Physics 507 : Problem set 9
Due : April 13, 2015
Please work together in a group of two or three. Please turn in your solutions separately with all the
work explicitly written.
1
Problem 1
Let us consider the differential equation y 00 (z) + P (z)y 0 (z) + Q(z)y(z) = 0. The differential equation
has three regular singular points, ξ, η, ζ, which is reflected in
B
C
1
D
E
F
A
+
+
, Q(z) =
+
+
.
P (z) =
z−ξ
z−η z−ζ
(z − ξ)(z − η)(z − ζ) z − ξ
z−η z−ζ
By changing a variable from z to x as z = x1 , verify that the ∞ is a regular point if A + B + C = 2
and D + E + F is finite.
Sol: The change of variable from z to x changes the differential equation to
Q(x)
2x3 − x2 P (x)
ẏ(x) + 4 y(x) = 0.
x4
x
We can examine the coefficients of ẏ(x) and y(x)
2x3 − x2 P (x)
1
B
C
A
P̃ =
=
+
+
2−
,
x4
x
1 − ξx 1 − ηx 1 − ζx
1
E
F
Q(x)
D
=
+
+
Q̃ =
.
x4
(1 − ξx)(1 − ηx)(1 − ζx) 1 − ξx 1 − ηx 1 − ζx
ÿ(x) +
To be regular at x = 0, P̃ and Q̃ need to be regular. This can be achieved by imposing A + B + C = 2
and D + E + F = finite.
2
Problem 2
As we discussed in the class, a general solution for the Hypergeometric equation is given by the integral
(a, b are not integer)
I
dz z a (1 − z)b
I=
C
with an appropriate contour enclosing both the branch points z = 0, 1 twice (shown in class). Rewrite
the integral in terms of Beta functions with appropriate phases using the notation z − 1 = r1 eiθ1 , z =
r2 eiθ2 . One can deform the contour close to the branch cut, then r1 + r2 = 1, and thus r2 can be
expressed 1 − r1 . (With this we can write the full solution for the Hypergeometric equation in terms
of infinite series.)
Sol: We can take the contour as in the figure. We assign z − 1 = r1 eiθ1 , z = r2 eiθ2 , then z a (1 − z)b =
r1b r2a eibθ1 +iaθ2 . The contour can be deformed to be close to x− axis along the branch cut. Then the
small circle circling around the branch points cancel out and the four segments, two above and two
below the branch cut, are left. Their integrals are determined by the arguments of these four segments
depending on θ1 , θ2 . Their values are
1) θ1 = π, θ2 = 0 for the segment from z = 1 → z = 0 above the branch cut,
2) θ1 = π, θ2 = 2π for the segment from z = 0 → z = 1 below the branch cut,
3) θ1 = −π, θ2 = 2π for the segment from z = 1 → z = 0 below the branch cut,
4) θ1 = −π, θ2 = 0 for the segment from z = 0 → z = 1 above the branch cut.
R1
Thus the integral is (one can use the notation B(a + 1, b + 1) = 0 dz ra (1 − r)b )
I
Z 1
I=
dz z a (1 − z)b = −eiπb + eiπ(b+2a) − eiπ(−b+2a) + e−iπb
dz ra (1 − r)b .
C
0
3
Problem 3
We would like to write the solution of the Associated Legendre equation
m2
2 00
0
(1 − x )y − 2xy + n(n − 1) −
y=0
1 − x2
in terms of the Riemann P-symbol we discussed in the class.
a) Discuss the singularities of the differential equation.
b) Solve the indicial equations for each regular singular points. Write down the solution using the
Riemann P-symbol.
c) Use the variable z = 1−x
2 to change the location of the singularities.
d) Extract out the factor (1 − x)m/2 (1 + x)m/2 so that the Riemann P-symbol looks familiar to the
standard form of the Hypergeometric solution discussed in the class.
e) Finally, write the solution in the form 2 F1 .
Sol: a) The singularities, x = ±1, ∞, are all regular singular. This can be checked as in problem 1.
b) Let us try to solve the indicial equation
x = ∞ as the other point were done in the class. We
Pat
∞
can try to the following solution, y = x−γ k=0 ck x−k . The indicial equation gives the relation
(γ + k − n)(γ + k + n − 1) = 0 .
Thus we have γ = n, −n + 1. Along with the information we got from the class, we have the following
Riemann P solution.


−1
∞

 1
m
m
n
x
.
y=P
2

 2m
− 2 −m
−n
+
1
2
c) We can change the variable from x to z, we change the locations of the singularities to 0, 1, ∞. Thus
we get


1
∞
 0

m
m
1−x
n
y=P
.
2
2
2
 m

− 2 −m
−n
+
1
2
c) By extracting out the factor (1 − x)m/2 (1 + x)m/2 , we get

1
∞
 0
0
0
m+n
y = (1 − x)m/2 (1 + x)m/2 P

−m −m m − n + 1
1−x
2


.

e) Thus the solution has the standard form. We identify a = m + n, b = m − n + 1, c = 1 + m. Thus
we have
y(x) = c(1 − x2 )m/2 2 F1 (m + n, m − n + 1; m + 1;
4
1−x
).
2
Problem 4
Consider the Bessel function
1 0
n2
y + y + 1− 2 y =0.
x
x
00
By changing a variable y = xn eix F (x) followed by z = −2ix, rewrite the equation to the form of
Confluent Hypergeometric equation. Write the solution in the standard form with 1 F1 .
Sol: This is straight forward problem. By plugging in the expression, we get
xF 00 + (2n + 1 + 2ix)F 0 + i(2n + 1)F = 0 .
In terms of z, we get
xF̈ + (2n + 1 − z)Ḟ − (n + 1/2)F = 0 .
Thus we have the expressions
Jn (x) = c
x n
2
eix 1 F1 (n + 1/2, 2n + 1, −2ix) .