and The inequality (3.5) implies that The application of the analogue of the Gronwall’s inequality for piecewise continuous functions yields Since the equation is valid and (1 + NL$_{J}$ ) i (( a,b $^{))}$e NL$_{f}$ ( $^{b}$−a ) ≤ (1 + NL$_{J}$ ) e ( $^{ω}$−α )( $^{b}$−a ) for any real numbers a and b with a < b, one can confirm that 6 u ( t ) ≤ c$_{1}$ (1 + NL$_{J}$ ) e ( $^{ω}$−α )( $^{t}$−θ$_{k}$$_{0}$$_{−}$$_{1}$ ) + c$_{2}$e ωt + ∫ t θ$_{k}$$_{0}$$_{−}$$_{1}$ c$_{2}$NL$_{f}$ (1 + NL$_{J}$ ) e ( $^{ω}$−α )( $^{t}$−s $^{)}$e $^{ωs}$ds + ∑ θ$_{k}$$_{0}$$_{−}$$_{1}$<θ$_{k}$<t c$_{2}$NL$_{J}$ (1 + NL$_{J}$ ) e ( $^{ω}$−α )( $^{t}$−θ$_{k}$ $^{)}$e ωθ$_{k}$ ≤ c$_{1}$ (1 + NL$_{J}$ ) e ( $^{ω}$−α )( $^{t}$−θ$_{k}$$_{0}$$_{−}$$_{1}$ ) + c$_{2}$e ωt + c$_{2}$NL$_{f}$ (1 + NL$_{J}$ ) α e ωt ( 1 − e − α ( $^{t}$−θ$_{k}$$_{0}$$_{−}$$_{1}$ ) ) + c$_{2}$NL$_{J}$ (1 + NL$_{J}$ ) 1 − e − αθ e ωt 1 − $_{e}$− α ( $^{t}$−θ$_{k}$$_{0}$$_{−}$$_{1}$ + θ ) ) . 1 + ∫ t θ$_{k}$$_{0}$$_{−}$$_{1}$ NL$_{f}$ (1 + NL$_{J}$ ) i (( s,t $^{))}$e NL$_{f}$ ( $^{t}$−s $^{)}$ds + ∑ θ$_{k}$$_{0}$$_{−}$$_{1}$<θ$_{k}$<t NL$_{J}$ (1 + NL$_{J}$ ) i (( θ$_{k}$,t $^{))}$e NL$_{f}$ ( $^{t}$−θ$_{k}$ ) = (1 + NL$_{J}$ ) i (( θ$_{k}$$_{0}$$_{−}$$_{1}$,t $^{))}$e NL$_{f}$ ( $^{t}$−θ$_{k}$$_{0}$$_{−}$$_{1}$ ) u ( t ) ≤ h ( t ) + ∫ t θ$_{k}$$_{0}$$_{−}$$_{1}$ NL$_{f}$ (1 + NL$_{J}$ ) i (( s,t $^{))}$e NL$_{f}$ ( $^{t}$−s $^{)}$h ( s ) ds + ∑ θ$_{k}$$_{0}$$_{−}$$_{1}$<θ$_{k}$<t NL$_{J}$ (1 + NL$_{J}$ ) i (( θ$_{k}$,t $^{))}$e NL$_{f}$ ( $^{t}$−θ$_{k}$ $^{)}$h ( θ$_{k}$ ) . u ( t ) ≤ h ( t ) + ∫ t θ$_{k}$$_{0}$$_{−}$$_{1}$ NL$_{f}$u ( s ) ds + ∑ θ$_{k}$$_{0}$$_{−}$$_{1}$<θ$_{k}$<t NL$_{J}$u ( θ$_{k}$ ) . c$_{2}$ = Nϵ γ ( 1 ω + 1 1 − e − ωθ ) .