- 9 23 Show That The Roots Of The Equation 9 41 In Example 9 12 Are Given By K 1 S 2t 1 S 21 4 S 1 S 1 (18.87 KiB) Viewed 32 times
- 9 23 Show That The Roots Of The Equation 9 41 In Example 9 12 Are Given By K 1 S 2t 1 S 21 4 S 1 S 2 (56.72 KiB) Viewed 32 times
reference alongside equations (9.41) and (9.42)
9.23 Show that the roots of the equation (9.41) in Example 9.12 are given by k¯¯ = (1 + s-2t + √(1+s−21)² − 4(s − 1)) (s-1+√√(s− 1)² +4(s − 1)). k+ = Hence verify that s(t) = t is a solution of the initial value problem (9.42).
Example 9.12 (Example 9.9 revisited) Determine the shock speed for Example 9.9 and describe how a shock wave can be fitted so as to avoid multivalued solutions while conserving the mean value of the solution u at all points in time t. The shock speed is, as in the previous example, given by s'(t) = {(u+ + u¯). Characteristics first intersect at x = t = 1 and so s(1) = 1 (the tip of the cusp in Fig. 9.11). In order to ascertain appropriate values of u and u, we suppose that the characteristics with k = k¯(t) and k = k+ (t) intersect at x = s(t) at time t. t t = 1.5 1.5 t = 1 1.0 t = 0.5 0,5 t = 0 x -2 -1 0 1 2 3 -2 -1 0 1 2 3 Fig. 9.13 The weak solution of the problem in Example 9.8 amended to include a shock wave at s(t) = 2t for t > 1 as discussed in Example 9.11 shock line
Note that, since the characteristic through (1, 1) also passes through the origin, we know that k < 0 <k+. The equation of the characteristics is given by (9.30) so the values of k* can be found in terms of t and s, by inverting the relation s(t) = k + g(k*)t (9.41) for the specific initial condition g(x) = 1 -x/(1+x). Moreover, since u = = g(k), the location of the shock can be determined by solving the initial value problem s' (t) = (g(k+)+ g(k¯)), s(1)= 1, (9.42) using the values of k obtained from (9.41). The details are worked out in Exercise 9.23.