68
1 Mathematical Physics
=
λ A λ B N
0
A
(λ B − λ A )
1
λ A
1
s
−
1
s + λ A
−
1
λ B
1
s
−
1
s + λ B
= N
0
1
1
s
−
λ B
(λ B − λ A )
1
(s + λ A )
+
λ A
(λ B − λ A )
1
(s + λ B )
∴ N c = N
0
A
1 +
1
λ B − λ A
(λ A exp (−λ B t) − λ B exp (−λ A t))
1.74 (a) L{e
ax
} =
∞
0
e
−sx e
ax dx =
∞
0
e
−(s−a)x dx
=
1
s − a
, if s > a
(b) and (c). From part (a), L(e
ax ) =
1
s−a
Replace a by ai
L(e
iax ) = L{cos ax + i sin ax}
= L{cos ax} + iL{sin ax}
=
1
s − ai
=
s + ai
s 2 + a 2 =
s
s 2 + a 2 +
ia
s 2 + a 2
Equating real and imaginary parts:
L{cos ax} =
s
s 2 + a 2 ; L{sin ax} =
a
x 2 + a 2
1.3.10 Special Functions
1.75 Express H n in terms of a generating function T (ξ, s).
T (ξ, s) = exp[ξ
2
− (s − ξ )
2 ] = exp[−s
2
+ 2sξ ]
=
∞
n=0
H n (ξ )s
n
n!
(1)
Differentiate (1) first with respect to ξ and then with respect to s.
∂ T
∂ξ
= 2s exp(−s
2
+ 2sξ ) =
n
2s
n+1 H n (ξ )
n!
=
n
s
n H
n (ξ )
n!
(2)
Equating equal powers of s
H
n = 2n H n−1
(3)
∂ T
∂s
= ξ (−2s + 2ξ ) exp(−s
2
+2sξ ) =
n
(−2s +2ξ )s
n H n (ξ ) =
n
s
n−1 H n (ξ )
(n − 1)!
(4)
Equating equal powers of s in the sums of equations
H n+1 = 2ξ H n − 2n H n−1
(5)
It is seen that (5) satisfies the Hermite’s equation
1 Mathematical Physics
=
λ A λ B N
0
A
(λ B − λ A )
1
λ A
1
s
−
1
s + λ A
−
1
λ B
1
s
−
1
s + λ B
= N
0
1
1
s
−
λ B
(λ B − λ A )
1
(s + λ A )
+
λ A
(λ B − λ A )
1
(s + λ B )
∴ N c = N
0
A
1 +
1
λ B − λ A
(λ A exp (−λ B t) − λ B exp (−λ A t))
1.74 (a) L{e
ax
} =
∞
0
e
−sx e
ax dx =
∞
0
e
−(s−a)x dx
=
1
s − a
, if s > a
(b) and (c). From part (a), L(e
ax ) =
1
s−a
Replace a by ai
L(e
iax ) = L{cos ax + i sin ax}
= L{cos ax} + iL{sin ax}
=
1
s − ai
=
s + ai
s 2 + a 2 =
s
s 2 + a 2 +
ia
s 2 + a 2
Equating real and imaginary parts:
L{cos ax} =
s
s 2 + a 2 ; L{sin ax} =
a
x 2 + a 2
1.3.10 Special Functions
1.75 Express H n in terms of a generating function T (ξ, s).
T (ξ, s) = exp[ξ
2
− (s − ξ )
2 ] = exp[−s
2
+ 2sξ ]
=
∞
n=0
H n (ξ )s
n
n!
(1)
Differentiate (1) first with respect to ξ and then with respect to s.
∂ T
∂ξ
= 2s exp(−s
2
+ 2sξ ) =
n
2s
n+1 H n (ξ )
n!
=
n
s
n H
n (ξ )
n!
(2)
Equating equal powers of s
H
n = 2n H n−1
(3)
∂ T
∂s
= ξ (−2s + 2ξ ) exp(−s
2
+2sξ ) =
n
(−2s +2ξ )s
n H n (ξ ) =
n
s
n−1 H n (ξ )
(n − 1)!
(4)
Equating equal powers of s in the sums of equations
H n+1 = 2ξ H n − 2n H n−1
(5)
It is seen that (5) satisfies the Hermite’s equation
