and uniformly in t ∈ [0 , ∞ ) . Denote by L$_{1}$ = 4 K ( j ) ( 2 α − 1 1 − exp( − αθ/ 2) ) . Next, by using mathematical induction, we are going to show that z ( j ) s ( t, a ) , s = 1 , 2 , . . . , are defined for t ∈ [0 , ∞ ) and satisfy if L < L$_{1}$. Utilizing Lemma 4.2 and inequalities (4.40), (4.44), (4.45) and θ$_{i}$$_{+1}$ − θ$_{i}$ ≥ θ, i ∈ $^{Z}$, one can verify that As a consequence of (4.46), the sequence z ( j ) k $_{+1}$( t, a ) converges uniformly on t ∈ [0 , ∞ ) , | a | < δ/ 2 K ( j ) 1 , and Therefore, the limit function z ( j $^{)}$( t, a ) exists on the same domain, it is piecewise continuous, satisfies (4.41) and the following estimate Denote by z ( t ) = z ( j $^{)}$( t, a ) , for j = 1 , 2 , . . . , m. Next, we will verify that z ( j $^{)}$( t, a ) satisfies (4.35). For it, differentiate (4.41) Fix θ$_{k}$, k ∈ $^{Z}$, then | q$_{i}$ ( z$_{1}$ ) − q$_{i}$ ( z$_{2}$ ) | ≤ L | z$_{1}$ − z$_{2}$ | , (4.45) | z ( j ) s $_{+1}$( t, a ) − z ( j ) s ( t, a ) | ≤ K ( j ) 1 | a | exp( − αt/ 2) / 2 $^{s}$, s = 0 , 1 , 2 , . . . , (4.46) | z ( j ) k $_{+1}$( t, a ) − z ( j ) k ( t, a ) | ≤ K ( j ) 1 | a | L$_{1}$ exp( − αt/ 2) / (2 $^{k}$α ) . (4.47) | z ( j ) s ( t, a ) | ≤ 2 K ( j ) 1 | a | exp( − αt/ 2) , s = 1 , 2 , . . . . | z ( j $^{)}$( t, a ) | ≤ 2 K ( j ) 1 | a | exp( − αt/ 2) . (4.48) z ′ ( t ) = Y ′ $_{j}$( t ) a + G ( j ) 1 ( t, t ) r ( t, z ( t )) + G j $_{2}$( t, t ) r ( t, z ( t )) + t ∫ 0 G ( j ) 1 t ( t, s ) r ( s, z ( s )) ds − ∞ ∫ t G ( j ) 2 t ( t, s ) r ( s, z ( s )) ds + ∑ 0 <θ$_{k}$<t G ( j ) 1 t ( t, θ$_{k}$ +) q$_{k}$ ( z ( θ$_{k}$ )) − ∑ t<θ$_{k}$$^{<}$∞$^{G}$$^{(}$$^{j}$$^{)}$ 2 t ( t, θ$_{k}$ +) q$_{k}$ ( z ( θ$_{k}$ )) = A ( t ) Y$_{j}$ ( t ) a + G ( j $^{)}$( t, t ) r ( t, z ( t )) + ∞ ∫ 0 A ( t ) G ( j $^{)}$( t, s ) r ( s, z ( s )) ds + ∑ 0 <θ$_{i}$<t A ( t ) G ( j $^{)}$( t, θ$_{k}$ +) q$_{k}$ ( z ( θ$_{k}$ )) = A ( t ) z ( t ) + r ( t, z ( t )) . 25