Laplace transform

Pierre-Simon Marquis de Laplace (1749-1827) and William Rowan Hamilton (1805-1865)
Sometimes we might feel some difficulties to keep the pace with the theories covered. That would be no surprise since they were discovered by geniuses of their age.

Pierre-Simon Marquis de Laplace (1749-1827) France (estimated IQ of 190)

Laplace advanced the nebular hypothesis of solar system origin, and was the first to conceive the concept of black holes. His other accomplishments in physics include theories about the speed of sound and surface tension. Laplace viewed mathematics as just a tool for developing his physical theories.

William Rowan Hamilton (1805-1865) Ireland (estimated IQ of 160—170)

Hamilton was a childhood prodigy. At age of 5 mastered Latin, Greek and Hebrew. Home-schooled and self-taught, he started as a student of languages and literature. By the time he was 13, the future mathematician knew 13 different languages, including Sanskrit, Persian, Italian, Arabic, Syriac and Indian dialects. At age of 15, he found errors in Laplace's work. Hamilton's Principle of Least Action, and its associated equations and concept of configuration space, led to a revolution in mathematical physics. Hamilton also made revolutionary contributions to dynamics, differential equations, the theory of equations, numerical analysis, fluctuating functions, and graph theory.

Definition of Laplace transform

The Laplace transform is a method for solving differential equations. It has some advantages over the other methods, e.g.

Like operator of differentiation converts function $f(x)$ into other function $f'(x)$, so the Laplace operator converts functions. By means of Laplace operator discontinuous function can be converted into other function.

The process of solving DE consists of 3 main steps:

  1. Transform the given hard to solve problem into simple problem.
  2. Solve the simple problem.
  3. Transform the solution of simple problem back to have a solution of given problem.

The Laplace transform is closely related to Fourier transform. If Laplace transform is defined on interval from negative infinity to infinity (referred to as bilateral transform) then Fourier transform can be seen as a special case of Laplace transform.

Improper integral

Note: improper Riemann integral is such, that has a domain involving infinity (example: $\int_0^3 f(x)\ dx$ is proper, $\int_0^{\infty} f(x)\ dx$ is improper integral).

Let $f(x)$ be a continuous function on the interval $I: 0 \leq x \leq h$. If, as $h \to \infty$, the definite integral $\int_0^{h} f(x)\ dx$ approaches a finite limit $K$, we say that improper integral $\int_0^{\infty}f(x)\ dx$ exists and converges to the value $K$ (K is the area under the graph).

$$ \int_0^{\infty}f(x)\ dx = \lim_{h \to \infty}\int_0^{h}f(x)\ dx= K $$

If the limit on the right does not exist, we say the improper integral on the left diverges and does not exist. Sometimes the result of improper integral is obvious, but sometimes we have to work with limits.


Determine whether the integral $\int_0^{\infty}1/(x+1)\ dx$ exists.

$$ \int_0^{h}\frac1 {x+1}\ dx = \left.\log (x+1)\right|_{x=0}^h = \log(h+1) $$

As $h \to \infty,\ \log(h+1) \to \infty$. Hence the improper integral diverges and does not exist.


Determine whether the integral $\int_0^{\infty}1/(x^2+1)\ dx$ exists.

$$ \begin{align} \int_0^{h}\frac1 {x^2+1}\ dx &= \arctan h - \arctan 0 \nonumber \\ \lim_{h \to \infty} (\arctan h) &= \frac{\pi}{2} \nonumber \end{align} $$

Since $\arctan(0)= 0$ the integral exists and converges to $\pi/2$.

Laplace transform

If the improper integral

$$ \begin{equation} \color{red}{\int_{0}^{\infty}e^{-st}f(t)\ dt},\hskip1em 0 \leq t \leq \infty, \label{ref:laplace1} \end{equation} $$

converges for a value $s=s_0$, then it converges for every $s\gt s_0$. The integral, if exists, is called the Laplace transform of $f(t)$ and is written as ${\scr L}[f(t)]$. The variable $t$ in $(\ref{ref:laplace1})$ is substituted by limits and the result of Laplace operator is another function of $\boldsymbol s$, e.g. ${\scr L}[f(t)] = F(s)$ or ${\scr L}[g(t)] = G(s)$.

An operator changes function $f(t)$ into $g(t)$. Here transform changes $f(t)$ into $F(s)$. The Laplace transform replaces power series: the power series (with $n$ from zero to infinity) has to converge and converts given function into series. And in here we have improper integral for that task. The function $e^t$ within integral is our favorite one: brings us convergence of the integral for positive $s$ and is easy to integrate/differentiate.

Because we will often work with $\int e^{-st}\ dt$ it is useful to list following limits:

$$ \begin{align} \lim_{t \to \infty} \frac{e^{-st}}{s} &= 0,\hskip1em \text{if}\ s \gt 0 \label{ref:lapl10} \\ \lim_{t \to \infty} \frac{e^{-st}}{s^2} &= 0,\hskip1em \text{if}\ s \gt 0 \label{ref:lapl11} \\ \lim_{t \to \infty} \frac{te^{-st}}{s} &= 0,\hskip1em \text{if}\ s \gt 0 \label{ref:lapl12} \\ \lim_{t \to \infty} \frac{t^ne^{-st}}{s} &= 0,\hskip1em \text{if}\ s \gt 0,\ n\ \text{real} \label{ref:lapl13} \end{align} $$

Note: function $e^t$ is known to be one of the fastest growing function. Here is $e^{-t}$ as one of the fastest decreasing function.


Find the Laplace transform of the function $f(t) = 1$.

$$ \begin{align} &{\scr L}[f(t)] = \int_0^{\infty}e^{-st}\ dt = \lim_{h\to\infty}\int_0^he^{-st}\ dt = \lim_{h\to\infty}\left.\frac{-e^{-st}}{s}\right|_{t=0}^h = \lim_{h\to\infty}\left(\frac{-e^{-st} }{s} + \frac 1 s\right) \nonumber \\ &\underline{\underline{{\scr L}[f(t)] = \frac 1 s}} \nonumber \end{align} $$

The result is valid for $s \gt 0$ (otherwise the integral does not converge—check numerator). The right branch of hyperbola is the Laplace transform of $f(t)=1$.


Find the Laplace transform of the function $f(t) = t$.

$$ \begin{align} &{\scr L}[f(t)] = \int_0^{\infty}e^{-st}t\ dt = \lim_{h\to\infty}\int_0^he^{-st}t\ dt = \lim_{h\to\infty}\left.\frac{-e^{-st}t}{s} - \frac{e^{-st}}{s^2}\right|_{t=0}^h = (0-0)-\big(0-\frac{e^0}{s^2}\big) \nonumber \\ &\underline{\underline{{\scr L}[f(t)] = \frac 1 {s^2}}} \nonumber \end{align} $$

The result is valid for $s \gt 0$ (otherwise the integral does not converge) and the values from $(\ref{ref:lapl11}),\ (\ref{ref:lapl12})$ were used. The integral itself comes from integrating by parts, steps are not shown above.


Find the Laplace transform of the function $f(t) = e^{-2t}$.

$$ \begin{align} &{\scr L}[f(t)] = \int_0^{\infty}e^{-st}e^{-2t}\ dt = \lim_{h\to\infty}\int_0^he^{-(s+2)t}\ dt = \lim_{h\to\infty}\left.\frac{-e^{-(s+2)t}}{s+2}\right|_{t=0}^h = 0-\big(\frac{-e^0}{s + 2}\big) \nonumber \\ &\underline{\underline{{\scr L}[f(t)] = \frac 1 {s+2}}} \nonumber \end{align} $$

The result is valid for $s \gt -2$ (otherwise the integral does not converge). From the second integral above you can see that the Laplace transform of $e^{-2t}$ is again hyperbola, shifted to the left (exponential shift).


Find the Laplace transform of the function $f(t) = e^{3t}$.

$$ \begin{align} &{\scr L}[f(t)] = \int_0^{\infty}e^{-st}e^{3t}\ dt = \lim_{h\to\infty}\int_0^he^{-(s-3)t}\ dt = \lim_{h\to\infty}\left.\frac{-e^{-(s-3)t}}{s-3}\right|_{t=0}^h = 0-\big(\frac{-e^0}{s - 3}\big) \nonumber \\ &\underline{\underline{{\scr L}[f(t)] = \frac 1 {s-3}}} \nonumber \end{align} $$

The result is valid for $s \gt 3$ (otherwise the integral does not converge).

Linearity of the Laplace transform

We like the fact, that if the Laplace transform of $f_1(t)$ converges for $s_1$ and the Laplace transform of $f_2(t)$ converges for $s_2$ then for $s$ greater than both $s_1,\ s_2$

$$ \begin{align} {\scr L}[ f_1 + f_2] & = {\scr L}[f_1] + {\scr L} [f_2], \nonumber \\ {\scr L}[ c f ] & = c{\scr L}[f], \nonumber \\ {\scr L}[ c_1f_1 + c_2f_2] & = c_1{\scr L}[f_1] + c_2{\scr L} [f_2], \nonumber \end{align} $$

where $c_1,\ c_2$ are constants, i.e. the Laplace transform is a linear operator.

Example: for $s\gt 0:\hskip1em {\scr L} [1+5t] = {\scr L}[1] + 5{\scr L}[t] = 1/s + 5/ s^2$.

Example: for $s\gt 3:\hskip1em {\scr L} [4e^{3t}-5t] = 4{\scr L}[e^{3t}] - 5{\scr L}[t] = 1/(s-3) - 5/ s^2$.

Transforms of some basic functions

$$ \begin{array}{|ll|} \textbf{(1)}\ {\scr L}[1] = \frac 1 s \hskip1em &\textbf{(2)}\ {\scr L}[t^n] = \frac {n!}{s^{n+1}},\ n = 1,\ 2,\ 3,\ \dots \\ \textbf{(3)}\ {\scr L}[e^{at}] = \frac {1}{s-a} \hskip1em &\textbf{(4)}\ {\scr L}[e^{at}t^n] = \frac {n!}{(s-a)^{n+1}},\ n=0,\ 1,\ \dots \\ \textbf{(5)}\ {\scr L}[\sin kt] = \frac {k}{s^2+k^2} \hskip1em &\textbf{(6)}\ {\scr L}[\cos kt] = \frac {s}{s^2+k^2} \\ \textbf{(7)}\ {\scr L}[\sinh kt] = \frac {k}{s^2-k^2} \hskip1em &\textbf{(8)}\ {\scr L}[\cosh kt] = \frac {s}{s^2-k^2} \\ \end{array} $$

Sufficient conditions for existence of Laplace transform

Some functions do not posses Laplace transform. Transform ${\scr L}[e^{t^2}]$ does not exist as integral $\int_0^{\infty} e^{-st} e^{t^2}\ dt$ does not exist ($e^{t^2}$ grows much faster than $e^{-st}$ decreases); similarly ${\scr L}[1/t]$ does not exist as the integral $(\ref{ref:laplace1})$ diverges (e.g. on the interval $(0,1)$).

To the definition $(\ref{ref:laplace1})$ we add sufficient conditions for existence of ${\scr L}[f(t)]$:

These conditions are not necessary. E.g. $f(t)= 1/\sqrt t$ is not piecewise continuous but its transform does exist.

Transform of piecewise continuous functions


Find the Laplace transform of the function

$$ f(t) = \left \{ \begin{align} 1-t, \hskip1em & 0 \leq t \leq 1 \nonumber \\ 0, \hskip1em & 1 \lt t \leq 3 \nonumber \\ e^{2t}, \hskip1em & 3 \lt t \lt \infty \nonumber \end{align} \right. $$
$$ \begin{align} {\scr L}[f(t)] &= \int_0^{\infty} e^{-st}f(t)\ dt = \int_0^{1}e^{-st}(1-t)\ dt + \int_1^3e^{-st}\cdot 0\ dt + \int_3^{\infty}e^{-st}e^{2t}\ dt = \nonumber \\ &= \int_0^1e^{-st}\ dt - \int_0^1te^{-st}\ dt + \int_3^{\infty} e^{(2-s)t}\ dt = \nonumber \\ &= \left. -\frac 1 s e^{-st} \right|_{t=0}^1 - \left.\frac{-(st+1)e^{-st}}{s^2}\right|_0^1 + \left.\frac{e^{(2-s)t}}{2-s}\right|_3^{\infty} = \nonumber \\ &= \left. -\frac 1 s e^{-st} \right|_0^1 - \left.\frac{-(st+1)e^{-st}}{s^2}\right|_0^1 + \left.\frac{e^{(2-s)t}}{2-s}\right|_3^{\infty} = \nonumber \\ &= \left(-\frac 1 s e^{-s}-(-\frac 1 s)\right) - \left(\frac{-(s+1)e^{-s}}{s^2} - \frac{-1}{s^2}\right) + \left(0-\frac{e^{(2-s)\cdot 3}}{2-s}\right) = \nonumber \\ &= \underline{\underline{\frac{e^{-s} + s - 1}{s^2} + \frac{e^{6-3s}}{s-2}}} \nonumber \end{align} $$

For $s \gt 2$. Note that the integral $\int_3^{\infty} e^{(2-s)t}\ dt$ does not exist if $s \leq 2$. Note also that the result is not piecewise defined.

Inverse transforms

If $F(s)$ represents the Laplace transform of $f(t)$, i.e. ${\scr L}[f(t)] = F(s)$, then $f(t)$ is the inverse Laplace transform of $F(s)$.

$$ f(t) = {\scr L}^{-1}[F(s)] \iff F(s) = {\scr L}[f(t)] $$

Inverse transform is the hardest part of computing DE by Laplace transform. Finally the function $F(s)$ has to be converted back to $f(t)$. For such task, we can not use tables unless they would be too long to be useful. The work mostly consists of partial fractions decomposition.

Inverse transforms of some basic functions

$$ \begin{array}{|ll|} \textbf{(1)}\ 1 = {\scr L}^{-1}\left[\frac 1 s\right] \hskip1em &\textbf{(2)}\ t^n = {\scr L}^{-1}\left[\frac {n!}{s^{n+1}}\right],\ n = 1,\ 2,\ 3,\ \dots \\ \textbf{(3)}\ e^{at} = {\scr L}^{-1}\left[\frac {1}{s-a}\right] \hskip1em &\textbf{(4)}\ e^{at}t^n = {\scr L}^{-1}\left[\frac {n!}{(s-a)^{n+1}}\right],\ n=0,\ 1,\ \dots \\ \textbf{(5)}\ \sin kt = {\scr L}^{-1}\left[\frac {k}{s^2+k^2}\right] \hskip1em &\textbf{(6)}\ \cos kt= {\scr L}^{-1}\left[\frac {s}{s^2+k^2}\right] \\ \textbf{(7)}\ \sinh kt = {\scr L}^{-1}\left[\frac {k}{s^2-k^2}\right] \hskip1em &\textbf{(8)}\ \cosh kt = {\scr L}^{-1}\left[\frac {s}{s^2-k^2}\right] \\ \end{array} $$


Find $\scr{L}^{-1}\left[1/(s-1)\right],\ s\gt 1$

From the above table we know that

$$ e^{at} = {\scr L}^{-1}\left[\frac {1}{s-a}\right] $$

We find that for $a=1\ e^{t} = {\scr L}^{-1}\left[ {1}/{s-1}\right]$. Hence ${\scr L}^{-1}\left[ {1}/{s-1}\right] = e^t$ for $t \gt 0$.


Find $\scr{L}^{-1}\left[(5s-2)/(s^2+4)\right]$.

$$ \scr{L}^{-1}\left[\frac{5s - 2}{s^2+4}\right] = \scr{L}^{-1}\left[\frac{5s}{s^2+4}\right] - \scr{L}^{-1}\left[\frac{2}{s^2+4}\right] = 5 \cos 2t - \sin 2t. $$

From the table of transforms:

$$ \textbf{(5)}\ \sin kt = {\scr L}^{-1}\left[\frac {k}{s^2+k^2}\right],\hskip2em \textbf{(6)}\ \cos kt= {\scr L}^{-1}\left[\frac {s}{s^2+k^2}\right] $$

The match for the member $\scr{L}^{-1}\left[{5s}/{(s^2+4)}\right]$ is $\cos kt$. Then $k=2$.

$$ k=2:\ {\scr L}[\cos 2t] = \frac{s}{s^2+4} $$

The match for the member $\scr{L}^{-1}\left[{2}/{(s^2+4)}\right]$ is $\sin kt$. Then $k=2$.

$$ k=2:\ {\scr L}[\sin 2t] = \frac{2}{s^2+4} $$

Decomposition to partial fractions

The differential equation is solved in the domain $s$ where it is easy to solve. Finaly inverse Laplace transform on solution $Y(s) = p(s)/q(s)$, where $p(s),\ q(s)$ are polynomials, has to be applied. The polynomial $Y(s)$ is usually too complicated for inverse transform and has to be splitted into simple fraction according to rules shown below.

CasePolynomial observed Rule for decomposition
Real roots$(s-a)$$\frac{A}{s-a}$
Repeated roots$(s-a)^2$$\frac{A}{s-a} + \frac{A}{(s-a)^2}$
Imaginary roots $(s^2+a^2)$$\frac{As + B}{s^2 + a^2}$

Example: Case I (real roots)

Find ${\scr{L}}^{-1}[F],\ F =(6s^2+5s-3)/(s^3+2s^2-3s)$.

$F(s) = (6s^2+5s-3)/(s^3+2s^2-3s)$ can be rewritten into $F(s) = \frac 1 s + \frac 3 {s+3} + \frac2 {s-1}$.

To convert $(6s^2+5s-3)/(s^3+2s^2-3s)$ into $\frac 1 s + \frac 3 {s+3} + \frac2 {s-1}$ we have to employ partial fraction decomposition:

$$ \frac{6s^2+5s-3}{s^3+2s^2-3s} = \frac{6s^2+5s-3}{s(s+3)(s-1)} = \frac{A}{s} + \frac{B}{s+3} + \frac{C}{s-1} $$

By multipliyng both sides by $s(s+3)(s-1)$ we have

$$ \begin{align} 6s^2+5s-3 =& A(s+3)(s-1) + Bs(s-1) + Cs(s+3) \implies \nonumber \\ \implies 6s^2+5s-3 =& As^2 +2As -3A + Bs^2 -Bs + Cs^2 + 3Cs \nonumber \\ 6s^2+5s-3 =& (A+B+C)s^2 + (2A-B+3C)s +1 (-3A)\cdot 1 \implies \nonumber \\ \implies A=&1,\ B =3,\ C=2\nonumber \\ \end{align} $$

We used the method of undetermined coefficients and here comes

$$ \frac{6s^2+5s-3}{s^3+2s^2-3s} = \frac{1}{s} + \frac{3}{s+3} + \frac{2}{s-1}. $$


$$ \begin{align} & {\scr L}^{-1}\left[\frac 1 s + \frac 3 {s+3} + \frac2 {s-1}\right] = {\scr L}^{-1}\left[\frac 1 s\right] + {\scr L}^{-1}\left[\frac 3 {s+3}\right] + {\scr L}^{-1}\left[\frac 2 {s-1}\right] \nonumber \\ & {\scr L}^{-1}\left[\frac 1 s\right] = 1 \nonumber \\ & {\scr L}^{-1}\left[\frac 3 {s+3}\right] = 3e^{-3t}\nonumber \\ & {\scr L}^{-1}\left[\frac 2 {s-1}\right] = 2e^{t}\nonumber \end{align} $$

And finally

$$ \underline{\underline{f(t) = 1+3e^{-3t}+2e^t}}. $$

Example: Case II (repeated roots)

Find ${\scr{L}}^{-1}[F],\ F ={s^2}/{(s+1)^3}$.

$F(s) = {s^2}/{(s+1)^3}$ can be rewritten into $F(s) = \frac{1}{(s+1)} - \frac{2}{(s+1)^2} + \frac{1}{(s+1)^3}$.

To convert ${s^2}/{(s+1)^3}$ into $\frac{1}{(s+1)} - \frac{2}{(s+1)^2} + \frac{1}{(s+1)^3}$ we have to employ partial fraction decomposition:

$$ \frac{s^2}{(s+1)^3} = \frac{A}{s+1} + \frac{B}{(s+1)^2} + \frac{C}{(s+1)^3} $$

By multipliyng both sides by $(s+1)^3$ we have

$$ \begin{align} s^2 =& A(s+1)^2 + B(s+1) + C \nonumber \implies \\ \implies s^2 =& As^2 + 2As +A + Bs+B + C \nonumber \\ s^2 =& As^2 + (2A+B)s +(A +B + C) \implies \nonumber \\ \implies A=&1,\ B=-2,\ C=1\nonumber \\ \end{align} $$

We used the method of undetermined coefficients and here comes that

$$ \frac{s^2}{(s+1)^3} = \frac{1}{s+1} - \frac{2}{(s+1)^2} + \frac{1}{(s+1)^3} $$


$$ \begin{align} {\scr L}^{-1}\left[\frac{1}{s+1} - \frac{2}{(s+1)^2} + \frac{1}{(s+1)^3}\right] =& {\scr L}^{-1}\left[\frac 1 {s+1}\right] - {\scr L}^{-1}\left[\frac 2 {(s+1)^2}\right] + \nonumber \\ & + {\scr L}^{-1}\left[\frac 1 {(s+1)^3}\right] \nonumber \\ {\scr L}^{-1}\left[\frac 1 {s+1}\right] =& e^{-t} \nonumber \\ {\scr L}^{-1}\left[\frac 2 {(s+1)^2}\right] =& 2e^{-t}t\nonumber \\ {\scr L}^{-1}\left[\frac 1 {(s+1)^3}\right] =& \frac 1 2 e^{-t}t^2\nonumber \end{align} $$

And finally

$$ \underline{\underline{f(t) = e^{-t}-2e^{-t}t + \frac 1 2 e^{-t}t^2}}. $$

Example: Case III (imaginary quadratic roots)

Find ${\scr{L}}^{-1}[F],\ F =(1+\pi s)/[s^2(s^2+1)]$.

$F =(1+\pi s)/[s^2(s^2+1)]$ can be rewritten into $\frac{\pi}{s} + \frac{1}{s^2} -\frac{\pi s + 1}{s^2+1}$.

To convert $(1+\pi s)/[s^2(s^2+1)]$ into $\frac{\pi}{s} + \frac{1}{s^2} -\frac{\pi s + 1}{s^2+1}$ we can employ partial fraction decomposition:

$$ \begin{equation} \frac{1+\pi s}{s^2\color{red}{(s^2+1)}} = \frac{A}{s} + \frac{B}{s^2} + \color{red}{\frac{Cs+D}{s^2 +1}} \label{ref:lapla_partial3} \end{equation} $$

By multiplying both sides by $s^2(s^2+1)$ we get

$$ \begin{align} 1+ \pi s &= As(s^2+1) + B(s^2+1)+ Cs(s^2) + Ds^2 \implies \nonumber \\ \implies 1+\pi s &= As^3 + As +Bs^2 + B + Cs^3 + Ds^2 \nonumber \\ 1+\pi s &= s^3(A+C) + s^2 (B+D) + s(A) + 1(B) \implies \nonumber \\ \implies A &= \pi,\ B=1,\ C = -\pi,\ D= -1 \nonumber \end{align} $$

Now we can place the found constants back to $(\ref{ref:lapla_partial3})$:

$$ \frac{1+\pi s}{s^2{(s^2+1)}} = \frac{\pi}{s} + \frac{1}{s^2} + {\frac{-\pi s-1}{s^2 +1}} $$

That suggests we have to find inverse transform of below terms in order to collect ${\scr{L}}^{-1}[F]$.

$$ \begin{align} & {\scr L}^{-1}\left[\frac {\pi} {s}\right] = \pi \nonumber \\ & {\scr L}^{-1}\left[\frac {1} {s^2}\right] = t \nonumber \\ & {\scr L}^{-1}\left[\pi \frac {s} {s^2 + 1}\right] = \pi \cos t \nonumber \\ & {\scr L}^{-1}\left[\frac {1} {s^2 + 1}\right] = \sin t \nonumber \\ \end{align} $$

The solution we are looking for is then

$$ \underline{\underline{y(t) = \pi + t - \pi \cos t - \sin t}}. $$

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