ) cos(

lkct).

Imposing the second IC we get

Alk=

a

0

dr

0

dF(r) sin2 r3/2 sin Jl+ 12

(lkr)Pl(cos )a

0

dr0

d rsin (Jl+ 12

(

lkr))2(Pl (cos ))2

Now lets comment on the solution outside of a sphere. In this case the solution for fwill be

flk(r) =r1/2Jl+ 1

2(

lkr) + r1/2Yl+ 1

2(

lkr),

wherelkis now defined as

Jl+ 12

(

lka) +Yl+ 12

(

lka) =0.

Due to the general properties of the SL problem there is the following orthogonality condition

a

d r r f lk(r)fln(r) =0, k=n.

S 4.1 First we have to switch to the homogeneous BC. We can do it by the following change

v(x, t) =u(x, t)1 + x

.

vsatisfies the following PDEv

t =

2v

x2+ sin(7x)e3t

on the interval[0, ]subject to

v(0, t) =0, v(, t) =0, v(x, 0) = x

1.

We look for the solution in the form

v(x, t) =

n=1

an(t) sin(nx)

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which satisfies BC. Substituting this solution into the PDE we get the following set of ODEs

dan

dt=

n2an, n=7,

da7

dt = 49a7+ e3t.

These ODEs have the solutions in the form

an(t) =en2tan(0), n =7

a7(t) =e49t

a7(0) 1

46

+

1

46e3t.

Imposing the IC

v(x, 0) =

n=1

an(0) sin(nx) = x

1we get

an(0) = 2

0

sin(nx)x

1

dx.

Thus the final solution is

u(x, t) =1 x

+ 1

46

e3t e49tsin(7x) +

n=1

an(0)en2tsin(nx)

30

Mathematical Methods of Physics-Collection of Problems and Solutions

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Transcript of Mathematical Methods of Physics-Collection of Problems and Solutions