3150 Sample Final Exam S2013 Name. Partial Differential Equations 3150 Sample Final Exam Final Exam Date: Thursday, 2 May 2013 Instructions: This exam is timed for 120 minutes. You will be given 30 extra minutes to complete the exam. No calculators, notes, tables or books. Problems use only chapters 1, 2, 3, 4, 7 of the textbook. No answer check is expected. Details count 3/4, answers count 1/4. Problems 1,2,3: Selected problem types from the first two midterm exams. See the sample exams and exam solutions for midterms 1 and 2, for specific examples. 4. (CH4. Poisson Problem) Solve for i(x, y) in the Poisson problem ( ‘ +u u(x,0) u(x, y) p_ Cfr- V 41 (c-k tL & = sin(rx) sin(iry), 0 = 4sin(7rr), 0 < x < 1, 0 <y < 1, 0 <x < 1, on the other 3 boundary edges. 46 p1efe ‘r f2 i 7L IQ I Q__ 3 - , rj Le1 S/ ) 1 ( (Y( b A A.o L 0 FlJ 7 = c ept-4D V) 0 -fLA 2 çi U bo 42. 4 r1i l( L( txty C V44 fr) 6 - - I A— c/n hrlr) rL4 z prour Q(tc — — ( Yni ;)1- (mit) 2 (ii ‘((= . then staple. Use this 1 n o-4 th’tc = A4 ((i-)) .j aáA t c<. ckci-4 &4-D T 1 M 2 ‘fl=, Name. KE / 3150 Exam 3 F2009 5. (CH7. Fourier Transform: Infinite Rod) Let f(x) = pulse(x, —2,2), that is, f(x) = 1 for x S < cc. 2 and < f(x) = 0 elsewhere on —oc Aar (a) [75%] Solve the insulated rod heat conduction problem f u(x,t) = u(x,0) = ft) = FT L 3 5ac;k’c 1 —cc<x<oo, t>0, (x,t), 1 u —cc <x < cc. f(s), (b) [25%] Evaluate the limit of u(x,t) at x cc and x = < X () 7.Lt, —cc. ‘41 LflQr (L P-Ilk] q de (Lo Fi ç4 r3 /j . r-t = — C F’ 7i (ti feft ôcr / (&) (x) ()_ 2. -c) 5-$)c I \......L. e -cs/(2’ ; =(/(J;-)) 2h. 2 :2 (S X2.. - (ev 1- Use this page to start your solution. Attach extra pages as needed, then staple. rç(co) 0 j Er((—X’) “-I Name. 3150 Exam 3 F2009 1’.. { 6. (CH4. Heat Equation and Gauss’ Heat Kernel) Solve the insulated rod heat conduction problem < x < tLt(X,t) = x,t), u ( 1 —00 v(x,0) = f(x), —1<x<1, , t > 0, 2o Ix<1. otherwise 0 2 dz, and Fourier Hint: Use convolutions, the heat kernel, the error function erf(x) = ’ e_z 0 f transform rules to solve the problem. The answer is expressed in terms of the error function. —= t 1 l l .( . C ç 00 = ( ( c ç 20 - 2o x) _?>j e dc Lef x—l ;_ Use this page to start your solution. Attach extra pages as needed, then staple.