Singularity at the upper limit

**donifan** · May 13th 2008, 06:33 AM

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?

**CaptainBlack** · May 13th 2008, 06:42 AM

Originally Posted by donifan

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

**mr fantastic** · May 13th 2008, 06:48 AM

Originally Posted by donifan

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?

**donifan** · May 13th 2008, 07:11 AM

Originally Posted by CaptainBlack

$\frac{u}{1-u}=-1+\frac{1}{1-u}$

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.

**mr fantastic** · May 13th 2008, 04:35 PM

Originally Posted by donifan

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 .......

Thread: Singularity at the upper limit

LinkBack

Thread Tools

Display

Singularity at the upper limit

Similar Math Help Forum Discussions

Upper confidence limit value problem.

Finding the upper limit of the integral

Upper Lower Limit and intermediate value

having 2pi as the upper limit of an interval

Solve for upper limit of integral?

Search Tags