The inverse of the Laplace transform by contour integration

September 1, 2010

The inverse of the Laplace transform by contour integration

Filed under: Giải tích 7 (MA4247) — Ngô Quốc Anh @ 17:23

Usually, we can find the inverse of the Laplace transform $\mathcal L[\cdot](s)$ by looking it up in a table. In this entry, we show an alternative method that inverts Laplace transforms through the powerful method of contour integration.

Consider the piece-wise differentiable function $f(x)$ that vanishes for $x < 0$ . We can express the function $e^{-cx}f(x)$ by the complex Fourier representation of

$\displaystyle f(x){e^{ - cx}} = \frac{1}{{2\pi }}\int_{ - \infty }^\infty {{e^{i\omega x}}\left[ {\int_0^\infty {{e^{ - ct}}f(t){e^{ - i\omega t}}dt} } \right]d\omega }$

for any value of the real constant $c$ , where the integral

$\displaystyle I = \int_0^\infty {{e^{ - ct}}|f(t)|dt}$

exists. By multiplying both sides of first equation by $e^{cx}$ and bringing it inside the first integral

$\displaystyle f(x) = \frac{1}{{2\pi }}\int_{ - \infty }^\infty {{e^{(c + i\omega )x}}\left[ {\int_0^\infty {f(t){e^{ - (c + i\omega )t}}dt} } \right]d\omega }$ .

With the substitution $z = c+\omega i$ , where $z$ is a new, complex variable of integration,

$\displaystyle f(x) = \frac{1}{{2\pi }}\int_{c - \infty i}^{c + \infty i} {{e^{zx}}\left[ {\int_0^\infty {f(t){e^{ - zt}}dt} } \right]d\omega }$ .

The quantity inside the square brackets is the Laplace transform $\mathcal L[f](z)$ . Therefore, we can express $f(t)$ in terms of its transform by the complex contour integral

Theorem (Bromwich’s integral). The following line integral runs along the line $x = c$ parallel to the imaginary axis and $c$ units to the right of it, the so-called Bromwich contour

$\displaystyle f(t) = \frac{1}{{2\pi i}}\int_{c - \infty i}^{c + \infty i} {\mathcal L[f](z){e^{tz}}dz}$ .

Example. Let us invert

$\displaystyle L(s) = \frac{{{e^{ - 3s}}}}{{{s^2}(s - 1)}}$ .

Solution. From Bromwich’s integral,

$\displaystyle\begin{gathered} f(t) = \frac{1}{{2\pi i}}\int_{c - \infty i}^{c + \infty i} {\frac{{{e^{(t - 3)s}}}}{{{z^2}(z - 1)}}dz} \hfill \\ \qquad= \frac{1}{{2\pi i}}\oint_C {\frac{{{e^{(t - 3)s}}}}{{{z^2}(z - 1)}}dz} - \frac{1}{{2\pi i}}\int_{{C_R}} {\frac{{{e^{(t - 3)s}}}}{{{z^2}(z - 1)}}dz} \hfill \\ \end{gathered}$

where $C_R$ is a semicircle of infinite radius in either the right or left half of the $z$ -plane and $C$ is the closed contour that includes $C_R$ and Bromwich’s contour.

Our first task is to choose an appropriate contour so that the integral along $C_R$ vanishes. By Jordan’s lemma, this requires a semicircle in the right half-plane if $t-3 < 0$ and a semicircle in the left half-plane if $t- 3>0$ . Consequently, by considering these two separate cases, we force the second integral to zero and the inversion simply equals the closed contour.

Consider the case $t < 3$ first. Because Bromwich’s contour lies to the right of any singularities, there are no singularities within the closed contour and $f(t) = 0$ .

Consider now the case $t > 3$ . Within the closed contour in the left half-plane, there is a second-order pole at $z = 0$ and a simple pole at $z = 1$ . Therefore,

$\displaystyle f(t) = \mathop {\rm Res}\limits_{z = 0} \left[ {\frac{{{e^{(t - 3)s}}}}{{{z^2}(z - 1)}}} \right] + \mathop {\rm Res}\limits_{z = 1} \left[ {\frac{{{e^{(t - 3)s}}}}{{{z^2}(z - 1)}}} \right]$

where

$\displaystyle\mathop {\rm Res}\limits_{z = 0} \left[ {\frac{{{e^{(t - 3)s}}}}{{{z^2}(z - 1)}}} \right] = \mathop {\lim }\limits_{z \to 0} \frac{d}{{dz}}\left[ {{z^2}\frac{{{e^{(t - 3)s}}}}{{{z^2}(z - 1)}}} \right] = 2 - t$

and

$\displaystyle\mathop {\rm Res}\limits_{z = 1} \left[ {\frac{{{e^{(t - 3)s}}}}{{{z^2}(z - 1)}}} \right] = \mathop {\lim }\limits_{z \to 1} \frac{d}{{dz}}\left[ {(z - 1)\frac{{{e^{(t - 3)s}}}}{{{z^2}(z - 1)}}} \right] = {e^{t - 3}}$ .

Taking our earlier results into account, the inverse equals

$\displaystyle f(t) = \left[ {{e^{t - 3}} - (t - 3) - 1} \right]H(t - 3)$ .

Source: Green’s Functions with Applications by Dean G. Duffy

Comments (3)

3 Comments »

Excuse me, I have a question.
the first equality on the solution of the example, I think z have to be put instead of s.
like this
$\displaystyle f(t) = \frac{1}{2\pi i}\int_{c-\infty i}^{c+\infty i}\frac{e^{(t-3)z}}{z^2 (z-1)}dz$
Am I right?

Comment by zariski — September 10, 2010 @ 19:41

Reply
- Hi zariski.
  
  You are right, it also appears that $s$ in the second line need to be changed, i.e., the whole stuff reads as follows
  
  $\displaystyle\begin{gathered} f(t) = \frac{1}{{2\pi i}}\int_{c - \infty i}^{c + \infty i} {\frac{{{e^{(t - 3)z}}}}{{{z^2}(z - 1)}}dz} \hfill \\ \qquad= \frac{1}{{2\pi i}}\oint_C {\frac{{{e^{(t - 3)z}}}}{{{z^2}(z - 1)}}dz} - \frac{1}{{2\pi i}}\int_{{C_R}} {\frac{{{e^{(t - 3)z}}}}{{{z^2}(z - 1)}}dz}. \hfill \\ \end{gathered}$
  
  Thanks for visiting my blog and also pointing out the misprint.
  
  Comment by Ngô Quốc Anh — September 10, 2010 @ 21:27
  
  Reply
Thanks for this post. It was a useful overview.

Comment by Sunil — August 12, 2012 @ 9:10

Reply

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Ngô Quốc Anh

September 1, 2010