Differential Equations Lecture Work Solutions 333

Differential Equations Lecture Work Solutions 333 - 2....

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Unformatted text preview: 2. Apply the Beam-Warming scheme with Euler implicit time diﬀerencing to the linearized Burgers’ equation on the computational grid given in Figure 61 (use c = 2, µ = 2, ∆x = 1) and determine the steady state values of u at j = 2 and j = 3. the boundary conditions are u n = 1, 1 u 1 = 0, 2 and the initial conditions are un = 4 4 u1 = 0 3 Do not use a computer to solve this problem. un+1 − un j j + cux ∆t Let ν = c n+1 j n+1 = µuxx j ∆t ∆t and r = µ 2 then ∆x ∆x ν n+1 +1 +1 +1 u − un−1 + r un−1 − 2un+1 + un+1 un+1 = un − j j j j j j 2 j +1 or − ν ν +1 +1 + r un−1 + (2r + 1)un+1 + − r un+1 = un j j j j 2 2 In our case c = 2, µ = 2, ∆x = 1 and so if we let ∆t = 1, then r = ν = 2. Using these values in the above equation, we get +1 +1 −3un−1 + 5un+1 − un+1 = un j j j j For n = 1 (don’t forget to employ the boundary conditions) −3 · 1 + 5u2 − u2 = 0 2 3 −3u2 + 5u2 − 4 2 3 =0 The solution of this system of two equations is u2 = 2 19 22 u2 = 3 29 22 Now go to the next time step n = 2 = 19 22 −3u3 + 5u3 − 4 = 2 3 29 22 −3 + 5u3 − u3 2 3 The solution of this system of two equations is 333 ...
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This note was uploaded on 12/22/2011 for the course MAP 2302 taught by Professor Bell,d during the Fall '08 term at UNF.

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