Non-Linear First Order ODE: Critical Point Linearization

[Mike86]

Homework Statement

dx/dt = x - y + (x^2) - xy

dy/dt = -y + (x^2)

- Determine the critical points for the equation,
- Determine the linearized system for each critical point and discuss whether it can be used
to approximate the behaviour of the non-linear system. What is the type and stability of
each critical point?

Homework Equations

Critical points occur when:

dx/dt = 0 and dy/dt = 0

The Attempt at a Solution

Hey guys! First post, hope all goes well and there will me many more to come :)

Ok I am doing this problem and this is the first time I've come across non-linear first order ODE's, usually I have been doing linear ones!

Basically I have established the critical points occur at (0,0) and (1,1).

I am not sure at all how to determine the linearized system for each critical point. I have looked in a couple of textbooks and online but haven't found too much unfortunately.

Any advice on how to go about this step would be great. If anyone knows of a good place to find an worked example of a similar question that would be great too (I have found this is the best way for me to learn, personally works really well for me!)

Thanks in advance :)

 

[HallsofIvy]

dx/dt = x - y + (x^2) - xy
dy/dt = -y + (x^2)

So the critical points are where $x- y+ x^2- xy= 0$ and $-y+ x^2= 0$. From the second equation, $y= x^2$ and then the first equation becomes $x- x^2+ x^2- x^3$$= x- x^3= x(1- x^2)= x(1- x)(1+ x)= 0$. x can have three values, 0, 1, and -1.

If x= 0, then, from the second equation, y= 0 so, yes, (0, 0) is a critical point. If x= 1,then, from the second equation, y= 1 so, yes, (1, 1) is a critical point.

But you missed one critical point! Do you see what it is?

The most general way of "linearizing" equations is to look at the Jacobian: if dx/dt= f(x,y) and dy/dt= g(x,y) then the Jacobian is the matrix
$$\begin{bmatrix}\frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} \\ \frac{\partial g}{\partial x} & \frac{\partial g}{\partial y}\end{bmatrix}$$

Here, f(x,y)= x - y + (x^2) - xy and g(x,y)= -y + (x^2) so the Jacobian is
$$\begin{bmatrix}1+ 2x- y & -1- x\\ -2x & -1\end{bmatrix}$$

At (0, 0) that is
$$\begin{bmatrix}1+ 2(0)- 0 & -1- 0 \\-2(0) & -1\end{bmatrix}= \begin{bmatrix}1 & -1 \\ 0 & -1\end{bmatrix}$$
so the linearization of the system of equations is
$$\begin{bmatrix} \frac{dx}{dt}\\ \frac{dy}{dt}\end{bmatrix}= \begin{bmatrix}1 & -1 \\ 0 & -1\end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}$$
or dx/dt= x- y, dy/dt= -y.
Do the same at the other two points.

In this simple case, since the functions are polynomials we don't need to be that sophisticated. We can argue that, close to 0, higher powers are much smaller than first power terms and drop them:
dropping both $x^2$ and xy from x - y + (x^2) - xy gives x- y and dropping $x^2$ from -y + (x^2) gives -y so that we have dx/dt= x- y and dy/dt= -y as before.

At (1, 1), of course, x and y are not close to 0 but x- 1 and y- 1 are: let u= x- 1 and v= y- 1 so that x= u+ 1 and y= v+ 1. Then $x- y+ x^2- xy=$$u+1- (v+1)+ (u+1)^2- (u+1)(v+1)$$= u+ 1- v- 1+ u^2+ 2u+ 1- uv- u- v- 1$$= u^2- uv+ 2u- 2v$ and $-y+ x^2=$$-(v+1)+ (u+1)^2=$$-v- 1+ u^2+ 2u+ 1= u^2+ 2u- v$. Now linearize by dropping higher powers of u and v: dx/dt= 2u- 2v= 2(x- 1)- 2(y- 1)= 2x- 2y and dy/dt= 2u- v= 2(x- 1)- (y- 1)= 2x- y+ 1.

But the more "sophisticated" method of evaluating the Jacobian at the point is more general and simpler.

 

[Mike86]

Thank you so much for that!

The other critical point is at (-1, 1). I totally overlooked that one.

I have found an example about non-linear ordinary ODE's and for the point (0,0) they have used the same approach.

I have found the Jacobi matrix for (1,1) and (-1,1), being [2, -2; 2, -1] and [-2, 0; -2, -1] respectively.

Based on these linearizations I am now trying to find the type and stability of each point. For (0,0) it was simple enough. I found the eigenvalues as -1 and 1. And as -1 < 0 < 1, I concluded the critical point is a unstable saddle.

The example I have seen for points away from the origin took a similar approach to finding the linearization. But at one point the example used anti-differentiation.

The approach you used for the linearization of (1,1) is the first time I have see something like that. I understand what you have done but the final step of dropping higher powers kinda confused me.

Also with the final linearization, I am not sure how to establish the type and stability of the point! From your concluded values of dx/dt and dy/dt I can't find a way to establish what type and stability the critical point has. I have graphed the point on Phase Plane and it looks like a stable centre with orbits resembling circles. I just can't see this with the linearization.

Thanks again for that HallsofIvy, appreciate it soo much. I am guessing you are a math teacher or lecturer! :)

 

[Mike86]

Any ideas? Still stuck on this one :( Thanks!

 

[HallsofIvy]

The Jacobian is
$$\begin{bmatrix}1+ 2x- y & -1- x\\ -2x & -1\end{bmatrix}$$
and at (1, 1) that is
$$\begin{bmatrix}2& -2\\ -2 & -1\end{bmatrix}$$

We can determine what kind of point that is by finding the eigenvalues. The characteristic equation is
$$\left|\begin{array}{cc}2-\lambda & -2\\ -2 & -1-\lambda\end{array}\right|= (2-\lambda)(-1-\lambda)- 4= \lambda^2- \lambda- 6= (\lambda- 3)(\lambda+ 2)= 0$$
which has roots $\lambda= 3$ and $\lamba= -2$
Since those are of different sign, (1, 1) is a saddle point- there is an "inflow" along one line and an "outflow" along another.

In fact, we can see that an eigenvector corresponding to eigenvalue 3 is <-2, 1> so the line y= -2x is the "outflow" line. An eigenvector corresponding to eigenvalue -2 is <1, 2> so 2y= x, or y= (1/2)x, is the "inflow" line.

For this particular problem, we could have just looked at the determinant.
$$\left|\begin{array}{cc} 2 & -2 \\ -2 & -1\end{array}\right|= (2)(-1)- (-2)(-1)= -2- 4= -6$$
Since that is negative, and the determinant is always the product of the eigenvalues, the eigenvalues are of opposite sign and so we have a saddle point. If the determinant had been positive, we would know that the eigenvalues are of the same sign but would not know if they were both positive (a "source") or both negative (a "sink").

A less sophisticated way of doing that, without using matrices, is to look for "straight line solutions". The linearized equations, around (1, 1) are x'= 2x- 2y and y'= -2x- y. A linear solution would be of the form y= mx. Then y'= mx' so -2x- y= m(2x- 2y).
Since y= mx, that is \(\displaystyle -2x- mx= m(2x- 2mx)\) or \(\displaystyle x(-2- m)= x(2m- 2m^2)\). In order that that be true for all x we must have \(\displaystyle -2- m= 2m- 2m^2\) or \(\displaystyle 2m^2-3m- 2= (2m+1)(m-2)= 0 so m= -1/2 or 2.

That is, linear solutions are y= -(1/2)x and y= 2x, the lines we got from the eigenvectors.

At (2, -1), for example, which is on the line y= -(1/2)x, x'= 2x- 2y= 2(2)- 2(-1)= 4+ 2= 6> 0 and y'= -2x- y= -2(2)- (-1)= -2+ 1= -1< 0 so that x is increasing while y is decreasing- we have motion away from the point.

At (1, 2), which is on the line y= 2x, x'= 2x- 2y= 2(1)- 2(2)= 2- 4= -2< 0 and y'= -2x- y= -2(1)-(2)= -4< 0 so we have motion toward the point. Again, a saddle point.\)