# Thread: Some simple DiffEQ problems.

1. ## 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?

2. Let's write the first DE as...

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

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

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

... where...

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

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

No we have three possibilities...

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

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

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

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

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

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

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

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

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

3. 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]]]]]

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# consider the differential equation dx/dt = kx-x^3

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