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Piecewise dened functions and the Laplace transform We look at how to represent piecewise dened functions using Heavised functions, and use the Laplace transform to solve dierential equations with piecewise dened forcing terms. We repeatedly will use the rules: assume thatL(f(t)) =F(s), andc0. Then L uc(t)f(tc)=ecsF(s);L1ecsF(s)=uc(t)f(tc); where u c(t) =(

0; t < c

1; tc;Luc(t)=ecs:

For example, consider

f(t) =( 0;0t2 t2;22<1 which is plotted as follows:012345024 tf(t)It is pretty easy to see that f(t) =u2(t)(t2) Let's calculate its Laplace transform. We know from formula 3 withn= 1 thatL(t) = 1=s2, so

L(f(t)) =e2ss

2:

Consider the following function:

f(t) =8 >>:t;0t1 1;1t2

3t;2t3

0 3t <1

which is plotted as follows:

0123451012

tf(t)To write this using the Heaviside functions, lets go step by step from left to right. We start o with the functiont. Att= 1, we subtractt1 to gett(t1) = 1. At time 2, we subtractt2 to get 1(t2) = 3t. Then att= 3 we addt3 to get (3t)+(t3) = 0. So, f(t) =tu1(t)(t1)u2(t)(t2) +u3(t)(t3); and

L(f(t)) =1s

2ess 2e2ss

2+e3ss

2=1s 2

1ese2s+e3s

Sometimes you need to rewrite the function a bit to recognize that it is in delayed form.

For example, consider

u

2(t)(t3):

We want to write this in the formu2(t)f(t2) in order to use rule 13. To avoid problems, let's temporarily think offas a function ofu. We then want to ndf(u) so thatf(t2) =t3.

Lettingu=t2 is the same ast=u+ 2, so we see that

f(u) =u1

In short, we are writing

u

2(t)(t3) =u2(t)(t2)1=u2(t)(t2)u2(t):

Then

Lu2(t)(t3)=e2ss

2e2ss
where we use

L(t) =1s

2;L(1) =1s

As a more complicated example, consider the problem of writing u

1(t)sin(t) =u1(t)f(t1)

Introducing a variableu=t1; t=u+ 1, we wantf(u) = sin(u+ 1). To calculate the Laplace transform ofu1(t)sin(t), we have the problem that there is not an entry in the tables for the Laplace transform of sin(u+ 1). So we need to use trig identities to write sin(u+ 1) = cos(1)sin(u) + sin(1)cos(u) leading to the formula sin(t) = cos(1)sin(t1) + sin(1)cos(t1):

We then obtain the nal answer to our problem

L u1(t)sin(t)=Lu1(t)cos(1)sin(t1) +u1(t)sin(1)cos(t1) = cos(1)Lu1(t)sin(t1)+ sin(1)Lu1(t)cos(t1) = cos(1) ess

2+ 1+ sin(1)sess

2+ 1

Inverse Laplace transforms

Now let's do some inverse Laplace transforms. Consider

F(s) =sess

2+ 4:

We recognize thatL1

1s 2+4 = sin(2t), hence L 1ess 2+ 4 =u(t)sin(2(t)) =u(t)sin(2t2): Since sin(2t2) = sin(2t), this also equalsu(t)cos(2t), but in most examples things won't work out so nicely. We can plot this0234521012 tf(t)Now let's work outL1(F(s)) where

F(s) =se3s(s2+ 4)(s2+ 2s+ 2):

We carry out partial fractions on the rational part (at this step ignore the exponential) s(s2+ 4)(s2+ 2s+ 2)=16 s+23 s

2+ 4+16

s13 s

2+ 2s+ 2

We then splitF(s) into the two terms. For the rst term, we calculate L 1(16 s+23 )e3ss 2+ 4 =16 L1 e 3sss 2+ 4 +13 L1 e 3s2s 2+ 4 16 u3(t)cos(2(t3)) +13 u3(t)sin(2(t3)): For the second term, we need to complete the square and simplify to use the tables, 16 s13 s

2+ 2s+ 2=16

s13 (s+ 1)2+ 1=16 (s+ 1)(s+ 1)2+ 1+16 (s+ 1)2+ 1 We multiply bye3sand take the inverse Laplace transform, L 1(16 s13 )e3ss

2+ 2s+ 1

=16 L1 e

3ss+ 1(s+ 1)2+ 1

16 L1 e

3s1(s+ 1)2+ 1

=16 u3(t)e(t3)cos(t3)16 u3(t)e(t3)sin(t3):

A nice long example!

We will now carry out a problem, start to nish, with a function we looked at in lecture. The problem is long, but you should be able to follow each step and see how to get from one to the next. Recall the following function from lecture: f(t) =8 >>>>:0; t <1

1;1t <2;

2;2t <3;

3;3t <4;

0;4t <1

which is plotted as follows:01234567024 tf(t)We writef(t) =u1(t) +u2(t) +u3(t)3u4(t). To understand this formula, note that at timet= 1 the functionf(t) increases by 1, then at timet= 2 it increases by 1 more, then at timet= 3 it increases by 1 more, then at timet= 4 it decreases by 3. By the linearity of the Laplace transform and the formulaLuc(t)=ecs, we calculate L f(t)=ess +e2ss +e3ss 3e4ss
=1s e s+e2s+e3s3e4s

Now let's solve the following equation:

y

00+ 2y03y=f(t); y(0) = 1; y0(0) = 2:

We take the Laplace transform of both sides, using rule 18. AssumeL(y(t)) =Y(s). Then the equation becomes s2Y(s)sy(0)y0(0)+ 2sY(s)y(0)3Y(s) =1s e s+e2s+e3s3e4s

Inserting the values fory(0) andy0(0) we get

(s2+ 2s3)Y(s)s4 =1s e s+e2s+e3s3e4s

We solve forY(s) to obtain

Y(s) =s+ 4s

2+ 2s3+1s(s2+ 2s3)

e s+e2s+e3s3e4s We next apply partial fractions to simplify the rational functions we see. For the rst term, s+ 4s

2+ 2s3=s+ 4(s+ 3)(s1)=54

1(s1)14

1(s+ 3)

Using formula 2, we then obtain

L

1s+ 4s

2+ 2s3

=54 et14 e3t: You might note that this is the homogeneous solution satisfyingy(0) = 1; y0(0) = 2. We next nd the inverse Laplace transform of the second term. We need to apply partial fractions only to the rational part; the exponentials in the parentheses don't come up at this step. So, we write

1s(s2+ 2s3)=14

1(s1)+112

1(s+ 3)13

1s Then L

11s(s2+ 2s3)

=et4 +e3t12 13

We then get the following (somewhat long) formula

L

11s(s2+ 2s3)es+e2s+e3s3e4s

u

1(t)et14

+e3(t1)12 13 +u2(t)et24 +e3(t2)12 13 +u3(t)et34 +e3(t3)12 13

3u4(t)et44

+e3(t4)12 13 The solution to the original IVP equals this plus the rst term 54
et14 e3t.

Advanced example: square wave forcing.

This is an advanced example to illustrate the power of using the Laplace transform (and no, it won't be on the nal exam). We consider driving an undampened harmonic oscillator by a square wave that has the same period as the homogeneous solution. The square wave is a step function approximation to cos(t):02345678921012 tf(t)We can represent the square wave as a sum of Heaviside functions: f(t) =u0(t)2u(t) + 2u2(t)2u3(t) + 2u4(t)2u5(t) + 2u6(t)2u7(t) + You can imagine that the wave stops after a nite number of terms so we don't have to worry about innite sums, but the following steps don't require that. We rst calculate the

Laplace transform off(t):

L(f(t)) =L(u0(t))2L(u(t)) + 2L(u2(t))2L(u3(t)) + 2L(u4(t))2L(u5(t)) 1s 2ess +2e2ss
2e3ss
+2e4ss
2e5ss
1s

12es+ 2e2s2e3s+ 2e4s2e5s+

Now suppose we want to solve the equation

y

00+y=f(t); y(0) = 0; y0(0) = 0:

Taking the Laplace transform of both sides leads to the equation forY(s) =L(y(t)), (s2+ 1)Y(s) =1s

12es+ 2e2s2e3s+ 2e4s2e5s+

We divide bys2+ 1 and apply partial fractions to write 1s

2+ 11s

=1s ss 2+ 1; and obtain the formula forY(s):

Y(s) =1s

12es+ 2e2s2e3s+ 2e4s2e5s+

ss 2+ 1

12es+ 2e2s2e3s+ 2e4s2e5s+

If we take the inverse Laplace transform, we get the following formula fory(t): y(t) =u0(t)2u(t) + 2u2(t)2u3(t) + cos(t) + 2u(t)cos(t)2u2(t)cos(t2) + 2u3cos(t3) We next use cos(t) =cos(t), cos(t2) = cos(t), et cetera to write this as y(t) =u0(t)2u(t) + 2u2(t)2u3(t) + cos(t) u

0(t) + 2u(t) + 2u2(t) + 2u3(t) +

This is a little complicated, but you can see that the rst line is justf(t), and bounces between 1 and1. The second line is cos(t) times a function that jumps by 2 each timet increases by, so it will grow proportional tot, just like resonant driving. On the following page we compare square wave forcing to cos(t) forcing (both forcing terms at the same frequency as the homogeneous frequency). They look qualitatively the same, but a close inspection shows some dierences. The physical explanation for why the solution for the square wave force grows faster is that the square wave contains more energy than the cosine wave.

Square wave forcing:

Solution toy00+y=f(t); y(0) = 0; y0(0) = 0, wheref(t) is the square wave.0234567891510505101520 ty(t)Resonant forcing: The solution toy00+y= cos(t); y(0) = 0; y0(0) = 0, is given byy=12 tsint.023456789105051015 t:5tsin(t)quotesdbs_dbs6.pdfusesText_11

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