Lesson 5 Method of Weighted Residuals

Classical Solution Technique The fundamental problem in calculus of variations is to obtain a function f(x) such that small variations in the function  f(x) will not change the original function The variational function can be written in general form for a second-order governing equation (no first derivatives) as Where  and  are prescribed values

Classical solution (continued) An equation containing first-order derivatives may not have a corresponding variational function. In some cases, a pseudovariational function can be used where C=C(x)

Classical solution (continued) Example: Consider a rod of length L. The equation defining heat transfer in the rod is with boundary conditions Integrating twice, one obtains Applying boundary conditions, the final result is

Rayleigh-Ritz Method FEM variational approach attributed to Lord Rayleigh ( ) & Walter Ritz ( ) Let Assume a quadratic function with boundary conditions

R-R Method (continued) Hence, Now integrate Thus

R-R Method (continued) which becomes To find the value of C 3 that makes J a minimum, Therefore,

Variation Methods Given a function u(x), the following constraints must be met –(1) satisfies the constraints u(x 1 )=u 1 and u(x 2 )=u 2 –(2) is twice differentiable in x 1 <x<x 2 –(3) minimizes the integral

Variational methods (continued) Then it can be shown that u(x) is also the solution of the Euler-Lagrange equation where

Variational methods (continued) For higher derivatives of u, Hence,

Variational methods (continued) In 2-D, the constraints are –(1) satisfies the constraint u = u 0 on  –(2) is twice differentiable in domain A(x,y) –(3) minimizes the functional and

Variational methods (continued) Example: Find the functional statement for the 2-D heat diffusion equation Applying the Euler-Lagrange relation

Variational methods (continued) we find that which yields the final functional form

A Rayleigh-Ritz Example Begin with the equation with boundary conditions First find the variational statement (J)

R-R example (continued) The variational statement is Assume a quadratic approximation with boundary conditions

R-R example (continued) Thus, Now, To be a minimum, Finally

The Weak Statement Method of Weighted Residuals - one does not need a strong mathematical background to use FEM. However, one must be able to integrate. To illustrate the MWR, let us begin with a simple example - Conduction of heat in a rod of length L with source term Q.

weak statement (continued) Integrating, or This analytical solution serves as a useful benchmark for verifying the numerical approach.

weak statement (continued) There are basically two ways to numerically solve this equation using the FEM: Rayleigh – Ritz Method and the Galerkin Method ( which produces a “weak” statement) Consider

weak statement (continued) Since u = c i  i (x,y) is an approximate function, substitution into the above equation may not satisfy the equation. We set the equation equal to an error (  ) We now introduce a set of weighting functions (test functions) W i, and construct an inner product (W i,  ) that is set to zero – this forces the error of the approximate differential equation to zero (average).

weak statement (continued) Hence, Example: The inner product becomes

weak statement (continued) Integrating by parts (Green-Gauss Theorem) Problem: Use Galerkin’s Method to solve

weak statement (continued) The inner product is or The weak statement becomes

weak statement (continued) Since we obtain This is the same as the Rayleigh-Ritz Method but no variational principle is required.

Weighting Function Choices Galerkin Least Squares Method of Moments

Weighting Function Choices (continued) Collocation Sub domain

Least Squares Example Let For the previous example problem

Least Squares Example (continued) Thus, Some observations: (1) no integration by parts required – not a weak form (2) the Neumann boundary conditions do not appear naturally (3) in this case  i must satisfy global boundary conditions, which is difficult in most problems