Verification of Steady and Unsteady Heat Equation Solutions on a Finite Interval

[mathfied]

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,

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

<br /> \theta (x,t) = \frac{T}<br /> {2}\left[ {1 - \frac{x}<br /> {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)}}<br /> {{n\pi }}} \right]}




My attempt:

Part (A)::
-----------------------
Using separation of variables:
\begin{gathered}<br /> Let:\theta = (x,t) = X(x)T(t) \hfill \\<br /> X(x) = Ax + B \hfill \\<br /> T(t) = C \hfill \\<br /> so:X(x)T(t) = (Ax + B)(C) \hfill \\ <br /> \end{gathered}

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


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


\begin{gathered}<br /> so: \hfill \\<br /> \theta _0 (x) = \frac{T}<br /> {{2L}}(x + L) = \frac{{T(1 + \frac{x}<br /> {L})}}<br /> {2} \hfill \\ <br /> \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}<br /> unsteady - solution: - \rho ^2 &lt; 0 \hfill \\<br /> X(x) = Ae^{i\rho x} + Be^{ - ipx} = A\cos px + B\sin px \hfill \\<br /> T(t) = Ce^{ - \alpha ^2 \rho ^2 t} \hfill \\<br /> so:X(x)T(t) = (A\cos \rho x + B\sin \rho x)(Ce^{ - \alpha ^2 \rho ^2 t} ) \hfill \\ <br /> \end{gathered}

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


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


<br /> \begin{gathered}<br /> \theta (x,0) = \theta _0 (x) = \frac{{T(1 + \frac{x}<br /> {L})}}<br /> {2}:so: \hfill \\<br /> (A\cos \rho (x + L) + B\sin \rho (x + L))(e^{ - \alpha ^2 \rho ^2 (0)} ) = \frac{{T(1 + \frac{x}<br /> {L})}}<br /> {2} \hfill \\<br /> (A\cos \rho (x + L) + B\sin \rho (x + L))(1) = \frac{{T(1 + \frac{x}<br /> {L})}}<br /> {2} \hfill \\ <br /> \end{gathered}

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

Thanks so much.

 

[HallsofIvy]

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,

<br /> \frac{{\partial ^2 \theta }}<br /> {{\partial x^2 }} = \frac{1}<br /> {{\alpha ^2 }}\frac{{\partial \theta }}<br /> {{\partial t}}<br /> where \alpha is a constant, on the interval:
<br /> - L \leqslant x \leqslant L<br /> with conditions: \begin{gathered} \theta ( - L,t) = 0 \hfill \\<br /> \theta (L,t) = T \hfill \\ <br /> \end{gathered} is: \theta = \theta _0 (x) = \frac{{T(1 + \frac{x}<br /> {L})}}<br /> {2}
Why? You were not asked to solve the equation, only to show that that \theta_0 is the solution. Since it is linear in x, its second derivative with respect to x is 0 and since it is "steady state" (no dependence on t) its second derivative with respect to t is trivallly 0: the two partial derivatives are equal so it satisfies the differential equation.
By setting x= L and -L it is easy to see that it satisfies the boundary conditions.

(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}}:

<br /> \theta (x,t) = \frac{T}<br /> {2}\left[ {1 - \frac{x}<br /> {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)}}<br /> {{n\pi }}} \right]}




My attempt:

Part (A)::
-----------------------
Using separation of variables:
\begin{gathered}<br /> Let:\theta = (x,t) = X(x)T(t) \hfill \\<br /> X(x) = Ax + B \hfill \\<br /> T(t) = C \hfill \\<br /> so:X(x)T(t) = (Ax + B)(C) \hfill \\ <br /> \end{gathered}

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


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


\begin{gathered}<br /> so: \hfill \\<br /> \theta _0 (x) = \frac{T}<br /> {{2L}}(x + L) = \frac{{T(1 + \frac{x}<br /> {L})}}<br /> {2} \hfill \\ <br /> \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}<br /> unsteady - solution: - \rho ^2 &lt; 0 \hfill \\<br /> X(x) = Ae^{i\rho x} + Be^{ - ipx} = A\cos px + B\sin px \hfill \\<br /> T(t) = Ce^{ - \alpha ^2 \rho ^2 t} \hfill \\<br /> so:X(x)T(t) = (A\cos \rho x + B\sin \rho x)(Ce^{ - \alpha ^2 \rho ^2 t} ) \hfill \\ <br /> \end{gathered}

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


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


<br /> \begin{gathered}<br /> \theta (x,0) = \theta _0 (x) = \frac{{T(1 + \frac{x}<br /> {L})}}<br /> {2}:so: \hfill \\<br /> (A\cos \rho (x + L) + B\sin \rho (x + L))(e^{ - \alpha ^2 \rho ^2 (0)} ) = \frac{{T(1 + \frac{x}<br /> {L})}}<br /> {2} \hfill \\<br /> (A\cos \rho (x + L) + B\sin \rho (x + L))(1) = \frac{{T(1 + \frac{x}<br /> {L})}}<br /> {2} \hfill \\ <br /> \end{gathered}

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

In general you can't write the solution as you have shown. What you can do is write the solutions as a sum (possibly infinite) of such things- a Fouriere series.

I would recommend a standard method: define a new function
\phi(x,t)= \theta(x,t)+ \frac{T(1+ \frac{x}{L}}{2}- T[/itex]<br /> Since the added function, \theta_0(x,t)- T has second derivatives with respect to both x and t 0, this function satisfies the same differential equation as \theta(x,t) but satisifies the boundary conditions \phi(L, t)= 0, \phi(-L,t)= 0. Then, taking \rho so that \rho(-L)= -\pi and \rho(L)= \pi, the solution can be written as a sum in sin(n\rho x)<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Thanks so much. </div> </div> </blockquote>

 

[mathfied]

Hi. Thanks.
I understand that I have to get the general solution in a Summation form so that I can apply the Fourier series.

But I have a little difficulty on applying that new function you told me to introduce.


This is how I've been taught to work out the solution: I'll try to explain in as much detail as possible. Hopefully we can take it from there..

- 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}<br /> \theta ( - L) = T:so: \hfill \\<br /> (A\cos \rho (x + L) + B\sin \rho (x + L))(e^{ - \alpha ^2 \rho ^2 t} ) = T \hfill \\<br /> (A\cos \rho ( - L + L) + B\sin \rho ( - L + L))(e^{ - \alpha ^2 \rho ^2 t} ) = T \hfill \\<br /> (A\cos (0) + B\sin (0))(e^{ - \alpha ^2 \rho ^2 t} ) = T \hfill \\<br /> \left[ {A(e^{ - \alpha ^2 \rho ^2 t} )} \right] = T \hfill \\<br /> \end{gathered}
A = \frac{T}<br /> {{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}<br /> {{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:
-------------------------------------------------------------
<br /> \begin{gathered}<br /> \theta (L) = 0:so: \hfill \\<br /> (\frac{T}<br /> {{e^{ - \alpha ^2 \rho ^2 t} }}\cos (2\rho L) + B\sin (2\rho L))(e^{ - \alpha ^2 \rho ^2 t} ) = 0 \hfill \\ <br /> \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}<br /> {{e^{\frac{{ - \alpha ^2 n^2 \pi ^2 t}}<br /> {{4L^2 }}} }}_n \cos \left[ {\frac{{n\pi x}}<br /> {{2L}}} \right]} \right]} + \sum\limits_{{\text{even n}}}^{} {\left[ {B_n \sin \left[ {\frac{{n\pi x}}<br /> {{2L}}} \right]} \right]} } \right)e^{\frac{{ - \alpha ^2 n^2 \pi ^2 t}}<br /> {{4L^2 }}}=0

Now using the 3rd and final condition:
--------------------------------

\theta (x,0) = \theta _0 (x) = \frac{{T(1 + \frac{x}<br /> {L})}}<br /> {2}:so:

substituting t=0 into the general form:

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


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

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


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

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

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}<br /> {2}\left[ {1 - \frac{x}<br /> {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)}}<br /> {{n\pi }}} \right]}

any idea where to take this ? thanks so much..