Remark 1.

• 1.

  In Axiom (A), (a) − (b) is similar to $$ for each $$.

• 2.

  Two elements $$ can satisfy $$ without requiring ξ(θ) = δ(θ) for all θ ≤ 0.

• 3.

  We can see from the equivalence indicated in Part 1 of this statement that we need that ξ(0) = δ(0) for every $$ such that $$.

The space of bounded (D) uniformly continuous (C) functions defined from (−∞, 0] to $$ is denoted by CUD. The Banach space of all continuous, bounded functions from (−∞, T] into $$ with the standard norm is referred to as CD = CD(−∞, T].

The Banach space of continuous and bounded functions mapping [0, T] into $$, endowed with the standard norm, is denoted as CD^′ = CD^′[0, T] at this juncture. Its norm, represented as $$, is defined as the supremum over the interval [0, T] of the absolute values of these functions, denoted as |δ(t)|.

Definition 1 (see [17].)Let Q be a mapping from Ω into 2^Φ. A mapping T : (ϖ, δ) : ϖ ∈ Ω∧δ ∈ Q(ϖ)⟶Φ is called random operator with stochastic domain Q if Q is measurable and for every open set $$ and all δ ∈ Φ, $$. We will say that T is continuous if every T(ϖ) is continuous, where F is a σ-Borel algebra of Ω.

Definition 2 (see [17].)A mapping δ : Ω⟶Φ is known as a random (stochastic) fixed point of a random operator T if, for p-almost all ϖ ∈ Ω, it satisfies the properties δ(ϖ) ∈ Q(ϖ), T(ϖ)δ(ϖ) = δ(ϖ), and $$ for every open set $$. (i.e., δ is measurable).

Remark 2. If Q(ϖ) ≡ Φ, the definitions of a random operator with a stochastic domain and a random operator coincide.

Lemma 1 (see [17].)Let Q : Ω⟶2^Φ be measurable for any ϖ ∈ Ω such that Q(ϖ) is closed, convex, and solid (i.e., int (ϖ) ≠ ∅). Consider there is a measurable random variable named δ₀ : Ω⟶Φ with δ₀ ∈ intQ(ϖ) for every ϖ ∈ Ω. Assume T is a continuous random operator with the stochastic domain Q such that {δ ∈ Q(ϖ) : T(ϖ)δ = δ} ≠ ∅ for every ϖ ∈ Ω. Following that, T has a stochastic fixed point.

Definition 3. If δ₀ = ∅ and the continuous function $$ and $$ solves the integral equation then it is referred to as a mild solution to equation (1).

Now, we have listed the hypotheses that will be discussed in the following section.

•  

  (G₁): There exists a continuous function $$ such that for all $$ and $$, the inequality holds:

  $$ ()

•  

  (G₂): For every $$, the function $$ is continuous, and for all $$, the function $$ is strongly measurable.

•  

  (G₃): For each positive integer m, there exists $$ such that for almost every $$, and for ℵ, δ satisfying ‖ℵ‖ ≤ m and ‖δ‖ ≤ m, the inequality holds:

  $$ ()

•  

  (G₄): There exist a random funtion $$ such that

  $$ ()

•  

  (G₅): T₁(t), t > 0 is compact.

Theorem 1. Given that conditions (G₁)–(G₅) are satisfied, problem (1) possesses a mild random solution over the interval (−∞, T].

Proof 1. Consider the set $$, defined as the collection of functions u in $$ such that u(0, ϖ) = ϕ₁(0, ϖ) = 0. $$ is equipped with the uniform convergence topology. Additionally, let $$ be the random operator.

We will demonstrate that the mapping defined by (11) constitutes a random operator. To achieve this, we must establish that for any $$, $$ is a random variable. Initially, observe that $$ is measurable, given that the mapping ϕ(t, ε, δ) for $$ and $$ is measurable according to conditions (G₁)–(G₅).

Let $$ be defined as $$. Then, for all $$, D(ϖ) is bounded, closed, convex, and solid. Moreover, we know that D is measurable. Utilizing the hypotheses, for each $$, we have:

This implies that F is a random operator with a stochastic domain D.

Now, we need to demonstrate that the operator defined in (11) is completely continuous. Let $$ for k ≥ 1.

First, we need to show that F maps $$ into an equicontinuous family. Consider $$ and $$ with 0 ≤ t₁ ≤ t₂ ≤ ω. Then, we have

As far as we are aware, T₁(t), T₂(t) are uniformly continuous for $$ and their compactness for t > 0 proves the uniform operator topology is continuous. Since $$ the righthand side of the above inequalities are independent. Therefore, ‖(F(ϖ)δ)(t₁) − (F(ϖ)δ)(t₂)‖⟶0 as (t₁ − t₂)⟶0. Hence, F maps $$ into an equicontinuous family of functions. The equicontinuity for the cases t₁ ≤ t₂ ≤ 0 and t₁ ≤ 0 ≤ t₂ follows from the uniform continuity of F on (−∞, 0] and from the relation,

Since proving that $$ is uniformly bounded is straightforward, certain details will be omitted. Having established that $$ forms an equicontinuous family, the task now is to demonstrate that $$ is compact. To establish this, we need to demonstrate that the set,

Since T₁(t), T₂(t) are compact operators the set $$ is relatively compact in $$ for every ϵ, 0 < ϵ < t, Therefore by our hypothesis and for all $$, we have

Hence, there exist precompact sets arbitrarily close to the set W(t). Thus, W(t) is also precompact in $$.

Now, our aim is to prove the continuity of $$. Let δⁿ⟶δ in $$, where $$. We have

Now, we apply by dominance convergence theorem

Hence, $$ as n⟶∞, indicating that F is continuous. Since $$ is uniformly bounded, continuous, and compact, it is assured by Schauder’s fixed point theorem that F(ϖ) has a fixed point in $$. This completes the proof of the theorem.

Definition 4. Problem (3) is said to be controllable over the interval (−∞, k], provided that for every final state δ¹(ϖ), there exists a control y(t, ϖ) in $$, such that the solution δ(t, ϖ) of (3) satisfies δ(k, ϖ) = δ¹(ϖ).

Definition 5. If δ₀ = ∅ and the continuous function $$ and $$ solves the integral equation then it is referred to as a mild solution to equation (1).

Now, we have listed the additional hypotheses that will be discussed in the following section.

Let ϑ be the supremum of $$, and ϑ^′ be the supremum of $$.

(G₇) The linear operator $$, defined as $$, has a pseudo-inverse operator k⁻¹ in $$, and there exists a positive constant G such that ‖Bk⁻¹‖ ≤ G.

(G₈) There exists a random function $$ such that:

Theorem 2. If (G₁)–(G₈) are satisfied, then problem (3) is controllable on $$. Proof: Let us define the control: