By Jichun Li

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

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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.

Solving such a system is quite laborious if we consider that ﬁrst it is not easy to form the coeﬃcient matrix, and second it is quite expensive to store and solve the matrix directly. More details will be discussed when we come to the elliptic problems in a later chapter. A simple and eﬃcient method, the so-called Alternate Direction Implicit (ADI) method, for solving 2-D parabolic problems was ﬁrst proposed by 26 Computational Partial Diﬀerential Equations Using MATLAB Peaceman and Rachford in 1955 [11].

L=⎢ ⎥ ⎥, U = ⎢ ⎢ ⎢ ⎥ ⎥ . . . ⎥ ⎢ ⎢ ⎥ ⎣ ⎣ ⎦ 0 αn−1 cn−1 ⎦ βn−1 1 0 0 βn 1 0 0 αn Prove that the αi and βi can be computed by the relations α1 = a1 , βi = bi /αi−1 , αi = ai − βi ci−1 , i = 2, · · · , n. , y1 = f1 , yi = fi − βi yi−1 , i = 2, · · · , n, xn = yn /αn , xi = (yi − ci xi+1 )/αi , i = n − 1, · · · , 1. Prove that the total number of multiplication/division is 5n − 4. 3. 1. 1 on a 31 × 51 grid for (x, t) domain. Compare the numerical solution to its analytical solution: u(x, t) = 1 − 2 ∞ k=1 2 2 4 e−(2k−1) π t cos(2k − 1)πx.