Powers D.L. Boundary Value Problems and Partial Differential Equations

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238 Chapter 3 The Wave Equation

Example.

Estimate the ﬁrst eigenvalue of



+λ

φ = 0, 0 < x < 1,

φ(0) = φ(1) =0.

Let us try y(x) = x(1 − x), which satisﬁes the boundary conditions. Then



(x) = 1 −2x and

N(y) =







(x)



dx =



(1 −2x)

dx =

D(y) =



(x) dx =



(1 −x)

dx =

Therefore, N(y)/D(y) = 10. We know, of course, that φ

(x) = sin(π x),and

N(φ

) =



cos

(πx) dx =

, D(φ

) =



sin

(πx) dx =

so N(φ

)/D(φ

) =λ

=π

< 10, conﬁrming Eq. (4).



Example.

Estimate the ﬁrst eigenvalue of

(xφ



)



+λ

φ = 0, 1 < x < 2,

φ(1) = φ(2) =0.

The integrals to be calculated are

N(y) =



x(y



)

dx, D(y) =



dx.

The tabulation gives results for several trial functions. It is known that the ﬁrst

eigenvalue and eigenfunction are



ln 2



∼

20.5423,

(x) = sin



π ln x

ln 2



The error for the best of the trial functions is about 1.44%

y(x)

√

x(2 −x)(x −1) (2 −x)(x −1)

(2 −x)(x − 1)

N(y)

D(y)

23.7500 22.1349 20.8379



3.6 Wave Equation in Unbounded Regions 239

This method of estimating the ﬁrst eigenvalue is called Rayleigh’s me thod,

and the ratio N(y)/D(y) is called the Rayleigh quotient.Insomemechanical

systems, the Rayleigh quotient may be interpreted as the ratio between poten-

tial and kinetic energy. There are many other methods for estimating eigenval-

ues and for systematically improving the estimates.

EXERCISES

1. Using Eq. (3), show that if q ≥ 0, then λ

≥0also.

2. Verify the results for at least one of the trial functions used in the second

example.

3. Estimate the ﬁrst eigenvalue of the problem



+λ

(1 +x)φ = 0, 0 < x < 1,

φ(0) = φ(1) =0.

4. Verify that the general solution of the following differential equation is

ax cos(λ/x) +bx sin(λ/x), and then solve the eigenvalue problem.



φ = 0, 1 < x < 2,

φ(1) = 0,φ(2) = 0.

5. Estimate the lowest eigenvalue of the problem in Exercise 4 using y =

(x − 1)(2 − x).

6. Estimate the lowest eigenvalue of the problem



+λ

xφ =0, 0 < x < 1,

φ(0) = 0,φ(1) = 0.

Use the trial function x

(1 −x), and minimize the Rayleigh quotient with

respect to b.

3.6 Wave Equation in Unbounded Regions

When the wave equation is to be solved for 0 < x < ∞ or for −∞ <

x < ∞, we can proceed as we did for the solution of the heat equation in these

unbounded regions. That is to say, we separate variables and use a Fourier

integral to combine the product solutions.

240 Chapter 3 The Wave Equation

Consider the problem

∂

∂x

∂

∂t

, 0 < t, 0 < x, (1)

u(x, 0) = f (x), 0 < x, (2)

∂u

∂t

(x, 0) = g(x), 0 < x , (3)

u(0, t) = 0, 0 < t. (4)

We require in addition that the solution u(x, t) be bounded as x →∞.

On separating variables, we make u(x, t) = φ(x)T(t) and ﬁnd that the fac-

tors satisfy



+λ

T =0, 0 < t,



+λ

φ = 0, 0 < x,

φ(0) = 0, |φ(x)| bounded.

The solutions are easily found to be

φ(x;λ) = sin(λx), T(t;λ) = A cos(λct) +B sin(λct),

and we combine the products φ(x;λ)T(t;λ) in a Fourier integral

u(x, t) =



∞



A(λ) cos(λct) +B(λ) sin(λct)



sin(λx) dλ. (5)

The initial conditions become

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



∞

A(λ) sin(λx) dλ, 0 < x,

∂u

∂t

(x, 0) = g(x) =



∞

λcB(λ) sin(λx) dλ, 0 < x.

Both of these equations are Fourier integrals. Thus the coefﬁcient functions

are given by

A(λ) =



∞

f (x) sin(λx) dx, B(λ) =

πλc



∞

g(x) sin(λx) dx.

It is sufﬁcient to demand that



∞

|f (x)|dx and



∞

|g(x)|dx both be ﬁnite in

order to guarantee the existence of A and B.

The deﬁciency of the Fourier integral form of the solution given in Eq. (5)

is that the formula gives no idea of what u(x, t) looks like. The d’Alembert

3.6 Wave Equation in Unbounded Regions 241

solution of the wave equation can come to our aid again here. We know that

the solution of Eq. (1) has the form

u(x, t) = ψ(x +ct) +φ(x − ct).

The two initial conditions boil down to

ψ(x) +φ(x) = f (x), 0 < x,

ψ(x) −φ(x) = G(x) + A, 0 < x.

As in the ﬁnite case, we have deﬁned

G(x) =



g(y) dy

and A is any constant.

From the two initial conditions we obtain

ψ(x) =



f (x ) +G(x) +A



, x > 0,

φ(x) =



f (x) −G(x) −A



, x > 0.

Both f and G are known for x > 0. Thus

ψ(x + ct) =



f (x +ct) + G(x +ct) + A



is deﬁned for all x > 0andt ≥0. But φ(x −ct) is not yet deﬁned for x −ct < 0.

That means that we must extend the functions f and G in such a way as to

deﬁne φ for negative arguments and also satisfy the boundary condition. The

sole boundary condition is Eq. (4), which becomes

u(0, t) = 0 =ψ(ct) +φ(−ct).

In terms of

f and

G, extensions of f and G,thisis

0 =f (ct) +G(ct) +A +

f (−ct) −

G(−ct) −A.

Since f and G are not dependent on each other, we must have individually

f (ct) +

f (−ct) = 0, G(ct) −

G(−ct) = 0.

That is,

f is f

, the odd extension of f ,and

G is G

, the even extension of G.

Finally, we arrive at a formula for the solution:

u(x, t) =



(x + ct) +G

(x +ct)





(x −ct) −G

(x −ct)



. (6)

242 Chapter 3 The Wave Equation

Figure 5 Solution of Eqs. (1)–(4) with g(x) ≡ 0. On the left are graphs of

(x + ct) (solid) and of f

(x − ct) (dashed) at the times shown. On the right are

the graphs of u(x, t) for 0 < x, made by averaging the graphs on the left.

Now, given the functions f (x) and g(x), it is a simple matter to construct f

and G

and thus to graph u(x, t) as a function of either variable or to evaluate it

for speciﬁc values of x and t. By way of illustration, Fig. 5 shows the solution of

Eqs. (1)–(4) as a function of x at various times, for f (x) as shown and g(x) ≡ 0.

Another interesting problem that can be treated by the d’Alembert method

is one in which the boundary condition is a function of time. For simplicity,

we take zero initial conditions. Our problem becomes

∂

∂x

∂

∂t

, 0 < t, 0 < x, (7)

u(x, 0) = 0, 0 < x, (8)

∂u

∂t

(x, 0) = 0, 0 < x, (9)

3.6 Wave Equation in Unbounded Regions 243

u(0, t) = h(t), 0 < t. (10)

As u is to be a solution of the wave equation, it must have the form

u(x, t) = ψ(x +ct) +φ(x − ct). (11)

The two initial conditions, Eqs. (8) and (9), can be treated exactly as in the

ﬁrst problem. Of course, G(x) ≡ 0, and the constant A, being arbitrary, may

be taken as 0. The conclusion is that

ψ(x) = 0,φ(x) = 0, 0 < x.

Because both x and t arepositiveinthisproblem,weseethatψ(x + ct) = 0

always, so Eq. (11) may be simpliﬁed to

u(x, t) = φ(x −ct). (12)

The boundary condition Eq. (10) will tell us how to evaluate φ for negative

arguments. The equation is

u(0, t) = φ(−ct) = h(t), 0 < t. (13)

We now put together what we know of the function φ:

φ(q) =







0, q > 0,



−



, q < 0.

(14)

The argument q is a dummy, used to avoid association with either x or t.Equa-

tions (12) and (14) now specify the solution u(x, t) completely.

Example.

Ta ke h(t) as shown in Fig. 6. The major steps to construct the graph of φ(q).

Note that the graph of φ for negative argument is that of h,reﬂected.Inother

words, to make the graph of φ(q), start from the graph of h(t):(1)Graphthe

even extension h

(t); (2) replace the right half (from 0 up) with 0; (3) adjust

scales so that q =−c where t =−1, etc.

The graphs in Fig. 7 show u(x, t) as a function of x forvariousvaluesoft.It

is clear from both the graphs and the formula that the disturbance caused by

the variable boundary condition arrives at a ﬁxed point x at time x/c.Thusthe

disturbance travels with the velocity c, the wave speed. An example is animated

on the CD.



A wave equation accompanied by nonzero initial conditions and time-

varying boundary conditions can be solved by breaking it into two prob-

lems, one like Eqs. (1)–(4) with zero boundary condition, and the other like

Eqs. (7)–(10) with zero initial conditions.

244 Chapter 3 The Wave Equation

Figure 6 Graphs of h(t) and φ(q) for semi-inﬁnite string with time-varying

boundary condition.

Figure 7 Graphs of u(x, t) versus x for the semi-inﬁnite string under the

time-varying boundary condition u(0, t) = h(t),withh(t) asshowninFig.6.The

shape seen in the last drawing will continue to travel to the right.

3.6 Wave Equation in Unbounded Regions 245

EXERCISES

1. Derive a formula similar to Eq. (6) for the case in which the boundary con-

dition Eq. (4) is replaced by

∂u

∂x

(0, t) = 0, 0 < t.

2. Derive Eq. (6) from Eq. (5) by using trigonometric identities for the prod-

uct sin(λx) · cos(λct), and so forth, and recognizing certain Fourier inte-

grals.

3. Sketch the solution of Eqs. (1)–(4) as a function of x at times t = 0, 1/2c,

1/c,2/c,3/c,ifg(x) = 0everywhereandf (x) is the rectangular pulse

f (x) =



0, 0 < x < 1,

1, 1 < x < 2,

0, 2 < x.

4. Same as Exercise 3, but f (x) =0andg(x) is the rectangular pulse

g(x) =



0, 0 < x < 1,

c, 1 < x < 2,

0, 2 < x.

5. Sketch the solution of Eqs. (7)–(10) at times t = 0, π/2, π ,3π/2, 2π ,5π/2,

if h(t) = sin(t).Takec = 1.

6. Sketch the solution of Eqs. (7)–(10) at times t = 0, 1/2, 3/2, and 5/2, if

c = 1and

h(t) =



0, 0 < t < 1,

1, 1 < t < 2,

0, 2 < t.

7. Use the d’Alembert solution of the wave equation to solve the problem

∂

∂x

∂

∂t

, −∞ < x < ∞, 0 < t,

u(x, 0) = f (x), −∞ < x < ∞,

∂u

∂t

(x, 0) = g(x), −∞ < x < ∞.

8. The solution of the problem stated in Exercise 7 is sometimes written

u(x, t) =



f (x +ct) + f (x − ct)





x+ct

x−ct

g(z) dz.

246 Chapter 3 The Wave Equation

Show that this is correct. You will need Leibniz’s rule (see the Appendix) to

differentiate the integral.

3.7 Comments and References

Thewaveequationisoneoftheoldestequations of mathematical physics.

Euler, Bernoulli, and d’Alembert all solved the problem of the vibrating string

about 1750, using either separation of variables or what we called d’Alembert’s

method. This latter is, in fact, a very special case of the method of characteris-

tics, in essence a way of identifying new independent variables having special

signiﬁcance. Street’s Analysis and Solution of Partial Differential Equations has

a chapter on characteristics, including their use in numerical solutions. Wan’s

Mathematical Models and Their Analysis givesapplicationstotrafﬁcﬂowand

also discusses other wave phenomena. (See the Bibliography.)

Because many physical phenomena described by the wave equation are

part of our everyday experience — the sounds of musical instruments, for

instance — they are often featured in popular expositions of mathematical

physics. The book of Davis and Hersch (The Mathematical Experience)ex-

plains standing waves (product solutions) and superposition in an elemen-

tary way. Of course, many other phenomena are described by the wave equa-

tion. Among the most important for modern life are electrical and mag-

netic waves, which are solutions of special cases of the Maxwell ﬁeld equa-

tions. These and other kinds of waves (including water waves) are studied in

Main’s Vibrations and Waves in Physics; both exposition and ﬁgures are ﬁrst

rate.

The potential difference V between the interior and exterior of a nerve axon

can be modeled approximately by the Fitzhugh–Nagumo equations,

∂V

∂t

∂

∂x

+V −

−R,

∂R

∂t

=k(V +a −bR).

Here R represents a restoring effect and a, b,andk are constants. At ﬁrst glance,

one would expect V to behave like the solution of a heat equation. But a trav-

eling wave solution,

V(x, t) = F(x −ct), R(x, t) = G(x −ct),

of these equations can be found that shows many important features of nerve

impulses. This system and many other exciting biological applications of

mathematics are reported by Murray’s excellent book, Mathematical Biology.

Miscellaneous Exercises 247

More information about the Rayleigh quotient and estimation of eigenval-

ues is in Boundary and Eigenvalue Problems in Mathematical Physics, by Sagan.

The classic reference for eigenvalues, and indeed for the partial differential

equations of mathematical physics in general, is the work by Courant and

Hilbert, Methods of Mathematical Physics.

Chapter Review

See the CD for Review Questions.

Miscellaneous Exercises

Exercises1–5refertotheproblem

∂

∂x

∂

∂t

, 0 < x < a, 0 < t,

u(0, t) = 0, u(a, t) = 0, 0 < t,

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

∂u

∂t

(x, 0) = g(x), 0 < x < a.

1. Ta ke f (x) = 1, 0 < x < a,andg(x) ≡ 0.(Thisisratherunrealis-

tic if u(x , t) is the displacement of a vibrating string.) Find a series

(separation-of-variables) solution.

2. Sketch u(x, t) of Exercise 1 as a function of x at various times throughout

one period.

3. The solution u(x, t) of Exercise 1 takes on only the three values 1, 0, and

−1. Make a sketch of the region 0 < x < a,0< t, and locate the places

where u takesoneachofthevalues.

4. Ta ke g(x) = 0andf (x) to be this function:

f (x ) =











3hx

, 0 < x <

3h(a −x)

< x < a.

The graph of f is triangular, with peak at x = 2a/3. Find a series solution

for u(x, t).

5. Sketch u(x, t) of Exercise 4 as a function of x at times 0 to a/c in steps of

a/6c.