Higher Series Solutions of Second Order Linear Equations
Ordinary Differential Equations
Dr. Marco A Roque Sol
12/01/2015
Dr. Marco A Roque Sol
Ordinary Differential Equations
Series Solutions Near an Ordinary Point, Part I
Series Solutions Near an Ordinary Point, Part II
Higher Series Solutions of Second Order Linear Equations
Series Solutions Near an Ordinary Point, Part I
Now, we say that x = x0 is an ordinary point if provided both
Q(x)
P(x)
and
R(x)
P(x)
are analytic at x = x0 . That is, these two quantities have Taylor
series around x = x0 . Since we are dealing with coefficients that
are polynomials so this will be equivalent to saying that P(x0 ) 6= 0.
Since P is continuous, it follows that there is an interval about x0
in which P(x) is never zero
Dr. Marco A Roque Sol
Ordinary Differential Equations
Higher Series Solutions of Second Order Linear Equations
Series Solutions Near an Ordinary Point, Part I
Series Solutions Near an Ordinary Point, Part II
Series Solutions Near an Ordinary Point, Part I
Hence, according to the existence and uniqueness Theorem, there
exists in that interval a unique solution of the ODE that also
satisfies the initial conditions y (x0 ) = y0 , y 0 (x0 ) = y00 for arbitrary
values of y0 and y00 .
If a point is not an ordinary point (P(x0 ) = 0) we call it a singular
point. In this case at least one of Q(x0 ) and R(x0 ) is not zero and
the Theorem of Existence and uniqueness is not applied.
The basic idea to finding a series solution to a differential equation
is to assume that we can write the solution as a power series in the
form
Dr. Marco A Roque Sol
Ordinary Differential Equations
Higher Series Solutions of Second Order Linear Equations
Series Solutions Near an Ordinary Point, Part I
Series Solutions Near an Ordinary Point, Part II
Series Solutions Near an Ordinary Point, Part I
y (x) =
∞
X
an (x − x0 )n
n=0
and then try to determine what the an s need to be. We will only
be able to do this if the point x = x0 , is an ordinary point. We will
usually say that y , given above, is a series solution around x = x0 .
Example 5.3
Find a series solution of the equation
y 00 + y = 0;
Dr. Marco A Roque Sol
−∞ < x < ∞
Ordinary Differential Equations
Higher Series Solutions of Second Order Linear Equations
Series Solutions Near an Ordinary Point, Part I
Series Solutions Near an Ordinary Point, Part II
Series Solutions Near an Ordinary Point, Part I
Solution
We look for a solution in the form of a power series about x0 = 0
y = a0 + a1 x + ... + an x n + ... =
∞
X
an x n
n=0
and assume that the series converges in some interval |x| < ρ.
Differentiating the above equation term by term, we obtain
y0 =
∞
X
nan x n−1 =
n=1
Dr. Marco A Roque Sol
∞
X
(n + 1)an+1 x n
n=0
Ordinary Differential Equations
Series Solutions Near an Ordinary Point, Part I
Series Solutions Near an Ordinary Point, Part II
Higher Series Solutions of Second Order Linear Equations
Series Solutions Near an Ordinary Point, Part I
y 00 =
∞
X
n(n − 1)an x n−2 ==
n=2
(n + 2)(n + 1)an+2 x n
n=0
Substituting the series for y and
y + y 00 =
∞
X
an x n +
n=0
=
∞
X
∞
X
y 00
∞
X
in the ODE gives
(n + 2)(n + 1)an+2 x n
n=0
[an + (n + 2)(n + 1)an+2 ] x n = 0
n=0
For this equation to be satisfied for all x, the coefficient of each
power of x must be zero; hence we conclude that
an + (n + 2)(n + 1)an+2 = 0;
Dr. Marco A Roque Sol
n = 0, 1, 2, ...
Ordinary Differential Equations
Higher Series Solutions of Second Order Linear Equations
Series Solutions Near an Ordinary Point, Part I
Series Solutions Near an Ordinary Point, Part II
Series Solutions Near an Ordinary Point, Part I
The above equation is called a recurrence relation. The successive
coefficients can be evaluated one by one by writing the recurrence
relation first for n = 0, then for n = 1, and so forth. In this case
we obtain
a0
a2
a0
a0
= − , a4 = −
= , ...,
a2 = −
2·1
2!
4·3
4!
a2k =
(−1)k
a0 ,
(2k)!
k = 1, 2, 3, ...
Similarly, for the odd-numbered coefficient
a3 = −
a1
a1
=− ,
3·2
3!
a2k+1 =
a5 = −
(−1)k
a1 ,
(2k + 1)!
Dr. Marco A Roque Sol
a3
a1
= , ...,
4·3
5!
k = 1, 2, 3, ...
Ordinary Differential Equations
Higher Series Solutions of Second Order Linear Equations
Series Solutions Near an Ordinary Point, Part I
Series Solutions Near an Ordinary Point, Part II
Series Solutions Near an Ordinary Point, Part I
Substituting these coefficients into the equation, we have
y = a0 + a1 x −
(−1)n a0 2n
(−1)n a1 2n
a0 2 a1 3
x − x + ... +
x +
x + ...
2!
3!
(2n)!
(2n + 1)!
x2 x4
(−1)n 2n
y = a0 1 −
+
+ ... +
x + ... +
2!
4!
(2n)!
x3 x5
(−1)n 2n+1
+
+ ... +
x
+ ...
a1 x −
3!
5!
(2n)!
y = a0
∞
X
(−1)n
n=0
(2n)!
x
2n
Dr. Marco A Roque Sol
∞
X
(−1)n 2n+1
+ a1
x
(2n + 1)!
n=0
Ordinary Differential Equations
Higher Series Solutions of Second Order Linear Equations
Series Solutions Near an Ordinary Point, Part I
Series Solutions Near an Ordinary Point, Part II
Series Solutions Near an Ordinary Point, Part I
The first series is the Taylor expansion of cos(x) around a = 0 and
the second series is the Taylor expansion of sin(x) around a = 0.
Thus, the solution is
y = a0 cos(x) + a1 sin(x)
Now, let’s suppose that we don’t know the series that we found
and called them
C (x) =
∞
X
(−1)n
n=0
(2n)!
x 2n ;
S(x) =
∞
X
(−1)n 2n+1
x
(2n + 1)!
n=0
and applying the ratio test we find that both series converge
absolutely for all x in −∞ < x < ∞. Moreover we have
Dr. Marco A Roque Sol
Ordinary Differential Equations
Higher Series Solutions of Second Order Linear Equations
Series Solutions Near an Ordinary Point, Part I
Series Solutions Near an Ordinary Point, Part II
Series Solutions Near an Ordinary Point, Part I
C 0 (x) =
∞
X
n=1
−
∞
2n
(−1)n 2n−1 X (−1)n 2n−1
x
=
x
=
(2n)!
(2n − 1)!
n=1
∞
X
(−1)n 2n+1
x
= −S(x)
(2n + 1)!
n=0
Similarly we have
S 0 (x) =
∞
X
n=0
∞
(2n + 1)
(−1)n 2n X (−1)n 2n
x =
x = C (x)
(2n + 1)!
(2n)!
n=0
And therefore the Wronskian is given by
C (0) S(0) 1 0
=
=1
W = 0
C (0) S 0 (0) 0 1
and they form a fundamental solution.
Dr. Marco A Roque Sol
Ordinary Differential Equations
Higher Series Solutions of Second Order Linear Equations
Series Solutions Near an Ordinary Point, Part I
Series Solutions Near an Ordinary Point, Part II
Series Solutions Near an Ordinary Point, Part I
Example 5.4
Find a series solution of the Airy’s equation ( One can find it in the
analysys of Schrdinger’s equation for a particle confined within a
triangular potential well and for a particle in a one-dimensional
constant force field )
y 00 − xy = 0;
−∞ < x < ∞
Solution
For this equation P(x) = 1, Q(x) = 0, and R(x) = −x; hence
every point is an ordinary point. We assume that
y = a0 + a1 x + ... + an x n + ... =
∞
X
an x n
n=0
Dr. Marco A Roque Sol
Ordinary Differential Equations
Higher Series Solutions of Second Order Linear Equations
Series Solutions Near an Ordinary Point, Part I
Series Solutions Near an Ordinary Point, Part II
Series Solutions Near an Ordinary Point, Part I
and assume that the series converges in some interval |x| < ρ. The
series for y 00 is given
y 00 =
∞
X
n(n − 1)an x n−2 ==
n=2
(n + 2)(n + 1)an+2 x n
n=0
Substituting the series for y and
y 00 − xy =
∞
X
∞
X
y 00
in the ODE gives
(n + 2)(n + 1)an+2 x n − x
n=0
Dr. Marco A Roque Sol
∞
X
an x n = 0
n=0
Ordinary Differential Equations
Higher Series Solutions of Second Order Linear Equations
Series Solutions Near an Ordinary Point, Part I
Series Solutions Near an Ordinary Point, Part II
Series Solutions Near an Ordinary Point, Part I
y 00 − xy =
∞
X
(n + 2)(n + 1)an+2 x n −
n=0
y 00 − xy =
∞
X
an x n+1 = 0
n=0
(n + 2)(n + 1)an+2 x n −
n=0
∞
X
an−1 x n = 0
(n + 2)(n + 1)an+2 x n −
n=1
∞
X
∞
X
n=1
y 00 − xy = (2)(1)a2 +
(2)(1)a2 +
∞
X
∞
X
an−1 x n =
n=1
[(n + 2)(n + 1)an+2 − an−1 ] x n = 0
n=1
Again, for this equation to be satisfied for all x in some interval,
the coefficients of like powers of x must be equal; hence a2 = 0,
and we obtain the recurrence relation
Dr. Marco A Roque Sol
Ordinary Differential Equations
Higher Series Solutions of Second Order Linear Equations
Series Solutions Near an Ordinary Point, Part I
Series Solutions Near an Ordinary Point, Part II
Series Solutions Near an Ordinary Point, Part I
(n + 2)(n + 1)an+2 = an−1 ;
a3 = −
a0
2·3
a6 = −
a2 = 0;
a5
a0
=
,
5·6
2·3·5·6
a0
,
2 · 3 · 5 · 6 . . . (3n − 1)(3n)
Similarly, we have
a3n =
a4 = −
a1
3·4
a3n+1 =
a7 =
n = 1, 2, ...
a9 =
a6
a0
=
.
8·9
2·3·5·6·8·9
n = 1, 2, 3, ...
a4
a1
a7
a1
=
, a10 =
=
...
6·7
3·4·6·7
9 · 10
3 · 4 · 6 · 7 · 9 · 10
a1
,
3 · 4 · 6 · 7 . . . 9 · 10 . . . · (3n)(3n + 1)
Dr. Marco A Roque Sol
n = 1, 2, 3, ...
Ordinary Differential Equations
Higher Series Solutions of Second Order Linear Equations
Series Solutions Near an Ordinary Point, Part I
Series Solutions Near an Ordinary Point, Part II
Series Solutions Near an Ordinary Point, Part I
Thus the general solution of Airy’s equation is
y = a0
x3
x6
x 3n
1+
+
+ ... +
+ ... +
2·3 2·3·5·6
2 · · · 3 . . . (3n − 1)(3n)
x4
x7
x 3n+1
a1 x +
+
+ ... +
+ ...
3·4 3·4·6·7
3 · · · 4 . . . (3n)(3n + 1)
Let’s define
x6
x 3n
x3
+
+ ... +
+ ...
Ai(x) = 1 +
2·3 2·3·5·6
2 · · · 3 . . . (3n − 1)(3n)
x4
x7
x 3n+1
Bi(x) = x +
+
+ ... +
+ ...
3·4 3·4·6·7
3 · · · 4 . . . (3n)(3n + 1)
Dr. Marco A Roque Sol
Ordinary Differential Equations
Higher Series Solutions of Second Order Linear Equations
Series Solutions Near an Ordinary Point, Part I
Series Solutions Near an Ordinary Point, Part II
Series Solutions Near an Ordinary Point, Part I
The solution can be written as
y = a0 Ai(x) + a1 Bi(x)
where Ai(x) and Bi(x) are called the Airy functions of the first
and second kind respectively.
Notice that Ai satisfies the initial conditions Ai(0) = 1, Ai 0 (0) = 0
and that Bi satisfies the initial conditions Bi(0) = 0, Bi 0 (0) = 1.
Thus W (Ai, Bi)(0) = 1 6= 0, and consequently Ai and Bi are a
fundamental set of solutions. Hence the general solution of Airys
equation is
Dr. Marco A Roque Sol
Ordinary Differential Equations
Higher Series Solutions of Second Order Linear Equations
Series Solutions Near an Ordinary Point, Part I
Series Solutions Near an Ordinary Point, Part II
Series Solutions Near an Ordinary Point, Part I
y = a0 Ai(x) + a1 Bi(x)
Dr. Marco A Roque Sol
Ordinary Differential Equations
Higher Series Solutions of Second Order Linear Equations
Series Solutions Near an Ordinary Point, Part I
Series Solutions Near an Ordinary Point, Part II
Series Solutions Near an Ordinary Point, Part I
Example 5.5
Find the first four terms in each portion of the series solution
around x0 = 0 for the following differential equation
(x 2 + 1)y 00 − 4xy 0 + 6y = 0
Solution
We have a differential equation that doesn’t have a constant
coefficient for the second derivative. In this case we have
P(x) = x 2 + 1;
Dr. Marco A Roque Sol
P(0) = 1 6= 0
Ordinary Differential Equations
Higher Series Solutions of Second Order Linear Equations
Series Solutions Near an Ordinary Point, Part I
Series Solutions Near an Ordinary Point, Part II
Series Solutions Near an Ordinary Point, Part I
So x0 = 0 is an ordinary point for this differential equation. We
first need the solution and its derivatives,
y=
∞
X
y0 =
an x n ;
n=0
∞
X
nan x n−1 ;
y 00 =
n=1
∞
X
(n)(n − 1)an x n−2
n=2
Now, substituting
∞
∞
∞
X
X
X
n−2
n−1
(x + 1)
(n)(n − 1)an x
− 4x
nan x
+6
an x n = 0
2
n=2
n=1
Dr. Marco A Roque Sol
n=0
Ordinary Differential Equations
Series Solutions Near an Ordinary Point, Part I
Series Solutions Near an Ordinary Point, Part II
Higher Series Solutions of Second Order Linear Equations
Series Solutions Near an Ordinary Point, Part I
∞
X
+6
n
(n)(n − 1)an x +
n=2
∞
X
∞
X
(n)(n − 1)an x
n−2
− 4x
n=2
∞
X
nan x n−1
n=1
an x n = 0
n=0
We will only need to shift the second series down by two to get all
the exponents the same in all the series and start the first and
third seies at zero
∞
∞
X
X
(n)(n − 1)an x n +
(n + 2)(n + 1)an+2 x n
n=0
n=0
−4
∞
X
nan x n + 6
n=0
∞
X
Dr. Marco A Roque Sol
∞
X
an x n = 0
n=0
Ordinary Differential Equations
Higher Series Solutions of Second Order Linear Equations
Series Solutions Near an Ordinary Point, Part I
Series Solutions Near an Ordinary Point, Part II
Series Solutions Near an Ordinary Point, Part I
Solving this gives,
an+2 = −
(n − 2)(n − 3)an
;
(n + 2)(n + 1)
n = 0, 1, 2, ...
n = 0 a2 = −3a0
1
n = 1 a3 = − a1
3
n = 2 a3 = −
0
a2 = 0
12
n = 3 a3 = −
0
a3 = 0
20
Dr. Marco A Roque Sol
Ordinary Differential Equations
Higher Series Solutions of Second Order Linear Equations
Series Solutions Near an Ordinary Point, Part I
Series Solutions Near an Ordinary Point, Part II
Series Solutions Near an Ordinary Point, Part I
Now, from this point on all the coefficients are zero. In this case
both of the series in the solution will terminate. The solution in
this case is,
1
y (x) = a0 (1 − 3x 2 ) + a1 (x − x 3 )
3
Now, the two functions y0 = (1 − 3x 2 );
y0 (0) = 1;
y1 (0) = 0;
1
y1 = (x − x 3 ) satisfy
3
y00 (0) = 0;
y10 (0) = 1
W (y0 , y1 )(0) = 1
Dr. Marco A Roque Sol
Ordinary Differential Equations
Higher Series Solutions of Second Order Linear Equations
Series Solutions Near an Ordinary Point, Part I
Series Solutions Near an Ordinary Point, Part II
Series Solutions Near an Ordinary Point, Part I
Therefore {y0 , y1 } form a fundamental set. Hence the general
solution of the equation is
y = a0 y0 (x) + a1 y1 (x)
Dr. Marco A Roque Sol
Ordinary Differential Equations
Higher Series Solutions of Second Order Linear Equations
Series Solutions Near an Ordinary Point, Part I
Series Solutions Near an Ordinary Point, Part II
Series Solutions Near an Ordinary Point, Part II
Let’s considered the problem of finding solutions of
P(x)y 00 + Q(x)y 0 + R(x)y = 0
where P, Q, and R are polynomials. Remember, we say that a
point x0 , in an ordinary point if the functions
p(x) =
Q(x)
;
P(x)
q(x) =
R(x)
P(x)
are analytical at x0 , that is, they have Taylor series about the point
x0 .
p(x) =
∞
X
pn (x − x0 )n ;
n=0
Dr. Marco A Roque Sol
q(x) =
∞
X
qn (x − x0 )n
n=0
Ordinary Differential Equations
Higher Series Solutions of Second Order Linear Equations
Series Solutions Near an Ordinary Point, Part I
Series Solutions Near an Ordinary Point, Part II
Series Solutions Near an Ordinary Point, Part II
Otherwise, we say that the point x0 is singular. For regular points
we have the following result
Theorem 1
If x0 is an ordinary point of the differential equation
P(x)y 00 + Q(x)y 0 + R(x)y = 0
that is, if p = Q/P and q = R/P are analytic at x0 , then the
general solution of the above equation is
y (x) =
∞
X
an (x − x0 )n ; y (x) = a0 y1 (x) + a1 y2 (x)
n=0
Dr. Marco A Roque Sol
Ordinary Differential Equations
Higher Series Solutions of Second Order Linear Equations
Series Solutions Near an Ordinary Point, Part I
Series Solutions Near an Ordinary Point, Part II
Series Solutions Near an Ordinary Point, Part II
where a0 and a1 are arbitrary; y1 and y2 are two power series
solutions that are analytic at x0 . The solutions y1 and y2 form a
fundamental set of solutions. Further, the radius of convergence
for each of the series solutions y1 and y2 is at least as large as the
minimum of the radii of convergence of the series for p and q .
Example 5.6
What is the radius of convergence of the Taylor series for
(1 + x 2 )−1 about x = 0 ?
Dr. Marco A Roque Sol
Ordinary Differential Equations
Higher Series Solutions of Second Order Linear Equations
Series Solutions Near an Ordinary Point, Part I
Series Solutions Near an Ordinary Point, Part II
Series Solutions Near an Ordinary Point, Part II
Solution
One way to proceed is to find the Taylor series in question, namely,
1
= 1 − x 2 + x 6 − x 6 + ... + (−1)n x 2n + ...
1 + x2
Then it can be verified by the ratio test that ρ = 1. Another
approach is to note that the zeros of 1 + x 2 are x = ±i. Since the
distance in the complex plane from 0 to i or to i is 1, the radius of
convergence of the power series about x = 0 is 1.
Dr. Marco A Roque Sol
Ordinary Differential Equations