Solution January 22 2018

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ADV ANCED MA THEMA TICAL ANAL YSIS Jan 23, 2018 Surname: Name: Code: LM — Chemical Engineering E1 E2 E3 E4 Q Lab T ot These problems and relative solutions are protected by copyright, therefore they cannot be exploited for any commercial purpose or editorial publication. Any abuse or infringement of copyright will be prosecuted to the full extent of the law.c⃝ • W rite solutions below the corresponding text in the blank spaces of these sheets and, if needed, on the reverse side. Drafts will not be acknowledged. • Provide motivations for your answers. Exercise 1. (7 points) Solve the transport problem with damping { ut + 3ux + 2u = 0 x∈ R, t> 0 u(x, 0) =e−|x| x∈ R . Answer: Letting v =e2tu, v solves the following transport problem (without damping) { vt + 3vx = 0 x∈ R,t> 0, v(x, 0) =e−|x| x∈ R. After setting g(x) =v(x, 0) =e−|x|, we get v(x,t ) =g(x− 3t), thus v(x,t ) =e−|x−3t|, therefore u(x,t ) = e− ( 2t+|x−3t| ) = { ex−5t if x< 3t, et−x if x≥ 3t. Exercise 2. (9 points) Find1 the solution of the following heat equation:    ut−uxx +u = 4 sin3(x) t> 0, 0<x<π, u(0,t ) =u(π,t ) = 0 t≥ 0, u(x, 0) = 0 0 ≤x≤π. Hint: since initial and boundary conditions are homogeneous, look for a solution of the form u(x,t ) = +∞∑ n=1 cn(t) sin(nx), Answer: Differentiating formally the series of the solution, we get ut(x,t ) = +∞∑ n=1 c′ n(t) sin(nx), u xx(x,t ) = +∞∑ n=1 −n2cn(t) sin(nx), 1It may be useful to recall that sin 3(α) = 3 4 sin(α) − 1 4 sin(3α). plugging these expressions in the equation, collecting similar terms and writing the right-hand side as 4 sin3x = 3 sinx− sin 3x, we obtain +∞∑ n=1 [ c′ n(t) + (n2 + 1)cn(t) ] sin(nx) = 3 sinx− sin 3x; this leads us to infinitely many ordinary differential equations about the functions cn: • n = 1 : the equation is c′ 1(t) + 2c1(t) =…