INVERSE PROBLEM FOR A TWO-DIMENSIONAL STRONGLY

Electronic Journal of Differential Equations, Vol. 2018 (2018), No. 77, pp. 1–17. ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu

INVERSE PROBLEM FOR A TWO-DIMENSIONAL STRONGLY DEGENERATE HEAT EQUATION MYKOLA IVANCHOV, VITALIY VLASOV Communicated by Ludmila S. Pulkina

Abstract. This article concerns the existence and uniqueness of solutions in the problem of identifying the leading coefficient in a two-dimensional heat equation. We suppose that unknown coefficient depends on the time variable and the equation is strongly degenerate. Applying Schauder fixed-point theorem, we find conditions for existence of a classical solution.

1. Introduction We consider an inverse problem for a two-dimensional degenerate heat equation in a rectangular domain. Direct problems of this type are mathematical models of various processes such as seawater desalination, movement of liquid in porous medium, financial market behavior, etc. [1, 2, 4]. Such problems are comparatively well studied (see, for instance, [3]). Inverse problems arise when certain parameters of these processes are unknown. Various types of inverse problems for non-degenerate equations are well investigated and some results may be found in monographs and references therein [7, 13, 15]. The beginning of the research of inverse problems for one-dimensional degenerate parabolic equations was made in [8, 9, 10, 12, 16]. There the conditions of existence and uniqueness of solution for these problems were established in the cases of week and strong power degeneration. Later some results were obtained for parabolic equations with arbitrary types of degeneration [17, 18] and for free-boundary domains [5, 6]. Inverse problem for a two-dimensional weakly degenerate heat equation in a rectangular domain was considered in [11]. In this paper we establish conditions for existence and uniqueness of solution to an inverse problem for a two-dimensional strongly degenerate heat equation. 2. Statement of the problem and main result In the domain QT := {(x, y, t) : 0 < x < h, 0 < y < l, 0 < t < T } we consider a two-dimensional heat equation with unknown leading coefficient depending on the time variable. We suppose that the equation degenerates at the initial moment 2010 Mathematics Subject Classification. 35R30, 35K10, 35A09, 35A02. Key words and phrases. Inverse problem; strongly degenerate heat equation; Green function; classical solution. c

2018 Texas State University. Submitted November 29, 2017. Published March 20, 2018. 1

2

M. IVANCHOV, V. VLASOV

EJDE-2018/77

as a power with a given exponent β ≥ 1. We choose the case of mixed DirichletNeumann boundary conditions. The additional condition (so-called overdetermination condition) is taken accordingly to the physical sense and it presents the value of the heat flux on the part of the boundary of domain. So, the problem consists of finding a pair of functions (a(t), u(x, y, t)), a(t) > 0, t ∈ [0, T ] that satisfy the degenerate heat equation ut = tβ a(t)∆u + f (x, y, t),

(x, y, t) ∈ QT ,

(2.1)

the initial condition u(x, y, 0) = ϕ(x, y, 0),

(x, y) ∈ D := [0, h] × [0, l],

(2.2)

the boundary conditions u(0, y, t) = µ1 (y, t), u(h, y, t) = µ2 (y, t), uy (x, 0, t) = ν1 (x, t), uy (x, l, t) = ν2 (x, t),

(y, t) ∈ [0, l] × [0, T ], (x, t) ∈ [0, h] × [0, T ]

(2.3) (2.4)

and the overdetermination condition a(t)ux (0, y0 , t) = κ(t),

t ∈ (0, T ],

(2.5)

where y0 ∈ (0, l) is some arbitrary fixed point. Our goal is to determine conditions of the existence and uniqueness of solution to problem (2.1)–(2.5) in the spaces of continuously differentiable functions. The corresponding result is presented in a theorem, for which we use the following assumptions: (A1) β ≥ 1, ϕ ∈ C 2 (D), µi ∈ C 2,1 ([0, l] × (0, T ]) ∩ C 1,1 ([0, l] × [0, T ]), νi ∈ C 1,0 ([0, h] × (0, T ]) ∩ C([0, h] × [0, T ]) for i = 1, 2, f ∈ C 2,2,0 (QT ), κ(t) = 1−β κ0 (t)t 2 with κ0 ∈ C[0, T ]; (A2) ϕx (x, y) ≥ 0 for (x, y) ∈ D; µ1t (y, t) − f (0, y, t) < 0, µ2t (y, t) − f (h, y, t) ≥ 0, µ1yy (y, t) ≥ 0, µ2yy (y, t) ≤ 0 for (y, t) ∈ [0, l] × (0, T ]; ν1x (x, t) ≤ 0, ν2x (x, t) ≥ 0 for (x, t) ∈ [0, h] × [0, T ]; fx (x, y, t) ≥ 0 for (x, y, t) ∈ QT ; κ0 (t) > 0 for t ∈ [0, T ]; (A3) µ1 (y, 0) = ϕ(0, y), µ2 (y, 0) = ϕ(h, y) for y ∈ [0, l]; ν1 (x, 0) = ϕy (x, 0), ν2 (x, 0) = ϕy (x, l) for x ∈ [0, h]; µ1y (0, t) = ν1 (0, t), µ1y (l, t) = ν2 (0, t), µ2y (0, t) = ν1 (h, t), µ2y (l, t) = ν2 (h, t) for t ∈ [0, T ]. Theorem 2.1. Under assumptions (A1)–(A3), there exists unique solution (a, u) to (2.1)–(2.5), which belongs to the space C[0, T ]×(C 2,2,1 (QT )∩C 0,1,1 (QT )), where a(t) > 0 and t ∈ [0, T ]. Here we have the usual notation [14] for the Banach spaces C k,m,n (QT ) consisting of functions that are continuous in QT , with their derivatives of k-th order with respect to x, m-th order with respect to y, and n-th order with respect to t. The spaces C k,m ([0, h] × (0, T ]) and others are defined similarly. 3. Existence of solutions There are many approaches to study existence of solutions to inverse problems. Among them, we choose the method of reducing the problem (2.1)–(2.5) to an equation with respect to the coefficient a(t) and applying the Schauder fixed-point theorem.

EJDE-2018/77

INVERSE PROBLEM FOR A DEGENERATE HEAT EQUATION

3

Assume temporarily that a = a(t) > 0 from class C[0, T ] is known and find the solution to the direct problem (2.1)–(2.4) [7]: Z lZ

h

G12 (x, y, t, ξ, η, 0)ϕ(ξ, η)dξdη

u(x, y, t) = 0

Z

0 tZ l

+ 0

0

Z tZ

l

G12ξ (x, y, t, h, η, τ )τ β a(τ )µ2 (η, τ ) dη dτ

− 0

G12ξ (x, y, t, 0, η, τ )τ β a(τ )µ1 (η, τ ) dη dτ

0

Z tZ

(3.1)

h β

−

G12 (x, y, t, ξ, 0, τ )τ a(τ )ν1 (ξ, τ )dξdτ 0

0

Z tZ + 0

h

G12 (x, y, t, ξ, l, τ )τ β a(τ )ν2 (ξ, τ )dξdτ

0

Z tZ lZ +

h

G12 (x, y, t, ξ, η, τ )f (ξ, η, τ ) dξ dη dτ. 0

0

0

Here we denote by Gij (x, y, t, ξ, η, τ ), i ∈ {1, 2} the Green functions for the heat equation (2.1) with boundary conditions of i-th kind with respect to x and j-th kind with respect to y. They are defined the equality [7] Gij (x, y, t, ξ, η, τ ) ∞ (x − ξ + 2nh)2 X 1 = exp − 4π(θ(t) − θ(τ )) m,n=−∞ 4(θ(t) − θ(τ )) (x + ξ + 2nh)2 (y − η + 2ml)2 + (−1)i exp − exp − 4(θ(t) − θ(τ )) 4(θ(t) − θ(τ )) (y + η + 2ml)2 + (−1)j exp − , 4(θ(t) − θ(τ )) Z t θ(t) = τ β a(τ )dτ, i, j ∈ {1, 2}.

(3.2)

0

Note that Gij may be expressed as a product of two Green functions for onedimensional heat equations: Gij (x, y, t, ξ, η, τ ) = Gi (x, t, ξ, τ )Gj (y, t, η, τ ). Using Green function’s properties and integrating by parts, find the derivative ux (x, y, t): ux (x, y, t) Z lZ h = G22 (x, y, t, ξ, η, 0)ϕξ (ξ, η)dξdη 0

0

Z tZ

l

G22 (x, y, t, 0, η, τ )(µ1τ (η, τ ) − τ β a(τ )µ1ηη (η, τ ) − f (0, η, τ )) dη dτ

− 0

0

Z tZ

l

G22 (x, y, t, h, η, τ )(µ2τ (η, τ ) − τ β a(τ )µ2ηη (η, τ ) − f (h, η, τ )) dη dτ

+ 0

0

Z tZ − 0

0

h

G22 (x, y, t, ξ, 0, τ )τ β a(τ )ν1ξ (ξ, τ )dξdτ

4


EJDE-2018/77

h

Z tZ

G22 (x, y, t, ξ, l, τ )τ β a(τ )ν2ξ (ξ, τ )dξdτ

+ 0

0

h

Z tZ lZ

G22 (x, y, t, ξ, η, τ )fξ (ξ, η, τ ) dξ dη dτ.

+ 0

0

0

Substituting expression in (2.5) we obtain the equation Z l Z h G22 (0, y0 , t, ξ, η, 0)ϕξ (ξ, η)dξdη a(t) = κ(t) Z tZ

0

0 l

G22 (0, y0 , t, 0, η, τ )(µ1τ (η, τ ) − τ β a(τ )µ1ηη (η, τ )

− 0

0

− f (0, η, τ )) dη dτ Z tZ l G22 (0, y0 , t, h, η, τ )(µ2τ (η, τ ) − τ β a(τ )µ2ηη (η, τ ) + 0

0

(3.3)

− f (h, η, τ )) dη dτ Z tZ h − G22 (0, y0 , t, ξ, 0, τ )τ β a(τ )ν1ξ (ξ, τ )dξdτ 0

0

Z tZ

h

G22 (0, y0 , t, ξ, l, τ )τ β a(τ )ν2ξ (ξ, τ )dξdτ

+ 0

0

Z tZ lZ

h

+

G22 (0, y0 , t, ξ, η, τ )fξ (ξ, η, τ ) dξ dη dτ 0

0

−1

,

t ∈ (0, T ].

0

It is easy to see this equation and problem (2.1)–(2.5) are equivalent. Indeed, if (a(t), u(x, y, t)) is a solution to (2.1)–(2.5), then the function a(t) satisfies (3.3) as it is shown above. On the other hand, if a(t) is a solution to (3.3), then we substitute it into (2.1) and find the solution to problem (2.1)–(2.4) under the form (3.1). Condition (2.5) follows from (3.3). To establish estimates for solutions of (3.3),we denote amax := max[0,T ] a(t) and consider the following integral from (3.3), Z tZ l G22 (0, y0 , t, 0, η, τ )(f (0, η, τ ) − µ1τ (η, τ )) dη dτ 0

0

Z tZ ≥

min (f (0, y, t) − µ1t (y, t))

[0,l]×[0,T ]

Z 0

≥√

C2 amax

G22 (0, y0 , t, 0, η, τ ) dη dτ 0

t

≥ C1

l

0

dτ p

(3.4)

θ(t) − θ(τ )

Z 0

t

√

dτ C3 ≥ β−1 √ . β+1 −τ t 2 amax

tβ+1

Here we used the equality Z

l

G2 (y, t, η, τ )dη = 1,

(3.5)

0

that can be verified by direct computation. The constants Ci , i ∈ {1, 2, 3} depend on given data. It follows from (A2) that all other summands in denominator of

EJDE-2018/77


5

(3.3) are non-negative, therefore κ(t)t

a(t) ≤

β−1 2

√

amax

C3

. 1−β

Taking into account the assumptions κ(t) = κ0 (t)t 2 , κ0 (t) > 0, t ∈ [0, T ], we obtain √ κ0 (t) amax a(t) ≤ , t ∈ [0, T ]. C3 Hence, we get the estimate a(t) ≤ A1 < ∞,

t ∈ [0, T ],

(3.6)

where A1 is a known constant depending on the problem data. Now we estimate a(t) from below. Denote integrals in the denominator of (3.3) as Ii , i = 1, 6. Taking into account (3.5), we find I1 ≤ max ϕx (x, y) := C4 . D

Denote amin := min[0,T ] a(t). Next we have the estimate Z tZ l I2 = G22 (0, y0 , t, 0, η, τ )(f (0, η, τ ) − µ1τ (η, τ ) + τ β a(τ )µ1ηη (η, τ )) dη dτ 0

0

Z tZ ≤

max (f (0, y, t) − µ1t (y, t))

[0,l]×[0,T ]

+

max [0,l]×[0,T ]

G22 (0, y0 , t, 0, η, τ ) dη dτ 0

tβ µ1yy (y, t)

Z tZ

l

0

l

G22 (0, y0 , t, 0, η, τ )a(τ ) dη dτ. 0

0

Using (3.5), (3.6) and the known estimates of the Green function [7], C5 + C6 , G2 (0, t, 0, τ ) ≤ p θ(t) − θ(τ ) we obtain I3 ≤

t

C7 + C8 . √ amin

β−1 2

Taking into account estimates [7] G2 (0, t, h, τ ) ≤ C7 and (3.5), we have I2 ≤ C9 . To estimate I4 we use equality (3.5) and estimates (3.6), (3.7): Z t I4 ≤ C10 G2 (y0 , t, 0, τ )τ β a(τ )dτ 0 ∞ (y − η + 2ml)2 X 1 0 p exp − 4(θ(t) − θ(τ )) θ(t) − θ(τ ) m=−∞ 0 (y + η + 2ml)2 0 + exp − τ β a(τ )dτ 4(θ(t) − θ(τ )) Z θ(t) ∞ (y − η + 2ml)2 C10 1 X 0 √ = √ exp − 4z 2 π 0 z m=−∞ (y + η + 2ml)2 0 + exp − dz 4z

C10 = √ 2 π

Z

t

(3.7)

6


≤ C11

p

EJDE-2018/77

θ(t) ≤ C12 .

The integral I5 is estimated similarly. From (3.5) we obtain I6 ≤ C13 . The above estimates lead us to the inequality a(t) ≥

κ(t) C14 β−1 √ t 2 amin

+ C15

.

Taking into account assumptions on κ(t), we transform it to √ κ0 (t) amin a(t) ≥ , t ∈ [0, T ] β−1 √ C14 + C15 t 2 amin or √ C16 amin ≥ . √ C14 + C17 amin Solving this inequality we obtain an estimate for a(t) from below: a(t) ≥ A0 > 0,

t ∈ [0, T ],

(3.8)

where A0 is determined by given data. Therefore, the a priori estimates of solutions to the equation (3.3) are established. Now we put equation (3.3) in the form a(t) = P a(t),

t ∈ [0, T ].

(3.9)

It follows from (3.6), (3.8) that the operator P maps the set N := {a ∈ C[0, T ] : A0 ≤ a(t) ≤ A1 } into itself. By the Arzela-Ascoli theorem and the above estimates, it is clear that the compactness of P on N will be established if we prove that ∀ε > 0 exists such a δ > 0, that inequality |P a(t2 ) − P a(t1 )| < ε holds if |t1 − t2 | < δ for all t1 , t2 ∈ [0, T ] and for any a ∈ N . Firstly, let show that there exists limt→0 a(t). It follows from the above estimates and assumptions (A2) that the limit of denominator in (3.3) is equal to limt→0 I2 (t). Taking into account the equalities Z l lim G2 (y0 , t, η, τ )(f (0, η, τ ) − µ1τ (η, τ ))dη = f (0, y0 , t) − µ1t (y0 , t), τ →t 0 (3.10) Z t Z 1 1−β dτ dz √ √ =t 2 tβ+1 − τ β+1 1 − z β+1 0 0 and using the mean-value theorem, from (3.3) we have p r κ0 (0) limt→0 a(t∗ ) π lim a(t) = R t→0 β + 1 (f (0, y0 , 0) − µ1t (y0 , 0)) 1 √ 0 where t∗ ∈ [0, t]. From this we obtain κ0 (0) π lim a(t) = R t→0 β + 1 (f (0, y0 , 0) − µ1t (y0 , 0)) 1 √ 0

dz 1−z β+1

2 dz 1−z β+1

Consider the difference |P a(t1 ) − P a(t2 )| =

κ0 (t1 ) β−1 2

t1

ux (0, y0 , t1 )

−

κ0 (t2 ) β−1 2

t2

,

ux (0, y0 , t2 )

.

EJDE-2018/77

INVERSE PROBLEM FOR A DEGENERATE HEAT EQUATION β−1

β−1

|κ0 (t1 )t2 2 ux (0, y0 , t2 ) − κ0 (t2 )t1 2 ux (0, y0 , t1 )|

=

β−1

7

β−1

.

t1 2 ux (0, y0 , t1 )t2 2 ux (0, y0 , t2 ) It follows from (3.4) that β−1

β−1

t1 2 ux (0, y0 , t1 )t2 2 ux (0, y0 , t2 ) ≥ C18 > 0. β−1

β−1

Thus, it is sufficient to estimate the difference ux (0, y0 , t1 )t2 2 − ux (0, y0 , t2 )t1 2 . Consider for example the expression Z Z β−1 t2 l 2 G22 (0, y0 , t2 , 0, η, τ )µ(η, τ ) dη dτ ∆ := t2 0 t1 Z l

0 β−1 2

Z

− t1

G22 (0, y0 , t1 , 0, η, τ )µ(η, τ ) dη dτ ,

0

0

where µ(y, t) is some continuous on [0, l] × [0, T ] function. To find the limit Z tZ l β−1 lim t 2 G22 (0, y0 , t, 0, η, τ )µ(η, τ ) dη dτ, t→0

0

0

we decompose the Green function into two summands (y0 − η)2 1 exp − + G022 (0, y0 , t, 0, η, τ ). G22 (0, y0 , t, 0, η, τ ) = π(θ(t) − θ(τ )) 4(θ(t) − θ(τ )) It follows from the Green function estimates (3.7) that Z tZ l β−1 lim t 2 G022 (0, y0 , t, 0, η, τ )µ(η, τ ) dη dτ = 0. t→0

0

0

On the other hand, using (3.10) we find Z tZ l β−1 1 (y0 − η)2 exp − lim t 2 t→0 4(θ(t) − θ(τ )) 0 0 π(θ(t) − θ(τ )) (y0 + η)2 + exp − µ(η, τ ) dη dτ 4(θ(t) − θ(τ )) r Z β + 1 µ(y0 , 0) 1 dz p √ . = π 1 − z β+1 a(0) 0 It means that ∀ε > 0 there exists such t0 ∈ (0, T ] that r Z Z Z β−1 t l β + 1 µ(y0 , 0) 1 dz t 2 < ε, p √ G22 (0, y0 , t, 0, η, τ )µ(η, τ ) dη dτ − π 1 − z β+1 a(0) 0 0 0 when 0 < t < t0 . Therefore, if ti < t0 , i ∈ {1, 2}, then ∆ < 2ε. Consider the case when ti ≥ t0 , i ∈ {1, 2} and suppose for the definiteness that t1 < t2 . Present ∆ as follows: Z Z β−1 t2 l G22 (0, y0 , t2 , 0, η, τ )µ(η, τ ) dη dτ ∆ = t2 2 t1

Z

t1

Z

+ 0

0 l

β−1 β−1 t2 2 G22 (0, y0 , t2 , 0, η, τ ) − t1 2 G22 (0, y0 , t1 , 0, η, τ ) µ(η, τ ) dη dτ .

0

(3.11)

8


EJDE-2018/77

Using (3.7), we obtain Z

β−1 t 2

t2

Z

2

G22 (0, y0 , t2 , 0, η, τ )µ(η, τ ) dη dτ

0 t2 Z l

t1 β−1 2

l

Z

≤ t2

C19

+ C20 dτ

p

θ(t2 ) − θ(τ ) Z t2 β−1 2 q ≤ C21 |t2 − t1 | + C22 t2 t1

0

t1

Z = C21 |t2 − t1 | + C22

√

dz 1 − z β+1

√

dz 1−z

t1 t2

Z ≤ C21 |t2 − t1 | + C22

1

1 t1 t2

dτ tβ+1 2

− τ β+1

√ ≤ C21 |t2 − t1 | + C23 t2 − t1 . Consider the second summand from (3.11): Z Z β−1 t1 l β−1 t2 2 G22 (0, y0 , t2 , 0, η, τ ) − t1 2 G22 (0, y0 , t1 , 0, η, τ ) µ(η, τ ) dη dτ 0

0

β−1 β−1 ≤ t2 2 − t1 2

Z

t1

Z

0

Z

β−1 2

t1

l

G22 (0, y0 , t2 , 0, η, τ )µ(η, τ ) dη dτ

0

l

Z

|G22 (0, y0 , t2 , 0, η, τ ) − G22 (0, y0 , t1 , 0, η, τ )||µ(η, τ )| dη dτ

+ t1

0

0

:= ∆1 + ∆2 . Substituting the product of two one-dimensional Green functions instead of the Green function G22 (0, y0 , t2 , 0, η, τ ) and taking into account (3.5), (3.7), we find Z t1 Z l β−1 β−1 2 2 G2 (0, t2 , 0, τ )G2 (y0 , t2 , η, τ ) dη dτ ∆1 ≤ C24 t2 − t1 0 0 Z t1 β−1 β−1 ≤ C24 t2 2 − t1 2 G2 (0, t2 , 0, τ )dτ 0 β−1 1−β β−1 t2 β−1 2 − 1 < , ≤ C25 t2 2 − t1 2 t1 2 = C25 t1

when |t2 − t1 | < δ1 . By the same reasoning we obtain Z t1 Z l β−1 ∆2 ≤ C24 t1 2 |G2 (0, t2 , 0, τ ) − G2 (0, t1 , 0, τ )|G2 (y0 , t2 , η, τ ) dη dτ 0

Z

t1

Z

0

l

G2 (0, t1 , 0, τ )|G2 (y0 , t2 , η, τ ) − G2 (y0 , t1 , η, τ )| dη dτ 0 0 Z t1 β−1 2 |G2 (0, t2 , 0, τ ) − G2 (0, t1 , 0, τ )|dτ ≤ C24 t1 +

0

Z

t1

Z

l

G2 (0, t1 , 0, τ )|G2 (y0 , t2 , η, τ ) − G2 (y0 , t1 , η, τ )| dη dτ

+ 0

0

:= ∆2,1 + ∆2,2 .

EJDE-2018/77


9

Now we put the Green function G2 (0, t, 0, τ ) in the form G2 (0, t, 0, τ ) = p

1 π(θ(t) − θ(τ ))

+p

2

∞ X

π(θ(t) − θ(τ )) n=1

exp −

n2 h2 . θ(t) − θ(τ )

Then 1 Z t1 1 1 √ p −p dτ π 0 θ(t1 ) − θ(τ ) θ(t2 ) − θ(τ ) Z t1 X ∞ n2 h2 2 p exp − + θ(t1 ) − θ(τ ) π(θ(t1 ) − θ(τ )) 0 n=1 2 2 n h 2 dτ . exp − −p θ(t2 ) − θ(τ ) π(θ(t2 ) − θ(τ )) β−1

∆2,1 ≤ C24 t1 2

(3.12)

Now we transform and estimate the first summand: Z t1 β−1 1 1 2 p C26 t1 −p dτ θ(t1 ) − θ(τ ) θ(t2 ) − θ(τ ) 0 Z t1 β−1 dτ θ(t ) − θ(t1 ) p 2 p p = C26 t1 2 (θ(t ) − θ(τ )) (θ(t ) − θ(τ ))( θ(t ) θ(t2 ) − θ(τ ) 0 1 2 1 − θ(τ ) + Z β+1 β+1 t 1 β−1 (t2 − t1 )dτ q q ≤ C27 t1 2 q β+1 β+1 0 β+1 + β+1 tβ+1 − τ t − τ t1 − τ β+1 tβ+1 − τ β+1 1 2 2 Z t1 β−1 1 1 q = C27 t1 2 −q dτ. β+1 0 β+1 β+1 tβ+1 t − τ − τ 1 2 After the change of variable z = τ /t1 we obtain Z t1 β−1 1 1 p −p dτ C26 t1 2 θ(t1 ) − θ(τ ) θ(t2 ) − θ(τ ) 0 Z 1 1 √ ≤ C27 1 − z β+1 0 1 t2 −q dz < , when | − 1| < δ2 , t2 β+1 t β+1 1 ( ) −z t1

as the function Z I(σ) := 0

1

√

dz σ − z β+1

is continuous for σ ≥ 1. Now we estimate the second summand from (3.12): Z t1 X β−1 ∞ 2 n2 h2 p C26 t1 2 exp − θ(t1 ) − θ(τ ) π(θ(t1 ) − θ(τ )) 0 n=1 2 2 2 n h dτ −p exp − θ(t2 ) − θ(τ ) π(θ(t2 ) − θ(τ )) Z t1 Z θ(t1 )−θ(τ ) ∞ n2 h2 ∂ 1 X 2C26 β−1 2 √ = √ t1 exp − dz dτ z π z n=1 0 θ(t2 )−θ(τ ) ∂z

(3.13)

10


EJDE-2018/77

) < , − tβ+1 ≤ C28 (θ(t2 ) − θ(t1 )) ≤ C29 (tβ+1 1 2

when |t2 − t1 | < δ3 .

To estimate ∆2,2 , we extract the main term from the Green function, Z t1 β−1 ∆2,2 ≤ C24 t1 2 G2 (0, t1 , 0, τ )dτ 0 l

(η − y0 )2 4(θ(t2 ) − θ(τ )) θ(t2 ) − θ(τ ) 0 1 (η − y0 )2 −p exp − dη 4(θ(t1 ) − θ(τ )) θ(t1 ) − θ(τ ) Z t1 Z l + G2 (0, t1 , 0, τ )dτ |G02 (y0 , t2 , η, τ ) − G02 (y0 , t1 , η, τ )|dη . Z

×

p

1

exp −

0

(3.14)

0

G02 (y0 , t, η, τ )

Taking into account that has no singularity, we find Z l Z l Z t2 0 0 |G2 (y0 , t2 , η, τ ) − G2 (y0 , t1 , η, τ )|dη ≤ G02t (y0 , t, η, τ )dt dη 0

0

t1

≤ C30 |t1 − t2 |. We consider the expression Z l 1 (η − y0 )2 p exp − 4(θ(t2 ) − θ(τ )) θ(t2 ) − θ(τ ) 0 1 (η − y0 )2 −p dη exp − 4(θ(t1 ) − θ(τ )) θ(t1 ) − θ(τ ) Z l (η − y0 )2 1 p ≤ exp − 4(θ(t2 ) − θ(τ )) θ(t2 ) − θ(τ ) 0 2 (η − y0 ) − exp − dη 4(θ(t1 ) − θ(τ )) Z l 1 1 (η − y0 )2 p −p dη . + exp − 4(θ(t1 ) − θ(τ )) θ(t1 ) − θ(τ ) θ(t2 ) − θ(τ ) 0 After the change of variable z = √ 2

l

η−y0 θ(t2 )−θ(τ )

in the first summand we have

(η − y0 )2 4(θ(t2 ) − θ(τ )) θ(t2 ) − θ(τ ) 0 1 (η − y0 )2 −p exp − dη 4(θ(t1 ) − θ(τ )) θ(t1 ) − θ(τ ) Z √ l−y0 2 θ(t2 )−θ(τ ) θ(t2 ) − θ(τ ) =2 exp(−z 2 ) − exp − z 2 dz θ(t1 ) − θ(τ ) √ −y0 2 θ(t2 )−θ(τ ) Z ∞ θ(t2 ) − θ(τ ) ≤2 exp(−z 2 ) − exp − z 2 dz θ(t1 ) − θ(τ ) −∞ p √ θ(t1 ) − θ(τ ) =2 π 1− p . θ(t2 ) − θ(τ )

Z

p

1

exp −

(3.15)

EJDE-2018/77


The change of variable z = √ 2

l

η−y0 θ(t1 )−θ(τ )

11

in the second summand gives

(η − y0 )2 1 1 p −p dη 4(θ(t1 ) − θ(τ )) θ(t1 ) − θ(τ ) θ(t2 ) − θ(τ ) 0 p l−y0 Z θ(t1 ) − θ(τ ) 2√θ(t2 )−θ(τ ) −z2 =2 1− p e dz θ(t2 ) − θ(τ ) √ −y0 2 θ(t2 )−θ(τ ) p √ θ(t1 ) − θ(τ ) ≤2 π 1− p . θ(t2 ) − θ(τ )

Z

exp −

Now we return to the estimate of the first summand in (3.14), using (3.7) and (3.13), Z t1 Z l β−1 1 (η − y0 )2 p G2 (0, t1 , 0, τ )dτ exp − C24 t1 2 4(θ(t2 ) − θ(τ )) θ(t2 ) − θ(τ ) 0 0 2 1 (η − y0 ) dη −p exp − 4(θ(t1 ) − θ(τ )) θ(t1 ) − θ(τ ) Z t1 β−1 1 1 2 p ≤ C31 t1 −p dτ θ(t1 ) − θ(τ ) θ(t2 ) − θ(τ ) 0 p Z t1 β−1 θ(t1 ) − θ(τ ) 2 + C32 t1 dτ < , 1− p θ(t2 ) − θ(τ ) 0 when | tt21 − 1| < δ4 . So, we obtained the estimate of ∆. Other expressions from P a(t) are estimated by a similar way. Therefore, the conditions of Schauder fixed-point theorem are satisfied for (3.9), and the existence of the solution to the problem (2.1)–(2.4) is proved. 4. Uniqueness of the solution Consider the function H(t) := √

√ πκ(t) R t µ0 (τ )dτ β+1 0 √

,

tβ+1 −τ β+1

where Z

l

G2 (y0 , t, η, τ )(f (0, η, τ ) − µ1τ (η, τ ))dη.

µ0 (t) :=

(4.1)

0

It is easy to see that there exists the limit √ πκ0 (0) lim H(t) = √ Rt t→+0 β + 1µ0 (0) 0 √

dσ 1−σ β+1

> 0.

It means that H(t) is continuous function on [0, T ]. Using the function H(t), we will establish the estimates of solutions for the equation (3.3). Taking into account assumption (A2), from (3.3) we have a(t) ≤ R t R l 0

0

κ(t) G22 (0, y0 , t, 0, η, τ )(f (0, η, τ ) − µ1τ (η, τ )) dη dτ

12


EJDE-2018/77

p κ(t) π˜ amax (t) = Rt ≤√ R t µ0 (τ )dτ G2 (0, t, 0, τ )µ0 (τ )dτ β + 1 0 √ β+1 β+1 0 t −τ p ≤ a ˜max (t)Hmax (t), κ(t)

where a ˜max (t) := max[0,t] a(τ ), Hmax (t) := max[0,t] H(τ ). From this follows that 2 a(t) ≤ Hmax (t),

t ∈ [0, T ].

(4.2)

To estimate a(t) from below, we put the denominator from (3.3) as a sum R t µ0 (τ )dτ √ + S(t). Using the notation a ˜min (t) := min[0,t] a(τ ), Hmin (t) := 0

√1 π

θ(t)−θ(τ )

min[0,t] H(τ ) we obtain √ Z t Z t µ (τ )dτ µ0 (τ )dτ β+1 κ(t) 1 p √ p √ 0 ≤p . = β+1 β+1 π 0 t −τ θ(t) − θ(τ ) π˜ amin (t) 0 H(t) a ˜min (t) Repeating the reasons of the estimation of the denominator of (3.3) from above, we establish that S(t) ≤ C33 . Then p ˜min (t) H(t) a κ(t) √ a(t) ≥ = . κ(t) C H(t) a ˜min (t) √ + C33 1 + 33 H(t)

a ˜min (t)

κ(t)

It follows from the assumptions (A2) and the estimate (4.2) that p β−1 C33 H(t) a ˜min (t) ≤ C34 t 2 . κ(t) Therefore, a(t) ≥

p ˜min (t) Hmin (t) a 1 + C34 t

β−1 2

,

t ∈ [0, T ],

or a(t) ≥

2 Hmin (t)

(1 + C34 t

β−1 2

)2

,

t ∈ [0, T ].

(4.3)

Having the estimates (4.2) and (4.3) of solutions of equation (3.3), we can pass to the proof of the uniqueness of solution. Put the equation (3.3) in the form a(t) =

√1 π

Rt 0

κ(t) , √µ0 (τ )dτ + S(t, a(t))

t ∈ [0, T ].

(4.4)

θ(t)−θ(τ )

Suppose that this equation has two solutions a1 (t) and a2 (t). Denote b(t) := a1 (t)− a2 (t). The function b(t) satisfies the equation 1 Z t 1 1 p b(t) = κ(t) √ −p µ0 (τ )dτ + S(t, a2 (t)) π 0 θ2 (t) − θ2 (τ ) θ1 (t) − θ1 (τ ) 1 Z t −1 µ (τ )dτ p 0 − S(t, a1 (t)) √ + S(t, a1 (t)) π 0 θ1 (t) − θ1 (τ ) −1 1 Z t µ (τ )dτ p 0 × √ + S(t, a2 (t)) . π 0 θ2 (t) − θ2 (τ ) (4.5)

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13

It is clear from above that S(t, ai (t)) ≥ 0, i ∈ {1, 2}. Then it follows from (4.5) that |b(t)| 1 Z t 1 1 p µ0 (τ )dτ + |S(t, a2 (t)) ≤ κ(t) √ −p π 0 θ2 (t) − θ2 (τ ) θ1 (t) − θ1 (τ ) Z t 1 Z t −1 µ0 (τ )dτ µ (τ )dτ p p 0 − S(t, a1 (t))| . π 0 θ1 (t) − θ1 (τ ) 0 θ2 (t) − θ2 (τ )

(4.6)

Transform the expression 1 1 −p θ2 (t) − θ2 (τ ) θ1 (t) − θ1 (τ ) θ2 (t) − θ2 (τ ) − (θ1 (t) − θ1 (τ )) 1 p p =p . (θ1 (t) − θ1 (τ ))(θ2 (t) − θ2 (τ )) θ1 (t) − θ1 (τ ) + θ2 (t) − θ2 (τ )

p

Using estimate (4.3), we find p

1 θ2 (t) − θ2 (τ )

−p

√

1 θ1 (t) − θ1 (τ )

≤

β−1

β + 1(1 + C34 t 2 )3 √ b (t), 3 (t) tβ+1 − τ β+1 max 2Hmin

where bmax (t) := max[0,t] |b(τ )|. Therefore, we obtain Z 1 κ(t) t 1 µ0 (τ )dτ √ p −p π 0 θ2 (t) − θ2 (τ ) θ1 (t) − θ1 (τ ) κ2 (t)(1 + C34 t ≤ 4 (t) 2Hmin

β−1 2

)3

(4.7)

bmax (t).

Now we estimate the difference S(t, a2 (t)) − S(t, a1 (t)). Consider Z Z ∞ (ξ + 2nh)2 X 1 h l 1 ∆1 R := exp − 2π 0 0 θ2 (t) m,n=−∞ 4θ2 (t) (y − η + 2ml)2 (y + η + 2ml)2 0 0 × exp − + exp − 4θ2 (t) 4θ2 (t) ∞ 2 X (ξ + 2nh) 1 (y0 − η + 2ml)2 exp − exp − − θ1 (t) m,n=−∞ 4θ1 (t) 4θ1 (t) (y + η + 2ml)2 0 + exp − ϕξ (ξ, η)dηdξ 4θ1 (t) Z h Z l Z θ2 (t) ∞ (ξ + 2nh)2 ∂ 1 X ≤C35 exp − 4z 0 0 θ1 (t) ∂z z m,n=−∞ (y + η + 2ml)2 (y − η + 2ml)2 0 0 × exp − + exp − dz dηdξ. 4z 4z After differentiating and using (3.5) we obtain ∞ Z θ1 (t) Z h Z l 1 (ξ + 2nh)2 X ∆1 R ≤ C36 dz exp − z 2 m,n=−∞ 4z θ2 (t) 0 0 (y − η + 2ml)2 (y + η + 2ml)2 0 0 × exp − + exp − 4z 4z

14


EJDE-2018/77

∞ (ξ + 2nh)2 X 1 2 (ξ + 2nh) exp − z 3 m,n=−∞ 4z (y − η + 2ml)2 0 × (y0 − η + 2ml)2 exp − + (y0 + η + 2ml)2 4z (y + η + 2ml)2 0 × exp − dηdξ 4z Z θ1 (t) dz ≤ C37 |θ1 (t) − θ2 (t)| . ≤ C37 min{θ1 (t), θ2 (t)} θ2 (t) z

+

Applying (4.3), we arrive at the estimate (1 + C t β−1 2 34 2 ) ∆1 R ≤ C38 + 1 tβ−1 bmax (t) ≤ C39 bmax (t). 2 (t) tβ−1 Hmin Now we estimate the expression Z ∞ 1 n2 h2 1 1 t X p exp − −p ∆2 R := √ θ2 (t) − θ2 (τ ) π 0 n=1 θ2 (t) − θ2 (τ ) θ1 (t) − θ1 (τ ) 2 2 n h × exp − µ0 (τ )dτ θ1 (t) − θ1 (τ ) Z t Z ∞ n2 h2 θ2 (t)−θ2 (τ ) ∂ 1 X dz ≤ C40 dτ exp − ∂z z n=1 z 0 θ1 (t)−θ1 (τ ) ≤ C41 |θ1 (t) − θ2 (t)| ≤ C43 bmax (t). Other summands from S(t, a2 (t)) − S(t, a1 (t)) are estimated in a similar way. Hence, |S(t, a1 (t)) − S(t, a2 (t))| ≤ C42 bmax (t). (4.8) To estimate the denominator in (4.6), note that p κ(t) amax (t) κ(t) √ ≤ R t µ0 (τ )dτ R β+1 t √ µ0 (τ )dτ √1 √ √ π 0 0 θi (t)−θi (τ )

π

2 ≤ Hmax (t),

(4.9)

tβ+1 −τ β+1

for i ∈ {1, 2}. By applying estimates (4.7)–(4.9) to (4.6), we obtain (1 + C t β−1 3 β−1 34 2 ) 4 2 + C t Hmax (t)bmax (t) 43 4 (t) 2Hmin (1 + C t β−1 3 4 β−1 34 2 ) Hmax (t) 4 ≤ + C42 t 2 Hmax (t) bmax (t). 4 2Hmin (t)

|b(t)| ≤

(4.10)

As limt→0 Hmin (t) = limt→0 Hmax (t), there exists such a number t0 ∈ (0, T ] that the following inequality holds: (1 + C t β−1 3 4 β−1 34 2 ) Hmax (t) 2 H4 + C t (t) < 1, t ∈ [0, t0 ]. (4.11) 42 max 4 (t) 2Hmin Then we conclude from (4.10) that bmax (t) ≤ 0, t ∈ [0, t0 ], that is impossible. Therefore, b(t) ≡ 0, t ∈ [0, t0 ]. It follows from the uniqueness of solution to problem (2.1)–(2.5) that the function u(x, y, t) is uniquely defined in Qt0 .

EJDE-2018/77


15

Now let us show that the solution to (2.1)–(2.5) is unique in the whole domain QT . Suppose that there exist two solutions (ai (t), ui (x, y, t)), i ∈ {1, 2} to the problem. For their difference a := a1 − a2 , u := u1 − u2 we obtain the problem ut = tβ a1 (t)∆u + tβ a(t)∆u2 (x, y, t), u(x, y, 0) = 0, u(0, y, t) = u(h, y, t) = 0, uy (x, 0, t) = uy (x, l, t) = 0,

(x, y, t) ∈ QT ,

(x, y) ∈ D,

(4.12) (4.13)

(y, t) ∈ [0, l] × [0, T ],

(4.14)

(x, t) ∈ [0, h] × [0, T ],

(4.15)

a1 (t)ux (0, y0 , t) = −a(t)u2x (0, y0 , t),

t ∈ [0, T ].

(4.16)

˜ 12 (x, y, t, ξ, η, τ ) of the problem (4.12)–(4.15), we find Using the Green function G that Z tZ ˜ 12 (x, y, t, ξ, η, τ )τ β a(τ )∆u2 (ξ, η, τ ) dξ dη dτ. G (4.17) u(x, y, t) = D

0

Substituting it into the overdetermination condition (4.16): Z tZ lZ h a1 (t) ˜ 12 (0, y0 , t, ξ, η, τ )τ β a(τ )∆u2 (ξ, η, τ ) dξ dη dτ, G a(t) = − x u2x (0, y0 , t) 0 0 0 (4.18) for t ∈ [0, T ]. Thus, we obtain a homogeneous Volterra integral equation of the second kind for a(t). Put it as Z t a(t) = K(t, τ )a(τ )dτ, t ∈ [0, T ]. (4.19) 0

To study the behavior of the kernel K(t, τ ), we find Z lZ h ˆ 12 (x, y, t, ξ, η, 0)ϕ(ξ, η)dξdη u2 (x, y, t) = G 0

0

Z tZ

l

ˆ 12 (x, y, t, 0, η, τ )τ β a(τ )µ1 (η, τ ) dη dτ G ξ

+ 0

0

Z tZ

l

ˆ 12 (x, y, t, h, η, τ )τ β a(τ )µ2 (η, τ ) dη dτ G ξ

− 0

0

Z tZ

ˆ 12 (x, y, t, ξ, 0, τ )τ β a(τ )ν1 (ξ, τ )dξdτ G

− 0

0

Z tZ + 0

(4.20)

h

h

ˆ 12 (x, y, t, ξ, l, τ )τ β a(τ )ν2 (ξ, τ )dξdτ G

0

Z tZ lZ + 0

0

h

ˆ 12 (x, y, t, ξ, η, τ )f (ξ, η, τ ) dξ dη dτ, G

0

ˆ 12 (x, y, t, ξ, η, τ ) is the Green function of problem (2.2)–(2.4) for the equawhere G tion ut = tβ a2 (t)∆u + f (x, y, t). By differentiating (4.20) twice by x, y and integrating by parts, we obtain u2xx (x, y, t)

16


Z lZ

h

ˆ 12 (x, y, t, ξ, η, 0)ϕξξ (ξ, η)dξdη G

=

0 tZ l

0

Z

EJDE-2018/77

+

ˆ 12 (x, y, t, 0, η, τ )(µ1 (η, τ ) − τ β a2 (τ )µ1 (η, τ ) − f (0, η, τ )) dη dτ G τ ηη ξ

0

0

Z tZ

l

ˆ 12 (x, y, t, h, η, τ )(µ2 (η, τ ) − τ β a2 (τ )µ2 (η, τ ) − f (h, η, τ )) dη dτ G τ ηη ξ

− 0

0

Z tZ

h

ˆ 12 (x, y, t, ξ, 0, τ )tβ a2 (τ )ν1 (ξ, τ )dξdτ G ξξ

− 0

0

Z tZ

h

ˆ 12 (x, y, t, ξ, l, τ )τ β a2 (τ )ν2 (ξ, τ )dξdτ G ξξ

+ 0

0

h

Z tZ lZ

ˆ 12 (x, y, t, ξ, η, τ )fξξ (ξ, η, )τ ) dξ dη dτ, G

+ 0

0

0

u2yy (x, y, t) Z lZ h ˆ 12 (x, y, t, ξ, η, 0)ϕηη (ξ, η)dξdη = G 0

0

Z tZ

l

ˆ 12 (x, y, t, 0, η, τ )τ β a2 (τ )µ1 (η, τ ) dη dτ G ηη ξ

+ 0

0

Z tZ

l

ˆ 12 (x, y, t, h, η, τ )τ β a2 (τ )µ2 (η, τ ) dη dτ G ηη ξ

− 0

0

Z tZ

h

ˆ 12 (x, y, t, ξ, 0, τ )(ν1 (η, τ ) − tβ a2 (τ )ν1 (ξ, τ ) − f (ξ, 0, τ ))dξdτ G τ ξξ

− 0

0

Z tZ + 0

h

ˆ 12 (x, y, t, ξ, 0, τ )(ν2 (η, τ ) − tβ a2 (τ )ν2 (ξ, τ ) − f (ξ, l, τ ))dξdτ G τ ξξ

0

Z tZ lZ + 0

0

h

ˆ 12 (x, y, t, ξ, η, τ )fηη (ξ, η, )τ ) dξ dη dτ. G

0

From the above expression, we establish the following estimate for the kernel of (4.19)): C44 |K(t, τ )| ≤ p . t(t − τ ) This means that 1 K(t, τ ) ≡ p K1 (t, τ ), t(t − τ ) and the equation (4.19)) can be presented as Z t 1 p a(t) = K1 (t, τ )a(τ )dτ, t(t − τ ) 0

t ∈ [0, T ],

(4.21)

where |K1 (t, τ )| ≤ C44 . It was proved that there exists such an interval [0, t0 ] where a(t) ≡ 0. Then the equation (4.21)) can be transformed to the form Z t 1 p a(t) = K1 (t, τ )a(τ )dτ, t ∈ [t0 , T ], (4.22) t(t − τ ) t0

EJDE-2018/77


and the following estimate holds: C45 p 1 K1 (t, τ ) ≤ √ , t−τ t(t − τ )

17

t ∈ [t0 , T ].

Then the properties of the integral Volterra equations of the second kind imply that equation (4.22)) has only the trivial solution. Hence, a(t) ≡ 0, t ∈ [0, T ]. Using it in the equation (4.12)), we obtain u(x, y, t) ≡ 0, (x, y, t) ∈ QT because of the the uniqueness of solution of the problem (4.12)–(4.14) [14]. The proof of Theorem 2.1 is complete. References [1] H. Berestycki, J. Busca, I. Florent; An inverse parabolic problem arising in finance, C. R. Acad. Sci. Paris, 2000, 331, 965-969. [2] L. Caffarelli, A. Friedman; Continuity of the density of a gas flow in a porous medium, Trans. Amer. Math. Soc., 1979, 252, 99-113. [3] E. DiBenedetto; Degenerate parabolic equations, Springer, New York, 1993. [4] V. Grebenev; On a system of degenerate parabolic equations that arises in fluid dynamics, Sib. Mat. J., 1994, 35, 753-767. [5] N.Hryntsiv; Inverse problem for a degenerate heat equation in a free boundary domain, Math. Methods and Phys.-Mech. Fields, 2006, 49, 28-40. (Ukranian) [6] N. Hryntsiv; Inverse problem for a strongly degenerate parabolic equation in a free boundary domain, Visnyk Lviv. Univer. Ser. Mech.-Math., 2007, 67, 84-97. (Ukranian) [7] M. Ivanchov; Inverse problems for equations of parabolic type, Lviv, VNTL Publishers, 2003. [8] M. Ivanchov, N. Saldina; Inverse problem for a degenerate heat equation, Ukrainian Math. Journal, 2005, 57, 1563-1570. (Ukranian) [9] M. Ivanchov, N. Saldina; Inverse problem for a parabolic equation with strong power degeneration, Ukrainian Math. Journal, 2006, 58, 1487-1500. (Ukranian) [10] M. Ivanchov, N. Saldina; Inverse problem for strongly degenerate heat equation , J. Inv. Ill-Posed Problems, 2006, 14, 465-480. [11] M. Ivanchov, V. Vlasov; Inverse problem for a weakly degenerate two-dimensional heat equation, Visnyk Lviv. Univer. Ser. Mech.-Math., 2009, 70, 91-102. (Ukranian) [12] M. Ivanchov, A. Lorenzi, N. Saldina; Solving a scalar degenerate multidimensional identification problem in Banach space, J. Inv. Ill-Posed Problems, 2008, 16, 397-415. [13] V. Isakov; Inverse problems for partial differential equations. Third edition, Applied Mathematical Sciences, 127, Springer, 2017. [14] O. A. Ladyzhenskaya, V. A. Solonnikov, N. N. Uraltseva; Linear and quasilinear equations of the parabolic type, Moscow, Nauka, 1967. [15] A. I. Prilepko, D. G. Orlovsky, I. A. Vasin; Methods for solving inverse problems in mathematical physics, Marcel Dekker, New York, 1999. [16] N. V. Saldina; Inverse problem for a weakly degenerate parabolic equation, Math. Methods and Phys.-Mech. Fields, 2006, 49, 7-7. (Ukranian) [17] N. V. Saldina; Strongly degenerate problem with arbitrary behavior of coefficients, Visnyk Lviv. Univer. Ser. Mech.-Math., 2006, 66, 186-202. (Ukranian) [18] N. V. Saldina; Strongly degenerate inverse parabolic problem with arbitrary behavior of minor coefficients of the equation, Nauk. Visnyk Uzhgorod. Univ. Ser. Math and Informat, 2007, 12-13, 109-122. (Ukranian) Mykola Ivanchov Department of Differential Equations, Ivan Franko National University of Lviv, Universytetska st., 1, 79000, Ukraine E-mail address: [email protected] Vitaliy Vlasov Department of Differential Equations, Ivan Franko National University of Lviv, Universytetska st., 1, 79000, Ukraine E-mail address: [email protected]