Calculus of Variations

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Calculus of variations is the study of extrema of functionals. This page provides a brief introduction to some elements of the theory.


Contents

Variation and differentiability

Consider a linear functional \ J:V \to \mathbb R, where \ V is a normed linear space.

We define the differential of \ J at y \in V as

\ \Delta J[h] = J[y + h] - J[y] for all h \in F

We say that \ J is differentiable at \ y if there exist a linear functional \ \delta J:V \to \mathbb R and \ \epsilon so that

\Delta J[h] = \delta J[h] + \epsilon \|h\|

and \ \epsilon \to 0 as \|h\| \to 0.

Here, \ \delta J is called the variation or differential of \ J at \ y.

Uniqueness of variation

Lemma: Let \ \phi:V \to \mathbb R denote a linear functional, where \ V is defined as above.

If \ {\phi[h] \over \|h\|} \to 0 as \|h\| \to 0, then \ \phi is identically zero for all \ h \in V.

Proof: Suppose \phi[y_0] \neq 0 for some y_0 \in V. Then, for \ n = 1, 2, 3, ...

{\phi[\frac {h_0}n] \over \|\frac {h_0}n\|} = {\frac 1n \phi[h_0] \over \frac 1n \|h_0\|} = {\phi[h_0] \over \|h_0\|} \neq 0

and hence does not approach zero as \frac {h_0}n \to 0, a contradiction. QED


Theorem: For \ J defined as above, \ \delta J is unique.

Proof: Suppose, if possible, that there exist two distinct variations \ \delta J_1 and \ \delta J_2. Then,

\ \Delta J[h] = \delta J_1[h] + \epsilon_1 \|h\|
\ \Delta J[h] = \delta J_2[h] + \epsilon_2 \|h\|

Comparing the two equations, we get

\ (\delta J_1 - \delta J_2)[h] = (- \epsilon_1 + \epsilon_2)\|h\|

or

{(\delta J_1 - \delta J_2)[h] \over \|h\|} = - \epsilon_1 + \epsilon_2 \to 0 as \ h \to 0.

Hence, by the lemma, \ \delta J_1 - \delta J_2 = 0 or \ \delta J_1 = \delta J_2, a contradiction. QED

Function space norm

Consider a space \ C^0[a,b] consisting of all functions \ f:[a,b] \to \mathbb R continuous on \ [a,b]. We define the norm as

\|y\|_0 = \max_{x \in [a,b]}|y(x)|

Generally, a space \ C^n[a,b] consists of all functions \ f:[a,b] \to \mathbb R that are continuously differentiable \ n times. In this case, we define the norm as

\|y\|_n = \sum_{i=0}^n{\max_{x \in [a,b]}|y^{(i)}(x)|}

It can be shown that \|y\|_n satisfies the conditions for a norm.

Strong and weak extrema

Consider some linear functional \ J:F \to \mathbb R, where \ F is a normed function space.

Then, we say that \ J has a strong minimum at some \ y_0 \in F if there exists some \ \delta > 0 such that

\ J[y_0] \le J[y]

for all \|y - y_0 \|_0 < \delta, where y \in F.

Similarly, we define a weak minimum at some \ y_0 \in F with \|\cdot\|_1 in place of \|\cdot\|_0.

Notice that a strong minimum is also a weak minimum.

Strong and weak maximum are defined similarly with \ge in place of \le.

We may also define extrema using norm \| \cdot \|_n for \ n > 1.

A necessary condition for an extremum

Theorem: If a differentiable functional \ J has an extremum for some \ y_0 \in F, then

\ \delta J[h] = 0 for all h \in F.

Proof: Suppose, if possible, that \ \delta J \neq 0 for some \ h_0 \in F, where \ h_0 \neq 0.

Let \ \alpha > 0 and consider {\Delta J[\alpha h_0] \over {\| \alpha h_0\|}} = {\delta J[\alpha h_0] \over {\| \alpha h_0 \|}} + \epsilon = {\delta J[h_0] \over {\| h_0 \|}} + \epsilon.

As \alpha \to 0, we have \|\alpha h_0\| = \alpha \|h_0\| \to 0 and hence \ \epsilon \to 0.

That is, if \ \alpha is sufficiently small, then \ \Delta J[\alpha h_0] and \ \delta J[\alpha h_0] share the same sign.

However, \ \delta J[-\alpha h_0] = - \delta J[\alpha h_0]. Hence, within any neighbourhood of \ h = 0, we have \ \Delta J taking on either sign, a contradiction. QED

References

  1. Gelfand, I.M. and Fomin, S.V.: Calculus of Variations, Dover Publ., 2000.
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