Finding the Inverse Laplace Transform (Bromwich Contour)

In summary: I would have to use integration by parts twice each time it seems. I'm not sure if that will work out so easily. But I'll try some more...thanks for the help. ^^In summary, the inverse Laplace transform of e^{-\sqrt{s-a}c} where a and c are constants and s is a complex variable can be found using the Bromwich integral and the residue theorem. By choosing a keyhole contour centered at a with an outer radius of R and inner radius of \varepsilon, and using Jordan's Lemma, the integral along the outer circle and inner circle can be shown to approach zero as R and \varepsilon approach infinity and zero, respectively. This leads to the equation \frac{
  • #1
myshadow
30
1

Homework Statement



Find the inverse laplace transform: [tex] {\cal L}^{-1}\{e^{-\sqrt{s-a}c}\} [/tex]

where [itex] a [/itex] and [itex] c [/itex] are constants. [itex] s [/itex] is the complex variable.

Homework Equations



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

Residue theorem: [tex] \oint f(s) ds=\sum Res(f(s)) [/tex]

The Attempt at a Solution



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

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

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

Therefore,

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

[tex] \frac{1}{2\pi{}i}\lim_{T\rightarrow \infty}\int_{\gamma-Ti}^
{\gamma+Ti}e^{st-\sqrt{s-a}c}ds =

\frac{-1}{2\pi{}i}\int_{branch cut}

=\frac{-1}{2\pi{}i}\lim_{\substack{R\rightarrow \infty\\\varepsilon\rightarrow 0}}

(\int_{-R+a}^{-\varepsilon+a}e^{st-\sqrt{s-a}c}ds+

\int_{-\varepsilon+a}^{-R+a}e^{st-\sqrt{s-a}c}ds) [/tex]

[tex] = \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) [/tex]

[tex] = \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)

=\frac{e^{ta}}{\pi{}}\int_{a}^{\infty}e^{-xt}\sin{(\sqrt{x}c)}dx [/tex]

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:

[tex] \frac{ce^{\frac{-c^2}{4t}+at}}{2\sqrt{\pi{}}}t^{\frac{3}{2}} [/tex]

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

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  • #2
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
 
  • #3
Try using the substitution [itex]u=s-a[/itex] first to get[tex]\int e^{st}e^{-c\sqrt{s-a}}\,ds = e^{at}\int e^{ut}e^{-c\sqrt{u}}\,du[/tex]Then you could complete the square
[tex]ut-c\sqrt{u} = t\left(\sqrt{u}-\frac{c}{2t}\right)^2-t\left(\frac{c}{2t}\right)^2[/tex]to get
[tex]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[/tex]
 
  • #4
Ray,

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

[tex] f(t)= {\cal L}^{-1}\{e^{-\sqrt{s}}\}[/tex]

[tex] e^{-\sqrt{rs}}=e^{-c\sqrt{s}}=(\int_{0}^{\infty}f(t)e^{-t{s}}dt)^c [/tex]

I don't see how to get that to [tex] \int_{0}^{\infty}f(t)e^{-trs}dt [/tex] :confused:

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. ^^
 
  • #5
Your analysis is correct as I see it except you're not integrating over the correct bounds. If [itex]z=a+re^{\pi i}[/itex], what must the lower limit on r be in order to reach a?
 
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  • #6
myshadow said:
Ray,

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

[tex] f(t)= {\cal L}^{-1}\{e^{-\sqrt{s}}\}[/tex]

[tex] e^{-\sqrt{rs}}=e^{-c\sqrt{s}}=(\int_{0}^{\infty}f(t)e^{-t{s}}dt)^c [/tex]

I don't see how to get that to [tex] \int_{0}^{\infty}f(t)e^{-trs}dt [/tex] :confused:

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
 
  • #7
From similar steps as in the first post and letting [itex] u=s-a [/itex] I get the final integral
[tex] \frac{1}{2\pi{}i}\lim_{T\rightarrow \infty}\int_{\gamma-Ti}^
{\gamma+Ti}e^{st-\sqrt{s-a}c}ds =\frac{e^{ta}}{\pi{}}\int_{0}^{\infty}e^{-xt}\sin{(\sqrt{x}c)}dx [/tex]

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 [itex] z=a+Re^{\pi i} [/itex]...[itex] z [/itex] goes from [itex] -R+a [/itex] to [itex] -\varepsilon+a [/itex]. 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.
 
  • #8
myshadow said:
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 [itex] z=a+Re^{\pi i} [/itex]...[itex] z [/itex] goes from [itex] -R+a [/itex] to [itex] -\varepsilon+a [/itex]. 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 [itex](-\infty,a)[/itex]. Ok then, if I let [itex]z=a+re^{\pi i}[/itex], and [itex]z=a+re^{-\pi i}[/itex] like you and I both did for the analysis, and let r go from [itex](\infty,0)[/itex], does not z traverse over that entire branch-cut except for the indentation around the branch-point?
 
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  • #9
myshadow said:
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.
 
  • #10
vela said:
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.
 
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  • #11
jackmell said:
Assume a is on the real axis for simplicity, then you're integrating over a branch-cut [itex](-\infty,a)[/itex]. Ok then, if I let [itex]z=a+re^{\pi i}[/itex], and [itex]z=a+re^{-\pi i}[/itex] like you and I both did for the analysis, and let r go from [itex](\infty,0)[/itex], 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? :confused:
 
  • #12
myshadow said:
So what happens at t=0? Is the solution not valid? :confused:

What happens to the expression for t>0 as t tends to zero?
 
  • #13
jackmell said:
What happens to the expression for t>0 as t tends to zero?

[tex] \frac{ce^{\frac{-c^2}{4t}+at}}{2\sqrt{\pi{}}t^{\frac{3}{2}}} [/tex] 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.
 

FAQ: Finding the Inverse Laplace Transform (Bromwich Contour)

1. What is the purpose of finding the inverse Laplace transform using the Bromwich contour?

The Bromwich contour is a mathematical tool used to find the inverse Laplace transform of a function. Its purpose is to provide a way to calculate the inverse Laplace transform of a function that may not have a closed form solution. It is particularly useful in solving differential equations and understanding the behavior of systems in the time domain.

2. How is the Bromwich contour related to the Laplace transform?

The Bromwich contour is used to find the inverse Laplace transform of a function. It is related to the Laplace transform through the Bromwich integral, which is a complex integral used to calculate the inverse Laplace transform. The contour is a path along which the integral is evaluated, and its shape is chosen to ensure the integral converges.

3. What are the key steps in finding the inverse Laplace transform using the Bromwich contour?

The key steps in finding the inverse Laplace transform using the Bromwich contour are:

  • Choose an appropriate contour that ensures convergence of the integral.
  • Evaluate the integral using the appropriate manipulations and substitutions.
  • Take the limit as the contour approaches infinity to remove the contribution from the branch cut.
  • Use the residue theorem to calculate the inverse Laplace transform.

4. What are some common challenges in finding the inverse Laplace transform using the Bromwich contour?

Some common challenges in finding the inverse Laplace transform using the Bromwich contour include:

  • Choosing an appropriate contour that ensures convergence of the integral.
  • Manipulating the integral to make it suitable for evaluation.
  • Taking the limit as the contour approaches infinity to remove the contribution from the branch cut.
  • Calculating the residues of the function.

5. Can the inverse Laplace transform be calculated without using the Bromwich contour?

Yes, there are other methods for finding the inverse Laplace transform, such as using partial fraction decomposition, convolution, or the Heaviside expansion formula. However, the Bromwich contour is particularly useful for functions that do not have a closed form solution and for understanding the behavior of systems in the time domain.

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