5.5 Series Solutions Near a Regular Singular Point, Part I

Theorem 5.3.1: If p(x) andq(x) are analytic at x_{0} (i.e., x_{0} is an ordinary point of the ODE y′′+p(x)y′+q(x)y = 0), then the general solution to this ODE is

y = Σ_{n=0}^{∞}a_{n}(x − x_{0})^{n} = a_{0}ϕ_{0}(x) + a_{1}ϕ_{1}(x)

where ϕ_{i} are power series solutions that are analytic at x_{0}. The solutions ϕ_{0}, ϕ_{1} form a fundamental set of solutions. The radius of convergence for each of these series solutions is at least as large as the minimum radii of convergence of the series for \frac{Q}{P} and \frac{R}{P}.

If you prefer a power series expansion about 0, use u-substitution: let u = x − x_{0}. Then p(u + x_{0}) and q(u + x_{0}) are analytic at 0

(Semi-failed) attempt to transform 5.5 problem into 5.4 problem:

5.5: y′′ + p(x)y′ + q(x)y = 0

x^{2}y′′ + x^{2}p(x)y′ + x^{2}q(x)y = 0

x^{2}y′′ + x[xp(x)]y′ + [x^{2}q(x)]y = 0 where xp(x) and x^{2}q(x) are functions of x.

5.4: x^{2}y′′ + αxy′ + βy = 0 where α, β are constants.

Combine 5.3/5.4 methods.

Defn: x_{0} is a regular singular value if x_{0} is a singular value and xp(x) and x^{2}q(x) are analytic at x_{0}. A singular value which is not regular is called irregular.

Examples:

y′′ + \frac{y′}{x} + \frac{y}{x^{2}} = 0, regular singular value: x = 0.

y′′ + \frac{y′}{x^{2}} + \frac{y}{x} = 0, irregular singular value: x = 0.

y′′ + y′ + \frac{y}{x^{3}} = 0, irregular singular value: x = 0.

If p(x) and q(x) are rational functions, then xp(x) and x^{2}q(x) are analytic iff lim_{x→0}xp(x) and iff lim_{x→0}x^{2}q(x) are finite. (i.e., after reducing fractions, x is not in the denominator.

Ex: p(x) = \frac{1}{x} implies xp(x) = \frac{x}{x} = 1

Ex: p(x) = \frac{1}{x^{2}} implies xp(x) = \frac{x}{x^{2}} = \frac{1}{x}

If x_{0} = 0 is a regular singular value of the linear homogeneous DE, x^{2}y′′ + x[xp(x)]y′ + x^{2}q(x)y = 0 (*), then xp(x) = Σ_{n=0}^{∞}p_{n}x^{n} and x^{2}q(x) = Σ_{n=0}^{∞}q_{n}x^{n} for constants p_{n}, q_{n}.

If y = x^{r}Σ_{n=0}^{∞}a_{n}x^{n} = Σ_{n=0}^{∞}a_{n}x^{n+r} is a solution to (*) where r ≠ 0.

y′ = Σ_{n=0}^{∞}(n+r)a_{n}x^{n+r−1} and y′′ = Σ_{n=0}^{∞}(n+r)(n+r−1)a_{n}x^{n+r−2}

x^{2}Σ_{n=0}^{∞}(n + r)(n + r − 1)a_{n}x^{n+r−2} + x[xp(x)]Σ_{n=0}^{∞}(n + r)a_{n}x^{n+r−1} +[x^{2}q(x)]Σ_{n=0}^{∞}a_{n}x^{n+r}

Σ_{n=0}^{∞}(n + r)(n + r − 1)a_{n}x^{n+r} + [xp(x)]Σ_{n=0}^{∞}(n + r)a_{n}x^{n+r} +[x^{2}q(x)]Σ_{n=0}^{∞}a_{n}x^{n+r}

Σ_{n=0}^{∞}(n + r)(n + r − 1)a_{n}x^{n+r} + (Σ_{n=0}^{∞}p_{n}x^{n}) (Σ_{n=0}^{∞}(n + r)a_{n}x^{n+r}) + (Σ_{n=0}^{∞}q_{n}x^{n}) (Σ_{n=0}^{∞}a_{n}x^{n+r})

Thus the coefficient of x^{r} is r(r − 1)a_{0} + p_{0}ra_{0} + q_{0}a_{0} = 0

We can take a_{0} ≠ 0. Thus r(r − 1) + p_{0}r + q_{0} = 0

Thus we can solve for r using the quadratic formula.

Case 1: r_{1} > r_{2} both real and r_{1} − r_{2} is not an integer. Case 2: r_{1} > r_{2} both real and r_{1} − r_{2} = p, p an integer. Case 3: one repeated root. Case 4: two complex roots.