Fixed points and Stability (Nonlinear System)

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1
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$$dot x = -x + x^3$$
$$dot y = x + y$$
Where $(x,y) in mathbbR^2$




I found the fixed points to be:



$$(0,0),(0,1),(0,-1),(1,0),(1,1),(1,-1),(-1,0),(-1,1),(-1,-1)$$



The Jacobian Matrix to be:



beginbmatrix
-1+3x^2 & 0\
1 & 1 \
endbmatrix



But every point is unstable.



What am i doing wrong? I feel like:



1) I have too many fixed points and



2) My Jacobian matrix is wrong



Thanks.






nonlinear-system







share|cite|improve this question



















  • 1




    i have edited the equations in LaTeX .
    – Ahmad Bazzi
    Sep 11 at 4:20






  • 2




    @Jon: I only find three critical points at $$(x, y) = (-1, 1), (0, 0), (1, -1)$$. The Jacobian is correct and all three critical points are unstable. Draw a phase portrait to see these.
    – Moo
    Sep 11 at 4:32







  • 2




    @Jon: Stationary points occur when $dotx=0$ and $doty=0$. You should get $(0,0)$, $(1, -1)$ and $(-1, 1)$. The Jacobian looks correct.
    – Winter Soldier
    Sep 11 at 4:33










  • Thanks you both. I see what I did wrong. I appreciate the feedback.
    – Jon
    Sep 11 at 4:46














up vote
1
down vote

favorite













$$dot x = -x + x^3$$
$$dot y = x + y$$
Where $(x,y) in mathbbR^2$




I found the fixed points to be:



$$(0,0),(0,1),(0,-1),(1,0),(1,1),(1,-1),(-1,0),(-1,1),(-1,-1)$$



The Jacobian Matrix to be:



beginbmatrix
-1+3x^2 & 0\
1 & 1 \
endbmatrix



But every point is unstable.



What am i doing wrong? I feel like:



1) I have too many fixed points and



2) My Jacobian matrix is wrong



Thanks.






nonlinear-system







share|cite|improve this question



















  • 1




    i have edited the equations in LaTeX .
    – Ahmad Bazzi
    Sep 11 at 4:20






  • 2




    @Jon: I only find three critical points at $$(x, y) = (-1, 1), (0, 0), (1, -1)$$. The Jacobian is correct and all three critical points are unstable. Draw a phase portrait to see these.
    – Moo
    Sep 11 at 4:32







  • 2




    @Jon: Stationary points occur when $dotx=0$ and $doty=0$. You should get $(0,0)$, $(1, -1)$ and $(-1, 1)$. The Jacobian looks correct.
    – Winter Soldier
    Sep 11 at 4:33










  • Thanks you both. I see what I did wrong. I appreciate the feedback.
    – Jon
    Sep 11 at 4:46












up vote
1
down vote

favorite









up vote
1
down vote

favorite












$$dot x = -x + x^3$$
$$dot y = x + y$$
Where $(x,y) in mathbbR^2$




I found the fixed points to be:



$$(0,0),(0,1),(0,-1),(1,0),(1,1),(1,-1),(-1,0),(-1,1),(-1,-1)$$



The Jacobian Matrix to be:



beginbmatrix
-1+3x^2 & 0\
1 & 1 \
endbmatrix



But every point is unstable.



What am i doing wrong? I feel like:



1) I have too many fixed points and



2) My Jacobian matrix is wrong



Thanks.






nonlinear-system







share|cite|improve this question
















$$dot x = -x + x^3$$
$$dot y = x + y$$
Where $(x,y) in mathbbR^2$




I found the fixed points to be:



$$(0,0),(0,1),(0,-1),(1,0),(1,1),(1,-1),(-1,0),(-1,1),(-1,-1)$$



The Jacobian Matrix to be:



beginbmatrix
-1+3x^2 & 0\
1 & 1 \
endbmatrix



But every point is unstable.



What am i doing wrong? I feel like:



1) I have too many fixed points and



2) My Jacobian matrix is wrong



Thanks.






nonlinear-system




nonlinear-system






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Sep 11 at 4:25









Ananth Kamath

1,500824




1,500824










asked Sep 11 at 4:16









Jon

345




345







  • 1




    i have edited the equations in LaTeX .
    – Ahmad Bazzi
    Sep 11 at 4:20






  • 2




    @Jon: I only find three critical points at $$(x, y) = (-1, 1), (0, 0), (1, -1)$$. The Jacobian is correct and all three critical points are unstable. Draw a phase portrait to see these.
    – Moo
    Sep 11 at 4:32







  • 2




    @Jon: Stationary points occur when $dotx=0$ and $doty=0$. You should get $(0,0)$, $(1, -1)$ and $(-1, 1)$. The Jacobian looks correct.
    – Winter Soldier
    Sep 11 at 4:33










  • Thanks you both. I see what I did wrong. I appreciate the feedback.
    – Jon
    Sep 11 at 4:46












  • 1




    i have edited the equations in LaTeX .
    – Ahmad Bazzi
    Sep 11 at 4:20






  • 2




    @Jon: I only find three critical points at $$(x, y) = (-1, 1), (0, 0), (1, -1)$$. The Jacobian is correct and all three critical points are unstable. Draw a phase portrait to see these.
    – Moo
    Sep 11 at 4:32







  • 2




    @Jon: Stationary points occur when $dotx=0$ and $doty=0$. You should get $(0,0)$, $(1, -1)$ and $(-1, 1)$. The Jacobian looks correct.
    – Winter Soldier
    Sep 11 at 4:33










  • Thanks you both. I see what I did wrong. I appreciate the feedback.
    – Jon
    Sep 11 at 4:46







1




1




i have edited the equations in LaTeX .
– Ahmad Bazzi
Sep 11 at 4:20




i have edited the equations in LaTeX .
– Ahmad Bazzi
Sep 11 at 4:20




2




2




@Jon: I only find three critical points at $$(x, y) = (-1, 1), (0, 0), (1, -1)$$. The Jacobian is correct and all three critical points are unstable. Draw a phase portrait to see these.
– Moo
Sep 11 at 4:32





@Jon: I only find three critical points at $$(x, y) = (-1, 1), (0, 0), (1, -1)$$. The Jacobian is correct and all three critical points are unstable. Draw a phase portrait to see these.
– Moo
Sep 11 at 4:32





2




2




@Jon: Stationary points occur when $dotx=0$ and $doty=0$. You should get $(0,0)$, $(1, -1)$ and $(-1, 1)$. The Jacobian looks correct.
– Winter Soldier
Sep 11 at 4:33




@Jon: Stationary points occur when $dotx=0$ and $doty=0$. You should get $(0,0)$, $(1, -1)$ and $(-1, 1)$. The Jacobian looks correct.
– Winter Soldier
Sep 11 at 4:33












Thanks you both. I see what I did wrong. I appreciate the feedback.
– Jon
Sep 11 at 4:46




Thanks you both. I see what I did wrong. I appreciate the feedback.
– Jon
Sep 11 at 4:46















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