Kaprekar's Constant for two-digit numbers

Suppose we take a two-digit number a b, so the number actually is (10a+b). We also assume that a>b.
So, by the general subtraction process:

(10a+b) - (10b+a)
= 9(a-b)

let (a-b)=p, so the number becomes...

9(a-b)
= 9p
= 10p - 10 + 10 - p
= 10(p-1) + (10-p)
= (p-1) (10-p)
= (a-b-1) (10-a+b)

so, the possible cases are:

case I -
10-a+b = a
a-b-1 = b

which gives, 2a-b=10
a-2b=1

solving, we obtain a=19/3 which is not possible. So this case is discarded.

case II -
10-a+b = b
a-b-1 = a

which directly gives a=10 and b=-1, another absurd result. So, none of the cases satifies our pre-assumed existence of a constant. So, we can conclued that there does not exist any Kaprekar's Constant for two digit numbers.
Actually, if we work with any such two digit number, say 35, we will observe an interesting thing.

53 - 35 = 18
81 - 18 = 63
63 - 36 = 27
72 - 27 = 45
54 - 45 = 09
90 - 09 = 81
81 - 18 = 63
63 - 36 = 27
............
............

and thus it carries on...so basically we obtain a cycle of numbers, 63->27->45->9->81->63.....this occurs when there is no specific Kaprekar's Constant for that digit numbers.