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13 weightedresidual
This document discusses numerical methods for solving boundary value problems using weighted residual methods. It introduces the weighted residual method and describes how it works by minimizing the residual over the domain. The document provides examples of using the collocation method, including setting up the system of equations in matrix form and choosing appropriate trial functions that satisfy boundary conditions. It also includes an example problem of using the weighted residual method to solve the bar tensile problem.
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13 weightedresidual
- 1. 1 ENME 602 Spring2007 Dr. Eng. Mohammad Tawfik Numerical Solution of Boundary Value Problems Weighted Residual Methods ENME 602 Spring 2007 Dr. Eng. Mohammad Tawfik Objectives • In this section we will be introduced to the general classification of approximate methods • Special attention will be paid for the weighted residual method • Derivation of a system of linear equations to approximate the solution of an ODE will be presented using different techniques
- 2. 2 ENME 602 Spring2007 Dr. Eng. Mohammad Tawfik Classification of Approximate Solutions of D.E.’s • Discrete Coordinate Method – Finite difference Methods – Stepwise integration methods • Euler method • Runge-Kutta methods • Etc… • Distributed Coordinate Method ENME 602 Spring 2007 Dr. Eng. Mohammad Tawfik Distributed Coordinate Methods • Weighted Residual Methods – Interior Residual • Collocation • Galrekin • Finite Element – Boundary Residual • Boundary Element Method • Stationary Functional Methods – Reyligh-Ritz methods – Finite Element method
- 3. 3 ENME 602 Spring2007 Dr. Eng. Mohammad Tawfik Basic Concepts • A linear differential equation may be written in the form: ( )( ) ( )xgxfL = • Where L(.) is a linear differential operator. • An approximate solution maybe of the form: ( ) ( )∑= = n i ii xaxf 1 ψ ENME 602 Spring 2007 Dr. Eng. Mohammad Tawfik Basic Concepts • Applying the differential operator on the approximate solution, you get: ( )( ) ( ) ( ) ( ) ( )( ) ( ) 0 1 1 ≠−= −⎟ ⎠ ⎞ ⎜ ⎝ ⎛ =− ∑ ∑ = = xgxLa xgxaLxgxfL n i ii n i ii ψ ψ ( )( ) ( ) ( )xRxgxLa n i ii =−∑=1 ψ Residue
- 4. 4 ENME 602 Spring2007 Dr. Eng. Mohammad Tawfik Handling the Residue • The weighted residual methods are all based on minimizing the value of the residue. • Since the residue can not be zero over the whole domain, different techniques were introduced. ENME 602 Spring 2007 Dr. Eng. Mohammad Tawfik General Weighted Residual Method
- 5. 5 ENME 602 Spring2007 Dr. Eng. Mohammad Tawfik Objective of WRM • As any other numerical method, the objective is to obtain of algebraic equations, that, when solved, produce a result with an acceptable accuracy. • If we are seeking the values of ai that would reduce the Residue (R(x)) allover the domain, we may integrate the residue over the domain and evaluate it! ENME 602 Spring 2007 Dr. Eng. Mohammad Tawfik Evaluating the Residue ( )( ) ( ) ( )xRxgxLa n i ii =−∑=1 ψ ( )( ) ( )( ) ( )( ) ( ) ( )xRxgxLaxLaxLa nn =−+++ ψψψ ...2211 n unknown variables ( ) ( )( ) ( ) 0 1 =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −= ∫ ∑∫ =Domain n i ii Domain dxxgxLadxxR ψ One equation!!!
- 6. 6 ENME 602 Spring2007 Dr. Eng. Mohammad Tawfik Using Weighting Functions • If you can select n different weighting functions, you will produce n equations! • You will end up with n equations in n variables. ( ) ( ) ( ) ( )( ) ( ) 0 1 =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −= ∫ ∑∫ =Domain n i iij Domain j dxxgxLaxwdxxRxw ψ ENME 602 Spring 2007 Dr. Eng. Mohammad Tawfik Collocation Method • The idea behind the collocation method is similar to that behind the buttons of your shirt! • Assume a solution, then force the residue to be zero at the collocation points
- 7. 7 ENME 602 Spring2007 Dr. Eng. Mohammad Tawfik Collocation Method ( ) 0=jxR ( ) ( )( ) ( ) 0 1 =− = ∑= j n i jii j xFxLa xR ψ ENME 602 Spring 2007 Dr. Eng. Mohammad Tawfik Example Problem
- 8. 8 ENME 602 Spring2007 Dr. Eng. Mohammad Tawfik The bar tensile problem ( ) 0/ 00 ' 02 2 =⇒= =⇒= =+ ∂ ∂ dxdulx ux sBC xF x u EA ENME 602 Spring 2007 Dr. Eng. Mohammad Tawfik Bar application ( ) 02 2 =+ ∂ ∂ xF x u EA ( ) ( )∑= = n i ii xaxu 1 ψ ( ) ( ) ( )xRxF dx xd aEA n i i i =+∑=1 2 2 ψ Applying the collocation method ( ) ( ) 0 1 2 2 =+∑= j n i ji i xF dx xd aEA ψ
- 9. 9 ENME 602 Spring2007 Dr. Eng. Mohammad Tawfik In Matrix Form ( ) ( ) ( )⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ ⎪ ⎪ ⎩ ⎪ ⎪ ⎨ ⎧ −= ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ ⎪ ⎪ ⎩ ⎪ ⎪ ⎨ ⎧ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ nnnnnn n n xF xF xF a a a kkk kkk kkk MMMOMM 2 1 2 1 21 22212 12111 ... ... ... Solve the above system for the “generalized coordinates” ai to get the solution for u(x) ( ) jxx i ij dx xd EAk = = 2 2 ψ ENME 602 Spring 2007 Dr. Eng. Mohammad Tawfik Notes on the trial functions • They should be at least twice differentiable! • They should satisfy all boundary conditions! • Those are called the “Admissibility Conditions”.
- 10. 10 ENME 602 Spring2007 Dr. Eng. Mohammad Tawfik Using Admissible Functions • For a constant forcing function, F(x)=f • The strain at the free end of the bar should be zero (slope of displacement is zero). We may use: ( ) ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = l x Sinx 2 π ψ ENME 602 Spring 2007 Dr. Eng. Mohammad Tawfik Using the function into the DE: • Since we only have one term in the series, we will select one collocation point! • The midpoint is a reasonable choice! ( ) ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −= l x Sin l EA dx xd EA 22 2 2 2 ππψ { } { }faSin l EA −= ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − 1 2 42 ππ
- 11. 11 ENME 602 Spring2007 Dr. Eng. Mohammad Tawfik Solving: • Then, the approximate solution for this problem is: • Which gives the maximum displacement to be: • And maximum strain to be: ( ) ( ) EA fl EA fl SinlEA f a 2 2 2 21 57.0 24 42 ≈== πππ ( ) ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ≈ l x Sin EA fl xu 2 57.0 2 π ( ) ( )5.057.0 2 =≈ exact EA fl lu ( ) ( )0.19.00 =≈ exact EA lf ux ENME 602 Spring 2007 Dr. Eng. Mohammad Tawfik Homework #11 • Solve the beam bending problem, for beam displacement, for a simply supported beam with a load placed at the center of the beam using – Any weighting function – Collocation Method • Use three term Sin series that satisfies all BC’s • Write a program that produces the results for n-term solution. )(4 4 xF dx wd = 0 )()0( 0)()0( 2 2 2 2 == == dx lwd dx wd lww
- 12. 12 ENME 602 Spring2007 Dr. Eng. Mohammad Tawfik Exact Solution 12/1 10 3 15 7 412 2/10 60 13 12 )( 23 3 <<+−+−= <<+= x xxx x xx xw