Chap 5: Laplace Transform 1

Question 1

What do we use the Laplace transform for? What is its structure?

Answer 1

Laplace transform allows us to solve ODE’s. We use it to convert Initial Value Problems (integrals, derivatives) into algebraic formulas which can be solved, then convert them back to get the solution to the IVP

F(s) = L(f(t)) = int[0,inf] (e^(-st) * f(t) dt)

Question 2

Theorem 1: What is a useful property of the Laplace transform?

How can it be applied?

Answer 2

1. The Laplace transform is linear, meaning
   L(af(t) + bg(t)) = aL(f(t)) + bL(g(t))
2. There are pre-written lists of Laplace transforms of various kinds. Through linearity, we can chain these together to solve difficult ODE´s.

Question 3

For what values of t is the Laplace transformed concerned?

When does it not exist? (Graphical)

Answer 3

1. It is only concerned with t >= 0. Hence f(t) = 0, t < 0.
2. The Laplace transform does not exist when it goes to infinity. Meaning for F(s-a), when (s-a) > 0, it exists
   (e^-x = 0 for x->inf ). When (s-a) < 0, it doesn’t exist.

Question 4

Theorem 2: What is first-shifting of Laplace transforms?

Answer 4

Normally, we have F(s). However, we can shift something to F(s-a), the shift being -a. In short, replace all the s’s with s-a. Alt: t = t~ + a. Solve for t~ first, then swap with t to get the solution. A very neat trick that can be used when necessary.

Question 5

Theorem 3: When does the Laplace transform exist?

And what does its uniqueness mean?

Answer 5

1a) f(t) should satisfy the growth restriction:
E: M, k, s.t. A: t >= 0, then |f(t)| <= Me^(kt), and L(s) exists for all s > k
Which means that there exists an M and a K for all t-values, for which |f(t)| is always small than Me^(kt)
1b) f(t) should be piecewise continuous, meaning that the function can jump and be broken, but not go to infinity.

2) Two functions that have the same transform cannot differ over an interval that is larger than 0. They can, however, differ in points. Essentially, they’re the same.

Question 6

Theorem 4: What is the Laplace transform of the derivative?

How do we solve these?

Answer 6

If f, f’, (…), f (d/dt)^(n-1) are continous for all t>=0 and satisfies the growth restriction, and f (d/dt)^n is piecewise continous for all t>=0:

lap(f(t) (d/dt)^(n)) = s^n lap(f(t) - s^(n-1)f(0) - s^(n-2)f’(0) - (…) - f(0) (d/dt)^(n-1)

We solve by swapping all the derivatives with their respective s^n * lap(f(t)) - (…), solving algebraically for lap(f(t)), then use partial fraction decomposition, finishing with the Laplace transform table.

Question 7

Theorem 5: What is the Laplace transform of the integral? How does this correlate to the inverse?

How do we solve these?

Answer 7

Given f(t) being piecewise continuous and satisfying the growth restriction, for s > 0 and t > 0:

lap{ int0,t} = 1/s * F(s). Also,
int0,t = lap^(-1) {1/s * F(s)}

We solve by algebraically manipulating F(s) until we have several 1/s * F(s) = (A/(s-a) + B/(s^2-1) etc) *1/s (partial fraction decomposition). that can be transformed using the Laplace transform table

Question 8

How do we solve ODE’s using Laplace transform?

What if the y(0) values are shifted to y(a)?

Answer 8

First, replace all the derivatives (or alternatively, integrals) with their respective Laplace transformation. Solve for lap(y(t)), use partial fraction decomposition on the RHS, then inverse Laplace. You need the y(0), y’(0) … y(0) (d/dt)^(n-1).

If the y(0) values are shifted to y(a), set t = t~ + a and solve for t~, then replace t~ with t - a at the end.

Question 9

What are the main points of this chapter? (Chap 5:1)

Answer 9

The Laplace transform

1. The formula itself (Theorem 1)
2. Linearity
3. First shifting (Theorem 2)
4. It’s existence and uniqueness (Theorem 3)
5. Laplace of the derivative (Theorem 4)
6. Laplace of the integral (Theorem 5)
7. Laplace in ODE’s