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Unformatted text preview: ck forms and the solution in this region is an expansion fan.
Several characteristics are shown in Figure 10.1.13 and the expansion solution is given
< uL if x=t < uL
u(x t) = : x=t if uL x=t < uR
uR if x=t uR
1/uR 1/uL 1/uL x x Figure 10.1.13: Shock (left) and expansion (right) wave characteristics of the Riemann
problem of Example 10.1.8.
We conclude this example by examining the solution of the Riemann problem along
the line x = 0. Characteristics for several choices of initial data are shown in Figure
10.1.14 and, by examining these and (10.1.26), we see that
> uL if uL uR > 0
> u if u u < 0
u(0 t) = > 0 if uL < 0 uR > 0
> uL if uL > 0 uR < 0 (uL + uR )=2 > 0
: u if u > 0 u < 0 (u + u )=2 < 0
This data will be useful when constructing numerical schemes based on the solution of
Riemann problems. Problems 18 Hyperbolic Problems
t t x x
t t x x t t x x Figure 10.1.14: Characteristics of Riemann problems for Burgers' equation when uL uR >
0 (top) uL uR < 0 (center) uL > 0, uR < 0, (uL + uR)=2 > 0 (bottom left) and
uL < 0 uR > 0 (bottom right).
1. Show that the solution of the Riemann problem (10.1.16, 10.1.25) is given by
(10.1.26). You may begin by solving a problem with continuous initial data, e.g.,
if x < ;
u ( ; x) + u ( + x) if ; < x
u(x 0) = : 2
if < x
L and take the limit as ! 0. R R 10.2. Discontinuous Galerkin Methods 19 10.2 Discontinuous Galerkin Methods
In Section 9.3, we examined the use of the discontinuous Galerkin method for time
integration. We'll now examine it as a way of performing spatial discretization of conservation laws (10.1.1). The method might have some advantages when solving problems
with discontinuous solutions. The discontinuous Galerkin method was rst used for to
solve an ordinary di erential equation for neutron transport 21]. At the moment, it
is very popular and is being used to solve...
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- Spring '14