Necessary and Sufficient Condition for Quadratic Functional to be Positive Definite/Lemma 2
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Define the domain and codomain of $h$ -- presumably it's real, but make it clear
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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:
- $\displaystyle \int_a^b \paren {\paren {P h'^2 + Q h^2} t + \paren {1 - t} h'^2} \rd x = 0$
Define the domain and codomain of $h$ -- presumably it's real, but make it clear
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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 $\int_a^b \paren {P h'^2 + Q h^2} \rd x$