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In this paper, we study the quasi-coordinated search technique for a lost target assumed to move randomly on one of two disjoint lines according to a random walk motion, where there are two searchers beginning their search from the origin on the first line and other two searchers begin their search from the origin on the second line. But the motion of the two searchers on the first line is independent from the motion of the other two searchers on the second line. Here we introduce a model of search plan and investigate the expected value of the first meeting time between one of the searchers and the lost target. Also, we prove the existence of a search plan which minimizes the expected value of the first meeting time between one of the searchers and the target.

The searching for a lost target either located or moved is often a time-critical issue, that is, when the target is very important. The primary objective is to find and search for the lost target as soon as possible. The searching for lost targets has recently applications such as the search for a goldmine underground, the search for Landmines and navy mines, the search for the cancer cells in the human body, the search for missing black box of a plane crash in the depth of the sea of ocean, the search for a damaged unit in a large linear system such as telephone lines, and mining system, and so on [

One of the most complicate problems when a mother loses her son in a way of multiple ways, here the primary objective is finding the lost son, as soon as possible in a minimum time. The survival rate of the son in this region gradually decreases, so the search team must organize itself quickly to begin the mission of the searching for the lost son immediately. Also, when the target is serious as a car, which filled by explosives, and it moves on one road from disjoint roads, and then the search effort must be unrestricted and we can use more than searcher to detect the target at right time.

The search team which consists of 4 searchers will organize itself on 2 straight lines to find the lost target as soon as possible. We clarify a modern technique by collaboration between each two searchers to find the lost person in minimum time. This problem can be characterized as follows.

The space of search: 2 disjoint lines.

The target: The target moves with a random walk motion on one of 2 disjoint straight lines.

The means of search: Looking for the lost target performed by two searches on each line. The searchers start searching for the target from the origins of the two lines with continuous paths and with equal speeds. In addition, the search spaces (2 straight lines) are separated into many distances.

Assume that we have two searchers S_{1} and S_{2} that start together looking for the lost target from O_{1} on L_{1}. The two searchers coordinate their search about the lost target, where the searcher S_{1} searches to the right and goes from the O_{1} to H_{1}, and the searcher S_{2} searches to the left and goes from O_{1} to −H_{1}, the two searchers S_{1} and S_{2} reach to H_{1} and −H_{1} in the same time of G_{1}. Then they come back to O_{1} again in the same time of G_{2}. If one of the two searchers do not find the lost target, then the two searchers S_{1} and S_{2} begin the new cycle search for the lost target, where they go from O_{1} to H_{2} and −H_{2}, respectively and they will reach to H_{2} and −H_{2} in the same time of G_{3}. Then they come back to O_{1} again in the same time of G_{4} and so on. Also, we have two other searchers S_{3} and S_{4} start together looking for the lost target from O_{2} on the second line L_{2}, the searcher S_{3} searchers to the right and goes from O_{2} to H ¯ 1 , and the searchers S_{4} searches to the left and goes to the left and goes from O_{2} to − H ¯ 1 , the two searchers S_{3} and S_{4} reach to H ¯ 1 and − H ¯ 1 in the same time of G ¯ 1 . Then they come back to O_{2} again in the same time of G ¯ 2 . If one of the two searchers not find the lost target, then the two searchers S_{3} and S_{4} begin the new cycle search for the lost target, where they go from O_{2} to H ¯ 2 and − H ¯ 2 , respectively and they will reach to H ¯ 2 and − H ¯ 2 in the same time of G ¯ 3 , then they come back to O_{2} again in the same time of G ¯ 4 , and so on. The four searchers return to the O_{1} and O_{2} after searching successively common distances until the target is found.

A target is assumed to move randomly on one of two disjoint lines according to a stochastic process { S ( t ) , t ∈ I + } , I + = { 0 , 1 , 2 , ⋯ } . Assume that { Z i } i ≥ 0 is a sequence of independent identically distributed random variables such as for any i ≥ 1 : p ( Z i = 1 ) = p and p ( Z i = − 1 ) = 1 − p = q , where p , q > 0 . For t > 0 , t ∈ I + ,

S ( t ) = ∑ i = 1 t Z i , S ( 0 ) = 0 .

We assume the searchers S_{1} and S_{2} begin their search path from O_{1} on L_{1} with speeds V_{1}, and the searchers S_{3} and S_{4} begin their search path from O_{2} on L_{2} with speeds V_{2}, following the search paths which are functions ϕ 1 : R + → R and ϕ ¯ 1 : R + → R on L_{1} and ϕ 2 : R + → R and ϕ ¯ 2 : R + → R on L_{2}, respectively, such that:

| ϕ 1 ( t 1 ) − ϕ 1 ( t 2 ) | = | ϕ ¯ 1 ( t 1 ) − ϕ ¯ 1 ( t 2 ) | ≤ V 1 | t 1 − t 2 | , (1)

and

| ϕ 2 ( t 1 ) − ϕ 2 ( t 2 ) | = | ϕ ¯ 2 ( t 1 ) − ϕ ¯ 2 ( t 2 ) | ≤ V 2 | t 1 − t 2 | , ∀ t 1 , t 2 ∈ I + , (2)

where V_{1} and V_{2} are constants in R + and ϕ 1 ( 0 ) = ϕ ¯ 1 ( 0 ) = ϕ 2 ( 0 ) = ϕ ¯ 2 ( 0 ) = 0 . Let the set of all search paths of the two searchers S_{1} and S_{2}, which satisfy condition (1), be respectively by Φ v 1 and Φ ¯ v 1 respectively and the set of all search paths of the searchers S_{3} and S_{4} which satisfy condition (2), be represented by Φ v 2 and Φ ¯ v 2 , respectively. we represented to the path of S_{1} and S_{2} by ϕ 0 = ( ϕ 1 , ϕ ¯ 1 ) ∈ Φ 0 where ϕ ¯ 0 = ( ϕ 2 , ϕ ¯ 2 ) ∈ Φ ¯ 0 , where

Φ ¯ 0 = { ( ϕ 2 , ϕ ¯ 2 ) : ϕ 2 ∈ Φ v 2 , ϕ ¯ 2 ∈ Φ ¯ v 2 } .

The search plan of the four searchers be represented by ϕ ^ = ( ϕ 0 , ϕ ¯ 0 ) ∈ Φ ^ , where Φ ^ = { ( ϕ 0 , ϕ ¯ 0 ) : ϕ 0 ∈ Φ 0 , ϕ ¯ 0 ∈ Φ ¯ 0 } is the set of all search plan.

We assume that Z 0 = X if the target moves on L_{1} and Z 0 = Y if the target moves on L_{2} such that P ( Z 0 = X ) + P ( Z 0 = Y ) = 1 . There is a known probability measure v 1 + v 2 = 1 on L 1 ∪ L 2 which describes the location of the target, where v_{1} is probability measure induced by the position of the target on L_{1}, while v_{2} on L_{2}. The first meeting time valued in I + defined as

τ ϕ ^ = inf { t : ϕ 1 ( t ) = X + S ( t ) or ϕ ¯ 1 ( t ) = X + S ( t ) or ϕ 2 ( t ) = Y + S ( t ) or ϕ ¯ 2 ( t ) = Y + S ( t ) } ,

where Z_{0} is a random variable representing the initial position of the target and valued in 2 I (or 2 I + 1 ) and independent of S ( t ) , t > 0 .

At the beginning of the search suppose that the lost target is existing on any integer point on L_{1} but more than H_{1} or less than − H 1 or the lost target is existing on an integer point on L_{2} but more than H ¯ 1 or less than − H ¯ 1 . Let τ ϕ 1 be the first meeting time between S_{1} and the target and τ ϕ ^ 1 be the first meeting time between S_{2} and the target and τ ϕ 2 be the first meeting time between S_{3} and the target and τ ϕ ^ 2 be the first meeting time between S_{4} and the target. The main objective is to find the search plan ϕ ^ = ( ϕ 0 , ϕ ¯ 0 ) ∈ Φ ^ such that E ( τ ϕ ^ ) < ∞ . In this case ϕ ^ is said to be a finite search plan, and if E ( τ ϕ ^ * ) < E ( τ ϕ ^ ) , ∀ ϕ ^ ∈ Φ ^ , where E terms to expectation value, then we call ϕ ^ * is an optimal search plan. Given n > 0 , if z is: 0 ≤ k 1 ≤ n + z 2 ≤ n , where k 1 is integer, then

p ( S ( n ) = k 1 ) = { ( n k 1 ) p k 1 q n − k 1 0 , if k 1 doesnotexist

Let λ 1 , λ 2 , ζ 1 , ζ 2 be positive integers such that ζ 1 , ζ 2 > 1 , λ 1 = k θ 1 , λ 2 = k θ 2 , where k = 1 , 2 , ⋯ and θ 1 , θ 2 are the least positive integers and V 1 = V 2 = 1 .

We shall define the sequences { G i } i ≥ 0 , { H i } i ≥ 0 for the searcher S_{1} on the first line L_{1} and { G ¯ i } i ≥ 0 , { H ¯ i } i ≥ 0 for the searcher S_{3} on the second line L_{2} and the search plans with speeds 1 as follows:

G i = 2 1 2 [ 1 − ( − 1 ) i − 1 ] λ 1 ( ζ 1 i 2 + 1 4 − ( − 1 ) i 1 4 − 1 ) , H i = G 2 i − 1 , i ≥ 1 on L_{1},

G ¯ i = 2 1 2 [ 1 − ( − 1 ) i − 1 ] λ 2 ( ζ 2 i 2 + 1 4 − ( − 1 ) i 1 4 − 1 ) , H ¯ i = G ¯ 2 i − 1 , i ≥ 1 on L_{2}.

We shall define the search path as follows:

for any t ∈ I + , if G i ≤ t < G i + 1 , then

ϕ 1 ( t ) = ( 1 2 H i + 1 2 ) + ( − 1 ) i + 1 ( 1 2 H i + 1 2 ) + ( − 1 ) i ( t − G i ) ,

and

ϕ ¯ 1 ( t ) = − ϕ 1 ( t ) .

Also, if G ¯ i ≤ t < G ¯ i + 1 , then

ϕ 2 ( t ) = ( 1 2 H ¯ i + 1 2 ) + ( − 1 ) i + 1 ( 1 2 H ¯ i + 1 2 ) + ( − 1 ) i ( t − G ¯ i ) ,

and

ϕ ¯ 2 ( t ) = − ϕ 2 ( t ) .

We define the notion

φ 1 ( t ) = S ( t ) − t , φ ˜ 1 ( t ) = S ( t ) + t on L_{1},

φ 2 ( t ) = S ( t ) − t , φ ˜ 2 ( t ) = S ( t ) + t on L_{2},

the searchers S_{1} and S_{2} return to the origin of L_{1} after searching successively common distances H 1 , H 2 , H 3 , ⋯ , and − H 1 , − H 2 , − H 3 , ⋯ , respectively and the searchers S_{3} and S_{4} return to the origin of L_{2} after searching successively common distances H ¯ 1 , H ¯ 2 , H ¯ 3 , ⋯ , and − H ¯ 1 , − H ¯ 2 , − H ¯ 3 , ⋯ , respectively until the target is found.

Theorem 1: If ϕ ^ = ( ϕ 0 , ϕ ¯ 0 ) ∈ Φ ^ is a search plan defined above, then the expectation E ( τ ϕ ^ ) if finite if

w 1 ( x ) = ∑ i = 1 ∞ ( ζ 1 i − 1 ) p ( φ ˜ 1 ( G 2 i − 1 ) < − x ) ,

w 2 ( x ) = ∑ i = 1 ∞ ( ζ 1 i − 1 ) p ( φ 1 ( G 2 i − 1 ) > − x ) ,

w 3 ( x ) = ∑ i = 1 ∞ ( ζ 1 i ( ζ 1 i − 2 ) + 1 ) p ( φ ˜ 1 ( G 2 i ) < − x ) ,

w 4 ( y ) = ∑ i = 1 ∞ ( ζ 1 i ( ζ 1 i − 2 ) + 1 ) p ( φ 1 ( G 2 i ) > − x ) ,

w 5 ( y ) = ∑ i = 1 ∞ ( ζ 2 i − 1 ) p ( φ ˜ 2 ( G 2 i − 1 ) < − y ) ,

w 6 ( y ) = ∑ i = 1 ∞ ( ζ 2 i − 1 ) p ( φ 2 ( G ˜ 2 i − 1 ) > − y ) ,

w 7 ( y ) = ∑ i = 1 ∞ ( ζ 2 i ( ζ 2 i − 2 ) + 1 ) p ( φ ˜ 2 ( G ˜ 2 i ) < − y ) ,

and

w 8 ( y ) = ∑ i = 1 ∞ ( ζ 2 i ( ζ 2 i − 2 ) + 1 ) p ( φ 2 ( G ˜ 2 i ) > − y ) . (3)

are finite.

Proof: Assume that X and Y are independent of S ( t ) , t > 0 , if X > 0 , then X + S ( t ) > ϕ 1 ( t ) until the first meeting between S_{1} and the target on L_{1}, also if X < 0 , then X + S ( t ) < ϕ ^ 1 ( t ) until the first meeting between S_{2} and the target on L_{2}. We can apply this assumption on the second line by replacing X by Y and ϕ 1 , ϕ ¯ 1 by ϕ 2 , ϕ ¯ 2 respectively. Hence, for any i ≥ 0

p ( τ ϕ ^ > t ) = p ( τ ϕ 0 > t or τ ϕ ^ 0 > t ) ,

hence

E ( τ ϕ ^ ) = ∫ 0 ∞ p ( τ ϕ ^ > t ) d t ≤ ∑ i = 0 ∞ [ ∫ G i G i + 1 p ( τ ϕ 0 > G i ) d t + ∫ G ˜ i G ˜ i + 1 p ( τ ϕ ^ 0 > G ˜ i ) d t ] = ∑ i = 0 ∞ ( 2 1 2 [ 1 − ( − 1 ) i + 2 ] λ 1 ( ζ 1 i + 1 2 + 1 4 − ( − 1 ) i + 1 1 4 − 1 )

− 2 1 2 [ 1 − ( − 1 ) i + 1 ] λ 1 ( ζ 1 i 2 + 1 4 − ( − 1 ) i 1 4 − 1 ) ) p ( τ ϕ ^ 0 > G i ) + ( 2 1 2 [ 1 − ( − 1 ) i + 2 ] λ 2 ( ζ 2 i + 1 2 + 1 4 − ( − 1 ) i + 1 1 4 − 1 )

− 2 1 2 [ 1 − ( − 1 ) i + 1 ] λ 2 ( ζ 2 i 2 + 1 4 − ( − 1 ) i 1 4 − 1 ) ) p ( τ ϕ ^ 0 > G ˜ i ) = λ 1 [ ( ( ζ 1 − 2 ) + 1 ) p ( τ ϕ 0 > 0 ) + ( ζ 1 − 1 ) p ( τ ϕ 0 > G 1 ) + ( ζ 1 ( ζ 1 − 2 ) + 1 ) p ( τ ϕ 0 > G 2 ) + ( ζ 1 2 − 1 ) p ( τ ϕ 0 > G 3 ) + ( ζ 1 2 ( ζ 1 − 2 ) + 1 ) p ( τ ϕ 0 > G 4 ) + ( ζ 1 3 − 1 ) p ( τ ϕ 0 > G 5 ) + ( ζ 1 3 ( ζ 1 − 2 ) + 1 ) p ( τ ϕ 0 > G 6 ) + ⋯ ]

+ λ 2 [ ( ( ζ 2 − 2 ) + 1 ) p ( τ ϕ ¯ 0 > 0 ) + ( ζ 2 − 1 ) p ( τ ϕ ¯ 0 > G ¯ 1 ) + ( ζ 2 ( ζ 2 − 2 ) + 1 ) p ( τ ϕ ¯ 0 > G ¯ 2 ) + ( ζ 2 2 − 1 ) p ( τ ϕ ¯ 0 > G ¯ 3 ) + ( ζ 2 2 ( ζ 2 − 2 ) + 1 ) p ( τ ϕ ¯ 0 > G ¯ 4 ) + ( ζ 2 3 − 1 ) p ( τ ϕ ¯ 0 > G ¯ 5 ) + ( ζ 2 3 ( ζ 2 − 2 ) + 1 ) p ( τ ϕ ¯ 0 > G ¯ 6 ) + ⋯ ] (4)

to solve Equation (4) we shall find the value of p ( τ ϕ 0 > G 2 i − 1 ) , p ( τ ϕ ¯ 0 > G ¯ 2 i − 1 ) , p ( τ ϕ 0 > G 2 i ) and the value of p ( τ ϕ ¯ 0 > G ¯ 2 i ) as the following

p ( τ ϕ 0 > G 2 i − 1 ) ≤ ∫ − ∞ 0 p ( x + S ( G 2 i − 1 ) < − H i / X = x ) v 1 ( d x ) + ∫ 0 ∞ p ( x + S ( G 2 i − 1 ) > H i / X = x ) v 1 (dx)

We get

p ( τ ϕ 0 > G 2 i − 1 ) ≤ ∫ − ∞ 0 p ( φ ˜ 1 ( G 2 i − 1 ) < − x ) v 1 ( d x ) + ∫ 0 ∞ p ( φ 1 ( G 2 i − 1 ) > − x ) v 1 ( d x ) (5)

also,

p ( τ ϕ ¯ 0 > G ¯ 2 i − 1 ) ≤ ∫ − ∞ 0 p ( φ ˜ 2 ( G ¯ 2 i − 1 ) < − y ) v 2 ( d y ) + ∫ 0 ∞ p ( φ 2 ( G ¯ 2 i − 1 ) > − y ) v 2 ( d y ) (6)

p ( τ ϕ 0 > G 2 i ) ≤ ∫ − ∞ 0 p ( X + S ( G 2 i ) < − 2 H i ) v 1 ( d x ) + ∫ 0 ∞ p ( x + S ( G 2 i ) > 2 H i ) v 1 ( d x ) (7)

We get

p ( τ ϕ 0 > G 2 i ) ≤ ∫ − ∞ 0 p ( φ ˜ 1 ( G 2 i ) < − x ) v 1 ( d x ) + ∫ 0 ∞ p ( φ 1 ( G 2 i ) > − x ) v 1 ( d x ) (8)

p ( τ ϕ 0 > G ¯ 2 i ) ≤ ∫ − ∞ 0 p ( φ ˜ 1 ( G ¯ 2 i ) < − y ) v 2 ( d y ) + ∫ 0 ∞ p ( φ 1 ( G ¯ 2 i ) > − y ) v 2 ( d y ) (9)

substituting by (5), (6), (7) and (8) in (4) we can get

E ( τ ϕ ^ ) ≤ λ 1 [ ( ( ζ 1 − 2 ) + 1 ) p ( τ ϕ 0 > 0 ) + ( ζ 1 − 1 ) p ( τ ϕ 0 > G 1 ) + ( ζ 1 ( ζ 1 − 2 ) + 1 ) p ( τ ϕ 0 > G 2 ) + ( ζ 1 2 − 1 ) p ( τ ϕ 0 > G 3 ) + ( ζ 1 2 ( ζ 1 − 2 ) + 1 ) p ( τ ϕ 0 > G 4 ) + ( ζ 1 3 − 1 ) p ( τ ϕ 0 > G 5 ) + ( ζ 1 3 ( ζ 1 − 2 ) + 1 ) p ( τ ϕ 0 > G 6 ) + ⋯ ]

+ λ 2 [ ( ( ζ 2 − 2 ) + 1 ) p ( τ ϕ ¯ 0 > 0 ) + ( ζ 2 − 1 ) p ( τ ϕ ¯ 0 > G ¯ 1 ) + ( ζ 2 ( ζ 2 − 2 ) + 1 ) p ( τ ϕ ¯ 0 > G ¯ 2 ) + ( ζ 2 2 − 1 ) p ( τ ϕ ¯ 0 > G ¯ 3 ) + ( ζ 2 2 ( ζ 2 − 2 ) + 1 ) p ( τ ϕ ¯ 0 > G ¯ 4 ) + ( ζ 2 3 − 1 ) p ( τ ϕ ¯ 0 > G ¯ 5 ) + ( ζ 2 3 ( ζ 2 − 2 ) + 1 ) p ( τ ϕ ¯ 0 > G ¯ 6 ) + ⋯ ]

hence

E ( τ ϕ ^ ) ≤ λ 1 [ ( ( ζ 1 − 2 ) + 1 ) p ( τ ϕ 0 > 0 ) + { ∫ − ∞ 0 w 1 ( x ) v 1 ( d x ) + ∫ 0 ∞ w 2 ( x ) v 1 ( d x ) } + { ∫ − ∞ 0 w 3 ( y ) v 2 ( d y ) + ∫ 0 ∞ w 4 ( y ) v 2 ( d y ) } ] + λ 2 [ ( ( ζ 2 − 2 ) + 1 ) p ( τ ϕ ¯ 0 > 0 ) + { ∫ − ∞ 0 w 5 ( y ) v 2 ( d y ) + ∫ 0 ∞ w 6 ( y ) v 2 ( d y ) } + { ∫ − ∞ 0 w 7 ( y ) v 2 ( d y ) + ∫ 0 ∞ w 8 ( y ) v 2 ( d y ) } ]

where,

w 1 ( x ) = ∑ i = 1 ∞ ( ζ 1 i − 1 ) p ( φ ˜ 1 ( G 2 i − 1 ) < − x ) ,

w 2 ( x ) = ∑ i = 1 ∞ ( ζ 1 i − 1 ) p ( φ 1 ( G 2 i − 1 ) > − x ) ,

w 3 ( x ) = ∑ i = 1 ∞ ( ζ 1 i ( ζ 1 − 2 ) + 1 ) p ( φ ˜ 1 ( G 2 i ) < − x ) ,

w 4 ( x ) = ∑ i = 1 ∞ ( ζ 1 i ( ζ 1 − 2 ) + 1 ) p ( φ 1 ( G 2 i ) > − x ) ,

w 5 ( y ) = ∑ i = 1 ∞ ( ζ 2 i − 1 ) p ( φ ˜ 2 ( G ¯ 2 i − 1 ) < − y ) ,

w 6 ( y ) = ∑ i = 1 ∞ ( ζ 2 i − 1 ) p ( φ 2 ( G ¯ 2 i − 1 ) > − y ) ,

w 7 ( y ) = ∑ i = 1 ∞ ( ζ 2 i ( ζ 2 − 2 ) + 1 ) p ( φ ˜ 2 ( G ¯ 2 i ) < − y ) ,

and

w 8 ( y ) = ∑ i = 1 ∞ ( ζ 2 i ( ζ 2 − 2 ) + 1 ) p ( φ 2 ( G ¯ 2 i ) > − y ) .

Lemma 1: For any k ≥ 0 , let a n ≥ 0 for n ≥ 0 , and a n + 1 ≤ a n . Let { d n } n ≥ 0 be a strictly increasing sequence of integers with d 0 = 0 ,

∑ n = k ∞ ( d n + 1 − d n ) a d n + 1 ≤ ∑ k = d k ∞ a k ≤ ∑ n = k ∞ ( d n + 1 − d n ) a d n ,

For more details see [

Theorem 2: The chosen search plan satisfies

w 1 ( x ) ≤ w 9 ( | x | ) , w 2 ( x ) ≤ w 10 ( | x | ) ,

w 2 ( x ) ≤ w 11 ( | x | ) , w 4 ( x ) ≤ w 12 ( | x | ) ,

w 5 ( y ) ≤ w 13 ( | y | ) , w 6 ( y ) ≤ w 14 ( | y | ) ,

w 7 ( y ) ≤ w 15 ( | y | ) , and w 8 ( y ) ≤ w 16 ( | y | ) ,

where, w 9 ( | x | ) , w 10 ( | x | ) , w 11 ( | x | ) , w 12 ( | x | ) , w 13 ( | y | ) , w 14 ( | y | ) , w 15 ( | y | ) , and w 16 ( | y | ) are linear function.

Proof: This theorem will prove for w 2 ( x ) and w 6 ( y ) , and by similar way we can prove the other cases

w 2 ( x ) = ∑ i = 0 ∞ ( ζ 1 i − 1 ) p ( φ 1 ( G 2 i − 1 ) > − x )

and

w 6 ( y ) = ∑ i = 0 ∞ ( ζ 2 i − 1 ) p ( φ 2 ( G ¯ 2 i − 1 ) > − y )

1) if x ≤ 0 , then

w 2 ( x ) ≤ w 2 (0)

and if y ≤ 0 , then

w 6 ( y ) ≤ w 6 ( 0 ) ,

2) if x > 0 , then

w 2 ( x ) = w 2 ( 0 ) + ∑ i = 0 ∞ ( ζ 1 i − 1 ) p ( − x < φ 1 ( G 2 i − 1 ) ≤ 0 ) ,

and if y > 0 , then

w 6 ( y ) = w 6 ( 0 ) + ∑ i = 0 ∞ ( ζ 2 i − 1 ) p ( − y < φ 2 ( G ¯ 2 i − 1 ) ≤ 0 ) ,

from Theorem (2), see (Mohamed [

w 2 ( 0 ) = ∑ i = 0 ∞ ( ζ 1 i − 1 ) p ( φ 1 ( G 2 i − 1 ) > 0 ) ≤ ∑ i = 1 ∞ ( ζ 1 i − 1 ) ε G 2 i − 1 , 0 < ε < 1

and

w 6 ( 0 ) = ∑ i = 0 ∞ ( ζ 2 i − 1 ) p ( φ 2 ( G ¯ 2 i − 1 ) > 0 ) ≤ ∑ i = 1 ∞ ( ζ 2 i − 1 ) ε G ¯ 2 i − 1 , 0 < ε < 1

Let us define the following

1) V ( n ) = φ 1 ( n θ 1 ) / 2 = ∑ i = 1 n W i , where { W i } is a sequence of (i. i. d. r. v.) V ¯ ( n ) = φ 2 ( n θ 2 ) / 2 = ∑ i = 1 n W ¯ i , where { W ¯ i } is a sequence of (i. i. d. r. v.).

2) d n = G 2 n − 1 / θ 1 = k ( ζ 1 n − 1 ) , d ¯ n = G ¯ 2 n − 1 / θ 2 = k ( ζ 2 n − 1 ) .

3) a ( n ) = n n + k p ( − x / 2 < V ( n ) ≤ 0 ) = ∑ i = 0 | x | / 2 p [ − ( j + 1 ) < V ( n ) ≤ ( − j ) ] , a ¯ ( n ) = n n + k p ( − y / 2 < V ¯ ( n ) ≤ 0 ) = ∑ i = 0 | y | / 2 p [ − ( j + 1 ) < V ¯ ( n ) ≤ ( − j ) ] ,

4) m 1 is an integer such that d m 1 = b 1 | x | + b 2 , and m 2 is an integer such that d m 2 = b ¯ 1 | y | + b ¯ 2 ,

5) α 1 = ζ 1 ( ζ 1 − 1 ) k , and α 2 = ζ 2 ( ζ 2 − 1 ) k ,

and

6) U 1 ( j , j + 1 ) = ∑ n = 0 ∞ p [ − ( j + 1 ) < V ( n ) < ( − j ) ] , U ¯ 1 ( j , j + 1 ) = ∑ n = 0 ∞ p [ − ( j + 1 ) < V ¯ ( n ) ≤ ( − j ) ] ,

then U 1 ( j , j + 1 ) and U ¯ 1 ( j , j + 1 ) satisfies the condition of the renewal equation, for more details see [

If n > d m 1 and n > d m 2 then by Theorem (2) see (Mohamed [

w 2 ( x ) − w 2 ( 0 ) = ∑ i = 1 ∞ ( ζ 1 i − 1 ) p ( − x < φ 1 ( G 2 i − 1 ) ≤ 0 ) = ∑ n = 1 n 1 ζ 1 n a ( d n ) + ∑ n = n 1 + 1 ∞ ζ 1 n a ( d n ) ≤ ∑ n = 1 n 1 ζ 1 n + α 1 ∑ n = n 1 + 1 ∞ ( d n − d n − 1 ) a ( d n ) ≤ ∑ n = 1 n 1 ζ 1 n + α 1 ∑ n = d m 1 ∞ a ( n ) ≤ ∑ n = 1 n 1 ζ 1 n + α 1 ∑ n = d m 1 ∞ ∑ i = 0 | x | / 2 p [ − ( j + 1 ) < V ( n ) ≤ ( − j ) ] ≤ ∑ n = 1 n 1 ζ 1 n + α 1 ∑ j = 0 | x | / 2 U 1 ( j , j + 1 )

and

w 6 ( x ) − w 6 ( 0 ) = ∑ i = 0 ∞ ( ζ 2 i − 1 ) p ( − y < φ 2 ( G 2 i − 1 ) ≤ 0 ) = ∑ n = 1 n 2 ζ 2 n a ( d ¯ n ) + ∑ n = n 2 + 1 ∞ ζ 2 n a ( d ¯ n ) ≤ ∑ n = 1 n 2 ζ 2 n + α 2 ∑ n = n 2 + 1 ∞ ( d ¯ n − d ¯ n − 1 ) a ( d ¯ n )

≤ ∑ n = 1 n 2 ζ 2 n + α 2 ∑ n = d m 2 ∞ a ¯ ( n ) ≤ ∑ n = 1 n 2 ζ 2 n + α 2 ∑ j = 0 | y | / 2 U ¯ 1 ( j , j + 1 )

Since U 1 ( j , j + 1 ) and U ¯ 1 ( j , j + 1 ) satisfied the condition of the renewal equation, hence U 1 ( j , j + 1 ) and U ¯ 1 ( j , j + 1 ) is bounded for all j by a constant, so

w 2 ( x ) ≤ w 2 ( 0 ) + N 1 + N 2 | x | = w 10 ( | x | ) ,

and

w 6 ( x ) ≤ w 6 ( 0 ) + N ¯ 1 + N ¯ 2 | x | = w 14 ( | y | ) .

Theorem 3: If ϕ ^ = ( ϕ 0 , ϕ ¯ 0 ) ∈ Φ ^ is a finite search plan, then E | Z 0 | is finite.

Proof: If E ( τ ϕ ^ ) < ∞ , then p ( τ ϕ ^ isfinite ) = 1 and so

p ( τ ϕ 0 is finite ) + p ( τ ϕ ^ 0 isfinite ) = 1 ,

then, we conclude that

p ( τ ϕ 0 is finite ) = 1 and p ( τ ϕ ¯ 0 isfinite ) = 0 ,

or

p ( τ ϕ 0 is finite ) = 0 and p ( τ ϕ ¯ 0 isfinite ) = 1 .

On the first line L_{1} if p ( τ ϕ 0 is finite ) = 1 , then X 0 = ϕ ( τ ϕ 0 ) − S ( τ ϕ 0 ) with probability one and hence

E | X 0 | ≤ E ( τ ϕ 0 ) + E | S ( τ ϕ 0 ) | .

If E ( τ ϕ 0 ) < ∞ , but | S ( τ ϕ 0 ) | ≤ τ ϕ 0 , then E | S ( τ ϕ 0 ) | ≤ E ( τ ϕ 0 ) and E | X 0 | < ∞ .

On the second line L_{2} if p ( τ ϕ ¯ 0 isfinite ) = 1 , then Y 0 = ϕ ( τ ϕ ¯ 0 ) − S ( τ ϕ ¯ 0 ) with probability one, by the same way we can get E | Y 0 | is finite on the second line L_{2}.

Theorem 4: Let for any t ∈ I + , let S ( t ) be a process. The mapping ϕ ^ → E ( τ ϕ ^ ) ∈ R + is lower semi-continuous on Φ ^ ( t ) .

Proof: Let I ( ϕ ^ , t ) be the indicator function of the set { τ ϕ ^ ≥ t } by the Fatou Lebesque theorem see (Stone [

E ( τ ϕ ^ ) = E [ ∑ t = 1 ∞ I ( ϕ ^ , t ) ] = E [ ∑ t = 1 ∞ lim i → ∞ inf I ( ϕ ^ n , t ) ] ≤ lim i → ∞ inf E ( τ ϕ ^ ) ,

for any sequence ϕ ^ n → ϕ ^ in Φ ^ ( t ) is sequentially compact [

We have described a new kind of search technique to find a lost moving target on one of two disjoint lines. The motion of the four searchers on the two lines in the quasi-coordinated search technique is independent, and this helps us to find the lost target without waste of time and cost, especially if this target is valuable as the search for lost children. Actually we calculated the finite search plan. Also; we proved the existence of an optimal search plan which minimizes the expected value of the first meeting time between one of the searchers and the target.

In the future work, we will introduce an important search problem, looking for a randomly moving target as a general case and the searchers will begin their mission from any point on the line.

The authors declare no conflicts of interest regarding the publication of this paper.

Teamah, A.A.M. and Afifi, W.A. (2019) Quasi-Coordinate Search for a Randomly Moving Target. Journal of Applied Mathematics and Physics, 7, 1814-1825. https://doi.org/10.4236/jamp.2019.78124