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 $t>0$ $y(t) = y(0) + \int_0^t y'(s) ds = \int_0^t 1 ds = t.$

Let $P(D)$ be a differential operator, where $D=d/dt$ and $P(x)$ is a degree $n$ polynomial. Suppose we are facing an equation of the type $P(D) y(t) = g(t)$ with some homogeneous boundary condition (meaning the function $y(t)$ vanishes on the boundary of the domain).

Suppose we know the solution for $P(D) G(t; s) = \delta(t-s).$

Then, we can solve the original equation by doing an integral $y(t) = \int G(t; s) g(s) ds.$ Indeed, we have $P(D) y(t) = \int P(D) G(t; s) g(s) ds = \int \delta(t-s)g(s) ds = g(t).$

Consider the domain being $[0,\infty)$, and we have $d/dt y(t) = g(t), \quad g(t)=1_{[1,2]}(t). y(0) = 0$ In this case, we can first solve for the Green's function $d/dt G(t; s) = \delta(t-s), \quad G(0;s)=0$ we get $G(t;s) = 1$ for $t>s$, and 0 else.

Thus, we can get get $y(t) = \int g(s) G(t;s) ds = \int_{1}^2 1_{t>s} ds.$ if $t>2$, then $y(t) = \int_1^2ds = 1$, if $1<t<2$, we get $\int_1^t ds = (t-1)$. if $t<1$, we get $y(t)=0$.

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 $\int_{(1-\epsilon, 1+\epsilon)} y(x) dx = O(\epsilon) \to 0$) hence we had a discontinuity of the slope of $y(x)$ when $x=1$. We may write down the general solution over the interval $(0,1)$ that vanishes on $x=0$ $y_-(x) = a \sin(x) , \quad x \in [0,1]$ and the general solution over $x \in (1,2)$ that vanishes on $x=2$ $y_+(x) = b \sin(x-2), \quad x \in [1,2]$ then we can solve the condition that $y_-(1) = y_+(1), \quad y_-'(1) - y'_+(1) = 1$ this gives $a \sin(1) = b \sin(-1), \quad a\cos(1) - b \cos(-1) = 1$ so $a=-b$ and $a = 1 / (2 \cos(1))$.