The inverse Laplace transform

Differential equations and Laplace transforms: Laplace-transformations

The inverse Laplace transform

The Laplace transform is injective, at least with the right choice of functions in the time domain. Therefore, we first discuss a special class of functions.

A collection $J$ of real points is called discrete if each closed interval of finite length has a finite number of points of $J$ .

A real function $f$ called piecewise continuous on an interval $I$ if there is a discrete subset $J$ of $I$ , such that $f$ is defined and continuous at each point of $I$ outside $J$ and if at each point $j$ of $J$ the left-hand limit $\lim_{x\uparrow j}f(x)$ and the right limit $\lim_{x\downarrow j}f(x)$ of the function $f$ exist and are distinct.

The collection $J$ is uniquely determined for a piecewise continuous function. It consists of all jumps of $f$ .

Let $c$ be a positive real number. We say that a function $f$ has exponential order $c$ if there are numbers $M\gt0$ and $t_0\gt 0$ such that $\left|f(t)\right|\le M\cdot\e^{c\,t}$ for $t\gt t_0$ . We say that $f$ has exponential order if there is a positive real number $c$ such that $f$ has exponential order $c$ .

The set of integers is a discrete set of real points. The set of rational numbers is not.

Continuous functions are piecewise functions without jumps.

The unit step function, also called the Heaviside function, is the real function $u_a$ defined by

$u_a(x) =\begin{cases} 0 &\text{if } x\lt a\\ 1&\text{if }x\ge a\end{cases}$

This is a piecewise continuous function with one jump, namely $x=a$ . This function has exponential order $c$ for any positive value of $c$ .

The Laplace transform is injective on the space of the above functions:

Uniqueness theorem

Let $f$ and $g$ two piecewise continuous functions on $\ivco{0}{\infty}$ having exponential order $c$ .

The Laplace-transform $\laplace{(f)}(s)$ is defined for $s\gt c$ and satisfies $\lim_{s\to\infty}\laplace{(f)}(s) = 0$
If $\laplace{(f)}(s)=\laplace{(g)}(s)$ for all $s\gt c$ then $f(t) = g(t)$ for all $t$ that are no jump of $f$ or $g$ .

Compare the Heaviside step function $u_c$ with the function $v_c$ defined by $v_c(x) =\begin{cases} 0 &\text{if } x\le c\\ 1&\text{if }x\gt c\end{cases}$ As real functions their values differ at $c$ , but as functions defined outside all jumps, they are equal.

The Laplace transforms of the functions $u_c$ and $v_c$ are equal. If we want to conclude from equality of their Laplace transforms that two piecewise continuous functions are equal, we should not take into consideration the values of the functions at jumps.

Just like the Laplace transform also the inverse Laplace transform can be defined by means of an integral: $\mathcal{L}^{-1}(F(s))(t) = \frac{1}{2\pi\ii}\int_{\gamma-\ii R}^{\gamma+\ii R}\ee^{st}\cdot F(s)\dd s$

Here $\gamma$ and $R$ are sufficiently large real numbers. This integral should be evaluated with complex variables; integrating variable $s$ runs through the complex numbers on the vertical line from $\gamma-\ii R$ to $\gamma+\ii R$ .

We will not calculate any inverse by use of this formula.

In view of this statement the Laplace transform is an injective mapping on the domain $E$ of piecewise continuous functions on $\ivco{0}{\infty}$ of exponential order. Here we take the domain of a piecewise continuous function $f$ to be the set of all real points in $\ivco{0}{\infty}$ that are no jumps of $f$ .

Because $E$ is a linear space (check!) and the Laplace transform is linear (see the linearity theorem), $\laplace{}$ is an invertible linear transformation from $E$ to the image of $E$ under $\laplace{}$ . As a consequence, the inverse Laplace transform is also linear.

According to the first statement of the uniqueness theorem, the image of $E$ under $\laplace{}$ lies within the vector space of those functions on an interval of the shape $\ivoo{a}{\infty}$ that approach $0$ for $s\to\infty$ . This property does not suffice for an exact description of the image. However, in case of a rational function $y(s)$ , this condition means that the degree of the numerator of $y(s)$ is less than the degree of the denominator of $y(s)$ ; below this turns out be a sufficient condition for $y(s)$ to lie in the image of $\laplace{}$ .

Finding the inverse Laplace transform can be laborious. We consider the case of a rational function for an impression of the calculation of the inverse Laplace transform $s$ . By partial fraction decomposition, any rational function is a sum of fractions whose denominators are powers of linear or irreducible quadratic polynomials. We describe how to find the inverse Laplace transform of such a function in $s$ .

Inverse Laplace transforms of rational functions

If $y(s)=\frac{p(s)}{q(s)}$ is a rational function with $p(s)$ and $q(s)$ polynomials, such that $p(s)$ has degree less than the degree of $q(s)$ , then $y(s)$ is the Laplace-transform of a piece-wise continuous function $f(t)$ of exponential order. This inverse Laplace-transform $f(t)$ of $y(s)$ can be found as follows.

First apply partial fraction decomposition to $y(s)$ ; then $f(t)$ is a linear combination of the inverse Laplace transforms of the terms of the partial fraction decomposition.
Each term of the fraction decomposition is of the form $\frac{b}{(s-a)^{m}}$ or $\frac{p\cdot s + q}{((s-a)^2+b^2)^{m}}$ , where $m$ is a natural number and $p$ , $q$ , $a$ , $b$ are real numbers with $b\ne0$ . In the first case, the inverse Laplace transform can be found with the aid of the first line of the table below.
In the second case (where the denominator is the $m$ -th power of an irreducible quadratic polynomial) the calculation uses the frequency shift to reduce to the case $a=0$ and next time scaling to reduce further to the case $b=1$ ; in the second line of the table below, the case $m=1$ is taken care of.
In the remaining cases, the term is of the form $\frac{p\cdot s + q}{(s^2+1)^{m}}$ with $m\ge2$ . Its inverse Laplace transform can be found by use of the for differentiation rule in the frequency domain applied to $t^{k}\cdot\cos(t)$ and $t^{k}\cdot\sin(t)$ for $k\lt m$ .

$\begin{array}{|c|c|}\hline y(s)&\mathcal{L}^{-1}y(t)\\ \hline\dfrac{1}{(s-a)^{m}}& \dfrac{t^{m-1}\ee^{a\cdot t}}{(m-1)!} \\ \hline\dfrac{p\cdot s + q}{s^2+1}& p\cdot\cos(t)+q\cdot\sin(t) \\\hline \end{array}$

Example

The general formula for the case $m=1$ with irreducible quadratic denominator is $\mathcal{L}^{-1}\left(\dfrac{p\cdot s+ q }{(s-a)^2+b^2}\right)=\e^{a\cdot t}\cdot\left(p\cdot \cos(b\cdot t)+ \frac{q+p\cdot a}{b} \cdot\sin(b\cdot t)\right)$

This formula can be found as indicated:

$\begin{array}{rcl}\mathcal{L}^{-1}\left(\dfrac{p\cdot s+ q }{(s-a)^2+b^2}\right)&=& \mathcal{L}^{-1}\left(\dfrac{p\cdot (s-a)+ q+p\cdot a }{(s-a)^2+b^2}\right)\\ &&\phantom{x}\color{blue}{\text{rewritten to argument }s-a }\\ &=& \e^{a\cdot t}\cdot \mathcal{L}^{-1}\left(\dfrac{p\cdot s+ q+p\cdot a }{s^2+b^2}\right)\\&&\phantom{x}\color{blue}{\text{frequency shift: }\mathcal{L}^{-1}( y(s-a)) =\e^{a\cdot t}\mathcal{L}^{-1}(y(s))} \\ &=& \e^{a\cdot t}\cdot \dfrac{1}{b}\cdot \mathcal{L}^{-1}\left(\dfrac{p\cdot \frac{s}{b}+ \frac{q+p\cdot a}{b} }{\left(\frac{s}{b}\right)^2+1}\right)\\&&\phantom{x}\color{blue}{\text{rewritten to argument }\frac{s}{b} }\\ &=& \e^{a\cdot t}\cdot \left(\mathcal{L}^{-1}\left(\dfrac{p\cdot s+ \frac{q+p\cdot a}{b} }{s^2+1}\right)(b\cdot t)\right)\\&&\phantom{x}\color{blue}{\text{time scaling: }\frac{1}{b}\mathcal{L}^{-1}( y(\frac{s}{b})) (t)=\mathcal{L}^{-1}(y(s))(b\cdot t) }\\ &=& \e^{a\cdot t}\cdot\left(p\cdot \cos(b\cdot t)+ \frac{q+p\cdot a}{b} \cdot\sin(b\cdot t)\right)\\&&\phantom{x}\color{blue}{\text{formula from the table} }\end{array}$

With the first formula of the table we can handle all terms of a partial fraction decomposition of $y(s)$ of the form $\frac{b}{(s-a)^m}$ .

The remaining terms have the form $\frac{p\cdot s + q}{((s-a)^2+b^2)^m}$ for real numbers $p$ , $q$ , $a$ , $b$ with $b\ne0$ . As we have seen in the example, with the aid of a frequency shift we can reduce the calculation of the inverse Laplace transform to the case where $a=0$ , and next, by means of a time scaling to the case where $b=1$ . For $m=1$ , the inverse Laplace transform is in the table. For $m=2$ , the inverse Laplace transform of $\frac{p\cdot s + q}{(s^2+1)^m}$ can be found by use of the following four known Laplace transforms:
$\begin{array}{rclcrcl}\laplace{(\cos(t))} &=&\dfrac{s}{s^2+1}&\qquad& \laplace{(\sin(t))} &=&\dfrac{1}{s^2+1} \\ \laplace{(t\cdot \cos(t))} &=&\dfrac{s^2-1}{(s^2+1)^2}&\qquad& \laplace{(t\cdot \sin(t))} &=&\dfrac{2s}{(s^2+1)^2} \end{array}$
We use these formulas, in order to determine, for any real numbers $p$ and $q$ , the inverse Laplace transform of the function $y(s) = \frac{p\cdot s+q}{(s^2+1)^2}$

From $\laplace{(t\cdot \cos(t))} =\frac{s^2-1}{(s^2+1)^2} =\frac{1}{s^2+1} -\frac{2}{(s^2+1)^2}=\laplace{(\sin(t))}-\frac{2}{(s^2+1)^2}$ we deduce, thanks to the linearity of the Laplace transform,
$\frac{2}{(s^2+1)^2} = \laplace{\left( \sin(t)-t\cdot\cos(t) \right)}$
Using this and $\frac{2s}{(s^2+1)^2} =\laplace{(t\cdot \sin(t))}$ we find
$\begin{array}{rcl} \dfrac{p\cdot s+ q }{(s^2+1)^2}&=&\displaystyle\frac{p}{2}\cdot \frac{2s}{(s^2+1)^2}+\frac{q}{2}\cdot \frac{2}{(s^2+1)^2}\\ &=&\displaystyle\frac{p}{2}\cdot\laplace{(t\cdot \sin(t))}+\frac{q}{2}\cdot \left(\laplace{(\sin(t))} -\laplace{(t\cdot \cos(t))} \right)\\ &=&\displaystyle\laplace{\left(\left(\frac{p}{2}\cdot t+\frac{q}{2}\right)\cdot \sin(t)-\frac{q}{2}\cdot t\cdot \cos(t) \right)}\\ \end{array}$
This shows that
$\mathcal{L}^{-1}\left(\dfrac{p\cdot s+ q }{(s^2+1)^2}\right) =\frac{p\cdot t+q}{2}\cdot \sin(t)-\frac{q\cdot t}{2}\cdot\cos(t)$

For $m\gt 2$ we can proceed similarly, first determining $\laplace{\left(t^{m-1}\cdot \cos(t)\right)}$ and $\laplace{\left(t^{m-1}\cdot \sin(t)\right)}$ by means of the rule for the derivative in the frequency domain.

Later we will see that, in order to calculate the inverse Laplace transform of a product of two functions in the frequency domain, a so-called convolution can be used. This provides another method for determining $\mathcal{L}^{-1}\left(\frac{p\cdot s + q}{(s^2+1)^m}\right)$ .

The fact that $\mathcal{L}^{-1}y$ is the sum of the inverse Laplace transforms of the terms of the partial fraction decomposition, of course, follows from the linearity of $\mathcal{L}^{-1}$ . The remaining steps in the calculation are all based on the properties of the Laplace transform dealt with before.

Compute the inverse Laplace transform of the function
$y(s) = {{1}\over{s^3-s^2+4\cdot s-4}}$

$\mathcal{L}^{-1}(y)(t) =$ ${{-\sin \left(2\cdot t\right)-2\cdot \cos \left(2\cdot t\right)+2\cdot \euler^{t}}\over{10}}$

In order to find the solution, we first determine the partial fraction decomposition of $y(s)$ :
$\begin{array}{rcl} \displaystyle y(s) &=&\displaystyle {{1}\over{s^3-s^2+4\cdot s-4}}\\ &&\phantom{xx}\color{blue}{\text{function rule for }y}\\ &=&\displaystyle {{1}\over{\left(s-1\right)\cdot \left(s^2+4\right)}}\\ &&\phantom{xx}\color{blue}{\text{denominator factored into irreducible factors}}\\ &=&\displaystyle {{1}\over{5\cdot \left(s-1\right)}}-{{s+1}\over{5\cdot \left(s^2+4\right)}}\\ &&\phantom{xx}\color{blue}{\text{partial fraction decomposition completed}}\\ \end{array}$
We now use linearity of $\mathcal{L}^{-1}$ :
$\begin{array}{rcl} \mathcal{L}^{-1}(y) &=&\displaystyle \mathcal{L}^{-1}\left({{1}\over{5\cdot \left(s-1\right)}}-{{s+1}\over{5\cdot \left(s^2+4\right)}}\right)\\ &&\phantom{xx}\color{blue}{\text{above function rule for }y}\\ &=&\displaystyle {{1}\over{5}}\cdot \mathcal{L}^{-1} \left({{1}\over{s-1}}\right)-{{1}\over{5}}\cdot \mathcal{L}^{-1} \left({{s}\over{s^2+4}}\right)-{{1}\over{10}}\cdot\mathcal{L}^{-1} \left({{2}\over{s^2+4}}\right) \\ &&\phantom{xx}\color{blue}{\text{linearity of the inverse Laplace transform}}\\ &=&\displaystyle {{1}\over{5}}\cdot \left(\euler^{t}\right)-{{1}\over{5}}\cdot \left(\cos \left(2\cdot t\right)\right)-{{1}\over{10}}\cdot \left(\sin \left(2\cdot t\right)\right) \\ &&\phantom{xx}\color{blue}{\text{calculation rule $\mathcal{L}^{-1}\left(\dfrac{p\cdot s+ q }{(s-a)^2+b^2}\right)=\e^{a\cdot t}\cdot\left(p\cdot \cos(b\cdot t)+ \frac{q+p\cdot a}{b} \cdot\sin(b\cdot t)\right)$}}\\ &=& \displaystyle {{-\sin \left(2\cdot t\right)-2\cdot \cos \left(2\cdot t\right)+2\cdot \euler^{t}}\over{10}} \end{array}$

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