UNKNOWN COEFFICIENT PROBLEM FOR MIXED EQUATION OF PARABOLIC-HYPERBOLIC TYPE WITH NON-LOCAL BOUNDARY CONDITIONS ON CHARACTERISTICS Текст научной статьи по специальности «Математика»

ISSN 2074-1871 Уфимский математический журнал. Том 16. № 2 (2024). С. 82-89.

UNKNOWN COEFFICIENT PROBLEM FOR MIXED EQUATION OF PARABOLIC-HYPERBOLIC TYPE WITH NON-LOCAL BOUNDARY CONDITIONS ON CHARACTERISTICS

D.K. DURDIEV

Abstract. For an equation of a mixed parabolic-hyperbolic type with a characteristic line of type change, we study the inverse problem associated with the search for an unknown coefficient at the lowest term of the parabolic equation. In the direct problem, we consider an analog of the Tricomi problem for this equation with a nonlocal condition on the characteristics in the hyperbolic part and the Dirichlet condition in the parabolic part of the domain. In order to determine the unknown coefficient by the solution on the parabolic-part of the domain, the integral overdetermination condition is proposed. Global results on the unique solvability of the inverse problem in the sense of the classical solution are proved.

Keywords: parabolic-hyperbolic equation, characteristic, Green's function, inverse problem, contraction principle mapping.

Mathematics Subject Classification: 35A01; 35A02; 35L02; 35L03; 35R03.

1. Formulation of problem

Let be a domain in the plane of variables x, y, consisting of the union of two subdomains, i.e. Qit = Qht U Q2i, where

Пит ={(x,y) : 0 <x<l, 0 <y ^ T},

П21 = [(x,y) : -y<x ^ у + I, — 1^<У< 0},

and I, T are fixed positive numbers. In this domain we consider the equation д2и 1 — sign у д2и 1 + sign уди 1 + sign у

dx2 2 dy2 2 dy 2 = (L1)

Equation (1.1) is of a mixed parabolic-hyperbolic type. For this equation, the line of change of type y = 0 is a characteristic (parabolic degeneration of the second kind [ ]),

Direct problem. In the domain QlT find the solution of equation ( ) satisfying the following boundary conditions:

u(°,y) = My), u(l,y) = <p2(y), y e [°,n (1.2)

U (f, -1) + U , = MX), X e [°, I] , (1.3) where = pi(y), = ^2(y), ^ = ip(x) are given functions.

D.K. Durdiev, Unknown coefficient problem for mixed equation of parabolic-hyperbolic

type with non-local boundary conditions on characteristics.

@ Durdiev D.K.. Submitted September 12, 2023.

A classical solution to direct problem ( )-( ) is a function u(x, y) in the class С (Pit) П С1 (PtT) nCj;2 (PizT) ПС2 (П2г), which solves equation (1,1) and satisfies conditions (1.2), (1.3).

An inverse problem is on finding a pair of functions и = u(x, y), q = q(x), in the classes

и ЕС (P,T ) П С1 (Pit ) П С^ (PUt ) П С2 (^), q Е С [0, /],

such that these functions solve equation (1.1) and satisfy boundary conditions (1.2), (1.3) and the following overdetermination condition:

T

/ftfoM^d^/M, x Е [ML (i.4)

0

where h = h(y), f = f(x) are given sufficiently smooth functions.

Direct and inverse problems for mixed type equations are not studied in so many details as similar problems for classical equations. Nevertheless, such problems are relevant from the point of view of applications. The importance of considering equations of mixed type, where the equation is of parabolic type in one part of the domain and hyperbolic in the other, was first pointed out by GePfand in his work |2|, Another example is the following phenomenon in electrodynamics: a mathematical study of the tension of an electromagnetic field in an inhomogeneous medium consisting of a dielectric and a conducting medium leads to a system consisting of a wave equation and a heat equation, see |3|. There are many examples of such kind.

For the first time, an analogue of the Tricomi problem for a hyperbolic-parabolic equation was studied in |4|, Further, such problems with different boundary and non-local conditions for parabolic-hyperbolic equations with both characteristic and non-characteristic type change lines were formulated and studied in |5|-|8|.

Methods for solving direct and inverse problems for finding the solution of an initial boundary value problem for equations of the parabolic-hyperbolic type and the unknown right-hand side (linear problem) of the equation in a rectangular domain were proposed in |9|-|11|, In this direction, we also point out work |12|, in which such problems were studied for equations of mixed parabolic-hyperbolic type with the time fractional derivative in the parabolic part of the equation.

Various inverse problems for particular second order equations of hyperbolic and parabolic types can be found in monographs |13|-|16|, see also the references therein.

For equations of mixed parabolic-hyperbolic type, inverse coefficient problems were not studied before. This article continues the study of the author |17|, in which the local unique solvability of the inverse problem on determining the variable coefficient at the lowest term of a hyperbolic equation for a mixed hyperbolic-parabolic equation with a noncharacteristic line of type change was investigated. Note that the problems considered below, in addition to their independent interest, are also of interest from the point of view of studying the solvability of inverse coefficient problems for parabolic equations.

Throughout this paper we shall assume that the following conditions are satisfied:

(Bl) ( ^(y), My)) e С 1[0, T], ф(х) Е С3 [0, A];

(B2) ыо) = Ы0), ^i(o) - Ы0) = Ф(о) - Ш;

(B3) h(y) Е С 1[0, T], h(0) = h(T) = 0, f(x) e С3[0,I], f h(y)tpi(y)dy = /(0),

0

T

J h(y)My)dy = /(/), f(x) = 0 for all x e [0, /].

0

2. Direct problem Assume that the function q(x) is known.

Theorem 2.1. Let conditions Bl, B2, q(x) E C1 [0,/] be .satisfied. Then there exists an unique solution to direct problem ( )-( ) in the domain Qit-

We denote t(x) := u(x, 0) v(x) = u(x, 0). Due to the unique solvability of the Cauchv problem for the wave equation, the solution to equation ( ) in the domain Q2i can be written using the d'Alembert formula:

1 1 rx-y

u(x,y) = -[T(x + y) + T(x — y)] - - u(s) ds. (2.1)

2 2 J x+y

Taking into account condition ( ) and the identities t(0) = ^1(0), t(/) = ^2(0) (a consequence of the definition of the classical solution), we obtain the identity

i

2r(x) + ^(0) + p2(0) - J v(s)ds = 2^(x), x E [0, /]. (2.2)

0

It follows from (1.3) at x = I, y = —I that

«(-,—-) = m — M0).

Comparing this with (2.1) at x = -2 ,y = — |, we have

i

J u(s) ds = ^1(0) + 3^2(0) — 2^(1). 0

Substituting this into identity (2.2), we find

T (x) = M) — ^2(0)+ ^(x). (2.3)

Thus, we have found the function t(x).

In order to find v(x) we use equation ( ) in domain and calculate limy^+0. Then we easily obtain v(x) = t"(x) — q(x)r(x) and the same

v(x) = $\x) — q(x) (^(l) — ^2(0) + ^(x)) .

It is clear that for known t(x) and v(x) the solution to direct problem ( )-( ) in Q2i is given bv formula ( ). Under the assumptions of Theorem we have u E C2 (Q2z).

It is known |1| that the Green's function of the first initial boundary value problem for the equation

Uxx — Uy = 0, X E (0, I), y > 0,

is of the form

G(x,C,y) =- V

v 2,/Wy ^

V V n= — cc

exp(—■— exp( —

In view of this, we rewrite equation ( ) in the domain Q1tT with conditions ( ) to an integral equation

I y

u(x, y)=J G(x,C, y)T(£)d£, + j Gi(x, 0,y - rj)(pi(rj) drj

0 0 (2.4)

y y l

- j Gi (x,l,y - V)MV) drj- J j G(x,C,y - V)Q(Ou(C, V)d£ drj. 0 0 0

We also note that integral equation ( ) is of Volterra type since r(x) is known. In view of the conditions imposed on tpu, in (Bl), this equation determines the function u G C^ (Put) , that is, the solution to problem ( ), ( ) in the domain Put.

Thus, the constructed functions in Put and P2 is the classical solution to direct problem ( )-( ) in the domain Pit. This completes the proof of Theorem

3. Inverse problem

Assume that conditions (B3) are satisfied. Multiplying equation ( ) in the domain Put by the function h(y), integrating over the segment [0,T] and taking into consideration ( ), we

t

f(x) 1 f

^(x) = + ^ h(y)u(x, У)dУ, x G [0, l]. (3.1)

( x) ( x)

0

( x)

the operator form:

u(x, y) = U[u](x, y), (x, y) G Put, (3.2)

where the operator U is defined by the identity

y i

Uu(x, y)=uo(x, y)+ G(x,^y - v)

0 0

f + i h'( s №,s )ds

, rj)d^drj, (3.3)

( ) ( )

0 u0

function:

i y y

uo(x, y) := J G(x,C, y^(OdC + J GC(x, 0,y - V)Vi (rn)drj - J G? (x,l,y - ^M^drj. 0 0 0 ( x)

The main result of this section is the following theorem.

Theorem 3.1. Let conditions (Bl)-(BS) be satisfied. Then, there exist positives numbers T* such, that equation ( ) has an unique continuous solution in the domain Put for T G (0,T*).

Proof. Owing to ( ) it is clear that under the assumptions of the theorem the operator U maps a function u G C (Put) into a function belonging to the same space. We define the norm in C (Put) as follows:

\\v\\lT = max |u(x,y)l . (x,y)eUuT

For the sake of brevity, we introduce notations

/0 := mn 1 /(x)|, fi := max |/"(x)|, h0 := max |h(y)l.

xe[0,z] xe[0,z] xe[0,T]

T U

5 (u0, r) := {u GC (Put) : \\u - u0\\lT ^ r) cC (Put )

into itself, where a radius r is a known number and a center is an element u0 = u0(x, y) of the functional space C (Put)- This fact will prove that in the domain P1iT equation ( ) has an unique continuous solution satisfying the inequality ||u — u0|it ^ r. It is obvious that each element u G S (u0, r) satisfies an estimate

MllT <

where R denotes a known positive number.

IT

+ r =: R,

Let us estimate

IT

.

functions G, G^ in the definitions of the function u0(x, y). In this case, we use the identity

J G(x,C, y)dC =1, 0

which follows from the definition of the function G. Taking this into account and explicit form ( ) of the function r(;r), the first term of u0(x, y) can be easily estimated:

G(x,C, y)r(OdC

<

C [0,T ]

+ 2

C[0,z]

(3.4)

G

G(x,î, y) = y ^ exP — (

n=1

2

In view of this expression we have the identities

2 ^

Gn(x, 0,y — v) = y 2^exP

n=1

(n^ \ 2

T) (y —

. nirx . nn£ sin —-— sin —-—.

I

nji mix 1 f

sin —:— = 7 I Gv(x, ty — V)(1 — 0 d^

0

which can be confirmed straightforwardly. Using these relations, we transform the following integral:

y i y

J Gc(x, 0,y — 'q)^1('q)d'q = ^ j (l — Q J Gv(x,^,y — y)'^1(y) drjd^ 0 0 0

iy

G(x,^,y — y)'^1(y) q — G(x,^,y — ri)^1(ri)dy J> d£

0

1 — x

~T

Vi(y)

— i ( — «

G(x,Ç, y)'M0)+ G(x,i,y — vWMdv

Here in the intermediate calculations we have used the relation limv^y G(x, y — rj) = 5(x — £), where 5(-) is the Dirac delta function and

i

/ a(^8(x — £)d£ = a(x),

which is valid for each continuous function a(x) on the interval (0,1).

|

2

y

These relations for (x, y) G C (Put) yield the estimate

G^(x, 0,y — (7]) dr]

^ (2 + T) ||^||cl[o;T]

(3.5)

For Gç(x,l,y — rj) we then have

Gç(x,^y — V) = 22 ^exP

n=1

(nn \ 2

T) (y —

,nn . nnx

(—1)n_ sin—= — -l Gv (x,Ç,y — V)Ç d£.

Using this, we transform the last term of u0(x, y) as follows:

1

i y

Gç (x,^y — V)lf2(V)dV = — J î Gv (x,î,y — V)lf2(V)dVdî

00

(y

G(x,^y — V)^2(rri) y — j G(x,^y — V)^2(rn) dV } dC

0

— ~^2(y) + T A

G(x,C, y)M0) + J G(x,C,y — v)v2(v)dv 0

d .

For (x, y) G C t) we also have the estimate

Gç (x,hy — rn)^2('ri)d'ri

^ (2 + T) ||Hlci[o,T]

(3.6)

Then inequalities (3.4)-(3.6) imply the estimate

^ (2 + T) \\^i\c![0;T] + (3 + T) \\^\\c![0;T] + 2 |M\CM . (3.7)

.

We now proceed to obtaining conditions for T, under which it is possible to apply the fixed point theorem to the operator U. Let u G S (u0, r), then, for all (x, y) G Put we have the inequalities

y

Wu — uo| ^J J G(x,i,y — v) 00

inoi + 1

T

| ( )| | ( )|

ih'(s)| |u(e, S)i ds

| u( , )|

<

^2h0rr2 , flRrri (rr\

0 0

Condition ||u — u0||iT ^ f (that is, Uu G S(u0, r)) te satisfied if T is chosen by the condition mi(T) < r. Let Ti be a positive root of the quadratic equation mi(T) — r = 0, that is

Tl

1

2 Rho

y/fi + 4r foho — fi

It is easy to see that mu(T) increases monotonicallv in T G (0,Tu) , mu(T) ^ ^t T ^ 0 and ml(T) < r for T G (0, Ti). This means that Uu G S(u0, r) for T <Tu.

U

S(u0, r). To prove this, we take any two elements (ul, u2) G S(u0, r) and estimate the norm

y

y

y

y

of the difference between their images Uu1, Uu2. For this purpose, using ( ) we have the inequality

y i

|Uu1 - Uu2\\lT ^

G{x, ^ y - y)

0 0

no

no

K(f, v) — u2(C,

T

(3.8)

+

I №1

Ih' (s )||u1(e, s )u1(e, y)-u2(C, s)u2(C, y)l ds

d^dy.

Here to estimate the expression [u1^, s)u1(£, y) — u2(£, s)u2(£, y)I, we use inequality | u1^, S)u1(e, y) — u2(e, s)u2(Z, y) l ^ lu1^, s) l lu1^, y) — u2(C, y) l

+ | u2(£, v) ll u1^, S) — u2(£, s)l

R\\u1-u2\

IT

(sy) e [0, T] x [0,1] x [0,y],

which holds for all (u1, u2) e S (u0, r). Continuing estimating in ( ), we get

(Uu1 — Uu2\\iT ^

2 Rh0 T 2 + J±T

0 0

lu1 — u2\

IT

-: m2(T) Wu1 — u2

\it '

We choose T so that the inequality m2 (T) < 1 holds, then the operator U contracts the distance between elements of the ball S(u0; r). It is clear that this condition is satisfied bv T e (0,T2), where

T2

1

2 —

4 Rh0

^ff[+m0ho — h

is a positive root of the quadratic equation m2(T) — 1 — 0.

Let T* — min(T1, T2). Then the operator U with T e (0,T*) is a contraction mapping of the ball S(u0, r) into itself. Hence, according to the contraction mapping principle, equation ( ) possesses a unique solution u(x, y) e S(u0, r). The proof is complete. □

u( x, ) ( x)

In this formula T e (0,TI > 0 is an arbitrary fixed number. We also note that T* does .

Thus the following theorem is valid.

Theorem 3.2. Let conditions (Bl)-(BS) be satisfied and T e (0,T*). Then formula ( ) determines q(x) e C 1[0,1] for each fixed I > 0.

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Durdimurod Kalandarovich Durdiev,

Bukhara Branch of Romanovskii Institute of Mathematics,

Uzbekistan Academy of Sciences

M. Ikbal Str. 11,

200100, Bukhara, Uzbekistan,

Bukhara State University,

M. Ikbal Str. 11,

200100, Bukhara, Uzbekistan

E-mail: [email protected], [email protected]