Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 9 docx

31.5 Systems of Constant Coefficient Differential EquationsThe Laplace transform can be used to transform a system of constant coefficient differential equations into a system of algebra

We use the convolution theorem to find the inverse Laplace transform of ˆy(s).

y(t) =

Z t 0

1

2sin(2τ ) cos(t − τ ) dτ + cos t

= 14

Z t 0

sin(t + τ ) + sin(3τ − t) dτ + cos t

= 14



− cos(t + τ ) − 1

3cos(3τ − t)

t 0

+ cos t

= 14

= −1

3cos(2t) +

4

3cos(t)Alternatively, we can find the inverse Laplace transform of ˆy(s) by first finding its partial fraction expansion

ˆy(s) = s/3

3cos(2t) +

4

3cos(t)Example 31.4.3 Consider the initial value problem

y00+ 5y0 + 2y = 0, y(0) = 1, y0(0) = 2

Without taking a Laplace transform, we know that since

y(t) = 1 + 2t + O(t2)the Laplace transform has the behavior

ˆy(s) ∼ 1

s +2

s2 + O(s−3), as s → +∞

31.5 Systems of Constant Coefficient Differential Equations

The Laplace transform can be used to transform a system of constant coefficient differential equations into a system

of algebraic equations This should not be surprising, as a system of differential equations can be written as a singledifferential equation, and vice versa

Example 31.5.1 Consider the set of differential equations

y10 = y2

y20 = y3

y30 = −y3− y2− y1+ t3with the initial conditions

We substitute this into the third equation.

1 − s2(s2+ 1)

We then take the inverse Laplace transform of ˆy1

=

Z ∞ s

Exercise 31.8

The function f (t) t ≥ 0, is periodic with period 2T ; i.e f (t + 2T ) ≡ f (t), and is also odd with period T ; i.e

f (t + T ) = −f (t) Further,

Z T 0

f (t) e−st dt = ˆg(s)

Show that the Laplace transform of f (t) is ˆf (s) = ˆg(s)/(1 + e−sT) Find f (t) such that ˆf (s) = s−1tanh(sT /2)

Hint, Solution

Exercise 31.9

Find the Laplace transform of tν, ν > −1 by two methods

1 Assume that s is complex-valued Make the change of variables z = st and use integration in the complex plane

2 Show that the Laplace transform of tν is an analytic function for <(s) > 0 Assume that s is real-valued Makethe change of variables x = st and evaluate the integral Then use analytic continuation to extend the result tocomplex-valued s

e−tln t dt

[ γ = 0.5772 is known as Euler’s constant.]

Hint, Solution

Exercise 31.11

Find the Laplace transform of tνln t Write the answer in terms of the digamma function, ψ(ν) = Γ0(ν)/Γ(ν) What

is the answer for ν = 0?

Hint, Solution

1 Expand ˆf (s) using partial fractions and then use the table of Laplace transforms.

2 Factor the denominator into (s − 2)(s2+ 1) and then use the convolution theorem

Hint, Solution

Exercise 31.14

Solve the problem,

y00− ty0+ y = 0, y(0) = 0, y0(0) = 1,with the Laplace transform

where c is to the right of the singularities of ˆf (s).

t Hint: cut the s-plane along the negative real axis and deform the contour onto the cut.Remember that R0∞e−ax 2

cos(bx) dx =pπ/4a e−b 2 /4a

y(τ ) cos(t − τ ) dτ, y(0) = 0

and the ‘initial condition’ u(t) = u0(t), −1 ≤ t ≤ 0, where u0(t) is given Show that the Laplace transform ˆu(s) ofu(t) satisfies

ˆu(s) = u0(0)

y(τ ) dτ = e−t, y(0) = 1

Hint, Solution

dt = i1− i2with initial conditions

i1(0) = i2(0) = E0

2R, q(0) = 0.

Solve the system by Laplace transform methods and show that

i1 = E02R +

E02ωLe

−αt

sin(ωt)where

α = R2L and ω

Z b a

g(s, t) dt =

Z b a

∂

∂sg(s, t) dtHint 31.5

Z ∞ s

e−stf (t) dt =

Z ∞ n=0

(n+1)T

X

nT

e−stf (t) dt

The sum can be put in the form of a geometric series.

e−steat dt

=

Z ∞ 0

for <(s) > <(a)

Leat = 1

s − a for <(s) > <(a)2

L[sin(at)] =

Z ∞ 0

e−stsin(at) dt

= 1ı2

Z ∞ 0

e(−s+ıa)t− e(−s−ıa)t
dt

= 1ı2

, for <(s) > 0

= 1ı2

1

L[cos(at)] = L d

dt

sin(at)a

L[sinh(at)] =

Z ∞ 0

e−stsinh(at) dt

= 12

Z ∞ 0

e(−s+a)t− e(−s−a)t
dt

= 12

for <(s) > |<(a)|

= 12

1

6 First note that

L sin(at)t

(s) =

Z ∞ s

L[sin(at)](σ) dσ

Now we use the Laplace transform of sin(at) to compute the Laplace transform of sin(at)/t

L sin(at)t



=

Z ∞ s

a

σ2+ a2 dσ

=

Z ∞ s

1(σ/a)2 + 1

dσa

=

harctan

σa

i∞ s



= arctan

as



7

L

Z t 0

L[f (t)] = 1

s(1 + e−πs)Solution 31.2

L[af (t) + bg(t)] =

Z ∞ 0

e−st af (t) + bg(t)
dt

= a

Z ∞ 0

e−stf (t) dt + b

Z ∞ 0

e−stectf (t) dt

=

Z ∞ 0

Now consider the Laplace transform of tnf (t) for n ≥ 1.

L[tnf (t)] =

Z ∞ 0

e−sttnf (t) dt

= − dds

Z ∞ 0

If R0β f (t)t dt exists for positive β and f (t) is of exponential order α then the Laplace transform of f (t)/t is defined for

s > α

L f (t)t



=

Z ∞ 0

e−st 1

tf (t) dt

=

Z ∞ 0

Z ∞ s

e−σt dσ f (t) dt

=

Z ∞ s

Z ∞ 0

e−σtf (t) dt dσ

=

Z ∞ s

ˆ

f (σ) dσ

Solution 31.6

L

Z t 0

f (τ ) dτ



=

Z ∞ 0

e−st

Z t 0

f (τ ) dτ

∞ 0

−

Z ∞ 0

−e

−st

s

ddt

Z t 0

f (τ ) dτ

dt

= 1s

Z ∞ 0

e−stf (t) dt

= 1sˆ

f (s)

Solution 31.7

f (t) is periodic with period T

L[f (t)] =

Z ∞ 0

e−stf (t) dt

=

Z T 0

e−stf (t) dt +

Z 2T T

e−stf (t) dt

=

Z T 0

Solution 31.8

ˆ

f (s) =

Z ∞ 0

e−st(−1)nf (t) dt

=

Z T 0

Z T 0

f (t) e−st dt = 1 − e

−st

By inspection we see that this is satisfied for f (t) = 1 for 0 < t < T We conclude:

e−sttνdt

Assume s is complex-valued The integral converges for <(s) > 0 and ν > −1

Method 1 We make the change of variables z = st

C is the path from 0 to ∞ along arg(z) = arg(s) (Shown in Figure 31.4)

Since the integrand is analytic in the domain  < r < R, 0 < θ < arg(s), the integral along the boundary of thisdomain vanishes

Z R



+

Z R eı arg(s)R

Re(z) arg(s)

Figure 31.4: The Path of Integration

integral bound

Z

C R

e−zzνdz

C 

e−zzνdz

estπs

1/2

e−2(as)1/2

(2α)

= eαt

√π

Z π π/2

eR e ıθ t

√π

≤

Z π π/2

eR e ıθ t

√π

≤

√π

√R

Z π π/2

eR cos(θ)t dθ

≤

√π

√R

Z π/2 0

e−Rt sin(φ) dφ

<

√π

√R

π2Rt

→ 0 as R → ∞

We could show that the integral along CR− vanishes by the same method

Now we have

1ı2π

Z

B

+Z

L +

+Z

C

+Z

L −



estπs

1/2

e−2(as)1/2 ds = 0

We show that the integral along C vanishes as  → 0 with the maximum modulus bound.

Z

C 

estπs

1/2

e−2(as)1/2 ds

...

e−sttνdtexists for <(s) > It converges uniformly for <(s) ≥ c > On this domain of uniform convergence we caninterchange differentiation and integration

d ˆf

ds =...

This integral defines ˆf (s) for <(s) > Note that the integral converges uniformly for <(s) ≥ c > On this domain

we can interchange differentiation and integration

ˆ0...

= √2π

r π4te

−4a/(4t)

= e

−a/t

√tThus the inverse Laplace transform is

f (t) = e