Solving Fixed Point Problems and Differential Equations using Iterative Methods, Assignments of Mathematical Methods for Numerical Analysis and Optimization

Download Solving Fixed Point Problems and Differential Equations using Iterative Methods and more Assignments Mathematical Methods for Numerical Analysis and Optimization in PDF only on Docsity! Jeffrey Hellrung Friday, November 23, 2005 Math 269A, Assignment 08 1. Consider the task of solving the fixed point problem x = G(x) using fixed point iteration, e.g., xn+1 = G(xn) for n = 0, 1, 2, . . .. (a) Give conditions on G and the initial iterate x0 that will insure that the fixed point iteration converges. (b) Assume one is using Backward Euler to solve the differential equation dy/dt = f(y). If the implicit equation that arises is solved using fixed point iteration, what is the restriction on the timestep that will ensure that the iteration converges? (c) Assume one is using the trapezoidal method to solve the differential equation dy/dt = f(y). If the implicit equation that arises is solved using fixed point iteration, what is the restriction on the timestep that will ensure that the iteration converges? (d) What conclusion can you draw about the utility of using implicit ODE methods for stiff systems when the implicit equations that arise are solved using fixed point iteration? Solution (a) Let x∗ be a fixed point of G (supposing one exists), i.e., G(x∗) = x∗. Then fixed point iteration converges if G is a contraction mapping within a neighborhood of x∗ and the initial iterate x0 lies in this neighborhood. Explicitly, we want |xn+1 − x ∗| ≤ α|xn − x ∗| for some 0 < α < 1. In this case, we have ∣ ∣ ∣ ∣ G(xn) − G(x ∗) xn − x∗ ∣ ∣ ∣ ∣ ≤ α < 1, and by the Mean Value Theorem, the difference quotient is the value of |G′| somewhere between xn and x ∗. It follows that if G′ is bounded in absolute value by some α < 1 in some neighborhood of x∗, then if the initial iterate x0 lies in this neighborhood, fixed point iteration will converge, i.e., xn → x ∗. Indeed, if |G′| < α uniformly, then there exists a unique fixed point, and fixed point iteration will converge for any initial iterate. (b) The Backward Euler method takes the form yk+1 = yk + hf(yk+1), hence we wish to solve x = G(x) = yk + hf(x). By the comments above, this will have a unique fixed point, and fixed point iteration will converge for any initial iterate, if |G′(x)| = h ∣ ∣ ∣ ∣ ∂f ∂y (x) ∣ ∣ ∣ ∣ ≤ α < 1 for all x ∈ R. 1 (c) The Trapezoidal method takes the form yk+1 = yk + h 2 f(yk) + h 2 f(yk+1), hence we wish to solve x = G(x) = yk + h 2 f(yk) + h 2 f(x). Thus we require that |G′(x)| = h 2 ∣ ∣ ∣ ∣ ∂f ∂y (x) ∣ ∣ ∣ ∣ < 1. (d) Using implicit methods for solving stiff ODEs by fixed point iteration provides little advantage, if any, over explicit methods, since similar time step restrictions are necessary. 2. Consider the system of differential equations given by du dt = 1 100 − ( 1 100 + u + v ) (1 + (u + 1000)(u + 1)) , dv dt = 1 100 − ( 1 100 + u + v ) ( 1 + v2 ) . (a) Give an expression for the Jacobian of this system. (b) What are the eigenvalues of the Jacobian when u = 0.0 and v = 0.5? (c) Assuming that u is close to 0.0 and v is close to 0.5, estimate the timestep required to obtain a qualitatively correct solution when a fourth order Runge-Kutta method is used. (d) Assume one is using Backward Euler to solve this system. Write out, in component form, the iteration that one must implement if one is going to use Newton’s method to solve the non-linear equations that must be solved to advance the solution. Solution (a) We have F (u, v) = ( 1 100 − ( 1 100 + u + v ) (1 + (u + 1000)(u + 1)) 1 100 − ( 1 100 + u + v ) ( 1 + v2 ) ) , so JF(u,v) = ( −3u2 − 2uv − 100101u/50 − 1001v − 101101/100 −u2 − 1001u − 1001 −v2 − 1 −3v2 − 2uv − v/50 − 1 ) . (b) At (u, v) = (0.0, 0.5), the Jacobian is JF(0.0,0.5) = ( −1511.51 −1001. −1.25 −1.76 ) , which has eigenvalues λ1 = −1512.34, λ2 = −0.931675. (c) We require both hλ1 and hλ2 to be in the region of absolute stability, hence h ∈ (0, 0.00184171). 2