## Laplace Transform

It is known from quantum mechanics that the law of conservation of momentum can
be derived from translation symmetry. If it is possible to move around the
physical system in space without its properties being altered, then in that
case conservation of momentum is guaranteed. However, the foundation of this
theorem is purely mathematical, and can be understood as follows.
The series expansion of a function $ f(x+\xi) $ around $x$ will be set up:
$$ f(x+\xi) = \sum_{k=0}^\infty \frac{ \xi^k }{ k ! } f^{(k)}(x)
= \left[ \sum_{k=0}^\infty \frac{1}{k!}
\left( \xi \ \frac{d}{dx} \right)^k \right] f(x) $$
In the expression between square brackets the series expansion of $ e^x $ is
recognized. Therefore we can write symbolically:
$$ f(x+\xi) = e^{ \, \xi \frac{d}{dx} } f(x) $$
With this formula in mind, consider an arbitrary convolution-integral:
$$ \int_{- \infty}^{+ \infty} \! h(\xi) \phi(x-\xi) \, d\xi $$
Convolution integrals do frequently occur. With a linear system, the response at
a disturbance is the convolution-integral of the disturbance with the so-called
unity-response. The unity-response is the way in which the system reacts upon
the simplest of all disturbances, that is a steep peak of very short duration
at time zero, a *Dirac delta*. Convolution integrals can be rewritten with
help of the operator-expression for $ f(x-\xi) $ as follows:
$$ \int_{- \infty}^{+ \infty} \! h(\xi)
e^{ \, - \xi \frac{d}{dx} } \, d\xi \quad \phi(x) $$
The integral in this expression should be well known. Quite "incidentally"
namely it is the (double-sided) Laplace transform:
$$ H(p) = \int_{-\infty}^{+\infty} \! h(\xi) e^{\, - \xi p} \, d\xi $$
It seems that Laplace's integral shows up quite spontaneously with elementary
considerations about convolution-integrals in combination with Operator
Calculus; the formula for the convolution-integral can be written as:
$$ H(\frac{d}{dx}) \, \phi(x) $$ The fact that
Laplace transforms are a very powerful means for solving differential
equations can now be understood without much effort. Suppose we have a linear
inhomogeneous differential equation. In general it has the form:
$$ D( \, \frac{d}{dx} ) \, \phi(x) = f(x) $$
Then with help of Operator Calculus we can immediately write the solution as:
$$ \phi(x) = \frac{1}{ D( \, \frac{d}{dx} ) } f(x) $$
Put $ H(d/dx) = 1/D(d/dx) $ , then the excercise becomes: find the inverse
Laplace transform of $ H(p) $. Call this inverse function $ h(x) $.
Finding the solution then follows entirely the abovementioned pattern:
$$ \phi(x) = \int_{- \infty}^{+ \infty} \! h(\xi) f(x-\xi) \, d\xi $$