Math 2280 - Assignment 12
Dylan Zwick
Spring 2014
Section 8.4 - 1, 6, 8, 9, 14
Section 8.5 - 1, 5, 6, 13, 16
1
Section 8.4 - Method of Frobenius: The Exceptional
Cases
8.4.1 - Either apply the method from Example 1 in the textbook to find
two linearly independent Frobenius series solutions, or find one such
solution and show (as in Example 2 from the textbook) that a second
such solution does not exist for the differential equation:
xy ′′ + (3 − x)y ′ − y = 0.
Solution - We can rewrite the differential equation as:
y ′′ +
3−x ′ 1
y − y = 0.
x
x
1
3−x
and Q(x) = − are singular at x = 0. The funcx
x
tions p(x) = xP (x) = 3 − x and q(x) = x2 Q(x) = −x are analytic at
x = 0. So, x = 0 is a regular singular point.
Both P (x) =
The leading terms of p(x) and q(x) are p(0) = p0 = 3, q(0) = q0 = 0.
So, the indicial equation is:
r(r − 1) + 3r = r 2 + 2r = r(r + 2) = 0.
The roots of the indicial equation are r = 0, −2.
Plugging the Frobenius series and its derivatives into the ODE we
get:
∞
X
(n + r)(n + r − 1)cn xn+r−1 +
n=0
∞
X
3(n + r)cn xn+r−1 −
n=0
r)cn xn+r −
∞
X
n=0
2
cn xn+r = 0.
∞
X
n=0
(n +
The xn+r−1 coefficient is a multiple of the indicial equation, so it must
be 0 and c0 is arbitrary. Taking out this term and shifting the first two
series by 1 we get:
∞
X
[((n + r + 1)(n + 4) + 3(n + r + 1))cn+1 − (n + r + 1)cn ]xn+r = 0.
n=0
From this we get the recursion relation:
cn+1 =
(n + r + 1)cn
.
(n + r + 1)(n + r + 3)
cn+1 =
cn
(n + 1)cn
=
.
(n + 1)(n + 3)
n+1
For r = 0 we get:
The first few terms of this sequence are:
c0 = c0
c2 =
c1
c0
=
4
3·4
c1 =
c3 =
c0
3
c2
c0
=
,
5
3·4·5
and, in general,
cn =
So, we get one solution:
3
2c0
.
(n + 2)!
y1 (x) = c0
1+2
∞
X
n=1
!
xn
.
(n + 2)!
For r = −2 we get the recursion relation:
cn+1 =
(n − 1)cn
.
(n − 1)(n + 1)
For c0 , c1 all is good c0 = c0 , c1 = c0 . For c2 we get the relation:
c2 (1 − 1)(1 + 1) − (1 − 1)c1 = 0.
This is true for any c2 , and so c2 is arbitrary. Note this c2 corresponds
with the c0 in our r = 0 solution.
So, we get a second solution:
−2
y2 (x) = c0 x (1 + x) = c0
1
1
.
+
x2 x
So, our final solution (replacing the c0 term in our first solution by
c2 ) is:
y(x) = c0
1
1
+
2
x
x
+ c2
4
1+2
∞
X
n=1
!
xn
.
(n + 2)!
8.4.6 - Either apply the method from Example 1 in the textbook to find
two linearly independent Frobenius series solutions, or find one such
solution and show (as in Example 2 from the textbook) that a second
such solution does not exist for the differential equation:
2xy ′′ − (6 + 2x)y ′ + y = 0.
Solution - We can rewrite the differential equation as:
y ′′ −
6 + 2x ′
1
y + y = 0.
2x
2x
The functions
P (x) = −
6 + 2x
,
2x
Q(x) =
1
,
2x
are both singular at x = 0. The corresponding functions:
p(x) = xP (x) = −
6 + 2x
,
2
q(x) = x2 Q(x) =
x
,
2
are nonsingular at x = 0. Therefore, x = 0 is a regular singular point.
We have p(0) = p0 = −3, and q(0) = q0 = 0. So, the indicial equation
is:
r(r − 1) − 3r = r(r − 4) = 0.
5
The roots of this equation are r = 0, 4. Our Frobenius series solution,
if we plug it ino the ODE, will be:
∞
X
n+r−1
2cn (n + r)(n + r − 1)x
n=0
−
∞
X
2cn (n + r)xn+r +
n=0
−
∞
X
6(n + r)cn xn+r−1
n=0
∞
X
cn xn+r = 0.
n=0
The xr−1 coefficient gives us the indicial equation, so c0 is arbitrary,
and for the rest of the terms we get (after a shift):
∞
X
[(2n + 2r − 6)(n + r + 1)cn+1 − (2n + 2r − 1)cn ]xn+r = 0.
n=0
From this we get the recurrence relation:
cn+1 =
(2n + 2r − 1)cn
.
(2n + 2r − 6)(n + r + 1)
For r = 4 we have the relation
cn+1 =
(2n + 7)cn
.
(2n + 2)(n + 5)
From this we get the coefficients:
c0 = c0 ,
c2 =
c1 =
7c0
,
2·5
9 · 7c0
9c1
=
,
4·6
(6 · 4 · 2)(7 · 6 · 5)
6
and in general,
8
(2n + 5)!!
cn =
c0
.
5
2n n!(n + 4)!
So, one solution is:
y1 (x) = c0 x4
!
∞
8 X (2n + 5)!!xn
.
1+
5 n=1 2n n!(n + 4)!
For r = 0 our recursion relation is:
cn+1 =
(2n − 1)cn
.
(2n − 6)(n + 1)
For n = 3 we have a problem. Indeed, our relation is:
(6 − 6)(4)c4 − 5c3 = 0 ⇒ 0c4 − 5c3 = 0.
As we cannot assume c3 = 0 necessarily, this cannot work. So, for
r = 0 we do not get a second Frobenius series solution.
7
8.4.8 - Either apply the method from Example 1 in the textbook to find
two linearly independent Frobenius series solutions, or find one such
solution and show (as in Example 2 from the textbook) that a second
such solution does not exist for the differential equation:
x(1 − x)y ′′ − 3y ′ + 2y = 0.
Solution - We can rewrite the differential equation as:
y ′′ −
3
2
y′ +
y = 0.
x(1 − x)
x(1 − x)
Both of the functions:
P (x) = −
3
;
x(1 − x)
and
Q(x) =
2
x(1 − x)
are singular at x = 0. The corresponding functions:
p(x) = xP (x) = −
3
;
1−x
and
q(x) = x2 Q(x) =
2x
1−x
are nonsingular at x = 0. The initial values are p(0) = p0 = 3, and
q(0) = q0 = 0. So, the indicial equation is:
8
r(r − 1) − 3r = r(r − 4).
The roots of the indicial equation are r = 0, 4.
If we plug the Frobenius series solution into our ODE we get:
∞
X
n+r−1
(n + r)(n + r01)cn x
−
n=0
∞
X
(n + r)(n + r − 1)cn xn+r
n=0
−
∞
X
n+r−1
3(n + 1)cn x
+
∞
X
2cn xn+r = 0.
n=0
n=0
The xr−1 coefficient is a multiple of the indicial equation, so it’s automatically 0, and c0 is arbitrary. For the rest of the coefficients we get
the recurrence relation:
[(n + r + 1)(n + r) − 3(n + r + 1)]cn+1 − [(n + r)(n + r − 1) − 2]cn = 0
⇒ cn+1 =
[(n + r)(n + r − 1) − 2]
cn .
(n + r + 1)(n + r − 3)
For r = 4 we get the recurrence relation:
cn+1 =
n+2
n+1
cn .
The general formula from this currence relation is cn = (n + 1)c0 , and
the Frobenius series solution is:
y1 (x) = c0 x4
1+
∞
X
n=1
9
(n + 1)xn
!
=
c0 x4
.
(1 − x)2
For r = 0 we get the recurrence relation:
(n − 2)
cn .
(n − 3)
cn+1 =
The first few terms are:
c0 = c0 ,
2
c1 = c0 ,
3
1
1
c2 = c1 = c0 ,
2
3
c3 = 0c2 = 0.
At n = 3 we get:
(3 − 3)c4 + c3 = 0.
As c3 = 0 this is true for any c4 . The arbitrary c4 just gives us our
r = 4 solution. So, our second solution is:
y2 (x) = c0
2
1 2
1+ x+ x ,
3
3
and our general solution is (replacing c0 in the first solution with c4 ):
c4 x4
1 2
2
.
y(x) = c0 1 + x + x +
3
3
(1 − x)2
10
8.4.9 - For the differential equation
xy ′′ + y ′ − xy = 0,
first find the first four nonzero terms in a Frobenius series solution.
Then use the reduction of order technique to find the logarithmic
term and the first three nonzero terms in a second linearly independent solution.
Solution - We can rewrite our differential equation as:
1
y ′′ + y ′ − y = 0.
x
The functions:
P (x) =
1
,
x
and
Q(x) = −1
are not both analytic at x = 0. In particular, P (x) is not. The corresponding functions:
p(x) = xP (x) = 1,
and
q(x) = x2 Q(x) = −x2
are both analytic at x = 0. The leading terms are p(0) = p0 = 1, and
q(0) = q0 = 0. The corresponding indicial equation is:
11
r(r − 1) + r = r 2 = 0.
The roots of the indicial equation are r = 0, repeated twice. Plugging
the Frobenius series solution into the ODE we get:
∞
X
n+r−1
(n + r)(n + r − 1)cn x
∞
∞
X
X
n+r−1
cn xn+r+1 = 0.
(n + r)cn x
−
+
n=0
n=0
n=0
For r = 0, the x−1 term gives us the indicial equation. The x0 term is
(1 · (0)c1 + c1 )x0 = 0,
which gives us c1 = 0. For the higher-order terms we get:
cn+1 [(n + 2)(n + 1) + (n + 2)]xn+1 − cn xn+1 = 0
⇒ cn+2 =
cn
.
(n + 2)2
The first 4 non-zero terms are:
c0 = c0
c2 =
c0
4
c4 =
c2
c0
=
16
64
c6 =
c4
c0
=
.
36
2304
So,
x2 x4
x6
y1 (x) = c0 1 +
+
+
+··· .
4
64 2304
Now, for the next solution we have
12
Z
P (x)dx =
⇒ e−
R
Z
P (x)dx
1
dx = ln x
x
= e− ln x =
1
.
x
So,
y2 (x) = y1 (x)
Z
1
x
[y1 (x)]2
dx.
Now,
x2 x4
x6
x2 x4
x6
[y1 (x)] = 1 +
+
+
+···
1+
+
+
+···
4
64 2304
4
64 2304
2
= 1+
x2 3x4 5x6
+
+
+ ···.
2
32
576
Its inverse is:
1
x2 5x4 23x6
=1−
+
−
+ ···.
[y1 (x)]2
2
32
576
So,
Z 1 x 5x3 23x5
− +
−
+···
x 2
32
576
= ln x −
x2 5x4 23x6
+
−
+ ···.
4
128 3456
Therefore,
x2 5x4 23x6
y2 (x) = y1 (x) ln x −
+
−
+··· .
4
128 3456
13
8.4.14 - For the differential equation
x2 y ′′ + x(1 + x)y ′ − 4y = 0,
first find the first four nonzero terms in a Frobenius series solution.
Then use the reduction of order technique to find the logarithmic
term and the first three nonzero terms in a second linearly independent solution.
Solution - We can rewrite the differential equation as:
y ′′ +
1+x ′
4
y − 2 y = 0.
x
x
The functions
P (x) =
1+x
,
x
Q(x) = −
4
x2
are both singular at x = 0. The corresponding functions
p(x) = xP (x) = 1 + x,
q(x) = x2 Q(x) = −4,
are both nonsingular at x = 0. The initial values are p(0) = p0 = 1,
and q(0) = q0 = −4. The indicial equation is:
r(r − 1) + r − 4 = r 2 − 4 = (r − 2)(r + 2).
14
The roots of the indicial equation are r = ±2. Plugging the Frobenius
series into the ODE we get:
∞
X
(n + r)(n + r − 1)cn xn+r +
n=0
∞
X
(n + r)cn xn+r
n=0
∞
∞
X
X
n+r+1
+
(n + r)cn x
−4
cn xn+r = 0
n=0
n=0
This gives us the recurrence relation:
([(n + r + 1)(n + r) + (n + r + 1) − 4]cn+1 + (n + r)cn )xn+r+1 = 0
⇒ cn+1 = −
(n + r)
cn .
(n + r + 1)2 − 4
For r = 2 we get
(n + 2)
cn .
(n + 5)(n + 1)
cn+1 = −
The first few terms are:
c0 = c0 ,
2
c1 = − c0 ,
3
c2 = −
3
1
c1 = c0 ,
12
10
c3 = −
4
2
c2 = −
c0 .
21
105
From this we get:
2
1 2
2 3
y1 (x) = c0 x 1 − x + x −
x +··· .
5
10
105
2
For the next solution we have:
15
P (x) =
Z
1
1+x
= + 1,
x
x
P (x)dx = ln x + x,
e−
R
P (x)dx
=
e−x
.
x
Now,
4
9 2
62 3
[y1 (x)] = x 1 − x + x −
x +··· ,
5
25
525
7 2
142 3
1 + 45 x + 25
x + 2625
x +···
1
=
.
[y1 (x)]2
x4
2
4
Therefore,
e−
R
P (x)dx
=
y12
=
e−x 1 + 45 x +
7 2
x
25
x5
+
142 3
x
2625
+···
1
1
13
1
− 4−
+
+ ···.
5
3
x
5x
50x
1750x2
Integrating this we get
−
1
1
1
13
+
+
−
+ ···.
4x4 15x3 100x2 1750x
So,
1
1
1
13
y2 (x) = y1 (x) − 4 +
+
−
+··· .
4x
15x3 100x2 1750x
16
Section 8.5 - Bessel’s Equation
8.5.1 - Differentiate termwise the series for J0 (x) to show directly that
J0′ (x) = −J1 (x) (another analogy with the cosine and sine functions).
Solution - The series for J0 (x) is:
∞
X
(−1)m x 2m
J0 (x) =
.
2
(m!)
2
m=0
Its termwise derivative is:
J0′ (x)
∞
X
(−1)m 2m x2m−1
=
(m!)2
22m
m=1
∞
(−1)m x 2m−1 X (−1)m+1 x 2m+1
=
=
m!(m
−
1)!
2
m!(m + 1)! 2
m=0
m=1
∞
X
= −J1 (x).
The move from the third term to the fourth is just a shift of the series
by 1.
17
8.5.5 - Express J4 (x) in terms of J0 (x) and J1 (x).
Solution - This just requires a number of applications of the formula:
Jp+1 (x) =
2p
Jp (x) − Jp−1 (x).
x
We have:
6
6 4
2
J4 (x) = J3 (x) − J2 (x) =
J2 (x) − J1 (x) −
J1 (x) − J0 (x)
x
x x
x
2
6 4 2
J1 (x) − J0 (x) − J1 (x) −
J1 (x) − J0 (x)
=
x x x
x
24
48 8
J1 (x) −
−
− 1 J0 (x).
=
x3 x
x2
18
8.5.6 - Derive the recursion formula:
[(m + r)2 − p2 ]cm + cm−2 = 0
for Bessel’s equation.
Solution - Bessel’s equation of order p ≥ 0 is:
x2 y ′′ + xy ′ + (x2 − p2 )y = 0.
We can rewrite this equation as:
1
x2 − p2
y ′′ + y ′ +
y = 0.
x
x2
The coefficient functions:
P (x) =
Q(x) =
1
,
x
x2 − p2
x2
are both singular at x = 0. The corresponding functions:
p(x) = xP (x) = 1,
q(x) = x2 Q(x) = x2 − p2
are both nonsingular at x = 0. The initial values are:
19
p0 = p(0) = 1,
q0 = q(0) = −p2 .
The corresponding indicial equation is:
r(r − 1) + r − p2 = r 2 − p2 = (r − p)(r + p).
This indicial equation has roots r = ±p.
If we plug the Frobenius series solution y =
∞
X
cm xm+r into Bessel’s
m=0
equation of order p we get:
∞
X
(m + r)(m + r − 1)cm xm+r +
∞
X
(m + r)cm xm+r
m=0
m=0
+
∞
X
cm xm+r+2 −
m=0
∞
X
p2 cm xm+r = 0.
m=0
We can rewrite this as:
[r(r − 1) + r − p2 ]c0 xr + [(r + 1)r + (r + 1) − p2 ]c1 xr+1
∞
∞
X
X
(m + r)cm xm+r
(m + r)(m + r − 1)cm xm+r +
+
m=2
m=2
+
∞
X
cm xm+r+2 −
m=0
∞
X
p2 cm xm+r = 0.
m=2
The term c0 is multiplied by the indicial equation, and so is arbitrary,
while for r = ±p we must have c1 = 0 for the xr+1 coefficient to be 0.
For the remaining terms we can shift all the sums starting at m = 2
by 2 to get:
20
∞
X
[((m + r + 2)(m + r + 1) + (m + r + 2) − p2 )cm+2 + cm ]xm+p+2 = 0.
m=0
Simplifying this, and applying the identity principle, we get:
[(m + r + 2)2 − p2 ]cm+2 + cm = 0
for m ≥ 0. This is equivalent to :
[(m + r)2 − p2 ]cm + cm−2 = 0
for m ≥ 2. Which is what we wanted to derive.
21
Z
xm Jn (x)dx can be evaluated in terms
Z
of Bessel functions and the indefinite integral J0 (x)dx. The latter
Z x
integral cannot be simplified further, but the function
J0 (t)dt is
8.5.13 - Any integral of the form
0
tabulated in Table 11.1 of Abramowitz and Stegun. Use the identities:
d
ds dx
[xp Jp (x)] = xp Jp−1 (x);
d −p
[x Jp (x)] = −x−p Jp+1(x);
dx
to evaluate the integral
Z
x2 J0 (x)dx.
Solution - From the first identity we get:
Z
If we integrate
Z
Z
xJ0 (x)dx = xJ1 (x).
x2 J0 (x)dx by parts we get:
2
2
x J0 (x)dx = x J1 (x) −
Z
xJ1 (x)dx.
From the second identity we get:
Z
J1 (x)dx = −J0 (x).
22
Integrating
Z
xJ1 (x)dx by parts we get:
Z
xJ1 (x)dx = −xJ0 (x) +
Z
J0 (x)dx.
So, our final solution is:
Z
2
2
x J0 (x)dx = x J1 (x) + xJ0 (x) −
23
Z
J0 (x)dx + C.
8.5.16 - Same instructions as 8.5.13, only with the integral:
Z
xJ1 (x)dx.
Solution - We note the integral:
Z
Integrating
Z
J1 (x)dx = −J0 (x).
xJ1 (x)dx by parts we get:
Z
xJ1 (x)dx = −xJ0 (x) +
24
Z
J0 (x)dx + C.