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Ex. 1 Ex. 2 Ex. 3 Ex 4 Total Mathematical Methods in for Materials Engineering January 10, 2019 Surname: Name: Id. code: • All answers and calculations must be clearly justified. You have to write your answers on these sheets only. You are not allowed to use or even have with you notes, texts or any electronic device, including mobiles. To get a positive evaluation you have to get at least 10/20 points in the first two items, at least 5/12 points in the remaining two. 1. (10 points) Consider the following Cauchy problem: ut +u2ux = 0, x∈ R,t> 0 u(x, 0) =g(x) :=    0 x≤ 0 −1 0 <x< 1 0 x≥ 1 Find the solution by the method of characteristics, specifying where it is a continuous function. Sketch of the solution. It is a conservation law ut +q(u)x =ut +q′(u)ux = 0 where q′(u) =u2 and q(u) = 1 3u3. The characteristic lines that start at the point ( x0, 0) satisfy the equation x =x0 + (g(x0))2t The families of characteristic lines which transport the initial data g(x) are x =x0 + (0)2t =x0 x0 < 0, x =x0 + (−1)2t =x0 +t 0<x 0 < 1, x =x0 + (0)2t =x0 x0≥ 1. We see that there is a region where the characteristic lines are missing. In this region we can defineu as a rarefaction wave that links with continuity the values from u = 0, transported by the x = x0 lines, to u =−1, transported by the x = x0 +t characteristics. We can formally compute the equation of the rarefaction wave, originated in (x0,t 0) = (0, 0), as u(x,t ) =r (x−x0 t−t0 ) =r (x t ) where r = (q′)−1 (inverse function). Since q′ =u4 we obtain u =± √x t To ensure the continuity of the solution u in the rarefaction fan, we have to choose the negative square root. The characteristic lines have the equation x =ht, where 0 <h< 1. A shock line leaves from the point ( x,t ) = (1, 0) . The RH condition is x′(t) = 1 3, so…