Question

1. When considering the heat conduction in a rod (of length L) with zero temperature at both ends, we encounter eigenvalue problem ψ't λψ = 0, ψ(0) = ψ(L) = 0. Show that in this problem, all eigenvalues λ are real and positive [Remember: eigenfunction (x) can be complex when eigenvalue λ is complex.]

Answer 1

EX Psi'' + lambda Psi = 0 rightarrow if lambda = 0 Rightarrow Psi'' = 0 Rightarrow Psi (t) = c_1 t + c_2 Psi (0) = Psi (L) = 0 0 = c_2 0 = L c_1 + 0 Rightarrow L notequalto 0 Rightarrow c_1 = 0 Therefore Psi (t) = c_1 t + c_2 is trivial solution Therefore lambda notequalto 0 if lambda < 0 Rightarrow Psi'' + lambda Psi = 0 Therefore D^2 Psi + lambda Psi = 0 D^2 Psi = d^2/dt^2 Psi (D^2 + lambda) Psi = 0 A.E D^2 + lambda = 0 lambda < 0 Rightarrow lambda = -m^2 Rightarrow D^2 = + m^2 D = plusminus m Therefore Psi (t) = c_1 e^mt + c_2 e^- mt Psi (0) = Psi (t) = 0 0 = c_1 + c_2 0 = c_1 e^m L + c_2 e^- mL Rightarrow c_1 = c_2 = 0 Therefore lambda < 0 is also trivial sol \r\n