Questions tagged [dirac-delta]

Mathematically, the delta function is not a function, because it is too singular. Instead, it is said to be a “distribution.” It is a generalized idea of functions, but can be used only inside integrals.

In fact, $~\int \delta~(x)~dx~$ can be regarded as an “operator” which pulls the value of a function at zero. Put it this way, it sounds perfectly legitimate and well-defined. But as long as it is understood that the delta function is eventually integrated, we can use it as if it is a function.

Definition: The Dirac delta function or, delta function $~(~\delta~(x)~)~$ is defined by the properties $$\delta~(x) = \begin{cases} 0 \quad \text{if} \ x\not=0\\ \infty \quad \text{if} \ x=x\end{cases}\qquad \text{and}\qquad \int_0^1 \delta~(x)~dx=1$$

This function is very useful as an approximation for a tall narrow spike function, namely an impulse. For example, to calculate the dynamics of a baseball being hit by a bat, approximating the force of the bat hitting the baseball by a delta function is a useful device. The delta function not only enables the equations to be simplified, but it also allows the motion of the baseball to be calculated by only considering the total impulse of the bat against the ball, rather than requiring the details of how the bat transferred energy to the ball.

There are three main properties of the Dirac Delta function that we need to be aware of. These are,

$1.\quad$ $$~\delta \left( {t - a} \right) = 0~~~~~~~t \ne a~$$ $2.\quad$ $$~\displaystyle \int_{{a - \varepsilon }}^{{ a + \varepsilon }}{{\delta \left( {t - a} \right)~dt}} = 1,\hspace{0.25in}\varepsilon > 0~$$ $3.\quad$ $$~\displaystyle \int_{{ a - \varepsilon }}^{{ a + \varepsilon }}{{f\left( t \right)\delta \left( {t - a} \right)~dt}} = f\left( a \right),\hspace{0.25in}~~~~~~~~~~\varepsilon > 0~$$

References:

https://en.wikipedia.org/wiki/Dirac_delta_function

http://mathworld.wolfram.com/DeltaFunction.html