16.920J/SMA 5212 Numerical Methods for PDEs (Sc"-2hSc"-c")=0 (S2-2hS-1)=0 characteristic polynomial p(S)=(S2-2hS-1)=0 EXAMPLE 2 Leapfrog Time Discretization: Time Shift Operator The solution to the characteristic polynomial is a(Ah)=s=Ah+Vi+A*h <HG, and G2 are the two roots The complementary solution to the modal equation would then be Bσ1"+B2O2 ghe The particular solution to the modal equation is D'( /e Combining the two components of the solution together, r=(c)+(m) 1(+(+(=2 Ah-2hleh-1 Slide 27 EXAMPLE 2 Leapfrog Time Discretization: Stability Criterion For the solution to be stable. the transient (complementary) solution must not be allowed to grow indefinitely with time, thus implying that =(++b)<1 -√1+h2<116.920J/SMA 5212 Numerical Methods for PDEs 17 2 2 1 ( 2 ) 0 ( 2 1) 0 n n n n S c h Sc c S c S h S S λ λ − − = − − = Slide 26 EXAMPLE 2 Leapfrog Time Discretization: Time Shift Operator The solution to the characteristic polynomial is 2 2 σ (λh) = S = λh ± 1+ λ h The complementary solution to the modal equation would then be 1 1 2 2 n n n c = β σ + β σ The particular solution to the modal equation is 2 2 2 1 hn h n h h ahe e p e h e µ µ µ µ λ = − − . Combining the two components of the solution together, n u ( ) ( ) n n = c + p ( ) ( ) 2 2 2 2 1 2 2 2 1 1 2 1 n n hn h h h ahe e h h h h e h e µ µ µ µ β λ λ β λ λ λ ￾ ✁ ￾ ✁ = ✂ + + + − + ✄ + ✂ ✄ − − ☎ ✆ ☎ ✆ Slide 27 EXAMPLE 2 Leapfrog Time Discretization: Stability Criterion For the solution to be stable, the transient (complementary) solution must not be allowed to grow indefinitely with time, thus implying that ( ) ( ) 2 2 1 2 2 2 1 1 1 1 h h h h σ λ λ σ λ λ = + + < = − + < characteristic polynomial 2 p(S) = (S − 2hλS −1) = 0 σ1 and σ2 are the two roots