Selected HW Solutions

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1 Selected HW Solutions HW1 1 & See web page notes Derivative Approximations. For example: df f i+1 f i 1 = dx h i 1 f i + hf i + h h f i + h3 6 f i + f i + h 6 f i + 3 a realmax b realmin = 1 18 = c with a phantom, ɛ M = f i hf i + h f i h3 6 f i + = HW Show that the bisection method converges linearly. Suppose the true root is x. Define the error e n = x n x. Then bisection guarantees that e n b a/ n since we are always dividing the interval in half. Thus e n+1 = C e n where 0 C 1 for all n. Typically we don t hit the root and 0 < C < 1. This is linear convergence. 3 Complete the proof of the quadratic convergence of Newton s method. e n+1 = e n fe n + x f e n + x = e n e nf x + e nf x/ + f x + e n f x + e n e n + e nf /f e 1 + e n f /f n e n + f e n e n + f f e n f f e n f e n f f e n 1 f f e n

2 HW3 3 The parabola px through the three points x 1 h, f 0, x 1, f 1, x 1 + h, f is px = x x 1x x h f 0 x x 0x x h f 1 + x x 0x x 1 h f p x = x x 1 + x x h f 0 x x 0 + x x h f 1 + x x 0 + x x 1 h f HW4 p x 1 = h h f 0 + h h f = f f 0 h p x = h f 0 h f 1 + h f = f f 1 + f 0 h = p x 1 1 Derive Simpson s rule by Richardson extrapolation of TR: 4Ah Ah 3 = 1 3 In the TR integration formula, 4 h f 0 + f 1 + f h f 0 + f A = h fx 0 + fx 0 + h hf 0 + h f 0 + h3 4 f 0 From the formula for the exact integral setting x = x 0 + h, = h 3 f 0 + 4f 1 + f The local error is I hf 0 + h f 0 + h3 6 f 0 e l = I A h3 1 f 0 3 Exact answers are i cos0, ii lncos , iii Siπ tabulated

3 HW5 1 For du/dt = fu, prove that the backward Euler method is first-order accurate, using the definition of the local truncation error. Taylor expanding, we get ut + t = u + t u t + t + t τ u + t u + t u / + = u + t u + t u + + t τ or τ = t u / + O t. Prove that backward Euler is A-stable and L-stable. To analyze stability, we consider the model problem u = au, a < 0. For backward Euler we have 1 u n+1 = u n + a t u n+1, u n+1 = 1 a t u n Gu n 1 G = 1 1 a t = a t 1 which is always true, so the method is A-stable. L-stable, since it is A-stable and lim t G = 0. 3 The growth factors are a = 1 G BE = t, G T R = 1 t 1 + t Backward Euler is also a At t =, G T R = 0. G BE = 1/3, which is fine but not very accurate. The exact growth factor H = e t b For t =.1, TR oscillates between positive and negative values because G T R = 0.05/.05 < 0. BE has the correct monotonically decaying behavior since G BE = 1/3.1 though not very accurate since H HW6 1 For the linear ODE du/dt = au, prove that the fourth-order Runge-Kutta method is in fact fourth-order accurate. The exact growth factor H = e a t = 1 + a t + a t! + a3 t 3 3! 3 + a4 t 4 4! + a5 t 5 5! +

4 H agrees with the numerical growth factor G through fourth order, since from the formulas for RK4 with fu = au, G = 1 + a t + a t + a3 t a4 t 4 4 No Taylor series expansion is involved in finding G. Use the facts that K 1 = 1 au, K = au + 1 a t u, K 3 = au + 1 a t u a3 t u, K 4 = 1 au + 1 a t u a3 t u a4 t 3 u Show that the local error e l LT E for the backward Euler method for du/dt = au. The backward Euler method applied to the linear IVP du/dt = au is u n+1 = u n + a t u n+1 The LTE is defined by ut n+1 = ut n + a t ut n+1 + LT E Calculating the local error and assuming exact initial data ut n = u n, we have e l = ut n+1 u n+1 = a t ut n+1 u n+1 + LT E which implies to leading order in t. e l = a t e l + LT E LT E HW7 1 Note that an extremely small change in the initial conditions leads to opposite behavior in the solutions at large t 3, since the Lorenz equations are a chaotic dynamical system. Using different numerical methods or error tolerances effectively produces small changes in the initial conditions at later timesteps, leading to opposite behaviors in the solutions at large t, here at t 18 and 4. 4

5 HW8 1 Verify the TRBDF divided difference formula for the local error: 1 k t γ f 1 n γ1 γ f n+γ γ f n+1 = using and k t When does 1 1 γ u γ1 γ u t + γ t γ u t + t k t 3 u = e l u t + t = u t + t u t + t! u t + u t + γ t = u t + γ t u t + γ t u t +! d 3γ + 4γ dγ 1 γ = 0? 3 Cf. Molers NCM Figures 7.4 and 7.5. Try the nonstiff solvers ode45f,[0 1./delta],delta; ode3f,[0 1./delta],delta; ode113f,[0 1./delta],delta; vs. the stiff solvers ode3tbf,[0 1./delta],delta; ode3tf,[0 1./delta],delta; ode3sf,[0 1./delta],delta; ode15sf,[0 1./delta],delta; HW9 Solve the nonlinear BVP y = y 1, y0 = 0, y1 = 1. The residual F is F = D*y - y.^ b; b = [zerosn-,1; 1/h^]; 5

6 where b incorporates the BCs. The Jacobian J = F/ y is See bvp.m. J = D - spdiags*y,0,n-1,n-1; HW10 5n 4 5n function x = tridisolvea,b,c,d x = d; n = lengthx; % forward elimination for j = 1:n-1 mu = aj/bj; % n-1 flops bj+1 = bj+1 - mu*cj; % n-1 flops xj+1 = xj+1 - mu*xj; % n-1 flops end % back solve xn = xn/bn; % 1 flop for j = n-1:-1:1 end end xj = xj-cj*xj+1/bj; % n-1 flops % Total = 5n-4 flops HW11 1 Show that the Jacobi spectral radius µ = cosπh for Laplace s equation on the unit square with second-order accurate central differences only 1D version. In 1D, A = tridiag[ 1 1]. For Jacobi, M = I, M 1 = 1 I B = I M 1 A = 1 tridiag[1 0 1] Bv j = 1 sinj + 1πh + sinj 1πh = cosπh sinjπh = cosπhv j 6

7 a Derive the equation for the SOR ω opt, assuming Young s formula. λ = ω 1 4ω 1 = ω µ ω = 4 ± 16 16µ = 1 1 µ µ µ b For Laplace s equation on the unit square, the SOR spectral radius is λ = 1 1 cos πh 1 = cos πh 1 sinπh + sin πh 1 + sinπh1 sinπh = 1 sinπh 1 + sinπh HW1 1 Show that the TR method for the heat equation u t = u xx is second-order accurate. Hint: Use the facts that and ux + h ux + ux h h = u xx + h 1 u xxxx + u xx t + t = u xx + t u xxt + t u xxtt + The LTE is given by ut + t = u + t [ux + h ux + ux h + h ux + h, t + t ux, t + t + ux h, t + t] + t τ Taylor expanding, we get u + t u t + t u tt + t3 6 u ttt + = u + t [ u xx + h 1 u xxxx + + ] u xx + t u xxt + t u xxtt + h 1 u xxxx + + t τ 7

8 Then using u xxt = u tt and u xxtt = u ttt, τ = t 1 u ttt h 1 u xxxx + Show that the Lax-Friedrichs method is CFL stable for the wave equation u t +cu x = 0 by deriving the modified equation u t +cu x = D n u xx and requiring D n 0. The LF scheme is ut + t = 1 Taylor expanding, we get c t ux h + ux + h ux + h ux h h u + t u t + t u tt + = u + h u xx + c t u x + h 6 u xxx + The modified PDE u t + cu x = h t 1 r u xx D n u xx, r = c t h shows that LF is diffusive. Note that D num 0 when r 1. 8