# 29 i f u p1 pu f u f u 10210b 10210c split

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Unformatted text preview: ng 26]. Van Leer 27] found an improvement that provided better performance near sonic and stagnation points of the ow. The split uxes are evaluated by upwind techniques. Thus, at an interface x = xj , f + is evaluated using Uj (xj t) and f ; is evaluated using Uj+1(xj t). Calculating uxes based on the solution of Riemann problems is another popular way of specifying numerical uxes for vector systems. To this end, let w(x=t uL uR ) be the solution of a Riemann problem for (10.1.1a) with the peicewise-constant initial data (10.1.25). The solution of a Riemann problem \breaking" at (xj tn) would be w((x ; xj )=(t ; tn) Uj (xj tn) Uj (xj+1 tn)). Using this, we would calculate the numerical ux at (xj t), t > tn , as F(Uj (xj tn) Uj+1(xj tn)) = f (w(0 Uj (xj tn) Uj+1(xj tn)): (10.2.11) 32 Hyperbolic Problems Example 10.2.4. Let us calculate the numerical ux based on the solution of a Riemann problem for Burgers' equation (10.1.16). Using the results of Example 10.1.8) we know that the solution of the appropriate Riemann problem is 8 if Uj Uj+1 > 0 > Uj > >U > j +1 if Uj Uj +1 < 0 < if Uj < 0 Uj+1 > 0 w(0 Uj Uj+1) = > 0 : > Uj if Uj > 0 Uj+1 < 0 (Uj + Uj+1)=2 > 0 > > :U j +1 if Uj > 0 Uj +1 < 0 (Uj + Uj +1 )=2 < 0 (The arguments of Uj and Uj+1 are all (xj tn). These have been omitted for clarity.) With f (u) = u2=2 for Burgers' equation, we nd the numerical ux 82 > Uj =2 if Uj Uj +1 > 0 >2 > U =2 if U U < 0 > j +1 < j j +1 if Uj < 0 Uj+1 > 0 F (Uj Uj+1) = > 0 : > Uj2 =2 if Uj > 0 Uj +1 < 0 (Uj + Uj +1 )=2 > 0 > >2 : U =2 if U > 0 U < 0 (U + U )=2 < 0 j j +1 j j +1 j +1 Letting u+ = max(u 0) u; = min(u 0) we can write the numerical ux more concisely as F (Uj Uj+1) = max (Uj+)2 =2 (Uj; )2=2]: +1 When used with a piecewise-constant basis and forward Euler time integration, the resulting discontinuous Galerkin scheme is identical to Godunov's nite di erence scheme 18]. This was the rst di erence sche...
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