Question

Problem 3: Solve the following initial value / Neumann problem by separation of variables: (8 points) Utt – 9uze = 0, (t, x) ER [0, 2], u(0,2) = cos? (*), u(0, 2) = [1 - COS s()], uz(t,0) = uz(t, 2) = 0.

Answer 1

Edution: Given Utt gUxoc (toJERX[012] U101_) = cos? () V (0,4)= ['- cos()] Uslo) Ux (t, 2) so By wing Sepration of variable: let Maat) = XOTI+) Usc (tox) XGUTIES Usc (t,0) = 0 810)T(H) x'loj so 반 Uzltj2) = 0 V X (0) T(+20 x'(2) 20 rta XlX) Tit) = It 2012 od X1%) "14) اق scloc) Tt) Dr ozu dap2 x"(sv) 717) Hence UH guxa x(x T1+) = 9x11x) T1+) x"(2) TH(+) Xla) -la k is is constant = 1 TIH)

X I TIH) 9 TH) Y"1+) TA = 9 TI+) K and X!! (OL) - K x"(x) k X (a) = 0 T XL) Card let k=1?, 170 x" () - 12 X 1212 -O XX - A é -AX X (2) A ex t Ben + ta bene dx X'(x) ine 40) =0 A At EXIB (130) A B x'(2) = 0 24 -22 23 À A é + Edsbé A= (A=B) XE -21 = Be (: AzB ) To de 24 -21 ri trivial solution case I Kao X! -- x' =A X = AX+B

x' = A 0 *: X(0) trivial solution Cause ( let Kari , Azə X + 12x - 0 x 100) A cos (an) + B sin(aus X'lov =tA sint 101) + A B Cos (10) ..' xlo) 0 - A sinlo) + AB coslo) 113 Cos (0) = Αγο BEO X512) 50 0 = - TA gin(21) + IB cas (21) 0 -AA sin(2x) for non rivial solution sin(20) - regina o 2. TT sin (24) Sinina) nal, 2, 3, -- - NTT 2

86 X(X) = Ancosala) Ancos (T3) i''(+) =.9THIK 11 (kamy?) 9K (2127173 -91 TH) 7'"+) = -9 (nyt - 1"1+) + 91 TBT TL4) 17 ho T Renee "T p cor (Z1T *) + 2 sin Int*) P

Ultix) = X () Tit) VH »)= An cos (nt x) [ Pn cox (3 MT) + An sin ( 317 )) (np9 x) cos (21) + Bo Bin (3 m+T+) cos(max)] Ultin) ŽI An cos nal [ An U10, 20) = CoR2 (72) Hence U lonx) = COSP (1T30 cos? (N) An con (192) + 0] = n=1 *(1+ Con ( 2172)) = An C05( ME x) n II 30 tu. A los A, cos + I cos 2 ( 22 ) x) + V (2T«) + Ą = 1 by comparing both side 2 नि A = Az = Adnan कि Hence Anat 2 14 lon) = [ 1--os (59 mux) (3 mT) - sin ( 39n+ ) + Bn Core COR (WITX) Ent)cos (Innt) Ut(t, x) = 8 h con A Cornilla N21

(40) 04101%) = 2 E [o + 14 [! - C08 (] en benut) con ("x)] [ + cosº (112) – 2 C06 ()] = (131) 8n cou ( ) ] 1 +40* 200*2)) - 20(") = 2(8) Br.com (16) 2009 (272) - TT) B. CO3(2x) + B2 ( 600(2.Tv) + B3 ( 3 371).col 3.TR 2 cos 32T 2 + (sete) By composing both side (210)B1 (221)B, = ) = -2 BIF $ --4 3T B2 = 6TT Bn= 14. + 37T 6T Hence COS UH) T (12) con({ht) 1 6 TT 1. sin ( 30 eux) cor (ht 14