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Given this dynamical system: $$\begin{cases}x'=-4y+4y^3\\y'=-4x\end{cases}$$

I have seen that the system has a saddle at $(0,0)$. Now I am asked to approximate (using a third order Taylor expansion) the equation of the stable manifold near this point. I know that what I have to do is analogous to what the user @John B answered here. However my case is more complicated since I can't solve the system required to know the coefficients $a,b,...$

If I take $\varphi(x)=a+bx+cx^2+dx^3+\cdots$ then I get the relations $$-4ba+4ba^3=0\\-4=-4b^2+12a^2b^2-8ca+8ca^3\\ 0=-4bc+12a^2cb+12ab^3-8cb+24a^2b^2c-12da+12da^3 \\...$$

My problem is that every time I add a new relation a new coefficient appears, having always one more variable than equations in the system. How can I solve it?

John Keeper
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1 Answers1

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$$\begin{cases}\frac{dx}{dt} =-4y+4y^3\\\frac{dy}{dt}=-4x\end{cases}$$ $$\frac{dy}{dx}=\frac{x}{y-y^3}$$ This ODE is separable and then integrable which leads to : $$y(x)=\pm\sqrt{1\pm\sqrt{C-2x^2}}$$ Since in the wording of the question the point $(0,0)$ is mentioned as belonging to the solution, this condition implies $C=1$ and : $$y(x)=\pm\sqrt{1-\sqrt{1-2x^2}}$$ Series expansion around $(0,0)$ : $$y(x)=\pm\left(x+\frac14 x^3+\frac{7}{32}x^5+...\right)$$

Nevertheless, the above calculus is probably not the expected method. $$ $$

OTHER METHOD : A series of the form $y(x)=a_0+a_1x+a_2x^2+a_3x^3+...$ is expected.

Preliminary inspection of $\frac{dy}{dx}=\frac{x}{y-y^3}=\frac{x}{y}\:\frac{1}{(1-y^2)}$ makes understand that $y(x)$ is an odd function. Thus : $$y(x)=a_1x+a_3x^3+...$$ $$\frac{dy}{dx}=a_1+3a_3x^2+...=\frac{x}{ (a_1x+a_3x^3+...)-(a_1x+a_3x^3+...)^3}$$ $$-x+(a_1+3a_3x^2+...)\big((a_1x+a_3x^3+...)-(a_1x+a_3x^3+...)^3 \big)=0$$ $$(a_1^2-1)x+(4a_1a_3-a_1^4)x^3+...=0$$ $$\begin{cases} a_1^2-1=0 \quad\implies\quad a_1=\pm 1\\ 4a_1a_3-a_1^4=0 \quad\implies\quad a_3=\frac{a_1^3}{4} \end{cases}$$ $$y(x)=a_1\big(x+\frac14a_1^2 x^3+...\big)$$ $$y(x)=\pm\big(x+\frac14 x^3+...\big)$$

JJacquelin
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  • Why $\frac{dy}{dx}=\frac{x}{y-y^3}=\frac{x}{y}:\frac{1}{(1-y^2)}$ makes understand that $y(x)$ is an odd function? Also, since $x$ is a function, shouldn't it be $\frac{dy}{dx}=a_1+3a_3x^2x'+...$ (implicit differentiation) – John Keeper Mar 24 '19 at 14:29
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    Obviously if $x$ is changed to $-x$ and $y$ is changed to $-y$ then $\frac{dy}{dx}$ remains the same and $\frac{x}{y}\frac{1}{(1-y^2)} $ remains the same. The equality continue to be valid. Thus if $(x,y)$ is solution also $(-x,-y)$ is solution. This proves that $y(x)$ is an odd function. Thus the series for $y(x)$ is made only with odd powers of $x$ : So, $a_0=a_2=a_4=…=0$. – JJacquelin Mar 24 '19 at 16:52
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    The derivative of $a_3x^2$ with respect to $x$ (NOT with respect to $t$) is $3a_3x^2$ which is not $3a_3x^2\frac{dx}{dt}$ . Take care to not confuse the derivative with respect to one variable with the derivative with respect to another variable. The use of notation apostrophe is confusing in some cases. – JJacquelin Mar 24 '19 at 17:02
  • What would you have done if the function was not odd or even? – John Keeper Apr 06 '19 at 14:17
  • If the function wax nor odd or even the series would be with all tems $y(x)=a_0+a_1x+a_2x^2+a_3x^3+...$ Same method but more boring calculus. – JJacquelin Apr 06 '19 at 15:44
  • I'm trying to follow the same steps you did to solve this post with the function $\frac{dy}{dx}=\frac{-sin(x)+0.5}{y}$. Using that this ode is separable and integrable I got $y(x)=\pm \sqrt{2(cos(x)-cos(x_0))+x-x_0}$ (the critical saddle point is $(x_0,0)$ but now I can't approximate this using the taylor expansion, maybe I should create a new post? – John Keeper Apr 06 '19 at 16:24
  • Let $x=x_0+t$ . Expand to series at $(0,0)$. – JJacquelin Apr 06 '19 at 17:17