Preface

This tutorial was made solely for the purpose of education and it was designed for students taking Applied Math 0330. It is primarily for students who have very little experience or have never used Mathematica before and would like to learn more of the basics for this computer algebra system. As a friendly reminder, don't forget to clear variables in use and/or the kernel.

Finally, the commands in this tutorial are all written in bold black font, while Mathematica output is in normal font. This means that you can copy and paste all commands into Mathematica, change the parameters and run them. You, as the user, are free to use the scripts for your needs to learn the Mathematica program, and have the right to distribute this tutorial and refer to this tutorial as long as this tutorial is accredited appropriately.

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Inverse Laplace transform

Let F(λ) be the Laplace transform of some function. This means that there exists a function f(t), defined on half-line, such that

\[ F(\lambda ) = f^L (\lambda ) = \left[ {\cal L} f \right] \left(\lambda \right) = \int_0^{\infty} f(t)\, e^{-\lambda\,t} \,{\text d}t . \]
The objective of this section is to determine f(t) when only F(λ) is given. It is known that the inverse Laplace transform can be found using the Bromwich integral:
\begin{align*} f(t) &= \frac{1}{2\pi{\bf j}} \, \int_{s-{\bf j}\infty}^{s+{\bf j}\infty} \, F(\lambda )\, e^{\lambda \,t} \,{\text d} \lambda \\ &= \frac{f(t+0) + f(t-0)}{2} = \frac{e^{st}}{2\pi} \left[ \int_0^{\infty} e^{{\bf j}\xi t} \,F(s+ {\bf j}\xi )\,{\text d} \xi + \int_0^{\infty} e^{{\bf j}\xi t} \,F(s- {\bf j}\xi )\,{\text d} \xi \right] , \end{align*}
where s is the abscissa of convergence for the function f(t). The above indefinite Bromwich integral is hard to use in practical applications. Instead, some kind of approximation formulas are available (for instance, fast Fourier transform). We demonstrate how to find the inverse Laplace transform using three approaches: partial fraction decomposition, the convolution integral, and the residue method.

We observe that application of the Laplace transformation to ordinary differential equations lead to functions F(λ) of a complex variable λ that are represented either by the ratio of two polynomials or by such ratio times the exponential function:

\[ F(\lambda ) = \frac{P(\lambda )}{Q(\lambda )} \qquad\mbox{or} \qquad F(\lambda ) = \frac{P(\lambda )}{Q(\lambda )} \, e^{-a\lambda} , \]
where \( P(\lambda ) = p_n \lambda^n + p_{n-1} \lambda^{n-1} + \cdots + p_0 \quad\mbox{and} \quad Q(\lambda ) = q_m \lambda^m + q_{m-1} \lambda^{m-1} + \cdots + q_0 . \) In applications, the degree m of the denominator always exceeds n, the degree the numerator. Later, we will consider a special case when m = n.

It should be noted that the latter case (when ratio of two polynomials is multiplied by the exponential term) can be reduced to the former one (without exponential multiple) with the aid of the shift rule:

\[ {\cal L}^{-1} \left[ \frac{P(\lambda )}{Q(\lambda )} \, e^{-a\lambda} \right] = f(t-a)\, H(t-a) , \qquad \mbox{where} \quad f(t) = {\cal L}^{-1} \left[ \frac{P(\lambda )}{Q(\lambda )} \right] . \]
Therefore, we concentrate our attention on determination of the inverse Laplace transform of the rational functions
\[ F(\lambda ) = \frac{P(\lambda )}{Q(\lambda )} . \]
The first observation is that the Bromwich integral is zero for negative t. Therefore, the inverse Laplace transform, if it exists, should be multiplied by the Heaviside function.

The second general observation is that the Bromwich integral restores the function in such a way that at the point of discontinuity t0 of the function f(t), the Bromwich integral assigns its mean value:

\[ {\cal L}^{-1} \left[ F(\lambda ) \right] \left( t_0 \right) = \frac{f(t_0 +0) + f(t_0 -0)}{2} = \frac{1}{2}\,\lim_{\epsilon \to 0, \ \epsilon \ge 0} \, \left[ f(t_0 + \epsilon ) + f(t_0 - \epsilon ) \right] . \]

One of the methods to find inverse Laplace transform is to use partial fraction decomposition and then apply inverse Laplace transform to each term (usually using a table of given Laplace transforms). However, Mathematica allows one to find the inverse Laplace transform straight forward:

InverseLaplaceTransform[4/s/(4 + s^2), s, t]
Out[8]= 4 (1/4 - 1/4 Cos[2 t])
Simplify[%]
Out[9]= 2 Sin[t]^2
Apart[4/s/(4 + s^2), s]
Out[10]= 1/s - s/(4 + s^2)
LaplaceTransform[2 Sin[t]^2, t, lambda]
Out[11]= 4/(lambda (4 + lambda^2))

The general approach:

InverseLaplaceTransform[%, lambda, t]
Out[7]= {{y[t] -> Cos[2 t] + Sin[2 t]}}

 

 

 

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