Motivation/Definition of Laplace Transform
The Laplace Transform of a function \( f(x) \) (or \(f(t)\), just replace all \(x\) below by \(t\)) is defined as
\begin{align}
\mathscr{L}[f](s) = \lim_{b\to\infty} \int_0^{b} e^{-sx} f(x) dx = \int_0^{\infty} e^{-sx} f(x) dx \tag{1}
\end{align}
To emphasize, such an integral transform \( \mathscr{L} \) is applied to the function \( f(x) \), \( x > 0 \), to make it become a new function in another parameter \( s > 0 \) now. It exists when the improper integral in (1) converges. We are interested in Laplace Transform because of its effect of converting an ODE in the time/space (\(t\) or \(x\)) domain to a simple algebraic equation in the complex frequency domain \( s \) that may be easier to solve, and then we can just apply an inverse transform to retrieve the physical solution for the ODE. This will be explained in greater detail in later posts.
The form of the Laplace Transform (1) can be motivated by rewriting it with \( z = e^{-s} \) (and using \(t\) instead of \(x\) for clarity):
\begin{align}
\int_0^{\infty} f(t) z^t dt \tag{2}
\end{align}
that very much looks like a generalized power series of \( \sum_{t=0}^{\infty} f(t) z^t \) if we move from a continuous integration to a discrete summation. We use a negative exponential so that it is easier for computation and the integral can converge. Essentially, by using the Laplace Transform, we are employing an extended version of the power series solution method to solve ODEs (see more in this MathOverflow post).
The simplest but often needed properties of the Laplace Transform are its additivity and scalar multiplication. For two functions \(f\) and \(g\) with both their Laplace Transforms existing, from (1):
\begin{align}
\mathscr{L}[f+g](s) &= \int_0^{\infty} e^{-sx} [f(x) + g(x)] dx \\
&= \int_0^{\infty} e^{-sx} f(x) dx + \int_0^{\infty} e^{-sx} g(x) dx \\
&= \mathscr{L}[f](s) + \mathscr{L}[g](s) \tag{3} \\
\mathscr{L}[cf](s) &= \int_0^{\infty} e^{-sx} cf(x) dx \\
&= c \int_0^{\infty} e^{-sx} f(x) dx = c\mathscr{L}[f](s) \tag{4}
\end{align}
where \(c\) can be any constant. This is mainly due to the additivity and scalar multiplication for integrals themselves.
Table of Notable Laplace Transform Values
Here we list some of the commonly seen Laplace Transform pairs. First, it is obvious that for a constant function:
\begin{align}
\mathscr{L}[1](s) &= \int_0^{\infty} e^{-sx} (1) dx = \int_0^{\infty} e^{-sx} dx \\
&= [-\frac{1}{s}e^{-sx}]_0^{\infty} \\
&= -[(0)-(\frac{1}{s})] = \frac{1}{s} \tag{5}
\end{align}
For positive integer powers of \(x\), i.e. \(x^n\), \( n\geq 1\), recursively using integration by parts leads to
\begin{align}
\mathscr{L}[x^n](s) &= \int_0^{\infty} e^{-sx} x^n dx \\
&= \int_0^{\infty} -\frac{1}{s} x^n d(e^{-sx}) \\
&= [-\frac{1}{s}x^ne^{-sx}]_{0}^{\infty} + \int_0^{\infty} \frac{1}{s} e^{-sx} d(x^n) \\
&= [-(0-0)] + \int_0^{\infty} \frac{1}{s} e^{-sx} nx^{n-1} dx \\
&= \frac{n}{s} \int_0^{\infty} e^{-sx} x^{n-1} dx \\
&= \frac{n}{s} \mathscr{L}[x^{n-1}](s) \\
&= \frac{n(n-1)}{s^2} \mathscr{L}[x^{n-2}](s) \\
&= \frac{n!}{s^n} \mathscr{L}[1](s) = \frac{n!}{s^{n+1}} \tag{6}
\end{align}
where in the last step we just use (5). For the exponential function \( e^{ax} \), we have
\begin{align}
\mathscr{L}[e^{ax}](s) &= \int_0^{\infty} e^{-sx} e^{ax} dx \\
&= \int_0^{\infty} e^{-(s-a)x} dx \\
&= [-\frac{1}{s-a} e^{-(s-a)x}]_0^{\infty} \\
&= -\frac{1}{s-a} [0-1] = \frac{1}{s-a} \tag{7}
\end{align}
where we need \( s > a \) for this integral to converge. Meanwhile, for sines (and cosines) like in the form of \( \sin(ax) \), its Laplace Transform is inferred by doing I.B.P. twice:
\begin{align}
\mathscr{L}[\sin(ax)](s) &= \int_0^{\infty} e^{-sx} \sin(ax) dx \\
&= \int_0^{\infty} -\frac{1}{a}e^{-sx} d(\cos(ax)) \\
&= [-\frac{1}{a}e^{-sx}\cos(ax)]_0^{\infty} + \int_0^{\infty} \frac{1}{a}\cos(ax)d(e^{-sx}) \\
&= \frac{1}{a} -\int_0^{\infty} \frac{s}{a}e^{-sx}\cos(ax)dx \\
&= \frac{1}{a} -\int_0^{\infty} \frac{s}{a^2}e^{-sx}d(\sin(ax))\\
&= \frac{1}{a} -\frac{s}{a^2}[e^{-sx}\sin(ax)]_0^{\infty} + \frac{s}{a^2} \int_0^{\infty} \sin(ax)d(e^{-sx}) \\
&= \frac{1}{a} -\frac{s}{a^2}(0) -\frac{s^2}{a^2} \int_0^{\infty}e^{-sx} \sin(ax)dx \\
&= \frac{1}{a} -\frac{s^2}{a^2}\mathscr{L}[\sin(ax)](s) \tag{8}
\end{align}
So
\begin{align}
\mathscr{L}[\sin(ax)](s) &= \frac{1}{a} -\frac{s^2}{a^2}\mathscr{L}[\sin(ax)](s) \\
(1+\frac{s^2}{a^2})\mathscr{L}[\sin(ax)](s) &= \frac{1}{a} \\
\mathscr{L}[\sin(ax)](s) &= \frac{1}{a} \frac{1}{1+\frac{s^2}{a^2}} = \frac{1}{a} \frac{a^2}{s^2+a^2} = \frac{a}{s^2+a^2} \tag{9}
\end{align}
In the same fashion, we should also get
\begin{align}
\mathscr{L}[\cos(ax)](s) &= \frac{s}{s^2+a^2} \tag{10}
\end{align}
For the hyperbolic functions, we simply apply (7), (3), (4):
\begin{align}
\mathscr{L}[\sinh(ax)](s) &= \mathscr{L}[\frac{e^{ax}-e^{-ax}}{2}](s) \\
&= \frac{1}{2} \mathscr{L}[e^{ax}](s) -\frac{1}{2} \mathscr{L}[e^{-ax}](s) \\
&= \frac{1}{2} \frac{1}{s-a} -\frac{1}{2}\frac{1}{s+a} \\
&= \frac{a}{s^2-a^2} \tag{11}
\end{align}
and similarly
\begin{align}
\mathscr{L}[\cosh(ax)](s) &= \mathscr{L}[\frac{e^{ax}+e^{-ax}}{2}](s) = \frac{s}{s^2-a^2} \tag{12}
\end{align}
It is not easy to compute an inverse Laplace transformation directly, which involves complex contour integration using the Bromwich formula. Rather, we usually consult a table of Laplace transform pairs just like (5)-(12) above and recognize the expressions.
Finally, we note that the Laplace Transform \( \mathscr{L}[f](s) \) of a function, if it exists, will always tend to \( 0 \) as \( s \to \infty \) under some mild condition imposed. This StackExchange post shows how to prove this. Here we give a slightly more polished version. If (1) indeed converges, and we further make a slightly stronger assumption (notice the absolute sign below) that there will be some \( K > 0 \) so that
\begin{align}
\mathscr{L}[f](K) = \int_0^{\infty} e^{-Kx} |f(x)| dx < \infty \tag{13}
\end{align}
is integrable and finite. By the Dominated Convergence Theorem in Real Analysis, since for \(s > K\) large enough, \( |e^{-sx} f(x)| < |e^{-sx}| |f(x)| < e^{-sx} |f(x)| < e^{-Kx} |f(x)| \), then we may put the limit \(s \to \infty\) inside the integral:
\begin{align}
\lim_{s \to \infty} \mathscr{L}[f](s) &= \lim_{s \to \infty} \int_0^{\infty} e^{-sx} f(x) dx \\
&= \int_0^{\infty} \lim_{s \to \infty} e^{-sx} f(x) dx \\
&= \int_0^{\infty} (0) f(x) dx = 0 \tag{14}
\end{align}
Exercise
Find the Laplace Transforms of the following functions (in \(t\)):
- \( t^3 -4t^2 + 1\)
- \( 5\sin(3t) -3\cos(5t) \)
- \( e^{-2t} + \sinh(2t) \)
as well as the inverse Laplace Transforms of:
- \( 5/s^4 + 3/s^2 -2/(s-1) \)
- \( 4/(s^2 + 9) \)
- \( 1/(s^4 -1) \) (Hint: Partial Fractions)
Answer
For the first part,
- \( 3!/s^4 -4(2!/s^3) + 1/s = 6/s^4 -8/s^3 + 1/s \)
- \( 5(3/(s^2+3^2)) -3(s/(s^2+5^2)) = 15/(s^2+9) -3s/(s^2+25) \)
- \( 1/(s-(-2)) + 2/(s^2-2^2) = 1/(s+2) + 2/((s-2)(s+2)) = s/(s^2-4) \) (Notice that the original expression is just \( \cosh(2t) \))
For the second part,
- \( 5(t^3/3!) + 3(t/1!) -2e^t = 5t^3/6 + 3t -2e^t \)
- Notice that it is just \( 4/(s^2 + 3^2) \), so the inverse Laplace transform is \( (4/3) \sin(3t) \).
- Following the hint, we decompose the expression into partial fractions:
\begin{align}
\frac{1}{s^4 -1} &= \frac{1}{2}\frac{1}{s^2-1} -\frac{1}{2}\frac{1}{s^2+1} \\
&= \frac{1}{4}\frac{1}{s-1} -\frac{1}{4}\frac{1}{s+1} -\frac{1}{2}\frac{1}{s^2+1}
\end{align}
Hence the desired inverse Laplace transform is \( (1/4)e^t -(1/4)e^{-t} -(1/2)\sin t \).
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