Past exam of the mathematics course of the University of Cambridge 2006 ii Paper 4 39C b Solution Created 2026-09-24 Updated 2026-10-14
Use a Fourier mode . Put . Its amplification polynomial of a multistep method isThe two roots are real, one nonnegative and one nonpositive. For , evaluation at gives and at zero gives , so the positive root lies below one. The negative root is at least exactly when the value at , namely , is nonnegative. Thus both roots have modulus at most one exactly for . At they are , and at they are ; all unit-modulus roots are simple, satisfying the multistep root condition for a multistep method. Since the largest available is , the Adams-Bashforth stability for centered diffusion condition isFor , the highest-frequency mode has a root less than and grows, proving necessity as well as sufficiency.