Review / Part III — Series & Numerical Methods / Ch 15

Power Series Solutions

Ch 15 — Part III — Series & Numerical Methods

When coefficients vary with \(t\) or equations lack elementary solutions, we can often represent \(y(t) = \sum a_n (t - t_0)^n\) as a power series and solve for the coefficients.

Recipe

  1. Substitute \(y = \sum a_n (t-t_0)^n\) and its derivatives into the ODE.
  2. Shift indices so all sums line up as \(\sum (\text{stuff}) (t-t_0)^n = 0\).
  3. Set coefficients of each power of \((t-t_0)^n\) to zero.
  4. Solve the resulting recurrence to get \(a_n\) in terms of \(a_0, a_1, \ldots\).
  5. Determine radius of convergence (usually dictated by nearest singular point).

Airy equation

\(y'' - t y = 0\) has series solution \(y = a_0 + a_1 t + \ldots\) with recurrence \((n+2)(n+1) a_{n+2} = a_{n-1}\) — two linearly independent series solutions Ai(\(t\)) and Bi(\(t\)).

Caveats

At regular singular points, use Frobenius (shifted exponent). At irregular singular points, power series may diverge.

Practice Problems

Problem 1easy
Find the first four nonzero terms of the power series about 0 solving \(y' = y\) with \(y(0) = 1\).
Hint
Recurrence \(a_{n+1} = a_n/(n+1)\).
Solution

\(a_0 = 1, a_1 = 1, a_2 = 1/2, a_3 = 1/6, a_4 = 1/24\). So \(y = 1 + t + t^2/2 + t^3/6 + t^4/24 + \ldots = e^t\).

Answer: \(y = 1 + t + t^2/2 + t^3/6 + \ldots\) (i.e. \(e^t\)).

Problem 2medium
Solve \(y'' + y = 0\) by power series with \(y(0)=1,\; y'(0)=0\).
Hint
\((n+2)(n+1) a_{n+2} = -a_n\).
Solution

\(a_0 = 1,\; a_1 = 0\). Then \(a_2 = -1/2,\; a_4 = 1/24,\; a_6 = -1/720,\ldots\); odd coefficients zero. This is \(\cos t\).

Answer: \(y = \cos t.\)

Problem 3hard
Derive the recurrence for \(y'' - t y = 0\).
Hint
Expand and shift.
Solution

\(y = \sum a_n t^n\), \(y'' = \sum n(n-1) a_n t^{n-2}\), \(ty = \sum a_n t^{n+1}\). Re-indexing yields \(\sum [(n+2)(n+1) a_{n+2} - a_{n-1}] t^n = 0\) for \(n\ge 1\), with \(2 a_2 = 0\). Recurrence: \(a_{n+2} = a_{n-1}/((n+2)(n+1))\) for \(n\ge 1\), \(a_2 = 0\).

Answer: See solution.