Laplace transform III - The Shifting Theorems

#differential_equations

With the Fourier transform, we had the stretch-and-shift property that made computing Fourier transforms of slightly modified functions very easy, namely:

F (f (a t + b)) = \frac{1}{| a |} e^{\frac{2 π i s b}{a}} (F f) (\frac{s}{a}) .

As a special case of this, when $a = 1$ we have the following "stretch theorem":

F (f (t + b)) = e^{2 π i s b} \cdot F f .

In other words, shifting the input of $f$ by $b$ simply "costs" a factor of $e^{2 π i s b}$ on the transform of $f$ . For example, we can instantly compute

F (Π (t + 1)) = e^{2 π i s} sinc (s) .

This property can potentially save us from lots of tedious integration.

There's also the corresponding property for the inverse Fourier transform, namely

F^{- 1} (F (s + b)) = e^{- 2 π i t b} \cdot F^{- 1} F .

How about the Laplace transform? Let's take a look and see!

The first shift theorem

Let's start by computing

\begin{aligned} L (f (t + b)) & = \int_{0}^{\infty} f (t + b) e^{- t s} d t \\ = \int_{b}^{\infty} f (u) e^{- (u - b) s} d u (using u = t + b) \\ = e^{b s} \int_{b}^{\infty} f (u) e^{- u s} d u \\ = e^{b s} \int_{b}^{\infty} f (t) e^{- t s} d t \end{aligned}

This last integral almost looks like the Fourier transform of $f$ , except that the lower bound is $b$ and not $0$ . In order to fix this issue, let's introduce the function

u_{b} (t) = {\begin{cases} 0, & 0 \leq t < b \\ 1, & t \geq b \end{cases}

We call this the unit step function at time $t = b$ . It's graph is simply

coming soon

We can think of $u_{b} (t)$ as like a "switch" that "turns on" at time $t = b$ . In particular, note that

\int_{0}^{\infty} u_{b} (t) f (t) e^{- t s} d t = \int_{b}^{\infty} f (t) e^{- t s} d t .

Based on our work above, we can thus conclude:

L (f (t + b)) = e^{b s} \cdot L (u_{b} (t) f (t)) .

As currently written, this property doesn't look like it would be very helpful to us. After all, none of our "Fab Four" Laplace transforms involves a unit step function, so we are unlikely to ever know the Laplace transform on the right-hand side.

However, we can flip things around a bit and turn this into a more useful equality, namely:

L (u_{b} \cdot f) = e^{- b s} L (f (t + b)) .

Example

Consider the piecewise-defined function

f (t) = {\begin{cases} 0, & 0 \leq t < 1 \\ t - 1, & 1 \leq t < 2 \\ 1, & 2 \leq t \end{cases}

We can rewrite $f (t)$ as a combination of its "component" functions using step functions to "turn on" and "turn off" the various components. Indeed, we claim that

f (t) = (u_{0} (t) - u_{1} (t)) \cdot 0 + (u_{1} (t) - u_{2} (t)) \cdot (t - 1) + u_{2} (t) \cdot 1.

To verify this, simply note that:

$u_{0} (t) - u_{1} (t)$ takes the value $1$ on the interval $[0, 1)$ and the value $0$ everywhere else
$u_{1} (t) - u_{2} (t)$ takes the value $1$ on the interval $[1, 2)$ and the value $0$ everywhere else
$u_{2} (t)$ takes the value $1$ on the interval $[2, \infty)$ and the value $0$ everywhere else

We can now simplify our expression for $f (t)$ , gathering terms according to step functions. We eventually find

f (t) = u_{1} (t) \cdot (t - 1) + u_{2} (t) \cdot (2 - t) .

It then follows that

\begin{aligned} L f & = L (u_{1} \cdot (t - 1)) + L (u_{2} \cdot (2 - t)) \\ = e^{- 1 \cdot s} L ((t + 1) - 1) + e^{- 2 s} L (2 - (t + 2)) \\ = e^{- s} L (t) + e^{- 2 s} L (- t) \\ = e^{- s} \frac{1}{s^{2}} - e^{- 2 s} \frac{1}{s^{2}} . \end{aligned}

We can also reverse this shifting property to give a formula for inverse Laplace transforms. First, let's express the original shifting property a little more suggestively, as

L (u_{b} \cdot f) = e^{- b s} L (f ∣_{t \mapsto t + b})

I like to read this as, "If you are transforming a step function times a known function, you can trade out the step function at the cost of a negative exponential term, and then transform the original function after first stepping forward by the step."

We can now reverse this rule to obtain the following:

L^{- 1} (e^{- b s} F) = u_{b} \cdot (L^{- 1} F) ∣_{t \mapsto t - b}

In words, "If you are transforming back and see a negative exponential function, you can trade out the exponential for a step function and then transform back the original function, but remember to step back after you do."

Example

Using the above reverse shifting property, we can compute

\begin{aligned} L^{- 1} (\frac{2 e^{- s}}{s^{2} + 4}) & = L^{- 1} (e^{- s} \cdot \frac{2}{s^{2} + 4}) \\ = u_{1} (t) \cdot (L^{- 1} (\frac{2}{s^{4} + 4})) ∣_{t \mapsto t - 1} \\ = u_{1} (t) \cdot (\sin (2 t)) ∣_{t \mapsto t - 1} \\ = u_{1} (t) \cdot \sin (2 (t - 1)) . \end{aligned}

The second shift theorem

Recall that the shifting property for the inverse Fourier transform is

F^{- 1} (F (s + b)) = e^{- 2 π i t b} \cdot F^{- 1} F .

We claim that we similarly have

L^{- 1} (F (s + b)) = e^{- b t} \cdot L^{- 1} F

To verify this, let $f (t) = L^{- 1} F$ , so that $L f = F$ , and observe

L (e^{- b t} f (t)) = \int_{0}^{\infty} e^{- b t} f (t) e^{- t s} d t = \int_{0}^{\infty} e^{- t (s + b)} d t = (L f) (s + b) = F (s + b) .

This proves $e^{- b t} f (t) = L^{- 1} (F (s + b))$ , as claimed.

Before seeing an example, I would like to rewrite this shifting property in a way I find more helpful in real life, using similar notation to the previous one:

L^{- 1} F = e^{- b t} L^{- 1} (F ∣_{s \mapsto s - b}) .

I read this as "If I'm doing an inverse transform, I can step back by $b$ before transforming if I pay the cost of a negative exponential."

Example

Consider the function $F (s) = \frac{1}{s^{2} + 2 s + 2}$ . In this form, we don't have much hope of computing its inverse transform. However, if we complete the square in the denominator, then we see that

F (s) = \frac{1}{(s + 1)^{2} + 1},

and so if we "stepped back" by $1$ we would have the function

F (s - 1) = \frac{1}{s^{2} + 1},

whose inverse transform we do know.

With this idea in mind, using the above shifting property we can compute

\begin{aligned} L^{- 1} F & = L^{- 1} (\frac{1}{s^{2} + 2 s + 2}) \\ = L^{- 1} (\frac{1}{(s + 1)^{2} + 1}) \\ = e^{- 1 \cdot t} L^{- 1} (\frac{1}{s^{2} + 1}) \\ = e^{- t} \sin (t) \end{aligned}

Just as with the first shifting theorem, we can "reverse" this theorem to give another shifting problem for the forward transform, namely:

L (e^{- b t} \cdot f) = (L f) ∣_{s \mapsto s + b} .

In word, "If you are computing a forward transform and see a negative exponential, you can just transform the function next to it, so long as you step forward after you do."

For example, we can use this to compute

L (e^{- 3 t} t^{4}) = (L (t^{4})) ∣_{s \mapsto s + 3} = (\frac{4!}{s^{5}}) ∣_{s \mapsto s + 3} = \frac{4!}{(s + 3)^{5}} .

In summary: the four shifting results

The Shifting Theorems

We have the following shifting theorems:

$L (u_{b} \cdot f) = e^{- b s} L (f ∣_{t \mapsto t + b}) \Leftrightarrow L^{- 1} (e^{- b s} F) = u_{b} \cdot (L^{- 1} F) ∣_{t \mapsto t - b}$
$L^{- 1} F = e^{- b t} L^{- 1} (F ∣_{s \mapsto s - b}) \Leftrightarrow L (e^{- b t} \cdot f) = (L f) ∣_{s \mapsto s + b}$

Notice that the forward transforms always involve a step forward, the inverse transforms a step back, and everything always "costs" a negative exponential.

Suggested next notes

Laplace transform IV - Convolution redux