Question

boundam ualue prodlem *Solue the fMlawing u(o,y) ay) iy)o Co<y<T) > uylz,o)O UyCa, )O * The answer 75 UCi,4) = H (n Cash n+ Bn snh ha)Cos nu PM M gy1 Cos ny du Cn 0,1,2 BM=-Qnh n Tcosh n (n=1,2, Jogupcs ny dy (n,2, please Show the sduing process n

Answer 1

Given boundary value problem is Uxxt ilyy so foro 5x21,ocych ... (1) 460, y) = gey), ux C1, Y)=0 for Bo <ylt .. (2) Ung (8,0) =0 , My (x, ) o for ocR41... (3) We start looking for solutions of the PDE (1) that satisfy the boundorty condition () and have the special form uck, y) = x(x) Y CY) ...(4) Differentiating (9) on t wice with respect to x and twice with respect to y and substitute these derivative into the PPE G). We then obtain x"(x) Y (4) + X(X) Y "(y)=0 - - x = =-X, where x is the seperating constant The above equation implies 1x2 = XX, 0<x<1 ...(5) and dy , osyl ... (6) Now, the boundary condition to (3) for u imply My (1,0) = xcx). Y Coco, My CX, 1) = x(x).Y (11) 50 Since u is non trivial ie uzo, so we can not take x(x)=0 hence Y'(o)Do and Y CA 20 ...(7) VV There fore, the funcion Y is a solution of the eigen value problem +Y 20 dyz , ocy<T Y'(o)50, Y Ct) Do Here three case artise, as follows - case-1: When x40 The solution is given by Yey)= & cash (Fiy) + ß sinh any) Now, appling the boundarly condition, we have Y()co 1 do YCA) to 400 Hence we get ycy)=o, a trivial solution. Therefore, this eigen value problem does not admit negative eigen values.

Casea: xDol the solution is given by cya at By Now Yeo) 20 Y'(I) = 0 pro Hence, x o is an eigenvalue with eigen function or rosa he Yocy El Csince the eigenfunction is unique upto a multiplicali ove factor) Case-3: 220 The solution is given by r cy)= a caste ) + sin(ray) Now, Yo) 20 8=0 YCH) TO & Sinan) 20. > Venttinen consequently, yo is an eigenvalues if and only if *- (OT) ie, aan², n. 1, 2, 3,... The associated eigen function is Y(Y) = osny Therefore, the solution of eigen value problem is an intinite sequence of non-negative simple eigen values and their associated eigenfunctions. we use the convinient nostation: Yn (Y) = cos(ny), Anan, n=0, 12,... Now considering the ODE (5) for Daxn. dk - mx = 0 The solutions are given by X (X) = 8 + 8x Xn(x)= Y cơ bx + Sp 3 n h ( x ) Thus the product solution of the intial boundary valuo problem given by . (x, y) = hot Box 2, C = 2A, Cash (n) + By Sinh (2) x (Pa) , 2 , 3 The superposition principle implies, U (x, y) = to + Box 7 Ancash (nx) + Bn sin are . 7 7 2

com 2 iny) -2% no where I an cos(ny) no is the Now the initial condition (2) imply. 120,4)= gly) Ao + EAC (ny) = 710 + 2 Ancolny) = costing), where an cos (ny) is the o comparing the coefficient of Tourier cosine series of Hy). cos coscny) in both side we get To = An - Anaan, an is the fourier coefficient, which is given by an=2 / 5 Hence, an= 2 2 y cony dy, n=0,1,2,.. gey) cos (myjay too na 0, 1, 2, ... nai Ako, ux Cl, y co → + È An Sinhen) at Bn cashin)?csony) so comparing the coefficient of cos(ny), we get Boro and An sinh (0) + Bn Cash (n)=0 Bna - sinh (n) An cosh (n) By = – 2 sinh (n) [ glycos (ny) ay n=1, 2, 3 ... IT Cash(n) Jo Hence the solution of the given problem is n2-1 where, u (x, y) = 40 + Elan cash (nx) + Bsinh (nx)} cocky) 'An= 2 2 8 gig cashyjay, 3=0, 1, 2, 3 ... Bm = -2.smh.com S rey) casco yay, n=1,2;3,- teosh (n) Jo