Math 308
Worksheet 3: Section 5.4
11 July 2012
Definition. Start with a generic second–order, linear differential equation R(x)y 00 +
P (x)y 0 + Q(x)y = 0 and divide through by R(x) to put the equation in normal form
y 00 + p(x) y 0 + q(x) y = 0 .
(1)
Question.
Pick a initial point xo . When can we find an analytic solution y(x) =
P∞
a
(x
−
xo )n ?
n=0 n
Definition. A function f (x) is analytic at xo if we can find positive number r > 0 so
P
that the Taylor series f (x) = ∞
n=0 an (x − xo ) converges when |x − xo | < r.
If both functions p(x) and q(x) are analytic at xo , then xo is an ordinary point of
the differential equation; otherwise, xo is a singular point.
Answer. Theorem 5.3.1 says we can find an analytic solution y(x) at xo if both
p(x) andPq(x) are analytic at xo . (That is, if xo is ordinary, then we can solve for
n
y(x) = ∞
n=0 an (x − xo ) .) Also, the radius of convergence for y(x) is at least as
big as the smallest radius of convergence for p(x) and q(x). (Use Theorem 5.3.1 to
answer HW (A) below.)
Summary. AtP
an ordinary point, we can find analytic solutions. The will be of the
n
form y(x) = ∞
n=0 an (x − xo ) . To find y(x) we proceed in two steps: (i) use the
initial values to determine a0 and a1 ; and (ii) use the differential equation (1) to find
a recursive relationship for the other an .
Example. Fix two numbers (parameters) α and β. For the Euler equation (a second–
order, linear, homogeneous differential equation)
x2 y 00 + α x y 0 + β y = 0 ,
(2)
we have
α
β
and q(x) = 2 .
x
x
So xo = 0 is a singular point of the
Euler
equation.
(So,
we won’t always be able to
P∞
n
find solutions of the form y(x) = n=0 an x .)
p(x) =
What about singular points? Can we still find solutions to (1) at a singular point?
Yes, if the singularity isn’t too bad. (But it probably won’t be analytic!) What we
need is for the singular point xo to look a lot like a singularity of the Euler equation.
Mathematically, this means both limits
lim (x − xo )p(x) and
x→xo
lim (x − xo )2 q(x)
x→xo
exist. When both these limits exist, we say xo is a regular singular point; if either
one of the limits does not exist, we say xo is a irregular singular point.
Beware! While we can construct solutions to (1) at a regular singular point, they
will be more complicated than mere power series.
Example. For the singular point xo = 0 of the Euler equation (2), we have
lim xp(x) = lim α = α
x→0
x→0
2
lim x q(x) = lim β = β .
x→0
x→0
Since both limits exist, xo is a regular singular point.
Homework. As part of your preparation for Quiz 2 (Wednesday, July 18) and the
Final Exam (Tuesday, August 07), I recommend the following:
(A) Section 5.3: Problems 5, 6.
(B) Section 5.4: Problems 2, 4, 6, 9.
(C) Section 5.4: Problems 18, 20, 24, 34.
(D) Section 5.4: Problems 35, 37, 39.
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Colleen Robles
robles@math.tamu.edu