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Heat flow in semi inifinite wall: Dirichlet problem or separate variables?

  1. Feb 4, 2010 #1
    1. The problem statement, all variables and given/known data

    The plane region defined by [tex]\[A = \left\{ {(x,y)/0 < x < 1,0 < y < \infty } \right\}\][/tex] defines the profile of a semi infinite wall, where there's a stationary regime heat flow. The left and right sides are isolated at 15º, and the base is isolated at 0º. Find T(x,y) in each point of the wall.

    2. Relevant equations

    Two-dimensional Heat equation: [tex]\[\frac{{{\partial ^2}T}}{{\partial {x^2}}} + \frac{{{\partial ^2}T}}{{\partial {y^2}}} = \frac{{\partial T}}{{\partial t}}\][/tex]

    Image

    ico65l.jpg

    3. The attempt at a solution

    This problem got me thinking, because it's done either one of two ways: either I treat it as a problem of separate variables and use Fourier Series, or I treat it as a differential geometry problem (Dirichlet problem, since in stationary regime the heat equation becomes Laplace's equation). I tried thinking of it as a Dirichlet problem.

    If I treat it as a differential geometry problem, I have to use conform transformations. First I have to multiply the whole thing by [tex]\[\pi \][/tex], so that then the wall has its sides at x = 0 and x = [tex]\[\pi \][/tex] (the base remains at y = 0, and it remains semi infinite). Then I have to traslate it [tex]\[ - \frac{\pi }{2}\][/tex], so that now the sides are at x = [tex]\[ - \frac{\pi }{2}\][/tex] and x = [tex]\[\frac{\pi }{2}\][/tex].

    So far, I transformed [tex]\[x + iy\][/tex] into [tex]\[\left( {\frac{\pi }{2}x - \frac{\pi }{2}} \right) + i\frac{\pi }{2}y\][/tex]. Now, I've only seen this kind of problem in which there are 3 contour conditions being solved by applying the transform sin(z). If I do that, then I transform the strip into Im(z) > 0, now I have transformed [tex]\[x + iy\][/tex] into [tex]\[\sin \left( {\frac{\pi }{2}x - \frac{\pi }{2}} \right)\cosh \left( {\frac{\pi }{2}y} \right) + i\cos \left( {\frac{\pi }{2}x - \frac{\pi }{2}} \right)\sinh \left( {\frac{\pi }{2}y} \right)\][/tex].

    The image is then transformed into:

    242y9mp.jpg

    Now, using polar coordinates, T doesn't depend of the radius, [tex]\rho \[/tex], so [tex]\[{\nabla ^2}T = \frac{1}{\rho }\frac{{{\partial ^2}T}}{{\partial {\varphi ^2}}}\][/tex]. This means that [tex]\[T(x,y) = A\theta + B\][/tex].

    But I can't go pass that. I don't know how to find A and B using the countour conditions.

    Any ideas?
     
  2. jcsd
  3. Feb 4, 2010 #2
    It compiled awfully! If you don't understand what it says, I repost it:

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

    The plane region defined by [tex]\[A = \left\{ {(x,y)/0 < x < 1,0 < y < \infty } \right\}\][/tex] defines the profile of a semi infinite wall, where there's a stationary regime heat flow. The left and right sides are isolated at 15º, and the base is isolated at 0º. Find T(x,y) in each point of the wall.

    2. Relevant equations

    Two-dimensional Heat equation: [tex]\[\frac{{{\partial ^2}T}}{{\partial {x^2}}} + \frac{{{\partial ^2}T}}{{\partial {y^2}}} = \frac{{\partial T}}{{\partial t}}\][/tex]

    Image

    ico65l.jpg

    3. The attempt at a solution

    This problem got me thinking, because it's done either one of two ways: either I treat it as a problem of separate variables and use Fourier Series, or I treat it as a differential geometry problem (Dirichlet problem, since in stationary regime the heat equation becomes Laplace's equation). I tried thinking of it as a Dirichlet problem.

    If I treat it as a differential geometry problem, I have to use conform transformations. First I have to multiply the whole thing by [tex]\[\pi \][/tex], so that then the wall has its sides at x = 0 and x = [tex]\[\pi \][/tex] (the base remains at y = 0, and it remains semi infinite). Then I have to traslate it [tex]\[ - \frac{\pi }{2}\][/tex], so that now the sides are at x = [tex]\[ - \frac{\pi }{2}\][/tex] and x = [tex]\[\frac{\pi }{2}\][/tex].

    So far, I transformed [tex]\[x + iy\][/tex] into [tex]\[\left( {\frac{\pi }{2}x - \frac{\pi }{2}} \right) + i\frac{\pi }{2}y\][/tex]. Now, I've only seen this kind of problem in which there are 3 contour conditions being solved by applying the transform sin(z). If I do that, then I transform the strip into Im(z) > 0, now I have transformed [tex]\[x + iy\][/tex] into [tex]\[\sin \left( {\frac{\pi }{2}x - \frac{\pi }{2}} \right)\cosh \left( {\frac{\pi }{2}y} \right) + i\cos \left( {\frac{\pi }{2}x - \frac{\pi }{2}} \right)\sinh \left( {\frac{\pi }{2}y} \right)\][/tex].

    The image is then transformed into:

    242y9mp.jpg

    Now, using polar coordinates, T doesn't depend of [tex]\rho [/tex], so then [tex]\[{\nabla ^2}T = \frac{1}{\rho }\frac{{{\partial ^2}T}}{{\partial {\varphi ^2}}}\][/tex]. This means that [tex]\[T(x,y) = A\theta + B\][/tex].

    But I can't go pass that. I don't know how to find A and B using the countour conditions.

    Any ideas?
     
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