TeachingDifferential equations – Laplace transformation
The Laplace transform is an integral transform of the form \(F(s)=\displaystyle\int_{\alpha}^{\beta}K(s,t)f(t)\,dt\) that transforms a function \(f(t)\) into another function \(F(s)\). The function \(K(s,t)\) is called the kernel of the integral transform.
For the Laplace transform we have: \(K(s,t)=e^{-st}\) and \((\alpha,\beta)=(0,\infty)\).
Definition: If the function \(f:[0,\infty)\to\mathbb{R}\) satisfies certain conditions, then
\[F(s)=\int_0^{\infty}e^{-st}f(t)\,dt\]is called the Laplace transform of the function \(f\). Notation: \(F(s)=\mathcal{L}\{f(t)\}(s)=\displaystyle\int_0^{\infty}e^{-st}f(t)\,dt\).
This is an improper integral which does not exist for every function \(f\). So the Laplace transform only exists for those function \(f\) for which this integral
\[\int_0^{\infty}e^{-st}f(t)\,dt=\lim\limits_{A\to\infty}\int_0^Ae^{-st}f(t)\,dt\]exists.
Theorem: If \(f\) is piecewise continuous on every subinterval \([0,A]\) with \(A>0\) and \(|f(t)|\leq Ke^{at}\) for all \(t\geq M>0\), then the Laplace transform
\[F(s)=\int_0^{\infty}e^{-st}f(t)\,dt\]of \(f\) exists for \(s>a\geq0\).
So the Laplace transform exists for the class of functions that are piecewise continuous and of exponential order.
Proof: Note that
\[\int_0^{\infty}e^{-st}f(t)\,dt=\int_0^Me^{-st}f(t)\,dt+\int_M^{\infty}e^{-st}f(t)\,dt.\]The first integral on the right-hand side exists for every piecewise continuous function \(f\). For the second integral we use that for \(t\geq M\)
\[|e^{-st}f(t)|\leq Ke^{-st}e^{at}=Ke^{(a-s)t}.\]Since \(\displaystyle\int_0^{\infty}e^{(a-s)t}\,dt\) converges if \(a-s<0\), we conclude that the second integral also converges for \(s>a\geq0\).
Last modified on April 25, 2021
