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2D heat equation bounday conditions for different intervals

  1. May 10, 2012 #1
    1. The problem statement, all variables and given/known data

    I have boundary conditions on my heat equation ## \dot{T}(x,t) = T''(x,t) ##

    ## T(0,t) = T(L,t), ##

    ## \frac{\partial T(0,t)}{\partial x} = \frac{\partial T(L,t)}{\partial x} ##

    Then

    at ## T= 0##

    ## T(x,0) = 1 ## for ## 0<x<L/4 ##
    ## T(x,0) = 0 ## for ## L/4<x<L ##

    2. Relevant equations

    None.

    3. The attempt at a solution

    I have done the usual separation of variables, then getting two ODEs equal to a constant. I've called it ## \alpha^2 ## I have eliminated the possibility that ##\alpha^2 = 0. ## Will this always happen?

    Now do I need to consider ## \alpha^2 < 0 ## and ## \alpha^2 >0 ## rather than just ##\alpha^2\ne 0 ##?

    I find the fact that the boundary conditions run over an interval to be off putting. Am I to use the differential boundary conditions instead for now? I have deduced that:

    ##T(x,t) = c_1(e^{\alpha x} - e^{-\alpha x})c_2 e^{\alpha^2 t} ##

    Of course if ##T_i(x,t) ## is a solution then ##T(x,t) = \displaystyle \sum_i T_i(x,t) ##

    I have found that

    ## T(x,t) = \sum_{n=1}^{\infty} C_n \sin ({n \pi x \over l}) e^{n^2 \pi^2 t \over l^2} ##

    So I need to find the Fourier coefficients of that function that is constant on the two different intervals. This is where I get stuck.

    ##C_n = \frac{1}{l} \int_0^{2l} f(x) \sin ({n \pi x \over l}) dx = \frac{1}{l} \int_0^{l/4} 0\cdot dx + \int_{l/4}^{l} \sin ({n \pi x \over l}) dx+ \frac{1}{l} \int_{l}^{5l/4} 0\cdot dx + \int_{5l/4}^{2l} \sin ({n \pi x \over l}) dx ##

    I don't think this the correct form for the coefficient:

    ## C_n = \frac{1}{n \pi} \left[ -(-1)^n + 1 +0 + \cos({4 n \pi \over 5})\right] ##

    So is it just integration over ## [0,l] ## ?

    This would give

    ## T(x,t) = \sum_{n=1}^{\infty} 2 \sin ({(2n-1) \pi x \over l}) e^{n^2 \pi^2 t \over l^2} ##
     
    Last edited: May 10, 2012
  2. jcsd
  3. May 11, 2012 #2
    I don't think your solution is correct. I don't like using T for the solution so I will write it as u(x,t)

    Let ##u(x,t)=\varphi(x)T(t)##. Then ##\frac{\varphi''}{\varphi}=\frac{T'}{T}=-\lambda##.

    So we have ##\varphi''+\lambda\varphi = 0\Rightarrow \varphi(x)=A\cos x\sqrt{\lambda} + B\sin x\sqrt{\lambda}## and ##T(t)=\exp(-\lambda t)##.

    Next we have to solve ##\varphi## for
    ##\varphi_1(0) = 1\quad\quad\varphi_2'(0) = 0##
    ##\varphi_1(0) = 0\quad\quad\varphi_2'(0) = 1##

    ##\varphi(x) = A\cos x\sqrt{\lambda} + B\frac{\sin x\sqrt{\lambda}}{\sqrt{\lambda}}##

    Using your initial conditions, we get
    $$
    A(1 - \cos L\sqrt{\lambda}) = B\frac{\sin L\sqrt{\lambda}}{\sqrt{\lambda}} \quad \text{and} \quad B(1 - \cos L\sqrt{\lambda}) = -A\sqrt{\lambda}\sin L\sqrt{\lambda}
    $$
    I solved for B and obtained
    $$
    B = \frac{-A\sqrt{\lambda}\sin L\sqrt{\lambda}}{1 - \cos L\sqrt{\lambda}}.
    $$
    From this, I can substitute and get
    $$
    A\left[(1 - \cos L\sqrt{\lambda})^2 + \sin^2 L\sqrt{\lambda}\right] = 0
    $$
    That means ##\lambda_n = \frac{4\pi^2n^2}{L^2}##.
    This leads to
    $$
    u(x,t) = \sum_{n=1}^{\infty}a_n\left(\cos \frac{2x\pi n}{L} + \frac{L}{2\pi n}\sin\frac{2x\pi n}{L}\right)\exp\left(-\frac{4t\pi^2 n^2}{L^2}\right)
    $$
     
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