the laplace transform and initial value problems part 1

7/29/2019 The Laplace Transform and Initial Value Problems Part 1

1/7Copyright Ren Barrientos Page 1

THE LAPLACE TRANSFORM AND INITIAL VALUE PROBLEMS (Part I)

We are ready to apply the transform to initial value problems. The transform converts expressions involving

derivatives to algebraic ones involving the transform of the dependent variable because, as we learned earlier,

0 and

0 0 And by an inductive argument we also established the formula for higher derivatives:

Observe that the right hand side of this identity does not involve derivatives except for the initial conditions0which are just real numbers. Hence, the Laplace transforms converts derivatives of an unknownfunction into an algebraic equation in its transform.

It will be convenient to use the notation in place of. The formula for the transform of thenbecomes:

(3

)

and similarly for the transform of higher derivatives:

0 0 0

Here is how the transform is used to solve initial value problems:

Let us solve a very simple equation to see how this works.

Example 1 Solve the IVP 0; 0 2Solution solves the differential equation and the initial condition requires that 2. Thus, the

solution is 2 and this is something that we have seen several times before. Let us see what theLaplace transform gives us:

, , ,

Given the initial value problem:

1) Apply the Laplace transform to both sides of the differential equation to obtain an algebraic equation in.

2) Solve the algebraic equation forY(s).3) Find 1{Y(s)} to reproducey(t).

0

0 0 0 0

This side of the equation involvesderivatives of the unknown function

This side of the equation is an algebraicequation in



Taking the transform of both sides of the differential equation:

0Applying linearity: 0Applying (3),

0 0

where . Since 0 2, after collecting like terms we have: 1 2 0Solving for :

2 1Since Y(s) = {y}, we can reproduce by taking the inverse transform of both sides:

2 1 2 1

1

as expected.

Clearly this is not a method that we would use for such a simple equation, but it works and it is speciallyuseful for situations in which the forcing function is piecewise defined or periodic.

Example 2 Solve the IVP 2 ; 0 1Solution

This is a first order linear equation and may easily be solved with an integrating factor (the

integrating factor in this case is ). Let us see what the Laplace transform method gives us:Taking the transform of both sides of the differential equation:

2 By linearity, 2 Applying (3),

2 0 1 1Solving for

2 2 1 12 1 1 1 2

Hence,

1 12 1 2

2 1Decomposing the first term into partial fractions,

1 12 1 1/3 1 2/32 1Thus,

1/3 1 2/3

2 1 2

2 1 1/3 1

4/32 1

We are ready for :



13 1

1 43

12 1

13 1

1 43

12 1/2

13

1

1 23

1

1/2

Example 3 [a second order equation] Solve the IVP 1; 0 0, 0 2Solution

Once again, taking the Laplace transform of both sides, we obtain:

1or

1

0 0 1

Substituting the initial conditions 0 0, 0 2, 0 2 1

Simplifying and collecting : 1 2 1

Solving for : 1

1 2

1

Thus,

1 1 2

1It is now a matter of computing this inverse transform. The first term requires partial fractions and

the second is clearly identifiable with the sine function. Verify that

1 1

1

1

Thus,

1 1 2

1

1

1 2

1

1 Example 4 [a typical spring-mass system] Solve the IVP 8cos3; 0 1, 0 0Solution

Taking the Laplace transform of both sides, we obtain:

8cos3or



0 0 8 9

Substituting the initial conditions 0 1, 0 0,

1 0 8 9Simplifying and collecting :

1 8 9Solving for :

8 1 9

1Thus,

8 1 9

1Verify that

8 1 9 1 9Thus, 1

9

1

2 1

9

Example 5 [over-damped spring-mass system] Solve the IVP 3 4 0; 0 0, 0 2Solution


3

4

0

Thus, 3 0 0 4 0 0Substituting the initial conditions,

3 6 4 03 4 1 6

Hence,

63 4 1 6

3 1 1By partial fractions,

3

1 9

3 1

Applying the inverse transform,

3 1 1 93

1 1/3

we obtain the typical solution associated with of over-damped systems:

/ Example 6 [critically damped system] Solve the IVP 4 4 0; 0 4, 0 0.



SolutionTaking the Laplace transform of both sides, we obtain:

4 4 0or

0 04 0 4 0Substituting the initial conditions, 4 4 4 4 0

Solving for : 4 4 4 16or

4 16 4 4 4 2

16 2

We recognize the second term on the right hand side of this equation to be of the form

1/ with the shift

2 so we expect to use the First Shifting Theorem at some point. The first term needs to be treatedwith partial fractions:4

2

2

2 4 and 8Thus,

4 2 8

2 16

2 4 2

8 2

Therefore,

4

2 8

2 4 1 2 8 1

2

4 8This is a typical solution of critically damped systems.

Example 7 [resonant system] Solve the IVP 2 32 5 sin 4 ; 0 0, 0 0Solution


2 32 5sin4Thus,2 0 032 20 16

Substituting the initial conditions,

2 32 20 16 16 10 16

First shifting Theorem

with 1/



Hence,

10 16In order to find , we inverse transform:

10

16

However, a quick glance to a basic table of transforms reveals that there is no entry that can help us

to find this inverse transform. However, a more extensive table has an entry that states:

sin cos 2 Using this formula in the reverse direction with 4,

10 16 10128

128 4

This example illustrates a fairly common problem that arises in the study of the transform: how do we find the

inverse transform of the product of two functions and ? That is, how do we find ?The answer to this question comes in the form of the convolution Theorem, which states:

where and . The integral operation is called the convolution of the

functions and , denoted by :

In the previous example, we may use the convolution theorem to obtain by realizing that

and identifying the functions

1/

16and

1/

16. Thus,

10 16 10 1

16 1

16

10 14 sin4 14

sin4

1016 sin4

sin4

Exercise: Evaluate the integral and obtain .Example 8 [piecewise defined forcing functions] Solve the IVP 49 98 0 0 51 5 ; 0 2,

0 0.SolutionUsing the unit step function, we may write the forcing functions as 98 5. Hence theequation can be written as 49 98 5Taking the Laplace transform of both sides, we obtain:

49 98 5Thus,



0 0 49 98 Substituting the initial conditions:

2 49 98 Solving for :

98 49 2 49The first term needs to be decomposed by partial fractions. The second term may be identified with

the transform of a hyperbolic cosine:

98 49

2

1 7

1 7

Hence,

1 7 1

7 2

2 49

To obtain we use the inverse transform:

1 7 1 7 2 2 49 5 2| 2cosh7 2 5 2cosh7

The effect of the forcing function in the previous example can be more appreciated if we write the solution

without the unit step function:

If 5, 2cosh7If 5, 2 2cosh7

2 2cosh7 35 2 cosh7 2Hence,

2 cosh7 52cosh7 35 cosh7 2 5

is continuous at the interface 5: 2 cosh35 2cosh0 cosh 35 2 1.58 10.2 4 6 8 10

21024

41024

61024

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the laplace transform and initial value problems part 1

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