During the transitional phase spanning from the realm of fuzzy logic to the realm of neutrosophy, a multitude of hybrid models have emerged, each surpassing its predecessor in terms of superiority. Given the pervasive presence of indeterminacy in the world, a higher degree of precision is essential for effectively handling imprecision. Consequently, more sophisticated variants of neutrosophic sets (NSs) have been conceived. The key objective of this paper is to introduce yet another variant of NS, known as the q-rung orthopair fuzzy-valued neutrosophic set (q-ROFVNS). By leveraging the extended spatial range offered by q-ROFS, q-ROFVNS enables a more nuanced representation of indeterminacy and inconsistency. Our endeavor commences with the definitions of q-ROFVNS and q-ROFVN numbers (q-ROFVNNs). Then, we propose several types of score and accuracy functions to facilitate the comparison of q-ROFVNNs. Fundamental operations of q-ROFVNSs and some algebraic operational rules of q-ROFVNNs are also provided with their properties, substantiated by proofs and elucidated through illustrative examples. Drawing upon the operational rules of q-ROFVNNs, the q-ROFVN weighted average operator (q-ROFVNWAO) and q-ROFVN weighted geometric operator (q-ROFVNWGO) are proposed. Notably, we present the properties of these operators, including idempotency, boundedness and monotonicity. Furthermore, we emphasize the applicability and significance of the q-ROFVN operators, substantiating their utility through an algorithm and a numerical application. To further validate and evaluate the proposed model, we conduct a comparative analysis, examining its accuracy and performance in relation to existing models.
Citation: Ashraf Al-Quran, Faisal Al-Sharqi, Atiqe Ur Rahman, Zahari Md. Rodzi. The q-rung orthopair fuzzy-valued neutrosophic sets: Axiomatic properties, aggregation operators and applications[J]. AIMS Mathematics, 2024, 9(2): 5038-5070. doi: 10.3934/math.2024245
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During the transitional phase spanning from the realm of fuzzy logic to the realm of neutrosophy, a multitude of hybrid models have emerged, each surpassing its predecessor in terms of superiority. Given the pervasive presence of indeterminacy in the world, a higher degree of precision is essential for effectively handling imprecision. Consequently, more sophisticated variants of neutrosophic sets (NSs) have been conceived. The key objective of this paper is to introduce yet another variant of NS, known as the q-rung orthopair fuzzy-valued neutrosophic set (q-ROFVNS). By leveraging the extended spatial range offered by q-ROFS, q-ROFVNS enables a more nuanced representation of indeterminacy and inconsistency. Our endeavor commences with the definitions of q-ROFVNS and q-ROFVN numbers (q-ROFVNNs). Then, we propose several types of score and accuracy functions to facilitate the comparison of q-ROFVNNs. Fundamental operations of q-ROFVNSs and some algebraic operational rules of q-ROFVNNs are also provided with their properties, substantiated by proofs and elucidated through illustrative examples. Drawing upon the operational rules of q-ROFVNNs, the q-ROFVN weighted average operator (q-ROFVNWAO) and q-ROFVN weighted geometric operator (q-ROFVNWGO) are proposed. Notably, we present the properties of these operators, including idempotency, boundedness and monotonicity. Furthermore, we emphasize the applicability and significance of the q-ROFVN operators, substantiating their utility through an algorithm and a numerical application. To further validate and evaluate the proposed model, we conduct a comparative analysis, examining its accuracy and performance in relation to existing models.
Modern problems of natural science lead to the need to generalize the classical problems of mathematical physics, as well as to the formulation of qualitatively new problems, which include non-local problems for differential equations. Among nonlocal problems, problems with integral conditions are of great interest. Integral conditions are encountered in the study of physical phenomena in the case when the boundary of the process flow region is inaccessible for direct measurements. Inverse problems arise in various fields of human activity, such as seismology, mineral exploration, biology, medical visualization, computed tomography, earth remote sensing, spectral analysis, nondestructive control, etc. Various inverse problems for certain types of partial differential equations have been studied in many works. A more detailed bibliography and a classification of problems are found in [1,2,3,4,5]. Inverse problems for one-dimensional pseudo-parabolic equations of third-order were studied in [6]. The existence and uniqueness of the solution of the inverse problem for the third order pseudoparabolic equation with integral over-determination condition is studied in [7]. Khompysh [8] investigated the reconstruction of unknown coefficient in pseudo-parabolic inverse problem with the integral over determination condition and studied the uniqueness and existence of solution by means of method of successive approximations. Studies of wave propagation in cold plasma and magnetohydrodynamics also reduce to the partial differential equations of fourth-order. To the study of nonlocal boundary value problems (including integral conditions) for partial differential equations of the fourth-order are devoted large number of works, see, for example, [9,10]. It should be noted that boundary value problems with integral conditions are of particular interest. From physical considerations, the integral conditions are completely natural, and they arise in mathematical modelling in cases where it is impossible to obtain information about the process occurring at the boundary of the region of its flow using direct measurements or when it is possible to measure only some averaged (integral) characteristics of the desired quantity.
In this article, we study the an inverse boundary value problem for a fourth order pseudo parabolic equation with periodic and integral condition to identify the time-dependent coefficients along with the solution function theoretically, i.e. existence and uniqueness.
Statement of the problem and its reduction to an equivalent problem. In the domain DT={(x,t):0≤x≤1,0≤t≤T}, we consider an inverse boundary value problem of recovering the timewise dependent coefficients p(t) in the pseudo-parabolic equation of the fourth-order
ut(x,t)−butxx(x,t)+a(t)uxxxx(x,t)=p(t)u(x,t)+f(x,t) | (1.1) |
with the initial condition
u(x,0)+δu(x,T)=φ(x)(0≤x≤1), | (1.2) |
boundary conditions
u(0,t)=u(1,t),ux(0,t)=ux(1,t),uxx(0,t)=uxx(1,t)(0≤t≤T), | (1.3) |
nonlocal integral condition
∫10u(x,t)dx=0(0≤t≤T) | (1.4) |
and with an additional condition
u(0,t)=∫t0γ(τ)u(1,τ)dτ+h(t)(0≤t≤T), | (1.5) |
where b>0, δ≥0-given numbers, a(t)>0,f(x,t),φ(x),γ(τ),h(t) -given functions, u(x,t) and p(t) - required functions.
Denote
ˉC4,1(DT)={u(x,t):u(x,t)∈C2,1(DT),utxx,uxxxx∈C(DT)}. |
Definition.By the classical solution of the inverse boundary value problem (1.1)-(1.5)we mean the pair {u(x,t),p(t)} functions u(x,t)∈ˉC4,1(DT), p(t)∈C[0,T] satisfying equation (1.1) in DT, condition (1.2) in [0, 1] and conditions (1.3)-(1.5) in [0, T].
Theorem 1. Let be b>0,δ≥0,φ(x)∈C[0,1],f(x,t)∈C(DT), ∫10f(x,t)dx=0, 0<a(t)∈C[0,T], h(t)∈C1[0,T], h(t)≠0(0≤t≤T), γ(t)∈C[0,T],δγ(t)=0 (0≤t≤T) and
∫10φ(x)dx=0,φ(0)=h(0)+δh(T). |
Then the problem of finding a solution to problem (1.1)-(1.5) is equivalent to the problem of determining the functions u(x,t)∈ˉC4,1(DT) and p(t)∈C[0,T], from (1.1)-(1.3) and
uxxx(0,t)=uxxx(1,t)(0≤t≤T), | (1.6) |
γ(t)u(1,t)+h′(t)−butxx(0,t)+a(t)uxxxx(0,t)= |
=p(t)(∫t0γ(τ)u(1,τ)dτ+h(t))+f(0,t)(0≤t≤T). | (1.7) |
Proof. Let be {u(x,t),p(t)} is a classical solution to problem (1.1)-(1.5). Integrating equation (1.1) with respect to x from 0 to 1, we get:
ddt∫10u(x,t)dx−b(utx(1,t)−utx(0,t))+a(t)(uxxx(1,t)−uxxx(0,t))= |
=p(t)∫10u(x,t)dx+∫10f(x,t)dx(0≤t≤T). | (1.8) |
Assuming that ∫10f(x,t)dx=0, taking into account (1.3) and (1.4), we arrive at the fulfillment of (1.6).
Further, considering h(t)∈C1[0,T] and differentiating with respect to t (1.5), we get:
ut(0,t)=γ(t)u(1,t)+h′(t)(0≤t≤T) | (1.9) |
Substituting x=0 into equation (1.1), we have:
ut(0,t)−butxx(0,t)+a(t)uxxxx(0,t)=p(t)u(0,t)+f(0,t)(0≤t≤T). | (1.10) |
Now, suppose that {u(x,t),p(t)} is a solution to problem (1.1)-(1.3), (1.6), (1.7). Then from (1.8), taking into account (1.3) and (1.6), we find:
ddt∫10u(x,t)dx−p(t)∫10u(x,t)dx=0(0≤t≤T). | (1.11) |
Due to (1.2) and ∫10φ(x)dx=0, it's obvious that
∫10u(x,0)dx+δ∫10u(x,T)dx=∫10φ(x)dx=0. | (1.12) |
Obviously, the general solution(1.11) has the form:
∫10u(x,t)dx=ce−∫t0p(τ)dτ(0≤t≤T). | (1.13) |
From here, taking into account (1.12), we obtain:
∫10u(x,0)dx+δ∫10u(x,T)dx=c(1+δe−∫T0p(τ)dτ)=0. | (1.14) |
By virtue of δ≥0, from (1.14) we get that c=0, and substituting into (1.13) we conclude, that ∫10u(x,t)dx=0(0≤t≤T). Therefore, condition (1.4) is also satisfied.
Further, from (1.7) and (1.10), we obtain:
ddt[u(0,t)−(∫t0γ(τ)u(1,τ)dτ+h(t))]= |
=p(t)[u(0,t)−(∫t0γ(τ)u(1,τ)dτ+h(t))](0≤t≤T). | (1.15) |
Let introduce the notation:
y(t)≡u(0,t)−(∫t0γ(τ)u(1,τ)dτ+h(t))(0≤t≤T) | (1.16) |
and rewrite the last relation in the form:
y′(t)+p(t)y(t)=0(0≤t≤T). | (1.17) |
From (1.16), taking into account (1.2), δγ(t)=0 (0≤t≤T) and φ(0)=h(0)+δh(T), it is easy to see that
y(0)+δy(T)=u(0,0)−h(0)+δ[u(0,T)−(∫T0γ(τ)u(1,τ)dτ+h(T))]=u(0,0)+ |
+δu(0,T)−(h(0)+δh(T))−δ∫T0γ(τ)u(1,τ)dτ=φ(0)−(h(0)+δh(T))=0. | (1.18) |
Obviously, the general solution (1.17) has the form:
y(t)=ce−∫t0p(τ)dτ(0≤t≤T). | (1.19) |
From here, taking into account (1.18), we obtain:
y(0)+δy(T)=c(1+δe−∫T0a0(τ)a1(τ)dτ)=0. | (1.20) |
By virtue of δ≥0, from (1.20) we get that c=0, and substituting into (1.19) we conclude that y(t)=0(0≤t≤T). Therefore, from (1.16) it is clear that the condition (1.5). The theorem has been proven.
It is known [5] that the system
1,cosλ1x,sinλ1x,...,cosλkx,sinλkx,... | (2.1) |
forms the basis of L2(0,1), where λk=2kπ(k=0,1,...).
Since system (2.1) forms a basis in L2(0,1), it is obvious that for each solution {u(x,t),a(t)} problem (1.1)–(1.3), (1.6), (1.7):
u(x,t)=∞∑k=0u1k(t)cosλkx+∞∑k=1u2k(t)sinλkx(λk=2πk), | (2.2) |
where
u10(t)=∫10u(x,t)dx,u1k(t)=2∫10u(x,t)cosλkxdx(k=1,2,...), |
u2k(t)=2∫10u(x,t)sinλkxdx(k=1,2,...). |
Applying the formal scheme of the Fourier method, to determine the desired coefficients u1k(t)(k=0,1,...) and u2k(t)(k=1,2,...) functions u(x,t) from (1.1) and (1.2) we get:
u″10(t)=F10(t;u,p)(0≤t≤T), | (2.3) |
(1+bλ2k)u′ik(t)+a(t)λ4kuik(t)=Fik(t;u,p)(i=1,2;0≤t≤T;k=1,2,...), | (2.4) |
u10(0)+δu10(T)=φ10, | (2.5) |
uik(0)+δuik(T)=φik(i=1,2;k=1,2,...), | (2.6) |
where
F1k(t;u,a,b)=p(t)u1k(t)+f1k(t)(k=0,1,...), |
f10(t)=∫10f(x,t)dx,f1k(t)=2∫10f(x,t)cosλkxdx(k=1,2,...), |
φ10=∫10φ(x)dx,φ1k=2∫10φ(x)cosλkxdx(k=1,2,...), |
F2k(t;u,a,b)=p(t)u2k(t)+f2k(t), |
f2k(t)=2∫10f(x,t)sinλkxdx(k=1,2,...),φ2k=2∫10φ(x)sinλkxdx(k=1,2,...). |
Solving problem (2.3)-(2.6), we find:
u10(t)=(1+δ)−1(φ10−δ∫T0F0(τ;u,p)dτ)+∫t0F10(τ;u,p)dτ(0≤t≤T), | (2.7) |
uik(t)=e−∫t0a(s)λ4k1+bλ2kds1+δe−∫T0a(s)λ4k1+bλ2kdssφik+11+bλ2k∫t0Fik(τ;u,p)e−∫tτa(s)λ4k1+bλ2kdsdτ− |
−δe−∫T0a(s)λ4k1+bλ2kds1+δe−∫T0a(s)λ4k1+bλ2kds11+bλ2k∫T0Fik(τ;u,p)e−∫tτa(s)λ4k1+bλ2kdsdτ(i=1,2;0≤t≤T;k=1,2,...). | (2.8) |
After substituting the expression u1k(t)(k=0,1,...), u2k(t)(k=1,2,...) in (2.2), to define a component u(x,t) solution of problem (1.1)-(1.3), (1.6), (1.7), we obtain:
u(x,t)=(1+δ)−1(φ0−δ∫T0F0(τ;u,p)dτ)+∫t0F0(τ;u,p)dτ+ |
+∞∑k=1{e−∫t0a(s)λ4k1+bλ2kds1+δe−∫T0a(s)λ4k1+bλ2kdssφ1kk+11+bλ2k∫t0F1k(τ;u,p)e−∫tτa(s)λ4k1+bλ2kdsdτ− |
−δe−∫T0a(s)λ4k1+bλ2kds1+δe−∫T0a(s)λ4k1+bλ2kds11+bλ2k∫T0F1k(τ;u,p)e−∫tτa(s)λ4k1+bλ2kdsdτ}cosλkx+ |
+∞∑k=1{e−∫t0a(s)λ4k1+bλ2kds1+δe−∫T0a(s)λ4k1+bλ2kdssφ2kk+11+bλ2k∫t0F2k(τ;u,p)e−∫tτa(s)λ4k1+bλ2kdsdτ− |
−δe−∫T0a(s)λ4k1+bλ2kds1+δe−∫T0a(s)λ4k1+bλ2kds11+bλ2k∫T0F2k(τ;u,p)e−∫tτa(s)λ4k1+bλ2kdsdτ}sinλkx. | (2.9) |
Now from (1.7), taking into account (2.2), we have:
p(t)=[h(t)]−1{h′(t)−f(0,t)+γ(t)u10(t)−p(t)∫t0γ(τ)u10(τ)dτ+ |
+∞∑k=1(bλ2ku′1k(t)+a(t)λ4ku1k(t)+γ(t)u1k(t)−p(t)∫t0γ(τ)u1k(τ)dτ). | (2.10) |
Further, from (2.4), taking into account (2.8), we obtain:
bλ2ku′1k(t)+a(t)λ4ku1k(t)+γ(t)u1k(t)=F1k(t;u,p)−u′1k(t)+γ(t)u1k(t)= |
=bλ2k1+bλ2kF1k(t;u,p)+(a(t)λ4k1+bλ2k+γ(t))u1k(t)= |
=bλ2k1+bλ2kFk(t;u,p)+(a(t)λ4k1+bλ2k+γ(t))[e−∫t0a(s)λ4k1+bλ2kds1+δe−∫T0a(s)λ4k1+bλ2kdssφ1k+ |
+11+bλ2k∫t0F1k(τ;u,p)e−∫tτa(s)λ4k1+bλ2kdsdτ− |
−δe−∫T0a(s)λ4k1+bλ2kds1+δe−∫T0a(s)λ4k1+bλ2kds11+bλ2k∫T0F1k(τ;u,p)e−∫tτa(s)λ4k1+bλ2kdsdτ](0≤t≤T;k=1,2,...). | (2.11) |
p(t)=[h(t)]−1{h′(t)−f(0,t)+ |
+γ(t))[(1+δ)−1(φ10−δ∫T0F0(τ;u,p)dτ)+∫t0F10(τ;u,p)dτ]− |
−p(t)∫t0γ(τ)u10(τ)dτ+∞∑k=1[bλ2k1+bλ2kF1k(t;u,p)+ |
+(a(t)λ4k1+bλ2k+γ(t))[e−∫t0a(s)λ4k1+bλ2kds1+δe−∫T0a(s)λ4k1+bλ2kdssφ1k+11+bλ2k∫t0F1k(τ;u,p)e−∫tτa(s)λ4k1+bλ2kdsdτ− |
+11+bλ2k∫t0F1k(τ;u,p)e−∫tτa(s)λ4k1+bλ2kdsdτ− |
−δe−∫T0a(s)λ4k1+bλ2kds1+δe−∫T0a(s)λ4k1+bλ2kds11+bλ2k∫T0F1k(τ;u,p)e−∫tτa(s)λ4k1+bλ2kdsdτ]+ |
+p(t)∫t0γ(τ)u1k(τ)dτ]}. | (2.12) |
Thus, the solution of problem (1.1)–(1.3), (1.6), (1.7)is reduced to the solution of system (2.9), (2.12) with respect to unknown functions u(x,t) and p(t).
To study the question of the uniqueness of the solution of problem (1.1)–(1.3), (1.6), (1.7) the following plays an important role.
Lemma 1. If {u(x,t),p(t)}-any solution of problem (1.1)–(1.3), (1.6), (1.7), then the functions
u10(t)=∫10u(x,t)dx,u1k(t)=2∫10u(x,t)cosλkxdx(k=1,2,...), |
u2k(t)=2∫10u(x,t)sinλkxdx(k=1,2,...) |
satisfy the system consisting of equations (27), (28) on [0,T].
It is obvious that if u10(t)=∫10u(x,t)dx, u1k(t)=2∫10u(x,t)cosλkxdx(k=1,2,...), u2k(t)=2∫10u(x,t)sinλkxdx(k=1,2,...) is a solution to system (2.7), (2.8), then the pair {u(x,t),p(t)} functions u(x,t)=∑∞k=0u1k(t)cosλkx+∑∞k=1u2k(t)sinλkx(λk=2πk) and p(t) is a solution to system (2.9), (2.12).
Consequence. Let system (29), (32) have a unique solution. Then problem (1.1)–(1.3), (1.6), (1.7) cannot have more than one solution, i.e. if problem (1.1)-(1.3), (1.6), (1.7) has a solution, then it is unique.
In order to study the problem (1.1)–(1.3), (1.6), (1.7) consider the following spaces.
Denote by Bα2,T [6] the set of all functions of the form
u(x,t)=∞∑k=0u1k(t)cosλkx+∞∑k=1u2k(t)sinλkx(λk=2πk), |
considered in DT, where each of the functions u1k(t)(k=0,1,...), u2k(t)(k=1,2,...) continuous on [0,T] and
J(u)=‖u10(t)‖C[0,T]+{∞∑k=1(λαk‖u1k(t)‖C[0,T])2}12+{∞∑k=1(λαk‖u2k(t)‖C[0,T])2}12<+∞, |
α≥0. We define the norm in this set as follows:
‖u(x,t)‖Bα2,T=J(u). |
Through EαT denote the space Bα2,T×C[0,T] vector - functions z(x,t)={u(x,t),p(t)} with norm
‖z(x,t)‖EαT=‖u(x,t)‖Bα2,T+‖p(t)‖C[0,T]. |
It is known that Bα2,T and EαT are Banach spaces.
Now consider in space E5T operator
Φ(u,p)={Φ1(u,p),Φ2(u,p)}, |
operator
Φ1(u,p))=˜u(x,t)≡∞∑k=0˜u1k(t)cosλkx+∞∑k=1˜u2k(t)sinλkx,Φ2(u,p)=˜p(t), |
˜u10(t),˜uik(t)(i=1,2;k=1,2,...),˜p(t) are equal to the right-hand sides of (2.7), (2.8) and (2.12), respectively.
It is easy to see that
1+bλ2k>bλ2k,1+δ≥1,.1+δe−∫T0a(s)λ4k1+bλ2kds≥1. |
Then, we have:
‖˜u0(t)‖C[0,T]≤|φ10|+(1+δ)√T(∫T0|f10(τ)|2dτ)12+(1+δ)T‖p(t)‖C[0,T]‖u10(t)‖C[0,T], | (2.13) |
(∞∑k=1(λ5k‖˜uik(t)‖C[0,T])2)12≤√3(∞∑k=1(λ5k|φik|)2)12+√3(1+δ)b√T(∫T0∞∑k=1(λ3k|fik(τ)|)2dτ)12+ |
+√3(1+δ)bT‖p(t)‖C[0,T](∞∑k=1(λ5k‖uik(t)‖C[0,T])2)12(i=1,2), | (2.14) |
‖˜p(t)‖C[0,T]≤‖[h(t)]−1‖C[0,T]{‖h′(t)−f(0,t)‖C[0,T]+ |
+‖γ(t)‖C[0,T][|φ0|+(1+δ)√T(∫T0|f0(τ)|2dτ)12+(1+δ)T‖p(t)‖C[0,T]‖u0(t)‖C[0,T]]+ |
+T‖γ(t)‖C[0,T]‖p(t)‖C[0,T]‖u10(t)‖C[0,T]+ |
+(∞∑k=1λ−2k)12[(∞∑k=1(λk‖f1k(t)‖)2C[0,T])12+‖p(t)‖C[0,T](∞∑k=1(λ3k‖u1k(t)‖C[0,T])2)12+ |
+(‖γ(t)‖C[0,T]+1b‖a(t)‖C[0,T])[(∞∑k=1(λ3k|φ1k|)2)12+√T(1+δ)b(∫T0∞∑k=1(λk|f1k(τ)|)2dτ)12+ |
+T(1+δ)b‖p(t)‖C[0,T](∞∑k=1(λ5k‖u1k(t)‖C[0,T])2)12]++T‖γ(t)‖C[0,T]‖p(t)‖C[0,T](∞∑k=1(λ5k‖u1k(t)‖C[0,T])2)12]}. | (2.15) |
Let us assume that the data of problem (1.1)–(1.3), (1.6), (1.7) satisfy the following conditions:
1.φ(x)∈W2(5)(0,1),φ(0)=φ(1),φ′(0)=φ′(1), |
φ″(0)=φ″(1),φ‴(0)=φ‴(1),φ(4)(0)=φ(4)(1); |
2.f(x,t),fx(x,t),fxx(x,t)∈C(DT),fxxx(x,t)∈L2(DT), |
f(0,t)=f(1,t),fx(0,t)=fx(1,t),fxx(0,t)=fxx(1,t)(0≤t≤T); |
3.b>0,δ≥0,γ(t),a(t)∈C[0,T],h(t)∈C1[0,T],h(t)≠0(0≤t≤T). |
Then from (2.10)–(2.12), we have:
‖˜u(x,t)‖B52,T≤A1(T)+B1(T)‖p(t)‖C[0,T]‖u(x,t)‖B52,T, | (2.16) |
‖˜p(t)‖C[0,T]≤A2(T)+B2(T)‖p(t)‖C[0,T]‖u(x,t)‖B52,T, | (2.17) |
where
A1(T)=‖φ(x)‖L2(0,1)+(1+δ)√T‖f(x,t)‖L2(DT)+2√3‖φ(5)(x)‖L2(0,1)+ |
+2√3b(1+δ)√T‖fxxx(x,t)‖L2(DT),B1(T)=(1+δ)(1+√3b)T, |
A2(T)=‖[h(t)]−1‖C[0,T]{‖h′(t)−f(0,t)‖C[0,T]+ |
+‖γ(t)‖C[0,T](‖φ(x)‖L2(0,1)+(1+δ)√T‖f(x,t)‖L2(DT))+ |
+(∞∑k=1λ−2k)12[‖‖fx(x,t)‖C[0,T]‖L2(0,1)+ |
+(‖γ(t)‖C[0,T]+1b‖a(t)‖C[0,T])(‖φ(3)(x)‖L2(0,1)+√T(1+δ)b‖fx(x,t)‖L2(DT))]}, |
B2(T)=‖[h(t)]−1‖C[0,T](∑∞k=1λ−2k)12[(‖γ(t)‖C[0,T]+1b‖a(t)‖C[0,T])T(2+δ)b+ |
+T‖γ(t)‖C[0,T]+1]. |
From inequalities (2.16), (2.17) we conclude:
‖u(x,t)‖B52,T+‖˜p(t)‖C[0,T]≤A(T)+B(T)‖p(t)‖C[0,T]‖u(x,t)‖B52,T, | (2.18) |
A(T)=A1(T)+A2(T),B(T)=B1(T)+B2(T). |
We can prove the following theorem.
Theorem 2. Let conditions 1-3 be satisfied and
(A(T)+2)2B(T)<1. | (2.19) |
Then problem (1.1)–(1.3), (1.6), (1.7) has in K=KR(‖z‖E5T≤R=A(T)+2) in the space E5T only one solution.
Proof. In space E5T consider the equation
z=Φz, | (2.20) |
where z={u,p}, components P Φ1(u,p),Φ2(u,p) of operators Φ(u,p) are defined by the right-hand sides of equations (2.9) and (2.12).
Consider the operator Φ(u,p) in a ball K=KR from E5T. Similarly to (2.18) we obtain that for any z={u,p}, z1={u1,p1}, z2={u2,p2}∈KR :
‖Φz‖E5T≤A(T)+B(T)‖p(t)‖C[0,T]‖u(x,t)‖B52,T, | (2.21) |
‖Φz1−Φz2‖E5T≤B(T)R(‖p1(t)−p2(t)‖C[0,T]+‖u1(x,t)−u2(x,t)‖B52,T). | (2.22) |
Then from estimates (2.21), (2.22), taking into account (2.19), it follows that the operator Φ acts in a ball K=KR and is contractive. Therefore, in the ball K=KR operator Φ has a single fixed point {u,p}, which is the only one in the ball K=KR solution of equation (2.20), i.e. is the only one solution in the ball K=KR of system (2.9), (2.12) in the ball.
Functions u(x,t), as an element of space B52,T is continuous and has continuous derivatives ux(x,t),uxx(x,t), uxxx(x,t),uxxxx(x,t) in DT.
From (2.4), it is easy to see that
(∞∑k=1(λk‖u′ik(t)‖C[0,T])2)12≤√2b‖a(t)‖C[0,T](∞∑k=1(λ5k‖uik(t)‖C[0,T])2)12+ |
+√2b‖‖fx(x,t)+p(t)ux(x,t)‖C[0,T]‖L2(0,1)(i=1,2). |
Hence it follows that ut(x,t) and utxx continuous in DT.
It is easy to check that equation (1.1) and conditions (1.2), (1.3), (1.6), (1.7) are satisfied in the usual sense. Consequently, {u(x,t),p(t)} is a solution to problem (1.1)–(1.3), (1.6), (1.7). By the corollary of Lemma 1, it is unique in the ball K=KR. The theorem has been proven.
With the help of Theorem 1, the unique solvability of the original problem (1.1)–(1.5) immediately follows from the last theorem.
Theorem 3. Let all the conditions of Theorem 1 be satisfied, ∫10f(x,t)dx=0(0≤t≤T), δγ(t)=0 (0≤t≤T) and the matching condition is met:
∫10φ(x)dx=0,φ(0)=h(0)+δh(T). |
Then problem (1.1)–(1.5) has in the ball K=KR(‖z‖E5T≤R=A(T)+2) from E5T the only classical solution.
The article considered an inverse boundary value problem with a periodic and integral condition, when the unknown coefficient depends on time for a linear pseudoparabolic equation of the fourth order. An existence and uniqueness theorem for the classical solution of the problem is proved.
The authors have declared no conflict of interest.
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