MATH201/content/Dirak δ-function (lec 21).md

#start of lec 21
From newton's second law: $ma=F$
$m\frac{dv}{dt}=f(t)$
integrate both sides:
$m\int_{t_{0}} ^{t_{1}} \frac{dv}{dt}dt =\int _{{t_{0}}} ^{t_{1}}f(t) \, dt$
$mv(t_{1})-mv(t_{0})=\int _{t_{0}}^{t_{1}}f(t) \, dt$
that is, change in momentum on the LHS equates to an impulse on the RHS.
(picture shown, you can have the same impulse, the same area under the graph if you squish down $f(t)$ to be narrower, as long as you make it taller. If we take it to the extreme we get the Dirak delta function.)

# Dirak delta function
#dirak_delta
The definition of the Dirak delta function is:
$\delta(t-a)=\begin{cases}0,  & t\ne a \\''\infty'',  & t=a\end{cases}$
however, a more useful definition is:
$\int _{-\infty} ^{\infty} \delta(t-a)f(t)\, dt=f(a)$

properties:
$\int_{{-\infty}}^{\infty} \delta(t-a)\, dt=1$
![draw](drawings/Drawing-2023-10-25-13.16.20.excalidraw.png)
$\int _{-\infty} ^t \delta(t-a)\, dt=\begin{cases}0, & t<a \\ 1,  & t\geq a\end{cases}=u(t-a)$
$u'(t-a)=\delta(t-a)$

What is $\mathcal{L}\{\delta(t-a)\}$?
$\mathcal{L}\{\delta(t-a)\}=\int _{0} ^\infty \delta(t-a)e^{-st} \, dt$ for $a>0$
Using the definition of Dirak $\int _{-\infty} ^{\infty} \delta(t-a)f(t)\, dt=f(a)$:
$$\mathcal{L}\{\delta(t-a)\}=\int _{-\infty} ^{\infty} e^{-st} \delta(t-a) \, dt=e^{-as}$$
## Examples of DE's with Dirak
#ex #second_order_nonhomogenous #dirak_delta #IVP
Solve for $w(t)$:
$$w''+6w'+5w=e^t\delta(t-1) \qquad w(0)=0 \qquad w'(0)=4$$
Solving this using something like #voparam would be very difficult, using LT should be very easy!
Hit it with the LT!
$s^2W-\cancel{ sw(0) }-\cancelto{ 4 }{ w'(0) }+6sW-\cancel{ 6w(0) }+5W=\int _{0} ^\infty e^{-st}e^t\delta(t-1)\, dt$
Now we try to isolate for $W$
$=s^2W-4+6sW+5W=\int _{-\infty} ^\infty e^{-st}e^t\delta(t-1)\, dt$
(we can extend the range of the integral as the integrand is 0 for all t<1)
this allows us to use the definition of Dirak on the RHS, namely: $\int _{-\infty} ^{\infty} \delta(t-a)f(t)\, dt=f(a)$
> Alternatively, you could have taken the laplace of the RHS by using properties 2 and 14 from the [big LT table](drawings/bigLTtable.png).

$W(s^2+6s+5)=4+ee^{-s}$
$W(s)=\frac{4}{(s+1)(s+5)}+\frac{ee^{-s}}{(s+1)(s+5)}$
Using partial fractions:
$w(t)=\mathcal{L}^{-1}\left\{ \frac{1}{s+1} - \frac{1}{s+5} \right\}+\frac{e}{4}\mathcal{L}^{-1}\left\{ e^{-s}\left( \frac{1}{s+1} - \frac{1}{s+5} \right) \right\}$
$=\underbrace{(e^{-t}-e^{ -5t }) }_{ \text{this came from initial conditions} }+\frac{e}{4}\mathcal{L}^{-1}\{\dots\}$
for second term use property:
$\mathcal{L}^{-1}\{e^{as}F(s)\}=f(t-a)u(t-a)$
$$y(t)= (e^{-t}-e^{ -5t }) +\frac{e}{4}u(t-1)(e^{ -(t-1) }-e^{ -5(t-1) })$$
notice that the right most term came from the impulse and the effect it had on the system.
side note: delta functions are useful for quantum physics.