In 1801 Leonhard Euler gave the solution to the hypergeometric differential equation. It is closely related to the Gaussian hypergeometric function , which was first systematically investigated by Carl Friedrich Gauß .
Hypergeometric differential equation
The hypergeometric function , where the gamma function denotes, satisfies the linear differential equation of the 2nd order :
2
F.
1
(
a
,
b
;
c
;
z
)
=
∑
k
=
0
∞
Γ
(
a
+
k
)
Γ
(
b
+
k
)
Γ
(
c
)
Γ
(
a
)
Γ
(
b
)
Γ
(
c
+
k
)
z
k
k
!
{\ displaystyle \ textstyle {} _ {2} F_ {1} (a, b; c; z) = \ sum _ {k = 0} ^ {\ infty} {\ frac {\ Gamma (a + k) \ , \ Gamma (b + k) \, \ Gamma (c)} {\ Gamma (a) \, \ Gamma (b) \, \ Gamma (c + k)}} {\ frac {z ^ {k}} {k!}}}
Γ
(
⋅
)
{\ displaystyle \ Gamma (\ cdot)}
z
(
1
-
z
)
d
2
d
z
2
2
F.
1
(
a
,
b
;
c
;
z
)
+
[
c
-
(
a
+
b
+
1
)
z
]
d
d
z
2
F.
1
(
a
,
b
;
c
;
z
)
-
a
b
2
F.
1
(
a
,
b
;
c
;
z
)
=
0
{\ displaystyle z (1-z) {\ frac {\ rm {d ^ {2}}} {{\ rm {d}} z ^ {2}}} \; {} _ {2} F_ {1} (a, b; c; z) + \ left [c- (a + b + 1) z \ right] {\ frac {\ rm {d}} {{\ rm {d}} z}} \; { } _ {2} F_ {1} (a, b; c; z) -ab \; {} _ {2} F_ {1} (a, b; c; z) = 0}
.
Singularities
The differential equation of the 2nd order has three liftable singularities , the values of which are determined in the following.
Based on the hypergeometric differential equation in the illustration
d
2
d
z
2
2
F.
1
(
a
,
b
;
c
;
z
)
+
p
(
z
)
d
d
z
2
F.
1
(
a
,
b
;
c
;
z
)
-
q
(
z
)
2
F.
1
(
a
,
b
;
c
;
z
)
=
0
{\ displaystyle {\ frac {\ rm {d ^ {2}}} {{\ rm {d}} z ^ {2}}} \; {} _ {2} F_ {1} (a, b; c ; z) + p (z) {\ frac {\ rm {d}} {{\ rm {d}} z}} \; {} _ {2} F_ {1} (a, b; c; z) -q (z) \; {} _ {2} F_ {1} (a, b; c; z) = 0}
With
p
(
z
)
=
c
-
(
a
+
b
+
1
)
z
z
(
1
-
z
)
=
c
-
c
z
+
(
c
-
a
-
b
-
1
)
z
z
(
1
-
z
)
=
c
z
+
c
-
a
-
b
-
1
1
-
z
{\ displaystyle p (z) = {\ frac {c- (a + b + 1) z} {z (1-z)}} = {\ frac {c-cz + (cab-1) z} {z ( 1-z)}} = {\ frac {c} {z}} + {\ frac {cab-1} {1-z}}}
and
q
(
z
)
=
a
b
z
(
1
-
z
)
{\ displaystyle q (z) = {\ frac {ab} {z (1-z)}}}
we get the two liftable singularities at and .
z
=
0
{\ displaystyle z = 0}
z
=
1
{\ displaystyle z = 1}
The third liftable singularity is obtained through the substitution . First, the derivatives of the hypergeometric function are substituted as follows:
t
=
1
z
,
d
t
d
z
=
-
1
z
2
=
-
t
2
{\ displaystyle \ textstyle t = {\ frac {1} {z}}, {\ frac {{\ rm {d}} t} {{\ rm {d}} z}} = {\ frac {-1} {z ^ {2}}} = - t ^ {2}}
d
d
z
2
F.
1
(
a
,
b
;
c
;
z
)
=
d
d
t
2
F.
1
(
a
,
b
;
c
;
t
)
⋅
d
t
d
z
=
-
t
2
⋅
d
d
t
2
F.
1
(
a
,
b
;
c
;
t
)
{\ displaystyle {\ frac {\ rm {d}} {{\ rm {d}} z}} \; {} _ {2} F_ {1} (a, b; c; z) = {\ frac { \ rm {d}} {{\ rm {d}} t}} \; {} _ {2} F_ {1} (a, b; c; t) \ cdot {\ frac {{\ rm {d} } t} {{\ rm {d}} z}} = - t ^ {2} \ cdot {\ frac {\ rm {d}} {{\ rm {d}} t}} \; {} _ { 2} F_ {1} (a, b; c; t)}
and
d
2
d
z
2
2
F.
1
(
a
,
b
;
c
;
z
)
=
d
d
t
(
-
t
2
⋅
d
d
t
2
F.
1
(
a
,
b
;
c
;
t
)
)
⋅
d
t
d
z
=
-
t
2
(
-
2
t
⋅
d
d
t
2
F.
1
(
a
,
b
;
c
;
t
)
-
t
2
⋅
d
2
d
t
2
2
F.
1
(
a
,
b
;
c
;
t
)
)
=
t
4th
⋅
d
2
d
t
2
2
F.
1
(
a
,
b
;
c
;
t
)
+
2
t
3
⋅
d
d
t
2
F.
1
(
a
,
b
;
c
;
t
)
{\ displaystyle {\ begin {aligned} {\ frac {\ rm {d ^ {2}}} {{\ rm {d}} z ^ {2}}} \; {} _ {2} F_ {1} (a, b; c; z) & = {\ frac {\ rm {d}} {{\ rm {d}} t}} {\ Big (} -t ^ {2} \ cdot {\ frac {\ rm {d}} {{\ rm {d}} t}} \; {} _ {2} F_ {1} (a, b; c; t) {\ Big)} \ cdot {\ frac {{\ rm {d}} t} {{\ rm {d}} z}} \\ & = - t ^ {2} {\ Big (} -2t \ cdot {\ frac {\ rm {d}} {{\ rm {d}} t}} \; {} _ {2} F_ {1} (a, b; c; t) -t ^ {2} \ cdot {\ frac {\ rm {d ^ {2}} } {{\ rm {d}} t ^ {2}}} \; {} _ {2} F_ {1} (a, b; c; t) {\ Big)} \\ & = t ^ {4 } \ cdot {\ frac {\ rm {d ^ {2}}} {{\ rm {d}} t ^ {2}}} \; {} _ {2} F_ {1} (a, b; c ; t) + 2t ^ {3} \ cdot {\ frac {\ rm {d}} {{\ rm {d}} t}} \; {} _ {2} F_ {1} (a, b; c ; t) \ end {aligned}}}
Thus the hypergeometric differential equation, after division by , takes on the following form:
t
4th
{\ displaystyle t ^ {4}}
d
2
d
t
2
2
F.
1
(
a
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b
;
c
;
t
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+
p
~
(
t
)
⋅
d
d
t
2
F.
1
(
a
,
b
;
c
;
t
)
-
q
~
(
t
)
2
F.
1
(
a
,
b
;
c
;
t
)
=
0
{\ displaystyle {\ frac {\ rm {d ^ {2}}} {{\ rm {d}} t ^ {2}}} \; {} _ {2} F_ {1} (a, b; c ; t) + {\ tilde {p}} (t) \ cdot {\ frac {\ rm {d}} {{\ rm {d}} t}} \; {} _ {2} F_ {1} ( a, b; c; t) - {\ tilde {q}} (t) \; {} _ {2} F_ {1} (a, b; c; t) = 0}
With
p
~
(
t
)
=
2
t
+
1
t
2
p
(
z
=
1
t
)
=
2
t
+
1
t
2
(
c
t
+
c
-
a
-
b
-
1
1
-
1
t
)
=
c
+
2
t
+
c
-
a
-
b
-
1
t
(
t
-
1
)
{\ displaystyle {\ tilde {p}} (t) = {\ frac {2} {t}} + {\ frac {1} {t ^ {2}}} p (z = {\ tfrac {1} { t}}) = {\ frac {2} {t}} + {\ frac {1} {t ^ {2}}} {\ Big (} ct + {\ frac {cab-1} {1 - {\ frac {1} {t}}}} {\ Big)} = {\ frac {c + 2} {t}} + {\ frac {cab-1} {t (t-1)}}}
and
q
~
(
t
)
=
1
t
4th
q
(
z
=
1
t
)
=
1
t
4th
a
b
1
t
(
1
-
1
t
)
=
a
b
t
2
(
t
-
1
)
{\ displaystyle {\ tilde {q}} (t) = {\ frac {1} {t ^ {4}}} q (z = {\ tfrac {1} {t}}) = {\ frac {1} {t ^ {4}}} {\ frac {ab} {{\ frac {1} {t}} (1 - {\ frac {1} {t}})}} = {\ frac {ab} {t ^ {2} (t-1)}}}
Accordingly, the hypergeometric differential equation also has a lifting singularity at.
z
=
1
t
=
∞
{\ displaystyle z = {\ tfrac {1} {t}} = \ infty}
Solution of the hypergeometric differential equation
With the power series approach with complex coefficients , the hypergeometric differential equation reads:
u
(
z
)
=
∑
k
=
0
∞
u
k
z
k
{\ displaystyle \ textstyle u (z) = \ sum _ {k = 0} ^ {\ infty} u_ {k} z ^ {k}}
u
k
{\ displaystyle u_ {k}}
z
(
1
-
z
)
d
2
d
z
2
∑
k
=
0
∞
u
k
z
k
+
[
c
-
(
a
+
b
+
1
)
z
]
d
d
z
∑
k
=
0
∞
u
k
z
k
-
a
b
∑
k
=
0
∞
u
k
z
k
=
0
{\ displaystyle z (1-z) {\ frac {\ rm {d ^ {2}}} {{\ rm {d}} z ^ {2}}} \ sum _ {k = 0} ^ {\ infty } u_ {k} z ^ {k} + \ left [c- (a + b + 1) z \ right] {\ frac {\ rm {d}} {{\ rm {d}} z}} \ sum _ {k = 0} ^ {\ infty} u_ {k} z ^ {k} -ab \ sum _ {k = 0} ^ {\ infty} u_ {k} z ^ {k} = 0}
.
After executing the derivations, the representation results
z
(
1
-
z
)
∑
k
=
2
∞
k
(
k
-
1
)
u
k
z
k
-
2
+
[
c
-
(
a
+
b
+
1
)
z
]
∑
k
=
1
∞
k
u
k
z
k
-
1
-
a
b
∑
k
=
0
∞
u
k
z
k
=
0
{\ displaystyle z (1-z) \ sum _ {k = 2} ^ {\ infty} k (k-1) u_ {k} z ^ {k-2} + \ left [c- (a + b + 1) z \ right] \ sum _ {k = 1} ^ {\ infty} ku_ {k} z ^ {k-1} -ab \ sum _ {k = 0} ^ {\ infty} u_ {k} z ^ {k} = 0}
.
Summarizing the powers of leads to
z
{\ displaystyle z}
∑
k
=
2
∞
k
(
k
-
1
)
u
k
z
k
-
1
-
∑
k
=
2
∞
k
(
k
-
1
)
u
k
z
k
+
c
∑
k
=
1
∞
k
u
k
z
k
-
1
-
(
a
+
b
+
1
)
∑
k
=
1
∞
k
u
k
z
k
-
a
b
∑
k
=
0
∞
u
k
z
k
=
0
{\ displaystyle \ sum _ {k = 2} ^ {\ infty} k (k-1) u_ {k} z ^ {k-1} - \ sum _ {k = 2} ^ {\ infty} k (k -1) u_ {k} z ^ {k} + c \ sum _ {k = 1} ^ {\ infty} ku_ {k} z ^ {k-1} - (a + b + 1) \ sum _ { k = 1} ^ {\ infty} ku_ {k} z ^ {k} -ab \ sum _ {k = 0} ^ {\ infty} u_ {k} z ^ {k} = 0}
.
The index shift results in
∑
k
=
0
∞
(
k
+
1
)
k
u
k
+
1
z
k
-
∑
k
=
0
∞
k
(
k
-
1
)
u
k
z
k
+
c
∑
k
=
0
∞
(
k
+
1
)
u
k
+
1
z
k
-
(
a
+
b
+
1
)
∑
k
=
0
∞
k
u
k
z
k
-
a
b
∑
k
=
0
∞
u
k
z
k
=
0
{\ displaystyle \ sum _ {k = 0} ^ {\ infty} (k + 1) ku_ {k + 1} z ^ {k} - \ sum _ {k = 0} ^ {\ infty} k (k- 1) u_ {k} z ^ {k} + c \ sum _ {k = 0} ^ {\ infty} (k + 1) u_ {k + 1} z ^ {k} - (a + b + 1) \ sum _ {k = 0} ^ {\ infty} ku_ {k} z ^ {k} -ab \ sum _ {k = 0} ^ {\ infty} u_ {k} z ^ {k} = 0}
.
This equation is obviously true if:
(
k
+
1
)
k
u
k
+
1
-
k
(
k
-
1
)
u
k
+
c
(
k
+
1
)
u
k
+
1
-
(
a
+
b
+
1
)
k
u
k
-
a
b
u
k
=
0
{\ displaystyle (k + 1) ku_ {k + 1} -k (k-1) u_ {k} + c (k + 1) u_ {k + 1} - (a + b + 1) ku_ {k} -abu_ {k} = 0}
.
The following recursion has thus been found for the coefficient :
u
k
{\ displaystyle u_ {k}}
u
k
+
1
=
k
(
k
-
1
)
+
(
a
+
b
+
1
)
k
+
a
b
(
k
+
1
)
k
+
c
(
k
+
1
)
u
k
=
k
2
-
k
+
k
a
+
k
b
+
k
+
a
b
(
c
+
k
)
(
1
+
k
)
u
k
=
k
2
+
k
a
+
k
b
+
a
b
(
c
+
k
)
(
1
+
k
)
u
k
=
(
a
+
k
)
(
b
+
k
)
(
c
+
k
)
(
1
+
k
)
u
k
=
(
a
,
k
)
(
b
,
k
)
(
c
,
k
)
(
1
,
k
)
u
k
{\ displaystyle {\ begin {aligned} u_ {k + 1} & = {\ frac {k (k-1) + (a + b + 1) k + ab} {(k + 1) k + c (k +1)}} u_ {k} \ qquad = {\ frac {k ^ {2} -k + ka + kb + k + ab} {(c + k) (1 + k)}} u_ {k} \ \ & = {\ frac {k ^ {2} + ka + kb + ab} {(c + k) (1 + k)}} u_ {k} \ qquad \ qquad \ qquad \; \; = {\ frac {(a + k) (b + k)} {(c + k) (1 + k)}} u_ {k} & = {\ frac {(a, k) (b, k)} {(c, k) (1, k)}} u_ {k} \ end {aligned}}}
Here the Pochhammer symbol denotes .
(
x
,
n
)
≡
Γ
(
x
+
n
)
Γ
(
x
)
{\ displaystyle (x, n) \ equiv {\ tfrac {\ Gamma (x + n)} {\ Gamma (x)}}}
If the initial value is set, the first basic solution of the hypergeometric differential equation is:
u
0
=
1
{\ displaystyle u_ {0} = 1}
u
(
z
)
=
2
F.
1
(
a
,
b
;
c
;
z
)
=
∑
k
=
0
∞
(
a
,
k
)
(
b
,
k
)
(
c
,
k
)
(
1
,
k
)
z
k
=
∑
k
=
0
∞
Γ
(
a
+
k
)
Γ
(
b
+
k
)
Γ
(
c
)
Γ
(
a
)
Γ
(
b
)
Γ
(
c
+
k
)
z
k
k
!
{\ displaystyle u (z) = {} _ {2} F_ {1} (a, b; c; z) = \ sum _ {k = 0} ^ {\ infty} {\ frac {(a, k) (b, k)} {(c, k) (1, k)}} z ^ {k} = \ sum _ {k = 0} ^ {\ infty} {\ frac {\ Gamma (a + k) \ , \ Gamma (b + k) \, \ Gamma (c)} {\ Gamma (a) \, \ Gamma (b) \, \ Gamma (c + k)}} {\ frac {z ^ {k}} {k!}}}
.
For is obtained as a second linearly independent basic solution
c
∉
Z
{\ displaystyle c \ notin \ mathbb {Z}}
v
(
z
)
=
z
1
-
c
2
F.
1
(
a
-
c
+
1
,
b
-
c
+
1
;
2
-
c
;
z
)
{\ displaystyle v (z) = z ^ {1-c} {} _ {2} F_ {1} (a-c + 1, b-c + 1; 2-c; z)}
Both together span the entire solution space of the hypergeometric differential equation:
y
(
z
)
=
C.
1
u
(
z
)
+
C.
2
v
(
z
)
{\ displaystyle y (z) = C_ {1} u (z) + C_ {2} v (z)}
With
C.
1
,
C.
2
∈
C.
{\ displaystyle C_ {1}, C_ {2} \ in \ mathbb {C}}
See also
literature
Individual evidence
^ Leonhard Euler: Transformationis Singularis, Nova Acta Academiae Scientiarum Imperialis Petropolitanae, Volume 12, 1801, Pages 58-70, online at books.google.de
↑ Erwin Kreyszig: Advanced Engineering Mathematics, John Wiley & Sons 1988, page 204f.
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