Convolution Theorem/Proof 1
Theorem
Let $\GF \in \set {\R, \C}$.
Let $f: \R \to \GF$ and $g: \R \to \GF$ be functions.
Let their Laplace transforms $\laptrans {\map f t} = \map F s$ and $\laptrans {\map g t} = \map G s$ exist.
Then:
- $\map F s \map G s = \ds \laptrans {\int_0^t \map f u \map g {t - u} \rd u}$
Proof
\(\ds \laptrans {\int_0^t \map f u \map g {t - u} \rd u}\) | \(=\) | \(\ds \int_{t \mathop = 0}^\infty e^{-s t} \paren {\int_{u \mathop = 0}^t \map f u \map g {t - u} \rd u} \rd t\) | Definition of Laplace Transform | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_{t \mathop = 0}^\infty \int_{u \mathop = 0}^t e^{-s t} \map f u \map g {t - u} \rd u \rd t\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{M \mathop \to \infty} \int_{t \mathop = 0}^M \int_{u \mathop = 0}^t e^{-s t} \map f u \map g {t - u} \rd u \rd t\) | ||||||||||||
\(\text {(1)}: \quad\) | \(\ds \) | \(=\) | \(\ds \lim_{M \mathop \to \infty} s_M\) |
where $s_M$ is defined to be:
- $\ds \int_{t \mathop = 0}^M \int_{u \mathop = 0}^t e^{-s t} \map f u \map g {t - u} \rd u \rd t$
The region in the plane over which $(1)$ is to be integrated is $\mathscr R_{t u}$ below:
Setting $t - u = v$, that is $t = u + v$, the shaded region above is transformed into the region $\mathscr R_{u v}$ the $u v$ plane:
Thus:
\(\ds s_M\) | \(=\) | \(\ds \iint_{\mathscr R_{t u} } e^{-s t} \map f u \map g {t - u} \rd u \rd t\) | from $(1)$ above | |||||||||||
\(\text {(2)}: \quad\) | \(\ds \) | \(=\) | \(\ds \iint_{\mathscr R_{u v} } e^{-s \paren {u + v} } \map f u \map g v \size {\dfrac {\map \partial {u, t} } {\map \partial {u, v} } } \rd u \rd v\) | Change of Variables Theorem (Multivariable Calculus) |
where $\dfrac {\map \partial {u, t} } {\map \partial {u, v} }$ is the Jacobian of the transformation:
\(\ds J\) | \(=\) | \(\ds \dfrac {\map \partial {u, t} } {\map \partial {u, v} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \begin{vmatrix} \dfrac {\partial u} {\partial u} & \dfrac {\partial u} {\partial v} \\ \dfrac {\partial t} {\partial u} & \dfrac {\partial t} {\partial v} \end{vmatrix}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \begin{vmatrix} 1 & 0 \\ 1 & 1 \end{vmatrix}\) |
Thus the right hand side of $(2)$ is:
- $(3): \quad \ds s_M = \int_{v \mathop = 0}^M \int_{u \mathop = 0}^{M - v} e^{-s \paren {u + v} } \map f u \map g v \rd u \rd v$
Let $\map K {u, v}$ be the function defined as:
- $\map K {u, v} = \begin{cases} e^{-s \paren {u + v} } \map f u \map g v & : u + v \le M \\ 0 & : u + v > M \end{cases}$
This function is defined over the square region in the diagram below:
but is zero over the lighter shaded portion.
Now we can write $(3)$ as:
\(\ds s_M\) | \(=\) | \(\ds \int_{v \mathop = 0}^M \int_{u \mathop = 0}^M \map K {u, v} \rd u \rd v\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \lim_{M \mathop \to \infty} s_M\) | \(=\) | \(\ds \int_{v \mathop = 0}^\infty \int_{u \mathop = 0}^\infty \map K {u, v} \rd u \rd v\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_{v \mathop = 0}^\infty \int_{u \mathop = 0}^\infty e^{-s \paren {u + v} } \map f u \map g v \rd u \rd v\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\int_{v \mathop = 0}^\infty e^{-s u} \map f u \rd u} \paren {\int_{v \mathop = 0}^\infty e^{-s v} \map g u \rd v}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map F s \map G s\) |
Hence the result.
$\blacksquare$
Sources
- 1965: Murray R. Spiegel: Theory and Problems of Laplace Transforms ... (previous) ... (next): Chapter $2$: The Inverse Laplace Transform: Solved Problems: The Convolution Theorem: $20$