16.920J/SMA 5212 Numerical Methods for PDEs the characteristic polynomial becomes P(a)=P(S)=[(1-b)S-1=0 The principal root is therefore Ah+2h2 which, upon comparison with e"=1+hh+Ah+., is only first-order accurate The solution is 1-th IMPLICIT TIME-MARCHING SCHEME Euler-Backward For the Parabolic PDE, A is always real and <0 Therefore, the transient component will always tend towards zero for large n irregardless of h(≡△ The time-marching scheme is always numerically stable In this way, the implicit Euler/Euler-backward time discretization scheme will allow us to resolve different time-scaled events with the use of different time-step sizes. A small time-step size is used for the short time scaled events, and then a large time-step size used for the longer time-scaled events there is no constraint on hm Slide 47 IMPLICIT TIME-MARCHING SCHEME Euler-Backward However, numerical solution of u requires the solution of a set of simultaneous algebraic equations or matrix inversion, which is computationally much more16.920J/SMA 5212 Numerical Methods for PDEs 30 the characteristic polynomial becomes (σ ) (S ) (1 hλ ) S 1 0 ￾ ✁ Ρ = Ρ = − − = ✂ ✄ The principal root is therefore 1 2 2 1 .... 1 h h h σ λ λ λ = = + + + − 1 2 2 which, upon comparison with 1 .... , is only 2 first-order accurate. h e h h λ = + λ + λ + The solution is ( ) ( ) 1 1 1 1 1 n u h n h ahe U h h e µ µ β λ λ + ☎ ✆ = + ✝ ✞ − − − ✟ ✠ Slide 46 IMPLICIT TIME-MARCHING SCHEME Euler-Backward For the Parabolic PDE, λ is always real and < 0. Therefore, the transient component will always tend towards zero for large n irregardless of h (≡ ∆t). The time-marching scheme is always numerically stable. In this way, the implicit Euler/Euler-backward time discretization scheme will allow us to resolve different time-scaled events with the use of different time-step sizes. A small time-step size is used for the short time￾scaled events, and then a large time-step size used for the longer time-scaled events. There is no constraint on hmax. Slide 47 IMPLICIT TIME-MARCHING SCHEME Euler-Backward However, numerical solution of u requires the solution of a set of simultaneous algebraic equations or matrix inversion, which is computationally much more