The Laplace Transform: Introduction

The Laplace transform is an integral transform that allows us to solve a linear differential equation by converting it to algebraic expressions such as rational functions.

The Laplace transform $\mathcal{L}\{F(t)\}$ of a function $F(t)$ is defined by
\mathcal{L}\{F(t)\}=\int_0^\infty e^{-st}F(t)dt
Since the improper integral in \eqref{eq:laplace} is a function of $s$, we also write $\mathcal{L}\{F(t)\}=f(s)$.

Example. Let $F(t)=1$, $t>0$.
$$\mathcal{L}\{F(t)\}=\int_0^\infty e^{-st}dt=\frac{1}{2}$$
provided $s>0$.

Example. Let $F(t)=e^{kt}$, $t>0$ and $k$ a constant.
\mathcal{L}\{F(t)\}&=\int_0^\infty e^{-st}e^{kt}dt\\
&=\int_0^\infty e^{-(s-k)}tdt\\
provided $s>k$.

\mathcal{L}\{t\}&=\int_0^\infty te^{-st}dt\\
for $s>0$.

Example. Again by the definition of the Laplace transform one obtains
by simply calculating the required improper integrals. There appears to be a pattern and one may expect the formula
This is indeed true and we will prove this later.

Example. By the definition of the Laplace transform one obtains
\mathcal{L}\{\sin kt\}&=\frac{k}{s^2+k^2}\\
\mathcal{L}\{\cos kt\}&=\frac{s}{s^2+k^2}

The Laplace transform is linear, namely

Example. $\sinh kt=\frac{e^{kt}-e^{-kt}}{2}$ so by the linearity of the Laplace transform
\mathcal{L}\{\sinh kt\}&=\frac{1}{2}(\mathcal{L}\{e^{kt}\}-\mathcal{L}\{e^{-kt}\}\\
Hence we have
\mathcal{L}\{\sinh kt\}=\frac{k}{s^2-k^2}
Similarly we also obtain
\mathcal{L}\{\cosh kt\}=\frac{s}{s^2+k^2}

One thought on “The Laplace Transform: Introduction

  1. Pingback: The Laplace Transform: Transforms of Derivatives | MathPhys Archive

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