Some inverse problems for strong parabolic systems
The questions of well-posedness and approximate solution of inverse problems of finding unknown functions on the right-hand side of a system of parabolic equations are investigated. For the problems considered, theorems on the existence, uniqueness, and stability of a solution are proved and example...
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Ukrains’kyi Matematychnyi Zhurnal| _version_ | 1860509529581551616 |
|---|---|
| author | Akhundov, A. Ya. Ахундов, А. Я. |
| author_facet | Akhundov, A. Ya. Ахундов, А. Я. |
| author_sort | Akhundov, A. Ya. |
| baseUrl_str | https://umj.imath.kiev.ua/index.php/umj/oai |
| collection | OJS |
| datestamp_date | 2020-03-18T19:54:30Z |
| description | The questions of well-posedness and approximate solution of inverse problems of finding unknown functions on the right-hand side of a system of parabolic equations are investigated. For the problems considered, theorems on the existence, uniqueness, and stability of a solution are proved and examples that show the exactness of the established theorems are given. Moreover, on the set of well-posedness, the rate of convergence of the method of successive approximations suggested for the approximate solution of the given problems is estimated. |
| first_indexed | 2026-03-24T02:42:33Z |
| format | Article |
| fulltext |
K O R O T K I P O V I D O M L E N N Q
UDC 517.5
A. Y. Akhundov (Inst. Math. and Mech. Nat. Acad. Sci. Azerbaijan, Baku)
SOME INVERSE PROBLEMS
FOR STRONG PARABOLIC SYSTEMS
DEQKI OBERNENI ZADAÇI
DLQ SYL\NO PARABOLIÇNYX SYSTEM
The questions of correctness and approximate solution of the inverse problems of finding unknown functions
on the right-hand side of the system of parabolic equations are investigated in the work. For the considered
problems, the theorems on the uniqueness, existence, and stability of solution have been proved and examples,
which show the exactness of the established theorems are given.
Moreover, on the set of correctness, the rate of convergence of the method of successive approximation,
suggested for approximate solution of the given problems has been estimated.
DoslidΩeno korektnist\ ta nablyΩene rozv’qzuvannqobernenyx zadaç vyznaçennq nevidomyx funkcij
u pravij çastyni systemy paraboliçnyx rivnqn\. Dlq cyx zadaç dovedeno teoremy [dynosti, isnuvannq
ta stabil\nosti rozv’qzkui navedeno pryklady, wo pokazugt\ toçnist\ vstanovlenyx teorem.
TakoΩ na mnoΩyni korektnosti vstanovleno ocinku ßvydkosti zbiΩnosti metodu poslidovnoho
nablyΩennq, wo zaproponovanyj dlq rozv’qzuvannqdanyx zadaç.
1. Statement of the problems. Let D′ ⊂ Rn−1, D ⊂ Rn, and Q = D′ × (a, b) ⊂ Rn
(a, b are some numbers) be bounded domains with boundaries ∂D′, ∂D, and ∂Q ∈ C2+α
let x′ = (x1, . . . , xn−1) and x = (x′, xn) be arbitrary points of the domainsD′ andD or
Q, respectively. The spaces Cl+α,(l+α)/2(·), l = 0, 1, 2, 0 < α < 1, and norms in these
spaces were defined in [1, p. 16], ‖vk‖l =
∑m
k=1
‖vk‖Cl , 0 < T, is a given number.
For simplicity, without loss of generality, the following system of parabolic equations
is taken as a model:
ukt − ∆uk = Φk (x, t, u) , k = 1,m, (1)
(x, t) ∈ D × (0, T ] ((x, t) ∈ Q× (0, T ]) ,
where ukt =
∂uk(x, t)
∂t
, ∆· =
∑n
i=1
∂2·
∂x2
i
is a Laplace operator, u = (u1, . . . , um) ,
uk (x, 0) = ϕk(x), x ∈ D̄ = D ∪ ∂D
(
x ∈ Q̄ = Q ∪ ∂Q
)
, (2)
uk(x, t) = ψk(x, t), (x, t) ∈ ∂D × [0, T ] ((x, t) ∈ ∂Q× [0, T ]) . (3)
From system (1) – (3) under the corresponding conditions on the input data, it is possible
to define exactly or approximately the functions uk(x, t), k = 1,m. The questions of
solvability of problem (1) – (3) in more general statement were considered, for example,
in the works [1, 2].
Let the right-hand side of equation (1) contain unknown functions and have one of
the following forms: 1) Φk(·) = fk(t)gk (x, t, u) ; 2) Φk(·) = fk (x′, t) gk (x, t, u) ;
3) Φk(·) = fk (uk) gk (x, t, u) (gk (x, t, u) are the given functions).
c© A. Y. AKHUNDOV, 2006
ISSN 1027-3190. Ukr. mat. Ωurn., 2006, t. 58, # 1 115
116 A. Y. AKHUNDOV
Then, in system (1) – (3), it is necessary to join some additional conditions. Depend-
ing on the structure of the right-hand side of Φk(·), the following inverse problems are
considered:
Problem I. It is required to define the functions
{
fk(t), uk(x, t), k = 1,m
}
from
conditions (1) – (3) and
uk (x∗, t) = pk(t), x∗ ∈ D is fixed point, t ∈ [0, T ] . (4)
Problem II. It is required to define the functions
{
fk (x′, t) , uk(x, t), k = 1,m
}
from conditions (1) – (3) and
uk (x′, c, t) = qk (x′, t) , c ∈ (a, b) is fixed point, (x′, t) ∈ D′ × (0, T ]. (5)
Problem III. It is required to define the functions
{
fk (uk) , uk(x, t), k = 1,m
}
from conditions (1) – (3) and
uk (x∗, t) = h(t), x∗ ∈ D is fixed point, t ∈ [0, T ] . (6)
Relative to the input data of Problems I – III, we suppose that:
10) gk (x, t, w) ∈ Lip(loc)B, |gk(·)| ≥ ν1 > 0, (x, t, w) ∈ B (in Problems I, III —
B = D̄ × [0, T ] ×R1, and in Problem II — B = Q̄× [0, T ] ×R1);
20) ϕk(x) ∈ C2+α
(
D̄
)
, ψk(x, t) ∈ C2+α,1+α/2 (∂D × [0, T ]) , ϕk(x) = ψk (x, 0) ,
x ∈ ∂D (in Problem II — ϕk(x) ∈ C2+α
(
Q̄
)
, ψk(x, t) ∈ C2+α,1+α/2 (∂Q× [0, T ]) ,
ϕk(x) = ψk (x, 0) , x ∈ ∂Q);
30) [ψkt (x, 0) − ∆ϕk(x)] gk (x, T, ψ (x, T )) =
[
ψkt(x, T ) − ∆ψk(x, T )
]
gk(x, 0,
ϕ(x)), x ∈ ∂D;
40) pk(t) ∈ C1+α [0, T ] , pk(0) = ϕk (x∗);
50) qk (x′, t) ∈ C2+α,1+α/2
(
D̄′ × [0, T ]
)
, qk (x′, 0) = ϕk (x′, c) , x′ ∈ D̄′;
60) hk(t) ∈ C1+α [0, T ] , hk(0) = ϕk (x∗) , ν2 ≤ hk(t) ≤ ν3, t ∈ [0, T ];
70) [ψkt (x, 0) − ∆ϕk(x)] gk (x∗, 0, ϕ(x)) = [pkt(0) − ∆ϕk(x)|x=x∗ ] gk(x, 0,
ϕ(x)), x ∈ ∂D;
80) [ψkt (x, 0) − ∆ϕk(x)] gk (x, 0, ϕ(x)) |xn=c =
[
qkt(x′, 0)−∆ϕk(x)|xn=c
]
gk(x,
0, ϕ(x)), x ∈ ∂Q;
90) [ψkt (x, 0) − ∆ϕk(x)] gk (x∗, 0, h(t)) = [hkt(0) − ∆ϕk(x)|x=x∗ ] gk(x, 0,
ϕ(x)), x ∈ ∂D.
Definition 1. The functions
{
fk(t), uk(x, t), k = 1,m
}
, are called the solution of
Problem I, if :
1) fk(t) ∈ C ([0, T ]), 2) uk(x, t) ∈ C2,1
(
D̄ × [0, T ]
)
, 3) correlations (1) – (3), (5)
are satisfied.
Definition 2. The functions
{
fk
(
x′, t
)
, uk(x, t), k = 1,m
}
are called the solu-
tion of Problem II, if: 1) fk (x′, t) ∈ C
(
D̄′ × [0, T ]
)
, 2) uk(x, t) ∈ C2,1
(
Q̄× [0, T ]
)
,
3) correlations (1) – (3), (6) are satisfied.
Definition 3. The functions
{
fk(uk), uk(x, t), k = 1,m
}
are called the solution of
Problem III, if : 1) fk (uk) ∈ C(R1), 2) uk(x, t) ∈ C2,1
(
D̄ × [0, T ]
)
, 3) correlations
(1) – (3), (7) are satisfied.
Consider Problems I – III relatively the class of incorrect Hadamard’s problems. The
solution of these problems does not always exist, and if it exists, then it can be nonunique
and unstable.
ISSN 1027-3190. Ukr. mat. Ωurn., 2006, t. 58, # 1
SOME INVERSE PROBLEMS FOR STRONG PARABOLIC SYSTEMS 117
The inverse problems of finding the right-hand side of the scalar equation of parabolic
type were considered earlier in the works [3 – 13] (see also the bibliography in these
works).
2. Uniqueness and estimation of the solutions stability. It is know that the unique-
ness theorem and also estimation of solution’s stability of the inverse problems take cen-
tral place in the investigation of their correctness questions [13]. Under the fairly general
suppositions, the following theorems of solution’s uniqueness of Problems I – III have
been proved and the estimations of solution’s stability have been established.
Theorem 1. Let conditions 10), 20), 40), 70) be fulfilled. Then if the solution of
Problem I exists and belongs to the set K1 = {
(
fk, uk, k = 1,m
)
/fk(t) ∈ Cα [0, T ] ,
uk(x, t) ∈ C2+α,1+α/2
(
D̄ × [0, T ]
)
}, then it is unique and the estimation of stability is
true:
‖u− ū‖0 +
∥∥f − f̄∥∥
0
≤M1
[
‖g − ḡ‖0 + ‖ϕ− ϕ̄‖2 +
∥∥ψ − ψ̄
∥∥
2,1
+ ‖p− p̄‖1
]
,
where M1 > 0 depends on data of Problem I and the set K1,
{(
f̄k(t), ūk(x, t), k =
= 1,m
)}
is a solution of Problem I with data ḡk(·), ϕ̄k(·), ψ̄k(·), p̄k(·), which satisfy
conditions 10), 20), 40), 70), respectively.
Theorem 2. Let conditions 10), 20), 50), 80) be fulfilled. Then if the solution of
Problem II exists and belongs to the set
K2 =
{(
fk, uk, k = 1,m
)
/fk (x′, t) ∈ Cα,α/2
(
D̄′ × [0, T ]
)
,
uk(x, t) ∈ C2+α,1+α/2
(
Q̄× [0, T ]
)}
,
then it is unique and the estimation of stability is true:
‖u− ū‖0 +
∥∥f − f̄∥∥
0
≤M2
[
‖g − ḡ‖0 + ‖ϕ− ϕ̄‖2 +
∥∥ψ − ψ̄
∥∥
2,1
+ ‖q − q̄‖0,1
]
,
where M2 > 0 depends on data of Problem II and the set K2,
{
f̄k(x′, t), ūk(x, t),
k = 1,m
}
is a solution of Problem II with data ḡk(·), ϕ̄k(·), ψ̄k(·), q̄k(·), which sat-
isfy conditions 10), 20, 50), 80), respectively.
Theorem 3. Let conditions 10), 20), 60), 90) be fulfilled. Then if the solution of
Problem III exists and belongs to the set
K3 =
{(
fk, uk, k = 1,m
)
|fk(·) ∈ Cα
(
R1
)
, ‖f‖C(R′) ≤
≤ ‖f‖C[ν2,ν3]
, uk(x, t) ∈ C2+α,1+α/2
(
D̄ × [0, T ]
)}
,
then it is unique and the estimation of stability is true:
‖u− ū‖0 +
∥∥f − f̄∥∥
0
≤M3
[
‖g − ḡ‖0 + ‖ϕ− ϕ̄‖2 +
∥∥ψ − ψ̄
∥∥
2,1
+
∥∥h− h̄∥∥
1
]
, (7)
where M3 > 0 depends on data of Problem III and the set K3,
{
f̄k(·), ūk(x, t),
k = 1,m
}
is a solution of Problem III with data ḡk(·), ϕ̄k(·), ψ̄k(·), h̄k(·), which satisfy
conditions 10), 20), 60), 90), respectively.
Theorems 1 – 3 are proved with close method. Let us show the proof of Theorem 3.
ISSN 1027-3190. Ukr. mat. Ωurn., 2006, t. 58, # 1
118 A. Y. AKHUNDOV
Proof. From equation (1) as x = x∗ taking into account the conditions of Theorem 3
for the function fk (uk) , we obtain:
fk (hk(t)) = [hkt(t) − ∆uk|x=x∗ ] /gk (x∗, t, h(t)) , (8)
k = 1,m, (x, t) ∈ Ω = D × (0, T ],
Define the function [2, p. 87]
ρk(x, t) ∈ C2+α,1+α/2
(
Ω̄
)
, ρk (x, 0) = ϕk(x), x ∈ D̄,
ρk(x, t) = ψk(x, t), k = 1,m, (x, t) ∈ S = ∂D × [0, T ] .
(9)
Let zk(x, t) = uk(x, t) − ūk(x, t), λk (uk, ūk) = fk (uk) − f̄k (uk) , δ1k (x, t, u) =
= gk (x, t, u) − ḡk (x, t, u) , δ2k(x) = ϕk(x) − ϕ̄k(x), δ3k(x, t) = ψk(x, t) − ψ̄k(x, t),
δ4k(x) = hk(t) − h̄k(t), δ5k(x, t) = ρk(x, t) − ρ̄k(x, t), k = 1,m.
It is easy to check that the functions
{
λk (ukūk) , ϑk(x, t) = zk(x, t) − δ5k(x, t),
k = 1,m
}
satisfy the system:
ϑkt − ∆ϑk = λk (uk, ūk) gk (x, t, u) + Fk(x, t), (x, t) ∈ Ω, (10)
ϑk (x, 0) = 0, x ∈ D̄; ϑk(x, t) = 0, (x, t) ∈ S, (11)
λk
(
hk, h̄k
)
= Hk(t) − ∆zk|x=x∗/gk (x∗, t, h(t)) , t ∈ [0, T ] , (12)
where
Fk(x, t) = f̄k (ūk) [gk (x, t, u) − gk (x, t, ū) + δ1k (x, t, ū)] − δ5kt(x, t) + ∆δ5k,
Hk(t) = δ4k(t)/gk (x∗, t, h)−
−[δ1k
(
x∗, t, h̄
)
+ gk (x∗, t, h) − gk
(
x∗, t, h̄
)
]/
[
ḡk (x∗, t, h) · gk
(
x∗, t, h̄
)]
.
Under the conditions of Theorem 3 and from the definition of the setK3 it follows that
coefficients and right-hand side of equation (10) satisfy the Hölder condition. It means
that there exists classical solution of definition problem of ϑk(x, t) from conditions (10),
(11) and it can be represented in the form [1, p. 468]
ϑk(x, t) =
t∫
0
∫
D
Gk (x, t; ξ, τ)
[
λk (uk, ūk) gk (ξ, τ, u) + Fk (ξ, τ)
]
dξdτ, (13)
where dξ = dξ1 . . . dξn, Gk (x, t; ξ, τ) is Green’s function of Problem (10), (11), for
which the following estimations [1] (Chapter IV) are true:
|Gk (x, t; ξ, τ)| ≤ N1 (t− τ)−n/2 exp
(
−N2|x− ξ|2/ (t− τ)
)
,∫
D
∣∣Dl
xGk (x, t; ξ, τ)
∣∣ dξ ≤ N3 (t− τ)−(l−α)/2
, l = 0, 1, 2, (14)
here, Dl
x· are various derivatives in xi of order l and Ni > 0, i = 1, 2, 3, depend on data
of Problem III.
Taking into account that ϑk(x, t) = zk(x, t) − δ5k(x, t), k = 1,m, from (13) we
obtain
ISSN 1027-3190. Ukr. mat. Ωurn., 2006, t. 58, # 1
SOME INVERSE PROBLEMS FOR STRONG PARABOLIC SYSTEMS 119
zk(x, t) = δ5k(x, t) +
t∫
0
∫
D
Gk (x, t; ξ, τ) [λk (uk, ūk) gk (ξ, τ, u) + Fk (ξ, τ)] dξdτ.
(15)
Assume that
æ = ‖u− ū‖0 +
∥∥f − f̄∥∥
0
.
Under the conditions of theorem and from definition of the setK3, taking into account
estimation (14), we obtain
|zk(x, t)| ≤M4
[
‖δ5‖2,1 + ‖δ1‖0
]
+M5æ t, (x, t) ∈ Ω̄, (16)
∣∣λk
(
hk, h̄k
)∣∣ ≤M6
[
‖δ1‖0 + ‖δ4‖1 + ‖δ5‖2,1
]
+M7æ tα/2, t ∈ [0, T ] . (17)
Inequalities (16), (17) are satisfied for any values (x, t) ∈ Ω̄. Therefore, they must be
also satisfied for maximum values of the left parts.
Consequently,
æ ≤M8
[
‖δ1‖0 + ‖δ4‖1 + ‖δ5‖2,1
]
+M9æ tα/2. (18)
Let T1, 0 < T1 ≤ T, be such number that M9T
α/2
1 < 1. Then we obtain from (17)
that if (x, t) ∈ D̄× [0, T1] , then the estimation of stability (18) for solution of Problem III
is true.
By induction method, we show that estimation (18) is true for all t ∈ [0, T ].
So, it is proved that the estimation of stability (7) holds for all (x, t) ∈ D̄ × [0, T ].
The uniqueness of the solution of Problem III follows from estimation (7) if gk (x, t, w) =
= ḡk (x, t, w) , ϕk(x) = ϕ̄k(x), ψk(x, t) = ψ̄k(x, t), hk(t) = h̄k(t).
So, Theorem 3 is completely proved.
3. The method of successive approximations. The method of successive approxi-
mations is applied for approximate solution of the considered inverse problems.
The method of successive approximations with reference to Problem I consists of the
following:
Let
{
f
(s)
k (t), u(s)
k (x, t), k = 1,m
}
∈ K1 be already found. Consider the problem on
definition of u(s+1)
k (x, t), k = 1,m, from the conditions
u
(s+1)
kt − ∆u(s+1)
k = f (s)
k (t)gk
(
x, t, u(s)
)
, (x, t) ∈ Ω = D × (0, T ], (19)
u
(s+1)
k (x, 0) = ϕk(x), x ∈ D̄; u
(s+1)
k (x, t) = ψk(x, t), (20)
(x, t) ∈ S = ∂D × (0, T ].
This problem has a unique classical solution (if input data will satisfy conditions 10),
20), 40), 70)) belonging to C2+α,1+α/2
(
Ω̄
)
[1, p. 364].
Then, under the functions u(s+1)
k (x, t), k = 1,m, from the condition
f
(s+1)
k (t) =
[
pkt(t) − ∆u(s+1)|x=x∗
]
/gk
(
x, t, u(s+1)
)
|x=x∗ , t ∈ [0, T ] , (21)
f
(s+1)
k (t) ∈ Cα [0, T ] , k = 1,m, are defined and these functions are used for the next
step of iteration. So, if we choose f (0)
k (t) ∈ Cα [0, T ] , u(0)
k (x, t) ∈ C2+α,1+α/2
(
Q̄
)
,
ISSN 1027-3190. Ukr. mat. Ωurn., 2006, t. 58, # 1
120 A. Y. AKHUNDOV
k = 1,m, from system (19) – (21) for s = 0, 1, 2, . . . , we consequently find the functions
f
(s)
k (t) ∈ Cα [0, T ] , u(s)
k (x, t) ∈ C2+α,1+α/2
(
Ω̄
)
, k = 1,m.
Let us show uniform boundedness of the sequences
{
f
(s)
k (t)
}
and
{
u
(s)
k (x, t)
}
, k =
= 1,m, which we need below.
Lemma 1. Let conditions 10), 20), 40), 70) be fulfilled. Then if f (0)
k (t) ∈ Cα [0, T ] ,
u
(0)
k ∈ C2+α,1+α/2
(
Ω̄
)
, then the functions
{
f
(s)
k (t), u(s)
k (x, t), k = 1,m
}
found from
system (19) – (21) for s = 1, 2, . . . are uniformly bounded (by sup norm) at (x, t) ∈ Ω̄.
Proof. As stated above, if we choose
{
f
(0)
k (t), u(0)
k (x, t), k = 1,m
}
∈ K1, then
under the conditions of Lemma 1 and by virtue of statement of the theorem proved in [1,
p. 364] it follows that
{
f
(s)
k (t), u(s)
k (x, t), k = 1,m
}
∈ K1 for any s = 1, 2, . . .. Then,
by Green’s function [1, p. 468], we find the expressions for solution of the problem of
definition of u(s+1)
k (x, t), k = 1,m, form (19), (20):
u
(s+1)
k (x, t) = ρk(x, t) +
+
t∫
0
∫
D
Gk (x, t; ξ, τ)
[
f
(s)
k (τ) gk
(
ξ, τ, u(s)
)
− ρkτ (ξ, τ) + ∆ρk
]
dξdτ, (22)
where ρk(x, t) is defined in (9), Gk (x, t; ξ, τ) is Green’s function of problem (19), (20)
for which estimations (14) are true.
Taking into account estimations (14) and the conditions of Lemma 1, from (22) and
(21) we obtain ∣∣∣Dl
xu
(s+1)
k (x, t)
∣∣∣ ≤ N4 ‖ρ‖2,1 +N5
∣∣∣f (s)
k (t)
∣∣∣ t(2+α−l)/2,
k = 1,m, l = 0, 1, 2, (x, t) ∈ Ω̄,∣∣∣f (s+1)
k (t)
∣∣∣ ≤ N6 ‖p‖1 +N7
∣∣∣D2
xu
(s+1)
k (x∗, t)
∣∣∣ tα/2, t ∈ [0, T ] ,
or
γ(s+1) ≤ N8
[
‖ρ‖2,1 + ‖p‖1
]
+N9t
α/2γ(0),
where
γ(s) =
2∑
l=0
∥∥∥Dl
xu
(s)
∥∥∥
0
+
∥∥∥f (s)
∥∥∥
0
.
From the last inequality, we have
γ(s+1) ≤ N10
[
‖ρ‖2,1 + ‖p‖1
]
(1 − σs) / (1 − σ) + σsγ(0), σ = N9t
α/2.
Let T2, 0 < T2 ≤ T, be such a number that N9T
α/2
2 < 1. Then we obtain that the
sequences
{
f
(s)
k (t)
}
,
{
Dl
xu
(s)
k (x, t)
}
, l = 0, 1, 2, k = 1,m, are uniformly (on sup norm)
bounded if (x, t) ∈ D̄ × [0, T2].
Considering problem (19) – (20) is turn for the intervals (T2, 2T2) , (2T2, 3T2) and
etc., for finite number of steps we shall obtain the uniform boundedness of the sequences{
f
(s)
k (t)
}
,
{
Dl
xu
(s)
k (x, t)
}
, l = 0, 1, 2, k = 1,m, for all (x, t) ∈ D̄ × [0, T ].
Theorem 4. Let: 1) conditions 10), 20), 40), 70) be fulfilled, 2) Problem I have a
unique solution belonging to the set K1. Then the functions
{
f
(s)
k (t), u(s)
k (x, t), k =
ISSN 1027-3190. Ukr. mat. Ωurn., 2006, t. 58, # 1
SOME INVERSE PROBLEMS FOR STRONG PARABOLIC SYSTEMS 121
= 1,m
}
, found from system (19) – (21) uniformly tend to the solution of Problem I with
rate of geometric progression.
Proof. From equation (1) with x = x∗, taking into account the conditions of Theo-
rem 4 for the functions fk(t), we obtain
fk(t) = [pkt(t) − ∆uk|x=x∗ ] /gk (x∗, t, p(t)) , t ∈ [0, T ] . (23)
Subtracting from the correlations of system (1), (2), (23) the corresponding correla-
tions of system (19) – (21), we obtain that the functions
{
λ
(s)
k (t) = fk(t) − f (s)
k (t), z(s)k (x, t) = uk(x, t) − u(s)
k (x, t), k = 1,m
}
satisfy the conditions of the system
z
(s+1)
kt − ∆z(s+1)
k = λ(s)
k (t)gk (x, t, u) + f (s)
k (t)
[
gk (x, t, u) − gk
(
x, t, u(s)
)(s)
]
, (24)
(x, t) ∈ Ω,
z
(s+1)
k (x, 0) = 0, x ∈ D̄; z
(s+1)
k (x, t) = 0, (x, t) ∈ S, (25)
λ
(s+1)
k (t) = −∆z(s+1)
k |x=x∗/gk (x∗, t, p(t)) + [
(
pkt(t) − ∆u(s+1)|x=x∗
)
×
×(gk
(
x∗, t, u(s+1) (x∗, t)
)
− gk (x∗, t, p(t)))]/
gk
(
x∗, t, u(s+1) (x∗, t)
)
gk (x∗, t, p(t)) . (26)
It follows from the assumptions of Theorem 4 and statement of Lemma 1, that the
right-hand side of (24) belongs to the class Cα,α/2
(
Ω̄
)
and it is uniformly bounded.
Therefore, there exists classical solution of problem of definition of z(s+1)
k (x, t) from
conditions (24), (25) and it can be represented in the form [1, p. 468]
z
(s+1)
k (x, t) =
t∫
0
∫
D
Gk (x, t; ξ, τ) ×
×
[
λ
(s)
k (τ) gk (ξ, τ, u) + f (s) (τ)
(
gk (ξ, τ, u) − gk
(
ξ, τ, u(s)
))]
dξdτ. (27)
For Green’s function of problem (24), (25) Gk (x, t; ξ, τ), estimations (14) are true.
Acting as in the proof of Lemma 1, we obtain∣∣∣Dl
xz
(s+1)
k (x, t)
∣∣∣ ≤ N11
[ ∣∣∣λ(s)
k (t)
∣∣∣ +
∣∣∣z(s)k (x, t)
∣∣∣
]
t(2+α−l)/2,
l = 0, 1, 2, (x, t) ∈ Ω̄,∣∣∣λ(s+1)
k (t)
∣∣∣ ≤ N12
[ ∣∣∣z(s+1)
k (x∗, t)
∣∣∣ +
∣∣∣D2
xz
(s+1)
k (x∗, t)
∣∣∣
]
tα/2, t ∈ [0, T ] ,
or
æ(s+1) ≤ N13æ(s)tα/2,
where æ(s) =
∑2
l=0
∥∥Dl
xz
(s)
∥∥
0
+
∥∥λ(s)
∥∥
0
.
Applying the last inequality, we obtain
æ(s+1) ≤ θs+1æ(0), θ = N13t
α/2. (28)
ISSN 1027-3190. Ukr. mat. Ωurn., 2006, t. 58, # 1
122 A. Y. AKHUNDOV
Let now T3, 0 < T3 ≤ T, be such a number that N13T
α/2
3 < 1. Then it follows
that the sequence
{
æ(s)
}
is majorized by decreased geometric progression. Acting as
indicated above, we obtain that inequality (28) is true for all t ∈ [0, T ]. It means that
æ(s) → 0 as s→ ∞ not slower than geometric progression.
So, we obtain that the functions
{
f
(s)
k (t), u(s)
k (x, t), k = 1,m
}
found from (19) –
(21) uniformly tend to the solution of problem (1) – (4) as s → ∞ with rate of conver-
gence, which is not slower than rate of convergence of the geometric progression.
As in Theorem 4, the following convergence theorems on the method of successive
approximations used in Problems II are proved.
Theorem 5. Let: 1) conditions 10), 20), 50), 80) be fulfilled; 2) Problem II have
a unique solution belonging to the set K2. Then the functions
{
f
(s)
k (x′, t), u(s)
k (x, t),
k = 1,m
}
found from the system
u
(s+1)
kt − ∆u(s+1)
k = f (s)
k (x′, t) gk
(
x, t, u(s)
)
, (x, t) ∈ Q× (0, T ], (29)
u
(s+1)
k (x, 0) = ϕk(x), x ∈ Q̄;
u
(s+1)
k (x, t) = ψk(x, t), (x, t) ∈ ∂Q× [0, T ],
(30)
f
(s+1)
k (x′, t) =
[
qkt (x′, t) − ∆u(s+1)|x=c
]
/g
(
x, t, u(s+1)
)
|xn=c,
(x′, t) ∈ D̄′ × [0, T ] ,
(31)
uniformly tend to the solution of Problem II with rate of geometric progression.
4. Existence of solution. The existence of solution of Problems I, II is proved by the
method of successive approximations used in Section 3.
Theorem 6. Let conditions 10), 20), 40), 70) be fulfilled. Then Problem I has at least
one solution in the sense of Definition 1.
Theorem 7. Let conditions 10), 20), 50), 80) be fulfilled. Then Problem II has at
least one solution in the sense of Definition 2.
Theorems 6, 7 are proved with close method. Let prove below Theorem 7.
Proof. Note that if we choose f (0)
k (x′, t) ∈ Cα,α/2
(
D̄′ × [0, T ]
)
, u
(0)
k (x, t) ∈
∈ C2+α,1+α/2
(
Q̄× [0, T ]
)
, k = 1,m, then, under the conditions of Theorem 7,
u(s)(x, t) ∈ C2+α,1+α/2
(
Q̄× [0, T ]
)
for all s = 1, 2, . . . [1, p. 364]. Then, under the
conditions of Theorem 7, it follows from (29) that f (s)
k (x′, t) ∈ Cα,α/2
(
D̄′ × [0, T ]
)
,
k = 1,m. Using the functions ρk(x, t), k = 1,m, defined in (9) and representations of
solution through Green’s function [1, p. 468], let us find the expressions for solution of
problem of definition u(s+1)
k (x, t) from conditions (29), (30):
u
(s+1)
k (x, t) = ρk(x, t) +
t∫
0
∫
Q
Gk (x, t; ξ, τ)F (s)
k (ξ, τ) dξdτ,
where F (s)
k (x, t) = f (s)
k (x′, t) gk
(
x, t, u(s)
)
− ρkt + ∆ρk.
By analogy with Lemma 1, we prove the following lemma:
Lemma 2. Let the conditions of Theorem 7 be fulfilled. Then the sequences{
f
(s)
k (x′, t)
}
,
{
Dl
xu
(s)
k (x, t)
}
, l = 0, 1, 2, k = 1,m, are uniformly bounded (on
sup norm) if (x, t) ∈ Q̄× [0, T ] .
The equipotential continuity of the sequences
{
Dl
xu
(s)
k (x, t)
}
, l = 0, 1, 2, k = 1,m,
follows from the inequality
ISSN 1027-3190. Ukr. mat. Ωurn., 2006, t. 58, # 1
SOME INVERSE PROBLEMS FOR STRONG PARABOLIC SYSTEMS 123
∣∣∣Dl
xu
(s+1)
k (x, t) −Dl
xu
(s)
k (x̄, t̄)
∣∣∣ ≤
∣∣∣Dl
xu
(s+1)
k (x, t) −Dl
xu
(s+1)
k (x̄, t)
∣∣∣ +
+
∣∣∣Dl
xu
(s+1)
k (x̄, t) −Dl
xu
(s+1)
k (x̄, t̄)
∣∣∣ ≤ ∣∣Dl
xρk(x, t) −Dl
xρk (x̄, t)
∣∣ +
+
∣∣Dl
xρk (x̄, t) −Dl
xρk (x̄, t̄)
∣∣ +
t∫
0
∫
Q
∣∣Dl
xGk (x, t; ξ, τ) −Dl
xGk (x̄, t; ξ, τ)
∣∣×
×
∣∣∣F (s)
k (ξ, τ)
∣∣∣ dξdτ +
t̄∫
0
∫
Q
∣∣Dl
xGk (x̄, t; ξ, τ) −Dl
xGk (x̄, t̄; ξ, τ)
∣∣×
×
∣∣∣F (s)
k (ξ, τ)
∣∣∣ dξdτ +
t∫
t̄
∫
Q
∣∣Dl
xGk (x̄, t; ξ, τ)
∣∣ ∣∣∣F (s)
k (ξ, τ)
∣∣∣ dξdτ
taking into account the uniform boundedness of
{
f
(s)
k (x′, t)
}
, the continuity and bound-
edness of input data, estimations (14), and the following [1, p. 469] relations:
∣∣Dl
xG (x, t; ξ, τ) −Dl
xG (x̄, t; ξ, τ)
∣∣ ≤
≤ N14 |x− x̄|α |t− τ |−(n+2+α)/2 exp
(
−N15 |x− ξ|2 / (t− τ)
)
,
∣∣Dl
xG (x, t; ξ, τ) −Dl
xG (x̄, t̄; ξ, τ)
∣∣ ≤
≤ N16 |t− t̄|(2+α−l)/2 (t̄− τ)−(n+2+α)/2 exp
(
−N17 |x− ξ|2 / (t− τ)
)
.
The equipotential continuity of the sequence
{
f
(s)
k (x′, t)
}
, k = 1,m, follows from
the inequality
∣∣∣f (s)
k (x′, t) − f (s)
k (x̄′, t̄)
∣∣∣ ≤
≤
∣∣∣f (s)
k (x′, t) − f (s)
k (x̄′, t)
∣∣∣ +
∣∣∣f (s)
k (x̄′, t) − f (s)
k (x̄′, t̄)
∣∣∣ ≤
≤
[∣∣∣qkt(x′, t) − qkt(x̄′, t)
∣∣∣ +
∣∣∣∆u(s+1)
k (x, t)
∣∣∣
xn=c
−
−
∣∣∣∆u(s+1)
k (x̄, t)
∣∣∣
x̄n=c
]
/
∣∣∣gk(x, t, u(s))|xn=c
∣∣∣+
+
∣∣∣gk(x, t, u(s))|xn=c − gk(x̄, t, u(s))|x̄n=c
∣∣∣
(∣∣qkt(x̄′, t)
∣∣+
+
∣∣∣∆u(s+1)
k (x̄, t)
∣∣
x̄n=c
∣∣∣
)
/
∣∣∣gk(x, t, u(s))
∣∣
xn=c
gk(x̄, t, u(s))
∣∣
x̄n=c
∣∣∣+
+
[∣∣∣qkt(x̄′, t) − qkt
(
x̄′, t̄
)∣∣∣+
+
∣∣∣∆u(s)
k (x̄, t)
∣∣
x̄n=c
− ∆u(s)
k (x̄, t̄)
∣∣
x̄n=c
∣∣∣
]
/
∣∣∣gk(x̄, t, u(s) )
∣∣
x̄n=c
∣∣∣+
+
∣∣∣gk(x̄, t, u(s))
∣∣
x̄n=c
− gk(x̄, t̄, u(s))
∣∣
x̄n=c
∣∣∣×
ISSN 1027-3190. Ukr. mat. Ωurn., 2006, t. 58, # 1
124 A. Y. AKHUNDOV
×
(∣∣qkt(x̄′, t̄)
∣∣ +
∣∣∆u(s+1)
k (x̄, t̄)
∣∣
x̄n=c
)/∣∣∣qk(x̄, t, u(s))
∣∣
x̄n=c
gk(x̄, t̄, u(s))
∣∣
x̄n=c
∣∣∣,
taking into account the uniform boundedness and equipotential continuity of the sequence{
Dl
xu
(s)
k (x, t)
}
, l = 0, 1, 2, the continuity and boundedness of input datas.
The uniform boundedness and equipotential continuity of the sequence
{
u
(s)
kt (x, t)
}
follows from (29).
By Arcela’s theorem [2, p. 84], from the sequences
{
u
(s)
kt
}
,
{
Dl
xu
(s)
k
}
,
{
f
(s)
k
}
,
l = 0, 1, 2, k = 1,m, it is possible to choose sequences convergent to some functions
{u∗kt} ,
{
Dl
xu
∗
k
}
, l = 0, 1, 2, {f∗k} , respectively, and u∗k (x, t) ∈ C2,1
(
Q̄× [0, T ]
)
,
f∗k (x′, t) ∈ C
(
D̄′ × [0, T ]
)
.
Then, passing to the limit as s→ ∞ in correlations (29) – (31), we obtain relations
u∗kt − ∆u∗k = f∗k (x′, t) gk (x, t, u∗) , (x, t) ∈ Q× (0, T ],
u∗k (x, 0) = ϕk (x) , x ∈ Q̄ u∗k (x, t) = ψk (x, t) , (x, t) ∈ ∂Q× [0, T ] ,
f∗k (x′, t) = [qkt (x′, t) − ∆u∗k|xn=c] /gk (x, t, u∗) |xn=c, (x′, t) ∈ D̄′ × [0, T ] .
This implies that
u∗kt − ∆u∗k = gk (x, t, u∗) [qkt (x′, t) − ∆u∗k|xn=c] /gk (x, t, u∗) |xn=c.
By using the last equality with xn = c and taking into account conditions ϕk(x′, c,
0) = q (x′, 0) , we obtain u∗k (x′, c, t) = qk (x′, t) .
Thus, the existence of solution of Problem II is proved in terms of Definition 2.
1. Ladizhenskaya O. A., Solonnikov V. A., Uralceva N. I. Linear and quasilinear equations of parabolic type
(in Russian). – M., 1967.
2. Fridman A. Parabolic type partial equations (in Russian). – M., 1968.
3. Iskenderov A. D. Some inverse problems on definition of the right-hand sides of differential equations //
Izv. AN AzSSR. Fis.-Tech. and Math. Sci. Ser. – 1976. – # 2. – P. 58 – 63 (in Russian).
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# 6. – P. 1296 – 1299 (in Russian).
5. Cannon J. R., Duchateau P. An inverse problem for an unknown source in a heat equation // J. Math. Anal.
and Appl. – 1980. – 75. – P. 465 – 485.
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Received 16.05.2005
ISSN 1027-3190. Ukr. mat. Ωurn., 2006, t. 58, # 1
|
| id | umjimathkievua-article-3438 |
| institution | Ukrains’kyi Matematychnyi Zhurnal |
| keywords_txt_mv | keywords |
| language | English |
| last_indexed | 2026-03-24T02:42:33Z |
| publishDate | 2006 |
| publisher | Institute of Mathematics, NAS of Ukraine |
| record_format | ojs |
| resource_txt_mv | umjimathkievua/b4/1f21ca0b25bf8d8f7fc957ba91404fb4.pdf |
| spelling | umjimathkievua-article-34382020-03-18T19:54:30Z Some inverse problems for strong parabolic systems Деякі обернені задачі для сильно параболічних систем Akhundov, A. Ya. Ахундов, А. Я. The questions of well-posedness and approximate solution of inverse problems of finding unknown functions on the right-hand side of a system of parabolic equations are investigated. For the problems considered, theorems on the existence, uniqueness, and stability of a solution are proved and examples that show the exactness of the established theorems are given. Moreover, on the set of well-posedness, the rate of convergence of the method of successive approximations suggested for the approximate solution of the given problems is estimated. Досліджено коректність та наближене розв'язування обернених задач визначення невідомих функцій у правій частині системи параболічних рівнянь. Для цих задач доведено теореми єдиності, існування та стабільності розв'язкуі наведено приклади, що показують точність встановлених теорем. Також на множині коректності встановлено оцінку швидкості збіжності методу послідовного наближення, що запропонований для розв'язування даних задач. Institute of Mathematics, NAS of Ukraine 2006-01-25 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/3438 Ukrains’kyi Matematychnyi Zhurnal; Vol. 58 No. 1 (2006); 115–124 Український математичний журнал; Том 58 № 1 (2006); 115–124 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/3438/3615 https://umj.imath.kiev.ua/index.php/umj/article/view/3438/3616 Copyright (c) 2006 Akhundov A. Ya. |
| spellingShingle | Akhundov, A. Ya. Ахундов, А. Я. Some inverse problems for strong parabolic systems |
| title | Some inverse problems for strong parabolic systems |
| title_alt | Деякі обернені задачі для сильно параболічних систем |
| title_full | Some inverse problems for strong parabolic systems |
| title_fullStr | Some inverse problems for strong parabolic systems |
| title_full_unstemmed | Some inverse problems for strong parabolic systems |
| title_short | Some inverse problems for strong parabolic systems |
| title_sort | some inverse problems for strong parabolic systems |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/3438 |
| work_keys_str_mv | AT akhundovaya someinverseproblemsforstrongparabolicsystems AT ahundovaâ someinverseproblemsforstrongparabolicsystems AT akhundovaya deâkíobernenízadačídlâsilʹnoparabolíčnihsistem AT ahundovaâ deâkíobernenízadačídlâsilʹnoparabolíčnihsistem |