In this section, we present our major results under the general Lipschitz condition defined above. We will establish the result by employing similar argument employed in the proof of Theorems 3.1 in [1] and 3.1 in [8] by highlighting the major changes due to condition (iii).

Proof.
Let a∈A0 and x0(·,a):I→A~ be the unique weak solution of the Cauchy problem(15)ddtη,Xtξ∈-P1t,Xtη,ξ,X0=a.For Kηξ and β defined by (iii) and (iv), we define b:A0(η,ξ)→Lloc1(I) by(16)baηξt=βηξt+KηξtWx0t,aηξ.By remark 2.1 in [1], the map η,aξ→η,x0(·,a)ξ is weakly continuous from A0(η,ξ) to C(I,D⊗_E). Hence from (16), it follows that b(·) is continuous from A0(η,ξ) to Lloc1(I). And we have(17)d0,P2t,x0t,aη,ξ≤baηξt,a.e. in Ifor each aηξ∈A0(η,ξ).

As in [8] we fix ϵ>0 and set ϵ=ϵ/2n+1, n∈N. Define Φ0:A0(η,ξ)→2L(I,D⊗_E) and ϕ0:A0(η,ξ)→2L(I,D⊗_E) by (14) and (15) in [1]. Using (16) and Lemma 2, ϕ0(·) is lower semicontinuous (l.s.c.) and for each aη,ξ∈A0(η,ξ), ϕ0(aηξ)≠0, and W(t)≠t. Again by Lemma 3, there exists φ0(η,ξ)→L(I,D⊗_E), a continuous selection of ϕ0(·). Set p0(t,a)(η,ξ)=φ0(aηξ)(t) as in [1]; then p0(·,a)(η,ξ) is continuous, p0(t,a)(η,ξ)∈P2(t,a)(η,ξ), and(18)p0t,aη,ξ≤baηξt+ϵ0,a.e. t∈I.If we set Mηξ=∫0tKηξ(s)ds, where aηξ is as defined, then, for each a∈A0, we can define βn(aηξ)(t), n≥1 as follows:(19)βnaηξt=∫0taηξsMηξt-Mηξsn-1dsn-1!+WT∑i=0nϵiMηξtn-1n-1!,t∈I.Thus by (19) βn(·) is continuous from A0(η,ξ) to Lloc1(I,R) since b(·) is continuous. Now if x1(·,a):I→A~ is the unique solution of the Cauchy problem(20)ddtη,Xtξ∈η,p0tξ-P1t,Xtη,ξ,X0=athen, by (11) in [1], we have(21)η,x1tξ-η,x0tξ≤∫0tp0s,aη,ξds≤∫0tbaηξsds+ϵ0T<β1aηξt,for each aηξ∈A0(η,ξ) and t∈I∖{0}. Now set η,pn(s,a)ξ≡pn(s,a)(η,ξ) and assume that there exist sequences {pn(·,a)}n∈N and {xn(·,a)}n∈N such that, for each n≥1, (a), (b), and (d) in [1] hold in this case while (c) becomes(22)η,p1t,aξ-η,pn-1t,aξ≤WKηξtβnt,a.e. t∈I,where W is due to the Lipschitz function K. We now obtain the following by (22) and (11) in [1], t∈I∖{0}(23)η,xn+1t,aξ-η,xnt,aξ≤∫0tpns,aη,ξ-pn-1s,aη,ξds≤W∫0tKηξsβnaηξsds=W∫0tbaηξsMηξt-Mηξsnn!ds+WT∑i=0nϵiMηξtnn!<Wβ1aηξt.Since P is maximal monotone and hence hypermaximal monotone, we get(24)dη,pt,aξ,P2t,xn+1t,aη,ξ≤KηξtWxn+1t,a-xnt,aηξ<KηξtWβn+1aηξt.By (24) and Lemma 2, the multivalued map ϕn+1:A0(η,ξ)→2L1(I,D⊗_E) defined by (19) in [1] is l.s.c. with decomposable closed nonempty values. Then by Lemma 3, the sesquilinear form valued map φn+1(aηξ)(t) still admits a continuous selection of Φn+1(·).

If we set η,pn+1(t,a)ξ=φn+1(aηξ)(t) for aηξ∈A0(η,ξ), t∈I, we have that pn+1 satisfies the properties (a), (b) in [1] and (22); hence by (24), we obtain(25)Φn+1aηξ=clvηξ∈ϕn+1aηξ:vηξ-η,pnkt,aξ<KηξtWβn+1aηξt,t∈I.Again by (22) and (23), we have(26)η,pn·,aξ-η,pn-1·,aξ=∫0Tpns,aη,ξ-pn-1s,aη,ξds≤W∫0TbaηξsMηξT-Mηξsnn!ds+WT∑i=0nϵiMηξTnn!≤WKηξt1nn!baηξ+Tϵ.Since aηξ→b(aηξ)(t) is continuous, then it is locally bounded. It follows by (26) that the sequence {pn(·,a)}n∈N satisfies the Cauchy condition uniformly. If p(·,a) is the limit of the given sequence, then aηξ→η,p(·,a)ξ is also weakly continuous from A0(η,ξ) into L1(I,D⊗_E).

Now if we use (23) and (26), we get(27)η,xn+1·,aξ-η,xn·,aξ≤η,pn·,aξ-η,pn-1·,aξ≤WKηξt1nn!baηξ1+Tϵ.Hence {η,xn(·,a)ξ} is Cauchy in C(I,D⊗_E) with respect to a. Then the map aηξ→η,xn(·,a)ξ is weakly continuous from A0(η,) to C(I,D⊗_E) and so also the map η,xn(·,a)ξ→η,x(·,a)ξ uniformly and(28)dη,pnt,aξ,Pt,xt,aη,ξ≤KηξtWxn·,a-x·,aηξ.Therefore, the result (22) in [1] holds here. If we let p0=p and x1(·,a) be the unique weak solution of the Cauchy problem (20), we obtain by (11) in [1](29)η,xnt,aξ-η,x1t,aξ≤∫0tη,pns,aξ-η,ps,aξds.If n→∞, then η,x1(·,a)ξ≡η,x(·,a)ξ. Therefore, x(·,a) is the weak solution of (20), and the result(30)η,x·,aξ∈STaη,ξholds here under the general Lipschitz condition.