Sufficient Condition for Twice Differentiable Functional to have Minimum
This article needs to be linked to other articles. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding these links. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{MissingLinks}} from the code. |
Theorem
Let $J$ be a twice differentiable functional.
Let $J$ have an extremum for $y=\hat y$.
Let the second variation $\delta^2 J \sqbrk {\hat y; h}$ be strongly positive with respect to $h$.
Then $J$ acquires the minimum for $y = \hat y$ .
Proof
By assumption, $J$ has an extremum for $y = \hat y$:
- $\delta J \sqbrk {\hat y; h} = 0$
The increment is expressible then as:
- $\Delta J \sqbrk {\hat y; h} = \delta^2 J \sqbrk {\hat y; h} + \epsilon \size h^2$
where $\epsilon \to 0$ as $\size h \to 0$.
By assumption, the second variation is strongly positive:
- $\delta^2 J \sqbrk {\hat y; h} \ge k \size h^2, \quad k \in \R_{>0}$
Hence:
- $\Delta J \sqbrk {\hat y; h} \ge \paren {k + \epsilon} \size h^2$
What remains to be shown is that there exists a set of $h$ such that $\epsilon$ is small enough so that right hand side is always positive.
Since $\epsilon \to 0$ as $\size h \to 0$, there exist $c \in \R_{>0}$, such that:
- $\size h < c \implies \size \epsilon < \dfrac 1 2 k$
Choose $h$ such that this inequality holds.
Then
\(\ds \frac 1 2 k\) | \(>\) | \(\ds \epsilon > -\frac 1 2 k\) | $\big \vert + k$, by Membership is Left Compatible with Ordinal Addition | |||||||||||
\(\ds \frac 3 2 k\) | \(>\) | \(\ds k + \epsilon > \frac 1 2 k\) | $\big \vert \cdot \size h^2$, by Membership is Left Compatible with Ordinal Multiplication | |||||||||||
\(\ds \frac 3 2 k \size h^2\) | \(>\) | \(\ds \paren {k + \epsilon} \size h^2 > \frac 1 2 k \size h^2\) |
This article, or a section of it, needs explaining. In particular: What does this mean? : $\big \vert + k$ and $\big \vert \cdot \size h^2$ You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code. |
Therefore:
- $\Delta J \sqbrk {\hat y; h} \ge \paren {k + \epsilon} \size h^2 > \dfrac 1 2 k \size h^2 $
For $k \in \R_{>0}$ and $\size h \ne 0$ right hand side is always positive.
Thus, there exists a neighbourhood around $y = \hat y$ where the increment is always positive:
- $\exists c \in \R_{>0}: \size h < c \implies \Delta J \sqbrk {\hat y; h} > 0$
and $J$ has a minimum for $y = \hat y$.
$\blacksquare$
Sources
- 1963: I.M. Gelfand and S.V. Fomin: Calculus of Variations ... (previous) ... (next): $\S 5.24$: Quadratic Functionals. The Second Variation of a Functional