1-D Finite-Element Methods with Poisson’s Equation

1. 1-D Finite-Element Methodswith Poisson’s Equation

2. Outline This topic discusses an introduction to finite-element methods • Review of Poisson’s equation • Defining a new kernel V(x) • Approximate solutions using uniform test functions

3. Outcomes Based Learning Objectives By the end of this laboratory, you will: • Understand how to approximate the heat-conduction/diffusion and wave equations in two and three dimensions • You will understand the differences between insulated and Dirichlet boundary conditions

4. The Target Equation Recall the first of Maxwell’s equations (Gauss’s equation): If we are attempting to solve for the underlying potential function, under the assumption that it is a conservative field, we have This is in the form of Poisson’s equation

5. The Target Equation In one dimension, this simplifies to: Define: and thus we are solving for

6. The Integral If , it follows that for any test function f(x) and therefore Substituting the alterative definition of V(x) into this equation, we get

7. The Integral Consider the first test function f1(x) :

8. Integration by Parts Again, take but before we apply integration by parts, expand the integral: Everything in the second integral is known: bring it to the right:

9. Integration by Parts The left-hand integral is no different from before, and performing integration by parts, we have

10. Integration by Parts First, substituting in the first test function: which yields 1 1 1

11. The System of Linear Equations Again, recall that we approximated the solution by unknown piecewise linear functions: where we define on

12. Unequally Spaced Points Recall that we approximated the left-hand integral by substituting the piecewise linear functions to get:

13. Integration by Parts That is, our linear equations are: This time, let’s take an actual example

14. Integration by Parts Consider the equation with the boundary conditions u(0) = 0 u(1) = 0

15. Integration by Parts The solution to the boundary value problem u(0) = 0 u(1) = 0 is the exact function

16. Integration by Parts We know in one dimension if the right-hand side is close to zero, the solution is a straight line Thus, choose 9 interior points with a focus on the centre: >> x_uneq = [0 0.16 0.25 0.33 0.41 0.5 0.59 0.67 0.75 0.84 1];

17. Integration by Parts We will compare this approximation with the approximation found using 9 equally spaced interior points • The finite difference approximation >> x_eq = 0:0.1:1;

18. The Test Functions The function to find the approximations is straight-forward: function [ v ] = uniform1d( x, uab, rho ) n = length( x ) - 2; idx = 1./diff(x); M = diag( -(idx( 1:end - 1 ) + idx( 2: end )) ) + ... diag( idx( 2:end - 1 ), -1 ) + ... diag( idx( 2:end - 1 ), +1 ); b = zeros ( n, 1 ); for k = 1:n b(k) = 0.5*int( rho, x(k), x(k + 2) ); end b(1) = b(1) - idx(1)*uab(1); b(end) = b(end) - idx(end)*uab(end); v = [uab(1); M \ b; uab(2)]; end int( rho, a, b ) approximates

19. The Test Functions The right-hand function of is also straight-forward: function [ u ] = rho( x ) u = sin( pi*x ).^4; end

20. The Equations Thus, we can find our two approximations: >> x_eq = (0:0.1:1)'; >> plot( x_eq, uniform1d( x_eq, [0, 0], @rho ), 'b+' ); >> hold on >> x_uneq = [0 0.16 0.25 0.33 0.41 0.5 0.59 0.67 0.75 0.84 1]'; >> plot( x_uneq, uniform1d( x_uneq, [0, 0], @rho ), 'rx' );

21. The Equations It is difficult to see which is the better function, therefore create a function storing the actual solution (as found in Maple): function u = u(x) u = -1/16*( ... (cos(pi*x).^4 - 5*cos(pi*x).^2 + 4)/pi^2 + ... 3*x.*(x - 1) ... ); end

22. The Equations Instead, plotting the errors: >> x_eq = 0:0.1:1; >> plot( x_eq, uniform1d( x_eq, [0, 0], @rho ) - u(x_eq), 'b+' ); >> hold on >> xnu = [0 0.16 0.25 0.33 0.41 0.5 0.59 0.67 0.75 0.84 1]; >> plot( x_uneq, uniform1d( x_uneq, [0, 0], @rho ) - u(x_uneq), 'rx' );

23. Integration by Parts Understanding that the right-hand side has a greater influence in the centre, appropriately changing the sample points yielded a significantly better approximation

24. Summary In this topic, we have generalized Laplace’s equation to Poisson’s equation • Used the same uniform test functions • We looked at a problem for which there is an exact solution • Changing the points allowed us to get better approximations

25. What’s Next? The impulse function (the derivative of a step function) is difficult to deal with… We will next consider test functions that avoid this… • The test functions will be tents • This generalizes to higher dimensions

26. References [1] Glyn James, Advanced Modern Engineering Mathematics, 4th Ed., Prentice Hall, 2011, §§9.2-3.

27. Usage Notes • These slides are made publicly available on the web for anyone to use • If you choose to use them, or a part thereof, for a course at another institution, I ask only three things: • that you inform me that you are using the slides, • that you acknowledge my work, and • that you alert me of any mistakes which I made or changes which you make, and allow me the option of incorporating such changes (with an acknowledgment) in my set of slides Sincerely, Douglas Wilhelm Harder, MMath dwharder@alumni.uwaterloo.ca