# trying to understanding applications of Eigenvals to euler spirals?

The excerpts in this post are from Gilbert Strang's Introduction To Linear Algebra Book, Page 315 and 316.

What we're solving for

I did the math, and i saw that he was just writing y'' in terms of the position y. But i still don't understand how he came up with the formulas for Yn+1 and Zn+1 and what he was trying to explain. Any help is much appreciated, thanks in advance.

Steps he took to solve it

A visual representation of a Forward Euler Spiral

$$y'' = -y$$

define $$z = y'$$ or $$y'=z$$

Then $$z' = y'' = -y$$

If we let some small amount of time go by.

$$y(t + \Delta t) = y(t) + y'(t) \delta t\\ z(t + \Delta t) = z(t) + z'(t) \delta t$$

But we know that $$y',z'$$ can be described in terms of $$y,z$$

$$y(t + \Delta t) = y(t) + z(t) \Delta t\\ z(t + \Delta t) = z(t) - y(t) \Delta t$$

$$\begin{bmatrix} y(t+\Delta t)\\z(t + \Delta t)\end{bmatrix} = \begin{bmatrix} 1&\Delta t\\-\Delta t&1\end{bmatrix}\begin{bmatrix} y(t)\\z(t)\end{bmatrix}$$

If we start at $$t = 0$$

$$\begin{bmatrix} y(\Delta t)\\z(\Delta t)\end{bmatrix} = \begin{bmatrix} 1&\Delta t\\-\Delta t&1\end{bmatrix}\begin{bmatrix} y(0)\\z(0)\end{bmatrix}$$

$$\begin{bmatrix} y(2\Delta t)\\z(2\Delta t)\end{bmatrix} = \begin{bmatrix} 1&\Delta t\\-\Delta t&1\end{bmatrix}\begin{bmatrix} y(\Delta t)\\z(\Delta t)\end{bmatrix} = \begin{bmatrix} 1&\Delta t\\-\Delta t&1\end{bmatrix}^2\begin{bmatrix} y(0)\\z(0)\end{bmatrix}$$

$$\begin{bmatrix} y(n\Delta t)\\z(n\Delta t)\end{bmatrix} = \begin{bmatrix} 1&\Delta t\\-\Delta t&1\end{bmatrix}^n\begin{bmatrix} y(0)\\z(0)\end{bmatrix}$$

It might be simpler to say:

$$\begin{bmatrix} y\\z\end{bmatrix}' = \begin{bmatrix} 0&1\\-1&0\end{bmatrix}\begin{bmatrix} y\\z\end{bmatrix}$$

Turning a second degree diff eq. into a system of 1st degree equations

If it wasn't a system of equations, you would say:

$$y' = ay\\ y = y(0) e^{at}$$

Just because you have vectors and matrices, no reason to change that...

$$\mathbf x' = A\mathbf x\\ \mathbf x = e^{At}\mathbf x_0$$

and

$$e^{at} = \sum \frac {a^nt^n}{n!}$$

If $$A$$ can be diagonalized, then

$$A = P^{-1}\Lambda P\\ A^n = P^{-1}\Lambda^nP\\ e^{At} = P^{-1}(\sum \frac {\Lambda^nt^n}{n!})P\\ \mathbb x' = P^{-1} e^{\Lambda t} P \mathbb x_0$$

or

$$P\mathbb x' = e^{\Lambda t} P \mathbb x_0$$

The eigenvalues of $$A$$ have most of the information that describes your differential equation.

• omg, thanks so much, I've been sitting on this question for days. I wish i could +1 you but the least I could do is reply for how grateful I am Nov 16, 2018 at 4:49