Let F(λ) be the Laplace transform of some function. This
means that there exists a function f(t), defined on half-line,
such that
\begin{equation} \label{EqInverse.1}
F(\lambda ) = f^L (\lambda ) = \left[ {\cal L} f \right] \left(\lambda \right) =
\int_0^{\infty} f(t)\, e^{-\lambda\,t} \,{\text d}t .
\end{equation}
For a given function
F(λ) of a complex variable λ that is analytic in a half plane Reλ >
s for some real
s ∈ ℝ, determination of a function
f(
t) whose Laplace transform is
F(λ) is called the
inverse Laplace transform; it is usually denoted as
\( {\cal L}^{-1} \left[ F(\lambda ) \right] = f(t) , \) so
F(λ) and
f(
t) are related via Eq.\eqref{EqInverse.1}. Actually, determination of the inverse Laplace transformation is equivalent to solving the
integral equation of the first kind \eqref{EqInverse.1}. It is known that such integral equation is an ill-posed problem, and its solution usually requires a regularization. It was the English mathematician
Thomas John I'Anson Bromwich (1875--1929)
who gave a regularization for the inverse Laplace transform that can be expressed as the
contour integral:
\begin{align}
f(t) &= {\cal L}^{-1} \left[ F(\lambda ) \right] = \mbox{P.V. } \frac{1}{2\pi{\bf j}} \, \int_{s-{\bf j}\infty}^{s+{\bf
j}\infty} \, F(\lambda )\, e^{\lambda \,t} \,{\text d} \lambda = \lim_{N\to \infty} \frac{1}{2\pi{\bf j}} \, \int_{s-{\bf j}N}^{s+{\bf
j}N} \, F(\lambda )\, e^{\lambda \,t} \,{\text d} \lambda
\label{EqInverse.2}
\\
& \notag
\\
&= \frac{f(t+0) + f(t-0)}{2} = \frac{e^{st}}{2\pi} \left[
\int_0^{\infty} e^{{\bf j}\xi t} \,F(s+ {\bf j}\xi )\,{\text d} \xi +
\int_0^{\infty} e^{{\bf j}\xi t} \,F(s- {\bf j}\xi )\,{\text d} \xi
\right] \qquad ({\bf j}^2 = -1), \notag
\end{align}
where the abbreviation P.V. stands for the
Cauchy principal value, which indicates that the symmetrical regularization of the
improper integration is done along the vertical line Reλ
=
s in the
complex plane such that
s is greater than the
real part of all singularities of
F(λ)
and
F(λ) is bounded on the line of integration.
This integral is usually referred to as the
Bromwich integral. The real number
s in the Bromwich integral should exceed
the abscissa of convergence for the
function
f(
t). Formula \eqref{EqInverse.2} allows us to make the following observations.
Theorem 1:
Let
f(
t) be a function of
bounded variation on interval [0, ∞) that satisfies the
Dirichlet conditions on every finite subinterval. Suppose that
f(
t) is of exponential order then its Laplace transform
\( F(\lambda ) = f^L (\lambda ) = \int_0^{\infty} e^{-\lambda t} f(t)\,{\text d}t \) exists and has the abscissa of convergence σ∈ℝ. Then for any
s > σ, we have
-
The inverse Laplace transform is identically zero for negative t:
\[
{\cal L}^{-1} \left[ F(\lambda ) \right] (t) \equiv 0 \quad \mbox{ for } \quad t < 0.
\]
-
The Bromwich integral with s > σ restores
the function in such a way that at the point of
discontinuity t0 of the function f(t), the
Bromwich integral assigns its mean value:
\begin{equation} \label{EqInverse.3}
{\cal L}^{-1} \left[ F(\lambda ) \right] \left( t_0 \right) =
\frac{f(t_0 +0) + f(t_0 -0)}{2} = \frac{1}{2}\,\lim_{\epsilon \to 0, \ \epsilon > 0} \,
\left[ f(t_0 + \epsilon ) + f(t_0 - \epsilon ) \right] .
\end{equation}
-
\[
\frac{1}{2}\, f(0^{+} ) = \frac{1}{2}\, f(0+) = \mbox{P.V.}\,\frac{1}{2\pi{\bf j}} \, \int_{s - {\bf j}\infty}^{s + {\bf j}\infty} f^L (\lambda )\,{\text d}\lambda = \lim_{N\to +\infty} \frac{1}{2\pi{\bf j}} \, \int_{s - {\bf j}N}^{s + {\bf j}N} f^L (\lambda )\,{\text d}\lambda \qquad ({\bf j}^2 =-1).
\]
Remark:
There are known three possible notations for a right-sided limit: f(t+) or f(t+0) or f(t+); correspondingly, the left-sided limit is denoted as f(t-0) or f(t-) or f(t-).
The inverse Laplace transform, if it
exists, should be multiplied by the Heaviside function.
Example 1:
The inverse Laplace transform always assign the half of the limit value at the origin. For example, the Laplaced transform of constant unit function is
\[
{\cal L} \left[ \, 1 \,\right] = \int_0^{\infty} 1 \cdot e^{-\lambda t}{\text d} t = \int_0^{\infty} e^{-\lambda t}{\text d} t = \frac{1}{\lambda} .
\]
However,when inverse Laplace transform is applied to 1/λ, it results in
\[
{\cal L}^{-1}_{\lambda \to t} \left[ \, \frac{1}{\lambda} \,\right] = H(t) =
\begin{cases}
1 , & \ \mbox{ for} \quad t > 0 , \\
½ , & \ \mbox{ for} \quad t = 0, \\
0, & \ \mbox{ for} \quad t < 0.
\end{cases}
\]
As another example, we consider a familiar cosine function. Its Laplace transform is
\[
{\cal L}_{t\to\lambda} \left[ \, \cos t \,\right] = \int_0^{\infty} \cos t \cdot e^{-\lambda t}{\text d} t = \int_0^{\infty} \cos t\,e^{-\lambda t}{\text d} t = \frac{\lambda}{\lambda^2 + 1} .
\]
Application of the Inverse Laplace transform yields another function
\[
{\cal L}^{-1}_{\lambda \to t} \left[ \, \frac{\lambda}{\lambda^2 + 1} \,\right] = f(t) =
\begin{cases}
\cos t , & \ \mbox{ for} \quad t > 0 , \\
½ , & \ \mbox{ for} \quad t = 0, \\
0, & \ \mbox{ for} \quad t < 0.
\end{cases}
\]
So we see that the inverse laplace transform actually multiplies the cosine function by the Heaviside function.
End of Example 1
Although the inverse Laplace transform formula via indefinite Bromwich integral may seem intimidating, we will be showing that there are much easier methods to obtain the result for the inverse transform of a function.
We focus on the practical approaches of finding the inverse Laplace transformation.
We observe that applications of the Laplace transformation to
ordinary differential equations lead to functions F(λ)
of a complex variable λ that are represented either by the ratio of two
polynomials or by such ratio times the exponential function:
\[
F(\lambda ) = \frac{P(\lambda )}{Q(\lambda )} \qquad\mbox{or} \qquad
F(\lambda ) = \frac{P(\lambda )}{Q(\lambda )} \, e^{-a\lambda} ,
\]
where
\( P(\lambda ) = p_n \lambda^n + p_{n-1}
\lambda^{n-1} + \cdots + p_0 \quad\mbox{and} \quad Q(\lambda ) = q_m
\lambda^m + q_{m-1} \lambda^{m-1} + \cdots + q_0 . \) In
applications, the degree
m of the denominator always
exceeds
n, the degree the numerator. Later, we will consider a
special case when
m = n.
It should be noted that the latter case (when ratio of two polynomials
is multiplied by the exponential term) can be reduced to the former
one (without exponential multiple) with the aid of the shift
rule:
\begin{equation} \label{EqInverse.4}
{\cal L}^{-1} \left[ \frac{P(\lambda )}{Q(\lambda )} \, e^{-a\lambda} \right] = f(t-a)\, H(t-a) , \qquad \mbox{where} \quad
f(t) = {\cal L}^{-1} \left[ \frac{P(\lambda )}{Q(\lambda )} \right] .
\end{equation}
Therefore, we concentrate our attention on determination of the
inverse Laplace transform of the rational functions
\begin{equation} \label{EqInverse.5}
F(\lambda ) = \frac{P(\lambda )}{Q(\lambda )} .
\end{equation}
Partial Fraction Decomposition
One of the methods to find the inverse Laplace transform of a given fracion \eqref{EqInverse.5} is to use
partial fraction decomposition (or expansion), a familair topic from calculus. This allows one to represent a ratio of two polynomials, Eq.\eqref{EqInverse.5}, as a sum of simple fractions, with denominators having either a single zero or a pair of complex conjugate zeroes. Every term in the sum has a simple form because the degree of numerator is strictly less than degree of corresponding denominator. Since the inverse Laplace transform of each term in the decomposition sum is known and tabulated, the restoration of the inverse Laplace transform of the required fraction \eqref{EqInverse.5} becomes straight forward---it is expressed as a sum of known functions.
The advantage of this apporach is obvious---we avoid application of the Bromwich integral \eqref{EqInverse.2} and its regularization because finding the inverse Laplace transform is an ill-posed problem. Application of partial fraction decomposition has a drawback of performing tedious calculations and being supplied with a table of Laplace transforms of elementary functions. Fortunately, a computer algebra system such as matlab weakens the majority of these inconveniences.
Theorem 2:
Suppose that the ratio of two relatively prime polynomials is
\[
F(\lambda ) = \frac{P(\lambda )}{\left( \lambda - \lambda_1 \right)\left( \lambda - \lambda_2 \right) \cdots \left( \lambda - \lambda_n \right)}
\]
where λ
1, λ
2, …, λ
n are distinct real numbers and
P(λ) is a polynomial of degree less than
n. Then
\[
F(\lambda ) = \frac{a_1}{\lambda - \lambda_1} + \frac{a_2}{\lambda - \lambda_2} + \cdots + \frac{a_n}{\lambda - \lambda_n} ,
\]
where 𝑎
k,
k = 1, 2, …,
n, can be computed from
F(λ) by ignoring the factor λ−λ
k and setting λ = λ
k elsewhere.
Example 2:
Find the inverse Laplace transform of
\[
F(\lambda ) = \frac{6 + 2\lambda -3 \lambda^2}{\lambda^3 + 5\lambda^2 + 2 \lambda -8} .
\tag{2.1}
\]
First, we factor the denominator
\[
\lambda^3 + 5\lambda^2 + 2 \lambda -8 = \left( \lambda +4 \right) \left( \lambda +2 \right) \left( \lambda -1 \right) .
\tag{2.2}
\]
syms s
factor(s^3 + 5*s^2 + 2*s - 8)
>> ans =
[s - 1, s + 4, s + 2]
Then using partial fraction decomposition, we represent the function (2.1) as a sum of simple terms
\[
F(\lambda ) = \frac{6 + 2\lambda -3 \lambda^2}{\lambda^3 + 5\lambda^2 + 2 \lambda -8} = \frac{a}{\lambda +4} + \frac{b}{\lambda +2} + \frac{c}{\lambda -1} .
\tag{2.3}
\]
To find the values of coefficients 𝑎, b, and c, we have two options. First, we utilize Theorem 2. So we ignore the factor λ+4 in the denominator and set λ = −4. This yields
\[
a = \frac{6 + 2(-4) -3(-4)^2}{(-4+2)(-4-1)} = \frac{-50}{10} =5.
\]
Similarly, the other coefficients are given by
\[
b = \frac{6 +2(-2) -3(-2)^2}{(4-2)(-1-2)} = \frac{-10}{-6} = \frac{5}{3},
\]
and
\[
c = \frac{6 +2(1) -3(1)^2}{(1+4)(1+2)} = \frac{5}{15} = \frac{1}{3}
\]
This gives
\[
F(\lambda ) = \frac{6 + 2\lambda -3 \lambda^2}{\lambda^3 + 5\lambda^2 + 2 \lambda -8} = \frac{-5}{\lambda +4} + \frac{5/3}{\lambda +2} + \frac{1/3}{\lambda -1} .
\tag{2.4}
\]
We check with
matlab
syms t s
partfrac((6 + 2*s - 3*s^2)/(s + 4)/(s + 2)/(s - 1))
>> ans =
1/(3*(s - 1)) + 5/(3*(s + 2)) - 5/(s + 4)
Another option to find the values of these coefficients is to multiply Eq.(2.3) by the denominator:
\[
6 + 2\lambda -3 \lambda^2 = a\left( \lambda +2 \right)\left( \lambda -1 \right) + b \left( \lambda +4 \right)\left( \lambda -1 \right) + c \left( \lambda +2 \right)\left( \lambda +4 \right) .
\]
Upon setting in this equation λ = −4, we get
\[
6 + 2(-4) -3 (-4)^2 = a\left( -4 +2 \right)\left( -4 -1 \right) \qquad \Longrightarrow \qquad a=5.
\]
In a similar way, we determine the values of coefficients
b and
c.
Using
matlab, we find the inverse Laplace transform of each simple term:
syms t s
ilaplace(1/(s+4), s, t)
ilaplace(1/(s+2), s, t)
ilaplace(1/(s-1), s, t)
Collecting all terms, we find
\[
{\cal L}^{-1} \left[ \frac{6 + 2\lambda -3 \lambda^2}{\lambda^3 + 5\lambda^2 + 2 \lambda -8} \right] = \left[ \frac{1}{3}\, e^t + \frac{5}{3}\,e^{-2t} - 5\, e^{-4t} \right] H(t) ,
\]
where
H(
t) is the Heaviside function.
Finally, we demonstrate
matlab capability to determine the inverse Laplace transform with a one-line command:
syms s t
ilaplace((6 + 2*s - 3*s^2)/(s + 4)/(s + 2)/(s - 1), s, t)
>> ans =
(5*exp(-2*t))/3 - 5*exp(-4*t) + exp(t)/3
■
Theorem 3:
Let Q(λ) be a polynomial of degree n ≥ 1, and P(λ) be a polynomial of degree < n. Then the number of arbitrary constants in the partial fraction expansion of P(λ)/Q(λ) equals degree n of the denominator.
Example 3:
Find the inverse Laplace transform of
\[
F(\lambda ) = \frac{6 + 2\lambda -3 \lambda^2}{ \left( \lambda +4 \right) \left( \lambda +2 \right) \left( \lambda -1 \right)^3} .
\tag{3.1}
\]
We expand this function into the sum of simple terms:
\[
F(\lambda ) = \frac{a}{\lambda +4} + \frac{b}{\lambda +2} + \frac{c_1}{\lambda -1} + \frac{c_2}{(\lambda -1)^2} + \frac{c_3}{(\lambda -1)^3} ,
\tag{3.2}
\]
which has five constants. Using the common denominator, we get
\[
\frac{6 + 2\lambda -3 \lambda^2}{ \left( \lambda +4 \right) \left( \lambda +2 \right) \left( \lambda -1 \right)^3} = \frac{a \left( \lambda +2 \right) \left( \lambda -1 \right)^3 + b \left( \lambda +4 \right) \left( \lambda -1 \right)^3 + c_1 \left( \lambda +4 \right) \left( \lambda +2 \right) \left( \lambda -1 \right)^2 + c_2 \left( \lambda +4 \right) \left( \lambda +2 \right) \left( \lambda -1 \right) + c_3 \left( \lambda +4 \right) \left( \lambda +2 \right)}{\left( \lambda +4 \right) \left( \lambda +2 \right) \left( \lambda -1 \right)^3} .
\]
Two sides have the same denominators, so to be equal they should have the same numerators:
\[
6 + 2\lambda -3 \lambda^2 = a \left( \lambda +2 \right) \left( \lambda -1 \right)^3 + b \left( \lambda +4 \right) \left( \lambda -1 \right)^3 + c_1 \left( \lambda +4 \right) \left( \lambda +2 \right) \left( \lambda -1 \right)^2 + c_2 \left( \lambda +4 \right) \left( \lambda +2 \right) \left( \lambda -1 \right) + c_3 \left( \lambda +4 \right) \left( \lambda +2 \right) .
\]
Since this is a polynomial equation, it should be valid for all real or complex λ. So both sides are the same whatever value of λ is chosen. The right-hand side suggests that if we take λ = −4, then −2, and finally λ = 1, we get
\begin{align*}
6 + 2(-4) -3(-4)^2 &= -50 = 10\,a \qquad \Longrightarrow \qquad a = -1/5,
\\
6 + 2(-2) -3(-2)^2 &= -10 = 2\,b \qquad \Longrightarrow \qquad b = 5/27,
\\
6 + 2(1) -3(1)^2 &= 5 = 15\,c_3 \qquad \Longrightarrow \qquad c_3 = 1/3.
\end{align*}
The remaining coefficients (or all coefficients) may be obtained by equating like powers of λ:
\begin{align*}
&\lambda^0 : \qquad \phantom{-}6 = -2a -4b + 2 c_1 - 2c_2 + 2 c_3 ,
\\
&\lambda^1 : \qquad \phantom{-}2 = 5a + 11 b - c_1 - c_2 +3 c_3 ,
\\
&\lambda^2 : \qquad -3 = -3 a - 9b -3 c_1 + 2 c_2 + c_3 ,
\\
&\lambda^3 : \qquad \phantom{-}0 = -a + b + c_1 + c_2 ,
\\
&\lambda^4 : \qquad \phantom{-}0 = a + b + c_1 .
\end{align*}
In the following matlab codes, we use s instead of λ because I am too lazy to type six letters. Other people who are not like me can enjoy typing lambda.
Solving the coresponding system of algebraic equations (of course, with the aid of matlab), we obtain
syms a b c1 c2 c3 s
x = expand((s + 2)*a*(s - 1)^3 + b*(s + 4)*(s - 1)^3 + c1*(s + 4)*(s + 2)*(s - 1)^2 + c2*(s + 4)*(s + 2)*(s - 1) + c3*(s + 4)*(s + 2))
coeffs(x, s)
eqn1 = -2*a - 4*b + 8*c1 - 8*c2 + 8*c3 == 6;
eqn2 = 5*a + 11*b - 10*c1 + 2*c2 + 6*c3 == 2;
eqn3 = -3*a - 9*b - 3*c1 + 5*c2 + c3 == -3;
eqn4 = -a + b + 4*c1 + c2 == 0;
eqn5 = a + b + c1 == 0;
[X,Y] = equationsToMatrix([eqn1, eqn2, eqn3, eqn4, eqn5], [a, b, c1, c2, c3]);
linsolve(X,Y)
>> ans =
x = 8*c1 - 4*b - 2*a - 8*c2 + 8*c3 + 5*a*s + 11*b*s - 10*c1*s + 2*c2*s + 6*c3*s - 3*a*s^2 - a*s^3 + a*s^4 - 9*b*s^2 + b*s^3 + b*s^4 - 3*c1*s^2 + 4*c1*s^3 + 5*c2*s^2 + c1*s^4 + c2*s^3 + c3*s^2
>> ans =
[8*c1 - 4*b - 2*a - 8*c2 + 8*c3, 5*a + 11*b - 10*c1 + 2*c2 + 6*c3, 5*c2 - 9*b - 3*c1 - 3*a + c3, b - a + 4*c1 + c2, a + b + c1]
>> ans =
-1/5
5/27
2/135
-4/9
1/3
Therefore,
\[
F(\lambda ) = \frac{6 + 2\lambda -3 \lambda^2}{ \left( \lambda +4 \right) \left( \lambda +2 \right) \left( \lambda -1 \right)^3} = \frac{-1/5}{\lambda +4} + \frac{5/27}{\lambda +2} + \frac{2/135}{\lambda -1} + \frac{-4/9}{(\lambda -1)^2} + \frac{1/3}{(\lambda -1)^3} .
\tag{3.3}
\]
We check this expansion with matlab:
syms t s
partfrac((6 + 2*s - 3*s^2)/(s + 4)/(s + 2)/(s - 1)^3)
>> ans =
2/(135*(s - 1)) - 4/(9*(s - 1)^2) + 5/(27*(s + 2)) + 1/(3*(s - 1)^3) - 1/(5*(s + 4))
The inverse Laplace transforms of each simple fraction in Eq.(3.3) are known. It is confirmed by matlab:
syms t s
ilaplace(1/(s+4),s, t)
ilaplace(1/(s+2),s, t)
ilaplace(1/(s-1),s, t)
ilaplace(1/(s-1)^2 ,s, t)
ilaplace(1/(s-1)^3 ,s, t)
>> ans =
exp(-4*t)
>> ans =
exp(-2*t)
>> ans =
exp(t)
>> ans =
t*exp(t)
>> ans =
(t^2*exp(t))/2
Then the inverse Laplace transform of
F(λ) is the sum of all components:
\[
{\cal L}^{-1} \left[ F(\lambda ) \right] = \left[ -\frac{1}{5}\, e^{-4t} + \frac{5}{27} \, e^{-2t} + \frac{2}{135}\, e^{t} -\frac{4}{9}\,t\, e^t + \frac{t^2}{6}\, e^{t} \right] H(t) .
\]
syms t s
ilaplace((6 + 2*s - 3*s^2)/(s + 4)/(s + 2)/(s - 1)^3, s, t)
>> ans =
(5*exp(-2*t))/27 - exp(-4*t)/5 + (2*exp(t))/135 + (t^2*exp(t))/6 - (4*t*exp(t))/9
■
Example 4:
Find the inverse Laplace transform of
\[
F(\lambda ) = \frac{6 + 2\lambda -3 \lambda^2}{ \left( \lambda +2 \right) \left( \lambda -1 \right) \left( \lambda^2 +6\lambda + 13 \right)} .
\tag{4.1}
\]
We expand this function into the sum of simple fractions:
\[
F(\lambda ) = \frac{6 + 2\lambda -3 \lambda^2}{ \left( \lambda +2 \right) \left( \lambda -1 \right) \left( \lambda^2 +6\lambda + 13 \right)} = \frac{a}{\lambda +2} + \frac{b}{\lambda -1} + \frac{c_1 + c_2 \lambda}{\lambda^2 +6\lambda + 13} .
\tag{4.2}
\]
So we need to determine the values of coefficients
in Eq.(4.2). A usual way to acomplish this goal is to add all simple fractions and equate the result to the given function
F(λ):
\[
\frac{6 + 2\lambda -3 \lambda^2}{ \left( \lambda +2 \right) \left( \lambda -1 \right) \left( \lambda^2 +6\lambda + 13 \right)} =
\frac{a\left( \lambda -1 \right) \left( \lambda^2 +6\lambda + 13 \right) + b \left( \lambda +2 \right) \left( \lambda^2 +6\lambda + 13 \right) + c \left( \lambda +2 \right) \left( \lambda -1 \right) }{ \left( \lambda +2 \right) \left( \lambda -1 \right) \left( \lambda^2 +6\lambda + 13 \right)} .
\]
Equating numerators, we get a system of algebraic equations
syms s a b c1 c2
x = expand(a*(s - 1)*(s^2 + 6*s + 13) + b*(s + 2)*(s^2 + 6*s + 13) + (c1 + c2*s)*(s^2 + s - 2))
coeffs(x, s)
>> ans =
[26*b - 13*a - 2*c1, 7*a + 25*b + c1 - 2*c2, 5*a + 8*b + c1 + c2, a + b + c2]
\begin{align*}
\lambda^3 : \ & a + b + c_2 &= 0 , \\
\lambda^2 : \ & 5a + 8b + c_1 + c_2 &= -3 , \\
\lambda^1 : \ & 7a + 25 b + c_1 -2 c_2 &= 2 , \\
\lambda^0 : \ & -13 a + 26 b - 2c_1 &= 6 .
\end{align*}
Solving this system of algebraic equations, we obtain
syms s a b c1 c2
eqn1 = 26*b - 13*a - 2*c1 == 6;
eqn2 = 7*a + 25*b + c1 - 2*c2 == 2;
eqn3 = 5*a + 8*b + c1 + c2 == -3;
eqn4 = a + b + c2 == 0;
[X,Y] = equationsToMatrix([eqn1, eqn2, eqn3, eqn4], [a, b, c1, c2]);
linsolve(X,Y)
>> ans =
2/3
1/12
-25/4
-3/4
\[
a = \frac{2}{3}, \quad b= \frac{1}{12} , \quad c_1 = - \frac{25}{4}, \quad c_2 = - \frac{3}{4} .
\]
So the expansion of
F(λ) into simple fractions becomes
\[
F(\lambda ) = \frac{6 + 2\lambda -3 \lambda^2}{ \left( \lambda +2 \right) \left( \lambda -1 \right) \left( \lambda^2 +6\lambda + 13 \right)} = \frac{2/3}{\lambda +2} + \frac{1/12}{\lambda -1} - \frac{1}{4} \cdot \frac{25 + 3 \lambda}{\lambda^2 +6\lambda + 13} .
\]
We check with
matlab:
syms s
partfrac((6 + 2*s - 3*s^2)/(s + 2)/(s - 1)/(s^2 + 6*s + 13))
>> ans =
1/(12*(s - 1)) + 2/(3*(s + 2)) - ((3*s)/4 + 25/4)/(s^2 + 6*s + 13)
Then we apply the inverse Laplace transform to each term and obtain the required inverse Laplace value:
\[
{\cal L}^{-1} \left[ \frac{1}{\lambda +2} \right] = e^{-2t} H(t) , \qquad {\cal L}^{-1} \left[ \frac{1}{\lambda - 1} \right] = e^{t} H(t)
\]
syms t s
ilaplace(1/(s+2), s, t)
ilaplace(1/(s-1), s, t)
>> ans =
exp(-2*t)
>> ans =
exp(t)
and
\[
{\cal L}^{-1} \left[ \frac{c_1 + c_2 \lambda}{\lambda^2 +6\lambda + 13} \right] = \frac{1}{2} \,e^{-3t} \left[ c_1\, \sin 2t + c_2 \left( \cos 2t - 3\,\sin 2t \right) \right] H(t) ,
\tag{4.3}
\]
where
H(
t) is the Heaviside function.
syms t s c1 c2
ilaplace((c1+c2*s)/(s^2 + 6*s+13), s, t)
>> ans =
c2*exp(-3*t)*(cos(2*t) + sin(2*t)*(c1/(2*c2) - 3/2))
To justify formula (4.3), we make substitution s = λ + 3 and obtain
\[
\frac{c_1 + c_2 \lambda}{\lambda^2 +6\lambda + 13} = \frac{c_1 - 3\, c_2 + c_2 \left( \lambda + 3 \right)}{\left( \lambda + 3 \right)^2 + 4} = \frac{c_1 - 3\, c_2 + c_2 s}{s^2 + 2^2} .
\]
The inverse Laplace transforms of the following fractions are well-known
\[
{\cal L}^{-1}_{s\to t} \left[ \frac{1}{s^2 + 2^2} \right] = \frac{1}{2}\,\sin 2t \,H(t) \qquad\mbox{and} \qquad {\cal L}^{-1}_{s\to t} \left[ \frac{s}{s^2 + 2^2} \right] = \cos 2t\,H(t) .
\]
Now the inverse Laplace transforms (4.3) follow from the
attenuation rule:
\[
{\cal L} \left[ e^{-at} f(t) \right] = f^L \left( \lambda + 3 \right) .
\tag{4.4}
\]
Therefore, we get
\[
{\cal L}^{-1} \left[ \frac{6 + 2\lambda -3 \lambda^2}{ \left( \lambda +2 \right) \left( \lambda -1 \right) \left( \lambda^2 +6\lambda + 13 \right)} \right] = \left[ \frac{1}{12}\, e^t + \frac{2}{3}\, e^{-2t} - \frac{25}{8}\, e^{-3t} \sin 2t - \frac{3}{8}\, e^{-3t} \left( \cos 2t -3\,\sin 2t \right) \right] H(t) .
\]
syms t s
ilaplace((6+2*s-3*s^2)/(s+2)/(s-1)/(s^2 +6*s+13),s,t)
>> ans =
(2*exp(-2*t))/3 + exp(t)/12 - (3*exp(-3*t)*(cos(2*t) + (8*sin(2*t))/3))/4
■
Example 5:
Find the inverse Laplace transform of
\[
F(\lambda ) = \frac{6 + 2\lambda -3 \lambda^2}{ \left( \lambda +2 \right) \left( \lambda -1 \right) \left( \lambda^2 +6\lambda + 13 \right)^2} .
\tag{5.1}
\]
We expand this function into the sum of simple terms:
\[
F(\lambda ) = \frac{6 + 2\lambda -3 \lambda^2}{ \left( \lambda +2 \right) \left( \lambda -1 \right) \left( \lambda^2 +6\lambda + 13 \right)^2} = \frac{a}{\lambda +2} + \frac{b}{\lambda -1} + \frac{c_1 + c_2 \lambda}{\lambda^2 +6\lambda + 13} + \frac{c_2 + c_4 \lambda}{\left( \lambda^2 +6\lambda + 13 \right)^2} .
\tag{5.2}
\]
syms s
partfrac((6+2*s-3*s^2)/(s+2)/(s-1)/(13+6*s + s^2)^2, s)
>> ans =
1/(240*(s - 1)) + 2/(15*(s + 2)) - ((3*s)/4 + 25/4)/(s^2 + 6*s + 13)^2 - ((11*s)/80 + 9/16)/(s^2 + 6*s + 13)
In order to find the values of coefficients in Eq.(5.2), we combine its right-hand side into one fraction and equate it to the given one. This yields the polynomial equation
\[
6 + 2\lambda -3 \lambda^2 = a \left( \lambda -1 \right) \left( \lambda^2 +6\lambda + 13 \right)^2 + b \left( \lambda + 2 \right) \left( \lambda^2 +6\lambda + 13 \right)^2 + \left( c_1 + c_2 \lambda \right) \left( \lambda^2 + \lambda -2 \right) \left( \lambda^2 +6\lambda + 13 \right) + \left( c_3 + c_4 \right) \left( \lambda^2 + \lambda -2 \right) .
\]
Upon expansion and equating the coefficients of like powers of λ, we come to the system of algebraic equations
syms s a b c1 c2 c3 c4
x =expand(a*(s - 1)*(s^2 + 6*s + 13)^2 + b*(s + 2)* (s^2 + 6*s + 13)^2 + (c1 + c2*s)*(s^2 + s - 2)*(s^2 + 6*s + 13) + (c3 + c4*s)*(s^2 + s - 2))
coeffs(x, s)
>> ans =
[338*b - 169*a - 26*c1 - 2*c3, 13*a + 481*b + c1 - 26*c2 + c3 - 2*c4, 94*a + 280*b + 17*c1 + c2 + c3 + c4, 50*a + 86*b + 7*c1 + 17*c2 + c4, 11*a + 14*b + c1 + 7*c2, a + b + c2]
\begin{align*}
&\lambda^0 : \qquad \phantom{-}6 = -169 a + 338 b -26 c_1 -2 c_3 ,
\\
&\lambda^1 : \qquad \phantom{-}2 = 13 a + 481 b + c_1 -26 c_2 + c_3 -2 c_4 ,
\\
&\lambda^2 : \qquad -3 = 94 a + 280 b + 17 c_1 + c_2 + c_3 + c_4 ,
\\
&\lambda^3 : \qquad \phantom{-}0 = 50 a + 86 b+ 7 c_1 + 17 c_2 + c_4 ,
\\
&\lambda^4 : \qquad \phantom{-}0 = 11 a + 14 b + c_1 + 7c_2 ,
\\
&\lambda^5 : \qquad \phantom{-}0 = a + b + c_2 .
\end{align*}
The right-hand side expressions are sorted by powers of λ (here,
s). For convenience, these were determined with
matlab:
syms s a b c1 c2 c3 c4
eqn1 = -169*a + 338*b - 26*c1 - 2*c3 == 6;
eqn2 = 13*a + 481*b + c1 - 26*c2 + c3 - 2*c4 == 2;
eqn3 = 94*a + 280*b + 17*c1 + c2 + c3 + c4 == -3;
eqn4 = 50*a + 86*b + 7*c1 + 17*c2 + c4 == 0;
eqn5 = 11*a + 14*b + c1 + 7*c2 == 0;
eqn6 = a + b + c2 == 0;
[C,D] = equationsToMatrix([eqn1, eqn2, eqn3, eqn4, eqn5, eqn6], [a, b, c1, c2, c3, c4]);
linsolve(C,D)
>> ans =
2/15
1/240
-9/16
-11/80
-25/4
-3/4
We solve this system of equations and obtain
\[
a = \frac{2}{15}, \quad b= \frac{1}{240} , \quad c_1 = - \frac{9}{16} , \quad c_2 = - \frac{11}{80} , \quad c_3 = - \frac{25}{4} , \quad c_4 = - \frac{3}{4} .
\]
This allows us to rewrite the given fraction as
\[
F(\lambda ) = \frac{6 + 2\lambda -3 \lambda^2}{ \left( \lambda +2 \right) \left( \lambda -1 \right) \left( \lambda^2 +6\lambda + 13 \right)^2} = \frac{2/15}{\lambda +2} + \frac{1/240}{\lambda -1} - \frac{45 + 11 \lambda}{80 \left( \lambda^2 +6\lambda + 13 \right)} - \frac{25 + 3 \lambda}{4 \left( \lambda^2 +6\lambda + 13 \right)^2} .
\tag{5.3}
\]
We check with
matlab:
syms t s
partfrac((6 + 2*s - 3*s^2)/(s + 2)/(s - 1)/(s^2 + 6*s + 13)^2)
>> ans =
1/(240*(s - 1)) + 2/(15*(s + 2)) - ((3*s)/4 + 25/4)/(s^2 + 6*s + 13)^2 - ((11*s)/80 + 9/16)/(s^2 + 6*s + 13)
The inverse Laplace transforms of first two simple fractions are known from the previous example.
\[
{\cal L}^{-1} \left[ \frac{1}{\lambda +2} \right] = e^{-2t} H(t) \qquad {\cal L}^{-1} \left[ \frac{1}{\lambda -1} \right] = e^t H(t) ,
\]
The third fraction can be written as
\[
\frac{45 + 11 \lambda}{80 \left( \lambda^2 +6\lambda + 13 \right)} = \frac{1}{80} \cdot \frac{12 + 11 \left( \lambda + 3 \right)}{\left( \lambda + 3 \right)^2 + 4} .
\]
We set
s = λ + 3, and consider an auxiliary fraction
\[
F_1 (s) = \frac{12 + 11\,s}{s^2 + 4} .
\]
Its inverse Laplace transform is known
\[
{\cal L}^{-1}_{s\to t} \left[ \frac{12 + 11\,s}{s^2 + 4} \right] = \left[ 11\,\cos 2t + 6 \sin 2t \right] H(t) .
\]
syms t s
ilaplace((12 + 11*s)/(s^2 + 4), s , t)
>> ans =
11*cos(2*t) + 6*sin(2*t)
Then using the attenuation rule,
\( {\cal L} \left[ e^{-at} f(t) \right] = f^L \left( \lambda + 3 \right) , \)
we get
\[
{\cal L}^{-1} \left[ \frac{45 + 11 \lambda}{80 \left( \lambda^2 +6\lambda + 13 \right)} \right] = \frac{1}{80} \cdot \left[ 11\,\cos 2t + 6\,\sin 2t \right] e^{-3t} H(t) .
\]
In order to find the inverse Laplace transform of the last term in right-hand side of Eq.(5.3), we rewrite it as
\[
\frac{25 + 3 \lambda}{4 \left( \lambda^2 +6\lambda + 13 \right)^2} = \frac{1}{4}\cdot \frac{16 + 3 \left( \lambda + 3 \right)}{\left[ \left( \lambda + 3 \right)^2 + 2^2 \right]^2}
\]
Again, setting
s = λ + 3, we consider another auxiliary fraction
\[
F_2 (s) = \frac{16 + 3s}{\left( s^2 + 2^2 \right)^2} ,
\]
which has the inverse Laplace transform
\[
{\cal L}^{-1}_{s\to t} \left[ F_2 (s) \right] = {\cal L}^{-1}_{s\to t} \left[ \frac{16 + 3s}{\left( s^2 + 2^2 \right)^2} \right] = \frac{1}{4} \left[ 3t
\,\sin 2t -8t\,\cos 2t + 4\,\sin 2t \right] H(t) .
\]
syms t s
ilaplace((16+3*s)/(s^2 +4)^2, s, t)
>> ans =
ilaplace((16+3*s)/(s^2 +4)^2, s, t)
Using the attenuation rule, we obtain
\[
{\cal L}^{-1}_{\lambda\to t} \left[ \frac{25 + 3 \lambda}{4 \left( \lambda^2 +6\lambda + 13 \right)^2} \right] = \frac{1}{16} \left[ 3t
,\sin 2t -8t\,\cos 2t + 4\,\sin 2t \right] e^{-3t} H(t) .
\]
Collecting all inverse Laplace transformations of simple fractions into one expression, we obtain
\[
{\cal L}^{-1} \left[ \frac{6 + 2\lambda -3 \lambda^2}{ \left( \lambda +2 \right) \left( \lambda -1 \right) \left( \lambda^2 +6\lambda + 13 \right)^2} \right] = \left[ \frac{2}{15}\, e^{-2t} + \frac{1}{240}\, e^t - \frac{1}{16}\,e^{-3t} \left( 4\,\sin 2t -8t\,\cos 2t + 3t\,\sin 2t \right) - \frac{1}{80}\,e^{-3t} \left( 11\,\cos 2t + 6\,\sin 2t \right) \right] H(t) .
\]
syms t s
ilaplace((6+2*s-3*s^2)/(s+2)/(s-1)/(s^2 +6*s+13)^2,s,t)
>> ans =
(2*exp(-2*t))/15 + exp(t)/240 - (exp(-3*t)*(4*sin(2*t) - 8*t*cos(2*t) + 3*t*sin(2*t)))/16 - (11*exp(-3*t)*(cos(2*t) + (6*sin(2*t))/11))/80
■
Example 6:
matlab
allows one to find the inverse Laplace transform in a straight forward way. For example, consider the rational function
\[
F(\lambda ) = \frac{4}{\lambda \left( 4 + \lambda^2 \right)} .
\tag{6.1}
\]
Since
matlab has a build-in command to determine the inverse Laplace transform, we apply it first and obtain
syms t s
ilaplace(4/s/(4 + s^2), s, t)
>> ans =
1 - cos(2*t)
\[
{\cal L}^{-1}_{\lambda\to t} \left[ \frac{4}{\lambda \left( 4 + \lambda^2 \right)} \right] = \left[ 1 - \cos 2t \right] H(t) = 2\,\sin^2 t \,H(t) .
\tag{6.2}
\]
We check the answer by transforming the function
F(λ) into the sum of simple terms:
syms t s
partfrac(4/s/(4 + s^2), s)
>> ans =
1/s - s/(s^2 + 4)
\[
F(\lambda ) = \frac{4}{\lambda \left( 4 + \lambda^2 \right)} = \frac{1}{\lambda} + \frac{\lambda}{4 + \lambda^2} .
\]
We recognize immediately that
syms t s
ilaplace(1/s,s)
ilaplace(s/(s^2 + 4),s,t)
>> ans =
1
>> ans =
cos(2*t)
\[
{\cal L}^{-1} \left[ \frac{1}{\lambda} \right] = H(t) \qquad\mbox{and} \qquad {\cal L}^{-1} \left[ \frac{\lambda}{4 + \lambda^2} \right] = \cos 2t \,H(t) .
\]
Adding these expressions we obtain the required inverse Laplace transform of function (6.2).
We check our answer by taking the Laplace transform of the obtained function
syms t s
x = laplace(2*sin(t)^2, t, s)
ilaplace(x, s, t)
>> ans =
4/(s*(s^2 + 4))
>> ans =
1 - cos(2*t)
■
-
Doetsch, G., Guide to the Applications of the Laplace and Z-Transforms, 1971, Van Nostrand Reinhold Company, 2nd Edition, London.
-
Doetsch, G., Introduction to the Theory and Application of the Laplace Transformation, 1974, Springer, Berlin.
-
Man, Y-K., An improved Heaviside approach to partial fraction expansion and its applications, International Journal of Mathematical Education in Science and Technology,
Volume 40, 2009, Issue 7, pp. 808--814. https://doi.org/10.1080/00207390902825310
-
Man, Y-K., Introducing the improved Heaviside approach to partial fraction decomposition to undergraduate students: results and implications from a pilot study, International Journal of Mathematical Education in Science and Technology,
Volume 43, 2012 - Issue 7, pp. 911--922. https://doi.org/10.1080/0020739X.2012.662292
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