Engineering Letter, 16:4, EL_16_4_03

Transcription

1

2 3 Exisence In his secion we reduce he problem (5)-(8) o an equivalen problem of solving a linear inegral equaion of Volerra ype for C(s). For his purpose we firs consider following free boundary problem: u xx = u for < x < s(), >, () u(, ) = C(s) where C(s), >, () u(x, ) = ϕ(x) where ϕ(x), < x b, and ϕ(b) =, b >, (2) u(s(), ) = for >, and s() = b, (3) u x (s(), ) = ds() d for >, (4) x = s() is he free boundary which is no given and is o be found ogeher wih u(x, ). Definiion. We say ha u(x, ), s() form a soluion of ()-(4) for all < σ, ( < σ ) if (i) 2 u x and u 2 are coninuous for < x < s(), < < σ; (ii)u and u x are coninuous for x s(), < < σ; (iii)u(x, ) is coninuous also for =, < x b, and lim inf u(x, ) lim sup u(x, ) < as, x (if ϕ() = f() hen u is required o be coninuous a x = = ); R (iv)s() is coninuously differeniable for < σ, and (v)he equaions ()-(4) are saisfied. Theorem. Assume ha C(s) ( < ) and ϕ(x) ( x b) are coninuously differeniable funcions. Then here exis a unique soluion u(x, ), s() of he sysem (2..)-(2..4) for all <. Furhermore, he funcion x = s() is monoone nondecreasing in and he funcion u(x, ) we find for following inegral represenaion. u(x, ) = C(s)G ξ (x, ;, τ)dτ u ξ (s(), )G(x, ; s(τ), τ)dτ (5) b ϕ(ξ)g(x, ; ξ, )dξ Where G(x, ; ξ, τ) is Green s funcion for he half-plan x > and where G(x, ; ξ, τ) = K(x, ; ξ, τ) K( x, ; ξ, τ), (x ξ)2 K(x, ; ξ, τ) = exp{ 4π( τ) 4( τ) }. Proof. See []. We shall now reduce he problem of solving ()-(4) o a problem of solving an inegral equaion. By inroducing and suppose ha u, s form a soluion of ()-(4) and ϕ() = C(s) =, we can reduce he problem of solving ()-(4) o a problem of solving an following inegral equaion [], υ() = b and by (4),(), we have Where ϕ (ξ)n(s(), ; ξ, )dξ (7) C (s)n(s(), ;, τ)dτ υ(τ)g x (s(), ; s(τ), τ)dτ, s() = b υ(τ)dτ. (8) N(x, ; ξ, τ) = K(x, ; ξ, τ) K( x, ; ξ, τ). Thus for every soluion u, s of he sysem ()-(4) for all < σ, he funcion υ() defined by (7) saisfies he nonlinear inegral equaion of Volerra ype (7) (for < < ), where s() is given by (8) coninuous for σ. Suppose conversely ha for some σ >, υ() is a coninuous soluion of he inegral equaion (7) for < σ, wih s() given by (8). We prove ha u(x, ), s() hen form a soluion of ()-(4) for all < σ, where u(x, ) is defined by (5) wih u ξ (s(τ), τ) replaced by υ(τ), []. Now we consider he following Inverse problem. By above verifying, we noe ha, if for some σ >, C(s) is a coninuously differeniable soluion of he linear Volerra inegral equaion of firs kind (7) (forc(s)), where s() is given, u(x, ) hen form a soluion of Inverse problem. By (7), we can wrie where C (s)n(s(), ;, τ)dτ =, (9) = b ϕ (ξ)n(s(), ; ξ, )dξ (2) υ(τ)g x (s(), ; s(τ), τ)dτ /2υ(), where s(), herefore υ() are given. We wan o solve he inegral equaion (9) where is given by (2). For his purpose we wrie he equaion (9) as following form, where u(τ)k(, τ)dτ = [, b], (2) υ() = u x (s(), ), () u(τ) = C (s), K(, τ) = N(s(), ;, τ),

3 and is given by (2). Proposiion. Assume he following : a) The funcion K : [, b] [, b] R is coninuous. b)we define, L = max K(, τ).,τ [,b] c) We se X = C[, b] and M = {u X : u < r} for fixed r >. Then, he original inegral equaion (2) has a leas one soluion u M. Proof. Define he operaor (T u)() = u() u(τ)k(, τ)dτ for all [, b]. Then, he inegral equaion (2) corresponds o he following fixed-poin problem : u = T u, u M. (22) We claim ha he equaion (22) has a soluion. For his purpose, we need o prove he following : ) The se M is a bounded, closed, convex, nonempy subse of Banach space X. 2) The operaor T : M M is compac. Then, he Schauder fixed-poin heorem ells us he equaion (22) has a soluion. Lemma. The se M is a bounded, closed, convex and nonempy subse of Banach space X. Proof. The bounded and nonempy propery subse of Banach space M of X are clear. We know ha, he se M is convex iff u, v M and α imply αu ( α)v M. Le u, v M and α, hen αu ( α)v αu ( α)v Hence, αu ( α)v M. = α u ( α) v αr ( α)r = r. Now we wan o show ha he se M is closed. To his end, le u n be a sequence in M such ha u n u as n. By he definiion of M, for each u n, we have Thus we can wrie u n r, n =, 2,... u = u u n u n u u n u n u u n r = r, as n. Hence, u M. Thus, he se M is closed. Lemma 2. Le us consider he inegral operaor (T u)() = u() Where Se u(τ)k(, τ)dτ for all [, b]. b min{ ε (r h ε k (2r h ))δ rl h, r h ε k (2r h ) }. Q = {(, τ, u) R 3 : (, τ) [, τ] and u r} for fixed r >. Suppose ha he funcion F : Q R F (, τ, u) = u(τ)k(, τ) is coninuous. Se X = C[, b] and M = {u X : u < r} for fixed r >. Then, he operaor T : M M is compac. Proof. Since Q and R are normed spaces and F : Q R is a coninuous operaor on he compac se Q, hence, F is uniformly coninuous on Q. This implies ha, for each ε >, here is a number δ > such ha F (, τ, u) F (, τ, v) < ε, (23) for all (, τ, u), (, τ, v) Q wih u v < δ. We firs show ha he operaor u : [, b] R is coninuous. In fac, if u M, hen he funcion u : [, b] R is coninuous, and u(τ) r for all τ [, b] and h : [, b] R is coninuous and h(τ) for all τ [, b]. Hence he funcion T u : [, b] R is also coninuous. Le u v = max u(τ) v(τ) < δ τ b implies max u() b = max ( b T u T v = u(τ)k(, τ)dτ v() = max ( b v(τ)k(, τ)dτ (u()u(τ) v()v(τ))k(, τ)dτ ) (u()u(τ) u()v(τ) u()v(τ) v()v(τ))k(, τ)dτ ) = max ( b (u()(u(τ) v(τ))

4 v(τ)(u() v()))k(, τ)dτ ) h max b ( u() (u(τ) v(τ))k(, τ) v(τ) (u() v())k(, τ) )dτ (rε rε)b = 2rbε h h by (23). Hence, T : M M is coninuous. We now show ha T (M) M. If u M, hen T u = u() u h max b u(τ)k(, τ)dτ u(τ) K(, τ) dτ r 2 Lb r, h for b rl. h Hence, T u M. Thus T (M) M for b rl. h We now show ha T : M M is compac. Since he se M is bounded, i suffices o show ha he se T (M) is relaively compac. By he Arzela-Ascolli heorem i remains o show ha T (M) is equiconinuous. Le < δ and, [, b]. Then by (23) u() = (T u)() (T u)( ) = u(τ)k(, τ)dτ u( ) h( ) = u(τ)k(, τ)dτ [ u() F (, τ, u) u( ) h( ) F (, τ, u)]dτ [ u() F (, τ, u) u( ) h( ) F (, τ, u)]dτ [ u() (F (, τ, u) F (, τ, u)) F (, τ, u)( u() u( ) h( ) )]dτ [ u() (F (, τ, u) F (, τ, u)) F (, τ, u)( u() u( ) h( ) )]dτ (r h ε k (2r h ))b (r h ε k (2r h ))δ ε for b ε (r h ε k (2r h ))δ r h ε k (2r h ). Where k = 4 Numerical Resuls max F (, τ, u).,τ [,b] In his secion we apply he Collocaion mehod o some examples in order o compare numerical soluion wih exac soluion. Example. In Inverse problem, suppose ha s() =. Then we obain he following inegral equaion : 4 C (s) exp{ 2 4( τ) } π dτ = τ 2 [exp{ τ τ 4 } τ ( τ)2 exp{ τ 4( τ) }]dτ wih exac soluion C(s) = e. Suppose ha (, ). The resul of applying Collocaion mehod wih 2 nodes of inerval (,) and wih 2 base funcions ϕ j () = j, j =,, 2,..., for above inegral equaion is in he following form: C collo (s) = , which he righ hand inegral is approximaed by Gaussian hree poins rule. Comparing of numerical soluion and exac soluion is given in figure. Example 2. For s() = 3/2, we obain he following inegral equaion : 3/8 3 4( τ) } C (s) exp{ τ [ 3/2 τ 3/2 τ τ 3/2 τ 3/2 τ Wih exac soluion dτ = 3/4 π exp{ (3/2 τ 3/2 ) 2 } 4( τ) exp{ (3/2 τ 3/2 ) 2 }]dτ. 4( τ) C(s) =

5 Figure : ( ) exac soluion, (*) approximaed poins Figure 2: ( ) exac soluion, (*) approximaed poins Figure 3: ( ) exac soluion, (*) approximaed poins

6 Suppose ha (, ). The resul of applying Collocaion mehod wih 2 nodes of inerval (,) and wih 2 base funcions ϕ j () = j, j =,, 2,..., for above inegral equaion is in he following form: C collo (s) = , which he righ hand inegral is approximaed by Gaussian hree poins rule. Comparing of numerical soluion and exac soluion is given in figure2. Example 3. Suppose ha s() = 2, hen we obain he following inegral equaion : π /2 4 C (s) exp{ 4( τ) } dτ = τ [( τ) exp{ ( τ)(2 τ 2 ) } τ 4 2 τ 2 τ exp{ (2 τ 2 ) 2 4( τ) }]dτ, wih exac soluion C(s) = Suppose ha (, ). The resul of applying Collocaion mehod wih 2 nodes of inerval (,) and wih 2 base funcions ϕ j () = j, j =,, 2,..., for above inegral equaion is in he following form: C collo (s) = , which he righ hand inegral is approximaed by Gaussian hree poins rule. Comparing of numerical soluion and exac soluion is given in figure 3 (lef). Also for s() = 2 by asympoic approximaion given in [2], we can obain upper and lower bounds for C(s) in he following form: exp{2 3 } C(s) exp{3 3 }. Figure 3 (righ) shows ha numerical approximaion lies beween upper and lower bounds. References [] Ablowiz M J and Delillo S,J. Phys.A: Mah. Gen.35(23),. [2] Cannon, J.R.The One-dimensional Hea Equaion, Cambridge Universiy Press, (984). [3] Colon D and Reemsen R 984 The soluion of inverse Sefan problem in wo space variables SIAM J. Appl.Mah [4] D. D. ANG, A. PHAM NGOC DINH, D. N. THANH, A bidimensional inverse Sefan problem: idenificaion of boundary value J.Compu.Appl.Mah.8 (997) [5] D. D. ANG, A. PHAM NGOC DINH, D. N. THANH, Regularizaion of a wo-dimensional wophase inverse Sefan problem Inverse Probl.3 (997) 7-9. [] D. COLTON, The inverse Sefan problem for he hea equaion in wo space variables Mahemaika 2 (974) [7] Delillo and Salvaori M C, J. Nonlinear Mah. Phys. 9(22),44. [8] Delillo, Salvaori M C and Sanchini G, Phys. Le. A3(23),25. [9] E. BOBULA, K. TWARDOWSKA,On a cerain inverse Sefan problem, Bull. POL.Acad.Sci.Tech. Sci.33 (985) [] Friedman,A.Prial differenial equaions of Parabolic ype. Prenice Hall, New Jersey.(94). [] Gold man N L 997 Inverse Sefan Problems(Dordrech:Kluwer). [2] Jochum P 98 The inverse Sefan Problem as a problem of nonlinear approximaion heory J.Approx [3] Jochum P 982 The numerical soluion of he inverse Sefan problem Numer. Mah [4] M.B. STAMPELLA, D.A. TARZIA, Deerminaion of one or wo unknown hermal coefficiens of a semiinfinie maerial hrough a wo-phase Sefan problem In.J.Eng.Sci.27 (989) [5] Reemsen R and Kirsch A 984 A mehod for he numerical soluion of he one-dimensional inverse Sefan Problem Numer.Mah [] Tien R H and Geiger GE. A hea- ransfer analysis os he solidificaion of a binary euecic sysems, J. Hea Trans ASME 97;89;23-4. [7] Voller V R, Developmen and applicaion of a hea balance inegral mehod for analysis of meallurgical solodificaion. Appl. Mah. Model 989;3:3-.