Contour Integration

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TeeEm
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#1
Report Thread starter 6 years ago
#1
Hello clever purists.

I spent the entire day on this PDE today and most of the time was wasted trying to make a contribution over an arc vanish and failed.



I posted the whole question for completeness, which is definitely correct on the assumption that this arc does not contribute.

For the time being I had to put in my solution the classic line "It can be shown ..." which greatly annoys me. (page 3 about 1/3 of the way down)

Can someone look at the contribution of γ1 and γ2.
It is a standard bromwich contour with a branch cut.

(problem is of course the centre of this arc is at (c,0) combined with the square roots)

Square roots are not a problem on the hole because its centre is at O.

Any pointers/solutions or if you know a textbook for me to look at ...
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Blazy
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#2
Report 6 years ago
#2
(Original post by TeeEm)
-
Rough argument (I think this works but I'm no expert - someone can fill this in I'm sure. Or try Stackexchange!):

Define a branch cut of  s^{-\frac{1}{2}} about the negative real axis by letting  s = Re^{i\theta} where  -\pi < \theta < \pi

Along Gamma 1, you can show that:
 \displaystyle \left |e^{st}e^{-\frac{x}{\sigma}\sqrt{s}}  \right | = e^{tR\cos\theta}e^{\frac{-x\sqrt{R}}{\sigma}\cos\frac{\theta}{2}}
(with some other terms outside the integral which we won't need to worry...and for some reason, it's not showing  \cos\frac{\theta}{2} ...)

Now on Gamma 1, we have  \frac{\pi}{2} < \theta < \pi , so  \cos\theta < 0 and  \cos\frac{\theta}{2} > 0 so the integrand vanishes as  R\rightarrow \infty  .

Similarly on Gamma 2,  -\pi< \theta < -\frac{\pi}{2} and so by symmetry, the integrand also vanishes.
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TeeEm
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#3
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#3
(Original post by Blazy)
Rough argument (I think this works but I'm no expert - someone can fill this in I'm sure. Or try Stackexchange!):

Define a branch cut of  s^{-\frac{1}{2}} about the negative real axis by letting  s = Re^{i\theta} where  -\pi < \theta < \pi

Along Gamma 1, you can show that:
 \displaystyle \left |e^{st}e^{-\frac{x}{\sigma}\sqrt{s}}  \right | = e^{tR\cos\theta}e^{\frac{-x\sqrt{R}}{\sigma}\cos\frac{\theta}{2}}
(with some other terms outside the integral which we won't need to worry...and for some reason, it's not showing  \cos\frac{\theta}{2} ...)

Now on Gamma 1, we have  \frac{\pi}{2} < \theta < \pi , so  \cos\theta < 0 and  \cos\frac{\theta}{2} > 0 so the integrand vanishes as  R\rightarrow \infty  .

Similarly on Gamma 2,  -\pi< \theta < -\frac{\pi}{2} and so by symmetry, the integrand also vanishes.
Thanks for looking at this.

The problem here is that the centre of the arc is not at the origin in a Bromwich Contour but at (c,0) so the arc parameterize to c+Re, which gets very messy with the square roots. I can intuitively see it vanishes by translating it to the left by c.

(I have argued like you suggested for the arc around the branch point further down)
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Blazy
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#4
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#4
(Original post by TeeEm)
Thanks for looking at this.

The problem here is that the centre of the arc is not at the origin in a Bromwich Contour but at (c,0) so the arc parameterize to c+Re, which gets very messy with the square roots. I can intuitively see it vanishes by translating it to the left by c.

(I have argued like you suggested for the arc around the branch point further down)
I understand, but I was wondering whether if you could take this contour?
Image

Only thing is I'm not so sure whether or not the straight part connecting the arcs with the C1 bit vanishes or not.
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TeeEm
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#5
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#5
(Original post by Blazy)
I understand, but I was wondering whether if you could take this contour?
Image

Only thing is I'm not so sure whether or not the straight part connecting the arcs with the C1 bit vanishes or not.
I guess you could

The Laplace inversion is the value of the contour C1 which in this case is equal to the negative of C3 + C5 so the extra parts from everything else would definitely vanish but I do not know how easy they will be to show with the square roots.

Thanks for the suggestion.
Will try it later
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