Lesson 34, Derivatives, Integrals, and Products of Laplace Transforms

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Handout Lesson 34, Derivatives, Integrals, and Products of Laplace Transforms

Textbook Section(s).

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This lesson is based on Section 7.4 of your textbook by Edwards, Penney, and Calvis.

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Products of Laplace transforms.

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We start this section with an example.

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Example 228. Inverse Laplace transforms of products.

Compare

\begin{gather*} \mathscr{L}^{-1}\left\{\frac{1}{s-2}\cdot\frac{1}{s+1} \right\} \quad \text{ and } \quad \mathscr{L}^{-1}\left\{\frac{1}{s-2}\right\} \mathscr{L}^{-1}\left\{\frac{1}{s+1}\right\} \end{gather*}

What are you supposed to learn from this comparison?

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In the last example, we were able to compute \(\mathscr{L}^{-1}\{\text{a product of Laplace transforms}\}\) because the product was a rational function, we were able to use the partial fractions tool to turn the product into a sum, and then we could use the linearity of \(\mathscr{L}^{-1}\) to finish the computation.

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Question: Is there a way to compute \(\mathscr{L}^{-1}\{F(s)\cdot G(s)\}\) in general, especially when \(F(s)\cdot G(s)\) is not a rational function?

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We have not yet encountered Laplace transforms that are not rational functions, so the examples that we are doing today will not really show you the power of this new technique. That being said, they will help to bolster our confidence in the new technique and help us to develop skills that will be required later.

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Definition 229.

Let \(f(t)\) and \(g(t)\) be piecewise continuous functions for \(t \geq 0\text{.}\) We define the convolution of \(f\) and \(g\text{,}\) denoted \(f * g\text{,}\) for \(t\geq 0\) by

\begin{gather} (f*g)(t) = \int_0^t f(w)g(t-w) \, dw\tag{✶} \end{gather}

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NOTE: Your textbook and the table of Laplace transforms use \(\tau\) instead of \(w\) for the variable of integration in (✶). I am choosing to use \(w\) because it is easy to distinguish from \(t\) when writing. Sometimes it is difficult to distinguish between \(t\) and \(\tau\) when writing, and all of us will be writing on the exam.

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Lemma 230.

When \(f*g\) is defined,

\begin{gather*} f*g=g*f \end{gather*}

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This is Theorem 1 from Section 7.4 of your textbook by Edwards, et. al.

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Theorem 231.

If \(f\) and \(g\) are piecewise continuous for \(t \geq 0\) and of exponential order with constant \(c\text{,}\) then \(\mathscr{L}\{f(t)*g(t)\}\) exists for \(s \gt c\) and

\begin{gather*} \mathscr{L}\{f(t)*g(t)\} = F(s)G(s)\\ \mathscr{L}^{-1}\{F(s)G(s)\} = f(t)*g(t) \end{gather*}

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Example 232. Inverse Laplace transforms of products.

In Example 228, we showed that \(\mathscr{L}^{-1}\left\{\frac{1}{s-2}\cdot\frac{1}{s+1} \right\} = \frac{1}{3}e^{2t}-\frac{1}{3}e^{-t} \text{.}\) Verify that

\begin{gather*} \mathscr{L}^{-1}\left\{\frac{1}{s-2}\right\} * \mathscr{L}^{-1}\left\{\frac{1}{s+1}\right\}= \frac{1}{3}e^{2t}-\frac{1}{3}e^{-t} \end{gather*}

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Example 233. Laplace transforms of convolutions.

Let \(f(t)=t^2\) and \(g(t)=\cos(t)\text{.}\)

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Compute \(f(t)*g(t)\text{.}\)
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Verify that \(\mathscr{L}\{f(t)*g(t)\}= \mathscr{L}\{f(t)\}\mathscr{L}\{g(t)\}\text{.}\)
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Example 234. Inverse Laplace transforms. Product or Partial Fractions?

Compute \(\mathscr{L}^{-1}\left\{\frac{2}{s^3(s-4)}\right\}\text{.}\)

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Derivative and Integrals of Transforms.

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This is Theorem 2 from Section 7.4 of your textbook by Edwards, et. al.

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Theorem 235.

If \(f\) is piecewise continuous for \(t \geq 0\) and of exponential order with constant \(c\text{,}\) then

\begin{gather*} F'(s) = \mathscr{L}\{-tf(t)\} \qquad \text{for } s \gt c\\ f(t) = -\frac{1}{t}\mathscr{L}^{-1}\{F'(s)\} \qquad \text{for } t \gt 0\\ \mathscr{L}\{t^nf(t)\} = (-1)^nF^{(n)}(s) \qquad \text{for } s \gt c \text{ and } n=1,2,3, \dots \end{gather*}

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