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# Open Mathematics

### formerly Central European Journal of Mathematics

Editor-in-Chief: Gianazza, Ugo / Vespri, Vincenzo

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Volume 15, Issue 1

# Determining of right-hand side of higher order ultraparabolic equation

Nataliya Protsakh
• Corresponding author
• National Forestry Engineering University of Ukraine, L’viv, Generala Chuprynky, 103, 79057, Ukraine
• Pidstryhach Institute for Applied Problems of Mechanics and Mathematics NAS of Ukraine, L’viv, Naukova, 3b, 79060, Ukraine
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Published Online: 2017-08-29 | DOI: https://doi.org/10.1515/math-2017-0086

## Abstract

In the paper the conditions of the existence and uniqueness of the solution for the inverse problem for higher order ultraparabolic equation are obtained. The equation contains two unknown functions of spatial and time variables in its right-hand side. The overdetermination conditions of the integral type are used.

MSC 2010: 35K70; 35R30

## 1 Introduction

The mathematical modelling of many applied problems of physics, biology and finance (such as processes of the diffusion with inertia, grows of population etc.) leads to the initial-boundary value problems for linear and nonlinear ultraparabolic equations (see [13]). The results of investigation of the unique solvability for the direct initial-boundary value problems for nonlinear ultraparabolic equations could be find here [4, 5]. If in the equation one or several coefficients or the right-hand side function are unknown, then the problems of their identification with the use of some additional information are called inverse problems.

In this paper we investigate the inverse problem of determination of two unknown functions of spatial and time variables in the right-hand side of the nonlinear higher order ultraparabolic equation. As an additional information we use the overdetermination conditions of the integral type. Using the known results for the direct problem for nonlinear ultraparabolic equations [5] and the method of successive approximations, we determine the sufficient conditions of the unique solvability of the inverse problem. Note that in [6] the right-hand side function of the higher order ultraparabolic equation contains the several unknown functions that depend only on time variable.

The problems of identifying of one right-hand side function in parabolic, hyperbolic or ultraparabolic equations were considered in [717], of the several unknown functions in [6, 11, 18, 19]. The investigation of those problems is based on the method of integral equations and the Shauder principle [1114, 18], the iterative and regularization methods [17] or the method of successive aproximation [6, 7, 15, 16].

## 2.1 Statement of the problem

Let Ωx ⊂ ℝn and Ωy ⊂ ℝl be bounded domains with boundaries ΩxCm0 and ΩyC1;T ∈ (0, ∞), x ∈ Ωx, y ∈ Ωy, t ∈ (0, T);G = Ωx × Ωy, Π = Ωy × (0, T), QT = Ωx × Ωy × (0, T) and ΣT = Ωx × Ωy × (0, T), ST = Ωx × Ωy × (0, T); {n, l, m0} ⊂ ℕ, {α, γ} ⊂ ℕn, ν is the outward unit normal vector to the surface ST, t, yi are partial derivatives on t and yi respectively, $\begin{array}{}{D}^{\alpha }=\frac{{\mathrm{\partial }}^{|\alpha |}}{\mathrm{\partial }{x}_{1}^{{\alpha }_{1}},\dots ,\mathrm{\partial }{x}_{n}^{\alpha n}},|\alpha |={\alpha }_{1}+\cdots +{\alpha }_{n}.\end{array}$

We shall use the following spaces:

• L(QT) := {w : QT → ℝ; w is measurable and there is a constant C such that |w(x, y, t)| ≤ C a.e. on QT} with ∥w;L(QT)∥ = inf{C : |w(x, y, t)| ≤ C a.e. on QT};

• Lx) := {w : Ωx → ℝ; w is measurable and there is a constant C such that |w(x)| ≤ C a.e. on Ωx} with ∥w;Lx)∥ = inf{C : |w(x)| ≤ C a.e. on Ωx};

• $\begin{array}{}{L}^{2}\left(G\right):=\left\{w:G\to \mathbb{R};\underset{G}{\int }|w\left(x,y\right){|}^{2}dx\phantom{\rule{thinmathspace}{0ex}}dy<\mathrm{\infty }\right\}\text{\hspace{0.17em}with\hspace{0.17em}}\parallel w;{L}^{2}\left(G\right)\parallel =\left(\underset{G}{\int }|w\left(x,y\right){|}^{2}dx\phantom{\rule{thinmathspace}{0ex}}dy{\right)}^{\frac{1}{2}};\end{array}$

• $\begin{array}{}{L}^{2}\left({\mathrm{\Omega }}_{x}\right):=\left\{w:{\mathrm{\Omega }}_{x}\to \mathbb{R};\underset{{\mathrm{\Omega }}_{x}}{\int }|w\left(x\right){|}^{2}dx<\mathrm{\infty }\right\}\text{\hspace{0.17em}with\hspace{0.17em}}\parallel w;{L}^{2}\left({\mathrm{\Omega }}_{x}\right)\parallel =\left(\underset{{\mathrm{\Omega }}_{x}}{\int }|w\left(x\right){|}^{2}dx{\right)}^{\frac{1}{2}};\end{array}$

• $\begin{array}{}{L}^{2}\left(0,T\right):=\left\{w:\left[0,T\right]\to \mathbb{R};\underset{0}{\overset{T}{\int }}|w\left(t\right){|}^{2}dt<\mathrm{\infty }\right\}\text{\hspace{0.17em}with\hspace{0.17em}}\parallel w;{L}^{2}\left(0,T\right)\parallel =\left(\underset{0}{\overset{T}{\int }}|w\left(t\right){|}^{2}dt{\right)}^{\frac{1}{2}};\end{array}$

• $\begin{array}{}{L}^{2}\left({Q}_{T}\right):=\left\{w:{Q}_{T}\to \mathbb{R};\underset{{Q}_{T}}{\int }|w\left(x,y,t\right){|}^{2}\mathit{d}\mathit{x}\phantom{\rule{thinmathspace}{0ex}}\mathit{d}\mathit{y}\phantom{\rule{thinmathspace}{0ex}}\mathit{d}\mathit{t}<\mathrm{\infty }\right\}\end{array}$ with $\begin{array}{}\parallel w;{L}^{2}\left({Q}_{T}\right)\parallel =\left(\underset{{Q}_{T}}{\int }|w\left(x,y,t\right){|}^{2}\mathit{d}\mathit{x}\phantom{\rule{thinmathspace}{0ex}}\mathit{d}\mathit{y}\phantom{\rule{thinmathspace}{0ex}}\mathit{d}\mathit{t}{\right)}^{\frac{1}{2}};\end{array}$

• Wk,2(⋅) is a set of functions wL2(⋅) such that Dα u ∈ L2(⋅), |α| ≤ k, with $∥w;Wm0,2(Ωx)∥=(∫Ωx∑|α|≤m0|Dαw(x)|2dx)12,∥w;W1,2(0,T=(∫0T[|w(t)|2+|w′(t)|2]dt)12;$

• $\begin{array}{}{W}_{0}^{{m}_{0},2}\left({\mathrm{\Omega }}_{x}\right)\end{array}$ is a closure of $\begin{array}{}{C}_{0}^{\mathrm{\infty }}\left({\mathrm{\Omega }}_{x}\right)\text{\hspace{0.17em}in\hspace{0.17em}}{W}^{{m}_{0},2}\left({\mathrm{\Omega }}_{x}\right)\end{array}$ under the norm of this space, where $\begin{array}{}{C}_{0}^{\mathrm{\infty }}\left({\mathrm{\Omega }}_{x}\right)\end{array}$ is a set of functions with compact support, all infinitely differentiable on Ωx;

• Ck(O) is the space of all k – times continuously differentiable functions on O;

• C([0, T];L2(G)) is the set of continuous functions ([0, T] → L2(G));

• C1([0, T];C2y)) is the set of continuously differentiable functions ([0, T] → C2y));

• Cx;L2(Π)) is the set of continuous functions (ΩxL2(Π));

• C1y;C1x)) is the set of continuously differentiable functions (ΩyC1x)).

We also introduce the following spaces:

• $\begin{array}{}{V}_{1}\left({Q}_{T}\right):=\left\{w:{D}^{\alpha }w\in {L}^{2}\left({Q}_{T}\right)\left(|\alpha |\le {m}_{0}\right),\frac{{\mathrm{\partial }}^{i}w}{\mathrm{\partial }{v}^{i}}{|}_{{\mathrm{\Sigma }}_{T}}=0\left(i\in \left\{0,1,\dots ,{m}_{0}-1\right\}\right)\right\};\end{array}$

• $\begin{array}{}{V}_{2}\left(G\right):=\left\{w:{D}^{\alpha }w,\phantom{\rule{thinmathspace}{0ex}}{\mathrm{\partial }}_{yj}\phantom{\rule{thinmathspace}{0ex}}w\in {L}^{2}\left(G\right)\left(|\alpha |\le {m}_{0},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}j\in \left\{1,\dots ,l\right\}\right),\phantom{\rule{thinmathspace}{0ex}}\frac{{\mathrm{\partial }}^{i}w}{\mathrm{\partial }{\nu }^{i}}{|}_{\mathrm{\partial }{\mathrm{\Omega }}_{x}×{\mathrm{\Omega }}_{y}}=0\left(i\in \left\{0,1,\dots ,{m}_{0}-1\right\}\right),\phantom{\rule{thinmathspace}{0ex}}w{|}_{{\mathrm{\Omega }}_{x}×{\mathrm{\Gamma }}_{1}}=0\right\};\end{array}$

• $\begin{array}{}{V}_{3}\left({Q}_{T}\right):=\left\{w:{D}^{\alpha }w,\phantom{\rule{thinmathspace}{0ex}}{\mathrm{\partial }}_{{y}_{j}}w,\phantom{\rule{thinmathspace}{0ex}}{\mathrm{\partial }}_{t}w\in {L}^{2}\left({Q}_{T}\right)\left(|\alpha |\le {m}_{0},\phantom{\rule{thinmathspace}{0ex}}j\in \left\{1,\dots ,l\right\}\right),\phantom{\rule{thinmathspace}{0ex}}w{|}_{{S}_{T}^{1}}=0,\frac{{\mathrm{\partial }}^{i}w}{\mathrm{\partial }{\nu }^{i}}{|}_{{\mathrm{\Sigma }}_{T}}=0\left(i\in \left\{0,1,\phantom{\rule{thinmathspace}{0ex}}\dots ,\phantom{\rule{thinmathspace}{0ex}}{m}_{0}-1\right\}\right)\right\};\end{array}$

• V4(QT):= {w : wV3(QT), DαwL2(QT) (|α| ≤ 2m0).

We consider the equation $∂tu+∑i=1lλi(x,y,t)∂yiu−∑0<|α|=|γ|≤m0(−1)|γ|Dγ(aαγ(x)Dαu)+c(x,y,t)u+g(x,y,t,u)==f1(x,y,t)q1(x)+f2(x,y,t)q2(t)+f0(x,y,t),(x,y,t)∈QT$(1) with the initial condition $u(x,y,0)=u0(x,y),(x,y)∈G,$(2) the boundary conditions $∂iu∂νi|ΣT=0(i∈{0,1,…,m0−1});u|ST1=0$(3) and the overdetermination conditions $∫ΠK1(y,t)u(x,y,t)dydt=E1(x),x∈Ωx,$(4) $∫GK2(x,y)u(x,y,t)dxdy=E2(t),t∈[0,T],$(5) where u(x, y, t), q1(x), q2(t) are unknown functions, $\begin{array}{}{S}_{T}^{1}=\left\{\left(x,\phantom{\rule{thinmathspace}{0ex}}y,\phantom{\rule{thinmathspace}{0ex}}t\right)\in {S}_{T}\phantom{\rule{thinmathspace}{0ex}}:\phantom{\rule{thinmathspace}{0ex}}\sum _{i=1}^{l}{\lambda }_{j}\left(x,\phantom{\rule{thinmathspace}{0ex}}y,\phantom{\rule{thinmathspace}{0ex}}t\right)\mathrm{cos}\left(\nu ,\phantom{\rule{thinmathspace}{0ex}}{y}_{i}\right)<0\right\}.\end{array}$

Assume that the following conditions hold:

(A): aαγLx), 0 < |α| = |γ| ≤ m0, $∑0<|α|=|γ|≤m0∫Ωxaαγ(x)DαwDγwdx≥a0∫Ωx∑|α|=m0|Dαw|2dx$ for all w$\begin{array}{}{W}_{0}^{{m}_{0},2}\left({\mathrm{\Omega }}_{x}\right)\end{array}$, a0 > 0;

(C): cL(QT), c(x, y, t) ≥ c0 for almost all (x, y, t) ∈ QT, where c0 is a constant;

(E): E1$\begin{array}{}{W}_{0}^{{m}_{0},2}\left({\mathrm{\Omega }}_{x}\right)\end{array}$, E2W1,2(0, T);

(F): f1Cx;L2(Π)), f2C([0, T];L2(G)), f0L2(QT);

(G): g(x, y, t, ξ) is measurable with respect to the variables (x, y, t) in the domain QT for all ξ ∈ ℝ1 and is continuous with respect to ξ for almost all (x, y, t) ∈ QT, moreover, there exists a positive constant g0, such that $|g(x,y,t,ξ)−g(x,y,t,η)|≤g0|ξ−η| for almost all (x,y,t)∈QT and all {ξ,η}⊂R1;$

(S): there exists Γ1Ωy ⊂ ℝl − 1, such that the surface $\begin{array}{}{S}_{T}^{1}\end{array}$ = Ωx × Γ1 × (0, T);

(K): K1C1([0, T];C1y)), K1(y, 0) = K1(y, T) = 0 for all y ∈ Ωy, $K1|Γ2×(0,T)=0, where Γ2=∂Ωy∖Γ1,K2∈C1(Ωy;C1(Ω¯x)),DαK2|∂Ωx×Ωy=0(|α|≤m0−1),K2|Ωx×Γ2=0;$

(L): λiC(QT), yiλiL(QT) for all i ∈ {1,…, l};

(U): u0V2(G).

## 2.2 Direct problem

First we assume that in Eq. (1) $\begin{array}{}{q}_{1}\left(x\right)={q}_{1}^{\ast }\left(x\right),\phantom{\rule{thinmathspace}{0ex}}{q}_{2}\left(t\right)={q}_{2}^{\ast }\left(t\right),\text{\hspace{0.17em}where\hspace{0.17em}}{q}_{1}^{\ast }\in {L}^{2}\left({\mathrm{\Omega }}_{x}\right),\phantom{\rule{thinmathspace}{0ex}}{q}_{2}^{\ast }\in {L}^{2}\left(0,\phantom{\rule{thinmathspace}{0ex}}T\right),\end{array}$ are known functions; consider the initial-boundary value problem for the Eq. (1) with the initial condition (2) and with the boundary conditions (3).

The results presented in [5] yield the following statements.

#### Theorem 2.1

Suppose that the conditions (A), (C), (G), (L), (F), (U), (S) hold, and, besides:

1. $\begin{array}{}{D}^{\alpha }{a}_{\alpha \gamma }\in {L}^{\mathrm{\infty }}\left({\mathrm{\Omega }}_{x}\right),{\mathrm{\partial }}_{{y}_{k}}c\in {L}^{\mathrm{\infty }}\left({Q}_{T}\right)\left(0<|\alpha |=|\gamma |\le {m}_{0},\phantom{\rule{thinmathspace}{0ex}}k\in \left\{1,...,l\right\}\right),{q}_{1}^{\ast }\in {L}^{2}\left({\mathrm{\Omega }}_{x}\right),{q}_{2}^{\ast }\in {L}^{2}\left(0,T\right),{\mathrm{\partial }}_{{y}_{k}}{f}_{s}\in {L}^{2}\left({Q}_{T}\right)\left(s\in \left\{0,1,2\right\},k\in \left\{1,\dots ,l\right\}\right);\end{array}$

2. $\begin{array}{}|{\mathrm{\partial }}_{{y}_{i}}g\left(x,y,t,\xi \right)|\le {g}^{1}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\left(i\in \left\{1,...,l\right\}\right),\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}g\left(x,y,t,0\right){|}_{{S}_{T}^{1}}=0\end{array}$ for almost all (x, y, t) ∈ QT and all ξ ∈ ℝ1, where g1 is a positive function;

3. $\begin{array}{}{f}_{0}{|}_{{S}_{T}^{1}}=0,\phantom{\rule{thinmathspace}{0ex}}{f}_{1}{|}_{{S}_{T}^{1}}=0,\phantom{\rule{thinmathspace}{0ex}}{f}_{2}{|}_{{S}_{T}^{1}}=0,\end{array}$ then there exists a unique function u*V3(QT) ∩ C([0, T];L2(G)), that satisfies the condition (2) and the equality $∫QT[∂tu∗v+∑i=1lλj(x,y,t)∂yiu∗v+∑0<|α|=|γ|≤m0aαγ(x)Dαu∗Dγv+c(x,y,t)u∗v++g(x,y,t,u∗)v]dxdydt=∫QT[f1(x,y,t)q1∗(x)+f2(x,y,t)q2∗(t)+f0(x,y,t)]vdxdydt$(6) for all functions νV1(QT).

Moreover, u*V4(QT) ∩ C([0, T];L2(G)), u* satisfies the condition (2) and the Eq. (1) for almost all (x, y, t) ∈ QT (so, u* is a solution to the problem (1)(3)).

The proof is carried out according to the scheme of proof of Theorem 1, 2 [5].

#### Remark

The solution u* of the problem (1)(3) has such appraisals: $∫G|u∗(x,y,τ)|2dxdy≤M1(∫QT(|f1(x,y,t)|2|q1∗(x)|2+|f2(x,y,t)|2|q2∗(t)|2++|f0(x,y,t)|2)dxdydt+∫G|u0(x,y)|2dxdy),τ∈[0,T],$(7) $∫G∑i=1l|∂yiu∗(x,y,τ)|2dxdy≤M2(∫QT(|f1(x,y,t)|2|q1∗(x)|2+|f2(x,y,t)|2|q2∗(t)|2++∑i=1l|∂yif1(x,y,t)|2|q1∗(x)|2+∑i=1l|∂yif2(x,y,t)|2|q2∗(t)|2+|f0(x,y,t)|2++∑i=1l|∂yif0(x,y,t)|2)dxdydt+∫G(|u0(x,y)|2+∑i=1l|∂yiu0(x,y)|2dxdy)),τ∈[0,T],$(8) $∫QT|∂tu∗|2dxdydt≤M3(∫QT(|f1(x,y,t)|2|q1∗(x)|2+|f2(x,y,t)|2|q2∗(t)|2+∑i=1l|∂yif1(x,y,t)|2|q1∗(x)|2++∑i=1l|∂yif2(x,y,t)|2|q2∗(t)|2+|f0(x,y,t)|2+∑i=1l|∂yif0(x,y,t)|2)dxdydt++∫G(|u0(x,y)|2+∑i=1l|∂yiu0(x,y)|2dxdy+∑α|=m0|Dαu0(x,y)|2dxdy)),τ∈[0,T],$(9) where the constants M1, M2, M3 are independent of $\begin{array}{}{q}_{1}^{\ast },{q}_{2}^{\ast }\end{array}$.

## 2.3 Inverse problem

#### Definition

A triple of functions (u(x, y, t), q1(x), q2(t)) is a solution to the problem (1)(5), if uV4(QT) ∩ C([0, T];L2(G)), q1L2x), q2L2(0, T), it satisfies Eq. (1) for almost all (x, y, t) ∈ QT and the conditions (2), (4), (5) hold.

Denote: $Δ1(x):=∫ΠK1(y,t)f1(x,y,t)dydt,Δ2(t):=∫GK2(x,y)f2(x,y,t)dxdy,A1(x):=−∑0<|α|=|γ|≤m0Dγ(aαγ(x)DαE1(x))−∫ΠK1(y,t)f0(x,y,t)dydt,A2(t):=E2′(t)−∫GK2(x,y)f0(x,y,t)dxdy,B1(x,y,t):=−∑i=1l∂yi(λi(x,y,t)K1(y,t))−∂tK1(y,t)+K1(y,t)c(x,y,t),B2(x,y,t):=−∑i=1l∂yi(λi(x,y,t)K2(x,y))+K2(x,y)c(x,y,t),F12(x,y,t):=K1(y,t)f2(x,y,t),F21(x,y,t):=K2(x,y)f1(x,y,t).$

Let $\begin{array}{}\left({u}^{\ast }\left(x,y,t\right),{q}_{1}^{\ast }\left(x\right),{q}_{2}^{\ast }\left(t\right)\right)\end{array}$ be a solution to the problem (1) — (5). Multiplication Eq. (1) on K1(y, t) and its integration on Π imply the equality $∫ΠK1(y,t)(∂tu∗+∑i=1lλi(x,y,t)∂yiu∗−∑0<|α|=|γ|≤m0Dγ(aαγ(x)Dαu∗)+c(x,y,t)u∗++g(x,y,t,u∗))dydt=q1∗(x)Δ1(x)+∫ΠF12(x,y,t)q2∗(t)dydt+∫ΠK1(y,t)f0(x,y,t)dydt$(10) for almost all x ∈ Ωx. Since the equality $∫ΠK1(y,t)∑0<|α|=|γ|≤m0Dγ(aαγ(x)Dαu∗)dydt=∑0<|α|=|γ|≤m0Dγ(aαγ(x)DαE1(x))$(11) follows from (4), then from (10), hypothesis (K) and (11) we obtain $Δ1(x)q1∗(x)=A1(x)+∫ΠB1(x,y,t)u∗dydt+∫ΠK1(y,t)g(x,y,t,u∗)dydt−−∫ΠF12(x,y,t)q2∗(t)dydt$(12) for almost all x ∈ Ωx.

On the basis of (1) and (5), we receive $∫GK2(x,y)(f1(x,y,t)q1∗(x)+f2(x,y,t)q2∗(t)+f0(x,y,t)−∑j=1lλi(x,y,t)uyi∗++∑0<|α|=|γ|≤m0Dγ(aαγ(x)Dαu∗)−c(x,y,t)u∗−g(x,y,t,u∗))dxdy=E2′(t)$(13) for almost all t ∈ [0, T]. Integrating (13) by parts and using the hypothesis (K), we obtain $Δ2(t)q2∗(t)=A2(t)+∫G(B2(x,y,t)u∗+∑0<|α|=|γ|≤m0DγK2(x,y)aαγ(x)Dαu∗++K2(x,y)g(x,y,t,u∗))dxdy−∫GF21(x,y,t)q1∗(x)dxdy,$(14) for almost all t ∈ [0, T].

Then let us show that if functions $\begin{array}{}{u}^{\ast }\in {V}_{4}\left({Q}_{T}\right)\cap C\left(\left[0,T\right];{L}^{2}\left(G\right)\right),\phantom{\rule{thinmathspace}{0ex}}{q}_{1}^{\ast }\in {L}^{2}\left({\mathrm{\Omega }}_{x}\right),\phantom{\rule{thinmathspace}{0ex}}{q}_{2}^{\ast }\in {L}^{2}\left(0,T\right)\end{array}$ satisfy (2), (12), (14) and Eq. (1) for almost all (x, y, t) ∈ QT, then they are the solution to the problem (1) — (5) with the right-hand side function $\begin{array}{}{f}_{1}\left(x,y,t\right){q}_{1}^{\ast }\left(x\right)+{f}_{2}\left(x,y,t\right){q}_{2}^{\ast }\left(t\right)+{f}_{0}\left(x,y,t\right)\end{array}$ in the Eq. (1).

Let $\begin{array}{}{E}_{1}^{\ast }\left(x\right)=\underset{\mathrm{\Pi }}{\int }{K}_{1}\left(y,t\right){u}^{\ast }\left(x,y,t\right)dy\phantom{\rule{thinmathspace}{0ex}}dt,\phantom{\rule{thinmathspace}{0ex}}x\in {\mathrm{\Omega }}_{x},\phantom{\rule{thinmathspace}{0ex}}{E}_{2}^{\ast }\left(t\right)=\underset{G}{\int }{K}_{2}\left(x,y\right){u}^{\ast }\left(x,y,t\right)dx\phantom{\rule{thinmathspace}{0ex}}dy,\phantom{\rule{thinmathspace}{0ex}}t\in \left[0,T\right].\end{array}$ Then it is obvious, that $Δ1(x)q1∗(x)=−∑0<|α|=|γ|≤m0Dγ(aαγ(x)DαE1∗(x))+∫Π(−K1(y,t)f0(x,y,t)−F12(x,y,t)q2∗(t)−+B1(x,y,t)u∗+K1(y,t)g(x,y,t,u∗))dydt,$(15) $Δ2(t)q2∗(t)=(E2∗(t))′+∫G(−K2(x,y)f0(x,y,t)−F21(x,y,t)q1∗(x)++B2(x,y,t)u∗+∑0<|α|=|γ|≤m0DγK2(x,y)aαγ(x)Dαu∗+K2(x,y)g(x,y,t,u∗))dxdy.$(16)

On the other hand, $\begin{array}{}{q}_{1}^{\ast }\left(x\right),{q}_{2}^{\ast }\left(t\right)\end{array}$ and u*(x, y, t) satisfy the equalities (12), (14). Then from (12), (14) – (16) we have $∑0<|α|=|γ|≤m0Dγ(aαγ(x)DαE1∗(x))=∑0<|α|=|γ|≤m0Dγ(aαγ(x)DαE1(x)),x∈Ωx,$(17) $(E2∗(t))′=E2′(t),t∈[0,T].$(18)

Since u*(x, y, t) satisfies the condition (3), we get Dα $\begin{array}{}{E}_{1}^{\ast }\end{array}$(x) = 0 (|α| ≤ m0 − 1), xΩx, thus from (17) it follows that $\begin{array}{}{E}_{1}^{\ast }\end{array}$(x) = E1(x), x ∈ Ωx. Integrating (18) by parts, we get $E2∗(t)−E2∗(0)=E2(t)−E2(0),t∈[0,T].$

Besides, (5) implies the equality $\begin{array}{}{E}_{2}\left(0\right)=\underset{G}{\int }{K}_{2}\left(x,y\right){u}_{0}\left(x,y\right)dx\phantom{\rule{thinmathspace}{0ex}}dy.\end{array}$ Thus, $\begin{array}{}{E}_{2}^{\ast }\end{array}$(0) = E2(0). So, $\begin{array}{}{E}_{2}^{\ast }\end{array}$ (t) = E2(t), t ∈ [0, T] and u*(x, y, t) satisfies the overdetermination conditions (4), (5). Thus, we proved the following lemma

#### Lemma 2.2

The triple offunctions $\begin{array}{}{u}^{\ast }\left(x,y,t\right),{q}_{1}^{\ast }\left(x\right),{q}_{2}^{\ast }\left(t\right)\right),\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}where\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{u}^{\ast }\in {V}_{4}\left({Q}_{T}\right)\cap C\left(\left[0,T\right];{L}^{2}\left(G\right)\right),\phantom{\rule{thinmathspace}{0ex}}{\mathrm{\partial }}_{t}{u}^{\ast }\in {L}^{2}\left({Q}_{T}\right),\phantom{\rule{thinmathspace}{0ex}}{q}_{1}^{\ast }\in {L}^{2}\left({\mathrm{\Omega }}_{x}\right),\phantom{\rule{thinmathspace}{0ex}}{q}_{2}^{\ast }\in {L}^{2}\left(0,T\right),\end{array}$ is a solution to the problem (1)(5) if and only if it satisfies Eq. (1) for almost all (x, y, t) ∈ QT, and (2), (12), (14) hold.

Denote $C1:=(mes Ωy)2infΩx|Δ1(x)|2inf[0,T]|Δ2(t)|2∫QT(F12(x,y,t))2dxdydt∫QT(F21(x,y,t))2dxdydt.$

#### Lemma 2.3

Let C1 < 1 and Δ1(x) ≠ 0 for all x ∈ Ωx, Δ2(t) ≠ 0 for all t ∈ (0, T). Suppose that the conditions (A), (C), (L), (U), (G), (E), (K), (F) hold, and $\begin{array}{}{D}^{\alpha }{a}_{\alpha \gamma }\in {L}^{\mathrm{\infty }}\left({\mathrm{\Omega }}_{x}\right),\phantom{\rule{thinmathspace}{0ex}}{\mathrm{\partial }}_{{y}_{k}}c\in {L}^{\mathrm{\infty }}\left({Q}_{T}\right),\phantom{\rule{thinmathspace}{0ex}}{\mathrm{\partial }}_{{y}_{k}}{f}_{s}\in {L}^{2}\left({Q}_{T}\right),\phantom{\rule{thinmathspace}{0ex}}{f}_{s}{|}_{{S}_{T}^{1}}=0\left(s\in \left\{0,1,2\right\},\phantom{\rule{thinmathspace}{0ex}}0<|\alpha |=|\gamma |\le {m}_{0},\phantom{\rule{thinmathspace}{0ex}}k\in \left\{1,\dots ,l\right\}\right),\end{array}$ Then for each fixed function u*V3(QT) ∩ C([0, T];L2(G)) there exists a unique solution (q1(x), q2(t)) of the system of equations (12), (14), where q1L2x), q2L2(0, T).

#### Proof

We shall use the method of successive approximations. We construct an approximation $\begin{array}{}\left({q}_{1}^{m}\left(x\right),{q}_{2}^{m}\left(t\right)\right)\end{array}$ to the solution of system of equations (12), (14), where $\begin{array}{}{q}_{1}^{m}\left(x\right),{q}_{2}^{m}\left(t\right),m\in \mathbb{N},\end{array}$ are such that $q11(x):=0,q21(t):=0,q1m(x)=1Δ1(x)[A1(x)+∫Π(B1(x,y,t)u∗+K1(y,t)g(x,y,t,u∗)−−F12(x,y,t)q2m−1(t))dydt],x∈Ωx,m≥2,$(19) $q2m(t)=1Δ2(t)[A2(t)+∫G(B2(x,y,t)u∗+∑0<|α|=|γ|≤m0DγK2(x,y)aαγ(x)Dαu∗++K2(x,y)g(x,y,t,u∗)−F21(x,y,t)q1m−1(x))dxdy],t∈[0,T],m≥2.$(20)

Now we show that the sequences $\begin{array}{}\left\{{q}_{1}^{m}\left(x\right){\right\}}_{m=1}^{\mathrm{\infty }},\phantom{\rule{thinmathspace}{0ex}}\left\{{q}_{2}^{m}\left(t\right){\right\}}_{m=1}^{\mathrm{\infty }},\end{array}$ are fundamental in L2x), L2(0, T) respectively and they converge to a solution of the system of equations (12), (14). Denote $q~1m(x)=q1m(x)−q1m−1(x),q~2m(t)=q2m(t)−q2m−1(t).$

According to (19), (20) $q~1m(x)=−1Δ1(x)∫ΠF12(x,y,t)q~2m−1(t)dydt,x∈Ωx,m≥3,$(21) $q~2m(t)=−1Δ2(t)∫GF21(x,y,t)q~1m−1(x)dxdy,t∈[0,T],m≥3.$(22)

Raising both sides of (21), (22) to the second power and integrating them on x, t respectively, by the using of the Hölder’s inequality, we get $∫Ωx|q~1m(x)|2dx≤mes ΩyinfΩx|Δ1(x)|2∫QT(F12(x,y,t))2dxdydt∫0T|q~2m−1(t)|2dt,m≥3,$(23) $∫0T|q~2m(t)|2dt≤mes Ωyinf[0,T]|Δ2(t)|2∫QT(F21(x,y,t))2dxdydt∫Ωx|q~1m−1(x)|2dx,m≥3.$(24)

It follows from (23), (24) that $∫Ωx|q~1m(x)|2dx≤C1∫Ωx|q~1m−2(x)|2dx≤C1m−22max∫Ωx|q~11(x)|2dx;∫Ωx|q~12(x)|2dx,m≥3,∫0T|q~2m(t)|2dt≤C1∫0T|q~2m−2(t)|2dt≤C1m−22max∫0T|q~21(t)|2dt;∫0T|q~22(t)|2dt,m≥3,$(25)

Since C1 < 1, then in view of (25), we conclude that for all k, m ∈ ℕ, m ≥ 3, the inequality $∫Ωx|q1m+k(x)−q1m(x)|2dx≤∑i=m+1m+k∫Ωx|q~1i(x)|2dx≤≤∑i=m+1m+kC1i−22max∫Ωx|q~11(x)|2dx;∫Ωx|q~12(x)|2dx≤≤C1(m−1)/2(1−C1k/2)1−C11/2max∫Ωx|q~11(x)|2dx;∫Ωx|q~12(x)|2dx≤≤C1(m−1)/21−C11/2max∫Ωx|q~11(x)|2dx;∫Ωx|q~12(x)|2dx$(26) is true. Similarly we obtain $∫0T|q2m+k(t)−q2m(t)|2dt≤C1(m−1)/21−C11/2max∫0T|q~21(t)|2dt;∫0T|q~22(t)|2dt.$(27)

It follows from (26), (27) that for any ε > 0 there exists $\stackrel{~}{m}$, such that for all k, m ∈ ℕ, m > $\stackrel{~}{m}$ the inequalities $\begin{array}{}\parallel {q}_{1}^{m+k}\left(x\right)-{q}_{1}^{m}\left(x\right);{L}^{2}\left({\mathrm{\Omega }}_{X}\right)\parallel \le \epsilon ,\parallel {q}_{2}^{m+k}\left(t\right)-{q}_{2}^{m}\left(t\right);{L}^{2}\left(0,T\right)\parallel \le \epsilon \end{array}$ hold. Thus, the sequence $\begin{array}{}\left\{{q}_{1}^{m}{\right\}}_{m=1}^{\mathrm{\infty }}\end{array}$ is fundamental in L2x), the sequence $\begin{array}{}\left\{{q}_{2}^{m}{\right\}}_{m=1}^{\mathrm{\infty }}\end{array}$ is fundamental in L2(0, T) and $\begin{array}{}{q}_{1}^{m}\to {q}_{1}\text{\hspace{0.17em}in\hspace{0.17em}}{L}^{2}\left({\mathrm{\Omega }}_{x}\right),{q}_{2}^{m}\to {q}_{2}\end{array}$ in L2(0, T).

Passing to the limit in (19), (20) as m → ∞, we conclude the statement of Lemma 2.3.

Assume that $\begin{array}{}\left({q}_{1}^{\left(1\right)}\left(x\right),\phantom{\rule{thinmathspace}{0ex}}{q}_{2}^{\left(1\right)}\left(t\right)\right),\phantom{\rule{thinmathspace}{0ex}}\left({q}_{1}^{\left(2\right)}\left(x\right),\phantom{\rule{thinmathspace}{0ex}}{q}_{2}^{\left(2\right)}\left(t\right)\right)\end{array}$ are two solutions of the system of equations (12), (14). Then their difference $\begin{array}{}\left({\stackrel{~}{q}}_{1}\left(x\right),{\stackrel{~}{q}}_{2}\left(t\right)\right),\text{\hspace{0.17em}where\hspace{0.17em}}{\stackrel{~}{q}}_{1}\left(x\right)={q}_{1}^{\left(1\right)}\left(x\right)-{q}_{1}^{\left(2\right)}\left(x\right),{\stackrel{~}{q}}_{2}\left(t\right)={q}_{2}^{\left(1\right)}\left(t\right)-{q}_{2}^{\left(2\right)}\left(t\right),\end{array}$ satisfies the system of equations: $q~1(x)=−1Δ1(x)∫ΠF12(x,y,t)q~2(t)dydt,x∈ΩX,$(28) $q~2(t)=−1Δ2(t)∫GF21(x,y,t)q~1(x)dxdy,t∈[0,T].$(29)

As in the case of proof of the existence of solution, from (28)–(29) we conclude $∫Ωx|q~1(x)|2dx≤C1∫Ωx|q~1(x)|2dx,∫0T|q~2(t)|2dt≤C1∫0T|q~2(t)|2dt.$

Therefore, $\begin{array}{}{q}_{1}^{\left(1\right)}\left(x\right)\equiv {q}_{1}^{\left(2\right)}\left(x\right),\phantom{\rule{thinmathspace}{0ex}}{q}_{2}^{\left(1\right)}\left(t\right)\equiv {q}_{2}^{\left(2\right)}\left(t\right).\end{array}$ The lemma is proved. □

We shall use Friedrichs’ inequality $∫ΩX∑|α|=j|Dαw|2dx≤γk,j∫Ωx∑|α|=k|Dαw|2dx,j∈{0,1,…,k},w∈W0k,2(ΩX),$ where the constant γk,j depends on Ωx, k, j. Denote $\begin{array}{}{\mathrm{\Gamma }}_{k}=\sum _{j=1}^{k}\end{array}$γk,j.

Denote $f1=supΩX∫Π(f1(x,y,t))2dydt,f2=sup[0,T]∫G(f2(x,y,t))2dxdy,λ1=maxiesssupΩy|∂yiλj(x,y,t)|,ϰ1=lλ1−2c0+2g0+1+2/T,C2:=3 mes Ωyinf[0,T]|Δ2(t)|2∫QT(F21(x,y,t))2dxdydt,C3:=2 mes ΩyinfΩx|Δ1(x)|2∫QT(F12(x,y,t))2dxdydt,C4:=max{2infΩx|Δ1(x)|2∫QT(B1(x,y,t)+K1(y,t)g0)2dxdydt++3inf[0,T]|Δ2(t)|2∫QT(B2(x,y,t)+K2(x,y)g0)2dxdydt;3m02Γm0inf[0,T]|Δ2(t)|2∫QT∑0<|α|=|γ|≤m0(DγK2(x,y)aαγ(x))2dxdydt},C5:=Teϰ1TC4min{1,2a0}(f11−C2+f21−C3).$

Let a number T satisfies the inequalities $ϰ1>0 and C5<1.$(30)

#### Theorem 2.4

Let C2 < 1, C3 < 1, C5 < 1, Δ1(x)≠ 0 for all x ∈ Ωx, Δ2(t)≠ 0 for all t∈(0, T), a number T satisfies the condition (30), and let the hypotheses (A), (C), (L), (U), (G), (E), (K), (F), (S) hold, and DαaαγLx), yk cL(QT), ykfsL2(QT), ${f}_{s}{|}_{{S}_{T}^{1}}$ = 0 (0 < |α| = |γ| ≤ m0, k ∈ {1, …, l}, s ∈ {0, 1, 2}). Then a solution to the problem (1)(5) exists.

#### Proof

We use the method of successive approximations. As in [7], we construct an approximation (um(x, y, t), $\begin{array}{}{q}_{1}^{m}\left(x\right),\phantom{\rule{thinmathspace}{0ex}}{q}_{2}^{m}\left(t\right)\right)\end{array}$ to the solution of problem (1)–(5), where the functions um(x, y, t) and $\begin{array}{}{q}_{1}^{m}\left(x\right),\phantom{\rule{thinmathspace}{0ex}}{q}_{2}^{m}\left(t\right),\end{array}$ m ∈ ℕ, satisfy the system of equalities $q11(x):=0,q21(t):=0,q1m(x)=1Δ1(x)[A1(x)+∫Π(B1(x,y,t)um−1+K1(y,t)g(x,y,t,um−1)−−F12(x,y,t)q2m(t))dydt],x∈ΩX,m≥2,$(31) $q2m(t)=1Δ2(t)[A2(t)+∫G(B2(x,y,t)um−1+∑0<|α|=|γ|≤m0DγK2(x,y)aαγ(x)Dαum−1++K2(x,y)g(x,y,t,um−1)−F21(x,y,t)q1m(x))dxdy],t∈[0,T],m≥2.$(32) um satisfies the equality $∫QT[∂tumv+∑i=1lλj(x,y,t)∂Yiumv+∑0<|α|=|γ|≤m0aαγ(x)DαumDγv+c(x,y,t)umv++g(x,y,m,t,um)v]dxdydt=∫QT(f1(x,y,t)q1m(x)+f2(x,y,t)q2m(t)+f0(x,y,t))vdxdydt,m≥1,$(33) for all vV1(QT), and the condition $um(x,y,0)=u0(x,y),(x,y)∈G.$(34)

It follows from Theorem 2.1 that for each m ∈ ℕ there exists a unique function umV3(QT) ∩ C([0, T]; L2(G)) that satisfies the equalities (33), (34) when $\begin{array}{}{q}_{1}^{m}\left(x\right),\phantom{\rule{thinmathspace}{0ex}}{q}_{2}^{m}\left(t\right)\end{array}$ are known, and due to Lemma 2.3 for each function um−1V3(QT) ∩ C([0, T]; L2(G)) there exists a unique pair of functions $\begin{array}{}\left({q}_{1}^{m}\left(x\right),\phantom{\rule{thinmathspace}{0ex}}{q}_{2}^{m}\left(t\right)\right),\end{array}$ such that $\begin{array}{}{q}_{1}^{m}\in {L}^{2}\left({\mathrm{\Omega }}_{x}\right),\phantom{\rule{thinmathspace}{0ex}}{q}_{2}^{m}\in {L}^{2}\left(0,\phantom{\rule{thinmathspace}{0ex}}T\right)\end{array}$ and it satisfies (31), (32).

Now we show that $\begin{array}{}\left\{\left({u}^{m}\left(x,\phantom{\rule{thinmathspace}{0ex}}y,\phantom{\rule{thinmathspace}{0ex}}t\right),{q}_{1}^{m}\left(x\right),{q}_{2}^{m}\left(t\right)\right){\right\}}_{m=1}^{\mathrm{\infty }}\end{array}$ converges to the solution of the problem (1)–(5). Denote $zm:=zm(x,y,t)=um(x,y,t)−um−1(x,y,t),r1m(x)=q1m(x)−q1m−1(x),r2m(t)=q2m(t)−q2m−1(t),m≥2.$

Using (33) with v = zmeϰt, ϰ ≥ 0, we find $12∫G|zm|2e−ϰTdxdy+∫QT[ϰ2|zm|2+∑i=1lλi(x,y,t)∂yizmzm+∑0<|α|=|γ|≤m0aαγ(x)DαzmDγzm++c(x,y,t)|zm|2+(g(x,y,t,um)−g(x,y,t,um−1))zm]e−ϰtdxdydt==∫QT(r1m(x)f1(x,y,t)+r2m(t)f2(x,y,t))zme−ϰtdxdydt,m≥2.$(35)

Let us estimate terms of the equality (35). $I1:=∫QT∑i=1lλi(x,y,t)∂yizmzme−ϰtdxdydt≥12∫ST2∑i=1lλi(x,y,t)|zm|2cos⁡(ν,yi)e−ϰtdxdσdt++lλ12∫QT|zm|2e−ϰtdxdydt,I2:=∫QT∑0<|α|=|γ|≤m0aαγ(x)DαzmDγzme−ϰtdxdydt≥a0∫QT∑|α|=m0|Dαzm|2e−ϰtdxdydt,I3:=∫QTc(x,y,t)|zm|2e−ϰtdxdydt≥c0∫QT|zm|2e−ϰtdxdydt,I4:=∫QT(g(x,y,t,um)−g(x,y,t,um−1))zme−ϰtdxdydt≤g0∫QT|zm|2e−ϰtdxdydt;I5:=∫QT(r1m(x)f1(x,y,t)+r2m(t)f2(x,y,t))zme−ϰtdxdydt≤δ2f1∫ΩX|r1m(x)|2dx++δ2f2∫0T|r2m(t)|2dt+1δ∫QT|zm|2e−ϰtdxdydt,δ>0.$

Taking into account 𝓘1 – 𝓘5, from (35) we obtain $∫G|zm|2e−ϰTdxdy+∫ST2∑i=1lλi(x,y,t)|zm|2cos⁡(ν,yi)e−ϰtdxdσdt++∫QT[(x−lλ1+2c0−2δ−2g0)|zm|2+2a0∑|α|=m0|Dαzm|2]e−ϰtdxdydt≤≤δf1∫Ωx|r1m(x)|2dx+δf2∫0T|r2m(t)|2dt,m≥2.$(36)

After choosing x = x1, δ = T in (36), for m ≥ 2 we get inequalities $∫QT[|zm|2+∑|α|=m0|Dαzm|2]dxdydt≤Teϰ1Tmin{1;2a0}[f1∫Ωx|r1m(x)|2dx+f2∫0T|r2m(t)|2dt].$(37)

According to (31), (32) we have $r1m(x)=1Δ1(x)∫Π[B1(x,y,t)zm−1+K1(y,t)(g(x,y,t,um−1)−g(x,y,t,um−2))−−F12(x,y,t)r2m(t)]dydt,x∈Ωx,m≥3,$(38) $r2m(t)=1Δ2(t)∫G[B2(x,y,t)zm−1+∑0<|α|=|γ|≤m0DγK2(x,y)aαγ(x)Dαzm−1++K2(x,y)(g(x,y,t,um−1)−g(x,y,t,um−2))−F21(x,y,t)r1m(x)]dxdy,t∈[0,T],m≥3.$(39)

Raising both sides of the equalities (38), (39) to the second power and integrating them on x, t with the use of Hölder’s and Friedrichs’ inequalities, we obtain $∫Ωx|r1m(x)|2dx≤2infΩx|△1(x)|2[∫QT(B1(x,y,t)+K1(y,t)g0)2dxdydt∫QT|zm−1|2dxdydt++mes Ωy∫QT(F12(x,y,t))2dxdydt∫0T|r2m(t)|2dt],m≥3,$(40) $∫0T|r2m(t)|2dt≤3inf[0,T]|△2(t)|2[∫QT(B2(x,y,t)+K2(x,y)g0)2dxdydt∫QT|zm−1|2dxdydt++m02Γm0∫QT∑0<|α|=|γ|≤m0Dγ(K2(x,y)aαγ(x))2dxdydt∫QT∑|α|=m0|Dαzm−1|2dxdydt++mesΩy∫QT(F21(x,y,t))2dxdydt∫Ωx|r1m(x)|2dx],m≥3.$(41)

The sum of (40) and (41) gives the estimates $∫Ωx|r1m(x)|2dx≤C41−C2∫QT[|zm−1|2+∑|α|=m0|Dαzm−1|2]dxdydt,m≥3,$(42) $∫0T|r2m(t)|2dt≤C41−C3∫QT[|zm−1|2+∑|α|=m0|Dαzm−1|2]dxdydt,m≥3.$(43)

Moreover, (37), (42) and (43) imply the inequalities $\begin{array}{}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{2em}{0ex}}{f}_{1}\underset{{\mathrm{\Omega }}_{x}}{\int }|{r}_{1}^{m+1}\left(x\right){|}^{2}dx+{f}_{2}\underset{0}{\overset{T}{\int }}|{r}_{2}^{m+1}\left(t\right){|}^{2}dt\phantom{\rule{1em}{0ex}}\le \\ {C}_{4}\left(\frac{{f}_{1}}{1-{C}_{2}}+\frac{{f}_{2}}{1-{C}_{3}}\right)\underset{{Q}_{T}}{\int }\left[|{z}^{m}{|}^{2}+\sum _{|\alpha |={m}_{0}}|{D}^{\alpha }{z}^{m}{|}^{2}\right]dx\phantom{\rule{thinmathspace}{0ex}}dy\phantom{\rule{thinmathspace}{0ex}}dt\le {C}_{5}\left[{f}_{1}\underset{{\mathrm{\Omega }}_{x}}{\int }|{r}_{1}^{m}\left(x\right){|}^{2}dx+{f}_{2}\underset{0}{\overset{T}{\int }}|{r}_{2}^{m}\left(t\right){|}^{2}dt\right],\end{array}$ m ≥ 3. Therefore $f1∫Ωx|r1m(x)|2dx+f2∫0T|r2m(t)|2dt≤C5m−2[f1∫Ωx|r12(x)|2dx+f2∫0T|r22(t)|2dt],m≥3.$(44)

Taking into account (44) and the condition C5 < 1, we obtain the inequality $f1∫Ωx|q1m+k(x)−q1m(x)|2dx+f2∫0T|q2m+k(t)−q2m(t)|2dt≤≤∑i=m+1m+k[f1∫Ωx|r1i(x)|2dx+f2∫0T|r2i(t)|2dt]≤∑i=m+1m+k(C5)i−2[f1∫Ωx|r12(x)|2dx++f2∫0T|r22(t)|2dt]≤(C5)m−1(1−(C5)k)1−C5[f1∫Ωx|r12(x)|2dx+f2∫0T|r22(t)|2dt]≤≤(C5)m−11−C5[f1∫Ωx|r12(x)|2dx+f2∫0T|r22(t)|2dt],$(45) for each k, m ∈ ℕ, m ≥ 3.

Then, for each ε > 0 there exists $\stackrel{~}{m}$, such that for all k, m ∈ ℕ, m > $\stackrel{~}{m}$, the inequalities ∥$\begin{array}{}{q}_{1}^{m+k}\end{array}$(x) − $\begin{array}{}{q}_{1}^{m}\left(x\right);{L}^{2}\left({\mathrm{\Omega }}_{x}\right)\parallel \le \epsilon ,\phantom{\rule{thinmathspace}{0ex}}\parallel {q}_{2}^{m+k}\left(t\right)-{q}_{2}^{m}\left(t\right);{L}^{2}\left(0,T\right)\parallel \end{array}$ε hold. Thus, the sequence $\begin{array}{}\left\{{q}_{1}^{m}{\right\}}_{m=1}^{\mathrm{\infty }}\end{array}$ is fundamental in L2x), $\begin{array}{}\left\{{q}_{1}^{m}{\right\}}_{m=1}^{\mathrm{\infty }}\end{array}$ is fundamental in L2(0, T). Then from (37) we obtain that $\begin{array}{}\left\{{u}^{m}{\right\}}_{m=1}^{\mathrm{\infty }}\end{array}$ is fundamental in L2(QT) ∩ C([0, T]; L2(G)) and the sequence $\begin{array}{}\left\{{D}^{\alpha }{u}^{m}{\right\}}_{m=1}^{\mathrm{\infty }}\end{array}$ is fundamental in L2(QT). Therefore, $um→u in L2(QT)∩C([0,T];L2(G)),Dαum→Dαu in L2(QT),|α|≤m0,$(46) $q1m→q1 in L2(Ωx),q2m→q2 in L2(0,T),$(47) as m → ∞.

Moreover, the functions yium(i ∈ {1, …, l}) and tum fulfill the estimates (8), (9) with yium, tum, $\begin{array}{}{q}_{1}^{m}\left(x\right),\phantom{\rule{thinmathspace}{0ex}}{q}_{2}^{m}\left(t\right)\end{array}$ instead of yiu*, tu*, $\begin{array}{}{q}_{1}^{\ast }\left(x\right),\phantom{\rule{thinmathspace}{0ex}}{q}_{2}^{\ast }\left(t\right)\end{array}$ respectively. From (46) and (47) we conclude that the right-hand sides of (8), (9) are bounded by the constants M4, M5, that are independent of m. Thus, we obtain $∫QT∑i=1l|∂yium|2dxdydt≤M4,∥∂tum;L2(QT)∥≤M5.$(48)

It follows from (48) that we can choose a subsequence of the sequence $\begin{array}{}\left\{{u}^{m}{\right\}}_{m=1}^{\mathrm{\infty }}\end{array}$ (for this subsequence, we preserve the same notation) such that $∂yium→∂yiu weakly in L2(QT)(i∈{1,…,l}),∂tum→∂tu weakly in L2(QT).$(49)

By using relations (46), (47), (49), (31), (32), (33) and Lemma 2.2, we conclude that (u(x, y, t), q1(x), q2(t)) satisfies the condition (2) and the equality $∫QT[∂tuv+∑i=1lλi(x,y,t)∂yiuv+∑0<|α|=|γ|≤m0aαγ(x)DαuDγv+c(x,y,t)uv+g(x,y,t,u)v]dxdydt==∫QT(f1(x,y,t)q1(x)+f2(x,y,t)q2(t)+f0(x,y,t))vdxdydt,m≥1,$(50) for all vV1(QT).

Let us show that (u(x, y, t), q1(x), q2(t)) is a solution for the problem (1)–(5).

The equality $∫Ωx[∂tuv1+∑i=1lλi(x,y,t)∂yiuv1+∑0<|α|=|γ|≤m0aαγ(x)DαuDγv1+c(x,y,t)uv1+g(x,y,t,u)v1]dx==∫Ωx[f1(x,y,t)q1(x)+f2(x,y,t)q2(t)+f0(x,y,t)]v1dx$(51) for all (y, t) ∈ Π and each function v1$\begin{array}{}{W}_{0}^{{m}_{0},2}\end{array}$x) follows from (50). So, from (51) we get that the function u is a weak solution of the Dirichlet problem for the elliptic equation $∑0<|α|=|γ|≤m0Dγ(aαγ(x)Dαu)=F(x,y,t),x∈Ωx;∂iu∂νi|Ωx=0(i∈{0,1,…,m0−1}),$(52) where F(x, y, t) = f1(x, y, t)q1(x) + f2(x, y, t)q2(t)+ f0(x, y, t) − tu$\begin{array}{}\sum _{i=1}^{l}\end{array}$ λi(x, y, t)yiuc(x, y, t)ug(x, y, t, u). Since function F(x, y, t) ∈ L2x) for almost all (y, t) ∈ Π, and the condition (2) is fulfilled, there exists a unique weak solution u of the problem (52). Then, from [21] and the condition ΩxCm0 we conclude that Dαu*(., y, t) ∈ L2x), |α| ≤ m0. Thus, u* y, t) ∈ $\begin{array}{}{W}_{0}^{{m}_{0},2}\end{array}$x) ∩ W2m0,2x). Using the scheme [20, p. 219] we prove that the function u*(x, y, t) satisfies the equation (1) for almost all (x, y, t) ∈ QT.

From (46), (47), (49), (31), (32), (33) and Lemma 2.2, we obtain that (u(x, y, t), q1(x), q2(t)) is a solution for the problem (1)–(5). The theorem is proved. □

Let us show an example of the problem (1)–(5), for which the conditions Ci < 1 (i ∈ {1, 2, 3}) are true.

#### Example 2.5

Let n = l = 1, Ωx = (0, x0), Ωy = (0, y0), x ∈ Ωx, y ∈ Ωy, t ∈(0, T), QT = (0, x0) × (0, y0)×(0, T). Assume that the coefficients of Eq. (1) satisfy the conditions (A), (C), (L), (G) and λ1(x, y, t)> 0 for all (x, y, t) ∈ QT, functions Ki and fi, i ∈ {1, 2}, are as follows: $K1(y,t):=(y0−y)8(T−t)5t,K2(x,y):=(y0−y)2(x0−x)2x2,f1(x,y,t):=y,f2(x,y,t):=y5.$(53)

Note that functions (53) satisfy conditions (K), (F) and f1(x, 0, t) = f2(x, 0, t) = 0, (x, t) ∈ (0, x0) × (0, T), yf1, yf2L2(QT). Then for any triple of numbers (x0, y0, T) the constants $C1=0.04458896297,C2=1152Tx08y0,C3=35280151915621x08y0T.$

If, for instance, y0 = T and x0 ∈ (2.413690382; 2.846159484) then C2 = 0.75497472, C3 = 0.3543612398.

#### Theorem 2.6

Let the conditions of Theorem 2.4 hold. Then the problem (1)(5) has at most one solution.

#### Proof

Assume that $\begin{array}{}\left({u}^{\left(1\right)}\left(x,\phantom{\rule{thinmathspace}{0ex}}y,\phantom{\rule{thinmathspace}{0ex}}t\right),{q}_{1}^{\left(1\right)}\left(x\right),{q}_{2}^{\left(1\right)}\left(t\right)\right),\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\left({u}^{\left(2\right)}\left(x,\phantom{\rule{thinmathspace}{0ex}}y,\phantom{\rule{thinmathspace}{0ex}}t\right),{q}_{1}^{\left(2\right)}\left(x\right),{q}_{2}^{\left(2\right)}\left(t\right)\right)\end{array}$ are two solutions of the problem (1)–(5). Then their difference $\begin{array}{}\left(\stackrel{~}{u}\left(x,\phantom{\rule{thinmathspace}{0ex}}y,\phantom{\rule{thinmathspace}{0ex}}t\right),{\stackrel{~}{q}}_{1}\left(x\right),{\stackrel{~}{q}}_{2}\left(t\right)\right),\phantom{\rule{thinmathspace}{0ex}}\text{\hspace{0.17em}where\hspace{0.17em}}\phantom{\rule{thinmathspace}{0ex}}\stackrel{~}{u}\left(x,\phantom{\rule{thinmathspace}{0ex}}y,\phantom{\rule{thinmathspace}{0ex}}t\right)\end{array}$ = u(1)(x, y, t) − u(2)(x, y, t), $\begin{array}{}{\stackrel{~}{q}}_{1}\left(x\right)={q}_{1}^{\left(1\right)}\left(x\right)-{q}_{1}^{\left(2\right)}\left(x\right),\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\stackrel{~}{q}}_{2}\left(t\right)={q}_{2}^{\left(1\right)}\left(t\right)-{q}_{2}^{\left(2\right)}\left(t\right),\end{array}$ satisfies the condition $\stackrel{~}{u}$(x, y, 0) ≡ 0, the equality $∫QT[∂tu~v+∑i=1lλi(x,y,t)∂yiu~v+∑0<|α|=|γ|≤m0aαγ(x)Dαu~Dγv+c(x,y,t)u~v+(g(x,y,t,u(1))−−g(x,y,t,u(2)))v]dxdydt=∫QT[f1(x,y,t)q~1(x)+f2(x,y,t)q~2(t)]vdxdydt$(54) for all vV1(QT), and the system of equalities $q~1(x)=1Δ1(x)∫Π[B1(x,y,t)u~+K1(y,t)(g(x,y,t,u(1))−g(x,y,t,u(2)))−−F12(x,y,t)q~2(t)]dydt,x∈Ωx,$(55) $q~2(t)=1Δ2(t)∫G[B2(x,y,t)u~+∑0<|α|=|γ|≤m0DγK2(x,y)aαγ(x)Dαu~+K2(x,y)(g(x,y,t,u(1))−−g(x,y,t,u(1)))−F21(x,y,t)q~1(x)]dxdy,t∈[0,T].$(56)

According to (54) with x = 1−2C0+2/δ + 1 − 2g0, δ > 0, for the triple of functions $\begin{array}{}\left(\stackrel{~}{u}\left(x,y,t\right),\phantom{\rule{thinmathspace}{0ex}}{\stackrel{~}{q}}_{1}\left(x\right),\phantom{\rule{thinmathspace}{0ex}}{\stackrel{~}{q}}_{2}\left(t\right)\right)\end{array}$ we have the equality $∫G|u~|2e−ϰTdxdy+∫QT[ϰ|u~|2+2∑i=1lλi(x,y,t)∂yiu~u~+2∑0<|α|=|γ|≤m0aαγ(x)Dαu~Dγu~+2c(x,y,t)|u~|2++2(g(x,y,t,u(1))−g(x,y,t,u(2)))u~]e−ϰtdxdydt==2∫QT[f1(x,y,t)q~1(x)+f2(x,y,t)q~2(t)]u~e−ϰtdxdydt.$

From here, similarly as from (35) we obtain the estimate $∫G|u~|2e−ϰτdxdy+∫ST2∑i=1lλi(x,y,t)|u~|2cos⁡(ν,yi)e−ϰtdxdσdt+∫QT[(ϰ−lλ1+2c0−2δ−2g0)|u~|2++2a0∑|α|=m0|Dαu~|2]e−ϰtdxdydt≤δf1∫Ωx|q~1(x)|2dx+δf2∫0T|q~2(t)|2dt$(57)

Choosing δ = T, ϰ = ϰ1 in (57) gives the equality $∫QT[|u~|2+∑|α|=m0|DαU~|2]dxdydt≤Teϰ1Tmin{1;2a0}[f1∫Ωx|q~1(x)|2dx+f2∫0T|q~2(t)|2dt].$(58)

Taking into account (55), (56) we get $∫Ωx|q~1(x)|2dx≤C41−C2∫QT[|u~|2+∑|α|=m0|Dαu~|2]dxdydt;∫0T|q~2(t)|2dt≤C41−C3∫QT[|u~|2+∑|α|=m0|Dαu~|2]dxdydt,$ and using (58), we obtain $(1−C5)[f1∫Ωx|q~1(x)|2dx+f2∫0T|q~2(t)|2dt]≤0.$(59)

Since C5 < 1, then $\begin{array}{}{\stackrel{~}{q}}_{1}\left(x\right)\equiv 0,\phantom{\rule{thinmathspace}{0ex}}{\stackrel{~}{q}}_{2}\left(t\right)\equiv 0,\phantom{\rule{thinmathspace}{0ex}}\text{\hspace{0.17em}thus\hspace{0.17em}}{q}_{1}^{\left(1\right)}\left(x\right)\equiv {q}_{1}^{\left(2\right)}\left(x\right),\phantom{\rule{thinmathspace}{0ex}}{q}_{2}^{\left(1\right)}\left(t\right)\equiv {q}_{2}^{\left(2\right)}\left(t\right).\end{array}$

Then from (58) it follows that $\begin{array}{}\underset{{Q}_{T}}{\int }|\stackrel{~}{u}{|}^{2}\end{array}$ dx dy dt ≤ 0, so u(1) = u(2) in QT. The proof of the theorem is complete. □

## 3 Conclusions

In this paper we showed the results of the investigation of the unique solvability of the inverse problem with the integral overdetermination conditions of identifying of two functions of different arguments in the right-hand side of the higher order ultraparabolic equation.

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Accepted: 2017-07-06

Published Online: 2017-08-29

Citation Information: Open Mathematics, Volume 15, Issue 1, Pages 1048–1062, ISSN (Online) 2391-5455,

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