By Jichun Li

Short assessment of Partial Differential Equations The parabolic equations The wave equations The elliptic equations Differential equations in broader areasA speedy assessment of numerical tools for PDEsFinite distinction equipment for Parabolic Equations advent Theoretical concerns: balance, consistence, and convergence 1-D parabolic equations2-D and 3D parabolic equationsNumerical examples with MATLAB codesFiniteRead more...

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**Example text**

108] or [3, p. 237]): n+ 1 1 n t (uj + unj+1 ) − [f (unj+1 ) − f (unj )], 2 2 x t n+ 1 n+ 1 [f (uj+ 12 ) − f (uj− 12 )]. 26) can be easily constructed: un+1 − un−1 f (unj+1 ) − f (unj−1 ) j j + = 0. 31) Finally, we want to mention that similar schemes can be developed for conservation laws in two and three space dimensions (cf. [7] or [3]). 34) 46 Computational Partial Diﬀerential Equations Using MATLAB where the constant a > 0. Let ζ = x + at, η = x − at. 35) where φ(ζ, η) = u(x, t). 35) twice gives u(x, t) = ψ1 (x + at) + ψ2 (x − at), where ψ1 and ψ2 are arbitrary twice diﬀerentiable functions.

40) whose truncation error is O(( t)2 + ( x)2 ). As for stability, using the von Neumann technique, we obtain λ−2+ or 1 4 1 1 1 = μ2 [−4λ sin2 k x − sin2 k x] λ 2 2 λ 2 1 1 (1 + 2μ2 sin2 k x)λ2 − 2λ + (1 + 2μ2 sin2 k x) = 0. 40) is unconditionally stable. 45) which holds true for any (x, y, t) ∈ (0, 1)2 × (0, tF ). 45). 46) where for simplicity we assume that x = y = h. Denote μ = ht . By von Neumann stability analysis, the ampliﬁcation factor λ satisﬁes the equation 1 1 λ2 − 2λ + 1 = μ2 [−4λ sin2 kx x − 4λ sin2 ky y] 2 2 or 1 1 λ2 − 2[1 − 2μ2 (sin2 kx x + sin2 ky y)]λ + 1 = 0.

Xj = j x, 0 ≤ j ≤ J, x = 1/J, tn = n t, 0 ≤ n ≤ N, x = tF /N. First let us consider a central diﬀerence scheme: un+1 − unj unj+1 − unj−1 j +a = 0. 7) is unconditionally unstable. Hence we have to be very careful in choosing proper diﬀerence approximations. , when a > 0, we approximate ux |nj by backward diﬀerence. , in this case by forward diﬀerence. To check its stability, let us denote μ = a unj = λn eik(j t x. 8), we see that the ampliﬁcation factor λ(k) = 1 − μ(1 − e−ik x ) = 1 − μ + μ cos k x − iμ sin k x.