satisfied where and Now, we are ready to prove the following result. Lemma 2.1 If conditions ( C 1) − ( C 6) hold, then for each x ( t ) ∈ A$_{x}$ there exists a unique solution φ β x ( t $_{)}$( t ) of (2.9) such that sup t ∈ R ∥ φ β x ( t $_{)}$( t ) ∥ ≤ r$_{1}$ for a constant r$_{1}$ = 4 KP$_{2}$ α M$_{h}$ 1 + √ 1 − 16 K $^{2}$P 2 2 α 2 M$_{h}$ . Proof. Consider the set C$_{0}$ ( R ) of continuous functions y ( t ) satisfying ‖ y ‖$_{0}$ ≤ r$_{1}$, where ‖ y ‖$_{0}$ = sup t ∈ R ‖ y ( t ) ‖ . Define the operator Π on C$_{0}$ as Using the mean value theorem we obtain that Hence, the inequality is valid so that Let 6 H$_{2}$ ( y$_{1}$, y$_{2}$, y$_{3}$, s ) = f ( y$_{1}$ + y$_{2}$, s ) − f ( y$_{3}$ + y$_{2}$, s ) − D$_{u}$f ( y$_{2}$, s )( y$_{1}$ − y$_{3}$ ) . ‖ Π y ‖$_{0}$ ≤ 2 K ( P$_{2}$ ‖ y ‖ 2 0 + M$_{h}$ ) α . ‖ Π y ( t ) ‖ ≤ 2 K ( P$_{2}$ ‖ y ‖ 2 0 + M$_{h}$ ) α ‖ H$_{1}$ ( y ( s ) , ν$_{β}$ ( s ) , s ) ‖ ≤ ∫ 1 0 ‖ D$_{u}$f ( θy$_{1}$ + y$_{2}$, s ) − D$_{u}$f ( y$_{2}$, s ) ‖ dθ ‖ y$_{1}$ ‖ ≤ ∫ 1 0 ∫ 1 0 ‖ D$_{uu}$f ( τθy$_{1}$ + y$_{2}$, s ) ‖ dτdθ ‖ y$_{1}$ ‖ 2 ≤ P$_{2}$ ‖ y$_{1}$ ‖ $^{2}$. Π y ( t ) = ∞ ∫ −∞ G$_{β}$ ( t, s )( H$_{1}$ ( y ( s ) , ν$_{β}$ ( s ) , s ) + h ( x ( s ))) ds. H$_{1}$ ( y$_{1}$, y$_{2}$, s ) = f ( y$_{1}$ + y$_{2}$, s ) − f ( y$_{2}$, s ) − D$_{u}$f ( y$_{2}$, s ) y$_{1}$. G$_{β}$ ( t, s ) =     V$_{β}$ ( t ) P$_{β}$V$_{β}$ ( s ) − 1 t ≥ s, V$_{β}$ ( t )( I − P$_{β}$ ) V$_{β}$ ( s ) − 1 t ≤ s, y ( t ) = ∞ ∫ −∞ G$_{β}$ ( t, s )( H$_{1}$ ( y ( s ) , ν$_{β}$ ( s ) , s ) + h ( x ( s ))) ds, (2.10)