Chapter 5 Series Solutions

Consider the following second order initial value problem

\{ \begin{matrix}P (x)y′′ + Q(x)y′ + R(x)y = 0 \\ y(x_{0}) = y_{0} \\ y′(x_{0}) = y′_{0} \end{matrix}

We want to find the solution of the form y(x) = ∑_{n=0}^{∞} a_{n}(x − x_{0})^{n} as a convergence power series. Since P (x_{0}) ≠ 0, x_{0} is a regular point, we write

y′′ + p(x)y′ + q(x)y = 0, where p = \frac{Q}{P}, and q = \frac{R}{P}.

If p(x) = ∑ p_{n}(x − x_{0})^{n} q(x) = ∑ q_{n}(x − x_{0})^{n}

y′_{n}(x) = ∑_{n=1}^{∞} na_{n}(x − x_{0})^{n−1}
y′′_{n}(x) = ∑_{n=2}^{∞} n(n − 1)a_{n}(x − x_{0})^{n−2}
p(x)y′(x) = p_{0}a_{1} + (p_{1}a_{1} + p_{0}2a_{2})(x − x_{0}) + ⋯
q(x)y′(x) = q_{0}a_{0} + (q_{1}a_{0} + q_{0}a_{1})(x − x_{0}) + ⋯

Comparing the Coefficients 2a_{2} + p_{0}a_{1} + q_{0}a_{0} = 0 6a_{3} + p_{1}a_{1} + 2p_{0}a_{2} + a_{0}q_{1} + q_{0}a_{1} = 0 y(x_{0}) = y_{0} ⇒ a_{0} = y_{0} y′(x_{0}) = y′_{0} ⇒ a_{1} = y′_{0} Thus, we can solve a_{n} inductively.

Theorem 1 : If p(x), q(x) are analytic on (x_{0}−ε, x_{0}+ε), then ∑ a_{n}(x−x_{0})^{n} obtained by the procedure also converges on (x_{0} − ε, x_{0} + ε).

Example : Solve \{ \begin{matrix}y′′ + y = 0 \\ y(0) = a_{0}, y′(0) = a_{1} \end{matrix} by series expansion. sol : To find y(x) = ∑_{n=0}^{∞} a_{n}(x − x_{0})^{n}, for x_{0} = 0.

Since y′′(x) = ∑_{n=2}^{∞} n(n − 1)a_{n}(x − x_{0})^{n−2},

we have ∑_{n=2}^{∞} n(n − 1)a_{n}(x − x_{0})^{n−2} + ∑_{n=0}^{∞} a_{n}(x − x_{0})^{n} = 0.

Let n − 2 = m, then

∑_{m=0}^{∞}(m + 2)(m + 1)a_{m+2}(x − x_{0})^{m} + ∑_{m=0}^{∞} a_{m}(x − x_{0})^{m} = 0.

Therefore (m + 2)(m + 1)a_{m+2} = −a_{m}, or a_{m+2} = \frac{−a_{m}}{(m + 2)(m + 1)}.

For example

a_{2} = \frac{−a_{0}}{2 !} , a_{4} = \frac{a_{0}}{4 !} , a_{6} = \frac{−a_{0}}{6 !} , a_{3} = \frac{−a_{1}}{3 !} , a_{5} = \frac{a_{1}}{5 !} , a_{7} = \frac{−a_{1}}{7 !} .

Therefore

y(x) = a_{0}(1 − \frac{x^{2}}{2 !} + \frac{x^{4}}{4 !} − \frac{x^{6}}{6 !} + ⋯) + a_{1}(1 − \frac{x^{3}}{3 !} + \frac{x^{5}}{5 !} − \frac{x^{7}}{7 !} + ⋯)

converges. Remark : We can obtain an ”formally” form recurrence relation. We need to determine where these power series converge.

\underline{Method 1} : Please check the radius of convergence directly.

cos x = 1 − \frac{x^{2}}{2!} + \frac{x^{4}}{4!} − \frac{x^{6}}{6!} + ⋯, a_{2n} = \frac{x^{2n}}{(2n)!}

\sqrt[2n]{|a_{2n}|} = \frac{|x|}{\sqrt[2n]{(2n)!}} < \frac{|x|}{R} where (2n)! >> R^{2n} ∀R > 0

For n large , lim_{n→∞} \sqrt[2n]{|a_{2n}|} = 0

∴ radius of convergence = ∞

Similarly, sin x = x − \frac{x^{3}}{3!} + \frac{x^{5}}{5!} − \frac{x^{7}}{7!} + ⋯ converges for all x ∈ ℝ. \underline{Method 2} : y′′ + py′ + qy = 0 p, q are analytic ( where p = 0, q = 1) For p, q are constant, p, q are analytic. So radius of convergence is infinite for p and q. By Theorem, y(x) = a_{0} + a_{1}x + a_{2}x^{2} + ⋯ converges for all x ∈ ℝ. Example : (Airy Function) y′′ − xy = 0 y(1) = a_{0} y′(1) = a_{1} y′′ − [(x − 1) + 1]y = 0 Since (x − 1) + 1 is analytic and radius of convergence is infinite, therefore y(x) = ∑_{n=0}^{∞} a_{n}(x − 1)^{n} converges for all x ∈ ℝ

y′′(x) = ∑_{n=2}^{∞} n(n − 1)a_{n}(x − 1)^{n−2} = ∑_{m=0}^{∞}(m + 2)(m + 1)a_{m+2}(x − 1)^{m}

[(x − 1) + 1]y = ∑_{n=0}^{∞} a_{n}(x − 1)^{n+1} + ∑_{n=0}^{∞} a_{n}(x − 1)^{n}

= ∑_{m=1}^{∞} a_{m−1}(x − 1)^{m} + ∑_{m=0}^{∞} a_{m}(x − 1)^{m}

⇔ \begin{matrix}m = 0 & , & 2a_{2} = a_{0} \\ m = 1 & , & 6a_{3} = a_{0} + a_{1} \\ m = 2 & , & 12a_{4} = a_{1} + a_{2} \end{matrix}

Two linearly independent solution given respectively by y_{3}(x) ⇔ a_{0} = 1 , a_{1} = 0 ; y_{4}(x) ⇔ a_{0} = 0 , a_{1} = 1 y_{1}(x) is a power series around x = 0 with y(0) = 1, y′(0) = 0. y_{2}(x) is a power series around x = 0 with y(0) = 0, y′(0) = 1.

Determine the radius of convergence

x − \frac{x^{3}}{3!} + \frac{x^{5}}{5!} − \frac{x^{7}}{7!} + ⋯

a_{n} = \{ \begin{matrix}(−1)^{n} \frac{1}{n!} & , & if n is odd number \\ 0 & , & if n is even number \end{matrix}

f(z) = ∑_{n=0}^{∞} a_{n}(z − z_{0})^{n}

The equality is valid when z = z_{0}. In general, there exists R > 0

s.t. ∑_{n=0}^{∞} a_{n}(z − z_{0})^{n} \{ \begin{matrix}convergent for |z − z_{0}| < R \\ divergent for |z − z_{0}| > R \\ no conclusion on |z − z_{0}| = R \end{matrix}

How to Determine R ?

R = \frac{1}{\overline{lim_{n→∞}}|a_{n}|^{\frac{1}{n}}}

Sufficient condition to determine R :

(1) If lim_{n→∞} |a_{n}|^{\frac{1}{n}} exists, then lim_{n→∞} |a_{n}|^{\frac{1}{n}} = \frac{1}{R}

(2) If lim_{n→∞} \frac{|a_{n+1}|}{|a_{n}|} exists, then lim_{n→∞} \frac{|a_{n+1}|}{|a_{n}|} = \frac{1}{R}

Remark : If f(x) = ∑ a_{n}(x − x_{0})^{n} is valid in (x_{0} − ε, x_{0} + ε), then f ∈ C^{∞} and

a_{0} = f(x_{0}), a_{1} = f′(x_{0}), a_{2} = \frac{f′′(x_{0})}{2!},

(1 + x^{2})y′′ + 2xy′ + 4x^{2}y = 0

⇒ y′′ + \frac{2x}{1 + x^{2}} y′ + \frac{4x^{2}}{1 + x^{2}} y = 0

The radius of convergence for p(x), q(x) around x = \frac{1}{2} is exactly \frac{\sqrt{5}}{2} Remark : From local existence and unique theorem Then y(x) on −∞ < x < ∞ for given y(\frac{1}{2}), y′(\frac{1}{2}) Example : What is the radius of convergence for given series solution of (1 + x^{2})y′′ + 2xy′ + 4x^{2}y = 0 ? Sol. :

(1) around x = 0

y′′ + \frac{2x}{1 + x^{2}} y′ + \frac{4x^{2}}{1 + x^{2}} y = 0

p(x) = \frac{2x}{1 + x^{2}} = 2x(1 − x^{2} + x^{4}) converges for |x^{2}| < 1

diverges for |x^{2}| ≥ 1

some for g(x) = \frac{4x^{2}}{1 + x^{2}} radius of convergence = 1 ∴ y(x) = ∑ a_{n}x^{n} converges at least on |x| < 1.

(2) around x = \frac{1}{2}

p(x) = \frac{2x}{1 + x^{2}} = p_{n}(x − \frac{1}{2})^{n}, radius of convergence = \frac{\sqrt{5}}{2} Some for g(x) ∴ y(x) = ∑ b_{n}(x − \frac{1}{2})^{n} converges at least on |x − \frac{1}{2}| < \frac{\sqrt{5}}{2} Facts : (complex analytic) \frac{2z}{1 + z^{2}} is differentiable (∴ analytic) on ℂ\{±i\}. If B_{ρ}(\frac{1}{2}) ⊆ ℂ\{±i\} then radius of convergence for ∑_{n=0}^{∞} a_{n}(x − \frac{1}{2})^{n} is at least e In above example, radius of convergence = e. Remark : The solution exists for all x ∈ ℝ. But the power series only converges

on (\frac{1}{2} − \frac{\sqrt{5}}{2}, \frac{1}{2} + \frac{\sqrt{5}}{2}).

Remark : \frac{1}{1 + x^{2}} exists for all x ∈ ℝ. But \frac{1}{1 + x^{2}} = ∑_{n=0}^{∞} a_{n}(x − \frac{1}{2})^{n} only on

(\frac{1}{2} − \frac{\sqrt{5}}{2}, \frac{1}{2} + \frac{\sqrt{5}}{2}).

P(x)y′′ + Q(x)y′ + R(x)y = 0 , P. Q. R : polynomials of x. x = x_{0} is a ordinary point if lim_{x→x_{0}} \frac{Q(x)}{P(x)} and lim_{x→x_{0}} \frac{R(x)}{P(x)} exists. Example : P(x) = x, Q(x) = sinx, R(x) = x^{2}, x_{0} = 0

⇒ \{ \begin{matrix}y′′(x) + \frac{sin x}{x}y′ + xy = 0 \\ y(0) = a_{0} y′(0) = a_{1} x ≠ 0 ℝ = ∞ \end{matrix}