On convolution theorem of Laplace transform: Schiff

In summary, the paper discusses the convolution theorem related to the Laplace transform, as presented by Schiff. It examines the mathematical framework and implications of the theorem, highlighting how convolution in the time domain corresponds to multiplication in the Laplace domain. The study emphasizes the applications of this theorem in solving differential equations and analyzing linear systems, providing insights into its significance in engineering and applied mathematics.
  • #1
psie
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TL;DR Summary
I'm reading Schiff's book The Laplace Transform. On page 92-93 he proves the convolution theorem. It's a very self-contained proof, with no reference to any prior results. However, I have some integration-related questions that I struggle with.
Here follows the theorem and proof:

Theorem (Convolution Theorem). If ##f## and ##g## are piecewise continuous on ##[0,\infty)## and of exponential order ##\alpha##, then $$\mathcal{L}\left[(f*g)(t)\right]=\mathcal{L}\big(f(t)\big)\cdot\mathcal{L}\big(g(t)\big)\quad \Big(Re(s)>\alpha\Big).$$

Proof. Let us start with the product \begin{align} \mathcal{L}\big(f(t)\big)\cdot\mathcal{L}\big(g(t)\big)&=\left(\int_0^\infty e^{-s\tau}f(\tau)d\tau\right)\left(\int_0^\infty e^{-su}g(u)du\right) \nonumber \\
&=\int_0^\infty \left(\int_0^\infty e^{-s(\tau+u)}f(\tau)g(u)du\right)d\tau \nonumber .
\end{align}
Substituting ##t=\tau+u##, and noting that ##\tau## is fixed in the interior integral, so that ##du=dt##, we have $$\mathcal{L}\big(f(t)\big)\cdot\mathcal{L}\big(g(t)\big)=\int_0^\infty \left(\int_\tau^\infty e^{-st}f(\tau)g(t-\tau)dt\right)d\tau .\tag 1$$ If we define ##g(t)=0## for ##t<0##, then ##g(t-\tau)=0## for ##t<\tau## and we can write ##(1)## as $$\mathcal{L}\big(f(t)\big)\cdot\mathcal{L}\big(g(t)\big)=\int_0^\infty \int_0^\infty e^{-st}f(\tau)g(t-\tau)dtd\tau .$$ Due to the hypotheses on ##f## and ##g##, the Laplace integrals of ##f## and ##g## converge absolutely and hence, in view of the preceding calculation, $$\int_0^\infty \int_0^\infty |e^{-st}f(\tau)g(t-\tau)|dtd\tau$$ converges. This fact allows us to reverse the order of integration,* so that \begin{align} \mathcal{L}\big(f(t)\big)\cdot\mathcal{L}\big(g(t)\big)&=\int_0^\infty \int_0^\infty e^{-st}f(\tau)g(t-\tau)d\tau dt \nonumber \\
&=\int_0^\infty \left(\int_0^t e^{-st}f(\tau)g(t-\tau)d\tau\right)dt \nonumber \\
&=\int_0^\infty e^{-st} \left(\int_0^t f(\tau)g(t-\tau)d\tau\right)dt \nonumber \\ &=\mathcal{L}[(f*g)(t)]. \nonumber
\end{align}

*Let $$a_{mn}=\int_n^{n+1}\int_m^{m+1} |h(t,\tau)|dtd\tau,\quad b_{mn}=\int_n^{n+1}\int_m^{m+1} h(t,\tau) dtd\tau,$$ so that ##|b_{mn}|\leq a_{mn}##. If $$\int_0^\infty\int_0^\infty |h(t,\tau)|dtd\tau <\infty,$$ then ##\sum_{n=0}^\infty\sum_{m=0}^\infty a_{mn}<\infty##, implying ##\sum_{n=0}^\infty\sum_{m=0}^\infty |b_{mn}|<\infty##. Hence, by a standard result on double series, the order of summation can be interchanged $$\sum_{n=0}^\infty\sum_{m=0}^\infty b_{mn}=\sum_{m=0}^\infty\sum_{n=0}^\infty b_{mn},$$
i.e., $$\int_0^\infty\int_0^\infty h(t,\tau) dtd\tau =\int_0^\infty\int_0^\infty h(t,\tau) d\tau dt.$$

Questions:

1. I do not understand the following part "...and hence, in view of the preceding calculation, ##\int_0^\infty \int_0^\infty |e^{-st}f(\tau)g(t-\tau)|dtd\tau## converges".

We know that ##\mathcal{L}\big(f(t)\big)## and ##\mathcal{L}\big(g(t)\big)## converge absolutely. So does their product converge absolutely (I assume this is the statement he is making)? If yes, how come?

The definition of absolute convergence given in the book is that ##\int_0^\infty |e^{-st}f(t)|dt## converges for a given real or complex parameter ##s##.

2. Regarding the footnote, if ##I=\int_0^\infty\int_0^\infty f(x,y)dxdy##, can we then always write ##I## as a double series, i.e. ##I=\sum_{n=0}^\infty\sum_{m=0}^\infty c_{mn}## where ##c_{mn}=\int_n^{n+1}\int_m^{m+1} f(x,y)dxdy## (I assume this is what Schiff is doing)? If not, what justifies that we can in this case and how?
 
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  • #2
psie said:
1. I do not understand the following part "...and hence, in view of the preceding calculation, ##\int_0^\infty \int_0^\infty |e^{-st}f(\tau)g(t-\tau)|dtd\tau## converges".
You are given that ##|f(\tau)|\le K_1 e^{\alpha\tau}## and ##|g(t-\tau)|\le K_2 e^{\alpha(t-\tau)}##, hence

##|e^{-st}f(\tau)g(t-\tau)|\le K_1K_2 e^{-Re(s)t} e^{\alpha\tau} e^{\alpha(t-\tau)} =K_1K_2 e^{(\alpha-Re(s))t} ##

Which is integrable for ##Re(s) > \alpha##.
 
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  • #3
martinbn said:
Which is integrable for ##Re(s) > \alpha##.
Hmm, I get then that $$\int_0^\infty \int_0^\infty |e^{-st}f(\tau)g(t-\tau)|dtd\tau\leq \int_0^\infty \int_0^\infty K_1K_2e^{(\alpha-Re(s))t}dtd\tau,$$ under the assumption that improper integrals depending on a parameter respect monotonicity. Are you sure the above converges?

By the way, in his book the definition of exponential order ##\alpha## means that it's supposed to hold eventually, so we have ##|f(\tau)|\le K_1 e^{\alpha\tau}## for ##\tau\geq\tau_0## where ##\tau_0\geq 0##.
 
  • #4
psie said:
Hmm, I get then that $$\int_0^\infty \int_0^\infty |e^{-st}f(\tau)g(t-\tau)|dtd\tau\leq \int_0^\infty \int_0^\infty K_1K_2e^{(\alpha-Re(s))t}dtd\tau,$$ under the assumption that improper integrals depending on a parameter respect monotonicity. Are you sure the above converges?

Recall that [itex]g(u) = 0 [/itex] for [itex]u < 0[/itex], so the lower limit of the inner integral is really [itex]\tau[/itex]. Also note that convergence requires [itex]\alpha - \operatorname{Re}(s) < 0[/itex], so we can replace it by [itex]-|\alpha - \operatorname{Re}(s)|[/itex]. Then [tex]\begin{split}
\int_0^\infty \int_0^\infty |e^{-st}f(\tau)g(t-\tau)|\,dt\,d\tau &\leq \int_0^\infty \int_\tau^\infty K_1K_2e^{-|\alpha-\operatorname{Re}(s)|t}\,dt\,d\tau \\
&= K_1K_2\frac1{|\alpha - \operatorname{Re}(s)|}\int_0^\infty e^{-|\alpha-\operatorname{Re}(s)|\tau}\,d\tau.\end{split}[/tex]
 
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  • #5
Thank you.

I think I can answer my second question. Let ##\int_0^\infty f(x) dx=\lim_{t\to\infty}\int_0^t f(x)dx## be an improper Riemann integral. Then, if we let ##t=n\in\mathbb N##, we get ##\int_0^n f(x)dx = \sum_{i=0}^{n-1} \int_i^{i+1} f(x)dx##, and in the limit ##n\to\infty##, we get $$\int_0^\infty f(x)dx=\sum_{i=0}^\infty \int_i^{i+1} f(x)dx.$$ So if the integral converges, then the sum converges. However, consider ##f(x)=\sin(2\pi x)##. Then the integral over the interval ##[i,i+1]## is ##0##, so the sum converges, but the improper integral does not. Hence the converse does not hold.
 

FAQ: On convolution theorem of Laplace transform: Schiff

What is the convolution theorem in the context of the Laplace transform?

The convolution theorem states that the Laplace transform of the convolution of two functions is equal to the product of their individual Laplace transforms. Mathematically, if \( f(t) \) and \( g(t) \) are two functions with Laplace transforms \( F(s) \) and \( G(s) \), respectively, then the Laplace transform of the convolution \( (f * g)(t) \) is given by \( \mathcal{L}\{f * g\}(s) = F(s)G(s) \). This theorem is particularly useful in solving differential equations and analyzing linear systems.

How is the convolution of two functions defined?

The convolution of two functions \( f(t) \) and \( g(t) \) is defined as the integral of their product over time, with one of the functions being time-reversed and shifted. It is mathematically expressed as: \[(f * g)(t) = \int_0^t f(\tau) g(t - \tau) d\tau\]This operation combines the effects of both functions over a specified interval, providing a new function that reflects their interaction.

What are the applications of the convolution theorem in engineering and physics?

The convolution theorem has numerous applications in engineering and physics, particularly in the fields of signal processing, control systems, and communications. It allows for the analysis of linear time-invariant systems by simplifying the process of determining system responses to inputs. For example, it is used in filtering signals, analyzing circuit behavior, and solving differential equations that model physical systems.

Are there any limitations to the convolution theorem?

Yes, the convolution theorem has certain limitations. It is applicable primarily to functions that are piecewise continuous and of exponential order, meaning they do not grow too quickly as \( t \) approaches infinity. Additionally, the theorem assumes that the functions involved are defined for \( t \geq 0 \). If these conditions are not met, the convolution may not yield meaningful results or may not be computable using the Laplace transform.

How does Schiff's work contribute to the understanding of the convolution theorem?

Schiff's work provides a comprehensive treatment of the convolution theorem in the context of the Laplace transform, offering detailed proofs and applications. His contributions help clarify the mathematical foundations of the theorem, ensuring that it is well understood and correctly applied in various scientific and engineering scenarios. Schiff emphasizes the importance of the theorem in solving complex problems and highlights its relevance in both theoretical and practical contexts.

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