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## Homework Statement

A uniform rod of length l has an initial (at time t = 0) temperature distribution given by u(x, 0) = sin([itex]\frac{πx}{l}[/itex]), 0 [itex]\leq[/itex] x [itex]\leq[/itex] l.

The temperature u(x, t) satisfies the classical one-dimensional diffusion equation, u

_{t}= ku

_{xx}

The ends of the rod are perfectly insulated, so u

_{x}(0, t) = u

_{x}(l, t) = 0.

Find general solution u(x, t) by the method of separation of variables. What happens to the solution as t -> ∞ ? (Do not find the Fourier coefficients other than the one required to answer the last part.)

## Homework Equations

## The Attempt at a Solution

We have method of Seperation of Variables so set:

u(x, t) = X(x)T(t) so we have XT' = kX''T (by using this with u

_{t}= ku

_{xx})

This gives us: [itex]\frac{X''}{X}[/itex] = [itex]\frac{T'}{kT}[/itex] = λ (by separating the variables where λ is a constant)

__Looking at X:__we have X'' = Xλ so X'' - Xλ = 0

this means we have [itex]\alpha[/itex]

^{2}= λ and so [itex]\alpha[/itex] = [itex]\pm[/itex][itex]\sqrt{λ}[/itex]

we require λ < 0, so let λ = -[itex]\omega[/itex]

^{2}and so we now have: [itex]\alpha[/itex] = i[itex]\omega[/itex]

this gives us X(x) = Asin([itex]\omega[/itex]x) + Bcos([itex]\omega[/itex]x)

Applying our boundary conditions to X(x) we have:

X' = A[itex]\omega[/itex]cos([itex]\omega[/itex]x) - B[itex]\omega[/itex]sin([itex]\omega[/itex]x)

thus X'(0) = A[itex]\omega[/itex] = 0 so we know A=0

and X'(l) = -B[itex]\omega[/itex]sin([itex]\omega[/itex]l) = 0 so we know sin([itex]\omega[/itex]l) = 0 and so [itex]\omega[/itex]l = nπ

this means [itex]\omega[/itex] = [itex]\frac{nπ}{l}[/itex]

then finally for X we have that: X(x) = Bcos([itex]\frac{nπx}{l}[/itex]), n=0,1,2,3,....

__Looking at T:__we have [itex]\frac{T'}{T}[/itex] = kλ and integrate to find:

ln|T| = kλt + C so T = Ce

^{kλt}

and we know that λ = -[itex]\omega[/itex]

^{2}= (-n

^{2}π

^{2})/(l

^{2})

Hence we know our General Solution is

**u(x,t) = X(x)T(t)**so

__Our General Solution is:__

u(x,t) = [itex]\sum[/itex]B

_{n}cos([itex]\frac{nπx}{l}[/itex])exp{(-n

^{2}π

^{2}kt)/(l

^{2})}

Also given an initial condition:

u(x,0) = sin([itex]\frac{πx}{l}[/itex])

using this, we set t=0 in u(x,t) so we have:

u(x,0) = [itex]\sum[/itex]B

_{n}cos([itex]\frac{nπx}{l}[/itex]) = sin([itex]\frac{πx}{l}[/itex])

so by inspection when n=0, B

_{0}= sin([itex]\frac{πx}{l}[/itex])

and when n=1,2,3,... we have B

_{1}, B

_{2}, B

_{3}, ... = 0.

Thus our full solution for these boundary conditions and initial conditions is:

u(x,t) = sin([itex]\frac{πx}{l}[/itex])

__________________________________________________________

**Is this all correct what I've done? The reason I'm unsure if I'm correct is due to the last part of the question asking "What happens to the solution when t --> ∞?" as you can see there is no t in the final solution. It also notes "Do not find the Fourier coefficients other than the one required to answer the last part" and I'm not sure what this means or how it effects the question really.**

ANY help is very much appreciated! Thanks very much!

ANY help is very much appreciated! Thanks very much!