Engineering Math

24.2 Laplace Transform Methods

There is a popular algebraic technique which may be the fastest way to find closed form solutions to the initial value problem and to find the fundamental matrix. This method of Laplace¹ transforms succeeds so well because of the algebraic technique of partial fractions and the fact that the Laplace transform is a linear mapping.

This presentation will emphasize the algebraic procedures. The analytical questions are not trivial and are given a discussion in Section A.2 of the appendix.

Definition 24.2.1 Let f be a function defined on [0,∞) which has exponential growth, meaning that

|f (t)| \leq Ceλt

for some real λ. Then the Laplace transform of f, denoted by ℒ

is defined as

\int ℒf (s) = \infty e-tsf (t)dt 0

for all s sufficiently large. It is customary to write this transform as F

or ℒf

and the function as f

instead of f. In other words, t is considered a generic variable as is s and you tell the difference by whether it is t or s. It is sloppy but convenient notation.

Lemma 24.2.2 ℒ is a linear mapping in the sense that if f,g have exponential growth, then for all s large enough and a,b scalars,

ℒ (af (t)+ bg(t))(s) = aℒf (s)+ bℒg (s)

Proof: Let f,g be two functions having exponential growth. Then for s large enough,

\int \infty -ts ℒ (af (t)+ bg(t)) \equiv 0 e (af (t)+ bg(t))dt \int \infty \int \infty = a e-tsf (t)dt+ b e- tsg(t)dt = aℒf (s) + bℒg (s) ■ 0 0

The usefulness of this method in solving differential equations, comes from the following observation.

' \int \infty ' - ts -st\infty \int \infty -st ℒ (x (t)) = 0 x (t) e dt = x (t)e |0 + 0 se x (t)dt = - x(0)+ sℒx (s).

In the table, Γ

denotes the gamma function

\int \infty Γ (p+ 1) = e-ttpdt 0

The function u_c

denotes the step function which equals 1 for t > c and 0 for t < c.

The expression in Formula 20.) is defined as follows

\int δ(t- c) f (t)dt = f (c)

It models an impulse and is sometimes called the Dirac delta function. There is no such function but it is called this anyway. In the following, n will be a positive integer and f ∗g

≡∫ ₀^tf

du. Also, F

will denote ℒ

the Laplace transform of the function t → f

Table of Laplace Transforms



f (t)	F (s)


1.) 1	1∕s

2.) e^at	1∕ (s − a)

3.) tⁿ	n! sn+1-

4.) t^p,p > −1	Γ (p+1) -sp+1-

5.) sinat	a s2+a2-

6.) cosat	s s2+a2-

7.) e^ibt	ss2++ibb2-

8.) sinhat	s2a−a2-

9.) coshat	s2s−a2-

10.) e^at sinbt	(s−ab)2+b2-

11.) e^at cosbt	(s−sa−)a2+b2-


f (t)	F (s)

12.) e^at sinhbt	(s−ab)2−b2

13.) e^at coshbt	---s−2a-2 (s−a)−b

14) tⁿe^at	---n!-- (s−a)n+1

15.) u_c (t)	e−cs --s-

16.) u_c (t) f (t− c)	e^−csF (s)

17.) e^ctf (t)	F (s− c)

18.) f (ct)	1 c F (s) c

19.) f ∗ g (t)	F (s) G (s)

20.) δ (t− c)	e^−cs

21.) f^{^′} (t)	sF (s) − f (0)

22.) (− t) ⁿf (t)	dnFn ds (s)

You should verify the claims in this table. It is best if you do it yourself. The fundamental result in using Laplace transforms is this. If you have F

= G

then aside from finitely many jumps on each bounded interval, it follows that f

= g

. Thus you just go backwards in the table to find the desired functions. To see this shown, see Section A.2 on Page 1323. I will illustrate with a second order differential equation having constant coefficients. Of course you can change to a first order system and this will be the emphasis next, but you can also use the method directly. Note

\int \infty \int \infty y''(t)e- stdt = y'(t)e-st|\infty0 + s y'(t)e- stdt 0 \int \infty 0 = - y'(0)+ s y'(t)e-stdt [0 \int ] ' -st\infty \infty -st = - y (0)+ s y (t)e |0 + s 0 y(t)e dt = - y'(0)- sy(0)+ s2Y (s)

A similar formula holds for higher derivatives. You can also get this by iterating 21.

Example 24.2.3 Find all solutions to the equation y^′′− 2y^′ + y = e^−t.

From the table, first go to y^′′. This gives −y^′

−sy

+ s²Y

then you go to the next term which gives −2sY

+ 2y

and finally, you get Y

from the y. On the right you get from formula 2. 1∕

. Therefore, you have

1 s2Y (s)- 2sY (s) +Y (s)- y'(0)- sy (0) + 2y(0) = s+-1

(s2 - 2s + 1) Y (s) = y'(0)+ (s- 2)y(0)+-1-- s + 1

Thus we find the Laplace transform of the function desired.

-1- Y (s) = y'(0) ----1-----+ y(0)--s---2---+ ----s+1---- s2 - 2s+ 1 s2 - 2s+ 1 (s2 - 2s+ 1) ' 1 s - 2 s1+1 = y (0) s2 --2s+-1-+ y(0)s2 --2s+-1-+ (s2---2s+-1)

Now you go backwards in the table. This typically involves doing partial fractions to get something which is in the table. It may be tedious, but is completely routine. You can also get this from a computer algebra system. More on this later. Thus we need

( 1 ) ( s- 2 ) ( 1 ) y'(0)ℒ -1 -2-------- + y(0)ℒ- 1 -2-------- +ℒ -1 -------2--------- s - 2s+ 1 s - 2s + 1 (s+ 2)(s - 2s+ 1)

-1- 1 1 1 -2--s+1-----= ------- + -------2 - ------- (s - 2s + 1) 4 (s + 1) 2(s- 1) 4(s- 1)

Now you go backwards in the table to find that this comes from

1 - t 1 t 1 t 4 e + 2te - 4e .

Next consider the other two terms.

s- 2 1 1 s2---2s+-1 = -------2 + s---1 (s- 1)

These are in the table.

-1 ( s- 2 ) t t ℒ s2 --2s-+1 = - te + e

( ) -1 ----1----- t ℒ s2 - 2s+ 1 = te

Therefore, our solution is

( ) 1 1 1 y'(0)tet + y(0) - tet + et + 4e-t + 2tet - 4et

If you specify y^′

= y

= 1, then you will find the unique solution to the differential equation with initial conditions. It is

y (t) = 1tet + 3et + 1e-t 2 4 4

You can check that this satisfies the initial conditions and the equation.

To present this in a unified manner write as a first order system as described above.

( )' ( ) ( ) ( ) ( ) ( ) x1 = 0 1 x1 + 0 , x1(0) = a x2 - 1 2 x2 e- t x2(0) b

taking the Laplace transform in terms of the entries of the matrices, gives

( ) ( ) ( )( ) ( ) X1 (s) x1 (0) 0 1 X1 (s) 0 s X2 (s) - x2 (0) = - 1 2 X2 (s) + 1∕ (s - 1)

Thus

( ( 1 0 ) ( 0 1 )) ( X (s) ) ( a ) ( 0 ) s - 1 = + 0 1 - 1 2 X2 (s) b 1∕ (s +1)

We can solve for

( ) ( ( ) ( ) )-1 (( ) ( )) X1(s) = s 1 0 - 0 1 a + 0 X2(s) 0 1 - 1 2 b 1∕(s+ 1)

( ) as2(-sa2-s-2s2+a1+)bs(s++b1+)1- = -as-2-a+bs2+bs+s- (s- 2s+1)(s+1)

Now you use partial fractions. It equals

( 1--1+4a + 11+2b--2a-+ --1-- ) 4 s-1 -as-2a+(sb-s12)+2bs+s 4(s+1) (s2--2s+1)(s+1)-

Now the solution to our problem is x₁ and so, we only need to go backwards in the table with the top line. Thus,

y (t) = 1 (- 1 +4a)et + 1(1+ 2b- 2a)tet + 1e-t 4 2 4

As before, say both a,b are 1. Then this reduces to

1( ) y (t) = 4 3e2t + 2te2t + 1 e-t

which is the same answer as before.

Using the table as just described really is a pretty good way to solve these kinds of equations, but there is a much easier way to do it. You let the computer algebra system do the tedious work for you. Here is the general idea for a first order system. Be patient. I will consider specific examples a little later. However, if you are looking for something which will solve all first order systems in closed form using known elementary functions, then you are looking for something which is not there. You can indeed speak of it in general theoretical terms but the only problems which are completely solvable in closed form are those for which you can exactly find the eigenvalues of the matrix. Unfortunately, this involves solving polynomial equations and none of us can do these in general.