The inverse problem of the heat equation with periodic boundary and integral overdetermination conditions
Tam metin
(2) Kanca Journal of Inequalities and Applications 2013, 2013:108 http://www.journalofinequalitiesandapplications.com/content/2013/1/108. Page 2 of 9. The parameter identification in a parabolic differential equation from the data of integral overdetermination condition plays an important role in engineering and physics [–]. This integral condition in parabolic problems is also called heat moments are analyzed in []. Boundary value problems for parabolic equations in which one or two local classical conditions are replaced by heat moments [–]. In [], a physical-mechanical interpretation of the integral conditions was also given. Various statements of inverse problems on determination of this coefficient in a one-dimensional heat equation were studied in [–, , , , ]. In the papers [, , ], the coefficient is determined from heat moment. Boundary value problems and inverse problems for parabolic equations with periodic boundary conditions are investigated in [, ]. In the present work, one heat moment is used with a periodic boundary condition for the determination of a source coefficient.The existence and uniqueness of the classical solution of the problem ()-() is reduced to fixed point principles by applying the Fourier method. The paper organized as follows. In Section , the existence and uniqueness of the solution of the inverse problem ()-() is proved by using the Fourier method. In Section , the continuous dependence upon the data of the inverse problem is shown. In Section , the numerical procedure for the solution of the inverse problem using the Crank-Nicolson scheme combined with the iteration method is given. Finally, in Section , numerical experiments are presented and discussed.. 2 Existence and uniqueness of the solution of the inverse problem We have the following assumptions on the data of the problem ()-(). (A ) E(t) ∈ C [, T], E(t) < , E (t) ≥ , ∀t ∈ [, T]; (A ) ϕ(x) ∈ C [, ]; () ϕ() = ϕ(), ϕ () = ϕ (), ϕ () = ϕ (), xϕ(x) dx = E(); () ϕn ≥ , n = , , . . . ; (A ) F(x, t) ∈ C(DT ); F(x, t) ∈ C [, ] for arbitrary fixed t ∈ [, T]; () F(, t) = F(, t), Fx (, t) = Fx (, t), Fxx (, t) = Fxx (, t); () Fn (t) > , n = , , . . . ; T ∞ () n= πn(ϕn + Fn (τ ) dτ ) ≤ E (t), ∀t ∈ [, T]; where ϕn = ϕ(x) sin(πnx) dx, Fn (t) = F(x, t) sin(πnx) dx, n = , , , . . . . Theorem Let the assumptions (A )-(A ) be satisfied. Then the following statements are true: () The inverse problem ()-() has a solution in QT ; () The solution of the inverse problem ()-() is unique in QT , where the number T ( < T < T) is determined by the data of the problem. Proof By applying the standard procedure of the Fourier method, we obtain the following representation for the solution of ()-() for arbitrary p(t) ∈ C[, T]:. u(x, t) =. ∞ n=. ϕn e–(π n). t– t p(s) ds . . t. Fn (τ )e–(π n). + . (t–τ )– t p(s) ds τ. dτ sin(πnx).. ().
(3) Kanca Journal of Inequalities and Applications 2013, 2013:108 http://www.journalofinequalitiesandapplications.com/content/2013/1/108. Page 3 of 9. Under the conditions () of (A ) and () of (A ), the series () and its x-partial derivative are uniformly convergent in QT since their majorizing sums are absolutely convergent. Therefore, their sums u(x, t) and ux (x, t) are continuous in QT . In addition, the tpartial derivative and the xx-second order partial derivative series are uniformly convergent in QT . Thus, we have u(x, t) ∈ C , (QT ) ∩ C , (QT ). In addition, ut (x, t) is continuous in QT . Differentiating () under the assumption (A ), we obtain . . xut (x, t) dx = E (t),. (). . and this yields. p(t) = K p(t) ,. (). where.
(4) . K p(t) = –E (t) +. ∞ n=. –. t –(π n) t– t p(s) ds –(π n) (t–τ )– τt p(s) ds πn ϕn e + Fn (τ )e dτ . . ∞ Fn (t) /E(t). πn n=. (). Denote – maxt∈[,T] E (t) – E() – maxt∈[,T] ( ∞ n= π n Fn (t)) C = , maxt∈[,T] E(t) T ∞ – mint∈[,T] E (t) + ∞ k= πn(ϕn + Fn (τ ) dτ ) – mint∈[,T] ( n= C = mint∈[,T] E(t). F (t)) π n n. .. Using the representation (), the following estimate is true: < C ≤ p(t) ≤ C . Introduce the set M as follows: M = p(t) ∈ C[, T] : C ≤ p(t) ≤ C . It is easy to see that K : M → M. Compactness of K is verified by analogy to []. By virtue of the Schauder fixed point theorem, we have a solution p(t) ∈ C[, T] of equation (). Now let us show that there exists QT ( < T ≤ T) for which the solution (p, u) of the problem ()-() is unique in QT . Suppose that (q, v) is also a solution pair of the problem.
(5) Kanca Journal of Inequalities and Applications 2013, 2013:108 http://www.journalofinequalitiesandapplications.com/content/2013/1/108. Page 4 of 9. ()-(). Then, from the representations () and () of the solution, we have u(x, t) – v(x, t) =. ∞ . t t ϕn e–(π n) t– p(s) ds – e–(π n) t– q(s) ds sin πn(x). n=. +. ∞ n=. t. –(π n) (t–τ )– t p(s) ds –(π n) (t–τ )– t q(s) ds τ τ dτ sin πn(x), Fn (τ ) e –e. (). . p(t) – q(t) = K p(t) – K q(t) ,. (). where. K p(t) – K q(t) t ∞ –(π n) t– t p(s) ds – e–(π n) t– q(s) ds ) n= πnϕn (e = E(t) t t ∞ –(π n) (t–τ )– τt p(s) ds – e–(π n) (t–τ )– τ q(s) ds ) dτ n= πn Fn (τ )(e . – E(t) Using the estimates –(π n) t– t p(s) ds –(π n) t– t q(s) ds ≤ (πn) T max p(t) – q(t), e –e ≤t≤T. –(π n) (t–τ )– t p(s) ds –(π n) (t–τ )– t q(s) ds e ≤ (πn) T max p(t) – q(t), τ τ –e ≤t≤T. we obtain . . max K p(t) – K q(t) ≤ α max p(t) – q(t).. ≤t≤T. ≤t≤T. Let α ∈ (, ) be an arbitrary fixed number. Fix a number T , < T ≤ T, such that C + C T ≤ α, C where. C = min E(t), t∈[,T]. C =. ∞ .
(6) πnϕn ,. n=. C = T max. t∈[,T]. ∞ . πnFn (t) .. n=. Then, from equality (), we obtain. p – q C[,T ] ≤ α p – q C[,T ] , which implies that p = q. By substituting p = q in (), we have u = v.. .
(7) Kanca Journal of Inequalities and Applications 2013, 2013:108 http://www.journalofinequalitiesandapplications.com/content/2013/1/108. Page 5 of 9. 3 Continuous dependence of (p, u) upon the data Theorem Under assumptions (A )-(A ), the solution (p, u) of the problem ()-() depends continuously upon the data for small T . The proof of the theorem is verified by analogy to [].. 4 Numerical method We use the finite difference method with a predictor-corrector type approach that is suggested in []. Apply this method to the problem ()-(). We subdivide the intervals [, ] and [, T] into Nx and Nt subintervals of equal lengths h = Nx and τ = NTt , respectively. Then we add two lines x = and x = (Nx + )h to generate the fictitious points needed for dealing with the boundary conditions. We choose the Crank-Nicolson scheme, which is absolutely stable and has a second-order accuracy in both h and τ . [] The Crank-Nicolson scheme for ()-() is as follows: j j+ j j j ui – ui = ui– – ui + ui+ τ h. j+ j+ j+ + ui– – ui + ui+ +. j+ j+ j+ j j p + pj ui + ui + Fi + Fi , . (). ui = φi ,. (). j u. (). j = uNx + ,. j. j. u + uNx . j. = uNx + ,. (). where ≤ i ≤ Nx and ≤ j ≤ Nt are the indices for the spatial and time steps respectively, j j ui = u(xi , tj ), φi = ϕ(xi ), Fi = F(xi , tj ), xi = ih, tj = jτ . At the t = level, adjustment should be made according to the initial condition and the compatibility requirements. Equations ()-() form an Nx × Nx linear system of equations AU j+ = b,. (). where j j j tr U j = u , u , . . . , uNx , ≤ j ≤ Nt , b = (b , b , . . . , bNx )tr , ⎡ – R + R + ⎢ – R + R + – ⎢ ⎢ ⎢ – R + R + · · · ⎢ ⎢ . ⎢ .. A=⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ – R + R + ⎣ – – R=. h , τ. R =. h j+ p + pj , . j = , , . . . , Nt ,. ⎤. – . – R + R +. ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥, ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ .
(8) Kanca Journal of Inequalities and Applications 2013, 2013:108 http://www.journalofinequalitiesandapplications.com/content/2013/1/108. Page 6 of 9. . j+ j j j j u + u + uNx + h F + F , j = , , . . . , Nt , b = – –R + R + . j+ j j j j bNx = uNx – – –R + R + uNx + u + h FNx + FNx , j = , , . . . , Nt , j+ j j j j bi = ui– – (–R + R + )ui + ui+ + h Fi + Fi , i = , , . . . , Nx – , j = , , . . . , Nt . Now, let us construct the predicting-correcting mechanism. First, integrating equation () with respect to x from to and using () and (), we obtain. p(t) =. –E (t) +. . xF(x, t) dx + ux (, t) . E(t). (). The finite difference approximation of () is j. pj =. Ej. where Ej = E(tj ), (Fin)j = For j = , p =. j. –((Ej+ – Ej )/τ ) + (Fin)j + (uNx + – uNx )/h . ,. xF(x, tj ) dx, j = , , . . . , Nt .. –((E – E )/τ ) + (Fin) + (φNx + – φNx )/h , Ej j. and the values of φi allow us to start our computation. We denote the values of pj , ui at j(s) the sth iteration step pj(s) , ui , respectively. In numerical computation, since the time step j+() j is very small, we can take pj+() = pj , ui = ui , j = , , , . . . , Nt , i = , , . . . , Nx . At each (s + )th iteration step, we first determine pj+(s+) from the formula j+(s). pj+(s+) =. j+(s). –((Ej+ – Ej + )/τ ) + (Fin)j+ + (uNx + – uNx )/h Ej. .. Then from ()-() we obtain j+(s+) j+(s+) j+(s) j+(s+) j+(s+) u = ui– – ui – ui + ui+ τ i h. j+(s) j+(s) j+(s) + ui– – ui + ui+ + j+(s). = uNx + ,. j+(s). + uNx. u. u. j+(s). j+(s+) j+(s+) j+ j+(s) j + Fi + Fi , p + pj+(s) ui + ui . () (). j+(s). . j+(s). = uNx + .. (). The system of equations ()-() can be solved by the Gauss elimination method, and j+(s+) is determined. If the difference of values between two iterations reaches the preui scribed tolerance, the iteration is stopped, and we accept the corresponding values pj+(s+) , j+(s+) j+ ui (i = , , . . . , Nx ) as pj+ , ui (i = , , . . . , Nx ), on the (j + )th time step, respectively. By virtue of this iteration, we can move from level j to level j + ..
(9) Kanca Journal of Inequalities and Applications 2013, 2013:108 http://www.journalofinequalitiesandapplications.com/content/2013/1/108. Page 7 of 9. Figure 1 The analytical and numerical solutions of p(t) when T = 1/2. The numerical solution is shown by a dashed line.. 5 Numerical example and discussions In this section, we present examples to illustrate the efficiency of the numerical method described in the previous section. Example Consider the inverse problem ()-() with + sin(πx) + exp(t) + (π) sin(πx) exp(t), . – exp(t), x ∈ [, ], t ∈ [, T]. ϕ(x) = + sin(πx), E(t) = π. F(x, t) =. . It is easy to check that the analytical solution of this problem is p(t), u(x, t) = + exp(t), + sin(πx) exp(t) .. (). Let us apply the scheme which was explained in the previous section for the step sizes h = ., τ = h . In the case when T = /, the comparisons between the analytical solution () and the numerical finite difference solution are shown in Figures and .. 6 Conclusions The inverse problem regarding the simultaneous identification of the time-dependent coefficient of heat capacity together with the temperature distribution in a one-dimensional heat equation with periodic boundary and integral overdetermination conditions has been considered. This inverse problem has been investigated from both theoretical and numerical points of view. In the theoretical part of the article, the conditions for the existence, uniqueness and continuous dependence upon the data of the problem have been.
(10) Kanca Journal of Inequalities and Applications 2013, 2013:108 http://www.journalofinequalitiesandapplications.com/content/2013/1/108. Figure 2 The analytical and numerical solutions of u(x, t) at the T = 1/2. The numerical solution is shown by a dashed line.. established. In the numerical part, a numerical example using the Crank-Nicolson finitedifference scheme combined with the iteration method is presented.. Competing interests The author declares that they have no competing interests. Acknowledgements Dedicated to Professor Hari M Srivastava. Received: 26 November 2012 Accepted: 24 February 2013 Published: 18 March 2013 References 1. Cannon, J, Lin, Y, Wang, S: Determination of a control parameter in a parabolic partial differential equation. J. Aust. Math. Soc. Ser. B, Appl. Math 33, 149-163 (1991) 2. Cannon, J, Lin, Y, Wang, S: Determination of source parameter in a parabolic equations. Meccanica 27, 85-94 (1992) 3. Fatullayev, A, Gasilov, N, Yusubov, I: Simultaneous determination of unknown coefficients in a parabolic equation. Appl. Anal. 87, 1167-1177 (2008) 4. Ivanchov, M, Pabyrivska, N: Simultaneous determination of two coefficients of a parabolic equation in the case of nonlocal and integral conditions. Ukr. Math. J. 53, 674-684 (2001) 5. Ismailov, M, Kanca, F: An inverse coefficient problem for a parabolic equation in the case of nonlocal boundary and overdetermination conditions. Math. Methods Appl. Sci. 34, 692-702 (2011) 6. Kanca, F, Ismailov, M: Inverse problem of finding the time-dependent coefficient of heat equation from integral overdetermination condition data. Inverse Probl. Sci. Eng. 20, 463-476 (2012) 7. Ionkin, N: Solution of a boundary-value problem in heat conduction with a nonclassical boundary condition. Differ. Equ. 13, 204-211 (1977) 8. Vigak, V: Construction of a solution of the heat-conduction problem with integral conditions. Dokl. Akad. Nauk Ukr. SSR 8, 57-60 (1994) 9. Ivanchov, N: Boundary value problems for a parabolic equation with integral conditions. Differ. Equ. 40, 591-609 (2004) 10. Choi, J: Inverse problem for a parabolic equation with space-periodic boundary conditions by a Carleman estimate. J. Inverse Ill-Posed Probl. 11, 111-135 (2003) 11. Ashyralyev, A, Erdogan, A: On the numerical solution of a parabolic inverse problem with the Dirichlet condition. Int. J. Math. Comput. 11, 73-81 (2011) 12. Sakinc, I: Numerical solution of the quasilinear parabolic problem with periodic boundary condition. Hacet. J. Math. Stat. 39, 183-189 (2010) 13. Samarskii, AA: The Theory of Difference Schemes. Dekker, New York (2001). Page 8 of 9.
(11) Kanca Journal of Inequalities and Applications 2013, 2013:108 http://www.journalofinequalitiesandapplications.com/content/2013/1/108. doi:10.1186/1029-242X-2013-108 Cite this article as: Kanca: The inverse problem of the heat equation with periodic boundary and integral overdetermination conditions. Journal of Inequalities and Applications 2013 2013:108.. Page 9 of 9.
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