Inverse Laplace Transforms

Section 5.5 Inverse Laplace Transforms

I would like to define an operation \(\calL^{-1}\) which undoes the Laplace transform.

\begin{equation*} \calL^{-1} \left\{ \calL \{ f(t) \} \right\} = f(t) \end{equation*}

This is very difficult to calculate directly, and the construction of the inverse transform is beyond the scope of this course. (It relies on complex ananlysis.) I’ll state the result for your interest, but it’s not a result I’m going to work with at all.

Theorem 5.5.1.

Let \(\alpha\) be a real number which is larger than all real singularities of \(F(s)\text{.}\) Let \(\gamma\) be a path in \(\CC\) which goes in a straight line from \(\alpha - \imath T\) to \(\alpha + \imath T\text{.}\) Then the inverse Laplace transform is defined by the following limit and integral.

\begin{equation*} \calL^{-1} (F(s)\} = \frac{1}{2\pi \imath} \lim_{T \rightarrow \infty} \int_{\gamma} e^{st} F(s) ds \end{equation*}

This is a contour integral in \(\CC\text{,}\) a common object in the study of complex variables. The use of the distance to singularities makes sense in that discipline, since contour integrals depend on the location of singularities. For those with a background in complex variables, this is not a closed curve and the homotopy implications depend a great deal on the function \(F\text{.}\)

For our purposes, I’ll just use tables to find inverse Laplace transforms, using the functions we already know. I’ll try to turn the \(s\) domain answers into familiar forms and invert known functions. This is a bit tricky, but can often be done. The use of shifts will be particularly inportant. I should note that \(\calL^{-1}\) is also linear.

Here is a table of the most important inverse transforms.

\begin{align*} \calL^{-1} \left\{ \frac{n!}{s^{n+1}} \right\} \amp = t^n\\ \calL^{-1} \left\{ \frac{1}{s-a} \right\} \amp = e^{at}\\ \calL^{-1} \left\{ \frac{k}{s^2 + k^2} \right\} \amp = \sin kt\\ \calL^{-1} \left\{ \frac{s}{s^2 + k^2} \right\} \amp = \cos kt\\ \calL^{-1} \left\{ \frac{k}{s^2 - k^2} \right\} \amp = \sinh kt\\ \calL^{-1} \left\{ \frac{s}{s^2 -k^2} \right\} \amp = \cosh kt\\ \calL^{-1} \left\{ e^{-sa} \right\} \amp = \delta_a(t) \end{align*}

The shift in an inverse transform is captured by the rule

\begin{equation*} \calL^{-1} \left\{ e^{-sa} F(s) \right\} = u_a(t) f(t-a)\text{.} \end{equation*}

This shifts of trigonometric functions are common enough that it is useful to mention them here.

\begin{align*} \calL^{-1} \left\{ \frac{\beta}{(s+ \alpha)^2 + \beta^2} \right\} \amp = e^{-\alpha t} \sin \beta t\\ \calL^{-1} \left\{ \frac{s + \alpha}{(s+ \alpha)^2 + \beta^2} \right\} \amp = e^{-\alpha t} \cos \beta t \end{align*}

Finally, the last rule from Section 5.4 is useful for dealing with \(s\)-domain functions which I can identity as derivatives of functions previous listed in the tables. Inverting that rule gives the following rule for inverse transforms.

\begin{equation*} \calL^{-1} \left\{ \frac{d^n}{ds^n} F(s) \right\} = (-1)^n t^n f(t) \end{equation*}

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