FINITE ELEMENT METHOD: APPROXIMATE SOLUTIONS

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1 FINITE ELEMENT METHOD: APPROXIMATE SOLUTIONS I Introduction: Most engineering problems which are expressed in a differential form can only be solved in an approximate manner due to their complexity. The best known techniques are the finite difference and the finite element methods. These two techniques although apparently different achieve similar objectives by reducing the infinite degrees of freedom of a continuous system to a finite set. By doing this the problem can be solved numerically and becomes amenable to computer solving. The finite difference technique defines a series if nodes at which the discrete version of the differential equation is satisfied. In finite elements the differential equation or its inner product version is satisfied in an average sense over a region or element. Two categories of methods are widely used to achieve this objective:. Methods of weighted residuals including: a Collocation method b Least square method c Galerkin s method 2. Variational methods including a Rayleigh-Ritz method II Methods of Weighted Residuals The main steps of the this category of methods is as follow. Assume a trial function that satisfies the boundary conditions and contains unknown coefficients to be determined 2. The residual is then computed by substituting the trial function into the differential equation. The unknowns in the trial function are determined such that the chosen trial function best approximate the exact solution. To this end a weighting function is selected and the weighted average of the residual over the problem domain is set to zero. Consider the following sample problem d 2 u 2 u = x, < x < u =, and u = Let ũ = ax x be the trial function. The residual R is then calculated as R = d2 ũ ũ + x = 2a ax x + x 2 A weighting function ω is then chosen such as the average of the residual over the problem domain is zero.i.e ω R = ω { 2a ax x + x} =

2 The next step is to decide about the weighting test function to be used since the resultant approximate solution will differs depending on this choice. The methods of weighted residuals can be classified based on how the test function is determined. For instance. In the Collocation Method ω = δx where < <. Direct substitution with =.5 leads to ũ =.2222x x 2. In the Least Square method the test function is determined from the residual such that Direct substitution leads to ω = dr da = 2 x x ũ =.25x x. In the Galerkin method, the test function comes from the chosen trial function. that is Direct substitution leads to ω = = x x da ũ =.2272x x Notes: In order to improve the approximate solutions, we can add more terms to the previously selected trial function. For example, another trial function is ũ = a x x + a 2 x 2 x This trial function has two unknown constants to be determined. Computation of the residual using the present trial function yields R = d2 ũ 2 ũ + x = a 2 x + x 2 + a 2 2 x x 2 + x + x We need the same number of test functions as that of the unknown so that the constants can be determined properly. The table below summarizes how to determine test functions for a chosen trial function with n unknowns coefficients. 2

3 Table : Weighting function for the methods of weighted residuals Collocation ω i = δx, i =, 2,...n where is a point within the domain Least Squares Galerkin ω i = R a i, i =, 2,...n where R is the residual and a i is an unknown coefficient in the trial function ω i = ũ a i, i =, 2,...n where ũ is the selected trial function Application to the present trial function results in the following test functions for each method.. Collocation Method ω = δx x ω 2 = δx x 2 x and x 2 must be selected such that the resultant weighted residual can produce two independent equations to determine unknowns a and a 2 uniquely. 2. Least Squares Method ω = 2 x + x 2 ω 2 = 2 x x 2 + x Note that the least squares method produces a symmetric matrix regardless of a chosen trial function. As a matter of fact Consider the general differential function written as Lu = f where L is a linear differential operator. A trial solution is chosen such that ũ = a i g i i= in which g i is a known function in terms of the spatial coordinates system and it is assumed to satisfy boundary conditions. Substitution leads to the following residual R = a i + p i= and p are functions in terms of the spatial coordinates system. In the least square method the weighting functions are ω j = h j j =, 2,..., n The weighted average of the residual over the domain yields the matrix equation I = ω j R dω = A ij a i b j = Ω i= j =, 2,..., n

4 . Galerkins Method A ij = h j dω = A ji Ω II. ω = x x ω 2 = x 2 x Galerkins method does not result in a symmetric matrix when applied. However, Galerkins method may produce a symmetric matrix under certain conditions. Weak Formulation The formulation described in the preceeding section is called the strong formulation of the weighted residual method. The strong formulation requires evaluation of ω d2ũ 2 whicncludes the highest order of derivative term in the differential equation. The integral must have a non-zero finite value to yield a meaningful approximate solution to the differential equation. This means that the trial function should be differentiable twice and its second derivative should not vanish. To reduce the requirement for a trial function in terms of order of differentiability, integration by parts is applied to the strong formulation. I = d2ũ ω 2 ũ + x = dω ωũ + xω [ + ω ] = In the last equation the trial function needs the first order differentiation instead of the second order differentiation. As a result, the requirement for the trial function is reduced. This formulation is called the weak formulation. Weak formulation has an advantage for Galerkins method where test functions are obtained directly from the selected trial function. II.2 Piecewise continuous trial function Regardless of the weak or strong formulation, the accuracy of an approximate solution so much depends on the chosen trial function. However, assuming a proper trial function for the unknown exact solution is not an easy task. This is especially true when the unknown exact solution is expected to have a large variation over the problem domain, the domain has a complex shape in two-dimensional or three-dimensional space, and/or the problem has complicated boundary conditions. In order to overcome these problems, a trial function can be described using piecewise continuous functions.i.e. splines. Consider solving the same problem from the previous section using the weak formulation d 2 u 2 u = x, < x < u =, and u = Take the trial function as Let ũ = a Φ x + b Φ 2 x where x x / Φ x = 2 x / x 2/ 2/ x x / Φ 2 x = x / x 2/ x 2/ x 4

5 II. Galerkin s finite element formulation As seen in the previous section, use of piecewise continuous functions for the trial function has advantages. As we increase the number of sub-domains for the piecewise functions, we can represent a complex function by using sum of simple piecewise linear functions. Later, the sub-domains are called finite elements. This section shows how to compute weighted residual in a systematic manner using finite elements and piecewise continuous functions. In the previous section, the piecewise continuous functions were defined in terms of the generalised coefcients i.e. a, a 2, etc.. For a systematic formulation, the piecewise continuous functions are dened in terms of nodal variables. We proceed by dividing the solution interval [, ] into a number of subintervals [, + ] and i = to n. Consider a sub-domain with two nodes, one at each end. At each node, the corresponding coordinate value ; or + and the nodal variable ũ i ; or ũ i+ are assigned. Let us assume the unknown trial function to be Since ũ = c x + c 2 ũ i = c + c 2 then ũ i+ = c + + c 2 ũ = H xũ i + H 2 xũ i+ where H x = + x H 2 x = x = + H x and H 2 x are called linear shape functions. They satisfy the following conditions. The shape function associated with node i has a unit value at node i and vanishes at other nodes The weak formulation can be transformed to or i= xi+ H = H 2 + = H + = H 2 = dω ωũ + xω I + u u = [ + ω ] = where I = i= xi+ dω ωũ + xω + 5

6 over eacnterval [, + ] we have ũ = H xũ i + H 2 xũ i+ H x = + x H 2 x = x = + leading to two contributions from two weighting functions I = I 2 = xi+ xi+ ω = du i = H x ω = dh du i+ = H 2 x H ũ + xh dh 2 H 2ũ + xh 2 Careful calculation leads to the following quadratic form I = + h i u i + h i u i I 2 = + h i u i + h i u i or or I I 2 = u i u i+ u i u i

7 III The Variational Method In the variational method, the original boundary value problem is transformed into finding the extremum maximun/minimum of the functional of the problem f y d f x2 y = F inding Extremum of J = fx, y, y = dy Consider the following sample problem d 2 u 2 u = x, < x < u =, and u = To solve this problem using the variational method we consider finding the extremum of the following integral x where J = fx, u, du since III. fx, u, du = 2 2 du + 2 u2 u x f u d f u = u x d2 u 2 = It is generally difficult to to derive the functional of the problem. The Rayleigh-Ritz Method The Rayleigh-Ritz method obtains an approximate solution to a differential equation with given boundary conditions using the functional of the equation. The procedure of this technique can be summarized in two steps as given below:. Assume an admissible solution which satises the Dirichlet boundary condition or essential boundary condition and contains unknown coefficients. 2. Substitute the assumed solution into the functional and nd the unknown coefcients that minimize/maximize the functional. In the previous problem, assume the approximate solution to be of the form Inserting this in the last integral leads to ũ = ax x J = 2 a2 [ 2x 2 + x 2 x 2 ] a x 2 x minimizing the functional with respect to the unknown coefficient leads to a =

8 dj da = Additional terms can be added to improve the approximate solutions. In the previous problem, assume the approximate solution to be of the form ũ = a x x + a 2 x 2 x where a and a 2 are two unknown coefficients the value of whics determined by setting J a = and J a 2 = III.2 The Rayleigh-Ritz finite element method The Rayleigh-Ritz method can be applied to a problem domain using continuous piecewise functions. As s result, the problem domain is divided into sub-domains of nite elements. For elements with two nodes apiece, the linear shape functions can be used for the Rayleigh-Ritz method. Consider the following sample problem d 2 u 2 u = x, < x < u =, and u = or J = 2 2 du + 2 u2 u x is maximum/minimum now can be expanded as J = i= xi+ 2 2 du + 2 u2 u x Using linear shape functions we get over eacnterval [, + ] ũ = u i H x + u i+ H 2 x and xi+ 2 2 du + 2 u2 u x 8