# Some simple DiffEQ problems.

• Sep 17th 2009, 04:49 PM
qtpipi
Some simple DiffEQ problems.
The differential equation dx/dt = (1/10)*x*(10-x)-h models a logistic population with with harvesting at a rate h. Determine the dependence of the critical points on the paramater h, and then construct a bifurcation diagram.

Consider the differential equation dx/dt = kx-x^3. (a) If k<=0, show that the only critical value c=0 of x is stable. (b) If k>0, show that the critical point c=0 is now unstable, but that the critical points c= +/- sqrt(k) are stable. Thus the qualitative nature of the solutions changes at k=0 as the parameter k increases, and so k=0 is a bifurcation point for the differential equation with parameter k. The plot of all points of the form (k,c) where c is a critical point of the equation is the "pitchfork diagram" (here the book shows a c vs. k plot of a parabola that opens to the right)

Any help?
• Sep 18th 2009, 12:11 AM
chisigma
Let's write the first DE as...

$\frac{dx}{dt} = x - .1\cdot x^{2} - h$ (1)

... and suppose to know the 'initial value' $x(0)= x_{0}$. It is important to establish the sign of second member of (1) and to do that we have to factorize it as follows...

$-.1 \cdot (x^{2} -10\cdot x +10\cdot h) = -.1 (x-a) (x-b)$ (2)

... where...

$a= 5 + \sqrt {25 -10\cdot h}$

$b= 5 - \sqrt{ 25 -10\cdot h}$ (3)

No we have three possibilities...

1) $25 -10\cdot h <0 \rightarrow h>2.5$

In this case $x^{'} (t)$ is everywhere negative and $\forall x_{0}$ is...

$\lim_{x \rightarrow \infty} x(t) = - \infty$

2) $25 -10\cdot h = 0 \rightarrow h=2.5$

In this case $x^{'} (t)$ is everywhere negative with the only exception of $x= 5$ where is null. The situation is the same as 1) unless you have $x_{0} = 5$ and in this case you have the constant solution $x=5$

3) $25 -10\cdot h > 0 \rightarrow h<2.5$

In this case $x^{'}(t)$ is negative for $x>a$ or $x , is null for $x=a$ or $x=b$ and is positive for $a>x>b$. We have three possibilities...

3a) $x_{0}>b \rightarrow \lim_{t \rightarrow \infty} = a$

3b) $x_{0} = b \rightarrow \lim_{t \rightarrow \infty} = b$

3b) $x_{0} < b \rightarrow \lim_{t \rightarrow \infty} = - \infty$

Kind regards

$\chi$ $\sigma$
• Sep 18th 2009, 06:18 AM
shawsend
Quote:

Originally Posted by qtpipi
and then construct a bifurcation diagram.
Any help?

Hey, you into programming? Mathematica is a big help with this stuff. No I don't work for them; just know a good thing when I see it. You know what a bifurcation diagram is right? It illustrates the dependence of the critical points on the parameter. So just calculate the zeros of the right side as a function of h and then just plot the results. If you're interested, try interpreting the code below. It first finds the roots, selects only the real ones, then arranges them in a suitable order to connect the points and draw a smooth plot. Now, modify my code to construct the pitch-fork bifurcation for the second one.

Also, in the news yesterday was an article about tipping points in global warming:

Is There a Climate-Change Tipping Point? - TIME

Here's something fun: Explain what that has to do with this thread.

Code:

Clear[h] myList =   Table[myRoots = x /. Solve[1/10 x (10 - x) - h == 0, x];   {h, #} & /@ myRoots,   {h, -5, 5, 0.1}]; myRealRoots =   Flatten[Select[myList, Element[#[[2]], Reals] &], 1]; curve = FindCurvePath[myRealRoots] ListLinePlot[myRealRoots[[curve[[1]]]]]