13.8: Laplace inverse

Up to now we have computed the inverse Laplace transform by table lookup. For example, \(\mathcal{L}^{-1} (1/(s - a)) = e^{at}\). To do this properly we should first check that the Laplace transform has an inverse.

We start with the bad news: Unfortunately this is not strictly true. There are many functions with the same Laplace transform. We list some of the ways this can happen.

1. If \(f(t) = g(t)\) for \(t \ge 0\), then clearly \(F(s) = G(s)\). Since the Laplace transform only concerns \(t \ge 0\), the functions can differ completely for \(t < 0\).
2. Suppose \(f(t) = e^{at}\) and

\[g(t) = \begin{cases} f(t) & \text{ for } t \ne 1 \\ 0 & \text{ for } t = 1. \end{cases} \nonumber \]

That is, \(f\) and \(g\) are the same except we arbitrarily assigned them different values at \(t = 1\). Then, since the integrals won’t notice the difference at one point, \(F(s) = G(s) = 1/(s - a)\). In this sense it is impossible to define \(\mathcal{L}^{-1} (F)\) uniquely.

The good news is that the inverse exists as long as we consider two functions that only differ on a negligible set of points the same. In particular, we can make the following claim.

Theorem \(\PageIndex{1}\)

Suppose \(f\) and \(g\) are continuous and \(F(s) = G(s)\) for all \(s\) with \(\text{Re} (s) > a\) for some \(a\). Then \(f(t) = g(t)\) for \(t \ge 0\).

This theorem can be stated in a way that includes piecewise continuous functions. Such a statement takes more care, which would obscure the basic point that the Laplace transform has a unique inverse up to some, for us, trivial differences.

We start with a few examples that we can compute directly.

Example \(\PageIndex{1}\)

Let

\[f(t) = e^{at}. \nonumber \]

So,

\[F(s) = \dfrac{1}{s - a}. \nonumber \]

Show

\[f(t) = \sum \text{Res} (F(s) e^{st}) \nonumber \]

\[f(t) = \dfrac{1}{2\pi i} \int_{c - i\infty}^{c + i \infty} F(s) e^{st}\ ds \nonumber \]

The sum is over all poles of \(e^{st}/(s - a)\). As usual, we only consider \(t > 0\).

Here, \(c > \text{Re} (a)\) and the integral means the path integral along the vertical line \(x = c\).

Solution

Proving Equation 13.8.4 is straightforward: It is clear that

\[\dfrac{e^{st}}{s -a} \nonumber \]

has only one pole which is at \(s = a\). Since,

\[\sum \text{Res} (\dfrac{e^{st}}{s - a}, a) = e^{at} \nonumber \]

we have proved Equation 13.8.4.

Proving Equation 13.8.5 is more involved. We should first check the convergence of the integral. In this case, \(s = c + iy\), so the integral is

\[\dfrac{1}{2\pi i} \int_{c - i \infty}^{c + i \infty} F(s) e^{st} \ ds = \dfrac{1}{2\pi i} \int_{-\infty}^{\infty} \dfrac{e^{(c + iy)t}}{c + iy - a} i \ dy = \dfrac{e^{ct}}{2 \pi} \int_{-\infty}^{\infty} \dfrac{e^{iyt}}{c + iy - a} \ dy. \nonumber \]

The (conditional) convergence of this integral follows using exactly the same argument as in the example near the end of Topic 9 on the Fourier inversion formula for \(f(t) = e^{at}\). That is, the integrand is a decaying oscillation, around 0, so its integral is also a decaying oscillation around some limiting value.

Now we use the contour shown below.

We will let \(R\) go to infinity and use the following steps to prove Equation 13.8.5.

1. The residue theorem guarantees that if the curve is large enough to contain \(a\) then
   \[\dfrac{1}{2\pi i} \int_{C_1 - C_2 - C_3 + C_4} \dfrac{e^{st}}{s - a}\ ds = \sum \text{Res} (\dfrac{e^{st}}{s - a}, a) = e^{at}. \nonumber \]
2. In a moment we will show that the integrals over \(C_2, C_3, C_4\) all go to 0 as \(R \to \infty\).
3. Clearly as \(R\) goes to infinity, the integral over \(C_1\) goes to the integral in Equation 13.8.5 Putting these steps together we have

\[e^{at} = \lim_{R \to \infty} \int_{C_1 - C_2 - C_3 + C_4} \dfrac{e^{st}}{s - a} \ ds = \int_{c - i\infty}^{c + i\infty} \dfrac{e^{st}}{s - a} \ ds \nonumber \]

Except for proving the claims in step 2, this proves Equation 13.8.5.

To verify step 2 we look at one side at a time.

\(C_2\): \(C_2\) is parametrized by \(s = \gamma (u) = u + iR\), with \(-R \le u \le c\). So,

\[|\int_{C_2} \dfrac{e^{st}}{s - a} \ ds| = \int_{-R}^{c} |\dfrac{e^{(u + iR)t}}{u + iR - a}| \le \int_{-R}^{c} \dfrac{e^{ut}}{R} \ du = \dfrac{e^{ct} - e^{-Rt}}{tR}. \nonumber \]

Since \(c\) and \(t\) are fixed, it’s clear this goes to 0 as \(R\) goes to infinity.

The bottom \(C_4\) is handled in exactly the same manner as the top \(C_2\).

\(C_3\): \(C_3\) is parametrized by \(s = \gamma (u) = -R + iu\), with \(-R \le u \le R\). So,

\[|\int_{C_3} \dfrac{e^{st}}{s - a} \ ds| = \int_{-R}^{R} |\dfrac{e^{(-R + iu)t}}{-R + iu - a}| \le \int_{-R}^{R} \dfrac{e^{-Rt}}{R + a} \ du = \dfrac{e^{-Rt}}{R + a} \int_{-R}^{R} \ du = \dfrac{2\text{Re}^{-Rt}}{R+a}. \nonumber \]

Since \(a\) and \(t > 0\) are fixed, it’s clear this goes to 0 as \(R\) goes to infinity.

Example \(\PageIndex{2}\)

Repeat the previous example with \(f(t) = t\) for \(t > 0\), \(F(s) = 1/s^2\).

This is similar to the previous example. Since \(F\) decays like \(1/s^2\) we can actually allow \(t \ge 0\)

Theorem \(\PageIndex{2}\) Laplace inversion 1

Assume \(f\) is continuous and of exponential type \(a\). Then for \(c > a\) we have

\[f(t) = \dfrac{1}{2\pi i} \int_{c - i\infty}^{c + i\infty} F(s) e^{st}\ ds. \nonumber \]

As usual, this formula holds for \(t > 0\).

Proof

The proof uses the Fourier inversion formula. We will just accept this theorem for now. Example 13.8.1 above illustrates the theorem.

Theorem \(\PageIndex{3}\) Laplace inversion 2

Suppose \(F(s)\) has a finite number of poles and decays like \(1/s\) (or faster). Define

\[f(t) = \sum \text{Res} (F(s) e^{st}, p_k), \text{ where the sum is over all the poles } p_k. \nonumber \]

Then \(\mathcal{L} (f; s) = F(s)\)

Proof

Proof given in class. To be added here. The basic ideas are present in the examples above, though it requires a fairly clever choice of contours.

The integral inversion formula in Equation 13.8.13 can be viewed as writing \(f(t)\) as a ‘sum’ of exponentials. This is extremely useful. For example, for a linear system if we know how the system responds to input \(f(t) = e^{at}\) for all \(a\), then we know how it responds to any input by writing it as a ‘sum’ of exponentials.