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

  • #1

Main Question or Discussion Point

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.
 

Answers and Replies

  • #2
81
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)"

Are you sure about this solution? the solutions for ODE's are in the form of e^(rx), so e^(ix)... is correlating to cos(x)? correct? Check that

Alternatively you could solving this using laplace transforms, which much easier, and evaluate the poles accordingly, s, functions, using the residual formulature and going from there.

YS
 

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