Laplace Transforms - Rhea

Sometimes an ODE can be really complex and hard to solve by any basic methods we have looked at in previous tutorials. For example, it may involve exponential functions, trigonometric functions, Heaviside functions, and anything else you can imagine. Hence, a famous French physicist Pierre-Simon Laplace found a transform method, which converts the functions in "time-domain" to "complex number-domain", to overcome the problem. It is great for transforming the calculation of differentiation and integration to the simple algebraic calculation.

7.1 Concept

Here is a basic concept map from my course note when first learning Laplace transform:

$ \begin{bmatrix} ODE & in & y(t) \\ "Time-domain" \end{bmatrix} → \begin{bmatrix} Laplace & Transform \\ L=[y(t)] \end{bmatrix} → \begin{bmatrix} Algebraic & equation & in & Y(s) \\ "Complex & number-domain" \end{bmatrix} $;

$ \begin{bmatrix} Algebraic & equation & in & Y(s) \\ "Complex & number-domain" \end{bmatrix} → \begin{bmatrix} Inverse & Laplace & Transform \\ L^{-1}[Y(s)] \end{bmatrix} → \begin{bmatrix} ODE & in & y(t) \\ "Time-domain" \end{bmatrix} $.

That means for some complicated ODEs, Laplace Transform is used to convert them to some algebraic equations. Then inverse Laplace Transform is used to convert the solution back.

Laplace Transform is defined as $ Y(s)=L[y(t)]=\int_{0}^{\infty} y(t) e^{-st} dt $， which is also time function $ y(t) $ expressed in the "complex frequency" domain.

Let's do a simple example. Consider the time function $ y(t)=e^{2t} $. Follow the definition above, we have $ Y(s)=L[e^{2t}]=\int_{0}^{\infty} e^{2t} e^{-st} dt=\int_{0}^{\infty} e^{(2-s)t} dt=\lim_{b \to \infty} {\int_{0}^{b} e^{2-s}t dt}=\lim_{b \to \infty} {\frac{e^{2-s}t}{2-s} |_{0}^{b}}=\lim_{b \to \infty} {\frac{1}{2-s} (e^{(2-s)b}-e^0)} $

$ =\begin{cases} \frac{1}{s-2}, & if & s＞2 \\ undefined, & if & s≤2 \end{cases} $.

If doing Laplace Transform to a derivative $ \frac{dy}{dt} $:

$ L[\frac{dy}{dt}]=\int_{0}^{\infty} \frac{dy}{dt} e^{-st} dt $, by the definition of Laplace Transform,

$ =[ye^{-st} - \int (-s e^{-st} y dt)]_{0}^{\infty} $, integration by part,

$ =[ye^{-st} + s \int y e^{-st} dt]_{0}^{\infty} $,

$ =[ye^{-st}]_{0}^{\infty} + sY(s), again by the definition of Laplace Transform, <math>=sY(s) - y(0) (= sL[y] - y(0)) $,

This includes the initial condition $ t=0 $, which is also an important difference with Fourier Transform.

7.2 Linearity of Laplace Transform

7.3 Heaviside Unit Step Function for Discontinuous Functions

7.4 Exercises

7.5 References

Institute of Natural and Mathematical Science, Massey University. (2017). 160.204 Differential Equations I: Course materials. Auckland, New Zealand.