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Need help with deriving the eigen-value and eigen-func for a S-L PDE (1dim Heat EQ)

  1. Nov 1, 2011 #1
    Hi,
    I am struggling with the heat equation
    ut = kuxx
    with the boundary conditions
    u(0,t) = u'(L,t) = 0
    and initial condition
    u(x,0) = f(x)
    0 ≤ x ≤ L
    0 ≤ t

    I want to derive it's eigenvalue using complex analysis.

    After separating the variables into u(x,t) = X(x)T(t) = XT and getting
    T' + λkT = 0 (1)
    X'' + λX = 0 (2)

    I start with (2);

    It is easily shown that λ ≠ 0 because it only yields elementary solutions.

    Equation (2)s Auxiliary Equation (AE) is
    r = ± i*√(λ)

    So
    X = Acosh(rx) + Bsinh(rx)
    and
    X' = Asinh(rx) + Bcosh(rx)

    Using the Boundrary Conditions I get
    X(0)=0 [itex]\Rightarrow[/itex] A = 0 and B ≠ 0
    X'(L) = 0 [itex]\Rightarrow[/itex] cosh(rx) = 0
    so
    X = Bsinh(rx)

    Now comes my first question:

    My textbook says that cosh(rx) only is zero when
    λ = (((2n -1)[itex]\pi[/itex])/(2*L))2 n = 1,2,3,... (3)
    Can't it also be
    λ = (((1 + 2n)[itex]\pi[/itex])/(2*L))2 n = 0,1,2,... ? (4)
    Why is it as (3) instead of (4) and will (4) cause problems if I want to expand u(x,t) in a sine series?

    Now for my second and primary question.

    Using (3) I get
    i*λL = n*[itex]\pi[/itex] - 0.5[itex]\pi[/itex] [itex]\Rightarrow[/itex]
    √(λ) = ((2n -1)[itex]\pi[/itex]) / (2Li) = -i((2n -1)[itex]\pi[/itex])/(2L) [itex]\Rightarrow[/itex]
    λ = (-((2n -1)[itex]\pi[/itex])/(2L))2
    That means λ < 0 so √(-λ) is real.
    How can I get the eigen function of λ equal Csin(√(-λ)x) when X = sinh(√(-λ)x) ?

    Thanks in advance.
     
  2. jcsd
  3. Nov 6, 2011 #2
    Re: Need help with deriving the eigen-value and eigen-func for a S-L PDE (1dim Heat E

     
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