The Laplace transform has been in recent years of great importance in engineering studies, mathematics, physics, among other scientific areas, as well as being of great interest in theory, provides a simple way to solve problems that come from science and engineering.
Originally the Laplace transform was presented by Pierre-Simón Laplace in his study on probability theory and was initially treated as a mathematical object of purely theoretical interest.
Current applications arise when various mathematicians tried to give a formal justification to the “operational rules” used by Heaviside in the study of equations of electromagnetic theory.
Let f be a function defined for t ≥ 0. The Laplace transform is defined as follows:
The Laplace transform is said to exist if the previous integral converges, otherwise the Laplace transform is said to not exist.
In general, lowercase letters are used to denote the function to be transformed, and the capital letter corresponds to its transform. In this way we will have:
Consider the constant function f (t) = 1. We have that its transform is:
Whenever the integral converges, that is, whenever s> 0. Otherwise, s <0, the integral diverges.
Let g (t) = t. Its Laplace transform is given by
By integrating by parts and knowing that te -st tends to 0 when t tends to infinity and s> 0, together with the previous example we have:
The transform may or may not exist, for example for the function f (t) = 1 / t the integral that defines its Laplace transform does not converge and therefore its transform does not exist.
Sufficient conditions to guarantee that the Laplace transform of a function f exists are that f is piecewise continuous for t ≥ 0 and is of exponential order.
A function is said to be piecewise continuous for t ≥ 0, when for any interval [a, b] with a> 0, there is a finite number of points t k, where f has discontinuities and is continuous in each subinterval [t k-1 , t k ].
On the other hand, a function is said to be of exponential order c if there are real constants M> 0, c and T> 0 such that:
As examples we have that f (t) = t 2 is of exponential order, since | t 2 | <e 3t for all t> 0.
In a formal way we have the following theorem
Theorem (Sufficient conditions for existence)
If f is a piecewise continuous function for t> 0 and of exponential order c, then the Laplace transform exists for s> c.
It is important to note that this is a sufficiency condition, that is, it could be the case that there is a function that does not meet these conditions and even so its Laplace transform exists.
An example of this is the function f (t) = t -1/2 which is not piecewise continuous for t ≥ 0 but its Laplace transform exists.
Laplace transform of some basic functions
The following table shows the Laplace transforms of the most common functions.
The Laplace transform owes its name to Pierre-Simon Laplace, a French mathematician and theoretical astronomer who was born in 1749 and died in 1827. His fame was such that he was known as the Newton of France.
In 1744 Leonard Euler devoted his studies to integrals with the form
as solutions of ordinary differential equations, but he quickly abandoned this investigation. Later, Joseph Louis Lagrange, who greatly admired Euler, also investigated these types of integrals and related them to probability theory.
In 1782 Laplace began to study these integrals as solutions to differential equations and according to historians, in 1785 he decided to reformulate the problem, which later gave rise to the Laplace transforms as they are understood today.
Having been introduced into the field of probability theory, it was of little interest to scientists at the time and was only seen as a mathematical object of only theoretical interest.
It was in the mid-nineteenth century when the English engineer Oliver Heaviside discovered that differential operators can be treated as algebraic variables, thus giving Laplace transforms their modern application.
Oliver Heaviside was an English physicist, electrical engineer and mathematician who was born in London in 1850 and died in 1925. While trying to solve differential equations problems applied to the theory of vibrations and using Laplace’s studies, he began to shape the Modern applications of Laplace transforms.
The results presented by Heaviside quickly spread throughout the scientific community of the time, but as his work was not rigorous he was quickly criticized by the more traditional mathematicians.
However, the usefulness of Heaviside’s work in solving equations in physics made his methods popular with physicists and engineers.
Despite these setbacks and after some decades of failed attempts, at the beginning of the 20th century a rigorous justification could be given to the operational rules given by Heaviside.
These attempts bore fruit thanks to the efforts of various mathematicians such as Bromwich, Carson, van der Pol, among others.
Among the properties of the Laplace transform, the following stand out:
Let c1 and c2 be constants and f (t) and g (t) are functions whose Laplace transforms are F (s) and G (s) respectively, then we have:
Due to this property the Laplace transform is said to be a linear operator.
First translation theorem
If it happens that:
And ‘a’ is any real number, so:
Since the Laplace transform of cos (2t) = s / (s ^ 2 + 4) then:
Second translation theorem
If f (t) = t ^ 3, then F (s) = 6 / s ^ 4. And therefore the transform of
is G (s) = 6e -2s / s ^ 4
And ‘a’ is a nonzero real, we have to
Since the transform of f (t) = sin (t) is F (s) = 1 / (s ^ 2 + 1) we have
Laplace transform of derivatives
If f, f ‘, f’ ‘, …, f (n) are continuous for t ≥ 0 and are of exponential order and f (n) (t) is piecewise continuous for t ≥ 0, then
Laplace transform of integrals
Multiplication by t n
If we have to
Division by t
If we have to
Let f be a periodic function with period T> 0, that is f (t + T) = f (t), then
Behavior of F (s) as s tends to infinity
If f is continuous in parts and of exponential order and
When we apply the Laplace transform to a function f (t) we obtain F (s), which represents said transform. In the same way we can say that f (t) is the inverse Laplace transform of F (s) and is written as
We know that the Laplace transforms of f (t) = 1 and g (t) = t are F (s) = 1 / s and G (s) = 1 / s 2 respectively, therefore we have that
Some common inverse Laplace transforms are as follows
Furthermore, the inverse Laplace transform is linear, that is, it is true that
To solve this exercise we must match the function F (s) with one of the previous table. In this case if we take an + 1 = 5 and using the linearity property of the inverse transform, we multiply and divide by 4! Getting
For the second inverse transform we apply partial fractions to rewrite the function F (s) and then the property of linearity, obtaining
As we can see from these examples, it is common that the function F (s) that is evaluated does not agree precisely with any of the functions given in the table. For these cases, as can be seen, it is enough to rewrite the function until it reaches the appropriate form.
Applications of the Laplace transform
The main application of Laplace transforms is to solve differential equations.
Using the property of the transform of a derivative it is clear that
Y of the n-1 derivatives evaluated at t = 0.
This property makes the transform very useful for solving initial value problems where differential equations with constant coefficients are involved.
The following examples show how to use the Laplace transform to solve differential equations.
Given the following initial value problem
Use the Laplace transform to find the solution.
We apply the Laplace transform to each member of the differential equation
By the property of the transform of a derivative we have
By developing all the expression and clearing Y (s) we are left
Using partial fractions to rewrite the right side of the equation we get
Finally, our goal is to find a function y (t) that satisfies the differential equation. Using the inverse Laplace transform gives us the result
As in the previous case, we apply the transform on both sides of the equation and separate term by term.
In this way we have as a result
Substituting with the given initial values and solving for Y (s)
Using simple fractions we can rewrite the equation as follows
And applying the inverse Laplace transform gives us the result
In these examples, you might wrongly conclude that this method is not much better than traditional methods for solving differential equations.
The advantages of the Laplace transform is that you don’t need to use parameter variation or worry about the various cases of the indeterminate coefficient method.
Also when solving initial value problems by this method, from the beginning we use the initial conditions, so it is not necessary to perform other calculations to find the particular solution.
Systems of differential equations
The Laplace transform can also be used to find solutions to simultaneous ordinary differential equations, as the following example shows.
With the initial conditions x (0) = 8 and y (0) = 3.
If we have to
Solving gives us as a result
And applying the inverse Laplace transform we have
Mechanics and electrical circuits
The Laplace transform is of great importance in physics, it mainly has applications for mechanics and electrical circuits.
A simple electrical circuit is made up of the following elements
A switch, a battery or source, an inductor, a resistor, and a capacitor. When the switch is closed, an electric current is produced which is denoted by i (t). The charge on the capacitor is denoted by q (t).
By Kirchhoff’s second law, the voltage produced by source E in the closed circuit has to be equal to the sum of each of the voltage drops.
The electric current i (t) is related to the charge q (t) on the capacitor by i = dq / dt. On the other hand, the voltage drop in each of the elements is defined as follows:
The voltage drop across a resistor is iR = R (dq / dt)
The voltage drop across an inductor is L (di / dt) = L (d 2 q / dt 2 )
The voltage drop across a capacitor is q / C
With these data and applying Kirchhoff’s second law to the simple closed circuit, a second-order differential equation is obtained that describes the system and allows us to determine the value of q (t).
An inductor, a capacitor, and a resistor are connected to a battery E, as shown in the figure. The inductor is 2 henries, the capacitor is 0.02 farads and the resistance is 16 ohms. At time t = 0 the circuit is closed. Find the charge and the current at any time t> 0 if E = 300 volts.
We have that the differential equation that describes this circuit is the following
Where the initial conditions are q (0) = 0, i (0) = 0 = q ‘(0).
Applying the Laplace transform we get that
And solving for Q (t)
Then, applying the inverse Laplace transform we have
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