Why series solutions · Differential Equations

Intuition

The methods so far need constant coefficients, or a lucky substitution. Many equations of physics have variable coefficients — Airy’s equation for light near a rainbow’s edge, Legendre’s and Bessel’s equations for spheres and drums — and their solutions are not built from exponentials, sines or polynomials. They can still be found as power series: guess that the solution is an infinite polynomial, and let the equation decide its coefficients one by one.

Building a curve from its Taylor polynomials, one term at a time. Each new term makes the approximation better near the starting point, and the equation itself says what every term must be.

Airy's equation $y''=xy$ with $y(0)=1$ , $y'(0)=0$ . Its solution, $1+\tfrac{x^{3}}{6}+\tfrac{x^{6}}{180}+\cdots$ , is no combination of familiar functions. Two terms of the series fit it only near $0$ ; four fit it much further out; the full series converges for every $x$ .

A solution as a power series

A power series $\sum_{n\ge0}a_{n}(x-x_{0})^{n}$ converges on an interval about , and inside that interval it may be differentiated term by term — both are proved in the analysis course and used here. The method of this chapter assumes a solution of that form, substitutes it into the equation and reads off equations for the coefficients. It is needed when the coefficients vary: for with or not constant there is no characteristic equation.

y=\sum_{n=0}^{\infty}a_{n}x^{n}=a_{0}+a_{1}x+a_{2}x^{2}+\cdots

What a series solution is

$y(0)=a_{0}$ and $y'(0)=a_{1}$ : the first two coefficients are the initial value and the initial slope.

The coefficients are derivatives at the centre

y=\sum_{n\ge0}a_{n}x^{n}\ \Longrightarrow\ a_{n}=\frac{y^{(n)}(0)}{n!}

Differentiate the series k times, term by term, which the analysis course allows inside the interval of convergence. At zero every power of x vanishes except the constant term, which is k factorial times the k-th coefficient. So each coefficient is fixed by a derivative of the function at zero, and two series for the same function have the same coefficients.

Proof steps

y^{(k)}(x)=\sum_{n\ge k}n(n-1)\cdots(n-k+1)\,a_{n}x^{n-k}

Differentiate k times, term by term.

y^{(k)}(0)=k(k-1)\cdots1\cdot a_{k}=k!\,a_{k}

At zero every term but the first vanishes.

a_{k}=\frac{y^{(k)}(0)}{k!}

So the coefficients are fixed by the function: it has at most one power series about a point.

Applications

Practice

When Series Are Needed

Constant coefficients have a characteristic equation. Variable coefficients usually do not, and a power series takes its place.

y''=xy

Try it

Which equation needs a series solution rather than a characteristic equation?