Lecture 10

Heat equation with zero-flux endpoint conditions (cont’d)

Section 2.4 of text by Haberman

In the last lecture, we considered the 1D heat equation

∂u

∂t= k

∂2u

∂x2, (1)

with zero-flux boundary conditions

∂u

∂x(0, t) = 0,

∂u

∂xu(L, t) = 0, (2)

and initial condition

u(x, 0) = f(x), 0 ≤ x ≤ L. (3)

Recall that the method of separation of variables produced a discretely infinite set of solutions to

this problem. They have the form

un(x, t) = φn(x)Gn(t) = cos(nπx

L

)

e−k(nπ/L)2t, n = 0, 1, 2, · · · , (4)

Any linear combination of these functions, for example,

u(x, t) =

M∑

n=0

cnun(x, t)

=M∑

n=0

cn cos(nπx

L

)

e−k(nπ/L)2t, (5)

is also a solution to the heat equation satisfying the zero-flux boundary conditions. All solutions

un(x, t), with the exception of u0(x, t), tend to zero as t→ ∞. In fact,

u(x, t) → c0 as t→ ∞. (6)

It now remains to accomodate the initial condition u(x, 0) = f(x). We state here, once again

without proof, that the set of spatial functions {φn}∞

n=0 forms a complete set or basis for the space of

functions L2[0, L]. (Note that the function φ0(x) = 1 is included in this set.) This implies that there

exists a unique expansion of the form

f(x) = a0 +

∞∑

n=1

an cos(nπx

L

)

. (7)

59

The coefficients an are easily obtained from the fact that the φn also form an orthogonal set, i.e.,

∫ L

0φm(x)φn(x) dx =

∫ L

0cos

(mπx

L

)

cos(nπx

L

)

dx =

0, m 6= n,

L, m = n = 0,

L2 , m = n ≥ 1.

(8)

To determine a0 in Eq. (7), we multiply both sides with φ0(x) = 1 and integrate w.r.t. x from 0 to L:

∫ L

0f(x) dx =

∫ L

0a0 dx+

∞∑

n=1

an

∫ L

0cos

(nπx

L

)

dx

= a0L, (9)

which implies that

a0 =1

L

∫ L

0f(x) dx. (10)

To determine ak, k ≥ 1, in Eq. (7), we multiply both sides with φk(x) and integrate w.r.t. x from 0

to L:

∫ L

0f(x) cos

(

kπx

L

)

dx =

∫ L

0a0 cos

(

kπx

L

)

dx+

∞∑

n=1

an

∫ L

0cos

(

kπx

L

)

cos(nπx

L

)

dx

= akL

2. (11)

which implies that

ak =2

L

∫ L

0f(x) cos

(

kπx

L

)

dx, k ≥ 1. (12)

From the an so determined, we now construct the solution to the heat equation with zero-flux boundary

conditions and initial condition given in Eq. (1):

u(x, t) = a0 +∞∑

n=1

anun(x, t), (13)

i.e.,

u(x, t) = a0 +

∞∑

n=1

an cos(nπx

L

)

e−k(nπ/L)2t. (14)

Notice that all terms in the solution, with the exception of the a0 term, contain decaying expo-

nential terms, implying that

u(x, t) → a0 =1

L

∫ L

0f(x) dx, t→ ∞. (15)

In other words, for any initial condition u(x, 0) = f(x), the solution u(x, t) to this problem evolves to

the constant solution u(x, t) = a0. There are two noteworthy points here:

60

1. Several lectures ago, we showed that constant solutions u(x) = C represent the equilibrium or

steady-state temperature distributions for the zero-flux endpoint boundary value condition.

2. The particular value C = a0 =1

L

∫ L

0f(x) dx represents the average value of the intial tem-

perature distribution. We also showed this result in that previous lecture – it follows from the

conservation of thermal energy. The zero-flux conditions imply that no thermal energy is allowed

to flow into or flow out of the rod. Whatever the initial temperature distribution, differences in

temperature are eventually damped out as heat is transferred throughout the rod to achieve an

equal distribution throughout the rod.

Heat equation – nonhomogeneous problems

Nonhomogeneous boundary conditions

Section 8.2 of text by Haberman

Up to now, we have used the separation of variables technique to solve the homogeneous (i.e., no

sources) 1D heat equation∂u

∂t= k

∂2u

∂x2, (16)

with homogeneous boundary conditions, i.e.

u(0, t) = 0, u(L, t) = 0 (zero-temperature endpoint conditions), (17)

or∂u

∂x(0, t) = 0,

∂u

∂x(L, t) = 0 (zero-flux endpoint conditions). (18)

We simply state here that the separation of variables method can also be used for mixed homogenous

boundary conditions such as, for example,

u(0, t) = 0,∂u

∂x(L, t) = 0. (19)

They are homogeneous because of the appearance of zeros on the right hand sides.

In all of these cases, the separation of variables technique works because of the following property:

If u1 and u2 satisfy homogeneous boundary conditions, then u = c1u1 + c2u2 also satisfies

these conditions for any c1, c2 ∈ R.

61

We now consider nonhomogeneous boundary conditions of the form

u(0, t) = A, u(L, t) = B (fixed temperature endpoint conditions). (20)

The method of separation of variables cannot be applied in this case because of the lack of zeros on

the right hand sides:

If u1 and u2 satisfy the above nonhomogeneous boundary conditions, it does not follow

that c1u1 + c2u2 satisfy them for arbitrary c1, c2 ∈ R.

With a little modification of the problem, we shall be able to apply the separation of variables method.

Let us first recall that there is a unique equilibrium temperature distribution associated with the

boundary conditions in Eq. (20), namely

u(x, t) = ueq(x) = A+B −A

Lx, 0 ≤ x ≤ L. (21)

This function is a linear interpolation of the temperatures at x = 0 and x = L.

Let us also recall that the fixed temperature endpoint condition in (20) is not a zero-flux condition

so that it is possible for heat to travel either out of or into the rod through the endpoints. In a previous

lecture devoted to the idea of steady-state temperature distributions, we argued, on physical grounds,

that all solutions u(x, t) to the heat equation will eventually approach the distribution ueq(x). We

shall now show this.

First, suppose that u(x, t) is the solution to the heat equation with fixed endpoint temperature

boundary conditions in Eq. (20). Then define the following displacement from equilbrium function,

v(x, t):

v(x, t) = u(x, t) − ueq(x). (22)

We now show that we shall be able to use separation of variables to solve for v(x, t) instead of trying

to solve for u(x, t). First rewrite Eq. (22) as

u(x, t) = v(x, t) + ueq(x). (23)

Now compute the following derivatives:

∂u

∂t(x, t) =

∂v

∂t(x, t) +

∂ueq(x)

∂t=∂v

∂t(x, t),

∂2u

∂x2(x, t) =

∂2v

∂x2(x, t) +

∂2ueq(x)

∂x2=∂2v

∂x2(x, t), (24)

62

where the final result follows from the fact that ueq(x) is linear in x.

Substitution into Eq. (16) implies that v(x, t) also satisfies the heat equation, i.e.,

∂v

∂t= k

∂2v

∂x2. (25)

But note the boundary conditions satisfied by v(x, t):

v(0, t) = u(0, t) − ueq(0)

= A−A

= 0, (26)

v(L, t) = u(L, t) − ueq(L)

= B −B

= 0. (27)

In other words, v(x, t) satisfies the heat equation with homogeneous boundary conditions. This means

that we can employ separation of variables to solve for v. There is one final detail to consider, however:

By definition, the initial condition satisfied by v(x, t) is given by

v(x, 0) = u(x, 0) − ueq(x)

= f(x) − ueq(x)

= g(x). (28)

We now proceed with the separation of variables method as was done earlier for the zero-

temperature endpoint boundary conditions, using the functions

un(x, t) = φn(x)Gn(t) = sin(nπx

L

)

e−k(nπ/L)2t, n = 1, 2, · · · . (29)

First, expand g(x) = f(x) − ueq(x) as a Fourier series in the φn(x) eigenfunctions:

g(x) =

∞∑

n=1

an sin(nπx

L

)

, (30)

where

an =2

L

∫ L

0g(x) sin

(nπx

L

)

dx =2

L

∫ L

0[f(x) − ueq(x)] sin

(nπx

L

)

dx. (31)

The solution to this boundary value/initial value problem in v(x, t) then becomes

v(x, t) =

∞∑

n=1

an sin(nπx

L

)

e−k(nπ/L)2t. (32)

63

Recalling that u(x, t) = v(x, t) − ueq(x), we have the final result for u(x, t):

u(x, t) = ueq(x) +

∞∑

n=1

an sin(nπx

L

)

e−k(nπ/L)2t. (33)

Now note that all terms in the summation involve decaying exponentials. It follows that

u(x, t) → ueq(x) = A+B −A

Lx, (34)

which proves the earlier “physical” argument.

Example: We consider the particular case A = 1, B = 2, k = L = 1, with initial condition

u(x, 0) = f(x) = 5 sin(πx). (35)

The equilibrium temperature distribution is

ueq(x) = 1 + x. (36)

We must first compute the Fourier series expansion of the initial condition function in terms of the

basis set φn(x) = sin(nπx):

v(x, 0) = f(x) − ueq(x)

= 5 sin(πx) − (1 + x)

=

∞∑

n=1

an sin(nπx). (37)

It is convenient to compute first the Fourier series coefficients of ueq(x) in this basis (Exercise):

bn = 2

∫ 1

0(1 + x) sin(nπx) dx =

1

nπ[1 − 2(−1)n]. (38)

From this, we obtain the an coefficients:

a1 = 5 −6

π, (the “5” comes from f(x)),

an = −2

nπ[1 − 2(−1)n], n ≥ 2. (39)

This yields the solution for v(x, t):

v(x, t) =

∞∑

n=1

an sin(nπx)e−n2π2t, (40)

64

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

u(x)

x

u(x,0)

Solutions u(x, t) to nonhomogeneous PDE problem described in text. Solutions are plotted for tn =

n∆t, n = 0, 1, · · · 20, ∆t = 0.05.

from which is constructed the solution u(x, t):

u(x, t) = ueq(x) +∞

∑

n=1

an sin(nπx)e−n2π2t (41)

Note that u(x, t) → ueq(x) as t→ ∞.

In the figure above, solutions u(x, t) are plotted at intervals of ∆t = 0.05. M = 1000 terms were

used in the Fourier expansion for v(x, t). At t = 0, u(x, 0) = 5 sin(πx), as prescribed. For t > 0, there

is an instantaneous jump in the endpoint values of the temperature function to those of the prescribed

boundary conditions, i.e., u(0, t) = 1 and u(1, t) = 2. As t increases, we see that the solution u(x, t)

approaches the linear steady-state temperature distribution function ueq(x).

65

Lecture 11

Heat equation – nonhomogeneous problems (cont’d)

Time-independent source terms

Section 8.2 of text by Haberman, p. 350

We now continue our examination of heat flow for nonhomogeneous problems by allowing for the

existence of time-independent, or “steady”, source terms. Following our notation from the first few

lectures, the heat equation with a source term (energy density per unit time) for a homogeneous rod

(constant c, ρ and K0) is given by

cρ∂u

∂t= K0

∂2u

∂x2+Q(x). (42)

Dividing by the factor cρ yields

∂u

∂t= k

∂2u

∂x2+

1

cρQ(x), k =

K0

cρ. (43)

Note: Unfortunately, the equation presented in the textbook, cf. Pages 350 and 352,

omits the 1/(cρ) factor multiplying the source term Q(x), representing an inconsistency

with its earlier notation. In what follows, we’ll keep the above notation. In order to do

the Exercises 8.2 on p. 352 and arrive at the answers given at the back of the textbook,

however, you will have to set cρ = 1.)

We also allow for nonhomogeneous boundary conditions, in this case, fixed-temperature endpoint

conditions of the form,

u(0, t) = A, u(L, t) = B. (44)

Because of the appearance of the source term Q(x), as well as the nonzero endpoint conditions, the

separation of variables method cannot be applied directly to this problem. Instead, we proceed as

before, first checking if there exists an equilibrium temperature distribution ueq(x) for this problem.

From Eq. (43), such a distribution would have to satisfy the ODE boundary value problem,

d2u

dx2= −

Q(x)

K0, u(0) = A, u(L) = B. (45)

If a solution to this problem exists, then we may proceed as in the previous lecture by defining a

displacement solution.

66

Note: We start the above sentence with “If a solution ... exists” because it is not al-

ways guaranteed that a steady-state temperature distribution exists for nonhomoegenous

problems – see Problem 1.4.7 of the text (p. 20).

Assuming that a (unique) steady-state temperature distribution ueq(x) exists, we define the dis-

placement from equilibrium solution,

v(x, t) = u(x, t) − ueq(x), (46)

as before. (And, as before, we expect that the solution u(x, t) will approach the equilibrium solution

ueq(x) as t→ ∞. We’ll show this.) Now rewrite the above equation as

u(x, t) = v(x, t) + ueq(x), (47)

and compute the necessary derivatives for the heat equation:

∂u

∂t=∂v

∂t+ 0, (48)

and

∂2u

∂x2=

∂2v

∂x2+∂2ueq

∂x2

=∂2v

∂x2−Q(x)

K0, (49)

where we have used Eq. (45).

We now substitute these results into our problem in Eq. (43) to obtain

∂v

∂t= k

∂2v

∂x2−

k

K0Q(x) +

1

cρQ(x) = k

∂2v

∂x2. (50)

The cancellation of the term in Q(x) follows from the fact that

1

cρ=

k

K0. (51)

Thus, v(x, t) satisfies a homogeneous heat equation with no sources. And as for the boundary condi-

tions that v(x, t) will satisfy:

v(0, t) = u(0, t) − ueq(0)

= A−A

= 0, (52)

67

v(L, t) = u(L, t) − ueq(L)

= B −B

= 0. (53)

As with our previous problem, v(x, t) satisfies the homogeneous heat equation with homogeneous

boundary conditions. This means that we can employ separation of variables to solve for v.

One final detail: The initial condition satisfied by v(x, t) is given by

v(0, t) = u(0, t) − ueq(x)

= f(x) − ueq(x)

= g(x). (54)

We now proceed in the same manner as for the previous problem and expand g(x) as a Fourier series

in the basis functions associated with the zero-temperature endpoint problem, i.e., φn(x) = sin(nπx):

g(x) =∞∑

n=1

an sin(nπx

L

)

, (55)

where

an =2

L

∫ L

0g(x) sin

(nπx

L

)

dx =2

L

∫ L

0[f(x) − ueq(x)] sin

(nπx

L

)

dx. (56)

The solution to this boundary value/initial value problem in v(x, t) then becomes

v(x, t) =

∞∑

n=1

an sin(nπx

L

)

e−k(nπ/L)2t. (57)

Recalling that v(x, t) = v(x, t) − ueq(x), we have the final result for u(x, t):

u(x, t) = ueq(x) +∞∑

n=1

an sin(nπx

L

)

e−k(nπ/L)2t. (58)

Now note that all terms in the summation involve decaying exponentials. It follows that

u(x, t) → ueq(x). (59)

which supports our “physical” argument.

This concludes our treatment of the one-dimensional heat equation – for now! We’ll return to

consider more general cases, and then move on to higher-dimensional heat flow. As a motivation for

further discussion, we ask the following question:

68

What can’t we do yet?

The answer is, “A great deal”, including:

1. Time-dependent problems, for example, source terms Q(x, t) that vary in time. We’ll come back

to these problems later in the course.

2. Nonuniform rods, where the physical parameters c(x), ρ(x) and K0(x) may vary with x. We’ll

address this case in the section on “Sturm-Liouville theory”.

69

Appendix: Some additional comments (not presented in lectures)

1. “Eigenvalues” and “eigenfunctions” associated with boundary value problems

In the previous discussions, we have referred to the discretized separation constants λn and their

associated spatial solutions φn(x) as, respectively, eigenvalues and eigenfunctions. You may certainly

be wondering why they are called so, in light of your earlier experiences in linear algebra.

First, let’s recall those basic facts of linear algebra. If A is an n × n matrix, and there exists a

vector v ∈ Rn such that

Av = λv, (60)

then we say that v is an eigenvector of A with eigenvalue λ. Or, simply, v is a λ-eigenvector of A.

The important point is that A is a linear operator on Rn that maps the eigenvector v to a multiple

of itself.

In the special case that A is symmetric (also called self-adjoint), it will have n eigenvalues λn

with associated eigenvectors vn, i.e.,

Avk = λkvk, k = 1, 2, · · · n. (61)

Moreover, it is a standard result that eigenvectors corresponding to different eigenvalues are orthogonal

to each other, i.e.,

vk · vl = 0, if k 6= l. (62)

In fact, the set of eigenvectors {vk} forms an orthogonal basis for Rn. This implies that any vector

u ∈ Rn may be written as a unique linear combination of the vk:

u = c1v1 + c2v2 + · · · cnvn. (63)

As discussed previously, the coefficients ck are given by

ck =u · vk

vk · vk. (64)

Let us now return to the boundary value problem examined earlier,

d2φ

dx2+ λφ = 0, φ(0) = φ(L) = 0. (65)

We may rewrite this BVP as follows,

Lφ = λφ, φ(0) = φ(L) = 0, (66)

70

where L is the linear differential operator L = −d2

dx2. (At first, you may be somewhat puzzled by

the appearance of the minus sign in the definition of L, but that is just how things work.) In other

words, the boundary value problem of Eq. (65) has been rewritten as an eigenvalue problem in Eq.

(66) – we are looking for eigenfunctions φ of the linear differential operator L. And, as we know, the

eigenvalues and associated eigenfunctions of L are:

λk =

(

kπ

L

)2

, φk(x) = sin

(

kπx

L

)

, k = 1, 2, · · · . (67)

In a manner quite analogous to the case of eigenvalues/eigenvectors in Rn, the eigenfunctions

φk(x) form an orthonormal set of functions on [0, L]:

∫ L

0φk(x)φl(x) dx = 0, if k 6= l. (68)

Moreover, as stated in a previous lecture (but not proved!) they form a complete basis set of functions

in the space L2[0, L]: Any function f ∈ L2[0, L] may be expressed uniquely in terms of the φk as

follows:

f = c1φ1 + c2φ2 + · · · . (69)

And the coefficients ck are determined by appropriate scalar products:

ck =

∫ L0 f(x)φk(x) dx

∫ L0 φk(x)φk(x) dx

, k = 1, 2, · · · . (70)

We’ll see in a later discussion of “Sturm-Liouville theory” that the above results may be extended

to a general class of second-order linear differential operators with associated boundary conditions.

This class of operators includes the differential operators that arise from the application of the sep-

aration of variables method to Laplace’s equation in one, two and higher dimensions. (The example

given above corresponds to one dimension.) This explains the importance of Sturm-Liouville theory.

2. The “separation of variables method”: Does it really work, and why?

Regarding the question, “Does it really work:” A few people have asked me if the separation of

variables method actually generates the entire solution of the PDE, i.e., is it possible that there are

solutions to the PDE that do not have the form of a separated solution, i.e., u(x, t) cannot be expressed

in the form φ(x)G(t).

Recall that the S of V method yields individual “separated” or product solutions of the form

un(x, t) = φn(x)Gn(t), n = (0), 1, 2, · · ·. Moreover, in the case of homogeneous boundary conditions,

71

any linear combination of these functions is also a solution. The fact that such linear combinations can

generate all solutions to the PDE is due to the completeness of the orthogonal set of eigenfunctions

φn(x), as discussed above. Given any initial condition

u(x, 0) = f(x), (71)

(well, not any one – the function f(x) must be an element of the space L2([0, L]), but that space will

do) there is a unique set of coefficients an such that the series

u(x, t) =∞

∑

n=1

anun(x, t) (72)

satisfies the initial condition. For this reason, we know that such series solutions can produce any

solution to the PDE BVP/IP.

Just as a final note: In general, the function u(x, t) defined by the series solution in (72) is not

a separable function, i.e., it cannot be expressed in the form u(x, t) = φ(x)G(t), even though its

components are separable functions.

3. Separation of variables and the Schrodinger equation of quantum mechanics

Some of you have already seen the Schrodinger equation (SE) of quantum mechanics. (To those of

you who haven’t: don’t worry, you can skip this section without any negative consequences!) For

simplicity, we consider only the one-dimensional, time-dependent, SE, having the form

−~

i

∂ψ

∂t= Hψ. (73)

Here, ψ(x, t) is the “wavefunction” of the system which, according to quantum mechanical orthdoxy,

contains all of the information of the system. The constant ~ = h/(2π), where h is Planck’s constant,

the fundamental unit of quantum mechanics. Finally, H is the “Hamiltonian” operator associated

with the quantum mechanical system of concern. For a quantum mechanical particle with mass m in

a potential energy field V (x), the Hamiltonian assumes the form

H = −~

2

2m

∂2

∂x2+ V (x, t). (74)

The first component of this operator, the second-order partial differential operator in x, represents

the kinetic energy operator. The second component, V (x, t), is the potential energy operator, a

multiplicative factor. The result is that Eq. (73) is a second order partial differential equation in the

function ψ(x, t).

72

Indeed, if we write out Eq. (73), using the form of the Hamiltonian in (74), we obtain

−~

i

∂ψ

∂t= −

~2

2m

∂2

∂x2ψ + V (x, t)ψ. (75)

The fact that the time derivative is first order and the spatial derivative is second order might suggest

that this is something like a heat equation. The problem is that there is an imaginary factor on the

LHS. As such, solutions will be oscillatory, instead of damped exponential, as we show below.

In order to provide solutions to Eq. (73) we employ – guess what? – a separation of variables

method. In what follows, to make things easier, we assume that the potential V is time-independent,

i.e., V = V (x). This implies that the Hamiltonian operator H in (74) is only position-dependent.

Now assume a solution of the form

ψ(x, t) = φ(x)G(t). (76)

Substitution into (73) yields

−~

iφ(x)G′(t) = [Hφ(x)]G(t). (77)

For the moment, we have not written out the action of H on φ

Now separate the variables, i.e.,

−~

i

G′(t)

G(t)=Hφ(x)

φ(x)= E, (78)

where E is the separation constant. This leads to the two equations,

Hφ(x) = Eφ(x), (79)

and

G′(t) +iE

~G(t) = 0. (80)

Eq. (79) is known as the “time-independent Schrodinger equation. It is an eigenvalue problem in φ(x)

and E. In general, there will be an infinite set of eigenfunctions φn(x) with associated eigenvalues En,

i.e.,

Hφn = Enφn. (81)

The eigenvalues En represent allowable energies of the system.

The time-dependent functions Gn(t) associated with these eigenvalues are

Gn(t) = e−iEnt

~ . (82)

As a result, the separation of variables method yields an infinite set of solutions of the form

ψn(x, t) = φn(x)e−iEnt

~ . (83)

73

Once again, any linear combination of these solutions will be a solution ... .

Finally, what does the wavefunction tell us? To make a long story short, according to “quantum

orthodoxy,” the quantity ‖ψ(x, t)‖dx represents the probability of finding the particle in an infinites-

imal element dx centered at x at time t.

Example: The “particle in a box” model. This is one of the simplest models that can be studied:

A particle moving inside an infinitely-high potential well of the form,

V (x) =

0, 0 ≤ x ≤ L

∞, otherwise.(84)

The particle is confined to the region [0, L] because of the infinitely-high value of the potential outside

this region. (If V (x) were finite outside [0, L], then there would still be a probability of finding the

particle there, since it could “tunnel” through the potential well.) This implies that ψ(x, t) = 0 outside

the interval [0, L]. By continuity, it must also vanish at the endpoints of the interval. In other words,

it must satisfy the boundary conditions,

ψ(0, t) = 0, ψ(L, t) = 0. (85)

(Sound familiar?)

From Eq. (74), the hamiltonian for this system is simply

H = −~

2

2m

∂2

∂x2, 0 ≤ x ≤ L, (86)

since V (x) = 0 on this interval. Proceeding with the separation of variables method, the time-

independent Schrodinger equation in (79) becomes

−~

2

2mφ′′(x) = Eφ(x), 0 ≤ x ≤ L, (87)

with boundary conditions

φ(0) = 0, φ(L) = 0. (88)

We may rearrange Eq. (87) as follows,

φ′′(x) +2mE

~2φ(x) = 0, (89)

This has the same form as Eq. (65), with λ =2mE

~2. As a result, the eigenvalues Ek are given by

λk =2mEk

~2=

(

kπ

L

)2

, k = 1, 2, · · · , (90)

74

implying that the eigenvalues/energies are given by

Ek =~

2

2m

(

kπ

L

)2

. (91)

And the associated eigenfunctions are given, as before, by

φk(x) = sin

(

kπx

L

)

. (92)

The net result of this exercise is that the “natural states” of the particle-in-a-box model are discrete

states with energies Ek. (They are identical to the normal modes of a clamped string, as you’ll see in

the next section.) This is in contrast to classical mechanics, where a particle can assume a continuum

of possible energies E as it from one wall to another, hitting the wall and bouncing back with the

same energy (assuming no friction). As mentioned earlier, the discrete nature of the eigenfunctions,

i.e., their “quantization,” is brought to you by the boundary value condition φ(0) = φ(L) = 0.

The particle-in-a-box model may seem like a quite contrived model. It is actually important in

problems involving confinement, e.g. the confinement of nucleons (protons and neutrons) in atomic

nuclei. Years ago, however, (for example, when this instructor was studying quantum mechanics as

an undergraduate), it didn’t seem likely that this model would be experimentally realizable. Since

those prehistoric days, the situation has changed immensely: It is possible to produce large potential

wells in materials that can be used to confine one or more quantum particles. These are known

as “quantum dots.” They are already important in applications and may also prove very useful to

“quantum computers.”

75

Wave Equation: The Vibrating String

Section 4.2 of text by Haberman

Consider a thin, tightly-stretched string that assumes a horizontal position at equilibrium. This

equilibrium position will define the x-axis in a coordinate system – a point P will be characterized by

its x-coordinate, say 0 ≤ x ≤ L. Now suppose that this string has been displaced from its equilibrium

position, perhaps by plucking. Here we seek to model the time-evolution of such displacements. We

shall assume that the displacement of the string is purely vertical – in other words, a point P on the

string moves only up and down. This is a reasonable assumption for small displacements.

In what follows, we let u(x, t) represent the vertical position at time t of a point P with position

coordinate x. We shall formulate PDE for the time evolution of u which is based upon an examination

of the forces that act on the string.

y = u(x, t)

xx x + ∆x

y

Consider a segment of the string that lies between positions x and x + ∆x, where ∆x is small.

The mass of this segment is

∆m ≈ ρ0(x)∆x, (93)

where ρ0(x) is the lineal mass density at x (mass per unit length). The forces that act on this segment

can be classified into two groups:

1. Body forces: These are external forces, for example, gravity. We’ll let Q(x, t) represent the

net external force per unit mass, i.e., the acceleration, acting on the string.

2. Internal forces: These are due to the stretching of the string.

We now examine the internal forces in more detail. Firstly, we assume perfect flexibility of the

string, i.e., there is no resistance to bending. This implies that the internal, stretching forces are

directed tangentially. If there were resistance to bending, then non-tangential forces would have to be

76

exerted in order for the string to assume a non-horizontal shape. The tangential force at a point is

known as the tension T(x, t) – it is a vector.

u(x, t)

x x + ∆x

T(x + ∆x, t)

T(x, t)

-T(x, t)

θ(x, t)

θ(x + ∆x, t)

At each point of the string, there are two tangential forces, T(x, t) and −T(x, t) that cancel each

other – we’ll adopt the convention that T(x, t) points to the right. The net force that acts on the

segment will be determined by the tensions at the endpoints:

F = T(x+ ∆x, t) − T(x, t). (94)

Since we are considering only vertical displacements, we assume that the horizontal component of this

vector difference is zero. The vertical component of this difference, i.e., the net vertical force on the

segment, will be given by

T (x+ ∆x, t) sin θ(x+ ∆x, t) − T (x, t) sin θ(x, t), (95)

where T denotes the magnitude of the tension and θ(x, t) denotes the angle between the tangent to

the string at x and the horizontal, −π/2 ≤ θ ≤ π/2.

Where there is a net nonzero force, there will be an acceleration according to Newton’s second

law, F = ma. The vertical acceleration of the segment will be given by∂2u

∂t2. Recalling that the mass

of the segment is given by Eq. (93), Newton’s equation ma = F translates to

ρ0(x)∆x∂2u

∂t2≈ T (x+ ∆x, t) sin θ(x+ ∆x, t) − T (x, t) sin θ(x, t) + ρ0(x)∆xQ(x, t), (96)

There is still a little work to be done to arrive at the wave equation.

77

Lecture 12

Wave equation: vibrating string (cont’d)

Sections 4.2-4.4 of text by Haberman

We pick up where we left off in the last lecture. Applying Newton’s second law, F = ma, to the

segment of the string situated at [x, x+ ∆x] led to the expression

ρ0(x)∆x∂2u

∂t2≈ T (x+ ∆x, t) sin θ(x+ ∆x, t) − T (x, t) sin θ(x, t) + ρ0(x)∆xQ(x, t), (97)

where we have added the external or body forces acting on the segment. Division by ∆x yields

ρ0(x)∂2u

∂t2≈T (x+ ∆x, t) sin θ(x+ ∆x, t) − T (x, t) sin θ(x, t)

∆x+ ρ0(x)Q(x, t). (98)

We now claim that taking the limit ∆x→ 0 yields the following PDE in u(x, t):

ρ0(x)∂2u

∂t2=

∂

∂x(T (x, t) sin θ(x, t)) + ρ0(x)Q(x, t). (99)

This is still not satisfactory, since we have two unknowns – u(x, t) and θ(x, t). They are connected,

however, as follows: The slope of the tangent at x is

tan θ =∂u

∂x. (100)

For small oscillations, cos θ ≈ 1, implying that

∂u

∂x= tan θ ≈ sin θ. (101)

Eq. (99) then becomes

ρ0(x)∂2u

∂t2=

∂

∂x

(

T∂u

∂x

)

+ ρ0(x)Q(x, t). (102)

This is our general one-dimensional wave equation, which allows for the variation in mass density and

tension along the string. We’ll now make some additional simplifications in order to come up with a

simpler PDE. First, we’ll assume homogeneity of the string, i.e., ρ0(x) = ρ0 constant.

Secondly, we’ll assume that T (x, t) = T0 constant, i.e., constant tension throughout the string.

This is due to homogeneity plus the additional assumption that the string is perfectly elastic and

tightly stretched so that variations in T are negligible. (It is the variations in the direction of the

tension vector that are responsible for motion.)

With these assumptions, the above PDE becomes

∂2u

∂t2=T0

ρ0

∂2u

∂x2+Q(x, t). (103)

78

We now assume that there are no external forces except gravity acting on the string. And if the string

is so tightly stretched that its equilibrium position is horizontal, then the gravity term can be ignored

– there are no net forces on the string when it is in its horizontal equilibrium position. Thus we set

Q = 0 and the above equation becomes

∂2u

∂t2= c2

∂2u

∂x2, c =

√

T0

ρ0. (104)

This is known as the one-dimensional wave equation.

Let’s examine the dimensionality of c: Since T0 is tension, i.e., force (mass × acceleration), and ρ

is mass per unit length, we have[

T0

ρ0

]

=ML

T 2·L

M=L2

T 2. (105)

This implies that c has the dimensions of velocity. We’ll see below that this velocity is important in

the solutions of the wave equation.

We now provide some solutions to the wave equation (104). First, however, we’ll need to specify

a sufficient number of conditions in order to be able to extract a unique solution. As with the heat

equation, the second-order derivative in x will require two conditions, normally boundary conditions

at the ends of the string. There are some very interesting physical possibilities here, and we refer the

reader to Section 4.3, “Boundary Conditions”, of the textbook for a discussion. In what follows, we

shall examine the simplest, and perhaps most common condition, that of a string of length L and fixed

ends (discussed in some detail in Section 4.4 of text). This would be the situation of a guitar or violin

string – ignoring the fret. These boundary conditions will then take the form

u(0, t) = 0, u(L, t) = 0. (106)

As for the time variable t, we now have a second-order derivative in t, implying that we shall need

two initial conditions on u(x, t). We shall assume that the initial position and initial velocity of each

segment of the string is prescribed, i.e.,

u(x, 0) = f(x),

∂u

∂t(x, 0) = g(x), 0 ≤ x ≤ L. (107)

Recall that the second-derivative in t came from the acceleration term in Newton’s equation. In

particle mechanics problems, a knowledge of the initial position and velocity (or momentum) of a

79

particle is sufficient to determine a unique trajectory of the particle as dictated by Newton’s second

law.

The wave equation and boundary conditions are linear and homogeneous, which means that we

can try to use the method of separation of variables. As for the heat equation, we’ll look for solutions

of the form

u(x, t) = φ(x)G(t). (108)

(For some reason, the textbook switches to using h(t) for the time-dependent part.) Substitution into

Eq. (104) yields

φ(x)d2G

dt2= c2G(t)

d2φ(x)

dx2, (109)

or simply

φ(x)G′′(t) = c2φ′′(x)G(t). (110)

We “separate the variables”, putting the c2 term with the t-dependent part:

G′′

c2G=φ′′

φ= µ = −λ. (111)

We’ve also introduced the separation constants µ and −λ since the LHS of the equation is solely t-

dependent while the other side is solely x-dependent. For convenience, we’ll use −λ for the separation

constant since the spatial equation for φ will be identical to that of the heat equation, and we already

know that those eigenvalues were positive. The resulting separated equations for φ and G are

φ′′ + λφ = 0, φ(0) = φ(L) = 0.

G′′ + λc2G = 0. (112)

We know that solutions to the boundary value problem for φ(x) exist only for λ > 0. In this case,

we have an infinite set of discrete eigenvalues λn,

λn =(nπ

L

)2, n = 1, 2, · · · , (113)

with associated eigenfunctions,

φn(x) = sin(nπx

L

)

. (114)

From these values of λn, we find the solutions to the corrresponding equations for G(t),

G′′ + λnc2G = 0, (115)

80

to be

Gn(t) = C1 cos(√

λnct) + C2 sin(√

λnct),

= C1 cos

(

nπct

L

)

+ C2 sin

(

nπct

L

)

. (116)

As a result, the product solutions u(x, t) yielded by the separation variables technique, cf. Eq. (108),

are given by

un(x, t) = φn(x)Gn(t) = sin(nπx

L

)

[

an cos

(

nπct

L

)

+ bn sin

(

nπct

L

)]

. (117)

Any finite linear combination of the un(x, t) is also a solution to the wave equation with fixed-end

boundary conditions. We’ll examine these solutions in more detail later.

In order to accomodate the initial two conditions, we shall generally have to resort to infinite

series in the un, i.e.,

u(x, t) =∞∑

n=1

un(x, t)

=

∞∑

n=1

[

an sin(nπx

L

)

cos

(

nπct

L

)

+ bn sin(nπx

L

)

sin

(

nπct

L

)]

. (118)

From the first initial condition u(x, 0) = f(x), we have

f(x) =∞∑

n=1

an sin(nπx

L

)

, (119)

which is the Fourier expansion of f(x), as encountered with the heat equation. From the second initial

condition∂u

∂t(x, 0) = g(x), we have

g(x) =∞∑

n=1

bnnπc

Lsin

(nπx

L

)

, (120)

which may also be viewed as a Fourier expansion of g(x) with coefficients bnnπc

L.

From Eq. (119), we have, as for the heat equation,

an =2

L

∫ L

0f(x) sin

(nπx

L

)

dx. (121)

From Eq. (120), we have

bnnπc

L=

2

L

∫ L

0g(x) sin

(nπx

L

)

dx, (122)

or

bn =2

nπc

∫ L

0g(x) sin

(nπx

L

)

dx. (123)

81

So, in principle, we have now solved the 1D wave equation with fixed ends and two initial conditions.

Let us now return to the individual product solutions

un(x, t) = φn(x)Gn(t) = sin(nπx

L

)

[

an cos

(

nπct

L

)

+ bn sin

(

nπct

L

)]

. (124)

Each of these solutions is called a normal mode of vibration. The spatial portion, sin(nπx/L) defines

the profile of the normal mode – a wave with nodes at the endpoints. The time-dependent portion is

oscillatory – the frequency of this oscillation (number of oscillations in 2π units of time) is

ωn =nπc

L=nπ

L

√

T0

ρ0, n = 1, 2, · · · . (125)

This oscillatory time-dependence modulates the profile of the wave.

The frequencies ωn are the natural frequencies of the vibrating string. In practial applications,

frequencies are expressed in cycles per second – in these units, the natural frequencies are

νn =ωn

2π=nc

2L=

n

2L

√

T0

ρ0, n = 1, 2, · · · cycles/second (cps) or “Hertz” (Hz). (126)

The lowest frequency ω1 = πc/L is called the first harmonic or fundamental. All others are

multiples of ω1. Note that ω1 may be increased/decreased by increasing/decreasing the tension T0.

Each normal mode may also be written in the form (Exercise),

un(x, t) = sin(nπx

L

)

An sin

(

nπct

L+ σn

)

, (127)

where the amplitude and phase of the time-oscillation are given by, respectively,

An =√

a2n + b2n, tanσn =

an

bn. (128)

The time-oscillation of a normal mode may be viewed as a standing wave – the wave is standing,

or stationary, because its nodes, including the ones at the endpoints are fixed in time. However, each

standing wave may be expressed as a sum of two travelling waves. This is possible from the addition

law for sin and cos:

sin(nπx

L

)

sin

(

nπct

L

)

=1

2cos

nπ

L(x− ct) −

1

2cos

nπ

L(x+ ct), (129)

and

sin(nπx

L

)

cos

(

nπct

L

)

=1

2sin

nπ

L(x− ct) +

1

2sin

nπ

L(x+ ct). (130)

82

The terms with (x− ct) represent waves that are travelling to the right with velocity c – to see this,

examine the position of x(t) for which the argument x− ct is zero:

x(t) = ct, (131)

so that x(t) is increasing linearly in time.

Likewise, the terms with (x+ ct) represent waves that are travelling to the left with velocity c.

In fact, it is rather easy to show (via Chain Rule – exercise) that

u(x, t) = f(x− ct) + g(x+ ct) (132)

is a solution to the 1D wave equation for any functions f(x) and g(x).

83