Stability of a fixed point

682 Views Asked by At

Determine the stability of all the fixed points of the following functions:

So basically I understand how to find fixed points and whether they are attracting/repelling...but I am confused on how to check their stability, and also confused about what to do if the derivative =1 (see part a)

a. $$S(x)=\sin x$$ $$x=\sin x$$

$1$ fixed point at $x=0$

$$\mbox{Fix}\{S\}=\{0\}$$

$$S(x)=\sin x$$ $$S'(x)=\cos x$$ $$S'(0)=\cos 0$$ $$S'(0)=1$$

And this is where I am stuck? Since $1$ is inconclusive right? Also I don't understand lyapunov's stability theorem/if I even need to use that here...

b. $$f(x)=2sin(x)$$ $$x=2sin(x)$$ $$\mbox{Fix}\{S\}=\{0,±a\}$$ $$a≈1.895$$ $$f(x)=2sin(x)$$ $$f'(x)=2cos(x)$$ $$f'(0)=2cos(0)$$ $$f'(0)=2>1$$ x=0 is repelling $$f'(±a)=2cos(±a)$$ $$f'(±a)=\sqrt{4-a^2}<1$$ x=±a is attracting

So...does that mean it is unstable?

c. $$T(x)=tan(x)$$ $$x=tan(x)$$ $$\mbox{Fix}\{S\}=\{0,±a\}$$ $$a≈4.49$$

$$T(x)=tan(x)$$ $$T'(x)=sec^2(x)$$ $$T'(0)=sec^2(0)$$ $$T'(0)=1$$ so??? it is inconclusive? or what is the next step?

$$T(x)=tan(x)$$ $$T'(x)=sec^2(x)$$ $$T'(±a)=sec^2(±a)$$ $$T'(±a)=sec^2(±4.49)$$ $$T'(±4.49)=20.556>1$$

x=a is repelling

d. $$E(x)=e^x$$

no fixed points

e. $$F(x)=\frac{1}{4}e^x$$ $$x=\frac{1}{4}e^x$$ $$\mbox{Fix}\{S\}=\{0.357,2.153\}$$

$$F(x)=\frac{1}{4}e^x$$ $$F'(x)=\frac{1}{4}e^x$$

$$F'(x)=\frac{1}{4}e^x$$ $$F'(0.357)=\frac{1}{4}e^{0.357}$$ $$F'(0.357)=0.357<1$$

x=0.357 is attracting

$$F'(x)=\frac{1}{4}e^x$$ $$F'(2.153)=\frac{1}{4}e^{2.153}$$ $$F'(2.153)=2.153>1$$

x=2.153 is repelling

So...one attracting fixed point and one repelling fixed point. As for the stability...I am not sure?