EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 16, No. 2, 2023, 1140-1153 ISSN 1307-5543 – ejpam.com Published by New York Business Global Inverse Boundary Value Problem Solution for Deflected Beams Joined Together by Elastic Medium Omar Alomari1,∗, Pavel B.Dubovski2 1 School of Basic Sciences and Humanities, German Jordanian University, Amman, Jordan 2 Department of Mathematical Sciences , Stevens Institute of Technology, Hoboken, NJ, USA Abstract. In this paper, we extend the Euler-Bernoulli beam theory for bending boundary value problem into mechanically coupled system. We follow the inverse approach to find the exerted force on two beams separated by elastic material. The theory was utilized in two ways: in the first approach, we calculate the force exerted on the beams using known values for the stiffness constant and measured values for the beam deflections. In the second method, we calculate the stiffness constant using a single known force and measured deflections. These problems are typically ill- posed problems whose solution does not depend continuously on the boundary data. To minimize the variational functional, we develop an iterative algorithm based on the system of three equations: the direct, adjoint, and control equations. Then, we present numerical examples to obtain the solutions. 2020 Mathematics Subject Classifications: 65L09, 31A25, 34A55, 35R30, 65M32 Key Words and Phrases: Adjoint equation, Euler-Bernoulli beam theory, Inverse problem, Tikhonov regularization, touchscreen, ill-posed 1. Introduction Devices with touchscreens capability such as smartphones, tablets, televisions, and medical equipment have become a necessity nowadays [6]. Their popular usage and daily need have made it crucial for manufacturers of such devices to understand how their glass screens react to stresses of various kinds, especially for hand-held devices because of their extensive use [1, 4, 6, 14, 17]. To minimize the lengthy and costly lab testing experiments of evaluating the sensitivity of these devices to accidents such as falling on varies surfaces and manually recording the data on the responses and stresses generated, bending models can be developed to predict the response and stress generated using numerical computations [7, 18]. The model can deliver scientific expectations and results in a more efficient and ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v16i2.4758 Email addresses: omar.alomari@gju.edu.jo (O. Alomari), pavel.dubovskiy@stevens.edu (P. B.Dubovski) https://www.ejpam.com 1140 © 2023 EJPAM All rights reserved. O. Alomari, Pavel B.Dubovski / Eur. J. Pure Appl. Math, 16 (2) (2023), 1140-1153 1141 cost-effective design for new gadgets. A general model to describe touchscreens is presented in Figure 1. A framework wherein two beams are joined by persistent elastic layer of stiffness k. The two beams are basically bolstered at the edges. The top beam is loaded by a point force f . In this paper, we develop and implement a numerical scheme to represent the model in Figure 1 in non-denominational approach using the one-dimensional Euler-Bernoulli beam theory for bending boundary value problem [5, 8, 11, 12]. We utilize this scheme in two ways: First we consider the inverse problem: Given the spring stiffness k and known finite measurements of the resulting deflections v1 and v2 for the upper and lower beams respectively, then we find the imposed force f . Secondly, we use given forces f(x) and finite measurements of the resulting deflections v1(x) and v2(x) to find the spring stiffness k. L f E1, v1, ρ1, I1 E2, v2, ρ2, I2 Figure 1: Representation of the physical problem Following reference [10] and to avoid the complication of the shear stress imposed by the two dimensional nature of the plates, we consider the one-dimensional cross-section of the full problem, where we assume that the deflection of the glass plates can be described by beams. Therefore, we modeled the edge of the plate with a one-dimensional fixable beam, assuming the same bending from all parts of the plate without any shear stress. We use Euler-Bernoulli beam theory for the transverse displacements v1(x, t) and v2(x, t) (vertical displacements) of two simply supported one-dimensional beams of length L, thickness h, density ρ, flexural rigidity, I the moment of inertia, and E is Young’s modulus, for the top beam and for the lower beam (as shown in Figure 1), Considering the two elastic beams joined by continuous elastic layer of stiffness k. Both beams are pinned to a rigid frame along the boundary. The top beam is loaded by a point force f . The assumption of point load is based on the fact the affected area by the force is much smaller than the whole area of the plates. The governing set of equations for such system in this case is given as: ρ1h1v1tt = −(E1I1v1xx)xx + k(v2 − v1)− f(x, t) (1) ρ2h2v2tt = −(E2I2v2xx)xx − k(v2 − v1), (2) Our interest will be in the response to external load f(x, t) on the upper beam. The O. Alomari, Pavel B.Dubovski / Eur. J. Pure Appl. Math, 16 (2) (2023), 1140-1153 1142 boundary conditions are based on the assumption that the plates are pinned to a rigid frame along the boundary which maintains a fixed separation between them: v1(0, t) = v1(L, t) = v2(0, t) = v2(L, t) = 0, (3) v1xx(0, t) = v1xx(L, t) = v2xx(0, t) = v2xx(L, t) = 0, (4) The problem of steady state solution of the one-dimensional, two beam system was discussed an analyzed by Adriazola et al. [10], they solved the forward problem using analysis of Green’s functions for system response to point loads, they also solved the inverse problem of recovering the spring stiffness using Fourier decomposition. In this paper we consider the same time-independent (stationary) problem, therefore the system above, (1),(2), (3), and (4) becomes: −E1I1v (4) 1 + k(v2 − v1) = f(x), (5) −E2I2v (4) 2 − k(v2 − v1) = 0. (6) Subtracting (6) from (5) and letting u = v1 − v2 and L = 1 reduces the system to u(4) −Ku = g(x) (7) u(0, t) = u(1) = 0 = uxx(0) = uxx(1) = 0 (8) where: K = k E1I1 + k E2I2 , and g = −f E1I1 Obtaining the solution u(x) for this system (7) and (8), we then substitute the relation v1 = u+ v2 into (6) to get a modified equation for v2, −E2I2v (4) 2 − ku = 0. or v (4) 2 = g̃. where g̃ = −u E2I2 is the forcing on the lower beam due to the top beam, After solving for v2, we can recover v1 from the relation v1 = u+ v2. First we consider the inverse problem: Given the spring stiffness k and measurements of the resulting deflections v1(x) and v2(x), find the imposed force f(x), this is the same O. Alomari, Pavel B.Dubovski / Eur. J. Pure Appl. Math, 16 (2) (2023), 1140-1153 1143 as finding the function g ∈ L2[0, 1] in (7),(8) if K is known and a finite observation is given about the function u. Second we consider the inverse problem: Given a single imposed force f(x) and mea- surements of the resulting deflections v1(x) and v2(x), find the spring stiffness k, this is the same as recovering the coefficient K, in (7),(8), if a continuous function g(x) is given. 2. Formulation of the two inverse problems The problem of recovering a source function g or a coefficient K in (7)(8) are typical ill-posed problems whose solution does not depend continuously on the boundary data. That is, a small error in the specified data may result in an enormous error in the numer- ical solution [13]. We employ Tikhonov regularization technique to restore the stability of the numerical solution [3, 15]. Stable and efficient numerical methods are of high impor- tance. We assume that the only available information is finite observations of the solution uobs at xi, i = 1, 2, 3, . . . n. Henceforth, uobs is the interpolating cubic spline of the finite observations of the function u. 2.1. Inverse problem 1 Suppose that the functions g = g0 and K are known in (7)(8), we solve the direct problem and obtain the exact solution u(0) ̸= uobs. Then we seek a solution u1 in the neighborhood of u(0) and g1 = g0+v in neighborhood of g0, where, u1 meets the boundary conditions (8). We define the operator A as follows: Au = u(4) −Ku, D(A) = {u ∈ W 2 4 [0, 1] : u(0) = u(1) = u′′(0) = u′′(1) = 0}, Au = g, g ∈ L2[0, 1], where W p k (Ω) := {u ∈ Lp(Ω) : Dαu ∈ Lp(Ω) ∀α, |α| ≤ k}, and define the inner product (u, v) = ∫ 1 0 uv dx The inverse problem is to find u1 ∈ W 2 4 [0, 1] and v ∈ L2 such that: Au1 = g1 = g0 + v. Along the above, we consider the variational problem: O. Alomari, Pavel B.Dubovski / Eur. J. Pure Appl. Math, 16 (2) (2023), 1140-1153 1144 inf v∈L2 { α∥v∥2L2 + ∥u1 − uobs∥2W 2 4 } . Now we can reformulate the problem above as the following inverse problem of the function u = u1 − u(0). For given uobs and u(0), find u ∈ W 2 4 [0, 1] and v ∈ L2 such that: Au = v, u(0) = u(1) = u′′(0) = u′′(1) = 0, inf v∈L2 J(u, v), where, J = inf v∈L2 { α∥v∥2L2 + ∥u− (uobs − u(0))∥2W 2 4 } . In the last expression we seek u and v such that the minimum is attained. Here α is a constant. To minimize the functional J , we consider its variational δJ and equals it to zero, we obtain: δJ = 2α(v, δv) + 2(u− (uobs − u(0)), δu) = 0, here, δu satisfies Au = v, we get: Aδu = δv, now we introduce the adjoint operator, A∗ and obtain: (A∗q, δu) = (q, Aδu) = (q, δv). Clearly, A∗ = A, if we set A∗q = u− (uobs − u(0)), we obtain the control equation: αv + q = 0. So, the algorithm can be written as follows: Aun = vn, un(0) = un(1) = u′′n(0) = u′′n(1) = 0, (9) A∗qn = un − (uobs − u(0)), qn(0) = qn(1) = q′′n(0) = q′′n(1) = 0, (10) O. Alomari, Pavel B.Dubovski / Eur. J. Pure Appl. Math, 16 (2) (2023), 1140-1153 1145 vn+1 = vn − τ(αvn + qn). (11) Where arbitrary initial value v0 ∈ L2 and τ is a constant (stabilizer). For fixed n iteration steps, the computer program of the algorithm (9)-(11) is ap- proximated by a finite difference scheme justified in [2, 9, 16], that can be written in the following form: Set ∆x = 1 N , where, N is the number of grid points. The fourth derivative is approxi- mated as: D(4)(z(xi)) ≈ z(xi+2)− 4z(xi+1) + 6z(xi)− 4z(xi−1) + z(xi−2) (∆x)4 , and the computer program is approximated as follows: D(4)(u(xi))−Ku(xi) = v(xi), i = 1, 2, . . . , N, D(4)(q(xi))−Kq(xi) = u(xi)− (uobs(xi)− u(0)(xi)), i = 1, 2, . . . , N, We find u and q from above, after that we solve the control equation (11), which is approximated as follows: vn+1(i) = vn(i)− α(τvn(i)− q(xi)), i = 1, 2, . . . , N. Then, we are ready to pass to the next, n+ 1 iteration step. 2.1.1. Numerical solution for Inverse Problem 1 The aim of Inverse Problem 1 is to recover the right hand side function of (7) in the neighborhood of known g0(x) over [0, 1]. Given the known functions, K = 1, and g0(x) = 1 2 sin(2πx), by solving the forward problem (7) we obtain the initial solution u(0). If we set v(x) = 1 4 sin(πx) and calculate the exact solution ue for K = 1 and ge = g0(x) + v(x), then we can compare this solution with the calculated solution using our algorithm. The exact solution ue and the initial starting solution u(0) are shown in Figure 7, we set uobs to be the interpolating cubic spline of n finite values of ue after adding noise. Figure 2 shows a side by side comparison of the exact solution ue and the noisy measurements (uobs). We then pretend that we know neither the value of v(x) nor ue(x), we run the algorithm to recover v(x) based on an initial guess of v0 = vguess = 0 shown in Figure 5. Figure 6 shows a side by side comparison of the functions ve (expected value of v), and vc, the calculated value (recovered) of v. Figure 8 shows a side by side comparison of the functions ue (expected value of u), and vc, the calculated value of u. As shown in the figure the graphs of both recovered functions, u and v, are almost identical to the expected values. It is worth mentioning that Figure 3 shows the expected value of ge based on the exact solution ue is as expected (Aue = ge), however, when we plug the noisy solution uobs in O. Alomari, Pavel B.Dubovski / Eur. J. Pure Appl. Math, 16 (2) (2023), 1140-1153 1146 Auobs, we get instability as shown in Figure 4. Our method still works well with such noisy measurements. This is clear in Figure 9 that shows side by side comparison with Auc and Aue. 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 x -0.0005 0 0.0005 0.001 0.0015 0.002 0.0025 0.003 u e u obs Figure 2: Inverse problem 1, the graph shows noisy uobs compare to exact ue . 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 x -0.4 -0.2 0 0.2 0.4 0.6 0.8 Au e Figure 3: Inverse problem 1, the graph shows Aue . O. Alomari, Pavel B.Dubovski / Eur. J. Pure Appl. Math, 16 (2) (2023), 1140-1153 1147 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 x -400 -300 -200 -100 0 100 200 300 400 500 Au obs Figure 4: Inverse problem 1, the graph shows Auobs . 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 x 0 0.05 0.1 0.15 0.2 0.25 v e v guess Figure 5: Inverse problem 1, the graph shows the initial process of expected value of v,(ve) and the starting value of v, (v0 = vguess). 2.2. Inverse problem 2 Suppose that the functions g in (7) is known, we use series expansion method of K and the subsequent transformation of this problem to the first problem (recovery of the right-hand side). let u(x) = r(x) + βs(x) , and K = k0 + βk1(x). From (7), O. Alomari, Pavel B.Dubovski / Eur. J. Pure Appl. Math, 16 (2) (2023), 1140-1153 1148 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 x 0 0.05 0.1 0.15 0.2 0.25 0.3 v e v c Figure 6: Inverse problem 1, the graph shows a side by side comparison of the functions ve (expected value of v), and the calculated value of of v, (vc) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 x -0.0005 0 0.0005 0.001 0.0015 0.002 0.0025 0.003 u e u (0) Figure 7: Inverse problem 1, the graph shows a side by side comparison of the functions ue (expected value of u), and the initial value of u, (u(0)) before we start the calculation. r(4) + βs(4) − (k0 + βk1)(r + βs) = g combining the terms: r(4) − k0r = g (12) r(0) = r(1) = r′′(0) = r′′(1) = 0. O. Alomari, Pavel B.Dubovski / Eur. J. Pure Appl. Math, 16 (2) (2023), 1140-1153 1149 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 x -0.0005 0 0.0005 0.001 0.0015 0.002 0.0025 0.003 u e u c Figure 8: Inverse problem 1,the graph shows a side by side comparison of the functions ue (expected value of u), and the calculated value of of u, (uc) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 x -0.4 -0.2 0 0.2 0.4 0.6 0.8 Au e Au c Figure 9: Inverse problem 1, the graph shows a side by side comparison of the functions Aue (expected value of g(x)), and the calculated value of of Auc s(4) − k0s = k1r (13) s(0) = s(1) = s′′(0) = s′′(1) = 0. sobs = uobs − r β Note that Equation (12) is a linear forward problem since g(x) and k0 are known, therefore it can be solved directly to find the solution r(x). Also, sobs in (13) is calculated based on the known finite observation uobs O. Alomari, Pavel B.Dubovski / Eur. J. Pure Appl. Math, 16 (2) (2023), 1140-1153 1150 We utilize the algorithm developed above to solve the inverse problem (13) to recover the right-hand side v = k1r. Note that the only unknown in this right-hand-side is k1, therefore, k1 will be derived from the recovered right-hand-side in every iteration of the algorithm developed and justified in the previous sections. 2.2.1. Numerical solution for Inverse Problem 2 Our goal is to recover the coefficient function K = 1 of (7) over [0, 1] for the given function g(x) = sin(πx). These values were used in the calculations: β = 0.05, k0 = 0.5, initial guess for v is vguess = 0. As before, we pretend that we don’t know the function K, we run the algorithm to recover v(x) = k1r based on the mentioned initial guess. Figure 10 shows side by side comparison of exact value of u = r+ βs (ue) and calculated value of u (uc) . Figure 11 shows a side by side comparison of the exact function K = k0 + βk1 and the calculated value of K, (Kc). As we can see the graphs of both functions are almost identical as expected. 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 x -0.002 0 0.002 0.004 0.006 0.008 0.01 0.012 u c u e Figure 10: Inverse problem 2 with input parameters: g(x) = sin(πx), β = 0.05, k0 = 0.5, initial guess for v is vguess = 0, the graph shows the expected value of u, (ue) and the calculated value of u, (uc) 3. Conclusion We developed and implemented a numerical scheme to represent coupled mechanical system. We considered a one-dimensional model for two beams joined together by elastic layer. First we considered the inverse problem: Given the spring stiffness k and finite measurements of the resulting deflections v1(x) and v2(x), find the imposed force f(x), second we considered the inverse problem: Given a single imposed force f(x) and finite measurements of the resulting deflections v1(x) and v2(x), find the spring stiffness k. This development includes a numerical algorithm, that takes as input parameters the physical REFERENCES 1151 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 x 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 K c K e Figure 11: Inverse problem 2 with input parameters: g(x) = sin(πx), β = 0.05, k0 = 0.5, initial guess for v is vguess = 0, the graph shows the expected value of K, (Ke) and the calculated value of K, (Kc) measurements of the resulting deflections of the beams and outputs the external forces or spring stiffness. This computational framework can serve as a preliminary tool in the product development process, allowing scientists and engineers to simulate desired physical situations without costly testing of prototypes. The algorithm recovered the right-hand side f(x) of Equations (7), (8), this method is based on minimizing the defect in the functional between the calculated data and the measured data. 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