Definition of Laplace transformation
To define the Laplace transformation, we first remember the definition of aIncorrectly integrated.Hvis \ (g \) is integrated during the interval \ ([a, t] \) for every \ (t> a \) then it isIncorrect integral of \ (g \)over\ ([a, \ inbound) \) is defined as
\ [\ etiket {ækv.: 8.1.1} \ int^\ infty_a g (t) \, dt = \ lim_ {t \ til \ innty} \ int^t_a g (t) \, dt.\]
We say that the wrong integralconvergerendIf the limit in comparison {equr.: 8.1.1} exists;Otherwise we say that the wrong integral divergererofdoes not exist.Here is the definition of the Laplace transformation of a function \ (F \).
Definition 8.1.1: Laplace -transformation
Let \ (f \) be defined for \ (t \ ge0 \) and let \ (s \) be a real number.Laplace -transformationof \ (f \) the function \ (f \) is defined by
\ [\ etiket {ækv.: 8.1.2} f (s) = \ int_0^\ infty e^{-st} f (t) \, dt, \]
For the values of \ (s \) that the wrong integrated converge.
It is important to remember that the variable of integration compared to {ec.: 8.1.2} (t \) is, while \ (s \) is a parameter that is independent of \ (t \).t \) If the independent variable for \ (F \) because in applications the Laplace transformation is usually used on time functions.
Laplace transformation can be seen as operator \ ({\ cal l} \) that transforms the function \ (f = f (t) \) into the function \ (f = f (s) \).{EQ Can: 8.1.2} expressed as
\ [F = {\ cal l} (f).\ Not -Number \]
The functions \ (F \) and \ (F \) form aTransformed par, as we sometimes indicate
\ [f (t) \ leftrightarrow f (s).\ Not -Number \]
It can be demonstrated that if \ (s) \) is defined for \ (s = s_0 \), this is defined for all \ (s> s_0 \) (Exercise 8.1.14B).
Calculation of some simple Laplace transformations
Example 8.1.1
Find the Laplace transformation of \ (f (t) = 1 \).
Solution
Of comparison \ ref {EME: 8.1.2} with \ (f (t) = 1 \),
\ 2
If \ (s \ ne 0 \) then
\ [\ label {equest} \ over s}.\]
Therefore
Ov, & s> 0, \\ \ infty, & s <0.\]
If \ (s = 0 \) integrates the integral with the constant \ (1 \), and
ovinfty.\ Noummer \]
That is why \ (F (0) \) is not defined, and
\ 2
This result can be written in Operator Note Die
\ [{\ cal l} (1) = {1 \ over s}, \ quad s> 0, \ numer \]
Or if the transformation savings
\ [1 \ leftrightarrow {1 \ over s}, \ quad s> 0. \ nonumber \]
Remark
It is useful to combine the steps to integrate \ (0 \) to \ (t \) and leave \ (t → ∞ \).: 8.1.4} As separate steps we write
ov = \ links \ {\ start {array} {cl} {\ frac {1} {s},} & {s> 0} \\ {\ innty,} & {s <0} \ end {array} \ Juist.
We follow this practice in this chapter.
Example 8.1.2
Find the Laplace transformation of \ (f (t) = t \).
From comparison \ ref {equr.: 8.1.2} with \ (f (t) = t \),
\ [\ etiket {ækv.: 8.1.5} f (s) = \ int_0^\ infty e^{-st} t \, dt.\]
If \ (s \ ne0 \), integrated after partial yield
ov \ int_0^\ infty e^{-st} \, dt =-\ links [{t \ over s}+{1 \ over s^2} \ rechts] e^{-st} \ bigg |_0^\ infty \\ & = \ links \ {\ start {array} {rr} {1 \ over s^2}, \ quad s> 0, \ inntty, \, s <0.\ end {array} \ naar rechts.
As \ (s = 0 \), the integral is in comparison \ ref {equ.: 8.1.5}
\ [\ int_0^\ Infty t \, dt = {t^2 \ over2} \ bigg | _0^\ other = \ other. \ Number \]
That is why \ (F (0) \) is not defined and
\ [F (s) = {1 \ over s^2}, \ quad s> 0. \ numer \]
This result can also be written as
\ [{\ cal l} (t) = {1 \ over s^2}, \ quad s> 0, \ nonumber \]
Or if the transformation savings
\ [T \ leftrightarrow {1 \ over S^2}, \ quad s> 0. \ nonumber \]
Example 8.1.3
Find the Laplace transformation of \ (f (t) = e^{at} \), where \ (a \) is a constant.
From comparison \ ref {Equin: 8.1.2} with \ (f (t) = e^{at} \),
\ 2
Combination of the exponential yields
\ 2
However, we know from example 8.1.1 that
\ [\ int_0^\ infty e^{-st} \, dt = {1 \ over s}, \ quad s> 0. \ nonumber \]
Replacement \ (s \) with \ (s-a \) is displayed here
\ [F (s) = {1 \ About s-a}, \ quad s> a. \ No number \]
This can also be written as
\ [{ov> a.
Example 8.1.4
[[Find the LaPlace transformations of \ (f (t) = \ Singa t \) and \ (g (t) = \ cos \ omega t \), where \ (\ \ Omega \) is a constant.
Define
\;
In
\ [\ etiket {ækv.: 8.1.7} g (s) = \ int_0^\ infty e^{-st} \ cos \ omega t \, dt.\]
If \ (s> 0 \), Integrate comparison \ ref {equr.: 8.1.6} After sharing yields
\ 2 COS \ OMEGA T \, DT, \ Number \]
So
\ [\ label {ækv.: 8.1.8} f (s) = {\ omega \ over s} g (s).\]
If \ (s> 0 \), Integrate comparison \ ref {equr.: 8.1.7} After sharing yields
\ 2 \ sin \ omega t \, dt, \ number \]
So
\ [G (s) = {1 \ About s} - {\ Omega \ About s} f (s). \ No number \]
Now replace comparison \ ref {equr.: 8.1.8} in this to get
\ [G (s) = {1 \ over s} - {\ omega^2 \ over s^2} g (s).\ Numer \]
This solving for \ (g (s) \) delivers
\ [G (s) = {s \ over s^2+\ omega^2}, \ quad s> 0. \ nonumber \]
This and comparison \ ref {EQur.: 8.1.8} suggests that
\ 2
Tables above Laplace transformations
Extensive tables with Laplace transformations are collected and are often used in applications.
Example 8.1.5
Use the table with Laplace transformations to find \ ({\ cal l} (t^3e^{4t}) \).
The table contains the transformation saving
\ [T^ne^{at} \ leftrightarrow {n!\ over (s-a)^{n+1}}.\ nummers \]
Setting \ (n = 3 \) and \ (a = 4 \) gives here
\ [\ cal l (t^3e^{4t}) = {3!\ over (S-4)^4} = {6 \ over (S-4)^4}.\ numer \]
We sometimes write Laplace transformations of specific functions without explicitly indicating how they are obtained.
Linearity of the Laplace transformation
The following sentence offers an important characteristic of the Laplace transformation.
Position 8.1.2 Linearity property
Suppose that \ (\ cal l} (f_i) \) is defined for \ (s> s_i, \) \ (1 \ le i \ le n). \) Let \ (s_0 \) the greatest of the numbers\ (s_1 \), \ (s_ {2}, \) ..., \ (s_n, \) and leave \ (c_1 \), \ (c_2 \), ..., \ (c_n \) be constantFerne
\ [{ovmbox {for} S> S_0. \ no number \]
- Bevis
-
We provide proof of the case in which \ (n = 2 \).
Ovipl} (F_1)+C_2 {\ Cal L} (F_2). \ Dan {adapted} \ Nonumber \]
Example 8.1.6
Use position 8.1.2 and the well -known Laplace transformation
\ [{\ cal l} (e^{at}) = {1 \ over s-a} \ nonumber \]
at FICE \ ({\ cal l} (\ cosh bt) \, (b \ Ne0) \).
Solution
By definition,
\ [\ cosh bt = {e^{bt}+e^{-bt} \ over 2}.
Therefore
\ [\ label {equ} + {1 \ over 2} e^{-bt} \ rechts) \\ [4pt] & = & {1 \ over 2} {\ cal l} (e^{bt}) + + {1 \ over 2} {ov2} \, {1 \ over s+b}, \ end {array} \]
where the first transformation on the right is defined for \ (s> b \) and the second for \ (s> -b \);That is why both are defined for \ (s> | B \).In comparison \ ref {EQur.: 8.1.9} Gives
\ [.
With the following sentence we can start with well -known transformation pairs and distract others.Exercises 8.1.6In8.1.13.)
Sentence 8.1.3 First changing sentence
When
\ [\ etiket {ækv.: 8.1.10} f (s) = \ int_0^\ infty e^{-st} f (t) \, dt \]
If the Laplace transformation of \ (f (t) \) is for \ (s> s_0 \), then \ (f (s-a) \) is the Laplace transformation of \ (E^{to} f (t) \) for \ (s> s_0+a \).
- Bevis
-
Replacement \ (s \) with \ (s-a \) in comparison \ ref {Equ.: 8.1.10} Delivered
\ [\ etiket {ækv.: 8.1.11} f (s-a) = \ int_0^\ infty e^{-(s-a) t} f (t) \, dt \]
If \ (s-a> s_0 \);
\ 2
What implies the conclusion.
Example 8.1.7
Use statement 8.1.3 and the well -known Laplace transformations of \ (1 \), \ (T \), \ (\ COS \ Omega T \) and \ (\ Omega T \) to find
\ [ov), \ mbox {og} {\ cal l} (e^{\ lambda t} \ omega t). \ _ no number \]
Solution
The following table states the known transformation pairs on the left and the required transformation pairs that are stated on the right are obtained using sentence 8.1.3.
| \ (f (t) \ leftrightarrow f (s) \) | \ (e^{at} f (t) \ leftrightarrow f (s-a) \) |
|---|---|
| \ (1 \ leftrightarrow {1 \ over s}, \ quad s> 0 \) | \ (e^{at} \ leftrightarrow {1 \ over (s-a)}, \ quad s> a \) |
| \ (t \ leftrightarrow \ frac {1} {s^{2}}, \ quad s> 0 \) | \ (te^{at} \ leftrightarrow \ frac {1} {(s-a)^{2}}, \ quad s> a \) |
| \) | OV |
| \) | ov \) |
Existence of Laplace transformations
Not every function has a Laplace transformation.Exercise 8.1.3) bee
OV
For every real number \ (s \).That is why the function \ (f (t) = e^{t^2} \) has no Laplace transformation.
Our next goal is to set conditions that ensure the existence of the Laplace transformation of a function.We first assess some relevant definitions of Calculus.
Remember that a limit
\ [\ lim_ {t \ til t_0} f (t) \ nonumber \]
exists like and only as the one -sided limits
\ [\ glue_ {t \ to t_0-} f (t) \ quad \ text {en} \ quad \ lijue_ {t \ to t_0+} f (t) \ nonumber \]
Both exist and are equal;in this case,
\ [\ lim_ {t \ til t_0} f (t) = \ lim_ {t \ til t_0-} f (t) = \ lim_ {t \ til t_0+} f (t).\ numer \]
Also remember that \ (f \) is continuous on a point \ (t_0 \) in an open interval \ ((a, b) \) if and only if only as if
\ [\ glue_ {t \ to t_0} f (t) = f (t_0), \ niet -Number \]
in accordance with
\;
For the sake of simplicity, we define
ov \]
So comparison \ ref {Equ.: 8.1.12} can be expressed as
\ [f (t_0+) = f (t_0-) = f (t_0).\ geen nummer \]
If \ (f (t_0+) \) and \ (f (t_0-) have limited but different values, we say that \ (f \) aSkip discontinuityby \ (t_0 \) and
\ [f (t_0+)-f (t_0-) \ nonumber \]
is calledhoppeI \ (f \) on \ (t_0 \) (Figure 8.1.1).
If \ (f (t_0+) \) and \ (f (t_0-) \) is limited and equal, but either \ (f \) is not defined by \ (t_0 \) or it is defined but is defined but is defined but
\ [F (t_0) \ hverken f (t_0 +) = f (t_0 -), \ nonumber \]
We say that \ (f \) aRemovable discrapinuityby \ (t_0 \) (Figure 8.1.2) .This terminology is appropriate as a function \ (f \) with a removable discontinuity at \ (t_0 \) can be made continuous at \ (t_0 \) by defining (or redefine ))
\ [f (t_0) = f (t_0+) = f (t_0-).\ geen nummer \]
Remark
We know from the Calculus that a certain integral is not influenced by changing the values of his Integrand on isolated points.\).
Definition 8.1.4: Pieces continuously
- It is said that a function is \ (f \)ancientOn a definitively closed interval \ ([0, t] \) if \ (f (0+) \) and \ (f (t-) \) is limited and is continuous with the open interval \ ((0, t) \) Except for finally on many points where \ (f \) jump -discontinuities or removable discontinuities can have.
- It is said that a function is \ (f \)ancientInsative interval \ ([0, \ Infty) \) If it is continuous on \ ([0, t] \) for every \ (t> 0 \).
Figure 8.1.3 shows the graph for a typical piece of continuous function.
It seems in the calculation that if a function is continuous with a limited closed interval, it is integrated into this interval.But if \ (f \) is continuously on \ ([0, \ Inny) \), this is also \ (E^{-st} f (t) \), and therefore
\ [\ int_0^t e^{-st} f (t) \, dt \ nonumber \]
is found for every \ (t> 0 \).
\ [\ label {equ (t) \, dt \]
converge for \ (s \) in one or the other interval \ ((s_0, \ inny) \).If \ (f (t) = e ^^ {t^2} \).The following definition offers a limitation of the growth of a function that guarantees the convergence of its Laplace transformation for \ (s \) in a certain interval \ ((S_0, \ Infty) \).
Definition 8.1.5: From exponential order
It is said that a function is \ (f \)of exponential order\ (S_0 \) If there are constants \ (m \) and \ (t_0 \)
\ [\ label {ækv.: 8.1.14} |f (t) |\ le me^{s_0t}, \ quad t \ ge t_0.\]
In situations where the specific value of \ (S_0 \) is not relevant, we simply say that \ (f \) isof exponential order.
The following sentence offers useful adequate conditions for a function \ (F \) to have a Laplace transformation.The certificate has been withdrawn insideExercise 8.1.10.
Zin 8.1.6
If \ (f \) piece is continuously on \ ([0, \ Inny) \) and of exponential order \ (s_0, \), defined \ ({\ cal l} (f) \) for \ (s> s_0\).
Remark
We emphasize that the conditions for ZENT 8.1.6 are sufficient, butnot necessary, for \ (f \) to have a Laplace transformation.For example,Exercise 8.1.14 (C)Show that \ (f \) may have a Laplace transformation, although \ (f \) is not of exponential order
Example 8.1.8
If \ (F \) is limited to one or
\ [|f (t) |\ le m, \ quad t \ ge t_0, \ numer \]
Then comparison \ ref {equr.is \ (\ ethe \ omega t \) and \ (\ cos \ omega t \) of exponential order zero, and sentence 8.1.6 implies that \ ({{{} (\ \ Omega t) and \ ({\ cal l} (\ cos \ omega t) \) is found for \ (s> 0 \).
Example 8.1.9
It can be demonstrated that if \ (\ glue_ {t \ to \ Infty} e^{-s_0t} f (t) \) is found and limited \ (f \) of exponential order \ (S_0 \) (Exercise 8.1.9) .If (\ alpha \) is a real number and \ (s_0> 0 \), is \ (f (t) = t^\ alpha \) of exponential order \ (s_0 \), since then
OV
From the rule of L'hôpital.Exercise 8.1.9and position 8.1.6 implies that \ (\ cal l} (t^\ alpha) \) is found for \ (s \ ge s_0 \).Really means that \ (\ cal l} (t^\ alpha) \) is found for all \ (s> 0 \).Exercises 8.1.6In8.1.8.
Example 8.1.10
Find the Laplace transformation of the Continuous Function
OV.\ Non -Number \]
Solution
Da \ (f \) is defined by different formulas on \ ([0.1) \) and \ ([1, \ Inny) \), we write
\ 2st} (-3e^{-t}) \, dt. \ No number \]
Since
OVS \ Neq 0} \\ {1} & {s = 0} \ End {array} \ on the right.
In
ov3e^{ -(s+1)} \ over s+1}, \ quad s> -1, \ noummer \]
It follows that
Yu 2} {C+1}} & {C> -1, with Yu 0} Yu;
This is in accordance with sentence 8.1.6, since then
\ [|F (t) |\ le 3e^{-t}, \ quad t \ ge 1, \ non-nummer \]
And that is why \ (f \) is of exponential order \ (s_0 = -1 \).
Remark
In section 8.4 we develop a more effective method for finding Laplace transformations of continuous functions.
Example 8.1.11
We said that before
OV
For all \ (s \) sentence 8.1.6 that \ (f (t) = e^{t^2} \) is not of exponential order
ovs_0t} = \ infty, \ noummer \]
So
\ [e^{t^2}> me^{s_0t} \ nonumber \]
For sufficiently large values of \ (t \), for every choice of \ (m \) and \ (s_ {0} \) (Exercise 8.1.3).
