In this case the equation above is reduced to: f''(x) + p_1(x) f'(x) + p_0(x) f(x) = 0. One distinguishes the following cases: • Point is an
ordinary point when functions and are analytic at . • Point is a
regular singular point if has a pole up to order 1 at and has a pole of order up to 2 at . • Otherwise point is an
irregular singular point. We can check whether there is an irregular singular point at infinity by using the substitution w = 1/x and the relations: \frac{df}{dx}=-w^2\frac{df}{dw} \frac{d^2f}{dx^2}=w^4\frac{d^2f}{dw^2}+2w^3\frac{df}{dw} We can thus transform the equation to an equation in , and check what happens at . If p_1(x) and p_2(x) are quotients of polynomials, then there will be an irregular singular point at infinite
x unless the polynomial in the denominator of p_1(x) is of
degree at least one more than the degree of its numerator and the denominator of p_2(x) is of degree at least two more than the degree of its numerator. Listed below are several examples from ordinary differential equations from
mathematical physics that have singular points and known solutions.
Bessel differential equation This is an ordinary differential equation of second order. It is found in the solution to
Laplace's equation in
cylindrical coordinates: x^2 \frac{d^2 f}{dx^2} + x \frac{df}{dx} + (x^2 - \alpha^2)f = 0 for an arbitrary real or
complex number (the
order of the
Bessel function). The most common and important special case is where is an
integer . Dividing this equation by
x2 gives: \frac{d^2 f}{dx^2} + \frac{1} {x} \frac{df}{dx} + \left (1 - \frac {\alpha^2} {x^2} \right )f = 0. In this case has a pole of first order at . When , has a pole of second order at . Thus this equation has a regular singularity at 0. To see what happens when one has to use a
Möbius transformation, for example x = 1 / w. After performing the algebra: \frac{d^2 f}{d w^2} + \frac{1}{w} \frac{df}{dw} + \left[ \frac{1}{w^4} - \frac{\alpha ^2}{w^2} \right ] f= 0 Now at p_1(w) = \frac{1}{w} has a pole of first order, but p_0(w) = \frac {1} {w^4} - \frac {\alpha ^2} {w^2} has a pole of fourth order. Thus, this equation has an irregular singularity at w = 0 corresponding to
x at ∞.
Legendre differential equation This is an ordinary differential equation of second order. It is found in the solution of
Laplace's equation in
spherical coordinates: \frac{d}{dx} \left[ (1-x^2) \frac{d}{dx} f \right] + \ell(\ell+1)f = 0. Opening the square bracket gives: \left(1-x^2\right){d^2 f \over dx^2} -2x {df \over dx } + \ell(\ell+1)f = 0. And dividing by : \frac{d^2 f}{dx^2} - \frac{2x}{1-x^2} \frac{df}{dx} + \frac{\ell(\ell+1)}{1-x^2} f = 0. This differential equation has regular singular points at ±1 and ∞.
Hermite differential equation One encounters this ordinary second order differential equation in solving the one-dimensional time independent
Schrödinger equation E\psi = -\frac{\hbar^2}{2m} \frac {d^2 \psi} {d x^2} + V(x)\psi for a
harmonic oscillator. In this case the potential energy
V(
x) is: V(x) = \frac{1}{2} m \omega^2 x^2. This leads to the following ordinary second order differential equation: \frac{d^2 f}{dx^2} - 2 x \frac{df}{dx} + \lambda f = 0. This differential equation has an irregular singularity at ∞. Its solutions are
Hermite polynomials.
Hypergeometric equation The equation may be defined as z(1-z)\frac {d^2f}{dz^2} + \left[c-(a+b+1)z \right] \frac {df}{dz} - abf = 0. Dividing both sides by gives: \frac {d^2f}{dz^2} + \frac{c-(a+b+1)z } {z(1-z)} \frac {df}{dz} - \frac {ab} {z(1-z)} f = 0. This differential equation has regular singular points at 0, 1 and ∞. A solution is the
hypergeometric function. ==References==