Question

4. Consider the boundary value problem defined by the partial differential equation д?и д?и = 0, ду? y > 0, да? with boundary conditions u(0, y) = u(T,y) = 0, u(x, 0) = 1 and limy-v00 |u(x, y)|< 0o. (a) Use separation of variables to find the eigenvalues and general series solution in terms of the normal modes. (b) Impose the inhomogeneous boundary condition u(x,0) = 1 to find the constants in the general series solution and hence the solution to the boundary value problem. Hint: { for n even, 0, sin nx dx for n 2р — 1 оdd, 2р-1» and sin2 nx dx T/2.

Answer 1

tu=0 By method of Sepeeation of valiables, Let ulx,y) = x(x) Yly) 24 = x'y, 24 = xy' and dy MO Substituting this in the Laplace equation x"y+xy"=0 -*" Y" = x"+,x=0 and y'_s y=0 Now there are two seperate ordinaly differential equations.

Sased on the boundary conditions u 10, y) = ulty)=0 de can say that u(x6) must contain peliosie teems in This happens if x''+sx=0 yields a periodic solution Therefore it must be positive het d=p2 x+p²x=0 and y_p² y =o. Geneeal solutions to these differential equations alle x = c. cos(p) + C₂ Sin (px) by - РУ Y = C3 Cosh (Py) + C4 Sinh (py) (hyperbolic functions) for some arbitrary Constants C,, C2, C3, C4. ux,y) = (s, Cos (px) + C2 Sin (px)) (Cz Cosh (by)+ C4 Sinh (py)) 4(0,4)=0 [C=0 UCty)=0 Sin (p) = 0 Pr=nt for n=0,1,2,3,- p= 1 [b=n Substituting colh (ty) = alter and Sinh lry Y = ((3 + (4) ePy + ((3 - (A) = ty giren Lim lulx,y)< Lim lx Yl Zoo yo Lim lyko This happens if the telm Containing y o epy in y Vanishes.

That is if Ca +(4=0. Cz=-(4 Y= Czepy Ulx,y) = C₂ C3 Sin (nx) eo for n=0,1,2, 3, . . . Let C₂ C3 = bn for a particular n. Then Ulx,y) = Ibn Sin (nx) eny nzo U(x,y) = 8 bn Sin (nx) e ny given u (26,0) = 1 = bn Sin (1x) which means bris a Foulier Selies of l. Coefficient n=1 n=1 TT Hence | (1) Sin (nx) dx. Day Sin n x) dx So Sin/nx) dx =) br=2 14 for k=1,2,... if nus even if nis odd bn Take n = 2K-1 0 4 Sin (@k_1)x) e (2k-1)y. Sin (2k-1) x) e ulxy) TT (2k-1) K=