Laplace Transforms and Derivatives

Section 5.4 Laplace Transforms and Derivatives

This is a course on differential equations; if Laplace transforms are useful, I will need them to relate to derivatives. I will calculate what happens to a derivative of a function (of exponential order) in a Laplace transform.

\begin{align*} \calL \left\{ \frac{df}{dt} \right\} \amp = \int_0^\infty \frac{df}{dt} e^{-st} dt\\ \amp = \left. fe^{-st} \right|_0^\infty - \int_0^\infty (-s) f(t) e^{-st} dt\\ \amp = \lim_{a \rightarrow \infty} f(a) e^{-sa} - f(0) + s \calL \{ f(t) \} (s) = -f(0) + s F(s) \end{align*}

I can summarize the result in this rule.

\begin{equation*} \calL \{ f^\prime(t) \} (s) = -f(0) + sF(s)\text{.} \end{equation*}

This is a lovely and convenient property. In the \(s\)-domain, there is no more differentiation. Differentiation is replaced by the much more approachable multiplication by the variable \(s\text{,}\) along with evaluation of the original function at zero.

I can do a similar calculation for second derivatives. The calculation will involve two uses of integration by parts, though I can make use of the previous calculation to simplify the work.

\begin{align*} \calL \left\{ \frac{d^2f}{dt^2} \right\} \amp = \int_0^\infty \frac{d^2f}{dt^2} e^{-st} dt\\ \amp = \left. \frac{df}{dt} e^{-st} \right|_0^\infty + s\int_0^\infty \frac{df}{dt} e^{-st} dt\\ \amp = -f^\prime(0) - sf(0) + s^2 F(s) \end{align*}

Again, this is very convenient. The derivatives are completely removed in the \(s\)-domain, replaced with simple algebraic operations. The general result for any order of differentiation is this rule.

\begin{equation*} \calL \left\{ \frac{d^n f}{dt^n} \right\} = s^n F(s) - s^{n-1} f(0) - s^{n-2} f^\prime(0) - s^{n-3} f^{\prime \prime}(0) - \ldots - f^{(n-1)}(0) \end{equation*}

The understanding of Laplace transforms of derivative leads to an amazing use of Laplace transforms in solving DEs. A Laplace transform changes differentiation into simple algebraic operations; therefore, it should change some DEs into algebraic equations. Since algebraic equations are much easier to solve than DEs, this is a huge advantage. However, I have one problem. If we want to solve a DE, I want to solve it in the \(t\)-domain. I can transform to the \(s\)-domain and solve the algebraic equation, but I need to get back to the \(t\)-domain to finish. For this, I need to invert the transform, which is covered in Section 5.5.

Before I move on, here are two other properties of the Laplace transform which relate to derivatives. The first shows that integrals are also changed into something algebraic.

\begin{equation*} \calL \left\{ \int_0^t f(u) du \right\} = \frac{1}{s} \calL \{ f(t) \} \end{equation*}

The second is a parallel identity which shows how differentiation in the \(s\)-domain also comes from an algebraic operation on the \(t\)-domain.

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

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