Tail of an Almost Linear System

[EugP]

Homework Statement

This isn't really homework help, I'm just trying to see if there is a "proper" way of doing this.

Given

\frac{dx}{dt} = x - y^2

\frac{dy}{dt} = x - 2y + x^2

show that the system is almost linear.

Homework Equations

The Attempt at a Solution

\left(\begin{array}{c}x&y\end{array}\right)' = \left(\begin{array}{cc}1&0\\1&-2\end{array}\right) \left(\begin{array}{c}x&y\end{array}\right) + \left(\begin{array}{c}-y^2&x^2\end{array}\right),

where \left(\begin{array}{c}-y^2&x^2\end{array}\right),
is the tail.

I was taught that the way you get the tail, is just separate the parts that are not just x or y. But there has to be a better, or I should say correct way of finding the tail. Just separating the equation seems too simple. And besides, I tried this method for:

\frac{dx}{dt} = (1 + x) \sin{y}

\frac{dy}{dt} = 1 - x - \cos{y}

And it didn't work.

 

[HallsofIvy]

First, never despise a method because it is "too easy"!

Second, to "show that the system is almost linear" you need to show that the definition of "almost linear" is satisfied. What is that?

Third, to handle non-polynomial problems, use Taylor's series.
The Taylor's series for sin(y) is, as I am sure you know,
\sum_{i=0}^\infty \frac{(-1)^i y^{2i+1}}{(2i+1)!}
and the Taylor's series for cos(y) is
\sum_{i=0}^\infty \frac{(-1)^i y^{2i)}{(2i)!}
Use that to convert (1+x) sin(y) and 1- x- cos(y) to power series and separate the linear and non-linear parts just as you did above.

 

[EugP]

First, never despise a method because it is "too easy"!

Second, to "show that the system is almost linear" you need to show that the definition of "almost linear" is satisfied. What is that?

Third, to handle non-polynomial problems, use Taylor's series.
The Taylor's series for sin(y) is, as I am sure you know,
\sum_{i=0}^\infty \frac{(-1)^i y^{2i+1}}{(2i+1)!}
and the Taylor's series for cos(y) is
\sum_{i=0}^\infty \frac{(-1)^i y^{2i)}{(2i)!}
Use that to convert (1+x) sin(y) and 1- x- cos(y) to power series and separate the linear and non-linear parts just as you did above.

1) Heh, I don't despise "easy ways", it's just that I don't like to rely on tricks, without understanding the basis for them.

2) I know a few ways to show an ALS (almost linear system):

If the tail is g(x),
then a system is ALS, if

\frac{||g(x)||}{r} \rightarrow 0

as

r \rightarrow 0

then it is ALS.

Or if each function is at least twice differentiable, it is an ALS.

3) THIS is what I was looking for. I KNEW there had to be a REAL method to finding the tail.

Thanks HallsofIvy!

 

[HallsofIvy]

1) Heh, I don't despise "easy ways", it's just that I don't like to rely on tricks, without understanding the basis for them.

2) I know a few ways to show an ALS (almost linear system):

If the tail is g(x),
then a system is ALS, if

\frac{||g(x)||}{r} \rightarrow 0

as

r \rightarrow 0

then it is ALS.

Do you understand that this makes no sense at all since you have not said what "r" is with a function of x?

[/quote]Or if each function is at least twice differentiable, it is an ALS.[/quote]
Okay, now this makes it easy!

3) THIS is what I was looking for. I KNEW there had to be a REAL method to finding the tail.

Thanks HallsofIvy!

Don't you just love Taylor's series? The man's birthday should be national holiday!

 

[EugP]

Do you understand that this makes no sense at all since you have not said what "r" is with a function of x?

I didn't write what r is because I know what it is. But anyway, here it is:

r = ||x||

Don't you just love Taylor's series? The man's birthday should be national holiday!

 

[EugP]

Another quick question. I found this I think on some site:

F(x,y) = F(x_0,y_0) + F_x(x_0,y_0)(x - x_0) + F_y(x_0,y_0)(y - y_0) + \eta_1(x,y)

G(x,y) = G(x_0,y_0) + G_x(x_0,y_0)(x - x_0) + G_y(x_0,y_0)(y - y_0) + \eta_2(x,y)

\frac{d}{dt}\left(\begin{array}{c}x - x_0&y - y_0\end{array}\right) = \left(\begin{array}{cc}F_x(x_0,y_0)&F_y(x_0,y_0)\\G_x(x_0,y_0)&F_y(x_0,y_0)\end{array}\right)\left(\begin{array}{c}x - x_0&y - y_0\end{array}\right) + \left(\begin{array}{c}\eta_1(x,y)&\eta_2(x,y)\end{array}\right)

So for my problem, letting

F(x,y) = \frac{dx}{dt} = x - y^2

and

G(x,y) = \frac{dy}{dt} = x - 2y + x^2


F_x = 1.....F_y = -2y

G_x = 1 + 2x.....G_y = -2

then:

\frac{d}{dt}\left(\begin{array}{c}x - x_0&y - y_0\end{array}\right) = \left(\begin{array}{cc}1&-2y\\1 + 2x&-2\end{array}\right)\left(\begin{array}{c}x - x_0&y - y_0\end{array}\right) + \left(\begin{array}{c}\eta_1(x,y)&\eta_2(x,y)\end{array}\right)

What I can't figure out is how to find the tail...

 

[HallsofIvy]

You have an error here:
\frac{d}{dt}\left(\begin{array}{c}x - x_0&y - y_0\end{array}\right) = \left(\begin{array}{cc}1&-2y\\1 + 2x&-2\end{array}\right)\left(\begin{array}{c}x - x_0&y - y_0\end{array}\right) + \left(\begin{array}{c}\eta_1(x,y)&\eta_2(x,y)\end{ array}\right)
You don't have the "derivative" matrix evaluated at x_0 and y_0.
It should be
\frac{d}{dt}\left(\begin{array}{c}x - x_0&y - y_0\end{array}\right) = \left(\begin{array}{cc}1&-2y_0\\1 + 2x_0&-2\end{array}\right)\left(\begin{array}{c}x - x_0&y - y_0\end{array}\right) + \left(\begin{array}{c}\eta_1(x,y)&\eta_2(x,y)\end{ array}\right)


Use the "two variable" tangent plane formula you have to determine the "linear part" of the F and G at (x_0,y_0). The tail is just the difference between F and its tangent plane formula and G and its tangent plane formula. In other words, multiply out those 2 matrices
\left(\begin{array}{cc}1&-2y_0\\1 + 2x_0&-2\end{array}\right)\left(\begin{array}{c}x - x_0&y - y_0\end{array}\right)
and subtract from
\left(\begin{array}{c}x- y^2\\x-2y+ x^2\end{array}\right)[/itex]

 

[EugP]

You have an error here:
\frac{d}{dt}\left(\begin{array}{c}x - x_0&y - y_0\end{array}\right) = \left(\begin{array}{cc}1&-2y\\1 + 2x&-2\end{array}\right)\left(\begin{array}{c}x - x_0&y - y_0\end{array}\right) + \left(\begin{array}{c}\eta_1(x,y)&\eta_2(x,y)\end{ array}\right)
You don't have the "derivative" matrix evaluated at x_0 and y_0.
It should be
\frac{d}{dt}\left(\begin{array}{c}x - x_0&y - y_0\end{array}\right) = \left(\begin{array}{cc}1&-2y_0\\1 + 2x_0&-2\end{array}\right)\left(\begin{array}{c}x - x_0&y - y_0\end{array}\right) + \left(\begin{array}{c}\eta_1(x,y)&\eta_2(x,y)\end{ array}\right)


Use the "two variable" tangent plane formula you have to determine the "linear part" of the F and G at (x_0,y_0). The tail is just the difference between F and its tangent plane formula and G and its tangent plane formula. In other words, multiply out those 2 matrices
\left(\begin{array}{cc}1&-2y_0\\1 + 2x_0&-2\end{array}\right)\left(\begin{array}{c}x - x_0&y - y_0\end{array}\right)
and subtract from
\left(\begin{array}{c}x- y^2\\x-2y+ x^2\end{array}\right)[/itex]

<br /> <br /> Thank you so much! Eveything is crystal clear now!