Liapunov’s Second Method proof

[member 731016]

Homework Statement

Please see below

Relevant Equations

Please see below

For this theorem,

I'm trying to prove why $a > 0$ and $4ac - b^2 > 0$ for the function to be positive definite.

My working is, $V(x,y) = x(ax + by) + cy^2$ (I try on write the function in alternative forms)

$V(x,y) = ax^2 + y(bx + cy)$

However, does anybody please know where to go from here?

Thanks!

 

[Orodruin]

I suggest that you define $k = x/y$ (the case $y = 0$ trivially gives $a > 0$ as a requirement) and require that $V(ky,y) \geq 0$ regardless of $k$.

 

[member 731016]

I suggest that you define $k = x/y$ (the case $y = 0$ trivially gives $a > 0$ as a requirement) and require that $V(ky,y) \geq 0$ regardless of $k$.

Thank you for your reply @Orodruin!

I think I also see another method we could use for the proof.

Wolfram Alpha tells me that another equivalent way of writing $V(x,y)$ other than the two previous ways I have done, is

$V(x,y) = \frac{(2ax + by)^2}{4a} - \frac{b^2y^2 - 4acy^2}{4a}$

However, does anybody please know to derive this expression? It seems to work when I expand it, however, I have never seen this sort of expression before derived from first principles. That expression also neatly solves the proof by making sure the second term is always positive $\frac{y^2(b^2 - 4ac)}{2a}$ so that negatives cancels which occurs when $b^2 - 4ac <0$ and $a > 0$.

Thanks!

 

[AR72]

looks like some sort application of the quadratic formula, 4ac -b^2 gives it away.
The algebra can be done by setting the equation equal to zero (subtract V(x,y) from both sides and treat as part of the c term), applying quadratic formula, then set it equal to V(x,y)

 

[pasmith]

Complete the square in $x$.

 

[billtodd]

The matrix representing this function is: [a,b/2];[b/2,c]
A matrix is positive definite whenever <Av,v>>0 when v is a vector different than zero.
And that happens when det(A)>0 and trace(A)>=0.

the determinant is ac-b^2/4>0 and the trace is a+c>0.

if a<0 then because ac>0, also c<0 which is impossible since then we would get: a+c<0, contrary to the fact that the trace is positive.

But in order to know all the above, you really need to know linear algebra 2.

 

[billtodd]

BTW, one should assume obviously that a,b,c are real numbers.