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2.3.4 Two special cases

We know from the discussion given towards the end of Section 2.3.1--see the formula (2.19) in particular--that the Euler-Lagrange equation (2.18) is a second-order differential equation for $ y(\cdot)$ ; indeed, it can be written in more detail as

$\displaystyle {L}_{ y}={L}_{{y'}{x}}+{L}_{{ y'}{ y}}y'+{L}_{{ y'}{y'}}y''.$ (2.23)

Note that we are being somewhat informal in denoting the third argument of $ L$ by $ y'$ , and also in omitting the $ x$ -arguments. We now discuss two special cases in which (2.24) can be reduced to a differential equation that is of first order, and therefore more tractable.

SPECIAL CASE 1 (``no $ y$ "). This refers to the situation where the Lagrangian does not depend on $ y$ , i.e., $ L=L(x,y')$ . The Euler-Lagrange equation (2.18) becomes $ \frac d{dx}{L}_{ y'}=0,
$ which means that $ {L}_{ y'}$ must stay constant. In other words, extremals are solutions of the first-order differential equation

$\displaystyle {L}_{ y'}(x,y'(x))=c$ (2.24)

for various values of $ c\in\mathbb{R}$ . We already encountered such a situation in Example 2.2, where we actually had $ L=L(y')$ . Due to the presence of the parameter $ c$ , we expect that the family of solutions of (2.25) is rich enough to contain one (and only one) extremal that passes through two given points.

The quantity $ {L}_{ y'}$ , evaluated along a given curve, is called the momentum. This terminology will be justified in Section 2.4.

SPECIAL CASE 2 (``no $ x$ "). Suppose now that the Lagrangian does not depend on $ x$ , i.e., $ L=L(y,y')$ . In this case the partial derivative $ {L}_{{y'}{x}}$ vanishes from (2.24), and the Euler-Lagrange equation becomes

$\displaystyle 0=%\frac d{dx}\dd{L}{ y'}-\dd{L}{ y}=
{L}_{{ y'}{y}}y'+{L}_{{y'}{y'}}y''-{L}_{ y}.
$

Multiplying both sides by $ y'$ , we have

$\displaystyle 0={L}_{{ y'}{ y}}(y')^2+{L}_{{ y'}{y'}}y'y''-{L}_{ y}y'=\frac
d{dx}\left({L}_{ y'}y'-L\right)
$

where the last equality is easily verified (the $ {L}_{ y'}y''$ terms cancel out). This means that $ {L}_{ y'}y'-L$ must remain constant. Thus, similarly to Case 1, extremals are described by the family of first-order differential equations

$\displaystyle {L}_{ y'}(y(x),y'(x))y'(x)-L(y(x),y'(x))=c
$

parameterized by $ c\in\mathbb{R}$ .

The quantity $ {L}_{ y'}y'-L$ is called the Hamiltonian. Although its significance is not yet clear at this point, it will play a crucial role throughout the rest of the book.


\begin{Exercise}
Use the above \lq\lq no $x$'' result to show that extremals for the ...
... indeed given by the
equations~\eqref{e-brachistochrone-solution}.\end{Exercise}


next up previous contents index
Next: 2.3.5 Variable-endpoint problems Up: 2.3 First-order necessary conditions Previous: 2.3.3 Technical remarks   Contents   Index
Daniel 2010-12-20