Lecture4 - fundamental equation of mechanics

\[\begin{eqnarray*} &mx''& = &-kx& - &cx'&\\ &\uparrow& &\uparrow& &\uparrow&\\ &\text{Force (Newton)}& &\text{spring (Hooke)}& &\text{dash pot/damping}& \end{eqnarray*}\]

\[\begin{eqnarray*} \Longrightarrow mx''+cx'+kx=0\\\\ \Longrightarrow x''+ \frac{c}{m}x'+ \frac{k}{m}x=0 \end{eqnarray*}\]

Second order linear (homogeneous) ODEs

• to find two independent solutions (homogeneous solutions)

• general form:

\[\begin{eqnarray*} \frac{d^{2}y}{dt^{2}}+A\frac{dy}{dt}+By=0, \text{ given } y(0)=1, y'(0)=0. \end{eqnarray*}\]

• basic method guess and plug in the exponential solution:

\[\begin{eqnarray*} \text{try } y(t)&=&e^{rt}\\\\ \rightarrow r^{2}e^{rt} + Are^{rt} &+& Be^{rt} = 0\\\\ \rightarrow r^{2} + Ar &+& B = 0 \text{ characteristic equation}. \end{eqnarray*}\]

• we can solve r:

\[\begin{eqnarray*} r=\frac{-A\pm \sqrt{A^{2}-4B}}{2} \end{eqnarray*}\]

Case1: two real roots (overdamping)

\[\begin{eqnarray*} r_{1} \neq r_{2} \rightarrow y(t)=c_{1}e^{r_{1}t}+c_{2}e^{r_{2}t} \end{eqnarray*}\]

Example 4

\[\begin{eqnarray*} y''+4y'+3y=0, y(0)=1, y'(0)=0 \end{eqnarray*}\]

The characteristic equation is

\[\begin{eqnarray*} r^{2}&+&4r+3=0\\\\ &\Longrightarrow& (r+3)(r+1)=0 \rightarrow r=-3, -1\\\\ &\Longrightarrow& y(t)=c_{1}e^{-3t}+c_{2}e^{-t} \end{eqnarray*}\]

Use initial conditions

\[\begin{eqnarray*} y(0)&=&1=c_{1}+c_{2}, \\\\ y'(0)&=&0=-3c_{1}-c_{2}\\\\ &\Longrightarrow& c_{1}=\frac{-1}{2}, c_{2}=\frac{3}{2}\\\\ &\Longrightarrow& y(t)=\frac{-1}{2}e^{-3t}+\frac{3}{2}e^{-t} \end{eqnarray*}\]

Note

Why the solutions are always “damped”? Can you link this result to the value of r?

• the plot for possible solutions y(t):

Case2: two equal roots (critical damping)

\[\begin{eqnarray*} r_{1} = r_{2} = -a \rightarrow (r+a)^{2}=0 \rightarrow r^{2}+2ar+a^{2}=0. \end{eqnarray*}\]

So the ODE can be expressed as:

\[\begin{eqnarray*} y''+2ay'+a^{2}y=0. \end{eqnarray*}\]

Now we only have one solution:

\[\begin{eqnarray*} y(t)=e^{-at}=y_{1}(t) \end{eqnarray*}\]

But we need one more. Let’s try this:

\[\begin{eqnarray*} \text{let } y_{2}(t)=u(t)y_{1}(t)\\\\ a^{2}&\times& (y_{2} = ue^{-at})\\\\ 2a&\times& (y_{2}'=u(-ae^{-at}) +u'e^{-at})\\\\ 1&\times& (y_{2}''=a^{2}e^{-at}u-ae^{-at}u'-ae^{-at}u'+e^{-at}u'')\\\\ &\Longrightarrow& 0 = 0 + 0 + e^{-at}u''\\\\ &\Longrightarrow& u'' = 0 \rightarrow u'=c_{1} \rightarrow u(t) = c_{1}t+c_{2}. \end{eqnarray*}\]

We take another solution and obtain the general solution:

\[\begin{eqnarray*} y_{2}(t) &=& te^{-at}\\\\ y(t) &=& Cy_{1}(t) + Dy_{2}(t) = Ce^{-at}+Dte^{-at} \end{eqnarray*}\]

Case 3: complex roots (undamped/underdamped)

We get the complex solution

\[\begin{eqnarray*} y_{c}(t) = e^{(a+bi)t} = u + iv, \text{ where } u, v \in \mathbb{R} \end{eqnarray*}\]

Theorem 1

If u+iv is the complex solution to

\[\begin{eqnarray*} y''+Ay'+By = 0, \end{eqnarray*}\]

where A and B are real. Then u and v are real solutions.

Proof. Plug the complex solution into the ODE:

\[\begin{eqnarray*} (u+iv)'' + A(u+iv)' + B(u+iv) = 0\\\\ \Longrightarrow (u''+Au'+Bu) + i(v''+Av'+Bv) = 0. \end{eqnarray*}\]

The only way that this equation holds is the real part equal to zero and imaginary part equal to zero. So the linear combination of u and v is the general solution:

\[\begin{eqnarray*} y(t) = c_{1}u(t) + c_{2}v(t) \end{eqnarray*}\]

So the complex solution is

\[\begin{eqnarray*} \tilde{y_c}(t) = e^{at+ibt}=e^{at}\cos{bt} + ie^{at}\sin{bt}. \end{eqnarray*}\]

And the general solution is

\[\begin{eqnarray*} y(t) = e^{at}(c_{1}\cos{bt} + c_{2}\sin{bt}) \end{eqnarray*}\]

Note

How about considering this complex solution

\[\begin{eqnarray*} y_{c}(t) = e^{at-ibt} \text{?} \end{eqnarray*}\]

Can we obtain the same general solution? And why?

Example 5

\[\begin{eqnarray*} y''+4y'+5y=0, y(0)=1, y'(0)=0 \end{eqnarray*}\]

The characteristic equation is

\[\begin{eqnarray*} r^{2}&+&4r+5=0\\\\ &\Longrightarrow& r=\frac{-4\pm \sqrt{-4}}{2} = -2\pm i\\\\ \end{eqnarray*}\]

So the complex solution is:

\[\begin{eqnarray*} e^{(-2\pm i)t} \rightarrow e^{-2t}(\cos{t}\pm \sin{t}) \end{eqnarray*}\]

The general solution is therefore:

\[\begin{eqnarray*} y(t) &=& e^{-2t}(c_{1}\cos{t}+c_{2}\sin{t})\\\\ y'(t) &=& -2e^{-2t}(c_{1}\cos{t}+c_{2}\sin{t}) + e^{-2t}(-c_{1}\sin{t}+c_{2}\cos{t}) \end{eqnarray*}\]

Use initial conditions

\[\begin{eqnarray*} y(0)&=&1=c_{1}, \\\\ y'(0)&=&0=-2c_{1}+c_{2} \rightarrow c_{2}=2\\\\ &\Longrightarrow& y(t)=e^{-2t}(\cos{t}+2\sin{t})=\sqrt{5}e^{-2t}\cos{}(t-\phi) \end{eqnarray*}\]

This is a solution underdamped.

Let us now revisit the problem with slightly modification on the coefficients:

\[\begin{eqnarray*} y'' &+& 2py'+\omega_{o}^{2}y=0,\\\\ \text{where } &\omega_{o}& \text{is the natural frequency and p represents the damping.} \end{eqnarray*}\]

The characteristic equation is

\[\begin{eqnarray*} r^{2} + 2pr + \omega_{o}^{2} = 0 \\\\ \rightarrow r = -p \pm \sqrt{p^{2}-\omega_{o}^{2}} \end{eqnarray*}\]

• if p=0, then there is no damping or called undamped.

\[\begin{eqnarray*} &&y''+\omega_{o}^{2}y=0\\\\ &\Longrightarrow& r=\pm i\omega_{o}\\\\ &\Longrightarrow& y(t) = c_{1}\cos{\omega_{o}t} + c_{2}\sin{\omega_{o}t} = A\cos{}(\omega_{o}t-\phi) \end{eqnarray*}\]

• if another case

\[\begin{eqnarray*} p^{2}-\omega_{o}^{2} < 0 \rightarrow p<\omega_{o}, \end{eqnarray*}\]

then we call the solution underdamped.

Definition 1

We define the pseudo-frequency as:

\[\begin{eqnarray*} \omega_{d}^{2} = \omega_{o}^{2} - p^{2} \end{eqnarray*}\]

\[\begin{eqnarray*} r = -p \pm \sqrt{-\omega_{d}^{2}} = -p \pm i\omega_{d} \end{eqnarray*}\]

So the general solution is

\[\begin{eqnarray*} y(t) &=& e^{-pt}(c_{1}\cos{}(\omega_{d}t) + c_{2}\sin{}(\omega_{d})) \\\\ &=& Ae^{-pt}\cos{}(\omega_{d}t-\phi) \end{eqnarray*}\]

Similarity between electrical and spring-dash pot system