View source for Titchmarsh-Weyl m-function

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A function arising in an attempt to properly determine which singular boundary-value problems are self-adjoint (cf. also [[Self-adjoint differential equation|Self-adjoint differential equation]]). Begin with a formally symmetric differential expression

$$
L y = \frac{-(p y')' + q y}{w} ,
$$

where $p\ne 0$, $q,w>0$ are measurable coefficients over $[a,b)$, and which is defined on a domain within $L^2(a,b;w)$. The Titchmarsh–Weyl $m$-function is defined as follows: For $\lambda = \mu + i \nu$, $\nu\ne 0$, let $\phi$ and $\psi$ be solutions of $L y = \lambda y$ satisfying

$$
\begin{aligned}
  \phi(a,\lambda) &= \sin\alpha, & \psi(a,\lambda) &= \cos\alpha, \\
  p\phi'(a,\lambda) &= -\cos\alpha, & p\psi'(a,\lambda) &= \sin\alpha .
\end{aligned}
$$

Now consider a real boundary condition at $b'$, $a<b'<b$, of the form

$$\cos\beta\, x(b')+\sin\beta\, px'(b')=0,$$

and let $\chi(x,\lambda)=\phi(x,\lambda)+\ell(\lambda)\psi(x,\lambda)$ satisfy it. Then

$$\ell(\lambda)=-\frac{\cot\beta\,\phi(b',\lambda)+p\phi'(b',\lambda)}{\cot\beta\,\psi(b'(\lambda)+p\psi'(b',\lambda)}.$$

If $z=\cot\beta$, $\ell$ is a [[Meromorphic function|meromorphic function]] in the complex $z$-plane; indeed, it is a [[Fractional-linear mapping|fractional-linear transformation]] of the $z$-plane into itself. From the well-known properties of fractional-linear transformations, as $\beta$ varies over real values $0\leq\beta\leq\pi$, $z$ varies over the real $z$-axis, and $\ell$ describes a circle in the $z$-plane.

It can be shown that if $b'$ increases, the circles become nested. Hence there is at least one point inside all. For such a point $\ell=m(\lambda)$,

$$\int\limits_a^b|\chi(x,\lambda)|^2w(x)dx<\infty.$$

There exists at least one solution of $Ly=\lambda y$, which is square-integrable.

If the limit of the circles is a point, then $m(\lambda)$ is unique and only $\chi(x,\lambda)$ is square-integrable. This is the limit-point case. If the limit of the circles is itself a circle, then $m(\lambda)$ is not unique and all solutions of $Ly=\lambda y$ are square-integrable. This is the limit-circle case.

Nonetheless, the differential operator

$$Ly=\frac{-(py')'+qy}{w}$$

whose domain satisfies

$$\sin\alpha\, y(a)-\cos\alpha\, py'(a)=0,$$

$$\lim_{x\to b}[p(x)(y(x)\chi'(\lambda,x)-y'(x)\chi(x,\lambda)]=0,$$

where $\ell=m$ on the limit circle or limit point, is a self-adjoint differential operator (cf. also [[Self-adjoint operator|Self-adjoint operator]]; [[Self-adjoint differential equation|Self-adjoint differential equation]]) on $L^2(a,b;w)$.

If the circle limit is a point, the second boundary condition (at $b$) is automatic.

The [[Spectral measure|spectral measure]] of $L$ is given by

$$\rho(\lambda)-\rho(\mu)=\frac1\pi\lim_{\epsilon\to0}\int\limits_\mu^\lambda\operatorname{Im}(m(\nu+i\epsilon))d\nu.$$

The [[Spectral resolution|spectral resolution]] of arbitrary functions in $L^2(a,b;w)$ is

$$f(x) = \lim_{(\mu,\nu) \to (-\infty,\infty)} \int_\mu^\nu g(\lambda) \psi(x,\lambda) d\rho(\lambda) , $$

where the limit is in the mean-square sense, and

$$g(\lambda)=\lim_{b'\to b}\int\limits_a^{b'}f(x)\psi(x,\lambda)dx.$$

====References====
<table><TR><TD valign="top">[a1]</TD> <TD valign="top">  E.A. Coddington,   N. Levinson,   "Theory of ordinary differential equations" , McGraw-Hill  (1955)</TD></TR><TR><TD valign="top">[a2]</TD> <TD valign="top">  A.M. Krall,   "$M(\lambda)$ theory for singular Hamiltonian systems with one singular point"  ''SIAM J. Math. Anal.'' , '''20'''  (1989)  pp. 644–700</TD></TR></table>