Chemical Engineering - Advanced Mathematical Analysis

Full exam

ADVANCED MATHEMATICAL ANALYSIS Jan 23, 2018 Surname: Name:Code: LM | Chemical EngineeringE1 E2 E3 E4 Q LabTotThese 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. c Write 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+ 3 u x+ 2 u= 0x2R; t >0 u(x;0) =ej xj x2R: Answer:Lettingv=e2 t u,vsolves the following transport problem (without damping) (vt+ 3 v x= 0 x2R; t >0; v(x;0) =ej xj x2R: After settingg(x) =v(x;0) =ej xj , we getv(x; t) =g(x3t), thus v(x; t) =ej x3tj ; thereforeu(x; t) =e 2t+jx3tj
= ex 5t ifx < > :u t u xx+ u= 4 sin3 (x)t >0;0< x <  ; u(0; t) =u(; t) = 0t0; u(x;0) = 0 0x : Hint: since initial and boundary conditions are homogeneous, look for a solution of the form u(x; t) =+ 1 X n=1c n( t) sin(nx); Answer:Dierentiating formally the series of the solution, we get ut( x; t) =+ 1 X n=1c 0 n( t) sin(nx); u xx( x; t) =+ 1 X n=1 n2 cn( t) sin(nx);1 It may be useful to recall that sin3 () =34 sin( )14 sin(3 ). plugging these expressions in the equation, collecting similar terms and writing the right-hand side as 4 sin3 x= 3 sinxsin 3x, we obtain +1 X n=1 c0 n( t) + (n2 + 1)c n( t) sin(nx) = 3 sinxsin 3x; this leads us to innitely many ordinary dierential equations about the functionsc n: n= 1 : the equation isc0 1( t) + 2c 1( t) = 3; its general solution is c1( t) =H 1e 2t +32 ; n= 3 : the equation isc0 3( t) + 10c 3( t) =1; its general solution is c3( t) =H 3e 10t 110 ; n6 = 1 andn6 = 3 : the equation is homogeneous:c0 n( t) + (n2 + 1)c n( t) = 0; its general solution iscn( t) =H ne (n2 +1)t ; notice also that every ordinary dierential equation in the list above solve a null initial value problem: sinceu(x;0) = 0 we can say thatc n(0) = 0 for all n, therefore H1= 32 ; H 3=110 ; H n= 0 8n6 = 1;3; the solution to our problem is then u(x; t) =32 1e 2t
sin(x)110 1e 10t
sin(3x): Exercise 3.(7 points) Evaluate the Fourier transform of f(x) =ej xj (1;1)( x) =( ej xj jxj < > :u tt u xx= 2 t2 x2R; t >0 u(x;0) =x2 x2R ut( x;0) = 0x2R Answer:It is a vibrating string problem, withc= 1. Using d'Alembert formula, the solution is u(x; t) =12  (x+t)2 + (xt)2  +12 Z t 0dZ x+(t) x(t)2 2 d =x2 +t2 + 2Z t 0( t)2 d=x2 +t2 +16 t 4 : Theory question (it will be evaluated only if the total score of previous exercises exceeds 24). Givenm2N, write the formula providing all distributional solutions of the division problem u2 D0 (R) :xm u= 0 inD0 (R) and prove it.