Ordinary Di erential Equations. Session 7roquesol/Math_308_Fall_2019_Session_7_Print.pdfLaplace...

Laplace Transform

Ordinary Differential Equations. Session 7

Dr. Marco A Roque Sol

10/08/2019

Dr. Marco A Roque Sol Ordinary Differential Equations. Session 7

Laplace TransformDefinition of The Laplace TransformSolution of Initial Value ProblemsStep Functions

Definition of The Laplace Transform

Laplace Transform

Among the tools that are very useful for solving linear differentialequations are integral transforms. An integral transform is arelation of the form

F (s) =

∫ β

αK (s, t)f (t)dt

where K (s, t) is a given function, called the kernel of thetransformation, and the limits of integration α and β are alsogiven. It is possible that α = −∞ or β =∞ or both. The relation,introduced above, transforms the function f into another functionF , which is called the transform of f .




There are several integral transforms that are useful in appliedmathematics, but we consider only the Laplace Transform (https://en.wikipedia.org/wiki/Pierre-Simon_Laplace )

Laplace Transform

Let f (t) be given for t ≥ 0. Then the Laplace transform of f ,which we will denote by L {f (t)} = F (s), is defined by theequation

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

∫ ∞0

e−st f (t)dt

whenever this improper integral converges.


https://en.wikipedia.org/wiki/Pierre-Simon_Laplace



The Laplace transform makes use of the kernel K (s, t) = e−st . Inparticular for linear second order differential equations withconstant coeficients is particular useful, since the solutions arebased on the exponential function.

The general idea in using the Laplace transform to solve adifferential equation is as follows:

1. Use the relation L {f (t)} = F (s) to transform an initial valueproblem for an unknown function f in the t-domain (time domain)into an algebraic problem for F in the s-domain (frequencydomain).




2. Solve this algebraic problem to find F .

3. Recover the desired function f from its transform F . This laststep is known as inverting the transform (which in general involvecomplex integration) and denoted by L −1{F (s)}(= limω→∞

12πi

∫ σ+iωσ−iω F (s)estds).

OBS The full power of Laplace Transform becomes available onlywhen we regard F (s) as a function of a complex variable. However,for our purposes it will be enough to consider only real values for s.

The Laplace transform F of a function f exists if f satisfies certainconditions:




Theorem

Suppose that

1. f is piecewise continuous on the interval 0 ≤ t ≤ A for anypositive A.

2. |f (t)| ≤ Keat when t ≥ M. In this inequality, K , a, and M arereal constants, K and M necessarily positive.

Then the Laplace transform L {f (t)} = F (s), defined by

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

∫ ∞0

e−st f (t)dt

exists for s > a.




Remember that a function, f (t), is piecewise continuous on theinterval α ≤ t ≤ β if the interval can be partitioned by a finitenumber of points α = t0 < t1 < . . . < tn = β so that

1. f is continuous on each open subinterval ti−1 < t < ti .

2. f approaches a finite limit as the endpoints of each subintervalare approached from within the subinterval.




Example 62

Find the Laplace transform for f (t) = 1, t ≥ 0

Solution

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

∫ ∞0

e−st f (t)dt

L {1} = F (s) =

∫ ∞0

e−stdt = − limA→∞

e−st

s

∣∣∣A0

=1

s; s > 0




Example 63

Find the Laplace transform for f (t) = eat , t ≥ 0

Solution

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

∫ ∞0

e−st f (t)dt

L {eat} = F (s) =

∫ ∞0

eate−stdt =

∫ ∞0

e−(s−a)tdt =

− limA→∞

e−(s−a)t

s − a

∣∣∣A0

=1

s − a; s > a




Example 64

Find the Laplace transform for

f (t) =

1 0 ≤ t < 1k t = 10 t > 1

where k is a constant. In engineering contexts f (t) oftenrepresents a unit pulse, perhaps of force or voltage.

Solution

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

∫ ∞0

f (t)e−stdt =

∫ 1

0e−stdt =

−e−st

s

∣∣∣10

=1− e−s

s




Example 65

Find the Laplace transform for f (t) = sin(at), t ≥ 0

Solution

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

∫ ∞0

e−st f (t)dt =

∫ ∞0

sin(at)e−stdt = Int by Parts

F (s) = limA→∞

[−e−stcos(at)

a

∣∣∣A0− s

a

∫ A

0e−stcos(at)dt

]= Int by Parts

F (s) =1

a− s2

a2

∫ ∞0

sin(at)e−stdt =1

a− s2

a2F (s)




Hence, solving for F(s), we have

F (s) =a

s2 + a2

Now, the Laplace transform is a linear operator, that is, supposethat f1 and f2 are two functions whose Laplace transforms exist fors > a1 and s > a2, respectively. Then, for s > max{a1, a2}

L {c1f1 + c2f2} =

∫ ∞0

e−st{c1f1 + c2f2}dt =

c1

∫ ∞0

e−st f1dt + c2

∫ ∞0

e−st f2dt = c1L {f1}+ c2L {f2}

Thus, we have

L {c1f1 + c2f2} = c1L {f1}+ c2L {f2}Dr. Marco A Roque Sol Ordinary Differential Equations. Session 7



OBS

L −1{d1F1 + d2F2} = d1L−1{F1}+ d2L

−1{F2}Example 66

Find the Laplace transform for f (t) = 5e−2t − 3sin(4t), t ≥ 0

Solution

L {5e−2t − 3sin(4t)} = 5L {e−2t} − 3L {sin(4t)} =

L {5e−2t − 3sin(4t)} =5

s + 2− 12

s2 + 16; s > 0

L −1{ 5

s + 2− 12

s2 + 16} = 5L −1{ 1

s + 2} − 3L −1{ 4

s2 + 42} =

5e−2t − 3sin(4t)Dr. Marco A Roque Sol Ordinary Differential Equations. Session 7


Solution of Initial Value Problems

To see how we can apply the method of Transform Laplace tosolve linear differential equations with constant coefficients. Weestablish the following results.

Theorem

Suppose that f is continuous and f ′ is piecewise continuous on anyinterval 0 ≤ t ≤ A. Suppose further that there exist constantsK , a, and M such that |f (t)| ≤ Keat for t ≥ M. Then

L {f ′} = sL {f } − f (0)

proof

L {f , (t)} =

∫ ∞0

e−st f ′(t)dt = limA→∞

∫ A

0e−st f ′(t)dt =




limA→∞

[∫ t1

0e−st f ′(t)dt +

∫ t2

t1

e−st f ′(t)dt +

∫ t3

t2

e−st f ′(t)dt + ...+

∫tn−1

tn = Ae−st f ′(t)dt

]=

and integrating by parts, we have

limA→∞

{e−st f (t)

∣∣∣t10

+ e−st f (t)∣∣∣t2t1

+ ...+ e−st f (t)∣∣∣tn=A

tn−1

+

s

[∫ t1

0e−st f ′(t)dt +

∫ t2

t1

e−st f ′(t)dt + ...+

∫ tn=A

tn−1

e−st f ′(t)dt

]}=




limA→∞

[e−sAf (A)− f (0) + s

∫ A

0e−st f (t)dt

]= s

∫ A

0e−st f (t)dt − f (0)

In his way we obtain

L {f ′} = sL {f } − f (0)

As a corollary, we have the following

Corollary

Suppose that the functions f , f ′, ..., f (n−1) are continuous and thatf (n) is piecewise continuous on any interval 0 ≤ t ≤ A.




Suppose further that there exist constants K , a, and M such that|f (t)| ≤ Keat , |f ′(t)| ≤ Keat , ..., |f (n−1)(t)|Keat for t ≥ M. ThenL {f (n)(t)} exists for s > a and is given by

L {f (n)(t)} = snL {f (t)} − sn−1f (0)

−sn−2f ′(0) . . .− sf (n−2)(0)− f (n−1)(0)

We use the previous results to solve IVP’s using LaplaceTransform. It is most useful for problems involvingnonhomogeneous differential equations. However, just to illustratethe method we will start with a homogeneus case




Example 67

Consider the IVP

y ′′ − y ′ − 2y = 0; y(0) = 1, y ′(0) = 0

Solution

Using the traditional method we find that the general solution is

y(x) = c1e−t + c2e

2t

and applying initial conditions we get c1 = 2/3 and c2 = 1/3.Hence, the particular solution is

y(x) =2

3e−t +

1

3e2t




Now, let us solve the same problem by using the Laplacetransform. We start off with the differential equation

y ′′ − y ′ − 2y = 0

Applying the Laplace Transform

L {y ′′ − y ′ − 2y = 0}because of linearity

L {y ′′} −L {y ′} − 2L {y} = 0

and using corollary 6.2.1

s2L {y} − sy(0)− y ′(0)− [sL {y} − y(0)]− 2L {y} = 0




Taking Y (s) = L {y} and applying initial coditions(y(0) = 1, y ′(0) = 0), we obtain

s2Y (s)− sy(0)− y ′(0)− [sY (s)− y(0)]− 2Y (s) = 0 =⇒

Y (s) =s − 1

s2 − s − 2=

s − 1

(s − 2)(s + 1)

The above can be written, using partial fractions, as

Y (s) =1/3

s − 2+

2/3

s + 1

Now, applying the inverse Laplace transform




y(t) = L −1{Y (s)} = L −1{

1/3

s − 2+

2/3

s + 1

}

y(t) = L −1{Y (s)} =1

3L −1

{1

s − 2

}+

2

3L −1

{1

s + 1

}but we know that L {eat} = 1

s−a or equivalently L −1{ 1s−a} = eat

y(t) =1

3e2t +

2

3e−t

The same procedure can be applied to the general second orderlinear equation with constant coefficients

ay ′′ + by ′ + cy = f (t)




obtaining

Y (s) =(as + b)y(0) + ay ′(0)

as2 + bs + c+

F (s)

as2 + bs + c; F (s) = L {f (t)}

The main difficulty that occurs in solving initial value problems bythe transform method lies in the problem of determining thefunction y = y(t), corresponding to the inverse transform of Y (s).

Since we will not deal with the formula for the inverse transform,because it requires complex integration, we will use a table ofcommon Laplace Transform for basic functions (http://www.math.tamu.edu/~roquesol/Laplace_Table.pdf)


http://www.math.tamu.edu/~roquesol/Laplace_Table.pdf

http://www.math.tamu.edu/~roquesol/Laplace_Table.pdf



Example 68

Consider the IVP

y ′′ + y = sin(2t); y(0) = 2, y ′(0) = 1

Solution

Let us solve the problem by using the Laplace transform. We startoff with the differential equation

y ′′ + y = sin(2t)


L {y ′′ + y = sin(2t)}




because of linearity

L {y ′′}+ L {y} = L {sin(2t)}

and using a previous corollary

s2L {y} − sy(0)− y ′(0) + L {y} =2

s2 + 4

taking Y (s) = L {y} and applying initial coditions(y(0) = 2, y ′(0) = 1), we obtain

s2Y (s)− 2s − 1 + Y (s) =2

s2 + 4=⇒

Y (s) =2s3 + s2 + 8s + 6

(s2 + 1)(s2 + 4)





Y (s) =as + b

s2 + 1+

cs + d

s2 + 4=

(as + b)(s2 + 4) + (cs + d)(s2 + 1)

(s2 + 1)(s2 + 4)=

(a + c)s3 + (b + d)s2 + (4a + c)s + (4b + d)

(s2 + 1)(s2 + 4)=

2s3 + s2 + 8s + 6

(s2 + 1)(s2 + 4)

Then, comparing coefficients of like powers of s, we have

a + c = 2; b + d = 1;

4a + c = 0; 4b + d = 6;




Therefore, a = 2, b = 5/3, c = 0, d = −2/3 and

Y (s) =2s

s2 + 1+

5/3

s2 + 1− 2/3

s2 + 4

Now, taking the inverse, we have

y(t) = 2cos(t) +5

3sin(t)− 1

3sin(2t)




Example 69

Consider the IVP

y (4) − y = 0; y(0) = 0, y ′(0) = 1, y ′′(0) = 0, y ′′′(0) = 0

Solution

Let us solve the problem by using the Laplace transform. We startoff with the differential equation

y (4) − y = 0


L {y (4) − y = 0}





L {y (4)} −L {y} = 0

and using corollary 6.2.1

s4L {y} − s3y(0)− s2y ′(0)− sy ′′(0)− y ′′′(0)− Y (s) = 0

taking Y (s) = L {y} and applying initial coditions(y(0) = 0, y ′(0) = 1, y ′′(0) = 0, y ′′′(0) = 0), we obtain

s4L {y} − s2y ′(0)− Y (s) = 0 =⇒

Y (s) =s2

s4 − 1=

s2

(s2 − 1)(s2 + 1)





Y (s) =as + b

s2 − 1+

cs + d

s2 + 1=

(as + b)(s2 + 1) + (cs + d)(s2 − 1)

(s2 − 1)(s2 + 1)=

(a + c)s3 + (b + d)s2 + (a− c)s + (b − d)

(s2 − 1)(s2 + 1)=

s2

(s2 − 1)(s2 + s)

Then, comparing coefficients of like powers of s, we have

a + c = 0; b + d = 1;

a− c = 0; b − d = 0;




therefore, a = 0, b = 1/2, c = 0, and d = 1/2, and

Y (s) =1/2

s2 − 1+

1/2

s2 + 1

and take the inverse, we have

y(t) =sinh(t) + sin(t)

2=

(et − e−t)/2 + sin(t)

2=

et

4− e−t

4+

sin(t)

2




Example 70

The Laplace transforms of certain functions can be foundconveniently from their Taylor series expansion. Using the Taylorseries for sin(t)

sin(t) =∞∑n=0

(−1)nt2n+1

(2n + 1)!

Let

f (t) =

{sin(t)

t t 6= 01 t = 0




Find the Taylor series for f about t = 0. Assuming that theLaplace transform of this function can be computed term by term,determine L {f (t)}

Solution

For t 6= 0 we have that f can written as

f (t) =sin(t)

t=∞∑n=0

(−1)nt2n+1

(2n + 1)!t=∞∑n=0

(−1)nt2n

(2n + 1)!


L {f (t)} = L

{ ∞∑n=0

(−1)nt2n

(2n + 1)!

}





L {f (t)} =∞∑n=0

(−1)n

(2n + 1)!L {t2n}

and using the table of Laplace Transforms, L {tm} = m!sm+1

L {f (t)} =∞∑n=0

(−1)n

(2n + 1)!

(2n)!

s2n+1

L {f (t)} =∞∑n=0

(−1)n(2n)!

(2n + 1)!s2n+1=∞∑n=0

(−1)n(2n)!

(2n + 1)(2n)!s2n+1




L {f (t)} =∞∑n=0

(−1)n

(2n + 1)s2n+1=∞∑n=0

(−1)n(1/s)2n+1

(2n + 1)

but, if we remember

tan−1(x) =∞∑n=0

(−1)nx2n+1

(2n + 1)

Therefore, we have

L {f (t)} = tan−1(1/s)



Step Functions

To deal effectively with functions having jump discontinuities, it isvery helpful to introduce a function known as the unit stepfunction or Heaviside function. This function will be denoted byuc and is defined by

uc(t) =

{0 t < c1 t ≥ c



Step Functions

The step can also be negative. For instance

u(t) = (1− uc(t)) =

{1 t < c0 t ≥ c



Step Functions

In fact, any piecewise-defined function can be written as a linearcombination of uc(t)’s functions. For instance consider thefunction

f (t) =

2 0 ≤ t < 11 1 ≤ t < 22 2 ≤ t < 30 3 ≤ t



Step Functions

We start with the function f1(t) = 2u0, which agrees with f (t) on[0, 1). To produce the jump down of one unit at t = 1, we add−u1(t) to f1(t), obtaining f2(t) = 2u0 − u1(t), which agrees withf (t) on [1, 2). The jump of one unit at t = 2 corresponds toadding u2(t), which gives f3(t) = 2u0 − u1(t) + u2(t). Thus weobtain

f (t) = f3(t) = 2u0 − u1(t) + u2(t)

The Laplace transform of uc for c ≥ 0 is easily determined:

L {uc} =

∫ ∞0

e−stucdt =

∫ ∞c

e−stdt =e−cs

s, s > 0



Step Functions

For a given function f defined for t ≥ 0, we will often want toconsider the related function g defined by

g(t) = uc f (t − c)

which represents a translation of f a distance c in the positive tdirection.



Step Functions

Theorem

If F (s) = L {f (t)} exists for 0 ≤ a < s, and if c is a positiveconstant, then

L {uc f (t − c)} = e−csL {f (t)} = e−csF (s),

Conversely, if f (t) = L −1{F (s)}, then

L −1{e−csF (s)} = uc f (t − c)



Step Functions

Example 71

If the function f is defined

f (t) =

{sin(t) 0 ≤ t < π/4sin(t) + cos(t − π/4) π/4 ≤ t

find L {f (t)}.



Step Functions

Solution

Note that f (t) = sint + g(t), where

g(t) =

{0 0 ≤ t < π/4cos(t − π/4) π/4 ≤ t

Thus

g(t) = uπ/4cos(t − π/4)



Step Functions

and

L {f (t)} = L {sin(t)}+ L {uπ/4cos(t − π/4)} =

L {sin(t)}+ e−πs/4L {cos(t)}

and using the table of Laplace Transforms

L {f (t)} =1

s2 + 1+ e−πs/4

s

s2 + 1=

1 + se−πs/4

s2 + 1



Step Functions

Let’s consider the following theorem

Theorem

If F (s) = L {f (t)} exists for 0 ≤ a < s, and if c is a constant, then

L {ect f (t)} = F (s − c), s > a + c

Conversely, if f (t) = L −1{F (s)}, then

L −1{F (s − c)} = ect f (t)



Step Functions

Example 72

Find the inverse transform of

H(s) =1

s2 − 4s + 5Solution

First of all the polynomial s2 − 4s + 5, has complex roots. Bycompleting the square in the denominator, we can write

H(s) =1

(s − 2)2 + 1= F (s − 2)

where F (s) = (s2 + 1)−1. L −1{F (s)} = sin(t). It follows fromthe previous theorem that

h(t) = L −1{H(s)} = L −1{F (s − 2)} = e2tsin(t)



Step Functions

Shift in the time − domain (t − domain)

If F (s) = L {f (t)} exists for 0 ≤ a < s, and if c is a positiveconstant, then

L {uc f (t − c)} = e−csF (s) L −1{e−csF (s)} = uc f (t − c)

Shift in the frequency − domain (s − domain)

If F (s) = L {f (t)} exists for 0 ≤ a < s, and if c is a constant, then

L {ect f (t)} = F (s − c), s > a + c L −1{F (s − c)} = ect f (t)


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