Necessary and Sufficient Condition for Quadratic Functional to be Positive Definite/Lemma

Theorem

If the function $\map h x$ satisfies the equation

$\displaystyle -\frac \d {\d x}\paren{Ph'}+Qh=0$

and the boundary conditions

$\map h a=\map h b=0$

then

$\displaystyle\int_a^b\paren{Ph'^2+Qh^2}\rd x=0$

Proof

 $\displaystyle 0$ $=$ $\displaystyle \int_a^b\sqbrk 0 h\rd x$ $\displaystyle$ $=$ $\displaystyle \int_a^b\sqbrk{-\frac \d {\d x} \paren {Ph'}+Qh}h\rd x$ $\displaystyle$ $=$ $\displaystyle \int_a^b Qh^2\rd x-\int_a^b\frac \d {\d x}\paren{Ph'}h\rd x$ $\displaystyle$ $=$ $\displaystyle \int_a^b Qh^2\rd x-Ph'h\bigg\vert_a^b+\int_a^b Ph'\rd h$ Integration by Parts $\displaystyle$ $=$ $\displaystyle \int_a^b Qh^2\rd x+\int_a^b Ph'^2\rd x$ $\displaystyle$ $=$ $\displaystyle \int_a^b\paren{Ph'^2+Qh^2}\rd x$

$\Box$

Sources

1963: I.M. Gelfand and S.V. Fomin: Calculus of Variations ... (previous) ... (next): $\S 5.26$: Analysis of the Quadratic Functional $\int_a^b \left ( { P h'^2 + Q h^2 } \right ) \mathrm d x$