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Two dimensional Heat equation of a semi infinite strip

  1. Jan 1, 2015 #1
    1. The problem statement, all variables and given/known data
    Consider
    [tex] \frac{\partial u}{\partial t} = k\left( \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} \right) \\
    0<x<L\\
    y>0 [/tex]
    subject to the initial condition
    IC: [itex] u(x,y,0) = f(x,y) [/itex]

    And solve with the following boundary conditions:
    BC1: [itex] \quad u(0,y,t) = 0 [/itex]
    BC2: [itex] \quad u(L,y,t) = 0[/itex]
    BC3: [itex] \quad u(x,0,t) = 0[/itex]

    2. Relevant equations
    Already stated above:

    3. The attempt at a solution


    A solution is given as follows:
    You can determine the Fourier sine transform in y:
    [tex] \frac{\partial \overline{U}}{\partial t} = k\left( \frac{\partial^2 \overline{U}}{\partial x^2} - \omega^2 \overline{U} \right) [/tex]
    Then this eqaution can be solved with Fourier sine series using BC1 and BC2:
    [tex] \overline{U} = \sum_{n=1}^{\infty} A_n sin(\frac{n \pi x}{L}) [/tex]
    And the previous equation can be used to get an expression for [itex] A_n [/itex]:
    [tex] \frac{\partial A_n}{\partial t} = -kA_n[({\frac{n\pi}{L}}^2 + \omega^2] \\
    A_n(\omega,t) = A_n(\omega,0) e^{-kt[({\frac{n\pi}{L}}^2 + \omega^2]} [/tex]
    With the initial condition IC1 you can get:
    [tex] \overline{U}(x,\omega,0) = F(x,\omega) = \sum_{n=1}^{\infty} A_n(\omega,0) sin(\frac{n \pi x}{L}) \\
    A_n(\omega,0) = \frac{2}{L} \int_0^L F(x,\omega) sin(\frac{n \pi x}{L}) dx [/tex]

    And now you can use the inverse Fourier sine transform to get [itex] u(x,y,t) [/itex]:
    [tex] u(x,y,t) = \int_0^\infty \overline{U}(x,\omega,t) sin(\omega y) d\omega [/tex]
    With [itex] \overline{U}(x,\omega,t) [/itex]:
    [tex] \overline{U}(x,\omega,t) = \sum_{n=1}^{\infty} A_n(\omega,t) sin(\frac{n \pi x}{L}) [/tex]
    With [itex] A_n(\omega,t) [/itex]:
    [tex] A_n(\omega,t) = A_n(\omega,0) e^{-kt[({\frac{n\pi}{L}}^2 + \omega^2]} [/tex]
    With [itex] A_n(\omega,0) [/itex]:
    [tex] A_n(\omega,0) = \frac{2}{L} \int_0^L F(x,\omega) sin(\frac{n \pi x}{L}) dx [/tex]
    With [itex] F(x,\omega) [/itex]:
    [tex] F(x,\omega) = \frac{2}{\pi} \int_0^\infty f(x,y) sin(\omega y) dy [/tex]


    I think I understand the solution, but the problem is, I don't see how BC3 comes in to play. I now have to do solve the same problem but with these conditions:
    IC: [itex] u(x,y,0) = f(x,y) [/itex]
    BC1: [itex] \quad u(0,y,t) = 0 [/itex]
    BC2: [itex] \quad u(L,y,t) = 0[/itex]
    BC3: [itex] \quad \frac{\partial u}{\partial y}(x,0,t) = 0[/itex]

    but in the derivation of the previous problem I only followed how IC1, BC1 and BC2 were used to derive the solution. How does BC3 matter?
     
  2. jcsd
  3. Jan 2, 2015 #2

    haruspex

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    I'm no expert on transforms, but I notice that your equation for the inverse is such that if you substitute y=0 on the right the integral will always be zero. It follows that the way ##\bar U## is defined encapsulates the assumption that u(x,0, t) = 0. How is it defined?
     
  4. Jan 2, 2015 #3
    Thanks, I just noticed also that the definition of the Fourier sine transform has a boundary term in it:
    [tex] S[\frac{d^2 f}{d x^2}] = \frac{2}{\pi} \omega f(0) - \omega^2 F(\omega) [/tex]
    And this of course explains the use of BC3.
    In my second case then it would be preferrable to use the Fourier cosine transform defined as:
    [tex] C[\frac{d^2 f}{d x^2}] = -\frac{2}{\pi} \frac{df}{dx}(0) - \omega^2 F(\omega) [/tex]
    And the rest would more or less follow just like the first problem

    In general, how would you decide what the best approach is to tackle problems like these, whether to use sine or cosine transform and whether in x or y etc.? What BC's would make a certain approach easy? Do the IC's matter? Does the region/domain matter?
    Could for example a double Fourier Transform be used in any of these two problems or would you run into practical problems?
    Maybe some others who know Fourier Transforms well can shed some light on this as well.
     
  5. Jan 2, 2015 #4
    On a side note, I said in the first post that with [itex] \overline{U}(x,\omega,t) [/itex] being:
    [tex] \overline{U}(x,\omega,t) = \sum_{n=1}^\infty A_n(\omega,t) sin(\frac{n\pi}{L}x) [/tex]
    you can get a differential equation for [itex] A_n(\omega,t) [/itex]:
    [tex] \frac{\partial A_n(\omega,t)}{\partial t} = -k A_n [ (\frac{n \pi}{L})^2 +\omega^2 ] [/tex]
    But how do you get from the sum:
    [tex] \sum_{n=1}^\infty \frac{\partial A_n(\omega,t)}{\partial t} = \sum_{n=1}^\infty -k A_n [ (\frac{n \pi}{L})^2 +\omega^2 ] [/tex]
    to a term by term equation? I know it should be correct, but I forgot the proper way to prove that.
     
  6. Jan 2, 2015 #5
    Never mind the last post, I just realized you have to use the orthogonality of the sine functions.
    But I haven't figured out a method for my general question:
     
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