Boyce/DiPrima 9 th ed, Ch 11.2: Sturm- Liouville Boundary Value Problems Elementary Differential...
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Transcript of Boyce/DiPrima 9 th ed, Ch 11.2: Sturm- Liouville Boundary Value Problems Elementary Differential...
Boyce/DiPrima 9th ed, Ch 11.2: Sturm-Liouville Boundary Value ProblemsElementary Differential Equations and Boundary Value Problems, 9th edition, by William E. Boyce and Richard C. DiPrima, ©2009 by John Wiley & Sons, Inc.
We now consider two-point boundary value problems of the type obtained in Section 11.1 by separating the variables in a heat conduction problem for a bar of variable material properties and with a source term proportional to temperature.
This kind of problem also occurs in many other applications.
These boundary value problems are commonly known as Sturm-Liouville problems.
They consist of a differential equation of the form
together with the boundary conditions
10,0)()(])([ xyxryxqyxp
0)1()1(,0)0()0( 2121 yyyy
Differential Operator Form
Consider the linear homogeneous differential operator L:
The differential equation
can then be written as
We assume that the functions p, p', q and r are continuous on0 x 1, and further, that p(x) > 0 and r(x) > 0 on [0,1].
The boundary conditions are said to be separated, since each one only involves one of the boundary points:
yxqyxpyL )(])([
0)()(])([ yxryxqyxp
yxryL )(
0)1()1(,0)0()0( 2121 yyyy
Lagrange’s Identity (1 of 3)
We next establish Lagrange’s identity, which is basic to the study of linear boundary value problems.
Let u and v be functions having continuous second derivatives on the interval 0 x 1. Then
Integrating the first term on the right twice by parts, we obtain
We thus obtain Lagrange’s identity:
1
0
1
0])([][ dxquvvupvdxuL
1
0
1
0
1
0
1
0
1
0
1
0
][)()()()()(
])([)()()()()()(][
dxvuLxvxuxvxuxp
dxquvuvpxvxuxpxvxuxpvdxuL
1
0
1
0)()()()()(][][ xvxuxvxuxpdxvuLvuL
Lagrange’s Identity Using Boundary Conditions (2 of 3)
Suppose that u and v satisfy the boundary conditions
If neither 2 nor 2 is zero, then
The result also holds if either 2 or 2 is zero.
Thus Lagrange’s identity reduces to
0
)0()0()0()0()0()1()1()1()1()1(
)0()0()0()0()0()1()1()1()1()1(
)()()()()(
2
1
2
1
1
0
vuvupvuvup
vuvupvuvup
xvxuxvxuxp
0)1()1(,0)0()0( 2121 yyyy
0][][1
0 dxvuLvuL
Lagrange’s Identity: Inner Product Form (3 of 3)
From the previous slide, we have Lagrange’s identity
In Section 10.2 we introduced the inner product (u,v) with
Thus Lagrange’s identity can be rewritten as
provided the previously mentioned assumptions are satisfied.
0][][1
0 dxvuLvuL
0)()(),(1
0 dxxvxuvu
,0][,],[ vLuvuL
Complex Inner Products
The inner product of two complex-valued functions on [0,1] is defined as
where is the complex conjugate of v.
Lagrange’s identity, as shown in the previous slides,
is still valid in this case since p(x), q(x), 1, 1, 2, and 2 are all real quantities.
0)()(),(1
0 dxxvxuvu
,0][,],[ vLuvuL
Sturm-Liouville Problems, Lagrange’s Identity
Consider again the Sturm-Liouville boundary value problem
This problem always has eigenvalues and eigenvectors.
In Theorems 11.2.1 – 11.2.4 of this section, we will state several of their important properties.
Each property is illustrated by the Sturm-Liouville problem
whose eigenvalues are n = n2 2 and corresponding eigenfunctions n(x) = sin n x.
These results rely on Lagrange’s identity.
,0)1(,0)0(,0 yyyy
0)1()1(,0)0()0(
10,0)()(])([
2121
yyyy
xyxryxqyxp
Theorem 11.2.1
Consider the Sturm-Liouville boundary value problem
All of the eigenvalues of this problem are real.
We give an outline of the proof on the next slide.
It can also be shown that all of the eigenfunctions are real.
0)1()1(,0)0()0(
10,0)()(])([
2121
yyyy
xyxryxqyxp
Theorem 11.2.1: Proof Outline
Suppose that is a (possibly complex) eigenvalue with corresponding eigenvector . Let = u + iv, (x) = U(x) + iV(x), where u, v, U, V are real.
Then it follows from Lagrange’s identity that
or
Since r(x) is real, we have
Recalling r(x) > 0 on [0,1], it follows that is real.
0,,0][,],[ rrLL
1
0
1
0)()()()()()( dxxrxxdxxrxx
0)()()(0)()()(1
0
221
0 dxxrxVxUdxxrxx
Theorem 11.2.2
Consider the Sturm-Liouville boundary value problem
If 1(x) and 2(x) are two eigenfunctions corresponding to the eigenvalues 1 and 2, respectively, and if 1 2, then
Thus 1(x) and 2(x) are orthogonal with respect to the weight function r(x).
0)1()1(,0)0()0(
10,0)()(])([
2121
yyyy
xyxryxqyxp
0)()()(1
0 21 dxxrxx
Theorem 11.2.2: Proof Outline
Let 1, 2, 1(x) and 2(x) satisfy the hypotheses of theorem.
Then
or
Since r(x), 2 and 2(x) are all real, we have
Since 1 2, we have
0,,0][,],[ 21212121 rrLL
1
0 21
1
0 211 )()()()()()( dxxrxxdxxrxx
0)()()(1
0 2121 dxxrxx
0)()()(1
0 21 dxxrxx
Theorem 11.2.3
Consider the Sturm-Liouville boundary value problem
The eigenvalues are all simple. That is, to each eigenvalue there corresponds one and only one linearly independent eigenfunction.
Further, the eigenvalues form an infinite sequence and can be ordered according to increasing magnitude so that
Moreover,
0)1()1(,0)0()0(
10,0)()(])([
2121
yyyy
xyxryxqyxp
n 321
n
nlim
Illustration of Sturm-Liouville Properties
The properties stated in Theorems 11.2.1 – 11.2.3 are again illustrated by the Sturm-Liouville problem
whose eigenvalues n = n2 2 are real, distinct and increasing,
such that n as n .Also, the corresponding real eigenfunctions n(x) = sin n x are orthogonal to each other with respect to r(x) = 1 on [0, 1].
,0)1(,0)0(,0 yyyy
Orthonormal Eigenfunctions
We now assume that the eigenvalues of the Sturm-Liouville problem are ordered as indicated in Theorem 11.2.3.
Associated with the eigenvalue n is a corresponding real eigenfunction n(x) determined up to a multiplicative constant.
It is often convenient to choose the arbitrary constant multiplying each eigenfunction so as to satisfy the condition
The eigenfunctions are said to be normalized, and form an orthonormal set with respect to the weight function r, since they are orthogonal and normalized.
,2,1,1)()(1
0
2 ndxxrxn
Kronecker Delta
For orthonormal eigenfunctions, it is useful to combine the orthogonal and normalization relations into one symbol.
To this end, the Kronecker delta mn is helpful:
Thus, for our orthonormal eigenfunctions, we have
nm
nmmn if,1
if,0
mnnm dxxrxx 1
0)()()(
Example 1: Orthonormal Eigenfunctions
Consider the Sturm-Liouville problem
In this case the weight function is r(x) = 1. The eigenvalues and eigenfunctions are n = n2 2 and yn(x) = sin n x.
To find the orthonormal eigenfunctions, we choose kn so that
Now
and hence the orthonormal eigenfunctions are
,0)1(,0)0(,0 yyyy
,2,1,1)sin(1
0
2 ndxxnkn
,2/2cos12/sin11
0
221
0
22 nnn kdxxnkxdxnk
,2,1,sin2)( nxnxn
Example 2: Boundary Value Problem (1 of 3)
Consider the Sturm-Liouville problem
In this case is r(x) = 1. The eigenvalues n satisfy
and the corresponding eigenfunctions are
To find the orthonormal eigenfunctions, we choose kn so that
or
0)1()1(,0)0(,0 yyyyy
,2,1,1sin1
0
2 ndxxk nn
,2,1,sin nxkxy nnn
,2,1,1sin1
0
22 nxdxk nn
0cossin
Example 2: Orthonormal Eigenfunctions (2 of 2)
We have
where in the last step we have used
Thus the orthonormal eigenfunctions are
,2
cos1
2
cossin
4
2sin2
4
2sin
22
2cos
2
1sin1
2222
1
0
21
0
21
0
22
nn
n
nnnn
n
nnn
n
nn
nnnn
kkk
xxkdx
xkxdxk
0cossin
,2,1,)cos1(
sin2)(
2/12
n
xx
n
nn
Eigenfunction Expansions (1 of 2)
We now investigate expressing a given function f as a series of eigenfunctions of the Sturm-Liouville boundary value problem
For example, if f is continuous and has a piecewise continuous derivative on [0, 1], and satisfies f (0) = f (1) = 0, then f can be expanded in a Fourier sine series of the form
which converges for each x in [0, 1].
This expansion of f is given in terms of the eigenfunctions of
dxxnxfbxnbxfn
nn
1
01
sin)(2,sin)(
0)1()1(,0)0()0(
10,0)()(])([
2121
yyyy
xyxryxqyxp
0)1(,0)0(,0 yyyy
Eigenfunction Expansions (2 of 2)
Suppose that a given function f, satisfying suitable conditions, can be expressed in a series of orthonormal eigenfunctions n(x) of the Sturm-Liouville boundary value problem.
Then
To obtain cm, we assume the series can be integrated term by term, after multiplying both sides by m(x)r(x) and integrating:
In inner product form, we have
dxxrxxfcxcxf nn
nnn
1
01
)()()(),()(
mn
mnnm cdxxrxxcdxxrxxf
1
1
0
1
0)()()()()()(
,2,1,, mrfc mm
Theorem 11.2.4
Consider the Sturm-Liouville boundary value problem
Let 1, 2, …, n,… be the normalized eigenfunctions for this problem, and let f and f ' be piecewise continuous on 0 x 1.
Then the series
converges to [ f (x+) + f (x-)]/2 at each point x in the open interval 0 < x < 1.
0)1()1(,0)0()0(
10,0)()(])([
2121
yyyy
xyxryxqyxp
dxxrxxfcxcxf nn
nnn
1
01
)()()(),()(
Theorem 11.2.4: Discussion
If f satisfies further conditions, then a stronger conclusion can be established. Consider again the Sturm-Liouville problem
Suppose f is continuous and f ' piecewise continuous on[0,1].
If 2 = 0, then assume that f(0) = 0. If 2 = 0, then assume that f(1) = 0. Otherwise no boundary conditions need to be prescribed for f. Then the series
converges to f (x) at each point in the closed interval [0, 1].
dxxrxxfcxcxf nn
nnn
1
01
)()()(),()(
0)1()1(,0)0()0(
10,0)()(])([
2121
yyyy
xyxryxqyxp
Example 3: Function f (1 of 3)
Consider the function
Recall from Example 2, for the Sturm-Liouville problem
the orthonormal eigenfunctions are
Then by Theorem 11.2.4, we have
,0)1()1(,0)0(,0 yyyyy
10,)( xxxf
,2,1,)cos1(
2,sin)(
2/12
nkxkx
n
nnnn
dxxrxxfcxcxf nn
nnn
1
01
)()()(),()(
Example 3: Coefficients (2 of 3)
Thus
Integrating by parts, we obtain
where in the last step we have used
Next, recall that
dxxxkdxxrxxfc nnnn 1
0
1
0sin)()()(
n
nn
n
n
n
nnn kkc
sin2cossin
0cossin
,2,1,)cos1(
22/12
nkn
n
Example 3: Eigenfunction Expansion (3 of 3)
We have
It follows that
Thus
,2,1,cos1
sin222
ncnn
nn
12
1 cos1
sinsin4sin)(
n nn
nn
nnnn
xxkcxf
n
n
nn
nnn kc
sin2
cos1
2sin22
Sturm-Liouville Problems and Algebraic Eigenvalue Problems
Sturm-Liouville boundary value problems are of great importance in their own right, but they can also be viewed as belonging to a much more extensive class of problems that have many of the same properties.
For example, there are many similarities between Sturm-Liouville problems and the algebraic system Ax = x, where the n x n matrix A is real symmetric or Hermitian.
Comparing the results mentioned in Section 7.3 with those of this section, in both cases the eigenvalues are real and the eigenfunctions or eigenvectors form an orthogonal set, and can be used as a basis for expressing an essentially arbitrary function or vector, respectively, as a sum.
Linear Operator Theory
The most important difference is that a matrix has only a finite number of eigenvalues and eigenvectors, while a Sturm-Liouville system has infinitely many.
It is of fundamental importance in mathematics that these seemingly different problems – the matrix problem Ax = x and the Sturm-Liouville problem,
which arise in different ways, are actually parts of a single underlying theory.
This theory is linear operator theory, and is part of the subject of functional analysis.
,0)1()1(,0)0()0(
10,0)()(])([
2121
yyyy
xyxryxqyxp
Self-Adjoint Problems (1 of 4)
Consider the boundary value problem consisting of the differential equation L[y] = r(x)y, where
and n homogeneous boundary conditions at the endpoints.
If Lagrange’s identity
is valid for every pair of sufficiently differentiable functions that satisfy the boundary conditions, then the problem is said to be self-adjoint.
,)()()( 01 yxPdx
dyxP
dx
ydxPyL
n
n
n
0][,],[ vLuvuL
Self-Adjoint Problems and Structure of Differential Operator L (2 of 4)
Lagrange’s identity involves restrictions on both the differential equation and the boundary conditions.
The differential operator L must be such that the same operator appears in both terms of Lagrange’s identity,
This requires L to be of even order. Further, a second order operator must have the form
and a fourth order operator must have the form
Higher order operators must have an analogous structure.
0][,],[ vLuvuL
yxqyxpyL )(])([
yxsyxqyxpyL )(])([])([
Self-Adjoint Problems and Boundary Conditions (3 of 4)
In addition, the boundary conditions must be such as to eliminate the boundary terms that arise during the integration by parts used in deriving Lagrange’s identity.
For example, in a second order problem this is true for the separated boundary conditions
and also in other cases, one of which is given in Example 4, as we will see.
0)1()1(,0)0()0( 2121 yyyy
Fourth Order Self-Adjoint Problems and Eigenvalue, Eigenvector Properties (4 of 4)
Suppose we have self-adjoint boundary value problem for L[y] = r(x)y, where L[y] is given by
We assume that p(x), q(x), r(x), s(x) are continuous on [0,1] and that the derivatives p', p'' and q' are also continuous.
Suppose also that p(x) > 0 and r(x) > 0 on [0,1].
Then there is an infinite sequence of real eigenvalues tending to , the eigenfunctions are orthogonal with respect to the weight function r, and an arbitrary function f can be expressed as a series of eigenfunctions.
However, the eigenfunctions may not be simple in these more general problems.
yxsyxqyxpyL )(])([])([
Sturm-Liouville Problems and Fourier Series
We have noted previously that Fourier sine (and cosine) series can be obtained using the eigenfunctions of certain Sturm-Liouville problems involving the differential equation
This raises the question of whether we can obtain a full Fourier series, including both sine and cosine terms, by choosing a suitable set of boundary conditions.
The answer is yes, as we will see in the following example.
This example will also serve to illustrate the occurrence of nonseparated boundary conditions.
0 yy
Example 4: Boundary Value Problem (1 of 3)
Consider the boundary value problem
This is not a Sturm-Liouville problem because the boundary conditions are not separated.
The boundary conditions above are called periodic boundary conditions since the require that y and y' assume the same values at x = L as at x = -L.
It is straightforward to show that this problem is self-adjoint.
0)()(,0)()(,0 LyLyLyLyyy
Example 4: Eigenvalues and Eigenfunctions (2 of 3)
Our boundary value problem is
It can be shown that 0 = 0 is an eigenvalue with corresponding eigenfunction 0(x) = 1.All other eigenvalues are given by
To each of these eigenvalues there corresponds two linearly independent eigenfunctions.
For example, the eigenfunctions corresponding to n are
This shows that the eigenfunctions may not be simple when the boundary conditions are not separated.
0)()(,0)()(,0 LyLyLyLyyy
,/,,/2,/ 222
21 LnLL n
LnxLnx nn /sin)(,/cos)(
Example 4: Eigenfunction Expansion (3 of 3)
Further, if we seek to expand a given function f of period 2L in a series of eigenfunctions for this problem, we obtain
which is just the Fourier series of f.
1
0 sincos2
)(n
nn L
xnb
L
xna
axf