Proof.
We decompose our proof into three steps for clarity.

Step 1. For any n∈N, we consider the existence of the weak solution for the following problem in B(0,n)≜Bn⊂RN:
(3.2)ut-Δut-Δu+f(x,u)=g(x), x∈Bn,u(x,0)=u0∈H1(Bn),u|∂Ω=0.
Choose a smooth function χn(x) with
(3.3)χn(x)={1,x∈Bn-1,0,x∉Bn.

Since Bn is a bounded domain, so the existence and uniqueness of solutions can be obtained by the standard Faedo-Galerkin methods; see [6, 8, 11, 16]; we have the unique weak solution
(3.4)un∈𝒞1([0,T];H1(Bn))∩Lp(0,T;Lp(Bn)), un(x,0)=χn(x)u0(x).

Step 2. According to Step 1, we denote (d/dt)un=unt; then un satisfies
(3.5)unt-Δunt-Δun+f(x,un)=g(x), x∈Bn,(3.6)un(x,0)=χn(x)u0(x),(3.7)un|∂Bn=0.
For the mathematical setting of the problem, we denote ∥·∥L2(Bn)≜∥·∥Bn, ∥·∥L1(RN)≜∥·∥1, ∥·∥L2(RN)≜∥·∥, ∥·∥L∞(RN)≜∥·∥∞.

Multiplying (3.5) by un in Bn, using f1(u)u≥0, (F2) and (A), we have
(3.8)12ddt(∥∇un∥Bn2+∥un∥Bn2)+∥∇un∥Bn2≤∫Bna(x)(β2-α2|u|p)dx+∫Bngundx≤β2∥a(x)∥1-∫Bnα2a(x)|u|pdx+∥g∥22λ+λ2∥un∥Bn2.
By the Poincaré inequality, for some ν>0, we conclude that
(3.9)12ddt(∥∇un∥Bn2+∥un∥Bn2)+ν(∥∇un∥Bn2+∥un∥Bn2)+∫Bnα2a(x)|u|pdx ≤β2∥a(x)∥1+∥g∥22λ.
Hence, it follows that
(3.10)∥∇un(T)∥Bn2+∥un(T)∥Bn2+2ν∫0T(∥∇un(T)∥Bn2+∥un(T)∥Bn2)+2∫0T∫Bnα2a(x)|u|pdx ≤(2β2∥a(x)∥1+∥g∥2λ)T.
We get the following estimate:
(3.11)supt∈[0,T]∥∇un(t)∥Bn2+∥un(t)∥Bn2≤C,∫0T(∥∇un(t)∥Bn2+∥un(t)∥Bn2)≤C,∫0T∫Bnα2a(x)|u(t)|pdx≤C.
Similar to (3.9), using (F1), (F2), and (A), we get
(3.12)∫0T∫Bn|u(t)|pdx≤C,
where C is independent of n.

(
F
1
)
and (F2) yield
(3.13)|f1(un)|≤C(|un|p-1+|un|),|f2(un)|≤C(|un|p-1+1).
Choose q such that (1/p)+(1/q)=1; then (p-1)q=p. Noting that p≥2, then 1<q≤2, and we have the embedding Lp(Bn)↪Lq(Bn). According to (3.12) and (3.13), we get
(3.14)∫0T∫Bn|f1(u)|q≤C∫0T∫Bn(|un|p-1+|un|)qdx dt≤C∫0T∫Bn|un|(p-1)qdx dt+C∫0T∫Bn|un|qdx dt≤C∫0T∫Bn|un|p+C∫0T∫Bn|un|pdx dt≤C,∫0T∫Bn|f2(u)|q≤C∫0T∫Bn|a(x)|q(|un|p-1+1)qdx dt≤C|a(x)|∞q-1∫0T∫Bna(x)(|un|(p-1)q+1)dx dt≤C|a(x)|∞q-1(C|a(x)|1+∫0T∫Bna(x)|un|pdx dt)≤C,
where C is independent of n.

Thanks to (3.14), f1(un) is bounded in Lp(0,T;Lq(Bn)), and af2(un) is bounded in Lp(0,T;Lq(Bn)).

For ∀v∈L2(0,T;H01(Bn)),
(3.15)∫0T∫Bn-Δunv=∫0T∫Bn∇un∇v≤(∫0T∥∇un∥Bn2)1/2(∫0T∥∇v∥Bn2)1/2≤(∫0T∥∇un∥2)1/2(∫0T∥∇v∥Bn2)1/2≤C∥∇v∥H01(Bn),
where C is independent of n. We can obtain that -Δun is bounded in L2(0,T;H-1(Bn)).

Since g(x)∈L2(ℝN),
(3.16)g(x)∈L2(0,T;ℝN).
Therefore, there exists s>0, such that L2(0,T;H-1(Bn)), L2(0,T;H01(Bn)), Lq(0,T;Lq(Bn)), and L2(0,T;L2(Bn)) are continuous embedding to Lq(0,T;H-s(Bn)).

According to (3.5) and (3.14)–(3.16), we obtain
(3.17)unt-Δunt∈Lq(0,T;H-s(Bn)).
So un has a subsequent (we also denote un) weak* convergence to u in L2(0,T;H-1(Bn)) and L2(0,T;L2(Bn)); unt-Δunt has a subsequent (we also denote unt-Δunt) weak* convergence to ut-Δut. Let un=0 outside of Bn; we can extend un to ℝN.

As introduced in [6, 20], Cc∞(ℝN) is dense in the dual space of H-1(Bn), L2(Bn), Lq(Bn), and H-s(Bn), so we can choose for all ϕ∈L2(0,T;Cc∞(ℝN))∩Lq(0,T;Cc∞(ℝN)) as a test function such that
(3.18)〈Δun,ϕ〉→〈Δu,ϕ〉,〈unt-Δunt,ϕ〉→〈ut-Δut,ϕ〉.

Since for all ϕ∈Cc∞(ℝN), there exists bounded domain Ω⊂ℝN such that ϕ=0, x∉Ω. It follows that un is uniformly bounded in L2(0,T;H01(Ω)), and unt-Δunt∈Lq(0,T;H-s(Ω)). Since H01(Ω)⊂⊂L2(Ω)⊂H-s(Ω), according to Lemma 2.8, there is a subsequence un (we also denote un) that converges strongly to u in L2(0,T;L2(Ω)).

Using the continuity of f1 and f2, we have
(3.19)〈f1(un),ϕ〉→〈f1(u),ϕ〉,〈a(x)f2(un),ϕ〉→〈a(x)f2(u),ϕ〉.

In line with (3.18) and (3.19), and let n→∞, we geting for all ϕ∈L2(0,T;Cc∞(ℝN))∩Lq(0,T;Cc∞(ℝN)):
(3.20)〈ut-Δut-Δu+f1(u)+a(x)f2(u),ϕ〉=〈g(x),ϕ〉.
Thus, u is the weak solution of (3.2) and satisfies
(3.21)u∈𝒞1([0,T];H1(ℝN))∩Lp(0,T;Lp(ℝN)).

Step 3 (uniqueness and continuous dependence). Let u0, v0 be in H1(ℝN), and setting w(t)=u(t)-v(t), we see that w(t) satisfies
(3.22)wt-Δwt-Δw+f1(u)-f1(v)+a(x)(f2(u)-f2(v))=0, x∈ℝN.
Taking the inner product with w of (3.22), using (F1), (F2), and (A), we obtain
(3.23)12ddt(∥∇w∥2+∥w∥2)+∥∇w∥2 ≤|∫(f1(u)-f1(v))w dx| +|∫a(x)(f2(u)-f2(v))w dx| ≤C(1+∥a∥∞)∥w∥2.
By the Gronwall Lemma, we get
(3.24)∥∇w(t)∥2+∥w(t)∥2≤eCt(∥∇w(0)∥2+∥w(0)∥2).
This is uniqueness and is continuous dependence on initial conditions.

Thanks to Theorem 3.1, and leting S(t)u0=u(t), S(t):H1(ℝN)→H1(ℝN) is a C0 semigroup.