Asymptotic Methods - Method of Steepest Descents

[Tangent87]

Hi, I'm hoping someone on here has some experience with using the method of steepest descents as I wasn't taught the course very well last term at university and I'm having trouble applying the method outlined in the books to my own examples. Specifically, I'm doing question B2/18 on page 73 here: http://www.maths.cam.ac.uk/undergrad/pastpapers/2001/Part_2/list_II.pdf

So far I've identified the saddle points of $$\phi(t)=\frac{t-t^{-1}}{2}$$ as $$t=\pm i$$ and then writing t=x+iy we have:

$$\phi(x+iy)=\left(\frac{x}{2}+\frac{x}{x^2+y^2}\right)+i\left(\frac{y}{2}-\frac{y}{x^2+y^2}\right)$$

The paths of steepest descent are given by $$Im\phi=\pm 1$$ which leads to the two paths being (I think):

$$x=-\sqrt{\frac{y(2+2y-y^2)}{y-2}}$$
(t=+i)

and $$x=-\sqrt{\frac{y(2-2y-y^2)}{y+2}}$$
(t=-i)

Where we take the negative roots so that $$Re\phi<0$$ which we require for steepest descent rather than ascent.

I'm now kind of stuck as these are pretty messy paths and even if I were to sketch them I'm not sure how to "deform" the current Hankel contour we have for the integral, onto them.

 

[mighty2000]

Hi there,

I am a scientist with experience using the method of steepest descents. I understand that you are having trouble applying the method to your specific example and I am happy to help.

Firstly, I want to commend you on correctly identifying the saddle points of your function \phi(t). This is an important step in the method of steepest descents.

Next, I want to address your concern about the paths of steepest descent being messy and difficult to deform onto your current Hankel contour. This is a common issue when using this method and it is important to remember that the paths of steepest descent are not necessarily the actual paths that the integral will follow. They are simply a tool to help us evaluate the integral.

In your case, it looks like you have correctly calculated the paths of steepest descent as x=-\sqrt{\frac{y(2+2y-y^2)}{y-2}} and x=-\sqrt{\frac{y(2-2y-y^2)}{y+2}}. However, these paths are only valid for y\neq 2 and y\neq -2 respectively. This means that you will need to consider different paths for these values of y.

Additionally, it may be helpful to consider the paths in terms of t rather than x and y. This could potentially simplify your calculations and make it easier to deform the contour onto the paths of steepest descent.

I hope this helps and please feel free to reach out with any further questions. Good luck with your problem!