# Necessary and Sufficient Condition for Quadratic Functional to be Positive Definite/Lemma 1

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## Theorem

Let the function $\map h x$ satisfy the equation:

- $-\map {\dfrac \d {\d x} } {P h'} + Q h = 0$

Let $\map h x$ have the boundary conditions:

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

Then:

- $\displaystyle \int_a^b \paren {P h'^2 + Q h^2} \rd x = 0$

## Proof

\(\ds 0\) | \(=\) | \(\ds \int_a^b \paren 0 h \rd x\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \int_a^b \paren {-\map {\frac \d {\d x} } {P h'} + Q h} h \rd x\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \int_a^b Q h^2 \rd x - \int_a^b \map {\frac \d {\d x} } {P h'} h \rd x\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \int_a^b Q h^2 \rd x - \bigintlimits {P h' h} a b + \int_a^b P h' \rd h\) | Integration by Parts | |||||||||||

\(\ds \) | \(=\) | \(\ds \int_a^b Q h^2 \rd x + \int_a^b P h'^2 \rd x\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \int_a^b \paren {P h'^2 + Q h^2} \rd x\) |

$\blacksquare$

## 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$