This happens to be somewhat involved for nonlinear

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Unformatted text preview: postpone it and, instead, note that an upwind ux for a scalar problem is k F (Uk (xk t)) Uk+1(xk t)) = f (Uk (x(xt))t)) iif a(Uk (xk t)) + a(Uk+1 (xk t)) > 0 f (Uk+1 k f a(Uk (xk t)) + a(Uk+1 (xk t)) 0 (10.2.5b) where a(u) = fu(u): (10.2.5c) 22 Hyperbolic Problems A simple numerical ux that is relatively easy to apply to vector systems and employs upwind information is the Lax-Friedrichs function 12] 1 F(Uk (xk t) Uk+1(xk t)) = 2 f (Uk (xk t)) + f (Uk+1(xk t)) ; max(Uk+1(xk t) ; Uk(xk t))] (10.2.5d) where max is the maximum absolute eigenvalue of the Jacobian matrix fu(u), u 2 Uk (xk t)) Uk+1(xk t)]. Example 10.2.1. The simplest discontinuous Galerkin scheme uses piecewise-constant (p = 0) solutions Uj ( t) = c0j (t)P0( ) = c0j : In this case, (10.2.4a) becomes _ hj c0j + f (U(xj t)) ; f (U(xj;1 t)) = 0: In this initial example, let's choose a scalar problem and evaluate the ux using the average (10.2.5a) F (Uk (xk t)) Uk+1(xk t)) = f (Uk (xk t)) +2f (Uk+1(xk t)) = f (c0 k ) +2f (c0 k+1) and upwind (10.2.5b) f (c0 k ) if a(c0 k ) + a(c0 k+1) > 0 f (c0 k+1) if a(c0 k ) + a(c0 k+1) 0 numerical uxes. With these ux choices, we have the ordinary di erential systems c0j + f (c0 j+1)2; f (c0 j;1) = 0 _ hj and c0j + (1 ; j )f (c0 j+1) + (1 + j )f (c0 j ) ;2(1 ; j;1)f (c0 j ) ; (1 + j;1)f (c0 j;1) = 0 _ hj where j = sgn(a(c0 j ) + a(c0 j +1)): In the (simplest) case when f (u) = au with a a positive constant, we have the two schemes ; c0j + a(c0 j+1 h c0 j;1) = 0 j = 0 1 : : : J _ 2 F (Uk (xk t)) Uk+1(xk t)) = and j c0j + a(c0 j ; c0 j;1) = 0 _ h j j = 0 1 : : : J: 10.2. Discontinuous Galerkin Methods 23 Initial conditions for c0j (0) may be speci ed by interpolating the initial data at the center of each interval, i.e., c0 j (0) = u0(xj ; hj =2), j = 1 2 : : : J . We use these two techniques to solve an initial value problem with a = 1 and u0(x t) = sin 2 x: Thus, the exact solution is u(x t) = sin 2 (x ; t): Piecewise-constant discontinuous Galerkin solutions with upwind and centered uxes are shown at t = 1 in Figure 10.2.2. A 16-element uniform mesh was used and time integration was performed using the...
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This document was uploaded on 03/16/2014 for the course CSCI 6860 at Rensselaer Polytechnic Institute.

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