Chapter 5

Boundary Value Problem

In this chapter I will consider the so-called boundary value problem (BVP), i.e., a differential equationplus boundary conditions. The properties of these problems in general are very different from theproperties of the IVP studied earlier.

5.1 Motivation

Boundary value problems appear is many different situations. For example, finding a heteroclinictrajectory connecting two equilibria x1 and x2 is actually a BVP, since I am looking for x such thatx(t) → x1 when t→ ∞ and x(t) → x2 when t→ −∞. Another example when BVP appears naturallyis the study of periodic solutions. Recall that the problem x = f(t,x), where f is T -periodic withrespect to t, has a T -periodic solution ϕ if and only if ϕ(0) = ϕ(T ). Some BVP may have a uniquesolution, some have no solutions at all, and some may have infinitely many solution.

Exercise 5.1. Find the minimal positive number T such that the boundary value problem

x′′ − 2x′ = 8 sin2 t, x′(0) = x′(T ) = −1

has a solution.

Arguably one of the most important classes of BVP appears while solving partial differentialequations (PDE) by the separation of the variables (Fourier method). To motivate the appearanceof PDE, I will start with a system of ordinary differential equations of the second order, which wasstudied by Joseph-Louis Lagrange in his famous “Analytical Mechanics” published first in 1788, wherehe used, before Fourier, the separation of variables1 technique.

Consider a system of k + 1 equal masses connected by the springs with the same spring constantc. Two masses at the ends of the system are fixed. Let uj(t) be the displacement of the j-th mass attime t. Using the second Newtons law and the Hook’s law for the springs, I end up with the system

muj = c(uj+1 − uj)− c(uj − uj−1) = c(uj+1 − 2uj + uj−1), j = 1, . . . , k − 1,

with the initial conditions

uj(0) = αj , u′j(0) = βj , j = 1, . . . , k − 1

1As far as I am aware this method appeared first in one of Daniel Bernoulli’s works on vibration of strings in 1753,together with “Fourier’s series.” “The Analytic Theory of Heat” by Joseph Fourier was published in 1822

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and the boundary conditions (the ends are fixed)

u0(t) = uk(t) = 0, t ∈ R.

This problem can be solved directly, by using the methods from Chapter 3. However, Lagrange,following Bernoulli, makes the following assumption: He assumes that the unknown function uj canbe represented as

uj(t) = T (t)J(j),

i.e., as the product of functions, the first of which depends only on t and the second one depends onlyon the integer argument j. On plugging this expression into the original system and rearranging I find

mT (t)

cT (t)=J(j + 1)− 2J(j) + J(j − 1)

J(j).

Since the left-hand side depends only on t whereas the right-hand side depends only on j then theyboth must be equal to the same constant, which I will denote −λ. Therefore, I get

T +cλ

mT = 0,

and−(J(j + 1)− 2J(j) + J(j − 1)) = λJ(j).

Moreover, from the boundary conditions for the original equation, it follows that

J(0) = J(k) = 0.

These boundary conditions plus the previous equation is actually an eigenvalue problem for a lineardifference operator, which acts on function, defined on integers. To emphasize this I denote

∆2 : f(j) 7→ −(f(j + 1)− 2f(j) + f(j − 1)),

and hence my problem can be rewritten as

∆2J = λJ, J(0) = J(k) = 0.

My goal is thus to determine those values of the constant λ for which my problem has non-trivial(different from zero) solutions.

The general theory of linear difference equations with constant coefficients is very similar to thetheory of linear ODE with constant coefficients. In particular, the general solution to the homogeneousequation is given as a linear combination of two linearly independent solutions. Whereas we use anansatz eqt for ODE, for the difference equations one should use qj . So, by plugging J(j) = qj into myequation and canceling I find

−(q2 − 2q + 1) = λq,

or, after rearranging,

q2 − 2

(1− λ

2

)q + 1 = 0.

Instead of directly solving this quadratic equation, I will start with a qualitative analysis.

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If my equation has two real roots q1 = q2 ∈ R then the general solution is C1qj1 + C2q

j2 and the

boundary conditions would imply that

C1 + C2 = 0

C1qk1 + C2q

k2 = 0,

which has a nontrivial solution C1, C2 if and only if the corresponding determinant is zero, which, bydirect calculation is impossible for q1 = q2. The case q1 = q2 ∈ R is analyzed similarly, only in thiscase the general solution is C1q

j1 +C2jq

j1. Now I assume that q1 = q2 ∈ C, i.e., q1,2 = re±iθ. But from

my quadratic equation by Vieta’s theorem I have q1q2 = 1 and hence r = 1. Now

C1 + C2 = 0

C1eiθk + C2e

−iθk = 0,

and the corresponding determinant is (check)

sin θk,

which is equal to zero only if

θl =πl

k, l = 1, 2, . . .

I determined my θl, but the original constant is λ, which should be expressed through θl. From Vieta’stheorem I have

1− λ

2=q1 + q2

2= cos θl,

hence

λl = 2(1− cos θl) = 4 sin2θl2

= 4 sin2πl

2k,

and hence, to have different constants, I should only take l = 1, . . . , k− 1. Therefore, I found that mydifference operators has k − 1 eigenvalues with k − 1 eigenfunctions.

Returning to the equation for the time dependent variable, I hence have

Tl+cλlmTl = 0 =⇒ Tl = Al sinωlt+Bl cosωlt = Cl sin(ωlt+φl), wl = 2

√c

msin

πl

2k, l = 1, . . . , k−1.

The full solution to the original system can be determined by noting that any linear combination ofTl(t)Jl(j) is also a solution, and hence

uj(t) =k−1∑l=1

Tl(t)Jl(j) =k−1∑l=1

Cl sin(ωlt+ φl)Jl(j), j = 1, . . . , k − 1.

These solutions have 2(k− 1) arbitrary constants which are uniquely determined by the initial condi-tions for k − 1 moving masses.

In a nutshell, an attempt to solve my original problem by the separation of variables method endedup with an eigenvalue problem for a linear difference operator, or, in a slightly different language, in aboundary value problem for a linear difference equation. Now, if I carefully take the limit k → ∞ in my

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problem, it can be shown (those who saw difference operators before should not find this surprising)that my system of ODE will turn into the so-called wave equation

ut = a2uxx, u(t, 0) = u(t, 1) = 0

for the unknown function u of two variables: time t and position x. I also must have the initialconditions

u(0, x) = f(x), ut(0, x) = g(x), 0 ≤ x ≤ 1.

Now I can do the same trick assuming that

u(t, x) = T (t)X(x).

After separating the variables I will find (the student should fill in all the details) that

X ′′ + λX = 0, X(0) = X(1) = 0.

This problem has a nontrivial solutions only if

λk = (πk)2, k = 1, 2, . . .

These are the eigenvalues of my differential operator. The corresponding eigenfunctions are

Xk(x) = sinπkx, k = 1, 2, . . .

The equation for Tk hence takes the form

T ′′ + (πka)2T = 0,

which has the solutionTk(t) = Ak sinπkat+Bk cosπkat.

Finally, using the superposition principle, I have that the infinite linear combination

u(t, x) =

∞∑k=1

Tk(t)Xk(x) =

∞∑k=1

(Ak sinπkat+Bk cosπkat) sinπkx

solves my problem and satisfies the boundary conditions. Now I can use the initial conditions todetermine Ak, Bk. For instance, using the first initial condition I find

∞∑k=1

Bk sinπkx = f(x),

which means that Bk are the Fourier coefficients of f of its Fourier sine series. To find these coefficientsI will use the orthogonality of sines:∫ 1

0sinπnx sinπmx dx = 0, m = n, =

1

2, n = m.

Therefore

Bk = 2

∫ 1

0f(x) sinπkx dx.

There are a lot of questions here about the convergence, completeness of the system of sines, etc. Iwill answer all these questions from a general and abstract point of view.

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Exercise 5.2. Recall that while solving the one-dimensional heat equation ut = α2uxx, we needed tosolve the boundary value problem U ′′ + λU = 0 with some boundary conditions. The result was thefamilies of eigenvalues and eigenfunctions that form an orthogonal family (actually, as we prove later,a basis). In this problem you are asked to solve the same problem for different boundary conditions.

1. U ′(0) = U ′(1) = 0

2. U(0) = U ′(1) = 0

3. U(−π) = U(π), U ′(−π) = U ′(π)

In both problems you are asked to find the values of λ’s (eigenvalues) for which the problem has anot-trivial solution, the corresponding solutions (eigenfunctions), and check that the set of all solutionsforms an orthogonal family with respect to the standard inner product in L2[0, 1] for the first andsecond problems and in L2[−π, π] for the second one.

Exercise 5.3. Consider the Laplace equation on the circular disk

∆u = 0, x2 + y2 < 1, u(x, y)|x2+y2=1 = h(x, y),

Introducing the polar coordinates x = r cos θ, y = r sin θ it can be proved that the Laplace operatorin the polar coordinates takes the form (I do not require a proof, but this is a good exercise in thechange of variables)

∆u =∂2u

∂x2+∂2u

∂y2=∂2v

∂r2+

1

r

∂v

∂r+

1

r2∂2v

∂θ2,

where v(r, θ) = u(r cos θ, r sin θ).Write the boundary conditions for the equation in polar coordinates:

v(1, θ) = H(θ), H(θ) = h(x, y)|x2+y2=1.

Note also that we should have v(r, θ) = v(r, θ + 2π), v′(r, θ) = v′(r, θ + 2π) (periodicity condition).

1. Separate the variables in polar coordinates and solve two ordinary differential equations, takinginto account the boundary conditions. Note that additionally we should request that the solutionmust be bounded and smooth to satisfy the original equation. A previous problem, part (c),may be useful.

2. Write down a formal solution to the original problem.

5.2 Inner products and Hilbert spaces

Definition 5.1. Let X be a complex vector space. An inner product on X is a function

⟨· , ·⟩ : X ×X −→ C,

for which the following properties hold for any x, y, z ∈ X and α ∈ C:

(1) ⟨x+ y , z⟩ = ⟨x , z⟩+ ⟨y , z⟩(2) ⟨αx , y⟩ = α ⟨x , y⟩(3) ⟨x , y⟩ = ⟨y , x⟩(4) ⟨x , x⟩ > 0 if x = 0.

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The immediate consequences are

(5) ⟨x , y + z⟩ = ⟨x , y⟩+ ⟨x , z⟩(6) ⟨x , αy⟩ = α ⟨x , y⟩(7) ⟨0 , x⟩ = ⟨x , 0⟩ = ⟨0 , 0⟩ = 0.

The properties (3) and (6) look somewhat easier if X is a real vector space. In the following, if notstated otherwise, I will assume that X is a complex vector space. Some proofs may be different forreal and complex cases. The properties of the inner product are motivated by the real dot product

x · y = ⟨x ,y⟩ =3∑j=1

xjyj , x,y ∈ R3,

which is actually equivalent to the dot product

x · y = |x||y| cos θ,

where θ is an angle between vectors x and y, and hence allows to measure angles. The complexconjugation is necessary to guarantee that if x ∈ C3 then

x · x ∈ R.

The definition I chose means that for any x, y ∈ C3

x · y =

3∑j=1

xjyj

is taken as the inner product. Sometimes an expression is used when the conjugation is applied to thefirst factors.

Definition 5.2. Pair (X, ⟨· , ·⟩) is an inner product space.

Each inner product induces a norm by the formula

∥x∥ := ⟨x , x⟩1/2 .

Lemma 5.3 (Cauchy–Bunyakovskii–Schwarz inequality).

| ⟨x , y⟩ | ≤ ∥x∥∥y∥

Proof. I will assume that X is a real vector space. The case of the complex vector space is left as anexercise. For any real constant α

0 ≤ ⟨x+ αy , x+ αy⟩ = ∥x∥2 + 2α ⟨x , y⟩+ α2∥y∥2,

which is a quadratic polynomial with respect to the variable α, and hence should have the discriminantnon-positive:

⟨x , y⟩2 − ∥x∥2∥y∥2 ≤ 0,

which concludes the proof.

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The lemma also shows that the equality is possible if and only if x and y are linearly dependent(prove), which in its turn implies that

∥x∥ = supy =0

| ⟨x , y⟩ |∥y∥

,

since

∥x∥ =⟨x , x⟩∥x∥

≤ supy =0

| ⟨x , y⟩ |∥y∥

≤ ∥x∥.

Moreover, now one can easily show that ∥x∥ is indeed a norm. Consider the expression

∥x+ y∥2 = ⟨x+ y , x+ y⟩ = ∥x∥2 + 2| ⟨x , y⟩ |+ ∥y∥2 ≤ ∥x∥2 + 2∥x∥∥y∥+ ∥y∥2 = (∥x∥+ ∥y∥)2,

which proves the triangle inequality, the other properties of the norm are straightforward.Having a norm on a vector space X means that we have a metric and a topology induces by this

metric, and therefore we can talk about convergence.

Definition 5.4. An inner product space is called a Hilbert space if it is complete (with respect to thenorm induced by the inner product).

The natural question whether any norm can be generated by some inner product. The answer isnegative, and the simplest way to check whether ∥ · ∥ can be generated by an inner product is to checkthe parallelogram law:

Lemma 5.5 (Parallelogram law). If X is an inner product space then

∥x+ y∥2 + ∥x− y∥2 = 2∥x∥2 + 2∥y∥2

for any x, y ∈ X.

The proof is left as an (easy) exercise.

Example 5.6. The real vector space Rk becomes a (finite dimensional) Hilbert space if the innerproduct is defined as

⟨Ax ,y⟩A =k∑

i,j=1

aijxiyj

for any positive definite real matrix A = (aij)k×k. The complex vector space Ck becomes a (finitedimensional) Hilbert space if the inner product is defined as

⟨Ax ,y⟩A =k∑

i,j=1

aijxiyj

for any Hermitian matrix A.

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Example 5.7. The complex space l2 consists of all infinite sequences x = (xi)∞i=1 that satisfy∑∞

i=1 |xi|2 <∞ (square summable sequences) with the inner product

⟨x , y⟩ =∞∑i=1

xiyi.

The inner product is well defined due to the Cauchy–Bunyakovskii–Schwarz inequality.The first thing to note about l2 is that it is not finite dimensional, opposite to Rk or Ck and

therefore is infinite dimensional. It is a Hilbert space, because every Cauchy sequence has a limit.Moreover, there exists a countable dense subset of l2, which means that any element of l2 is the limitof elements from this subset (can you construct this subset explicitly?) Such metric spaces, containingfinite or countable dense subset are called separable. We shall need this notion later.

Example 5.8. Consider the space of continuous real function C([a, b],R) with the inner product

⟨x , y⟩ =∫ b

ax(t)y(t) dt.

You should check that it is indeed an inner product. This inner product space is not Hilbert because itis quite easy to give examples of sequences of continuous functions which would lead to still integrablebut discontinuous functions. Fortunately, it is always possible to consider a completion of an innerproduct space, and the completion of the space of continuous functions, defined on the open set Ωis called L2(Ω). More precisely L2(Ω) is a Hilbert space of all (equivalence classes of) measurablefunctions x : Ω −→ [−∞,∞] that are square summable, i.e.,∫

Ω|x|2 <∞,

where the integral is Lebesgue’s integral. The inner product is given by

⟨x , y⟩ =∫Ωxy,

and the corresponding norm

∥x∥L2(Ω) =

∫Ω|x|2.

The Hilbert space L2(Ω) is also separable.

5.3 Orthogonality. Basis. Fourier series

In inner product spaces we can measure angles between vectors. In particular, we can specify whichtwo vectors are orthogonal.

Definition 5.9. Let (X, ⟨· , ·⟩) be an inner vector space. Vectors x, y ∈ X are orthogonal if

⟨x , y⟩ = 0.

An immediate consequence of this definition is the famous Pythagoras theorem:

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Lemma 5.10 (Pythagoras theorem). If x and y are orthogonal then

∥x+ y∥2 = ∥x∥2 + ∥y∥2.

Given any family of vectors (ei)i∈I , span(ei)i∈I denotes all the subspace of X formed by all finitelinear combinations in the family. A family (ei)i∈I of elements of an inner product space (X, ⟨· , ·⟩) iscalled orthonormal family if

⟨ei , ej⟩ = δij , for all i, j ∈ I.

Here δij is Kronecker’s symbol, which is equal to 1 is i = j and 0 otherwise. Note that any orthonormalfamily is linearly independent (prove it). Gram–Schmidt orthonormalization procedure allows toconstruct an orthonormal family our of linearly independent (finite or countably infinite) family ofvectors. Details are left to the textbooks on functional analysis.

An orthonormal family (ei)i∈I in an inner product space (X, ⟨· , ·⟩) is called maximal if the onlyvector x ∈ X that satisfies ⟨x , ei⟩ = 0 for all i ∈ I is x = 0. A convenient sufficient condition for anorthonormal family to be maximal is as follows: (ei)i=I is maximal if

span(ei)i∈I = X.

Any inner product set has a maximal orthonormal family, but in general the proof will require theaxiom of choice. However, in the case of a separable inner product space, we can say more withoutany heavy machinery. In particular,

Theorem 5.11. Let (X, ⟨· , ·⟩) be a separable infinite dimensional inner product space. Then(a) There exists a countable infinite orthonormal family (ei)

∞i=1 such that x ∈ X and ⟨x , ei⟩ = 0

for all i imply that x = 0.(b) Any orthonormal family is finite or countably infinite.

Proof.

Finally, I will need definition of a basis for an infinite dimensional space.

Definition 5.12. Let X be a Banach space. Then a countably infinite family (ei)∞i=1 is a (Schauder)

basis for X if for any x ∈ X there exist unique scalars (αi)∞i=1 such that

x =∞∑i=1

αiei.

Now my goal is to show that in a separable Hilbert space a maximal orthonormal family is exactlya (Schauder) basis.

Theorem 5.13. Let (X, ⟨· , ·⟩) be an infinite dimensional separable Hilbert space and let (ei)∞i=1 be a

maximal orthonormal family. Then(a) Any element x ∈ X can be expanded as a convergent series

x =∞∑i=1

⟨x , ei⟩ ei,

which is called the Fourier series of x.

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(b) The scalars ⟨x , ei⟩ ∈ C, i ≥ 1, which are called the Fourier coefficients of x relative to thefamily (ei)

∞i=1, satisfy Parseval’s formula

∥x∥2 = ⟨x , x⟩ =∞∑i=1

| ⟨x , ei⟩ |2.

Proof. First, let me show that∑∞

i=1 | ⟨x , ei⟩ | <∞. We have

0 ≤ ∥x−n∑i=1

⟨x , ei⟩ ei∥2 =

⟨x−

n∑i=1

⟨x , ei⟩ ei , x−n∑j=1

⟨x , ej⟩ ej

⟩=

= ∥x∥2 −n∑i=1

| ⟨x , ei⟩ |2 −n∑j=1

| ⟨x , ei⟩ |2 +n∑

i,j=1

⟨x , ei⟩ ⟨x , ej⟩ ⟨ei , ej⟩ =

= ∥x∥2 −n∑i=1

| ⟨x , ei⟩ |2.

Therefore,n∑i=1

| ⟨x , ei⟩ |2 ≤ ∥x∥2, for any n,

and hence∞∑i=1

| ⟨x , ei⟩ |2 ≤ ∥x∥2.

The last inequality is called Bessel’s inequality. Note that I did not use the fact that my space isHilbert. This inequality is true in any inner product space.

Next, I will show∑∞

i=1 ⟨x , ei⟩ ei converges. Consider the sequence of partial sums

sn =n∑i=1

⟨x , ei⟩ ei,

and

∥sn − sl∥2 =

⟨n∑

i=l+1

⟨x , ei⟩ ei ,n∑

j=l+1

⟨x , ej⟩ ej

⟩=

n∑i=l+1

| ⟨x , ei⟩ |2,

which proves that (sn) is a Cauchy sequence. Since the space is complete (Hilbert), then there is thelimit y :=

∑∞i=1 ⟨x , ei⟩ ei.

Finally, we need to show that x = y. It is equivalent to consider the condition ⟨x− y , ei⟩ for alli ≥ 1 since (ei)

∞i=1 is maximal by the assumptions. Using the continuity of the inner product, we have⟨

x−∞∑i=1

⟨x , ei⟩ ei , ej

⟩= lim

n→∞

⟨x−

n∑i=1

⟨x , ei⟩ ei , ej

⟩.

But ⟨x−

n∑i=1

⟨x , ei⟩ ei , ej

⟩= 0, for all n ≥ j.

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Therefore, x = y.Finally, from the previous we established that

0 = limn→∞

∥x−n∑i=1

⟨x , ei⟩ ei∥2 = limn→∞

(∥x∥2 −

n∑i=1

| ⟨x , ei⟩ |2),

which proves Parseval’s formula.

Example 5.14. Consider the space L2([0, 2π)],C) and the family, defined by

en(θ) :=1√2πeinθ, n ∈ Z, 0 ≤ θ ≤ 2π.

Note that the index n takes both positive and negative values. This is an orthonormal family in theHilbert space with the inner product

⟨x , y⟩ =∫ 2π

0x(θ)y(θ) dθ,

which can be checked directly.To prove that it is a maximal family will require several additional facts, in particular a complex

version of the Weierstrass trigonometric polynomial approximation theorem, and hence given withoutproof. The proved theorem asserts that the family (en) is a basis for our Hilbert space, and anyelement of this space can be uniquely represented by its Fourier coefficients.

The last, and probably most important question to ask is what is an ultimate source of maximalorthonormal families (note that the maximality is usually the property, which is most difficult toprove).

5.4 Self-adjoint operators

Recall that an operator L from a normed vector space X to a normed vector space Y is linear if

L(αx+ y) = αLx+ Ly, for any x, y ∈ X.

A linear operator is bounded if there is a constant C ≥ 0 such that ∥Lx∥ ≤ C∥x∥ for any x ∈ X.A basic fact that a linear operator is bounded if and only if it is continuous. For continuous (and,correspondingly, bounded linear operators) it is convenient to introduce the operator (or uniform)norm, which is given by the infimum among all possible constants C. Recall that

∥L∥ = supx=0

∥Lx∥∥x∥

= sup∥x∥=1

∥Lx∥ = sup∥x∥≤1

∥Lx∥.

Assume now that we deal with an inner product space (X, ⟨· , ·⟩).

Definition 5.15. A linear operator L : X −→ X is called self-adjoint if

⟨Lx , y⟩ = ⟨x , Ly⟩ , for all x, y ∈ X.

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It can be shown that if X is a Hilbert space then any self-adjoint operator is continuous and hencebounded.

For any linear operator L it is possible to consider an eigenvalue problem: To find all scalars λand non-zero vectors x ∈ X than Lx = λx. The scalars are called eigenvalues and the vectors areeigenvectors. This is a direct generalization of eigenvalue problem on finite dimensional vector spaces.

For the self-adjoint operators we have

Theorem 5.16. Let L be a self-adjoint operator. Then(a) ⟨Lx , x⟩ is real for any x ∈ X.(b) Let λ be an eigenvalue of L. Then λ is real. Moreover, λ ≥ 0 is L is positive semi-definite and

λ > 0 if L is positive definite.(c) Eigenvectors corresponding to distinct eigenvalues are orthogonal.(d)

∥L∥ = supx =0, y =0

| ⟨Lx , y⟩ |∥x∥∥y∥

= supx =0

| ⟨Lx , x⟩ |∥x∥2

= sup∥x∥=1

| ⟨Lx , x⟩ |.

Proof. (a) ⟨Lx , x⟩ = ⟨x , Lx⟩ = ⟨Lx , x⟩.(b) Let Lx = λx and x = 0. Then ⟨Lx , x⟩ = λ ⟨x , x⟩, and

λ =⟨Lx , x⟩⟨x , x⟩

,

which, by using (a), proves (b).(c) Let Lx1 = λ1x1 and Lx2 = λ2x2 such that λ1 = λ2. We have

⟨Lx1 , x2⟩ = λ1 ⟨x1 , x2⟩ = ⟨x1 , Lx2⟩ = λ2 ⟨x1 , x2⟩ =⇒ (λ1 − λ2) ⟨x1 , x2⟩ = 0.

(d) Let me start with the first expression for ∥L∥. Let

α = supx =0,y =0

| ⟨Lx , y⟩ |∥x∥∥y∥

= sup∥x∥=∥y∥=1

| ⟨Lx , y⟩ | .

To prove the formula, I will show that α ≤ ∥L∥ ≤ α. The first inequality is a consequence of theCauchy–Bunyakovskii–Schwarz inequality, since

| ⟨Lx , y⟩ | ≤ ∥Lx∥∥y∥ ≤ ∥L∥∥x∥∥y∥.

To prove the second inequality, start with ∥x∥ = 1

∥Lx∥2 = ⟨Lx ,Lx⟩ = ∥Lx∥⟨Lx ,Lx⟩∥Lx∥

≤ ∥Lx∥ sup∥x∥=1,∥y∥=1

| ⟨Lx , y⟩ | = ∥Lx∥α,

which, using∥L∥ = sup

∥x∥=1∥Lx∥ ≤ α,

finishes the proof.

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Now to the second part of (d), for which we introduce

β = supx =0

| ⟨Lx , x⟩ |∥x∥2

= sup∥x∥=1

| ⟨Lx , x⟩ |.

The goal is to show that β ≤ ∥L∥ ≤ β. The first follows from the Cauchy–Bunyakovskii–Schwarzinequality. For the second inequality we first note that

⟨Lx , x⟩ ≤ β∥x∥2.

Now, direct calculation together with the parallelogram law for two arbitrary x, y ∈ X for which∥x∥ = ∥y∥ = 1, yield

4Re ⟨Lx , y⟩ = ⟨L(x+ y) , x+ y⟩ − ⟨L(x− y) , x− y⟩≤ β

(∥x+ y∥2 + ∥x− y∥2

)= 2β

(∥x∥2 + ∥y∥2

)= 4β,

orRe ⟨Lx , y⟩ ≤ β.

Now,

⟨Lx , y⟩ = | ⟨Lx , y⟩ |eiθ =⟨Lx , ye−iθ

⟩.

Since ∥ye−iθ∥ = 1 then

Re⟨Lx , ye−iθ

⟩= Re eiθ| ⟨Lx , y⟩ | = | ⟨Lx , y⟩ |,

which implies∥L∥ = sup

∥x∥=∥y∥=1| ⟨Lx , y⟩ | ≤ β.

The big question in the last theorem is why we needed at all the last point (d), since it seems thatthe properties of the eigenvalues and eigenvectors of a self-adjoint operator are almost what we arelooking for. The answer is that we never proved that eigenvalues of L have to exist, or form a countableset. Actually, how simple examples show, it is possible to come up with self adjoint operators that donot have eigenvalues at all, or have an uncountable number of eigenvalues. To remedy this issue wewill have to choose only those self-adjoint operators, which are at the same time compact.

Exercise 5.4. Consider an operator M mapping L2[0, 1] to L2[0, 1] defined by

y(t) = tx(t)

(multiplication by t). Show that M is self-adjoint. Also show that M has no eigenvalues.

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5.5 Compact self-adjoint operators. Spectral theorem

Definition 5.17. A linear operator A : X −→ Y between normed vector spaces X and Y is compactif for any bounded sequence (xn)

∞n=1 in X from the sequence A(xn) in Y it is possible to choose a

convergent subsequence.

If an operator is compact than it is bounded and hence continuous.It is a remarkable fact that any compact and self-adjoint operator acting in an infinite dimensional

inner product space possesses similar properties. In particular, there are at most countably infinitenumber of nonzero real eigenvalues, each one of finite multiplicity, and the corresponding eigenvectorsform a maximal orthonormal family if the operator is in addition injective.

Theorem 5.18. Let (X, ⟨· , ·⟩) be an infinite dimensional inner product space and let A : X −→ X bea compact and self-adjoint operator with an infinite dimensional range. Then2

(a) There exists an infinite sequence (λn) of eigenvalues of A and an infinite sequence (pn) ofcorresponding eigenvectors that satisfy

|λ1| = ∥A∥, |λ1| ≥ |λ2| ≥ . . . ≥ |λn| ≥ . . . , λn = 0, limn→∞

λn = 0,

Apn = λnpn, ⟨pk , pl⟩ = δkl, k, l ≥ 1,

|λ1| =| ⟨Ap1 , p1⟩ |

∥p1∥2= sup

x =0

| ⟨Ax , x⟩ |∥x∥2

,

|λn| =| ⟨Apn , pn⟩ |

∥pn∥2= sup

x =0,⟨x ,pk⟩=0, 1≤k≤n−1

| ⟨Ax , x⟩ |∥x∥2

, n ≥ 2.

(b) For any vector x ∈ X,

Ax =

∞∑n=1

λn ⟨x , pn⟩ pn.

(c) Let λ be any nonzero eigenvalue of A. Then there exists n ≥ 1 such that λn = λ, moreover theset of such λn is finite.

(d) If L is injective (i.e., kerL = 0) then (pn) is a maximal orthonormal family, and hence abasis if X is a separable Hilbert space.

Proof. (i) First, I will prove that there exist λ1 and p1, ∥p1∥ = 1 such that 0 < |λ1| = ∥A∥ =| ⟨Ap1 , p1⟩ | and Ap1 = λ1p1.

Recall that ∥A∥ = sup∥x∥=1 | ⟨Ax , x⟩ | and ∥A∥ > 0 due to the assumption that the range of Ais infinite dimensional. Therefore, there exists a sequence (xn) such that | ⟨Axn , xn⟩ | → ∥A∥ with∥xn∥ = 1 for all n. Introduce λ1 = limn→∞ | ⟨Axn , xn⟩ | = ∥A∥ > 0. Since A is compact that from thesequence (Axn) it is possible to choose a convergent subsequence, say (Axl), which converges in X.

Consider

∥Axl − λ1xl∥2 = ∥Axl∥2 − 2λ1 ⟨Axl , xl⟩+ λ21 ≤≤ ∥A∥2 − 2λ1 ⟨Axl , xl⟩+ λ21 → 0, l → ∞.

2I am copying a very detailed proof from Ciarlet, Linear and Nonlinear Functional Analysis with Applications, SIAM,2013

127

Therefore Axl − λ1xl → 0 when l → ∞.Define p1 := liml→∞ xl. Then from the above and rearrangements

p1 = liml→∞

xl = liml→∞

(− 1

λ1(Axl − λ1xl) +

1

λ1Axl

)=

1

λ1liml→∞

Axl,

and ∥p1∥ = 1 since ∥xl∥ = 1 for all l.Now

Ap1 = A( liml→∞

xl) = liml→∞

Axl = λ1p1,

which proves that λ1 = ∥A∥ or λ1 = −∥A∥ is an eigenvalue of A with unit eigenvector p1.(ii) Now let me make a second step. Consider the subspace X2 := x ∈ X : ⟨x , p1⟩ = 0. Then

the image of A(X2) of X2 is again in X2, since

⟨Ax , p1⟩ = ⟨x ,Ap1⟩ = λ1 ⟨x , p1⟩ = 0.

Therefore, we can consider the restriction A2 of A to X2, which is again compact and self-adjointoperator. Moreover, A2 = 0. Indeed, assuming that A2 = 0 then

A(x− ⟨x , p1⟩ p1) = 0, x ∈ X,

because x− ⟨x , p1⟩ p1 ∈ X2. On the other hand, we find

Ax = λ1 ⟨x , p1⟩ p1

for all x ∈ X, which contradicts the fact that the image of A is infinite dimensional. Now thearguments from (i) are repeated verbatim for A2, and we find that there exists λ2 and p2, ∥p2∥ = 1,Ap2 = λ2p2, ⟨p1 , p2⟩ = 0,

0 ≤ |λ2| = ∥A2∥ = supx =0, x∈X2

| ⟨A2x , x⟩ |∥x∥2

= supx =0, x∈X2

| ⟨Ax , x⟩ |∥x∥2

≤ supx =0

| ⟨Ax , x⟩ |∥x∥2

= |λ1|.

We can continue. The only thing to check every step that An+1 = 0. This can be argued similarly,by considering x −

∑nm=1 ⟨x , pm⟩ pm, which belongs to Xn+1, and assuming that An+1 = 0, thus

getting a contradiction.(iii) Now let me show that limn→∞ λn = 0. Assume that it is not true, i.e., |λn| ≥ δ > 0 for all n.

Now consider a sequence (λ−1n pn), which is bounded and therefore has a convergent subsequence, say

λ−1m pm. Since A is compact, we can find yet another convergent subsequence (Aλ−1

l pl) = (pl). Butsince all pn are orthogonal then

∥pk − pl∥2 = ∥pk∥2 + ∥pl∥2 = 2,

and hence we cannot have a limit. Contradiction. Part (a) is done.(iv) Consider xn := x −

∑nk=1 ⟨x , pk⟩ pk. By construction, and using the notations introduced

above, we have that xn ∈ Xn+1 = x ∈ X : ⟨x , pk⟩ = 0, 1 ≤ k ≤ n. Since xn is orthogonal to∑nk=1 ⟨x , pk⟩ pk, then, by the Pythagoras theorem, ∥xn∥ ≤ ∥x∥. Moreover, Axn = An+1xn, and

∥An+1∥ = |λn+1|.

128

Now

∥Ax−n∑k=1

λk ⟨x , pk⟩ pk∥ = ∥Axn∥ = ∥An+1xn∥ ≤ ∥An+1∥∥xn∥ ≤ |λn+1|∥x∥,

which proves (b).(v) Here let me prove that we found all nonzero eigenvalues. Assume opposite, let Aλ = λp, and

λ = λn for all n. We know that p must be orthogonal to all pk. Using (iv)

Ap = limn→∞

n∑k=1

λk ⟨p , pk⟩ pk = 0,

which is a contradiction to λ = 0.(vi) Here I will show that each eigenvalue has a finite multiplicity. By part (ii) there exist orthog-

onal eigenvectors pn, where n is such that λn = λ, n ≥ 1 (note that the number of vectors is finitesince λn → 0), such that

Apn = λpn.

Therefore, span(pn) ⊆ p ∈ X : Ap = λp. To prove the claim we need to show that this inclusionis an equality. Assume contrary, there is p such that Ap = λp, and p /∈ span(pn). We can use theGram-Schmidt pronominalization to produce p ∈ span(pn), p and ⟨p , pn⟩ = 0 for any n. Moreover,Ap = λp, because p is a linear combination of eigenvectors with the eigenvalue λ. We also have that⟨p , pn⟩ = 0 for all n ≥ 1 because A is self adjoint. Now use the same argument as in (v) to reach acontradiction.

(vii) I will show that thatkerA = (span(pn))

⊥.

Here Z⊥ is the orthogonal complement of Z ⊆ X, defined as Z⊥ := x ∈ X : ⟨x , z⟩ = 0, z ∈ Z.Let x ∈ X be such that Ax = 0. Then for any n ≥ 1,

⟨x , pn⟩ =1

λn⟨x , λnpn⟩ =

1

λn⟨Ax , pn⟩ = 0,

which means that x ∈ (span(pn))⊥. Conversely, if x ∈ (span(pn))

⊥ then ⟨x , pn⟩ = 0 for all n ≥ 1 andthus Ax = 0 by (iv).

Remark 5.19.

• If the range of A is finite dimensional, of dimension N then we can show that there are N realeigenvalues that satisfy

|λ1| ≥ |λ2| ≥ . . . ≥ |λN | > 0,

with the corresponding eigenvectors (pn)Nn=1. Other statements should be modified to change ∞

for N .

• If A is not injective, i.e., kerA = 0, then the last point in the theorem should be replaced withthe statement that any vector y ∈ range (A) can be represented by a convergent Fourier series

y =

∞∑k=1

⟨y , pk⟩ pk.

129

We did not see many examples of compact or self adjoint operator (we saw none, to be precise),and here is an example, which is of paramount importance for us.

Example 5.20. Consider the space L2 (I;C) = L2(I) of square integrable complex valued functionon the interval I = (a, b) and the integral operator

K : x ∈ L2(I) 7→∫Ig(t, s)x(s) ds,

where g(t, s) is assumed to be an element of L2(I × I). The inner product is defined in the usual way:

⟨x , y⟩ =∫Ix(s)y(s) ds.

The claim is that K is compact and, moreover, if g(t, s) = g(s, t) then K is self-adjoint.To show that K is compact, consider the basis

ϕn(t) =1√b− a

exp

(2πni

t− a

b− a

),

which is a generalization of the standard complex Fourier basis for the interval (0, 2π) to the arbitraryinterval (a, b). Moreover, it can be shown that

ψn,m(t, s) = ϕn(t)ϕm(s)

is an orthonormal basis for L2(I × I). Then the Fourier series theorem says that k(t, s) can berepresented by a Fourier series, and

∥k − kN∥ → 0, N → ∞,

wherekN =

∑|m|,|n|≤N

⟨k , ψn,m⟩ψn,m.

Denoting KN the integral operator with the kernel kN , we have that

∥K −KN∥ ≤ ∥k − kN∥ → 0.

The dimension of the range of KN is 2N+1, and therefore KN is compact because any linear operatorwith finite dimensional range is compact (prove it). Now, we have a sequence of compact operatorsconverging to K, and by using the fact that the subspace of compact operators is closed (it requiresproof), we find that K is compact.

The proof that K is self adjoint is left as an exercise.

Exercise 5.5. Consider the integral operator y = Kx defined as

y(t) =

∫Ik(t, s)x(s) ds,

which maps L2(I) to L2(I). Assuming that k(t, s) = k(s, t), show that K is self-adjoint.

130

5.6 Sturm–Liouville operators. Symmetric operators

Finally, we return back to differential equations. I consider a problem of the following form

−(p(x)u′(x)

)′+ q(x)u(x) = f(x), x ∈ (a, b), (5.1)

with some prescribed boundary conditions (see below).Any second order linear equation of the form

p0(x)u′′(x) + p1(x)u

′(x) + p2(x)u(x) = g(x)

can be put into form (5.1) (see homework problems).In the following I will consider the eigenvalue problem in the operator form

Lu = λu, (5.2)

where

Lu =1

r(x)(−(p(x)u′)′ + q(x)u). (5.3)

I assume that p(x), p′(x), q(x), and r(x) are continuous real-valued functions on [a, b], and p(x) > 0

and r(x) > 0 for all x ∈ [a, b]; let H be a complex Hilbert space u :∫ ba |u(x)|

2r(x) dx <∞ with theinner product

⟨u , v⟩ =∫ b

au(x)v(x)r(x) dx.

I denote D the set of all twice continuously differentiable functions on [a, b], satisfying

B1u = β1u(a) + γ1u′(a) = 0, B2u = β2u(b) + γ2u

′(b) = 0, (5.4)

where |β1|+ |γ1| > 0 and |β2|+ |γ2| > 0. Clearly, D ⊂ H.

Definition 5.21. The eigenvalue problem (5.2), (5.3), (5.4) is called the regular Sturm–Liouvilleproblem, and operator L defined by (5.3) is called the Sturm–Liouville operator.

To emphasize that the set D is the set on which operator L is defined, I will denote DL = D.It is very tempting to use the spectral theorem for compact self-adjoint operators, and conclude

that L has a countable number of eigenvalues and eigenvectors that form a basis for H, however,note that L is not even defined on all H, since there are a lot of functions in H which are not twicecontinuously differential. For the following exposition, however, it is important to mention that L isdensely defined on H because the set DL is dense in H (I state this fact without proof). But eventhis is not all. A differential operator is an example of an unbounded operator (think of A := d

dxacting on the sequence (sinnx)). Therefore, the Sturm–Liouville operator is unbounded and none ofthe studied machinery will work for it.

However, Sturm–Liouville operator is symmetric.

Definition 5.22. Linear operator L : DL −→ RL is called symmetric if DL is dense in H, RL ⊆ H,and

⟨Lu , v⟩ = ⟨u , Lv⟩

for all u, v ∈ DL.

131

Carefully note that the definition depends on the domain DL.

Lemma 5.23. The Sturm–Liouville operator is symmetric.

Proof. Let u, v ∈ DL. Then we have

⟨Lu , v⟩ − ⟨u , Lv⟩ =∫ b

a

(−(pu′)′v + quv + (pv′)′u− quv

)dx

=

∫ b

a

(−(pu′)′v + (pv′)′u

)dx.

By integrating by parts, I find

⟨Lu , v⟩ − ⟨u , Lv⟩ = p(uv′ − u′v)|ba.

From (5.4) we have, since the constants are real, that if u, v ∈ DL then u, v ∈ DL:

β1u(a) + γ1u′(a) = 0,

β1v(a) + γ1v′(a) = 0,

and hence the determinant ∣∣∣∣u(a) u′(a)v(a) v′(a)

∣∣∣∣must vanish. The same is true for ∣∣∣∣u(b) u′(b)

v(b) v′(b)

∣∣∣∣ .Therefore,

⟨Lu , v⟩ − ⟨u , Lv⟩ = 0.

Remark 5.24. I actually showed that for any two u, v ∈ DL it is true that

⟨Lu , v⟩ − ⟨u , Lv⟩ = p(uv′ − u′v)|ba,

which is called Green’s formula, from which, it follows that

uLv − vLu =(p(uv′ − u′v)

)′,

which is called Lagrange’s identity. I thus have that L is symmetric if and only if p(uv′ − u′v)|ba = 0.

Exactly as for the bounded self-adjoint operators, it can be proved

Theorem 5.25. Let L be a symmetric operator on a Hilbert space H. Then if λ is an eigenvalueof L then λ ∈ R. If p1 and p2 are two eigenvectors of L corresponding to distinct λ1 and λ2 then⟨p1 , p2⟩ = 0. If (pn) is an orthonormal family of eigenvectors of L with the corresponding eigenvalues(µn), and (pn) forms a basis for H, then for every y ∈ DL one has

Ly =

∞∑n=1

µn ⟨y , pn⟩ pn.

132

The proof is left as an exercise. I remark that the fact that (pn) is a basis for H is included as anassumption in the last theorem, and therefore this theorem requires some other method to find suchbasis to be able to be applied.

Now assume that L is injective (this is equivalent to saying that 0 is not an eigenvalue of L). Thismeans that there exists the inverse operator L−1. In this case I have the following key statement.

Theorem 5.26. Let L be the Sturm–Liouville operator given by (5.3), (5.4) and 0 is not an eigenvalueof L. Then there is a continuous function g(x, y) defined for all a ≤ x, y ≤ b such that

(L−1u)(x) =

∫ b

ag(x, y)u(y)r(y) dy

for all u in the range RL. Furthermore, g(x, y) can be chosen so that it is real valued and g(x, y) =g(y, x). Consequently, L−1 is a compact self adjoint operator.

I will give a constructive proof of this theorem, which shows how to actually find such g(x, t).Function g(x, t) is called the Green function of problem (5.2)–(5.4). Note that if we know Green’sfunction than any solution to (5.1) with boundary conditions (5.3) can be written as

u(x) =

∫ b

ag(x, y)f(y) dy.

Proof. Consider second order differential equation Lu = λu. According to the general theory it has afundamental set of solutions u1(x, λ) and u2(x, λ), and the general solution is given by u = c1u1+c2u2.The boundary conditions imply that

c1B1u1 + c2B1u2 = 0, c1B2u1 + c2B2u2 = 0.

This system has a nontrivial solution if and only if

∆(λ) =

∣∣∣∣B1u1 B1u2B2u1 B2u2

∣∣∣∣ = 0.

Now consider the problem Lu = 0, and let uj , j = 1, 2 be the solutions to Lu = 0 with Bjuj = 0. Since,by assumption, λ = 0 is not an eigenvalue, then the determinant above cannot be zero, and henceBiuj = 0 for i = j. This means that uj , j = 1, 2 are linearly independent and form a fundamentalsolution set. Consider the Wronskian

W (x) = det

[u1 u2u′1 u′2

]and yet another, non-homogeneous differential equation Lu = v for some function v. According to thegeneral theory, the general solution to this equation is

u = c1u1 + c2u2 + up,

where up can be found by the variation of the parameter formula, namely

up(x) =

∫ x

a

u1(x)u2(y)− u2(x)u1(y)

p(x)W (x)v(y)r(y) dy.

133

We want to choose the coefficients c1 and c2 so that when v is continuous then u ∈ DL, i.e., it satisfiesthe boundary conditions (5.4). If we set c2 = 0 and

c1 = −∫ b

a

u2(y)

p(x)W (x)v(y)r(y) dt,

then the solutions becomes

u(x) = −∫ b

x

u1(x)u2(y)

p(x)W (x)v(y)r(y) dy −

∫ x

a

u2(x)u1(y)

p(x)W (x)v(y)r(y) dy =

∫ b

ag(x, y)v(y)r(y) dy,

where

g(x, t) =

−c−1u1(x)u2(y), a ≤ x ≤ y ≤ b,

−c−1u2(x)u1(y), a ≤ y ≤ x ≤ b,

where c = p(x)W (x), which does not depend on x due to Lagrange’s identity.One can check that u satisfies the boundary conditions, and the kernel g(x, y) is continuous.

Moreover, since u1, u2 are real valued then g(x, y) is real and g(x, y) = g(y, x), which implies that L−1

is compact and self-adjoint.

The next question is what to do if 0 actually an eigenvalue of L. In this case consider Sturm–Liouville operator L− λI, where λ is a real number and not an eigenvalue of L. Such λ always existsbecause Sturm–Liouville operator is symmetric, and symmetric operators have orthogonal eigenvec-tors. Our Hilbert space is separable, and hence can have at most a countable number of eigenvectorsand, correspondingly, eigenvalues. Hence we can always pick a real number which is not an eigenvalue.

Corollary 5.27. Let L be a Sturm–Liouville operator (5.3) and assume that a real number λ is notan eigenvalue of L. Then there is a Green’s function g(x, y, λ) continuous for all a ≤ x, y ≤ b andsuch that

((L− λI)−1u)(x) =

∫ b

ag(x, y, λ)u(y)r(y) dy,

for all u in the range of L− λI. Moreover, the operator (L− λI)−1 is compact and self-adjoint.

Why would be bother about the inverse? Because, the eigenvalue problem Lu = λu can berewritten as 1

λu = L−1u, i.e., the inverse operator has the same eigenvectors and eigenvalues inverseto the original one.

Finally, I can state and prove the main result of this last part of our course.

Theorem 5.28. Let L be a Sturm–Liouville operator. Then there exists a sequence of real numbers(µn) and an orthonormal basis (ϕn) of H such that each ϕn is a C(2) function, and Lϕn = µnϕn.Moreover, each real number µ appears at most twice in the sequence (µn). Furthermore, each functionv ∈ H can be written in terms of the Fourier series

v =

∞∑n=1

⟨v , ϕn⟩ϕn.

134

Proof. Pick a real number λ, which is not an eigenvalue of L. The operator (L−λI)−1 is then compactand self-adjoint, therefore it has the sequence of eigenvalues (µn) and the corresponding eigenvectors(ϕn), that forms a basis of H (this follows from the fact that (L− λI)−1 is injective). It follows fromthe last theorems that µn = −λ − µ−1

n is a sequence of eigenvalues of L with the same eigenvectors.Each ϕn is a C(2) function since it is a solution of the second order differential equation with continuouscoefficients. Also, each eigenvalue appears at most twice since second order ODE can have at mosttwo linearly independent solutions.

5.7 Green’s function. Properties. Examples

The constructive way to produce Green’s function can be used to infer the following properties ofGreen’s function:

• Green’s function g(x, y) for all x = y solves the homogeneous equation Lg = 0 and satisfies thehomogeneous boundary conditions B1g = B2g = 0.

• It is a continuous function of variable x.

• At the point x = y its derivative has a jump discontinuity whose value is given by

g′x|x=y+0 − g′x|x=y−0 = − 1

p(y).

Note that I use ”−” sign, because in my definition of the Sturm–Liouville operator I also usedminus. In some textbooks, this minus is omitted.

Exercise 5.6. Prove the stated properties.

These properties are enough to effectively construct Green’s function.

Example 5.29. Consider L = d2

dx2with the boundary conditions u(0) = u(1) = 0. To find Green’s

function, I write

g(x, y) =

C1x+ C2, x ≤ y

C3x+ C4, x ≥ y.

Using the boundary conditions, I find that C2 = 0 (from the left boundary condition) and C3 = −C4.Therefore,

g(x, y) =

C1x, x ≤ y

C3(x− 1), x ≥ y.

The continuity and the jump conditions together imply that

C1y = C2(y − 1),

C3 − C1 = 1,

from where we findC1 = y − 1, C3 = y.

135

Finally,

g(x, y) =

(y − 1)x x ≤ y

y(x− 1) x ≥ y.

Now any solution to the problem Lu = f(x) can be written as

u(x) =

∫ b

ag(x, y)f(y) dy.

Example 5.30. Let

L = − d2

dx2+ ω2, ω > 0, u(0) = u(1) = 0.

It is convenient to express the general solution to the homogeneous equation as

A sinhωx+B coshωx.

(Here the idea is to use such basis functions that satisfy the initial conditions u(0) = 1, u′(0) = 0 andu(0) = 0, u′(0) = 1, recall the notion of the principal matrix solution). Thus we have

g(x, y) =

C1 sinhωx+ C2 coshωx, x ≤ y

C3 sinhωx+ C4 coshωx, x ≥ y.

From the boundary conditions we find

g(x, y) =

C1 sinhωx, x ≤ y

C3 sinhω(1− x), x ≥ y.

The second line can be inferred directly, but also can be argued as follows. From the second boundarycondition we have that

C3 = −C4 coshω

sinhω.

Therefore,C4

sinhω(sinhω coshωx− coshω sinhωx) = A sinhω(1− x),

where I usedsinh(α± β) = sinhα coshβ ± sinhβ coshα.

Now the continuity and jump conditions becomes

C1 sinhωt = C3 sinhω(1− y),

−ωC3 coshω(1− y)− C1ω coshωy = −1.

Multiplying the first equation by ω coshω(1− y) and the second by sinhω(1− y) and adding, we find,after again using the formula for sinh(α+ β), that

C1 =sinhω(1− y)

ω sinhω, C3 =

sinhωy

ω sinhω.

136

Finally,

g(x, y) =

sinhωx sinhω(1− y)

ω sinhω, x ≤ y

sinhω(1− x) sinhωy

ω sinhω, y ≥ t.

Exercise 5.7. Consider the boundary value problem

xw′′ + (a− x)w′ = −λw, w(L) = w(R) = 0,

where a, L,R are real constants.Note that the differential operator on the left is not a Sturm–Liouville operator (it is not symmet-

ric). By choosing properly a multiplicative factor v(x) show that you can rewrite this problem in thesymmetric form

(−p(x)u′)′ + q(x)u = −λr(x)u.What are p(x), q(x) and r(x)?

Exercise 5.8. Consider the regular Sturm–Liouville problem

(−p(x)y′)′ + q(x)y = λr(x)y

with the boundary conditions

β1y(a) + γ1y′(a) = 0, β2y(b) + γ2y

′(b) = 0.

Prove that if y1, y2 are two eigenvectors corresponding to the same eigenvalue λ then y1 and y2 arelinearly dependent (i.e., all eigenvalues for this problem are simple).

Exercise 5.9. Find Green’s function for

x3y′′ + 3x2y′ + xy = f(x), y(1) = 0, y(2) + 2y′(2) = 0.

Exercise 5.10. For the boundary problem (note the boundary conditions!)

y′′ + y = f(x), y(0) = y(π), y′(0) = y′(π)

find Green’s function.

Exercise 5.11. The violin operator is defined to be the Sturm–Liouville operator

Ly = −τρy′′

on 0 ≤ x ≤ l with y(0) = y(l) = 0. τ denotes the tension in the violin string, ρ is the density of thestring, and l the length of the string. Show that λn = n2α are the eigenvalues of L for n = 1, 2, . . . andappropriate choice of α. The first eigenvalue λ1 is defined to be the pitch of the violin operator. Showthat one can increase the pitch be either increasing the tension, decreasing the length, or decreasingthe density.

5.8 Appendix

5.8.1 Delta-function

5.8.2 Singular Sturm–Liouville problem. Examples

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