“…Therefore we have that y m ⇀ y in L 2 (0, T ); H 1 0 (Ω) , y m ⇀ y in C [0, T ]; L 2 (Ω) . Now proceeding as in [23], one can prove by interpretation that the solution y is equivalent to the initial problem (1.1). Furthermore, setting Λ 1 = max (C 1 (α, λ 1 , T ), C 2 (α, λ 1 , T )) and Λ 2 = max (C 3 (α, λ 1 , T ), C 4 (α, λ 1 , T )), an using the estimates obtained at the beginning of the proof, we have that y L 2 ((0,T );H 1 0 (Ω)) = ˆT 0 y H 1 0 (Ω) dt…”