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Ex. 1 Ex. 2 Ex. 3 Ex 4 T otal Mathematical and Numerical Methods for Engineering September 10, 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. 1. (10 points) Consider the following Cauchy problem: ut +uux = 0, x∈ R,t> 0 u(x, 0) =g(x) := { 1 x≤ 0 x x> 0 Find the solution by the method of characteristics, specifying where it is a continuous function. Sketch of the solution . It is Burger’s equation, that is ut +q(u)x =ut +q′(u)ux = 0 with q′(u) =u and q(u) = u2 2 . The characteristic lines that start at the point ( x0, 0) satisfy the equation x =x0 +g(x0)t The families of characteristic lines which transport the initial data g(x) are x =x0 + 1·t =x0 +t x 0≤ 0, x =x0 +x0·t =x0(1 +t) x0 > 0. In the region spanned by the characteristic lines starting at x0 < 0 the solution is u(x,t ) = 1. Consider the region spanned by the characteristic lines starting at x0 > 0. The characteristic going through the point ( x,t ) has base point x0 = x 1+t, therefore in that region the solution is u(x,t ) = x 1+t. A shock line leaves from the point ( x,t ) = (0, 0) . The RH condition is x′(t) = 1 2 1− x2 (1+t)2 1− x 1+t = 1 2 1 +t +x 1 +t = 1 2 + 1 2 x 1 +t. This is a linear equation. The solution satisfying the initial condition x(0) = 0 is x(t) = 1 +t− √ 1 +t. The solution of the equation is u(x,t ) = 1 if x < 1 +t−√1 +t and u(x,t ) = x 1+t if x > 1 +t−√1 +t. The solution is continuous for all t≥ 0 and all x such that x̸= 1 +t−√1 +t. 2. Let Ω := (0,π )× (0, +∞). Solve the following problem    ∂2 ttu(x,t )−∂2 xxu(x,t ) = 0, (x,t )∈ Ω ∂xu(0,t ) =∂xu(π,t ) = 0, t> 0 u(x, 0) = 1 +x, x ∈…