Sturm-Liouville Problem/Unit Weight Function

Theorem
Let $ P, Q : \R \to \R $ be real mappings such that $ P $ is smooth and positive, while $ Q $ is continuous:


 * $ \displaystyle P \left ( { x } \right ) \in C^\infty $


 * $ \displaystyle P \left ( { x } \right ) > 0 $


 * $ \displaystyle Q \left ( { x } \right ) \in C^0 $

Let the Sturm-Liouville equation, with $ w \left ( { x } \right ) = 1 $, be of the form:


 * $ - \left ( { P y' } \right )' + Qy = \lambda y $

where $ \lambda \in \R $.

Let it satisfy the following boundary conditions:


 * $ y \left ( { a } \right ) = y \left ( { b } \right ) = 0 $

Then all solutions of the Sturm-Liouville equation, together with their eigenvalues, form infinite sequences $ \{ { y^{ \left ( { n } \right ) } } \} $ and $ \{ { \lambda^{ \left ( { n } \right ) } } \} $.

Furthermore, each $ \lambda^{ \left ( { n } \right ) } $ corresponds to an eigenfunction $ y^{ \left ( { n } \right ) } $, unique up to a constant factor.

Lemma
The given Sturm-Liouville equation is an Euler equation of the following functional:


 * $ \displaystyle J \left [ { y } \right ] = \int_a^b \left ( { P y'^2 + Q y^2 } \right ) \mathrm d x $

constrained by a subsidiary condition:


 * $ \displaystyle \int_a^b y^2 \mathrm d x = 1 $

Proof
According to Simplest Variational Problem with Subsidiary Conditions, the following equation must hold:


 * $ \displaystyle F_y - \frac{ \mathrm d }{ \mathrm d x } F_{ y' } + \lambda \left ( { G_y - \frac{ \mathrm d }{ \mathrm d x } G_{ y' } } \right) = 0 $

where:


 * $ F = P y'^2 + Q y^2 $


 * $ G = y^2 $

Then the Euler equation reads:


 * $ \displaystyle 2 Q y - 2 \left ( { P y' } \right )' + 2 \lambda y = 0 $

Division by $ 2 $ and rearrangement of terms yields the desired result.

By Necessary Condition for Integral Functional to have Extremum for given function, if $ y $ is an extremum of $ J $, it is also a solution of the Sturm-Liouville equation.

Lemma
$ J $ is bounded from below.

Proof
Since $ Q $ is continuous on an interval, it is bounded.

Since $ P > 0 $, it holds that:


 * $ \displaystyle \int_a^b \left ( { Py'^2 + Qy^2 } \right ) \mathrm d x > \int_a^b Qy^2 \mathrm d x \ge M \int_a^b y^2 \mathrm d x = M $

where


 * $ \displaystyle M = \min_{ a \le x \le b } Q \left ( { x } \right )$

Therefore, $ J $ is bounded from below.

Introduce a new variable $ \displaystyle t = \pi \frac{ x - a }{ b - a } $.

Then the interval of consideration $ \left [ { a \,. \,. \, b } \right ] $ is mapped onto $ \left [ { 0 \,. \,. \, \pi } \right ] $.

Choose Ritz sequence $ \{ { \phi_n \left ( { t } \right ) } \} = \{ { \sin nt  } \} $, where $ n \in \N $.

Lemma
The elements of the sequence $ \{ { \sin nt } \} $ are orthogonal on the interval $ \left [ { 0 \,. \,. \, \pi } \right ] $:


 * $ \displaystyle \int_0^\pi \sin \left ( { k t } \right ) \sin \left ( { l t } \right ) \mathrm d x = \frac{ \pi }{ 2 } \delta_{ k l } $

Proof
The product involves two elements of the sequence $ \{ { \sin nt } \} $.

Their indices either match each other or not.

Suppose $ k = l $.

Then:

Suppose $ k \ne l $.

Then:

By Proof by Cases, the statement is proved.

Let the trial solution be of the following form:


 * $ \displaystyle y \left ( { x } \right ) = \sum_{ k = 1 }^n \alpha_k \sin \left ( { k t \left ( { x } \right ) } \right ) $

Trial solution has to satisfy boundary and subsidiary conditions.

Boundary conditions are satisfied without further requirements.

Subsidiary condition results into an additional constraint on coefficients $ \alpha_k $:

All the points $ \boldsymbol \alpha $ constitute a set $ \sigma_n $ which is a surface of an $ n $-dimensional sphere, defined by the subsidiary condition.

For the assumed trial mapping the functional $ J_n \left ( { \boldsymbol \alpha } \right ) $ reads as:


 * $ \displaystyle J_n \left ( { \boldsymbol \alpha } \right ) = \frac{ \pi }{ b - a } \int_0^\pi \left [ { P \left ( { \sum_{ k = 1 }^n \alpha_k \sin k t } \right )'^2 + Q  \left ( { \sum_{ k = 1 }^n \alpha_k \sin k t  } \right )^2 } \right ] \mathrm d t $

The integrand is a second order polynomial the components of $ \boldsymbol \alpha $.

Hence, $ J $ is continuous the components of $ \boldsymbol \alpha $.

The components of $ \boldsymbol \alpha $ constitute a closed and bounded set.

By definition, $ \sigma_n $ is a compact set.

Thus, $ J_n \left ( { \boldsymbol \alpha } \right ) $ is continuous on $ \sigma_n $.

By Continuous Function on Compact Space is Bounded, $ J_n \left ( { \boldsymbol \alpha } \right ) $ has a minimum on $ \sigma_n $.

Let $ y_n^{ \left ( { 1 } \right ) } \left ( { x } \right ) $ be defined as:


 * $ \displaystyle y_n^{ \left ( { 1 } \right ) } \left ( { x } \right ) = \sum_{ k = 1 }^n \alpha_k^{ \left ( { 1 } \right ) } \sin k t \left ( { x } \right ) $

for which $ J_n \left ( { \boldsymbol \alpha } \right ) $ achieves the minimum $ \lambda_n^{ \left ( { 1 } \right ) } $.

Then the $ n $-th element of the sequence $ \{ { y_n^{ \left ( { 1 } \right ) } } \} $ corresponds to the $ n $-th element of the sequence of minima $ \{ { \lambda_n^{ \left ( { 1 } \right ) } } \} $ of $ J_n \left ( { \boldsymbol \alpha } \right ) $.

Since $ \sigma_n \subset \sigma_{ n + 1} $, where $ \sigma_n $ has $ \alpha_{ n + 1 } = 0 $, it holds that:


 * $ \displaystyle J_n \left ( { \alpha_1, \ldots, \alpha_n } \right ) = J_{ n + 1 } \left ( { \alpha_1, \ldots, \alpha_n, 0 } \right ) $

By Ritz Method implies Not Worse Approximation with Increased Number of Functions:


 * $ \displaystyle \lambda_{ n + 1 }^{ \left ( { 1 } \right ) } \le \lambda_n^{ \left ( { 1 } \right ) } $

By increasing the domain of definition of $ y_n^{ \left ( { 1 } \right ) } $, the minima of $ J_n \left ( { \boldsymbol \alpha } \right ) $ cannot increase.

From the last inequality and $ J $ being bounded from below it follows, that the following limit exists:


 * $ \displaystyle \lambda^{ \left ( { 1 } \right ) } = \lim_{ n \to \infty } \lambda_n^{ \left ( { 1 } \right ) } $