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Heat Equation Problem

  1. Jul 12, 2008 #1
    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,

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

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

    [itex]
    \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]}[/itex]




    My attempt:

    Part (A)::
    -----------------------
    Using separation of variables:
    [itex]\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} [/itex]

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


    [itex]\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}[/itex]


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

    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:
    [itex]Let:\theta = (x,t) = X(x)T(t)[/itex]

    [itex]\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}[/itex]

    now to use the conditions:
    ------------------------------
    [itex]\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}[/itex]


    [itex]\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}[/itex]


    [itex]
    \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}[/itex]

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

    Thanks so much.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 13, 2008 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

     
  4. Jul 13, 2008 #3
    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:
    -------------------------------------------------------
    [itex]\theta(x,t) = X(x)T(t) = (A\cos \rho x + B\sin \rho x)(Ce^{ - \alpha ^2 \rho ^2 t} )[/itex]

    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:
    --------------------------------------------------------------

    [itex]\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}[/itex]
    [itex]A = \frac{T}
    {{e^{ - \alpha ^2 \rho ^2 t} }}[/itex]


    - What usually happens with these questions is that the first condition says [itex] \theta ( - L) = 0[/itex] 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 [itex]A = \frac{T}
    {{e^{ - \alpha ^2 \rho ^2 t} }}[/itex] 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:
    -------------------------------------------------------------

    [itex]
    \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} [/itex]

    sin(2pl)=0 when [itex]2\rho L = n\pi[/itex] so [itex]\rho = \frac{n\pi}{2L}[/itex]

    so now subsituting "p" back into the general equation, we get:
    [itex]\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[/itex]

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


    [itex]\theta (x,0) = \theta _0 (x) = \frac{{T(1 + \frac{x}
    {L})}}
    {2}:so:[/itex]

    substituting t=0 into the general form:

    [itex]\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}[/itex]


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

    [itex]
    \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}[/itex]


    [itex]
    \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}[/itex]

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

    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:

    [itex]\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]}[/itex]

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