Solving Differential Equations Using the Laplace Transformation

7/29/2019 Solving Differential Equations Using the Laplace Transformation

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Chapter 4: Solving Differential Equations using the Laplace

Transform

In previous chapters we have investigated solving differential equations of the form:

)(' 011

1 tgyayayayan

n

n

n =++++

" (1)

(Note: An nth

-order linear non-homogeneous differential equation can be transformedinto a non-homogeneous system of n linear, 1

st-order differential equations).

We have seen that if the coefficients an, an-1, . . . , a0 are numbers, we can find the

general solution of the above, by first solving the corresponding homogeneous system

and then adding the particular solution of the non-homogeneous equation.The degree of difficulty in obtaining a solution for (1), depends on the nature of g(x). If

g(x) is a smooth continuous function we have investigated two techniques: i) Method ofundetermined coefficients; ii) Method of variation of parameters. On the other hand, if

g(x) is a non-smooth function, such as a piecewise-function, solving (1) using the known

techniques can become very difficult.

In this chapter, we investigate a technique that transforms (1) into an algebraic equationthat can often be easily solved, so that the solution to the differential equation can be

obtained. It works for smooth and non-smooth functions g(x). Furthermore, the

technique is also valid, even if the coefficients of the left had-side of (1) are functions ofthe independent variable t.

The new technique, classified as an integral transformation, is referred to as: LaplaceTransform.

4.0 Introduction and Background Information

Definition: Let f(t) be a function on the interval [0, ). The Laplace Transform off(t) is the function (in terms ofs ):

dttfetfL st )()}({0

=

Because the Laplace transform yields a function ofs, we often use the notation

L{ f(t) } = F(s)to denote the Laplace transform off(t).


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The Laplace transforms of functions f(t) can be computed by hand. Clearly the

calculations can become very complex and tedious, depending on the nature off(t). AComputer Algebra System, can be used to directly evaluate the Laplace transform of a

given function.

In Maple the laplace function is in the library inttrans. Lets use Maple to calculatesome Laplace transformations:

Example: Calculate the Laplace transformations of the functions:

i) f(t) = eat

ii) f(t) = sin( at); iii) f(t) = cos(at); iv) f(t) = tn

v) f(t) = 5t + e4t

sin3t

Solution: First we invoke the library inttrans and then use the function laplace:

> restart:>with(inttrans);addtable fourier fouriercos fouriersin hankel hilbert invfourier invhilbert invlaplace, , , , , , , , ,[

invmellin laplace mellin savetable, , , ]

>

> laplace( exp(a*t), t, s);1

s a

> laplace(sin(a*t), t, s);a

+s2 a2

> laplace( cos(a*t), t, s);s

+s2 a2

> assume( n>0);

> laplace(t^n, t, s);

s( ) n~ 1

( ) +n~ 1

The above can also be written as: 1

!+n

n

Note that : (n+ 1) = n!

> laplace(t^5, t, s);


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1201

s6

> laplace( 5 *t + exp(4*t) - sin(3*t), t, s);

+ 51

s21

s 43

+s2 9

The Laplace transformation obeys the linearity property which is stated by the following

theorem:

Theorem 1: Let a and b be constants. Then the Laplace transformation of a linearcombination of functions f(t) and g(t) is:

L{ a f(t) + b g(t) } = a L{ f(t) } + b L { g(t) }

In Calculus, we have seen that improper integrals may diverge. Thus the Laplace

tranform may not exist for some functions. The following definition and theorem provide

an understanding about the types of functions for which the Laplace transform exists.

Definition: A function f is of exponential order b if there are numbers b, M >0and T >0 such that:

|f(t) | M eb t

for t > T.

For example, the functions f(t) = 5t, f(t) = e-t and f(t) =10 sint are all of exponential

of order 1, since:

| 5 t| et

|e- t

| et and |10 sint | e

t for t > T

Definition: A function is piecewise continuous on interval [0, ) if in any intervala tb there are at most a finite number of points tk at which f has discontinuitiesand is continuous on each interval tk 1 < t < tk

Theorem 2: Suppose f is a piecewise continuous function on the interval [0, ) andthat is of exponential order b for t > T. Then the Laplace transform L { f(t) } exists

for s > b.


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Throughout the chapter we will only be concerned with functions that have a Laplace

transform.

Example: Find the Laplace transform of f(t) =

>

41

401

t

t

Solution: We first present a solution finding the Laplace transform by hand. Certainly

the given function is of exponential order (bounded by an exponential), so the Laplacetransform exists.

dtedtedtetftfL ststst

+==4

4

00

1)1()()}({

=

Mt

t

st

M

tst

s

e

s

e

t

=

=

=

+

= 4

4

lim0

= ( ) ( )sMsM

s ees

es

44

lim1

11

= ( )121 4 ses

Now lets use Maple to calculate the Laplace transform:

> restast:

>with(inttrans):

First we use the command "piecewise" to define the given function:

> f:= piecewise( t>=0 and t 4, 1);

:=f {-1 andt 0


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Let's get a visual idea how that function looks like:

>plot(f, t = 0..8);

> laplace(f,t,s);

laplace , ,{-1 andt 0


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Taking the Laplace transform of the above:

> laplace(Heaviside(t -4) - Heaviside(4 -t), t, s);

2 e( )4s

s1s

An important function in modeling many physical situations is the unit step function

Ua(t) which is defined as:

Definition: The unit step function U(t a) = Ua(t) is defined as:

U(t a) = Ua(t) =


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Definition: The unit impulse function a ( t t0) is defined as:

+

+ assume (t0 > 0):

> laplace(Dirac(t - t0), t, s);

e ( )s t0~


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> laplace(Dirac(t), t, s);

1

> laplace(Dirac(t - Pi), t,s);

e ( )s

Properties of the Laplace Transform:

A number of properties are stated to make the Laplace transform of various functionseasier to calculate. The properties are direct consequences of properties of integrals:

Shifting Property: IfL{ f(t) } = F(s), then L{ ea t

f(t)} = F( s a)

In order to be able to use the Laplace transform to solve differential equations, we need tobe able to compute the Laplace transform of the derivatives of arbitrary functions:

Laplace Transform of 1st

Derivative:

L{ f (t) } = s L{ f(t)} - f(0)

Laplace Transform of Higher Derivatives:

L{ f(n)

(t) } = sn

L{ f(t) } sn 1

f(0) - sn 2

f (0) - . . . s f(n 2)

(0) - f(n 1)

(0)

Derivatives of the Laplace Tranform:

L{ tn

f(t) } = n

nn

ds

sFd )()1(

Laplace transform involving the unit step function:

L{ F(t a) U(t a)} = e-a s

F(s), where F(s) = L{f(t)}


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Laplace transform of the Dirac delta function:

0)}({ 0st

ettL=

Inverse Laplace Transform:

So far we have seen how to find the Laplace transform of a function. In this section, we

reverse the process: given a function F(s) we want to find a function f(t) such that

L{ f(t) } = F(s)

Definition: The Inverse Laplace Transform of a function F(s) is the unique

continuous function f(t) on [0, ) that satisfies L{ f(t) } = F(s). We denote theinverse of the Laplace transform of F(s) as:

f(t) = L-1

{ F(s) }

The inverse Laplace transforms of functions F(s)) can be computed by hand. Clearly

the calculations can become very complex and tedious, depending on the nature ofF(s).A Computer Algebra System, can be used to directly evaluate the inverse Laplace

transform of a given function.In Maple the invlaplace function is in the library inttrans. Lets use Maple to calculate

some inverse Laplace transformations:

> restart:

>with(inttrans):

>

> invlaplace(s/(s^2 + 2* s + 5), s , t);

( )t

( )cos 2 t1

2e ( )t ( )sin 2 t

> invlaplace((3*s -4)/(s^2 - 4*s), s ,t);

+1 2 e ( )4 t >

invlaplace((2*s^3 - 4*s)/((s^2 - s)*(s^2 + 9)), s,t);

+ +1

5e t 11

5( )cos 3 t

11

15( )sin 3 t


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Properties of the inverse Laplace Transform:

Linearity Property: Let L-1

{ F(s) } and L-1

{ G(s) } be the inverse laplace transforms for

f(t) andg(t) respectively. Let a and b be constants. Then:

L-1

{a F(s) + b G(s) } = a L-1

{ F(s)} + b L-1

{ G(s) }

Note: The inverse Laplace transform does not always exists. The following statement

provides a condition that F(s) must satisfy in order forL-1

{ F(s) } to exist:

Existence of the inverse Laplace transform: If the limit:

0)}({lim)(lim ==

tfLsFss

then the inverse Laplace transform exists as a continuous or a piecewise continuousfunction. Otherwise, it does not exist.

Inverse Laplace transform involving the step function:

L-1

{e- a s

F(s) } = f(t a) U(t a), where L{ f(t) } = F(s)

4.1 Solving a Differential Equation using Laplace Transformations

Laplace transforms can be used to solve initial value problems, by transforming a

differential equation to an algebraic equation. The overall method can be summarized as

follows:

Method:

Compute the Laplace transform of each term of the differential equation Solve the resulting (algebraic) equation forL{ y(t)} = Y(s) Determine the functiony(t) by computing the inverse laplace transform of Y(s)


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Consider the initial value problem:

)(01

1

1

tgyadt

yda

dt

yda

n

n

nn

n

n

=+++

"

y(0) = y0, y(0) = y1, . . . , y(n-1)

(0) = yn-1

Taking the Laplace transform of each term, yields:

)}({}{01

1

1 tgLyLadt

ydLa

dt

ydLa

n

n

nn

n

n =++

+

"

Using the property of the laplace transformations of derivatives, we obtain:

++ )0()0()()0()0()( )2(211

)1(1 nnn

n

nnn

n yyssYsayyssYsa ""

)()(0 sGsYa =++"

We solve the resulting algebraic equation for Y(s) and then take the inverse laplace

transform to obtain the expression for Y(t).

Example: Solve the initial value problem, using the laplace transform:

y + y 6 y = sin4t, y(0) = 2, y(0) =0, y(0) = -1Plot the solution

Solution: We will use Maple to implement the steps of the solution methodology.

>with(inttrans):

We define the given differential equation:

> dfe:=diff(y(t),t$3) + diff(y(t), t$2) - 6*diff(y(t), t)=sin(4*t);

:=dfe =+

3

t3( )y t

2

t2( )y t 6

t

( )y t ( )sin 4 t

>


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Take the Laplace transform of both sides:

> step1:= laplace(dfe, t,s);

step1 s ( )s ( )s ( )laplace , ,( )y t t s ( )y 0 ( )( )D y 0 ( )( )( )D( )2

y 0:=

s ( )s ( )laplace , ,( )y t t s ( )y 0 ( )( )D y 0 6s ( )laplace , ,( )y t t s 6 ( )y 0+ + =

41

+s2 16

>

Note that Maple displays the terms of the left side, using the property of the laplace

transform of the derivatives. (multiply out the terms in the nested parentheses)

Let's substitute on the above the initial conditions (Be careful on the syntax of the

command):

> step2:= subs({y(0) = 2, D(y)(0) =0, (D@@2)(y)(0)=-1},step1);

step2 :=

+ + s2 ( )s ( )laplace , ,( )y t t s 2 13 s ( )s ( )laplace , ,( )y t t s 2 6s ( )laplace , ,( )y t t s

41

+s2 16=

>Recall that laplace(y(t),t,s) = Y(s). Thus we can solve the above algebraic equation for

Y(s):

> step3:=simplify(solve(step2, laplace(y(t), t,s)));

:=step3+ + +2s4 19s2 204 2s3 32s

s ( )+ + + s4 10s2 s3 16s 96

>

We now compute the inverse laplace ransform of the above:

> solution:= invlaplace(step3, s,t);

:=solution + 17

8

2

25e ( )2 t 7

125e ( )3 t 11

1000( )cos 4 t

1

500( )sin 4 t


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The solution can be easily plotted as:

>plot(solution, t = -2..2);

>

Example: Solve the initial value problem:

y + 8 y = f(t) where

with(inttrans):

>

> f:=t-> 1 - Heaviside(t -Pi);

:=f t 1 ( )Heaviside t

> dfe:= diff(y(t), t$2) + 8 * y(t) = f(t);

:=dfe =+

2

t2( )y t 8 ( )y t 1 ( )Heaviside t


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We take the Laplace transforms of both sides:

> step1:=laplace(dfe, t,s);

:=step1 = +s ( )s ( )laplace , ,( )y t t s ( )y 0 ( )( )D y 0 8 ( )laplace , ,( )y t t s 1

s

e ( )s s

We apply the initial conditions to the previous step:

> step2:=subs({y(0) =0, D(y)(0) =0}, step1);

:=step2 =+s2 ( )laplace , ,( )y t t s 8 ( )laplace , ,( )y t t s 1

s

e ( )s s

Solve the above for laplace(y(t), t,s) = Y(s) :

> step3:= simplify(solve(step2, laplace(y(t), t,s)));

:=step3 +1 e ( )s s ( )+s2 8

Find the solution by taking the inverse Laplace transfrom of the above:

> sol:= invlaplace(step3, s,t);

:=sol + +18

18

( )cos 2 2 ( ) +t ( )Heaviside t 18

18

( )cos 2 2 t

>plot(sol, t = 0..6);

>


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4.2 Solving Systems of Differential Equations using the Laplace

Transform

Laplace transforms can be used to solve initial value problems that involve a system oflinear differential equations. The method is applied in much the same way as

demonstrated for the case of a single initial value problem

Method:

Compute the Laplace transform of each term for each of the differential equations Solve the resulting (algebraic) system for the Laplace transform of each unknown

function

Determine the unknown functions by taking the inverse Laplace transform for each ofthe expressions resulting from the previous step.

Example: Solve the system: x = -2 x 4y - costy = -x 2y + sint

x(0) = 0, x(0) = 0, y(0) =0, y(0) =0

Solution: Lets use Maple, and follow the steps outlined in the above method:

>with(inttrans):

First we define the system of differential equations:

> sys:={diff(x(t), t$2) = - 2*x(t) - 4*y(t) - cos(t),

diff(y(t),t$2) = -x(t) - 2*y(t) + sin(t)};

:=sys { },=

2

t2( )x t 2 ( )x t 4 ( )y t ( )cos t =

2

t2( )y t +( )x t 2 ( )y t ( )sin t

Apply the Laplace transform to each equation:

> step1:=laplace(sys, t,s);

step1 s ( )s ( )laplace , ,( )x t t s ( )x 0 ( )( )D x 0 ={:=

2 ( )laplace , ,( )x t t s 4 ( )laplace , ,( )y t t s s

+s2 1,

s ( )s ( )laplace , ,( )y t t s ( )y 0 ( )( )D y 0 =

+( )laplace , ,( )x t t s 2 ( )laplace , ,( )y t t s1

+s2 1}


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Apply the initial conditions:

> step2:= subs({x(0)=0, D(x)(0) =0, y(0)=0, D(y)(0)=0},step1);

step2 =s2 ( )laplace , ,( )x t t s 2 ( )laplace , ,( )x t t s 4 ( )laplace , ,( )y t t s s+s2 1,{:=

=s2 ( )laplace , ,( )y t t s +( )laplace , ,( )x t t s 2 ( )laplace , ,( )y t t s1

+s2 1}

Solve the resulting algebraic system forX(s) and Y(s):

> step3:= solve(step2,{laplace(x(t), t,s), laplace(y(t),t,s)} );

:=step3 { },=( )laplace , ,( )y t t s + +s2 s 2

s2 ( )+ +4 s4 5s2=( )laplace , ,( )x t t s + +s

3 2s 4

s2 ( )+ +4 s4 5s2

Take the inverse Laplace transform of the each term of the above expression:

> sol:=invlaplace(step3, s,t);

sol =( )y t + + 1

2t

1

4

1

12( )cos 2 t

1

12( )sin 2 t

1

3( )cos t

1

3( )sin t ,{:=

=( )x t + + +t1

2

1

6( )cos 2 t

1

6( )sin 2 t

1

3( )cos t

4

3( )sin t }

>

4.3 Application Models using the Laplace Transform

Example: A double pendulum consists of two masses m1 and m2 attached to strings

of lengths l1 and l2 respectively. For oscillations, of small displacements, in a verticalplane, the system of differential equations describing the motion is:

0)()( 112122

2

2122

12

2

121 =++++

glmmdt

dllm

dt

dlmm and

022221

2

2122

2

22

22 =++

glmdt

dllm

dt

dlm


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Assuming that m1 =3, m2 = 1, l1 = l2 = 16, g = 32, solve the equations of motion to

determine 1(t) and 2(t). Are the motions periodic? Assume that the initial

conditions are: 1(0) = 1, 2(0) = 0, 1(0) = 0, 2(0) = -1Plot the solutions.

Solution: Since we assume small oscillations, we can use the approximations:

11sin and 22sin

Substituting the given values to the equations of motion, we obtain:

032164''16''164 122

1

2 =++

03216''16''16 222

1

2 =++

which simplifies to:

4 1 + 2 + 81 = 01 + 2 + 2 2 = 0

Lets use Maple, and solve the above system with the Laplace transforms:

>with(inttrans):

We define the system of equations of motion:

> sys:= {4*diff(theta1(t), t$2) + diff(theta2(t), t$2) +8*theta1(t) =0, diff(theta1(t), t$2) + diff(theta2(t), t$2)

+ 2*theta2(t) = 0};

sys :=

{ },=+ +4

2

t2( )1 t

2

t2( )2 t 8 ( )1 t 0 =+ +

2

t2( )1 t

2

t2( )2 t 2 ( )2 t 0

Taking the Laplce transorms of each equation:

> step1:= laplace(sys, t,s);

step1 s ( )s ( )laplace , ,( )1 t t s ( )1 0 ( )( )D 1 0{:=s ( )s ( )laplace , ,( )2 t t s ( )2 0 ( )( )D 2 0 2 ( )laplace , ,( )2 t t s+ + 0= ,

4s ( )s ( )laplace , ,( )1 t t s ( )1 0 4 ( )( )D 1 0 s ( )s ( )laplace , ,( )2 t t s ( )2 0 +

( )( )D 2 0 8 ( )laplace , ,( )1 t t s + 0= }


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Subtituting the initial conditions:

> step2:= subs({theta1(0) =1, D(theta1)(0) =0, theta2(0) =0,D(theta2)(0) = -1}, step1);

step2 {:==+ + +s ( )s ( )laplace , ,( )1 t t s 1 1 s2 ( )laplace , ,( )2 t t s 2 ( )laplace , ,( )2 t t s 0,

=+ + +4s ( )s ( )laplace , ,( )1 t t s 1 1 s2 ( )laplace , ,( )2 t t s 8 ( )laplace , ,( )1 t t s 0

}

Solving for the laplace transform of each of the unknown functions:

> step3:= solve(step2,{laplace(theta1(t), t,s),laplace(theta2(t), t,s) } );

step3 :=

{ },=( )laplace , ,( )2 t t s +8 8s 3s2

+ +3s4 16s2 16=( )laplace , ,( )1 t t s + +

2 3s3 8s

+ +3s4 16s2 16

Taking the inverse Laplace of each of the terms in the above expression:


sol =( )1 t + +1

2

cos2

33 t

1

83

sin2

33 t

1

2( )cos 2 t

1

8( )sin 2 t ,{:=

=( )2 t + ( )cos 2 t1

4 ( )sin 2 t cos

2

3 3 t1

4 3 sin

2

3 3 t }

>

Plotting each of the solutions as a function of time:

> assign(sol);

>plot(theta1(t), t = 0..44);


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>plot(theta2(t), t = 0..44);

Observe that the motion of the pendulums is not periodic.

We can also generate thephase plane portrait of the solution as:

>plot([theta1(t), theta2(t), t =0..45], numpoints=200);

The above plot is known as theLissajous curves.


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Example: Two tanks T1 and T2 are connected with two pipes, so fluid can flow fromone to another and vice versa. Tank T1 contains initially 100 gal of pure water. Tank T2

contains initially 100 gal or pure water in which 150 lb of salt are dissolved. The inflow

into T1 is 2 gal/min fm T2 and 6 gal/min containing 6 lb of salt from the outside. The

inflow into T2 is 8 gal / min from T1. The outflow from T2 is 8 gal/min. Set up a model for the concentrate of salt in each tank Solve the model Plot the solution(s)

Solution: Let y1(t) and y2(t) represent the solution concentrations of salt in T1 and

T2 tanks respectively.

Observe that: (Rate of change in concentration) = ( Inflow / min ) (Outflow / min)

Thus:

6100

2

100

8' 211 ++= yyy

212100

8

100

8' yyy =

The initial conditions for the model are: y1(0) = 0 and y2(0) = 150

Note that the above is linear 1st-order non-homogeneous system.

We can use the (power of) Laplace transform to solve the above system:

>

> restart:

>with(inttrans):

We define the system of equations of motion:

> sys:= {diff(y1(t), t) = -0.08*y1(t) + 0.02*y2(t) + 6,diff(y2(t), t) = 0.08*y1(t) - 0.08*y2(t)};

:=sys { },=t

( )y1 t + +.08 ( )y1 t .02 ( )y2 t 6 =t

( )y2 t .08 ( )y1 t .08 ( )y2 t


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Taking the Laplce transforms of each equation:

> step1:= laplace(sys, t,s);step1 s ( )laplace , ,( )y2 t t s 1. ( )y2 0. ={:=

.08000000000 ( )laplace , ,( )y1 t t s .08000000000 ( )laplace , ,( )y2 t t s ,

s ( )laplace , ,( )y1 t t s 1. ( )y1 0. =

+ +.08000000000 ( )laplace , ,( )y1 t t s .02000000000 ( )laplace , ,( )y2 t t s6.

s1.}

Substituting the initial conditions:

> step2:= subs({y1(0.) =0, y2(0.) =150}, step1);

step2 s ( )laplace , ,( )y2 t t s 150. ={:=

.08000000000 ( )laplace , ,( )y1 t t s .08000000000 ( )laplace , ,( )y2 t t s ,

s ( )laplace , ,( )y1 t t s =

+ +.08000000000 ( )laplace , ,( )y1 t t s .02000000000 ( )laplace , ,( )y2 t t s6.

s1.}

Solving for the laplace transfrom of each of the unknown functions:

> step3:= solve(step2,{laplace(y1(t), t,s), laplace(y2(t),t,s) } );

step3 =( )laplace , ,( )y1 t t s 75.+75.s 4.

s ( )+ +3. 625.s2 100.s,{:=

=( )laplace , ,( )y2 t t s 150. + +2. 50.s 625.s2

s ( )+ +3. 625.s2 100.s}

Taking the inverse Laplace of each of the terms in the above expression:


sol =( )y1 t 100. 62.50000000 e( ).1200000000 t

37.50000000 e ( ).04000000000 t ,{:==( )y2 t + 100. 125. e ( ).1200000000 t 75. e ( ).04000000000 t }

>


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Plotting each of the solutions as a function of time:

> assign(sol);

>plot(y1(t), t = 0..150);

>plot(y2(t), t = 0..150);

Superimposing the plots:

>plot([y1(t), y2(t)], t =0..150);

>We can observe that an equilibrium is "eventually" achieved where each concentration

levels off to the value of 100 lb. Clearly the same result is obtained by taking the limit, in

the expressions obtained for y1(t) andy(t), as t approaches infinity.


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4.4 Problems and Projects

Problems:

1) Find the following Laplace transforms using Maple. Rewrite each function using the

Heaviside . Make sure to verify that the re-definition of your function in terms of the

Heaviside function is correct:


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4) Rewrite the following function in terms of theHeaviside function. Show all steps to

demonstrate that your formula is correct:

>


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x = - 17 y + f(t), y = 0.25 x y f(t), where


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Documents

Solving Differential Equations Using the Laplace Transformation