Exponential Stability

Question

What is the value of k to make origin exponential stable on given system?

$$\begin{align} x'=-kx+x^3\\ \end{align} $$

An equilibrium point is exponential stable if condition shown below holds: $$∀t > 0, ||x(t)|| ≤α ||x(0)||e^{-λt} $$ $$α>0$$

Answer 1

Let $k\ne 0$ then we have$$\dfrac{x'}{x(x^2-k)}=1\\-\dfrac{1}{k}\left(\dfrac{1}{x}-\dfrac{x}{x^2-k}\right)dx=dt\\-\dfrac{1}{k}(\ln |x|-\dfrac{1}{2}\ln |x^2-k|)=t+c_1\\c_2e^{-kt}=\dfrac{|x|}{\sqrt{|x^2-k|}}\\c_2^2e^{2kt}x^2=|x^2-k|$$if $x^2(0)<k$ we get$$c_3e^{2kt}x^2=k-x^2\\x^2=\dfrac{k}{1+c_3e^{2kt}}$$which satisfies the stability with $\lambda=k$ and $\alpha>\dfrac{1}{x(0)}\sqrt{\dfrac{k}{c_3}}$. Now let $x^2(0)> k$ therefore$$x^2=\dfrac{k}{1-c_3e^{2kt}}$$which doesn't hold for large enough $t$. If $x^2(0)=k$ then $x^2=k$ which is stable but not exponentially. Also for $k=0$ the stability doesn't hold unless if $x(0)=0$.

Answer 2

Choose the Lyapunov function $V(x)=0.5x^2=0.5||x||^2$ which fulfils the following inequality.

$$\dfrac{1}{3}\cdot||x||^2 \leq V(x)\leq 1\cdot||x||^2$$

$$\implies \dot{V}=x\left[-kx+x^3 \right]=-kx^2+x^4=-kx^2(1-x^2)$$

For $||x||<1$ the time derivative is negative definite if $k>0$. We define a domain $D=\{x\in \mathbb{R}\,|\,||x||<r<1 \}$. As long as $x\in D$ we can write

$$\dot{V}=-kx^2(1-x^2)\leq -kx^2(1-r^2)=-\left[k(1-r^2)\right]\cdot\left[\dfrac{1}{2}x^2\right]<-\left[k(1-r^2)\right]V.$$

Or

$$\dot{V}\leq -\left[k(1-r^2)\right]V.$$

Hence, the exponential stability of the origin is verified as long as $r<1$ and $k>0$ (See Khalil Theorem 4.10 from Nonlinear Systems by Hassan Khalil).