We will solve the heat equation with boundary/initial conditions: 2uxx, 0<x<6, t≥0 u₁ = 2uxx ux (0, 1) = 0, ux (6,t) = 0, and u(x, 0) { 0, 0 < x <3 4, 3 ≤x≤6 This models the temperature in a thin rod of length L = 6 with thermal diffusivity α = 2 where the no heat is gained or lost through the ends of the rod (insulated ends) and the initial temperature distribution is u(x, 0). For extra practice we will solve this problem from scratch. The problem splits into cases based on the sign of 2. (Notation: For the cases below, use constants a and b) • Case 1:= 0 X(x) = a+bx Plugging the boundary values into this formula gives 0 = X'(0) = b 0 = X'(6) = 6* So X(x) = 0. We will deal with this case using A0 later.... • Case 2: λ = y² (In your answers below use gamma instead of lambda) X(x) = a*e^(gamma*x)+b*e^(-gamma*x) Plugging the boundary values into this formula gives 0 = X'(0) = 0 = X'(6) = So X(x) = 0 We can ingore this case. • Case 3:λ=2 (In your answers below use gamma instead of lambda) X(x)=a*cos(gamma*x)+b*sin(gamma*x) Plugging the boundary values into this formula gives 0 = X'(0) = b*gamma 0 X'(6) = -b*sin(gamma*6)*gamma Which leads us to the eigenvalues Yn = and eigenfunctions X,(x) = (Notation: Eigenfunctions should not include any constants a or b.) Plug the eigenvalues λ = y from Case 3 into the differential equation for T(t) and solve: Tn (t) = (Notation: use c for the unknown constant.) Combining all of the X, and T, we get that u(x,t) = Σ An ΣΑ n=0 n where A, are unknown constants. We compute A, by plugging t = 0 into the formula for u(x, t) and setting equal to the initial heat distribution given in the problem. 8 u(x, 0) = Σ An n=0 So the A, are Fourier coefficients. An = 2 6 6 0, 0 < x < 3 4, 3≤x≤6 dx (for n 0) Ao = = Remember that sin(n) = 0.