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

## Theorem

Let $\map h x : \closedint a b \to \R$ be continuously differentiable $\forall x \in \closedint a b$.

Suppose the function $\map h x$ satisfies the equation:

$-\map {\dfrac \d {\d x} } {\paren {t P + \paren {1 - t} } h'} + t Q h = 0$

subject to the boundary conditions:

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

Then:

$\ds \int_a^b \paren {\paren {P h'^2 + Q h^2} t + \paren {1 - t} h'^2} \rd x = 0$

## Proof

 $\ds 0$ $=$ $\ds \int_a^b \paren 0 h \rd x$ $\ds$ $=$ $\ds \int_a^b \paren {-\map {\frac \d {\d x} } {\paren {t P + \paren {1 - t} } h'} + t Q h } h \rd x$ $\ds$ $=$ $\ds \int_a^b t Q h^2 \rd x - \int_a^b h \map \d {\paren {t P + \paren {1 - t} } h'}$ $\ds$ $=$ $\ds \int_a^b t Q h^2 \rd x - h \bigintlimits {\paren {t P + \paren {1 - t} } h'} a b + \int_a^b \paren {\paren {t P + \paren {1 - t} } h'} \rd h$ Integration by Parts $\ds$ $=$ $\ds \int_a^b t Q h^2 \rd x + \int_a^b \paren {\paren {t P + \paren {1 - t} } h'} h' \rd x$ $\ds$ $=$ $\ds \int_a^b \paren {\paren {P h'^2 + Q h^2} t + \paren {1 - t} h'^2} \rd x$

$\blacksquare$

## Sources

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