D3 Lecture 10 IITB

MA-108 Ordinary Differential Equations

M.K. Keshari

Department of MathematicsIndian Institute of Technology Bombay

Powai, Mumbai - 76

6th April, 2015D3 - Lecture 10

M.K. Keshari D3 - Lecture 10

Recall: we defined an improper integral a

g(t) dt = limT

Ta

g(t) dt

its convergence and divergence.

The Laplace transform of f(t), t 0 is

F (s) =

0

estf(t) dt

defined for those values of s for which the improper integralconverges.

(Linearity) L(c1f1 + c2f2) = c1L(f1) + c2L(f2).

(First shifting theorem) If L(f(t)) = F (s) for s > s0, thenL(eatf(t)) = F (s a) for s > s0 + a.


Laplace Transforms

1 L(1) =1

s, L(t) =

1

s2, L(tn) =

n!

sn+1, s > 0.

2 L(sint) =

s2 + 2, L(cost) =

s

s2 + 2, s > 0.

3 L(sinht) =

s2 2 , L(cosht) =s

s2 2 , s > |b|.

4 L(tneat) =n!

(s a)n+1 , s > a.5 L(eat sint) =

(s a)2 + 2 , s > a.

6 L(eat cost) =s a

(s a)2 + 2 , s > a.

7 L(eat sinh bt) =b

(s a)2 b2 , s > a+ |b|.

8 L(eat cosh bt) =s a

(s a)2 b2 , s > a+ |b|.


Existence of Laplace transform

The function f(t) = et2

does not have a Laplace transform.We defined two kinds of discontinuities at a point t0, jumpdiscontinuity, and removable discontinuity.

We defined that a function f : [0, T ] R is piecewisecontinuous, if limit at end points 0, T exist, and f iscontinuous on (0, T ) except at finite many points, where wehave discontinuities of above two types.

We defined that a function f : [0,) R is piecewisecontinuous if it is so on [0, T ] for all T > 0.



If f is piecewise continuous on closed interval [a, b], then ba

f(t)dt

exists.If f is piecewise continuous on [0,), then so is estf(t).Hence T

0

estf(t)dt

exists for every T > 0. But piecewise continuity of f does notguarantee that the Laplace transform

0

estf(t)dt = limT

T0

estf(t)dt

exists for some s (s0,). For example, take f(t) = et2 .M.K. Keshari D3 - Lecture 10


The reason is that et2

grows too rapidly in comparison to est

for any fixed s.

Definition

A function f is of exponential order s0, if there existconstants M and t0 such that

|f(t)| Mes0t, t t0.

We say f is of exponential order, if f is so of order s0 s0.Ex. f(t) = et

2is not of exponential order.

limt

et2

Mes0t= lim

t1

Met

2s0t =

So et2> Mes0t for large t, for any fixed s0,M .



Theorem

If f is piecewise continuous on [0,) and of exponential orders0, then Laplace transform L(f) is defined for s > s0.

Proof. Assume |f(t)| Mes0t, t t0. We need to showthat the integral

0

estf(t)dt = t00

estf(t)dt+ t0

estf(t)dt

converges. The first integral exists and is finite, since f ispiecewise continuous. For t > t0,

|estf(t)| estMesot =Me(sso)t

Thus the second integral converges, since it is dominated by aconvergent integral. Therefore L(f) exists.



Ex. If f is bounded on [t0,), say|f(t)| M, t t0

then f is of exponential order s0 = 0. For example, sint andcost are of exponential order 0. Thus L(sint) andL(cost) exists for s > 0 (already seen).

Exercise. If limt es0tf(t) exists and is finite, then showthat f is of exponential order s0.

Ex. If R and s0 > 0, then limt es0tt = 0Hence t is of exponential order s0 for any s0 > 0.Q. Does this mean L(t) exists for any R. No. We needpiecewise continuity for t 0.If 0, then t is continuous on (0,), hence L(t) existsfor 0.


Ex. Find Laplace transform of piecewise continuous function

f(t) =

{1, 0 t < 1et, t 1

Solution.

L(f) = F (s) =

0

estf(t)dt

=

10

estdt+ 1

estetdt

= 1sest|10 +

1s+ 1

e(s+1)t|1

=

1 es

s+e(s+1)

s+ 1, s > 1, s 6= 0

1 +1

e, s = 0


1 140100001 PAREKH MANMATH KARTIKEY2 140100024 ABHISHEK KUMAR CHAUDHARY3 140100047 RAVI JAIN4 140100054 SUNNY SONI5 140100069 CHINMAY MAHESHWARI6 140100079 SANSKAR JAIN7 14B030007 ASHU AGRAWAL ABSENT8 14B030025 MANISH GODARA9 14D100007 AUTI GUNJAN SANJAY10 14D100015 ASHOK KUMAR CHAHIL11 14D260004 YASHPAL SINGH BRAR12 14D260012 AMRITESH SHARMA13 140100016 MONISH PATHARE14 140100031 MESHRAM AYUSH DEEPAK15 140100053 DHEERAJ KUMAR VERMA16 140100066 DINESH KUMAR17 140100081 CHETAN SINGH18 14D100004 CHIKHALIKAR AKASH KEDAR


Inverse Laplace Transform

If L(f(t)) = F (s) is the Laplace transform of f , then we sayf is an inverse Laplace transform of F , write f = L1(F ).In order to solve an IVP using Laplace transform, we need tofind inverse Laplace transforms. The formula for inverseLaplace transform uses complex function theory.

L1(F (s)) =1

2pilimT

TTeT (+i)F ( + i) d,

where is suitably defined. In this course, we will use thetable of Laplace transform to find inverse transform.

Linearity Property: If F1, . . . , Fr are Laplace transforms andci R, thenL1(c1F1 + . . .+ crFr) = c1L1(F1) + . . .+ crL1(Fr).


Inverse Laplace Transform

Ex. L1(

1

s2 1)= sinh t, L1

(s

s2 + 9

)= cos 3t.

Ex.

L1(

8

s+ 5+

7

s2 + 3

)= L1

(8

s+ 5

)+ L1

(7

s2 + 3

)Note: L(f) = F = L (eatf(t)) = F (s a), L(1) = 1/s.

= 8e5t +73sin(

3t).

Ex. L1(

3s+ 8

s2 + 2s+ 5

)= L1

(3(s+ 1) + 5

(s+ 1)2 + 4

)= etL1

(3s+ 5

s2 + 4

)= etL1

(3s

s2 + 4

)+etL1

(5

s2 + 4

)= et

[3 cos 2t+

5

2sin 2t

].


If P,Q are polynomials with deg P deg Q, then L1 ofP (s)/Q(s) is found, by finding partial fractions, usingHeaviside Method.

Ex. (Q is product of distinct linear factors)

Let F (s) =6 + (s+ 1)(s2 5s+ 11)s(s 1)(s 2)(s+ 1) . Find L

1(F (s)).

The partial fraction of F (s) is of the form

F (s) =A

s+

B

s 1 +C

s 2 +D

s+ 1.

A = F (s)s|s=0 = 6 + (s+ 1)(s2 5s+ 11)

(s 1)(s 2)(s+ 1) |s=0 =17

2,

B = F (s)(s 1)|s=1 = 6 + (s+ 1)(s2 5s+ 11)

s(s 2)(s+ 1) |s=1

=6 + 2.7

2 = 10,M.K. Keshari D3 - Lecture 10

C = F (s)(s 2)|s=2 = 6 + (s+ 1)(s2 5s+ 11)

s(s 1)(s+ 1) |s=2

=6 + 3.5

6=

7

2,

D = F (s)(s+ 1)|s=1 = 6 + (s+ 1)(s2 5s+ 11)

s(s 1)(s 2) |s=1

6

6 = 1.

L1 (F (s)) = L1(17

2s 10s 1 +

7

2(s 2) 1

s+ 1

)=

17

2+ 10et + 7

2e2t et


Ex. (Q is power of a linear factor)

Let F (s) =s2 5s+ 7(s+ 2)3

. Find L1(F (s)).


F (s) =A

s+ 2+

B

(s+ 2)2+

C

(s+ 2)3.

To find A,B,C, expand the numerator of F (s) in powers of(s+ 2).

s2 5s+ 7 = ((s+ 2) 2)2 5((s+ 2) 2) + 7= (s+ 2)2 9(s+ 2) + 21.

Thus A = 1, B = 9, C = 21.Therefore, L1(F (s)) = L1

(1

s+ 2 9

(s+ 2)2+

21

(s+ 2)3

)= e2tL1

(1

s 9s2

+21

s3

)= e2t

(1 9t+ 21

2t2).


Ex. Let F (s) =8 + 3s

(s2 + 1)(s2 + 4). Find L1(F (s)).


8 + 3s

(s2 + 1)(s2 + 4)=A+Bs

s2 + 1+Cs+D

s2 + 4.

Equate the powers of s in

8 + 3s = (A+Bs)(s2 + 4) + (C +Ds)(s2 + 1)

and solve to get A,B,C,D.

We have a simpler method in this particular case, here Q is apolynomial in s2, put x = s2 in

1

(x+ 1)(x+ 4)=

1

3

(1

x+ 1 1x+ 4

)M.K. Keshari D3 - Lecture 10

1(s2 + 1)(s2 + 4)=

1

3

(1

s2 + 1 1s2 + 4

)Hence

F (s) =8 + 3s

(s2 + 1)(s2 + 4)=

1

3

(8 + 3s

s2 + 1 8 + 3ss2 + 4

)Therefore,

L1(F (s)) = L1(

8

3(s2 + 1)+

s

s2 + 1 8

3(s2 + 4) ss2 + 4

)

=

(8

3sin t+ cos t 4

3sin 2t cos 2t

).


Laplace transform of Derivatives

Our goal is to apply Laplace transforms to differentialequations. So we want to know the Laplace transform ofderivative of a function. Consider T

0

estf (t) dt = f(t)est|T0 T0

(s)estf(t) dt

= f(T )esT f(0) + s T0

estf(t) dt

If f is of exponential order s0, then as T ,f(T )esT 0 and

0estf (t) dt = L(f).

Thus we have the following theorem.


Theorem

Let f be continuous on [0,) and of exponential order s0.Let f be piecewise continuous on [0,). Then the Laplacetransform for f exists for s > s0 and is given by

L(f ) = sL(f) f(0)

We do not need f to be of exponential order.

Proof : If f was continuous on [0,), the proof is done onlast slide. If f is only piecewise continuous witht1 < t2 < . . . < tn being the discontinuities in [0, T ], then T

0

estf(t) dt =ni=1

ti+1ti

estf (t) dt

=ni=1

[f(t)est|ti+1ti

ti+1ti

(s)estf(t) dt]


= f(tn)estn est0f(t0) + s

tnt0

estf(t)

Noting that t0 = 0 and tn = T and allowing T , we getL(f ) = sL(f) f(0)

Ex. Let us compute L(cost) using that

L(sint) =

s2 + 2.

For f(t) = sint, use L(f ) = sL(f) f(0). ThenL( cost) = s

s2 + 2 0

L(cost) = s

s2 + 2

L(cost) =s

s2 + 2


Q. How does this help us solve initial value problems?Consider the ODE

y + y = 0, y(0) = 5.

We already know that the solution is given by y = 5ex. Letus verify this using Laplace transform.

Let us assume that the given equation has a solution and itis of exponential order s0 for some s0.

Then L( + ) = L(0) = sL() (0) + L() = 0.This says that L() =

5

s+ 1.

Applying inverse Laplace transform, we get that (x) = 5ex.

Remark. Solving IVP with Laplace transform requires initialconditions at t = 0.


We have have the following result about L(f (n)).

Theorem

Let f, f , . . . , f (n1) be continuous and f (n) be piecewisecontinuous on [0,). Let f, f , . . . , f (n1) be of exponentialorder s0 for some s0. Then the Laplace transforms off, f , . . . , f (n1), f (n) exists and

L(f (n)) = snL(f) f (n1)(0) sf (n2) . . . sn1f(0).

We do not need that f (n) be of exponential order.

Ex: Consider y + 4y = 3 sin t, y(0) = 1, y(0) = 1.We know this equation has a unique solution on R. Assumeit is of exponential order s0 0. Applying Laplace transformon [0,), we get that for all s > s0

L() + 4L() = 31

s2 + 1M.K. Keshari D3 - Lecture 10

L() + 4L() =3

s2 + 1

(s2L() s(0) (0)) + 4L() = 3s2 + 1

(s2 + 4)L() s+ 1 = 3s2 + 1

L() =3

(s2 + 1)(s2 + 4)+

s 1s2 + 4

L() =1

s2 + 1 2s2 + 4

+s

s2 + 4

Therefore,

(t) = sin t sin 2t+ 12cos 2t.

From uniqueness theorem, this is the solution on all of R.


Ex: Solve y + 2y + 2y = 1, y(0) = 3, y(0) = 1.The equation has a unique solution defined on all of R.Assume is of exponential of order s0. Then for all s s0,

L() + 2L() + 2L() = L(1)

(s2L() s(0) (0)) + 2(sL() (0)) + 2L() = 1s

(s2 + 2s+ 2)L() (s+ 2)(0) (0) = 1s

((s+ 1)2 + 1)L() + 3(s+ 2) 1 = 1s

L() =1 (3s+ 5)s((s+ 1)2 + 1)s

= F (s)

We want to compute L1(F (s)). We use partial fractions.


F (s) =3s2 5s+ 1((s+ 1)2 + 1)s

=A

s+B(s+ 1) + C

(s+ 1)2 + 1

= 3s2 5s+ 1 = A((s+ 1)2 + 1) + (B(s+ 1) + C)sLet s = 0,1, 1, to get the following equations.

1 = 2A, 3 = A C, 7 = 5A+ 2B + C

This implies A = 1/2, C = 5/2 and B = 7/2.

= L() = 12s 7(s+ 1)

2((s+ 1)2 + 1) 5

2((s+ 1)2 + 1)

= (t) = 12 7

2et cos t 5

2et sin t


Ex. More generally, to solve a constant coefficient IVP

y + py + qy = r(t), y(0) = a, y(0) = b, p, q R

let be the unique solution, which has a Laplace transform forall s s0. Applying Laplace transform, we get

(s2L() s(0) (0)) + p(sL() (0)) + qL() = L(r)

= (s2 + ps+ q)L() = L(r) + sa+ b+ paWe can simply this to an equation L() = F (s) and computethe inverse Laplace transform of F , to get (t).

Remark. Although the unique solution exist on R, Laplacetransform gives solution only on [0,).


Unit Step Function

Let us consider IVP with constant coefficients, where theforcing function r(t) is piecewise continuous. To solve it usingLaplace transform, we need to find Laplace transform ofpiecewise continuous functions. To do this in a systematicway, let us begin with a definition.

Definition

The unit (or Heaviside) step function is defind as

u(t) =

{0, t < 01, t 0

Replacing t by t a, we get

u(t a) ={0, t < a1, t a


Ex. Express them in terms of unit step functions.

Ramp Function =

{0, 0 < t < a

t a, t > a = (t a)u(t a).

f(t) =

{sin t, 0 < t < t0t, t t0 = sin t+ u(t t0)(t sin t).

f(t) =

sin t, 0 < t < t0cos t, t0 t t1t, t > t1

= sin t +u(t t0)(cos t sin t) +u(t t1)(tcos t).

f(t) =

f1, 0 t < t1f2, t1 t < t2. .fn, tn1 t

= f1+u(tt1)(f2f1)+ . . .+u(ttn1)(fnfn1).M.K. Keshari D3 - Lecture 10

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D3 Lecture 10 IITB