Solving Eigenvalue & Eigenfunction of 1D Heat Equation

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SUMMARY

The discussion focuses on solving the 1D heat equation, represented as ut = kuxx, with boundary conditions u(0,t) = 0 and u'(L,t) = 0, and initial condition u(x,0) = f(x). The participant seeks to derive eigenvalues using complex analysis, specifically examining the separation of variables leading to the equations T' + λkT = 0 and X'' + λX = 0. The participant questions the validity of eigenvalue solutions derived from the auxiliary equation, particularly whether λ = (((1 + 2n)π)/(2L))^2 could also be valid. Additionally, they express confusion regarding the relationship between eigenfunctions and the derived eigenvalues.

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  • Understanding of partial differential equations (PDEs), specifically the heat equation.
  • Familiarity with boundary value problems and eigenvalue problems.
  • Knowledge of complex analysis and its application in solving differential equations.
  • Experience with separation of variables technique in PDEs.
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  • Explore the derivation of eigenvalues for the heat equation using complex analysis.
  • Learn about the implications of boundary conditions on eigenfunction solutions.
  • Investigate the Laplace transform method for solving PDEs and its advantages.
  • Study the relationship between hyperbolic functions and trigonometric functions in the context of differential equations.
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Mathematicians, physicists, and engineers dealing with heat transfer problems, as well as students studying partial differential equations and eigenvalue problems.

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 \Rightarrow A = 0 and B ≠ 0
X'(L) = 0 \Rightarrow 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)\pi)/(2*L))2 n = 1,2,3,... (3)
Can't it also be
λ = (((1 + 2n)\pi)/(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*\pi - 0.5\pi \Rightarrow
√(λ) = ((2n -1)\pi) / (2Li) = -i((2n -1)\pi)/(2L) \Rightarrow
λ = (-((2n -1)\pi)/(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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Powertravel said:
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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