Necessary and Sufficient Condition for Quadratic Functional to be Positive Definite/Dependent on N Functions

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Let $K$ be a functional, such that:

$\displaystyle K\sqbrk{\mathbf h}=\int_a^b\paren{\mathbf h'\mathbf P\mathbf h'+\mathbf h\mathbf Q\mathbf h}\rd x$

where $\mathbf h$ is an N-dimensional vector, $\mathbf Q$ is a $N\times N$ matrix, and $\mathbf P$ is a $N\times N$ symmetric positive definite matrix.

Suppose $\closedint a b$ does not contain a point conjugate to $a$.


$\forall\mathbf h:\map {\mathbf h} a=\map {\mathbf h} b=0:K\sqbrk{\mathbf h}>0$

if and only if the above holds.


Necessary Condition

Let $\mathbf W$ be an arbitrary differentiable symmetric matrix.


\(\displaystyle 0\) \(=\) \(\displaystyle \int_a^b \frac \d {\d x}\paren{\mathbf h\mathbf W\mathbf h}\rd x\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \int_a^b\mathbf h\mathbf W'\mathbf h\rd x+2\int_a^b\mathbf h'\mathbf W\mathbf h\rd x\) $\quad$ $\quad$

Suppose, $\mathbf W$ is such that:

$\displaystyle\mathbf Q+\mathbf W'=\mathbf W\mathbf P^{-1}\mathbf W$


\(\displaystyle \int_a^b\paren{\mathbf h'\mathbf P\mathbf h'+\mathbf h\mathbf Q\mathbf h}\rd x\) \(=\) \(\displaystyle \int_a^b\paren{\mathbf h'\mathbf P\mathbf h'+2\mathbf h'\mathbf W\mathbf h+\mathbf h\mathbf Q\mathbf h+\mathbf h\mathbf W'\mathbf h}\rd x\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \int_a^b\paren{\mathbf h'\mathbf P\mathbf h'+2\mathbf h'\mathbf W\mathbf h+\mathbf h\mathbf W\mathbf P^{-1}\mathbf W\mathbf h}\rd x\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \int_a^b\paren{\mathbf P^{1/2}\mathbf h'+\mathbf P^{1/2}\mathbf h'\mathbf P^{-1/2}\mathbf W\mathbf h+\mathbf P^{-1/2}\mathbf W\mathbf h }^2\rd x\) $\quad$ $\quad$

Note that:

$\displaystyle\mathbf P^{1/2}\mathbf h'+\mathbf P^{-1/2}\mathbf W\mathbf h\ne 0$


$\displaystyle\map {\mathbf h} x=0:\forall x\in\closedint a b$

However, this contradicts the absence of conjugate points.

Thus, $K>0$.


Sufficient Condition

Consider the following functional:

$\displaystyle\int_a^b\sqbrk{Kt+\mathbf h'\paren{1-t}\mathbf h'}\rd x$

The corresponding Euler's equations are:

$\displaystyle -\frac \d {\d x}\sqbrk{t\mathbf P\mathbf h'+\paren{1-t}\mathbf h'}+t\mathbf Q\mathbf h=0$

Suppose the interval $\closedint a b$ contains a point $\tilde a$ conjugate to $a$.

Hence the determinant $\size {h_{ij} }$ vanishes.

Therefore there exists a linear combination of $h_i$ not identically equal to zero such that $\map {\mathbf h} {\tilde a}=0$.

Furthermore, since the Euler's equations are continuous with respect to $t$, so is the solution of this equation.

Suppose, $\tilde a=b$.

By lemma, $K$ vanishes.

This contradicts the positive definiteness of $K$.

Therefore, $\tilde a\ne b$.

Thus, for $t=1$ the conjugate point may only reside in $\openint a b$.