Hi. Having problems with this tricky Heat Equation Question. Managed to do part (a) and would appreciate verification that it's right.

But I can't manage to finish off the second part. I've started it off so please do advice me. Thanks a lot!

QUESTION:

-----------------------------------------

(a)

Show that the steady solution (which is independent of t) of the heat equation,

$\displaystyle

\frac{{\partial ^2 \theta }}

{{\partial x^2 }} = \frac{1}

{{\alpha ^2 }}\frac{{\partial \theta }}

{{\partial t}}

$ where $\displaystyle \alpha$ is a constant, on the interval:

$\displaystyle

- L \leqslant x \leqslant L

$ with conditions: $\displaystyle \begin{gathered} \theta ( - L,t) = 0 \hfill \\

\theta (L,t) = T \hfill \\

\end{gathered} $ is: $\displaystyle \theta = \theta _0 (x) = \frac{{T(1 + \frac{x}

{L})}}

{2}$

(b)

Use methods of separation of variables to show that the unsteady solution for

$\displaystyle \theta = (x,t)$ withconditions: $\displaystyle \theta ( - L) = T,{\text{ }}\theta (L) = 0,{\text{ }}\theta (x,0) = \theta _0 (x){\text{ is}}$:

$\displaystyle

\theta (x,t) = \frac{T}

{2}\left[ {1 - \frac{x}

{L}} \right] - 2T\sum\limits_{n = 1}^\infty {\left[ {e^{ - \alpha ^2 n^2 \pi ^2 t/L^2 } \frac{{( - 1)^n \sin (n\pi x/L)}}

{{n\pi }}} \right]}$

My attempt:

Part (A)::

-----------------------

Using separation of variables:

$\displaystyle \begin{gathered}

Let:\theta = (x,t) = X(x)T(t) \hfill \\

X(x) = Ax + B \hfill \\

T(t) = C \hfill \\

so:X(x)T(t) = (Ax + B)(C) \hfill \\

\end{gathered} $

now to use the conditions:

----------------------------------

$\displaystyle \begin{gathered}

when:\theta ( - L,t) = 0, \hfill \\

A(x + L) + B = 0 \hfill \\

A(0) + B = 0,so:B = 0 \hfill \\

\end{gathered} $

$\displaystyle \begin{gathered}

when:\theta (L,t) = T \hfill \\

A(x + L) + 0 = T \hfill \\

A(2L) = T \hfill \\

A = \frac{T}

{{2L}} \hfill \\

\end{gathered}$

$\displaystyle \begin{gathered}

so: \hfill \\

\theta _0 (x) = \frac{T}

{{2L}}(x + L) = \frac{{T(1 + \frac{x}

{L})}}

{2} \hfill \\

\end{gathered}$

That's part (A) done. is my method to approach the final answer correct?

Part (B)::

-----------------------

I'm having problems here. Can't quite finish the question off. Please could you guide me out here. Thanks.

Using separation of variables:

$\displaystyle Let:\theta = (x,t) = X(x)T(t)$

$\displaystyle \begin{gathered}

unsteady - solution: - \rho ^2 < 0 \hfill \\

X(x) = Ae^{i\rho x} + Be^{ - ipx} = A\cos px + B\sin px \hfill \\

T(t) = Ce^{ - \alpha ^2 \rho ^2 t} \hfill \\

so:X(x)T(t) = (A\cos \rho x + B\sin \rho x)(Ce^{ - \alpha ^2 \rho ^2 t} ) \hfill \\

\end{gathered}$

now to use the conditions:

------------------------------

$\displaystyle \begin{gathered}

\theta ( - L) = T:so: \hfill \\

(A\cos \rho (x + L) + B\sin \rho (x + L))(e^{ - \alpha ^2 \rho ^2 t} ) = T \hfill \\

(A\cos \rho ( - L + L) + B\sin \rho ( - L + L))(e^{ - \alpha ^2 \rho ^2 t} ) = T \hfill \\

(A\cos (0) + B\sin (0))(e^{ - \alpha ^2 \rho ^2 t} ) = T \hfill \\

\left[ {A(e^{ - \alpha ^2 \rho ^2 t} )} \right] = T \hfill \\

\end{gathered}$

$\displaystyle \begin{gathered}

\theta (L) = 0:so: \hfill \\

(A\cos (2\rho L) + B\sin (2\rho L))(e^{ - \alpha ^2 \rho ^2 t} ) = 0 \hfill \\

\end{gathered}$

$\displaystyle

\begin{gathered}

\theta (x,0) = \theta _0 (x) = \frac{{T(1 + \frac{x}

{L})}}

{2}:so: \hfill \\

(A\cos \rho (x + L) + B\sin \rho (x + L))(e^{ - \alpha ^2 \rho ^2 (0)} ) = \frac{{T(1 + \frac{x}

{L})}}

{2} \hfill \\

(A\cos \rho (x + L) + B\sin \rho (x + L))(1) = \frac{{T(1 + \frac{x}

{L})}}

{2} \hfill \\

\end{gathered}$

I'm stuck here. Any ideas on how to approach the final answer (as shown on the question)????

Thanks so much.