17 02 20

Pagina 1

Ex. 1 Ex. 2 Ex. 3 Ex. 4 Total Mathematical Methods for Biomedical Engineering February 17, 2020 Surname: Name: Matricola: • 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 + 3u2ux = 0, x∈ R,t> 0 u(x, 0) =g(x) :=    0 x≤ 0 −1 0 <x< 1 0 x≥ 1. x∈ R, Find the solution by the method of characteristics. Solution. It is a conservation equation ut +q(u)x =ut +q′(u)ux = 0 where q′(u) = 3u2 and q(u) =u3. The characteristic lines that start at the point (x0, 0) satisfy the equation x =x0 + 3g2(x0)t. The three families of characteristic lines which transport the initial data g(x) are x =x0 x0≤ 0 x =x0 + 3t 0<x 0 < 1, x =x0 x0≥ 1, There is a rarefaction zone starting at (x,t ) = (0, 0). In that region the solution satisfies 3u2 = x t , sou(x,t ) =−√x 3t (observe the minus sign in the inverse of the square: it is necessary since the solution is u(x,t ) =−1 in the region to the right). At (x,t ) = (1, 0) a shock line starts. The RH equation is { s′(t) = 1 s(0) = 1, so the shock line has equation s(t) = 1 + t, and it intersects the rarefaction zone at ( x,t ) = (3/2, 1/2). At that point the RH equation is { s′(t) = s 3t s(1/2) = 3/2, so the shock line has equation s(t) = 2−2/33t1/3. The solution is u(x,t ) =    0 if x≤ 0 −√x 3t if x≥ 0 and x≤ 3t and t≤ 2−2/33t1/3 −1 if x> 3t and x<t + 1, 0 otherwise . 2. (10 points) Solve the following problem for the wave equation with mixed boundary conditions    (∂2…