added lec 18 and fixed broken link

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Sasserisop 2023-10-18 18:18:24 -06:00
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# (Heaviside) Unit step function
We got a joke! Heaviside (British electrician) was told during a restaurant that mathematicians are finally using his formula, he replied: I don't need the chef to tell me the food was good. (a lil bit cynical!)
Anyways, the Heaviside step function is defined as:
$u(t-a)=\begin{cases}0,\quad t<a \\1,\quad a\leq t\end{cases}$
graph it, it just "switches on" when $t=a$
if we use laplace transforms we will see that solving equations with this Heaviside function is very natural. Using #mouc wont work, #voparam might work but it's messy.
#ex
compute:
$\mathcal{L}\{u(t-a)f(t-a)\}$
$\int _{0} ^\infty e^{-st}u(t-a)f(t-a)\, dt$
$=\int _{a}^\infty e^{-st}\cdot 1f(t-a)\, dt$
let $x=t-a$
$=\int _{0} ^\infty e^{-s(x+a)}f(x)\, dx$
$=e^{-as}\int _{0} ^\infty e^{-sx}f(x)\, dx$
the right side is nothing but the LT of f. How nice!
$$\mathcal{L}\{u(t-a)f(t-a)\}=e^{-as}F(s)$$
Very useful formula! ^
$\mathcal{L}\{u(t-a)f(t)\}=\mathcal{L}\{u(t-a)f(t+a-a)\}=e^{-as}\mathcal{L}\{f(t+a)\}$
$\mathcal{L}^{-1}\{e^{-as}F(s)\}=u(t-a)f(t-a)$
#ex
multiple functions being switched:
$f(t)=\begin{cases} 1, \quad t<1 \\ t, \quad 1\leq t\leq \pi \\ \sin(t), \quad \pi\leq t\end{cases}$
then we express it using heaviside functions:
$f(t)=1-u(t-1)+tu(t-1)-tu(t-\pi)+\sin(t)u(t-\pi)$
think it about switching on certain parts of the equation at certain times t.
#ex
current in electric circuit I(t) is defined by:
$I''+I=g(t),\quad I(0)=I'(0)=0$
$g(t)=\begin{cases}1,\quad 0\leq t<\pi \\ -1,\quad \pi\leq t\end{cases}$
find $I(t)$
the capacitance here is huge, 1F! (I believe it comes from the coefficient of $y$ term)
and we are switching it instantaneously, lets imagine we don't blow anything up.
$I''+I=1-2u(t-\pi)$
$\mathcal{L}\{I(t)\}=J(s)$
$s\cdot I(0)=s\cdot I'(0)=0$
using formula we derived in earlier ex:
$s^2J(s)+J=\frac{1}{s}-\frac{2e^{-\pi s}}{s}$
$J=\underbrace{ \frac{1}{s(s^2+1)} }_{ =F(s) }-2 \frac{1}{s(s^2+1)}e^{-\pi s}$
$\mathcal{L}^{-1}\left\{ \frac{1}{s(s^2+1)} \right\}=\mathcal{L}^{-1}\{\frac{1}{s}-\frac{s}{s^2+1}\}=1-\cos t$
$=1-\cos(t)-2u(t-\pi)(1-\cos(t+\pi))$
$$=1-\cos(t)-2u(t-\pi)(1+\cos(t))$$
#end of lec 18

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@ -15,7 +15,7 @@ $v''+\left( \frac{2y_{1}'}{y_{1}}+p(x) \right)v'=\frac{f(x)}{y_{1}}$
substitute $v'=u$ substitute $v'=u$
$u'+\left( \frac{2y_{1}'}{y_{1}}+p(x) \right)u=\frac{f(x)}{y_{1}}$<- This is now a linear first order equation #de_L_type2 $u'+\left( \frac{2y_{1}'}{y_{1}}+p(x) \right)u=\frac{f(x)}{y_{1}}$<- This is now a linear first order equation #de_L_type2
This can be solved with prior tools now, We compute the integrating factor $\mu$ This can be solved with prior tools now, We compute the integrating factor $\mu$
$\mu=e^{\int(2y_{1}'/y_{1}+p)dx}=e^{\ln(y_{1}^2)}e^{\int P(x) \, dx}=y_{1}^2\cdot e^{\int p(x) \, dx}$ $\mu=e^{\int(2y_{1}'/y_{1}+p)dx}=e^{\ln(y_{1}^2)}e^{\int p(x) \, dx}=y_{1}^2\cdot e^{\int p(x) \, dx}$
From there, continue on as you would with any linear first order equation. From there, continue on as you would with any linear first order equation.
Isn't this nice? some kind of magic. We made some guesses and we arrived somewhere. Isn't this nice? some kind of magic. We made some guesses and we arrived somewhere.
## What you need to remember: ## What you need to remember:
@ -24,10 +24,10 @@ I know memorizing formulas robs the richness of mathematics, but that is just th
1) $y''+p(x)y'+q(x)y=f(x)$ 1) $y''+p(x)y'+q(x)y=f(x)$
2) $u'+\left( \frac{2y_{1}'}{y_{1}}+p(x) \right)u=\frac{f(x)}{y_{1}}$ <- Notice, if the coefficient for the $u$ term is $0$, you can treat the equation as a separable equation to minimize computation (integrate both sides to get u, then move on to step 5). Otherwise, move on to step 3. 2) $u'+\left( \frac{2y_{1}'}{y_{1}}+p(x) \right)u=\frac{f(x)}{y_{1}}$ <- Notice, if the coefficient for the $u$ term is $0$, you can treat the equation as a separable equation to minimize computation (integrate both sides to get u, then move on to step 5). Otherwise, move on to step 3.
3) $\mu(x)=y_{1}^2\cdot e^{\int p(x) \, dx}$ <- where $y_{1}$ is one of your homogenous solutions. 3) $\mu(x)=y_{1}^2\cdot e^{\int p(x) \, dx}$ <- where $y_{1}$ is one of your homogenous solutions.
4) $u(x)=\frac{1}{\mu(x)}\int \mu(x) \frac{f(x)}{y_{1}} \, dx$ <-don't confuse $q(x)$ with $\frac{f(x)}{y_{1}}$ 4) $u(x)=\frac{1}{\mu(x)}\int \mu(x) \frac{f(x)}{y_{1}} dx$ <-don't confuse $q(x)$ with $\frac{f(x)}{y_{1}}$
5) $y(x)=y_{1}(x)\int u(x) \, dx$ is your general solution. 5) $y(x)=y_{1}(x)\int u(x) dx$ is your general solution.
Notice the gotcha, we have three nested integrals. If you notice any of the integrals is difficult to compute, you might have to give up and use another method like #mouc or #voparam. No method is better than the other. For example, solving $y''+y'+y=xe^x$ is trivial using #mouc but practically impossible using #reduction_of_order Notice the gotcha, we have three nested integrals. If you notice any of the integrals is difficult to compute, you might have to give up and use another method like #mouc or #voparam . No method is better than the other. For example, solving $y''+y'+y=xe^x$ is trivial using #mouc but practically impossible using #reduction_of_order .
# Reduction of order example # Reduction of order example
#ex #second_order_nonhomogenous #reduction_of_order #ex #second_order_nonhomogenous #reduction_of_order

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#start of lec 17
summary of all the equations (separable linear exact, ...) available on eclass
(homogenous (2 forms), Bernoulli equation, linear coefficients,)
second order equations:
put it in standard form
solve the homogenous eq,
with constant coefficients
3 cases: overdamped critically damped under damped
now find particular solution to get general solution $y=y_{h}+y_{p}$
either use #mouc or #voparam
we also have Cauchy-Euler equations (two methods, either guess y=x^r or use the general formula)
reduction of order is a technique when you're given a solution and you have to find the other
mechanical systems $my''+by'+ky=F_{o}\cos(\gamma t)$ (he recommends solving with mouc) that gives you the transient solution.
the particular solution is the steady part.
recall $\mu$=amplitude $\gamma_{r}$=resonance freq $\mu(\gamma_{_{r}})$=amplitude at resonance
Laplace transforms: definition and 4 important properties
That's all that will be covered on the midterm. 25 minutes spent covering what will be on the midterm!
So, why did we learn all this stuff about laplace transforms? We will now see how its useful:
#ex #LT #second_order_nonhomogenous
$$y''+y=t^2+2 \qquad y(0)=1 \qquad y'(0)=-1$$
LHS=RHS, so the Laplace transforms of each side must also be equal:
$\mathcal{L}\{y''+y\}=\mathcal{L}\{t^2+2\}$
applying linearity:
$\mathcal{L}\{y''\}+\mathcal{L}\{y\}=\mathcal{L}\{t^2\}+\mathcal{L}\{2\}$
using properties:
$s^2Y(s)-s+1+Y(s)=\frac{2!}{s^3}+\frac{2}{s}$
factor out Y(s)
$Y(s)(s^2+1)=\frac{2(s^2+1)}{s^3}+s-1$
$Y(s)=\frac{2}{s^3}+\frac{s}{s^2+1}-\frac{1}{s^2+1}$
now we take the inverse LT to obtain y(s)
$$y(t)=t^2+\cos(t)-\sin(t)$$
Done!
second example:
#ex #LT #second_order_nonhomogenous
$$y''-y=(t-2)e^{t-2} \qquad y(2)=0 \qquad y'(2)=0$$
we need y at 0 but its as 2.
so we make a substitution: $x=t-2$
$\frac{d^2y}{dx^2}-y=xe^x$ <- notice the y here is not the same as the y above, lousy notation again :)
where $y(0)=0 \qquad \frac{dy}{dx}(0)=0$
hit it with the LT!
notice 1/s^2 is LT of x so we have to shift it
$s^2Y(s)-Y(s) =\frac{1}{(s-1)^2}$
$Y(s)=\frac{1}{(s-1)^3(s+1)}=\frac{A}{s-1}+\frac{B}{(s-1)^2}+\frac{C}{(s-1)^3}+\frac{D}{s+1}$
$\frac{{A(s-1)^2(s+1)+B(s-1)(s+1)+C(s+1)+D(s-1)^3}}{(s-1)^3(s+1)}$
$\begin{matrix}A+D=0 \\-A-B-3D=0 \\ -A+c+3D=0 \\ A-B+C-D=1\end{matrix}$
$A=\frac{1}{8}$
$B=-\frac{1}{4}$
$C=\frac{1}{2}$
$D=-\frac{1}{8}$
plug into expression then take inv LT to obtain y(t):
final solution: $y(x)=\frac{1}{8}e^x-\frac{1}{4}xe^x+\frac{1}{4}x^2e^x-\frac{1}{8}e^-x$
where x=t-2
all done!
#end of lec 17 #start of lec 18
#ex
$$y''+ty'-2y=2 \qquad y(0)=y'(0)=0$$
hit it with the LT!
$\mathcal{L}\{y''\}+\mathcal{L}\{ty'\}-2\mathcal{L}\{y\}=\frac{2}{s}$
$s^2Y(s)-\frac{d}{ds}\mathcal{L}\{y'\}-2Y(s)=\frac{2}{s}$
$s^2Y-\frac{d}{ds}(sY(s))-2Y=\frac{2}{s}$
apply product rule:
$s^2Y-Y-s\frac{dY}{ds}=\frac{2}{s}$
^ Boooo! another differential equation! :(
$\frac{dY}{ds}$ lies in the s "phase space"
$-s\frac{dY}{ds}+s\left( s-\frac{3}{5} \right)=\frac{2}{s}$
This is a linear equation!
divide by -s to get it in standard form
$\frac{dY}{ds}-\left( s-\frac{3}{5} \right)y=-\frac{2}{s^2}$
compute integrating factor:
$\mu(s)=e^{-\int (s-3/s) \, ds}=e^{-s^2/2}e^{\ln{s^3}}=s^3e^{-s^2/2}$
$\frac{d}{ds}(s^3e^{-s^2/2}Y)=-2se^{-s^2/2}$
$s^3e^{-s^2/2}Y=-2\int se^{-s^2/2} \, ds$
use u sub.
$u=\frac{s^2}{2}$
$-2\int e^{-u} \, du$
$=2e^{-s^2/2}+C$
$Y(s)=2s^3+C \frac{e^{s^2/2}}{s^3}$
is this even a legitimate thing to take an inverse of?
the lim of the expression approaches inf as s approaches inf
So what do we do? Well we have that C term. We have to set $C=0$
then:
$Y(s)=\frac{2}{s^3}$
$$y(t)=t^2$$
we just solved a new equation that hasn't fit into our previous equation types using LT. How cool is that!

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# This is the main index # This is the main index
I have written these notes for myself, I thought it would be cool to share them. These notes may be inaccurate, incomplete, or incoherent. No warranty is expressed or implied. Reader assumes all risk and liabilities. I have written these notes for myself, I thought it would be cool to share them. These notes may be inaccurate, incomplete, or incoherent. No warranty is expressed or implied. Reader assumes all risk and liabilities.
</br> </br>
Good luck on midterms! <3 -Oct 18 2023
</br>
[Separable equations (lec 1)](separable-equations-lec-1.html) [Separable equations (lec 1)](separable-equations-lec-1.html)
[Homogenous equations (lec 2)](homogenous-equations-lec-2.html) [Homogenous equations (lec 2)](homogenous-equations-lec-2.html)
[Linear equations (lec 2-3)](linear-equations-lec-2-3.html) [Linear equations (lec 2-3)](linear-equations-lec-2-3.html)
@ -15,6 +17,9 @@ I have written these notes for myself, I thought it would be cool to share them.
[Free vibrations (lec 11-12)](free-vibrations-lec-11-12.html) (raw notes, not reviewed or revised yet.) [Free vibrations (lec 11-12)](free-vibrations-lec-11-12.html) (raw notes, not reviewed or revised yet.)
[Resonance & AM (lec 13-14)](resonance-am-lec-13-14.html) [Resonance & AM (lec 13-14)](resonance-am-lec-13-14.html)
[Laplace transform (lec 14-16)](laplace-transform-lec-14-16.html) (raw notes, not reviewed or revised yet.) [Laplace transform (lec 14-16)](laplace-transform-lec-14-16.html) (raw notes, not reviewed or revised yet.)
[Solving IVP's using Laplace transform (lec 17-18)](solving-ivps-using-laplace-transform-lec-17-18.html) (raw notes, not reviewed or revised yet.)
[(Heaviside) Unit step function (lec 18)](unit-step-function) (raw notes, not reviewed or revised yet.)
</br> </br>
[How to solve any DE, a flow chart](Solve-any-DE.png) [How to solve any DE, a flow chart](Solve-any-DE.png)