Hamilton's principle/Mathematics/Several dependent variables

Plain-language version: More than one variable

A mechanical system rarely has one degree of freedom. This page extends the machinery to an integrand depending on $n$ unknown functions of a single independent variable. The result is a system of $n$ Euler-Lagrange equations, one per unknown, and this is the form in which Hamilton's principle is actually stated for mechanics.

Setup

Varying one degree of freedom at a time.

Let $y=(y_1,\dots ,y_n)$ be $n$ functions of $x$ on $[a,b]$, each in $C^2$, with all $2n$ endpoint values prescribed:

$y_i(a)=y_{i,a},y_i(b)=y_{i,b},i=1,\dots ,n\mathrm{(}\mathrm{1}\mathrm{)}$

Let $L(x,y_1,\dots ,y_n,{y_1}^{\prime },\dots ,{y_n}^{\prime })$ have continuous second partials in all $2n+1$ slots, and define

$J[y]={\int }_a^{b}L(x,y_i,{y_i}^{\prime })dx\mathrm{(}\mathrm{2}\mathrm{)}$

The variation space is now $n$ copies of the earlier one: each ${\eta }_i\in C^2[a,b]$ with ${\eta }_i(a)={\eta }_i(b)=0$, and the ${\eta }_i$ are independent of one another. That independence is the whole content of what follows.

Derivation

Form the family $y_i+\epsilon {\eta }_i$, differentiate under the integral, and set $\epsilon =0$:

$\delta J={\int }_a^b\sum _{i=1}^n[\frac{\partial L}{\partial y_i}{\eta }_i+\frac{\partial L}{\partial {y_i}^{\prime }}{{\eta }_i}^{\prime }]dx\mathrm{(}\mathrm{3}\mathrm{)}$

Integrate each term of the second kind by parts. Every boundary term vanishes, since every ${\eta }_i$ vanishes at both ends:

$\delta J={\int }_a^b\sum _{i=1}^n[\frac{\partial L}{\partial y_i}-\frac{d}{dx}\frac{\partial L}{\partial {y_i}^{\prime }}]{\eta }_{i}dx\mathrm{(}\mathrm{4}\mathrm{)}$

Stationarity requires (4) to vanish for every choice of the $n$ variations. Here is the step that needs care.

Select one index at a time. Fix $k$ and choose ${\eta }_i=0$ for all $i\ne k$, with ${\eta }_k$ arbitrary. This is legitimate precisely because the ${\eta }_i$ are independent. The sum in (4) collapses to its $k$-th term, and the fundamental lemma gives

$\frac{\partial L}{\partial y_k}-\frac{d}{dx}\frac{\partial L}{\partial {y_k}^{\prime }}=0\mathrm{(}\mathrm{5}\mathrm{)}$

Repeating for each $k$ yields $n$ equations. $◼$

The single-index trick fails under constraints. If the $y_i$ are related — a bead confined to a wire, a rigid rod between two masses — then the ${\eta }_i$ are not independent, ${\eta }_k$ cannot be varied alone, and (5) does not follow. That case is Holonomic constraints, and the repair is Lagrange multipliers. Everything in this page rests on independence, and it is worth knowing exactly where.

Counting

(5) is a system of $n$ coupled second-order equations, so its general solution carries $2n$ constants. The endpoint data (1) supplies exactly $2n$ conditions. The count matches again, as it did for one variable.

Coupling enters through the mixed partials ${\partial }^{2}L/\partial {y_i}^{\prime }\partial {y_k}^{\prime }$. Expanding the total derivative in (5),

$\sum _{k=1}^n\frac{{\partial }^{2}L}{\partial {y_i}^{\prime }\partial {y_k}^{\prime }}{y_k}^{″}+\sum _{k=1}^n\frac{{\partial }^{2}L}{\partial {y_i}^{\prime }\partial y_k}{y_k}^{\prime }+\frac{{\partial }^{2}L}{\partial {y_i}^{\prime }\partial x}-\frac{\partial L}{\partial y_i}=0\mathrm{(}\mathrm{6}\mathrm{)}$

The matrix $W_{ik}={\partial }^{2}L/\partial {y_i}^{\prime }\partial {y_k}^{\prime }$ is the Hessian with respect to the derivatives. The system can be solved for the ${y_k}^{″}$ if and only if $W$ is invertible. This is the $n$-variable non-degeneracy condition, and its definiteness version is the Legendre condition of the Legendre condition. A singular $W$ signals a constrained or gauge system, which is where Noether's second theorem applies.

First integrals

Both results from the one-variable case generalise, with the same proofs.

If $L$ does not contain $y_k$, then

$p_k=\frac{\partial L}{\partial {y_k}^{\prime }}=\text{const}\mathrm{(}\mathrm{7}\mathrm{)}$

and $y_k$ is called a cyclic or ignorable coordinate.

If $L$ does not contain $x$, then

$\sum _{i=1}^n{y_i}^{\prime }\frac{\partial L}{\partial {y_i}^{\prime }}-L=\text{const}\mathrm{(}\mathrm{8}\mathrm{)}$

which is the $n$-variable Beltrami identity of the Beltrami identity, and is the Hamiltonian.

Hamilton's own form

This is the form in which the historical statement is made. Hamilton writes the equations of motion of a system of free points as a single formula,

$\Sigma .m(x^{″}\delta x+y^{″}\delta y+z^{″}\delta z)=\delta U\mathrm{(}\mathrm{9}\mathrm{)}$

and after transforming to general marks of position ${\eta }_1,\dots ,{\eta }_{3n}$ he arrives at

$\frac{d}{dt}\frac{\delta T}{\delta {\eta }^{\prime }}-\frac{\delta T}{\delta \eta }=\frac{\delta U}{\delta \eta }\mathrm{(}\mathrm{10}\mathrm{)}$

Equation (10) is (5) with $L=T+U$, allowing for Hamilton's sign convention on the force-function. The correspondence between his notation and this volume's is tabulated on Notation.

Worked example: the free particle in the plane

Take $x$ as time $t$, coordinates $(y_1,y_2)$, and

$L=\frac{1}{2}m({\dot{y}}_1^2+{\dot{y}}_2^2)\mathrm{(}\mathrm{11}\mathrm{)}$

Neither coordinate appears in $L$, so both are cyclic and (7) gives $m{\dot{y}}_1=c_1$, $m{\dot{y}}_2=c_2$. Both momentum components are conserved and the path is a straight line traversed at constant speed.

$t$ does not appear either, so (8) applies as well and gives $\frac{1}{2}m({\dot{y}}_1^2+{\dot{y}}_2^2)=\text{const}$. That is energy conservation, and it is not independent of the momentum results here — it follows from them. Independent symmetries need not give independent first integrals, a point Noether makes explicitly and which is taken up on Noether's first theorem.

Next

Several independent variables takes the remaining step, from ordinary to partial differential equations.