ODE Chapter 5 Sturm-Liouville Theory

Chapter 5

Sturm-Liouville Theory

5.1 Oscillation and Separation Theory

Consider the differential equation

a2(x)y′′ + a1(x)y′ + a0(x)y = 0 (5.1.1)

where a2(x) is not zero for all x ∈ [a, b] , ai(x) ∈ C[a, b]. Rewrite (5.1.1) in the form

y′′ +a1

a2

y′ +a0

a2

y = y′′ + p(x)y′ + q(x)y = 0

Definek(x) = e

∫p(s)ds,

Thend

dx(k(x)y′) + k(x)q(x)y = 0

or

(ky′)′+ g(x)y = 0 (5.1.2)

Define the differential operators

L(y) = (ky′)′+ g(x)y

M(y) = a2y′′ + a1y

′ + a0y (5.1.3)

The adjoint of M is defined by

M(y) = (a2y)′′ − (a1y)′ + a0y

1

2 CHAPTER 5. STURM-LIOUVILLE THEORY

M(y) = a2y′′ + (2a′2 − a1)y′ + (a′′2 − a′1 + a0)y.

After some manipulation it is easy to show that

vM(u)− uM(v) = [(a1 − a′2)vu+ a2(vu′ − uv′)]′.This result is called LaGrange’s identity and we rewrite it as

vM(u)− uM(v) =d

dxP (u, v).

By an integration we obtain Green’s formula ,∫ b

a

[vM(u)− uM(v)

]dx = P (u, v)|ba

If M(u) = M(u), the equation M(u) = 0 is said to be self-adjoint. Hence M(u) = 0 isself-adjoint if

a′2 = a1.

In this case Lagrange’s identity becomes

vM(u)− uM(v) = [a2(vu′ − uv′)]′ = [a2(x)N(v, u)]′

andM(u) = (a2u

′)′ + a0u

Clearly the operator L defined by (5.1.3) is self-adjoint and the discussion preceding (5.1.2)shows every general linear equation can be put into self adjoint form.

5.1.1 Separation theorems

Theorem 5.1.1. [Sturm Separation Theorem]

1. A nontrivial solution of M(y) = 0 can have at most a finite number of zeros on [a, b].

2. All zeros of a solution are simple.

3. If u1(x), u2(x) are linearly independent solutions of M(y) = 0 then between any twozeros of u1(x) there is precisely one zero of u2(x).

Proof. 1. Suppose there exists infinitely many zeros,{zn}, select a subsequence {xn} suchthat xn → x̂. Then

0 = limn→∞

y(xn) = y(x̂).

Also,

y′(x̂) = limn→∞

y(xn)− y(x̂)

xn − x̂= 0

and so y(x) = 0 by uniqueness.

5.1. OSCILLATION AND SEPARATION THEORY 3

2. The proof of this was a much earlier exercise.

3. Suppose x0, x1 are consecutive zeros of u1(x), and assume thatx0 < x1. Then u2(x0) 6=0 and u2(x1) 6= 0, or else W (u1, u2)(xi) = 0, i = 0, 1. So without loss of generalityassume u1(x) > 0, x ∈ (x0, x1), u2(x0) > 0. Now

W (u1, u2)(x0) = −u′1(x0)u2(x0)

W (u1, u2)(x1) = −u′1(x1)u2(x1)

Since u′1(x0) > 0,W (x0) < 0. Because the Wronskian of u1, u2 cannot change sign,W (x1) < 0. But u′1(x1) < 0 so this would require that u2(x1) < 0. Hence u2(x) mustvanish in (x0, x1).

If we apply the argument with the roles of u1 and u2 interchanged, we see that betweentwo consecutive zeros of u2 there must be a zero of u1(x). Hence the zeros of u1 andu2 must interface.

Roughly speaking, the Sturm Separation theorem states that linearly independent solu-tions have the same number of zeros. If we consider two different equations, for example

y′′ + y = 0, y′′ + 4y = 0

then solutions of the second equation oscillate more rapidly than those of the first. Moregenerally, Sturm Comparison theorems address the rate of oscillation of solutions of differentequations.

5.1.2 Oscillation Theory

Here we shall consider equations of the form

L(y) = (ky′)′ + gy = 0, a < x < b

where k ∈ C1(a, b), g ∈ C0[a, b], and k > 0 on [a, b].

Theorem 5.1.2. Let Li(y) = (ky′)′ + giy = 0 where g2 > g1 and g2 is not identically equalto g1 on any subinterval of (a, b). If L1(u1) = 0 and L2(u2) = 0, then between any twoconsecutive zeros of u1(x) there is a zero of u2(x).


Proof. Suppose u1(x1) = u1(x2) = 0 and u2(x) 6= 0 on (x1, x2). Without loss of generalitytake u1, u2 > 0 on (x1, x2) Lagrange’s identity gives

u2L1(u1)− u1L1(u2) =d

dx(k(x)(u2u

′1 − u1u

′2).

We also have

L1(u2)− L2(u2) = (g1 − g2)u2.

Hence

u2L1(u1)− u1[L2(u2) + (g1 − g2)u2] = (k(x)(u2u′1 − u1u

′2))′

or

k(x)(u2u′1 − u1u

′2)|x2

x1=

∫ x2

x1

(g2 − g1)u1u2dx > 0.

However, the left hand side reduces to

k(x2)u2(x2)u′1(x2)− k(x1)u2(x1)u′1(x1). (*)

Since u2(x2) ≥ 0, u′1(x2) < 0 and u2(x1) ≥ 0, u′1(x1) > 0 the above expression is nonpositiveand hence we obtain a contradiction.

Suppose that we assumed u2(x1) = 0, then in the above proof expression (∗) becomes

k(x2)u′1(x2)u2(x2).

If u2(x2) ≥ 0, then again (∗) ≤ 0 and a contradiction would be obtained. Thus Theorem5.1.2 could be restated: If k ∈ C1(a, b), g ∈ C0[a, b], k > 0 on [a, b] and u1(a) = u2(a) = 0and u1(x1) = 0, a < x1 < b, then there exists z, a < z < x1 such that u2(z) = 0. Thus u2(x)has at least as many zeros as u1(x) on [a, b].

A more general version of this theorem is

Theorem 5.1.3. Assume p, q ∈ C0[a, b] and z(x) is a non trivial solution of

z′′ + q(x)z = 0

where

z(a) = z(b) = 0.

If ∫ b

a

(p− q)z2dx ≥ 0

5.1. OSCILLATION AND SEPARATION THEORY 5

then a nontrivial solution of

y′′ + p(x)y = 0

y(a) = 0

has a zero in the interval (a, b].

Proof. Suppose y(x) 6= 0 in (a, b]. Then

z(z′′ + qz) = 0

z2

y(y′′ + py) = 0

or

zz′′ − z2y′′

y= z2(p− q)

orz

y(yz′ − zy′)′ = z2(p− q).

Now note that

limx→a

z(x)

y(x)= lim

z→a

z′(a)

y′(a)

and since z′(a) 6= 0, y′(a) 6= 0, z′, y′ ∈ C[a, b], this limit exists and is finite. Hence it makessense to write ∫ b

a

z

y(yz′ − zy′)′dx =

∫ b

a

z2(p− q)dx ≥ 0.

Now integrate by parts and since z(b) = 0, y(b) 6= 0, z(a) = y(a) = 0,

z

y(yz′ − zy′)|ba −

∫ b

a

(yz′ − zy′)(zy

)′ =

∫ b

a

z2(p− q)dx ≥ 0, (∗)

or

−∫ b

a

(yz′ − zy′)2

y2dx ≥ 0

or

0 ≥∫ b

a

(yz′ − zy′)2

y2dx.

The right hand side is identically zero if y(x) = cz(x) in which case the result is triviallytrue. So if y(x) 6= cz(x), the right hand side is positive and we get a contradiction. Hencey(x) must vanish in (a, b].


The proof shows that if p(x) 6= q(x) then∫ b

a

z2(p− q)dx > 0.

In this case y(x) must have a zero in (a, b). If not, then just as before we could derive (*) bydividing by y(x) and the boundary term in (*) would vanish since y(b) = 0, and we wouldobtain ∫ b

a

(yz′ − zy′)2

y2dx < 0,

which is a contradiction.We conclude with a generalization of these results.

Theorem 5.1.4. Let ki ∈ C1[a, b], gi ∈ C[a, b] with ki > 0. If z is a nontrivial solution of

(k1z)′ + g1z = 0

z(a) = z(b) = 0,

and y is a non trivial solution of(k2y

′)′ + g2y = 0

y(a) = 0

and ∫ b

a

(k1 − k2)(z′)2 + (g2 − g1)z2dx ≥ 0,

then y(x) has a zero in (a, b]. If the inequality is strict, the zero is in the open interval (a, b).

Proof. Multiply the first equation by z and subtract the second multiplied by (z2/y) toobtain the Picone formula,∫ b

a

(k1 − k2)(z′)2 + (g2 − g1)z2dx+

∫ b

a

k2(yz′ − zy′

y)2dx =

z

y(k1yz

′ − k2y′z)|ba

and proceed as before.

As an immediate consequence of this theorem we obtain

Theorem 5.1.5. [Sturm-Picone Theorem] Suppose

(k1z′)′ + g1z = 0

(k2y′)′ + g2y = 0

where g2 ≥ g1 and k1 ≥ k2 > 0 and g2 6≡ g1,k2 6≡ k1 on [a, b] and

z(a) = z(b) = 0.

then y(x) has a zero in (a, b).

5.2. BOUNDARY VALUE PROBLEMS 7

5.2 Boundary Value Problems

We consider the problem of solving

M(y) = a2y′′ + a1y

′ + a0y = f(x), a < x < b (5.1.4)

subject to the boundary conditions

B1(y) = α11y(a) + α12y′(a) + β11y(b) + β12y

′(b) = γ1 (5.1.5)

B2(y) = α21y(a) + α22y′(a) + β21y(b) + β22y

′(b) = γ2.

Here a2, a1, a0 ∈ C[a, b], a2(x) 6= 0, αij, βij, γi are constants. Equations (5.1.4) and (5.1.5)constitute what is called a boundary value problem(BVP). If β11 = β12 = 0 and α21 =α22 = 0, the boundary conditions are separated. If α12 = β12 = 0 and α21 = β21 = 0, andγ1 = γ2 = 0 the boundary conditions are periodic. If α12 = β11 = β12 = α21 = β21 = β22 = 0we have the initial conditions y(a) = γ1

α11, y′(a) = γ2

α21. Note that the boundary operators Bi

are linear. We refer to {f(x), γ1, γ2} as the data of the BVP.

Theorem 5.1.1. The BVP (5.1.4)-(5.1.5) with data {0, γ1, γ2} has a unique solution iff theBVP with data {0, 0, 0} has only the trivial solution.

Proof. Assume first that we know that solutions to the BVP (5.1.4), (5.1.5) are unique andsuppose y1, y2 are linearly independent solutions of (5.1.4). If u is any solution of M(y) = 0,with data {0, γ1, γ2}, then, since every solution must be a linear combination of y1 and y2,there exist unique α1, α2 so that

u = α1y1 + α2y2.

Then

α1B1(y1) + α2B1(y2) = γ1

α1B2(y1) + α2B2(y2) = γ2,

or (B1(y1) B1(y2)B2(y1) B2(y2)

)(α1

α2

)=

(γ1

γ2

)(5.1.6)

Since α1, α2 are unique,

det

(B1(y1) B1(y2)B2(y1) B2(y2)

)6= 0. (5.1.7)


Now let us suppose that w solves the boundary value problem with data {0, 0, 0} and wewrite

w = ξ1y1 + ξ2y2 (5.1.8)

then as above we get (B1(y1) B1(y2)B2(y1) B2(y2)

)(ξ1

ξ2

)=

(00

)and since the coefficient matrix is nonsingular we must have ξ1 = ξ2 = 0 by (5.1.7).

Conversely, if the BVP with data{0, 0, 0} has only the trivial solution then from (5.1.8)we conclude that (5.1.7) holds and hence (5.1.6) has a unique solution.

Theorem 5.1.2. The BVP

M(y) = f, x ∈ (a, b)

B1(y) = γ1, B2(y) = γ2

has a unique solution if the BVP with data {0,0,0} has only the trivial solution.

Proof. Let u1, u2 be linearly independent solutions of M(y) = 0 and let yp be a particularsolution of M(y) = f. Seek a solution of the BVP in the form

y = α1u1 + α2u2 + yp.

Solve for α1, α2 such that

B1(y) = α1B1(u1) + α2B1(u2) +B1(yp) = γ1

B2(y) = α1B2(u1) + α2B2(u2) +B2(yp) = γ2

or (B1(u1) B1(u2)B2(u1) B2(u2)

)(α1

α2

)=

(γ1 −B1(yp)γ2 −B2(yp)

).

Since the homogeneous problem has only the trivial solution the matrix is invertible and so(α1

α2

)=

(B1(u1) B1(u2)B2(u1) B2(u2)

)−1(γ1 −B1(yp)γ2 −B2(yp)

).

5.2. STURM-LIOUVILLE BOUNDARY VALUE PROBLEMS 9

5.2 Sturm-Liouville Boundary Value Problems

In paractice one often encounters a second order differential equation in so-called self-adjointform and generally one finds that the most common boundary conditions are either separatedor periodic. A second order operator L is in self-adjoint form if

L(y) = (ky′)′+ g(x)y.

We are particularly interested in BVP’s of the form

L(y) + λp(x)y = 0, a < x < b, (5.2.9)

B1(y) = 0, B2(y) = 0, (5.2.10)

where k, k′, g, p are real and continuous on [a, b], and k, p > 0 on [a, b]. The correpondingseparated boundary conditions are given by

B1(y) = α1y(a) + α2y′(a) = 0 (5.2.11)

B2(y) = β1y(b) + β2y′(b) = 0. (5.2.12)

The BVP (5.2.9)-(5.2.12) is called a Regular Sturm-Liouville Eigenvalue Problem. The valuesof λ for which the BVP has a nontrivial solution are called eigenvalues.

For a general Ly = a0y′′ + a1y

′ + a2y, the BVP

L(y) + λp(x)y = 0

B1(y) = 0, B2(y) = 0

is said to be self-adjoint provided∫ b

a

[uL(v)− vL(u)]dx = 0

for all u, v that satisfy the above boundary conditions (5.2.11)-(5.2.12).

Theorem 5.2.1. The BVP corresponding to the regular SLBVP

L(u) + λpu = 0

α1y(a) + α2y′(a) = 0

β1y(b) + β2y′(b) = 0

is self-adjoint.


Proof. Integration by parts yields the so-called Green’s formula∫ b

a

[vL(u)− uL(v)]dx = k(x) (u′v − v′u) |ba ≡ P (u, v)|ba (5.2.13)

If u and v satisfy the boundary conditions at x = a, then(v(a) v′(a)u(a) u′(a)

)(α1

α2

)=

(00

).

Since α21 + α2

2 6= 0, we haveu′(a)v(a)− v′(a)u(a) = 0.

In the same way we see that if u and v satisfy the conditions at x = b, then

u′(b)v(b)− v′(b)u(b) = 0.

From this we see that P (u, v)|ba = 0 and the result follows.

Note that the above proof shows, in general, that if u, v satisfy separated B.C’s, thenP (u, v)|ba = 0. The next theorem states that “eigenfunctions corresponding to differenteigenvalues are orthogonal with respect to the weight p(x).”

Theorem 5.2.2. Let (λ, u), (µ, v) denote an eigenpair of the RSLBVP,

L(y) + λp(x)y = 0

α1y(a) + α2y′(a) = 0

β1y(b) + β2y′(b) = 0.

Then

(λ− µ)

∫ b

a

u(x)v(x)p(x)dx = 0,

i.e., u and v are orthogonal with respect to the weight p.

Proof. From Green’s formula (5.2.13), we know that

(λ− µ)

∫ b

a

uvp(x)dx =

∫ b

a

[u(L(v)− v(L(u)] dx =

= k(x) (u′v − v′u) |ba = 0.

Theorem 5.2.3. The eigenvalues of the RSLBVP are real.


Proof. IfL(u) + λpu = 0,

thenL(u) + λpu = 0,

orL(u) + λpu = 0

and clearlyB1(y) = 0 = B2(y).

Thus (λ, µ) is an eigenpair and so

(λ− λ)

∫ b

a

uupdx = 0,

or

(λ− λ)

∫ b

a

|u|2pdx = 0.

Since p > 0, and |u|2 6≡ 0 (since u is an eigenfunction), we must have λ = λ.

An eigenvalue λ is said to be simple if the dimension of the null space of Lλ is one, i.e.,the diemnsion of {ϕ : Lλϕ = 0} is one. Otherwise λ is a multiple eigenvalue.

Theorem 5.2.4. The eigenvalues of the RSLBVP are simple.

Proof. Suppose u, v are eigenfunctions corresponding to the same eigenvalue λ. Then(u(a) v(a)u′(a) v′(a)

)(α1

α2

)=

(00

)where u, v satisfy

L(y) + λpy = 0.

Since α21 + α2

2 6= 0, the determinant of the coefficient matrix must be zero (i.e., the ho-mogeneous equation has nonzero solutions) and hence u, v must be linearly dependent, i.e,v(x) = cu(x).

If k(a) = k(b) and instead of the separated boundary conditions, we consider the periodicboundary conditions,

y(a) = y(b), y′(a) = y′(b),

then Theorems 5.2.1-5.2.3 are still true Theorem 5.2.4 is no longer true.We have yet to address whether the RSLBVP has any eigenvalues. The following exam-

ples indicate that there are infinite, but a countable number, of eigenvalues.


Example 5.2.5. Find the eigenvalues and eigenfunctions for

y′′ + λy = 0, 0 < x < l

with the boundary conditions

1. y(0) = 0, y(l) = 0.

If λ = 0, y(x) = ax + b and the boundary conditions imply a = b = 0. If λ < 0, sayλ = −µ2, then

y(x) = a sinh(µx) + b cosh(µx).

Then y(0) = 0 only if b = 0 and y(l) = 0 only if a = 0.

If λ > 0, say λ = µ2, then

y(x) = a sin(µx) + b cos(µx).

and y(0) = 0 if b = 0. To satisfy the boundary condition at x = l we need

a sin(µl) = 0.

We get a nontrivial solution if

µl = nπ, n = ±1,±2,±3.....

Hence the eigenvalues and eigenfunctions are given by

λn = (nπ/l)2, yn(x) = sin(nπx/l), n = 1, 2, 3, ...

2. y(0) = 0, y′(l) = 0As above it is easy to verify that eigenvalues must be positive. If λ = µ2 > 0, then

y(x) = a sin(µx) + b cos(µx).

and y(0) = 0 if b = 0. The second boundary condition gives

aµ cos(µl) = 0

or

µl =

(n+

1

2

)π, n = 0,±1,±2, · · ·

or

λn =

((n+

1

2

)π

l

)2

, n = 0,±1,±2, · · ·

and

yn(x) = sin

((n+

1

2

)π

lx

).


3. y′(0) = y′(l) = 0If λ = 0, then y = ax + b and y′(0) = y′(l) = 0 if a = 0. Hence the constantfunction is an eigenfunction corresponding to the eigenvalue 0. It is easy to verifythan an eigenvalue for this problem cannot be negative. If λ = µ2 > 0, then y(x) =a cosµx+ b sinµx and y′(0) = 0 if b = 0. Then

y′(l) = −aµ sinµl = 0

ifµl = nπ, n = 0,±1,±2, · · ·

Hence eigenvalues and eigenfunctions are given by

λn =(nπl

)2

, yn(x) = sin(nπlx), n = 0, 1, 2, · · ·

4. y(0) + y′(0) = 0, y(l) = 0If λ = −µ2 < 0, µ > 0 then

y(x) = aeµx + be−µx

and the boundary conditions require that

y(0) + y′(0) = (a+ b) + µ(a− b) = 0

y(l) = aeµl + be−µl = 0

which implies b = exp(2µl)a and

a[(1− e2µl) + µ(1 + e2µl)] = 0.

This can be written as

e2µl =1− µ1 + µ

and by graphing each side it is easy to see that this equation is satisfied only whenµ = 0. Thus we have a = 0 and hence b = 0. If λ = 0, y = ax + b and the boundaryconditions require that

y(0) + y′(0) = b+ a = 0

a(l − 1) = 0

If l = 1, then a is arbitrary and an eigenfunction is y(x) = x − 1. If l 6= 1, thena = 0 and hence b = 0 and so λ = 0 is not an eigenvalue. If λ = µ2 > 0 theny = a cosµx+ b sinµx and to satisfy the boundary conditions we need

a+ µb = 0, a cosµl + b sinµl = 0


or

tanµl = µ.

It is easy to see that there are infinitely many eigenvalues λn that satisfy√λn = tan

√λnl

with corresponding eigenfunctions

yn(x) = sin√λnx−

√λn cos

√λnx.

5. y(0) = y(l), y′(0) = y′(l)

If λ = −µ2 < 0, then

y(x) = a sinhµx+ b coshµx

and

y(0) = b = y(l) = a sinhµl + b coshµl

while

y′(0) = aµ = y′(l) = aµ coshµl + bµ sinhµl

or (sinhµl coshµl − 1

µ(coshµl − 1) µ sinhµl

)(ab

)=

(00

).

The determinant of he coefficient matrix is 2µ(coshµl−1) which vanishes only if µ = 0and so a = b = 0. If λ = 0, an obvious eigenfunction is y = 1. If λ = µ2 > 0, then

y(x) = a sinµx+ b cosµx.

We see that

y(0)− y(l) = 0 if b− (a sinµl + b cosµl) = 0

and

y′(0)− y′(l) = 0 if aµ− (aµ cosµl − bµ sinµl) = 0

or (− sinµl 1− cosµl

µ(1− cosµl) µ sinµl

)(ab

)=

(00

).

We obtain a nontrivial solution if

−µ sin2 µl − µ(1− cosµl)2 = 0.

5.3. GREEN’S FUNCTIONS 15

Expanding the second term and simplifying we arrive at cosµl = 1 and so

µ =2nπ

l, n = ±1,±2, · · ·

In this case a, b are arbitrary and so to each eigenvalue

λn =

(2nπ

l

)n = ±1,±2, · · ·

there are two linearly independent eigenfunctions

yn(x) = an sin2nπ

lx+ bn cos

2nπ

lx

5.3 Green’s Functions

In this section we present an elementary introduction to the notion of a Green’s functionfor the class of regular Sturm Liouville systems studied in the last section. In particular weinterested in solving a RSLBVP when the differential equation has an extra nonhomogeneousright hand side. In order to simplify matters a bit, let us assume that the weight functionin the previous sections is p(x) ≡ 1, so we consider the BVPλ

Lλ(y) = (ky′)′ + g(x)y + λy = 0, a < x < b

B1(y) = 0

B2(y) = 0

whereBi(y) = αi1(a) + αi2y

′(a) + βi3y(b) + βi2y′(b)

and k ∈ C1(a, b), k(x) > 0, x ∈ [a, b].

Definition 5.3.1. A Green’s function for BVPλ is a function G(x, ξ, λ) for (x, ξ) ∈ [a, b]×[a, b] such that

1. The following hold

(a) G(·, ·, λ) is continuous on [a, b]× [a, b],

(b)∂G

∂x(·, ξ, λ) is continuous on [a, ξ)× (ξ, b], and,

(c)∂G(x, ξ, λ)

∂x

∣∣∣∣x=ξ+

x=ξ−≡ ∂G

∂x(ξ+, ξ, λ)− ∂G

∂x(ξ−, ξ, λ) =

1

k(ξ)


2. for all ξ ∈ [a, b], G(x, ξ, λ) solves Lλ(G) = 0, x 6= ξ.

3. for all ξ ∈ (a, b), Bi(G) = 0.

Theorem 5.3.2. If λ is not an eigenvalue of BVPλ, then the boundary value problem hasa unique Green’s function G(x, ξ, λ) and it is symmetric, i.e., G(x, ξ, λ) = G(x, λ, ξ).

Proof. We provide a proof by construction for two special cases, (I.) Separated BoundaryConditions, (II.) Periodic Boundary Conditions (which are unseparated):

I.) Separated boundary conditions,

B1(y) = α1y(a) + α2y′(a) = 0

B2(y) = β1y(b) + β2y′(b) = 0.

Choose ui such that Lλ(ui) = 0 and Bi(ui) = 0. This can be done as this amounts tosolving an initial value problem corresponding to initial conditions specified at x = aand x = b. The solutions u1, u2 must be linearly independent. Indeed, suppose w =c1u1 + c2u2 ≡ 0. Then

B1(w) = c2B1(u2) = 0B2(w) = c1B2(u1) = 0.

If B1(u2) = 0 then u2 would be an eigenfunction, but λ is not an eigenvalue. Hence wemust have c2 = 0. Similarly, c1 = 0.

Now seek G(x, ξ, λ) in the form

G(x, ξ, λ) =

{Au1(x) a ≤ x ≤ ξ

Bu2(x) ξ ≤ x ≤ b.

We need

Au1(ξ) = Bu2(ξ)

and

Bu′2(ξ)− Au′1(ξ) =1

k(ξ).

By Cramer’s rule, one obtains

A =u2(ξ)

k(ξ)W (u1, u2)(ξ)

B =u1(ξ)

k(ξ)W (u1, u2)(ξ)


and hence

G(x, ξ, λ) =

u1(x)u2(ξ)

k(ξ)W (u1, u2)(ξ)a ≤ x ≤ ξ

u1(ξ)u2(x)

k(ξ)W (u1, u2)(ξ)ξ ≤ x ≤ b.

To see that G is symmetric, recall that Abel’s formula states that

W (u1, u2)(ξ) = W (x0) exp

(−∫ ξ

x0

k′(x)

k(x)dx

)= W (x0)

k(x0)

k(ξ).

HenceW (u1, u2)(ξ)k(ξ) = constant.

Example 5.3.3. Find the Green’s function for

y′′ = 0, 0 < x < 1

y(0) = 0, y(1) = 0.

Here Lλ = L0 and it is easy to verify that λ = 0 is not an eigenvalue. Take u1(x) =x, u2(x) = x− 1 and

W (u1, u2) =

∣∣∣∣ x x− 11 1

∣∣∣∣ = 1

Hence

G(x, ξ, λ) =

{x(ξ − 1) 0 ≤ x ≤ ξ

ξ(x− 1) ξ ≤ x ≤ 1.

Example 5.3.4. Find the Green’s function for

y′′ + λy = 0, λ > 0, 0 < x < π

y(0) = 0, y(π) = 0.

Here we regard L(y) = y′′ and Lλ(y) = y′′+ λy. We know the eigenvalues are given by

λn = n2, n = 1, 2, 3 · · ·

If λ 6= n, take u1(x) = sin√λx, u2(x) = sin

√λ(x− π)x. Then

k(0)W (u1, u2)(0) =

∣∣∣∣ 0 − sin√λπ√

λ√λ cos

√λπ.

∣∣∣∣ = −√λ sin

√λπ


and so

G(x, ξ, λ) =

−sin

√λx sin

√λ(ξ − π)√

λ sin√λπ

0 ≤ x ≤ ξ

−sin√λξ sin

√λ(x− π)√

λ sin√λπ

ξ ≤ x ≤ 1.

II.) For our second example we consider the special case of Initial Value Problems. Thatis, in the special case in which the boundary conditions reduce to the initial values

B1(u) = u(a)

B2(u) = u′(a)

In this case we seek

G(x, ξ, λ) =

{0 a ≤ x ≤ ξ

Au1(x) +Bu2(x) ξ ≤ x

where u1, u2 are linearly independent solutions of Lλ = 0. In this case the continuityand jump condition give

Au1(ξ) +Bu2(ξ) = 0

and

Au′1(ξ) +Bu′2(ξ) =1

k(ξ).

and so

A = − u2(ξ)

k(ξ)W (u1, u2)(ξ), B =

u1(ξ)

k(ξ)W (u1, u2)(ξ).

Hence

G(x, ξ, λ) =

0 a ≤ x ≤ ξ

u1(ξ)u2(x)− u1(x)u2(ξ)

k(ξ)W (u1, u2)(ξ)ξ ≤ x.

Recall that the Heaviside function is defined by

H(x) =

{1 x > 0

0 x < 0

Let uξ(x) denote the solution of

Lλ(uξ(x)) = 0

uξ(ξ) = 0

u′ξ(ξ) =1

k(ξ).


Thus we see that the Green’s function for the initial value problem satisfies

G(x, ξ) = H(x− ξ)uλ(x).

Such a Green’s function is often referred to as the causal fundamental solution.

For more general boundary conditions, we might seek G(x, ξ) in the form

G(x, ξ) = H(x− ξ)uξ + Au1(x) +Bu2(x)

where u1, u2 are linearly independent solutions of Lλ = 0.

Example 5.3.5. Construct the Green’s function for

u′′ = 0, 0 < x < 1

u(0) + u(1) = 0

u′(0) + u′(1) = 0

SeekG(x, ξ) = H(x− ξ)uξ(x) + Ax+B ≡ E(x, ξ) + Ax+B

where

u′′ξ = 0, x > ξ > 0

uξ(ξ+) = 0, u′ξ(ξ

+) = 1.

Then

E(x, ξ)) =

{0 0 ≤ x ≤ ξ

x− ξ x ≤ ξ ≤ 1

and

B1(G) = (E(0, ξ) +B) + (E(1, ξ) + A+B)

= 2B + A+ (1 + ξ) = 0,

B2(G) = (0 + A) + (1 + A)

Solving, one obtainsA = −1/2, B = −1/4 + ξ/2

and hence

G(x, ξ) =

12x− 1

4+ ξ

2, 0 ≤ x ≤ ξ

(x− ξ)− 14

+ ξ2− x

2, x < ξ ≤ 1

or

G(x, ξ) = −1

4+|x− ξ|

2


We now turn to the main application of Green’s function in this section. Namely, weconsider the nonhomogeneous BVP.

Lλ(y) = (ky′)′ + g(x)y + λy = f(x), a < x < b

B1(y) = 0

B2(y) = 0

whereBi(y) = αi1(a) + αi2y

′(a) + βi3y(b) + βi2y′(b)

and k ∈ C1(a, b), k(x) > 0, x ∈ [a, b].First we recall a classical formula whose general counterpart has far reaching consequences

in the theory of ordinary and partial differential equations and the theory of weak solutions.At this point we will only consider a very special case. Namely, given any two functions uand v, a straightforward calculation gives the so-called Lagrange Identity:

vLλ(u)− uLλ(v) =d

dxP (u, v)

where (see (5.2.13))P (u, v) = k (u′v − uv′)

and we note that integration gives the Green’s formula∫ b

a

[vLλ(u)− uLλ(v)] = P (u, v)|x=bx=a

Let G(x, ξ) denote the Green’s function for the homogeneous BVPλ. From Lagrange’sidentity, for x 6= ξ

G(x, ξ)Lλ(y)− yLλ(G(x, ξ)) =d

dx[k(Gy′ − yG′)]

which implies ∫ ξ−

a

GLλ(y)dx = k(Gy′ −G′y)]ξ−a

and ∫ b

ξ+

GLλ(y)dx = k(Gy′ −G′y)]bξ+ .


Hence ∫ b

a

GLλ(y)dx = k(Gy′ −G′y)]ba − k(Gy′ −G′y)]ξ+

ξ− .

Suppose that B1, B2 are boundary conditions with the property that if u, v satisfy B1(u) =0 = B2(v), then

[k(Gy′ −G′y)]ba = 0

Let us refer to such boundary conditions as regular boundary conditions. We have seen forexample that separated boundary conditions are regular and regular boundary conditionsresult in a self-adjoint boundary value problem. Assume the the boundary conditions areregular and B(y) = B2(y) = 0. Then∫ b

a

GLλ(y)dx = −[k(Gy′ −G′y)]ξ+

ξ−

= k[∂G

∂x(ξ+, ξ)− ∂G

∂x(ξ−, ξ)]y(ξ)

= y(ξ).

Thus, formally at least, if y satisfies L(y) = f , then we should have y(x) =

∫ b

a

G(x, ξ, λ)f(ξ) dξ.

That is, again on a purely formal level, we have∫ b

a

G(x, ξ)Lλ(y)(x)dx =

∫ b

a

LλG(x, ξ)(y)(x)dx = y(ξ)

which suggest that LλG(x, ξ) = δ(x− ξ), i.e., the solution to

Lλ(y) = f

Bi(y) = 0

would be given by

y(x) =

∫ b

a

G(x, ξ)f(ξ)dξ

provided that λ is not an eigenvalue. This is indeed the case and we will argue this forseparated boundary conditions.

Theorem 5.3.6. If λ is not and eigenvalue of BVPλ where the boundary conditions areseparated, that is

B1(y) = αy(a) + αy′(a)B2(y) = βy(b) + βy′(b),


then the unique solution ofLλ(y) = f

B1(y) = γ1, B2(y) = γ2

is given by

y(x) =γ2

B2(y1)y1(x) +

γ1

B1(y2)y2(x) +

∫ b

a

G(x, ξ)f(ξ)dξ

where y1, y2 are (not identically zero) solutions of

L(y) = 0B1(y1) = 0, B2(y2) = 0.

(Note that since y1 and y2 are not identically zero, we must have B1(y2) 6= 0 and B2(y1) 6= 0.

Proof. Since B1(G) = B2(G) = 0,

B1(y) =γ1

B1(y2)B1(y2) = γ1

and similarly B2(y) = γ2

To see that the nonhomogeneous differential equation is satisfied, we consider

u(x) =∫ baG(x, ξ)f(ξ)dξ

=∫ xaG(x, ξ)f(ξ)dξ +

∫ bxG(x, ξ)f(ξ)dξ

Thenu′(x) =

∫ x−a

∂G∂xfdξ +G(x, x−)f(x−)

+∫ bx+

∂G∂xfdξ −G(x, x+)f(x+)

=∫ xaGx(x, ξ)f(ξ)dξ +

∫ bxGx(x, ξ)f(ξ)dξ.

Differentiating again we have

u′′(x) =∫ x−a

Gxx(x, ξ)f(ξ)dξ +Gx(x, x−)f(x−)

+∫ bx+ Gxx(x, ξ)f(ξ)dξ −Gx(x, x

+)f(x+).

We need the following observation,

∂G

∂x(x, x−) =

∂G

∂x(x+, x)

∂G

∂x(x, x+) =

∂G

∂x(x−, x).


For example, to verify the first of these claims, note that

∂G

∂x(x, x−) = lim

ε→0+

∂G

∂x(x, x− ε)

= limε→0+

limh→0

G(x+ h, x− ε)−G(x, x− ε)h

.

The partials exist because we are in the open region x > ξ away from the diagonal (x = ξ)where only one-sided derivatives exist. Moreover, because G is smooth when x > ξ we mayinterchange the order of limits to obtain

∂G

∂x(x, x−) = lim

ε→0+limh→0+


= limh→0+

limε→0+


= limh→0+

G(x+ h, x)−G(x, x)

h

=∂G

∂x(x+, x)

Similarly for the other statement. Hence we have

u′′(x) =

∫ b

a

Gxx(x, ξ)f(ξ)dξ + [Gx(x+, x)−Gx(x−, x)]f(x−)

=

∫ b

a

Gxx(x, ξ)f(ξ)dξ + f(x)/k(x).

Thus

Lλ(u) = ku′′ + k′u′ + gu+ λu

=

∫ b

a

[k(x)Gxx(x, ξ) + k′(x)Gx(x, ξ) + g(x)G(x, ξ)

+ λG(x, ξ)

]f(ξ)dξ +

k(x)f(x)

k(x)

=

∫ b

a

LλG(x, ξ)y(ξ) dξ + f(x)

= f(x)

since Lλ(G) = 0.


Recall that we previously argued that the nonhomogeneous BVP

L(y) = fB1(y) = γ1, B2(y) = γ2

has a unique solution if the BVP with the data 0, 0, 0 has only the trivial solution. In lightof the previous theorem we see that this is merely the statement that if λ = 0 is not aneigenvalue of BVPλ, the nonhomogeneous BVP is uniquely solvable.

Theorem 5.3.6 is often phrased in the following equivalent form: Either the BVP

Lλ(u) = fB1(u) = γ1, B2(u) = γ2

has a unique solution for each f or else the associated homogeneous problem has a nontrivialsolution. This statement is referred to as the Fredholm Alternative.

Our results do not say that if λ is an eigenvalue the nonhomogeneous BVP is not solvable,and we address this situation now. First consider the analogy from linear algebra.

Remark 5.3.7. Suppose A is an n× n matrix and we want to solve Ax = b. Then we haveTheorem: (uniqueness) The solution to Ax = b (if it exists) is unique if and only if Ax = 0implies that x = 0.

On the other hand the Fredholm Alternative gives an explicit criteria for existence.Theorem: (existence) The equation Ax = b has a solution if and only if < b, v >= 0 forevery v satisfying A∗v = 0.

Now suppose A = AT and let M = A − λI. Then My(λ) = f is solvable if and only iff ⊥ N(MT ) (the nullspace of MT ). Since MT = M and N(M) = {v|Av = λv}, we have(A−λI)y = f is solvable if and only if y ⊥ v, where Av = λv. That is, if λ is an eigenvalueof A, then (A − λI)y = f has a solution if and only if f is orthogonal to the eigenspacecorresponding to λ.

For the differential operator Lλ we have an analogous result. Again, we restrict ourattention to separated boundary conditions.

Theorem 5.3.8. Suppose that (µ, v) is an eigenpair for

Lλ(y) = 0

Bi(y) = 0, i = 1, 2.

Then the nonhomogeneous problem

Lµ(y) = f, Bi(y) = 0, i = 1, 2,

has a solution if and only if∫ baf(x)v(x)dx = 0.


Proof. Suppose u is a solution of the nonhomogeneous problem. Since u, v satisfy the ho-mogeneous boundary conditions, we have

P (u, v) = k(uv′ − u′v)∣∣ba

= 0

Green’s formula (5.2.13) gives∫ b

a

vfdx =

∫ b

a

[0− vf ] dx =

∫ b

a

[uLλ(v)− vLλ(u)]dx = 0.

Now suppose∫ bavfdx = 0 and let us choose u to the IVP solve Lλ(u) = f with the initial

conditionsu(a) = v(a), u′(a) = v′(a).

Then B1(u) = 0 since B1(v) = 0. We need to show that B2(u) = 0 = β1u(b) + β2u′(b).

Green’s formula gives∫ b

a

vfdx =

∫ b

a

[uLλ(v)− vLλ(u)]dx = k[uv′ − vu′]|ba.

But we already know that B1(u) = 0 so this gives

0 =

∫ b

a

vfdx = k(b)[u(b)v′(b)− v(b)u′(b)].

Since β21 + β2

2 6= 0, without loss of generality assume β1 6= 0. If β2 = 0, then β1v(b) = 0 sov(b) = 0 and hence u(b) = 0 and so B2(u) = 0. If β2 6= 0, then β1/β2v(b) = v′(b). Since vis a nontrivial solution of Lλ(y) = 0, we see from the previous relations that v(b) 6= 0 andv′(b) 6= 0. Thus is follows that

u(b) =v(b)u′(b)

v′(b)

and so

β1u(b) + β2u′(b) = β1

v(b)

v′(b)u′(b) + β2u

′(b)

= u′(b)

(β1v(b)

v′(b)+ β2

)= 0

and hence B2(u) = 0. Thus u solves the nonhomogeneous boundary value problem.


Exercises for Chapter 5

1. Considerk0y

′′ + g(x)y = 0 (∗)where

k0 > 0, 0 < g1 < g(x) < g2.

If z1, z2 are consecutive zeros of a solution to (*), show that

π

√k0

g2

≤ (z2 − z1) ≤ π

√k0

g1

.

2. Suppose that y(x) is a solution of

y′′ + a(x)y = 0.

(a) If a(x) ≥ m > 0, x0 ≤ x ≤ ∞, show that y(x) has infinitely many zeros but ifa(x) < 0, x0 ≤ x ≤ ∞, then y(x) has at most one zero.

(b) Iflimx→∞

a(x) = a0 > 0,

prove that the distance between consecutive zeros tends to π/√a0 as the zeros

tend to infinity.

3. Find the eigenvalues and eigenfunctions of u′′ + λu = 0 with the boundary conditions:

(a) u(0) = u(1) = 0,

(b) u′(0) = u′(1) = 0,

(c) u(0) = 0, u(1)− u′(1) = 0 .

4. Find the Green’s function for

y′′ − γ2y = 0, y′(0) = 0, y(1) = 0, γ > 0.

5. Find the eigenvalues and eigenfunctions for the following boundary value problem andthen construct the Green’s function.

d

dx

(xdy

dx

)+λ

xy = 0, 1 < x < e

u(1) = 0, u(e) = 0.


6. Do parts a) through c):

(a) Show that the eigenvalues and eigenfunctions of the boundary value problem

d

dx

((1 + x)2 d

dx(u)

)+ λu = 0, 0 < x < 1

u(0) = u(1) = 0

are given by

λn =( nπ

ln 2

)2

+1

4, n = 1, 2, 3, . . .

un(x) =sin ((nπ ln(1 + x))/(ln 2))√

1 + x.

(b) Construct the Green’s function for this problem.

(c) If

f(x) =1√

1 + x, λ =

1

4+( π

ln 2

)2

,

doesL(u) + λu = f(x)

u(0) = u(1) = 0

have a solution? If so, is it unique?


Bibliography

[1] D.A. Sanchez, Ordinay Differential Equations and Stability Theory: An Introduction,W.H. Freeman and Company, 1968.

[2] F. Brauer, J.A. Nohel, Qualitative Theory of Ordinary Differential Equtions, Dover,1969.

[3] I. Stakgold, “Green’s Functions and Boundary Value Problems,” Wiley-Interscience,1979.

[4] H. Sagan, “Boundary and Eigenvalue Problems in Mathematical Physics,” J. Wiley &Sons, Inc, New York, 1961.

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ODE Chapter 5 Sturm-Liouville Theory