Finding the Inverse Laplace Transform (Bromwich Contour)

[myshadow]

Homework Statement

Find the inverse laplace transform: $${\cal L}^{-1}\{e^{-\sqrt{s-a}c}\}$$

where $a$ and $c$ are constants. $s$ is the complex variable.

Homework Equations

Bromwich integral: $$\frac{1}{2\pi{}i}\lim_{T\rightarrow \infty}\int_{\gamma-Ti}^<br /> {\gamma+Ti}e^{st-\sqrt{s-a}c}ds$$

Residue theorem: $$\oint f(s) ds=\sum Res(f(s))$$

The Attempt at a Solution

There is a branch point at $s=a$. Also, there are no poles. Therefore, the sum of the residues is equal to zero. So I have a keyhole contour centered at $a$ with an outer radius of $R$ and inner radius of $\varepsilon$. I chose $\gamma=2a$. I attached a general image of the contour that I used. The contour is centered at $a$.

Therefore, $s=Re^{i\theta}+a$ along the the outer circle,
$s=xe^{\pi i}+a$ going from left to right along the branch cut,
$s=\varepsilon e^{i\theta}+a$ along the the inner circle, and
$s=xe^{-\pi i}+a$ going from right to left along the branch cut.

From Jordan's lemma the integral along the outer radius goes to zero as $R\rightarrow \infty$. Similarly, the integral along the inner circle (keyhole) goes to zero as $\varepsilon\rightarrow 0$

Therefore,

$$\frac{1}{2\pi{}i}\lim_{T\rightarrow \infty}\int_{\gamma-Ti}^<br /> {\gamma+Ti}e^{st-\sqrt{s-a}c}ds + \frac{1}{2\pi{}i}\int_{branch cut}=0$$

$$\frac{1}{2\pi{}i}\lim_{T\rightarrow \infty}\int_{\gamma-Ti}^<br /> {\gamma+Ti}e^{st-\sqrt{s-a}c}ds = <br /> <br /> \frac{-1}{2\pi{}i}\int_{branch cut}<br /> <br /> =\frac{-1}{2\pi{}i}\lim_{\substack{R\rightarrow \infty\\\varepsilon\rightarrow 0}}<br /> <br /> (\int_{-R+a}^{-\varepsilon+a}e^{st-\sqrt{s-a}c}ds+<br /> <br /> \int_{-\varepsilon+a}^{-R+a}e^{st-\sqrt{s-a}c}ds)$$

$$= \frac{-1}{2\pi{}i}(-\int_{a}^{\infty}e^{(a-x)t-\sqrt{x}ic}e^{\pi{}i}dx+\int_{a}^{\infty}e^{(a-x)t+\sqrt{x}ic}e^{-\pi{}i}dx)$$

$$= \frac{-1}{2\pi{}i}(\int_{a}^{\infty}e^{(a-x)t-\sqrt{x}ic}dx-\int_{a}^{\infty}e^{(a-x)t+\sqrt{x}ic}dx)<br /> <br /> =\frac{e^{ta}}{\pi{}}\int_{a}^{\infty}e^{-xt}\sin{(\sqrt{x}c)}dx$$

And this is where I'm stuck. I think this integral doesn't have a closed form. I can expand the integral into an infinite series but I don't know how to integrate the terms. I tried doing the Inverse Laplace transform in mathematica. The answer it gave me is:

$$\frac{ce^{\frac{-c^2}{4t}+at}}{2\sqrt{\pi{}}}t^{\frac{3}{2}}$$

So what I ended up getting doesn't agree with Mathematica. Also I know from tables that the inverse laplace transform of $$e^{-\sqrt{s}}$$ is $$\frac{a}{2\sqrt{\pi{}t^3}}e^{\frac{-a^2}{4t}}$$ I really don't know what I did wrong. I don't know what to do from here. Thanks in advance.

 

[Ray Vickson]

If you can find the inverse transform of exp(-c*sqrt(s)), you can use standard theorems to get the inverse transform of exp(-c*sqrt(s-a)). Now c*sqrt(s) = sqrt(r*s), where r = c^2, so if f(t) = inverse of exp(-sqrt(s)), we have exp(-sqrt(r*s)) = int_{t=0..inf} f(t) exp(-t*(r*s)) dt = int_{t=0..inf} f(t/r)*exp(-t*s) dt/r (r*t <---t), so inv of exp(-sqrt(r*s)) is f(t/r)/r = f(t/c^2)/c^2.

RGV

 

[vela]

Try using the substitution $u=s-a$ first to get$$\int e^{st}e^{-c\sqrt{s-a}}\,ds = e^{at}\int e^{ut}e^{-c\sqrt{u}}\,du$$Then you could complete the square
$$ut-c\sqrt{u} = t\left(\sqrt{u}-\frac{c}{2t}\right)^2-t\left(\frac{c}{2t}\right)^2$$to get
$$e^{at}\int e^{ut}e^{-c\sqrt{u}}\,du = e^{at-\frac{c^2}{4t}}\int e^{t(\sqrt{u}-\frac{c}{2t})^2}\,du$$

 

[myshadow]

Ray,

I follow everything you wrote except "exp(-sqrt(r*s)) = int_{t=0..inf} f(t) exp(-t*(r*s)) dt"

$$f(t)= {\cal L}^{-1}\{e^{-\sqrt{s}}\}$$

$$e^{-\sqrt{rs}}=e^{-c\sqrt{s}}=(\int_{0}^{\infty}f(t)e^{-t{s}}dt)^c$$

I don't see how to get that to $$\int_{0}^{\infty}f(t)e^{-trs}dt$$

Vela, that looks interesting. So with that substitution I end up with a branch point at u=0 right? I'll see what i get with the contour and get back to you. Thanks for the help guys. ^^

 

[jackmell]

Your analysis is correct as I see it except you're not integrating over the correct bounds. If $z=a+re^{\pi i}$, what must the lower limit on r be in order to reach a?

 

[Ray Vickson]

Ray,

I follow everything you wrote except "exp(-sqrt(r*s)) = int_{t=0..inf} f(t) exp(-t*(r*s)) dt"

$$f(t)= {\cal L}^{-1}\{e^{-\sqrt{s}}\}$$

$$e^{-\sqrt{rs}}=e^{-c\sqrt{s}}=(\int_{0}^{\infty}f(t)e^{-t{s}}dt)^c$$

I don't see how to get that to $$\int_{0}^{\infty}f(t)e^{-trs}dt$$

Vela, that looks interesting. So with that substitution I end up with a branch point at u=0 right? I'll see what i get with the contour and get back to you. Thanks for the help guys. ^^

It's just the statement exp(-sqrt(S)) = int_{0..inf} f(t) exp(-S*t) dt, with S replaced by r*s!

RGV

 

[myshadow]

From similar steps as in the first post and letting $u=s-a$ I get the final integral
$$\frac{1}{2\pi{}i}\lim_{T\rightarrow \infty}\int_{\gamma-Ti}^ <br /> {\gamma+Ti}e^{st-\sqrt{s-a}c}ds =\frac{e^{ta}}{\pi{}}\int_{0}^{\infty}e^{-xt}\sin{(\sqrt{x}c)}dx$$

I evaluated the integral with mathematica and it agrees with what i got by just taking the inverse directly with mathematica. The only thing it is a conditional expression valid with t>0. But doing the inverse directly it doesn't give that condition... thoughts?

jackmell, is this what you were referring to? that the limits should be from zero to infinity? I'm not sure how u figured that out without making the substitution. Without making the substitution $z=a+Re^{\pi i}$...$z$ goes from $-R+a$ to $-\varepsilon+a$. Then taking the limit as R goes to infinity and epsilon goes to zero aren't the limits from a to infinity?

ray, so what ur suggesting is to do a substitution...I saw the constant outside the square root and had immediately thought it didn't match with the table...lol guess i made this problem more complicated than it needed to be. -_-;

So out of curiosity how would I evaluate this final integral? I can do a series expansion...but i don't know how to integrate the individual terms.

 

[jackmell]

jackmell, is this what you were referring to? that the limits should be from zero to infinity? I'm not sure how u figured that out without making the substitution. Without making the substitution $z=a+Re^{\pi i}$...$z$ goes from $-R+a$ to $-\varepsilon+a$. Then taking the limit as R goes to infinity and epsilon goes to zero aren't the limits from a to infinity?

Assume a is on the real axis for simplicity, then you're integrating over a branch-cut $(-\infty,a)$. Ok then, if I let $z=a+re^{\pi i}$, and $z=a+re^{-\pi i}$ like you and I both did for the analysis, and let r go from $(\infty,0)$, does not z traverse over that entire branch-cut except for the indentation around the branch-point?

 

[vela]

I evaluated the integral with mathematica and it agrees with what i got by just taking the inverse directly with mathematica. The only thing it is a conditional expression valid with t>0. But doing the inverse directly it doesn't give that condition... thoughts?

By default, Mathematica uses the one-sided Laplace transform, so it's understood the inverse transform is valid only for t>0.

When you evaluate the Bromwich integral, you get to choose which way to close the contour. When t>0, the integral over the outer circle vanishes only when you close the contour in the left half-plane. When t<0, you have to close it in the right half-plane. There's nothing going on the right half-plane, so you get f(t)=0 for t<0.

 

[jackmell]

By default, Mathematica uses the one-sided Laplace transform, so it's understood the inverse transform is valid only for t>0.

When you evaluate the Bromwich integral, you get to choose which way to close the contour. When t>0, the integral over the outer circle vanishes only when you close the contour in the left half-plane. When t<0, you have to close it in the right half-plane. There's nothing going on the right half-plane, so you get f(t)=0 for t<0.

I don't think we have to Vela. I'll conjecture we can still close the contour over the left half-plane when t<0 if we take the limit over all the contours simultaneously because in the limit, Cauchy's Theorem still applies. I can't at the moment however, show this.

 

[myshadow]

Assume a is on the real axis for simplicity, then you're integrating over a branch-cut $(-\infty,a)$. Ok then, if I let $z=a+re^{\pi i}$, and $z=a+re^{-\pi i}$ like you and I both did for the analysis, and let r go from $(\infty,0)$, does not z traverse over that entire branch-cut except for the indentation around the branch-point?

ohhh, i get it. i didn't change my bounds when I did the substitution! thanks.

By default, Mathematica uses the one-sided Laplace transform, so it's understood the inverse transform is valid only for t>0.

When you evaluate the Bromwich integral, you get to choose which way to close the contour. When t>0, the integral over the outer circle vanishes only when you close the contour in the left half-plane. When t<0, you have to close it in the right half-plane. There's nothing going on the right half-plane, so you get f(t)=0 for t<0.

So what happens at t=0? Is the solution not valid?

 

[jackmell]

So what happens at t=0? Is the solution not valid?

What happens to the expression for t>0 as t tends to zero?

 

[myshadow]

What happens to the expression for t>0 as t tends to zero?

$$\frac{ce^{\frac{-c^2}{4t}+at}}{2\sqrt{\pi{}}t^{\frac{3}{2}}}$$ goes to zero as t goes to zero...well I'm almost 100% sure its valid for t=0, since other inverse laplace transforms are valid for t=0. I was wondering how that translates to the bromwich contour since mathematica gave the conditional existence of t>0 for the final integral i got from using residue theory.