Question

2. Solve the initial-boundary value problem One = 48m2 for 0 < x < 8, t > 0, u(0, t) = u(8,t) = 0 for t > 0, u(2,0) = 2e-4x for 0 < x < 8. (60 pts.)

Answer 1

given heat equation 4 d²u to for olna 8 ana tzo ulo, ti= ul8,t=0 -ur Unol = e . On 2 it we know that general heat equation du = karu For osach to and b-c. uco,ti= u(2 t o tzo un ol = F (1) then solution is given by -kn272 + m -knet Dn sin antenne Vln, tia n = 1 where one 2 , Fles sin ( h) da we get Compare k= 4 it with L=8 given F (n) equation = 2 ē the 42 чn'm E 6.4 Hence solution n= L n²t2 t 16 n - L -Yn antn now 2n = Uniti e an sin (AT) e Unit) - Ě on Sin (th) e now an= 2 j ze ve sin ("Th) de = + j en sin ( Tendre that sinbada = ean Casinon - blosbn) [a2+62) un a on t e , (-451 (nama) - na ces (ar)) 8 -4r o Now we know lan nt - n ht (16+ n²

- 4n an= - (-+) (32 sin (Atm ) + n cus (n yanl) 2 (16+ n²t2 nh- [ 32 sin (nn) + nT cos (open) + nT co 16 (16+ n²t² 64 - 42 [ a2 sin(nra) + nu cos (min) Dn= n Clo24+ n2² - 4 ona - 4 è un (n2H² + 1024] [32 sin (1 ) + nT cos(nm)] Hence complete solution ) 9 Uh, t1 = P n=L where { an sin (nn) e - - 4eun [ 32 sin (1T) + nT Cor (AT)] [n²t2 Horu)