### 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.

Piece-wise continuous functions of exponential order 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 Laplace transform is injective on the space of the above functions:

Uniqueness theorem

Let #c# be a positive real number and #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#.

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#.

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

In order to find the solution, we first determine the partial fraction decomposition of #y(s)#:

\[\begin{array}{rcl}

\displaystyle y(s) &=&\displaystyle {{s^2+2\cdot s+3}\over{s^4+4\cdot s^3+11\cdot s^2+14\cdot s+10}}\\

&&\phantom{xx}\color{blue}{\text{function rule for }y}\\

&=&\displaystyle {{s^2+2\cdot s+3}\over{\left(s^2+2\cdot s+2\right)\cdot \left(s^2+2\cdot s+5\right)}}\\

&&\phantom{xx}\color{blue}{\text{denominator factored into irreducible factors}}\\

&=&\displaystyle {{2}\over{3\cdot \left(s^2+2\cdot s+5\right)}}+{{1}\over{3\cdot \left(s^2+2\cdot s+2\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({{2}\over{3\cdot \left(s^2+2\cdot s+5\right)}}+{{1}\over{3\cdot \left(s^2+2\cdot s+2\right)}}\right)\\

&&\phantom{xx}\color{blue}{\text{above function rule for }y}\\

&=&\displaystyle {{1}\over{3}}\cdot \mathcal{L}^{-1} \left({{2}\over{\left(s+1\right)^2+4}}\right)+{{1}\over{3}}\cdot \mathcal{L}^{-1} \left({{1}\over{\left(s+1\right)^2+1}}\right) \\

&&\phantom{xx}\color{blue}{\text{linearity of the inverse Laplace transform}}\\

&=&\displaystyle {{1}\over{3}}\cdot \left(\euler^ {- t }\cdot \sin \left(2\cdot t\right)\right)+{{1}\over{3}}\cdot \left(\euler^ {- t }\cdot \sin \left(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 {{\euler^ {- t }\cdot \left(\sin \left(2\cdot t\right)+\sin \left(t\right)\right)}\over{3}}

\end{array}\]

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