Mathematical Tools for Physics - Department of Physics - University

4—Differential Equations 74

Examples of equations that show up in physics problems are

y′′+y= 0

(1−x^2 )y′′− 2 xy′+(+ 1)y= 0 regular singular points at± 1

x^2 y′′+xy′+ (x^2 −n^2 )y= 0 regular singular point at zero

xy′′+ (α+ 1−x)y′+ny= 0 regular singular point at zero

(4.18)

These are respectively the classical simple harmonic oscillator, Legendre equation, Bessel equation,
generalized Laguerre equation.
A standard procedure to solve these equations is to use series solutions, but not just the standard
power series such as those in Eq. (2.4). Essentially, you assume that there is a solution in the form of
an infinite series and you systematically compute the terms of the series. I’ll pick the Bessel equation

from the above examples, as the other three equations are done the same way. The parameternin that

equation is often an integer, but it can be anything. It’s common for it to be^1 / 2 or^3 / 2 or sometimes

even imaginary, but there’s no need to make any assumptions about it for now.
Assume a solution in the form :

Frobenius Series: y(x) =

∑∞

0

akxk+s (a 06 = 0) (4.19)

Ifs= 0or a positive integer, this is just the standard Taylor series you saw so much of in chapter two,

but this simple-looking extension makes it much more flexible and suited for differential equations. It

often happens thatsis a fraction or negative, but this case is no harder to handle than the Taylor

series. For example, what is the series expansion of(cosx)/xabout the origin? This is singular at

zero, but it’s easy to write the answer anyway because you already know the series for the cosine.

cosx

x

=

1

x

−

x

2

+

x^3

24

−

x^5

720

+···

It starts with the term 1 /xcorresponding tos=− 1 in the Frobenius series.

Alwaysassume thata 06 = 0, because that just defines the coefficient of the most negative power,

xs. If you allow it be zero, that’s just the same as redefiningsand it gains nothing except confusion.

Plug this into the Bessel differential equation.

x^2 y′′+xy′+ (x^2 −n^2 )y= 0

x^2

∑∞

k=0

ak(k+s)(k+s−1)xk+s−^2 +x

∑∞

k=0

ak(k+s)xk+s−^1 + (x^2 −n^2 )

∑∞

k=0

akxk+s= 0

∑∞

k=0

ak(k+s)(k+s−1)xk+s+

∑∞

k=0

ak(k+s)xk+s+

∑∞

k=0

akxk+s+2−n^2

∑∞

0

akxk+s= 0

∑∞

k=0

ak

[

(k+s)(k+s−1) + (k+s)−n^2

]

xk+s+

∑∞

k=0

akxk+s+2= 0

The coefficients of all the like powers ofxmust match, and in order to work out the matches efficiently,

and so as not to get myself confused in a mess of indices, I’ll make an explicit change of the index in
the sums.Do this trick every time. It keeps you out of trouble.