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3 Inhomogeneous Linear Differential Equations: The Forced Harmonic Oscillator

# 3 Inhomogeneous Linear Differential Equations: The Forced Harmonic Oscillator

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148

Mathematics for Physical Chemistry

TABLE 12.1 Particular Trial Solutions for the Variation of Parameters

Inhomogeneous term

Trial solution

1

A

tn

A 0 + A1 t + A2 t + · · · + An

e αt

Ae αt

t net

e αt (A

e αt sin (βt )

e αt [A cos (βt ) + B sin (βt )]

α,β

e αt

e αt [A cos (βt ) + B sin (βt )]

α,β

cos (βt )

Forbidden characteristic root

0

0

+ A1 t + A2 t + · · · + An

(ω2 − α 2 )A cos (αt) + ω2 − α 2 B sin (αt)

F0 sin (αt)

,

m

where we have replaced k/m by ω2 . This can be a valid

equation for all values of t only if A = 0 and if

F0

.

m(ω2 − α 2 )

The particular solution is

z p (t) =

α

F0 α

,

m(ω2 − α 2 )

F0 α

vz (0)

.

b2 =

ω

mω(ω2 − α 2 )

Substitution of this into the inhomogeneous equation gives

−α 2 A cos (αt) − α 2 B sin (αt)

k

A cos (αt) + B sin (αt)

+

m

F0 sin (αt)

=

,

m

t n)

vz (0) = b2 ω +

z p = A cos (αt) + B sin (αt).

B=

0

α

Use of Table 12.1 gives the particular solution

=

tn

Exercise 12.11. Find an expression for the initial velocity.

The solution in the previous example is a linear

combination of the natural motion and a motion proportional to the external force. If the frequencies of

these are not very different, a motion such as shown in

Figure 12.5, known as beating, can result. There is a

periodic variation of the amplitude of vibration with a

circular frequency equal to ω − α. You can hear this beating

when a piano is being tuned. There are two or three strings

for each note, and they are tuned separately. Each string

can excite a sympathetic vibration in the other, which acts

as an external force. When the frequencies of two strings

are slightly different, you can hear a pulsation like that in

Figure 12.5, which shows the case that α = 1.100ω.

F0

sin αt .

− α2 )

m(ω2

z

The general solution is

F0

sin αt ,

m(ω2 − α 2 )

(12.63)

where the constants b1 and b2 are to be determined by

the initial conditions. Let us assume that z(0) = 0, so that

b1 = 0.

z(t) = b1 cos(ωt) + b2 sin(ωt) +

t

F0

sin αt ,

− α2 )

dz

F0 α

= b2 ω cos(ωt) +

cos αt .

vz (t) =

dt

m(ω2 − α 2 )

z(t) = b2 sin(ωt) +

m(ω2

The constant b2 would be determined by another initial

condition on vz (0)

FIGURE 12.5 The position of a forced harmonic oscillator as a function

of time for the case α = 1.1ω.

CHAPTER | 12 Differential Equations

149

12.4 DIFFERENTIAL EQUATIONS WITH

SEPARABLE VARIABLES

In this section we discuss equations that can be manipulated

algebraically into the form

g(y)

dy

= f (x),

dx

(12.64)

where g(y) is a known integrable function of y, f(x) is a

known integrable function of x, and y(x) is the unknown

function. To solve Eq. (12.64), we symbolically multiply

both sides of the equation by dx and use the relation:

dy

dx = dy.

dx

(12.65)

(12.66)

If we have manipulated the equation into this form, we

say that we have separated the variables, because we have

no x dependence in the left-hand side of the equation and

no y dependence in the right-hand side. We can perform

an indefinite integration on both sides of this equation to

obtain

g(y)dy =

f (x)dx + C,

(12.67)

where C is the difference of two constants of integration.

We could do a definite integration

y2

y1

g(y)dy =

x2

f (x)dx,

1

dc = ln (c) = −k

c

ln (c) = −kt + C,

dt + C = −kt + C,

where C is a constant of integration. We take the exponential

of each side of this equation to obtain

eln (c) = c = eC e−kt = c(0)e−kt ,

c(t) = c(0)e−kt .

(12.71)

A definite integration can be carried out instead of an

indefinite integration:

c t1

We now have

g(y)dy = f (x)dx.

We perform an indefinite integration

c(0)

c(t1 )

1

dc = ln

c

c(0)

= −k

t1

0

dt = −kt1 .

The limits on the two definite integrations must be done

correctly. If the lower limit of the time integration is zero,

the lower limit of the concentration integration must be the

value of the concentration at zero time. The upper limit is

similar.

Exercise 12.12. In a second-order chemical reaction

involving one reactant and having no back reaction,

dc

= kc2 .

dt

Solve this differential equation by separation of variables.

Do a definite integration from t = 0 to t = t1 .

(12.68)

x1

where

y1 = y(x1 ),

(12.69)

y2 = y(x2 ).

(12.70)

The result of this integration is an algebraic equation

that can be solved for y as a function of x.

Example 12.9. In a ﬁrst-order chemical reaction with

no back reaction, the concentration of the reactant is

governed by

dc

− = kc,

dt

where c is the concentration of the single reactant, t is the

time, and k is a function of temperature called the rate

constant. Solve the equation to find c as a function of t.

We divide by c and multiply by dt to separate the

variables:

1

1 dc

dt = dc = −k dt.

c dt

c

If you are faced with a differential equation and if you

think that there is some chance that separation of variables

will work, try the method. If it doesn’t work, you haven’t

lost very much time.

12.5 EXACT DIFFERENTIAL EQUATIONS

Sometimes an equation can be manipulated into the Pfafﬁan

form:

M(x,y)dx + N (x,y,)dy = 0.

(12.72)

Some such differential forms are exact, which means that

they are differentials of functions, as explained in Chapter

8. Other differentials are inexact , which means that they are

not differentials of any function. If the differential is exact,

the equation is called an exact differential equation. The

test for exactness is based on the Euler reciprocity relation,

as in Eq. (8.29): If

∂M

∂y

=

x

∂N

∂x

,

y

(12.73)

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Mathematics for Physical Chemistry

then the differential is exact. If the differential equation is

exact, there is a function f(x,y) such that

d f = M(x,y)dx + N (x,y)dy = 0,

(12.74)

f (x,y) = x 2 y = k,

(12.75)

where k is a constant. We solve this algebraic equation to

obtain our general solution:

which implies that

f (x,y) = k,

where k is a constant, because a constant function has a

differential that vanishes. This equation can be solved for

y in terms of x, providing a solution to the differential

equation.

In Chapter 9 we discussed the procedure for finding the

function in Eq. (12.75) by using a line integral,

f (x1 ,y1 ) = f (x0 ,y0 ) +

df,

(12.76)

C

where C represents a curve beginning at (x0 ,y0 ) and ending

at (x1 ,y1 ). A convenient curve is the rectangular path from

(x0 ,y0 ) to (x1 ,y0 ) and then to (x1 ,y1 ). On the first part of

this path, y is constant at y0 , so the dy integral vanishes and

y is replaced by y0 in the dx integral. On the second part of

the path, x is constant at x1 , so the dx integral vanishes and

x is replaced by x1 in the dy integral:

x1

f (x1 ,y1 ) = f (x0 ,y0 ) +

M(x,y0 )dx

x0

+

y1

N (x1 ,y)dy.

where k is a constant. We drop the subscripts on x1 and y1

and consider x0 and y0 to be constants:

(12.77)

y0

y = k/x 2 .

Some condition would have to be specified to obtain the

value of k.

Exercise 12.13. Solve the equation (4x +y)dx +x dy = 0.

12.6 SOLUTION OF INEXACT

DIFFERENTIAL EQUATIONS USING

INTEGRATING FACTORS

If we have an inexact differential equation

M(x,y)dx + N (x,y)dy = 0,

we cannot use the method of the previous section. However,

some inexact differentials yield an exact differential when

multiplied by a function known as an integrating factor. If

the function g(x, y) is an integrating factor for the differential

in Eq. (12.78), then

g(x,y)M(x,y)dx + g(x,y)N (x,y)dy = 0

Both integrals are now ordinary integrals. If we can perform

the integrations, we will have an algebraic equation that

can be solved for y as a function of x. The solution will

contain an arbitrary constant, because the same constant can

be added to f (x1 ,y1 ) and f (x0 ,y0 ) in Eq. (12.77) without

changing the equality.

Example 12.11. Solve the differential equation

y

dy

= .

dx

x

We convert the equation to the Pfaffian form,

2x y dx + x 2 dy = 0.

y dx − x dy = 0.

The equation is exact, because

∂ 2

(2x y) = 2x and

(x ) = 2x,

∂y

∂x

so that the Pfaffian form is the differential of a function

f = f (x, y).We do a line integral from (x0 ,y0 ) to (x1 ,y0 ) and

then to (x0 ,y1 ):

=

=

x1

2x y0 dx +

x0

y0 x12

x12 y1

y0 x02

x02 y0

(12.79)

is an exact differential equation that can be solved by the

method of the previous section. A solution for Eq. (12.79)

will also be a solution for Eq. (12.78).

Example 12.10. Solve the differential equation

f (x1 ,y1 ) − f (x0 ,y0 ) =

(12.78)

+

y1

x12 dy = 0

y0

x12 y1

= 0,

− x12 y0 = 0

We test for exactness:

∂y

∂y

x

∂( − x)

∂x

y

= 1,

= −1.

The equation is not exact. We now show that 1/x 2 is an

integrating factor. Multiplication by this factor gives

y

dx −

x2

1

dy = 0.

x

CHAPTER | 12 Differential Equations

151

This is an exact differential of a function f = f (x, y), since

1

,

x2

1

= 2.

x

y

∂(y/x 2 )

∂y

∂( − 1/x)

∂x

String position

given by y = y (x , t)

=

x

y

We can solve this equation by the method of the previous

section:

x1

y1 1

y0

dx

dy = 0

f (x1 ,y1 ) − f (x0 ,y0 ) = s

2

x0 x

y0 x 1

y0

y0

y1

y0

= − +

+

=0

x1

x0

x1

x1

y0

y1

=

= 0.

x0

x1

We regard x0 and y0 as constants, so that

y

y0

=

= k,

x

x0

where k is a constant. We solve for y in terms of x to obtain

the solution

y = kx.

This is a general solution, since the original equation was

first order and the solution contains one arbitrary constant.

If an inexact differential has one integrating factor, it

has been shown that it has an infinite number of integrating

factors. Unfortunately, there is no general procedure for

finding an integrating factor except by trial and error.

1/y 2

Exercise 12.14. Show that

is an integrating factor

for the equation in the previous example and show that it

x

Equilibrium position

x=0

x=L

FIGURE 12.6 A flexible string.

2. Its motion is restricted to small vibrations, so that the

string is not appreciably stretched.

3. It has a finite length and both ends of the string are fixed

in position.

Figure 12.6 depicts the string. We use the equilibrium

(straight) position of the string as our x axis and let one end

of the string be at x = 0 and the other end of the string be at

x = L. The displacement of the string from its equilibrium

position in the y direction is denoted by y(x, t) and the

displacement in the z direction is denoted by z(x,t).

The equation of motion of the string is derived by

writing Newton’s second law for a small segment of the

string and taking a mathematical limit as the length of the

segment becomes infinitesimal.5 The motions in the y and z

directions are independent of each other and can be solved

separately. The equation of motion for the y direction is

∂2 y

∂t 2

=

T

ρ

∂2 y

∂x2

= c2

∂2 y

∂x2

,

(12.80)

where T is the magnitude of the tension force on the string

and ρ is the mass of the string per unit length. We let

12.7 PARTIAL DIFFERENTIAL EQUATIONS

c=

T

.

ρ

(12.81)

Differential equations that contain partial derivatives are

called partial differential equations. These equations

involve functions of several independent variables.

12.7.2 Solution by Separation of Variables

12.7.1 Waves in a String

We seek a solution that can be written as a product of factors,

each of which depends on only one variable:

We consider the classical equation of motion of a

ﬂexible string. There are important similarities between

this equation and the Schrödinger equation of quantum

mechanics. The flexible string is a model system designed

to provide an approximate representation of a real string

such as those found in musical instruments. This flexible

string is defined as follows:

1. It is completely flexible, so that no force is required to

bend the string.

y(x,t) = ψ(x)θ (t).

(12.82)

This is called a solution with the variables separated. We

regard it as a trial solution and substitute it into the

differential equation. This method of separation of variables

is slightly different from our previous version, since we are

5 Robert G. Mortimer, Physical Chemistry, 3rd ed., Elsevier, 2008, pp.

1268–1269.

152

Mathematics for Physical Chemistry

now separating two independent variables instead of one

independent variable and one dependent variable.

Since ψ does not depend on t and θ does not depend on

x, the result of substituting the trial solution into Eq. (12.80)

is

d2 ψ

d2 θ

2

=

c

.

(12.83)

θ

(t)

ψ(x)

dt 2

dx 2

We can evaluate the constants to make the solution apply

to a specific case. We have the boundary conditions that

y(0) = 0 and y(L) = 0. If y must vanish at x = 0 and at x = L,

then ψ must vanish at these points, since the factor θ does

not depend on x:

ψ(0) = 0

(12.91)

We write ordinary derivatives since we now have functions

of only one variable. We separate the variables by dividing

both sides of Eq. (12.83 ) by the product ψ(x)θ (t). We also

divide by c2 :

ψ(L) = 0.

1 d2 θ

1 d2 ψ

=

.

2

2

c θ (t) dt

ψ(x) dx 2

(12.84)

We now use the fact that x and t are mathematically

independent variables. If we temporarily keep t fixed at

some value, the left-hand side of Eq. (12.84) is equal to a

constant. We can still allow x to vary, so the right-hand side

of the equation must be a constant function of x. We equate

it to the constant −κ 2 :

1 d2 ψ

= −κ 2 = constant.

ψ(x) dx 2

θ (t) = b1 cos (κct) + b2 sin (κct).

a1 = 0,

(12.93)

since cos (0) = 1 and sin 0 = 0. Since sin 0 = 0,

the condition of Eq. (12.91) is satisfied. Equation (12.92)

requires that

sin (κ L) = 0.

(12.94)

We know that

sin (nπ ) = 0 (n = 0,1,2, . . . ),

(12.95)

where n is some integer. Therefore,

κ=

(12.86)

We denote the constant by the symbol −κ 2 because this

choice will make κ into a real quantity. We now multiply

Eq. (12.85) by ψ(x) and multiply Eq. (12.86) by c2 θ (t).

We obtain

d2 ψ

+ κ 2ψ = 0

(12.87)

dx 2

and

d2 θ

+ κ 2 c2 θ = 0.

(12.88)

dt 2

The separation of variables is complete, and we have two

ordinary differential equations. Except for the symbols

used, both of these equations are the same as Eq. (12.12). We

transcribe the real solution to that equation with appropriate

changes in symbols:

ψ(x) = a1 cos (κ x) + a2 sin (κ x),

(12.92)

Equation (12.91) requires that

(12.85)

For the same reason, the left-hand side must be a constant

function of t, and must equal the same constant:

1 d2 θ

= −κ 2 .

c2 θ (t) dt 2

and

(12.89)

(12.90)

These are general solutions to the ordinary differential

equations of Eqs. (12.87) and (12.88), but we do not

necessarily have a general solution to our partial differential

equation, because there can be solutions that are not of the

form of Eq. (12.82).

(n = 1,2,3, . . . ).

L

(12.96)

We are not interested in the case that n = 0, because this

corresponds to a stationary string at its equilibrium position.

Negative values of n simply change the sign of the function,

which is irrelevant.

The coordinate factor ψ in our solution is now

ψ(x) = a2 sin

nπ x

.

L

(12.97)

To determine the time-dependent factor θ , we apply initial

conditions. Let us assume that at t = 0, the string is passing

through its equilibrium position, which corresponds to y = 0

for all x. If so, then b1 = 0 since cos (0) = 1. We now have

θ = b2 sin

nπ ct

L

.

The complete solution is

y(x,t) = A sin

nπ x

nπ ct

sin

L

L

,

(12.98)

where we write A = a2 b2 . The maximum amplitude

(maximum value of y) is equal to A, and another initial

condition would be required to specify its value.

We have a set of solutions, one for each value of the

integer n. Figure 12.7 shows the function ψ(x) for several

values of n. Each curve represents the shape of the string

at an instant when θ (t) = 1. At other times, the string

is vibrating between such a position and a position given

CHAPTER | 12 Differential Equations

153

When a string in a musical instrument is struck

or bowed, it will usually not vibrate according to a

single harmonic. The following Fourier series is a linear

combination that satisfies Eq. (12.80) and can represent

any possible motion of the string:

n=5

n=4

y(x,t) =

n=3

sin

n=1

+bn sin

n=2

nπ x

L

an cos

nπ ct

L

nπ ct

L

.

(12.102)

n=1

FIGURE 12.7 Standing waves in a flexible string.

by −ψ(x). There are fixed points at which the string is

stationary. These points are called nodes, and the number

of nodes other than the two nodes at the ends of the string

is equal to n − 1. A wave with stationary nodes is called a

standing wave.

We let λ represent the wavelength, or the distance for

the sine function in ψ to go through a complete period:

nλ = 2L.

(12.99)

The period of oscillation is the time required for the

sine function in the factor θ (t) to go through a complete

oscillation and return the string to its original position and

velocity, which requires the argument of the sine function

to range through 2π . If τ represents the period,

nπ cτ

2L

= 2π ; τ =

.

L

nc

This is a Fourier series in time with coefficients that

depend on position. The fact that a linear combination of

solutions can be a solution to the equation is an example

of the principle of superposition. We can regard the linear

combination as a physical representation of constructive

and destructive interference of the different harmonics.

The strengths of the different harmonics are represented by

the values of the coefficients an and bn . Different musical

instruments have different relative strengths of different

harmonics.

For a string of finite length with fixed ends, only standing

waves can occur. For an infinitely long string, traveling

waves can also occur. In a traveling wave, nodes move along

the string. A traveling wave does not generally correspond

to a solution with the variables separated. The following

wave function represents a traveling wave:

y(x,t) = A sin k(x − ct) .

Example 12.12. Show that the function in Eq. (12.103)

satisfies the wave equation:

(12.100)

The frequency ν is the reciprocal of the period:

ν=

n

nc

=

2L

2L

T

ρ

1/2

.

(12.101)

In musical acoustics, the oscillation corresponding to n = 1

is called the fundamental, that for n = 2 is the ﬁrst overtone,

and so on. The fundamental is also called the ﬁrst harmonic,

the first overtone is called the second harmonic, and

so on.

Exercise 12.15. A certain violin string has a mass per unit

length of 20.00 mg cm−1 and a length of 55.0 cm. Find the

tension force necessary to make it produce a fundamental

tone of A above middle C (440 oscillations per second =

440 s−1 = 440 Hz).

(12.103)

c2

∂2 y

∂t 2

= c2

∂2 y

∂x2

∂2 y

∂t 2

= −A(kc)2 sin k(x − ct) ,

∂2 y

∂x2

= −Ac2 k 2 sin k(x − ct) .

,

We can show that the function of Eq. (12.103) represents

a traveling wave by showing that a node in the wave moves

along the string. When t = 0, there is a node at x = 0. At

a later time, this node is located at a value of x such that

k(x − ct) is still equal to zero. At a time t, x = ct at the node,

so that the speed of the wave is equal to c. Using Eq. (14)

of Appendix B, we can show that

A sin k x − ct

= A sin kx cos (kct) − cos kx sin kct

.

(12.104)

154

Mathematics for Physical Chemistry

This equation exhibits the fact that a traveling wave is

equivalent to the sum of two standing waves that interfere

with each other constructively and destructively.

Exercise 12.16. Find the speed of propagation of a

traveling wave in an infinite string with the same mass per

unit length and the same tension force as the violin string

in the previous exercise.

12.7.3 The Schrödinger Equation

The time-dependent Schrödinger equation for a particle

moving only in the x direction is

2

∂2

+ V (x)

2m ∂ x 2

=i

,

∂t

(12.105)

where = h/2π , where h is Planck’s constant, equal to

6.6260755 × 10−34 J s, and where V (x) represents the

potential energy. We separate the variables by assuming

the product wave function

(x,t) = ψ(x)θ (t).

(12.106)

Substitution of this trial solution into the differential

equation gives

2

2m

θ (t)

∂ 2ψ

∂θ

+ V (x)θ (t)ψ(x) = i ψ(x) . (12.107)

∂x2

∂t

1 ∂ 2ψ

1 ∂θ

.

+ V (x)ψ(x) = i

2m ψ(x) ∂ x 2

θ (t) ∂t

2

(12.108)

Each side of this equation must be a constant function of

its independent variable, which we denote by E:

∂ 2ψ

2m(E − V∝

= −

ψ,

2

∂x2

iE

∂θ

= − θ.

∂t

(12.109)

(12.110)

Both of these equations are similar to the equation that

we solved for the flexible string. Equation (12.109) can be

rewritten

− 2 ∂ 2ψ

+ V (x)ψ = Eψ.

2m ∂ x 2

12.8 SOLUTION OF DIFFERENTIAL

EQUATIONS USING LAPLACE

TRANSFORMS

Some differential equations can be solved by taking the

Laplace transforms of the terms in the equation, applying

some of the theorems presented in Section 11.3 to obtain

an expression for the Laplace transform of the unknown

function, and then finding the inverse transform. We

illustrate this procedure with the differential equation for

the less than critically damped harmonic oscillator,6 Eq.

(12.43), which can be rewritten

d2 z

dt 2

ζ dz

k

ζ

k

+ z = z (2) + z (1) + z = 0,

m dt

m

m

m

(12.112)

where we use the notation z (2) for the second derivative

d2 z/dt 2 and z (1) for the first derivative dz/dt. We take the

Laplace transforms of the terms this equation, applying Eq.

(11.48) and the n = 2 version of Eq. (11.49), to express

the Laplace transforms of the first and second derivatives.

We let Z(s) be the Laplace transform of z and obtain the

algebraic equation

+

k

ζ

[s Z − z(0)] + Z = 0.

m

m

(12.113)

We solve this equation for Z:

s 2 Z − sz(0) − z (1) (0) +

Division by ψ(x)θ (t) separates the variables:

boundary conditions impose conditions such that E can take

on only specific discrete values, and we say that the energy

is quantized.

(12.111)

This is the time-independent Schrödinger equation for

motion in the x direction. It is called an eigenvalue equation,

and the quantity E is called the eigenvalue . This eigenvalue

E is a possible value of the energy. In many cases, the

Z=

sz(0) + z (1) (0) + (ζ /m)z(0)

.

s 2 + (ζ /m)s + k/m

(12.114)

When we find the inverse transform of Z, we will have our

answer. In order to match an expression for a transform

in Table 11.1, we complete the square in the denominator.

(That is, we add a term so that we have a perfect square plus

another term:)

sz(0) + z (1) (0) + (ζ /m)z(0)

s 2 + (ζ /m)s + ζ 2 /4m 2 − ζ 2 /4m 2 + k/m

(12.115)

(1)

z(0)(s + ζ /m) + z (0)

=

.

(12.116)

(s + ζ /2m)2 − ζ 2 /4m 2 + k/m

Z =

6 Erwin Kreyszig, Advanced Engineering Mathematics, 3rd ed.,

pp. 156–157, Wiley, New York, 1972.

CHAPTER | 12 Differential Equations

155

We have also expressed the numerator in terms of the

quantity that is squared in the denominator. We now make

the substitutions,

a=

ζ2

k

k

ζ

and ω2 =

− a2,

=

2

2m

m

4m

m

12.9.1 Euler’s Method

Euler’s method is simple to understand and implement, but

it is not very accurate. Consider a differential equation for

a variable x as a function of t that can be schematically

represented by

dx

= f (x,t)

dt

so that Eq. (12.115) can be written

z(0)(s + a)

z (1) (0)

+

(s + a)2 + ω2

(s + a)2 + ω2

z(0)(s + a)

az(0) + z (1) (0)

=

+

. (12.117)

(s + a)2 + ω2

(s + a)2 + ω2

Z =

We assume the case of less than critical damping, so that

ω2 is positive.

From Table 11.1, we have the inverse transforms,

L−1

L−1

s

+ ω2

ω

2

s + ω2

s2

= cos(ωt),

= sin(ωt)

with the initial condition that x(0) = x0 , a known value.

A formal solution can be written

L e

f (t) = F(s + a),

(12.118)

so that

z (1) (0) + az(0)

sin(ωt) e−at .

ω

(12.119)

Except for the symbols used for constants, this agrees with

the solution obtained earlier in the chapter.

z(t) = z(0) cos(ωt) +

Exercise 12.17. Obtain the solution of Eq. (12.112) in the

case of critical damping, using Laplace transforms.

Our discussion of the Laplace transform method for

solving differential equations suffices only to introduce the

method. The book by Kreyszig in the list at the end of the

book is recommended for further study.

12.9 NUMERICAL SOLUTION OF

DIFFERENTIAL EQUATIONS

Many differential equations occur for which no solution

can be obtained with pencil and paper. With the use of

programmable computers, it is now frequently possible to

obtain numerical approximations to the solutions of these

equations to any desired degree of accuracy.

t

x(t ) = x0 +

0

f (x,t)dt.

(12.121)

Like any other formal solution, this cannot be used in

practice, since the variable x in the integrand function

depends on t in some way that we do not yet know.

Euler’s method assumes that if t is small enough, the

integrand function in Eq. (12.121) can be replaced by its

value at the beginning of the integration. We replace t by

the symbol t and write

and from the shifting theorem of Eq. (11.47)

−at

(12.120)

x( t) ≈ x0 +

t

0

f (x0 ,0)dt = x0 + f (x0, 0) t.

(12.122)

A small value of t is chosen, and this process is repeated

until the desired value of t is reached. Let xi be the value

of x obtained after carrying out the process i times, and let

ti equal i t, the value of t after carrying out the process i

times. We write

xi+1 ≈ xi +

t f (xi, ti ).

(12.123)

Euler’s method is analogous to approximating an

integral by the area under a bar graph, except that the height

of each bar is obtained by starting with the approximate

height of the previous bar and using the known slope of the

tangent line.

Example 12.13. The differential equation for a first-order

chemical reaction without back reaction is

dc

= −kc,

dt

where c is the concentration of the single reactant and k is

the rate constant. Set up an Excel spreadsheet to carry out

Euler’s method for this differential equation. Carry out the

calculation for the initial concentration 1.000 mol l−1, k =

1.000 s−1 for a time of 2.000 s and t = 0.100 s.

156

Mathematics for Physical Chemistry

where c is the concentration of the single reactant and k

is the rate constant. Set up an Excel spreadsheet to carry

out Euler’s method for this differential equation. Carry out

✩ the calculation for the initial concentration 1.000 mol l−1 ,

k = 1.000 l mol−1 s−1 for a time of 2.000 s and for t =

Here are the numbers from the spreadsheet, using:

Time

Concentration

0.0

1

0.1

0.9

0.2

0.81

0.3

0.729

0.4

0.6561

0.5

0.59049

0.6

0.531441

0.7

0.4782969

0.8

0.43046721

0.9

0.387420489

1.0

0.34867844

1.1

0.313810596

1.2

0.282429536

1.3

0.254186583

1.4

0.228767925

1.5

0.205891132

F1 =

1.6

0.185302019

F2 =

1.7

0.166771817

1.8

0.150094635

1.9

0.135085172

2.0

0.121576655

12.9.2 The Runge–Kutta Method

Since Euler’s method is not accurate except for very

small values of t, more sophisticated methods have been

devised. One such widely used method is the Runge–Kutta

method, which is somewhat analogous to using Simpson’s

method for a numerical integration, as discussed in

Chapter 7.7

Consider again the differential equation represented by

dx

= f (x,t).

dt

In the Runge–Kutta method, Eq. (12.123) is replaced by

1

xi+1 ≈ xi + (F1 + 2F2 + 2F3 + F4 ),

6

(12.124)

where

F3 =

F4 =

t f (xi, ti ),

1

F1, ti +

2

1

t f xi + F2, ti +

2

t f xi + F3, ti + t

tf

xi +

(12.125)

t

2

t

2

.

,

(12.126)

,

(12.127)

(12.128)

✪ We do not present the derivation of this method, which is

discussed in the book by Burden and Faires listed at the end

The result of the spreadsheet calculation is

of the book.

There are also other numerical methods for solving

c(2.00 s) ≈ 0.1216 mol l−1

differential equations, which we do not discuss. The

numerical methods can be extended to sets of simultaneous

differential equations such as those that occur in the analysis

−kt

of chemical reaction mechanisms. Many of these sets of

c(2.00 s) = c(0)e

equations have a property called stiffness that makes them

−1

= (1.000 mol l )

difficult to treat numerically. Techniques have been devised

× exp −(1.000 s−1 ) 2.000 s

to handle this problem, which is beyond the scope of this

book.8

−1

= 0.1353 mol l

Using a value of

calculation is

t = 0.05 s, the result of the spreadsheet

c(2.00 s) ≈ 0.1258 mol l−1

Exercise 12.18. The differential equation for a secondorder chemical reaction without back reaction is

dc

= −kc2 ,

dt

12.9.3 Solution of Differential Equations

with Mathematica

Mathematica can solve differential equations both symbolically and numerically.

7 See the book by Burden, Faires, and Reynolds and the book by Hornbeck

listed at the end of the book.

8 C. J. Aro, Comput. Phvs. Comm. 97, 304 (1996).

CHAPTER | 12 Differential Equations

Symbolic Solution

The statement DSolve is used to carry out a symbolic

solution of a differential equation. We illustrate this with

an example.

Example 12.14. Use Mathematica to solve the differential

equation

dy

= ay(x).

dx

We enter the Mathematica statement

DSolve[y’[x]= =a y[x], y[x], x]

and press the “Enter” key. Notice how the statement is

written inside the brackets. First comes the equation, with

the first derivative denoted by y’. The double equal sign

must be used to let Mathematica know that an equation

is to be solved. We have used a blank space between the

a and the y[x] to indicate multiplication. After a comma

comes the specification of the dependent variable, y[x].

Note the use of brackets, not parentheses. The independent

variable must be included inside the brackets. After another

comma comes the statement of the independent variable.

Mathematica returns the output

Out[1]={{y[x] →

ea x

C[1]}}

Note the space between the a and the x and the space

between the exponential and the constant C[1] in the output.

The constant C[1] is to be determined by initial conditions.

An initial condition can be included in the original input

statement. For example, if y(0) = 2 , we would enter

DSolve[{y’[x]==a y[x],y[0]==2},y[x],x]

and press the “Enter” key or a “Shift-Return.” The output

would be

Out[1]={{y[x] → 2

ea x }}

Numerical Solution

Mathematica carries out numerical solutions of differential

equation for which no exact solution can be written. The

solution is given in terms of an interpolating function,

which is a table of values of the unknown function for

different values of the independent variable. The program

finds a numerical value of the function for a specific value

of the independent variable by interpolation in this table.

The statement NDSolve is used to solve the differential

equation, as in the next example:

Example 12.15. Obtain the numerical solution to the

differential equation

dy

= 2 sin (x)

dx

157

for the interval 0 < x < π with the initial condition

y(0) = 1.

We type the input

NDSolve[{y’[x]==2 sin[x], y[0]==1}, y, {x, 0, Pi}]

and press the “Enter” key. The output appears

Out[1]={{y → InterpolatingFunction[{0,3.141593}, <

>]}}

To obtain the value of the function at some value of x,

say x = 2, we type the input

y[2] /.%1

and press the “Enter” key or a “Shift-return”. The /. is the

replacement operator in Mathematica and is typed as two

characters, a forward slash and a period. The %1 means that

the output line number 1 is referred to. If the interpolating

function had been in line 3, we would have typed %3. The

output result now appears

Out[2]={3.83229}

To obtain a graph of the solution, we enter

Plot[Evaluate[y[x] /. %1], {x, 0, Pi}]

and press the “Enter” key or a “Shift-Return”. The graph

appears as the output.

PROBLEMS

1. An object moves through a fluid in the x direction. The

only force acting on the object is a frictional force that

is proportional to the negative of the velocity:

Fx = −ζ υx = −ζ

dx

dt

.

Write the equation of motion of the object. Find the

general solution to this equation and obtain the particular

solution that applies if x(0) = 0 and vx (0) = v0 =

constant. Construct a graph of the position as a function

of time.

2. A particle moves along the z axis. It is acted upon by a

constant gravitational force equal to −kmg, where k is

the unit vector in the z direction. It is also acted on by a

frictional force given by

F f = −kζ

dz

dt

,

where ζ is a constant called a “friction constant.” Find

the equation of motion and obtain a general solution.

Find z as a function of time if z(0) = 0 and vx (0) = 0.

Draw a graph of z as a function of time.

158

Mathematics for Physical Chemistry

3. An object sliding on a solid surface experiences a

frictional force that is constant and in the opposite

direction to the velocity if the particle is moving, and is

zero it is not moving. Find the position of the particle as

a function of time if it moves only in the x direction and

the initial position is x(0) = 0 and the initial velocity is

vx (0) = v0 = constant. Proceed as though the constant

force were present at all times and then cut the solution

off at the point at which the velocity vanishes. That is,

just say that the particle is fixed after this time. Construct

a graph of x as a function of time for the case that

v0 = 10.00 m s−1 .

4. A harmonic oscillator has a mass m = 0.300 kg and a

force constant k = 155 N m−1 :

(a) Find the period and the frequency of oscillation.

(b) Find the value of the friction constant ζ necessary

to produce critical damping with this oscillator.

Find the value of the constant λ1 .

(c) Construct a graph of the position of the oscillator

as a function of t for the initial conditions

z(0) = 0,vz (0) = 0.100 m s−1 .

5. A less than critically damped harmonic oscillator has a

mass m = 0.3000 kg, a force constant k = 98.00 N m−1 ,

and a friction constant ζ = 1.000 kg s−1 .

concentration in the tank, write and solve the differential

equation that governs the number of moles of solute in

the tank. The inlet pipe allows A moles per hour to flow

in and the overflow pipe allows Bn moles per hour to flow

out, where A and B are constants and n is the number

of moles of solute in the tank. Find the values of A and

B that correspond to a volume in the tank of 100.0 l, an

input of 1.000 l h−1 of a solution with 1.000 mol l−1 ,

and an output of 1.000 l h−1 of the solution in the tank.

Find the concentration in the tank after 5.00 h, if the

initial concentration is zero.

9. An nth-order chemical reaction with one reactant obeys

the differential equation

dc

= −kcn ,

dt

where c is the concentration of the reactant and k is a

constant. Solve this differential equation by separation

of variables. If the initial concentration is c0 moles per

liter, find an expression for the time required for half of

the reactant to react.

10. Find the solution to the differential equation

d2 y

dy

− 2y = −xe x .

dx 2

dx

(a) Find the circular frequency of oscillation ω and 11. Test the following equations for exactness and solve the

exact equations:

compare it with the frequency that would occur if

there were no damping.

(a) (x 2 + x y + y 2 )dx + 4x 2 − 2x y + 3y 2 dy = 0.

(b) Find the time required for the real exponential

(b) ye x dx + e x dy = 0.

factor in the solution to drop to one-half of its

(c) 2x y − cos (x) dx + x 2 − 1 dy = 0.

value at t = 0.

6. A forced harmonic oscillator with a circular frequency 12. Use Mathematica to solve the differential equation

symbolically

ω = 6.283 s−1 (frequency ν = 1.000 s−1 ) is exposed

to an external force F0 sin (αt) with circular frequency

dy

+ y cos (x) = e− sin (x) .

α = 7.540 s−1 such that in the solution of Eq. (12.63)

dx

becomes

13. Use Mathematica to obtain a numerical solution to the

z(t) = sin(ωt) + 0.100 sin αt .

differential equation in the previous problem for the

range 0 < x < 10 and for the initial condition y(0) = 1.

Using Excel or Mathematica, make a graph of z(t) for a

Evaluate the interpolating function for several values of

time period of at least 20 s.

x and make a plot of the interpolating function for the

7. A forced harmonic oscillator with mass m = 0.300 kg

range 0 < x < 10.

and a circular frequency ω = 6.283 s−1 (frequency ν =

14.

Solve

the differential equation

1.000 s−1 ) is exposed to an external force F0 exp ( −

βt) sin (αt) with α = 7.540 s−1 and β = 0.500 s−1 .

d2 y

− 4y = 2e3x + sin (x).

Find the solution to its equation of motion. Construct a

dx 2

graph of the motion for several values of F0 .

8. A tank contains a solution that is rapidly stirred, so that 15. Radioactive nuclei decay according to the same

differential equation which governs first-order chemical

it remains uniform at all times. A solution of the same

reactions, Eq. (12.71). In living matter, the isotope 14 C

solute is flowing into the tank at a fixed rate of flow,

is continually replaced as it decays, but it decays without

and an overflow pipe allows solution from the tank to

replacement beginning with the death of the organism.

flow out at the same rate. If the solution flowing in has

The half-life of the isotope (the time required for half of

a fixed concentration that is different from the initial

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