A loaded strut in equilibrium satisfies the force-balance condition for a…

PHYSICS · Multi-part · Fermi University Physics · Mathematical Methods for Physics · Mechanics & Dynamical Systems · Nonlinear dynamics and chaos

Error, Chance and the Machine, Ch Iteration: nonlinear systems and eigenvalues

Problem

A loaded strut in equilibrium satisfies the force-balance condition f(x)=0f(x) = 0 for a dimensionless displacement xx measured from the unloaded configuration. At the critical buckling load the two equilibrium branches merge into a single degenerate equilibrium, so ff has a double root at x∗x^*: f(x∗)=f′(x∗)=0f(x^*) = f'(x^*) = 0 but f′′(x∗)≠0f''(x^*) \neq 0. Write f(x)=(x−x∗)2h(x)f(x) = (x-x^*)^2 h(x) with hh smooth and h(x∗)≠0h(x^*) \neq 0. There Newton's method converges only linearly, with ratio 12\frac12. Consider instead the doubled step xn+1=xn−2f(xn)/f′(xn)x_{n+1} = x_n - 2f(x_n)/f'(x_n). ((a)) With e=x−x∗e = x - x^*, show that the doubled step satisfies en+1≈Cen2e_{n+1} \approx C e_n^2 with C=h′(x∗)/(2h(x∗))C = h'(x^*)/\bigl(2h(x^*)\bigr), so it is quadratic at a double root. ((b)) For one strut the force-balance condition is f(x)=x3−x2−x+1=(x−1)2(x+1)f(x) = x^3 - x^2 - x + 1 = (x-1)^2(x+1). From x0=2x_0 = 2, take three steps of plain Newton and three of the doubled step. Give the errors, and compare them with the ratio 12\frac12 and with the constant CC of part (a).

Browse the problem library