Question

Assignment 0220 Marks) Solve the following IVBP: PDE : Uxx = (1/25) utt ICs: u (x,0) = x2 (nt - x), ut (x,0) = sin(x) BCs: u(0,t) = 0, u(nt,t) = 0 for 0<x<, t> 0. for 0<x<T. for t>0.

Answer 1

The given IVBP is Vax (+) utt for 02x2 MT, tyo initial conditions are: u(,0) = x(TT-x) for "+(x,o) sina OLXCIT Boundary conditions are: ulo, t) = 0; u (TI, U)=0 Step-D :- The solution for the wave equation will be in the form of u(a, ) x(). TCU) By substituting it in the it in the given equation, we get x(n), T"() (as) *"(x), 7 (4) T"CE) x"(x) 25x(+) s -) 0 T(E) CS Scanned with CamScanner

From eg , we get 2DE's; 1 x"(a) + 25XX(X) = 0 2) T"(t) + XT(L) =D Step- We have boundary conditions, a(0,t) u(IT ,t) 3) u (0,5) = x(0). I(t) = 0 » u ( 2 ) x(), T(t) = =0 which means WC get X(0)=0} x ()=0 Step - The value 't' have three cases. Case - il x=0 Then the generat solutions for x(x) & r(t) are X(x) = atba t(t) = ct de CS Scanned with CamScanner

So, for s=0 X(x) = atba we have , x 10) a 20 X(T) = at 6 1 - 0 => b=0 Therefore, we get azo, b=0 Here, we got the only trivial solution. So, we can ignore it. Case - ii, i- For ato The general solution is X (X) a Cos (Vix) + 6 sin (5x) T(+) Ccos ( qvae) + d sin (evat) [since I - 1 5 we have, x(0) - 0 & X(T) = *(0) - a =D a=0 X(T) = b sin (FX (A)) = 0 = b =0 (1) sin(x) = 0 get trivial solution. So, we can CS for bio, we Scanned with Camsigmome it.

For sin (TXT) = 0 VT NTT (nea) nl (n (n = 1,2,3 ..) So, u(a,) = x(se). TCH) Up (2,6) Xn(x), To(t) got Xn(x) = bn sin (no) then, Tnct) eneos ( (ant) + dnsin (ant) So, un (x, t) = brisin (nx) encos ( + dosin (nt Here the coefficient bn can be absorbed into en & dn Un (x, t) sin (nu) [cn eos (mt) + dm sin (95)] CS Scanned with CamScanner

Case-it, i sco The general solution is Fax -v-ax X («) + be Brat • Sext T(t) = ce ae t de we have X (0) -0 X(T) = 20 x(0) a+b=0 TEX ae -TF + be =0 expressing them in terms of matrix equation, 16.11.12 FAT -TA e e The derterminant => AFX TX -e e we have sto , e the only solution possible is (a,b=0) which is a trivial solution So, we can ignore it. CS Scanned with CamScanner

Step - : Now, the superposition of all cases of a we get, u(x,t) E un (x, t) Therefore, u (x, t) E È sin(nx) [ encos (mt) + dnsin (mt) 021 Step- Applying intial conditions, U (4,0) = a( - x) ut(x,0) = sin x - > E sim (na) [ n(o) +0]: x"(1-2) 2) E cnsin (mx) = x (TT-x) Therefore, we can see that on is the fourier sine coefficients of x (1-x). TT en = % (+(1-2) sin ( nx) da ㅠ CS Scanned with CamScanner 0

Cn = # [4* (6 - 17 ") sin(am) - 41n cos (mt) – 2Tn a)-cm ny en - [urincos cos (nm) + 20n] [ since sin (OTT) = 0 .] TTY en un cos (717) + en 14 en 4 cos (MT) +2' n3 Now, u(,0) = sin a "t(a,0) 2 sin (na) [ da(3)] sinx Here, can see that do (7) is the fourier sine coefficient of sina. ㅠ du (3) f sma). (sin (ma)) da res CS Scanned with CamScanner

dn (1/2) Sin (m) ท-| do 5 sin (0) Tn (1-n") rol since Sin(n) :o) Therefore dn 20 So, our final solution is u(x,t) § sin (nn) -14cos (03)+2) cos(at) nii n3 when no even u(,t) a sin (nx) 도 n zeven when neodd ulx, t), E sim (mu) ( 2 COS (ht ng hrodd CS Scanned with CamScanner

Therefore the solution to the given wave equation is clearly explained step by step.