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Chapter 9
Series solutions of O.D.E.
(Frobenius’ method)

Let us look at the a very simple (ordinary) differential equation,

y''(t) = t\kern 1.66702pt y(t),
(9.1)

with initial conditions y(0) = a, y'(0) = b. Let us assume that there is a solution that is analytical near t = 0. This means that near t = 0 the function has a Taylor’s series

y(t) = {c}_{0} + {c}_{1}t + \mathop{\mathop{…}} ={ \mathop{∑ }}_{k=0}^{∞}{c}_{ k}{t}^{k}.
(9.2)

(We shall ignore questions of convergence.) Let us proceed

\eqalignno{ y'(t) & = &{c}_{1} + 2{c}_{2}t + \mathop{\mathop{…}} & = &{\mathop{∑ }}_{k=1}^{∞}k{c}_{ k}{t}^{k−1}, & & & & & & & & \cr y''(t) & = &2{c}_{2} + 3 ⋅ 2t + \mathop{\mathop{…}} & = &{\mathop{∑ }}_{k=2}^{∞}k(k − 1){c}_{ k}{t}^{k−2}, & & & & & & & & \cr ty(t) & = &{c}_{0}t + {c}_{1}{t}^{2} + \mathop{\mathop{…}} & = &{\mathop{∑ }}_{k=0}^{∞}{c}_{ k}{t}^{k+1}. & & & & \text{(9.3)} }

Combining this together we have

\begin{eqnarray} y'' − ty& =& [2{c}_{2} + 3 ⋅ 2t + \mathop{\mathop{…}}] − [{c}_{0}t + {c}_{1}{t}^{2} + \mathop{\mathop{…}}] %& \\ & =& 2{c}_{2} + (3 ⋅ 2{c}_{3} − {c}_{0})t + \mathop{\mathop{…}} %& \\ & =& 2{c}_{2} +{ \mathop{∑ }}_{k=3}^{∞}\left \{k(k − 1){c}_{ k} − {c}_{k−3}\right \}{t}^{k−2}.%&(9.4) \\ \end{eqnarray}

Here we have collected terms of equal power of t. The reason is simple. We are requiring a power series to equal 0. The only way that can work is if each power of x in the power series has zero coefficient. (Compare a finite polynomial....) We thus find

{c}_{2} = 0,\kern 2.77695pt \kern 2.77695pt k(k − 1){c}_{k} = {c}_{k−3}.
(9.5)

The last relation is called a recurrence of recursion relation, which we can use to bootstrap from a given value, in this case {c}_{0} and {c}_{1}. Once we know these two numbers, we can determine {c}_{3},{c}_{4} and {c}_{5}:

{c}_{3} = {1\over 6}{c}_{0},\kern 2.77695pt \kern 2.77695pt \kern 2.77695pt {c}_{4} = {1\over 12}{c}_{1},\kern 2.77695pt \kern 2.77695pt \kern 2.77695pt {c}_{5} = {1\over 20}{c}_{2} = 0.
(9.6)

These in turn can be used to determine {c}_{6},{c}_{7},{c}_{8}, etc. It is not too hard to find an explicit expression for the c’s

\begin{eqnarray} {c}_{3m}& =& {3m − 2\over (3m)(3m − 1)(3m − 2)}{c}_{3(m−1)} %& \\ & =& {3m − 2\over (3m)(3m − 1)(3m − 2)} {3m − 5\over (3m − 3)(3m − 4)(3m − 5)}{c}_{3(m−1)} %& \\ & =& {(3m − 2)(3m − 5)\mathop{\mathop{…}}1\over (3m)!} {c}_{0}, %& \\ {c}_{3m+1}& =& {3m − 1\over (3m + 1)(3m)(3m − 1)}{c}_{3(m−1)+1} %& \\ & =& {3m − 1\over (3m + 1)(3m)(3m − 1)} {3m − 4\over (3m − 2)(3m − 3)(3m − 4)}{c}_{3(m−2)+1}%& \\ & =& {(3m − 2)(3m − 5)\mathop{\mathop{…}}2\over (3m + 1)!} {c}_{1}, %& \\ {c}_{3m+1}& =& 0. %&(9.7) \\ \end{eqnarray}

The general solution is thus

y(t) = a\left [1 +{ \mathop{∑ }}_{m=1}^{∞}{c}_{ 3m}{t}^{3m}\right ] + b\left [1 +{ \mathop{∑ }}_{m=1}^{∞}{c}_{ 3m+1}{t}^{3m+1}\right ].
(9.8)

The technique sketched here can be proven to work for any differential equation

y''(t) + p(t)y'(t) + q(t)y(t) = f(t)
(9.9)

provided that p(t), q(t) and f(t) are analytic at t = 0. Thus if p, q and f have a power series expansion, so has y.

 9.1 Singular points
 9.2 *Special cases
  9.2.1 Two equal roots
  9.2.2 Two roots differing by an integer
  9.2.3 Example 1
  9.2.4 Example 2

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