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Lecture 6 18.086 Info • Project proposals…! Leapfrog scheme • Easiest numerical scheme for 2nd order problem: Leapfrog n t) Notation: Uj,n = U (j x,CFL criterion for Leapfrog scheme Uj,n+1 • 2Uj,n + Uj,n 1 !" 2!"Uj+1,n u =0 Equation: =+ c 2 !t !x t Stability: |r| ≤ 1 (equiv. CFL condition!) Lecture t n+1! Accuracy: 2nd order n-2! with physical domain of dependence n! n-1! • 2Uj,n +nU j 1,n +1 = !in –1 – " (!in+1 # !in–1 ) Scheme: !i x2 numerical domain of dependence unstable |uΔt/Δx| >1 c stable |uΔt/Δx| ≤1 c Lecture / see Mathematica notebook leapfrog_stability.nb i-3! i-2! i-1! i! i+1! i+2! i+3! x Accuracy of FD schemes: Alternative way (using modified equations) • Approach is general and gives GLOBAL error directly (not local error) Using wave equation and leapfrog as an example: • Uj,n+1 2Uj,n + Uj,n t2 1 =c 2 Uj+1,n 2Uj,n + Uj x2 1,n • Plug in analytical u and expand in Taylor on LHS and RHS of FD equation E.g.: 1 uj,n+1 = u(x, t) + ux (x, t) t + utt (x, t) t2 + O( t3 ) 2 • Simplifying LHS and RHS gives 1 1 2 2 utt + t utttt + . . . = c (uxx + x2 uxxxx + . . .) 12 12 • 2 4 u = c u = c uxxxx , so the global error is Note that tttt ttxx 1 ⇥ 2 4 t c 12 x 2 ⇤ uxxxx => p=2 in time and space Staggered grids • • • Remember: @ Equivalent 1st order problem: @t with v1 = ut , v2 = cux ✓ v1 v2 ◆ = ✓ 0 c c 0 ◆ @ @x ✓ There are a number of “physical” wave equations that are of this form, most importantly the Maxwell equations of electrodynamics (in vacuum): Leads to a 1st order system as above! v1 v2 ◆ correct typo in one-way LF scheme! We could try this…. • Discretize 1st order problem: with 1st order “leapfrog” @ @t ✓ v w ◆ = ✓ 0 c c 0 ◆ @ @x ✓ v w ◆ This would give us 2nd order accuracy, because we use two centered differences for space and time in leapfrog! Uj,n+1 Uj,n Uj+1,n Uj =c 2 t 2 x • 1way wave eq leapfrog: • 1-way wave eq. leapfrog: dispersion-free |G|=1 for all k, as long as CFL condition r<=1 • Here: Vj,n+1 Vj,n Wj+1,n Wj 1,n =c 2 t 2 x Wj,n+1 Wj,n Vj+1,n Vj 1,n =c 2 t 2 x 1,n Lecture: interesting numerical “dispersion”! 1way-Leapfrog • “Coupled” equations decouple in current leapfrog formulation! Diffusion and Advection (6.5) • The heat equation is another famous PDE ut = Duxx • D is the diffusion constant (units length^2/time) Dk2 t • Growth factor: G(k, t) = e • Solution to IV u(x,0) = δ(x) is given by: x2 1 u(x, t) = p e 4Dt 4⇡Dt Numerical scheme (FD) • Heat equation is dissipative, so why not try Forward Euler: Uj,n+1 Uj,n t • • • • = Uj+1,n 2Uj,n + Uj x2 1,n Expected accuracy: O(Δt) in time, O(Δx2) in space. t 1 Stability in the usual way gives R = 2 x 2 We can use previous ODE methods using the old method of lines, of course… -2 Or implicit methods (see book, 6.5). Notice that matrix contains O(Δx ) terms and is stiff! Incorporating boundary conditions into FD matrices • Writing the heat equation using method of lines (with centered differences in x), we have position j=i ~ut = A~u • Away from boundaries, Ai = (0, . . . , • Boundary conditions determine values of A near boundaries, e.g. 0 B B B B B B @ 0 1 0 0 0 0 0 2 1 0 0 0 0 1 2 1 0 0 0 0 1 2 1 0 0 0 0 1 2 0 0 0 0 0 1 0 10 u0 u1 u3 ... CB CB CB CB CB CB A @ uN 1 uN 1 C C C C C C A 1, 2, 1, 0, . . . , 0) BCs: u(0,t)=v1 u(X,t)=v2 see book, 1.2