# Thread: Heat Equation Tricky Question

1. ## Heat Equation Tricky Question

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.

2. 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, $f(x) = \tfrac{T}{2}\left( 1+\tfrac{x}{L}\right)$.

3. 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.

4. 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 .

5. 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!