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Powertravel
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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.
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.