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Chapter 5
Boundary Value Problems
A boundary value problem for a given differential equation consists of finding a solution of the
given differential equation subject to a given set of boundary conditions. A boundary condition is
a prescription some combinations of values of the unknown solution and its derivatives at more
than one point.
Let I = (a, b) ⊆ R be an interval. Let p, q, r : (a, b) → R be continuous functions.
Throughout this chapter we consider the linear second order equation given by
y
′′
′
- q(x)y = r(x), a < x < b. (5.1)
Corresponding to ODE (5.1), there are four important kinds of (linear) boundary conditions.
They are given by
Dirichlet or First kind : y(a) = η 1 , y(b) = η 2 ,
Neumann or Second kind : y′(a) = η 1 , y′(b) = η 2 ,
Robin or Third or Mixed kind : α 1 y(a) + α 2 y′(a) = η 1 , β 1 y(b) + β 2 y′(b) = η 2 ,
Periodic : y(a) = y(b), y′(a) = y′(b).
Remark 5.1 (On periodic boundary condition) If the coefficients of ODE (5.1) are periodic
functions with period l = b − a and if φ is a solution of ODE (5.1) (note that this solution exists on
R), then ψ defined by ψ(x) = φ(x + l) is also a solution. If φ satisfies the periodic boundary
conditions, then ψ(a) = φ(a) and ψ
′ (a) = φ
′ (a). Since solutions to IVP are unique in the present
case, it must be that ψ ≡ φ. In other words, φ is a periodic function of period l.
Boundary Value Problems do not behave as nicely as Initial value problems. For, there are
BVPs for which solutions do not exist; and even if a solution exists there might be many more.
Thus existence and uniqueness generally fail for BVPs. The following example illustrate all the
three possibilities.
Example 5.2 Consider the equation
y
′′
(i) The BVP for equation (5.2) with boundary conditions y(0) = 1, y(π 2 ) = 1 has a unique solution. This solution is given by sin x + cos x.
(ii) The BVP for equation (5.2) with boundary conditions y(0) = 1, y(π) = 1 has no solutions.
(iii) The BVP for equation (5.2) with boundary conditions y(0) = 1, y(2π) = 1 has an infinite
number of solutions.
42 5.1. Adjoint forms, Lagrange identity
5.1 Adjoint forms, Lagrange identity
In mathematical physics there are many important boundary value problems corresponding to
second order equations. In the studies of vibrations of a membrane, vibrations of a structure one
has to solve a homogeneous boundary value problem for real frequencies (eigen values). As is
well known in the case of symmetric matrices that there are only real eigen values and
corresponding eigen vectors form a basis for the underlying vector space and thereby all
symmetric matrices are diagonalisable. Self-adjoint problems can be thought of as
corresponding ODE versions of symmetric matrices, and they play an important role in
mathematical physics.
Let us consider the equation
L[y] ≡ l(x)y
′′
′
- q(x)y = 0, a < x < b. (5.3)
Integrating zL[y] by parts from a to x, we have
Z (^) x a zL[y] dx = (zl)y
′ − (zl)
′ y +
(zp)y
x
a+Z^ x a
(zl)
′′ − (zp)
′
If we define the second order operator L
∗ by
L∗[z] ≡ (zl)′′^ − (zp)′^ + (zq) = l(x)z′′^ + (2l′^ − p)(x)z′^ + (l′′^ − p′^ + q)(x)z = 0, (5.5) then the
equation (5.4) becomes
Z (^) x
a
z L[y] − y L
∗ [z] dx = l(y
′ z − yz
′ ) + (p −
l′ ) y z
x
a. (5.6)
The operator L
∗
is called the adjoint operator corresponding to the operator L. It can be easily
verified that adjoint of L∗is L itself. If L and L∗^ are the same, then L is said to be self-adjoint.
Thus, the necessary and sufficient condition for L to be self-adjoint is that p = 2l′^ − p, and q = l′′^ −
p′^ + q, (5.7)
which is satisfied if
p = l
′
. (5.8)
Thus if L is self-adjoint, we have
L[y] ≡ l(x)y
′′
′
A general operator L may not be self-adjoint but it can always be converted into a self-adjoint by
suitably multiplying L with a function.
Lemma 5.3 (self-adjointisation) The operator h(x)L[y] is self-adjoint, where h is given by
h(x) =
l(x)exp Z^
x
p(t) l(t)dt (^). (5.10)
In fact h(x)L[y] is given by
d
dx
α(x)
dy dx + β(x) y = 0, (5.11)
where
α(x) = exp
Z
x
p(t) l(t)dt , β(x) = q(x)^
l(x)exp^ Z
x
p(t) l(t)dt
Sivaji Ganesh Sista MA 417: Ordinary Differential Equations 44 5.2. Two-point boundary value problem
Note that the boundary conditions are in the most general form, and they include the first three
conditions given at the beginning of our discussion on BVPs as special cases.
Let us introduce some nomenclature here.
Definition 5.5 Assume hypothesis (HBVP). A nonhomogeneous boundary value problem consists
of solving
L[y] = f, U 1 [y] = η 1 , U 2 [y] = η 2 , (5.23)
for given constants η 1 and η 2 , and a given continuous function f on the interval [a, b]. Definition
5.6 The associated homogeneous boundary value problem is then given by L[y] = 0, U 1 [y] = 0,
U 2 [y] = 0. (5.24)
Let us list some properties of the solutions for BVP that are consequences of the linearity of the
differential operator L.
Lemma 5.7 (i) A linear combination of solutions of the homogeneous BVP (5.24) is also a
solution of the homogeneous BVP (5.24).
(ii) If u, v are two solutions of the nonhomogeneous BVP (5.23), then their difference u − v is a
solution of the homogeneous BVP (5.24).
(iii) If y solves the nonhomogeneous BVP (5.23) and z solves the homogeneous BVP (5.24),
then the function y + z nonhomogeneous BVP (5.23).
(iv) Let u be a (fixed) solution of the nonhomogeneous BVP (5.23). Then any solution y of the
nonhomogeneous BVP (5.23) is given by y = u + z for some function z that solves the homogeneous BVP (5.24).
Given a fundamental pair of solutions to the ODE L[y] = 0, it is possible to say whether (and
when) the homogeneous BVP has only trivial solution and is characterised in terms of the given
fundamental pair. Recall that a fundamental pair of solutions to L[y] = 0 always exists.
Lemma 5.8 Let φ 1 , φ 2 be a fundamental pair of solutions to the ODE L[y] = 0. Then the following
are equivalent.
(1) The nonhomogeneous boundary value problem has a unique solution for any given
constants η 1 and η 2 , and a given continuous function f on the interval [a, b].
(2) The associated homogeneous boundary value problem has only trivial solution.
(3) The determinant U 1 [φ 1 ] U 1 [φ 2 ] U 2 [φ 1 ]
U 2 [φ 2 ]
Before we give a proof of this result, let us make an observation concerning the condition (5.25).
Remark 5.9 Though (3) above depends on a given fundamental pair, due to its equivalence with
the (1) and (2), it is indeed the case that the condition (5.25) is independent of the choice of
fundamental pair.
Also note that there is a subtle difference between equivalence of (1) and (2), and the
statement(s) of Lemma 5.7. What is it? In effect, equivalence of (1) and (2) does not really
follow from Lemma 5.7. Therefore we will have to prove the equivalence of (1) and (2).
MA 417: Ordinary Differential Equations Sivaji Ganesh Sista Chapter 5 : Boundary Value Problems 45
Proof :
Our strategy for the proof is to prove (i). (1)⇐⇒(3) (ii). (2)⇐⇒(3). In fact, it is enough to prove
(1)⇐⇒(3), since (2)⇐⇒(3) follows by taking f = 0, η 1 = η 2 = 0.
We already illustrated how to find solutions of L[y] = f starting from a fundamental pair of solu
tions to the ODE L[y] = 0 and we gave an expression for a general solution of L[y] = f. Let us
pick any one of such solutions, let us denote it by z.
Since φ 1 , φ 2 is a fundamental pair of solutions to the ODE L[y] = 0, general soution y of L[y] = f
is given by
y = z + c 1 φ 1 + c 2 φ 2 , c 1 , c 2 ∈ R. (5.26)
Now y is a solution of nonhomogeneous BVP (5.23) if and only if we can solve for c 1 and c 2 from
the algebraic equations
U 1 [y] = U 1 [z] + c 1 U 1 [φ 1 ] + c 2 U 1 [φ 2 ] = η 1 , U 2 [y] = U 2 [z] + c 1 U 2 [φ 1 ] + c 2 U 2 [φ 2 ] = η 2 , (5.27) The
above system (5.27) has a solution for every η 1 and η 2 if and only if (5.25) holds. Exercise 5.
Consider the ODE
y′′^ + y = f(x), 0 ≤ x ≤ π. (5.28)
Determine if the following BVPs for the ODE (5.28) have unique solution for every f, η 1 , η 2 by
applying the above result.
(i) U 1 [y] ≡ y(0) + y
′ (0) = η 1 , U 2 [y] ≡ y(π) = η 2.
(ii) U 1 [y] ≡ y(0) = η 1 , U 2 [y] ≡ y(π) = η 2.
Exercise 5.11 Prove that the nonhomogeneous BVP (posed on [0, π])
y
′′
- y = 0, y(0) = 0, y(π) = 1 (5.29)
has no solution. Comment on the solutions of the associated homogeneous BVP.
5.3 Fundamental solutions, Green’s functions
Fundamental solution of an ODE gives rise to a representation (integral) formula for solution of
the nonhomogeneous equation. When we want to take care of boundary conditions, we impose
boundary conditions on fundamental solutions and get Green’s functions. Thus Green’s
functions give rise to a representation formula for solution of the nonhomogeneous BVP. The
concepts of a fundamental solution as well as a Green’s function are defined in terms of the
homogeneous BVPvassociated to the nonhomogeneous BVP.
Let Q denote the square Q := [a, b] × [a, b] in the xξ-plane. Let us partition Q by the line x = ξ
and call the two resulting triangles Q 1 and Q 2. Let
Q 1 = {(x, ξ) : a ≤ ξ ≤ x ≤ b },
Q 2 = {(x, ξ) : a ≤ x ≤ ξ ≤ b }.
Note that the diagonal x = ξ belongs to both the triangles.
∂x(x, x−) −∂γ ∂x(x, x+) =^1
p(x), a < x < b. (5.31)
Lemma 5.14 Assume hypothesis (HBVP). A fundamental solution exists but is not unique.
Proof :
For each a ≤ ξ ≤ b, let y(x; ξ) be the solution of the IVP
L[y] = 0, y(ξ) = 0, y′ (ξ) = 1
p(ξ). (5.32)
Then
γ(x, ξ) :=
(
0 for (x, ξ) ∈ Q 2
y(x; ξ) for (x, ξ) ∈ Q 1 (5.33)
is a fundamental solution, and can be verified easily.
Let z be a solution of L[z] = 0, and let h ∈ C[a, b]. Then the function
γ 1 (x, ξ) := γ(x, ξ) + z(x)h(ξ) (5.34)
is also a fundamental solution.
MA 417: Ordinary Differential Equations Sivaji Ganesh Sista Chapter 5 : Boundary Value Problems 47
Example 5.15 With the definition a+ := max(0, a),
(i) a fundamental solution for L[y] ≡ y′′^ = 0 is (x − ξ)+.
(ii) a fundamental solution for L[y] ≡ y′′^ + λ^2 y = 0 is (^1) λsin λ(x − ξ)+.
Given a fundamental solution for the homogeneous equation L[y] = 0, one can easily construct
a solution of nonhomogeneous equation L[y] = f, as asserted by the following result.
Theorem 5.16 Assume hypothesis (HBVP). Let γ(x, ξ) be a fundamental solution. Then the
function v defined by
v(x) =
Z (^) b a
γ(x, ξ) f(ξ) dξ (5.35)
is twice continuously differentiable and is a solution of L[v] = f on the interval [a, b].
Proof :
We split the integral in (5.35) at the point x and write
v(x) =
Z (^) x a γ(x, ξ) f(ξ) dξ
Z (^) b x γ(x, ξ) f(ξ) dξ. (5.36)
Differentiating (5.36) w.r.t. x yields
v
′ (x) = γ(x, x) f(x)
Z (^) x a
∂γ
∂x(x, ξ) f(ξ) dξ − γ(x, x) f(x) +
Z (^) b x
∂γ
∂x(x, ξ) f(ξ) dξ. (5.37)
Since γ(x, ξ) is continuous, the above equation becomes
v′(x) =
Z (^) x a
∂γ
∂x(x, ξ) f(ξ) dξ
Z (^) b x
∂γ
∂x(x, ξ) f(ξ) dξ. (5.38)
Differentiating (5.38) w.r.t. x yields
v′′ (x) = ∂γ
∂x(x, x−) f(x) +
Z (^) x a
∂^2 γ
∂x
(x, ξ) f(ξ) dξ
−∂γ^
∂x(x, x+) f(x) + Z (^) b x
∂^2 γ
∂x
(x, ξ) f(ξ)
dξ.(5.39)
By Exercise 5.13, above equation (5.39) reduces to
v
′′ (x) =^1
p(x)f(x) + Z (^) x a
2 γ
∂x
(x, ξ) f(ξ) dξ
Z (^) b x
2 γ
∂x
(x, ξ) f(ξ) dξ.
(5.40)
Since L[γ(., ξ)] = 0 (by definition of a fundamental solution), thanks to the above calculations,
we get
L[v] = pv′′^ + p′v′^ + qv =
Z (^) b a
L[γ(., ξ)](x, ξ) f(ξ) dξ + f(x) = f(x). (5.41)
Definition 5.17 (Green’s function) A Green’s function G(x, ξ) for the homogeneous boundary
value problem (5.24)
L[y] = 0, U 1 [y] = 0, U 2 [y] = 0. (5.42)
is defined by the following two properties:
(i) G(x, ξ) is a fundamental solution of L[y] = 0.
(ii) U 1 [G(., ξ)] = 0, U 2 [G(., ξ)] = 0 for each a < ξ < b.
Sivaji Ganesh Sista MA 417: Ordinary Differential Equations 48 5.3. Fundamental solutions, Green’s functions
5.3.1 Construction of Green’s functions
In this paragraph we are going to construct Green’s functions under the assumption that the
homogeneous BVP
L[y] = 0, U 1 [y] = 0, U 2 [y] = 0. (5.43)
has only trivial solution.
Let (λ 1 , λ 2 ) 6= (0, 0) be such that a 1 λ 1 + a 2 λ 2 = 0 and let φ 1 be solution of L[y] = 0 satisfying φ 1 (a)
= λ 1 and φ′ 1 (a) = λ 2. Choose another solution φ 2 of L[y] = 0 similarly. This way of choosing φ 1
and φ 2 make sure that both are non-trivial solutions.
Note that φ 1 and φ 2 form a fundamental pair of solutions of L[y] = 0, since we assumed that
homogeneous BVP has only trivial solutions.
By Lagrange’s identity (5.20), we get
d dx p(φ
′ 1 φ 2 − φ 1 φ
′ (^2) ) = 0. This implies p(φ
′ 1 φ 2 − φ 1 φ
′ 2 ) ≡ c, a constant and non-zero, (5.44)
as a consequence of (φ′ 1 φ 2 − φ 1 φ′ 2 ) being the wronskian corresponding to a fundamental pair of
solutions.
Green’s function is then given by
(
G(x, ξ) :=
c
φ 1 (x)φ 2 (ξ) for (x, ξ) ∈ Q 2
φ 1 (ξ)φ 2 (x) for (x, ξ) ∈ Q (^1) .(5.45)
Verifying that G(x, ξ) has the required properties for it to be a Green’s function is left as an
exercise.
Exercise 5.18 Verify that G(x, ξ) is indeed a Green’s function.
Z (^) b a
Z (^) b
(G(x, ξ) − Γ(ξ, x)) f(ξ)g(x) dξ dx = 0. (5.53) a
Since the relation holds for arbitrary f and g, we must have
G(x, ξ) = Γ(ξ, x). (5.54)
Setting Γ = G, we get symmetry of Green’s function G; and now once again we get G = Γ and
thus establishing uniqueness of Green’s function.
Example 5.20 Green’s function for
y
′′
- y = 0 in [0, 1], y(0) = y(1) = 0 (5.55)
is given by
G(x, ξ) := (
ξ(x − 1) for 0 ≤ ξ ≤ x ≤ 1,
x(ξ − 1) for 0 ≤ x ≤ ξ ≤ 1.(5.56)
Note that the homogeneous BVP in this case, has only trivial solution.
Exercise 5.21 How to solve general nonhomogeneous BVP? It is necessary to ask this question
since Green’s functions give rise to solutions of nonhomogeneous equations but with homogeneous boundary conditions. (Hint: Use linearity of the differential operator) Solve, for instance, the non-homogeneous BVP
y
′′ = f(x) in [0, 1], y(0) = 0, y(1) + y
′ (1) = 2. (5.57) Exercise 5.22 Solve the
nonhomogeneous BVP
y′′^ + y = exin [0, 1], y(0) = y(1) = 0 (5.58)
by (i) using a fundamental pair of solutions and a special solution of the nonhomogeneous differ ential equation; (ii) using Green’s function.
Exercise 5.23 Find Green’s function for
y
′′ = 0 in [0, 1], y
′ (0) = y(1) = 0 (5.59)
Exercise 5.24 Find Green’s function for
y′′ +
4x
= 0 in [1, 2], y(1) = y(2) = 0 (5.60)
(Hint: Try a change of variable x = e
t )
Sivaji Ganesh Sista MA 417: Ordinary Differential Equations 50 5.4. Generalised Green’s function
5.4 Generalised Green’s function
In Section 5.3 we constructed Green’s function for the homogeneous BVP when the latter
problem had only trivial solutions. Now we investigate the case where the homogeneous BVP
has non-trivial solutions.
Let us look at what happens in the case of linear system of equations of size k in k variables.
Let A be a k × k matrix, and b ∈ R
k
. Suppose that Ax = 0 has non-trivial solutions. In this case
we know that Ax = b does not have solution for every b ∈ R
k
. We also know that Ax = b has a
solution if and only if b belongs to orthogonal complement of a certain subspace of Rk, i.e., b
must satisfy a compatibility condition. Note that, in this case, if Ax = b has one solution, then
there are infinitely many solutions.
We are going to see that there exists an analogue of Green’s function, called Generalised
Green’s function. We also prove that a situation similar to that of matricial analogy given above
occurs here too; and the compatibility condition for the existence of a solution to
nonhomogeneous BVP is a kind of Fredholm alternative.
Let us recall the notations Q 1 and Q 2.
Q 1 = {(x, ξ) : a ≤ ξ ≤ x ≤ b },
Q 2 = {(x, ξ) : a ≤ x ≤ ξ ≤ b }.
Definition 5.25 (Generalised Green’s function) Let φ 0 be a solution of L[y] = 0 with homo
geneous boundary conditions, such that R^
b
aφ
2 0 (x) dx = 1. A function Γ(x, ξ) defined in Q is
called a generalised Green’s function if Γ has the following properties:
(i) The function Γ(x, ξ) is continuous in Q.
(ii) The first and second order partial derivatives w.r.t. variable x of the function Γ(x, ξ) exist and
continuous up to the boundary on Q 1 and Q 2.
(iii) Let ξ ∈ [a, b] be fixed. Then γ(x, ξ), considered as a function of x, satisfies L[Γ(., ξ)] =
φ 0 (x)φ 0 (ξ) (5.61)
at every point of the interval [a, b], except at ξ, satisfying the homogeneous boundary
conditions U 1 [y] = 0 and U 2 [y] = 0.
(iv) The first derivate has a jump across the diagonal x = ξ, of magnitude 1/p, i.e.,
∂Γ ∂x
x=ξ
∂x (ξ+, ξ) −∂Γ^
∂x (ξ−, ξ) =^1
p(ξ), a < ξ < b, (5.62)
∂x (ξ+, , ξ) is defined as the limit of^
∂Γ
where
∂Γ
∂Γ
∂x (ξ−, , ξ) is defined as the limit of^
∂Γ
∂x (x, , ξ) as x → ξ+, i.e., (x, , ξ)^ ∈^ Q 1 ; and
∂x (x, , ξ) as x → ξ−, i.e., (x, , ξ)^ ∈^ Q 2. Note that plus and minus signs indicate that limits are taken from right and left sides of the diagonal x = ξ.
(v) The function Γ(x, ξ) satisfies the condition
Z (^) b
Γ(x, ξ)φ 0 (x) dx = 0 (5.63) a
The following result characterises the class of functions f, for which the nonhomogeneous
equation L[y] = f has a solution satisfying homogeneous boundary conditions.
MA 417: Ordinary Differential Equations Sivaji Ganesh Sista Chapter 5 : Boundary Value Problems 51
Theorem 5.26 The non-homogenous BVP
L[y] = f, U 1 [y] = 0, U 2 [y] = 0 (5.64)
has a solution φ(x) =
Z (^) b a
Step 3B: Γ must be continuous.
Γ 1 (ξ, ξ) = Γ 2 (ξ, ξ). (5.78)
This gives
a 2 (ξ) = a 1 (ξ) + 1. (5.79)
Step 3C: Γ must satisfy (5.63). In our case, this condition takes the form
Z (^1) Γ(x, ξ)xdx = 0 (5.80) 0
This gives us
a (^1) (ξ) = ξ^3 2−
5 ξ. (5.81)
Last Step : Formula for Generalised Green’s function. Generalised Green’s function is given by (
Γ(x, ξ) :=
Expression for solution (5.65).
Γ 1 (x, ξ) for 0 ≤ x ≤ ξ ≤ 1,
Γ (^2) (x, ξ) for 0 ≤ ξ ≤ x ≤ 1.(5.82)
φ(x) =
Z 1 0
Γ(x, ξ) f(ξ)
dξ =
Z (^) x 0
Γ 2 (x, ξ) f(ξ)
dξ +
Z 1 x
Γ 1 (x, ξ) f(ξ)
dξ (5.83)
Note that we did not check the jump condition and it is satisfied automatically since φ 0 is nor
malised solution. This finishes our discussion on Green’s functions.
MA 417: Ordinary Differential Equations Sivaji Ganesh Sista
