Differential Equations Z Transform

[Chiquita Palafox]

In mathematics, the Laplace transform is a powerful integral transform used to switch a function from the time domain to the s-domain. The Laplace transform can be used in some cases to solve linear differential equations with given initial conditions.

Laplace transform is named in honour of the great French mathematician, Pierre Simon De Laplace (1749-1827). Like all transforms, the Laplace transform changes one signal into another according to some fixed set of rules or equations. The best way to convert differential equations into algebraic equations is the use of Laplace transformation.

Laplace transformation plays a major role in control system engineering. To analyze the control system, Laplace transforms of different functions have to be carried out. Both the properties of the Laplace transform and the inverse Laplace transformation are used in analyzing the dynamic control system. In this article, we will discuss in detail the definition of Laplace transform, its formula, properties, Laplace transform table and its applications in a detailed way.

A function is said to be a piecewise continuous function if it has a finite number of breaks and it does not blow up to infinity anywhere. Let us assume that the function f(t) is a piecewise continuous function, then f(t) is defined using the Laplace transform. The Laplace transform of a function is represented by Lf(t) or F(s). Laplace transform helps to solve the differential equations, where it reduces the differential equation into an algebraic problem.

The Laplace transform is a well established mathematical technique for solving a differential equation. Many mathematical problems are solved using transformations. The idea is to transform the problem into another problem that is easier to solve. On the other side, the inverse transform is helpful to calculate the solution to the given problem.

The Laplace transform can also be defined as bilateral Laplace transform. This is also known as two-sided Laplace transform, which can be performed by extending the limits of integration to be the entire real axis. Hence, the common unilateral Laplace transform becomes a special case of Bilateral Laplace transform, where the function definition is transformed is multiplied by the Heaviside step function.

In pure and applied probability theory, the Laplace transform is defined as the expected value. If X is the random variable with probability density function, say f, then the Laplace transform of f is given as the expectation of:

Before proceeding into differential equations we will need one more formula. We will need to know how to take the Laplace transform of a derivative. First recall that \(f^(n)\) denotes the \(n^\mboxth\) derivative of the function \(f\). We now have the following fact.

That was a fair amount of work for a problem that probably could have been solved much quicker using the techniques from the previous chapter. The point of this problem however, was to show how we would use Laplace transforms to solve an IVP.

There are a couple of things to note here about using Laplace transforms to solve an IVP. First, using Laplace transforms reduces a differential equation down to an algebra problem. In the case of the last example the algebra was probably more complicated than the straight forward approach from the last chapter. However, in later problems this will be reversed. The algebra, while still very messy, will often be easier than a straight forward approach.

Second, unlike the approach in the last chapter, we did not need to first find a general solution, differentiate this, plug in the initial conditions and then solve for the constants to get the solution. With Laplace transforms, the initial conditions are applied during the first step and at the end we get the actual solution instead of a general solution.

Notice that we also had to factor a 2 out of the denominator of the first term and fix up the numerator of the last term in order to get them to match up to the correct entries in our table of transforms.

The first thing that we will need to do here is to take care of the fact that initial conditions are not at \(t = 0\). The only way that we can take the Laplace transform of the derivatives is to have the initial conditions at \(t = 0\).

This means that we will need to formulate the IVP in such a way that the initial conditions are at \(t = 0\). This is actually fairly simple to do, however we will need to do a change of variable to make it work. We are going to define

Note that unlike the previous examples we did not completely combine all the terms this time. In all the previous examples we did this because the denominator of one of the terms was the common denominator for all the terms. Therefore, upon combining, all we did was make the numerator a little messier and reduced the number of partial fractions required down from two to one. Note that all the terms in this transform that had only powers of \(s\) in the denominator were combined for exactly this reason.

The examples worked in this section would have been just as easy, if not easier, if we had used techniques from the previous chapter. They were worked here using Laplace transforms to illustrate the technique and method.

We are going to be given a transform, \(F(s)\), and ask what function (or functions) did we have originally. As you will see this can be a more complicated and lengthy process than taking transforms. In these cases we say that we are finding the Inverse Laplace Transform of \(F(s)\) and use the following notation.

The denominator of the third term appears to be #3 in the table with \(n = 4\). The numerator however, is not correct for this. There is currently a 7 in the numerator and we need a \(4! = 24\) in the numerator. This is very easy to fix. Whenever a numerator is off by a multiplicative constant, as in this case, all we need to do is put the constant that we need in the numerator. We will just need to remember to take it back out by dividing by the same constant.

So, probably the best way to identify the transform is by looking at the denominator. If there is more than one possibility use the numerator to identify the correct one. Fix up the numerator if needed to get it into the form needed for the inverse transform process. Finally, take the inverse transform.

Recall that in completing the square you take half the coefficient of the \(s\), square this, and then add and subtract the result to the polynomial. After doing this the first three terms should factor as a perfect square.

We needed an \(s + 4\) in the numerator, so we put that in. We just needed to make sure and take the 4 back out by subtracting it back out. Also, because of the 3 multiplying the \(s\) we needed to do all this inside a set of parenthesis. Then we partially multiplied the 3 through the second term and combined the constants. With the transform in this form, we can break it up into two transforms each of which are in the tables and so we can do inverse transforms on them,

The last part of this example needed partial fractions to get the inverse transform. When we finally get back to differential equations and we start using Laplace transforms to solve them, you will quickly come to understand that partial fractions are a fact of life in these problems. Almost every problem will require partial fractions to one degree or another.

In order for these two to be equal the coefficients of the \(s^2\), \(s\) and the constants must all be equal. So, setting coefficients equal gives the following system of equations that can be solved.

Here is a simplified example.
I have two structs VariableA & VariableB, and I use them to generate A, B.
Finally, I want to use A, B to generate the ODE function.
A has a main differential equation, and I want to add B into A dynamically.
It means that there can be 0 or 1 or 2 or more B components in A.
I have no idea where to start with.
Can I accomplish the idea? Is there any suggestion?

You can then define new variables that are the combined expressions and use those in the derivative equations. This library is under continued development (current day: 4/27/2019) and future features will make it easier to combine pre-built differential equation models to more easily build large systems of differential equations.

I'm learning about laplace tranform method to solve lineair differential equations but i'm wondering if laplace transformations can be used to solve every linear differential equations there is.Or are there some limitations?

One can say without any problem that almost no equation can be solved using Laplace transforms. There are two main reasons for this. One is that in general Laplace transforms cannot be computed, unless for functions with some prescribed growth. Otherwise nothing can be done. Even worse, one often writes a symbol for the Laplace transform of the solution without knowing whether the solution has that prescribed growth (still one can say that if at the hand we find a solution of which one can compute the Laplace transform, the reasoning is correct).The other main reason is of course that the method expects that we are able to compute explicitly, and like finding primitives this is more art than mathematics.

But I'm really not sure if I'm on the right track. I'm not terribly concerned with solving the problem completely, but rather I'm held up on the syntax of converting the partial differential equation to "Mathematica form", and how to proceed with Mathematica.

In the second term, we actually can interchange the order of integration and differentiation to see that it's just D[LaplaceTransform[T[x, t], t, s], x, 2]. Therefore, we replace the transformed function with a dummy:

A Fourier Transform is a mathematical operation that decomposes a function into its constituent frequencies. In other words, it converts a function from its original domain (such as time or position) to its frequency domain.

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