Laplace transforms are important tools for us to use when solving linear differential equations. The Laplace transform is a relation of the form

\mathcal{L}{f(x)} = \bar{f}(t) = \int limits_{0}^{\infty} e^{-tx} f(x) dx

As we can see, the Laplace transformation converts the function f into another function \bar{f} , which is called the Laplace transform of f . The general pattern for using Laplace transformations to solve linear differential equations is as follows: first, apply the Laplace transform to both sides of the differential equation to turn a problem to an algebraic equation for \bar{f} ; second, solve this algebraic equation to find \bar{f} ; and finally, recover the solution f(x) from its Laplace transform \bar{f} . This last step is known as the problem of finding the inverse Laplace transform.

In these notes, we’ll review some basic facts about inverse Laplace transforms and investigate their properties. You’ll learn how to calculate inverse Laplace transforms using the fraction decomposition and how to make use of Laplace transforms in differential equations.

Inverse Laplace Transforms

If \bar{f}(t) is the Laplace transform of f(x) , then we can introduce an operation \mathcal{L}^{-1}{\bar{f}(t)} , which returns the function whose Laplace transform is \bar{f}(t) . In other words,

f(x) = \mathcal{L}^{-1}{\bar{f}(t)}

We say that f(x) is the inverse Laplace transform of \bar{f}(t) . If f(x) is continuous in the interval [0, \infty) , the Laplace transform is unique. Thus, the simplest way for finding the inverse Laplace transforms is to recognize them from a table of Laplace transforms. Some elementary Laplace and inverse Laplace transforms are presented below:

\mathcal{L}{ 1 } = \dfrac{1}{t},,\quad t>0,, \quad \mathcal{L}^{-1}{ \dfrac{1}{t} } = 1 \ \ \mathcal{L}{ x^n } = \dfrac{n!}{t^{n+1}},,\quad t>0,, \quad n=0,1,2,\ldots

\mathcal{L}^{-1}{ \dfrac{1}{t^{n+1}} } = \dfrac{x^n}{n!} \ \ \mathcal{L}{ x^n e^{\alpha x} } = \dfrac{n!}{(t - \alpha)^{n+1}},,\quad t > \alpha ,, \quad n=0,1,2,\ldots

\mathcal{L}^{-1}{ \dfrac{1}{(t - \alpha)^{n+1}} } = \dfrac{x^n e^{\alpha x}}{n!} \ \ \mathcal{L}{ sin (\alpha x) } = \dfrac{\alpha}{t^2 + \alpha^2},,\quad t>0,,

\mathcal{L}^{-1}{ \dfrac{1}{t^2 + \alpha^2} } = \dfrac{sin (\alpha x)}{\alpha} \ \ \mathcal{L}{ cos (\alpha x) } = \dfrac{t}{t^2 + \alpha^2},,\quad t>0,,

\mathcal{L}^{-1}{ \dfrac{t}{t^2 + \alpha^2} } = cos (\alpha x) \ \ \mathcal{L}{ sinh (\alpha x) } = \dfrac{\alpha}{t^2 - \alpha^2},,\quad t>|\alpha|,,

\mathcal{L}^{-1}{ \dfrac{1}{t^2 - \alpha^2} } = \dfrac{sinh (\alpha x)}{\alpha} \ \ \mathcal{L}{ cosh (\alpha x) } = \dfrac{t}{t^2 - \alpha^2},,\quad t>|\alpha|,,

\mathcal{L}^{-1}{ \dfrac{t}{t^2 - \alpha^2} } = cosh (\alpha x)

Sometimes, we need to find the inverse Laplace transform for \bar{f}(t) which is not in a recognizable form. In this case, we can try to transform \bar{f}(t) into such a form using simple algebraic manipulations and the property of linearity. In the following notes, we’ll illustrate the use of the partial fraction decomposition for the derivation of inverse Laplace transforms.

Finding Inverse Laplace Transforms via the Partial Fraction Decomposition

Suppose \bar{f}(t) = P(t)/Q(t) , where P(t) and Q(t) are some polynomials, and the degree of P is less than the degree of Q . If the denominator Q(t) can be presented in the form Q(t) = (t-q_1) (t-q_2) \ldots (t-q_n) , where q_k are all distinct constants, then the expansion for \bar{f}(t) is as follows:

\bar{f}(t) = \dfrac{A_1}{t-q_1} + \dfrac{A_2}{t-q_2} + \cdots + \dfrac{A_n}{t-q_n}

Here, A_k are constants which can be determined by substitution. If one of the roots, q_k , is repeated

\bar{f}(t) = \dfrac{A_{k_1}}{t-q_k} + \dfrac{A_{k_2}}{(t-q_k)^2} + \cdots + \dfrac{A_{k_m}}{(t-q_k)^m}

To each factor of Q(t) of the form (t^2 + bt + c)^p , we have to assign a sum of p fractions of the form

\dfrac{B_1 t + C_1}{t^2 + bt + c} + \dfrac{B_2 t + C_2}{(t^2 + bt + c)^2} + \cdots + \dfrac{B_p t + C_p}{(t^2 + bt + c)^p}

The following examples illustrate the practical use of the partial fraction decomposition.

Example 1

Let’s find the inverse Laplace transform of the following function:

\bar{f}(t) = \dfrac{t+3}{(t-2)(t+1)}

Using the partial fraction decomposition, we set

\dfrac{t+3}{(t-2)(t+1)} = \dfrac{A_1}{t-2} + \dfrac{A_2}{t+1}

This equality holds under condition

t+3 = A_1 (t+1) + A_2 (t-2) = (A_1 + A_2)t + (A_1 - 2 A_2)

Thus, constants A_1 and A_2 are determined from the following system:

\begin{array}{l} A_1 + A_2 = 1 \ A_1 - 2A_2 = 3 \end{array}

We can easily find a solution: A_1 =5/3 , and A_2 = - 2/3 . Thus, using partial fractions, we have obtained that

\bar{f}(t) = \dfrac{t+3}{(t-2)(t+1)} = \dfrac{5}{3(t-2)} - \dfrac{2}{3(t+1)}

The inverse Laplace transform of the first term here is known:

\mathcal{L}^{-1}\dfrac{5}{3(t-2)} = \dfrac{5}{3},\mathcal{L}^{-1}\dfrac{1}{t-2} = \dfrac{5}{3}, e^{2x}

For the second term, we have:

\mathcal{L}^{-1}-,\dfrac{2}{3(t+1)} = -,\dfrac{2}{3} ,\mathcal{L}^{-1} \dfrac{1}{t+1} = -,\dfrac{2}{3}, e^{-x}

Due to the linearity of inverse Laplace transforms, we conclude that

\mathcal{L}^{-1}\dfrac{t+3}{(t-2)(t+1)} = \dfrac{5}{3}, e^{2x} -,\dfrac{2}{3}, e^{-x}

Example 2

Let’s find \mathcal{L}^{-1}{ 1/t(t-1)^3} . Due to the partial fraction decomposition, \bar{f}(t) can be presented in the form

\bar{f}(t) = \dfrac{1}{t(t-1)^3} = \dfrac{A}{t} + \dfrac{B_1}{t-1} + \dfrac{B_2}{(t-1)^2} + \dfrac{B_3}{(t-1)^3}

Constants A , B_1 , B_2 , and B_3 are determined from the following condition:

A(t-1)^3 + B_1 t (t-1)^2 + B_2 t (t-1) + B_3 t = 1

This condition is valid in the case

A = -1 ,,\quad B_1 = 1 ,,\quad B_2 = -1 ,,\quad B_3 = 1

Consequently, the partial fraction decomposition of \bar{f}(t) takes form:

\bar{f}(t) = \dfrac{1}{t(t-1)^3} = -,\dfrac{1}{t} + \dfrac{1}{t-1} -, \dfrac{1}{(t-1)^2} + \dfrac{1}{(t-1)^3}

To calculate the inverse Laplace transform, we use the property of linearity and reference expression:

\mathcal{L}^{-1}\dfrac{1}{(t - \alpha)^{n+1}} = \dfrac{x^n e^{\alpha x}}{n!} ,,\qquad n=1,2,3,\ldots

The result is as follows:

\mathcal{L}^{-1}{ \dfrac{1}{t(t-1)^3}} = -1 + e^x - x e^x + \dfrac{x^2}{2},e^x

Example 3

Find the inverse Laplace transform of

\bar{f}(t) = \dfrac{t^2}{t^4-1}

Using the partial fraction decomposition, we get

\dfrac{t^2}{t^4-1} = \dfrac{1}{2(t^2+1)} + \dfrac{1}{2(t^2-1)}

We know that

\mathcal{L}^{-1}{ \dfrac{1}{t^2 + 1}} = sin (x) ,,\quad \text{and} \quad \mathcal{L}^{-1}{ \dfrac{1}{t^2 - 1}} = sinh (x)

Thus, due to the linearity of the inverse Laplace transforms, we obtain

\mathcal{L}^{-1}{ \dfrac{t^2}{t^4-1}} = \dfrac{sin(x) + sinh(x)}{2}

Solutions of Linear Differential Equations with Constant Coefficients

Laplace transforms are so useful for solving linear differential equations because the Laplace transform of the n-th derivative f^{(n)}(x) can be related to the transform of f(x) in a simple manner. This relationship is expressed in the following way:

\mathcal{L}{f^{(n)}(x)} = t^n \bar{f}(t) - f^{(n-1)}(0) - tf^{(n-2)}(0) - \cdots - t^{n-1}f(0)

Suppose, the initial conditions on f(x) at x=0 are given by

f(0) = C_0 ,,\quad f'(0) = C_1 ,,\ldots ,, \quad f^{(n-1)}(0) = C_{n-1}

Then, we can write

\mathcal{L}{f^{(n)}(x)} = t^n \bar{f}(t) - C_{n-1} - tC_{n-2} - \cdots - t^{n-1}C_0

For the special cases n=1 and n=2 , this relation simplifies to

\begin{array}{l} \mathcal{L}{f'(x)} = t \bar{f}(t) - C_0 \quad \mathcal{L}{f''(x)} = t^2 bar{f}(t) - tC_0 - C_1 \end{array}

Thus, the Laplace transform converts a linear differential equation with constant coefficients into an algebraic equation. Once the inverse Laplace transform of the solution of this algebraic equation is known, the solution of the differential equation is found. Let’s illustrate this method by working through some examples.

Example 4

Consider the initial value problem for a differential equation of the fourth-order:

y^{IV} - y = 0 ,;\qquad y(0) = y''(0) = y'''(0) = 0 ,,\quad y'(0) = 1

If \bar{y}(t) is the Laplace transform of y(x) , it follows that

\mathcal{L}{y^{IV}} = t^4 \bar{y}(t) - y'''(0) - t y''(0) - t^2 y'(0) - t^3 y(0) = t^4 \bar{y}(t) - t^2

Solving an algebraic equation for \bar{y}(t) , we obtain

\bar{y}(t) = \dfrac{t^2}{t^4-1}

Using the result of Example 3, we get a solution:

y(x) = \mathcal{L}^{-1}{ \bar{y}(t)} = \dfrac{sin(x) + sinh(x)}{2}

Example 5

Let’s find a solution to the initial value problem

y'' + w^2 y = 1 ,;\qquad y(0) = a ,,\quad y'(0) = b

In performing a Laplace transform of the differential equation, we use the following relations:

\mathcal{L}{ y(x)} = \bar{y}(t) ,,\qquad \mathcal{L}{ y''(x)} = t^2\bar{y}(t) - at - b ,,\qquad \mathcal{L}{ 1 } = \dfrac{1}{t}

Next, we expand the solution for \bar{y}(t) in partial fractions:

\bar{y}(t) = \dfrac{at^2 + bt + 1}{t(t^2+w^2)} = \dfrac{1}{w^2 t} + \dfrac{b}{t^2+w^2} + \left(a - \dfrac{1}{w^2} \right) \dfrac{t}{t^2+w^2}

The inverse Laplace transform of the first term is simple: \mathcal{L}^{-1}{ 1/t } = 1 . For the second term and third term, we can refer to the table of elementary transforms:

\mathcal{L}^{-1}{ \dfrac{1}{t^2 + w^2} } = \dfrac{sin (w x)}{w} ,,\qquad \mathcal{L}^{-1}{ \dfrac{t}{t^2 + w^2}} = cos (w x)

Thus, using the property of linearity, we obtain

y(x) = \dfrac{1}{w^2} + \dfrac{b}{w},sin(wx) + \left(a - \dfrac{1}{w^2} \right)cos(wx)

Example 6

In this example, we’ll demonstrate the usefulness of Laplace transforms for solving systems of linear differential equations. Consider the following set of two differential equations with two unknown functions, y(x) and z(x) :

\begin{array}{l} y'' + y + z = 0 z' + y' = 0 \end{array}

For the initial values, we suppose

y(0) = 0 ,,\qquad y'(0) = 0 ,,\qquad z(0) = 1

Denote \mathcal{L}{ y(x) } = \bar{y}(t) and \mathcal{L}{ z(x) } = \bar{z}(t) . Then, taking the Laplace transforms of each differential equation of the system, we obtain a set of algebraic equations

\begin{array}{l} t^2 \bar{y} + \bar{y} + \bar{z} = 0 \ t \bar{z} + t \bar{y} - 1 = 0 \end{array}

Solving this last system for \bar{y}(t) and \bar{z}(t) , we find that

\bar{y}(t) = -,\dfrac{1}{t^3} ,,\qquad \bar{z}(t) = \dfrac{1}{t} + \dfrac{1}{t^3}

To find the inverse Laplace transforms, we use the following equality:

\mathcal{L}^{-1}{ \dfrac{1}{t^{n+1}}} = \dfrac{x^n}{n!} ,,\qquad n=1,2,3,\ldots

Thus, we conclude that

y(x) = -,\dfrac{x^2}{2} ,,\qquad z(x) = 1 + \dfrac{x^2}{2}

Wrapping Up Inverse Laplace Transforms

In this review article, we investigated the inversion problem for Laplace transformations and established its relationship with differential equations. After studying our examples, you’ll be able to use partial fraction decomposition and inverse Laplace transforms to solve linear differential equations with constant coefficients. Let us know in the comments if you have any other helpful applications for inverse Laplace transforms.

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