Definition:Sturm-Liouville Theory

Definition
Sturm-Liouville theory is the branch of mathematical physics concerned with the eigenvalues and eigenfunctions arising from the Sturm-Liouville equation:
 * $\map {\dfrac \d {\d x} } {\map p x \dfrac {\d y} {\d x} } + \map q x y = -\lambda \map w x y$

Sturm-Liouville Theory
It is assumed that the S-L problems is regular.

That is:
 * $\map p x$, $\map q x$ and $\map w x$ are real-valued Lebesgue integrable functions over the closed real interval $\closedint a b$
 * $\map p x > 0$ and $\map w x > 0$ over $\closedint a b$
 * $\map p x$, $\map q x$ and $\map w x$ have separated boundary conditions:


 * $(2): \quad \map y a \cos \alpha - \map p a \map {y'} a \sin \alpha = 0$


 * $(3): \quad \map y b \cos \beta - \map p b \map {y'} b \sin \beta = 0$

where $\alpha, \beta \in \closedint 0 \pi$.

The main tenet of Sturm-Liouville theory states that:


 * The eigenvalues $\lambda_1, \lambda_2, \lambda_3, \ldots$ of the regular Sturm-Liouville problem $(1)$ - $(2)$ - $(3)$ are real number and can be ordered such that:


 * $\lambda_1 < \lambda_2 < \lambda_3 < \cdots < \lambda_n < \cdots \to \infty$


 * Corresponding to each eigenvalue $\lambda_n$ is a unique (up to a normalization constant) eigenfunction $\map {y_n} x$ which has exactly $n - 1$ zeros in $\openint a b$.

The eigenfunction $\map {y_n} x$ is called the $n$th fundamental solution satisfying the regular Sturm-Liouville problem $(1)$ - $(2)$ - $(3)$.


 * The normalized eigenfunctions form an orthonormal basis:


 * $\displaystyle \int_a^b \map {y_n} x \map {y_m} x \map w x \rd x = \delta_{m n}$


 * in the Hilbert Space $\map {L_2} {\closedint a b, \map w x \rd x}$.

Here $\delta_{m n}$ is a Kronecker delta.

Since by assumption the eigenfunctions are normalized, the result is established by a proof of their orthogonality.

Note that, unless $\map p x$ is continuously differentiable and $\map q x$, $\map w x$ are continuous, the equation has to be understood in a weak sense.

Sturm-Liouville Form
The differential equation $(1)$ is said to be in Sturm-Liouville form or self-adjoint form.

All second-order linear ordinary differential equations can be recast in the form on the of $(1)$ by multiplying both sides of the equation by an appropriate integrating factor.

The same is not true of second-order partial differential equations, or if $y$ is a vector.

Examples
The Bessel equation:


 * $x^2 y'' + x y' + \paren {\lambda^2 x^2 - \nu^2} y = 0$

can be written in Sturm-Liouville form as:


 * $\paren {x y'}' + \paren {\lambda^2 x - \nu^2 / x} y = 0$

The Legendre equation:


 * $\paren {1 - x^2} y'' - 2 x y' + \nu \paren {\nu + 1} y = 0$

can easily be put into Sturm-Liouville form, since:
 * $\map D {1 - x}^2 = -2 x$

so, the Legendre equation is equivalent to:


 * $\paren {\paren {1 - x^2} y'}' + \nu \paren {\nu + 1} y = 0$

It takes more work to put the following differential equation into Sturm-Liouville form:


 * $x^3 y'' - x y' + 2 y = 0$

Divide throughout by $x^3$:


 * $y'' - \dfrac x {x^3} y' + \dfrac 2 {x^3} y = 0$

Multiplying throughout by an integrating factor of:


 * $\displaystyle e^{\int -x / x^3 \rd x} = e^{\int -1 / x^2 \rd x} = e^{1 / x}$

gives:


 * $e^{1 / x} y'' - \dfrac {e^{1 / x} } {x^2} y' + \dfrac {2 e^ {1 / x} } {x^3} y = 0$

which can be easily put into Sturm-Liouville form since:


 * $D e^{1 / x} = - \dfrac {e^{1 / x} } {x^2}$

so the differential equation is equivalent to:


 * $\paren {e^{1 / x} y'}' + \dfrac {2 e^{1 / x} } {x^3} y = 0$

In general, given a differential equation:


 * $\map P x y'' + \map Q x y' + \map R x y = 0$

dividing by $\map P x$, multiplying through by the integrating factor:


 * $\displaystyle e^{\int {\map Q x / \map P x} \rd x}$

and then rearranging gives the Sturm-Liouville form.

Sturm-Liouville Equations as Hermitian Differential Operators
Let us rewrite equation $(1)$ as:
 * $(1d): \quad \displaystyle \Lambda \map y x = \lambda \map w x \map y x$

with:
 * $\displaystyle \Lambda \equiv \paren {-\map {\dfrac \d {\d x} } {\map p x \dfrac \d {\d x} } + \map q x}$

The function $\map w x$ is positive-definite and hence equation $(1d)$ has the form of a generalized operator eigenvalue equation.

It can be transformed to a regular eigenvalue equation by substitution of:
 * $\map u x = \map w x^{1/2} \map y x$
 * $L = \map w x^{-1/2} \Lambda \map w x^{-1/2}$

Equation (1d) becomes:
 * $\paren {\map w x^{-1/2} \Lambda \map w x^{-1/2} }$
 * $\map w x^{1/2} \map y x = \lambda \map w x^{1/2} \map y x$

or:
 * $(1e): \quad L u = \lambda u$

The map $L$ can be viewed as a linear operator mapping a function $u$ to another function $L u$.

We may study this linear operator in the context of functional analysis.

Equation $(1e)$ is precisely the eigenvalue problem of $L$.

That is, we are trying to find the eigenvalues:
 * $\lambda_1, \lambda_2, \lambda_3, \ldots$

and the corresponding eigenvectors $u_1, u_2, u_3, \ldots$ of the $L$ operator.

The proper setting for this problem is the Hilbert space $\map {L_2} {\closedint a b, \map w x \rd x}$ with scalar product:


 * $\displaystyle \innerprod {u_i} {u_j} = \int_a^b \overline {\map {u_i} x} \map {u_j} x \rd x = \int_a^b \overline {\map {y_i} x} \map {y_j} x \map w x \rd x$

where:
 * $\map {u_k} x \equiv \map w x^{1/2} \map {y_k} x$

The functions $y$ solve the generalized eigenvalue problem $(1d)$ and the functions $u$ the ordinary eigenvalue problem $(1e)$.

In this space $L$ is defined on sufficiently smooth functions which satisfy the above boundary conditions.

Moreover, $L$ is a Hermitian operator.

This can be seen formally by using Integration by Parts twice, where the boundary terms vanish by virtue of the boundary conditions.

The functions $\map w x$, $\map p x$ and $\map q x$ are real.

From the vanishing of the boundary terms follows:
 * $\paren {\dfrac \d {\d x} }^* = -\dfrac \d {\d x}$

hence:
 * $L^* = \paren {w^{-1/2} \Lambda w^{-1/2} }^* = w^{-1/2} \Lambda^* w^{-1/2}$

and:
 * $\Lambda^* = \paren {-\map {\dfrac \d {\d x} } {\map p x \dfrac \d {\d x} } + \map q x}^* = \paren {-\map {\dfrac \d {\d x} } {\map p x \dfrac \d {\d x} } + \map q x}$

Both $L$ and $\Lambda$ are Hermitian.

It then follows that the eigenvalues $\lambda$ shared by $L$ and $\Lambda$ are real and that eigenfunctions of $L$ corresponding to different eigenvalues are orthogonal.

If:
 * $L u_k = \lambda_k u_k, \quad L u_\ell = \lambda_\ell u_\ell$ such that $\lambda_k \ne \lambda_\ell$

then:
 * $\displaystyle \int_a^b \overline {\map {u_k} x} \map {u_\ell} x \rd x = \int_a^b \overline {\map {y_k} x} \map {y_\ell} x \map w x \rd x = 0$

However, the operator $L$ is unbounded and hence existence of an orthonormal basis of eigenfunctions is not evident.

To overcome this problem one looks at the resolvent:


 * $\paren {L - z}^{-1}, \qquad z \in \C$

where $z$ is chosen to be some complex number which is not an eigenvalue.

Then, computing the resolvent amounts to solving the inhomogeneous equation, which can be done using the variation of parameters formula.

This shows that the resolvent is an integral operator with a continuous symmetric kernel (the Green's function of the problem).

As a consequence of the Arzelà-Ascoli Theorem, this integral operator is compact.

Existence of:
 * a sequence of eigenvalues $\alpha_n$ which converge to $0$

and:
 * eigenfunctions which form an orthonormal basis

follows from the spectral theorem for compact operators.

Finally, note that:
 * $\paren {L - z}^{-1} u = \alpha u$

is equivalent to:
 * $L u = \paren {z + \alpha^{-1} } u$

If the interval is unbounded, or if the coefficients have singularities at the boundary points, one calls $L$ singular.

In this case the spectrum no longer consists of eigenvalues alone and can contain a continuous component.

There is still an associated eigenfunction expansion (similar to Fourier series versus Fourier transform).

This is important in quantum mechanics, since the one-dimensional Schrödinger equation is a special case of a S-L equation.

Example
We wish to find a function $\map u x$ which solves the following Sturm-Liouville problem:


 * $L u = \dfrac {\d^2 u} {\d x^2} = \lambda u$

where the unknowns are $\lambda$ and $\map u x$.

As above, we must add boundary conditions.

We take for example:


 * $\map u 0 = \map u \pi = 0$

Observe that if $k$ is any integer, then the function:


 * $\map u x = \sin k x$

is a solution with eigenvalue $\lambda = -k^2$.

We know that:
 * the solutions of a S-L problem form an orthogonal basis
 * from the theory of Fourier series that this set of sinusoidal functions is an orthogonal basis.

Since orthogonal bases are always maximal (by definition) we conclude that the S-L problem in this case has no other eigenvectors.

Given the preceding, let us now solve the inhomogeneous problem:


 * $L u = x: x \in \openint 0 \pi$

with the same boundary conditions.

In this case, we must write $\map f x = x$ in a Fourier series.

By integrating $\displaystyle \int \map \exp {i k x} x \rd x$:


 * $(4): \quad \displaystyle L u = \sum_{k \mathop = 1}^\infty - 2 \frac {\paren {-1}^k} k \sin k x$

This particular Fourier series is troublesome because of its poor convergence properties.

It is not clear a priori whether the series converges pointwise.

Because of Fourier analysis, since the Fourier coefficients are "square-summable", the Fourier series converges in $L^2$.

This is all we need for this particular theory to function.

We mention for the interested reader that in this case we may rely on a result which says that Fourier's series converges at every point of differentiability, and at jump points (the function $x$, considered as a periodic function, has a jump at $\pi$) converges to the average of the left and right limits (see Convergence of Fourier Series).

Therefore, by using formula $(4)$, we obtain the solution:


 * $\displaystyle u = \sum_{k \mathop = 1}^\infty 2 \frac {\paren {-1}^k} {k^3} \sin k x$

In this case, we could have found the answer using antidifferentiation.

This technique yields $u = \dfrac {\paren {x^3 - \pi^2 x} } 6$, whose Fourier series agrees with the solution we found.

The antidifferentiation technique is not generally useful when the differential equation has many variables.

Application to Normal Modes
Suppose we are interested in the modes of vibration of a thin membrane, held in a rectangular frame, $0 < x < L_1, 0 < y < L_2$.

We know the equation of motion for the vertical membrane's displacement:
 * $\map W {x, y, t}$

is given by the wave equation:


 * $\dfrac {\partial^2 W} {\partial x^2} + \dfrac {\partial^2 W} {\partial y^2} = \dfrac 1 {c^2} \dfrac {\partial^2 W} {\partial t^2}$

The equation is separable, by substituting $W = \map X x \times \map Y y \times \map T t$.

The normal mode solutions that have harmonic time dependence and satisfy the boundary conditions $W = 0$ at $x = 0, L_1$ and $y = 0, L_2$ are given by:


 * $\map {W_{m n} } {x, y, t} = A_{m n} \map \sin {\dfrac {m \pi x} {L_1} } \map \sin {\dfrac {n \pi y} {L_2} } \map \cos {\omega_{m n} t}$

where:
 * $m$ and $n$ are non-zero integers
 * $A_{m n}$ is an arbitrary constant

and:
 * $\omega^2_{m n} = c^2 \paren {\dfrac {m^2 \pi^2} {L_1^2} + \dfrac {n^2 \pi^2} {L_2^2} }$

Since the eigenfunctions $W_{m n}$ form a basis, an arbitrary initial displacement can be decomposed into a sum of these modes, which each vibrate at their individual frequencies $\omega_{mn}$.

Infinite sums are also valid, as long as they converge.