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:
:$\map P x \in C^\infty$ 
:$\map P x > 0$
:$\map Q x \in C^0$
Let the Sturm-Liouville equation, with $\map w x = 1$, be of the form:
:$-\paren {P y'}' + Q y = \lambda y$
where $\lambda \in \R$.
Let it satisfy the following boundary conditions:
:$\map y a = \map y b = 0$
Then all solutions of the Sturm-Liouville equation, together with their eigenvalues, form infinite sequences $\sequence {y^{\paren n} }$ and $\sequence {\lambda^{\paren n} }$.
Furthermore, each $\lambda^{\paren n}$ corresponds to an eigenfunction $y^{\paren n}$, unique up to a constant factor.
== Proof ==
=== Outline ===
Firstly, an equivalence between the Sturm-Liouville equation and minimisation of functional $\ds J \sqbrk y = \int_a^b \paren {P y'^2 + Q y^2} \rd x$ problems is established.
Then, the lower bound of the functional $J$ is found, thus allowing $J$ to have a finite minimisation. 
Afterwards, a trial minimizing sequence is chosen, and $J$ becomes a function of expansion coefficients. The more coefficients, (possibly) the lower value.
Sequences of trial mappings $\sequence {y_n^{\paren 1} }$ and values $\sequence {\lambda_n^{\paren 1} }$ of $J$ are introduced. 
Convergence of $\sequence {\lambda_n^{\paren 1} }$ is shown at once.
As for $\sequence {y_n^{\paren 1} }$, convergence to $y^{\paren 1}$ is proved for its subsequence.
The rest of arguments rest upon this weaker result.
Furthermore, $\lambda^{\paren 1}$ and $y^{\paren 1}$ are observed to satisfy Sturm-Liouville equation.
Convergence of the original sequence $\sequence {y_n^{\paren 1} }$ is secured.
Finally, construction of the rest of eigenfunctions and eigenvalues is described. 
=== Lemma 1 ===
The given Sturm-Liouville equation is an Euler equation of the following functional:
:$\ds J \sqbrk y = \int_a^b \paren {P y'^2 + Qy^2} \rd x$
constrained by a subsidiary condition:
:$\ds \int_a^b y^2 \rd x = 1$
=== Proof ===
According to Simplest Variational Problem with Subsidiary Conditions, the following equation must hold:
:$F_y - \dfrac \rd {\rd x} F_{y'} + \lambda \paren {G_y - \dfrac \rd {\rd x} G_{y'} } = 0$
where:
:$F = P y'^2 + Q y^2$
:$G = y^2$
Then the Euler equation reads:
:$2 Q y - 2 \paren {P y'}' + 2 \lambda y = 0$
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 2 ===
$J$ is bounded from below.
=== Proof ===
Since $Q$ is continuous on an interval, it is bounded.
Since $P > 0$, it holds that:
:$\ds \int_a^b \paren {P y'^2 + Q y^2} \rd x > \int_a^b Q y^2 \rd x \ge M \int_a^b y^2 \rd x = M$
where:
:$\ds M = \min_{a \mathop \le x \mathop \le b} \map Q x$
Therefore, $J$ is bounded from below.
Introduce a new variable:
:$t := \pi \dfrac {x - a} {b - a}$
Then the interval of consideration $\closedint a b$ is mapped onto $\closedint 0 \pi$.
Choose Ritz sequence $\sequence {\map {\phi_n} t} = \sequence {\sin n t}$, where $n \in \N$.
=== Lemma 3 ===
The elements of the sequence $\sequence {\sin n t}$ are orthogonal on the interval $\closedint 0 \pi$:
:$\ds \int_0^\pi \map \sin {k t} \map \sin {l t} \rd x = \frac \pi 2 \delta_{k l}$
=== Proof ===
The product involves two elements of the sequence $\sequence {\map \sin {n t} }$.
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:
:$\ds \map y x = \sum_{k \mathop = 1}^n \alpha_k \map \sin {k \map t x}$
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 $\map {J_n} {\boldsymbol \alpha}$ reads as:
:$\ds \map {J_n} {\boldsymbol \alpha} = \frac \pi {b - a} \int_0^\pi \sqbrk {\map P {\sum_{k \mathop = 1}^n \alpha_k \sin {k t} }'^2 + \map Q {\sum_{k \mathop = 1}^n \alpha_k \sin {k t} }^2} \rd t$
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, $\map {J_n} {\boldsymbol \alpha}$ is continuous on $\sigma_n$.
By Continuous Function on Compact Subspace of Euclidean Space is Bounded, $\map {J_n} {\boldsymbol \alpha}$ has a minimum on $\sigma_n$. 
Let $\map {y_n^{\paren 1} } x$ be defined as:
:$\ds \map {y_n^{\paren 1} } x = \sum_{k \mathop = 1}^n \alpha_k^{\paren 1} \sin {k \map t x}$
for which $\map {J_n} {\boldsymbol \alpha}$ achieves the minimum $\lambda_n^{\paren 1}$, unrelated to $\lambda$.
Then the $n$-th element of the sequence $\sequence {y_n^{\paren 1} }$ corresponds to the $n$-th element of the sequence of minima $\sequence {\lambda_n^{\paren 1} }$ of $\map {J_n} {\boldsymbol \alpha}$.
Since $\sigma_n \subset \sigma_{n + 1}$, where $\sigma_n$ has $\alpha_{n + 1} = 0$, it holds that:
:$\map {J_n} {\alpha_1, \ldots, \alpha_n} = \map {J_{n + 1} } {\alpha_1, \ldots, \alpha_n, 0}$
By Ritz Method implies Not Worse Approximation with Increased Number of Functions:
:$\lambda_{n + 1}^{\paren 1} \le \lambda_n^{\paren 1}$
Therefore, by increasing the domain of definition of $y_n^{\paren 1}$ through additional summands, the minima of $\map {J_n} {\boldsymbol \alpha}$ cannot increase.
From the last inequality and $J$ being bounded from below it follows, that the following limit exists:
:$\ds \lambda^{\paren 1} = \lim_{n \mathop \to \infty} \lambda_n^{\paren 1}$
=== Lemma 4 ===
The sequence $\sequence {y_n^{\paren 1} }$ contains a uniformly convergent subsequence.
=== Proof ===
The sequence 
:$\ds \lambda_n^{\paren 1} = \frac \pi {b - a} \int_0^\pi \paren {P {y_n^{\paren 1} }'^2 + Q {y_n^{\paren 1} }^2} \rd t$
is convergent with its limit being $\lambda^{\paren 1}$.
Hence, it is bounded:
:$\ds \frac \pi {b - a} \int_0^\pi \paren {P {y_n^{\paren 1} }'^2 + Q {y_n^{\paren 1} }^2} \rd t \le M $
Furthermore:
Consequently:
:$\ds \frac \pi {b - a} \min_{a \mathop \le x \mathop \le b} \map P x \int_0^\pi \map { {y_n^{\paren 1} }'^2} t \rd t \le \frac \pi {b - a} \int_0^\pi P \map { {y_n^{\paren 1} }'^2} t \rd t \le M_1$
For positive $P$, division by $P$ does not affect the direction of inequality.
It follows that:
:$\ds \int_0^\pi \map { {y_n^{\paren 1} }'^2} t \rd t \le \frac {b - a} \pi \frac {M_1} {\min_{a \mathop \le x \mathop \le b} } \map P x = M_2$
Consider squared absolute value of $y_n^{\paren 1}$.
Then, for $0 \le t \le \pi$:
In other words:
:$\forall t \in \closedint 0 \pi, n \in \N : \size {\map {y_n^{\paren 1} } t - \map {y_n^{\paren 1} } 0} \le \sqrt {M_2 \pi}$
Thus $\sequence {y_n^{\paren 1} }$ is uniformly bounded.
In addition to this, for $0 \le t_1, t_2 \le \pi$:
Let $\epsilon$ be any strictly positive real number such that $\epsilon = \sqrt {M_2 \delta}$, where $\delta$ is a strictly positive real number.
Suppose $\delta$ is such that $\size {t_2 - t_1} < \delta$.
Then:
In other words:
:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall n \in \N: \forall t_1, t_2 \in \closedint 0 \pi: \size {t_2 - t_1} < \delta \implies \size {\map {y_n^{\paren 1} } {t_2} - \map {y_n^{\paren 1} } {t_1} } < \epsilon$
where metric is induced by norm.
Thus $\sequence {y_n^{\paren 1} }$ is uniformly equicontinuous.
By Arzela's Theorem, there exists a uniformly convergent subsequence $\sequence {y_{n_m}^{\paren 1} }$ from $\sequence {y_n^{\paren 1} }$.
Denote:
:$\ds \map {y^{\paren 1} } x = \lim_{m \mathop \to \infty} \map {y_{n_m}^{\paren 1} } x$
Now a proposition is established, needed for the upcoming Lemma 6.
=== Lemma 5 ===
Let $\map y t$ be continuous in $\closedint 0 \pi$.
Suppose:
:$\ds \forall h \in C^2 \openint 0 \pi : \map h 0 = \map h \pi = \map {h'} 0 = \map {h'} \pi = 0 : \int_0^\pi \sqbrk {-\paren {P h'}' + Q_1 h} y \rd t = 0$
Then $\map y t \in C^2 \openint 0 \pi$ and:
:$-\paren {P y'}' + Q_1 y = 0$
=== Proof ===
By Integration by parts, Product Rule for Derivatives, boundary conditions for $h$, and noticing that:
 
the previous integral can be rewritten as:
From the lemma:
:$\ds -P y + \int_0^t P' y \rd \zeta + \int_0^x \paren {\int_0^t Q_1 y \rd t} \rd \zeta = c_0 + c_1 t$
The as well as the second and third terms on the are differentiable $t$.
Thus $\paren {P y}'$ exists.
Differentiation $t$ leads to:
:$\ds -\paren {P y}' + P' y + \int_0^t Q_1 y \rd \zeta = c_1$
or:
:$\ds -P y' + \int_0^t Q_1 y \rd \zeta = c_1$
The , the second term on the and $P$ are continuous and differentiable $t$, while $P$ is also positive.
Therefore, $y'$ exists and is continuous.
Hence, $\paren {P y'}'$ exists and:
:$-\paren {P y'}' + Q_1 y = 0$
Furthermore, $P$ is continuous and differentiable, while $Q_1$ is continuous.
Then $y$ exists and is continuous.
=== Lemma 6 ===
$y^{\paren 1}$ together with $\lambda^{\paren 1}$ satisfy the Sturm-Liouville equation, where $\map w x = 1$:
:$-\paren {P {y^{\paren 1} }' }' + Q y^{\paren 1} = \lambda^{\paren 1} y^{\paren 1}$
=== Proof ===
Let $J_n$, together with its subsidiary condition $\ds \int_a^b y^2 \rd x = 1$, achieve a minimum for $\boldsymbol \alpha = \boldsymbol \alpha^{\paren 1}$.
Then the necessary condition for its minimum is:
Notice, that:
This leads to a system of equations:
:$\ds \int_0^\pi \paren {\map P t \sqbrk {\sum_{k \mathop = 1}^n \alpha_k^{\paren 1} \paren {\sin {k t} }'} \paren {\sin {r x} }' + \sqbrk {Q - \lambda_n^{\paren 1} } \sqbrk {\sum_{k \mathop = 1}^n \alpha_k^{\paren 1} \sin {k t} } \sin {r t} } \rd t = 0$
Multiplying each equation by an arbitrary constant $C_r^{\paren n}$ and summing over $r$ results in:
:$\ds \int_0^\pi \sqbrk {P y_n' h_n' + \paren {Q - \lambda_n^{\paren 1} y_n h_n} } \rd t = 0$
where:
:$\ds \map {h_n} t = \sum_{r \mathop = 1}^n C_r^{\paren n} \sin {r t}$
:$\ds y_n = \sum_{k \mathop = 1}^n \alpha_k \sin {k t}$
By Integration by parts:
Consider all real mappings $h$ such that:
:$\map h x \in C^2 \openint 0 \pi$
and satisfying the boundary conditions.
Then $C_r^{\paren n}$ can be chosen, such that:
:$\ds \lim_{n \mathop \to \infty} \int_0^\pi \size {\map {h_n} x - \map h x}^2 \rd x = 0$
:$\ds \lim_{n \mathop \to \infty} \int_0^\pi \size {\map {h_n'} x - \map {h'} x}^2 \rd x = 0$
:$\ds \lim_{n \mathop \to \infty} \int_0^\pi \size {\map {h_n} x - \map {h} x}^2 \rd x = 0$
Due to the existence of uniformly convergent subsequence, $y_n^{\paren 1}$ converges to $y^{\paren 1}$ uniformly on $\closedint 0 \pi$:
:$\ds \lim_{m \mathop \to \infty} \int_0^\pi \paren {-\paren {P h_{n_m}'}' + \paren {Q - \lambda_{n_m}^{\paren 1} } h_{n_m} } y_{n_m}^{\paren 1} \rd x = \int_0^\pi \paren { -\paren {P h'}' + \paren {Q - \lambda^{\paren 1} } h} y^{\paren 1} \rd x = 0$
By Lemma 5, where $Q_1 = Q - \lambda^{\paren 1}$, $y^{\paren 1} \in C^2 \closedint 0 \pi$ and satisfies Sturm-Liouville equation with $w = 1$.
=== Lemma 7 ===
$\sequence {\map {y_n^{\paren 1} } x}$ pointwise converges to $\map {y^{\paren 1} } x$.
=== Proof ===
By Existence and Uniqueness of Solution for Linear Second Order ODE with two Initial Conditions, where $\map R x = 0$, the Sturm-Liouville equation
:$-\paren {P y'}' + Q y = \lambda y$
satisfying the boundary conditions:
:$\map y 0 = \map y \pi = 0$
and the subsidiary condition:
:$\ds \int_0^\pi \map {y^2} t = 1$
is unique up to the sign of $y$. 
Let $\map {y^{\paren 1} } t$ be a solution corresponding to $\lambda = \lambda^{\paren 1}$
Due to the subsidiary condition, the condition $\map {y^{\paren 1} } t = 0$ cannot hold in the entire closed interval $\closedint 0 \pi$.
Hence, the set of roots to this condition is countable.
Then:
:$\exists t_0 \in \closedint 0 \pi: \map {y^{\paren 1} } {t_0} \ne 0$
Choose the sign so that $\map {y^{\paren 1} } {t_0} > 0$
Similarly, let $\map {y_n^{\paren 1} } t$ be a solution corresponding to $\lambda = \lambda_n^{\paren 1}$
Choose the signs so that:
:$\forall n \in \N : \map {y_n^{\paren 1} } {t_0} \ge 0$
Suppose $\map {y_n^{\paren 1} } t$ does not pointwise converge to $\map {y^{\paren 1} } t$.
By Arzela's Theorem there exists another subsequence from $\sequence {\map {y_n^{\paren 1} } t}$, converging to another solution $\overline y^{\paren 1}$, where $\lambda = \lambda^{\paren 1}$.
Because of the uniqueness of solutions, except for the sign, both solutions may differ only in their signs:
:$\map {\overline y^{\paren 1} } x = - \map {y^{\paren 1} } t$
Therefore:
:$\map {\overline y^{\paren 1} } {t_0} < 0$
This is impossible, since:
:$\forall n \in N : \map {y_n^{\paren 1} } {t_0} \ge 0$ 
Therefore $\map {y_n^{\paren 1} } t$ pointwise converges to $\map {y^{\paren 1} } t$, provided $\map {y_n^{\paren 1} } t$ is chosen with the correct sign.
=== Lemma 8 ===
Sequences $\sequence {y^{\paren n} }$ and $\sequence {\lambda^{\paren n} }$ are infinite.
=== Proof ===
Suppose, $y^{\paren r}$ and $\lambda^{\paren r}$ are known.
The next eigenfunction $y^{\paren {r + 1} }$ and the corresponding eigenvalue $\lambda^{\paren {r + 1} }$ can be found by minimising 
:$\ds J \sqbrk y = \int_0^\pi \paren {P y'^2 + Q y^2} \rd x $
where boundary and subsidiary conditions are supplied with orthogonality conditions:
:$\forall m \in \N : {1 \le m \le r} : \ds \int_0^\pi \map {y^{\paren m} } t \map {y^{\paren {r + 1} } } t \rd t = 0$
The new solution of the form:
:$\ds \map {y_n^{\paren {r + 1} } } t = \sum_{k \mathop = 1}^n \alpha_k^{\paren {r + 1} } \sin {k t}$
is now also orthogonal to mappings:
:$\ds \map {y_n^{\paren m} } t = \sum_{k \mathop = 1}^n \alpha_k^{\paren m} \sin {k t}$
This results into:
:$\ds \sum_{k \mathop = 1}^n \alpha_k^{\paren {r + 1} } \int_0^\pi \sin {k t} \paren {\sum_{l \mathop = 1}^n \alpha_l^{\paren m} \sin {l t} } \rd t = \frac \pi 2 \sum_{k \mathop = 1}^n \alpha_k^{\paren {r + 1} } \alpha_k^{\paren m} = 0$
These equations describe $r$ distinct $\paren {n - 1}$-dimensional hyperplanes, passing through the origin of coordinates in $n$ dimensions.
These hyperplanes intersect the sphere $\sigma_n$, resulting in an $\paren {n - r}$-dimensional sphere $\hat \sigma_{n - r}$.
By definition, it is a compact set.
By Continuous Function on Compact Subspace of Euclidean Space is Bounded, $\map {J_n} {\boldsymbol \alpha}$ has a minimum on $\hat {\sigma}_{n - r}$. 
Denote it as $\lambda_n^{\paren {r + 1} }$.
By Ritz Method implies Not Worse Approximation with Increased Number of Functions:
:$\lambda_{n + 1}^{\paren {r + 1} } \le \lambda_n^{\paren {r + 1} }$
This, together with $J$ being bounded from below, implies:
:$\ds \lambda^{\paren {r + 1} } = \lim_{n \mathop \to \infty} \lambda_n^{\paren {r + 1} }$
Additional constraints may or may not affect the new minimum:
:$\lambda^{\paren r} \le \lambda^{\paren {r + 1} }$
Let:
:$\ds \map {y_n^{\paren {r + 1} } } t = \sum_{k \mathop = 1}^n \alpha_k^{\paren {r + 1} } \sin {k t}$
$y^{\paren {r + 1} }$ satisfies Sturm-Liouville equation together with boundary, subsidiary and orthogonality conditions.
By Lemma 7, which is not affected by additional constraints, $\sequence {y_n^{\paren {r + 1} } }$ uniformly converges to $y^{\paren {r + 1} }$.
Thus, $y^{\paren {r + 1} }$ is an eigenfunction of Sturm-Liouville equation with an eigenvalue $\lambda^{\paren {r + 1} }$.
Orthogonal mappings are linearly independent.
Each eigenvalue corresponds only to one eigenfunction, unique up to a constant factor.
Thus:
:$\lambda^{\paren r} < \lambda^{\paren {r + 1} }$
== Sources ==
* : $\S 8.41 $: The Sturm-Liouville Problem
Category:Calculus of Variations