Difference between revisions of "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,

$ =sY(s) - y(0) (= sL[y] - y(0)) $,

Now consider the second order derivative $ \frac{d^2y}{dt^2} $,

$ L[\frac{d^2y}{dt^2}]=L[\frac{d}{dt} (\frac{dy}{dt})]=s L[\frac{dy}{dt}] - \frac{dy}{dx}(0) = s(sL[y]-y(0)) - \frac{dy}{dt}(0) = s^2 L[y] - sy(0) - \frac{dy}{dt}(0) $.

Here we can see the regular pattern and infer the general form $ L[\frac{d^ny}{dt^n}]=s^n Y(s) - s^{n-1}y(0) - ... - \frac{d^{n-1}y}{dt^{n-1}(0)} $.

Click here for a frequently used Laplace Transform table.

Inverse Laplace Transform converts the results back to a solution to ODE. The calculation may involve a method of partial fraction. In other words, when we solved the converted algebraic function (having $ Y(s) $), we need to convert it back, still referring to the Laplace Transform table, to have $ L^{-1}[Y(s)]=y(t) $. Refer here to see how to do partial fraction.

7.2 Linearity of Laplace Transform

By the definition of linear transformation, if a function $ f(x) $ is linear, then $ f(a+bx)=f(a)+bf(x) $. Test the linearity for the definition of Laplace Transform $ L[y(t)]=\int_{0}^{\infty} y(t) e^{-st} dt $ by this property,

$ L[f+g]=\int_{0}^{\infty} (f(t)+g(t)) e^{-st} dt=\int_{0}^{\infty} f(t) e^{-st} dt + \int_{0}^{\infty} g(t) e^{-st} dt = L[f]+L[g] $,

$ L[cf]=\int_{0}^{\infty} cf(t) e^{-st} dt = c \int_{0}^{\infty} f(t) e^{-st} dt = cL[f] $.

Therefore, it is linear.

We can do the same test for the inverse transform, which is also linear:

$ L^{-1} [X(s)+Y(s)]=L^{-1}[X(s)] + L^{-1}[Y(s)] $,

$ L^{-1} [cX(s)]=cL^{-1} [Y(s)] $.

7.3 Heaviside Unit Step Function

Heaviside unit step function is a discontinuous function named by a mathematician Oliver Heaviside. This function has a value $ 1 $ for positive arguments and a value $ 0 $ for negative arguments. In previous tutorials, all we solved are continuous ODEs, while Laplace Transform is able to convert discontinuous functions into continuous ones for further operations. Heaviside unit step function is a typical example of discontinuous functions: $ H(t-a)=\begin{cases} 0, & if & t＜a \\ 1, & if & t≥a \end{cases} $.

Refer to the above frequent Laplace Transform table for the transformation of Heaviside step function.

7.4 Example

Find the Laplace transform of the function $ y(t)=\begin{cases} 3t, & 0≤t＜1, \\ 1, & 1≤t＜2 \\ e^t, & 2≤t \end{cases} $.

In order to move the given function closer to the Heaviside style, we make a slight change:

$ y(t)=\begin{cases} 3t, & 0≤t＜1, \\ 3t+(1-3t), & 1≤t＜2 \\ 3t+(1-3t)+(e^t-1), & 2≤t \end{cases} $.

Add all them up to have $ y(t)=3t+H(t-1)(1-3t)+H(t-2)(e^t-1) $

$ =3t(1-H(t-1))+(H(t-1)-H(t-2))+e^t H(t-2) = 3t+(1-3t)H(t-1) + (e^t-1)H(t-2) $,

$ =\frac{3}{s^2}-e^{-s}(\frac{3}{s^2}+\frac{2}{s})+\frac{e^{2(1-s)}}{s-1}-\frac{2e^{-5}}{s} $, by the Laplace Transform table. Here the discontinuous function is converted to a continuous algebraic function, and we are ready for further analysis.

7.5 References

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