# Math Help - Singularity at the upper limit

1. ## Singularity at the upper limit

Hi,

I need to solve this integral from a physical model

$
\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?

2. Originally Posted by donifan
Hi,

I need to solve this integral from a physical model

$
\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

3. Originally Posted by donifan
Hi,

I need to solve this integral from a physical model

$
\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?

4. 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.

5. 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 .......