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Chapter 5 Sturm-Liouville Theory 5.1 Oscillation and Separation Theor y Consider the dierential equation a 2 (x)y + a 1 (x)y + a 0 (x)y  = 0 (5.1.1) where  a 2 (x) is not zero for all  x [a, b] ,  a i (x) C [a, b]. Rewrite (5.1.1) in the form y +  a 1 a 2 y +  a 0 a 2 y  =  y + p(x)y + q (x)y  = 0 Dene k(x) = e   p(s)ds , Then d dx (k(x)y ) + k (x)q (x)y  = 0 or (ky ) + g (x)y  = 0 (5.1.2) Dene the dierential operators L(y) = (ky ) + g (x)y M (y) = a 2 y + a 1 y + a 0 y  (5.1.3) The adjoint of M is dened by M (y) = (a 2 y) (a 1 y) + a 0 y 1

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2   CHAPTER 5. STURM-LIOUVILLE THEORY 

M (y) = a2y + (2a2 − a1)y + (a2 − a1 + 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   ,   ba

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 

a2 =  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 theoremsTheorem 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,{z n}, select a subsequence {xn} suchthat  xn → x. Then

0 = limn→∞ y(xn) = y( x).Also,

y( x) = limn→∞

y(xn) − y( x)

xn − x   = 0

and so y(x) = 0 by uniqueness.

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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)

=

0 and   u2(x1) = 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) = −u1(x0)u2(x0)

W (u1, u2)(x1) = −u1(x1)u2(x1)

Since   u1(x0)   >   0, W (x0)   <   0.   Because the Wronskian of   u1, u2   cannot change sign,W (x1)  < 0.  But  u1(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

 ∈C 1(a, b), g

 ∈C 0[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 equal to   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).

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4   CHAPTER 5. STURM-LIOUVILLE THEORY 

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

u2L1(u1) − u1L1(u2) =   ddx

(k(x)(u2u1 − u1u2).

We also haveL1(u2) − L2(u2) = (g1 − g2)u2.

Henceu2L1(u1) − u1[L2(u2) + (g1 − g2)u2] = (k(x)(u2u1 − u1u2))

or

k(x)(u2u1 − u1u2)|x2x1  =

   x2x1

(g2 − g1)u1u2dx > 0.

However, the left hand side reduces to

k(x2)u2(x2)u1(x2) − k(x1)u2(x1)u1(x1).   (*)

Since  u2(x2) ≥ 0, u1(x2) <  0 and u2(x1) ≥ 0, u1(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)u1(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 ∈  C 1(a, b), g ∈  C 0[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  ∈ C 0[a, b]  and  z (x)  is a non trivial solution of 

z  + q (x)z  = 0

where z (a) = z (b) = 0.

If     ba

( p − q )z 2dx ≥ 0

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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) = 0 in (a, b]. Then

z (z  + qz ) = 0

z 2

y (y + py) = 0

or

zz  − z 2 y

y  = z 2( p − q )

orz 

y(yz  − zy ) = z 2( p − q ).

Now note that

limx→a

z (x)

y(x) = lim

z→a

z (a)

y(a)

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

   ba

z y (yz  − zy )dx =    ba

z 2( p − q )dx ≥ 0.

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

y(yz  − zy )|ba −

   ba

(yz  − zy )(z 

y) =

   ba

z 2( p − q )dx ≥ 0,   (∗)

or

−   ba

(yz  − zy )2

y2  dx ≥ 0

or

0 ≥    ba

(yz  − zy

)2y2

  dx.

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

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6   CHAPTER 5. STURM-LIOUVILLE THEORY 

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

   b

a

z 2( 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

y2  dx < 0,

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

Theorem 5.1.4.   Let  ki

 ∈C 1[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     ba

(k1 − k2)(z )2 + (g2 − g1)z 2dx ≥ 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 (z 2/y) toobtain the Picone formula,   b

a

(k1 − k2)(z )2 + (g2 − g1)z 2dx +

   ba

k2(yz  − zy

y  )2dx =

 z 

y(k1yz  − k2yz )|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 ≡ g1,k2 ≡ k1  on   [a, b]  and 

z (a) = z (b) = 0.

then  y(x)  has a zero in  (a, b).

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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) + β 11y(b) + β 12y(b) = γ 1   (5.1.5)

B2(y) = α21y(a) + α22y(a) + β 21y(b) + β 22y(b) = γ 2.

Here   a2, a1, a0 ∈   C [a, b], a2(x) = 0, αij, β ij, γ i   are constants. Equations (5.1.4) and (5.1.5)constitute what is called a boundary value problem(BVP). If   β 11   =   β 12   = 0 and   α21   =

α22   = 0, the boundary conditions are separated. If  α12  =  β 12  = 0 and  α21  =  β 21  = 0, andγ 1 =  γ 2 = 0 the boundary conditions are periodic. If  α12 =  β 11 =  β 12  =  α21 =  β 21 =  β 22 = 0we have the initial conditions  y(a) =   γ 1

α11, y(a) =   γ 2

α21.  Note that the boundary operators  Bi

are linear. We refer to {f (x), γ 1, γ 2} as the data of the BVP.

Theorem 5.1.1.  The BVP (5.1.4)-(5.1.5) with data  {0, γ 1, γ 2} has a unique solution iff the BVP 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, γ 1, γ 2}, 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) = γ 1

α1B2(y1) + α2B2(y2) = γ 2,

or

  B1(y1)   B1(y2)

B2(y1)   B2(y2)   α1

α2  =   γ 1

γ 2   (5.1.6)

Since α1, α2  are unique,

det

  B1(y1)   B1(y2)B2(y1)   B2(y2)

= 0.   (5.1.7)

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8   CHAPTER 5. STURM-LIOUVILLE THEORY 

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

w =  ξ 1y1 + ξ 2y2   (5.1.8)

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

  ξ 1ξ 2

 =

  00

and since the coefficient matrix is nonsingular we must have  ξ 1  =  ξ 2 = 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) = γ 1, B2(y) = γ 2

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  y p   be a particularsolution of  M (y) = f. Seek a solution of the BVP in the form

y =  α1u1 + α2u2 + y p.

Solve for  α1, α2  such that

B1(y) = α1B1(u1) + α2B1(u2) + B1(y p) = γ 1

B2(y) = α1B2(u1) + α2B2(u2) + B2(y p) = γ 2

or   B1(u1)   B1(u2)B2(u1)   B2(u2)

  α1

α2

 =

  γ 1 − B1(y p)γ 2 − B2(y p)

.

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

  γ 1 − B1(y p)γ 2 − B2(y p)

.

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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-adjoint

form 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 correponding

separated boundary conditions are given by

B1(y) = α1y(a) + α2y(a) = 0 (5.2.11)

B2(y) = β 1y(b) + β 2y(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   ba

[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

β 1y(b) + β 2y(b) = 0

is self-adjoint.

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10   CHAPTER 5. STURM-LIOUVILLE THEORY 

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

   b

a

[vL(u)−

uL(v)]dx =  k(x) (uv−

vu)|b

a ≡P (u, v)

|b

a

  (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 = 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) = 0β 1y(b) + β 2y(b) = 0.

Then 

(λ − µ)

   ba

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

−µ)  

  b

a

uvp(x)dx =    b

a

[u(L(v)

−v(L(u)] dx =

= k(x) (uv − vu) |ba = 0.

Theorem 5.2.3.  The eigenvalues of the RSLBVP are real.

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5.2. STURM-LIOUVILLE BOUNDARY VALUE PROBLEMS    11

Proof.   If L(u) + λpu = 0,

then L(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)

   ba

|u|2 pdx = 0.

Since p > 0, and |u|2 ≡ 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 = 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 periodic

boundary 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.

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12   CHAPTER 5. STURM-LIOUVILLE THEORY 

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,   say λ = −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) = 0

As 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 + 12 π

l2

, n = 0, ±1, ±2, · · ·and 

yn(x) = sin

n +

 1

2

 π

l x

.

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5.2. STURM-LIOUVILLE BOUNDARY VALUE PROBLEMS    13

3.   y(0) = y (l) = 0If   λ   = 0, then   y   =   ax +  b   and   y(0) =   y(l) = 0   if   a   = 0. Hence the constant 

 function is an eigenfunction corresponding to the eigenvalue 0. It is easy to verify 

than an eigenvalue for this problem cannot be negative. If   λ  =  µ2 >  0, then   y(x) =a cos µx + b sin µx  and  y(0) = 0  if  b = 0. Then 

y(l) = −aµ sin µl = 0

if µl =  nπ, n = 0, ±1, ±2, · · ·

Hence eigenvalues and eigenfunctions are given by 

λn  =

l 2

, yn(x) = sin

l  x

, 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 − µ

1 + µ

and by graphing each side it is easy to see that this equation is satisfied only when µ = 0. Thus we have  a = 0   and hence  b = 0. If  λ = 0, y  =  ax + b  and the boundary conditions 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 = 1, then a   = 0   and hence   b   = 0   and so   λ   = 0   is not an eigenvalue. If   λ   =   µ2 >   0   then y =  a cos µx + b sin µx  and to satisfy the boundary conditions we need 

a + µb = 0, a cos µl + b sin µl = 0

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14   CHAPTER 5. STURM-LIOUVILLE THEORY 

or 

tan µl =  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 sinh µx + b cosh µx

and 

y(0) = b  =  y(l) = a sinh µl + b cosh µl

while 

y(0) = aµ  =  y (l) = aµ cosh µl + bµ sinh µl

or      sinh µl   cosh µl − 1

µ(cosh µl − 1)   µ sinh µl

  ab

 =

  00

.

The determinant of he coefficient matrix is  2µ(cosh µl−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 sin µx + b cos µx.

We see that 

y(0) − y(l) = 0 if  b − (a sin µl + b cos µl) = 0

and 

y(0) − y(l) = 0 if  aµ − (aµ cos µl − bµ sin µl) = 0

or      − sin µl   1 − cos µlµ(1 − cos µl)   µ sin µl

  ab =   0

0 .

We obtain a nontrivial solution if 

−µ sin2 µl − µ(1 − cos µl)2 = 0.

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5.3. GREEN’S FUNCTIONS    15

Expanding the second term and simplifying we arrive at  cos µl = 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 sin 2nπ

l  x + bn cos

 2nπ

l  x

5.3 Green’s FunctionsIn 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) + β i3y(b) + β i2y(b)

and  k ∈ C 1(a, b), k(x) >  0, x ∈ [a, b].

Definition 5.3.1.   A Green’s function for BVP λ   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 )

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16   CHAPTER 5. STURM-LIOUVILLE THEORY 

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

3. for all  ξ 

 ∈(a, b), Bi(G) = 0.

Theorem 5.3.2.   If  λ   is not an eigenvalue of BVP λ, then the boundary value problem has a 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) = 0B2(y) = β 1y(b) + β 2y(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

Bu2(ξ ) − Au1(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 )

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5.3. GREEN’S FUNCTIONS    17

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

−   ξx0

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 = 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

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18   CHAPTER 5. STURM-LIOUVILLE THEORY 

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

Au1(ξ ) + Bu2(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

uξ(ξ ) =  1

k(ξ ).

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5.3. GREEN’S FUNCTIONS    19

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) = 0Seek 

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

where 

uξ  = 0, x > ξ > 0

uξ(ξ +) = 0, uξ(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 obtains A = −1/2, B = −1/4 + ξ/2

and hence 

G(x, ξ ) = 1

2x −  1

4 +  ξ

2 ,   0 ≤ x ≤ ξ 

(x − ξ ) −   14 +   ξ

2 −   x

2 , x < ξ   ≤ 1

or 

G(x, ξ ) = −1

4 +

 |x − ξ |2

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20   CHAPTER 5. STURM-LIOUVILLE THEORY 

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

consider 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) + β i3y(b) + β i2y(b)

and  k ∈

C 1(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 (uv − uv)

and we note that integration gives the Green’s formula   ba

[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 = ξ 

G(x, ξ )Lλ(y) − yLλ(G(x, ξ )) =  d

dx[k(Gy − yG)]

which implies    ξ−a

GLλ(y)dx =  k(Gy − Gy)]ξ−

a

and    bξ+

GLλ(y)dx =  k(Gy − Gy)]bξ+.

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5.3. GREEN’S FUNCTIONS    21

Hence

   b

a

GLλ(y)dx =  k(Gy − Gy)]ba − k(Gy − Gy)]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 − Gy)]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 − Gy)]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) =

   ba

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) =

   ba

G(x, ξ )f (ξ )dξ 

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

separated boundary conditions.

Theorem 5.3.6.   If   λ   is not and eigenvalue of BVP λ   where the boundary conditions are separated, that is 

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

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22   CHAPTER 5. STURM-LIOUVILLE THEORY 

then the unique solution of Lλ(y) = f 

B1(y) = γ 1, B2(y) = γ 2

is given by 

y(x) =  γ 2B2(y1)

y1(x) +  γ 1

B1(y2)y2(x) +

   ba

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) = 0 and  B2(y1) = 0.

Proof.   Since  B1(G) = B2(G) = 0,

B1(y) =  γ 1B1(y2)

B1(y2) = γ 1

and similarly B2(y) = γ 2To see that the nonhomogeneous differential equation is satisfied, we consider

u(x) = ba

 G(x, ξ )f (ξ )dξ 

= xa

  G(x, ξ )f (ξ )dξ  + bx

 G(x, ξ )f (ξ )dξ 

Thenu(x) =  x−

a∂G∂x f dξ  + G(x, x−)f (x−)

+ bx+

∂G∂x

f dξ − G(x, x+)f (x+)

= xa

  Gx(x, ξ )f (ξ )dξ  + bx

 Gx(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).

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5.3. GREEN’S FUNCTIONS    23

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

∂G

∂x(x, x−) = lim

→0+

∂G

∂x(x, x

−)

= lim→0+

limh→0

G(x + h, x − ) − G(x, x − )

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

→0+limh→0+

G(x + h, x − ) − G(x, x − )

h

= limh→0+

lim→0+

G(x + h, x − ) − G(x, x − )h

= limh→0+

G(x + h, x) − G(x, x)

h

= ∂G

∂x(x+, x)

Similarly for the other statement. Hence we have

u(x) =

   ba

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

=   ba

Gxx(x, ξ )f (ξ )dξ  + f (x)/k(x).

Thus

Lλ(u) = ku + ku + gu + λu

=

   ba

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)

=   ba

LλG(x, ξ )y(ξ ) dξ  + f (x)

= f (x)

since Lλ(G) = 0.

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24   CHAPTER 5. STURM-LIOUVILLE THEORY 

Recall that we previously argued that the nonhomogeneous BVP

L(y) = f 

B1(y) = γ 1, B2(y) = γ 2has 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) = f B1(u) = γ 1, B2(u) = γ 2

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 have Theorem:  (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   for every  v   satisfying  A∗v = 0.

Now suppose  A  =  AT  and let  M   =  A−

λI . Then  M y

(λ) =  f   is solvable if and only if f  ⊥  N (M T )   (the nullspace of  M T ). Since  M T  =  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 eigenvalue of   A, then   (A − λI )y   =   f   has a solution if and only if   f   is orthogonal to the eigenspace corresponding 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  ba

 f (x)v(x)dx = 0.

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5.3. GREEN’S FUNCTIONS    25

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 − uv)ba = 0

Green’s formula (5.2.13) gives   ba

vfdx =

   ba

[0 − vf ] dx =

   ba

[uLλ(v) − vLλ(u)]dx = 0.

Now suppose ba

 vfdx = 0 and let us choose u  to the IVP solve  Lλ(u) = f  with the initialconditions

u(a) = v(a), u(a) = v (a).

Then   B1(u) = 0 since   B1(v) = 0. We need to show that   B2(u) = 0 =   β 1u(b) + β 2u

(b).Green’s formula gives   ba

vfdx =

   ba

[uLλ(v) − vLλ(u)]dx =  k[uv − vu]|ba.

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

0 =

   ba

vfdx =  k(b)[u(b)v(b) − v(b)u(b)].

Since  β 21  +  β 22 = 0, without loss of generality assume  β 1

 = 0. If  β 2  = 0, then  β 1v(b) = 0 so

v(b) = 0 and hence  u(b) = 0 and so  B2(u) = 0. If  β 2 = 0, then  β 1/β 2v(b) =  v(b). Since vis a nontrivial solution of  Lλ(y) = 0, we see from the previous relations that  v(b) = 0 andv(b) = 0. Thus is follows that

u(b) = v(b)u(b)

v(b)

and so

β 1u(b) + β 2u(b) = β 1v(b)

v(b)u(b) + β 2u(b)

= u

(b)β 1v(b)

v(b) +  β 2 = 0

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

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26   CHAPTER 5. STURM-LIOUVILLE THEORY 

Exercises for Chapter 5

1. Consider

k0y + g(x)y = 0 (∗)

wherek0 >  0,   0 < g1 < g(x) < g2.

If  z 1, z 2  are consecutive zeros of a solution to (*), show that

π

 k0

g2≤ (z 2 − z 1) ≤ 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 if a(x) <  0, x0 ≤ x ≤ ∞, then y(x) has at most one zero.

(b) If limx→∞

a(x) = a0 >  0,

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

a0  as the zerostend 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 − γ 2y = 0, y(0) = 0, y(1) = 0, γ > 0.

5. Find the eigenvalues and eigenfunctions for the following boundary value problem and

then construct the Green’s function.

d

dx

x

dy

dx

+

 λ

xy = 0,   1 < x < e

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

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5.3. GREEN’S FUNCTIONS    27

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?

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28   CHAPTER 5. STURM-LIOUVILLE THEORY 

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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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