Name. 3150 Sample Exam2S2Ol3 Partial Differential Equations 3150 Sample Midterm Exam 2 Exam Date: Monday, 22 April 2013 Instructions: This exam is timed for 50 minutes. You will be given double time to complete the exam. No calculators, notes, tables or books. Problems use only chapters 1 to 4 of the textbook. No answer check is expected. Details count 3/4, answers count 1/4. la. (CH3. Finite String: Fourier Series Solution) (a) [75%] Display the series formula without derivation details for the finite string problem I j u(x, t) u(0,t) u(L,t) u(x,0) ut(z,0) c u 2 (x, 1), 0, 0, 0, g(x), = = = = t > 0, t>0, t>0, < L, 0 < 0<z<L. (b) [25%] Display an explicit formula for the Fourier coefficients which contains the symbols L, g(x). 7l Lt _c.• j ‘141 )1 S4 ( 0 (ic) M’ pt QI j ( ii?TC) Co.f’ ‘)‘ I ‘7’- ‘Vhc,t)= ç (ii 1 fl fli ( 774..% 2er (r7rct )‘L 1I -n = ‘ 1lfl2.c(o) TC .. c t y1rrC — — I 4i- c) L)1 - 2-14 or#?3 ( 2-L h Use this page to start your solution. Attach extra pages as needed, then staple. 0 4.-) n/z’ )/ fy1T -p-) Name. frE 3150 Sample Exam 2 S2013 lb. (CH3. d’Alembert’s Solution: Finite String) Let = f 0.3z 0.3(1—x) 0<x0/5, 0.5<x< 1. and define g(x) = 0 on 0< x 1. Assumed is dAlemberts solution to the vibrating string problem on 0 < x < 1, t> 0, which is the formula u(x,t) Assume L = 1 and c (a) [25%] Define f, = (f(z - ct) + f(z + ct)) + 1 x+ct L gs)ds. 1/7r. g as appropriate periodic extensions of f and g, respectively. (b) [75%] Display a piecewise-defined formula for u(z, ir/3) on 0 <x < 1. c - —peM i,” os ‘ C €.AL1c rr P.. If -, (1÷2) ( ( ,, - -= j z -c—c 0 - tc ‘ c 2 X - ‘fl, €. - f C. L (+C c-&) pr, T’ ± ± (( ;t::v. 2= ) t ‘ P -4’ - -x i - U.. Use this page to start your solution. Attach extra pages as needed, then staple. 4LrPi trct 3150 Sample Exam 2 S2013 Name. 2. (CR3. Heat Conduction in a Bar) Consider the heat conduction problem in a laterally insulated bar of length 1 with one end at zero Celsius and the other end at 100 Celsius. The initial temperature along the bar is given by function f(x). u(0,t) u(l,t) () o<<i, t>o, = Ut U(X,0) = 0, 100, = f(x), = t>0, O<x<1. (a) [25%] Find the steady-state temperature ui(z). (b) [50%] Solve the bar problem with zero Celsius temperatures at both ends, but f(x) replacej (x,t). The answer has Fourier coefficients in integral form, 2 Qr). Call the answer u 1 by f(x) —u unevaluated, to save time. (c) [25%] Explain why (ta-) u(x) + u2(x,t). Cl’. 0 ‘ A U(x,t) = QQ4t4.t -it, çjL-.? 7 1 C (I) (z.) 1 r’J U n 2L*-= -’L(,,c 2 c C2 ‘.it41D. CAt” + CIO)C (O),U)lO L (c 1A qt)- 41 Use this page to start your solution. Attach extra pages as needed, then staple. E îI A v V N vv V V II I_I__c’ %. I L1 II E - 4&U ço r 0 cS I ( £ Q j1 E )h 4- N ! _l —s -E 4: — ::t— - 1’ I U) E L I I I I I I ‘‘1 i—I Ill F I - aj 4’- 4 C12 + V V ‘- II ...- VI II ‘ i ic.” ‘-1 II._ ..-_-c a) Z /- r •U (c F C ÷ ‘3 _j_ 1 ii + , :s .,- ‘ — cs r ‘ —4c-g Ii 4’. ‘ ? s ‘•Th — c 2 CU U, C a) 4.J U) U) U) C U, CU CU U, U) CU x U) c) CU 0 C 4-, D U, 0 I D