β
h ( t , 0 )
∂ h ∂ t ( t , 0 )
− ∞ < β < 1
h ( t , 0 ) → 0
t → ∞
∂ h ∂ t ( t , 0 ) < 0
β = 1
h ( t , 0 ) = h ( 0 , 0 )
0 ≤ t ≤ ∞
∂ h ∂ t ( t , 0 ) = 0
1 < β < 5 3
h ( t , 0 ) → ∞
∂ h ∂ t ( t , 0 ) > 0
β = 5 3
Exponentially
5 3 < β ≤ 2
t → t 1