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

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178 Chapter 2 The Heat Equation

is called a regular Sturm–Liouville problem if the following conditions are ful-

ﬁlled:

a. s(x), s



(x), q(x),andp(x) are continuous for l ≤x ≤r;

b. s(x)>0andp(x)>0 for l ≤ x ≤r;

c. The α’s and β’s are nonnegative, and α

+α

> 0, β

+β

> 0.

d. The parameter λ occurs only where shown.

Condition a and the ﬁrst condition b guarantee that the differential equation

has solutions with continuous ﬁrst and second derivatives. Notice that s(l) and

s(r) must both be positive (not zero). Condition c just says that there are two

boundary conditions: α

+ α

= 0onlyifα

= α

= 0, which would be no

condition. The other requirements contribute to the desired properties in ways

that are not obvious.

We are now ready to state the theorems that contain necessary information

about eigenfunctions.

Theorem 1. The regular Sturm–Liouville problem has an inﬁnite number of

eigenfunctions φ

,φ

,...,eachcorrespondingtoadifferenteigenvalueλ

,λ

,....

If n = m, the eigenfunctions φ

and φ

are o rthogonal with weight function p(x):



(x)φ

(x)p(x) dx =0, n =m.



The theorem is already proved, for the continuity of coefﬁcients and eigen-

functions makes our previous calculations legitimate. It should be noted that

any constant multiple of an eigenfunction is also an eigenfunction; but aside

from a constant multiplier, the eigenfunctions of a Sturm–Liouville problem

are unique.

A number of other properties of the Sturm–Liouville problem are known.

We summarize a few here.

Theorem 2. (a) The regular Sturm–Liouv i lle problem has an inﬁnite number of

eigenvalues, and λ

→∞as n →∞.

(b) Iftheeigenvaluesarenumberedinorder,λ

<λ

< ···, then the eigen-

function corresponding to λ

has exactly n − 1 zeros in the interval l < x < r

(endpoints excluded).

, α

, β

, are all greater than or equal to zero, then all

theeigenvaluesarenonnegative.



Examples.

1. We note that the eigenvalue problems in Sections 3–6 of this chapter are

all regular Sturm–Liouville problems, as is the problem in Eqs. (1)–(3) of

2.7 Sturm–Liouville Problems 179

this section. In particular, the problem



+λ

φ = 0, 0 < x < a,

φ(0) = 0, hφ(a) +κφ



(a) =0

is a regular Sturm–Liouville problem, in which

s(x) = p(x) = 1, q(x) = 0,α

=1,α

=0,

=h,β

=κ.

All conditions of the deﬁnition are met.

2. A less trivial example is

(xφ



)



+λ





φ = 0, 1 < x < 2,φ(1) = 0,φ(2) = 0.

We ide nti fy s(x) = x, p(x) = 1/x , q(x) = 0. This is a regular Sturm–

Liouville problem. The orthogonality relation is



(x)φ

(x)

dx = 0, n = m.

The conclusions of Theorems 1 and 2 hold for both examples.



EXERCISES

1. The general solution of the differential equation in Example 2 is

φ(x) = c

cos



λ ln(x)



sin



λ ln(x)



Find the eigenvalues and eigenfunctions, and verify the orthogonality rela-

tion directly by integration.

2. Check the results of Theorem 2 for the problem consisting of



+λ

φ = 0, 0 < x < a,

with boundary conditions

a. φ(0) = 0, φ(a) = 0; b. φ



(0) = 0, φ



(a) =0.

In case b, λ

=0.

3. Find the eigenvalues and eigenfunctions, and sketch the ﬁrst few eigenfunc-

tions of the problem



+λ

φ = 0, 0 < x < a,

180 Chapter 2 The Heat Equation

with boundary conditions

a. φ(0) = 0, φ



(a) =0,

b. φ



(0) = 0, φ(a) = 0,

c. φ(0) = 0, φ(a) + φ



(a) =0,

d. φ(0) −φ



(0) =0, φ



(a) =0,

e. φ(0) −φ



(0) =0, φ(a) + φ



(a) =0.

4. In Eqs. (1)–(3), take l =0, r = a, and show that

a. The eigenfunctions are φ

(x) = α

cos(λ

x) +α

sin(λ

x).

b. The eigenvalues must be solutions of the equation

−tan(λa) =

λ(α

+α

)

−α

5. Show by applying Theorem 1 that the eigenfunctions of each of the follow-

ing problems are orthogonal, and state the orthogonality relation.

a. φ



+λ

(1 +x)φ = 0, φ(0) = 0, φ



(a) =0;

b. (e



)



+λ

φ = 0, φ(0) −φ



(0) = 0, φ(a) = 0;

c. φ







φ = 0, φ(1) = 0, φ



(2) = 0;

d. φ



−sin(x)φ + e

φ = 0, φ



(0) =0, φ



(a) =0.

6. Consider the problem

(sφ



)



−qφ +λ

pφ = 0, l < x < r,

φ(r) = 0,

in which s(l) = 0, s(x)>0 for l < x ≤ r,butp and q satisfy the conditions of

a regular Sturm–Liouville problem. Require also that both φ(x) and φ



(x)

have ﬁnite limits as x →l+. Show that the eigenfunctions (if they exist) are

orthogonal.

7. The following problem is not a regular Sturm–Liouville problem. Why?

Solve, and show that the eigenfunctions are not orthogonal.



+λ

φ = 0, 0 < x < a,

φ(0) = 0,φ



(a) −λ

φ(a) = 0.

2.8 Expansion in Series of Eigenfunctions 181

8. Showthat0isaneigenvalueoftheproblem

(sφ



)



+λ

pφ = 0, l < x < r,



(l) = 0,φ



(r) = 0,

where s and p satisfy the conditions of a regular Sturm–Liouville problem.

9. Find all values of the parameter µ for which there is a nonzero solution of

this problem:



+µφ =0,

φ(0) +φ



(0) = 0,φ(a) +φ



(a) =0.

One solution is negative. Does this contradict Theorem 2?

2.8 Expansion in Series of Eigenfunctions

We have seen that the eigenfunctions that arise from a regular Sturm–Liouville

problem

(sφ



)



−qφ +λ

pφ = 0, l < x < r, (1)

φ(l) −α



(l) = 0, (2)

φ(r) + β



(r) = 0(3)

are orthogonal with weight function p(x):



p(x)φ

(x)φ

(x) dx = 0, n = m, (4)

and it should be clear, from the way in which the question of orthogonality

arose, that we are interested in expressing functions in terms of eigenfunction

series.

Suppose that a function f (x) is given in the interval l < x < r and that we

wish to express f (x) in terms of the eigenfunctions φ

(x) of Eqs. (1)–(3). That

is, we wish to have

f (x) =

∞



n=1

(x), l < x < r. (5)

The orthogonality relation Eq. (4) clearly tells us how to compute the coef-

ﬁcients. Multiplying both sides of the proposed Eq. (5) by φ

(x)p(x) (where

182 Chapter 2 The Heat Equation

m is a ﬁxed integer) and integrating from l to r yields



f (x )φ

(x)p(x) dx =

∞



n=1



(x)φ

(x)p(x) dx.

The orthogonality relation says that all the terms in the series, except that one

in which n = m, must disappear. Thus



f (x)φ

(x)p(x) dx =c



(x)p(x) dx

gives a formula for choosing c

We can now cite a convergence theorem for expansion in terms of eigen-

functions. Notice the similarity to the Fourier series convergence theorem. Of

course, the Fourier sine or cosine series are series of eigenfunctions on a regu-

lar Sturm–Liouville problem in which the weight function p(x) is 1.

Theorem. Let φ

,φ

,...be eigenfunctions of a regular Sturm–Liouville problem

Eqs. (1)–(3),inwhichtheα’s and β’s are not negative.

If f (x) is sectionally smooth on the interval l < x < r, then

∞



n=1

(x) =

f (x+) +f (x−)

, l < x < r, (6)

where



f (x)φ

(x)p(x) dx



(x)p(x) dx

Furthermore, if the series

∞



n=1





(x)p(x) dx



1/2

converges, then the series Eq. (6) converges uniformly, l ≤ x ≤ r.



EXERCISES

1. Ve r i f y t h at



nπ

ln(b)



,φ

=sin



ln(x)



are the eigenvalues and eigenfunctions of

2.8 Expansion in Series of Eigenfunctions 183

(xφ



)



+λ





φ = 0, 1 < x < b,

φ(1) = 0,φ(b) = 0.

Find the expansion of the function f (x) = x in terms of these eigenfunc-

tions. To what values does the series converge at x = 1andx = b?

2. If φ

,φ

,... are the eigenfunctions of a regular Sturm–Liouville problem

and are orthogonal with weight function p(x) on l < x < r and if f (x) is

sectionally smooth, then



(x)p(x) dx =

∞



n=1

where



(x)p(x) dx

and c

is the coefﬁcient of f as given in the theorem. Show why this should

be true, and conclude that c

√

→0asn →∞.

3. Verify that the eigenvalues and eigenfunctions of the problem



)



φ = 0, 0 < x < a,

φ(0) = 0,φ(a) = 0

are



nπ



,φ

(x) = exp



−



sin



nπx



Find the coefﬁcients for the expansion of the function f (x) =1, 0 < x < a,

in terms of the φ

4. If φ

,φ

,... are eigenfunctions of a regular Sturm–Liouville problem, the

numbers

√

are called normalizing c onstants, and the functions ψ

√

are called normalized eigenfunctions. Show that



(x)p(x) dx =1,



(x)ψ

(x)p(x) dx =0, n =m.

5. Find the formula for the coefﬁcients of a sectionally smooth function f (x)

in the series

f (x) =

∞



n=1

(x), l < x < r,

where the ψ

are normalized eigenfunctions.

184 Chapter 2 The Heat Equation

Show that, for the function in Exercise 5,



(x)p(x) dx =

∞



n=1

7. What are the normalized eigenfunctions of the following problem?



+λ

φ = 0, 0 < x < 1,



(0) = 0,φ



(1) = 0.

2.9 Generalities on the Heat Conduction Problem

On the basis of the information we have about the Sturm–Liouville problem,

we can make some observations on a fairly general heat conduction problem.

We take as a physical model a rod whose lateral surface is insulated. In order

to simplify slightly, we will assume that no heat is generated inside the rod.

Since material properties may vary with position, the partial differential

equation that governs the temperature u(x, t) in the rod will be

∂

∂x



κ(x)

∂u

∂x



=ρ(x)c(x)

∂u

∂t

, l < x < r, 0 < t. (1)

Any of the three types of boundary conditions may be imposed at either

boundary, so we use as boundary conditions

u(l, t) −α

∂u

∂x

(l, t) = c

, t > 0, (2)

u(r, t) + β

∂u

∂x

(r, t) = c

, t > 0. (3)

If the temperature is ﬁxed, the coefﬁcient of ∂u/∂x is zero. If the boundary is

insulated, the coefﬁcient of u is zero, and the right-hand side is also zero. If

there is convection at a boundary, both coefﬁcients will be positive, and the

signs will be as shown.

We already know that in the case of two insulated boundaries, the steady-

state solution has some peculiarities, so we set this aside as a special case. As-

sume, then, that either α

or β

or both are positive. Finally we need an initial

condition in the form

u(x, 0) = f (x), l < x < r. (4)

Equations (1)–(4) make up an initial value–boundary value problem.

2.9 Generalities on the Heat Conduction Problem 185

Assuming that c

and c

are constants, we must ﬁrst ﬁnd the steady-state

solution

v(x) = lim

t→∞

u(x, t).

The function v(x) satisﬁes the boundary value problem



κ(x)



=0, l < x < r, (5)

v(l) −α



(l) = c

, (6)

v(r) +β



(r) = c

. (7)

Since we have assumed that at least one of α

or β

is positive, this problem

can be solved. In fact, it is possible to give a formula for v(x) in terms of the

function (see Exercise 1)



dξ

κ(ξ)

=I(x). (8)

Before proceeding further, it is convenient to introduce some new functions.

Let ¯κ, ¯ρ,and¯c indicate average values of the functions κ(x), ρ(x),andc(x).

We shall deﬁne dimensionless functions s(x) and p(x) by

κ(x) =¯κs(x), ρ(x)c(x) =¯ρ¯cp(x).

Also, we deﬁne the transient temperature to be

w(x, t) = u(x, t) − v(x).

By direct computation, using the fact that v(x) is a solution of Eqs. (5)–(7),

we can show that w(x, t) satisﬁes the initial value–boundary value problem

∂

∂x



s(x)

∂w

∂x



p(x)

∂w

∂t

, l < x < r, 0 < t, (9)

w(l , t) −α

∂w

∂x

(l, t) = 0, 0 < t, (10)

w(r, t) +β

∂w

∂x

(r, t) = 0, 0 < t, (11)

w(x, 0) = f (x) −v(x) = g(x), l < x < r, (12)

which has homogeneous boundary conditions. The constant k is deﬁned to be

¯κ/ ¯ρ¯c.

Now we use our method of separation of variables to ﬁnd w.Ifw has the

form w(x, t) = φ(x)T(t), the differential equation becomes

T(t)



s(x)φ



(x)





p(x)φ(x)T



(t),

186 Chapter 2 The Heat Equation

and, on dividing through by pφT,weﬁndtheseparatedequation

(sφ



)



pφ



, l < x < r, 0 < t.

As before, the equality between a function of x and a function of t can hold

only if their common value is constant. Furthermore, we expect the constant

to be negative, so we put

(sφ



)



pφ



=−λ

and separate two ordinary equations



+λ

kT =0, 0 < t,

(sφ



)



+λ

pφ = 0, l < x < r.

The boundary conditions, being linear and homogeneous, can also be changed

into conditions of φ.Forinstance,Eq.(10)becomes



φ(l) −α



(l)



T(t) = 0, 0 < t,

and, because T(t) ≡ 0makesw(x, t) ≡ 0, we take the other factor to be zero.

We have, then, the eigenvalue problem

(sφ



)



+λ

pφ = 0, l < x < r, (13)

φ(l) −α



(l) = 0, (14)

φ(r) + β



(r) = 0. (15)

Since s and p are related to the physical properties of the rod, they should be

positive. We suppose also that s, s



,andp are continuous. Then Eqs. (13)–(15)

comprise a regular Sturm–Liouville problem, and we know the following.

1. There is an inﬁnite number of eigenvalues

0 <λ

<λ

< ···.

2. To each eigenvalue corresponds just one eigenfunction (give or take a con-

stant multiplier).

3. The eigenfunctions are orthogonal with weight p(x):



(x)φ

(x)p(x) dx =0, n =m.

The function T

(t) that accompanies φ

(x) is given by

(t) = exp



−λ



2.9 Generalities on the Heat Conduction Problem 187

We now begin to assemble the solution. For each n = 1, 2, 3,...,w

(x, t)

=φ

(x)T

(t) satisﬁes Eqs. (9)–(11). As these are all linear homogeneous equa-

tions, any linear combination of solutions is again a solution. Thus the tran-

sient temperature has the form

w(x, t) =

∞



n=1

(x) exp



−λ



The initial condition Eq. (12) will be satisﬁed if we choose the a

so that

w(x, 0) =

∞



n=1

(x) = g(x), l < x < r.

The convergence theorem tells us that the equality will hold, except possibly at

a ﬁnite number of points, if f (x) — and therefore g(x) — is sectionally smooth.

Thus w(x, t) is the solution of its problem, if we choose



g(x)φ

(x)p(x) dx



(x)p(x) dx

Finally, we can write the complete solution of Eqs. (1)–(4) in the form

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

∞



n=1

(x) exp



−λ



. (16)

Working from the representation Eq. (16) we can draw some conclusions

about the solution of Eqs. (1)–(4).

1. Since all the λ

are positive, u(x, t) does tend to v(x) as t →∞.

2. For any t

> 0, the series for u(x, t

) converges uniformly in l ≤ x ≤ r

because of the exponential factors; therefore u(x, t

) is a continuous func-

tion of x. Any discontinuity in the initial condition is immediately elimi-

nated.

3. For large enough values of t, we can approximate u(x, t) by

v(x) +a

(x) exp



−λ



(To judge how large t might be, we need to know something about the a

and the λ

.) Because φ

(x) is of one sign on the interval l < x < r (that

is, φ

(x)>0orφ

(x)<0 for all x between l and r), the graph of our

approximation will lie either above or below the graph of v(x) but will

not cross it (provided that a

=0).