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Orthogonality, Reality, and Uniqueness

Two eigenfunctions $ u_m(x)$ and $ u_n(x)$ are said to be orthogonal relative to the weight function $ \rho (x)$ if

$\displaystyle \int^b_a \overline{u}_m (x) u_n(x) \rho (x) dx=0~\quad~\textrm{whenever}~\quad~
m\not= n\,.
$

They are said to be orthonormal with respect to $ \rho (x)$ if

$\displaystyle \int^b_a \overline{u}_m(x) u_n(x) \rho (x) dx = \delta_{mn}\,.$ (312)

With these reminders at hand, one can now identify the two most important properties of a S-L system, the orthonormality of its eigenfunctions and the reality of its eigenvalues. The statement and the proof of these properties parallel those of the familiar eigenvalue problem from linear algebra,

$\displaystyle A\vec u = \lambda B\vec u
$

where $ A$ is a Hermitian and $ B$ is a positive definite matrix.


Theorem 1 (Orthogonality) Let $ \lambda_m$ and $ \lambda_n$ be any two distinct eigenvalues of the S-L problem 3.10 and 3.11 with corresponding eigenfunctions $ u_m$ and $ u_n$ . Then $ u_m$ and $ u_n$ are orthogonal with respect to the weight $ \rho (x)$ .

Orthogonality also holds in the following cases

  1. when $ p(a)=0$ and the first (1) of the boundary conditions 3.11 is dropped. This is equivalent to setting $ \alpha =\alpha '=0$
  2. when $ p(b)=0$ and the second (2) of the conditions 3.11 is dropped. This is equivalent to setting $ \beta =\beta '=0$
  3. when $ p(a)=p(b)$ and 3.11 are replaced by
    $\displaystyle u(a)$ $\displaystyle =$ $\displaystyle u(b)$ (313)
    $\displaystyle u'(a)$ $\displaystyle =$ $\displaystyle u'(b) ~~.$  

Remarks:

  1. In case (1.) or (2.), the S-L problem is said to be singular.
  2. The S-L problem with mixed Dirichlet-Neumann conditions at both ends is said to be regular.
  3. The same terminology, (``singular'') is also used when
    $\displaystyle \rho (x)$ $\displaystyle ~$ $\displaystyle \textrm{vanishes~at~an~endpoint},$  
    $\displaystyle q(x)$   $\displaystyle \textrm{is~singular~at~an~endpoint},$  
    $\displaystyle (a,b)$   $\displaystyle \textrm{is~unbounded}\,.$  

    In other words, we are not interested in the actual value of $ u(x)$ , just that it stays finite. This is sufficient to select (a multiple of the correct) one of two independent solutions to the differential equation.
  4. The boundary conditions 3.13 are those of a periodic S-L problem, for example, the one where $ x$ is the angle $ \varphi$ in cylindrical coordinates. (More on that later.)
  5. This theorem is analogous to the orthogonality $ \langle u_m,Bu_n\rangle =0$ , of the eigenvectors $ u_m$ and $ u_n$ of the familiar eigenvalue problem $ A\vec u=\lambda B\vec u$ .
  6. The physical significance of the orthogonality of the eigenfunctions is exemplified by the energy of a vibrating system governed by, say the wave equation, Eq.(3.2). Its total energy,

    $\displaystyle T.E. = \frac{1}{2}\int_a^b \left[ \rho(x) \left( \frac{\partial v...
...ht)^2 +T(x) \left( \frac{\partial v}{\partial x}\right)^2 +k(x) v^2 \right]dx~,$ (314)

    is the sum of its kinetic energy and its combined potential energies due to the tension in the string and due to the elasticity of the ambient medium in which the string makes its transverse excursions. Performing an integration by parts on the second term, dropping the endpoint terms due to the imposed homogeneous boundary conditions, and using the governing wave equation, Eq.(3.2), one finds that the total energy is
    $\displaystyle T.E.$ $\displaystyle =$ $\displaystyle \frac{1}{2}\int_a^b
\left[ \rho(x) \left( \frac{\partial v}{\part...
...artial}{\partial x} T(x)\frac{\partial v}{\partial x}
+k(x)v \right)v\right]
dx$  
      $\displaystyle =$ $\displaystyle \frac{1}{2}\int_a^b
\left[ \left( \frac{\partial v}{\partial t}\right)^2
-\frac{\partial^2 v}{\partial t^2} v\right] \rho(x)\, dx$  

    Suppose the total vibrational amplitude is a superposition of the amplitudes associated with with each eigenfrequency $ \omega_n$ ,

    $\displaystyle v(x,t)=\sum_n c_n u_n(x) \cos(\omega_n t+\delta_n)~.
$

    Then the total energy becomes
    $\displaystyle T.E.$ $\displaystyle =$ $\displaystyle \frac{1}{2} \sum_m\sum_n
[ \omega_m\omega_n
\sin(\omega_m t+\delta_m)\sin(\omega_n t+\delta_n)$  
      $\displaystyle +$ $\displaystyle \omega_m^2\cos(\omega_m t+\delta_m)\cos(\omega_n t+\delta_n) ]
\bar{c}_m c_n \int_a^b \bar{u}_m(x)u_n(x)\rho(x)\,dx$  

    The orthonormality, Eq.(3.12), implies that

    $\displaystyle \boxed{T.E.=\frac{1}{2} \sum_n \omega_n^2\vert c_n\vert^2}~.
$

    Thus we see that the orthonormality of the S-L eigenfunctions expresses the fundamental fact that the total energy, a constant independent of time, is composed of the mutually exclusive and constant energies residing in each normal mode (``vibratory degree of freedom'').
Proof in 3 Steps: In analogy to $ Au_m=\lambda_m Bu_m$ and $ Au_n=\lambda_n Bu_n$ one first considers

Step (1) $ \underbrace{-(p\overline{u}'_m)'-q\overline{u}_m}_{
\begin{array}{c}
\parallel\\
L\overline{u}_m
\end{array}} =\lambda_m
\rho \overline{u}_m$ ;     $ \underbrace{-(p u'_n)'-qu_n}_{
\begin{array}{c}
\parallel\\
Lu_n
\end{array}}=\lambda_n
\rho u_n$ .

Then multiply the equations respectively by $ u_n$ and $ \overline{u}_m$ and subtract them. The left hand side becomes

$\displaystyle l.h.s.$ $\displaystyle =$ $\displaystyle u_nL\overline{u}_m-\overline{u}_mLu_n\equiv \frac{d}{dx}p(\overline{u}_mu'_n-u_n \overline{u}'_m)$ (315)

We now interrupt the three-step proof to remark that this is an important identity known as Lagrange's Identity. We shall meet it and refer to it in several subsequent sections. This identity generalizes to higher dimensions by means of the vector identity $ u_n\nabla^2
\overline{u}_m-\overline{u}_m \nabla^2 u_n=\nabla\cdot (u_n\vec\nabla
\overline{u}_m- \overline{u}_m\vec\nabla u_n)$ .

The integral version of Lagrange's Identity is known as Green's identity

$\displaystyle \int^b_a (u_nL\overline{u}_m-\overline{u}_mLu_n)dx = p(x)(\overline{u}_mu'_n-u_n\overline{u}'_m)\Bigg\vert^b_a$ (316)

in 1 dimension. Observe the parallel of this with Green's Identity in three dimensions:

$\displaystyle \mathop{\int\!\!\int\!\!\int}\limits_{\textrm{volume}} (u_n\nabla...
...
(u_n\vec\nabla \overline{u}_m-\overline{u}_m\vec\nabla u_n)\cdot \vec{dS}\,.
$

We now continue the three-step proof by considering the right hand side of the above subtraction result,

$\displaystyle r.h.s. = (\lambda_m-\lambda_n)\rho \overline{u}_mu_n\,.$

Step (2) Both sides are equal. Upon integrating them, one obtains

$\displaystyle (\lambda_m -\lambda_n)\int^b_a \overline{u}_m u_n
\rho (x)dx=p(x)W[\overline{u}_m,u_n](x)\Bigg\vert^b_a
$

where

$\displaystyle W[\overline{u}_m,u_n]=\left\vert\begin{array}{cc}
\overline{u}_m &\overline{u}'_m\\
u_n &u'_n\end{array}\right\vert\,.
$

This would be called the Wronskian of $ \overline{u}_m$ and $ u_n$ if $ \lambda_m$ and $ \lambda_n$ were equal. The right hand side of this one-dimensional Green's identity depends only on the boundary (end) points. The idea is to point out that this right hand side vanishes for any one of the boundary conditions under consideration.

Step (3a) If one has D-N conditions

$\displaystyle \alpha u(a)+\alpha 'u'(a)$ $\displaystyle =$ 0  
$\displaystyle \beta u(b)+\beta 'u'(b)$ $\displaystyle =$ $\displaystyle 0\,,$  

then these D-N conditions imply
$\displaystyle W(a)$ $\displaystyle =$ 0  
$\displaystyle W(b)$ $\displaystyle =$ 0  

because $ 1^{\textrm{st}} = 2^{\textrm{nd}}$ columns are proportional. Thus for a regular S-L problem

$\displaystyle (\lambda_m -\lambda_n )\int^b_a \overline{u}_m(x)u_n(x) \rho (x)dx = 0\,,
$

i.e., one has orthogonality whenever $ \lambda_m \not= \lambda_n$ .

Step (3b) If one has a periodic S-L problem

  $\displaystyle p(b) = p(a)~~\qquad$ $\displaystyle ~~$  
  $\displaystyle \left.\begin{array}{l}
u(a) = u(b)\\
u'(a) = u'(b)\end{array}\right\}$ $\displaystyle \Rightarrow W(a)=W(b)\,.$  

i.e., one again has orthogonality whenever $ \lambda_m \not= \lambda_n$ .

Setp (3c) If one has a singular S-L problem

$\displaystyle \left.\begin{array}{l}
p(b) =0\\
W(b) = \textrm{finite}\end{array}\right\}\Rightarrow p(b)W(b) = 0\,.
$

Similar considerations at the other end point also yield zero. Once again one has orthogonality whenever $ \lambda_m \not= \lambda_n$ . To summarize, the eigenfunctions of different eigenvalues of regular, periodic, and singular Sturm-Liouville systems are orthogonal.


Lecture 21


Theorem 2 (Reality of Eigenvalues) For a regular, periodic, and singular S-L system the eigenvalues are real.

Proof: Step (1) Let $ u$ be an eigenfunction corresponding to the complex eigenvalue $ \lambda =\mu +i\nu$ . The eigenfunctions are allowed to be complex. Thus

$\displaystyle Lu = \lambda \rho (x) u~~\qquad~~$ $\displaystyle \textrm{and}$ $\displaystyle ~~\qquad~~L\overline{u} =
\overline{\lambda} \rho \overline{u}$  
$\displaystyle \alpha u(a)+\alpha ' u'(a)=0$ $\displaystyle ~~$ $\displaystyle \alpha \overline{u}(a)+\alpha '\overline{u'}
(a)=0$  
$\displaystyle \beta u(a)+\beta ' u'(a)=0$ $\displaystyle ~~$ $\displaystyle \beta \overline{u}(a)+\beta '\overline{u'}
(a)=0$  

because

\begin{displaymath}\left.
\begin{array}{l}
L = \overline{L},~\rho (x)=\overline{...
...a},\overline{\beta '}\end{array}\right\}~~\textrm{are~real}\,.
\end{displaymath}

Step (2) We have, therefore,

$\displaystyle \underbrace{\int^b_a (\overline{u} Lu-uL\overline{u})dx}_{
\displ...
...Bigg\vert^b_a} = (\lambda -\overline{\lambda})\int^b_a\overline{u} u\rho (x)dx
$

$\displaystyle 0=(\lambda -\overline{\lambda})\int^b_a\vert u\vert^2 \rho (x)dx\,.
$

This implies that $ \lambda = \overline{\lambda}$ , i.e., that $ \lambda $ is real.

We now inquire as to the number of independent eigenfunctions corresponding to each eigenvalue. This is a question of uniqueness. The examples on page [*] have only one such eigenfunction for each eigenvalue. Consider, however, the following

Example (Periodic S-L system)

$\displaystyle u'' +\lambda u$ $\displaystyle =$ $\displaystyle 0~~\qquad~~\qquad~~-1<x<1$  
$\displaystyle u(-1)$ $\displaystyle =$ $\displaystyle u(1)$  
$\displaystyle u'(-1)$ $\displaystyle =$ $\displaystyle u'(1)\,.$  

We note that $ p(-1)=p(1)$ . Consequently, this is a periodic S-L system.

The form of the solution can be written down by inspection. Letting $ \lambda =\alpha ^2$ , one obtains

$\displaystyle u(x)=c_1\cos \alpha x+c_2\sin\alpha x
$

without loss of generality we assume $ \alpha >0$ . The two boundary conditions imply

$\displaystyle 2c_2\sin\alpha =0
$

and

$\displaystyle -2\alpha c_1\sin\alpha =0\,.
$

Both conditions yield non-zero solutions whenever $ \alpha =0,\pi ,2\pi ,\dots$ . Consequently, the eigenvalues are

$\displaystyle \lambda_n=n^2\pi^2~~\qquad~~\qquad~~n=0,1,2,\cdots \,.
$

Note that for every eigenvalue (except $ \lambda_0$ ) there are two eigenfunctions
$\displaystyle \lambda_0$ $\displaystyle \colon$ $\displaystyle \frac{1}{2}$  
$\displaystyle \lambda_1$ $\displaystyle \colon$ $\displaystyle \cos\pi x\,,~\sin\pi x$  
$\displaystyle \vdots$      
$\displaystyle \lambda_n$ $\displaystyle \colon$ $\displaystyle \cos n\pi x\,,~\sin n\pi x\,.$  

Such nonuniqueness is expressed by saying that each of the eigenvalues $ \lambda_1,\lambda_2,\dots$ is degenerate, in this example doubly degenerate because there are two independent eigenfunctions for each eigenvalue.

The next theorem states that this cannot happen for a regular S-L system. Its eigenvalues are simple, which is to say they are nondegenerate.

Note that the theorem below uses Abel's Theorem, namely Theorem 4.

Theorem 3 (Uniqueness of solutions to the regular S-L system.) An eigenfunction of a regular Sturm-Liouville system is unique except for a constant factor, i.e., the eigenvalues of a regular S-L problem are simple.

Proof: For the same eigenvalue $ \lambda $ , let $ u_1$ and $ u_2$ be two eigenfunctions of the regular S-L system. For a regular S-L system the b.c. are

$\displaystyle \alpha u_1(a)+\alpha ' u'_1(a)$ $\displaystyle =$ 0  
$\displaystyle \alpha u_2 (a) +\alpha ' u'_2(a)$ $\displaystyle =$ $\displaystyle 0\,.$  

In other words, both solutions satisfy the D-N mixed boundary conditions at the left hand endpoint. The value of the Wronskian at $ x=a$ is

$\displaystyle W[u_1,u_2](a) = \left\vert\begin{array}{ll}
u_1 &u'_1\\
u_2 &u'_...
...x=a} = 0\,,~~\qquad~~{\textrm{columns~are}
\choose
\textrm{proportional}}\,.
$

Using Abel's Theorem: $ p(x)W[u_1,u_2](x)=$ constant, we obtain

$\displaystyle \frac{u'_1(x)}{u_1(x)} -\frac{u'_2(x)}{u_2(x)} = 0\Rightarrow u_1(x)=ku_2(x)\,.
$

This conclusion says that the solution $ u_1(x)$ is unique (up to a constant multiplicative factor).

NOTE: If the endpoint condition had been the periodic boundary condition, then one cannot conclude that the eigenvalues are simple. This is because

\begin{displaymath}
\begin{array}{l}
u(a)=u(b)\\
u'(a)=u'(b)\end{array}~~\textr...
...t}~\textrm{imply}~~[u_1(x)u'_2(x)-
u'_1(x)u_2(x)]_{x=a} = 0\,.
\end{displaymath}

The previous uniqueness theorem used Abel's theorem, which applies to a second order linear differential equation regardless of any boundary conditions imposed on its solutions.

Theorem 4 (Abel) If $ u_1$ and $ u_2$ are two solutions to the same differential equation

$\displaystyle \left[ -\frac{d}{dx} p\frac{d}{dx} - q\right] u=\lambda \rho u
$

(i.e., $ Lu = \lambda \rho u$ ), then

$\displaystyle p(x)[u_1(x)u'_2(x)-u_2(x)u'_1(x)] = \textrm{constant}\,.
$

Remark. The expression in square brackets,

$\displaystyle W=u_1u'_2-u_2u'_1
$

is called the ``Wronskian'' or the ``Wronskian determinant''.

Proof: Start with Lagrange's identity

$\displaystyle u_2 Lu_1-u_1Lu_2 = \frac{d}{dx} p(u_1u'_2-u_2u'_1)\equiv \frac{d}{dx}p(x) W
[u_1,u_2]\,.
$

Use the given differential equation to conclude that the left hand side vanishes, i.e.

$\displaystyle 0 = \frac{d}{dx} p(x) W[u_1,u_2]\,.
$

Thus $ p(x) W[u_1,u_2](x)$ is indeed a constant, independent of $ x$ .

A nice application of this theorem is that it gives us a way of obtaining a second solution to the given differential equation, if the first one is already known.

Using Abel's theorem, the Wronskian determinant can be rewritten in the form

$\displaystyle u^2_1\left(\frac{u'_2}{u_1} - u_2 \frac{u'_1}{u_1^2}\right) =
\frac{\textrm{const.}}{p}
$

or

$\displaystyle u^2_1\frac{d}{dx} \left(\frac{u_2}{u_1}\right) =
\frac{\textrm{const.}}{p}\,.
$

Integration yields the following

Corollary (Second solution)

$\displaystyle u_2=u_1(x)\int^x \frac{dx'}{p(x')u^2_1(x')} +c_1u_1\,.
$

Thus one is always guaranteed a second solution if a first solution is known.

Exercise 33.1 (NONDEGENERATE EIGENVALUES)
Consider the S-L eigenvalue problem

$\displaystyle [Lu_n](x)\equiv\left(-{d^2 \over dx^2} +x^2\right)u_n (x)=\lambda_nu_n(x)\,;~~ \lim_{x\to\pm\infty} u(x)=0\,;$ (317)

on the infinite interval $ (-\infty
,\infty)$ .

Show that the eigenvalues $ \lambda_n$ are nondegenerate, i.e. show that, except for a constant multiplicative factor, the corresponding eigenfunctions are unique.

Nota bene:

(i)
The eigenfunctions are known as the Hermite-Gaussian polynomials. They are known to professionals in Fourier optics who work with laser beams passing through optical systems. A laser beam which is launched one focal length away from a lens, passes through the lens, and then is observed (on, say, a screen) one focal length after that lens, has an amplitude pattern which is precisely $ u_n(x)$ , whenever the beam was launched with that such an amplitude pattern.
(ii)
These eigenfunctions are also known to physicists who work with simple harmonic oscillators (e.g. vibrating molecules), in which case the eigenfunctions are the quantum states of an oscillator and the eigenvalues are its allowed energies.

Exercise 33.2 (EVEN AND ODD EIGENFUNCTIONS)
Consider the ``parity'' operator $ P:L^2(-\infty,\infty) \to L^2(-\infty,\infty)$ defined by

$\displaystyle P\psi(x)\equiv \psi(-x)
$

(i)
For a given function $ \psi(x) $ , what are the eigenvalues and eigen functions of $ P$ ?
(ii)
Show that the eigenfunctions of the operator $ L$ defined by Eq.(3.17) are eigenfunctions of $ P$ . Do this by first computing

$\displaystyle P^{-1}LP\psi(x)
$

for $ \psi \in L^2(-\infty,\infty)$ and then pointing out how $ P^{-1}LP$ is related to $ L$ .

Next point out how this relationship applied to an eigenfunction $ u_n$ of the previous problem leads to the result $ P u_n=\mu u_n$ .

Exercise 33.3 (EIGENBASIS OF THE FOURIER TRANSFORM $ {\mathcal{F}}$ )

Consider the S-L eigenvalue problem

$\displaystyle [Lu_n](x)\equiv\left(-{d^2 \over dx^2} +x^2\right)u_n (x)=\lambda_nu_n(x)\,;~~ \lim_{x\to\pm\infty} u(x)=0\,;$

on the infinite interval $ (-\infty
,\infty)$ . We know that the eigenvalues are nondegenerate and are

$\displaystyle \lambda_n=2n+1\,,~~\qquad~~n=0,1,\dots\,.
$

Consider now the Fourier transform on $ L^2(-\infty ,\infty )$ :

$\displaystyle {\cal F}[u](k)\equiv \int^\infty_{-\infty}
\frac{e^{-ikx}}{\sqrt{2\pi}}u(x)dx\,.
$

(a)
By computing

$\displaystyle {\cal F}L{\cal F}^{-1}\hat{\psi}(k)
$

for arbitrary $ \hat{\psi} \in L^2(-\infty,\infty)$ , determine the Fourier representation

$\displaystyle {\cal F}L{\cal F}^{-1}\equiv \hat L~.
$

of the operator

$\displaystyle L=-\frac{d^2 }{dx^2} +x^2
$

(b)
By viewing $ {\cal F}$ as a map $ L^2(-\infty,\infty) \to
L^2(-\infty,\infty)$ , compare the operators $ \hat L$ and $ L$ .

State your result in a single English sentence and also as a mathematical equation.

(c)
Use the result obtained in (b) to show that each eigenfunction $ u_n$ of the S-L operator $ L$ is also an eigenfunction of $ {\cal F}$ :

$\displaystyle {\cal F}u_n=\mu ~u_n\,.
$

By applying the result (e) of the Fourier eigenvector Exercise on page [*] to the previous Exercise determine the only allowed values for $ \mu $ . What is the Fourier transform of a Hermite-Gauss polynomial $ u_n(x)$ ?

CONGRATULATIONS, you have just found an orthonormal eigenbasis of the Fourier transform operator $ {\cal F}$ (in terms of the eigenbasis of the S-L operator $ L$ )!

Exercise 33.4 (HOW TO NORMALIZE AN EIGENFUNCTION)
Consider the S-L system

$\displaystyle \left[{d\over {dx}}p {d\over {dx}} + q + \lambda \rho\right] u = 0
\qquad a < x < b
$

$\displaystyle \alpha u(a) + \alpha' u'(a) = 0\,; \quad \beta u(b) + \beta' u'(b) = 0~~.
$

Let $ w (x,\lambda)$ be that unique solution to $ {d\over {dx}}p
{dw\over {dx}} + (q + \lambda \rho) w = 0$ which satisfies $ \alpha
w (a,\lambda) + \alpha' w' (a,\lambda) = 0\!$ . i.e. it satisfies the left hand boundary condition. Then $ w_n (x) \equiv w(x,\lambda_n)$ is an eigenfunction of the above S-L system corresponding to the eigenvalue $ \lambda_n$ .

Calculate the normalization integral $ \int\limits^b_a w_n^2~\rho~dx$ as follows:

(a)
Obtain the preliminary formula

$\displaystyle (\lambda - \lambda_n) \int^b_a w_n (x) w(x,\lambda) \rho(x) dx = p(b)
W(w,w_n)\vert^{x=b} ~~.
$

(b)
By taking the limit $ \lambda\to\lambda_n$ show that

$\displaystyle \int^b_a w^2_n~\rho~dx = p(b) \left[ w'_n (b) {dw(b,\lambda)\over...
... {d\over {d\lambda}} w' (b,\lambda)
\Bigg\vert_{\lambda = \lambda_n}\right]~~,
$

where prime denotes differentiation w.r.t. $ x$ .

Exercise 33.5 (ORTHONORMALIZED BESSEL FUNCTIONS)
Consider the Sturm-Liouville (S-L) problem

$\displaystyle \left[-{d\over {dx}}~x~{d\over {dx}} + {\nu^2\over {x}}\right] u =
\lambda xu~~.
$

Here $ u$ , $ \displaystyle {du\over {dx}}$ bounded as $ x\to 0$ , $ u(1) =
0$ where $ \nu$ is a real number.
(a)
Using the substitution $ t = \sqrt {\lambda~} x$ , show that the above differential equation reduces to Bessel's equation of order $ \nu$ . One solution which is bounded as $ t\to 0$ is $ J_\nu (t)$ ; a second linearly independent solution, denoted by $ Y_\nu
(t)$ , is unbounded as $ t\to 0$ .
(b)
Show that the eigenvalues $ \lambda_1,\lambda_2,\dots$ of the given problem are the squares of the positive zeroes of $ J_\nu (\sqrt {\lambda\,})$ , and that the corresponding eigenfunctions are

$\displaystyle u_n (x) = J_\nu (\sqrt {\lambda_n~} x) ~~.
$

(c)
Show that the eigenfunctions $ u_n(x)$ satisfy the orthogonality relation

$\displaystyle \int\limits^1_0 x~u_m(x)~u_n (x)~dx = 0\quad m\ne n\,.
$

(d)
For the case $ \nu=0$ , apply the method of the previous problem to exhibit the set of orthonormalized eigenfunctions $ \{ u_0(x),u_1(x),u_2(x),\cdots \}$ .
(e)
Determine the coefficients in the Fourier-Bessel series expansion

$\displaystyle f(x) \doteq \sum\limits^\infty_{n=1} ~c_n~u_n (x)\,.
$

Exercise 33.6 (ORTHOGONALITY OF LEGENDRE POLYNOMIALS)
Consider the S-L problem

$\displaystyle \left[ -{d\over {dx}} (1-x^2) {d\over {dx}} + {m^2\over
{1-x^2}}\right] u = \lambda u
$

Here $ u$ , $ \displaystyle {du\over {dx}}$ bounded as $ x\to\pm 1$ . Here $ m=$ integer. The solutions to this S-L problem are $ u_n = P^m_n
(x)$ , the ``associated Legendre polynomials'', corresponding to $ \lambda_n = n(n+1)$ , $ n=$ integer. Show that

$\displaystyle \int\limits^1_{-1} P^m_n (x) P^m_{n'} (x) ~dx = 0\quad \lambda_n\ne
\lambda_{n'}\,.
$


Lecture 22



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Ulrich Gerlach 2007-04-05