- #1

Tangent87

- 148

- 0

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 [tex]\phi(t)=\frac{t-t^{-1}}{2}[/tex] as [tex]t=\pm i[/tex] and then writing t=x+iy we have:

[tex]\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)[/tex]

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

[tex]x=-\sqrt{\frac{y(2+2y-y^2)}{y-2}}[/tex]

(t=+i)

and [tex]x=-\sqrt{\frac{y(2-2y-y^2)}{y+2}}[/tex]

(t=-i)

Where we take the negative roots so that [tex]Re\phi<0[/tex] 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.

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

[tex]\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)[/tex]

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

[tex]x=-\sqrt{\frac{y(2+2y-y^2)}{y-2}}[/tex]

(t=+i)

and [tex]x=-\sqrt{\frac{y(2-2y-y^2)}{y+2}}[/tex]

(t=-i)

Where we take the negative roots so that [tex]Re\phi<0[/tex] 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.

Last edited by a moderator: