Inequality manipulation for a vector norm

Question

How can I manipulate the following inequality to reach from $$\dot{V}\leq -4x_1^2 +4x_1x_2 -2x_2^2 $$ to $$\dot{V}\leq -(3-\sqrt{5}) \|x\|^2 $$ where $x=[x_1 \;x_2]^T$ is a 2D vector and $\|x\|$ is the euclidean norm of $x?$ It is related to Lyapunov stability and I need to get something of the following form: $$\dot{V} \leq -k\|x\|^2,$$ where $k$ is some positive constant.

Answer 1

Let $x_1=a$ and $x_2=b$.

Thus, $$-4a^2+4ab-2b^2\leq-(3-\sqrt5)(a^2+b^2)$$ it's $$(1+\sqrt5)a^2-4ab+(\sqrt5-1)b^2\geq0$$ or $$(1+\sqrt5)^2-4(1+\sqrt5)ab+4b^2\geq0$$ or $$((1+\sqrt5)a-2b)^2\geq0.$$ We can get a value $-3+\sqrt5$ by the following way.

Let $$-4a^2+4ab-2b^2\leq k(a^2+b^2)$$ is true for all reals $a$ and $b$.

Thus, $$(4+k)a^2-4ab+(2+k)b^2\geq0,$$ for which we need $4+k>0$ and $$4-(4+k)(2+k)\leq0$$ or $$k^2+6k+4\geq0,$$ which since $k>-4$ (we see that even $k>-2$), gives $$k\geq-3+\sqrt5.$$ Id est, the minimal value of $k$ for which our inequality holds for all variables it's $k=-3+\sqrt5.$

Why we took a minimal value? Because we want that the equality will occur.

Answer 2

Your $\dot{V}$ can also be expressed as the following quadratic expression in $x=\begin{bmatrix}x_1 \;x_2\end{bmatrix}^\top$, namely

$$ \dot{V} = x^\top P\,x, $$

with

$$ P = \begin{bmatrix}-4 & 2 \\ 2 & -2\end{bmatrix}. $$

It can be shown that $P$ has the eigenvalues $\lambda_1 = -3 + \sqrt{5}$ and $\lambda_2 = -3 - \sqrt{5}$. Now when assuming that $x$ is real and by using that symmetric matrices have an orthogonal basis then $x$ can also be written as

$$ x = \alpha_1\,v_1 + \alpha_2\,v_2, $$

with $v_1,v_2$ the normalized eigenvectors corresponding to the eigenvalues $\lambda_1,\lambda_2$ respectively and $\alpha_1,\alpha_2 \in\,\mathbb{R}$. So $\dot{V}$ can also be written as

\begin{align} \dot{V} &= (\alpha_1\,v_1 + \alpha_2\,v_2)^\top P\,(\alpha_1\,v_1 + \alpha_2\,v_2) \\ &= (\alpha_1\,v_1 + \alpha_2\,v_2)^\top (\alpha_1\,\lambda_1\,v_1 + \alpha_2\,\lambda_2\,v_2) \\ &= \alpha_1^2\,\lambda_1 + \alpha_2^2\,\lambda_2 \end{align}

When looking for bounds of $\dot{V}$ in terms of some scalar times $\|x\|^2$ we might as well set $\|x\|^2$ to one. Since we have an orthonormal basis it holds that $\|x\|^2 = \alpha_1^2 + \alpha_2^2$. From this it can be observed that $\dot{V}$ is maximized when $\alpha_1^2 = 1$ and $\alpha_0^2 = 0$, since $\lambda_1$ is the largest eigenvalue of $P$. Similarly $\dot{V}$ is minimized when $\alpha_1^2 = 0$ and $\alpha_0^2 = 1$, since $\lambda_2$ is the largest eigenvalue of $P$.

Answer 3

I will add an alternative method. Introduce polar cooridnates $x_1=r\cos \alpha$ and $x_2 = r\sin \alpha$ (note that $x_1^2+x_2^2=||\boldsymbol{x}||^2 = r^2$) to obtain

$$\dot{V} \leq -4r^2\cos^2\alpha-2r^2\sin^2\alpha+4r^2\sin \alpha\cos\alpha$$ $$\implies \dot{V}\leq -r^2\left[2\cos^2\alpha+\sin^2\alpha-2\sin \alpha \cos\alpha \right]$$

Use $\cos^2 \alpha + \sin^2 \alpha = 1$ and $2\sin \alpha \cos\alpha=\sin 2\alpha$ $$\implies \dot{V}\leq r^2\left[\sin 2\alpha-\cos^2\alpha-1\right]$$

We want to find extrema of the expression in the bracket, hence we introduce $f(\alpha)=\sin 2\alpha -\cos^2 \alpha -1 \implies f'(\alpha)=2\cos2\alpha + 2\cos\alpha\left[-\sin \alpha \right]$ set this expression equal to $0$ and try to solve the resulting equation. Check which of the values results in the highest $f(\alpha)$ it should result in your upper bound.

Alternatively, you can apply $\cos^2 \alpha = \dfrac{1+\cos 2\alpha}{2}$ to obtain

$$f(\alpha) =\dfrac{2\sin 2\alpha -\cos 2\alpha -3}{2}.$$

In the last step, we use the following observation

$$C\sin(x+y) = C\sin x\cos y + C\cos x \sin y=A\sin x + B \cos x$$

By comparing these we obtain $$A=C\cos y$$ $$B=C\sin y$$ $$\implies C = \sqrt{A^2+B^2} \qquad \text{only this is important}$$ $$\implies \tan y = \dfrac{B}{A} \qquad \text{not necessary}$$

For

$$f(\alpha) =\dfrac{2\sin 2\alpha -\cos 2\alpha -3}{2}=\dfrac{\sqrt{5}\sin (2\alpha + y) -3}{2}$$

$$\implies \dfrac{-\sqrt{5}-3}{2} \leq f(\alpha) \leq \dfrac{+\sqrt{5}-3}{2}.$$

This method is sometimes fast because it is often possible to get a crude estimate for $f(\alpha)$ here $f(\alpha)\leq 1 - 0 -1 =0$ for the case that the sine is $1$ and the cosine is $0$ (unrealistic but crude enoght to get $\dot{V}\leq 0$.