Question

Problem 1. Consider the nonhomogeneous heat equation for u(r, t) subject to the nonhomogenoous boundary conditions u(0, t) 1, u(r, t) 0, t>o and the initial condition u(, 0)in() Find the solution u (z, t) by completing each of the following steps: (a) Find the equilibrium temperature distribution t) (b) Den ote u(x, t)-u(x, t)-ue(x). Derive the IBVP for the function u(x,t). (c) Find v(x, t) (d) Find u(x,t)

Answer 1

ut = uxx + sin 2x a) At equilibrium temperature, ut = 0, and then 0<x<π, t > 0 uxx + sin 2x = 0 or, uxx =-sin 2x Integrating twice w.r.t. "x", we get -sin 2x dx | dx= 4 b) Define r(x, t) = u(x, t)-ue (x) = u(x, t)-(-sin 2x + c, x + c2 To find c1 and c2 we define u(0, t) 0, u(n, t) 0 As u(0, t) = 1, thus r(0,t)-0-a(0, t) _ (1 sín 2(0) + Ci (0) + C2) =1-c2 Hence As u(T, t) = 0

Hence Thus r(x, t) = u(x, t) _ sin 2x--x + 1 | (@) Now the new IBVP for v(x, t), is (0, t) = 0 = u(n, t) Also the initial condition given u(x, 0) = 1--+ sin x thus initial condition for v(x,t) is r(x, 0)=u(x, 0)-Gsin 2x-1x+1) 1--+ sin x- sin 2x--x + 1 u(x, 0) = sin x-、-sin 2x (*) c) Thus the IBVP is

v(0, t)0 -v(t, t) r(x, 0) = sin x-、-sin 2x (*) Solution of the partial differential equation by method of separation of variables Let v(x, t) -X(x)T(t) ...(2) be solution to (1) Then The left side is totally a function of t and right side is totally a function of x. This can only happen if both sides reduce to a constant. Thus-k2 (say) TX Now consider the first equation or T' + k"T=0

-k2t It's a linear equation and its solution is T- ce Now consider the second equation --k2 or X" +k2X- o Eigen functions are cos kx and sin kx Its solution x(x) = (c2 cos kx + C3 sin kx) and hence the general solution is u(x, t) = (A cos kx + B sin kx)e-ft (4) Put x 0, then 0 0, t) - Ae-k't, we get A 0 (6) Then v(x,t (B sin kx)ek...(7) (B sin kn)e-k" t, -k2t 0 = v(n, t) Put x = π, then which gives sin kπ=0. Thus kπ= nT, n= 1,2 or, k= 1,2 (8)

Putting this value of k in (4), we get v(x, t) = (Bn sin nx)e-n°t (9) The above solution is true for every n 1,2, 3,... Taking all these answers together, we get 〉 (Bn sin nx)e-nzt v(x,t)= (9) n-1 Now to find the values of B, n-1,2,.3 we use the initial condition (), r(x,0) = sinx-Isin 2x = Σ (B" sin nxe-n2(0) 4 n=1 Σ(B, sin nx) 2 (Bn n-1

By comparison of coefficients of sin nx, we find that except for n-1, 2 rest of B,-0, and B1 = 1, B, 4 Thus the solution for p(x.t) becomes r(x, t) = (B, sin 1x)e-12t + (B2 sin 2x)e = (sin x)e-t-( sin 2x) e-4t -2 t d) Thus now using (@) v(x, t) + ( sin 2x-π x + 1 si2 u(x, t) = (sinxt sin (10) Please rate the solution as it is a very lengthy problem. Thanks