Answer Happy

Consider the following heat equation problem on 0s rs 1 in cylindrical geometry: at 2T 1 ar 1 2²T + at ar² rar 200² + +r² cos m 8, T(1, 0, t) = -5 Ir-1, T(r, 0, 0) = 0. at(r,e,t) ar (a) The solution will depend on 8 as cos m 8, so only a radial grid is necessary. While this simplifies the numerics, we now require a boundary condition at r=0.In Sec. 6.1 it is shown that the proper bound- ary condition is at lim, 0-²T = 0 ar
[see Eq. (6.1.37)]. For m=0 this implies T=0 at r = 0, but for m = 0 it implies T = 0 at r=0. Use the differencing techniques discussed in connection with Eqs. (6.2.23) and (6.2.25) to satisfy the mixed boundary condition at r = 1. (b) Use the Crank-Nicolson method to solve this problem for the case m = 2. Take the time step size At = 0.02, and take M = 20 (i.e. 21 grid points in r). Make an animation of the solution as a function of r at 8=0 for 0 ≤ t≤ 1/2.
Τ 2ατ Δx + (2κ - α Δx)Τ" 2x + a\x = (6.2.23)
TM = 2br n Ax+(2k-bAx) TM-1 2k+bAx (6.2.25)