Question

Solve the following partial differential equation using separation of variables method to determine the function 0 (x,t). Simplify the solution using Fourier series method. 2²0 2²0 (30 marks) Ox² at is Where: (x,0) = 0 0(0,t) = 0.21 0<t< 20 Q(x,20) = 0 do(0,1)= (1 - 2t) dx = (1-21)

Answer 1

Given PDE is ad = 9th with Bie (%,0) = 0 = 9 (20) ($(0,t) & 0.2 t = f(E) 18 (ort) = (1-at) = g(t) v hat the general solution of be (by separation Ø (art) = H (H) G(") variastes) or, pax = H(t) G (*) . Ptt a H"(t) G(X) Putting it we get H"(t) (x) Het = G(x) = (say) for do | H() > 0 ) H ( ) + 4 + 6+ using boundary conditions, we get trovial soln. (say) For doo (do ) H" (f) - 4²H(t) = o. 7 HCM - A ett Bedt Again we get a frivial non-frivial sol" ico sol" for hon het d=-42 Then "+ M²H=0 m² + M2 20 m = Mi HLt) = A cost & B sinpt H(O) = 0 = A 7 A=0 H(20) = 0 = B sin 20u = 0 Bto

Sin 20u = 0= sin na 20u = na 7 = 20 forn=0,112 - > d=- We also have => Ġ (x) + M+G(X) = 0 + G(x) = A cose + B sinux G60 - 0.2t O can. be written as polo,t) = f(t)} Tex(0,t) =o/ + Ploit) = 0 Pt(0,6) 20 g1t) e For (2) 9106) - HLAG(070 For (2) 8x(0,+) = H (+) Gi (0) = 0 » G (o) = 0 ; (*) = -AM Ersinux + BM Cs 4x > G (0) 20= Bu B=0 Gn (2) = An Cosu x = An as TX en Pn (2,t) a Hn (7) Gn (2) Fourier series solu 20 Edu(x,t) f(xit) & Dn sin nye ces momento

ی ی مینی & (ort) = f(t) = {on sin nato (fourder sine series) on = 20 { f(t) sin nate at For (3) we can similarly find that of (x, t) = 2 Fn sin (not) sin ( 2 ) qr(0,4) = g(4) > Zion on the sin egen In - 2 ) & (6) sin note at combining Co (W) & ② (we have pleit) - Ž Sinetes [ Da ws na +5, sinady Dua to se 0.2 t sinnat at ** 50 % + 24 (os may 10 TL 4] ای ی ی ی ی 12t 2 , 1, h +1

ad En 2 1 (1-2t) sinnat at nat (-4) 199)- 25 de in the east (19*4!) + et *(as ng )/" +2 13 l cos hoté) at ] - - - - - - - - - - - - at us n 20 De 780 (--)**' + 207 T on a to mor [ 1+ 39 (-1997 q(x, t) = 2 sin nato (ou ws na R + Fn sin nax]