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formerly Central European Journal of Mathematics

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Volume 15, Issue 1


Volume 13 (2015)

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


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.

Keywords: Higher order ultraparabolic equation; Inverse problem; Initial-boundary value problem; Unique solvability

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 Main results

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, Dα=|α|x1α1,,xnαn,|α|=α1++αn.

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};

  • L2(G):={w:GR;G|w(x,y)|2dxdy<} with w;L2(G)=(G|w(x,y)|2dxdy)12;

  • L2(Ωx):={w:ΩxR;Ωx|w(x)|2dx<} with w;L2(Ωx)=(Ωx|w(x)|2dx)12;

  • L2(0,T):={w:[0,T]R;0T|w(t)|2dt<} with w;L2(0,T)=(0T|w(t)|2dt)12;

  • L2(QT):={w:QTR;QT|w(x,y,t)|2dxdydt<} with w;L2(QT)=(QT|w(x,y,t)|2dxdydt)12;

  • 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;

  • W0m0,2(Ωx) is a closure of C0(Ωx) in Wm0,2(Ωx) under the norm of this space, where C0(Ωx) 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:

  • V1(QT):={w:DαwL2(QT)(|α|m0),iwvi|ΣT=0(i{0,1,,m01})};

  • V2(G):={w:Dαw,yjwL2(G)(|α|m0,j{1,,l}),iwνi|Ωx×Ωy=0(i{0,1,,m01}),w|Ωx×Γ1=0};

  • V3(QT):={w:Dαw,yjw,twL2(QT)(|α|m0,j{1,,l}),w|ST1=0,iwνi|ΣT=0(i{0,1,,m01})};

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

We consider the equation tu+i=1lλi(x,y,t)yiu0<|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 iuνi|ΣT=0(i{0,1,,m01});u|ST1=0(3) and the overdetermination conditions ΠK1(y,t)u(x,y,t)dydt=E1(x),xΩx,(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, ST1={(x,y,t)ST:i=1lλj(x,y,t)cos(ν,yi)<0}.

Assume that the following conditions hold:

(A): aαγLx), 0 < |α| = |γ| ≤ m0, 0<|α|=|γ|m0Ωxaαγ(x)DαwDγwdxa0Ωx|α|=m0|Dαw|2dx for all wW0m0,2(Ωx), a0 > 0;

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

(E): E1W0m0,2(Ωx), 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|ξη| for almost all (x,y,t)QT and all {ξ,η}R1;

(S): there exists Γ1Ωy ⊂ ℝl − 1, such that the surface ST1 = 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=ΩyΓ1,K2C1(Ωy;C1(Ω¯x)),DαK2|Ωx×Ωy=0(|α|m01),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) q1(x)=q1(x),q2(t)=q2(t), where q1L2(Ωx),q2L2(0,T), 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. DαaαγL(Ωx),ykcL(QT)(0<|α|=|γ|m0,k{1,...,l}),q1L2(Ωx),q2L2(0,T),ykfsL2(QT)(s{0,1,2},k{1,,l});

  2. |yig(x,y,t,ξ)|g1(i{1,...,l}),g(x,y,t,0)|ST1=0 for almost all (x, y, t) ∈ QT and all ξ ∈ ℝ1, where g1 is a positive function;

  3. f0|ST1=0,f1|ST1=0,f2|ST1=0, then there exists a unique function u*V3(QT) ∩ C([0, T];L2(G)), that satisfies the condition (2) and the equality QT[tuv+i=1lλj(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(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].


The solution u* of the problem (1)(3) has such appraisals: G|u(x,y,τ)|2dxdyM1(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) Gi=1l|yiu(x,y,τ)|2dxdyM2(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|2dxdydtM3(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 q1,q2.

2.3 Inverse problem


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=1lyi(λi(x,y,t)K1(y,t))tK1(y,t)+K1(y,t)c(x,y,t),B2(x,y,t):=i=1lyi(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 (u(x,y,t),q1(x),q2(t)) 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)yiu0<|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)udydt+ΠK1(y,t)g(x,y,t,u)dydtΠF12(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)ug(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))dxdyGF21(x,y,t)q1(x)dxdy,(14) for almost all t ∈ [0, T].

Then let us show that if functions uV4(QT)C([0,T];L2(G)),q1L2(Ωx),q2L2(0,T) 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 f1(x,y,t)q1(x)+f2(x,y,t)q2(t)+f0(x,y,t) in the Eq. (1).

Let E1(x)=ΠK1(y,t)u(x,y,t)dydt,xΩx,E2(t)=GK2(x,y)u(x,y,t)dxdy,t[0,T]. 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, q1(x),q2(t) 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)),xΩx,(17) (E2(t))=E2(t),t[0,T].(18)

Since u*(x, y, t) satisfies the condition (3), we get Dα E1(x) = 0 (|α| ≤ m0 − 1), xΩx, thus from (17) it follows that E1(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 E2(0)=GK2(x,y)u0(x,y)dxdy. Thus, E2(0) = E2(0). So, E2 (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 u(x,y,t),q1(x),q2(t)),whereuV4(QT)C([0,T];L2(G)),tuL2(QT),q1L2(Ωx),q2L2(0,T), 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)|2QT(F12(x,y,t))2dxdydtQT(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 DαaαγL(Ωx),ykcL(QT),ykfsL2(QT),fs|ST1=0(s{0,1,2},0<|α|=|γ|m0,k{1,,l}), 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).


We shall use the method of successive approximations. We construct an approximation (q1m(x),q2m(t)) to the solution of system of equations (12), (14), where q1m(x),q2m(t),mN, 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)q2m1(t))dydt],xΩx,m2,(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)q1m1(x))dxdy],t[0,T],m2.(20)

Now we show that the sequences {q1m(x)}m=1,{q2m(t)}m=1, 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)q1m1(x),q~2m(t)=q2m(t)q2m1(t).

According to (19), (20) q~1m(x)=1Δ1(x)ΠF12(x,y,t)q~2m1(t)dydt,xΩx,m3,(21) q~2m(t)=1Δ2(t)GF21(x,y,t)q~1m1(x)dxdy,t[0,T],m3.(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)|2dxmes ΩyinfΩx|Δ1(x)|2QT(F12(x,y,t))2dxdydt0T|q~2m1(t)|2dt,m3,(23) 0T|q~2m(t)|2dtmes Ωyinf[0,T]|Δ2(t)|2QT(F21(x,y,t))2dxdydtΩx|q~1m1(x)|2dx,m3.(24)

It follows from (23), (24) that Ωx|q~1m(x)|2dxC1Ωx|q~1m2(x)|2dxC1m22maxΩx|q~11(x)|2dx;Ωx|q~12(x)|2dx,m3,0T|q~2m(t)|2dtC10T|q~2m2(t)|2dtC1m22max0T|q~21(t)|2dt;0T|q~22(t)|2dt,m3,(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)|2dxi=m+1m+kΩx|q~1i(x)|2dxi=m+1m+kC1i22maxΩx|q~11(x)|2dx;Ωx|q~12(x)|2dxC1(m1)/2(1C1k/2)1C11/2maxΩx|q~11(x)|2dx;Ωx|q~12(x)|2dxC1(m1)/21C11/2maxΩx|q~11(x)|2dx;Ωx|q~12(x)|2dx(26) is true. Similarly we obtain 0T|q2m+k(t)q2m(t)|2dtC1(m1)/21C11/2max0T|q~21(t)|2dt;0T|q~22(t)|2dt.(27)

It follows from (26), (27) that for any ε > 0 there exists m~, such that for all k, m ∈ ℕ, m > m~ the inequalities q1m+k(x)q1m(x);L2(ΩX)ε,q2m+k(t)q2m(t);L2(0,T)ε hold. Thus, the sequence {q1m}m=1 is fundamental in L2x), the sequence {q2m}m=1 is fundamental in L2(0, T) and q1mq1 in L2(Ωx),q2mq2 in L2(0, T).

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

Assume that (q1(1)(x),q2(1)(t)),(q1(2)(x),q2(2)(t)) are two solutions of the system of equations (12), (14). Then their difference (q~1(x),q~2(t)), where q~1(x)=q1(1)(x)q1(2)(x),q~2(t)=q2(1)(t)q2(2)(t), satisfies the system of equations: q~1(x)=1Δ1(x)ΠF12(x,y,t)q~2(t)dydt,xΩX,(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)|2dxC1Ωx|q~1(x)|2dx,0T|q~2(t)|2dtC10T|q~2(t)|2dt.

Therefore, q1(1)(x)q1(2)(x),q2(1)(t)q2(2)(t). The lemma is proved. □

We shall use Friedrichs’ inequality ΩX|α|=j|Dαw|2dxγk,jΩx|α|=k|Dαw|2dx,j{0,1,,k},wW0k,2(ΩX), where the constant γk,j depends on Ωx, k, j. Denote Γk=j=1kγk,j.

Denote f1=supΩXΠ(f1(x,y,t))2dydt,f2=sup[0,T]G(f2(x,y,t))2dxdy,λ1=maxiesssupΩy|yiλj(x,y,t)|,ϰ1=lλ12c0+2g0+1+2/T,C2:=3 mes Ωyinf[0,T]|Δ2(t)|2QT(F21(x,y,t))2dxdydt,C3:=2 mes ΩyinfΩx|Δ1(x)|2QT(F12(x,y,t))2dxdydt,C4:=max{2infΩx|Δ1(x)|2QT(B1(x,y,t)+K1(y,t)g0)2dxdydt++3inf[0,T]|Δ2(t)|2QT(B2(x,y,t)+K2(x,y)g0)2dxdydt;3m02Γm0inf[0,T]|Δ2(t)|2QT0<|α|=|γ|m0(DγK2(x,y)aαγ(x))2dxdydt},C5:=Teϰ1TC4min{1,2a0}(f11C2+f21C3).

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), fs|ST1 = 0 (0 < |α| = |γ| ≤ m0, k ∈ {1, …, l}, s ∈ {0, 1, 2}). Then a solution to the problem (1)(5) exists.


We use the method of successive approximations. As in [7], we construct an approximation (um(x, y, t), q1m(x),q2m(t)) to the solution of problem (1)–(5), where the functions um(x, y, t) and q1m(x),q2m(t), m ∈ ℕ, satisfy the system of equalities q11(x):=0,q21(t):=0,q1m(x)=1Δ1(x)[A1(x)+Π(B1(x,y,t)um1+K1(y,t)g(x,y,t,um1)F12(x,y,t)q2m(t))dydt],xΩX,m2,(31) q2m(t)=1Δ2(t)[A2(t)+G(B2(x,y,t)um1+0<|α|=|γ|m0DγK2(x,y)aαγ(x)Dαum1++K2(x,y)g(x,y,t,um1)F21(x,y,t)q1m(x))dxdy],t[0,T],m2.(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,m1,(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 q1m(x),q2m(t) 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 (q1m(x),q2m(t)), such that q1mL2(Ωx),q2mL2(0,T) and it satisfies (31), (32).

Now we show that {(um(x,y,t),q1m(x),q2m(t))}m=1 converges to the solution of the problem (1)–(5). Denote zm:=zm(x,y,t)=um(x,y,t)um1(x,y,t),r1m(x)=q1m(x)q1m1(x),r2m(t)=q2m(t)q2m1(t),m2.

Using (33) with v = zmeϰt, ϰ ≥ 0, we find 12G|zm|2eϰTdxdy+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,um1))zm]eϰtdxdydt==QT(r1m(x)f1(x,y,t)+r2m(t)f2(x,y,t))zmeϰtdxdydt,m2.(35)

Let us estimate terms of the equality (35). I1:=QTi=1lλi(x,y,t)yizmzmeϰtdxdydt12ST2i=1lλi(x,y,t)|zm|2cos(ν,yi)eϰtdxdσdt++lλ12QT|zm|2eϰtdxdydt,I2:=QT0<|α|=|γ|m0aαγ(x)DαzmDγzmeϰtdxdydta0QT|α|=m0|Dαzm|2eϰtdxdydt,I3:=QTc(x,y,t)|zm|2eϰtdxdydtc0QT|zm|2eϰtdxdydt,I4:=QT(g(x,y,t,um)g(x,y,t,um1))zmeϰtdxdydtg0QT|zm|2eϰtdxdydt;I5:=QT(r1m(x)f1(x,y,t)+r2m(t)f2(x,y,t))zmeϰtdxdydtδ2f1ΩX|r1m(x)|2dx++δ2f20T|r2m(t)|2dt+1δQT|zm|2eϰtdxdydt,δ>0.

Taking into account 𝓘1 – 𝓘5, from (35) we obtain G|zm|2eϰTdxdy+ST2i=1lλi(x,y,t)|zm|2cos(ν,yi)eϰtdxdσdt++QT[(xlλ1+2c02δ2g0)|zm|2+2a0|α|=m0|Dαzm|2]eϰtdxdydtδf1Ωx|r1m(x)|2dx+δf20T|r2m(t)|2dt,m2.(36)

After choosing x = x1, δ = T in (36), for m ≥ 2 we get inequalities QT[|zm|2+|α|=m0|Dαzm|2]dxdydtTeϰ1Tmin{1;2a0}[f1Ωx|r1m(x)|2dx+f20T|r2m(t)|2dt].(37)

According to (31), (32) we have r1m(x)=1Δ1(x)Π[B1(x,y,t)zm1+K1(y,t)(g(x,y,t,um1)g(x,y,t,um2))F12(x,y,t)r2m(t)]dydt,xΩx,m3,(38) r2m(t)=1Δ2(t)G[B2(x,y,t)zm1+0<|α|=|γ|m0DγK2(x,y)aαγ(x)Dαzm1++K2(x,y)(g(x,y,t,um1)g(x,y,t,um2))F21(x,y,t)r1m(x)]dxdy,t[0,T],m3.(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)|2dx2infΩx|1(x)|2[QT(B1(x,y,t)+K1(y,t)g0)2dxdydtQT|zm1|2dxdydt++mes ΩyQT(F12(x,y,t))2dxdydt0T|r2m(t)|2dt],m3,(40) 0T|r2m(t)|2dt3inf[0,T]|2(t)|2[QT(B2(x,y,t)+K2(x,y)g0)2dxdydtQT|zm1|2dxdydt++m02Γm0QT0<|α|=|γ|m0Dγ(K2(x,y)aαγ(x))2dxdydtQT|α|=m0|Dαzm1|2dxdydt++mesΩyQT(F21(x,y,t))2dxdydtΩx|r1m(x)|2dx],m3.(41)

The sum of (40) and (41) gives the estimates Ωx|r1m(x)|2dxC41C2QT[|zm1|2+|α|=m0|Dαzm1|2]dxdydt,m3,(42) 0T|r2m(t)|2dtC41C3QT[|zm1|2+|α|=m0|Dαzm1|2]dxdydt,m3.(43)

Moreover, (37), (42) and (43) imply the inequalities f1Ωx|r1m+1(x)|2dx+f20T|r2m+1(t)|2dtC4(f11C2+f21C3)QT[|zm|2+|α|=m0|Dαzm|2]dxdydtC5[f1Ωx|r1m(x)|2dx+f20T|r2m(t)|2dt], m ≥ 3. Therefore f1Ωx|r1m(x)|2dx+f20T|r2m(t)|2dtC5m2[f1Ωx|r12(x)|2dx+f20T|r22(t)|2dt],m3.(44)

Taking into account (44) and the condition C5 < 1, we obtain the inequality f1Ωx|q1m+k(x)q1m(x)|2dx+f20T|q2m+k(t)q2m(t)|2dti=m+1m+k[f1Ωx|r1i(x)|2dx+f20T|r2i(t)|2dt]i=m+1m+k(C5)i2[f1Ωx|r12(x)|2dx++f20T|r22(t)|2dt](C5)m1(1(C5)k)1C5[f1Ωx|r12(x)|2dx+f20T|r22(t)|2dt](C5)m11C5[f1Ωx|r12(x)|2dx+f20T|r22(t)|2dt],(45) for each k, m ∈ ℕ, m ≥ 3.

Then, for each ε > 0 there exists m~, such that for all k, m ∈ ℕ, m > m~, the inequalities ∥q1m+k(x) − q1m(x);L2(Ωx)ε,q2m+k(t)q2m(t);L2(0,T)ε hold. Thus, the sequence {q1m}m=1 is fundamental in L2x), {q1m}m=1 is fundamental in L2(0, T). Then from (37) we obtain that {um}m=1 is fundamental in L2(QT) ∩ C([0, T]; L2(G)) and the sequence {Dαum}m=1 is fundamental in L2(QT). Therefore, umu in L2(QT)C([0,T];L2(G)),DαumDαu in L2(QT),|α|m0,(46) q1mq1 in L2(Ωx),q2mq2 in L2(0,T),(47) as m → ∞.

Moreover, the functions yium(i ∈ {1, …, l}) and tum fulfill the estimates (8), (9) with yium, tum, q1m(x),q2m(t) instead of yiu*, tu*, q1(x),q2(t) 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 QTi=1l|yium|2dxdydtM4,tum;L2(QT)M5.(48)

It follows from (48) that we can choose a subsequence of the sequence {um}m=1 (for this subsequence, we preserve the same notation) such that yiumyiu weakly in L2(QT)(i{1,,l}),tumtu 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,m1,(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 v1W0m0,2x) 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),xΩx;iuνi|Ωx=0(i{0,1,,m01}),(52) where F(x, y, t) = f1(x, y, t)q1(x) + f2(x, y, t)q2(t)+ f0(x, y, t) − tui=1l 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) ∈ W0m0,2x) ∩ 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):=(y0y)8(Tt)5t,K2(x,y):=(y0y)2(x0x)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.


Assume that (u(1)(x,y,t),q1(1)(x),q2(1)(t)),(u(2)(x,y,t),q1(2)(x),q2(2)(t)) are two solutions of the problem (1)–(5). Then their difference (u~(x,y,t),q~1(x),q~2(t)), where u~(x,y,t) = u(1)(x, y, t) − u(2)(x, y, t), q~1(x)=q1(1)(x)q1(2)(x),q~2(t)=q2(1)(t)q2(2)(t), satisfies the condition 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,xΩx,(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 (u~(x,y,t),q~1(x),q~2(t)) we have the equality G|u~|2eϰTdxdy+QT[ϰ|u~|2+2i=1lλi(x,y,t)yiu~u~+20<|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~]eϰtdxdydt==2QT[f1(x,y,t)q~1(x)+f2(x,y,t)q~2(t)]u~eϰtdxdydt.

From here, similarly as from (35) we obtain the estimate G|u~|2eϰτdxdy+ST2i=1lλi(x,y,t)|u~|2cos(ν,yi)eϰtdxdσdt+QT[(ϰlλ1+2c02δ2g0)|u~|2++2a0|α|=m0|Dαu~|2]eϰtdxdydtδf1Ωx|q~1(x)|2dx+δf20T|q~2(t)|2dt(57)

Choosing δ = T, ϰ = ϰ1 in (57) gives the equality QT[|u~|2+|α|=m0|DαU~|2]dxdydtTeϰ1Tmin{1;2a0}[f1Ωx|q~1(x)|2dx+f20T|q~2(t)|2dt].(58)

Taking into account (55), (56) we get Ωx|q~1(x)|2dxC41C2QT[|u~|2+|α|=m0|Dαu~|2]dxdydt;0T|q~2(t)|2dtC41C3QT[|u~|2+|α|=m0|Dαu~|2]dxdydt, and using (58), we obtain (1C5)[f1Ωx|q~1(x)|2dx+f20T|q~2(t)|2dt]0.(59)

Since C5 < 1, then q~1(x)0,q~2(t)0, thus q1(1)(x)q1(2)(x),q2(1)(t)q2(2)(t).

Then from (58) it follows that QT|u~|2 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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About the article

Received: 2016-10-24

Accepted: 2017-07-06

Published Online: 2017-08-29

Citation Information: Open Mathematics, Volume 15, Issue 1, Pages 1048–1062, ISSN (Online) 2391-5455, DOI: https://doi.org/10.1515/math-2017-0086.

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© 2017 Protsakh. This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 3.0 License. BY-NC-ND 3.0

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