AE  5332  –  Professor  Dora  E.  Musielak    

Lecture  2    

Introduction

A Boundary-Valued Problem (BVP) involving a second order ordinary differential equation (ODE)

𝑎! 𝑥 𝑦!! + 𝑎! 𝑥 𝑦! + 𝑎! 𝑥 𝑦 = 𝑓(𝑥)    Subject to boundary conditions is one such that

𝐿 ≡ 𝑎! 𝑥𝑑!

𝑑𝑥! + 𝑎! 𝑥𝑑𝑑𝑥 + 𝑎!(𝑥)  

then  

𝐿𝑦 = 𝑓    subject to the specified boundary conditions. In general, we can solve some equations of BVP using eigenvalue expansions. For example there are solutions that lead to trigonometric eigenfunctions. Such functions can be used to represent functions as Fourier series expansions. We wish to generalize some of these techniques in order to solve other BVPs. The Sturm-Liouville Problems (SLP) that we study here are a class of eigenvalue problems. Sturm-Liouville equations arise in applied mathematics and engineering. For example, they describe the vibrational modes of various systems, such as the vibratios of a string, or the energy eigenfunctions of a quantum mechanical oscillator, in which the eigenvalues correspond to the resonant frequencies of vibration or energy levels. Sturm-Liouville problems arise directly as eigenvalue problem in one-space dimensions. They also commonly arise from linear PDEs. In several space dimensions when the equations are separable in some coordinate system (Cartesian, cylindrical, or spherical coordinates).

What are eigenvalues? Eigenvalues ! Special set of scalar values associated with a linear system of equations (i.e., a matrix equation) that are sometimes also known as characteristic roots, characteristic values, or proper values.

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Sturm-Liouville Problems Definition: A BVP is called a Sturm-Liouville problem if it consists of 1) a second-order homogeneous linear differential equation

𝑑𝑑𝑥 𝑝(𝑥)

𝑑𝑦𝑑𝑥 + 𝑞 𝑥 + 𝜆  𝑤(𝑥) 𝑦 = 0

or

𝑝 𝑥 𝑦′ ! + 𝑞 𝑥 𝑦 + 𝜆  𝑤 𝑥 𝑦 = 0 where 𝑝 𝑥 , 𝑞(𝑥), and 𝑤(𝑥) are real functions such that p has a continuous derivative, q and w are continuous, and 𝑝(𝑥) > 0, 𝑤 𝑥 > 0, for all 𝑥 on a real interval 𝑎 ≤ 𝑥 ≤ 𝑏, and 𝜆 is a parameter of x; and 2) two supplementary conditions

𝛼  𝑦 𝑎 + 𝛽  𝑦! 𝑎 =  0

𝛾  𝑦 𝑏 + 𝛿  𝑦! 𝑏 = 0 where 𝛼,𝛽, 𝛾, 𝛿 are real constants such that 𝛼 and 𝛽 are not both zero, and 𝛾 and 𝛿 are not both zero. * * * * * Two important special cases are those in which the supplementary conditions are of the form

𝑦 𝑎 = 0                                        𝑦 𝑏 = 0 or

𝑦′ 𝑎 = 0                                        𝑦′ 𝑏 = 0 Example 1. The BVP

𝑑!𝑦𝑑𝑥! + 𝜆𝑦 = 0

𝑦 0 = 0,          𝑦 𝜋 = 0

is a S-L problem, since

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𝑑𝑑𝑥 1 ∙

𝑑𝑦𝑑𝑥 + 0+ 𝜆 ∙ 1 𝑦 = 0

and hence is of the form of the S-L problem in the definition, where 𝑝 𝑥 = 1, 𝑞 𝑥 = 0, and 𝑤 𝑥 = 1. The B.C.s are of the form in the definition where 𝛼 = 1,𝛽 = 0, 𝛾 = 1, 𝛿 = 0.

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Is the BVP defined by

𝑑𝑑𝑥 𝑥  

𝑑𝑦𝑑𝑥 + 2𝑥! + 𝜆  𝑥! 𝑦 = 0

3𝑦 1 + 4𝑦! 1 = 0

5𝑦 2 − 3𝑦! 2 = 0

a Sturm-Liouville problem?

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Determine if the following BVP a Sturm-Liouville problem.

𝑑!𝑦𝑑𝑥! + 𝜆𝑦 = 0

𝑦 0 = 0,              𝑦 𝜋 = 0

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Note that the value of 𝜆 is not specified in the previous equations. Thus, our job is to find values for 𝜆 for which there exists nontrivial solutions of the given differential equation satisfying the boundary conditions. Finding values of 𝜆 is part of the S-L problem. Such values of 𝜆 (when they exist) are called the eigenvalues of the BVP defined by the S-L problem (differential equation plus B.C.s). The corresponding solutions for such 𝜆 are the eigenfunctions of the problem.

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Sturm-Liouville Problems

Example 2. Find the eigenvalues and eigenfunctions of the S-L problem

𝑦!! − 𝑦! + 𝜆𝑦 = 0

𝑦 0 = 0,          𝑦 𝐿 = 0 Solution: To solve the given differential equation, evaluate its auxiliary equation

𝑟! − 𝑟 + 𝜆 = 0 It has roots

𝑟 =1± 1− 4𝜆

2

Note that the type of solutions will depend on whether (1− 4𝜆) is positive, negative, or zero. CASE 1: 1− 4𝜆 = 0 In this case, the roots of the auxiliary equation are identical, 𝑟 = !

!, !!, and for

repeated roots the general solution of the given differential equation is

𝑦 = 𝑐!𝑒!!! + 𝑐!  𝑥  𝑒

!!!

Apply the B.C. 𝑦 0 = 0,

𝑦 0 = 𝑐! = 0 Apply the B.C. 𝑦 𝐿 = 0,

𝑦 𝐿 = 𝑐!  𝐿  𝑒!!! = 0

𝑐! = 0

Thus, the solution is the trivial solution, 𝑦 = 0.

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CASE 2: 1− 4𝜆 > 0 In this case, we have two real roots

𝑟! =1+ 1− 4𝜆

2 ;                      𝑟! =1− 1− 4𝜆

2

and the general solution of the given differential equation is

𝑦 = 𝑐!  𝑒𝑥𝑝1+ 1− 4𝜆

2 𝑥 + 𝑐!  𝑒𝑥𝑝1− 1− 4𝜆

2 𝑥

Apply the first B.C.

𝑦 0 = 𝑐! + 𝑐! = 0 so

𝑐! = −𝑐! Apply the second B.C.

𝑦 𝐿 = 𝑐!  𝑒𝑥𝑝1+ 1− 4𝜆

2 𝐿 − 𝑐!  𝑒𝑥𝑝1− 1− 4𝜆

2 𝐿 = 0

This implies 𝑐! = 0,  and thus 𝑐! = 0. Again, we obtain the trivial solution. CASE 3: 1− 4𝜆 < 0 In this case, we have complex roots

𝑟 =12± 𝑖

4𝜆 − 12

and the general solution to the given differential equation is

𝑦 = 𝑐!𝑒!!! sin

4𝜆 − 12 𝑥 + 𝑐!𝑒

!!! cos

4𝜆 − 12 𝑥

Apply the first B.C.

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𝑦 0 = 𝑐! = 0 Apply the second B.C.

𝑦(𝐿) = 𝑐!𝑒!!! sin

4𝜆 − 12 𝐿 = 0

We know that for 𝑐! ≠ 0, this implies

sin4𝜆 − 12 𝐿 = 0

And we also know that the zeros of sine occur at integral multiples of 𝜋. Therefore, we must have

4𝜆 − 12  𝐿 = 𝑛𝜋

for 𝑛 = 1, 2, 3,… The n-values of 𝜆 that satisfy this equation are the eigenvalues

𝜆! =(2𝑛𝜋)! + 𝐿!

4𝐿!

and the eigenfunctions corresponding to the eigenvalues are

𝑦! = 𝑐!𝑒!!! sin

𝑛𝜋𝐿 𝑥

In summary, for the case 1− 4𝜆 < 0, we obtain Eigenvalues

𝜆! =(2𝑛𝜋)! + 𝐿!

4𝐿! ,                𝑛 = 1, 2, 3,…

Eigenfunctions

𝑦! = 𝑐!𝑒!!! sin

𝑛𝜋𝐿 𝑥,          𝑛 = 1, 2, 3,…

where 𝑐! are arbitrary non-zero constants.

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The functions defined by  

𝑐!𝑒!!! sin

𝜋𝐿 𝑥,           𝑐!𝑒

!!! sin

2𝜋𝐿 𝑥,…          

with non-zero constants are nontrivial solutions of the given S-L problem.

" Example 3. Find the eigenvalues and eigenfunctions of the Sturm-Liouville problem

𝑦!! + 𝜆  𝑦 = 0

𝑦 0 = 0,            𝑦 𝜋 = 0 Solution: CASE 1: 𝜆 = 0 In this case, the given differential equation reduces to

𝑑!𝑦𝑑𝑥! = 0

It has a general solution

𝑦 = 𝑐! + 𝑐!𝑥 Apply the B.C.s we can demonstrate that the only solution is the trivial solution. CASE 2: 𝜆 < 0 In this case, the roots of the characteristic or auxiliary equation are ± −𝜆 And the general solution of the given differential equation is

𝑦 = 𝑐!  𝑒 !!! + 𝑐!  𝑒! !!! and after applying the B.C.s we find only trivial solutions.

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CASE 3: 𝜆 > 0 In this case, the roots are conjugate complex numbers, ±  𝑖 𝜆, and the general solution of the given differential equation is

𝑦 = 𝑐! sin 𝜆 𝑥 +  𝑐! cos 𝜆 𝑥 Apply the B.C.s,

𝑦 0 = 𝑐! sin 𝜆 0+  𝑐! cos 𝜆 0 = 0

𝑐! = 0

𝑦 𝜋 = 𝑐! sin 𝜆 𝜋 = 0 If 𝑐! = 0, then we’d get a trivial solution. Thus,

sin 𝜆 𝜋 = 0 in order to obtain nontrivial solutions. So, we must have eigenvalues

𝜆 = 𝑛,          𝑛 = 1, 2, 3,…. or

𝜆 = 𝑛!,        for  𝑛 = 1, 2, 3,… The parameter 𝜆 in the S-L problem must be a member of the infinite sequence

1, 4, 9, 16,… . ,𝑛! The nontrivial solutions corresponding to the eigenvalues 𝜆 = 𝑛! are given by the eigenfunctions

𝜙! 𝑥 = 𝑐! sin𝑛𝑥 where 𝑐! are arbitrary nonzero constants. The functions defined by 𝑐! sin 𝑥 , 𝑐! sin 2𝑥 ,… are nontrivial solutions or eigenfunctions of the given problem.

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Eigenvalues are the characteristic values of S-L problems. Eigenfunctions are the characteristic nontrivial solutions of S-L problems. The values of the parameter 𝜆 in the S-L problem for which there exists nontrivial solutions are the eigenvalues that we must determine to solve the BVP. The corresponding nontrivial solutions are the eigenfunctions of the problem. The infinite set of eigenvalues can be arranged in a monotonic increasing sequence

𝜆! < 𝜆! < 𝜆! < ⋯ Note that 𝜆! → +∞ as 𝑛 → +∞

Example 4. Find the eigenvalues and eigenfunctions of the BVP

𝑑𝑑𝑥 𝑥

𝑑𝑦𝑑𝑥 +

𝜆𝑥  𝑦 = 0

𝑦! 0 = 0,              𝑦! 𝑒!! = 0

where 𝜆 is a nonnegative number. Solution: This is a S-L problem. CASE 1: 𝜆 = 0

𝑑𝑑𝑥 𝑥

𝑑𝑦𝑑𝑥 = 0

𝑥𝑑𝑦𝑑𝑥 = 𝑐             →    

𝑑𝑦𝑑𝑥 =

𝑐𝑥            →  𝑑𝑦 =

𝑐𝑥 𝑑𝑥

And the general solution of the given differential equation is

𝑦 = 𝑐 ln 𝑥 + 𝑐! where 𝑐, 𝑐! are arbitrary constants.

Is  it  possible  to  obtain    

𝜆 = 0  ?  

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Applying the B.C.s we find that both require 𝑐 = 0. However, neither of the conditions imposes restrictions on 𝑐!. Thus, for 𝜆 = 0 we obtain solutions

𝑦 = 𝑐! These are nontrivial solutions or eigenfunctions for all choices of 𝑐! ≠ 0. So, the answer to the question posed is yes, 𝜆 = 0 is an eigenvalue of the given S-L problem and 𝜑 = 𝑐! is an eigenfunction. CASE 2: 𝜆 > 0   In this case, with 𝑥 ≠ 0 the given equation

𝑑𝑑𝑥 𝑥

𝑑𝑦𝑑𝑥 +

𝜆𝑥  𝑦 = 0

is equivalent to the Euler-Cauchy equation

𝑥!𝑑!𝑦𝑑𝑥! + 𝑥

𝑑𝑦𝑑𝑥 + 𝜆  𝑦 = 0

How do we solve the above Euler-Cauchy equation?

Let

𝑥 = 𝑒! and the differential equation becomes

𝑑!𝑦𝑑𝑡! + 𝜆  𝑦 = 0

And since 𝜆 > 0, its general solution is

𝑦 = 𝑐! sin 𝜆  𝑡 +  𝑐! cos 𝜆  𝑡 Therefore, for 𝜆 > 0 and 𝑥 > 0, the general solution of the given differential equation may be written as

𝑦 = 𝑐! sin 𝜆 ln 𝑥 +  𝑐! cos 𝜆 ln 𝑥 since 𝑡 = ln 𝑥.

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Apply the B.C. 𝑦! 1 = 0  

𝑑𝑦𝑑𝑥 =

𝑐! 𝜆𝑥 cos 𝜆 ln 𝑥 −

𝑐! 𝜆𝑥  sin 𝜆 ln 𝑥

𝑦! 1 = 𝑐! 𝜆 cos 𝜆 ln 1 − 𝑐! 𝜆  sin 𝜆 ln 1 = 0

𝑐! 𝜆 = 0,               →   𝑐! = 0 Apply the B.C. 𝑦′(𝑒!!) = 0

𝑦! 𝑒!! = −𝑐! 𝜆  sin 𝜆 ln 𝑒!! = 0 and since ln 𝑒!! = 2𝜋,

𝑐! 𝜆  sin 2𝜋 𝜆 = 0 To obtain nontrivial solutions we must have

sin 2𝜋 𝜆 = 0 Thus

2𝜋 𝜆 = 𝑛𝜋,              𝑛 = 1, 2, 3,…. The eigenvalues of the given S-L problem are

𝜆! =𝑛!

4 ,              𝑛 = 1, 2, 3,…

And the corresponding nontrivial solutions or eigenfunctions are

𝜑! = 𝑐! cos𝑛 ln 𝑥2 ,            𝑛 = 1, 2, 3,…

where 𝑐! are arbitrary nonzero constants. The eigenvalues

𝜆 = 0,14 , 1,

94 , 4,

254 ,… ,

𝑛!

4

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for 𝑛 ≥ 0 are the eigenvalues of the S-L problem with corresponding eigenfunctions

 𝑐!, 𝑐! cosln 𝑥2 , 𝑐! cos

2 ln 𝑥2 ,…  

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