Prof. Dr. O. Junge, T. März Scientific Computing - M3 Center for Mathematical Sciences Munich University of Technology Name: NUMERICAL ANALYSIS 2 - FINAL EXAM Summer Term 2006 Matrikelnummer: I agree to the publication of the results of this examination on the course home page. Only the Matrikelnummer will be associated to the result (no names or other personal data). Signature Hints: Make sure that your copy of the exam is complete and that the copy is readable. This exam should consist of 10 (ten) pages. Make sure you filled the Name and Matrikelnummer fields. If appropriate, also sign the agreement to publish the results. You have 90 Minutes time to solve the assignments. The exam has 39 points altogether plus 4 extra points for the optional assignment 9. You need at least 17 points to pass. Please read the assignments carefully, before you start solving them. In places where you find a box of page width, we expect full answers. If the space provided is not sufficient, use the page s back side. In places where you find small boxes, you may mark the right solution(s) by a cross. In the multiple choice part (assignments 4-7), you should answer only if you know the solution, because you will get: +1 point for the right answer, 0 points for no answer, points for a wrong anser. 1 2 Assisting material is allowed, but limited to one DIN A4 sheet of paper with your own notes and reminders. Except for this sheet, no other utilities are permitted. 1
Assignment 1: 9 Point(s) Let f : ]0, [ R, f C 2 (]0, [), be convex (i.e. the tangents are always below the graph of f ) and strictly monotonously decreasing with a root in x > 0 (i.e. f (x ) = 0). (a) Draw a qualitative sketch of the graph of f and show that for each starting value 0 < x 0 < x the Newton iteration converges monotonously to x. 2
(b) What can happen for x 0 > x? Give an example (it is enough to draw a sketch). c) Using Newton s method, construct an iteration for computing x = 1/a without using the division operator. Here a > 0 is a given number. For which initial values x 0 does this method converge (maximal intervall)? Hint: reuse parts a) and b) 3
Assignment 2: 11 Point(s) In order to construct a one step method to approximate the solution of an initial value problem ẋ = f (t, x), x(t 0 ) = x 0, (1) we use the following ansatz: Let p(t) = kt + b be a linear polynomial, satisfying p(t) = x ṗ(t + τ 2 ) = f (t + τ 2, p(t + τ 2 )). The method is then defined by evaluating p at t + τ Ψ t+τ,t x := p(t + τ). a) Transform Ψ t+τ,t x into the general form of Runge-Kutta methods and write down its Butcher scheme. b) Derive the order of this method. c) To which subclass of RK-methods does the method above belong? explicit: implicit: d) Set up the stability function of this method. e) Is the method A-stable?................................................... yes: no: 4
Assignment 3: 4 Point(s) a) Consider the following initial value problem ẋ = x 2, x(1) = 1. Why has this initial value problem a unique solution? b) Here the solution of the IVP from part a) is approximated on the time intervall [1, 1.25] by three different RK-methods. Each method is applied on five different equidistant meshes k with stepsizes τ k = 10 k, k = 1, 2, 3, 4, 5. The figure below shows the maximal error on each mesh k. ɛ k = max t k x(t) x k (t) What is the order of each method? Name a representative method of this order: Method : Method : Method : 5
Assignment 4: 3 Point(s) The following properties of the (continuous) evolution Φ of an ODE are inherited by the discrete evolution Ψ: Φ t,t x = x.................................................................. true false Φ t,s Φ s,t 0 x = Φ t,t 0 x........................................................ true false d dτ Φt+τ,t x τ=0 = f (t, x)................................................... true false Assignment 5: Let an explicit Runge-Kutta method be given by the following Butcher-scheme: 1 Point(s) 0 1 1 2 2 1 1 2 0 2 1 0 0 1 1 8 1 2 1 2 1 8 This RK method is consistent.................................................... true Assignment 6: The intervalwise condition number κ[t 0, T] of an initial value problem satisfies false 1 Point(s) κ[t 0, T] W(T, t 0 ), where W(T, t 0 ) denotes the coresponding propagation matrix................... true false Assignment 7: 1 Point(s) In order to implement an RK-method with step size control one needs at least two methods of different order................................................. true false 6
Assignment 8: Consider the model problem for elliptic partial differential equations 9 Point(s) u = f in Ω, u Γ = g, with the one-dimensional domain Ω =]x 1, x 5 [ R and boundary Γ = {x 1, x 5 }. The domain is discretized using the following mesh: x 1 x 2 x 3 x 4 x 5 with equidistant stepsize h = x i+1 x i (i = 1 : 4). Our aim is to set up a linear system of equations to solve our model problem on the mesh by a finite difference scheme of second order. (a) Derive the coefficients of a 3-point-stencil such that the finite difference scheme is second order accurate. Hint: Use Taylor expansion of u(x h) and u(x + h). 7
(b) Set up the linear system of equations of the discretized problem for g 0. Give the entries of A h, u h and f h explicitly. (c) Which components in the discrete equation in (b) change, if we assume g 1? Give the changed component(s) explicitly. 8
(d) Describe a possibility to develop a finite difference method of fourth order based on the difference stencil of part (a). Hint: Assume f C 2 and substitute u in the Taylor expansion of u by a finite difference scheme for f. 9
The following assignment is optional Assignment 9: 4 Point(s) Derive the weak formulation of the problem div(d u) + bu = f in Ω R 2, u = 0 on Ω, where b : Ω R, b L (Ω), with b(x) > 0 x Ω, is a given scalar valued function, D : Ω R 2 2, D L (Ω), with ξ T D(x)ξ α > 0 ξ R 2 x Ω, is a given matrix valued function, Ω has Lipschitz boundary and f L 2 (Ω). Derive the bilinearform a. Derive the linear form f,.. State the weak form of the problem in terms of the bilinearform a and the linearform f,. on the correct function space. Hint: if D(x) I, then div(d u) = div( u) = u. 10