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Bulletin of TICMI Vol. 16, No. 1, 2012, 27–33 An Iteration Method for the Kirchhoff Static Beam Peradze Jemal I. Javakhishvili Tbilisi State University, Georgian Technical University (Received December 26, 2011; Accepted March 5, 2012) RL 00 u02 k−1 dx)uk = f , k = 1, 2, . . . , is used to solve the R boundary value problem for the nonlinear differential equation uiv −(λ+2/L 0L u02 dx)u00 = f . The approximation uk is expressed as well-defined integrals of the functions uk−1 and f . The method error uk − u is estimated. The iteration method uiv k − (λ + 2/L 0 Keywords: Kirchhoff equation, Static beam, Iteration method, Algorithm error. AMS Subject Classification: 65L10, 65L70. 1. Statement of the problem We consider the following boundary value problem ¶ µ Z 2 L 02 u (x) dx u00 (x) = f (x), u (x) − λ + L 0 ıv 0 < x < L, u(0) = u(L) = 0, (1.1) λ = const > 0, u00 (0) = u00 (L) = 0, (1.2) where f (x) is a given continuous function, and u(x) is the sought solution. Equation (1.1) is the stationary problem related to the equation µ Z L µ ¶2 ¶ 2 ∂2u ∂u ∂4u ∂ u + α − α + α = 0, dx 0 1 2 ∂t2 ∂x4 ∂x ∂x2 0 (1.3) which was proposed by Woinowsky–Krieger [9] in 1950 as a model for the deflection of an extensible dynamic beam with hinged ends. The nonlinear term of this equation was for the first time used by Kirchhoff [4] who generalized D’Alembert’s classical model. Therefore equations (1.1) and (1.3) are frequently called a Kirchhoff type equation for a dynamic and a static beam, respectively. The results of one of the initial mathematical studies of equations of (1.3) type are presented in [1] and [2]. For equation (1.1) and its generalizations, as well as for equations similar to (1.1), the problem of construction of numerical algorithms and estimation of their accuracy is studied in [3], [5]–[8]. Each of the algorithms used in these papers is a combination of two approximate methods, one of which reduces the problem to ISSN: 1512-0082 print c 2012 Tbilisi University Press ° 28 Bulletin of TICMI the finite-dimensional one and the other is some iterative process of solution of the discrete system. In the present paper, a technique somewhat different from the above-mentioned one is proposed to solve problem (1.1),(1.2). The differential equation (1.1) is solved by an iteration method. At each iteration step, a boundary value problem is obtained for a linear differential equation whose solution is written in integrals. The algorithm accuracy is estimated by the method of a priori inequalities. 2. The algorithm On choosing a function u0 (x), 0 ≤ x ≤ L, that together with its second derivative vanishes for x = 0 and x = L, we will seek for a solution of problem (1.1),(1.2) using the iteration process uıv k (x) ¶ µ Z 2 L 02 u (x) dx u00k (x) = f (x), − λ+ L 0 k−1 (2.4) 0 < x < L, u00k (0) = u00k (L) = 0, uk (0) = uk (L) = 0, (2.5) k = 1, 2, . . . , where uk (x) is the k-th approximation of the solution of problem (1.1),(1.2), k = 0, 1, . . . . The considered algorithm makes it possible to express uk (x) through the preceding approximation in the integral form. Indeed, on denoting 2 αk = λ + L Z 0 L u02 k (x) dx, we introduce the function vk (x) = u00k (x), k = 0, 1, . . . . Now, (2.4),(2.5) can be rewritten as relations u00k (x) = vk (x), 0 < x < L, uk (0) = uk (L) = 0 and vk00 (x) − αk−1 vk (x) = f (x), 0 < x < L, vk (0) = vk (L) = 0. For uk (x) we have µ ¶ Z x Z L 1 uk (x) = (x − L) ξvk (ξ) dξ + x (ξ − L)vk (ξ) dξ L 0 x (2.6) Vol. 16, No. 1, 2012 29 and vk (x) is representable in the form 1 vk (x) = √ √ αk−1 sinh( αk−1 L) µ Z ¡√ ¢ × sinh αk−1 (x − L) ¡√ ¢ + sinh αk−1 x Z x sinh 0 x L ¡√ ¢ αk−1 ξ f (ξ) dξ ¶ ¡√ ¢ sinh αk−1 (ξ − L) f (ξ) dξ . (2.7) Substituting (2.7) into (2.6) and applying the well-known equalities for hyperbolic functions, we obtain the desired formula uk (x) = − 1 αk−1 L µ Z (x − L) Z x ξf (ξ) dξ + x 0 ¶ L (ξ − L)f (ξ) dξ x µ Z ¡√ ¢ x ¡√ ¢ 1 + sinh αk−1 (x−L) sinh αk−1 ξ f (ξ) dξ √ √ αk−1 αk−1 sinh( αk−1 L) 0 ¶ Z L ¡√ ¢ ¡√ ¢ + sinh αk−1 x sinh αk−1 (ξ − L) f (ξ) dξ , k = 1, 2, . . . , (2.8) x which makes it possible to obtain approximations uk (x) evading the differential problem (2.4),(2.5). Therefore approximations uk (x) are found by means of (2.8). 3. Error of the algorithm and the equation for it Let us define the algorithm error as a difference ∆uk (x) = uk (x) − u(x), k = 0, 1, . . . , between an approximate and an exact solutions. Subtracting the respective relations in (1.1) and (1.2) from (2.4) and (2.5), we obtain ∆uıv k (x) ¶ µ Z ¢ 1 L ¡ 02 02 uk−1 (x) + u (x) dx ∆u00k (x) − λ+ L 0 µZ L ¶ ¡ 0 ¢ 1 0 0 − uk−1 (x) + u (x) ∆uk−1 (x) dx (u00k (x) + u00 (x)) = 0, L 0 ∆uk (0) = ∆uk (L) = 0, ∆u00k (0) = ∆u00k (L) = 0. (3.9) (3.10) We use (3.9), (10) to estimate the algorithm accuracy. For this, we need some a priori relations which are derived in the next paragraph. 30 4. Bulletin of TICMI Auxiliary inequalities For the well-defined functions, sufficiently smooth for 0 ≤ x ≤ L, we introduce the notation Z L (u(x), v(x)) = µZ ku(x)kp = ¶ Lµ p d u(x) 2 dxp 0 u(x)v(x) dx, 0 ¶1 2 dx , p = 0, 1, 2, ku(x)k = ku(x)k0 . Lemma 4.1: For a twice differentiable function u(x), 0 ≤ x ≤ L, that vanishes at x = 0 and x = L the inequalities √ 2 L ku(x)k ≤ ku(x)k1 ≤ √ ku(x)k2 L 2 (4.11) are valid. Proof : We have Z x u(x) = u0 (ξ) dξ. 0 Hence µZ |u(x)| ≤ ¶1 µ Z x 2 dξ 0 x ¶1 02 u (ξ) dξ 0 2 1 ≤ x 2 kuk1 . Therefore ku(x)k2 ≤ L2 kuk21 , 2 which implies the left inequality of (4.11). Using the latter and taking into account that ¯L ku(x)k21 = u(x)u0 (x)¯0 − (u(x), u00 (x)) = −(u(x), u00 (x)), we complete the proof. ¤ Lemma 4.2: For the solution of problem (1.1), (1.2) we have the inequality ku(x)k1 ≤ c1 , (4.12) where c1 is the constant calculated by the formula L c1 = 2 µ ¶− 1 4 1 2 kf (x)k 2 . + λL L (4.13) Vol. 16, No. 1, 2012 31 Proof : We multiply equation (1.1) by u(x) and then integrate the obtained equality with respect to x from 0 to L. Using (1.2), we get the relation ku(x)k22 ¶ µ 2 2 + λ + ku(x)k1 ku(x)k21 = (f (x), u(x)) , L which, together with (4.11), yields µ ¶ 2 2 L 2 λ + 2 + ku(x)k1 ku(x)k21 ≤ √ kf (x)k ku(x)k1 . L L 2 Thus we obtain the inequality 2 L2 ku(x)k41 ≤ L 8 µ ¶−1 2 +λ kf (x)k2 , L2 which implies estimate (4.12). ¤ Lemma 4.3: The solution of problem (2.4), (2.5) satisfies the relation kuk (x)k1 ≤ c2 , k = 1, 2, . . . , (4.14) where c2 is a constant independent of k and defined by L2 c2 = √ 2 µ ¶−1 2 + λL kf (x)k. L (4.15) Proof : We multiply equation (2.4) by uk (x) and integrate with respect to x from 0 to L. Taking (2.5) into account, we see kuk (x)k22 ¶ µ 2 2 + λ + kuk−1 (x)k1 kuk (x)k21 = (f (x), uk (x)) . L By applying (4.11) we find ¶ µ 2 L 2 2 λ + 2 + kuk−1 (x)k1 kuk (x)k21 ≤ √ kf (x)k kuk (x)k1 . L L 2 This implies (4.14). ¤ 32 5. Bulletin of TICMI Estimation of the algorithm error We multiply equation (3.9) by ∆uk (x) and integrate the obtained equality with respect to x from 0 to L. Applying (3.10) we obtain k∆uk (x)k22 ¶ µ ¢ 1¡ 2 2 kuk−1 (x)k1 + ku(x)k1 k∆uk (x)k21 + λ+ L + 1 ¢ 1 Y¡ 0 uk−p (x) + u0 (x), ∆u0k−p (x) = 0. L p=0 By (4.11) µ ¶ ¢ 2 1¡ 2 2 kuk−1 (x)k1 + ku(x)k1 k∆uk (x)k21 λ+ 2 + L L ≤ 1 1 Y (kuk−p (x)k1 + ku(x)k1 ) k∆uk−p (x)k1 . L p=0 Therefore µ ¶1 µ ¶1 ¢ 2 2 2 1¡ 2 2 k∆uk (x)k1 ≤ kuk−1 (x)k1 + ku(x)k1 L L ¶ µ ¢ −1 1¡ 2 2 2 kuk−1 (x)k1 + ku(x)k1 × λ+ 2 + L L × (kuk (x)k1 + ku(x)k1 ) k∆uk−1 (x)k1 . y 2 0≤y<∞ α+y Since max = 1 2 α−1/2 holds for α > 0, we have µ µ ¶¶− 1 2 2 (kuk (x)k1 + ku(x)k1 ) k∆uk−1 (x)k1 . k∆uk (x)k1 ≤ 2 + λL L (5.16) ¡ ¡ ¢¢− 1 2 Let the condition q = 2 L2 + λL (c1 + c2 ) < 1 be fulfilled, which, as follows from (4.13) and (4.15), is equivalent to the requirement Ã !p µ ¶− 3 2 √ 4 1 1X 2 < 1. q= L 2 + λL kf (x)k 2 4 L p=1 Then, by virtue of (5.16), (4.12), (4.14) and (4.11), we come to a conclusion that the iteration method (2.4),(2.5) or, which is the same, (2.8) reduces to the solution Vol. 16, No. 1, 2012 33 of problem (1.1),(1.2) and the estimate µ k∆uk (x)kp ≤ L √ 2 ¶1−p k = 1, 2, . . . , q k k∆u0 (x)k1 , p = 0, 1, holds for the error of the method. Acknowledgement The author acknowledges the support under grant 7353 of the Estonian Science Foundation. References [1] J.M. Ball, Initial-boundary value problems for an extensible beam, J. Math. Anal. Appl., 42 (1973), 61–90 [2] R.W. Dickey, Free vibrations and dynamic buckling of the extensible beam, J. Math. Anal. Appl., 29 (1970), 443–454 [3] T. Gudi, Finite element method for a nonlocal problem of Kirchhoff type, SIAM J. Numer. Anal., 11 pp., 2011 (in print) [4] G. Kirchhoff, Vorlesungen über mathematische Physik. I. Mechanik, Teubner, Leipzig, 1876 [5] T.F. Ma, Existence results and numerical solutions for a beam equation with nonlinear boundary conditions, Appl. Numer. Math., 47, no. 2 (2003), 189–196 [6] J. Peradze, A numerical algorithm for a Kirchhoff-type nonlinear static beam, J. Appl. Math., 2009, Art. ID 818269, 12 pp. [7] J. Peradze, On an iteration method of finding a solution of a nonlinear equilibrium problem for the Timoshenko plate, Z. Angew. Math. Mech., 91, no. 12 (2011), 993–1001 [8] S. Y. Tsai, Numerical Computation for Nonlinear Beam Problems, M. Sc. Thesis, National Sun YatSen University, Kaohsiung, Taiwan, 2005 [9] S. Woynowsky-Krieger, The effect of an axial force on the vibration of hinged bars, J. Appl. Mech., 17 (1950), 35–36