KindiakMath

The Shift Theorems

In our previous write-up, we encountered a new problem in solving our initial value problem; how do we evaluate the expression below?

\displaystyle \mathcal L^{-1} \left\{ e^{-2s} \cdot \frac{1}{s(s-2)(s-3)} \right\}

Well, let’s first define \mathcal L^{-1} properly.

Definition 1. For any function F, we write f(t) = \mathcal L^{-1}\{F(s)\} if there exists a function f such that

\mathcal L\{f(t)\} = F(s).

Example 1. Based on previous discussions,

\displaystyle \begin{aligned} \mathcal L^{-1}\{1/s\} &= 1, \\ \mathcal L^{-1}\{1/(s-a)\} &= e^{at}, \\ \mathcal L^{-1}\{e^{-as}/s\}&= U(t-a). \end{aligned}

We notice that setting a = 0 in the last case, 1 = U(t), which is false for t < 0, since 1 \neq 0 = U(t) in that case. Nevertheless, as far as Laplace inverse transforms are concerned, they are the same function. The reason is that 1 = U(t) for t \geq 0.

Hence, we will regard f = g whenever f(t) = g(t) for t \geq 0. (The formal identification can be described using equivalence relations). These investigations, however, raises an interesting observation. Writing f(t) = 1, we have

\displaystyle F(s) \equiv \mathcal L\{f(t)\} = \frac 1s.

Hence,

\displaystyle \mathcal L\{e^{at} f(t)\} = \frac 1{s-a} = F(s-a).

In fact, this result holds in general, and is known as the first of the shift theorems.

Theorem 1 (First Shift Theorem). For any real a and function f such that \mathcal L\{f\} exists, \mathcal L\{e^{at}f(t)\} exists and equals

\displaystyle \mathcal L\{e^{at} f(t)\} = \frac 1{s-a} = F(s-a).

Proof. For any s,

\displaystyle \begin{aligned} \mathcal L\{e^{at} f(t)\} &= \int_0^\infty e^{at}f(t) \cdot e^{-st}\, \mathrm dt \\ &= \int_0^\infty f(t) \cdot e^{-(s-a)t}\, \mathrm dt \\ &= F(s-a). \end{aligned}

This result is named the shift theorem because multiplying the exponential shifts the Laplace transform.

Can we reverse the question? That is, what would the expression \mathcal L^{-1}\{e^{-as} F(s) \} evaluate to? Let’s explore this question. Suppose there exists f,g such that \mathcal L\{f\} = F and

\mathcal L\{g(t)\} = e^{-as} F(s).

When in doubt, return to the definitions—that would be first-principles thinking. Unpacking the definition of Laplace transforms yields

\displaystyle \begin{aligned} \mathcal L\{g(t)\} &= e^{-as} \int_0^\infty f(t)e^{-st}\, \mathrm dt \\ &= \int_0^\infty f(t)e^{-s(a+t)}\, \mathrm dt \\ &= \int_a^\infty f(w-a)e^{-sw}\, \mathrm dw \\ &= \int_0^\infty f(w-a)U(w-a) e^{-sw}\, \mathrm dw \\ &= \int_0^\infty f(t-a)U(t-a) e^{-st}\, \mathrm dt \\ &= \mathcal L\{ f(t-a)U(t-a) \}.\end{aligned}

Therefore, we can define g(t) := f(t-a)U(t-a), so that we obtain the second shift theorem below:

Theorem 2 (Second Shift Theorem). For any real a and function f such that \mathcal L\{f\} exists, \mathcal L\{e^{at}f(t)\} exists and equals

\displaystyle \mathcal L^{-1} \{e^{-as} F(s)\} = f(t-a)U(t-a).

Returning to our original problem, we want to find an expression for

\displaystyle \mathcal L^{-1} \left\{ e^{-2s} \cdot \frac{1}{s(s-2)(s-3)} \right\}.

This means that if we allow \displaystyle F(s) = \frac{1}{s(s-2)(s-3)} and we can compute f(t) = \mathcal L^{-1}\{F(s)\}, then by the second shift theorem, we have the following result.

Lemma 1. \displaystyle \mathcal L^{-1} \left\{ e^{-2s} \cdot \frac{1}{s(s-2)(s-3)} \right\} = f(t-2)U(t-2).

The only problem is: what is f(t)? Since f(t) = \mathcal L^{-1}\{F(s)\}, we need to ask the question, how do we evaluate the following expression?

\displaystyle \mathcal L^{-1} \left\{ \frac{1}{s(s-2)(s-3)} \right\}

The denominator looks “separable”. Suppose we can, in fact, separate the fraction into a sum of partial fractions:

\displaystyle \frac{1}{s(s-2)(s-3)} = \frac As + \frac B{s-2} + \frac C{s-3}.

Then using the linearity of \mathcal L, we can show that \mathcal L^{-1} is linear too, so that

\displaystyle \begin{aligned} &\mathcal L^{-1}\left\{ \frac{1}{s(s-2)(s-3)} \right\}\\ &= \mathcal L^{-1}\left\{ \frac As + \frac B{s-2} + \frac C{s-3} \right\} \\ &= A \cdot \mathcal L^{-1} \left\{ \frac 1s \right\} + B \cdot \mathcal L^{-1} \left\{ \frac 1{s-2} \right\} + C \cdot \mathcal L^{-1} \left\{ \frac 1{s-3} \right\}\\ &= A \cdot 1 + B \cdot e^{2t} + C \cdot e^{3t} \\ &= A + B \cdot e^{2t} + C \cdot e^{3t} .\end{aligned}

Well, that is very convenient! All that remains is to compute A, B, C, and we are done! Of course, there are several more loose ends to tie, such as the various denominators that can arise, but that would become a matter of bookkeeping and exploration of more formulas. Perhaps we will explore those other formulas in an exercise.

Back to computing A, B, C. Beginning with the question at hand, we will clear denominators to obtain

\displaystyle \begin{aligned} \frac{1}{s(s-2)(s-3)} &= \frac As + \frac B{s-2} + \frac C{s-3} \\ 1 &= A(s-2)(s-3) + Bs(s-3) + Cs(s-2). \end{aligned}

Setting s = 0 yields 1 = A \cdot (-2) \cdot (-3) \Rightarrow A = 1/6.
Setting s = 2 yields 1 = B \cdot 2 \cdot (-1) \Rightarrow B = -1/2.
Setting s = 3 yields 1 = C \cdot 3 \cdot 1 \Rightarrow C = 1/3.

Back-substituting our results, we have the following expression for f.

Lemma 2. \displaystyle f(t) = \frac 16 - \frac 12 \cdot e^{2t} + \frac 13 \cdot e^{3t}.

Thus, we can finally answer our main question as follows.

Example 2. Solve the initial value problem

y'' - 5y' + 6y = U(t-2),\quad y(0) = 0, \quad y'(0) = 0.

Solution. Taking Laplace transforms on all sides,

\displaystyle \mathcal L\{y'' - 5 y' + 6y\} = \mathcal L\{ U(t-2) \}.

By linearity,

\displaystyle \mathcal L\{y''\} - 5 \mathcal L\{y'\} + 6 \mathcal L\{y\} = \mathcal L\{ U(t-2) \}.

By the Laplace transform of the second derivative,

\displaystyle \begin{aligned} \mathcal L\{y''\} &= s^2 Y(s) - s y(0) - y'(0) \\ &= s^2 Y(s) - 0 - 0 = s^2 Y(s). \end{aligned}

By the Laplace transform of the first derivative,

\displaystyle \begin{aligned} \mathcal L\{y'\} &= s Y(s) - y(0) \\ &= s Y(s) - 0 = s Y(s). \end{aligned}

By notation, \mathcal L\{y\} = Y(s). By the Laplace transform of unit step functions,

\displaystyle \mathcal L \{ U(t-2) \} = \frac{e^{-2s}}s.

Substituting all of these results,

\displaystyle s^2 Y(s) - 5sY(s) + 6Y(s) = \frac {e^{-2s}}s.

By basic algebruh,

\displaystyle Y(s) = \frac{e^{-2s}}{s(s-2)(s-3)}.

Taking inverse Laplace transforms and applying Lemmas 1 and 2 respectively,

\displaystyle \begin{aligned} y(t) &= \mathcal L^{-1}\{Y(s)\} \\ &= \mathcal L^{-1} \left\{ \frac{e^{-2s}}{s(s-2)(s-3)} \right\} \\ &= f(t-2) \cdot U(t-2) \\ &= \left(\frac 16 - \frac 12 \cdot e^{2(t-2)} + \frac 13 \cdot e^{3(t-2)} \right) \cdot U(t-2).\end{aligned}

—Joel Kindiak, 10 Feb 25, 2207H

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