Question

4. (5 marks) Consider the partial differential equation (1) for 1 € (0,2) and t > 0, with boundary conditions u(0, 1) = 0 ur(2, 1) = 0. Which of the following are solutions to the PDE and boundary conditions? In each case explain your answer. Note that initial conditions are not given. (Hint: it is not necessary to solve the problem above. (a) -3)*** e ular, 1) = Žen sin [(---) --] ~[(---) ;-)e-(1-1) e+(1-3)*(/2°1 u(3,t) - Cu COS u(1,t) = cos(naa) e n?-?1 (d) u(3,1) = nsin Een sin [(-->) 5+] c+(1-3)** (0) u(x,t) ensin » [(- - -) 3*) ** (1-3)*121*

Answer 1

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Partial differential PDE Answer: itial equation - U = Uxx Jos RECO,27 4770 with boundary conditions - U10,0) = 0 Uy (2 t) = 0 To check weather given solutions are of the above or not. So of ult) is a solution to the must satily the boundaryodna the PDE Ekle will check for each option of that satisfy the boundary conditions. (0) uer,t) = { an sin((n-13) ax] boundary conditions- ulo,t) = 8 ensin(n-1)x.o PDE 2. (n-1324 e -In-1, jx?4 e nel 1-sinoo Uto, O Os (n-1) Ux (t) = { cn cos((n-1, x x] •M-I) a ne1 -(m. 1 3.2t U, 10,+) = Ź en cospin-3)*.2).(n-1)*.e scos (n-1)2x cas Cannot) -(n-1₂124²7 & crane to ie uix+) does not eatingy given bound-15-cosina A cos (ca-2na) auy conditions. (171+1 - ulgt) is not & (cosx) -- sina La Isinx) a solution to PDE. da . (0X CS Scanned with CamScanner

USD = है Cnceof{n-1, x] e(-1) 2102377 (n-1) vast (0,+. E ancos (13.) 5.0] © { Che 11-12) ² (₂) ²7 ficoso - - U19 + 0 ie ulat) is not a solution to PDG. U13+) E Cncasinaxle 11 ont -nin't Ê en coslo) e nat Ecne U10,+) = Ulot) ²0 ie uixt is not a solution to the PDt. UYE): Ë ensinan-1.18 ] en mis)*(0,14 ulo, t) - Ensin [inne, og er 12214 in-3)*()*t lisino: o E Cnoo.e U10,0 - Ug (1,7) E cn colink]. (-3)$ (n-1) 449%+ Uy t9,+) 3 cn cos (m+)].(:1) 5 cm-kº (0,3'+ CS Scanned with CamScanner

cos(na - I 9 conacast + sinna sina cosna.o+00 101 Sinna o E en 0. nal (n-ba) I e In-1) Us (2,5) = 0 ps T Una =-E en (n-1) JE nal inisin flin- z) 5x) (n='s lengte É cn simfln-235x). Em D (n-1)) 바 . nol utoluxx ut 7 urx not satisfy the PDE = Unt) not a solution. les to x 9) - Liam sinf %) 5« Je Ink)+1%)? 110,t) = {chsin (n-1) 2.0) e{n- k)? (914 SO sinoo 1-433944 uz lst) = Ž contos (1) * 4x] . 6-4) the 2 e Uxx - 4x129+) o las solved in optiond) - In-13) (9) + E con sin (n. 12) 2 (n-1₂) -) Uux = t =) ulast) is a solution to the PDE. nal Ut = -Ecu sinfln-k)x] e In-k]?15)+ CS Scanned with CamScanner