Necessary and Sufficient Condition for Quadratic Functional to be Positive Definite/Lemma 2

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Theorem

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

$\displaystyle -\frac \d {\d x}\sqbrk{\paren{tP+\paren{1-t} }h'}+ tQh=0$

and the boundary conditions

$\map h {a,t}=\map h {b,t}=0$

then

$\displaystyle\int_a^b\sqbrk{\paren{Ph'^2+Qh^2}t+\paren{1-t} h'^2}\rd x=0$

Proof

\(\displaystyle 0\) \(=\) \(\displaystyle \int_a^b\sqbrk{0}h\rd x\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \int_a^b\sqbrk{-\frac \d{\d x} \sqbrk{\paren{tP+\paren{1-t} }h'}+tQh}h\rd x\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \int_a^b tQh^2\rd x-\int_a^b h\rd\sqbrk{\paren{tP+\paren{1-t} }h'}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \int_a^b tQh^2\rd x-h\sqbrk{\paren{tP+\paren{1-t} }h'}\bigg\vert_a^b+\int_a^b\sqbrk{\paren{tP+\paren{1-t} }h'}\rd h\) $\quad$ Integration by Parts $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \int_a^b tQh^2\rd x+\int_a^b\sqbrk{\paren{tP+\paren{1-t} }h'}h'\mathrm d x\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \int_a^b\sqbrk{\paren{Ph'^2+Qh^2}t+\paren{1-t}h'^2}\rd x\) $\quad$ $\quad$

$\Box$

Sources

1963: I.M. Gelfand and S.V. Fomin: Calculus of Variations ... (previous) ... (next): $\S 5.26$: Analysis of the Quadratic Functional $\int_a^b\paren{Ph'^2+Qh^2}\rd x$