Sometimes you want to model a short impulse:

• you give a swing a push, so that it started to swing up
• you plucked a string on the guitar, and it started to vibrate

This can be modeled by an inhomogenous equation of the form $$ D^2 u(t) + \omega^2 u(t) = g(t). $$ where $g(t)$ models the force (which varies over time). Sometimes we don't care about the precise shape of the function $g(t)$, but only the 'total effect' of the force given, then it is useful to use the delta function to model it.

Consider the following sequence of functions $$ g_n(t) = n \cdot 1_{[0,1/n]}(t). $$ where $1_{[a,b]}(t)$ means the value of the function is $1$ if $t \in [a,b]$ and $0$ otherwise. We have that $$ \int g_n(t) dt = 1 $$ for all $n$, but the graph of the function is getting narrower and taller.

We also have the property that, suppose we have a smooth function $f(t)$, then we have $$ \lim_{n \to \infty} \int g_n(t) f(t) dt = f(0). $$ consider the case that $f(t) = 1$ or $t$.

We will write $$ \delta(t) = \lim_{n \to \infty} g_n(t). $$ which is meaningful inside an integral.

Consider the equation $$ y'(t) = \delta(t), \quad t \geq 0 $$ and with initial condition $y(t)=0$ for $t<0$.

We can solve it by integration: for $b > 0$, we have $$ y(b) = y(-\epsilon) + \int_{-\epsilon}^b y'(t) dt = 0 + 1 = 1. $$ where $\epsilon$ is any positive number.

The shape of $y(t)$ is a 'step function', $$ y(t) = \begin{cases} 0 & t < 0 \cr 1 & t > 0 \end{cases} $$ the value of $y(0)$ is undefined (and doesn't matter)

Consider the equation $$ (d/dt)^2 y(t) = \delta(t), \quad t \geq 0 $$ and with initial condition $y(t)=0$ for $t<0$.

Let $g(t) = y'(t)$, then g(t)$ satisfies $$ (d/dt) g(t) = \delta(t) $$ which is just the case in Ex 1. We see $g(t) = 1$ only if $t>0$. Then, we get, for $t>0$ $$ y(t) = y(0) + \int_0^t y'(s) ds = \int_0^t 1 ds = t. $$

Consider the equation that, for $x \in [0,2]$ $$ (d/dx)^2 y(x) + y(x) = \delta (x-1) $$ with boundary condition $y(0) = y(2) = 0. $$

If we integrate this equation over the interval $(1-\epsilon, 1+\epsilon)$, across the location of the delta function, we find that $$ \lim_{\epsilon} y'(1+\epsilon) - y'(1-\epsilon) = 1 $$ (the term $\int_{¹⁾$.