Boundary Value Problems

MATH 467 Partial Differential Equations

J. Robert Buchanan

Department of Mathematics

Fall 2017

Objectives

In this lesson we will learn:

▶ to classify second order, two-point boundary value

problems, ▶ to place boundary value problems in self-adjoint form, ▶ properties of the eigenvalues and eigenfunctions of

boundary value problems.

This enables us to generalize the method of separation of variables.

Background

When solving the heat equation for 0 ≤ x ≤ L with homogeneous boundary conditions we had to solve the induced boundary value problem:

X′′(x) + λ X(x) = 0
X(0) = 0
X(L) = 0.

We saw that nontrival solutions existed only if

λ = λ_{n} = \frac{n^{2}π^{2}}{L^{2}}
X(x) = X_{n}(x) = sin \frac{nπx}{L} for n ∈ ℕ.

The set {λ_{n}}_{n=1}^{∞} is called the set of eigenvalues and {X_{n}(x)}_{n=1}^{∞} is the corresponding set of eigenfunctions.

Generalization

Consider the following second-order, linear homogeneous ODE of the form

P(x)y′′(x) + Q(x)y′(x) + R(x)y(x) = 0 for a < x < b.

Suppose that P(x), Q(x), and R(x) are continuous on interval [a, b] and that P(x) ≠ 0 for all x in [a, b].

▶ If Q(x) = P′(x) then the ODE can be written as

[P(x)y′(x)]′ + R(x)y(x) = 0.

▶ If Q(x) ≠ P′(x) multiply both sides of the ODE by

r(x) = \frac{1}{P(x)}e^{∫_{a}^{x} Q(s)/P(s) ds} = \frac{p(x)}{P(x)}.

Generalization

Consider the following second-order, linear homogeneous ODE of the form

P(x)y′′(x) + Q(x)y′(x) + R(x)y(x) = 0 for a < x < b.

Suppose that P(x), Q(x), and R(x) are continuous on interval [a, b] and that P(x) ≠ 0 for all x in [a, b].

▶ If Q(x) = P′(x) then the ODE can be written as

[P(x)y′(x)]′ + R(x)y(x) = 0.

▶ If Q(x) ≠ P′(x) multiply both sides of the ODE by

r(x) = \frac{1}{P(x)}e^{∫_{a}^{x} Q(s)/P(s) ds} = \frac{p(x)}{P(x)}.

Self-Adjoint Form

P(x)y′′(x) + Q(x)y′(x) + R(x)y(x) = 0
r(x)P(x)y′′(x) + r(x)Q(x)y′(x) + r(x)R(x)y(x) = 0
p(x)y′′(x) + p′(x)y′(x) + q(x)y(x) = 0
[p(x)y′(x)]′ + q(x)y(x) = 0

where q(x) = r(x)R(x). An ODE written in this last form is called self-adjoint.

Example

Suppose 0 < a ≤ x ≤ b and express the following ODE in self-adjoint form.

x y′′(x) + (1 − x)y′(x) + y(x) = 0

Note that

r(x) = \frac{1}{x}e^{∫_{a}^{x}(1−s)/s ds} = \frac{1}{a}e^{−x+a}.

Multiplying both sides of the ODE by r(x) produces,

x y′′(x) + (1 − x)y′(x) + y(x) = 0
\frac{x}{a}e^{−x+a} y′′(x) + \frac{1}{a}e^{−x+a}(1 − x)y′(x) + \frac{1}{a}e^{−x+a}y(x) = 0
[x e^{−x+a}y′(x)]′ + e^{−x+a}y(x) = 0.

Example

Suppose 0 < a ≤ x ≤ b and express the following ODE in self-adjoint form.

x y′′(x) + (1 − x)y′(x) + y(x) = 0

Note that

r(x) = \frac{1}{x}e^{∫_{a}^{x}(1−s)/s ds} = \frac{1}{a}e^{−x+a}.

Multiplying both sides of the ODE by r(x) produces,

x y′′(x) + (1 − x)y′(x) + y(x) = 0
\frac{x}{a}e^{−x+a} y′′(x) + \frac{1}{a}e^{−x+a}(1 − x)y′(x) + \frac{1}{a}e^{−x+a}y(x) = 0
[x e^{−x+a}y′(x)]′ + e^{−x+a}y(x) = 0.

Introduction of Operator Notation

Given that the ODE:

P(x)y′′(x) + Q(x)y′(x) + R(x)y(x) = 0

can be written in the self-adjoint form:

[p(x)y′(x)]′ + q(x)y(x) = 0

define the linear operator, L[y] = [p(x)y′]′ + q(x)y.

Sturm-Liouville BVP (1 of 2)

Consider a second-order linear ODE and boundary conditions:

[p(x)y′]′ + (q(x) + λr(x))y = 0 for a < x < b
α_{1}y(a) + β_{1}y′(a) = 0
α_{2}y(b) + β_{2}y′(b) = 0,

where α_{1}^{2} + β_{1}^{2} ≠ 0 and α_{2}^{2} + β_{2}^{2} ≠ 0. Remarks:

▶ This is known as a Sturm-Liouville BVP with separated

boundary conditions. ▶ The unknown constant λ is the eigenvalue parameter. ▶ The function r(x) is called the weight function. ▶ The solution y(x) is the eigenfunction corresponding to λ.

Sturm-Liouville BVP (2 of 2)

The boundary value problem can be concisely written as,

L[y] + λr(x)y = 0 for a < x < b
α_{1}y(a) + β_{1}y′(a) = 0
α_{2}y(b) + β_{2}y′(b) = 0.

Remarks: if ▶ p(a) = 0 or p(b) = 0, or ▶ p(a) = ±∞, p(b) = ±∞, q(a) = ±∞, q(b) = ±∞,

r(a) = ±∞, or r(b) = ±∞, or ▶ a = ±∞ or b = ±∞, then

the BVP is singular. Otherwise the BVP is regular.

Example

Find the eigenvalues and eigenfunctions of the regular Sturm-Liouville BVP:

y′′(x) + \frac{1}{x}y′(x) + \frac{λ}{x^{2}}y(x) = 0 for 1 < x < 2
y(1) = 0
y(2) = 0.

Hint: make the change of variable z = lnx.

Solution (1 of 5)

\frac{dy}{dx} = \frac{dy}{dz} \frac{dz}{dx} = \frac{1}{x} \frac{dy}{dz} = e^{−z} \frac{dy}{dz}
\frac{d^{2}y}{dx^{2}} = \frac{d}{dx} [e^{−z} \frac{dy}{dz}] = e^{−2z} \frac{dy}{dz} + e^{−2z} \frac{d^{2}y}{dz^{2}}

Substituting into the BVP yields:

e^{−2z} \frac{d^{2}y}{dz^{2}} + e^{−2z} \frac{dy}{dz} + e^{−z}e^{−z} \frac{dy}{dz} + λ(e^{−z})^{2}y = 0 for 0 < z < ln2
y′′(z) + 2y′(z) + λy(z) = 0 for 0 < z < ln2.

Solution (2 of 5)

If λ = 1 the general solution can be expressed as

y(z) = (C_{1} + C_{2}z)e^{−z} ⟹ y(x) = (C_{1} + C_{2} lnx)\frac{1}{x}. Using the boundary conditions:

y(1) = 0 = (C_{1} + C_{2} ln1)\frac{1}{1} ⟹ C_{1} = 0
y(2) = 0 = \frac{C_{2}}{2} ln2 ⟹ C_{2} = 0.

Solution (3 of 5)

If λ < 1 the general solution can be expressed as

y(z) = C_{1}e^{(−1−(1−λ)^{1/2})z} + C_{2}e^{(−1+(1−λ)^{1/2})z}
y(x) = C_{1}x^{(−1−(1−λ)^{1/2})} + C_{2}x^{(−1+(1−λ)^{1/2})}.

Using the boundary conditions:

y(1) = 0 = C_{1} + C_{2} ⟹ C_{1} = −C_{2}
y(2) = 0 = C_{1}2^{(−1−(1−λ)^{1/2})} − C_{1}2^{(−1+(1−λ)^{1/2})} ⟹ C_{1} = C_{2} = 0.

Solution (4 of 5)

If λ > 1 the general solution can be expressed as

y(z) = C_{1}e^{(−1−i(λ−1)^{1/2})z} + C_{2}e^{(−1+i(λ−1)^{1/2})z}
y(x) = \frac{1}{x} (C_{1} cos ((λ − 1)^{1/2} lnx) + C_{2} sin((λ − 1)^{1/2} lnx)) .

Using the boundary conditions:

y(1) = 0 = C_{1}
y(2) = 0 = \frac{C_{2}}{2} sin((λ − 1)^{1/2} ln2).

▶ If C_{2} = 0 then there are no nontrivial solutions.

▶ If C_{2} ≠ 0 then (λ − 1)^{1/2} ln2 = n π for n ∈ ℕ.

Solution (5 of 5)

Finally, the eigenvalues are

λ = λ_{n} = 1 + \frac{n^{2}π^{2}}{(ln2)^{2}}

with corresponding eigenfunctions

y(x) = y_{n}(x) = \frac{1}{x} sin (\frac{n π}{ln2} lnx)

for n ∈ ℕ.

Nontrivial Solutions

Let y_{1}(x) and y_{2}(x) be any two linearly independent solutions to the BVP:

L[y] = [p(x)y′]′ + q(x)y = 0 for a < x < b
α_{1}y(a) + β_{1}y′(a) = 0
α_{2}y(b) + β_{2}y′(b) = 0,

where α_{1}^{2} + β_{1}^{2} ≠ 0 and α_{2}^{2} + β_{2}^{2} ≠ 0.

If y(x) is any nontrivial solution to the BVP then there exist constants C_{1} and C_{2} such that

y(x) = C_{1}y_{1}(x) + C_{2}y_{2}(x).

Homework

▶ Read Section x.y ▶ Exercises: