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Math Help - Heat Equation Tricky Question

  1. #1
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    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,

    <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  < 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.
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  2. #2
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    Quote Originally Posted by mathfied View Post
    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}
    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).
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  3. #3
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    Quote Originally Posted by mathfied View Post
    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]}
    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.
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  4. #4
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    Quote Originally Posted by ThePerfectHacker View Post
    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}<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}

    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}<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 PerfectHacker? Thanks so much .
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  5. #5
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    Quote Originally Posted by ThePerfectHacker View Post
    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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