Laplace transform of derivatives - laplace-transform.com

1. Motivation

Our main goal to learn the Laplace Transform is to apply it in initial-value problems:

$\left\{\begin{array}{l} ay''+by'+cy=f(t), \\ y(0)=y_{0} \\ y'(0)=y_{1} \end{array}\right.$

From what we learned before, it is clear that we can evaluate the Laplace Transform of $f(t)$ and $y(t)$ above:

$F(s)=\mathscr{L}\{f(t)\}$

$Y(s)=\mathscr{L}\{y(t)\}$

Now, we want to be able to evaluate

$\mathscr{L}\{f''(t)\},\quad\mathscr{L}\{f'(t)\}$

such that we have an expression involving $F(s)$ .

2. Laplace transform derivatives theorem

3. Math behind Laplace transform of derivatives

a. First derivative

Let $f(t)$ be a function in the variable $t\geq0$ . Suppose $f'$ is continuous for $t\in[0,\infty)$ . By definition:

$\begin{aligned}\mathscr{L}\{f'(t)\} & = \int_{0}^{\infty}e^{-st}f'(t)dt \\ & = \lim_{b\to\infty}\int_{0}^{b}e^{-st}f'(t)dt \end{aligned}$

Let $u=e^{-st}$ and $dv=f'(t)dt$ . Then

$du=-se^{-st}dt$

and

$v=f(t)$

The integration by parts gives

$\begin{aligned} \int_{0}^{b}e^{-st}f'(t)dt & = \left.e^{-st}f(t)\right|_{t=0}^{t=b}-\int_{0}^{b}-se^{-st}f(t)dt \\ & = e^{-sb}f(b)-e^{-s\cdot0}f(0)+s\int_{0}^{b}e^{-st}f(t)dt \\ & = e^{-sb}f(b)-f(0)+s\int_{0}^{b}e^{-st}f(t)dt \end{aligned}$

We assume

$\displaystyle\lim_{b\to\infty}e^{-sb}f(b)=0$

then

$\begin{aligned} \mathscr{L}\{f'(t)\} & = \lim_{b\to\infty}\left[e^{-sb}f(b)-f(0)+s\int_{0}^{b}e^{-st}f(t)dt\right] \\ & = \lim_{b\to\infty}e^{-sb}f(b)-f(0)+s\lim_{b\to\infty}\int_{0}^{b}e^{-st}f(t)dt \\ & = 0-f(0)+s\lim_{b\to\infty}\int_{0}^{b}e^{-st}f(t)dt \end{aligned}$

Since

$\displaystyle F(s)=\lim_{b\to\infty}\int_{0}^{b}e^{-st}f(t)dt$

we concluded that

$\mathscr{L}\{f'(t)\}=sF(s)-f(0)$

b. Second derivative

In the same way, we can show a formula to the Laplace Transform of $\mathscr{L}\{f''(t)\}$ . Assuming $f'$ and $f''$ are both continuous on the interval $[0,\infty)$ , by the definition:

$\displaystyle\mathscr{L}\{f''(t)\}=\lim_{b\to\infty}\int_{0}^{b}e^{-st}f''(t)dt$

If $u=e^{-st}$ then $du=-se^{-st}dt$ and from $dv=f''(t)dt$ we have $v=f'(t)$ . Applying integration by parts:

$\begin{aligned} \int_{0}^{b}e^{-st}f''(t)dt & = \left.e^{-st}f'(t)\right|_{t=0}^{t=b}-\int_{0}^{b}-se^{-st}f'(t)dt \\ & = e^{-sb}f'(b)-e^{-s\cdot0}f'(0)+s\int_{0}^{b}e^{-st}f'(t)dt \\ & = e^{-sb}f'(b)-f'(0)+s\int_{0}^{b}e^{-st}f'(t)dt \end{aligned}$

Again, we assume that

$e^{-sb}f(b)\to0$

as $b\to\infty$ . Thus,

$\begin{aligned} \mathscr{L}\{f''(t)\} & = 0-f'(0)+s\cdot\overbrace{\lim_{b\to\infty}\int_{0}^{b}e^{-st}f'(t)d}^{\mathscr{L}\{f'(t)\}} \\ & = -f'(0)+s(sF(s)-f(0)) \end{aligned}$

Therefore,

$\mathscr{L}\{f''(t)\}=s^{2}F(s)-sf(0)-f'(0)$

As we repeat the same idea showed before, we can get the theorem results.

4. Illustrative example

Applying the Laplace Transform to ODE in the initial-value problem, we get:

$\mathscr{L}\{y''(t)+3y'(t)\}=\mathscr{L}\{0\}$

Due to the linearity of $\mathscr{L}$ and since

$\mathscr{L}\{0\}=0$

we can rewrite as follows

$\mathscr{L}\{y''(t)\}+3\mathscr{L}\{y'(t)\}=0$

Designing $\mathscr{L}\{y(t)\}=Y(s)$ and using the formulas above, we have

$\begin{aligned} \mathscr{L}\{y''(t)\} & = s^{2}Y(s)-sy(0)-y'(0) \\ & = s^{2}Y(s)-s\cdot1-(-1) \\ & = s^{2}Y(s)-s+1 \end{aligned}$

and

$\begin{aligned} \mathscr{L}\{y'(t)\} & = sY(s)-y(0) \\ & = sY(s)-1 \end{aligned}$

Thus, the ODE becomes

$\begin{aligned} & s^{2}Y(s)-s+1+3(sY(s)-1)=0 \\ & s^{2}Y(s)-s+1+3sY(s)-3=0 \\ & s^{2}Y(s)+3sY(s)-s-2=0 \end{aligned}$

where the solution $Y(s)$ is given by:

$\displaystyle Y(s)=\frac{s+2}{s^2+3s}$

Taking the inverse Laplace transform of $Y(s)$ , we can get the solution of the differential equation $y(t)$ .