Chemical Engineering - Advanced Mathematical Analysis

Full exam

ADVANCED MATHEMATICAL ANALYSIS Feb 8, 2016 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+ 2 u x+ 2 u y+ u= 0 (x; y)2R2 ; t >0 u(x; y;0) =e (jxj+jyj) (x; y)2R2: Answer:Lettingv=et u,vsolves the following transport problem (without damping) (vt+ 2 v x+ 2 v y= 0 ( x; y)2R2 ; t >0 v(x; y;0) =e (jxj+jyj) (x; y)2R2 after settingg(x; y) =v(x; y;0) =e (jxj+jyj) , we getv(x; y; t) =g(x2t; y3t) , thus v(x; y; t) =e jx2tj+jy2tj
; thereforeu(x; y; t) =e t+jx2tj+jy2tj2
: Exercise 2.(10 points) Solve1 the following boundary value problem for Laplace equation: 8 > < > :
u = 0x2 +y2 0 u(x;0) = 02x2 u(x; y) = 4y3 x2 +y2 = 4; y >0 where we look for a solutionC2 in the interior of the domain and continuous up to the boundary. Answer:We switch to polar coordinates: if we letu=u(%; #), thenusolves 8 > > < > > :
u =u %%+1% u %+1% 2u ##= 0 % < > :u tt 4u xx= 0 x2R; t >0 u(x;0) = 2e x2 x2R ut( x;0) = 8x e x2 x2R Answer:It is an initial value problem for the 1D wave equation withc= 2; the solution is given by d'Alembert's formula: u(x; t) =2 e (x2t)2 + 2e (x+2t)22 + 12
2Z x+2t x2t8 e 2 d = e (x2t)2 +e (x+2t)2  e (x+2t)2 e (x2t)2 = 2e (x2t)2 : Theory question (it will be evaluated only if the total score of previous exercises exceeds 24). State and prove the formula providing the Fourier transform of the derivative of a functionu:R!R. Answer: Theorem- Assumeu2C1 (R); u; u0 2L1 (R). Then (u0 )b() =i b u(): Proof -By the Fundamental Theorem of Integral Calculus, for anyu2C1 (R) andx2R, we have u(x) =u(0) +Z x 0u 0 (t)dt ; hence due to Lebesgue Theorem ( thanks tou0 2L1 ) there exist both limits ofu(x) asx! 1 and they are both nite, moreover (sinceu2L1 ) they are both equal to 0. Notice that there existvinL1 (lacking the propertyv0 2L1 ) s.t. these limits do not exist! e.g. +1 X n=1n  [n;n+1=n3 ]+ 1 X n=1n  [n;n+1=n3 ]: We conclude the proof: integrating by parts in (a;+a) the product ofC1 functionse ix u0 (x) we getZ+a ae ix u0 (x)dx= +i Z +a ae ix u(x)dx+h e ix u(x)i +a a; then passing to the limit, asa!+1, we obtain the conclusion.