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Implicit-Explicit Runge-Kutta schemes for hyperbolic systems ... - utenti

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Initial layer fix<br />

Next, we test the <strong>schemes</strong> with following two Riemann problems<br />

ρ(x,t)<br />

2.05<br />

2<br />

ρl = 2, ml = 1, zl = 1, x < 0.2,<br />

ρr = 1, mr = 0.13962, zr = 1, x > 0.2,<br />

ρl = 1, ml = 0, zl = 1, x < 0,<br />

ρr = 0.2, mr = 0, zr = 1, x > 0.<br />

ε=1e−008, t=0.50, N=200<br />

1.95<br />

−0.2 −0.15 −0.1 −0.05 0<br />

x<br />

0.05 0.1 0.15 0.2<br />

−1<br />

−2<br />

−3<br />

−4<br />

x<br />

ε=1e−008,<br />

10−3<br />

5<br />

4<br />

3<br />

2<br />

1<br />

0<br />

t=0.25, N=200<br />

−5<br />

0.2 0.25 0.3 0.35<br />

x<br />

0.4 0.45 0.5<br />

Initial layer <strong>for</strong> ρ in problem 1 (left) and departure from equilibrium z − zE in problem 2 (right) <strong>for</strong> ε =<br />

10 −8 . Left problem: ARS(2,2,2) (∗), ARSF(2,2,2) (◦), IMEX-SSP2(2,2,2) (×). Right problem: IMEX-<br />

SSP2(2,2,2) (+), IMEX-SSP2F(2,2,2) (×), ARS(2,2,2) (◦).<br />

z(x,t)−z E (x,t)<br />

29<br />

(1)<br />

(2)

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