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Laplace transforms
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Block 2 – Laplace Transforms
2. Laplace transforms
Contents
3.1 Introduction
3.2 The Laplace transform3.2.1 Definitions and notation3.2.2 Laplace transform of some simple functions3.2.3 Properties of the Laplace transform3.2.4 Table of Laplace transforms
3.3 The inverse Laplace transform3.3.1 Using partial fractions3.3.2 Finding inverses using the first shift theorem3.3.3 Completing the square
3.4 Solution of differential equations 13.4.1 Laplace transforms of derivatives3.4.2 Solution of 1st order differential equations3.4.3 Solution of 2nd order differential equations
3.5 Solution of differential equations3.5.1 Differential equations and Dirac functions3.5.2 Nonlinear differential equations
3.6 Systems of differential equations
3.7 Summary
Appendix A1 – Partial fractions
Appendix A2 – Completing the square
1
Block 2 – Laplace Transforms
Prerequisite knowledge
• 1st Year Math Modules or A-level mathematics/ HND.
• An understanding of improper integrals would be useful
Learning objectives
By the end of this block, you should be able to:
• Evaluate Laplace transforms
• Evaluate inverse Laplace transforms
• Solve linear constant coefficient differential equations
• Apply Laplace transforms to problems in vibration, control systems and oscillations in electrical circuits.
• Solve systems of differential equations
2
Block 2 – Laplace Transforms
3.1 Introduction
Laplace transforms are an interesting field of mathematics that can be used to solve problems involving differential equations and integro-differential equations. The technique is quite abstract in nature but it allows the solution of differential equations, without the need to do any integration or differentiation. The processes of integration and differentiation are replaced by algebraic manipulation, which is often considered easier to apply than concepts taken from calculus. A further advantage of the ‘Laplace transform method’ for solving initial value problems is the initial conditions are incorporated in an entirely natural way that does not require the solution of simultaneous equations.
The main reason for considering Laplace transforms at this stage is all engineering disciplines (except civil engineering) engage in a subject called control engineering, where Laplace transforms are used to analyse the response of engineering systems to changes in inputs to the system, whether it be a chemical reactor vessel or an auto-pilot in a plane.
Before we can attempt to solve differential equations using the Laplace transform, we need to introduce it and consider the Laplace transform and inverse Laplace transform for a number of simple functions and differential operators.
3.2 The Laplace transform
3.2.1 Definitions and notation
The Laplace transform of a function,
( )tf
is denoted as,
( ){ } ( )∫∞ −=
0dttfetfL st (3.1)
Alternative symbols used for the Laplace transform are given below,
( ){ } ( ) ( )sforsFtfL =
From (3.1) we see that the Laplace transform consists of an improper integral (one of the integration bounds is infinite). In the first instance with an improper integral you have to focus on whether the integral has a solution or not. For example the Laplace transform of the function,
( ) atetf =
only exists if
3
Block 2 – Laplace Transforms
0<− sa
otherwise the Laplace transform is
( ) ∫∫∞∞ − =
00dtedttfe btst
and b=a-s is positive. This integral is unbounded. Returning to the terminology used to describe the Laplace transform (3.1)
ste−
is called the kernel of the integral. In the original function,
( )tf
The independent variable is t, which can be considered a time variable. The function f(t) is considered to exist in the time domain. The Laplace transform F(s) exists in a frequency domain. s is the independent variable of the Laplace transform and strictly speaking is a complex variable although we will for the most part only consider its real part.
3.2.2 Laplace transforms of some simple functions
Consider the function,
( ) ctf = (3.2)
where c is a constant. The Laplace transform of (3.2) is given below.
( ){ } ∫∞ −=0
dtectfL st
As this is an improper integral it should be considered in the limit of the upper bound being finite and the limit to infinity taken once the integral is evaluated.
( ){ } ∫∫ −∞
∞→
− ==b st
b
st dtcedtectfL00
lim
−−−=
−= −−−∫ 0
00
es
ce
s
ce
s
cdtce sb
bstb st
( )sbes
c −−= 1
We can now let b tend to infinity,
( )s
ce
s
c sb
b=− −
∞→1lim
4
Block 2 – Laplace Transforms
To summarise,
{ }s
ccL =
This can be considered as a Laplace transform pair,
( ) ( )
==
s
csFctf ,
Worked Examples (Evaluating Laplace transforms by direct integration)
Let us consider another example in detail,
( ) ttf =
Solution
Consider the Laplace transform of t,
{ } ∫ ∫∞ −
∞→
− ==0 0
limb st
b
st dttedtettL
This can be integrated by parts,
∫0
bu
dvdt
dt=uv−∫0
bv
dudt
dt
u= t ⇒ du /dt=1
dv /dt= e−st ⇒ v=−1s
e−st
∫∫ −−− +
−=
b stb
stb st dtes
es
tdtte
00
0
1
∫∫ −−− +−=b stsbb st dte
se
s
bdtte
00
1
We could evaluate the integral on the right hand side but we can let b tend to infinity now and avoid the integration,
{ }200
11
11lim
sL
sdte
sdtte stb st
b==+= ∫∫
∞ −−
∞→
So the Laplace transform pair is,
5
Block 2 – Laplace Transforms
( ) ( )
==
2
1,
ssFttf
Exercise
Use the same approach (i.e. one integration by parts) to show that,
( ) ( )
==
32 2,
ssFttf
The main point of this block is not the integration process. In this section we are interested in growing our collection of functions that we ‘know’ the Laplace transform for although we will not do this by integrating the Laplace transform for every function we come across.
6
Block 2 – Laplace Transforms
For example, inspecting the three Laplace transform pairs given above a pattern is emerging for algebraic terms. The general rule is,
General Rule for Laplace transforms of algebraic terms
Consider the function,
( ) nttf =
Its Laplace transform is,
( )1
!+=
ns
nsF
One more example evaluated by direct integration,
Worked Example
Evaluate the Laplace transform of the function,
( ) ktetf =
Solution
{ } ( )∫ ∫∞ −−
∞→
− ==0 0
limb tks
b
stktkt dtedteeeL
( ) ( ) ( )( )0
00
11ee
kse
ksdte bks
btksb tks −
−−=
−−= −−−−−−∫
( )( )ks
eeks
bks
b −=−
−−= −−
∞→
11lim 0
provided
ksorks >>− 0
So the Laplace transform pair is,
( ) ( )
−==
kssFetf kt 1
,
As you will appreciate from the above examples deriving Laplace transforms by direct integration is quite a tedious process. To avoid direct integration many ingenious mathematical tricks and theorems are often used to find Laplace transforms. For example consider the following worked example.
7
Block 2 – Laplace Transforms
Worked Example
Evaluate the Laplace transform of the function,
( ) iatetf =
Where a is some real number and
1−=i
Solution
At first glance this looks almost the same as the preceding worked example and indeed the solution is the same.
( ) ( )
−==
iassFetf iat 1
,
So why have we done two worked examples that are so very similar? The answer is the second leads onto an additional useful result giving us two extra Laplace transforms. Three Laplace Transforms for the price of one!
Complex numbers given in exponential form can be represented using Euler’s formula
( ) atiatetf iat sincos +==
Reconsidering the Laplace transform of the previous example
{ } { } { } { }atiLatLatiatLeL iat sincossincos +=+= (3.3)
The separation of the Laplace transform into an application of the Laplace transform to the two terms is possible as the Laplace transform is a linear operator. This will be discuss further below, for now just accept it.
Again from the previous worked example,
{ }22
11
as
ias
ias
ias
iasiaseL iat
++=
++×
−=
−=
So from (3.3)
{ } { }2222
sincosas
ai
as
satiLatL
++
+=+ (3.4)
Therefore equating real and imaginary parts of (3.4) gives the following results.
8
Block 2 – Laplace Transforms
Laplace Transforms of Trigonometric Functions
The Laplace transform pair for sin at
( ) ( )
+==
22,sin
as
asFattf
The Laplace transform pair for cos at,
( ) ( )
+==
22,cos
as
ssFattf
3.2.3 Properties of the Laplace transform
The Laplace transform is a linear operator which means it has the property that,
( ) ( ){ } ( ){ } ( ){ }tgLtfLtgtfL βαβα +=+
where α and β are constants and f(t) and g(t) are functions.
Example
Determine,
{ } { } { }tt eLtLetL 33 2323 +=+
Solution
Using the linearity property of the Laplace transform,
{ } { } { }tt eLtLetL 33 2323 +=+
We can now apply the Laplace transforms derived in the previous section,
3
232 −
+=ss
{ }3
2323
23
−+=+
ssetL t
9
Block 2 – Laplace Transforms
Example
Determine,
{ }tettL 462sin435 −+−
Solution
Using the linearity property of the Laplace transform,
{ } { } { } { } { }tt eLtLtLLettL 44 62sin431562sin435 −+−=−+−
Applying the Laplace transforms derived or presented in the previous section,
4
6
4
24
3522 −
−
++−=
ssss
{ }4
6
4
83562sin435
224
−−
++−=−+−
ssssettL t
10
Block 2 – Laplace Transforms
Another mathematical tool that can be used to grow our collection of Laplace transforms is called the First Shift Theorem.
First Shift Theorem
If f(t) is a function having a Laplace transform, F(s) then the function,
( )tfeat
has a Laplace transform given by,
( ){ } ( )asFtfeL at −=
Sometimes,
( )asF − is denoted, ( )[ ] asssF −→
Here are a couple of worked examples using the first shift theorem to derive Laplace transforms.
Worked Example
Determine,
{ }tetL 2−
Solution
{ } ( )sFs
tL ==2
1
By the first shift theorem,
{ } ( )[ ] ( ) 222
2
1
+== +→
−
ssFetL ss
t
Worked Example
Determine,
{ }teL t 2sin3−
11
Block 2 – Laplace Transforms
Solution
{ } ( )sFs
tL =+
=4
22sin
2
By the first shift theorem,
{ } ( )[ ] ( ) 43
22sin
233
++== +→
−
ssFteL ss
t
3.2.4 Table of Laplace transforms
Deriving a Laplace transform every time a function crops up is a time consuming process. Remembering the Laplace transforms for all of the functions given above is also not a realistic proposition for most students.
Therefore in the exam the table of Laplace transforms presented below is handed out.
The table together with the first Shift theorem and the linearity property of the Laplace transform allows you to determine the Laplace transform of many functions.
12
Block 2 – Laplace Transforms
( )tf ( )sF
cs
c
t2
1
snt
1
!+ns
n
kteks −
1
atsin22 as
a
+atcos
22 as
s
+att sin
( ) 222
2
as
sa
+att cos
( ) 222
22
as
as
+−
( ) ( )tbgtaf + ( ) ( )sbGsaF +( )tf ′ ( ) ( )0fssF −
( )tf ′′ ( ) ( ) ( )002 fsfsFs ′−−( )at −δ sae−
( ) ( )
<≥−
=at
atatgtf
0
( )sGe sa−
Table of Laplace transforms.
The table above summarises the results presented in the previous two sections. In addition there are Laplace transforms for derivatives and something called the Dirac delta function. These are useful for the solution of differential equations and will be considered in detail in the coming sections.
13
Block 2 – Laplace Transforms
Challenge Exercise
Confirm by applying the first shift theorem to the Laplace transform of
{ }iatetL
that
{ } ( )222
22
cosas
asattL
+−=
and
{ } ( )222
2sin
as
asattL
+=
3.3 The inverse Laplace transform
If,
( ){ } ( )sFtfL =
Then the inverse Laplace transform is denoted, L-1 and,
( ){ } ( )tfsFL =−1
Examples
a) As
{ }ks
eL kt
−= 1
kteks
L =
−− 11
b) As
{ }22
sinas
aatL
+=
atas
aL sin
221 =
+−
14
Block 2 – Laplace Transforms
3.3.1 Using partial fractions
The most obvious way to find an inverse transformation is to use the table of Laplace transforms, however some manipulation of rational polynomials of the form,
( )( )sq
sp(3.5)
is often required. For example they might be represented as partial fractions, see Appendix A1 on partial fractions if you need your memory jogging.
Similar to the Laplace transformation the inverse Laplace transform is a linear operator, so
( ) ( ){ } ( ){ } ( ){ }sGLsFLsGsFL 111 −−− +=+ βαβα
Using partial fractions and the linear property given above we can calculate our first inverse Laplace transforms that do not appear in the Laplace transform table on page 11. See the worked example below.
Worked Example
Find
( )( )
−+−
23
11
ssL
Solution
We are ultimately going to use the first shift theorem and the table of Laplace transforms, but the first thing to do is represent the Laplace transform using partial fractions,
( )( )( ) ( )
( )( )23
32
2323
1
−+++−=
−+
+=
−+ ss
sBsA
s
B
s
A
ss
( )( )( )23
32
−++−+=
ss
BAsBA
Equating terms,
0: =+ BAs
132:1 =+− BA
These equations can be solved to give,
5
1−=A5
1=B
15
Block 2 – Laplace Transforms
( )( )
−+
+−=
−+−−−
2
1
5
1
3
1
5
1
23
1 111
sL
sL
ssL
By inspection of the table of Laplace transforms the inverse Laplace transform is,
( )( )tt ee
ssL 231
5
1
5
1
23
1 +−=
−+−−
Worked Example
Find
( )
++−
9
122
1
ss
sL
Solution
The first thing to do is represent the Laplace transform using partial fractions,
( )( ) ( )
( )9
99
99
122
2322
2222 ++++++=
++++=
++
ss
DsCssBsAs
s
DCs
s
B
s
A
ss
s
( ) ( )( )9
9922
23
++++++=
ss
BAssDBsCA
Equating terms,
0:3 =+CAs
0:2 =+ DBs
19: =As
19:1 =B
These equations can be solved to give,
9
1=A
9
1=B
9
1−=C
9
1−=D
( )
+−
+−
+
=
++ −−−−−
9
1
9
1
99
11
9
11
9
1
9
12
12
12
1122
1
sL
s
sL
sL
sL
ss
sL
16
Block 2 – Laplace Transforms
The first three terms are easy to evaluate using the table of Laplace transforms,
tts
sL
sL
sL 3cos
9
1
9
1
9
1
99
11
9
11
9
12
12
11 −+=
+−
+
−−−
The fourth term requires a little more work,
ts
Ls
Ls
L 3sin27
1
9
3
27
1
9
1
93
3
9
1
9
12
12
12
1 =
+=
+×=
+−−−
the key point being multiplying the rational function by 3 and dividing by 3 to give a Laplace transform that has the same form as one of the ‘standard’ transforms.
This is a common operation in the business of finding inverse Laplace transforms
Putting this all together,
( ) tttss
sL 3sin
27
13cos
9
1
9
1
9
1
9
122
1 −−+=
++−
3.3.2 Finding inverses using the 1st shift theorem
The first shift theorem applied to the inverse Laplace transform,
( ){ } ( )tfeasFL at=−−1
Sometimes the alternative notation is simpler to apply,
( )[ ]{ } ( )tfesFL atass =−→
−1
We will only do one example in this section, however you will have ample opportunity to see this technique throughout the rest of these notes.
Worked Example
Evaluate
( )
+−
21
2
1
sL
17
Block 2 – Laplace Transforms
Solution
( )t
ss
tes
Ls
L 2
22
12
1 1
2
1 −
+→
−− =
=
+
3.3.3 Completing the square
Sometimes a quotient of the form,
( )( )sq
sp
cannot be simplified using partial fractions. For example consider the quotient,
136
22 ++ ss
The quadratic term,
01362 =++ ss
has complex roots so cannot be factorised and therefore partial fractions does not offer a way forward in the next worked example. The alternative approach is to complete the square, see Appendix A2 if you have forgotten how to do this.
Worked Example
Evaluate
++−
136
22
1
ssL
Solution
As stated above the quotient cannot be represented as the sum of two partial fractions as the quadratic
01362 =++ ss
has no real roots. The answer is to complete the square.
( ) 43913)3(136 222 ++=−++=++ ssss
So,
18
Block 2 – Laplace Transforms
( )
++=
++−−
43
2
136
22
12
1
sL
ssL
tes
L t
ss
2sin4
2 3
32
1 −
+→
− =
+=
Worked Example
Evaluate
+++−
52
72
1
ss
sL
Solution
Completing the square on the denominator,
( ) 4115)1(52 222 ++=−++=++ ssss
So,
( )
+++=
+++ −−
41
7
52
72
12
1
s
sL
ss
sL
( )
++++= −
41
612
1
s
sL
++
+=
++=
+→
−
+→
−
+→
−
12
1
12
1
12
1
4
6
44
6
ssssss sL
s
sL
s
sL
The first term is ready for inversion,
tes
sL t
ss
2cos4 1
21 −
+→
− =
+
The second term requires a little more manipulation to see the inverse.
tes
Ls
L t
ssss
2sin34
23
4
6
12
1
12
1 −
+→
−
+→
− =
+=
+
Putting the two terms together,
19
Block 2 – Laplace Transforms
tetess
sL tt 2sin32cos
52
72
1 −−− +=
+++
Let us consider one more example, applying the same ideas to bed this in.
Worked Example
Find the inverse Laplace transform of
106
122 ++
−ss
s
Solution
Again we complete the square on the quadratic term,
( ) 13910)3(106 222 ++=−++=++ ssss
( )( )
( )
++−+=
++
−=
++− −−−
13
732
13
12
106
122
12
12
1
s
sL
s
sL
ss
sL
( ) ( )
++
−
++
+= −−
13
17
13
32
21
21
sL
s
sL
+−
+=
+→
−
+→
−
32
1
32
1
1
17
12
ssss sL
s
sL
tete tt sin7cos2 33 −− −=
This process at least initially has to be done step by step in a slow and methodical way otherwise silly mistakes will creep into the solution.
20
Block 2 – Laplace Transforms
3.4 Solution of differential equations 1
This to some degree is the end point of this block on Laplace transforms as we are studying Laplace transforms as another technique for solving differential equations.
3.4.1 Laplace transforms of derivatives
Consider the Laplace transform of the derivative of a function,
∫∞ −=
0dte
dt
df
dt
dfL st
Applying integration by parts,
∫0
∞ dfdt
e−st dt=∫ 0
∞u
dvdt
dt=uv−∫0
∞v
dudt
dt
steu −=
dtdt
dfdv =
dtsedu st−−=
fv =
( )[ ] ( )∫∫∞ −∞−∞ − +=
000dtetfstfedte
dt
df ststst
( )( ) ( ){ }tfsLfdtedt
df st +−=∫∞ − 00
0
Laplace transformation of the first derivative of a function f(t).
( ) ( )0fsFsdt
dfL −=
Similarly the Laplace transformation of a functions second derivative can be found by integration by parts,
Laplace transformation of the second derivative of a function f(t).
( ) ( ) ( )0022
2
fsfsFsdt
fdL ′−−=
21
Block 2 – Laplace Transforms
It has to be said that in the analysis above it is assumed that the function f(t) and its derivatives are sufficiently smooth for the integrals to exist.
Note the Laplace transforms of derivatives are included in the table of Laplace transforms.
3.4.2 Solution of first order differential equations
Now we have the Laplace transforms of derivatives and we can find the inverse Laplace transform we can now solve differential equations without having to integrate or differentiate anything!
Consider the following initial value problem,
teydt
dy 322 =−
(3.6)( ) 20 =y
Take the Laplace transform of both sides of the differential equation,
{ }teLydt
dyL 322 =
−
{ } { }teLyLdt
dyL 322 =−
( ) ( ) ( )3
220
−=−−
ssYysYs (3.7)
where Y(s) denotes the Laplace transform of the function, y(t).
We can now reorganise the above equation (3.7) to make the Laplace transform of y(t) the subject of the equation.
( ) ( ) 23
22 +
−=−
ssYs
( ) ( )( ) 2
2
32
2
−+
−−=
ssssY
( ) ( )( ) ( ) ( ) ( ) 3
2
32
42
32
322
−=
−−−=
−−−+=
sss
s
ss
ssY
Now all we have to do is find the inverse Laplace transform. In this case the inverse Laplace transform can be found directly from the table of Laplace transforms,
22
Block 2 – Laplace Transforms
( ){ } ( ) tes
LtysYL 311 23
2 =
−== −−
This is the solution to the initial value problem (3.6).
We have solved our first differential equation using the Laplace transform.
If you review the example given above you will see the Laplace transform method for solving differential equations can be separated into three steps,
The Laplace Transform Method
Step 1
Take the Laplace Transform of the given differential equation
Step 2
Make the transformed variable (Y(s) above) the subject of the transformed equation.
Step 3
Apply the inverse Laplace transform to find y(t).
As you can see from the example given above the initial value is included in the solution without having to introduce a constant A and then use the initial value to find the value of A.
The Laplace transform method of solving differential equations comes into its own when considering second order differential equations.
3.4.3 Solution of 2nd order differential equations
The nonhomogeneous second order constant coefficient differential equation reads,
( )tfcydt
dyb
dt
yda =++
2
2
Note using Laplace transforms we can solve this equation without the need to separate the solution into the complementary function (solution to the homogeneous problem (f(t)=0) and a particular integral to extend the solution to the nonhomogeneous differential equation. The solution of second order differential equations follows a similar line to the solution of first order differential equations using the Laplace transform method.
23
Block 2 – Laplace Transforms
Example
Solve the second order differential equation,
0652
2
=++ ydt
dy
dt
yd
Subject to the initial values,
( ) ( ) 00,10 ==dt
dyy
Solution
Apply the Laplace transform to both sides of the differential equation,
{ } 0652
2
=+
+
yLdt
dyL
dt
ydL
( ) ( ) ( ) ( ) ( )( ) ( ) 0605002 =+−+−− sYyssYdt
dysysYs
Substitute the initial values into the transformed equation,
( ) ( )( ) ( ) 06152 =+−+− sYssYssYs
Reorganise the transformed equation such that Y(s) is the subject of the equation,
( ) ( ) 5652 +=++ ssYss
( )65
52 ++
+=ss
ssY
We can now take the inverse Laplace transform of both sides,
( ){ }
+++= −−
65
52
11
ss
sLsYL (3.8)
( ){ } ( )tysYL =−1 (3.9)
Considering the right hand side of (3.8),
( )( ) 3232
5
65
52 +
++
=++
+=++
+s
B
s
A
ss
s
ss
s
24
Block 2 – Laplace Transforms
( ) ( )( )( )
( )( )( )32
23
32
23
+++++=
+++++=
ss
BAsBA
ss
sBsA
Equating terms
1: =+ BAs
523:1 =+ BA
These can be solved to give,
2,3 −== BA
32
12
1 12
13
3
2
2
3
65
5
+→+→
−−
−
=
+−
+=
+++
ssss ssssL
ss
sL
tt ee 32 23 −− −=
Putting the above together with (3.9),
( ) tt eety 32 23 −− −=
Let us do another worked example that includes a nonhomogeneous degeneracy!
Worked Example
Solve the second order differential equation,
teydt
dy
dt
yd 22
2
65 −=++
Subject to the initial values,
( ) ( ) 10,10 −==dt
dyy
Solution
Apply the Laplace transform to both sides of the differential equation,
{ } { }teLyLdt
dyL
dt
ydL 2
2
2
65 −=+
+
( ) ( ) ( ) ( ) ( )( ) ( )2
1605002
+=+−+−−
ssYyssY
dt
dysysYs
Substitute the initial values into the transformed equation,
25
Block 2 – Laplace Transforms
( ) ( )( ) ( )2
161512
+=+−++−
ssYssYssYs
Reorganise the transformed equation such that Y(s) is the subject of the equation,
( ) ( ) 42
1652 ++
+=++ s
ssYss
( ) ( ) ( ) ( ) ( )32
4
32
1
65
42
1
22 ++++
++=
++
+++=
ss
s
ssss
sssY
We can now take the inverse Laplace transform of both sides,
( ){ } ( ) ( ) ( )( )
++++
++= −−
32
4
32
12
11
ss
s
ssLsYL (3.10)
( ){ } ( )tysYL =−1
Considering the right hand side of the above equation we have two sets of partial fractions to calculate,
( ) ( ) ( ) 32232
122 +
++
++
=++ s
C
s
B
s
A
ss
( )( ) ( ) ( )( ) ( )32
23322
2
++++++++=
ss
sCsBssA(3.11)
This can be solved in the same way as the first worked example for 2nd order differential equations.
Alternatively you can initially set s=-2 and equating the LHS and RHS,
( ) 1321 =⇒+−= BB
Setting s=-3,
( ) 1231 2 =⇒+−= CC
A can now be found by equating the s2 terms on the RHS of (3.11),
10 −=⇒=+ ACA
Similarly (the details have been omitted),
26
Block 2 – Laplace Transforms
( )( ) 3
1
2
2
32
4
+−
+=
+++
ssss
s
Putting the partial fractions together,
( ) ( ) ( ) ( ) ( ) 3
1
2
2
3
1
2
1
2
1
32
4
32
122 +
−+
++
++
++
−=++
++++ sssssss
s
ss
( ) 22
1
2
1
++
+=
ss
We can now take the inverse Laplace transform of the RHS of (3.10).
( )tt
ssss
teessss
L 22
22
22
1 11
2
1
2
1 −−
+→+→
− +=
+
=
++
+
( ) ( )tety t += − 12
So we see that in principle solving homogeneous and nonhomogeneous second order differential equations can be treated in the same way using the Laplace transform method.
One last example before we move on.
Example
Solve the second order differential equation,
092
2
=+ ydt
yd
Subject to the initial values,
( ) ( ) 10,00 ==dt
dyy
Solution
Apply the Laplace transform to both sides of the differential equation,
{ } 092
2
=+
yLdt
ydL
( ) ( ) ( ) ( ) 09002 =+−− sYdt
dysysYs
Substitute the initial values into the transformed equation,
27
Block 2 – Laplace Transforms
( ) ( ) 0912 =+− sYsYs
Reorganise the transformed equation such that Y(s) is the subject of the equation,
( ) ( ) 192 =+ sYs
( )9
12 +
=s
sY
We can now take the inverse Laplace transform of both sides,
( ){ }
+= −−
9
12
11
sLsYL
( ){ } ( )tysYL =−1
Considering
ts
Ls
L 3sin3
1
9
3
3
1
9
12
12
1 =
+=
+−−
giving the solution,
( ) tty 3sin3
1=
28
Block 2 – Laplace Transforms
3.4.3.1 An LCR circuit example
Before we move on let us consider one last worked example. This is an initial value problem you have seen before, but solved using techniques considered in EPS Maths 3.
The worked example is a model for an LCR circuit, see the figure below. An LCR circuit is one that includes a resistor, a capacitor and an inductor, connected in series with a voltage source e(t).
Figure 3.1 – An LCR circuit.
Before closing the switch at time t=0 the charge on the capacitor, q and the resulting current in the circuit are zero. Applying Kirchhoff’s second law to the circuit gives a second order nonhomogeneous differential equation for the charge on the capacitor,
( )teqCdt
dqR
dt
qdL =++ 1
2
2
( ) ( ) 000 ==dt
dqq
In the circuit given above the different components have the following values,
( ) VoltsteFaradCHenryLOhmsR 2010,1,160 4 ==== −
Solution
Substituting the values of the electrical properties into the differential equation,
2010160 42
2
=++ qdt
dq
dt
qd
Taking Laplace transforms of both sides of the differential equation,
t=0R L
C q(t)e(t) i(t) q
29
Block 2 – Laplace Transforms
{ } { }2010160 42
2
LqLdt
dqL
dt
qdL =+
+
( ) ( ) ( ) ( ) ( )( ) ( )s
sQqssQdt
dqsqsQs
2010016000 42 =+−+−−
Substituting in the initial values,
( ) ( )s
sQss20
10160 42 =++
Making the Laplace transform of the solution of the differential equation the subject of the equation,
( ) ( )42 10160
20
++=
ssssQ
To take the inverse Laplace transform of the RHS it needs to be represented using partial fractions,
( )( )
( )42
242
4242 10160
10160
1016010160
20
++++++=
++++=
++ sss
CsBsssA
ss
CBs
s
A
sss
Equating terms,
BAs +=0:2
CAs +=1600:
A41020:1 =
These equations have the solution,
500
160,
500
1,
500
1 −=−== CBA
( )
+++−=
++ 4242 10160
1601
500
1
10160
20
ss
s
ssss
The quadratic term in the equation above has complex roots so the approach to take to find the inverse Laplace transform is to complete the square,
( ) ( ) ( )
++
−++
+−=
++
+−222222 6080
80
6080
801
500
1
6080
1601
500
1
ss
s
ss
s
s
The first two terms are now in a form that the inverse Laplace transform can be taken. The final term requires further manipulation,
30
Block 2 – Laplace Transforms
( ) ( )
++
−++
+−=2222 6080
60
3
4
6080
801
500
1
ss
s
s
Therefore,
( )
+−
+−
=
++ +→+→
−
8022
802242
1
60
60
3
4
60
1
500
1
10160
20
ssss ss
s
ssssL
( )
−−=
++−−− tete
sssL tt 60sin
3
460cos1
500
1
10160
20 808042
1
So the solution reads,
( )
−−= −− tetetq tt 60sin
3
460cos1
500
1 8080
3.5 Solution of Differential Equations 2
So far we have solved differential equations using the Laplace transform method without the direct application of techniques taken from calculus. It has offered a number of advantages compared to other techniques. We will consider the application of the Laplace transform method to a class of differential equations that are not easily solved any other way.
3.5.1 Differential equations and Dirac functions
Laplace transforms can be used to solve problems involving an impulsive force or current, where the impulse is delivered over a time interval, (t0, t1)
( ) ( )∫= 1
0
t
tdttftI
I(t) is the total momentum input to the system. For an electrical circuit the equivalent property is the applied voltage, V(t).
Dirac delta function
Suppose the applied force is given by the function.
( )
+<<=
otherwise
ttttf0
100 ε
εε
The above function should be interpreted as a constant force (1/ε) applied over a time interval of length ε. By construction,
31
Block 2 – Laplace Transforms
( ) ( ) 110
00=== ∫∫
+∞ εεε ε
t
tdtdttftI
Iε(t) is the area under the curve fε(t) and it is independent of ε. Taking the limit,
( ) ( )00 tttf −→→ δε ε
Where δ(t-t0) is called the Dirac delta function. This is a peculiar name for this construct as it does not have all of the properties of a function. For this reason it is a member of a class called generalised functions. The Dirac delta function is zero everywhere except at t=t0
where it is a singularity and therefore undefined. The important point is the Dirac delta function has the property,
( ) 10 0 =−∫∞
dtttδ
This is important as it represents an impulse of magnitude 1 acting over an infinitely short time interval. Another important property of the Dirac delta function is that,
( ) ( ) ( )00 0 tfdttftt =−∫∞δ
This means the Laplace transform of a Dirac delta function can be evaluated,
( ) sast edteat −∞ − =−∫0δ
This is called the sifting property of Dirac functions as it makes it possible to isolate a particular value of a function. To see this more clearly, consider,
( ){ } 10 == etL δ
and it thus follows that
{ } ( )tL δ=− 11
Therefore in principle any differential equation involving an impulse delivered over a very short time interval can be solved using Laplace transforms.
The last construct we need before we can start solving differential equations involving Dirac delta functions is the second shift theorem.
32
Block 2 – Laplace Transforms
The Second Shift Theorem
If
( ) ( )
<≥−
=at
atatgtf
0
then
( ) ( )sGesF sa−=
where( ) ( ){ }tgLsG =
This is pretty dry stuff. It will become a little clearer when you see an application of the second shift theorem.
Worked Example
Find the inverse Laplace transform of
( )2
3 1
sesF s−=
Solution
By inspection of the second shift theorem in this example,
3=aand
( )2
1
ssG =
hence,
( ){ } ts
LsGL =
= −−
211 1
then the 2nd shift theorem implies,
<≥−
=
−−
30
3312
31
t
tt
seL s
One more worked example to see how the second shift theorem can be used.
33
Block 2 – Laplace Transforms
Worked Example
Find the inverse Laplace transform of
( )7
12
+= −
sesF s
Solution
( )7
1
+=
ssG
hence,
( ){ } t
ss
ess
LsGL 7
7
11 1
7
1 −
+→
−− =
=
+=
Applying the 2nd shift theorem,
( )
<≥
=
+
−−−−
20
2
7
1 2721
t
te
seL
ts
We now have all of the tools in place to solve problems involving instantaneous impulses.
Consider a freely vibrating system without damping. The dynamics of the system is governed by a differential equation of the form,
022
2
=+ ydt
yd ω
See section 1.3.1 in the first semester of material. If the system is subjected to an instantaneous impulse of magnitude, a at a time, t=b, the second order differential equation is modified to,
( )btaydt
yd −=+ δω 22
2
Worked Example
Use Laplace transforms to solve the initial value problem,
( )342
2
−=+ tydt
yd δ
( ) ( ) 00,10 ==dt
dyy
34
Block 2 – Laplace Transforms
Solution
We follow the same solution strategy as previous examples looking at solving differential equations using the Laplace transform method.
Step 1
Take the Laplace transform of the differential equation,
{ } ( ){ }342
2
−=+
tLyLdt
ydL δ
( ) ( ) ( ) ( ) sesYdt
dysysYs 32 400 −=+−−
(See the table of Laplace transforms for the Laplace transform of a Dirac function)
Step 2
Substitute the initial values into the equation,
( ) ( ) sesYssYs 32 4 −=+−
Step 3
Make Y(s) the subject of the equation,
( ) ( ) sesYs s +=+ −32 4
( )44
122
3
++
+= −
s
s
sesY s
Step 4
Find the inverse Laplace transform for Y(s).
( ){ }
++
+= −−−−
44
12
12
311
s
sL
seLsYL s
ts
sL 2cos
421 =
+−
To find the inverse Laplace transform,
+−−
4
12
31
seL s
35
Block 2 – Laplace Transforms
we use the 2nd shift theorem. Consider,
ts
Ls
L 2sin2
1
4
2
2
1
4
12
12
1 =
+=
+−−
Therefore the 2nd shift theorem says,
( )( )
<
≥−=
+−−
30
332sin2
1
4
12
31
t
tts
eL s
Putting these results together gives the solution,
( ) ( )
≥−+
<=
362sin2
12cos
32cos
ttt
ttty
See the figure below for a plot of y(t).
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
0 2 4 6 8 10
x
y
Figure 3.2 – The displacement of a vibrating body subjected to an impulse (t=3.)
Let’s look at one last example before we move on.
36
Block 2 – Laplace Transforms
Example
Use Laplace transforms to solve the initial value problem,
( )72862
2
−=++ tydt
dy
dt
yd δ
(3.12)
( ) ( ) 60,00 ==dt
dyy
Solution
We follow the same solution strategy as in the previous example.
Step 1
Take the Laplace transform of the differential equation,
{ } ( ){ }72862
2
−=+
+
tLyLdt
dyL
dt
ydL δ
( ) ( ) ( ) ( ) ( )( ) ( ) sesYyssYdt
dysysYs 72 280600 −=+−+−−
Step 2
Substitute the initial values into the equation,
( ) ( ) ( ) sesYssYsYs 72 2866 −=++−
Step 3
Make Y(s) the subject of the equation,
( ) ( ) 686 72 +=++ − sesYss
( )86
6
86
222
7
+++
++= −
ssssesY s
Step 4
Find the inverse Laplace transform for Y(s). This is a long process in this example but provided you take your time you will see the approach is the same as the previous example.
( ){ }
+++
++= −−−−
86
6
86
22
12
711
ssL
sseLsYL s
37
Block 2 – Laplace Transforms
To find the inverses we have to represent the quotients using partial fractions as,
( )( )42862 ++=++ ssss
So,
( )( )( ) ( )
( ) ( )42
24
4242
2
+++++=
++
+=
++ ss
sBsA
s
B
s
A
ss
Equating terms,
BA
BAs
242:1
0:
+=
+=
This has solution,
1,1 −== BA
It follows that,
4286
2 77
27
+−
+=
++
−−−
s
e
s
e
sse
sss
and (steps omitted in the derivation)
4
3
2
3
86
62 +
−+
=++ ssss
42
112
1 13
13
4
13
2
13
86
6
+→+→
−−−
−
=
+−
+=
++ ssss sssL
sL
ssL
tt ee 42 33 −− −=
+−
+=
++−−−−−−
4
1
2
1
86
2 71712
71
seL
seL
sseL sss
t
ss
ess
L 2
2
1 1
2
1 −
+→
− =
=
+
t
ss
ess
L 4
4
1 1
4
1 −
+→
− =
=
+
38
Block 2 – Laplace Transforms
So by the second shift theorem,
( )
<≥
=
+
−−−−
70
7
2
1 7271
t
te
seL
ts
Similarly
( )
<≥
=
+
−−−−
70
7
4
1 7471
t
te
seL
ts
Putting this altogether, (i.e. the results in the boxes above) we get
( ) ( ) ( )
≥−+−<−
=−−−−−−
−−
733
733747242
42
teeee
teety
tttt
tt
See the figure below for a graphical solution.
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0 2 4 6 8 10
x
y
Figure 3.3 – Graphical solution of the initial value problem, (3.12).
39
Block 2 – Laplace Transforms
3.5.2 Nonlinear differential equations
Although Laplace transforms provide a good way of solving initial value problems without resorting to calculus based techniques they are restricted to linear differential equations with constant coefficients. This is differential equations of the form,
( )tfcydt
dyb
dt
yda =++
2
2
where a, b and c are constants. In this section we want to extend the range of application to nonlinear equations of the form,
( )tfydt
dyytc
dt
dy
dt
dyytb
dt
yd
dt
dyyta =
+
+
,,,,,,
2
2
For example
( ) 01 =−− yydt
dy
is a nonlinear first order differential equation. This is nonlinear as
1−= yc
There is no easy way of solving the above differential equation using Laplace transforms.
In its place we solve a linear differential equation that approximates the original nonlinear differential equation.
Approximating a nonlinear differential equation by a linear equation is called linearization. The derivation of an approximate linear differential equation uses a 1st order Taylor Series.
Recall that a Taylor series of a function f(y) centred on y0 is given by,
( ) ( ) ( ) ( ) ( ) ( ) +′′−+′−+= 02
0000 2
1yfyyyfyyyfyf
So for example consider the function,
( ) ( ) 21 yyyyyf −=−= (3.13)
( ) yyf 21−=′
Then a linear approximation to (3.13) centred on y0=2 is,
( ) ( ) ( ) ( )222 fyfyf ′−+≈
40
Block 2 – Laplace Transforms
( ) ( ) yyyf 34)3(22 −=−−+−≈ (3.14)
See the Figure below to compare the function and it’s linear approximation.
-14
-12
-10
-8
-6
-4
-2
0
2
4
6
0 1 2 3 4
y
fun
cti
on
y(1-y)
4-3y
Figure 3.4 – Comparison of a function with its linear approximation centred on y0=2.
We see that near the centre of the Taylor series expansion, y0=2, the approximation is very accurate and accuracy deteriates as you move away from the centre. Let us see how this impacts on the solution of an initial value problem,
Consider the initial value problem,
( ) 01 =−− yydt
dy
(3.15)( ) 20 =y
This is a separable differential equation (some revision) so can be solved by separating the dependent and independent variables,
( ) dtdyyy
=−1
1
and integrating both sides of the equation.
( ) ∫∫ =−
dtdyyy 1
1
CtdtRHS +== ∫41
Block 2 – Laplace Transforms
( ) ∫ ∫∫ −−=
−= dy
ydy
ydy
yyLHS
1
11
1
1 “From partial fractions”
( )
−
=−−=1
ln1lnlny
yyy
“I only need one integration constant”So
Cty
y +=
−1
ln
Rearranging the result such that y is the subject of the equation.
tCty
y
Aeey
ye ==
−= +
−
11
ln
tt
t
eA
A
Ae
Aey −−
=−
=1
the integration constant, A can be found using the initial value,
21
2 =⇒−
= AA
A
Giving the analytical solution
tey −−
=2
2
Quite a lot of work to produce an analytical solution to compare with the solution of the linear approximation to the linear problem,
Worked Example
Find an approximate solution to the initial value problem given below using a linear approximation to the function (3.13).
( ) 01 =−− yydt
dy
( ) 20 =y
42
Block 2 – Laplace Transforms
Solution
The linear approximation to the function has already been calculated, (3.14) giving the linear initial value problem,
034 =+− ydt
dy
( ) 20 =y
This is now in a form that the Laplace transform method can be applied to.
Step 1
Applying the Laplace transform
{ } { } 034 =+−
yLLdt
dyL
( ) ( ) ( ) 034
0 =+−− sYs
yssY
Step 2
Substitute the initial value and make Y(s) the subject of the equation,
( ) ( )s
s
ssYs
42423
+=+=+
( ) ( )3
42
++=
ss
ssY
Step 3
Find the inverse Laplace transform using partial fractions,
( )( )
( )3
3
33
42
+++=
++=
++
ss
BssA
s
B
s
A
ss
s
Equating terms in the numerator gives A and B,
3
2,
3
4 == BA
( ) 3
111 1
3
2
3
4
3
1
3
21
3
4
3
42
+→
−−−
+=
++
=
++
ssssL
sL
ss
sL
43
Block 2 – Laplace Transforms
te 3
3
2
3
4 −+=
Giving the result,
( ) tety 3
3
2
3
4 −+=
The figure below shows a comparison between the exact analytical solution and the approximate solution.
1
1.2
1.4
1.6
1.8
2
2.2
0 0.2 0.4 0.6 0.8 1
t
y
Exact Solution
Approximate Solution
Figure 3.5 – Comparison of the exact and approximate solution to (3.15).
From the figure above the two solutions are in close agreement for t<0.3 and the difference between the two curves grows with time. This is consistent with Figure 3.? Where as y decreases the linear approximation to the function (3.13) becomes less and less accurate.
This is not the first example of the use of linear approximation to produce a linear differential equation that can be solved that approximates a nonlinear differential equation. Recall last semester in the discussion of the oscillations of an undamped pendulum, the application of Newton’s second law (F=ma) gave the model equation,
θθsin
2
2
mgdt
dml −= (3.16)
Where θ is the angular displacement from the vertical, m is the pendulum mass, l is the length of the string and g is gravity. See Section 1.2.2.1 in the printed notes from last semester for more details. This is a second order nonlinear differential equation. Last semester the model was converted to an approximate linear model by noting that for small angular displacement,
θθ ≈sin
44
Block 2 – Laplace Transforms
Giving the linear model after a little simplification,
02
2
=+ θθl
g
dt
d(3.17)
which can be solved the conventional way or by using Laplace transforms. Now we see that the argument could run slightly differently! The linear approximation to sin θ centred on the origin is given by,
( ) ( ) ( ) ( )000 θθθθθ fff ′−+≈
( ) ( ) θθ =−+≈ 0cos00sinyf
giving the result.
The figure below shows a comparison of the analytical solution of (3.17) and a numerical solution of (3.16) calculated using MATLAB, showing in this case that introducing the linear approximation produces an error in the phase that grows progressively with time..
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
0 5 10 15 20
t
θ
Approximate Solution
Exact Solution
Figure 3.6 – Comparison of the nonlinear and linear undamped oscillator model.
45
Block 2 – Laplace Transforms
Exercise
Solve the initial value problem
02
2
=+ θθl
g
dt
d
( ) 00,2.00 === twhendt
dθπθ
using the Laplace transform method.
Engineering Application – Outflow from a water tank.
Volume conservation applied to a water tank emptying gives the equation,
outVdt
dV −=
where the right hand side is the volume flow rate out of the tank. See the schematic figure below.
Figure 3.7 Gravity driven outflow from a water tank.
It is more convenient to work with the water height, h rather than the volume, V, where,
hAV cros=
h
Acros
outV
46
Block 2 – Laplace Transforms
as if the outflow is due to gravity alone a model for the outflow is given by,
ghACV pipeDout ρ2=
There are a lot of symbols here that will not be defined as this is a mathematics module not a course in fluid mechanics, suffice it to say everything is a constant except h and V and these change with time, the remainder will soon be fixed to some convenient values so we can focus on the mathematics. Using the above the volume conservation equation becomes
ghACdt
dhA pipeDcros ρ2−=
Therefore given the initial height of the water, hmax the time the tank takes to empty can be estimated. To allow us to focus on the mathematics we group all of the constants together to give the initial value problem,
hCdt
dh −=
(3.18)( ) max0 hh =
where
gAA
CC pipe
cros
D ρ2=
Letting C=1 and hmax=1, the initial value problem can be solved (revision) to give the exact solution,
( ) 202
12
≤≤
−= tfor
tth (3.19)
Worked Example
Linearise the initial value problem (3.18) using a Taylor series of order one centred on h0=1 for the source term in (3.18) and solve using the Laplace transform method.
Solution (some steps have been omitted)
The first thing to do is derive the linear approximation to the source term. The general form of the truncated Taylor series is given below,
( ) ( ) ( ) ( )000 hfhhhfhf ′−+≈
Where in this case,
( ) hhf =
47
Block 2 – Laplace Transforms
and
( )h
hf2
1=′
thus
( ) ( ) hhhf2
1
2
1
12
111 +=−+=
The linear approximation to the original initial value problem is then,
+−= h
dt
dh
2
1
2
1
( ) 10 =h
Applying the Laplace transform method to both sides and substituting in the initial value,
+−=− H
ssH
2
1
2
11
Make H the subject of the equation,
( )12
12
+−=
ss
sH
Using partial fractions
( )2
11
21
12
12
++−=
+−=
ssss
sH
Taking inverse Laplace transforms
( ) t
ss
es
th 2
1
2
121
121
−
+→
+−=
+−=
Giving the result,
( ) 12 2
1
−=− t
eth (3.20)
The figure below shows a comparison between the approximate ( 3.20) and exact solution (3.19) to the initial value problem (3.18).
48
Block 2 – Laplace Transforms
0
0.2
0.4
0.6
0.8
1
1.2
0 0.5 1 1.5 2
t
h
Exact Solution
Approximate Solution
Figure 3.8 – Comparison of the exact and approximate solution to (3.18).
Exercise – Heating a slab up
The temperature of a slab being heated by an electric heater is given by the initial value problem,
( )44ambTTq
dt
dTC −−= α
( ) ambTT =0
where Tamb is the surrounding air temperature and has a value of 300K. C is the heat capacitance of the slab (a constant). q is the heat flux imposed by the heater (a constant) and α is called the coefficient of radiation (another constant).
Linearise the differential equation using a 1st order truncated Taylor Series centred on the initial temperature for the function,
44)( ambTTTf −=
Solve the resulting linear initial value problem using the Laplace transfer method given,
K
JC 000,1=
Wq 000,1=
49
Block 2 – Laplace Transforms
48101
K
W−×=α
Exercise – Flow across a distillation column tray
The flow dynamics for a tray in a distillation column is governed by volume conservation,
outintray VVdt
dhA −=
where the outflow term is given by a correlation for flow over a weir of the form,
5.1ChVout =
Note C is a constant. See the figure below for the configuration of the distillation tray.
Figure 3.9 – Flow over a tray in a distillation column.
The values for the physical properties in this scenario are given convenient values rather than physically correct values.
h
inV
Atray
50
Block 2 – Laplace Transforms
Given the initial value problem,
5.1232
1h
dt
dh −=
( ) 10 =h
Linearise the differential equation using a Taylor series centred on h0=1 and solve using the Laplace transform method.
3.6 Systems of differential equations
The extension of the Laplace transform method for solving differential equations is natural. Recall that for one differential equation the Laplace transform method is a three step process.
The Laplace Transform Method
Step 1
Take the Laplace Transform of the given differential equation
Step 2
Make the transformed variable (Y(s) above) the subject of the transformed equation.
Step 3
Apply the inverse Laplace transform to find y(t).
For a system of differential equations the second step is more complicated as it involves the solution of a system of (simultaneous) algebraic equations rather than one equation. By way of example consider the following initial value problem.
Example
Solve the following initial value problem using the Laplace transform method.
211 2xx
dt
dx +=
212 22 xx
dt
dx −=
( ) 201 =x , ( ) 102 =x
51
Block 2 – Laplace Transforms
Solution
The first step is to take the Laplace transform of the differential equations.
{ } { }211 2 xLxL
dt
dxL +=
{ } { }212 22 xLxL
dt
dxL −=
( ) 2111 20 XXxsX +=−
( ) 2122 220 XXxsX −=−
Substituting the initial values into the equations and reorganising to put all of the unknowns on one side of the equations,
( ) 221 21 =−− XXs (3.21)
( ) 122 21 =++− XsX (3.22)
This is two equations in two unknowns, X1 and X2 which can be solved. From (3.21)
1
12 2
1 −+=
s
XX (3.23)
Substituting for X1 in (3.22),
( ) 121
14 2
2 =++−+− Xs
s
X
which can be rearranged,
1
3
1
41
1
412
1
42 −
+=−
+−=−
+=
++
−−
s
s
s
s
sXs
s
( )( )1
3
1
1242 −
+=
−−++−
s
sX
s
ss
( )( ) 2
1
32
3
6
322 −
=+−
+=−+
+=sss
s
ss
sX
52
Block 2 – Laplace Transforms
Having found X2 we can substitute into (3.23) to give X1.
2
2
2
21
1
12
1
12
1
21 −=
−−+
−=
−
+−=
ss
s
sssX
Having solved the simultaneous equations taking the inverse Laplace transform gives the result,
{ } t
ss
esss
LXL 2
2
11
1 21
22
12
2
2 =
=
−=
−=
−→
−−
using the table of Laplace transforms and the 1st shift theorem.
Similarly
{ } tes
LXL 212
1
2
1 =
−= −−
Giving the result,
( ) tetx 21 2=
( ) tetx 22 =
What about 2nd order systems?
What about nonhomogeneous systems?
The method is exactly the same. A second order system example is given below, a nonhomogeneous first order system is left as an exercise.
Worked Example
Solve the following second order system using the Laplace transform method.
022
2
=−+ yxdt
xd
022
2
=+− yxdt
yd
( ) 40 =x , ( ) 20 =y
dt
dy
dt
dxt == 0
53
Block 2 – Laplace Transforms
Solution
Some of the details of the solution will be omitted as it is essentially the same steps as the last example.
Take the Laplace transform of the differential equations
( ) ( ) 02002 =−+
−− YX
dt
dxsxXs
( ) ( ) 02002 =+−
−− YX
dt
dysyYs
Substitute for the initial values and rearrange such that the “unknowns” are on one side
( ) sYXs 422 =−+ (3.24)
( ) sYsX 222 =++− (3.25)
From (3.24),
( ) sXsY 422 −+= (3.26)
Substituting for Y in (3.25) and rearranging such that X is the subject of the equation,
( )( )31
104
34
10422
3
24
3
+++=
+++=
ss
ss
ss
ssX
Substituting this into (3.26) and tidying up the equation gives,
( )( )31
8222
3
+++=ss
ssY
Representing X and Y using partial fractions,
( )( )( ) ( ) ( ) ( )
( )( )31
13
3131
10422
22
2222
3
+++++++=
+++
++=
+++=
ss
sDCssBAs
s
DCs
s
BAs
ss
ssX
Multiplying out the brackets on the numerator and equating terms gives the following equations.
4:3 =+CAs
0:2 =+ DBs
103: =+CAs
03:1 =+ DB
54
Block 2 – Laplace Transforms
These equations can be solved to give,
01,0,3 ==== DandCBA
Thus
31
322 +
++
=s
s
s
sX
Similarly,
31
322 +
−+
=s
s
s
sY
Taking the inverse Laplace transforms
{ }
++
+= −−−
313
21
211
s
sL
s
sLXL
Both of these inverse Laplace transforms can be found using the table of Laplace transforms,
( ) tttx 3coscos3 +=
Similarly
( ) ttty 3coscos3 −=
55
Block 2 – Laplace Transforms
There are many examples of systems of differential equations that are formulated as problems in vibration and circuit simulation, as an example application a mathematical model for the simulation of a circuit is given below.
Application
A two loop circuit is shown schematically in Figure 3.10 shown below.
Figure 3.10 – A double loop circuit.
Applying Kirchhoff’s 1st law at node X gives,
21 iii +=
Applying Kirchhoff’s 2nd law to the two loops in turn gives the differential equations,
( ) ( ) ViRiidt
dLiiR =++++ 12211211
012232
2 =−+ iRiRdt
diL
( ) ( ) 000 21 == ii
Given
OhmsROhmsRR 20,10 231 ===
HenrysLL 521 ==
VoltsV 200=
R1
R2 R
3
L1 L
2
V(t)
i1
i2
i
X
56
Block 2 – Laplace Transforms
Find the currents i1 and i2 using the Laplace transform method.
Solution
Firstly substitute the numerical values for the constant terms into the differential equations,
( ) 20020510 121
21 =+
+++ i
dt
di
dt
diii
020105 122 =−+ ii
dt
di
Dividing through by common factors in the differential equations,
( ) 4042 121
21 =++++ idt
di
dt
diii
042 122 =−+ ii
dt
di
and taking Laplace transforms
( )( ) ( )( )s
IisIisIII40
40022 1221121 =+−+−++
( )( ) 0420 1222 =−+− IIisI
Substituting in the initial values and tidying up,
( ) ( )s
IsIs40
26 21 =+++ (3.27)
( ) 024 21 =++− IsI (3.28)
From (3.28)
2
4 12 +
=s
II (3.29)
Substituting for I2 in (3.27) and making I1 the subject of the equation,
( )10
401 +
=ss
I
57
Block 2 – Laplace Transforms
From (3.29)
( ) ( )102
160
2
4 12 ++
=+
=ssss
II (3.30)
Using partial fractions,
( )( )
( )10
10
1010
401 +
++=+
+=+
=ss
BssA
s
B
s
A
ssI
equating terms in the numerators
4410
40 −=−=== ABandA
Therefore
+−
=
10
14
141 ss
I
Taking inverse Laplace transforms and applying the first shift theorem to the second term,
( ) ( )t
ss
ess
ti 10
101 14
144
10
144 −
+→
−=
−=
+−=
Similarly representing (3.30) using partial fractions and applying the 1st shift theorem (details omitted)
( ) tt eeti 1022 2108 −− +−=
3.7 Summary
This block has been organised into three separate sets of theory. The first looks at finding the Laplace transforms of functions. This can be done by evaluating the improper integral definition of the Laplace transform,
( ){ } ( )∫∞ −=
0dttfetfL st
This approach is not advocated here as a better approach is to use a table of Laplace transforms together with the first shift theorem and the linear property of the Laplace operator.
Having mastered Laplace transforms we focused on finding inverse Laplace transforms. This is a more difficult problem as it requires a certain ability to recognise patterns. The different
58
Block 2 – Laplace Transforms
techniques for finding inverse Laplace transforms, such as the use of partial fractions and completing the square are all about representing the Laplace transform in a way that it can be matched up with functions in the table of Laplace transforms. The first shift theorem in reverse can also be used to find inverse Laplace transforms.
We looked at the application of Laplace transforms to the solution of differential equations. This involves three steps,
Step 1 – Apply the Laplace transform to the differential equation
Step 2 – Rearrange the transformed equation to be explicit in the Laplace transform of the solution, Y(s)
Step 3 – Take the inverse Laplace transform to find the solution of the differential equation.
The Laplace transform method of solution for differential equations has a number of advantages over standard methods.
• The incorporation of initial values is simpler.• The solution technique requires no knowledge of calculus.• Certain problems involving near instantaneous disturbances to the modelled system
can be incorporated.
The last point above involves representing instantaneous disturbances using Dirac functions and solving the differential equation using the 2nd shift theorem.
The final part of the block looked at how this technique can be applied to nonlinear differential equations and systems of differential equations. Rather than solving a nonlinear differential equation the nonlinear term is replaced using a 1st order truncated Taylor Series centred on the initial value. This gives a linear differential equation that can be solved with Laplace transforms that is initially accurate and diverges from the exact solution for increasing time.
Linear systems of differential equations are relatively easy to solve using the Laplace transform method as you can follow the 3 step process given above. The second step for systems of differential equations includes the solution of simultaneous equations.
59
Block 2 – Laplace Transforms
Appendix A1- Partial Fractions (1st year notes)
In the same way you can add algebraic fractions together you can also separate them into combinations of algebraic fractions, such terms are called partial fractions.
Worked Example
Consider
23
1072 ++
+xx
x
Suppose we want to convert this into the sum of partial fractions.
Solution
Step 1. Factorise the denominator.
We have already used this as a worked example so we know
( )( )12
107
23
1072 ++
+=++
+xx
x
xx
x
Step 2. Assume a solution with unknown coefficients (A and B)
( )( ) 1212
107
++
+=
+++
x
B
x
A
xx
x
Step 3. Multiply both sides of the equation by the denominator (x+2)(x+1).
( )( )( )( )( )
( )( ) ( )( )1
12
2
12
12
10712
++++
+++=
+++++
x
Bxx
x
Axx
xx
xxx
and tidy up
( ) BxAxx )2(1107 +++=+
Step 4. Find A and B.
The above equation is true for all values of x. So let x be -1 and then -2.
1−=x
( ) ( )( ) ( ) BA )21(111017 +−++−=+−×
So 3=B
60
Block 2 – Laplace Transforms
2−=x
( ) ( )( ) ( ) BA )22(121027 +−++−=+−×
So 4=A
1
3
2
4
23
1072 +
++
=++
+xxxx
x
Where you have an algebraic fraction of the form,
( ) 21
2
+−
x
x
that is the denominator has a repeated root the above approach does not work you can try it if you like.
In this case you have to represent the algebraic fraction as the sum of two partial fractions in the form,
( ) ( ) 22 111
2
++
+=
+−
x
B
x
A
x
x
Multiply both sides by (x+1)2,
( ) ( )( )
( ) ( )( ) 2
22
2
2
1
1
1
1
1
21
+++
++=
+−+
x
Bx
x
Ax
x
xx
Tidy up,
( ) BAxx ++=− 12
You now equate terms to get two simultaneous equations in A and B to solve.
Ax =1:
BA +=−2:1
So
)(1 EasyA =
3−=B
61
Block 2 – Laplace Transforms
( ) ( ) 22 1
3
1
1
1
2
+−
+=
+−
xxx
x
You can also get algebraic fractions where the denominator includes quadratic terms (ax2+bx+c) that cannot be factorised into two linear factors as the roots are complex. This gives rise to a partial fraction of the form,
cbxax
BAx
+++
2
Where A and B are real constants.
Worked Example
Consider the algebraic fraction,
( )( )21
72 −++ xxx
x
Convert this into partial fractions.
Solution
Note the x2+x+1 has complex roots so cannot be factorised. The partial fraction representation is given below.
( )( ) 2121
722 −
+++
+=−++ x
C
xx
BAx
xxx
x
Now we have the partial fraction representation of the algebraic fraction we find the unknowns, A, B and C in the same way as the other examples.
62
Block 2 – Laplace Transforms
Step 1 Recombine the partial fractions over a common denominator
( ) ( ) ( )( )( )21
12
21 2
2
2 −+++++−+=
−+
+++
xxx
xxCxBAx
x
C
xx
BAx
Step 2 Multiply out the factors in the numerator, using F.O.I.L if required.
( ) ( ) ( )( )( ) ( )( )21
22
21
122
22
2
2
−+++++−+−=
−+++++−+
xxx
CCxCxBBxAxAx
xxx
xxCxBAx
Step 3 Tidy up
( )( )( ) ( )
( )( )21
22
21
72
2
2 −+++−++−++=
−++ xxx
CBxCBAxCA
xxx
x
Step 4 Equate terms
CAx +=0:2 (A1.1)
CBAx ++−= 27: (A1.2)
CB +−= 20:1 (A1.3)
Step 5 Solve system of equations to give A,B and C.
Three equations in three unknowns but the approach is the same as the other worked examples. From (A1.3),
AC −=
Substituting into the two remaining equations,
BAABA +−=⇒−+−= 3727 (A1.4)
AB −−= 20 (A1.5)
From (A1.5)
BA 2−=
Substituting for A in (A1.4) gives B.
( ) 177237 =⇒=⇒+−−= BBBB
Back substituting,
63
Block 2 – Laplace Transforms
22 −=−= BA
2=−= AC
Putting this all together gives the final result.
( )( ) 2
2
1
12
21
722 −
+++
+−=−++ xxx
x
xxx
x
The general rules for partial fractions are
Given a, b, c, l, m, n, p, q, r and s i.e. they are numbers then
( )( ) srx
B
qpx
A
srxqpx
nmx
++
+=
+++
( ) ( ) 22 qpx
B
qpx
A
qpx
nmx
++
+=
++
( )( ) qpx
C
cbxax
BAx
qpxcbxax
nmxlx
++
+++=
+++++
22
2
Multiply through by the denominator on the left hand side of the equation, tidy up and compare terms in the top line.
64
Block 2 – Laplace Transforms
Appendix A2 – Completing the square (1st year notes)
In any quadratic equation you can complete the square. Consider the quadratic function,
232 +−= xxy
You can complete the square by noting that
4
92
2
323
22 −+
−=+− xxx
This representation comes from noting that
42
222 bb
xbxx −
+=+
Where b is any number you like, such as -3. Once you have completed the square you can find the roots of the original quadratic although in the Laplace transform method that is not the point of the operation.
65