Prove that $V'(x,y)=-(x^2+y^2)+(2Ax+By)F_1(x,y)+(Bx+2Cy)G_1(x,y)$. Liapunov.

Question

We've got the following linear system:

$$\frac{dx}{dt}=a_{11}x+a_{12}y$$ $$\frac{dy}{dt}=a_{21}x+a_{22}y$$

The critical point $(0,0)$ is an assymptotically stable critical point of the system. I have proved that $a_{11}+a_{22}<0$ and $a_{11}a_{22}-a_{12}a_{21}>0.$ Then, we have constructed a Liapunov function $V(x,y)=Ax^2+Bxy+Cy^2$ such that $V$ is positive definite and $V'$ is negative definite (We have defined $A$, $B$ and $C$ so that $V'(x,y)=-x^2-y^2$ ).

Now, we have to show that the Liapunov function we have defined, is also a Liapunov function for the following system:

$$\frac{dx}{dt}=a_{11}x+a_{12}y+F_1(x,y)$$ $$\frac{dy}{dt}=a_{21}x+a_{22}y+G_1(x,y)$$

For that, first of all we have to prove that

$$V'(x,y)=-(x^2+y^2)+(2Ax+By)F_1(x,y)+(Bx+2Cy)G_1(x,y)$$

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I have rewrite it like this:

$$ V'(x,y) = \frac{\partial V}{\partial x}f(x,y)+\frac{\partial V}{\partial y}g(x,y)=(2Ax+By)(a_{11}x+a_{12}y)+(Bx+2Cy)(a_{21}x+a_{22}) $$

But now I don't know how to follow. Could anyone please help me?

Answer 1

Following from your question, I presume that you accepted my approach, though you didn't upvote or downvote my previous answer. In my previous answer, the Lyapunov function candidate was proposed:

$$ V = - \frac{1}{2} a_{21} x^{2} + \frac{1}{2} a_{12} y^{2}, $$

where $a_{11} < 0$, $a_{12} > 0$, $a_{21} < 0$, $a_{22} < 0$.

Comparing with your given Lyapunov function

$$ V = A x^{2} + B x y + C y^{2}, $$

you will find that

$$ A = - \frac{1}{2} a_{21} \quad \Rightarrow \quad 2 A = - a_{21}, $$ $$ B = 0, $$ $$ C = \frac{1}{2} a_{12} \quad \Rightarrow \quad 2 C = a_{12}. $$

Since $B = 0$, now you need to prove that

$$ \dot{V} = - \left(x^{2} + y^{2}\right) + 2 A x F + 2 C y G. $$

To maintain $a_{11} < 0$, and $a_{22} < 0$, it is proposed that

$$ a_{11} = \frac{1}{a_{21}} = - \frac{1}{2 A}, $$

$$ a_{22} = - \frac{1}{a_{12}} = - \frac{1}{2 C}. $$

Taking the time derivative for V, you get

$$ \dot{V} = - a_{21} x \left(a_{11} x + a_{12} y + F\right) + a_{12} y \left(a_{21} x + a_{22} y + G\right) $$

$$ \dot{V} = - a_{11} a_{21} x^{2} - a_{12} a_{21} x y - a_{21} x F + a_{12} a_{21} x y + a_{12} a_{22} y^{2} + a_{12} y G $$

Making substitutions and cancelling common factors to simplify the equation to

$$ \dot{V} = - \left(\frac{1}{a_{21}}\right) a_{21} x^{2} + a_{12} \left(- \frac{1}{a_{12}}\right) y^{2} - a_{21} x F + a_{12} y G $$

$$ \dot{V} = - x^{2} - y^{2} - a_{21} x F + a_{12} y G $$

Making further substitutions, you get

$$ \dot{V} = - \left(x^{2} + y^{2}\right) + 2 A x F + 2 C y G. $$

Answer 2

This becomes easier to show using matrix and vector notation. For this I define the following notation:

\begin{align} \vec{z} &= \begin{bmatrix} x \\ y \end{bmatrix}, \\ \textbf{A} &= \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}, \\ \textbf{P} &= \begin{bmatrix} A & \frac{1}{2} B \\ \frac{1}{2} B & C \end{bmatrix}, \\ \vec{f}(\vec{z}) &= \begin{bmatrix} F_1(x,y) \\ G_1(x,y) \end{bmatrix}. \end{align}

Using these definitions it can be shown that the initial dynamics and Lyapunov function can be written as

\begin{align} \frac{d\,\vec{z}}{dt} &= \textbf{A}\,\vec{z}, \\ V(\vec{z}) &= \vec{z}^\top \textbf{P}\,\vec{z}, \end{align}

with $\textbf{P}$ chosen such that

\begin{align} \frac{d\,V(\vec{z})}{dt} &= \frac{d\,\vec{z}^\top}{dt} \textbf{P}\,\vec{z} + \vec{z}^\top \textbf{P}\,\frac{d\,\vec{z}}{dt}, \\ &= \vec{z}^\top \textbf{A}^\top \textbf{P}\,\vec{z} + \vec{z}^\top \textbf{P}\,\textbf{A}\,\vec{z}, \\ &= \vec{z}^\top \left(\textbf{A}^\top \textbf{P} + \textbf{P}\,\textbf{A}\right)\vec{z}, \\ &= -\vec{z}^\top \vec{z}, \end{align}

so $\textbf{P}$ satisfies $\textbf{A}^\top \textbf{P} + \textbf{P}\,\textbf{A}=-\textbf{I}$ with $\textbf{I}$ the identity matrix.

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Your modified dynamics can be written as

$$ \frac{d\,\vec{z}}{dt} = \textbf{A}\,\vec{z} + \vec{f}(\vec{z}). $$

Substituting that in the derivative of the Lyapunov function yields

\begin{align} \frac{d\,V(\vec{z})}{dt} &= \frac{d\,\vec{z}^\top}{dt} \textbf{P}\,\vec{z} + \vec{z}^\top \textbf{P}\,\frac{d\,\vec{z}}{dt}, \\ &= \left(\vec{z}^\top \textbf{A}^\top + \vec{f}(\vec{z})^\top\right) \textbf{P}\,\vec{z} + \vec{z}^\top \textbf{P} \left(\textbf{A}\,\vec{z} + \vec{f}(\vec{z})\right), \\ &= \vec{z}^\top \left(\textbf{A}^\top \textbf{P} + \textbf{P}\,\textbf{A}\right)\vec{z} + \vec{f}(\vec{z})^\top \textbf{P}\,\vec{z} + \vec{z}^\top \textbf{P}\,\vec{f}(\vec{z}), \\ &= -\vec{z}^\top \vec{z} + 2\,\vec{z}^\top \textbf{P}\,\vec{f}(\vec{z}). \end{align}

Substituting back in the expressions for $\vec{z}$, $\textbf{P}$ and $\vec{f}(\vec{z})$ yields

$$ 2\,\vec{z}^\top \textbf{P}\,\vec{f}(\vec{z}) = (2\,A\,x+B\,y)\,F_1(x,y)+(B\,x+2\,C\,y)\,G_1(x,y). $$