Session 2
Series Solutions of ODE's
1 Introduction
A powerful and very general technique used to obtain
solutions of linear ordinary differential equations
is the power series method. One starts by
assuming that the solution can be expressed by a
simple power series then substitutes the series into the
differential equation and thus determines the required values of
the coefficients in the power series.
In its implementation due to Frobenius the method
allows the determination of solutions to a large number
of differential equations of great importance in
applications.
2 Power Series and Function Representations
For a real variable x an expression of the form
is called a power series. The series converges if it
has a limit as N ® ¥. Convergence
can be investigated by the ratio test which
evaluates the ratio between two subsequent terms in the
series and result in the
interval of convergence
where
Inside the interval of convergence, the series converges; outside
it does not.
Series are often useful to produce representations of other functions.
Examples of such series representations are
|
|
| cosx = |
¥
å
n = 0
|
(-1)n |
x2n
(2n!)
|
|
|
| |
|
|
|
| coshx = |
¥
å
n = 0
|
|
x2n
(2n!)
|
|
|
| |
|
|
|
| log |
1+x2
1-x2
|
= 2 x2 |
¥
å
n = 0
|
|
x4n
2n+1
|
|
|
| |
|
and
When converging series are used to represent functions the operations of analysis
(differentiation and integration) can be carried out on both sides of
the equation and the equality is maintained. For instance, the representation
|
|
| f(x) = |
¥
å
n = 0
|
An (x - x0)n |
|
| |
|
after differentiation leads to
and it is called the Taylor series expansion of f(x) near x0.
If x0 = 0 this becomes the Maclaurin series expansion.
3 Taylor Series for Functions of a Complex Variable
If the complex function of a complex variable f(z) is analytic in the
circular disk |z-a| £ r then it can be expanded in a Taylor series
converging inside the disk, i.e.
|
|
| f(z) = |
¥
å
n = 0
|
An (z - a)n = |
¥
å
n = 0
|
|
fn(a)
n!
|
(z-a)n |
|
| |
|
where
|
|
| An = |
1
2 pi
|
| ó
(ç)
õ
|
C
|
|
f(a) d a
(a-a)n+1
|
= |
fn(a)
n!
|
|
|
| |
|
where fn(a) is the n-th derivative of f(z) at z = a. Further, if f(z)
can be expanded in Taylor series in a circle around z = a then it is analytic
there.
4 Laurent Series
Laurent series generalize Taylor series by allowing negative powers. Consider the
annulus r1 < |z-a| < r2 inside which f(z) is analytic. By introducing
a crosscut Cauchy's integral formula can be applied on a single contour and the
Laurent expansion of f(z) becomes
|
|
| f(z) = |
n = ¥
å
n = -¥
|
cn (z-a)n |
|
| |
|
where
|
|
| cn = |
1
2 pi
|
| ó
(ç)
õ
|
C
|
|
f(a) da
(a-a)n+1
|
|
|
| |
|
The Laurent series expansion is unique.
5 Solving Differential Equations using Series
Consider the differential equation
Now assume the solution y(x) can be represented by a power series
around x = 0, i.e.
Differentiating
|
|
|
d2 y
d x2
|
= |
¥
å
k = 0
|
k (k-1) Ak xk-2 |
|
| |
|
and substituting in the original ODE
|
|
| Ly = |
¥
å
k = 0
|
k (k-1) Ak xk-2 - |
¥
å
k = 0
|
Ak xk = 0 |
|
| |
|
Term by term comparison produces the recurrence relation
for k = 2,3,.... Therefore only two of
the constant coefficients (A0 and A1) are independent
and the required solution of the problem is
|
|
|
y(x) = A0 (1 + |
1
2
|
x2 + |
1
24
|
x4 + ...) +A1 ( x + |
1
6
|
x3 + |
1
120
|
x5 + ...) = |
| | = A0 coshx + A1 sinhx = A0 u1(x) + A1 u2(x) |
|
| |
|
where u1(x) = coshx and u2(x) = sinhx are
two linearly independent solutions.
6 Singular Points
The standard form of writing a homogeneous second order linear ordinary
differential equation is
|
|
|
d2 y
dx2
|
+ a1(x) |
dy
dx
|
+ a2(x) y = 0 |
|
| |
|
We are interested in the behavior of the solution near x = x0.
This turns out to depend on the behavior of a1 and a2.
If both coefficients can be expanded in power series about x0
(i.e. they are regular), then x = x0 is an ordinary point of the
differential equation. If the expansion does not exist, then x0
is a singular point. But if the functions (x-x0) a1(x)
and (x-x0)2 a2(x) are both regular then x0 is
a regular singular point, otherwise, it is an irregular singular point.
The above is relevant because
1) If x0 is an ordinary point of a
differential equation then the equation possesses two linearly
independent solutions both expressible as power series.
2) If x0 is a regular singular point, at least one solution
exists of the form
|
|
| y = (x-x0)s |
¥
å
k = 0
|
Ak (x - x0)k |
|
| |
|
where s must be determined and can be real or imaginary. However,
a second independent solution may or may not exist.
3) If x0 is an irregular singular point then a non-trivial solution of this
type may or may not exist.
7 The Method of Frobenius
Conside the second order linear homogenoeus differential equation
written in the following standard form
|
|
| L y = R(x) |
d2 y
dx2
|
+ |
1
x
|
P(x) |
dy
dx
|
+ |
1
x2
|
Q(x) y = 0 |
|
| |
|
where R(x), P(x) and Q(x) are regular at x = 0 so that this point
is an ordinary point or at worst a regular singular point of the
differential equation and a nontrivial solution of the form
|
|
| y(x) = xs |
¥
å
k = 0
|
Ak xk |
|
| |
|
can be expected to exist.
Because of the regularity the following converging series exist
|
|
| P(x) = P0 + P1 x + P2 x2 + ... |
|
| |
|
|
|
| Q(x) = Q0 + Q1 x + Q2 x2 + ... |
|
| |
|
and
|
|
| R(x) = 1 + R1 x + R2 x2 + ... |
|
| |
|
where R(0) = 1 has been assumed without loss of generality.
The assumed nontrivial solution and the above three
series can now be substituted into the original differential
equation. Terms are rearranged in order of increasing powers of
x. Finally we verify that the coefficients of all powers of
x vanish independently.
Specifically, the vanishing of the
coefficient associated with the lowest power of x (xs-2) yields
|
|
| s (s-1) + P0 s + Q0 = s2 + (P0-1) s + Q0 = f(s) = 0 |
|
| |
|
which is called the indicial equation with corresponding roots
s1 and s2 which are the exponents of the leading terms of the
expected series solutions. These are called the exponents of the
differential equation at x = 0. If these exponents are not equal then,
by convention s1 > s2. The exponents can be real or complex but
in most cases of interest they are real.
For each such value of s, the vanishing of the
coefficient of the term associated with xs+k-2 yields an infinite
sequence of relationships, of which the first two are
|
|
| [(s+1) s + P0(s+1) + Q0] A1 = - [R1 s (s-1) + P1 s + Q1] A0 |
|
| |
|
|
|
|
[(s+2)(s+1) + P0(s+2) + Q0] A2 = - { [R1 (s+1) s + P1 (s+1) + Q1] A1 + |
| | + [R2 s (s-1) + P2 s + Q2] A0 } |
|
| |
|
These recurrence relationships can be more compactly rewritten as
the following recurrence formula
|
|
| f(s+k) Ak = - |
k
å
n = 1
|
gn(s+k) Ak-n |
|
| |
|
where
|
|
| gn(s) = Rn (s-n)2 + (Pn - Rn)(s-n) + Qn. |
|
| |
|
The recurrence formula allows determination of Ak from
the preceeding A's as long as f(s+k) ¹ 0.
If the roots s1 and s2 are distinct and if for each such s
|
|
| f(s+k) = (s+k)(s+k-1) + P0 (s+k) + Q0 ¹ 0 |
|
| |
|
for any k ³ 0, then one independent set of Ak's is obtained
for each value of s and the general
solution is of the form y = c1 u1(x) + c2 u2(x).
If s1 = s2 only one series solution can be obtained.
This is an exceptional case.
A second exceptional case arises if s1 ¹ s2 but
f(s1+k) or f(s2+k) vanishes for k = K where K is
a positive integer. If the
roots are real and s1 > s2 then f(s2+k) vanishes
only when s1-s2 is a positive integer.
Thus, if s1 - s2 = K, when k = K the recurrence
formula becomes
|
|
| (s-s2)(s-s2+K) AK = - |
K
å
n = 1
|
gn(s+K) AK-n |
|
| |
|
If s = s2, AK is necessarily arbitrary and a series
solution results corresponding to the smaller exponent s2
containing two arbitrary constants A0 and AK (i.e.
a complete solution) is obtained.
In summary,
- When x = 0 is an ordinary point or a regular singular point
for which s1 - s2 is not zero nor a
positive integer, two distinct series solution result.
- If s1 - s2 is a positive integer, a series solution
corresponding to s1 results but it may also be the case that
a solution for s2 exist and thus two distinct series solution result.
- If the exponents are equal, one series solution results.
Strategically, when s1-s2 = K ³ 1 one must first explore
the solution obtained with the smaller exponent s2 since if a
nontrivial solution is found, one can regard two constants in that solution,
A0 and AK as arbitrary and there is no need to obtain the solution
for s1.
If x is complex the solution has a pole at the origin when
the exponent is a negative integer and a branch point when
the exponent is nonintegral.
Example: Apply Frobenius method to find the general solution of the equation:
|
|
| L y = 2x2 |
d2 y
dx2
|
- x |
dy
dx
|
+ (1+x) y = 2 x2 y¢¢- x y¢+ (1+x)y = 0 |
|
| |
|
Since x = 0 is a regular singular point, we expect solutions of the form
|
|
| y(x) = xs |
¥
å
k = 0
|
Ak xk = |
¥
å
k = 0
|
Ak xk+s |
|
| |
|
Here
|
|
| y(x)¢ = |
¥
å
k = 0
|
(k+s) Ak xk+s-1 |
|
| |
|
and
|
|
| y(x)¢¢ = |
¥
å
k = 0
|
(k+s-1) (k+s) Ak xk+s-2 |
|
| |
|
Substituting into the given ODE yields
|
|
|
|
¥
å
k = 0
|
2 (k+s-1) (k+s) Ak xk+s - |
¥
å
k = 0
|
(k+s) Ak xk+s + |
| |
¥
å
k = 0
|
Ak xk+s + |
¥
å
k = 0
|
Ak xk+s+1 = 0 |
|
| |
|
and a change in the dummy index of the fourth sum readily makes
the x in all sums have the same power
|
|
|
|
¥
å
k = 0
|
2 (k+s-1) (k+s) Ak xk+s - |
¥
å
k = 0
|
(k+s) Ak xk+s + |
| |
¥
å
k = 0
|
Ak xk+s + |
¥
å
k = 1
|
Ak-1 xk+s = 0 |
|
| |
|
The above can also be written as
|
|
|
[2(s-1)s - s + 1]A0 xs + |
¥
å
k = 1
|
2 (k+s-1) (k+s) Ak xk+s |
| | - |
¥
å
k = 1
|
(k+s) Ak xk+s + |
¥
å
k = 1
|
Ak xk+s + |
¥
å
k = 1
|
Ak-1 xk+s = 0 |
|
| |
|
I.e.
|
|
|
[2(s-1)s - s + 1] A0 xs + |
| |
¥
å
k = 1
|
[{2(k+s-1) (k+s) - (k+s) + 1}Ak + Ak-1] xk+s = 0 |
|
| |
|
Collecting all terms who have as common factor the lowest power of x
(i.e. xs) yields the indicial equation
with roots s1 = 1 and s2 = 1/2.
And the recurrence relation is
|
|
| Ak = - |
1
2(s+k)2 - 3(s+k) + 1
|
Ak-1 |
|
| |
|
Finally the corresponding solutions are
|
|
| y1(x) = x [1 + |
¥
å
k = 1
|
|
(-1)k xk
[(2k+1)(2k-1) ... 5 ×3]k!
|
] |
|
| |
|
and
|
|
| y2(x) = x1/2 [1 + |
¥
å
k = 1
|
|
(-1)k xk
[(2k-1)(2k-3) ... 3 ×1]k!
|
] |
|
| |
|
Example: Apply Frobenius method to find the general solution of the equation:
|
|
| L y = x2 |
d2 y
dx2
|
+ (x2 + x) |
dy
dx
|
- y = 0 |
|
| |
|
Here the indicial equation is s2 - 1 = 0
with solutions s1 = +1, s2 = -1 and a
series solution is assured for s1.
The recurrence formula is
|
|
| (s+k-1)[(s+k+1)Ak + Ak-1] = 0 |
|
| |
|
and for s = s1 = 1 this is
so that the associated solution is
|
|
|
y1 = x [ A0 - |
A0
3
|
x + |
A0
3 ×4
|
x2 - |
A0
3 ×4 ×5
|
x3 + ...] = |
| |
| |
|
Now for s = s2 = -1 the recurrence formula becomes
which yields A1 = -A0 when k = 1, 0 = 0 when k = 2
(i.e. A2 can be anything we want, including zero) and
Ak = - Ak-1/k for k ³ 3. With this, the solution
associated with s = -1 becomes
8 Exceptional Cases
Consider first the exceptional case obtained when s1 = s2.
The series solution is
|
|
| y(x,s) = xs |
¥
å
k = 0
|
Ak(s) xk |
|
| |
|
A second solution is simply
|
|
| y2(x) = [ |
¶y(x,s)
¶s
|
]s = s1 |
|
| |
|
Next consider the exceptional case obtained when the exponents differ
by a positive integer K (i.e. s1 - s2 = K ³ 1). The
second solution here is
|
|
| y2(x) = [ |
¶
¶s
|
[(s-s2) y(x,s)]s = s2 |
|
| |
|
In summary, in all cases when the differential equation with x = 0
as a regular singular point with exponents s1 and s2 possesses
only one series solution of the form
|
|
| y1(x) = |
¥
å
k = 0
|
Ak xk+s1 = A0 u1(x) |
|
| |
|
any independent solution is of the form
|
|
| y2(x) = C u1(x) logx + |
¥
å
k = 0
|
Bk xk+s2 |
|
| |
|
9 A Particular Class of Equations
Consider the second order linear homogeneous differential equation
|
|
| (1+RM xM) |
d2 y
dx2
|
+ |
1
x
|
(P0 + PM xM) |
dy
dx
|
+ |
1
x2
|
(Q0 + QM xM) y = 0 |
|
| |
|
Special cases of this are Bessel's equation
|
|
| x2 |
d2 y
dx2
|
+ x |
d y
dx
|
+(x2 - p2) y = 0, |
|
| |
|
Legendre's equation
|
|
| (1-x2) |
d2 y
dx2
|
- 2x |
d y
dx
|
+p(p+1) y = 0, |
|
| |
|
Gauss equation
|
|
| x(1-x) |
d2 y
dx2
|
- [g- (a+ b+ 1)x] |
d y
dx
|
- aby = 0, |
|
| |
|
If for the general form
one assumes a series solution of the form
the indicial equation is
|
|
| f(s) = s2 + (P0 - 1) s + Q0 = 0 |
|
| |
|
while the recurrence formula is
for k = 1,2,...,M-1 and
for k ³ M.
Therefore for a solution one can write
|
|
| y(x) = |
¥
å
k = 0
|
Bk xMk+s |
|
| |
|
An exceptional case occurs only when s1 = s2 or when
s1 - s2 = K M.
10 Practice Problems
Obtain the solution of the following differential equation in terms of
Maclaurin series
Answer:
Let f(x) = ån = 0¥ An xn to obtain
|
|
| (2 A2 - A0) x0 + |
¥
å
n = 1
|
[(n+2)(n+1) An+2 + n An - An] xn = 0 |
|
| |
|
From the first coefficient A0 is an arbitrary constant;
then for n = 1 with A1 arbitrary constant, A3 = 0.
For n ³ 1
|
|
| An+2 = - |
(n-1)
(n+2)(n+1)
|
An |
|
| |
|
Must now consider two cases: n even and n odd. The general solution is
|
|
| f(x) = A1 x + A0 [1 + x2/2 - x4/24 + x6/240 + ...] |
|
| |
|
Use the method of Frobenius to obtain a general solution of the following
differential equation valid near x = 0
|
|
| x |
d2y
dx2
|
+ 2 |
dy
dx
|
+ x y = 0 |
|
| |
|
Answer:
Let y = f(x) = ån = 0¥ An xn+s. Thus
|
|
| x f(x) = x |
¥
å
n = 0
|
An xn+s = |
¥
å
n = 0
|
An xn+s+1 = |
¥
å
n = 1
|
An-1 xn+s |
|
| |
|
where instead of the dummy index n, the dummy n-1 was used.
Similarly
|
|
| 2 |
dy
dx
|
= 2 f(x)¢ = 2 |
¥
å
n = 0
|
(n+s) An xn+s-1 = 2 |
¥
å
n = -1
|
(n+1+s) An+1 xn+s |
|
| |
|
Where the dummy index has been changed from n to n+1 in order
to have x raised to the power n+s in the sum.
Finally,
|
|
| x |
d2 y
dx2
|
= x f(x)¢¢ = |
¥
å
n = 0
|
(n+s-1) (n+s) An xn+s-1 = |
¥
å
n = -1
|
(n+s) (n+1+s) An+1 xn+s |
|
| |
|
Where the dummy index has also been changed from n to n+1.
Now, substituting the above into the original ODE yields
|
|
| |
¥
å
n = -1
|
(n+s) (n+1+s) An+1 xn+s + |
¥
å
n = -1
|
2 (n+1+s) An+1 xn+s + |
¥
å
n = 1
|
An-1 xn+s = 0 |
|
| |
|
And division by xn+s yields the general recurrence relationship
|
|
| |
¥
å
n = -1
|
(n+1+s)(n+s) An+1 + |
¥
å
n = -1
|
2 (n+1+s) An+1 + |
¥
å
n = 1
|
An-1 = 0 |
|
| |
|
The indicial equation is now obtained from the above with n = -1 and it is
Assuming A0 is arbitrary but ¹ 0 one obtains
the roots s1 = 0 and s2 = -1. Since s1 - s2 = 1 we must
expect either no solution or a complete one for s2 = -1 and
of course, one solution for s1 = 0.
Now, when n = 0 the recurrence relation becomes
|
|
| (1+s) s A1 + 2 (1+s) A1 = A1 (1+s)(s+2) = 0 |
|
| |
|
which is always true when s = -1, regardless of the value of A1,
therefore A1 is also arbitrary
and we necessarily have a complete solution.
With n ³ 1
|
|
| An+1 = (-1) |
1
(n+1+s)(n+2+s)
|
An-1 |
|
| |
|
which yields, for s = -1,
|
|
| An+1 = (-1) |
1
n (n+1)
|
An-1 |
|
| |
|
Specifically, for n = 1,
for n = 2
for n = 3
|
|
| A4 = (-1) |
1
12
|
A2 = (-1) |
1
12
|
(-1) |
1
2
|
A0 |
|
| |
|
for n = 4
|
|
| A5 = (-1) |
1
20
|
A3 = (-1) |
1
20
|
(-1) |
1
6
|
A1 |
|
| |
|
The recurrence relationship can finally be represented by the
following two equations in more compact form
|
|
| A2l = (-1)l |
1
(2l)!
|
A0 , l ³ 0 |
|
| |
|
and
|
|
| A2l+1 = (-1)l |
1
(2l+1)!
|
A1 , l ³ 0 |
|
| |
|
Therefore, the general solution (corresponding to s = -1) is
|
|
|
y(x) = x-1 |
¥
å
n = 0
|
An xn = x-1 [ A0 + A1 x + A2 x2 + ...] = |
| |
= x-1 [A0 + A1 x - |
1
2
|
A0 x2 - |
1
6
|
A1 x3 + ..] = |
| |
= A0 [x-1 - |
1
2
|
x + |
1
24
|
x3 + ...] +A1 [1 - |
1
6
|
x2 + |
1
120
|
x4 + ...] = |
| | = A0 |
cos(x)
x
|
+ A1 |
sin(x)
x
|
|
|
| |
|
File translated from TEX by TTH, version 2.34.
On 6 Oct 2003, 17:15.