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Heat conduction (Fourier complex integral)

  1. Oct 14, 2012 #1
    1. The problem statement, all variables and given/known data

    A semi-infinite bar (0 < x < 1) with unit thermal conductivity is fully insulated
    at x = 0, and is constantly heated at x = 1 over such a narrow interval that the
    heating may be represented by a delta function:

    [itex]\frac{\partial U}{\partial t}=\frac{\partial^2 U}{\partial x^2}+\delta (x-1)[/itex]

    where U(x,t) is the temperature. Assuming the initial temperature is zero through-
    out, and that U -> 0 as x ->\infty 1, find U(x,t) using the appropriate Fourier trans-
    form in x. In particular, show that

    [itex]U(1,t)=\frac{(1+e^{-1/t})\sqrt{t}}{\sqrt{\pi }}+erf(\frac{1}{\sqrt{t}})-1[/itex]

    My main problem is that this is the first time I am solving this type of problem and every example I have tried to compare mine with never involves the dirac function so I can't be certain or not if I have the right solution.. Also I've never worked with the error function and have no idea how one would manipulate my final answer so that I can show U(1,t)

    2. Relevant equations

    If : [itex]f(\lambda )=\int_{-\infty }^{\infty }e^{-i\lambda x}F(u)du[/itex]

    Then : [itex]F(x)=\frac{1}{2\pi}\int_{-\infty }^{\infty }e^{i\lambda x}f(\lambda)d\lambda[/itex]

    Also : [itex]\int_{0}^{\infty }\delta (x-a)f(x)dx=f(a)[/itex]

    And for the final bit : [itex]\int_{0}^{\infty }\frac{cos\lambda x}{\lambda^2 }(1-e^{-\lambda^2 t})d\lambda =\sqrt{\pi t}e^{\frac{-x^2}{4t}}+\frac{\pi }{2}(xerf(\frac{x}{2\sqrt{t}}-\left | x \right |)[/itex]

    3. The attempt at a solution

    I begin by definining:

    [itex]u(\lambda ,t)=\int_{0}^{\infty }e^{-i\lambda x}U(x,t)d\lambda \\[/itex]

    Then taking the Fourier transform with respect to x:

    [itex]F[\frac{\partial U}{\partial t}]=F[\frac{\partial^2 U}{\partial x^2}+\delta (x-1)]\\[/itex]

    [itex]\frac{\partial U}{\partial t}=\int_{0}^{\infty }\frac{\partial^2 U}{\partial x^2}e^{-i\lambda x}dx+\int_{0}^{\infty}\delta (x-1)e^{-i\lambda x}dx\\[/itex]

    [itex]\frac{\partial U}{\partial t}=\int_{0}^{\infty }\frac{\partial^2 U}{\partial x^2}e^{-i\lambda x}dx+e^{-i\lambda }\\[/itex]

    Then using integration by parts:

    [itex]\frac{\partial U}{\partial t}=\left [ \frac{\partial U}{\partial x}e^{-i\lambda x} \right ]_{0}^{\infty }+i\lambda \int_{0}^{\infty }\frac{\partial U}{\partial x}e^{-i\lambda x}dx+e^{-i\lambda }\\[/itex]

    Here I used the fact that at x=0 the rod is "fully insulated" thus I believe the heat transfer here will remain 0?

    [itex]\frac{\partial U}{\partial t}=i\lambda \int_{0}^{\infty }\frac{\partial U}{\partial x}e^{-i\lambda x}dx+e^{-i\lambda }\\[/itex]

    Using integration by parts again:

    [itex]\frac{\partial U}{\partial t}=\left [ U(x,t)e^{-i\lambda x} \right ]_{0}^{\infty }+i^2\lambda^2 \int_{0}^{\infty }U(x,t)e^{-i\lambda x}dx+e^{-i\lambda }\\*[/itex]

    And again by the fact that at x=0 no heat should be exchanged:

    [itex]\frac{\partial U}{\partial t}=-\lambda^2 \int_{0}^{\infty }U(x,t)e^{-i\lambda x}dx+e^{-i\lambda }\\[/itex]

    [itex]\frac{\partial U}{\partial t}=-\lambda^2 u(\lambda ,t)+e^{-i\lambda }\\[/itex]

    Solving this simple ODE with the initial conditions I obtain:

    [itex]U(x,t)=\frac{e^{-i\lambda} (1-e^{-\lambda^2 t})}{\lambda^2 }\\[/itex]

    [itex]U(x,t)=\frac{1}{2\pi }\int_{0}^{\infty }e^{i\lambda x}\frac{e^{-i\lambda} (1-e^{-\lambda^2 t})}{\lambda^2 }d\lambda [/itex]

    Now I have absolutely NO idea how to get this into the integral they have given me for the final bit or if I have even done this right. Actually I also tried to expand this as an even series and you get exactly what they have given me.

    A little bit more work I've done.. again have no idea if its right because I don't know if this solution is correct but:

    [itex]U(1,t)=\frac{1}{2\pi }\int_{0}^{\infty }\frac{1-e^{-\lambda^2 t}}{\lambda^2 }d\lambda \\[/itex]

    [itex]U(1,t)=\frac{1}{2\pi }\int_{0}^{\infty }\frac{1}{\lambda^2 }d\lambda-\frac{1}{2\pi }\int_{0}^{\infty }\frac{e^{-\lambda^2 t}}{\lambda^2 }d\lambda \\[/itex]

    The integral on the left causes some problems, and I can see that the integral on the right is related to the error function but not sure what to do with the denominator in there..
     
    Last edited: Oct 14, 2012
  2. jcsd
  3. Oct 14, 2012 #2
    Nobody? :(

    It would help just to even know that I have come to the right final solution!
     
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