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I am struggling with the heat equation

u_{t}= ku_{xx}

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.

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

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