161 
This work is licensed under a Creative Commons Attribution 4.0 International License. 

 

  

 

 

The Optimal Classical Continuous Control Quaternary Vector of 

Quaternary Nonlinear Hyperbolic Boundary Value Problem 

 

 

 

 

 

 

 

 

 

 

Abstract 

       This work is concerned with studying the optimal classical continuous control quaternary 

vector problem. It is consisted of; the quaternary nonlinear hyperbolic boundary value problem 

and the cost functional. At first, the weak form of the quaternary nonlinear hyperbolic 

boundary value problem is obtained. Then under suitable hypotheses, the existence theorem of 

a unique state quaternary vector solution for the weak form where the classical continuous 

control quaternary vector is considered known is stated and demonstrated by employing the 

method of Galerkin and the compactness theorem. In addition, the continuity operator between 

the state quaternary vector solution of the weak form and the corresponding classical 

continuous control quaternary vector is demonstrated in three different infinite dimensional 

spaces (Hilbert spaces). Furthermore, with suitable hypotheses, the existence theorem of an 

optimal classical continuous control quaternary vector dominated by the weak form of the 

quaternary nonlinear hyperbolic boundary value problem is stated and demonstrated.     

 

Keywords: Optimal Classical Continuous Control Quaternary Vector, Quaternary Nonlinear 

Hyperbolic Boundary Value Problem, Weak form.  

 

1.  Introduction 

Different applications in real-life are classified as optimal control problems (OCPs). For 

example, in medicine [1], economics [2], robotics [3], Aircraft [4], and many other fields. Usually, this 

importance encouraged many researchers to be interested in studying OCPs in general and optimal 

classical continuous control problems (OCCCP) in particular. During the last decade, great attention 

has been made to studying the subject of OCCCP for a system dominated by nonlinear PDEs (NLPDEs) 

Ibn Al-Haitham Journal for Pure and Applied Sciences 

http://jih.uobaghdad.edu.iq/index.php/j/index: Journal homepage 

 

Doi: 10.30526/35.3.2833 

Article history: Received 13 March 2022, Accepted 17 May 2022, Published in July 2022. 

 

Jamil A. Ali Al-Hawasy 

Department of Mathematics, College of 

Sciences, Mustansiriyah University, 

Baghdad, Iraq. 

jhawassy17@uomustanriyah.edu.iq                        

 

Mayeada Abd Alsatar Hassan 

Baghdad  Directorate of Education, 

Baghdad, Iraq 

mayeadabd1989@uomustanriyah.edu.iq 

 

https://creativecommons.org/licenses/by/4.0/
file:///F:/العدد%20الثاني%202022/:%20http:/jih.uobaghdad.edu.iq/index.php/j/index
mailto:jhawassy17@uomustanriyah.edu.iq
mailto:jhawassy17@uomustanriyah.edu.iq
mailto:mayeadabd1989@uomustanriyah.edu.iq%7d


IHJPAS. 53 (3)2022 
 

162 

of the three types elliptic [5], hyperbolic [6], and parabolic [7]. Latter, the study of this subject 

expanded to include OCCCP for systems dominated by a couple of NLPDEs of their three 

types [8-10]; through recent years, these studies for these three types expanded to deal with 

OCCCP for systems dominated by triple NLPDEs [11-13]. All these studies encouraged us to 

investigate the OCCCP dominated by QNLHBVP.  
 

This article first concerns the mathematical formulation for the optimal classical 

continuous control quaternary vector problem. Then the existence theorem (ETH) of a unique 

state quaternary vector solution (SQVS) for the weak form (WF) “of the quaternary nonlinear 

hyperbolic boundary value problem (QNLHBVP)” is stated and demonstrated using the 

method of Galerkin (MGA) and the Aubin compactness theorem (ACTH) when the classical 

continuous control quaternary vector (CCCQV) is fixed under suitable hypotheses. 

Furthermore, the continuity operator between the SQVS of the WF for the QNLHBVP and the 

corresponding CCCQV is demonstrated. Lastly, the ETH of an optimal classical continuous 

control quaternary vector (OCCCQV) is stated and demonstrated with suitable hypotheses.   

 

2. Problem Description  

Let  𝐼 = [0, 𝑇], T < ∞, Ω ⊂ ℝ2, be an open bounded region with boundary Γ = 𝜕Ω ,𝑄 = Ω ×

𝐼 , Σ = Γ × 𝐼. The OCCCQV includes the quaternary state equations (QSEs) which are 

considered by the following QNLHBVP: 

𝑦1𝑡𝑡 − ∆𝑦1 + 𝑦1 − 𝑦2 + 𝑦3 + 𝑦4 = 𝑓1(𝑥, 𝑡, 𝑦1, 𝑢1), in 𝑄                                                      (1) 

𝑦2𝑡𝑡 − ∆𝑦2 + 𝑦1 + 𝑦2 − 𝑦3 − 𝑦4 = 𝑓2(𝑥, 𝑡, 𝑦2, 𝑢2), in 𝑄                                                      (2) 

𝑦3𝑡𝑡 − ∆𝑦3 − 𝑦1 + 𝑦2 + 𝑦3 + 𝑦4 = 𝑓3(𝑥, 𝑡, 𝑦3, 𝑢3), in 𝑄                                                      (3) 

𝑦4𝑡𝑡 − ∆𝑦4 − 𝑦1 + 𝑦2 − 𝑦3 + 𝑦4 = 𝑓4(𝑥, 𝑡, 𝑦4, 𝑢4), in 𝑄                                                       (4) 

With the following boundary conditions (BCs) and the initial conditions (ICs) 

𝑦𝑖 (𝑥, 𝑡) = 0, on Σ, for 𝑖 = 1,2,3,4                                                                                        (5) 

𝑦1(𝑥, 0) = 𝑦𝑖
0(𝑥),and 𝑦𝑖𝑡 (𝑥, 0) = 𝑦𝑖

1(𝑥), in Ω for 𝑖 = 1,2,3,4                                            (6) 

Where �⃗� = (𝑦1, 𝑦2, 𝑦3, 𝑦4) belongs to the Hilbert space (𝐻
2(Ω))4 is the SQVS, corresponding 

to the CCCQV �⃗⃗� = (𝑢1, 𝑢2, 𝑢3, 𝑢4) ∈ (𝐿
2(Q))4 and (𝑓1, 𝑓2, 𝑓3, 𝑓4) ∈ (𝐿

2(Q))4 is a vector of a 

given function on (𝑄 × ℝ × 𝑈1) × (𝑄 × ℝ × 𝑈2) × (𝑄 × ℝ × 𝑈3) × (𝑄 × ℝ × 𝑈4), with 

𝑈𝑖 ⊂ ℝ, ∀𝑖 = 1,2,3,4 .   

The quaternary controls constraints (QCCs) are 

�⃗⃗⃗⃗�𝐴= {�⃗⃗⃗� ∈ 𝑊 ⃗⃗⃗⃗⃗ ⊂ (𝐿
2(Q))4|�⃗⃗⃗� ∈ �⃗⃗⃗� ⊂ ℝ4 𝑎. 𝑒. 𝑖𝑛 𝑄},with �⃗⃗⃗�  ⊂ ℝ4is a convex. 

The cost function will is considered as 

 𝐺0(�⃗⃗�) = Σ
𝑖=1

4

∫
𝑄

𝑔0𝑖  (𝑥, 𝑡, 𝑦𝑖 , 𝑢𝑖 )𝑑𝑥𝑑𝑡                                                                                    (7) 

The OCCCP is to find  �⃗⃗� ∈ �⃗⃗⃗⃗�𝐴, s.t.  𝐺0(�⃗⃗�) = 𝑚𝑖𝑛
�⃗⃗⃗�∈�⃗⃗⃗⃗�𝐴 

𝐺0 (�⃗⃗⃗�)        

Let V⃗⃗⃗ = (𝑉)4; 𝑉 = 𝐻0
1(Ω), and �⃗⃗� = {�⃗�: �⃗� ∈ (𝐻1(Ω))4, 𝑣1 = 𝑣2 = 𝑣3 = 𝑣4 = 0 𝑜𝑛 𝜕Ω}. �⃗� =

(𝑣1, 𝑣2, 𝑣3, 𝑣4), we denote by (𝑣, 𝑣) and ∥ 𝑣 ∥0 the inner product (IP) and the norm in (𝐿
2(Ω))4, 

by (�⃗�, �⃗�)1 = Σ
𝑖=1

4

∥ 𝑣1 ∥1
2 the IP and the norm in V⃗⃗⃗, and V⃗⃗⃗∗ is the dual of V⃗⃗⃗. 

The WF of ((1)-(6)) when 𝑦⃗⃗⃗ ⃗ ∈ (𝐻0
1(Ω))4 is given a.e. on I and ∀𝑣𝑖 ∈ 𝑉𝑖 (∀𝑖 = 1,2,3,4 ) by : 

(𝑦1𝑡𝑡 , 𝑣1) + (∇𝑦1, ∇𝑣1) + (𝑦1, 𝑣1) − (𝑦2, 𝑣1) + (𝑦3, 𝑣1) + (𝑦4, 𝑣1) = (𝑓1, 𝑣1)                     (8) 

(𝑦1
0, 𝑣1) = (𝑦1(0), 𝑣1), and (𝑦1𝑡

1 , 𝑣1) = (𝑦1𝑡 (0), 𝑣1)                                                              (9) 



IHJPAS. 53 (3)2022 
 

163 

(𝑦2𝑡𝑡 , 𝑣2) + (∆𝑦2, ∇𝑣2) + (𝑦1, 𝑣2) + (𝑦2, 𝑣2) − (𝑦3, 𝑣2) − (𝑦4, 𝑣2) = (𝑓2, 𝑣2)                    (10) 

(𝑦2
0, 𝑣2) = (𝑦2(0), 𝑣2), and (𝑦2𝑡

1 , 𝑣2) = (𝑦2𝑡 (0), 𝑣2)                                                           (11) 

(𝑦3𝑡𝑡 , 𝑣3) + (∇𝑦3, ∇𝑣3) − (𝑦1, 𝑣3) + (𝑦2, 𝑣3) + (𝑦3, 𝑣3) + (𝑦4, 𝑣3) = (𝑓3, 𝑣3)                    (12) 

(𝑦3
0, 𝑣3) = (𝑦3(0), 𝑣3), and (𝑦3𝑡

1 , 𝑣3) = (𝑦3𝑡 (0), 𝑣3)                                                           (13) 

(𝑦4𝑡𝑡 , 𝑣4) + (∇𝑦4, ∇𝑣4) − (𝑦1, 𝑣4) + (𝑦2, 𝑣4) − (𝑦3, 𝑣4) + (𝑦4, 𝑣4) = (𝑓4, 𝑣4)                     (14) 

(𝑦4
0, 𝑣2) = (𝑦4(0), 𝑣4), and (𝑦4𝑡

1 , 𝑣4) = (𝑦4𝑡 (0), 𝑣4)                                                            (15) 

2.1. Assumptions (A): Suppose that  𝑓𝑖  is of the Carathéodory type on  𝑄 × (ℝ × 𝑈𝑖 ) and 

satisfies (for 𝑖 = 1,2,3,4): 

 (i)|𝑓𝑖 (𝑥, 𝑡, 𝑦𝑖 , 𝑢𝑖 )| ≤ 𝐹𝑖 (𝑥, 𝑡) + 𝛾𝑖 ∣ 𝑢𝑖 | + 𝛽𝑖 |𝑦𝑖 |, where 𝑦𝑖 , 𝑢𝑖 ∈ ℝ, 𝛽𝑖 , 𝛾𝑖 ∣> 0 and 𝐹𝑖 ∈ 𝐿
2(Q). 

 (ii) 𝑓𝑖  is satisfied Lipschitz condition (LIPC) w.r.t. 𝑦𝑖, i.e.  

       |𝑓𝑖 (𝑥, 𝑡, 𝑦𝑖 , 𝑢𝑖 ) − 𝑓𝑖 (𝑥, 𝑡, �̅�𝑖 , 𝑢𝑖 )| ≤ 𝐿𝑖 |𝑦𝑖 − �̅�𝑖 | , 𝑦𝑖 , �̅�𝑖 , 𝑢𝑖 ∈ ℝ, 𝐿𝑖 > 0, for (𝑥, 𝑡) ∈ 𝑄. 

2.2 Lemma1: (Gronwall inequality): Let 𝐾 be a nonnegative constant and let 𝑓 and 𝑔 be 

continuous nonnegative functions on [𝛼, 𝛽], satisfies: 𝑓(𝑡) ≤ 𝐾 + ∫ 𝑓(𝑠)𝑔(𝑠)𝑑𝑠
𝑡

𝛼
. Then,

𝑓(𝑡) ≤ 𝐾𝑒∫ 𝑔
(𝑠)𝑑𝑠

𝑡
𝛼 , for    𝛼 ≤ 𝑡 ≤ 𝛽. 

  

3. The Solution for the QSEs: 

3.1 Proposition [14]: Let 𝐷 is a measurable subset of ℝ𝑑  (𝑑 = 2,3), 𝑓: 𝐷 × ℝ𝑛 ⟶ ℝ𝑚 is of 

Carathéodory type satisfies ‖𝑓(𝑣, 𝑥)‖ ≤ (𝑣) + (𝑣)‖𝑥‖𝛼,∀(𝑣, 𝑥) ∈ 𝐷 × ℝ𝑛 , where 𝑥 ∈

𝐿𝑝 (𝐷 × ℝ𝑛 ), ∈ 𝐿1(𝐷 × ℝ ), ∈ 𝐿
𝑝

𝑝−𝛼(𝐷 × ℝ ), 𝛼 ∈ [0, 𝑝], if 𝑝 ≠ ∞. Then, the functional 

𝐹(𝑥) = ∫  
 

𝐷
𝑓(𝑣, 𝑥(𝑣))𝑑𝑣 is continuous.  

3.2 Theorem (ETH of a Unique SQVS): with Assumptions (A), for each given �⃗⃗� ∈ 𝐿2(Q), 

the WF ((8)-(15)) has a unique solution �⃗� = (𝑦1, 𝑦2, 𝑦3, 𝑦4) ∈ (𝐿
2(I × V))4 and �⃗�𝑡 =

(𝑦1𝑡 , 𝑦2𝑡 , 𝑦3𝑡 , 𝑦4𝑡 ) ∈ (𝐿
2(Q))4, �⃗�𝑡𝑡 = (𝑦1𝑡𝑡 , 𝑦2𝑡𝑡 , 𝑦3𝑡𝑡 , 𝑦4𝑡𝑡 ) ∈ (𝐿

2(I × V∗))4. 

Proof: Let 𝑉𝑛⃗⃗⃗⃗ = (𝑉𝑛)
4 ⊂ 𝑉 ⃗⃗⃗ (for each n) be the set of piecewise affine functions in Ω, let 

{𝑣𝑛 }𝑛=1
∞  be a sequence of subspaces of 𝑉 ⃗⃗⃗, s.t. ∀ �⃗� = (𝑣1, 𝑣2, 𝑣3, 𝑣4) ∈ 𝑉 ⃗⃗⃗, there is a sequence 

{𝑣𝑛 } 
  with �⃗�𝑛 = (𝑣1𝑛, 𝑣2𝑛 , 𝑣3𝑛 , 𝑣4𝑛 ) ∈ �⃗⃗�𝑛, ∀𝑛 and �⃗�𝑛 → �⃗� strongly (ST) in 𝑉 ⃗⃗⃗ then �⃗�𝑛 → �⃗� ST 

in (𝐿2(Ω))4. Let {𝑣𝑗⃗⃗⃗ ⃗ = (𝑣1𝑗 , 𝑣2𝑗 , 𝑣3𝑗 , 𝑣4𝑗 ): 𝑗 = 1,2, … , 𝑀(𝑛)} be a finite basis of �⃗⃗�𝑛 (where 𝑣𝑗⃗⃗⃗ ⃗  

is a piecewise affine function in Ω, with 𝑣𝑗⃗⃗⃗ ⃗(𝑥) = 0 on the boundary Γ) and let �⃗�𝑛 =

(𝑦1𝑛, 𝑦2𝑛, 𝑦3𝑛, 𝑦4𝑛) be the Galerkin approximate solution (GAS) to the exact solution �⃗� =

(𝑦1, 𝑦2, 𝑦3, 𝑦4) s.t.: 

𝑦𝑖𝑛 = Σ
𝑗=1

𝑛

𝑐𝑖𝑗 (𝑡)𝑣𝑖𝑗 (𝑥)                                                                                                         (16) 

𝑧𝑖𝑛 = Σ
𝑗=1

𝑛

𝑑𝑖𝑗 (𝑡)𝑣𝑖𝑗 (𝑥)                                                                                                         (17) 

Where 𝑐𝑖𝑗 (𝑡), 𝑑𝑖𝑗 (𝑡) are unknown functions, ∀𝑖 = 1,2,3,4, 𝑗 = 1,2, … , 𝑛. 

The MGA is utilized to approximate the WF ((8), (10), (12), (14))  w.r.t. 𝑥, they become after 

substituting 𝑦𝑖𝑛𝑡 = 𝑧𝑖𝑛 (∀𝑣𝑖 ∈ 𝑉𝑛, ∀𝑖 = 1,2,3,4): 

(𝑧1𝑛𝑡 , 𝑣1) + (∇𝑦1𝑛, ∇𝑣1) + (𝑦1𝑛, 𝑣1) − (𝑦2𝑛, 𝑣1) + (𝑦3𝑛, 𝑣1) + (𝑦4𝑛, 𝑣1) = (𝑓1, 𝑣1)        (18) 

(𝑦1𝑛
0 , 𝑣1) = (𝑦1

0, 𝑣1), and (𝑧1𝑛
1 , 𝑣1) = (𝑦1

1, 𝑣1)                                                                    (19) 

(𝑧2𝑛𝑡 , 𝑣2) + (∇𝑦2𝑛, ∇𝑣2) + (𝑦1𝑛, 𝑣2) + (𝑦2𝑛, 𝑣2) − (𝑦3𝑛, 𝑣2) − (𝑦4𝑛, 𝑣2) = (𝑓2, 𝑣2)       (20) 

(𝑦2𝑛
0 , 𝑣2) = (𝑦2

0, 𝑣2), and (𝑧2𝑛
1 , 𝑣2) = (𝑦2

1, 𝑣2)                                                                    (21) 

(𝑧3𝑛𝑡 , 𝑣3) + (∇𝑦3𝑛, ∇𝑣3) − (𝑦1𝑛, 𝑣3) + (𝑦2𝑛, 𝑣3) + (𝑦3𝑛, 𝑣3) + (𝑦4𝑛, 𝑣3) = (𝑓3, 𝑣3)       (22) 



IHJPAS. 53 (3)2022 
 

164 

(𝑦3𝑛
0 , 𝑣3) = (𝑦3

0, 𝑣3), and (𝑧3𝑛
1 , 𝑣3) = ((𝑦3

1, 𝑣3)                                                                  (23) 

(𝑧4𝑛𝑡 , 𝑣4) + (∇𝑦4𝑛, ∇𝑣4) − (𝑦1𝑛, 𝑣4) + (𝑦2𝑛, 𝑣4) − (𝑦3𝑛, 𝑣4) + (𝑦4𝑛, 𝑣4) = (𝑓4, 𝑣4)         (24) 

(𝑦4𝑛
0 , 𝑣2) = (𝑦4

0, 𝑣4), and (𝑧4𝑛
1 , 𝑣4) = ((𝑦4

1, 𝑣4)                                                                   (25) 

Where 𝑦𝑖𝑛
0 = 𝑦𝑖𝑛

0 (𝑥) = 𝑦𝑖𝑛
 (𝑥, 0) ∈ 𝑉𝑛 (respectively 𝑧𝑖𝑛

0 = 𝑦𝑖𝑛
1 = 𝑦𝑖𝑛

1 (𝑥) = 𝑦𝑖𝑛𝑡
 (𝑥, 0) ∈ 𝐿2(Ω) 

be the projection of 𝑦𝑖
0
 onto 𝑉 (be the projection of 𝑦𝑖

1 = 𝑦𝑖𝑡
  on to 𝐿2(Ω), ∀𝑖 = 1,2,3,4 ), i.e. 

𝑦𝑖𝑛
0 → 𝑦𝑖

0
 ST in 𝑉, with ∥ �⃗�𝑛

0 ∥1≤  𝑏0   and ∥ �⃗�𝑛
0 ∥0≤  𝑏0                                                    (26) 

𝑦𝑖𝑛
1 → 𝑦𝑖

1 ST in 𝐿2(Ω) , with ∥ �⃗�𝑛
1 ∥≤  𝑏1                                                                             (27) 

Substituting (16) & (17) with 𝑖 = 1,2,3,4 in )(18)-(25)( and setting 𝑣𝑖 = 𝑣𝑖𝑙  ,∀𝑙 = 1,2, … , 𝑛, 

then the obtained equations are equivalent to the following system of nonlinear ODEs of 1st 

order with ICs (which has a unique solution), i.e.  

𝐴1𝐷1(𝑡) + 𝐵1𝐶1(𝑡) − 𝐸𝐶2(𝑡) + 𝐹𝐶3(𝑡) + 𝐾𝐶4(𝑡) = 𝑏1(�̅�1
𝑇 (𝑥)𝐶1(𝑡))                                             

𝐴1𝐶1(0) = 𝑏1
0 and 𝐴1𝐷1̅̅ ̅(0) = 𝑏1

1                                                                                 

𝐴2𝐷2(𝑡) + 𝐵2𝐶2(𝑡) + 𝐻𝐶1(𝑡) − 𝐺𝐶3(𝑡) + 𝐷𝐶4(𝑡) = 𝑏2 (�̅�2
𝑇 (𝑥)𝐶1(𝑡))                                              

𝐴2𝐶2(0) = 𝑏2
0 and 𝐴2𝐷2̅̅ ̅(0) = 𝑏2

1                                                                                  

𝐴3𝐷3(𝑡) + 𝐵3𝐶3(𝑡) − 𝑅𝐶1(𝑡) + 𝑊𝐶2(𝑡) + 𝑍𝐶4(𝑡) = 𝑏3(�̅�3
𝑇 (𝑥)𝐶1(𝑡))                                               

𝐴3𝐶3(0) = 𝑏3
0 and 𝐴3𝐷3̅̅ ̅(0) = 𝑏3

1                                                                                  

𝐴4𝐷4(𝑡) + 𝐵4𝐶4(𝑡) − 𝑇𝐶1(𝑡) + 𝑀𝐶2(𝑡) − 𝑁𝐶3(𝑡) = 𝑏4(�̅�4
𝑇 (𝑥)𝐶1(𝑡))                                               

𝐴4𝐶4(0) = 𝑏4
0 and 𝐴4𝐷4̅̅ ̅(0) = 𝑏4

1                                                                                

where 𝐴𝑖 = (𝑎𝑖𝑙𝑗 )𝑛×𝑛, 𝑎𝑖𝑙𝑗 = (𝑣𝑖𝑗 , 𝑣𝑖𝑙 ), 𝐵𝑖 = (𝑏𝑖𝑙𝑗 )𝑛×𝑛, 𝑏𝑖𝑙𝑗 = (∇𝑣𝑖𝑗 , ∇𝑣𝑖𝑙 ) + (𝑣𝑖𝑗 , 𝑣𝑖𝑙 ), 𝐸 =

(𝑒𝑙𝑗 )𝑛×𝑛,𝑒𝑙𝑗 = (𝑣2𝑗 , 𝑣1𝑙 ),𝐹 = (𝑓𝑙𝑗 )𝑛×𝑛, 𝑓𝑙𝑗 = (𝑣3𝑗 , 𝑣1𝑙 ),𝐺 = (𝑔𝑙𝑗 )𝑛×𝑛, 𝑔𝑙𝑗 = (𝑣3𝑗 , 𝑣2𝑙 ), 𝐻 =

(ℎ𝑙𝑗 )𝑛×𝑛, ℎ𝑙𝑗 = (𝑣1𝑗 , 𝑣2𝑙 ),𝑅 = (𝑟𝑙𝑗 )𝑛×𝑛, 𝑟𝑙𝑗 = (𝑣1𝑗 , 𝑣3𝑙 ), 𝑊 = (𝑤𝑙𝑗 )𝑛×𝑛, 

 𝑤𝑙𝑗 = (𝑣2𝑗 , 𝑣3𝑙 ), 𝐾 = (𝑘𝑙𝑗 )𝑛×𝑛, 𝑘𝑙𝑗 = (𝑣4𝑗 , 𝑣1𝑙 ), 𝐷 = (𝑑𝑙𝑗 )𝑛×𝑛, 𝑑𝑙𝑗 = (𝑣4𝑗 , 𝑣2𝑙 ), 𝑍 =

(𝑧𝑙𝑗 )𝑛×𝑛, 𝑧𝑙𝑗 = (𝑣4𝑗 , 𝑣3𝑙 ), 𝑇 = (𝑡𝑙𝑗 )𝑛×𝑛, 𝑡𝑙𝑗 = (𝑣1𝑗 , 𝑣4𝑙 ), 𝑀 = (𝑚𝑙𝑗 )𝑛×𝑛, 𝑚𝑙𝑗 = (𝑣2𝑗 , 𝑣4𝑙 ), 

𝑁 = (𝑛𝑙𝑗 )𝑛×𝑛, 𝑛𝑙𝑗 = (𝑣3𝑗 , 𝑣4𝑙 ),𝑏𝑙 = (𝑏𝑙𝑖 ) = (𝑏𝑙𝑖 )𝑛×1, 𝑏𝑙𝑖 = (𝑓𝑖 (�̅�𝑙
𝑇 (𝑥)𝐶𝑙 (𝑡), 𝑢𝑖 ), 𝑣𝑙𝑖 ),   𝑏𝑙

𝑘 =

(𝑏𝑙𝑗
𝑘 ), 𝑏𝑙𝑗

0 = (𝑦𝑙
𝑘 , 𝑣𝑙𝑗 ), 𝐶𝑖 (𝑡) = (𝐶𝑖𝑗 (𝑡))𝑛×1, 𝐷𝑖 (0) = (𝑑𝑖𝑗 (0))𝑛×1, 𝐶𝑖 (0) = (𝐶𝑖𝑗 (0))𝑛×1, 

𝐷𝑖 (𝑡) = (𝑑𝑖𝑗 (𝑡))𝑛×1. 

Then corresponding to the sequence {�⃗�𝑛}, the following approximation problems are held, i.e. 

for each �⃗�𝑛 = (𝑣1𝑛, 𝑣2𝑛 , 𝑣3𝑛  , 𝑣4𝑛 ) ⊂ 𝑉𝑛⃗⃗⃗⃗  , and 𝑛 = 1,2, … 

(𝑦1𝑛𝑡𝑡 , 𝑣1𝑛) + (∇𝑦1𝑛, ∇𝑣1𝑛) + (𝑦1𝑛, 𝑣1𝑛) − (𝑦2𝑛, 𝑣1𝑛) + (𝑦3𝑛, 𝑣1𝑛) + (𝑦4𝑛, 𝑣1𝑛) 

= (𝑓1(𝑦1𝑛, 𝑢1), 𝑣1𝑛)                                                                                                                (28) 

(𝑦1𝑛
0 , 𝑣1𝑛) = (𝑦1

0, 𝑣1𝑛), and (𝑦1𝑛
1 , 𝑣1𝑛) = (𝑦1

1, 𝑣1𝑛)                                                              (29) 

(𝑦2𝑛𝑡𝑡 , 𝑣2𝑛 ) + (∇𝑦2𝑛, ∇𝑣2𝑛) + (𝑦1𝑛, 𝑣2𝑛) + (𝑦2𝑛, 𝑣2) − (𝑦3𝑛, 𝑣2𝑛 ) − (𝑦4𝑛, 𝑣2𝑛 ) =

(𝑓2(𝑦2𝑛, 𝑢2), 𝑣2𝑛 )                                                                                                                   (30) 

(𝑦2𝑛
0 , 𝑣2𝑛 ) = (𝑦1

0, 𝑣2𝑛 ), and (𝑦2𝑛
1 , 𝑣2𝑛 ) = (𝑦1

1, 𝑣2𝑛 )                                                             (31) 

(𝑦3𝑛𝑡𝑡 , 𝑣3𝑛 ) + (∇𝑦3𝑛, ∇𝑣3𝑛) − (𝑦1𝑛, 𝑣3𝑛) + (𝑦2𝑛, 𝑣3𝑛 ) + (𝑦3𝑛, 𝑣3𝑛 ) + (𝑦4𝑛, 𝑣3𝑛) =

(𝑓3(𝑦3𝑛, 𝑢3), 𝑣3𝑛 )                                                                                                                   (32) 

(𝑦3𝑛
0 , 𝑣3𝑛 ) = (𝑦3

0, 𝑣3𝑛 ), and (𝑦3𝑛
1 , 𝑣3) = (𝑦3

1, 𝑣3𝑛)                                                                

(33) 

(𝑦4𝑛𝑡𝑡 , 𝑣4𝑛 ) + (∇𝑦4𝑛, ∇𝑣4𝑛) − (𝑦1𝑛, 𝑣4𝑛 ) + (𝑦2𝑛, 𝑣4𝑛 ) − (𝑦3𝑛, 𝑣4𝑛) + (𝑦4𝑛, 𝑣4𝑛) =

(𝑓4(𝑦4𝑛, 𝑢4), 𝑣4𝑛 )                                                                                                                   (34) 

(𝑦4𝑛
0 , 𝑣4𝑛) = (𝑦4

0, 𝑣4𝑛), and (𝑦4𝑛
1 , 𝑣4𝑛 ) = (𝑦4

1, 𝑣4𝑛 )                                                              (35) 



IHJPAS. 53 (3)2022 
 

165 

Which have a sequence of unique solutions {�⃗�𝑛}. Substituting 𝑣𝑖𝑛 = 𝑦𝑖𝑛𝑡 , for 𝑖 = 1,2,3,4  in 

(28), (30), (32) and (34) resp. , using Lemma 1.2 in [15] for the 1st terms of the LHS of each 

equality, then adding the resulting equation, to get 
𝑑

𝑑𝑡
[∥ �⃗�𝑛𝑡 ∥0

2+
𝑑

𝑑𝑡
∥ �⃗�𝑛 ∥1

2] = 2[(𝑦2𝑛, 𝑦1𝑛𝑡 ) − (𝑦3𝑛, 𝑦1𝑛𝑡 ) − (𝑦4𝑛, 𝑦1𝑛𝑡 ) − (𝑦1𝑛, 𝑦2𝑛𝑡 ) +

(𝑦3𝑛, 𝑦2𝑛𝑡 ) + (𝑦4𝑛, 𝑦2𝑛𝑡 ) + (𝑦1𝑛, 𝑦3𝑛𝑡 ) − (𝑦2𝑛, 𝑦3𝑛𝑡 ) − (𝑦4𝑛, 𝑦3𝑛𝑡 ) + (𝑦1𝑛, 𝑦4𝑛𝑡 ) −
(𝑦2𝑛, 𝑦4𝑛𝑡 ) + (𝑦3𝑛, 𝑦4𝑛𝑡 )+(𝑓1(𝑦1𝑛, 𝑢1), 𝑦1𝑛𝑡 ) + (𝑓2(𝑦2𝑛, 𝑢2), 𝑦2𝑛𝑡 ) + (𝑓3(𝑦3𝑛, 𝑢3), 𝑦3𝑛𝑡 ) +
(𝑓4(𝑦4𝑛, 𝑢4), 𝑦4𝑛𝑡 )]                                                                                                               (36) 
Taking the absolute value for both sides, we get: 
𝑑

𝑑𝑡
[∥ �⃗�𝑛𝑡 ∥0

2+∥ �⃗�𝑛 ∥1
2] ≤ 2[∣ (𝑦2𝑛, 𝑦1𝑛𝑡 ) ∣ +∣ (𝑦3𝑛, 𝑦1𝑛𝑡 ) ∣ +∣ (𝑦4𝑛, 𝑦1𝑛𝑡 ) ∣ +∣(𝑦1𝑛, 𝑦2𝑛𝑡 )∣ 

+∣(𝑦3𝑛, 𝑦2𝑛𝑡 )∣+∣(𝑦4𝑛, 𝑦2𝑛𝑡 )∣+∣(𝑦1𝑛, 𝑦3𝑛𝑡 )∣+∣(𝑦2𝑛, 𝑦3𝑛𝑡 )∣+∣ (𝑦4𝑛, 𝑦3𝑛𝑡 ) ∣ +∣ (𝑦1𝑛, 𝑦4𝑛𝑡 ) ∣ 
+∣ (𝑦2𝑛, 𝑦4𝑛𝑡 ) ∣ +∣ (𝑦3𝑛, 𝑦4𝑛𝑡 ) ∣ +∣ (𝑓1(𝑦1𝑛, 𝑢1), 𝑦1𝑛𝑡 ) ∣ +∣ (𝑓2(𝑦2𝑛, 𝑢2), 𝑦2𝑛𝑡 ) ∣ + 
∣ (𝑓3(𝑦3𝑛, 𝑢3), 𝑦3𝑛𝑡 ) ∣ +∣ (𝑓4(𝑦4𝑛, 𝑢4), 𝑦4𝑛𝑡 ) ∣]                                                                   (37) 
 

Using Assumptions (A) for the R.H.L. of (37), integrating both sides (IBS) on [0, 𝑡], using  

∥ 𝑦𝑖𝑛 ∥0
 ≤∥ �⃗�𝑛 ∥1

 , ∥ �⃗�𝑖𝑛𝑡 ∥0
 ≤∥ �⃗�𝑖𝑛𝑡 ∥1

 , ∥ �⃗�𝑛𝑡 ∥0
 ≤∥ �⃗�𝑛𝑡 ∥1

 , to get   

∥ �⃗�𝑛𝑡 ∥0
2+∥ �⃗�𝑛 ∥1

2≤ 3 ∫  
0

𝑡

[∥ �⃗�𝑛𝑡 ∥0
2+∥ �⃗�𝑛 ∥1

2]𝑑𝑡 + (∥ 𝐹1 ∥𝑄
2 +∥ 𝐹2 ∥𝑄

2 +∥ 𝐹3 ∥𝑄
2 +∥ 𝐹4 ∥𝑄

2 ) +

𝛽5 ∫  
0

𝑡

∥ �⃗�𝑛 ∥1
2 𝑑𝑡 + (1 + 𝛽5 + 𝛾5) ∫  

0

𝑡

∥ �⃗�𝑛𝑡 ∥0
2 𝑑𝑡 + (𝛾5 + 𝑐5) ∥ �⃗⃗� ∥𝑄

2 + 𝑏0 + 𝑏1  

≤ 𝑑6 + 3 ∫  
0

𝑡

[∥ �⃗�𝑛 ∥1
2+∥ 𝑦𝑛𝑡 ∥0

2]𝑑𝑡 + 𝛽5 ∫  
0

𝑡

∥ 𝑦𝑛 ∥1
2 𝑑𝑡 + 𝛽6 ∫  

0

𝑡

∥ 𝑦𝑛𝑡 ∥0
2 𝑑𝑡  

≤ 𝑑6 + 𝛽7 ∫  
0

𝑡

[∥ �⃗�𝑛 ∥1
2+∥ 𝑦𝑛𝑡 ∥0

2]𝑑𝑡                                                                              (38) 

Where𝛽5 = max (𝛽1, 𝛽2, 𝛽3, 𝛽4), 𝛾5 = max (𝛾1, 𝛾2, 𝛾3, 𝛾4),∥ 𝑢𝑖 ∥𝑄
2 ≤ 𝑐𝑖(∀𝑖 = 1,2,3,4),  

∥ 𝐹𝑖 ∥𝑄
2 ≤ 𝑑𝑖,  𝑑5 = ∑  

𝑖=1

4

𝑑𝑖, 𝑐5 = max (𝑐1, 𝑐2, 𝑐3, 𝑐4),  𝑑6 = 𝛾5 + 𝑐5 + 𝑑5 + 𝑏0 + 𝑏1, 𝛽6 = 1 +

𝛽5 + 𝛾5 𝛽7 = max (3, 𝛽6). 

Using Lemma 2.2, ∀𝑡 ∈ [0, 𝑡] to get 

∥ �⃗�𝑛𝑡 (𝑡) ∥0
2+∥ �⃗�𝑛 (𝑡) ∥1

2≤ 𝑑6𝑒
𝛽7 ∫  

0

𝑇

𝑑𝑡
= 𝑏2(𝑐) ⟹ ∥ �⃗�𝑛𝑡 (𝑡) ∥0

2≤ 𝑏2(𝑐) and ∥ �⃗�𝑛(𝑡) ∥1
2≤ 𝑏2(𝑐). 

 Easily once can obtain that ∥ �⃗�𝑛𝑡 (𝑡) ∥𝑄
 ≤ 𝑏 1(𝑐) and ∥ �⃗�𝑛 (𝑡) ∥𝐿2(𝐼,𝑉)

 ≤ 𝑏 (𝑐). 

Then, by applying the Alaoglu’s theorem (ALTh), there is a subsequence of {�⃗�𝑛}𝑛∈𝑁, for 

simplicity say {�⃗�𝑛 }  s.t. �⃗�𝑛𝑡 → �⃗� weakly (WK) in (𝐿
2(𝑄))4 and �⃗�𝑛 → �⃗� WK in (𝐿

2(𝐼, 𝑉))4 . 

but                               

(𝐿2(ℝ, 𝑉))4 ⊂ (𝐿2(ℝ, Ω))4 ≅ ((𝐿2(ℝ, Ω))∗)4 ⊂ (𝐿2(ℝ, 𝑉∗))4                                         (39) 

Then the (ACTH)[15] can be employed here to get that �⃗�𝑛 → �⃗� ST in (𝐿
2(𝑄))4 . Now  

multiplying both sides (MBS) of ((28), (30),(32)&(34)) by 𝜙𝑖 (𝑇) ∈ 𝐶
2[0, 𝑇] s.t. 𝜙𝑖 (𝑇) =

 𝜙𝑖
′(𝑇) = 0, 𝜙𝑖 (0) ≠ 0, 𝜙𝑖

′(0) ≠ 0, ∀𝑖 = 1,2,3,4, IBS on [0, 𝑇], finally integrating by parts 

twice (IBPs2) the 1st term in each equation yield to 

− ∫  
0

𝑇
𝑑

𝑑𝑡
(𝑦1𝑛, 𝑣1𝑛 )𝜙1

′ 𝑑𝑡 + ∫  
0

𝑇

[(∇𝑦1𝑛, ∇𝑣1𝑛) + (𝑦1𝑛, 𝑣1𝑛 ) − (𝑦2𝑛, 𝑣1𝑛 ) + (𝑦3𝑛, 𝑣1𝑛 ) +

(𝑦4𝑛, 𝑣1𝑛)]𝜙1
 (𝑡)𝑑𝑡 =∫  

0

𝑇

(𝑓1(𝑦1𝑛, 𝑢1), 𝑣1𝑛)𝜙1
 (𝑡)𝑑𝑡 + (𝑦1𝑛

′  , 𝑣1𝑛 ) 𝜙1(0)                     (40) 



IHJPAS. 53 (3)2022 
 

166 

∫  
0

𝑇

(𝑦1𝑛, 𝑣1𝑛)𝜙1
′′𝑑𝑡 + ∫  

0

𝑇

[(∇𝑦1𝑛, ∇𝑣1𝑛) + (𝑦1𝑛, 𝑣1𝑛) − (𝑦2𝑛, 𝑣1𝑛) + (𝑦3𝑛, 𝑣1𝑛) +

(𝑦4𝑛, 𝑣1𝑛)]𝜙1
 (𝑡)𝑑𝑡=∫  

0

𝑇

(𝑓1(𝑦1𝑛, 𝑢1), 𝑣1𝑛)𝜙1
 (𝑡)𝑑𝑡 + (𝑦1𝑛

′  , 𝑣1𝑛 )𝜙1(0) − 

(𝑦1𝑛
0  , 𝑣1𝑛)𝜙1

′ (0)                                                                                                           (41) 

− ∫  
0

𝑇
𝑑

𝑑𝑡
(𝑦2𝑛, 𝑣2𝑛 )𝜙2

′ 𝑑𝑡 + ∫  
0

𝑇

[(∇𝑦2𝑛, ∇𝑣2𝑛) + (𝑦1𝑛, 𝑣2𝑛 ) + (𝑦2𝑛, 𝑣2𝑛) − (𝑦3𝑛, 𝑣2𝑛 ) −

(𝑦4𝑛, 𝑣2𝑛 )]𝜙2
 (𝑡)𝑑𝑡 =∫  

0

𝑇

(𝑓2(𝑦2𝑛, 𝑢2), 𝑣2𝑛 )𝜙2
 (𝑡)𝑑𝑡 + (𝑦2𝑛

′  , 𝑣2𝑛 ) 𝜙1(0)                           (42) 

∫  
0

𝑇

(𝑦2𝑛, 𝑣2𝑛 )𝜙2
′′(𝑡)𝑑𝑡 + ∫  

0

𝑇

[(∇𝑦2𝑛, ∇𝑣2𝑛) + (𝑦1𝑛, 𝑣2𝑛 ) + (𝑦2𝑛, 𝑣2𝑛 ) − (𝑦3𝑛, 𝑣2𝑛 ) −

(𝑦4𝑛, 𝑣2𝑛 )]𝜙2
 (𝑡)𝑑𝑡  

= ∫  
0

𝑇

(𝑓2(𝑦2𝑛, 𝑢2), 𝑣2𝑛)𝜙2
 (𝑡)𝑑𝑡 + (𝑦2𝑛

′  , 𝑣2𝑛 )𝜙2(0) − (𝑦2𝑛
0  , 𝑣2𝑛 )𝜙2

′ (0)                            (43) 

− ∫  
0

𝑇
𝑑

𝑑𝑡
(𝑦3𝑛, 𝑣3𝑛 )𝜙3

′ (𝑡)𝑑𝑡 + ∫  
0

𝑇

[(∇𝑦3𝑛, ∇𝑣3𝑛) − (𝑦1𝑛, 𝑣3𝑛 ) + (𝑦2𝑛, 𝑣3𝑛) + (𝑦3𝑛, 𝑣3𝑛 ) +

(𝑦4𝑛, 𝑣3𝑛 )]𝜙3
 (𝑡)𝑑𝑡 =∫  

0

𝑇

(𝑓3(𝑦3𝑛, 𝑢3), 𝑣3𝑛 )𝜙3
 (𝑡)𝑑𝑡 + (𝑦3𝑛

′  , 𝑣3𝑛 ) 𝜙3(0)                             (44) 

∫  
0

𝑇

(𝑦3𝑛, 𝑣3𝑛 )𝜙3
′′𝑑𝑡 + ∫  

0

𝑇

[(∇𝑦3𝑛, ∇𝑣3𝑛) − (𝑦1𝑛, 𝑣3𝑛) + (𝑦2𝑛, 𝑣3𝑛 ) + (𝑦3𝑛, 𝑣3𝑛 ) +

(𝑦4𝑛, 𝑣3𝑛 )]𝜙3
 (𝑡)𝑑𝑡  

=∫  
0

𝑇

(𝑓3(𝑦3𝑛, 𝑢3), 𝑣3𝑛)𝜙3
 (𝑡)𝑑𝑡 + (𝑦3𝑛

′  , 𝑣3𝑛 )𝜙3(0) − (𝑦3𝑛
0  , 𝑣3𝑛 )𝜙3

′ (0)               (45) 

− ∫  
0

𝑇
𝑑

𝑑𝑡
(𝑦4𝑛, 𝑣4𝑛 )𝜙4

′ (𝑡)𝑑𝑡 + ∫  
0

𝑇

[(∇𝑦4𝑛, ∇𝑣4𝑛) − (𝑦1𝑛, 𝑣4𝑛) + (𝑦2𝑛, 𝑣4𝑛 ) − (𝑦3𝑛, 𝑣4𝑛 ) +

(𝑦4𝑛, 𝑣4𝑛)]𝜙4
 (𝑡)𝑑𝑡 =∫  

0

𝑇

(𝑓4(𝑦4𝑛, 𝑢4), 𝑣4𝑛)𝜙4
 (𝑡)𝑑𝑡 + (𝑦4𝑛

′  , 𝑣4𝑛 ) 𝜙4(0)                    (46) 

∫  
0

𝑇

(𝑦4𝑛, 𝑣4𝑛)𝜙4
′′𝑑𝑡 + ∫  

0

𝑇

[(∇𝑦4𝑛, ∇𝑣4𝑛 ) − (𝑦1𝑛, 𝑣4𝑛 ) + (𝑦2𝑛, 𝑣4𝑛 ) − (𝑦3𝑛, 𝑣4𝑛 ) +

(𝑦4𝑛, 𝑣4𝑛)]𝜙4
 (𝑡)𝑑𝑡=∫  

0

𝑇

(𝑓4(𝑦4𝑛, 𝑢4), 𝑣4𝑛)𝜙4
 (𝑡)𝑑𝑡 + (𝑦4𝑛

′  , 𝑣4𝑛 )𝜙4(0) − 

(𝑦4𝑛
0  , 𝑣4𝑛)𝜙4

′ (0)                                                                                                         (47) 

First, since 𝑣𝑖𝑛 → 𝑣𝑖  ST in 𝐿
2(Ω) ⟹ {

𝑣𝑖𝑛 𝜙𝑖
 (𝑡) → 𝑣𝑖 𝜙𝑖

 (𝑡)

𝑣𝑖𝑛𝜙𝑖
′(𝑡) → 𝑣𝑖 𝜙𝑖

′(𝑡)
 ST in 𝐿2(I, V),  

𝑣𝑖𝑛 𝜙𝑖
 (0) → 𝑣𝑖 𝜙𝑖

 (0) ST in 𝐿2(Ω) and 𝑣𝑖𝑛 𝜙𝑖
′(0) → 𝑣𝑖 𝜙𝑖

′(0) ST in 𝐿2(Ω) for 𝑖 = 1,2,3,4 

Also since 𝑣𝑖𝑛 → 𝑣𝑖  ST in 𝑉 ⟹ {
𝑣𝑖𝑛 𝜙𝑖

′(𝑡) → 𝑣𝑖 𝜙𝑖
′(𝑡)

𝑣𝑖𝑛𝜙𝑖
′′(𝑡) → 𝑣𝑖 𝜙𝑖

′′(𝑡)
  ST in 𝐿2(Q). 

Second, we have  𝑦𝑖𝑛𝑡 → 𝑦𝑖𝑡 WK in 𝐿
2(Q) and 𝑦𝑖𝑛𝑡 → 𝑦𝑖𝑡 WK in 𝐿

2(I, V) and ST in 𝐿2(Q). 

Third and on the other hand, let 𝑤𝑖𝑛 = 𝑣𝑖𝑛 𝜙𝑖
  and 𝑤𝑖 = 𝑣𝑖  𝜙𝑖

  then 𝑤𝑖𝑛 → 𝑤𝑖 ST in 𝐿
2(Q) and 

then 𝑤𝑖𝑛 is measurable w.r.t. (𝑥, 𝑡), so using Assm (A-(i)), employing Proposition 1.3, the 

(𝑓𝑖 (𝑥, 𝑡, 𝑦𝑖𝑛, 𝑢𝑖 ), 𝑤𝑖𝑛)𝑑𝑥𝑑𝑡 is cont. w.r.t. (𝑦𝑖𝑛, 𝑢𝑖 , 𝑤𝑖𝑛), then   

∫  
0

𝑇

(𝑓𝑖 (𝑦𝑖𝑛, 𝑢𝑖 ), 𝑣𝑖𝑛)𝜙𝑖
 (𝑡)𝑑𝑡 → ∫  

0

𝑇

(𝑓𝑖 (𝑦𝑖 , 𝑢𝑖 ), 𝑣𝑖 )𝜙𝑖
 (𝑡)𝑑𝑡, ∀𝑖 = 1,2,3,4  

From these convergences,  (26) , and (27) we can passage the limits in ((40)-(47)), to get 



IHJPAS. 53 (3)2022 
 

167 

− ∫  
0

𝑇

(𝑦1𝑡 , 𝑣1)𝜙1
′ (𝑡)𝑑𝑡 + ∫  

0

𝑇

[(∇𝑦1, ∇𝑣1) + (𝑦1, 𝑣1) − (𝑦2, 𝑣1) + (𝑦3, 𝑣1) + (𝑦4, 𝑣1)]𝜙1
 (𝑡)𝑑𝑡 

=∫  
0

𝑇

(𝑓1(𝑦1, 𝑢1), 𝑣1)𝜙1
 (𝑡)𝑑𝑡 + (𝑦1

′  , 𝑣1) 𝜙1(0)                                                                    (48) 

∫  
0

𝑇

(𝑦1, 𝑣1)𝜙1
′′𝑑𝑡 + ∫  

0

𝑇

[(∇𝑦1, ∇𝑣1) + (𝑦1, 𝑣1) − (𝑦2, 𝑣1) + (𝑦3, 𝑣1) + (𝑦4, 𝑣1)]𝜙1
 (𝑡)𝑑𝑡  

=∫  
0

𝑇

(𝑓1(𝑦1, 𝑢1), 𝑣1)𝜙1
 (𝑡)𝑑𝑡 + (𝑦1

′  , 𝑣1)𝜙1(0) − (𝑦1
0 , 𝑣1)𝜙1

′ (0)                                         (49) 

− ∫  
0

𝑇

(𝑦2𝑡 , 𝑣2)𝜙2
′ 𝑑𝑡 + ∫  

0

𝑇

[(∇𝑦2, ∇𝑣2) + (𝑦1, 𝑣2) + (𝑦2, 𝑣2) − (𝑦3, 𝑣2) − (𝑦4, 𝑣2)]𝜙2
 (𝑡)𝑑𝑡 

=∫  
0

𝑇

(𝑓2(𝑦2, 𝑢2), 𝑣2)𝜙2
 (𝑡)𝑑𝑡 + (𝑦2

′  , 𝑣2) 𝜙1(0)                                                                    (50) 

∫  
0

𝑇

(𝑦2, 𝑣2)𝜙2
′′(𝑡)𝑑𝑡 + ∫  

0

𝑇

[(∇𝑦2, ∇𝑣2) + (𝑦1, 𝑣2) + (𝑦2, 𝑣2) − (𝑦3, 𝑣2) − (𝑦4, 𝑣2)]𝜙2
 (𝑡)𝑑𝑡  

=∫  
0

𝑇

(𝑓2(𝑦2, 𝑢2), 𝑣2)𝜙2
 (𝑡)𝑑𝑡 + (𝑦2

′  , 𝑣2)𝜙2(0) − (𝑦2
0 , 𝑣2)𝜙2

′ (0)                                       (51) 

− ∫  
0

𝑇

(𝑦3𝑡 , 𝑣3)𝜙3
′ (𝑡)𝑑𝑡 + ∫  

0

𝑇

[(∇𝑦3, ∇𝑣3) − (𝑦1, 𝑣3) + (𝑦2, 𝑣3) + (𝑦3, 𝑣3) + (𝑦4, 𝑣3)]𝜙3
 (𝑡)𝑑𝑡 

=∫  
0

𝑇

(𝑓3(𝑦3, 𝑢3), 𝑣3)𝜙3
 (𝑡)𝑑𝑡 + (𝑦3

′  , 𝑣3) 𝜙3(0)                                                                   (52) 

∫  
0

𝑇

(𝑦3, 𝑣3)𝜙3
′′𝑑𝑡 + ∫  

0

𝑇

[(∇𝑦3, ∇𝑣3) − (𝑦1, 𝑣3) + (𝑦2, 𝑣3) + (𝑦3, 𝑣3) + (𝑦4, 𝑣3)]𝜙3
 (𝑡)𝑑𝑡  

=∫  
0

𝑇

(𝑓3(𝑦3, 𝑢3), 𝑣3)𝜙3
 (𝑡)𝑑𝑡 + (𝑦3

′  , 𝑣3)𝜙3(0) − (𝑦3
0 , 𝑣3)𝜙3

′ (0)                                        (53) 

− ∫  
0

𝑇

(𝑦4𝑡 , 𝑣4)𝜙4
′ (𝑡)𝑑𝑡 + ∫  

0

𝑇

[(∇𝑦4, ∇𝑣4) − (𝑦1, 𝑣4) + (𝑦2, 𝑣4) − (𝑦3, 𝑣4) + (𝑦4, 𝑣4)]𝜙4
 (𝑡)𝑑𝑡 

=∫  
0

𝑇

(𝑓4(𝑦4, 𝑢4), 𝑣4)𝜙4
 (𝑡)𝑑𝑡 + (𝑦4

′  , 𝑣4) 𝜙4(0)                                                                     (54) 

∫  
0

𝑇

(𝑦4, 𝑣4)𝜙4
′′𝑑𝑡 + ∫  

0

𝑇

[(∇𝑦4, ∇𝑣4) − (𝑦1, 𝑣4) + (𝑦2, 𝑣4) − (𝑦3, 𝑣4) + (𝑦4, 𝑣4)]𝜙4
 (𝑡)𝑑𝑡  

= ∫  
0

𝑇

(𝑓4(𝑦4, 𝑢4), 𝑣4)𝜙4
 (𝑡)𝑑𝑡 + (𝑦4

′  , 𝑣4)𝜙4(0) − (𝑦4
0 , 𝑣4)𝜙4

′ (0)                                       (55) 

Case1: Choose 𝜙𝑖 ∈ 𝐶
2[0, 𝑇] s.t. 𝜙𝑖 (0) = 𝜙𝑖

′(0) =  𝜙𝑖
′(𝑇) = 𝜙𝑖

 (𝑇) = 0, ∀𝑖 = 1,2,3,4 

substituting in (49), (51), (53), (55) IBPs2 the 1st terms in the LHS, i.e. 

∫  
0

𝑇

(𝑦1𝑡𝑡 , 𝑣1)𝜙1
 𝑑𝑡 + ∫  

0

𝑇

[(∇𝑦1, ∇𝑣1) + (𝑦1, 𝑣1) − (𝑦2, 𝑣1) + (𝑦3, 𝑣1) + (𝑦4, 𝑣1)]𝜙1
 (𝑡)𝑑𝑡  

=∫  
0

𝑇

(𝑓1(𝑦1, 𝑢1), 𝑣1)𝜙1
 (𝑡)𝑑𝑡                                                                                                 (56) 

∫  
0

𝑇

(𝑦2𝑡𝑡 , 𝑣2)𝜙2
 (𝑡)𝑑𝑡 + ∫  

0

𝑇

[(∇𝑦2, ∇𝑣2) + (𝑦1, 𝑣2) + (𝑦2, 𝑣2) − (𝑦3, 𝑣2) − (𝑦4, 𝑣2)]𝜙2
 (𝑡)𝑑𝑡  

=∫  
0

𝑇

(𝑓2(𝑦2, 𝑢2), 𝑣2)𝜙2
 (𝑡)𝑑𝑡)                                                                                               (57) 

∫  
0

𝑇

(𝑦3𝑡𝑡 , 𝑣3)𝜙3
 𝑑𝑡 + ∫  

0

𝑇

[(∇𝑦3, ∇𝑣3) − (𝑦1, 𝑣3) + (𝑦2, 𝑣3) + (𝑦3, 𝑣3) + (𝑦4, 𝑣3)]𝜙3
 (𝑡)𝑑𝑡  



IHJPAS. 53 (3)2022 
 

168 

=∫  
0

𝑇

(𝑓3(𝑦3, 𝑢3), 𝑣3)𝜙3
 (𝑡)𝑑𝑡                                                                                                 (58) 

∫  
0

𝑇

(𝑦4𝑡𝑡 , 𝑣4)𝜙4
 𝑑𝑡 + ∫  

0

𝑇

[(∇𝑦4, ∇𝑣4) − (𝑦1, 𝑣4) + (𝑦2, 𝑣4) − (𝑦3, 𝑣4) + (𝑦4, 𝑣4)]𝜙4
 (𝑡)𝑑𝑡  

= ∫  
0

𝑇

(𝑓4(𝑦4, 𝑢4), 𝑣4)𝜙4
 (𝑡)𝑑𝑡                                                                                               (59) 

Hence �⃗� is a solution of (16),(18), (20) &(22) a.e. on I 

Case2: Choose 𝜙𝑖 ∈ 𝐶
2[0, 𝑇] s.t. 𝜙𝑖 (𝑇) = 0, &𝜙𝑖

 (0) = 0, ∀𝑖 = 1,2,3,4. MBS of (8), (10), 

(12) and (14) by 𝜙1(𝑡), 𝜙2(𝑡), 𝜙3(𝑡), and 𝜙4(𝑡)resp., IBS on [0, 𝑇],  then IBPs the 1
st term in 

the LHS of each equation, then subtracting each one of these obtained equations from those 

corresponding in (48),(50),(52) &(54) resp. to get 

(𝑦𝑖𝑡 (0), 𝑣𝑖 )𝜙𝑖
 (0) = (𝑦𝑖

′(0), 𝑣𝑖 )𝜙𝑖
 (0), ∀𝑖 = 1,2,3,4. 

Case 3: Choose 𝜙𝑖 ∈ 𝐶
2[0, 𝑇] s.t. 𝜙𝑖 (0) =  𝜙𝑖

′(𝑇) = 𝜙𝑖
 (𝑇) = 0, 𝜙𝑖

′(0) ≠ 0, ∀𝑖 = 1,2,3,4. 

MBS of (8), (10), (12) and (14) by 𝜙1(𝑡), 𝜙2(𝑡), 𝜙3(𝑡), and 𝜙4(𝑡)resp., IBS on [0, 𝑇] then 

IBPs2 the 1st term in the LHS of each equation, then subtracting each one of these obtained 

equations from those corresponding in (49),(51),(53), and (55) resp. to get 

(𝑦𝑖 (0), 𝑣𝑖 )𝜙𝑖
′(0) = (𝑦𝑖

0, 𝑣𝑖 )𝜙𝑖
′(0), ∀𝑖 = 1,2,3,4  

In the last two cases, the ICs (19), (11), (13) and (14) are held. 

To prove that �⃗�𝑛 → �⃗� ST in 𝐿
2(𝐼, 𝑉), we start by IBS (36) on [0, 𝑇], to get 

∫  
0

𝑇
𝑑

𝑑𝑡
∥ �⃗�𝑛 ∥0

2 𝑑𝑡 + 2 ∫  
0

𝑇

∥ �⃗�𝑛 ∥1
2 𝑑𝑡 = 2 ∫  

0

𝑇

[(𝑦2𝑛 , 𝑦1𝑛𝑡 ) − (𝑦3𝑛, 𝑦1𝑛𝑡 ) − (𝑦4𝑛, 𝑦1𝑛𝑡 ) −

(𝑦1𝑛, 𝑦2𝑛𝑡 ) + (𝑦3𝑛, 𝑦2𝑛𝑡 ) + (𝑦4𝑛, 𝑦2𝑛𝑡 ) + (𝑦1𝑛, 𝑦3𝑛𝑡 ) − (𝑦2𝑛, 𝑦3𝑛𝑡 ) − (𝑦4𝑛, 𝑦4𝑛𝑡 ) +

(𝑦1𝑛, 𝑦4𝑛𝑡 ) − (𝑦2𝑛, 𝑦4𝑛𝑡 ) + (𝑦3𝑛, 𝑦4𝑛𝑡 )]𝑑𝑡 + 2 ∫  
0

𝑇

[(𝑓1(𝑦1, 𝑢1), 𝑦1𝑛𝑡 ) + (𝑓2(𝑦2, 𝑢2), 𝑦2𝑛𝑡 ) +

(𝑓3(𝑦3, 𝑢3), 𝑦3𝑛𝑡 ) + (𝑓4(𝑦4, 𝑢4), 𝑦4𝑛𝑡 )]𝑑𝑡                                                                             (60) 

The same way applied to acquire (36)&(60), can be utilize here to obtain, i.e. 

∥ 𝑦𝑡 (𝑇) ∥0
2−∥ 𝑦𝑡 (0) ∥0

2+ 2 ∫  
0

𝑇

∥ �⃗�(𝑡) ∥1
2 𝑑𝑡=2 ∫  

0

𝑇

[(𝑦2, 𝑦1𝑡 ) − (𝑦3, 𝑦1𝑡 ) − (𝑦4, 𝑦1𝑡 ) − (𝑦1, 𝑦2𝑡 ) 

+(𝑦3, 𝑦2𝑡 ) +  (𝑦4, 𝑦2𝑡 ) + (𝑦1, 𝑦3𝑡 ) − (𝑦2, 𝑦3𝑡 ) − (𝑦4, 𝑦4𝑡 ) + (𝑦1, 𝑦4𝑡 ) − (𝑦2, 𝑦4𝑡 ) +

(𝑦3, 𝑦4𝑡 )]𝑑𝑡 +2 ∫  
0

𝑇

[(𝑓1(𝑦1, 𝑢1), 𝑦1𝑡 ) + (𝑓2(𝑦2, 𝑢2), 𝑦2𝑡 ) + (𝑓3(𝑦3, 𝑢3), 𝑦3𝑡 ) +

(𝑓4(𝑦4, 𝑢4), 𝑦4𝑡 )]𝑑𝑡                                                                                                               (61) 

Since 

∥ �⃗�𝑛𝑡 (𝑇) − �⃗�𝑡 (𝑇) ∥0
2−∥ �⃗�𝑛𝑡 (0) − �⃗�𝑡 (0) ∥0

2+ 2 ∫  
0

𝑇

∥ �⃗�𝑛 (𝑡) − 𝑦 ⃗⃗⃗(𝑡) ∥1
2 𝑑𝑡=(a)-(b)-(c)          (62) 

(a)= ∥ �⃗�𝑛𝑡 (𝑇) ∥0
2−∥ �⃗�𝑛𝑡 (0) ∥0

2+ 2 ∫  
0

𝑇

∥ �⃗�𝑛(𝑡) ∥1
2 𝑑𝑡 . 

(b)=(�⃗�𝑛𝑡 (𝑇), �⃗�𝑡 (𝑇)) − (�⃗�𝑛𝑡 (0), �⃗�𝑡 (0))) + 2 ∫  
0

𝑇

(�⃗�𝑛(𝑡), 𝑦 ⃗⃗⃗(𝑡))1
𝑑𝑡 

(c)=( �⃗�𝑡 (𝑇), �⃗�𝑛𝑡 (𝑇) − �⃗�𝑡 (𝑇)) − ( �⃗�𝑡 (0), �⃗�𝑛𝑡 (0) − �⃗�𝑡 (0)) + 2 ∫  
0

𝑇

(𝑦 ⃗⃗⃗(𝑡), �⃗�𝑛(𝑡) − 𝑦 ⃗⃗⃗(𝑡))1
𝑑𝑡 

Since �⃗�𝑛 → �⃗� ST in 𝐿
2(𝑄), and �⃗�𝑛𝑡 → �⃗� WK in 𝐿

2(𝑄), then from (60) and the Assm on 𝑓𝑖 , for 

𝑖 = 1,2,3,4, we get 



IHJPAS. 53 (3)2022 
 

169 

(𝑎) = 2 ∫  
0

𝑇

[(𝑦2𝑛, 𝑦1𝑛𝑡 ) − (𝑦3𝑛, 𝑦1𝑛𝑡 ) − (𝑦4𝑛, 𝑦1𝑛𝑡 ) − (𝑦1𝑛, 𝑦2𝑛𝑡 ) + (𝑦3𝑛, 𝑦2𝑛𝑡 ) + (𝑦4𝑛, 𝑦2𝑛𝑡 ) 

+(𝑦1𝑛, 𝑦3𝑛𝑡 ) − (𝑦2𝑛, 𝑦3𝑛𝑡 ) − (𝑦4𝑛, 𝑦4𝑛𝑡 ) + (𝑦1𝑛, 𝑦4𝑛𝑡 ) − (𝑦2𝑛, 𝑦4𝑛𝑡 ) + (𝑦3𝑛, 𝑦4𝑛𝑡 )]𝑑𝑡 +

2 ∫  
0

𝑇

[(𝑓1(𝑦1, 𝑢1), 𝑦1𝑛𝑡 ) + (𝑓2(𝑦2, 𝑢2), 𝑦2𝑛𝑡 ) + (𝑓3(𝑦3, 𝑢3), 𝑦3𝑛𝑡 ) + (𝑓4(𝑦4, 𝑢4), 𝑦4𝑛𝑡 )]𝑑𝑡 ⟶

2 ∫  
0

𝑇

[(𝑦2, 𝑦1𝑡 ) − (𝑦3, 𝑦1𝑡 ) − (𝑦4, 𝑦1𝑡 ) − (𝑦1, 𝑦2𝑡 ) + (𝑦3, 𝑦2𝑡 ) +  (𝑦4, 𝑦2𝑡 ) + (𝑦1, 𝑦3𝑡 ) −

(𝑦2, 𝑦3𝑡 ) − (𝑦4, 𝑦4𝑡 ) + (𝑦1, 𝑦4𝑡 ) − (𝑦2, 𝑦4𝑡 ) + (𝑦3, 𝑦4𝑡 )]𝑑𝑡 +2 ∫  
0

𝑇

[(𝑓1(𝑦1, 𝑢1), 𝑦1𝑡 ) +

(𝑓2(𝑦2, 𝑢2), 𝑦2𝑡 ) + (𝑓3(𝑦3, 𝑢3), 𝑦3𝑡 ) + (𝑓4(𝑦4, 𝑢4), 𝑦4𝑡 )]𝑑𝑡                                                  

By the same way that was employed to acquire (27), it used here to acquire  

�⃗�𝑛𝑡 (𝑇) → �⃗�(𝑇) ST in 𝐿
2(Ω)                                                                                                (63) 

 On the other hand, since �⃗�𝑛 → �⃗� in 𝐿
2(I, V), then using (27) & (63), yield to  

(b)⟶ RHS of (61)= 2 ∫  
0

𝑇

[(𝑦2, 𝑦1𝑡 ) − (𝑦3, 𝑦1𝑡 ) − (𝑦4, 𝑦1𝑡 ) − (𝑦1, 𝑦2𝑡 ) + (𝑦3, 𝑦2𝑡 ) +

 (𝑦4, 𝑦2𝑡 ) + (𝑦1, 𝑦3𝑡 ) − (𝑦2, 𝑦3𝑡 ) − (𝑦4, 𝑦4𝑡 ) + (𝑦1, 𝑦4𝑡 ) − (𝑦2, 𝑦4𝑡 ) + (𝑦3, 𝑦4𝑡 )]𝑑𝑡 + 

2 ∫  
0

𝑇

[(𝑓1(𝑦1, 𝑢1), 𝑦1𝑡 ) + (𝑓2(𝑦2, 𝑢2), 𝑦2𝑡 ) + (𝑓3(𝑦3, 𝑢3), 𝑦3𝑡 ) + (𝑓4(𝑦4, 𝑢4), 𝑦4𝑡 )]𝑑𝑡                      

All the term in (c) approach to zero, so as the 1st two terms in the LHS of (62), hence (62) gives 

∫  
0

𝑇

∥ �⃗�𝑛 (𝑡) − 𝑦 ⃗⃗⃗(𝑡) ∥1
2 𝑑𝑡 → 0 as 𝑛 → ∞, therefore �⃗�𝑛 → �⃗� ST in 𝐿

2(I, V) 

Uniqueness of the Solution: 

Let �⃗� = (𝑦1, 𝑦2, 𝑦3, 𝑦4) and �⃗̅� = (�̅�1, �̅�2, �̅�3, �̅�4) be two solutions of the SQVS of the WF ((8), 

(10), (12), and (14)), subtracting each equation from the other and replace 𝑣𝑖 = 𝑦𝑖 − �̅�𝑖 for 

each 𝑖 = 1,2,3,4, i.e. 

((𝑦1 − �̅�1)𝑡𝑡 , 𝑦1 − �̅�1) + ∇(𝑦1 − �̅�1, 𝑦1 − �̅�1) + (𝑦1 − �̅�1, 𝑦1 − �̅�1) − (𝑦2 − �̅�2, 𝑦1 − �̅�1) +

(𝑦3 − �̅�3, 𝑦1 − �̅�1) + (𝑦4 − �̅�4, 𝑦1 − �̅�1) = (𝑓1(𝑦1, 𝑢1) − 𝑓1(�̅�1, 𝑢1), 𝑦1 − �̅�1)               (64) 

((𝑦2 − �̅�2)𝑡𝑡 , 𝑦2 − �̅�2) + ∇(𝑦2 − �̅�2, 𝑦2 − �̅�2) + (𝑦1 − �̅�1, 𝑦2 − �̅�2) + (𝑦2 − �̅�2, 𝑦2 − �̅�2) −

(𝑦3 − �̅�3, 𝑦2 − �̅�2) − (𝑦4 − �̅�4, 𝑦2 − �̅�2) = (𝑓2(𝑦2, 𝑢2) − 𝑓2(�̅�2, 𝑢2), 𝑦2 − �̅�2)                (65) 

((𝑦3 − �̅�3)𝑡𝑡 , 𝑦3 − �̅�3) + ∇(𝑦3 − �̅�3, 𝑦3 − �̅�3) − (𝑦1 − �̅�1, 𝑦3 − �̅�3) + (𝑦2 − �̅�2, 𝑦3 − �̅�3) +

(𝑦3 − �̅�3, 𝑦3 − �̅�3) + (𝑦4 − �̅�4, 𝑦3 − �̅�3) = (𝑓3(𝑦3, 𝑢3) − 𝑓3(�̅�3, 𝑢3), 𝑦3 − �̅�3)                (66) 

((𝑦4 − �̅�4)𝑡𝑡 , 𝑦4 − �̅�4) + ∇(𝑦4 − �̅�4, 𝑦4 − �̅�4) + (𝑦1 − �̅�1, 𝑦4 − �̅�4) + (𝑦2 − �̅�2, 𝑦4 − �̅�4) −

(𝑦3 − �̅�3, 𝑦4 − �̅�4) + (𝑦4 − �̅�4, 𝑦4 − �̅�4) = (𝑓4(𝑦4, 𝑢4) − 𝑓4(�̅�4, 𝑢4), 𝑦4 − �̅�4)                (67) 

 

Collecting the above equalities for 𝑖 = 1,2,3,4 using Lemma 1.2 in ref. [15] for the 1st in LHS 

of above equations, to get  
1

2

𝑑

𝑑𝑡
∥ (�⃗� − �⃗̅�)

𝑡
(𝑡) ∥0

2+ 2 ∥ �⃗� − �⃗̅� ∥1
2= (𝑓1(𝑦1, 𝑢1) − 𝑓1(�̅�1, 𝑢1), 𝑦1 − �̅�1) +    

(𝑓2(𝑦2, 𝑢2) − 𝑓2(�̅�2, 𝑢2), 𝑦2 − �̅�2) + (𝑓3(𝑦3, 𝑢3) − 𝑓3(�̅�3, 𝑢3), 𝑦3 − �̅�3) +  

(𝑓4(𝑦4, 𝑢4) − 𝑓4(�̅�4, 𝑢4), 𝑦4 − �̅�4)                                                                                         (68) 

 

The LHS of (68) is positive, IBS of it w.r.t. 𝑡 from 0 to 𝑡, and using Assumptions (A-ii) of the 

RHS of it, yields to 

∫  
0

𝑑

𝑑𝑡

𝑡

∥ �⃗� − �⃗̅� ∥0
2  ≤ 2 ∫  

0

𝑡

𝐿 ∥ �⃗� − �⃗̅� ∥0
2 𝑑𝑡 , where 𝐿 = max (𝐿1, 𝐿2, 𝐿3, 𝐿4) 



IHJPAS. 53 (3)2022 
 

170 

Then, ∥ �⃗� − �⃗̅� ∥0
2≤ 2 ∫  

0

𝑇

𝐿 ∥ �⃗� − �⃗̅� ∥0
2,.  

By using Lemma (1.2), to acquire   

∥ �⃗� − �⃗̅� ∥0
2≤ 0𝑒

∫  
0

𝑇

2𝐿𝑑𝑡
= 0, ∀𝑡 ∈ 𝐼   

Again IBS of (68) w.r.t. 𝑡  from 0 to 𝑇, using the ICs and the above result for the RHS of the 

equations, to acquire 

∫  
0

𝑇
𝑑

𝑑𝑡
∥ �⃗� − �⃗̅� ∥0

2+ 2 ∥ �⃗� − �⃗̅� ∥1
2 𝑑𝑡 ≤ 𝐿 ∫  

0

𝑇

∥ (�⃗� − �⃗̅�) ∥0
2 𝑑𝑡 ⟹ ∫  

0

𝑇

∥ (�⃗� − �⃗̅�)
 
(𝑡) ∥1

2≤ 0 ⟹∥

(�⃗� − �⃗̅�)
 
(𝑡) ∥

𝐿2(I,V)
 = 0 ⟹ �⃗� = �⃗̅� i.e. The solution is unique  

3.1 Lemma: In addition to Assumptions (A), if the functions 𝑓𝑖  (for each 𝑖 = 1,2,3,4) is 
Lipschitz w.r.t. 𝑦𝑖   & 𝑢𝑖 , and if CCCQV is bounded, then the operator �⃗⃗� → �⃗��⃗⃗⃗� form (𝐿

2(𝑄))4 
to (𝐿∞(𝐼, 𝐿2(Ω)))4 or to (𝐿2(𝑄))4 or to (𝐿2(𝐼, 𝑉))4 is continuous. 

Proof: Let �⃗⃗� = (𝑢1, 𝑢2, 𝑢3, 𝑢4), �⃗⃗̅� = (�̅�1, �̅�2, �̅�3, �̅�4) ∈ (𝐿
2(𝑄))4, 𝛿�⃗⃗� = �⃗⃗̅� − �⃗⃗�, �⃗⃗�𝜀 =�⃗⃗� + 𝛿�⃗⃗� ∈

(𝐿2(𝑄))4, for > 0, then by Theorem 3.1, �⃗� = �⃗��⃗⃗⃗� = (𝑦1, 𝑦2, 𝑦3, 𝑦4) and �⃗�𝜀 = �⃗��⃗⃗⃗�𝜀 =

(𝑦1𝜀 , 𝑦2𝜀 , 𝑦3𝜀 , 𝑦4𝜀 ) are their corresponding SQVS which satisfy the WF ((8)-(15)). Setting 
𝛿�⃗�𝜀 = (𝛿𝑦1𝜀 , 𝛿𝑦2𝜀 , 𝛿𝑦3𝜀 , 𝛿𝑦4𝜀 )= �⃗�𝜀 − �⃗� , to obtain 
(𝛿𝑦1𝜀𝑡𝑡 , 𝑣1) + (∇𝛿𝑦1𝜀 , ∇𝑣1) + (𝛿𝑦1𝜀 , 𝑣1) − (𝛿𝑦2𝜀 , 𝑣1) + (𝛿𝑦3𝜀 , 𝑣1) + (𝛿𝑦4𝜀 , 𝑣1)  

= (𝑓1(𝑦1 + 𝛿𝑦1𝜀 , 𝑢1 + 𝛿𝑢1) − 𝑓1(𝑦1, 𝑢1), 𝑣1)                                                                    (69) 

𝛿𝑦1𝜀 (𝑥, 0) = 0 and 𝛿𝑦1𝜀𝑡 (𝑥, 0) = 0                                                                                     (70) 

(𝛿𝑦2𝜀𝑡𝑡 , 𝑣2) + (∆𝛿𝑦2𝜀 , ∇𝑣2) + (𝛿𝑦1𝜀 , 𝑣2) + (𝛿𝑦2𝜀 , 𝑣2) − (𝛿𝑦3𝜀 , 𝑣2) − (𝛿𝑦4𝜀 , 𝑣2)  

= (𝑓2(𝑦2 + 𝛿𝑦2𝜀 , 𝑢2 + 𝛿𝑢2) − 𝑓2(𝑦2, 𝑢2), 𝑣2)                                                                   (71) 

𝛿𝑦2𝜀 (𝑥, 0) = 0 and 𝛿𝑦2𝜀𝑡 (𝑥, 0) = 0                                                                                     (72) 

(𝛿𝑦3𝜀𝑡𝑡 , 𝑣3) + (∇δ𝑦3𝜀 , ∇𝑣3) − (𝛿𝑦1𝜀 , 𝑣3) + (𝛿𝑦2𝜀 , 𝑣3) + (𝛿𝑦3𝜀 , 𝑣3) + (𝛿𝑦4𝜀 , 𝑣3)  

= (𝑓3(𝑦3 + 𝛿𝑦3𝜀 , 𝑢3 + 𝛿𝑢3) − 𝑓3(𝑦3, 𝑢3), 𝑣3)                                                                   (73) 

𝛿𝑦3𝜀 (𝑥, 0) = 0 and 𝛿𝑦3𝜀𝑡 (𝑥, 0) = 0                                                                                     (74) 

(𝛿𝑦4𝜀𝑡𝑡 , 𝑣4) + (∇δ𝑦4𝜀 , ∇𝑣4) − (𝛿𝑦1𝜀, 𝑣4) + (𝛿𝑦2𝜀 , 𝑣4) − (𝛿𝑦3𝜀 , 𝑣4) + (𝛿𝑦4𝜀 , 𝑣4)  

= (𝑓4(𝑦4 + 𝛿𝑦4𝜀 , 𝑢4 + 𝛿𝑢4) − 𝑓4(𝑦4, 𝑢4), 𝑣4)                                                                    (75)  

𝛿𝑦4𝜀 (𝑥, 0) = 0 and 𝛿𝑦4𝜀𝑡 (𝑥, 0) = 0                                                                                     (76) 

Substituting 𝑣𝑖 = 𝛿𝑦𝑖𝜀𝑡 for 𝑖 = 1,2,3,4 in (69),(71),(73)and (75) resp., collecting the obtained 

equations. Using the same way that is used to get (37), a similar equation can be obtained but 

with 𝛿�⃗�𝜀  in position of �⃗�𝑛 , then IBS on [0, 𝑡], using Lip. on 𝑓𝑖  w.r.t. (𝑦𝑖 , 𝑢𝑖 ) resp. for (𝑖 =

1,2,3,4) to get   

∫
0

𝑡
𝑑

𝑑𝑡
[∥ 𝛿�⃗�𝜀𝑡 (𝑡) ∥0

2+∥ 𝛿�⃗�𝜀 ∥1
2]𝑑𝑡 ≤  

2 ∫  
0

[(
𝑡

∣ 𝛿𝑦2𝜀 ∣ +∣ 𝛿𝑦3𝜀 ∣ +∣ 𝛿𝑦4𝜀 ∣) ∣ 𝛿𝑦1𝜀𝑡 ∣ +�̅�1 ∣ 𝛿𝑦1𝜀𝑡 ∣
2+ �̿�1 ∣ 𝛿𝑢4 ∣∣ 𝛿𝑦1𝜀𝑡 ∣]𝑑𝑡 +  

+  2 ∫  
0

𝑡

[(∣ 𝛿𝑦1𝜀 ∣ +∣ 𝛿𝑦3𝜀 ∣ +∣ 𝛿𝑦4𝜀 ∣) ∣ 𝛿𝑦2𝜀𝑡 ∣ +�̅�2 ∣ 𝛿𝑦2𝜀𝑡 ∣
2+ �̿�2 ∣ 𝛿𝑢2 ∣∣ 𝛿𝑦2𝜀𝑡 ∣]𝑑𝑡 

2∫
0

𝑡

[(∣ 𝛿𝑦1𝜀 ∣ +∣ 𝛿𝑦2𝜀 ∣ +∣ 𝛿𝑦4𝜀 ∣) ∣ 𝛿𝑦3𝜀𝑡 ∣ +�̅�3 ∣ 𝛿𝑦3𝜀𝑡 ∣
2+  �̅̅�3 ∣ 𝛿𝑢3 ∣ ∣ 𝛿𝑦3𝜀𝑡 ∣]𝑑𝑡  

+2 ∫  
0

𝑡

[(∣ 𝛿𝑦1𝜀 +∣ 𝛿𝑦2𝜀 ∣ +∣ 𝛿𝑦3𝜀 ∣) ∣ 𝛿𝑦4𝜀𝑡 ∣ +�̅�4 ∣ 𝛿𝑦4𝜀𝑡 ∣
2 + �̅̅�4 ∣ 𝛿𝑢4 ∣∣ 𝛿𝑦4𝜀𝑡 ∣]𝑑𝑡  

Using the definitions of the norms and the relations between them, to get. 

∥ 𝛿�⃗�𝜀𝑡 ∥0
2+∥ 𝛿�⃗�𝜀 ∥1

2≤ 3 ∫  
0

𝑡

[∥ 𝛿�⃗�𝜀 ∥0
2+∥ 𝛿�⃗�𝜀𝑡 ∥1

2]𝑑𝑡 + �̃�3 ∫  
0

𝑡

[∥ 𝛿�⃗�𝜀 ∥0
2+∥ 𝛿�⃗�𝜀𝑡 ∥1

2]𝑑𝑡  



IHJPAS. 53 (3)2022 
 

171 

                                     +�̃�4 ∫  
0

𝑡

∥ 𝛿�⃗⃗� ∥0
2 𝑑𝑡 + �̃�4  ∫  

0

𝑡

∥ 𝛿�⃗�𝜀𝑡 ∥1
2 𝑑𝑡   

                                 ≤ �̃�4 ∥ 𝛿�⃗⃗� ∥𝑄
2 + 𝐿5 ∫  

0

𝑡

[∥ 𝛿�⃗�𝜀 ∥0
2+∥ 𝛿�⃗�𝜀𝑡 ∥1

2]𝑑𝑡  

Where �̃�3 = max (�̅�1, �̅�2, �̅�3, �̅�4), �̃�4 = max (�̅̅�1, �̅̅�2, �̅̅�3, �̅̅�4),  𝐿5 = max (3 + �̃�3, 3 + �̃�4 + �̃�3). 

Applying the BGI, with𝐿2 = �̃�4𝑒
𝐿5 ∫  

0

𝑡

𝑑𝑡
, to get 

∥ 𝛿�⃗�𝜀𝑡 ∥0
2+∥ 𝛿�⃗�𝜀 ∥1

2≤ 𝐿2 ∥ 𝛿�⃗⃗� ∥𝑄
2 , ∀𝑡 ∈ 𝐼 ⟹∥ 𝛿�⃗�𝜀 ∥1

2≤ 𝐿2 ∥ 𝛿�⃗⃗�(𝑡) ∥𝑄
2 , ∀𝑡 ∈ 𝐼 

∥ 𝛿�⃗�𝜀 ∥𝐿∞(𝐼,𝐿2(Ω))≤ 𝐿 ∥ 𝛿�⃗⃗� ∥𝑄, ∥ 𝛿�⃗�∈ ∥𝐿2(𝐼,𝑉)≤ 𝐿 ∥ 𝛿�⃗⃗� ∥𝑄 and ∥ 𝛿�⃗�𝜀 ∥𝑄 ≤ 𝐿 ∥ 𝛿�⃗⃗� ∥𝑄 .   

4. The Existence of an OCCCQV 

4.1. Assumptions (B): Consider 𝑔𝑙𝑖 for (𝑙 = 0 &𝑖 = 1,2,3,4) is of Carathéodory type  on 𝑄 ×

(ℝ × 𝑈), and satisfies the following sub quadratic condition w.r.t. 𝑦𝑖 ∈ ℝ and 𝑢𝑖 ∈ 𝑈𝑖  ,  

|𝑔𝑙𝑖 (𝑥, 𝑡, 𝑦𝑖 , 𝑢𝑖 )| ≤ 𝐺𝑙𝑖 (𝑥, 𝑡) + 𝐶𝑙𝑖1𝑦𝑖
2 + 𝐶𝑙𝑖2𝑢𝑖

2, where  𝐺𝑙𝑖 ∈ 𝐿
1(Q), for (𝑥, 𝑡) ∈ 𝑄,𝑙 = 0 

4.1 Lemma: With assum. (B), the functional �⃗⃗� → 𝐺0(�⃗⃗�) is continuous on  (𝐿
2(𝑄))4. 

Proof: Using Assumptions (B) and proposition 1.3, the integral∫
𝑄

𝑔0𝑖  (𝑥, 𝑡, 𝑦𝑖 , 𝑢𝑖 )𝑑𝑥𝑑𝑡 is 

continuous on 𝐿2(𝑄), ∀𝑖 = 1,2,3,4, hence 𝐺0(�⃗⃗�) is continuous on  (𝐿
2(𝑄))4. 

4.2 Lemma : let 𝑔: 𝑄 × ℝ → ℝ is of Carathéodory type  on 𝑄 × (ℝ × ℝ) and satisfies 

|𝑔 (𝑥, 𝑡, 𝑦, 𝑢)| ≤ 𝐺 (𝑥, 𝑡) + 𝑐 𝑦 
2 + �́�𝑢 

2, where  𝐺(𝑥, 𝑡) ∈ 𝐿1(Q),𝑢 ∈ 𝑈, 𝑐, 𝑐́ ≥ 0, 𝑈 ⊂ ℝ, is 

compact.. Then, ∫
𝑄

𝑔 (𝑥, 𝑦, 𝑢 )𝑑𝑥 is continuous on 𝐿
2(𝑄) w.r.t. 𝑦. 

4.1 Theorem: In addition to Assumptions (A&B), if the set �⃗⃗⃗� is convex and compact., �⃗⃗⃗⃗�𝐴 ≠

𝜙, the function 𝑓𝑖 , for (𝑖 = 1,2,3,4) have the form  

𝑓𝑖 (𝑥, 𝑡, 𝑦𝑖 , 𝑢𝑖 ) = 𝑓𝑖1(𝑥, 𝑡, 𝑦𝑖 ) + 𝑓𝑖2(𝑥, 𝑡)𝑢𝑖 ,where  |𝑓𝑖1(𝑥, 𝑡, 𝑦𝑖 )| ≤ 𝑖 (𝑥, 𝑡) + 𝑐𝑖 ∣ 𝑦𝑖 ∣, 

|𝑓𝑖2(𝑥, 𝑡)| ≤ 𝐾𝑖 , 𝑖 ∈ 𝐿
2(Q), 𝑐𝑖 ≥ 0, for 𝑖 = 1,2,3,4. Then there exists an OCCQV. 

Proof: From the Assumptions on 𝑈𝑖 ⊂ ℝ for 𝑖 = 1,2,3,4 and the Egorov’s theorem, once get 

that �⃗⃗⃗⃗� is weakly compact, since �⃗⃗⃗⃗�𝐴 ≠ 𝜙, there exists a minimum sequence {�⃗⃗�𝑘 } =

{(𝑢1𝑘 , 𝑢2𝑘 , 𝑢3𝑘 , 𝑢4𝑘 )} ∈ �⃗⃗⃗⃗�𝐴, ∀𝑘 s.t. 𝑙𝑖𝑚
𝑘→∞

𝐺0(�⃗⃗�𝑘 ) = 𝑖𝑛𝑓
�⃗⃗⃗�𝑘∈�⃗⃗⃗⃗�𝐴

𝐺0(�⃗⃗̅�). Since {�⃗⃗�𝑘 } ∈ �⃗⃗⃗⃗�𝐴, ∀𝑘 and �⃗⃗⃗⃗� 

is weakly compact, there exists a subsequence of {�⃗⃗�𝑘 } say again {�⃗⃗�𝑘 } s.t. . �⃗⃗�𝑘 → �⃗⃗�  WK in 

(𝐿2(𝑄))4 and ∥ �⃗⃗�𝑘 ∥𝑄 ≤ 𝑑,∀𝑘. From Theorem 3.1, for each control {�⃗⃗�𝑘 } the WF (18), (20), 

(22), (24) has a unique SQVS , {�⃗�𝑘 = �⃗�𝑢𝑘 } s.t. the norm ∥ �⃗�𝑘 ∥𝐿2(𝐼,𝑉), ∥ �⃗�𝑘𝑡 ∥𝐿2(𝑄) are bounded, 

then by ATH there exists a subsequence of {�⃗�𝑘 } and {�⃗�𝑘𝑡 }, say again {�⃗�𝑘 } and {�⃗�𝑘𝑡 }, s.t. �⃗�𝑘 →

�⃗�  WK in (𝐿2(𝐼, 𝑉))4, �⃗�𝑘𝑡 → �⃗�  WK in (𝐿
2(𝑄))4. Now for each 𝑘. and by applying the ACTH 

[15], we get that there exists a subsequence of{�⃗�𝑘 } say a gain {�⃗�𝑘 } s.t. �⃗�𝑘 → �⃗�  ST in (𝐿
2(𝑄))4. 

Now since for each 𝑘, �⃗�𝑘  is a SQVS of the WF ((18), (20), (22), (24)) resp. , substituting in 

these equations, MBSs of each equation by 𝜙𝑖 (𝑡), ∀𝑖 = 1,2,3,4 (with 𝜙𝑖 ∈ 𝐶
2[0, 𝑇], s.t. 

𝜙𝑖 (𝑇) = 𝜙𝑖
′(𝑇) = 0, 𝜙𝑖 (0) ≠ 0, 𝜙𝑖

′(0) ≠ 0). Rewriting the 1st term in the LHS of each one 

then IBS on [0, 𝑇], finally IBPs for the 1st terms, one has 

∫  
0

𝑇
𝑑

𝑑𝑡
(𝑦1𝑘𝑡 , 𝑣1)𝜙1

 𝑑𝑡 + ∫  
0

𝑇

[(∇𝑦1𝑘 , ∇𝑣1) + (𝑦1𝑘 , 𝑣1) − (𝑦2𝑘 , 𝑣1) + (𝑦3𝑘 , 𝑣1) + (𝑦4𝑘 , 𝑣1)]𝜙1
 𝑑𝑡 

=∫  
0

𝑇

(𝑓11(𝑥, 𝑡, 𝑦1𝑘 ), 𝑣1)𝜙1
 (𝑡)𝑑𝑡 + ∫  

0

𝑇

(𝑓12(𝑥, 𝑡)𝑢1𝑘 , 𝑣1)𝜙1
 (𝑡)𝑑𝑡                                        (77) 

∫  
0

𝑇
𝑑

𝑑𝑡
(𝑦2𝑘𝑡 , 𝑣2)𝜙2

 𝑑𝑡 + ∫  
0

𝑇

[(∇𝑦2𝑘 , ∇𝑣2) + (𝑦1𝑘 , 𝑣2) + (𝑦2𝑘 , 𝑣2) − (𝑦3𝑘 , 𝑣2) − (𝑦4𝑘 , 𝑣2)]𝜙2
 𝑑𝑡  



IHJPAS. 53 (3)2022 
 

172 

= ∫  
0

𝑇

(𝑓21(𝑥, 𝑡, 𝑦2𝑘 ), 𝑣2)𝜙2
 (𝑡)𝑑𝑡 + ∫  

0

𝑇

(𝑓22(𝑥, 𝑡)𝑢2𝑘 , 𝑣2)𝜙2
 (𝑡)𝑑𝑡                                       (78) 

∫  
0

𝑇
𝑑

𝑑𝑡
(𝑦3𝑘 , 𝑣3)𝜙3

 𝑑𝑡 + ∫  
0

𝑇

[(∇𝑦3𝑘 , ∇𝑣3) − (𝑦1𝑘 , 𝑣3) + (𝑦2𝑘 , 𝑣3) + (𝑦3𝑘 , 𝑣3) + (𝑦4𝑘 , 𝑣3)]𝜙3
 𝑑𝑡  

= ∫  
0

𝑇

(𝑓31(𝑥, 𝑡, 𝑦3𝑘 ), 𝑣3)𝜙3
 (𝑡)𝑑𝑡 + ∫  

0

𝑇

(𝑓32(𝑥, 𝑡)𝑢3𝑘 , 𝑣3)𝜙3
 (𝑡)𝑑𝑡                                       (79) 

∫  
0

𝑇
𝑑

𝑑𝑡
(𝑦4𝑘 , 𝑣4)𝜙4

 𝑑𝑡 + ∫  
0

𝑇

[(∇𝑦4𝑘 , ∇𝑣4) − (𝑦1𝑘 , 𝑣4) + (𝑦2𝑘 , 𝑣4) − (𝑦3𝑘 , 𝑣4) + (𝑦4𝑘 , 𝑣4)]𝜙4
 𝑑𝑡    

= ∫  
0

𝑇

(𝑓41(𝑥, 𝑡, 𝑦4𝑘 ), 𝑣4)𝜙4
 (𝑡)𝑑𝑡 + ∫  

0

𝑇

(𝑓42(𝑥, 𝑡)𝑢4𝑘 , 𝑣4)𝜙4
 (𝑡)𝑑𝑡                                       (80) 

The same steps that are utilized in the proof of Theorem 3.1, can also be used here to passage 

the limit in the LHS of ((77)-(80)). We persist in the passage of the limit in RHS of ((77))-

(80)) as follows. Let ∀𝑖 = 1,2,3,4, 𝑣𝑖 ∈ 𝐶[Ω̅], 𝑤𝑖 = 𝑣𝑖 𝜙𝑖
 (𝑡), then 𝑤𝑖 ∈ 𝐶[Q̅] ∈ 𝐿

∞(𝑄) ⊂
𝐿2(𝑄), set 𝑓�̅�1(𝑦𝑖𝑘 ) = 𝑓𝑖1(𝑦𝑖𝑘 )𝑤𝑖, then 𝑓�̅�1: 𝑄 × ℝ → ℝ is of Carathéodory type , utilizing 

proposition 3.1, to get the integral  ∫  
𝑄

 

𝑓𝑖1(𝑦𝑖𝑘 )𝑤𝑖  𝑑𝑥𝑑𝑡, is continuous w.r.t.  𝑦𝑖𝑘 ∀𝑖 = 1,2,3,4. 

but  𝑦𝑖𝑘 → 𝑦𝑖  ST in 𝐿
2(𝑄)  and 𝑢𝑖𝑘 → 𝑢𝑖 WK in 𝐿

2(𝑄), then    

∫  
𝑄

 

𝑓𝑖1(𝑦𝑖𝑘 )𝑤𝑖  𝑑𝑥𝑑𝑡 → ∫  
𝑄

 

𝑓𝑖1(𝑦𝑖 )𝑤𝑖  𝑑𝑥𝑑𝑡, ∀   𝑤𝑖 ∈ 𝐶[Q̅] , for 𝑖 = 1,2,3,4                           (81)    

∫  
𝑄

 

𝑓𝑖2(𝑥, 𝑡)𝑢𝑖𝑘 𝑤𝑖  𝑑𝑥𝑑𝑡 → ∫  
𝑄

 

𝑓𝑖2(𝑥, 𝑡)𝑢𝑖 𝑤𝑖  𝑑𝑥𝑑𝑡, ∀   𝑤𝑖 ∈ 𝐶[Q̅] , for 𝑖 = 1,2,3,4                (82)                    

Form the density of  𝐶(Ω̅) in 𝑉, (81) &(82) are satisfies for each 𝑣𝑖 ∈ 𝑉 for 𝑖 = 1,2,3,4, hence 
the following WF is obtained              

(𝑦1𝑡 , 𝑣1) + (∇𝑦1, ∇𝑣1) + (𝑦1, 𝑣1) − (𝑦2, 𝑣1) + (𝑦3, 𝑣1) + (𝑦4, 𝑣1) 
= (𝑓11(𝑥, 𝑡, 𝑦1), 𝑣1) + (𝑓12(𝑥, 𝑡)𝑢1, 𝑣1), ∀𝑣1 ∈ 𝑉   a.e. on I                                                  (83) 
(𝑦2𝑡 , 𝑣2) + (∆𝑦2, ∇𝑣2) + (𝑦1, 𝑣2) + (𝑦2, 𝑣2) − (𝑦3, 𝑣2) − (𝑦4, 𝑣2) 
= (𝑓21(𝑥, 𝑡, 𝑦2), 𝑣2) + (𝑓22(𝑥, 𝑡)𝑢2, 𝑣2), ∀𝑣2 ∈ 𝑉  a.e. on I                                                  (84) 
(𝑦3𝑡 , 𝑣3) + (∇𝑦3, ∇𝑣3) − (𝑦1, 𝑣3) + (𝑦2, 𝑣3) + (𝑦3, 𝑣3) + (𝑦4, 𝑣3) 
= (𝑓31(𝑥, 𝑡, 𝑦3), 𝑣3) + (𝑓32(𝑥, 𝑡)𝑢3, 𝑣3)∀𝑣3 ∈ 𝑉     a.e. on I                                                (85) 
(𝑦4𝑡 , 𝑣4) + (∇𝑦4, ∇𝑣4) − (𝑦1, 𝑣4) + (𝑦2, 𝑣4) − (𝑦3, 𝑣4) + (𝑦4, 𝑣4) 
= (𝑓41(𝑥, 𝑡, 𝑦4), 𝑣4) + (𝑓42(𝑥, 𝑡)𝑢4, 𝑣4) ∀𝑣4 ∈ 𝑉    a.e. on I                                                 (86) 
Also, the same steps employed in Theorem 3.1 can be employed here to obtain that the ICs are 

held, hence �⃗� is the SQVS.  

Now since ∀𝑖 = 1,2,3,4, 𝑔0𝑖 (𝑥, 𝑡, 𝑦𝑖 , 𝑢𝑖 ) is continuous w.r.t.  (𝑦𝑖 , 𝑢𝑖 ), and 𝑈𝑖 is compact. With 

𝑢𝑖 (𝑥, 𝑡) ∈ 𝑈𝑖 a.e. in 𝑄, then using lemma 4.2, to get  

∫  
𝑄

 

𝑔0𝑖 (𝑥, 𝑡, 𝑦𝑖𝑘 , 𝑢𝑖𝑘 ) 𝑑𝑥𝑑𝑡 → ∫  
𝑄

 

𝑔0𝑖  (𝑥, 𝑡, 𝑦𝑖 , 𝑢𝑖𝑘 )𝑑𝑥𝑑𝑡                                                         (87)    

But 𝑔0𝑖 (𝑥, 𝑡, 𝑦𝑖 , 𝑢𝑖 ) is continuous and convex w.r.t. 𝑢𝑖  then ∫  
𝑄

 

𝑔0𝑖  (𝑥, 𝑡, 𝑦𝑖 , 𝑢𝑖 )𝑑𝑥𝑑𝑡 is weakly 

lowe semi cont. (WLSC) w.r.t. 𝑢𝑖 , ∀𝑖 = 1,2,3,4, i.e. 

∫  
𝑄

 

𝑔0𝑖  (𝑥, 𝑡, 𝑦𝑖 , 𝑢𝑖 )𝑑𝑥𝑑𝑡 ≤ 𝑙𝑖𝑚
𝑘→∞

𝑖𝑛𝑓 ∫  
𝑄

 

[𝑔0𝑖  (𝑥, 𝑡, 𝑦𝑖 , 𝑢𝑖𝑘 ) − 𝑔0𝑖  (𝑥, 𝑡, 𝑦𝑖𝑘 , 𝑢𝑖𝑘 )]𝑑𝑥𝑑𝑡  

                                          + 𝑙𝑖𝑚
𝑘→∞

𝑖𝑛𝑓 ∫  
𝑄

 

𝑔0𝑖  (𝑥, 𝑡, 𝑦𝑖𝑘 , 𝑢𝑖𝑘 )𝑑𝑥𝑑𝑡,  

                                         ≤ 𝑙𝑖𝑚
𝑘→∞

𝑖𝑛𝑓 ∫  
𝑄

 

𝑔0𝑖  (𝑥, 𝑡, 𝑦𝑖𝑘 , 𝑢𝑖𝑘 )𝑑𝑥𝑑  

⟹ Σ
𝑖=1

4

∫  
𝑄

 

𝑔0𝑖  (𝑥, 𝑡, 𝑦𝑖 , 𝑢𝑖 )𝑑𝑥𝑑𝑡 ≤ Σ
𝑖=1

4

∫  
𝑄

 

𝑔0𝑖  (𝑥, 𝑡, 𝑦𝑖𝑘 , 𝑢𝑖𝑘 )𝑑𝑥𝑑𝑡, thus 



IHJPAS. 53 (3)2022 
 

173 

𝐺0(�⃗⃗�) ≤ 𝑙𝑖𝑚
𝑘→∞

𝑖𝑛𝑓
�⃗⃗⃗�𝑘∈�⃗⃗⃗⃗�𝐴

𝐺0(�⃗⃗�𝑘 ) = 𝑙𝑖𝑚
𝑘→∞

𝐺0(�⃗⃗�𝑘 ) = 𝑖𝑛𝑓
�⃗⃗⃗�𝑘∈�⃗⃗⃗⃗�𝐴

𝐺0(�⃗⃗̅�), then 

𝐺0(�⃗⃗�) ≤ 𝑖𝑛𝑓
�⃗⃗⃗�𝑘∈�⃗⃗⃗⃗�𝐴

𝐺0(�⃗⃗̅�) ⟹ 𝐺0(�⃗⃗�) ≤ 𝑚𝑖𝑛
�⃗⃗⃗�𝑘∈�⃗⃗⃗⃗�𝐴

𝐺0(�⃗⃗̅�), then �⃗⃗� is OQCCCV. 

 

5. Conclusion  

     The method of Galerkin with the Aubin compactness theorem is used successfully to 

demonstrate the existence theorem of a unique state quaternary vector solution for the weak 

form for the quaternary nonlinear hyperbolic boundary value problem where the classical 

continuous control quaternary vector is considered given under suitable hypotheses. 

The continuity operator between the state quaternary vector solution of the weak form 

for the quaternary nonlinear hyperbolic boundary value problem and the corresponding 

classical continuous control quaternary vector is demonstrated. The existence theorem of an 

optimal classical continuous control quaternary vector under suitable hypotheses is 

demonstrated. 

 

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