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Consider the differential equation

$$x''+(2a)x'-(b^2/2)x+x^3=0 \tag{1}$$

where $a,b >0$ are constants.

(i) Write differential equation (1) as a first order dynamical system.

(ii) Determine the equilibrium point(s) of the system found in part (i).

(iii) Classify the equilibrium point $$(b/\sqrt2,0).$$

For (i) I got $$x'=y$$ $$y'=-x^3+(b^2/2)x-(2a)y$$

For (ii) I got equilibrium points $$(0,0), (b/\sqrt2,0), (-b/\sqrt2,0)$$

Now for (iii) I moved the eq point $(b/\sqrt2,0)$ to the origin. So I let $X=x-b/\sqrt2$ and $Y=y$ and then found the associated linear system as $$X'=Y$$ $$Y'=(b^2)X-(2a),Y$$ and then found the eigenvalues $$\lambda=-a+\sqrt{a^2+b^2}$$ and $$\lambda=-a-\sqrt{a^2+b^2}$$ and found that the first one is $>0$ and second one is $<0$. This means that it is a saddle point right? Do we have to find anything else to determine that it is a saddle point? And more importantly, i was told that we need to think about the linearization theorem to finish the question but i don't know how. Please help and if i have done any mistakes, please let me know!!

yoknapatawpha
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snowman
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  • no we haven't learned that and i am a billion percent sure we aren't meant to use jacobian. we have learned the linearization theorem though. – snowman Nov 09 '14 at 14:03
  • here is what we were give about linearization: "the behaviour of the phase paths in the neighbourhood of an equilibrium point of the nonlinear system x'=P(x,y), y'=Q(x,y) is the same as that of the associated linearized system EXCEPT when the linearization system has a zero or purely imaginary eigenvalues." And for iii, we were always told to shift an eq point to the origin when its not already there. if you let X=x-b/sqrt2 and Y=0 and find X' and Y' in terms of X and y, you get quadratic variables like X^2 and X^3 but you can ignore them because we are near origin so they will be very small. – snowman Nov 09 '14 at 15:52
  • so i ended up with the X' and Y' in the first post. then i found the matrix A which is 2x2 and just is the coefficients of the equated X' and Y'. then found |A-lamda*I|=0 where I is the 2x2 identity matrix. and that's how i got the lamda values. if you didn't understand, please say and i will try to explain more. I reaaaally need help on this question!! – snowman Nov 09 '14 at 15:54

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