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2.4.1 Hamilton's canonical equations

In Section 2.3.4 we came across two quantities, which we recall here and for which we now introduce special symbols. The first one was the momentum

$\displaystyle p:={L}_{ y'}(x,y,y')$ (2.27)

which we will usually regard as a function of $ x$ associated to a given curve $ y=y(x)$ . The second object was the Hamiltonian

$\displaystyle H(x,y,y',p):=p\cdot y'-L(x,y,y')$ (2.28)

which is written here as a general function of four variables but also becomes a function of $ x$ alone when evaluated along a curve. The inner product sign $ \cdot$ in the definition of $ H$ reflects the fact that in the multiple-degrees-of-freedom case, $ y'$ and $ p$ are vectors.

The variables $ y$ and $ p$ are called the canonical variables. Suppose now that $ y$ is an extremal, i.e., satisfies the Euler-Lagrange equation (2.18). It turns out that the differential equations describing the evolution of $ y$ and $ p$ along such a curve, when written in terms of the Hamiltonian $ H$ , take a particularly nice form. For $ y$ , we have

$\displaystyle \frac{dy}{dx}=y'(x)={H}_{ p}(x,y(x),y'(x)).

For $ p$ , we have

$\displaystyle \frac{dp}{dx}=\frac d{dx}{L}_{
y'}(x,y(x),y'(x))={L}_{ y}(x,y(x),y'(x))=-{H}_{ y}(x,y(x),y'(x))

where the second equality is the Euler-Lagrange equation. In more concise form, the result is

$\displaystyle \fbox{$y'={H}_{p}, \quad p'=-{H}_{y}$}$ (2.29)

which is known as the system of Hamilton's canonical equations. This reformulation of the Euler-Lagrange equation was proposed by Hamilton in 1835. Since we are not assuming here that we are in the ``no $ y$ " case or the ``no $ x$ " case of Section 2.3.4, the momentum $ p$ and the Hamiltonian $ H$ need not be constant along extremals.

Confirm directly from the equations~\eqref{e-p-def}--\eqref{e-h...
...mals and in the \lq\lq no $x$'' case $H$\ is constant along extremals.

An important additional observation is that the partial derivative of $ H$ with respect to $ y'$ is

$\displaystyle {H}_{y'}(x,y,y',p)=p-{L}_{y'}(x,y,y')=0$ (2.30)

where the last equality follows from the definition (2.28) of $ p$ . This suggests that, in addition to the canonical equations (2.30), another necessary condition for optimality should be that $ H$ has a stationary point as a function of $ y'$ along an optimal curve. To make this statement more precise, let us plug the following arguments into the Hamiltonian: an arbitrary $ x\in [a,b]$ ; for $ y$ , the corresponding position $ y(x)$ of the optimal curve; for $ p$ , the corresponding value of the momentum $ p(x)={L}_{ y'}(x,y(x),y'(x))$ . Let us keep the last remaining argument, $ y'$ , as a free variable, and relabel it as $ z$ for clarity. This yields the function

$\displaystyle H^*(z):={L}_{ y'}(x,y(x),y'(x))\cdot z-L(x,y(x),z).$ (2.31)

Our claim is that this function has a stationary point when $ z$ equals $ y'(x)$ , the velocity of the optimal curve at $ x$ . Indeed, it is immediate to check that

$\displaystyle \frac{dH^*}{dz}(y'(x))=0.$ (2.32)

Later we will see that in the context of the maximum principle this stationary point is actually an extremum, in fact, a maximum. Moreover, the statement about the maximum remains true when $ H$ is not necessarily differentiable or when $ y'$ takes values in a set with a boundary and $ {H}_{y'}\ne 0$ on this boundary; the basic property is not that the derivative vanishes but that $ H$ achieves the maximum in the above sense.

Mathematically, the Lagrangian $ L$ and the Hamiltonian $ H$ are related via a construction known as the Legendre transformation. Since this transformation is classical and finds applications in many diverse areas (optimization, geometry, physics), we now proceed to describe it. However, we will see that it does not quite provide the right point of view for our future developments, and it is included here mainly for historical reasons.

next up previous contents index
Next: 2.4.2 Legendre transformation Up: 2.4 Hamiltonian formalism and Previous: 2.4 Hamiltonian formalism and   Contents   Index
Daniel 2010-12-20