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

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586 Second-Order Linear Diﬀerential Equations

Writing the complex numbers in polar form

C(ω) ≡ ρ(ω)e

iγ(ω)

,B= Ae

iα

,p(iω) ≡ P (ω)e

iθ(ω)

we obtain

ρ(ω)=

P (ω)

and γ(ω)=α −θ(ω).

The real solution, u(t)=Re[U (t)], will then be

u(t)=Re[C(ω)e

iωt

]=ρ(ω)cos[ωt + γ(ω)]

P (ω)

cos[ωt + α − θ(ω)]. (24.44)

The function C(ω) is called the transfer function associated with the lin-

transfer function

ear operator L. Equation (24.44) shows that the output u(t) has the same fre-

quency as the input. It also indicates that the amplitude of u(t) is frequency-

dependent, making it possible to obtain large output amplitudes by varying

the frequency until P (ω) is minimum. This is the phenomenon of resonance

in AC circuits.

Example 24.6.3.

Let us apply the analysis above to Example 24.6.1 and, for

deﬁniteness, take the underdamped case. In this case, 4b>a

;andω

≡

√

is called the natural frequency of the system. The characteristic polynomial isnatural frequency

p(λ)=λ

+ aλ + b.Thus,

p(iω)=−ω

+ iωa + b =(ω

− ω

)+iωa

and

P (ω)=



(ω

− ω

)

+ ω

,θ(ω)=tan

−1



ωa

− ω



The amplitude of the output signal, sometimes called the gain function,isgain function

ρ(ω)=

P (ω)



(ω

− ω

)

+ ω

The minimum of the denominator occurs at ω = ω

, that is, when the driving

frequency equals the natural frequency. In such a situation we have ρ(ω)=A/(ω

a),

showing that the output signal will have a large amplitude when a, the damping

coeﬃcient, is small.

We have considered only the particular solution, u(t), because the most general

solution

y(t)=Ke

−at/2

cos(ω

t + β)+u(t)

in which K and β are constants, eventually reduces to u(t). The ﬁrst term on the

RHS, the transient term, decays to zero. The rate of this decay is determinedtransient term

by the time constant 2/a, the time interval during which the amplitude of the

time constant

transient term drops to 1/e of its initial value.



The importance of the sinusoidal signal becomes clear when we recall that

any periodic signal can be expanded in a Fourier series, R(t)=



∞

n=−∞

inωt

24.7 Problems 587

where ω is the fundamental frequency. The linearity of L suggests the solution

u(t)=Re[U(t)], where

U(t)=

∞



n=−∞

(ω)e

inωt

Substituting in L[U ]=R(t)gives

∞



n=−∞

(ω)p(inω)e

inωt

∞



n=−∞

inωt

Since the e

inωt

are orthonormal, we get C

(ω)=b

/[p(inω)], and

u(t)=Re

∞



n=−∞

inωt

p(inω)

Thus, u(t) is also periodic and has the same fundamental frequency as r(t).

24.7 Problems

24.1. Let f and g be two diﬀerentiable functions that are linearly dependent.

Show that their Wronskian vanishes. (Note that f and g need not be solutions

of a homogeneous SOLDE.)

24.2. Show that if (f



)and(f



) are linearly dependent at one point,

then f

and f

are linearly dependent at all a ≤ x ≤ b.Heref

and f

are

solutions of the DE of (24.4). Hint: Derive the identity

W (f

; x

)=W (f

; x

)exp

(

−

p(t) dt

)

24.3. Show by direct substitution that f

of Equation (24.6) indeed satisﬁes

(24.4) no matter what K is.

24.4. Show that the solutions to the SOLDE y



+ q(x)y = 0 have a constant

Wronskian.

24.5. Find a general integral formula for G

(x), the linearly independent

“partner” of the Hermite polynomial H

(x) which satisﬁes the Hermite DE



− 2xy



+2ny =0.

Specialize this to n =0, 1. Is it possible to ﬁnd G

(x)andG

(x)intermsof

elementary functions?

588 Second-Order Linear Diﬀerential Equations

24.6. Use Theorem 23.3.1 to construct

y =

W (x)

(x)



C +

(t)r(t)

W (t)



asolutionof



p +





y =

24.7. Show that each pair of the following functions satisfy the DE next to it.

Calculate the Wronskian, and give a solution satisfying the initial conditions

y(0) = 2 and y



(0) = 1.

(a) cos x and sin x; y



+ y =0. (b) e

and e

; y



+4y



+3y =0.

; y



1 − x



−

1 − x

y =0.

24.8. For the HSOLDE y



+ py



+ qy = 0, show that

p = −



− f



W (f

)

and q =





− f





W (f

)

Thus, knowing two solutions of an HSOLDE allows us to reconstruct the DE.

24.9. Show that the HSOLDE y



+ py



+ qy =0canbecastintheform



+ S(x)u = 0. Hint: Deﬁne w(x)byy = wu, substitute in the DE, and

demand that the coeﬃcient of u



be zero to obtain

w(x)=C exp



−

p(t) dt



Now show that the original DE can be written as u



+ S(x)u =0with

S(x)=q + p





= q −

−



24.10. Show that the adjoint of M given in Equation (24.14) is the original L.

24.11. Show that S-L equation (24.18) can be transformed into

+[λ − Q(t)]v =0,

by the so-called Liouville substitution, which changes both independent

and dependent variables:

u(x)=v(t)[p(x)w(x)]

−1/4

,t=

w(s)

p(s)

ds.

Then

Q(t)=

q(x(t))

w(x(t))

+[p(x(t))w(x(t))]

−1/4

[(pw)

1/4

24.7 Problems 589

24.12. Show that

(a) the Liouville substitution (see Problem 24.11) transforms the Bessel DE

(xu



)



+(k

x − ν

/x)u =0into



−

− 1/4



v =0.

(b) Specialize to ν =

and show that

1/2

(kt)=A

sin kt

√

+ B

cos kt

√

(x) is an analytic function of x to show that

1/2

(kt)=A

sin kt

√

24.13. Show that the functions x

λx

,wherer =0, 1, 2,...,k, are linearly

independent. Hint: Starting with (D −λ)

, apply powers of D −λ to a linear

combination of x

λx

for all possible r’s.

24.14. Suppose λ is a root of the polynomial

p(x) ≡ x

+ a

n−1

+ ···+ a

x + a

where all coeﬃcients are real. Show that λ

∗

isalsoarootofp(x). Hint:

Complex conjugate p(λ) = 0. Does the same result hold if the coeﬃcients

were complex?

24.15. Write Equation (24.39) in the more familiar Cartesian coordinates

and show that e =0givesacircle,0<e<1 gives an ellipse, e =1givesa

parabola, and e>1 gives a hyperbola. Show that except for the case of a

parabola, the Cartesian equation of the conic section is

(1 −e

)



x −

Km(1 − e

)



(1 −e

=1.

24.16. Derive all the formulas in Equation (24.41).

24.17. Find a basis of real solutions for each DE:

(a) y



+5y



+6=0. (b) y



+6y



+12y



+8y =0.

(4)

= y. (d) y

(4)

= −y.

24.18. Solve the following DEs subject to the given initial conditions.

(a) y

(4)

= y, y(0) = y



(0) = y



(0) = 0,y



(0) = 1.

(b) y

(4)

+ y



=0,y(0) = y



(0) = y



(0) = 0,y



(0) = 1.

(4)

=0,y(0) = y



(0) = y



(0) = 0,y



(0) = 2.

590 Second-Order Linear Diﬀerential Equations

24.19. Solve y



− 2y



+ y = xe

subject to the initial conditions y(0) =

0,y



(0) = 1.

24.20. Find the general solution of each equation:

(a) y



= xe

. (b) y



− 4y



+4y = x



+ y =sinx sin 2x. (d) y



− y =(1+e

−x

)

(e) y



− y = e

sin 2x. (f) y

(6)

− y

(4)

= x

(g) y



− 4y



+4=e

+ xe

. (h) y



+ y = e

24.21. Consider the Euler equation

(n)

+ a

n−1

(n−1)

+ ···+ a



+ a

y = r(x).

Substitute x = e

and show that such a substitution reduces this to a DE

with constant coeﬃcients. In particular, solve x



− 4xy



+6y = x.

24.22. Show that v = C

cos θ + C

sin θ canbewrittenasv = A cos(θ −θ

Find A and θ

in terms of C

and C

24.23. (a) Show that the extremum (maximum or minimum) of the function

y(t)=c

−at/2

+ c

−at/2

occurs at t =2/a − c

(b) Prove that if c

> 0, the extremum is maximum and if c

< 0, it is

minimum.

24.24. Verify that, for any diﬀerentiable function f,wehave

(D − λ)[e

λx

f(x)] = e

λx



(x)

and if ν = λ,then

(D −ν)[e

λx

f(x)] = e

λx

[(λ − ν)f(x)+f



(x)].

24.25. Derive Equation (24.44).

Chapter 25

Laplace’s Equation:

Cartesian Coordinates

In Chapter 22 we discussed the technique of the separation of variables for the

most important PDEs encountered in introductory physics and engineering

courses. One such PDE deserving special attention is the Laplace equation

∇

Φ = 0 (25.1)

which shows up extensively in problems in electrostatics and steady-state heat

conduction. The latter arises in situations in which the temperature does not

change with time, so that the LHS of Equation (22.3) vanishes.

Aside from its signiﬁcance in applications, Laplace’s equation is important

because its solution leads naturally to some of the most famous functions

of mathematical physics. In fact, when separating this equation in various

coordinate systems, one obtains not only such elementary functions as sines

and cosines, but also the more advanced “special functions” such as Legendre

polynomials and the Bessel functions. At the heart of such functions is the

linearity of Laplace’s equation which allows summing a (inﬁnite) number of

solutions to get a new solution. This leads naturally to solutions of Laplace’s

equation in terms of inﬁnite series.

In a typical situation, Φ is given on some surfaces bounding a volume in

space and its value is sought for all points in the volume. When the bounding

surfaces are arbitrarily shaped, the solution can be found only by numerical

techniques; but when they are primary surfaces of a coordinate system, then

we can generally solve the problem by separating Laplace’s equation in the

appropriate coordinate system.

592 Laplace’s Equation: Cartesian Coordinates

25.1 Uniqueness of Solutions

We shall see many examples of solutions to Laplace’s equation in various

coordinate systems in this and the following chapters. All of these solutions

will be obtained in the form of inﬁnite series. So, we know that solutions to

Laplace’s equation indeed exist. What we want to do in this section is to

show that the solution which satisﬁes all the boundary conditions is unique.

In other words, no matter how we ﬁnd the solution, as long as it satisﬁes the

boundary condition, it is the solution of Laplace’s equation. In fact, we can

be more general and prove the uniqueness for the Poisson equation ∇

Φ=ρ.

Consider the volume V with some surfaces bounding it. Figure 25.1 shows

two such volumes. Assume that two functions Φ

and Φ

satisfy the Poisson

equation at every point of the volume, and that they both satisfy some other

conditions related to the surfaces which we shall look into shortly. Let Φ =

− Φ

and note that Φ satisﬁes Laplace’s equation because

∇

Φ=∇

(Φ

− Φ

)=∇

−∇

= ρ − ρ =0.

For any function f, we have [see Equation (14.11)]

∇ · (f∇f)=∇f · ∇f + f∇

f = |∇f|

+ f ∇

For Φ—since it satisﬁes Laplace’s equation—we get

∇ · (Φ∇Φ) = ∇Φ · ∇Φ+Φ∇





= |∇Φ|

Integrating both sides of the last equation over the volume V and using the

divergence theorem on the LHS yields

(Φ∇Φ) ·

da =

(Φ

− Φ

)(

·∇Φ

−

·∇Φ

) da

|∇Φ|

dV. (25.2)

(a) (b)

Figure 25.1: A volume (shaded region) with its bounding surface. (a) The volume is

“inside” the bounding surface. (b) The volume is “outside” the bounding surface(s).

25.1 Uniqueness of Solutions 593

Now suppose:

• Dirichlet boundary condition:Φ

and Φ

take on the same value at Dirichlet boundary

condition

every point of the bounding surface(s), i.e., Φ

− Φ

=0onS;or

• Neumann boundary condition: the so-called normal derivatives

Neumann

boundary

condition; normal

derivatives

· ∇Φ

and

· ∇Φ

take on the same value at every point of the

bounding surface(s), i.e.,

· ∇Φ

−

· ∇Φ

=0onS.

Then, in either case, the ﬁrst line of Equation (25.2) yields zero. Since the

integrand of the RHS is never negative, the integrand must vanish.

It follows

that

|∇Φ|

=0 ⇒ ∇Φ=0 ⇒ Φ=constant ⇒ Φ

− Φ

= constant

for all points in the volume V . Since Φ = 0 on the bounding surface, the

constant must be zero, i.e., Φ

=Φ

for all points in the volume V .Wethus

have

Box 25.1.1. Let V be a volume bounded by a (possibly disconnected) sur-

face S. Then there exists a unique function which satisﬁes both Laplace’s

equation (or the Poisson equation) at every point of V and either Dirichlet

or Neumann boundary conditions on S.

Historical Notes

Pierre Simon de Laplace was a French mathematician and theoretical astronomer

who was so famous in his own time that he was known as the Newton of France. His

main interests throughout his life were celestial mechanics, the theory of probability,

and personal advancement.

At the age of 24 he was already deeply engaged in the detailed application of

Newton’s law of gravitation to the solar system as a whole, in which the planets and

their satellites are not governed by the Sun alone, but interact with one another

in a bewildering variety of ways. Even Newton had been of the opinion that di-

vine intervention would occasionally be needed to prevent this complex mechanism

from degenerating into chaos. Laplace decided to seek reassurance elsewhere, and

succeeded in proving that the ideal solar system of mathematics is a stable dynam-

ical system that will endure unchanged for all time. This achievement was only

one of the long series of triumphs recorded in his monumental treatise M´ecanique

C´eleste (published in ﬁve volumes from 1799 to 1825), which summed up the work

on gravitation of several generations of illustrious mathematicians. Unfortunately

for his later reputation, he omitted all reference to the discoveries of his predecessors

and contemporaries, and left it to be inferred that the ideas were entirely his own.

Many anecdotes are associated with this work. One of the best known describes

the occasion on which Napoleon tried to get a rise out of Laplace by protesting that

The integral is the limit of a sum. If no term of this sum is negative, and the sum

equals zero, then each term of the sum must be zero.

594 Laplace’s Equation: Cartesian Coordinates

he had written a huge book on the system of the world without once mentioning

God as the author of the universe. Laplace is supposed to have replied, “Sire, I

had no need of that hypothesis.” The principal legacy of the M´ecanique C´eleste to

later generations lay in Laplace’s wholesale development of potential theory, with its

far-reaching implications for a dozen diﬀerent branches of physical science ranging

from gravitation and ﬂuid mechanics to electromagnetism and atomic physics. Even

though he lifted the idea of the potential from Lagrange without acknowledgment,

he exploited it so extensively that ever since his time the fundamental equation of

potential theory has been known as Laplace’s equation.

Pierre Simon de

Laplace 1749–1827

After the French Revolution, Laplace’s political talents and greed for position

came to full ﬂower. His compatriots speak ironically of his “suppleness” and “versa-

tility” as a politician. What this really means is that each time there was a change

of regime (and there were many), Laplace smoothly adapted himself by changing his

principles—back and forth between fervent republicanism and fawning royalism—

and each time he emerged with a better job and grander titles. He has been aptly

compared with the apocryphal Vicar of Bray in English literature, who was twice a

Catholic and twice a Protestant. The Vicar is said to have replied as follows to the

charge of being a turncoat: “Not so, neither, for if I changed my religion, I am sure

I kept true to my principle, which is to live and die the Vicar of Bray.”

To balance his faults, Laplace was always generous in giving assistance and

encouragement to younger scientists. From time to time he helped forward in their

careers such men as the chemist Gay–Lussac, the traveler and naturalist Humboldt,

the physicist Poisson, and—appropriately—the young Cauchy, who was destined to

become one of the chief architects of nineteenth-century mathematics.

25.2 Cartesian Coordinates

The separation of Laplace’s equation in Cartesian coordinates is obtained

from Equation (22.12) by setting the constant C equal to zero.

This leads

to the following three equations

− α

X =0,

− α

Y =0,

+(α

+ α

)Z =0, (25.3)

where the α’s could be any number (including zero and complex). The speciﬁc

value that each α takes on depends on the boundary conditions (BCs). We

consider bounding surfaces parallel to the planes of the Cartesian coordinates.

The most eﬀective way of learning how to solve Laplace’s equation is to

go into the details of the solution of a number of speciﬁc examples. We do

so in the following, hoping that the reader will examine these examples very

carefully, taking note of steps taken with an eye on how each step would

change in a diﬀerent situation (diﬀerent BCs, etc).

Example 25.2.1.

Two semi-inﬁnite conducting plates starting on the y-axis andsemi-inﬁnite

electrically

conducting plates

parallel to the x-axis are grounded (the potential Φ is zero on them) and separated by

Recall from Subsection 22.2 that Φ(x, y, z)=X(x)Y (y)Z(z).

We have changed the sign of the α’s to illustrate how the boundary conditions force on

us the correct functional form of X, Y ,andZ.

25.2 Cartesian Coordinates 595

(b)(a)

Figure 25.2: (a) The semi-inﬁnite plates, and (b) the cross section of the two

(grounded) plates and the strip maintained at potential V .

adistanceb [Figure 25.2(a)]. Both plates extend from −∞ to ∞ in the z-direction. A

conducting strip of width b—also inﬁnite in both directions of the z-axis—is located

between the two plates and separated from them by an inﬁnitesimal gap, so that

the strip can be maintained at a diﬀerent potential of Φ = V . Figure 25.2(b) shows

the cross section of the geometry of the problem. We want to ﬁnd the potential in

the region enclosed by the conductors.

The potential is independent of z because, as a small observer moves along the Symmetry tells us

that the potential

is independent

of z.

z-axis keeping the other two coordinates ﬁxed, his detectors and instruments will not

detect any change in the physics of the problem, because the physical environment

of the detectors remains unchanged. So, Z(z) is a constant which we absorb in X(x)

or Y (y). Furthermore, substituting Z = const. in the third equation of (25.3) yields

+ α

=0.

Thus the problem is reduced to ﬁnding X(x)andY (y) which satisfy the diﬀer-

ential equations of (25.3). First let us consider the Y equation. If α

= 0, then the

solution will be of the form

Y (y)=Ay + B.

Thecaseofα

= 0 is a SOLDE with constant coeﬃcients whose most general

solution can be written as

Y (y)=Ae

√

+ Be

−

√

. (25.4)

The vanishing of Φ at y =0andy = b means that

Φ(x, 0) = X(x)Y (0) = 0 for all x ⇒ Y (0) = 0,

Φ(x, b)=X(x)Y (b) = 0 for all x ⇒ Y (b)=0.

Therefore, for the case of α

= 0, this implies

Y (0) = A ×0+B =0 ⇒ B =0,

Y (b)=Ab + B = Ab +0=0 ⇒ A =0.

Thus, if α

=0,wegetY (y)=0andΦ(x, y)=X(x)Y (y)=0whichisthetrivial

solution.

It follows that if we are interested in nontrivial solutions, we had better assume

that α

= 0. Then, Equation (25.4) gives

Y (0) = A + B =0 and Y (b)=Ae

√

+ Be

−

√

=0.