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I am considering the dynamical system: $u'=v-0.25(v-u)^2$ $v'=u(1+v)+0.25(u+v)^2$

I have calculated the linear stable and unstable manifold as, $E^s=sp(1,1)$ and $E^u=sp(-1,1)$ for eigenvalues $-1$ and $1$ respectively. I did this by calculating the linear eigenvalues and vectors and then using the defintions for the linear stable and unstable manifolds:

$E^s$ is the eigenspace corresponding to the eigenvalues with negative real parts. Let $v_1, . . . , v_k ⊂ R^n$ be the eigenvectors corresponding to the eigenvalues of $D(\bar{x})f$ having negative real parts. $E^s = sp{v_1, . . . , v_k}$.

$E^u$ is equivalent for a positive eigenvalue.

I am now trying to calculate an approximation for the unstable manifold but believe I should transform my system so that the linear unstable manifold is of the form ${(x,0):x\in\Re} but I am unsure on how to this.

My initial idea was to let $y=\tilde{y}-1$ but that doesn't satisfy the above for all y, which is where I am confused.

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\begin{align} u'&=v-\frac14(v-u)^2\\ v'&=u(1+v)+\frac14(u+v)^2 \end{align} With the transformation $U=u+v$, $V=u-v$ (there is no real need to use rotations, the scaling here doesn't hurt), you get \begin{align} U'&=U-\frac12V^2+\frac12U^2\\ V'&=-V-\frac12U^2\\ \end{align} The unstable manifold $\mathcal{U}$ is, to linear approximation, $V=0$. So to obtain better approximations, we can impose an ansatz $$ V=a_2U^2+a_3U^3+a_4U^4+\dots $$ and see what invariance gives us. Differentiating, we get $V'=(2a_2U+3a_3U^2+4a_4U^3+\dots)U'$. Substituting for $U',V'$ from the differential equation (by invariance), we have $$ -V-\frac12U^2 =(2a_2U+3a_3U^2+4a_4U^3+\dots)(U-\frac12V^2+\frac12U^2) $$ and substituting our ansatz $V=a_2U^2+a_3U^3+a_4U^4+\dots$, \begin{multline} -(a_2U^2+a_3U^3+a_4U^4+\dots)-\frac12U^2 =(2a_2U+3a_3U^2+4a_4U^3+\dots)\times\\\left(U-\frac12(a_2U^2+a_3U^3+a_4U^4+\dots)^2+\frac12U^2\right) \end{multline} Equating coefficients, \begin{align} -a_2-\frac12&=2a_2&a_2&=-\frac16\\ -a_3&=a_2+3a_3 & a_3&=\frac1{24}\\ -a_4&=\frac32a_3+4a_4 & a_4&=-\frac1{80}\\&&\vdots \end{align} Finally, you might want to transform back to the original $u,v$ coordinates.

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  • $\begingroup$ Ok, I think I follow what you have done but why do you choose the transformation you do? $\endgroup$ – KieranSQ Nov 15 '18 at 9:27
  • $\begingroup$ Shouldn't for the coeff $U^3: \ -a_3=a_2^2+3a_3+a_2$? $\endgroup$ – KieranSQ Nov 15 '18 at 10:04
  • $\begingroup$ Where does the $a_2^2$ come from? $\endgroup$ – user10354138 Nov 15 '18 at 22:03

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