Module

for

Shifting Theorems and the Step Function

Chapter 12  Fourier Series and the Laplace Transform

12.7  Laplace Transform Shifting Theorems and the Step Function

This section is a continuation of our development of the Laplace Transforms in Section 12.5 and Section 12.6.

We have shown how to use the Laplace transform to solve linear differential equations. Familiar functions that arise in solutions to differential equations are    and  .  Theorem 12.15 (the first shifting theorem) shows how their transforms are related to those of     and    by shifting the variable s in .  A companion result, called the second shifting theorem, Theorem 12.16, shows how the transform of can be obtained by multiplying by  .  Loosely speaking, these results show that multiplication of by    corresponds to shifting ,  and that shifting corresponds to multiplication of the transform by  .

Theorem 12.15   (Shifting the Variable  s).  If    is the Laplace transform of  ,  then

.

Proof.

Definition 12.3   (The Unit Step Function).  Let .  Then, the unit step function is

Figure 12.22.  The graph of the unit step function .

Theorem 12.16   (Shifting the Variable  t).  If is the Laplace transform of , and , then

,

where   and    are illustrated in Figure 12.23.

Figure 12.23.  Comparison of the functions    and  .

Proof.

Example 12.17.  Show that  .

Solution.

If we let , then , and if we apply Theorem 12.15, we obtain the desired result:

.

Explore Solution 12.17.

Example 12.18.   Show that  .

Solution.

If set , and then set .  We apply Theorem 12.16 to get

Explore Solution 12.18.

Extra Example 1.  Use Theorem 12.16 and find  .

Explore Solution for Extra Example 1.

Example 12.19.  Find    if   is as given in Figure 12.24.

Figure 12.24.  The function .

Solution.

We represent   in terms of step functions  .  Using the result of Example 12.18 and linearity, we obtain

Explore Solution 12.19.

Example 12.20.  Solve the initial value problem

A graph of the solution.

Solution.

As usual, we let denote the Laplace transform of .  The right hand side of the D.E. is   and    .
Taking Laplace transforms we write  .  Using the initial conditions , and we get

.

Solving for yields

.

Use the facts that  ,   and get  , , respectively.   Then we will apply Theorem 12.16.  We compute the solution, , as

Then using the trigonometric identity we can write this in a more familiar form

Explore Solution 12.20.

Laplace Transform

The Next Module is

(c) 2012 John H. Mathews, Russell W. Howell