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Math Help - Non-Linear First Order ODE: Critical Point Linearization

  1. #1
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    Non-Linear First Order ODE: Critical Point Linearization

    Question:

    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?

    Answer:

    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) from:

    Critical points occur when:

    dx/dt = 0 and dy/dt = 0

    I am not sure at all how to determine the linearized system for each critical point. I have looked in a couple of text books and online but havent 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
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  2. #2
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    To linearize the system about critical point you need to find the Jacobian and then evaluate that Jacobian at the critical points.
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  3. #3
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    I think you missed one. (-1,1)
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    Thanks for that guys! Yes I missed that one. Thanks.

    I've found the Jacobian matrix and evaluated it at the different points.

    I am having trouble using the linearizations to evaluate the type and stability of each point. I can find the eigenvalues of (0,0) and have established it is a unstable saddle.

    But for say (1,1) I can't get it right it seems. I have substituted X = x -1 and Y = y - 1.

    I ended up with something like: 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 I cant see how I can use this information to conclude what the type and stability of the critical point is.

    Thanks again guys!
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  5. #5
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    If your system is

    \dot{x} = f(x,y)\;\;\;\dot{y} = g(x,y) then the Jacobian matrix is

    \left(\begin{array}{cc}<br />
f_x & f_y\\<br />
g_x & g_y<br />
\end{array}\right)

    Sub in your critical points and look at the eigenvalues of this matrix. I think it's really the easiest way.
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  6. #6
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    If you find the jacobian and then evaluate at the various fixed points. Then find the eigenvalues of the jacobian. If all the eigenvalues have negative real part the solutions is stable. If any of the eigenvalues have positive real part then it is unstable. If the eigenvalues are purely imaginary then the fixed point is a center, but then it is technically stable.

    So let's do the fixed point (1,1)
    [Math]J(x,y) = \begin{bmatrix}1 + 2x - y& -1 - x \\ 2x & -1\end{bmatrix}[/tex]
    [Math]J(1,1) = \begin{bmatrix}2& -2 \\ 2 & -1\end{bmatrix}[/tex]
    Thus, the linearized system at (1,1) is
    [Math]\begin{bmatrix}x'\\ y'\end{bmatrix}= \begin{bmatrix}2& -2 \\ 2 & -1\end{bmatrix}\begin{bmatrix}x\\ y\end{bmatrix}[/tex]
    The eigenvalues of J(1,1) are
    \lambda_1 = 1/2 (1+i \sqrt{7})
    \lambda_2 = 1/2 (1-i \sqrt{7}).
    The real parts are positive, so (1,1) is linearly unstable.
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  7. #7
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    Quote Originally Posted by lvleph View Post
    If you find the jacobian and then evaluate at the various fixed points. Then find the eigenvalues of the jacobian. If all the eigenvalues have negative real part the solutions is stable. If any of the eigenvalues have positive real part then it is unstable. If the eigenvalues are purely imaginary then the fixed point is a center, but then it is technically stable.
    Care must be taken when trying to use the linear system to predict the stability of the nonlinear system. If the real part of the eigenvalue of the linear system is zero, nothing can be said about the stability of the nonlinear system.
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    I suppose I should say I meant linearly stable.
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  9. #9
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    I thought because at (1, 1) x and y are not close to 0 but x- 1 and y- 1 are: so I let let u= x- 1 and v= y- 1 so that x= u+ 1 and y= v+ 1.

    I thought this was standard approach to this sort of non-linear question. Was I wrong?
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  10. #10
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    I have never heard of your approach and I don't see the point, but that doesn't mean it is necessarily wrong. The procedure that I showed is the standard way of determine the linear system and its stability.
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    Isn't the method you showed a analysis of the linearized system only. Is the type and stability for the linearized system at (1,1) same as the type and stability of that point in the complete non-linear system?

    I guess that is what is confusing me.
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  12. #12
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    Looking at stability for the linearized system applies in a small region around the critical point. This is the way I was taught to check for stability. As Danny pointed out we can't say anything about the stability of the non-linear system if we are discussing centers. However, when it comes to critical points that have eigenvalues with non-zero real parts then yes we can imply that the nonlinear system is stable or unstable in a small region about the critical point. This is nice because it is all we need for stability.

    However, if you want to know about stability further out from the critical points then the linearization method may not work. In this case you need to use Lyapunov Functions, which are quite difficult to find. Using the Lyapunov function you can check the direction of flow across the boundary, or something like that, and it will tell you the type of stability.

    EDIT: Link to Lyapunov Function Description
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  13. #13
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    A better explanation of the other method can be seen here:
    Non-Linear First Order ODE: Critical Point Linearization

    The other example I did actually see myself was with a zero real part, so that might be the difference! Thanks a lot for that mate, appreciate it greatly
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  14. #14
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    I see what you were attempting. You were shifting the variables so that the critical point would then be at (0,0) which then allows you to eliminate the higher order terms. Yeah, you could do that. The Jacobian is more straightforward in my opinion. The Jacobian is sort of like using a straight line to approximate a curve.
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