Question

4. [10] Find the solution to given initial-boundary value problem: 4uxx = ut 0<x<TI, t> 0 u(0,t) = 5, uit, t) = 10, t> 0 u(x,0) = sin 3x - sin 5x, 0<x<T

Answer 1

The given initial boundary value problem is, Aro Alan • Ut , OLMET too el, ult,A) = 10 uco, t) = 5 Осигт Uh, o) = sinar sinsn. Applying Laplane transform on 11), weget su(a.s) - U(210) [:: Lsc («,477 4 ū (Hrs) - & ut nisla 1 (singkasingu) LS 24 13 = 3 umus? -U(X) dūras) dar or (2). so the Gimplementary function of (2) is, ą (2, 3) = a los hran -xt C2 sinn s where als) and Contents to be determined. C2 (s) are uplars) (&(sinsnasinsx)) pa & Singu- sinza A prst D-5 D . t. [ sinsx. --3 -25-6 sinsa St100 sinax -9-04 sinar 5436

Therefore, ūns) = a Cash (rs a) ffa sinn (S) Sinox Sinan 5+36 832 sfloo Laplace transform on the boundary Now applying ondition, we get U(0,5) = 5 S ū (ET,S) - 10 ūco,s) - § genues, a = islu un (15,5) = 10 gives, ş GShl v n) +Cg sing = 0 09, alu 5 s 2- Cos h(ra) (2 sinnlifor) Therefore the equation (3) becomes 2-CGs host) sin als al) ů (nrs) & Cash(s n) + sin hl Bit) sinsa sin 3x Stivo S+36 fransform on (4), weget the Applying inverse Laplace required general solution of the given cos notar) Kinet) = LYS 8 [Cos n(ren) - sin sn e sinane 741. PPE, 2 sin news}} sin na -1004 ~364

Now, Cos hun)- 2-los h (VI) sinn (ru) sin his it) Cas n (ru) sin h it) + Cash (18 ) Sex (84)-2S A 8+) sin h( ) sinals a linth **))-2 sinn - sin h( t) 5 2 = A (X,H LYS sin h ( F (n+7)) - 2 sin h ( 13 ) sinn (1) (SAL). Now this inversion can be Calculated using Contour Integration of the haploce Inversion integral. Cape it. I f(1) = f(t) = 21 fisits. eff st Sum of the residues ofe if(s) at the poles of f(s). Cauchy residues R= lim [CS-z;" Gls)] day where SZ (ny)! dsnt g(s) = est is. Hare the solution is, -360 AlxH) sin sx e loot & sinza e (Argwer). Unit) =