17 01 20 1

Page 1

Ex. 1 Ex. 2 Ex. 3 Ex. 4 Total Mathematical Methods in for Materials Engineering January 27, 2020 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 +uux = 0, x∈ R,t> 0 u(x, 0) =g(x) :=    1 x≤ 0 1−x 0<x< 1 0 x≥ 1. x∈ R, Find the solution by the method of characteristics, draw a graph of the solution for t = 1 2 and t = 2. Solution. It is Burgers’ equation, that is ut +q(u)x =ut +q′(u)ux = 0 where q′(u) =u and q(u) = 1 2u2. The characteristic lines that start at the point ( x0, 0) satisfy the equation x =x0 +g(x0)t. The three families of characteristic lines which transport the initial data g(x) are x =x0 + (1)t =x0 +t x 0≤ 0 x =x0 + (1−x0)t 0<x 0 < 1, x =x0 + (0)t =x0 x0≥ 1, Since g(x) is continuous, there is no rarefaction zone, nor there is a shock for small t. In the region t≤x≤ 1, the characteristic line passing through the point ( x,t ) has base point x0 = x−t 1−t , therefore the solution is u(x,t ) = 1−x0 = 1−x 1−t . All characteristic lines starting at 0 ≤ x0≤ 1 pass through the point ( x,t ) = (1, 1), therefore a shock line starts from such point. The RH equation is { s′(t) = 1 2 s(1) = 1, so the shock line has equation s(t) = t+1 2 . The solution is u(x,t ) =    1 if x≤t and t≤ 1 1 if x< t+1 2 and t≥ 1 1−x 1−t if t≤x≤ 1 0 if x≥ 1 and t< 1 0 if x> t+1 2 and t≥ 1. 2. (10 points) i) Solve the following problem for the Laplace equation with Dirichlet boundary condition …