Hassani S. Mathematical Methods: For Students of Physics and Related Fields

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658 Laplace’s Equation: Cylindrical Coordinates

27.11. Derive the expansion of the sine function in terms of Bessel functions.

Hint: See Example 27.5.1.

27.12. The integral



∞

−ax

(bx) dx may look intimidating, but leads to a

very simple expression. To see this:

(a) Substitute for J

(bx) its series representation, and express the result of

the integration in terms of the gamma function (a factorial, in this case).

(b) Use one of the results of Problem 11.1 to show that

∞

−ax

(bx) dx =

√

∞



n=0

Γ(n +

)



−



function:

∞

−ax

(bx) dx =



, 1; 1; −



(d) Now use the result of Problem 11.4 to express the integral in a very simple

form.

27.13. By writing the series representation of the Bessel function as in the

previous problem, and using the result of Problem 11.2, show that for integer

∞

−ax

(bx) dx =

√





Γ(m/2+1)Γ((m +1)/2)

Γ(m +1)

· F



+1,

m +1

; m +1;−



27.14. Multiply both sides of Equation (27.40) by ρJ

a/ρ)cosjϕ and

integrate appropriately to obtain A

. Switch cosine to sine and do the same

to ﬁnd B

27.15. Use Equation (27.18) to show that

ρJ





dρ =

27.16. Use the Parseval relation (27.38) for f (ρ)=ρ

to obtain

∞



n=1

4(m +1)

for any m. Hint: See Example 27.5.2.

27.17. A long heat conducting cylinder of radius a is composed of two halves

(with semicircular cross sections) with an inﬁnitesimal gap between them.

The upper and lower halves of the cylinder are in contact with heat baths of

temperatures +T

and −T

, respectively. Find the temperature both inside

and outside the cylinder.

27.7 Problems 659

27.18. A long heat conducting cylinder of radius a is composed of two halves

(with semicircular cross sections) with an inﬁnitesimal gap between them.

The upper and lower halves of the cylinder are in contact with heat baths

of temperatures +T

and −T

, respectively. The cylinder is inside a larger

cylinder (and coaxial with it) held at temperature T

. Find the temperature

inside the inner cylinder, between the two cylinders, and outside the outer

cylinder.

27.19. A long conducting cylinder of radius a is kept at potential V

.The

cylinder is inside a larger cylinder (and coaxial with it) held at potential V

Find the potential inside the inner cylinder, between the two cylinders, and

outside the outer cylinder.

Chapter 28

Other PDEs

of Mathematical Physics

Chapters 25, 26, and 27 discussed one of the most important PDEs of mathe-

matical physics, Laplace’s equation. The techniques used in solving Laplace’s

equation apply to all PDEs encountered in introductory physics. Since we

have already spent a considerable amount of time on these techniques, we

shall simply provide some illustrative examples of solving other PDEs.

28.1 The Heat Equation

The heat equation, sometimes also called the diﬀusion equation,wasin- diﬀusion equation

troduced in Chapter 22 [see Equation (22.3)]. The separation of variables

T (t, r)=g(t)R(r) yields

∂

∂t

[g(t)R(r)] = k

∇

[g(t)R(r)] ⇒ R(r)

= k

g(t)∇

Dividing both sides by g(t)R(r), we obtain

= k

∇



 

≡−λ

The LHS is a function of t, and the RHS a function of r. The independence of

these variables forces each side to be a constant. Calling this constant −k

for later convenience, we obtain an ODE in time and a PDE in the remaining

variables:

+ k

λg =0 and ∇

R + λR =0. (28.1)

The general solution of the ﬁrst equation is

g(t)=Ae

−k

λt

(28.2)

662 Other PDEs of Mathematical Physics

and that of the second equation can be obtained precisely by the methods of

the last chapter. We illustrate this by some examples, but ﬁrst we need to

keep in mind that λ is to be assumed positive, otherwise the exponential in

λ of the heat

equation is always

positive

Equation (28.2) will cause a growth of g(t) (and, therefore, the temperature)

beyond bounds.

28.1.1 Heat-Conducting Rod

Let us consider a one-dimensional conducting rod with one end at the origin

x = 0 and the other at x = b. The two ends are held at T = 0. Initially,

heat-conducting

rod

at t = 0, we assume a temperature distribution on the rod given by some

function f (x). We want to calculate the temperature at time t at any point

x on the rod.

Due to the one-dimensionality of the rod, the y-andz-dependence can be

ignored, and the Laplacian is reduced to a second derivative in x.Thus,the

second equation in (28.1) becomes

+ λX =0, (28.3)

where X is a function of x alone. The general solution of this equation is

X(x)=B cos(

√

λx)+C sin(

√

λx).

Since the two ends of the rod are held at T = 0, we have the boundary

conditions T (t, 0) = 0 = T (t, b), which imply that X(0) = 0 = X(b). These

give B =0and

sin(

√

λb)=0 ⇒

√

λb= nπ for n =1, 2,....

With a label n attached to λ, the solution, and the constant multiplying it,

we can now write



nπ



and X

(x)=C

sin



nπ



for n =1, 2,....

The (subscripted) solution of the time equation is also simply obtained:

(t)=A

−k

(nπ/b)

This leads to a general solution of the form

T (t, x)=

∞



n=1

−(nπk/b)

sin



nπ



, (28.4)

where B

≡ A

. The initial condition f(x)=T (0,x) yields

f(x)=

∞



n=1

sin(nπx/b)

The reader may check that the only solution for λ = 0 is the trivial solution.

Consult Section 25.2.

28.1 The Heat Equation 663

which is a Fourier series from which we can calculate the coeﬃcients

sin



nπ



f(x) dx.

Thus if we know the initial temperature distribution on the rod [the function

f(x)], we can determine the temperature of the rod for all time. For instance,

if the initial temperature distribution of the rod is uniform, say T

,then

sin



nπ



dx =



−

nπ

cos



nπ





nπ

[1 − (−1)

It follows that the odd n’s survive, and if we set n =2m +1,weobtain

2m+1

π(2m +1)

and

T (t, x)=

∞



m=0

−[(2m+1)πk/b]

2m +1

sin



(2m +1)π



This distribution of temperature for all time can be obtained numerically for

any heat conductor whose k is known. Note that the exponential in the sum

causes the temperature to drop to zero (the ﬁxed temperature of its two end

points) eventually. This conclusion is independent of the initial temperature

distribution of the rod as Equation (28.4) indicates.

28.1.2 Heat Conduction in a Rectangular Plate

As a more complicated example involving a second spatial variable, consider

a rectangular heat-conducting plate with sides of length a and b all held at

T = 0. Assume that at time t = 0 the temperature has a distribution function

conduction of heat

in a rectangular

plate

f(x, y). Let us ﬁnd the variation of temperature for all points (x, y)atall

times t>0.

The spatial part of the heat equation for this problem is

∂

∂x

∂

∂y

+ λR =0.

A separation of variables, R(x, y)=X(x)Y (y), and its usual procedure leads

to the following equation:

  

≡−μ

  

≡−ν

+ λ =0.

This leads to the following two ODEs:

+ μX =0,

+ νY =0,λ= μ + ν.

664 Other PDEs of Mathematical Physics

Due to the periodicity of the BCs, the general solutions of these equations are

trigonometric functions. The four boundary conditions

T (0,y,t)=T (a, y, t)=T (x, 0,t)=T (x, b, t)=0

determine the speciﬁc form of the solutions as well as the indexed constants

of separation:



nπ



and X

(x)=A

sin



nπ



for n =1, 2,...,



mπ



and Y

(y)=B

sin



mπ



for m =1, 2,....

So, λ becomes a double indexed quantity:

λ ≡ λ

= μ

+ ν



nπ





mπ



The solution to the g equation can be expressed as g(t)=C

−k

Putting everything together, we obtain

T (x, y, t)=

∞



n=1

∞



m=1

−k

sin



nπ



sin



mπ



where A

= A

is an arbitrary constant. To determine it, we impose

the initial condition T (x, y, 0) = f (x, y). This yields

f(x, y)=

∞



n=1

∞



m=1

sin



nπ



sin



mπ



from which we ﬁnd the coeﬃcients A

(see Theorem 25.2.5):

dyf (x, y)sin



nπ



sin



mπ



28.1.3 Heat Conduction in a Circular Plate

In this example, we consider a circular plate of radius a whose rim is held

at T = 0 and whose initial surface temperature is characterized by a func-

tion f(ρ, ϕ). We are seeking the temperature distribution on the plate for

circular plate

all time. The spatial part of the heat equation in z-independent cylindrical

coordinates,

appropriate for a circular plate, is

∂

∂ρ



∂R

∂ρ



∂

∂ϕ

+ λR =0

See the discussion of Subsection 22.3.

28.1 The Heat Equation 665

which, after the separation of variables, R(ρ, ϕ)=R(ρ)S(ϕ), reduces to

S(ϕ)=A cos mϕ + B sin mϕ for m =0, 1, 2,...,

dρ



λ −



R =0.

The solution of the last (Bessel) equation, which is well deﬁned for ρ =0and

vanishes at ρ = a,is

R(ρ)=CJ





with

√

λ =

and n =1, 2,...,

where, as usual, x

is the nth root of J

.Weseethatλ is a double-indexed

quantity. The time equation (28.2) has a solution of the form

g(t)=D

−k

= D

−k

Multiplying the three solutions and summing over the two indices yields the

most general solution

T (ρ, ϕ, t)=

∞



m=0

∞



n=1





−(kx

/a)

cos mϕ + B

sin mϕ).

The coeﬃcients are determined from the initial condition

f(ρ, ϕ)=T (ρ, ϕ, 0) =

∞



m=0

∞



n=1





cos mϕ + B

sin mϕ).

Except for the hyperbolic sine term, this equation is identical to (27.40).

Therefore, the coeﬃcients are given by expressions similar to Equation (27.41).

In the case at hand, we get

πa

m+1

)

2π

dϕ

dρ ρf (ρ, ϕ)J





cos mϕ,

πa

m+1

)

2π

dϕ

dρ ρf (ρ, ϕ)J





sin mϕ.

In particular, if the initial temperature distribution is independent of ϕ,

then only the term with m = 0 contributes,

and we get

T (ρ, t)=

∞



n=1





−(kx

/a)

With f (ρ)representingtheϕ-independent initial temperature distribution,

the coeﬃcient A

is found to be

)

dρ ρf (ρ)J





Note that the temperature distribution does not develop any ϕ dependence

at later times.

See the discussion after Equation (27.41).

666 Other PDEs of Mathematical Physics

28.2 The Schr¨odinger Equation

Chapter 22 separated the time part of the Schr¨odinger equation from its space

part, and resulted in the following two equations:

∇

ψ +



[E −V (r)]ψ =0 and



T, (28.5)

where E, the energy of the quantum particle, is the constant of separation.

We have also used ψ instead of R, because the latter is usually reserved to

denote a function of the radial variable r (or ρ) when separating the variables

of the Laplacian in spherical (or cylindrical) coordinates.

The solution of the time part is easily obtained: It is simply

T (t)=Ae

iEt/

= Ae

iωt

where ω ≡



. (28.6)

It is the solution of the ﬁrst equation in (28.5), the time-independent

time-independent

Schr¨odinger

equation

Schr¨odinger equation that will take up most of our time in this section.

Historical Notes

Erwin Schr¨odinger was a student at Vienna from 1906 and taught there for ten

years from 1910 to 1920 with a break for military service in World War I. While

at Vienna he worked on radioactivity, proving the statistical nature of radioactive

decay. He also made important contributions to the kinetic theory of solids, studying

the dynamics of crystal lattices.

After leaving Vienna in 1920 he was appointed to a professorship in Jena, where

he stayed for a short time. He then moved to Stuttgart, and later to Breslau before

accepting the chair of theoretical physics at Zurich in late 1921. During these years

of changing from one place to another, Schr¨odinger studied physiological optics, in

particular the theory of color vision.

Zurich was to be the place where Schr¨odinger made his most important contribu-

tions. From 1921 he studied atomic structure. In 1924 he began to study quantum

statistics soon after reading de Broglie’s thesis which was to have a major inﬂuence

on his thinking.

Schr¨odinger published very important work relating to wave mechanics and the

general theory of relativity in a series of papers in 1926. Wave mechanics, proposed

by Schr¨odinger in these papers, was the second formulation of quantum theory, the

ﬁrst being matrix mechanics due to Heisenberg. For this work Schr¨odinger was

awarded the Nobel prize in 1933.

Erwin Schr¨odinger

1887–1961

Schr¨odinger went to Berlin in 1927 where he succeeded Planck as the chair of

theoretical physics and he became a colleague of Einstein’s.

Although he was a Catholic, Schr¨odinger decided in 1933 that he couldn’t live

in a country in which the persecution of Jews had become a national policy. He left,

spending time in Britain where he was at the University of Oxford from 1933 until

1936. In 1936 he went to Austria and spent the years 1936–1938 in Graz. However,

the advancing Nazi threat caught up with him again in Austria and he ﬂed again,

this time settling in Dublin, Ireland, in 1939.

We used α in place of E in Chapter 22.

28.2 The Schr¨odinger Equation 667

His study of Greek science and philosophy is summarized in Nature and the

Greeks (1954) which he wrote while in Dublin. Another important book written

during this period was What Is Life (1944) which led to progress in biology. He

remained in Dublin until he retired in 1956 when he returned to Vienna.

During his last few years Schr¨odinger remained interested in mathematical physics

and continued to work on general relativity, uniﬁed ﬁeld theory, and meson

physics.

28.2.1 Quantum Harmonic Oscillator

As an important example of the Schr¨odinger equation, we consider a particle quantum harmonic

oscillator

in a one-dimensional harmonic oscillator potential.

The one-dimensional time-independent Schr¨odinger equation for a particle

of mass μ in a potential V (x)is

2μ



[E − V (x)]ψ =0,

where E is the total energy of the particle.

For a harmonic oscillator (with the “spring” constant k),

V (x)=

≡

μω

and



−



ψ +

2μ



Eψ =0,ω≡

To simplify the equation, we make the change of variables x =(



/μω)y.

The equation then becomes



− y

ψ +

ω

ψ =0, (28.7)

where the primes indicate diﬀerentiation with respect to y.

We could solve this DE by the Frobenius power series method. However,

tradition suggests that we ﬁrst look at the behavior of the solution at y →∞.

In this limit, we can ignore the last term in (28.7), and the DE becomes



− y

ψ ≈ 0

which can easily be shown to have (an approximate) solution of the form

±y

. Since the positive exponent diverges at inﬁnity, we have to retain

only the solution with negative exponent. Following the traditional steps,

we consider a solution of the form ψ(y) ≡ H(y)exp(−y

/2) in which the

asymptotic function has been separated. Substitution of this separated form

of ψ in (28.7) results in



− 2yH



+ λH =0 where λ =

ω

− 1. (28.8)

668 Other PDEs of Mathematical Physics

This is the Hermite diﬀerential equation.Hermite

diﬀerential

equation

To solve the Hermite DE by the Frobenius method, we assume an expan-

sion of the form H(y)=



∞

n=0

with



(y)=

∞



n=1

n−1

∞



n=0

(n +1)c

n+1



(y)=

∞



n=1

n(n +1)c

n+1

n−1

∞



n=0

(n +1)(n +2)c

n+2

where in the last step of each equation, we changed the dummy index to

m = n − 1, and in the end, replaced m with n. Substituting in Equation

(28.8) gives

∞



n=0

[(n +1)(n +2)c

n+2

+ λc

  

≡S

−2

∞



n=0

(n +1)c

n+1

=0. (28.9)

Now separate the zeroth term of the ﬁrst sum to obtain

=2c

+ λc

∞



n=1

[(n +1)(n +2)c

n+2

+ λc

Changing the dummy index to m = n − 1 yields

=2c

+ λc

∞



m=0

[(m +2)(m +3)c

m+3

+ λc

m+1

whose dummy index can be switched back to n. Substitution of this last result

in (28.9) now yields

+ λc

∞



n=0

[(n +2)(n +3)c

n+3

+ λc

n+1

− 2(n +1)c

n+1

=0.

Setting the coeﬃcients of powers of y equal to zero, we obtain

= −

n+3

2(n +1)− λ

(n +2)(n +3)

n+1

for n ≥ 0,

or, replacing n with n −1 and noting that the resulting recursion relation is

true for n = 0 as well, we obtain

recursion relation

for Hermite DE

n+2

2n −λ

(n +1)(n +2)

,n≥ 0. (28.10)

The ratio test yields easily that the series is convergent for all values of y.

Physics dictates

mathematics!

However, on physical grounds, i.e., the demand that lim

x→∞

ψ(x)=0,the

series must be truncated. Let us see why.