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1 Math-Net.Ru All Russian mathematical portal Ekaterina N. Kriger, Igor V. Frolenkov, An identification problem of nonlinear lowest term coefficient in the special form for two-dimensional semilinear parabolic equation, J. Sib. Fed. Univ. Math. Phys., 016, Volume 9, Issue, DOI: Use of the all-russian mathematical portal Math-Net.Ru implies that you have read and agreed to these terms of use Download details: IP: April 7, 018, 00:18:14

2 Journal of Siberian Federal University. Mathematics & Physics 016, 9, УДК An Identification Problem of Nonlinear Lowest Term Coefficient in the Special Form for Two-Dimensional Semilinear Parabolic Equation Ekaterina N. Kriger Igor V. Frolenkov Institute of Mathematics and Computer Science Siberian Federal University Svobodny, 79, Krasnoyarsk, Russia Received , received in revised form , accepted In this paper we investigate an identification problem of a coefficient at the nonlinear lowest term in a D semilinear parabolic equation with overdetermination conditions given on a smooth curve. The unknown coefficient has the form of a product of two functions each depending on time and a spatial variable. We prove solvability of the problem in classes of smooth bounded functions. We present an example of input data satisfying the conditions of the theorem and the corresponding solution. Keywords: inverse problem, semilinear parabolic equation, Cauchy problem, lowest term coefficient, weak approximation method, local solvability, overdetermination conditions on a smooth curve. DOI: / We consider an identification problem of a special coefficient at the nonlinear term in a twodimensional semilinear parabolic equation with Cauchy data. The desired coefficient depends on all variables and has the form of a product of two functions each depending on time and a spatial variable. Overdetermination conditions are given on two smooth curves. By means of overdetermination conditions the given inverse problem is reduced to a nonclassical direct problem for a loaded parabolic equation. Solvability of the direct problem is proved by the weak approximation method [1 3]. Efficiency of the weak approximation method the splitting method at differential level is that, as a rule, on every fractional step the split problem is simpler than the original one. Therefore we can estimate the solution of the split problem more precisely, obtain a priori estimates and establish its solvability. For the original inverse problem we prove a theorem of the solution existence in classes of smooth bounded functions. Earlier, the problems of coefficient identification for semilinear parabolic equations with Cauchy data have been studied in papers [4, 5]. In [6] the problem of identification of two coefficients in a semilinear parabolic equation with overdetermination conditions on smooth curves has been considered. Solvability of the identification problem of a coefficient represented as a sum of two functions at a nonlinear term in a semilinear parabolic equation has been investigated in [7]. In paper [8] unique solvability of an inverse boundary-value problem for an one-dimensional semilinear parabolic equation with the sought for coefficient ft + gx at the lowest term has e_katherina@mail.ru igor@frolenkov.ru c Siberian Federal University. All rights reserved 180

3 been proved. The boundary-value problem of identification of the coefficient at ut, x depending on t, x in a parabolic equation has been considered in [9]. In [10] we proved existence and uniqueness of solution of the identification problem of the coefficient λt, x, z = λ 1 t, x λ t, z at the source function in a two-dimensional parabolic equation with Cauchy data. The identification problems of coefficients of other forms at the source function in parabolic equations have been studied in papers [11, 1]. 1. The problem definition In the strip G [0,T ] = { t, x, z 0 t T, x, z R } we consider a Cauchy problem for the parabolic equation u t = L ut, x, z + u p λt, x, z + ft, x, z, t, x, z G 0,T ], 1 L ut, x, z = α 1 t u xx t, x, z + α t u zz t, x, z + β 1 t u x t, x, z + β t u z t, x, z, with the initial condition u0, x, z = u 0 x, z, x, z R. 3 Here the degree p 1 is an integer constant. The coefficients α k t α 0 > 0, β k t, k = 1,, are real-valued, continuous and bounded in [0, T ] functions. Together with the function ut, x, z we also need to determine the function λt, x, z = λ 1 t, x λ t, z. 4 Let the function ut, x, z satisfies overdetermination conditions u t, x, at = φt, x, u t, bt, z = ψt, z, 5 where the curves at, bt C 1 [0, T ]. Suppose the compatibility conditions are fulfilled u 0 x, a0 = φ0, x, u0 b0, z = ψ0, z, φ t, bt = ψ t, at. 6 Assume also that the ft, x, z, u 0 x, z, φt, x, ψt, z defined in G [0,T ], R, [0, T ] R, [0, T ] R, respectively, are continuous, have continuous derivatives occurring in the relation 7 and satisfy it: l 1 t l 1 x k φt, x 1 + l k t l z k ψt, z + k x k 1 z k u 0x, z + + x k 1 k z k ft, x, z C, l 1, l = 0, 1, k 1, k = 0, 1,..., 6. 7 Let also the following constraints on the initial data hold φt, x δ 1 > 0, ψt, z δ > 0, t, x, z G [0,T ], 8 φ t t, bt α1 t φ xx t, bt α t ψ zz t, at β1 t φ x t, bt where δ 1, δ, δ 3 are constants. β t ψ z t, at 1 + a t f t, bt, at δ3, t [0, T ], 9 181

4 . Reduction to the direct problem Let us reduce the inverse problem 1 5 to the auxiliary direct problem. For this purpose in equation 1 we set x = bt to express the coefficient λ 1 t, bt λ t, z. After that in 1 we set z = at to express the coefficient λ 1 t, x λ t, at. In the product of these two expressions we substitute the expression for λ 1 t, btλ t, at obtained from 1 by simultaneous substitution of x = bt, z = at. By 5 we obtain the following expression for the unknown coefficient: λt, x, z = φp t, bt φ t t, x α 1 tφ xx t, x β 1 tφ x t, x ft, x, at æt φ p t, x α tu zz t, x, at + β tu z t, x, at 1 + a t ψ t t, z α tψ zz t, z φ p t, x ψ p t, z β tψ z t, z + ft, bt, z ψ p α 1tu xx t, bt, z + β 1 tu x t, bt, z 1 + b t t, z ψ p, 10 t, z where æt = φ t t, bt α 1 t φ xx t, bt α t ψ zz t, at β 1 t φ x t, bt β t ψ z t, at 1 + a t ft, bt, at. We transform relation 10 to the form λt, x, z = A 0 t A 1 t, x + A t, x u zz t, x, at + A 3 t, x u z t, x, at B 1 t, z + B t, z u xx t, bt, z + B 3 t, z u x t, bt, z, 11 where the functions A 0 t, A i t, x, B i t, z, i = 1,, 3, are known, depend on the initial data and given by the following formulas: A 0 t = φp t, bt æt ; A 1 t, x = φ tt, x α 1 tφ xx t, x β 1 tφ x t, x ft, x, at φ p ; t, x A t, x = α t φ p t, x ; B 1t, z = ψ tt, z α tψ zz t, z β tψ z t, z ft, bt, z ψ p ; t, z A 3 t, x = β t 1 + a t φ p ; B t, z = α 1t t, x ψ p t, z ; B 3t, z = β 1t 1 + b t ψ p. t, z Substituting the relation 11 to equation 1 we arrive at the direct problem for the equation u t t, x, z = Lut, x, z + λt, x, z u p t, x, z + ft, x, z, t, x, z G 0,T ], 13 with the initial condition 3 and the coefficient λt, x, z of the form Existence of solution of the direct problem To prove solvability of the direct problem 13, 3 we use the weak approximation method [1 3]. We split equation 13 into two fractional steps and make a shift by in nonlinear terms and traces of unknown functions. 1 u t = Lu t, x, z, n < t n + 1 ; 14 18

5 u t = u t, x, z pm t, x, z, u t, x, z +ft, x, z, n + 1 < t n+1; 15 u 0, x, z = u 0 x, z, 16 n = 0, 1,,..., N 1, N = T. Here the coefficient at the nonlinear term has the following form: M t, x, z, u t, x, z = B 1 t, z+b t, zu xxt, bt, z+b 3t, zu xt, bt, z A 1 t, x + A t, x u zzt, x, at + A 3t, x u zt, x, at A 0 t. 17 By the n-th complete step we mean the half-interval n, n+1 ], whereas the j-th fractional step of the n-th complete step is the half-interval n + j 1/, n + j/ ]. Let us introduce the notation k1 k U k1,k 0 = x,z R x k1 z u 0x, z k, U k 1,k t = n<ξ t x,z R k x k 1 z k u ξ, x, z, k 1, k = 0, 1,..., 6, 6 6 U0 = U k1,k 0, U t = k 1=0 k =0 6 k 1=0 k =0 n < t n + 1, 6 Uk 1,k t. The following statements are true k 1. x k 1 z k u ξ, x, z U k 1,k t U t, k 1, k = 0, 1,..., 6, 18 n < ξ t, n < t n + 1; 19. The functions U k 1,k t, U t are nonnegative and nondecreasing on each time step n, n + 1 ]. Let us prove a priori estimates guaranteeing the compactness of the family of solutions u t, x, z of the problem We consider the zeroth complete step n = 0. At the first fractional step with t 0, ] we study the equation u t = Lu t, x, z with initial condition 16. By the maximum principle for the Cauchy problem 14, 16 we find u ξ, x, z u 0 x, z, 0 < ξ t, t ] 0,. 0 x,z R Differentiating equation 14 and the initial condition 16 with respect to x and z from 1 to 6 times and using the maximum principle we obtain k1 k x k1 z u ξ, x, z k k1 k x k1 z u 0x, z k, x,z R 0 < ξ t, t 0, ], k1, k = 0, 1,...,

6 We apply x,z R first to the right parts of the inequalities 0, 1 and then to their left parts. After that we apply to both sides of the obtained relations and sum up the results. 0<ξ t In view of 18 and 19 we get In the second fractional step t, ] we have U t U0, t 0, ]. u t = u t, x, z p M t, x, z, u t, x, z + ft, x, z, u t = u, x, z. Then we integrate equation 15 with respect to the time variable. The following inequality holds: u ξ, x, z u, x, z + ξ u θ, x, z p M θ, x, z, u θ, x, z + + ξ fθ, x, z dθ, < ξ t, t ]., 3 From the notation 1 and the conditions 7 9, since at, bt C 1 [0, T ], it follows that the functions A 0 t, A i t, x, B i t, z i = 1,, 3, and their derivatives with respect to x and z of order up to 6, including 6th, are bounded and continuous in G [0,T ]. Applying first and then <ξ t to both parts of the inequality 3 we obtain the estimate <ξ t ξ + x,z R u ξ, x, z u x,z R <ξ θ x,z R u ξ, x, z ξ, x, z + p x,z R <ξ θ x,z R fξ, x, z dθ+ M ξ, x, z, u ξ <ξ θ, x, z dθ, x,z R < ξ t, t ]., 4 Taking into account notation 17 19, conditions 7 9 and the fact that at, bt C 1 [0, T ], from 4 we find t U0,0t U 0,0 0 + C dθ + C t U 0,0 θ p 1 + U0,θ + U 0,1θ 1 + U,0θ + U 1,0θ dθ. 5 Hereinafter, C denotes generally speaking, various constants depending on degree p, constants δ 1, δ, δ 3 from 8, 9 and the constant from 7 which bounds the input data, and independent of the parameter. For the sake of convenience, we assume that C 1. Let us denote the result of differentiating of function u t, x, z p k 1 times with respect to x 184

7 and k times with respect to z by k 1,k t, x, z, k 1, k = 0, 1,..., 6, that is 0,0t, x, z = u t, x, z p, 1,0t, x, z = p u t, x, z p 1 x u t, x, z, 0,1t, x, z = p u t, x, z p 1 z u t, x, z, 1,1t, x, z = p p 1 u t, x, z p x u t, x, z z u t, x, z+ + p u t, x, z p 1 x z u t, x, z, and so forth to 6,6t, x, z. According to 18, 19 for t, x, z G [0,T ] the following inequality is valid: k1,k t, x, z p k 1 +k U t p, k 1, k = 0, 1,..., 6. 6 We differentiate equation 15 with respect to x and z from 1 to 6 times and integrate the obtained expression with respect to the time variable. Using the notation 17 and the inequality 6 we apply first and then to both parts of the obtained inequality. In view of x,z R <ξ t the conditions 7 9, notations 18, 19 and the fact that at, bt C 1 [0, T ], we get the estimate U k 1,k t U k1,k 0 + C p k 1+k 6 k 1,k =0 t U θ p + 1 dθ + C p k 1+k t Uk 1,k θ dθ + C pk 1+k 6 U θ p k 1,k =0 k 1, k = 0, 1,..., 6, t U θ p U k 1,k θ dθ, Summing up the inequalities 5 and 7 and using the estimate we have U t U0 + C p 1 t P p+ U θ dθ, t, ]. 7 where P p+ χ = C χ p+ + χ p+1 + χ p χ + 1 is a polynomial of degree p +, C is a nonzero constant independent of. The latter inequality holds for all t ]., Therefore taking into account the properties of definite integrals and the function U t we find U t U0 + C p 1 0 P p+ U θ dθ, t 0, ]. 8 We consider the Cauchy problem for the ordinary differential equation dωt dt = P p+ ωt, ω0 = U0. 9 By the Cauchy theorem [13], a solution ωt of the problem 9 exists on some interval [0, t ], where t 0, T ] depends on constant C and the initial data U0. The function ωt increases monotonously on the interval [0, t ] and ωt C 1 [0, t ]. 185

8 By 8, 9, if U t 0 ωt 0 for some 0 < t 0 t, then U t ωt, t [t 0, t ]. Since U 0 ω0, the fact that ωt is monotone implies U t ω, t [0, ]. We consider the first complete time step, where t, ]. Arguing in the same way as at the zeroth complete step we obtain U t ω, t [0, ]. After a finite number of steps we have U t ωt C, t [0, t ]. Thus, we have proved the following uniform by estimates: k1 k x k1 z u t, x, z k C, k 1, k = 0, 1,..., 6, t, x, z G [0,t ]. 30 Estimates 30 imply that the right parts of equations 14, 15 are bounded uniformly by on any time step. Therefore the left parts of the equations are bounded uniformly by too: u t t, x, z C, t, x, z G [0,t ]. Differentiating equations 14, 15 with respect to x and z, by 30 we get k1 k x k1 z u k t t, x, z C, k 1, k = 0, 1,..., 4, t, x, z G [0,t ]. 31 Estimates 30, 31 guarantee that the assumptions of Arzela s compactness theorem are satisfied. By this theorem, some subsequence u k t, x, z of the sequence of solutions u t, x, z to the problem with corresponding derivatives converges to a function ut, x, z C 0,4,4 t,x,z G M [0,t ]. In view of the theorem about convergence of the weak approximation method, the function ut, x, z is a solution of the problem 13, 3, moreover ut, x, z C 1,4,4 t,x,z C l,l 1,l t,x,z G M [0,t ] { G M [0,t ] = ut, x, z, where M > 0 is an integer constant, G M [0,t = ] {t, x, z 0 t t, x M, z M k t k u, k1+k x k 1 z k u C G M [0,t ], }, k = 0, 1,..., l, k 1 = 0, 1,..., l 1, k = 0, 1,..., l }. We choose constant M arbitrarily, therefore the solution ut, x, z of the problem 13, 3 belongs to the class C 1,4,4 t,x,z G[0,t ]. Furthermore the following estimations are valid for t, x, z G [0,t ]: k x k 1 z k ut, x, z C, k 1, k = 0, 1,..., 4. 3 From continuous differentiability of at, bt, the conditions 7 9 and the estimation 3, from 10 and 13 it follows that ut, x, z, λt, x, z belong to the class { Zt = ut, x, z, λt, x, z u C 1,4,4 t,x,z G[0,t ], λt, x, z C 0,, t,x,z G[0,t ] }, and satisfy the following inequality: 4 k x k 1 z k ut, x, z + k 1,k =0 k x k 1 z k λt, x, z C. 33 k 1,k =0 186

9 4. Existence of solution of the inverse problem Now we prove that the pair of functions ut, x, z, λt, x, z, where λt, x, z is defined by 10 and satisfies the condition 4, is a solution of the inverse problem 1 5. Since ut, x, z is a solution of the direct problem 13, 3, substituting ut, x, z, λt, x, z into 1, 3 we obtain the identities. Let us prove that the overdetermination conditions 5 are fulfilled. In the strip P [0,T ] = {t, y 0 t T, y R} we consider the auxiliary Cauchy problem for the equation v t t, y = a 1 t v yy t, y + a t v y t, y + a 3 t, y vt, y a 4 t, y v yy t, dt a 5 t, y v y t, dt, 34 with the initial condition v0, y = ry. 35 Suppose that the functions a 1 t, a t, a 3 t, y, a 4 t, y, a 5 t, y, ry and all their derivatives are continuous and bounded in P [0,T ], dt is a smooth bounded curve. The following statement takes place. Lemma 1. If solution vt, y C 1,4 t,y P[0,T ] of the problem 34, 35 exists and satisfies the condition 4 k vt, y yk C, then it is unique. Here, { C 1,4 t,y P[0,T ] = vt, y k=0 v tt, y, k y k vt, y C } P [0,T ], k = 0, 1,..., 4. The proof of the Lemma 1 basically repeats the proof of Lemma in paper [7]. We substitute z = at into 13. Transforming the obtained relation and taking into consideration that ut, x, at p φt, x p we find = ut, x, atp φt, x p + φt, x p φt, x p = = ut, x, at φt, x up 1 + u p φ + u p 3 φ u φ p + φ p 1 φt, x p + 1, u t t, x, at + u z t, x, at a t φ t t, x = α 1 t u xx t, x, at φ xx t, x + + β 1 t u x t, x, at φ x t, x + D 1 t, x ut, x, at φt, x D t, xα 1 t u xx t, bt, at φ xx t, bt +D 3 t, xβ 1 t u x t, bt, at φ x t, bt, 187

10 where D 0 t = ψ t t, at α t ψ zz t, at β t ψ z t, at ft, bt, at α 1 t u xx t, bt, at β 1 t u x t, bt, at 1 + b t ; D 1 t, x = 1 D 0 t up 1 + u p φ + u p 3 φ u φ p + φ p 1 φt, x p φ t t, x α 1 t φ xx t, x β 1 t φ x t, x ft, x, at ψ t t, at α t ψ zz t, at β t ψ z t, at ft, bt, at + + φ t t, x α 1 t φ xx t, x β 1 t φ x t, x ft, x, at α 1 t u xx t, bt, at β 1 t u x t, bt, at 1 + b t + ψ t t, at α t ψ zz t, at β t ψ z t, at ft, bt, at + α 1 t u xx t, bt, at β 1 t u x t, bt, at 1 + b t α t u zz t, x, at β t u z t, x, at 1 + a t + α t u zz t, x, at β t u z t, x, at 1 + a t ; D t, x = u tt, x, at α 1 t u xx t, x, at β 1 t u x t, x, at α t u zz t, x, at D 0 t D 3 t, x = D t, x 1 + b t. We introduce the notation: ν 1 t, x = ut, x, at φt, x. Hence, ν 1 t t, x = u t t, x, at + u z t, x, at a t φ t t, x. From the compatibility conditions 6 it follows that β t u z t, x, at + ft, x, at ; D 0 t ν 1 0, x = u0, x, a0 φ0, x = u 0 x, a0 φ0, x = 0. Consequently, the function ν 1 t, x is a solution of the following problem: ν 1 t t, x = α 1 t ν 1 xxt, x + β 1 t ν 1 xt, x + D 1 t, x ν 1 t, x D t, x α 1 t ν 1 xxt, bt D 3 t, x β 1 t ν 1 xt, bt, 36 ν 1 0, x = In accordance with the constraints 7 9, the inequality 33 and the fact that at, bt C 1 [0, T ], the functions D 1 t, x, D t, x, D 3 t, x and their derivatives with respect to x to the second order, including nd, are continuous and bounded in G [0,t ]. Therefore, the problem 36, 37 satisfies the assumptions of Lemma 1 where a 1 = α 1 t, a = β 1 t, a 3 = D 1 t, x, a 4 = D t, x α 1 t, a 5 = D 3 t, x β 1 t, d = at, r = 0. From Lemma 1 it follows that the solution of problem 36, 37 is unique. This solution is the function ν 1 t, x = 0. This means that ut, x, at = φt, x. Hence the first of the overdetermination conditions 5 holds. Similarly, 188

11 using the compatibility conditions 6 we prove that the second of overdetermination conditions 5 is valid: ut, bt, z = ψt, z. Similarly, using the compatibility conditions 6 we prove that the second of overdetermination conditions 5 is valid: ut, bt, z = ψt, z. Thus the following theorem has been proved. Theorem. Let the conditions 6 9 be satisfied. Then there exists a constant 0 < t T depending on the degree p and on the constants δ 1, δ, δ 3, C from 7 9 and such that a solution ut, x, z, λt, x, z = λ 1 t, x λ t, z of the inverse problem 1 5 in the class Zt exists and satisfies Example For the problem under study there is example of the initial data satisfying the assumptions of the theorem and the corresponding solution. In the strip G [0,T ] = { t, x, z 0 t T, x, z R } we consider the Cauchy problem for the parabolic equation u t = u xx + u zz + u x + u z + u λt, x, z cos x cos z + A, 38 A > 1 is a constant, t 0, T ], x, z R, with the initial condition u0, x, z = u 0 x, z = e x sin z + A. In addition to the function ut, x, z it is required to find the function λt, x, z = λ 1 t, x λ t, z. Let at = t, bt = t + C 1 [0, T ]. Then the overdetermination conditions have the form: ut, x, at = φt, x = e t+x sin t + A, ut, bt, z = ψt, z = e t+ sin z + A. The compatibility conditions hold u 0 x, a0 = φ0, x = A e x, u 0 b0, z = ψ0, z = e sin z + A, φt, bt = ψt, at = e t+ sin t + A. The functions ft, x, z = cos x cos z + A, u 0 x, z, φt, x, ψt, z satisfy the smoothness and boundedness conditions 7. We verify fulfilment of the conditions 8: φt, x = e t+x sin t + A ex A 1 ε A 1 > 0, 0 < ε 1, t [0, T ], x R, ψt, z = e t+ sin z + A e A 1 > 0, t [0, T ], z R. After that we consider 9: e t+ sin t + cos t + A e t+ sin t + A + e t+ sin t e t+ sin t + A e t+ cos t+ + cost + cos t + A A + cos t e t+ cost + e A 1 t+ cost + A 1 e t+ cost + A 1 e > 0, t [0, T ]. Consequently the conditions 8, 9 are satisfied for δ 1 = ε A 1, δ = e A 1, δ 3 = A 1 e. 189

12 Then we can find the functions ut, x, z, λt, x, z: ut, x, z = e t+x sin z + A, λt, x, z = λ 1 t, x λ t, z = Indeed, the obtained solution satisfies equation 1: 1 cos x e t+x cos z + A sin z + A. e t+x sin z + A = e t+x sin z + A e t+x sin z + e t+x sin z + A + e t+x cos z e t+x sin z + A 1 cos x e t+x cos z + A cos x cos z + A, sin z + A e t+x A + e t+x cos z e t+x cos x cos z + A cos x cos z + A = 0, cos z + A e t+x e t+x + cos x cos x = 0, 0 = 0. This example shows that the set of solutions to the problem under study is not empty. The research for this paper was carried out in Siberian Federal University within the framework of the project «Multidimensional Complex Analysis and Differential Equations» funded by the grant of the Russian Federation Government to port scientific research under ervision of a leading scientist, no. 14.Y References [1] N.N.Yanenko, The Method of Fractional Steps for Solving Multi-Dimensional Problems of Mathematical Physics, Novosibirsk, Nauka, 1967 in Russian. [] Yu.Ya.Belov, Inverse Problems for Partial Differential Equations, Utrecht, VSP, 00. [3] Yu.Ya.Belov, On Estimates of Solutions of the Split Problems for Some Multi-Dimensional Partial Differential Equations, Journal of Siberian Federal University. Mathematics & Physics, 009, no. 3, [4] Yu.Ya.Belov, I.V.Frolenkov, Coefficient Identification Problems for Semilinear Parabolic Equations, Doklady Mathematics, 7005, no., [5] E.N.Kriger, I.V.Frolenkov, An Identification Problem of the Special Form Coefficient at a Nonlinear Term for a Two-Dimensional Parabolic Equation, Differential equations. Function spaces. Approximation theory, International Conference dedicated to the 105th anniversary of the birthday of Sergei L. Sobolev August 18 4, 013, Novosibirsk, Russia, Abstracts, Novosibirsk, Sobolev Institute of Mathematics, 013, 176 in Russian. [6] Yu.Ya.Belov, I.V.Frolenkov, On the Problem of Identification of Two Coefficients of a Parabolic Semi-Linear Overdetermined Equation Defined on a Smooth Curve, Special issue of journal Kompyuternye tehnologii devoted to N.N. Yanenko s 85-th anniversary, 11006, no. 1, in Russian. 190

13 [7] E.N.Kriger, I.V.Frolenkov, An Identification Problem of Coefficient for Two-Dimensional Semilinear Parabolic Equation with the Cauchy Data, Russian Mathematics, 59015, no. 5, [8] E.G.Savateev, On the Problem of Identification of a Coefficient in a Parabolic Equation, Siberskii Mat. Zhurnal, , no. 1, in Russian. [9] A.I.Kozhanov, Nonlinear Loaded Equations and Inverse Problems, Vych. Matematika i Mat. Physika, 44004, no. 4, in Russian. [10] I.V.Frolenkov, E.N.Kriger, Existence of a Solution of the Problem of Identification of a Coefficient at the Source Function, Journal of Mathematical Sciences, 03014, no. 4, [11] I.V.Frolenkov, E.N.Kriger, An Identification Problem of the Source Function of the Special Form in Two-Dimensional Parabolic Equation, Journal of Siberian Federal University. Mathematics & Physics, 3010, no. 4, in Russian. [1] I.V.Frolenkov, E.N.Kriger, An Identification Problem of Coefficient in the Special Form at Source Function for Multi-Dimensional Parabolic Equation with Cauchy Data, Journal of Siberian Federal University. Mathematics & Physics, 6013, no., [13] L.S. Pontryagin, Ordinary Differential Equations, Moscow, Nauka, 198 in Russian. Об одной задаче идентификации коэффициента специального вида при нелинейном младшем члене в двумерном полулинейном параболическом уравнении Екатерина Н. Кригер Игорь В. Фроленков Институт математики и фундаментальной информатики Сибирский федеральный университет Свободный, 79, Красноярск, Россия В статье исследована задача идентификации коэффициента при нелинейном члене в двумерном полулинейном параболическом уравнении с данными Коши и условиями переопределения, заданными на гладкой кривой. В предположении, что искомый коэффициент представим в виде произведения двух функций, каждая из которых зависит от временной и только одной пространственной переменной, доказана разрешимость рассмотренной задачи «в малом» в классе гладких ограниченных функций. Построен пример входных данных, удовлетворяющих условиям доказанной теоремы, и соответствующего им решения. Ключевые слова: обратная задача, полулинейное параболическое уравнение, задача Коши, коэффициент при младшем члене, метод слабой аппроксимации, локальная разрешимость, условия переопределения на гладкой кривой. 191