Answer Happy

Need help with EXERCISE 9.23, Example 9.12 is given for
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.