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I have an equation $$ y' = (2+y)(k-y^2) $$ and I am asked to find the equilibrium solutions and bifurcation values for all values of $k$.

My approach was that... there exists $2$ equlibrium solutions for $k>0$, and $1$ equilibrium soltuion for $k = 0$ and $0$ equilibrium solution for $k < 0$.

After differentiating the function to $$ f' = -4y - 3y^2 $$

I do not know how to continue. Do I plug in $\sqrt k$ and $\sqrt {-k}$, which is the equilibrium solution to the differentiated function...? Thanks.

mle
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Topstar
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2 Answers2

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Hints:

To find the critical points, we set up and solve $y' = 0$, hence:

$$y' = (2+y)(k-y^2) = 0 \implies y = -2, ~\pm ~\sqrt{k}$$

Now that we have these critical points, how do they behave for various values of $k$?

Can you determine how to find the bifurcation points from this?

Look at the direction field plot and what happens to number of critical points for $k = 4, 1, 0, -1$:

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Another point of interest is $k = 4$, as look at the direction field plot (collapses to two equilibrium points):

enter image description here

Amzoti
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  • Um. Do I draw a phase line for y = -2 and y = +-sqrt(k) and determine whether its property?(like node etc) next? – Topstar Mar 25 '14 at 23:18
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    We have three critical points. Do the number of equilibrium points change as a result of changing $k$? If so, how? See, For example: http://www.sosmath.com/diffeq/first/bifurcation/example1/answer/answer.html – Amzoti Mar 25 '14 at 23:20
  • Okay, I found that sqrt(k) is a sink, -sqrt(k) is a source – Topstar Mar 25 '14 at 23:35
  • @Topstar: See additional value of $k$ of interest, we have $k = 4, 1, 0, -1 $. – Amzoti Mar 26 '14 at 03:20
  • Well done, as usual, Amzoti! – amWhy Mar 29 '14 at 12:08
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Find the equilibrium points as pointed out above by solving for $y'=0$ . As for bifurcation, we have to find the values of parameter $k$ at which there will be a qualitative change in dynamics.For $k>0$ , there will be three equilibrium points.

Rewriting $y'$ as $y' =(2+y)(\sqrt(k)+y)(\sqrt(k)-y)$

Now if k >0 , there are three roots for this polynomial and two are negative. Also if k is not equal to 4, the roots will be distinct. When k=4 , two roots will be -2 and one will be 2. When roots are distinct , the smallest root will be a stable equilibrium point followed by an unstable and another stable equilibrium point in that order. But at K=4, the stable and unstable equilibrium points coalesce at y =-2 and the equilibrium will be stable in one direction and unstable in the other. Hence there is a qualitative change in dynamics at k=4.

Once again, as k becomes zero. the second and third equilibrium points will coalesce and produce a similar situation as discussed above but only with direction of stability reversed. Finally for k <0 , there will be only one equilibrium point and it will be stable . Hence there is a qualitative change in dynamics at $k =0$ too.

Hence my guess would be that the roots are -2, $\sqrt(k)$,$-\sqrt(k)$ and bifurcation points are $k=4 ,0$

anonlinear
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