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I need to find a differential equation $x'=f(x)$ that has three equilibrium points that are

  1. all sinks;
  2. all sources.

So two differential equations total. Help!

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Can you add a little detail yo tour question? Do you just need any solution? –  Sam DeHority Feb 9 '13 at 6:32
    
I need an example equation that has three equilibrium points, all of which are sinks, and also an example equation having three equilibrium points, all of which are sources. –  Sara Feb 9 '13 at 6:36

2 Answers 2

Just define a function, $f$, that has three zeros, and to ensure these zeros are all sinks of the differential equation, make sure that $\frac{df}{dx} < 0$ at each zero. Note this requires that $f$ is a discontinuous function. As a specific example, with zeros at $-1$, $0$ and $1$, $$ f(x) = \left\{ \begin{array}{lc} -x-1 \;, & x \le -\frac{1}{2} \\ -x \;, & -\frac{1}{2} < x \le \frac{1}{2} \\ -x+1 \;, & x > \frac{1}{2} \end{array} \right. \;. $$ With instead $-f(x)$, the zeros are sources.

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hint: pay attention root of this equation x'=f(x) is equilibrium points let $f(x_0)=0$ therefore$$if\quad f'(x_0)\lt0 \quad then\quad x_0 \quad is\quad sink $$ and $$if\quad f'(x_0)\gt0 \quad then\quad x_0 \quad is\quad source $$ now you can find $f(x)$ for "1" let $f(x)=(x-a)(x-b)(x)$ s.t $f'(x)=3x^2-2(a+b)+ab\lt0$ then calculate lower bound and upper bound for a,b also do it for second part $f(x)=(x-a)(x-b)(x)$ then find a,b such that $f'(x)=3x^2-2(a+b)+ab\gt0$

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