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Math Help - Singularity at the upper limit

  1. #1
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    Singularity at the upper limit

    Hi,

    I need to solve this integral from a physical model

     <br />
\int\limits_{a, 1} \frac{u}{u-1} du

    which is singular at the upper limit. The reason for the singularity is that the function reaches a maximum at u=1. The model is completely physically based so the integral must have a solution. I was thinking about using Cauchy's theorem but I don't know how to find a proper contour. Even if I get complex numbers in the solution I think I still can interpret them somehow (or make some simplification like dropping the imaginary part).
    Any ideas?
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  2. #2
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    Quote Originally Posted by donifan View Post
    Hi,

    I need to solve this integral from a physical model

     <br />
\int\limits_{a, 1} \frac{u}{u-1} du

    which is singular at the upper limit. The reason for the singularity is that the function reaches a maximum at u=1. The model is completely physically based so the integral must have a solution. I was thinking about using Cauchy's theorem but I don't know how to find a proper contour. Even if I get complex numbers in the solution I think I still can interpret them somehow (or make some simplification like dropping the imaginary part).
    Any ideas?
    \frac{u}{1-u}=-1+\frac{1}{1-u}

    So if a is a constant different from 1:

    \lim_{b \to 1}\int_a^b \frac{u}{1-u}~du=(a-1)+\lim_{b \to 1}[-\ln(1-u]_a^b=\infty

    RonL
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  3. #3
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    Quote Originally Posted by donifan View Post
    Hi,

    I need to solve this integral from a physical model

     <br />
\int\limits_{a, 1} \frac{u}{u-1} du

    which is singular at the upper limit. The reason for the singularity is that the function reaches a maximum at u=1. The model is completely physically based so the integral must have a solution. I was thinking about using Cauchy's theorem but I don't know how to find a proper contour. Even if I get complex numbers in the solution I think I still can interpret them somehow (or make some simplification like dropping the imaginary part).
    Any ideas?
    The function u/(u - 1) = 1 + 1/(u - 1) does not have a maximum at u = 1. It is undefined at u= 1. That's the reason for the singularity. Your question boils down to finding the value of ln |u - 1| as u --> 1 ...... your prospects are bleak.

    What is the physical model and what is the question asked of this model?
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  4. #4
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    Quote Originally Posted by CaptainBlack View Post
    \frac{u}{1-u}=-1+\frac{1}{1-u}

    So if a is a constant different from 1:

    \lim_{b \to 1}\int_a^b \frac{u}{1-u}~du=(a-1)+\lim_{b \to 1}[-\ln(1-u]_a^b=\infty

    RonL

    Thanks for the answer. I totally agree but let me explain better. My problem indeed is to solve

    u\frac{du}{dt}=1-u

    so it is the function u that has a maximum at u=1. It is part of a model with a negative feedback of u on itself. so what I am trying to find here is the time for the maximum. The straight answer is that it is infinite, but my question is: is there any other possibility, as for example, a branch point in u(t)? I can solve it using a numerical method but I need to do it analytically.

    Thanks for any help.
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  5. #5
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    Quote Originally Posted by donifan View Post
    Thanks for the answer. I totally agree but let me explain better. My problem indeed is to solve

    u\frac{du}{dt}=1-u

    so it is the function u that has a maximum at u=1. It is part of a model with a negative feedback of u on itself. so what I am trying to find here is the time for the maximum. The straight answer is that it is infinite, but my question is: is there any other possibility, as for example, a branch point in u(t)? I can solve it using a numerical method but I need to do it analytically.

    Thanks for any help.
    I see where your mistake is. You're saying that u has a maximum at u = 1 because du/dt = 0 when u = 1, right?

    That's not correct.

    In fact, u has a horizontal asymptote with equation u = 1 ....... Note that as t --> +oo, u --> 1 and du/dt --> 0.

    It's not hard to show that curve u does not cut its horizontal asymptote. Therefore it obviously takes an infinite amount of time for u to equal 1 ......


    By the way, I know that u has a horizontal asymptote because the solution to the DE (assuming the boundary condition u(a) = 0, where a < 1) is

    t = a - u + \ln \left( \frac{1-a}{1-u}\right) = a - u - \ln \left( \frac{1-u}{1-a}\right)

    and a graph of t versus u clearly has a vertical asymptote at u = 1 ....... Try drawing a graph corresponding to a = 1/2, say and you'll see what I mean. So a graph of u versus t clearly has a horizontal asymptote u = 1 .......
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