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.

Outline
Firstly, an equivalence between the Sturm-Liouville equation and minimisation of functional $\displaystyle J \sqbrk y = \int_a^b \paren {P y'^2 + Q y^2} \rd x$ 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:


 * $\displaystyle J \sqbrk y = \int_a^b \paren {P y'^2 + Qy^2} \rd x$

constrained by a subsidiary condition:


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

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:


 * $\displaystyle \int_a^b \paren {P y'^2 + Q y^2} \rd x > \int_a^b Q y^2 \rd x \ge M \int_a^b y^2 \rd x = M$

where:


 * $\displaystyle M = \min_{a \mathop \le x \mathop \le b} \map Q x$

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 $\closedint 0 \pi$:


 * $\displaystyle \int_0^\pi \map \sin {k t} \map \sin {l t} \rd x = \frac \pi 2 \delta_{k l}$

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:


 * $\displaystyle \map y x = \sum_{k \mathop = 1}^n \alpha_k \map \sin {k \map t x}$

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:


 * $\displaystyle \map {J_n} {\boldsymbol \alpha} = \frac \pi {b - a} \int_0^\pi \sqbrk { P \paren {\sum_{k = 1}^n \alpha_k \sin {k t} }'^2 + Q \paren {\sum_{k = 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:


 * $\displaystyle \map {y_n^{\paren 1} } x = \sum_{k = 1}^n \alpha_k^{\paren 1} \sin {k \map t x}$

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:


 * $\displaystyle \map {J_n} {\alpha_1, \ldots, \alpha_n} = \map {J_{n + 1} } {\alpha_1, \ldots, \alpha_n, 0}$

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


 * $\displaystyle \lambda_{n + 1}^{\paren 1} \le \lambda_n^{\paren 1}$

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:


 * $\displaystyle \lambda^{\paren 1} = \lim_{n \mathop \to \infty} \lambda_n^{\paren 1}$

Lemma 4
The sequence $\sequence {y_n^{\paren 1} }$ contains a uniformly convergent subsequence.

Proof
The sequence


 * $\displaystyle \lambda_n^{\paren 1} = \frac \pi {b - a} \int_0^\pi \paren {P { y_n^{\paren 1} }'^2 + Q {y_n^{\paren 1} }^2} \rd t$

is convergent with its limit being $\lambda^{\paren 1}$.

Hence, it is bounded:


 * $\displaystyle \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:


 * $\displaystyle \frac \pi {b - a} \min_{a \mathop \le x \mathop \le b} \map P x \int_0^\pi \map { {y_n^{\paren 1} }'^2} t \rd t \le \frac \pi {b - a} \int_0^\pi P \map { {y_n^{\paren 1} }'^2} t \rd t \le M_1$

For positive $P$, division by $P$ does not affect the direction of inequality.

It follows that:


 * $\displaystyle \int_0^\pi \map { {y_n^{\paren 1} }'^2} t \rd t \le \frac {b - a} \pi \frac {M_1} {\min_{a \mathop \le x \mathop \le b} } \map P x = M_2$

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:


 * $\displaystyle \forall \epsilon \in \R_{>0} : \exists \delta \in \R_{>0} : \forall n \in \N : \forall t_1, t_2 \in \closedint 0 \pi : \size {t_2 - t_1} < \delta \implies \size {\map {y_n^{\paren 1} } {t_2} - \map {y_n^{\paren 1} } {t_1} } < \epsilon$

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:


 * $\displaystyle \map {y^{\paren 1} } x = \lim_{m \mathop \to \infty} \map {y_{n_m}^{\paren 1} } x$

Now a proposition is established, needed for the upcoming Lemma 6.

Lemma 5
Let $\map y t$ be continuous in $\closedint 0 \pi$.

Suppose:


 * $ \displaystyle \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 lemma:


 * $\displaystyle - P y + \int_0^t P' y \rd \zeta + \int_0^x \paren {\int_0^t Q_1 y \rd t} \rd \zeta = c_0 + c_1 t$

The as well as the second and third terms on the  are differentiable  $t$.

Thus $\paren {P y}'$ exists.

Differentiation $t$ leads to:


 * $\displaystyle -\paren {P y}' + P' y + \int_0^t Q_1 y \rd \zeta = c_1$

or:


 * $\displaystyle - P y' + \int_0^t Q_1 y \rd \zeta = c_1$

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:


 * $\displaystyle - \paren {P y'}' + Q_1 y = 0$

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 $\map w x = 1$:


 * $\displaystyle - \paren {P {y^{\paren 1} }' }' + Q y^{\paren 1} = \lambda^{\paren 1} y^{\paren 1}$

Proof
Let $J_n$, together with its subsidiary condition $\displaystyle \int_a^b y^2 \rd x = 1$, 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:


 * $\displaystyle \int_0^\pi \paren {\map P t \sqbrk {\sum_{k = 1}^n \alpha_k^{\paren 1} \paren {\sin {k t} }'} \paren {\sin {r x} }' + \sqbrk {Q - \lambda_n^{\paren 1} } \sqbrk {\sum_{k = 1}^n \alpha_k^{\paren 1} \sin {k t} } \sin {r t} } \rd t = 0$

Multiplying each equation by an arbitrary constant $C_r^{\paren n}$ and summing over $r$ results in:


 * $\displaystyle \int_0^\pi \sqbrk {P y_n' h_n' + \paren {Q - \lambda_n^{\paren 1} y_n h_n} } \rd t = 0$

where:


 * $\displaystyle \map {h_n} t = \sum_{r = 1}^n C_r^{\paren n} \sin {r t}$


 * $\displaystyle y_n = \sum_{k = 1}^n \alpha_k \sin {k t}$

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:


 * $\displaystyle \lim_{n \mathop \to \infty} \int_0^\pi \size {\map {h_n} x - \map h x}^2 \rd x = 0$


 * $\displaystyle \lim_{n \mathop \to \infty} \int_0^\pi \size {\map {h_n'} x - \map {h'} x}^2 \rd x = 0$


 * $\displaystyle \lim_{n \mathop \to \infty} \int_0^\pi \size {\map {h_n} x - \map {h} x}^2 \rd x = 0$

Due to the existence of uniformly convergent subsequence, $y_n^{\paren 1}$ converges to $y^{\paren 1}$ uniformly on $\closedint 0 \pi$:


 * $\displaystyle \lim_{m \mathop \to \infty} \int_0^\pi \paren {- \paren {P h_{n_m}'}' + \paren {Q - \lambda_{n_m}^{\paren 1} } h_{n_m} } y_{n_m}^{\paren 1} \rd x = \int_0^\pi \paren { -\paren {P h'}' + \paren {Q - \lambda^{\paren 1} } h} y^{\paren 1} \rd x = 0$

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:


 * $\displaystyle \int_0^\pi \map {y^2} t = 1$

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 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


 * $\displaystyle 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:


 * $\forall m \in \N : {1 \le m \le r} : \displaystyle \int_0^\pi \map {y^{\paren m} } t \map {y^{\paren {r + 1} } } t \rd t = 0$

The new solution of the form:


 * $\displaystyle \map {y_n^{\paren {r + 1} } } t = \sum_{k = 1}^n \alpha_k^{\paren {r + 1} } \sin {k t}$

is now also orthogonal to mappings:


 * $\displaystyle \map {y_n^{\paren m} } t = \sum_{k = 1}^n \alpha_k^{\paren m} \sin {k t}$

This results into:


 * $\displaystyle \sum_{k = 1}^n \alpha_k^{\paren {r + 1} } \int_0^\pi \sin {k t} \paren {\sum_{l = 1}^n \alpha_l^{\paren m} \sin {l t} } \rd t = \frac \pi 2 \sum_{k = 1}^n \alpha_k^{\paren {r + 1} } \alpha_k^{\paren m} = 0$

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:


 * $\displaystyle \lambda_{n + 1}^{\paren {r + 1} } \le \lambda_n^{\paren {r + 1} }$

This, together with $J$ being bounded from below, implies:


 * $\displaystyle \lambda^{\paren {r + 1} } = \lim_{n \mathop \to \infty} \lambda_n^{\paren {r + 1} }$

Additional constraints may or may not affect the new minimum:


 * $\lambda^{\paren r} \le \lambda^{\paren {r + 1} }$

Let:


 * $\displaystyle \map {y_n^{\paren {r + 1} } } t = \sum_{k \mathop = 1}^n \alpha_k^{\paren {r + 1} } \sin {k t}$

$y^{\paren {r + 1} }$ 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:


 * $\lambda^{\paren r} < \lambda^{\paren {r + 1} }$