Heat Equation Tricky Question

**mathfied** · July 12th 2008, 05:00 PM

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,

$ \frac{{\partial ^2 \theta }} {{\partial x^2 }} = \frac{1} {{\alpha ^2 }}\frac{{\partial \theta }} {{\partial t}} $ where $\alpha$ is a constant, on the interval:
$ - L \leqslant x \leqslant L $ with conditions: $\begin{gathered} \theta ( - L,t) = 0 \hfill \\ \theta (L,t) = T \hfill \\ \end{gathered}$ is: $\theta = \theta _0 (x) = \frac{{T(1 + \frac{x} {L})}} {2}$

(b)
Use methods of separation of variables to show that the unsteady solution for
$\theta = (x,t)$ with conditions: $\theta ( - L) = T,{\text{ }}\theta (L) = 0,{\text{ }}\theta (x,0) = \theta _0 (x){\text{ is}}$ :

$ \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:
$\begin{gathered} Let:\theta = (x,t) = X(x)T(t) \hfill \\ X(x) = Ax + B \hfill \\ T(t) = C \hfill \\ so:X(x)T(t) = (Ax + B)(C) \hfill \\ \end{gathered}$

now to use the conditions:
----------------------------------
$\begin{gathered} when:\theta ( - L,t) = 0, \hfill \\ A(x + L) + B = 0 \hfill \\ A(0) + B = 0,so:B = 0 \hfill \\ \end{gathered}$

$\begin{gathered} when:\theta (L,t) = T \hfill \\ A(x + L) + 0 = T \hfill \\ A(2L) = T \hfill \\ A = \frac{T} {{2L}} \hfill \\ \end{gathered}$

$\begin{gathered} so: \hfill \\ \theta _0 (x) = \frac{T} {{2L}}(x + L) = \frac{{T(1 + \frac{x} {L})}} {2} \hfill \\ \end{gathered}$

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

$\begin{gathered} unsteady - solution: - \rho ^2 < 0 \hfill \\ X(x) = Ae^{i\rho x} + Be^{ - ipx} = A\cos px + B\sin px \hfill \\ T(t) = Ce^{ - \alpha ^2 \rho ^2 t} \hfill \\ so:X(x)T(t) = (A\cos \rho x + B\sin \rho x)(Ce^{ - \alpha ^2 \rho ^2 t} ) \hfill \\ \end{gathered}$

now to use the conditions:
------------------------------
$\begin{gathered} \theta ( - L) = T:so: \hfill \\ (A\cos \rho (x + L) + B\sin \rho (x + L))(e^{ - \alpha ^2 \rho ^2 t} ) = T \hfill \\ (A\cos \rho ( - L + L) + B\sin \rho ( - L + L))(e^{ - \alpha ^2 \rho ^2 t} ) = T \hfill \\ (A\cos (0) + B\sin (0))(e^{ - \alpha ^2 \rho ^2 t} ) = T \hfill \\ \left[ {A(e^{ - \alpha ^2 \rho ^2 t} )} \right] = T \hfill \\ \end{gathered}$

$\begin{gathered} \theta (L) = 0:so: \hfill \\ (A\cos (2\rho L) + B\sin (2\rho L))(e^{ - \alpha ^2 \rho ^2 t} ) = 0 \hfill \\ \end{gathered}$

$ \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.

**ThePerfectHacker** · July 12th 2008, 06:33 PM

Originally Posted by mathfied

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

$ \frac{{\partial ^2 \theta }} {{\partial x^2 }} = \frac{1} {{\alpha ^2 }}\frac{{\partial \theta }} {{\partial t}} $ where $\alpha$ is a constant, on the interval:
$ - L \leqslant x \leqslant L $ with conditions: $\begin{gathered} \theta ( - L,t) = 0 \hfill \\ \theta (L,t) = T \hfill \\ \end{gathered}$ is: $\theta = \theta _0 (x) = \frac{{T(1 + \frac{x} {L})}} {2}$

We look for a solution of the form $\theta(x,t) = \bar \theta(x,t) + f(x)$ .
Where $\bar \theta(x,t)$ solves the homogenous boundary value problem and $f(x)$ is the steady-state function.
We want to get,
$\frac{d^2f}{dx^2} =0$ and $\frac{1}{\alpha^2}\frac{\partial \bar \theta}{\partial t} = \frac{\partial^2 \bar \theta}{\partial x^2}$ .
With with conditions,
$f(-L) = 0$ and $f(L) = T$ .
Since $f(x) = ax+b$ we want $-La+b=0$ and $La+b=T$ .
Solving this system of equations we get, $f(x) = \tfrac{T}{2}\left( 1+\tfrac{x}{L}\right)$ .

**ThePerfectHacker** · July 12th 2008, 06:48 PM

Originally Posted by mathfied

Use methods of separation of variables to show that the unsteady solution for
$\theta = (x,t)$ with conditions: $\theta ( - L) = T,{\text{ }}\theta (L) = 0,{\text{ }}\theta (x,0) = \theta _0 (x){\text{ is}}$ :

$ \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]}$

The only thing you need to know here is that if you are solving an equation with non-homogenous boundary conditions and you seperate it into a sum of two solutions one of which is a steady-state solution then the full solution to the original problem would be the sum of the steady-state function and the solution to the equation with homogenous boundary conditions. In this case the steady-state function (similar to first part) is $f(x)=\tfrac{T}{2}(1-\tfrac{x}{L})$ . Then you need to solve the equation for $\bar \theta(x,t)$ . Where $\bar \theta_t = \alpha^2 \bar \theta_xx$ with $\bar \theta (-L,t) = \bar \theta (L,t) = 0$ and $\bar \theta (x,0) = \theta_0(x) - f(x)$ . But this is a homogoneous heat equation. Its solutions is $\bar \theta (x,t) = \Sigma_{n=1}^{\infty} A_n e^{-\pi^2 n^2 \alpha^2 t/L^2} \sin \tfrac{\pi n x}{L}$ . And the $A_n$ are determined by integral forumals while solving this problem.

**mathfied** · July 13th 2008, 02:12 PM

Originally Posted by ThePerfectHacker

But this is a homogoneous heat equation. Its solutions is $\bar \theta (x,t) = \Sigma_{n=1}^{\infty} A_n e^{-\pi^2 n^2 \alpha^2 t/L^2} \sin \tfrac{\pi n x}{L}$ . And the $A_n$ are determined by integral forumals while solving this problem.

Hi. Thanks.
I understand that once I get $\bar \theta (x,t) = \Sigma_{n=1}^{\infty} A_n e^{-\pi^2 n^2 \alpha^2 t/L^2} \sin \tfrac{\pi n x}{L}$ then I have to apply the fourier series to get A_n.

I can get close to the summation formula you have just stated, but I seem to have 2 terms inside the summation instead of JUST ONE.

Here is what I got, please shed some light on where to take this.

This is how I've been taught to work out the solution: I'll try to explain in as much detail as possible.

- First I try to simplify the general form of the equation and try and turn it into a "summation" kind of form so that it is ready to have the FOURIER SERIES applied to it. So here is what i have so far:

First I have been taught to always use the general equation form for an unsteady solution which is:
-------------------------------------------------------
$\theta(x,t) = X(x)T(t) = (A\cos \rho x + B\sin \rho x)(Ce^{ - \alpha ^2 \rho ^2 t} )$

We then apply the 3 conditions on the general form to try and convert the general solution to something that resembles a summation form, so that we can apply the fourier series to it.

now to use the First condition. Please note that I have absorbed the "C " from the general equation into the other constants:
--------------------------------------------------------------
$\begin{gathered} \theta ( - L) = T:so: \hfill \\ (A\cos \rho (x + L) + B\sin \rho (x + L))(e^{ - \alpha ^2 \rho ^2 t} ) = T \hfill \\ (A\cos \rho ( - L + L) + B\sin \rho ( - L + L))(e^{ - \alpha ^2 \rho ^2 t} ) = T \hfill \\ (A\cos (0) + B\sin (0))(e^{ - \alpha ^2 \rho ^2 t} ) = T \hfill \\ \left[ {A(e^{ - \alpha ^2 \rho ^2 t} )} \right] = T \hfill \\ \end{gathered}$
$A = \frac{T} {{e^{ - \alpha ^2 \rho ^2 t} }}$

- What usually happens with these questions is that the first condition says $\theta ( - L) = 0$ instead of = T.
- If it was = 0, then the whole solution would have resulted in A=0.
- So going back to the GENERAL SOLUTION, I would have been able to cancel the term with the A becauase "0" makes the whole term go to "0".
- Instead we have A as $A = \frac{T} {{e^{ - \alpha ^2 \rho ^2 t} }}$ so I can't cancel the term out and instead have to work with both terms of the general solution when applying the second condition.

Using 2nd Condition:
-------------------------------------------------------------
$ \begin{gathered} \theta (L) = 0:so: \hfill \\ (\frac{T} {{e^{ - \alpha ^2 \rho ^2 t} }}\cos (2\rho L) + B\sin (2\rho L))(e^{ - \alpha ^2 \rho ^2 t} ) = 0 \hfill \\ \end{gathered}$

sin(2pl)=0 when $2\rho L = n\pi$ so $\rho = \frac{n\pi}{2L}$

so now subsituting "p" back into the general equation, we get:
$\left( {\sum\limits_{{\text{odd n}}}^{} {\left[ {\frac{T} {{e^{\frac{{ - \alpha ^2 n^2 \pi ^2 t}} {{4L^2 }}} }}_n \cos \left[ {\frac{{n\pi x}} {{2L}}} \right]} \right]} + \sum\limits_{{\text{even n}}}^{} {\left[ {B_n \sin \left[ {\frac{{n\pi x}} {{2L}}} \right]} \right]} } \right)e^{\frac{{ - \alpha ^2 n^2 \pi ^2 t}} {{4L^2 }}}=0$

Now using the 3rd and final condition:
--------------------------------
$\theta (x,0) = \theta _0 (x) = \frac{{T(1 + \frac{x} {L})}} {2}:so:$

substituting t=0 into the general form:

$\left( {\sum\limits_{{\text{odd n}}}^{} {\left[ {\frac{T} {{e^{\frac{{ - \alpha ^2 n^2 \pi ^2 (0)}} {{4L^2 }}} }}_n \cos \left[ {\frac{{n\pi x}} {{2L}}} \right]} \right]} + \sum\limits_{{\text{even n}}}^{} {\left[ {B_n \sin \left[ {\frac{{n\pi x}} {{2L}}} \right]} \right]} } \right)e^{\frac{{ - \alpha ^2 n^2 \pi ^2 (0)}} {{4L^2 }}} = \frac{{T(1 + \frac{x} {L})}} {2}$

anything to the power of 0 becomes "1", so the e's turn into 1.
so we have:

$ \left( {\sum\limits_{{\text{odd n}}}^{} {\left[ {T_n \cos \left[ {\frac{{n\pi x}} {{2L}}} \right]} \right]} + \sum\limits_{{\text{even n}}}^{} {\left[ {B_n \sin \left[ {\frac{{n\pi x}} {{2L}}} \right]} \right]} } \right) = \frac{{T(1 + \frac{x} {L})}} {2}$

$ \sum\limits_{{\text{n = 1}}}^\infty {\left[ {T_n \cos \left[ {\frac{{n\pi x}} {{2L}}} \right] + \left[ {B_n \sin \left[ {\frac{{n\pi x}} {{2L}}} \right]} \right]} \right]} = \frac{{T(1 + \frac{x} {L})}} {2}$

This is a bit similiar to what you wrote but the sum has 2 terms inside it. I can't seem to get rid of the term with the cos...
-------------------------------------------------------------

I now seem to have the general solution in a form that can have the fourier series applied to it. but i've tried this all day and just can't seem to get to the final answer which is:

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

any idea where to take this PerfectHacker? Thanks so much .

**mathfied** · July 14th 2008, 05:12 PM

Originally Posted by ThePerfectHacker

The only thing you need to know here is that if you are solving an equation with non-homogenous boundary conditions and you seperate it into a sum of two solutions one of which is a steady-state solution then the full solution to the original problem would be the sum of the steady-state function and the solution to the equation with homogenous boundary conditions. In this case the steady-state function (similar to first part) is $f(x)=\tfrac{T}{2}(1-\tfrac{x}{L})$ . Then you need to solve the equation for $\bar \theta(x,t)$ . Where $\bar \theta_t = \alpha^2 \bar \theta_xx$ with $\bar \theta (-L,t) = \bar \theta (L,t) = 0$ and $\bar \theta (x,0) = \theta_0(x) - f(x)$ . But this is a homogoneous heat equation. Its solutions is $\bar \theta (x,t) = \Sigma_{n=1}^{\infty} A_n e^{-\pi^2 n^2 \alpha^2 t/L^2} \sin \tfrac{\pi n x}{L}$ . And the $A_n$ are determined by integral forumals while solving this problem.

i think i get this now. thanks a lot for the suggestion.

im trying it out now. big headache but i think im getting there now!

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