Question

The function u(x, t) satisfies the partial differential equation with the boundary conditions u(0,t) = 0 , u(1,t) = 0 and the initial condition u(x,0) = f(x) = 2x if 0<x<} 2(1 – x) if}<x< 1 . The initial velocity is zero. Answer the following questions. (1) Obtain two ODES (Ordinary Differential Equations) by the method of separation of variables and separating variable -k? (2) Find u(x, t) as an infinite series satisfying the boundary condition and the initial condition.

Answer 1

Andy We have given given 2 24 بالي det? by wing Variables wing the method let the of separation of solution of ② is . the form Glut) = X(2). T (t) where X a function of only and To only is function of t From (2) have 2 ²u and de XT" =TA" 2 ate du ² علما Put this in Eqh vet @ XT" TX" x" TU Х the n is L.H.S. function of i function of it only Note that only and this Can let say X R.HS happen are if bath Gnstanta only TH Two O.DE -k? 2 and Xpek Thoi on -k? T x

Now, we we jet from 3) 8 Their respective solution are u(at)b = (6, 6 kn + (q Sin kn) (23 Cos Kt + ly sin kt) to find the series solution of und given condition 4 (0,4)=0 Eqh (c, Cas KO) + (g Sin KG) (C3 Cos Kt + ly Sinkt) (a).(cs Caskt + Cu Sin Kt) = Tc Put Equ u(n,A) (Ca Sin Sin Kn) C3 Ces Kt + Cu sinkt) @ ここ © ih @ 2 , we yet C) Intial velocity is zero Ų (X,0) = 0 Up (2 t) = how (( Sin Ka) (-k C3 K C₂ Sinky + K Cy sos kt) (a,0) > ((2 Sin Ke) (-K Cg Sin KKG) + K Ly Cas Kld). (( sinka) (K Cy) [cy=0 Put in Equ @ get

© ulaut) Cq Sin Kr. Cas kt U (1,A) (iu) ( sink. Glaskst zo G Sinka Note that C 20 otherwise & Zero But we get only solution we want hour zero So Sink) zo Sin(K) Sin (h17 Kahit 4- n=1, 2, 3, - So Equ becomes ucnit) { Bn Sin (Nora) Cos (net nal u(a,0) ful & Bn sin (nord) ; ohne finza {ali gu 2 (2~ y hench h hal So, with olach olacl Compare So l where Bu = I fond. Sin (nom) din l

c 2 stas Sin (172) de سلیم s 21. Sin (now) dne 2 / 211->) Sin in ota) din 4 > Now integrate by integration by part. / 231. Cos mera) 2 Sin (nan) n² it NIT 0 211-0) Cas(nan) 2f-2400 a Sin Mita) n² 4 2 1 Y2 nit 2 Sin (MI) Cascht 'no + 2 Sin 2 Cas Center + 2 nit t ht = ) MI n²2 8 Sin (ht nza u Sin (4) n2o2 2 + u Sin MT) n² --) Bra 2 But in Eqh get , we 8 sin (2) n²tt2 Sin (hita). Cos (nett) A