Categories: Mathematics, Physics.

# Calculus of variations

The calculus of variations lays the mathematical groundwork for Lagrangian mechanics.

Consider a functional $J$, mapping a function $f(x)$ to a scalar value by integrating over the so-called Lagrangian $L$, which represents an expression involving $x$, $f$ and the derivative $f'$:

\begin{aligned} J[f] = \int_{x_0}^{x_1} L(f, f', x) \dd{x} \end{aligned}

If $J$ in some way measures the physical “cost” (e.g. energy) of the path $f(x)$ taken by a physical system, the principle of least action states that $f$ will be a minimum of $J[f]$, so for example the expended energy will be minimized. In practice, various cost metrics may be used, so maxima of $J[f]$ are also interesting to us.

If $f(x, \varepsilon\!=\!0)$ is the optimal route, then a slightly different (and therefore worse) path between the same two points can be expressed using the parameter $\varepsilon$:

\begin{aligned} f(x, \varepsilon) = f(x, 0) + \varepsilon \eta(x) \qquad \mathrm{or} \qquad \delta f = \varepsilon \eta(x) \end{aligned}

Where $\eta(x)$ is an arbitrary differentiable deviation. Since $f(x, \varepsilon)$ must start and end in the same points as $f(x,0)$, we have the boundary conditions:

\begin{aligned} \eta(x_0) = \eta(x_1) = 0 \end{aligned}

Given $L$, the goal is to find an equation for the optimal path $f(x,0)$. Just like when finding the minimum of a real function, the minimum $f$ of a functional $J[f]$ is a stationary point with respect to the deviation weight $\varepsilon$, a condition often written as $\delta J = 0$. In the following, the integration limits have been omitted:

\begin{aligned} 0 &= \delta J = \pdv{J}{\varepsilon} \Big|_{\varepsilon = 0} = \int \pdv{L}{\varepsilon} \dd{x} = \int \pdv{L}{f} \pdv{f}{\varepsilon} + \pdv{L}{f'} \pdv{f'}{\varepsilon} \dd{x} \\ &= \int \pdv{L}{f} \eta + \pdv{L}{f'} \eta' \dd{x} = \Big[ \pdv{L}{f'} \eta \Big]_{x_0}^{x_1} + \int \pdv{L}{f} \eta - \dv{}{x}\Big( \pdv{L}{f'} \Big) \eta \dd{x} \end{aligned}

The boundary term from partial integration vanishes due to the boundary conditions for $\eta(x)$. We are thus left with:

\begin{aligned} 0 = \int \eta \bigg( \pdv{L}{f} - \dv{}{x}\Big( \pdv{L}{f'} \Big) \bigg) \dd{x} \end{aligned}

This holds for all $\eta$, but $\eta$ is arbitrary, so in fact only the parenthesized expression matters:

\begin{aligned} \boxed{ 0 = \pdv{L}{f} - \dv{}{x}\Big( \pdv{L}{f'} \Big) } \end{aligned}

This is known as the Euler-Lagrange equation of the Lagrangian $L$, and its solutions represent the optimal paths $f(x, 0)$.

## Multiple functions

Suppose that the Lagrangian $L$ depends on multiple independent functions $f_1, f_2, ..., f_N$:

\begin{aligned} J[f_1, ..., f_N] = \int_{x_0}^{x_1} L(f_1, ..., f_N, f_1', ..., f_N', x) \dd{x} \end{aligned}

In this case, every $f_n(x)$ has its own deviation $\eta_n(x)$, satisfying $\eta_n(x_0) = \eta_n(x_1) = 0$:

\begin{aligned} f_n(x, \varepsilon) = f_n(x, 0) + \varepsilon \eta_n(x) \end{aligned}

The derivation procedure is identical to the case $N = 1$ from earlier:

\begin{aligned} 0 &= \pdv{J}{\varepsilon} \Big|_{\varepsilon = 0} = \int \pdv{L}{\varepsilon} \dd{x} = \int \sum_{n} \Big( \pdv{L}{f_n} \pdv{f_n}{\varepsilon} + \pdv{L}{f_n'} \pdv{f_n'}{\varepsilon} \Big) \dd{x} \\ &= \int \sum_{n} \Big( \pdv{L}{f_n} \eta_n + \pdv{L}{f_n'} \eta_n' \Big) \dd{x} \\ &= \Big[ \sum_{n} \pdv{L}{f_n'} \eta_n \Big]_{x_0}^{x_1} + \int \sum_{n} \eta_n \bigg( \pdv{L}{f_n} - \dv{}{x}\Big( \pdv{L}{f_n'} \Big) \bigg) \dd{x} \end{aligned}

Once again, $\eta_n(x)$ is arbitrary and disappears at the boundaries, so we end up with $N$ equations of the same form as for a single function:

\begin{aligned} \boxed{ 0 = \pdv{L}{f_1} - \dv{}{x}\Big( \pdv{L}{f_1'} \Big) \quad \cdots \quad 0 = \pdv{L}{f_N} - \dv{}{x}\Big( \pdv{L}{f_N'} \Big) } \end{aligned}

## Higher-order derivatives

Suppose that the Lagrangian $L$ depends on multiple derivatives of $f(x)$:

\begin{aligned} J[f] = \int_{x_0}^{x_1} L(f, f', f'', ..., f^{(N)}, x) \dd{x} \end{aligned}

Once again, the derivation procedure is the same as before:

\begin{aligned} 0 &= \pdv{J}{\varepsilon} \Big|_{\varepsilon = 0} = \int \pdv{L}{\varepsilon} \dd{x} = \int \pdv{L}{f} \pdv{f}{\varepsilon} + \sum_{n} \pdv{L}{f^{(n)}} \pdv{f^{(n)}}{\varepsilon} \dd{x} \\ &= \int \pdv{L}{f} \eta + \sum_{n} \pdv{L}{f^{(n)}} \eta^{(n)} \dd{x} \end{aligned}

The goal is to turn each $\eta^{(n)}(x)$ into $\eta(x)$, so we need to partially integrate the $n$th term of the sum $n$ times. In this case, we will need some additional boundary conditions for $\eta(x)$:

\begin{aligned} \eta'(x_0) = \eta'(x_1) = 0 \qquad \cdots \qquad \eta^{(N-1)}(x_0) = \eta^{(N-1)}(x_1) = 0 \end{aligned}

This eliminates the boundary terms from partial integration, leaving:

\begin{aligned} 0 &= \int \eta \bigg( \pdv{L}{f} + \sum_{n} (-1)^n \dvn{n}{}{x}\Big( \pdv{L}{f^{(n)}} \Big) \bigg) \dd{x} \end{aligned}

Once again, because $\eta(x)$ is arbitrary, the Euler-Lagrange equation becomes:

\begin{aligned} \boxed{ 0 = \pdv{L}{f} + \sum_{n} (-1)^n \dvn{n}{}{x}\Big( \pdv{L}{f^{(n)}} \Big) } \end{aligned}

## Multiple coordinates

Suppose now that $f$ is a function of multiple variables. For brevity, we only consider two variables $x$ and $y$, but the results generalize effortlessly to larger amounts. The Lagrangian now depends on all the partial derivatives of $f(x, y)$:

\begin{aligned} J[f] = \iint_{(x_0, y_0)}^{(x_1, y_1)} L(f, f_x, f_y, x, y) \dd{x} \dd{y} \end{aligned}

The arbitrary deviation $\eta$ is then also a function of multiple variables:

\begin{aligned} f(x, y; \varepsilon) = f(x, y; 0) + \varepsilon \eta(x, y) \end{aligned}

The derivation procedure starts in the exact same way as before:

\begin{aligned} 0 &= \pdv{J}{\varepsilon} \Big|_{\varepsilon = 0} = \iint \pdv{L}{\varepsilon} \dd{x} \dd{y} \\ &= \iint \pdv{L}{f} \pdv{f}{\varepsilon} + \pdv{L}{f_x} \pdv{f_x}{\varepsilon} + \pdv{L}{f_y} \pdv{f_y}{\varepsilon} \dd{x} \dd{y} \\ &= \iint \pdv{L}{f} \eta + \pdv{L}{f_x} \eta_x + \pdv{L}{f_y} \eta_y \dd{x} \dd{y} \end{aligned}

We partially integrate for both $\eta_x$ and $\eta_y$, yielding:

\begin{aligned} 0 &= \int \Big[ \pdv{L}{f_x} \eta \Big]_{x_0}^{x_1} \dd{y} + \int \Big[ \pdv{L}{f_y} \eta \Big]_{y_0}^{y_1} \dd{x} \\ &\quad + \iint \eta \bigg( \pdv{L}{f} - \dv{}{x}\Big( \pdv{L}{f_x} \Big) - \dv{}{y}\Big( \pdv{L}{f_y} \Big) \bigg) \dd{x} \dd{y} \end{aligned}

But now, to eliminate these boundary terms, we need extra conditions for $\eta$:

\begin{aligned} \forall y: \eta(x_0, y) = \eta(x_1, y) = 0 \qquad \forall x: \eta(x, y_0) = \eta(x, y_1) = 0 \end{aligned}

In other words, the deviation $\eta$ must be zero on the whole “box”. Again relying on the fact that $\eta$ is arbitrary, the Euler-Lagrange equation is:

\begin{aligned} 0 = \pdv{L}{f} - \dv{}{x}\Big( \pdv{L}{f_x} \Big) - \dv{}{y}\Big( \pdv{L}{f_y} \Big) \end{aligned}

This generalizes nicely to functions of even more variables $x_1, x_2, ..., x_N$:

\begin{aligned} \boxed{ 0 = \pdv{L}{f} - \sum_{n} \dv{}{x_n}\Big( \pdv{L}{f_{x_n}} \Big) } \end{aligned}

## Constraints

So far, for multiple functions $f_1, ..., f_N$, we have been assuming that all $f_n$ are independent, and by extension all $\eta_n$. Suppose that we now have $M < N$ constraints $\phi_m$ that all $f_n$ need to obey, introducing implicit dependencies between them.

Let us consider constraints $\phi_m$ of the two forms below. It is important that they are holonomic, meaning they do not depend on any derivatives of any $f_n(x)$:

\begin{aligned} \phi_m(f_1, ..., f_N, x) = 0 \qquad \mathrm{or} \qquad \int_{x_0}^{x_1} \phi_m(f_1, ..., f_N, x) \dd{x} = C_m \end{aligned}

Where $C_m$ is a constant. Note that the first form can also be used for $\phi_m = C_m \neq 0$, by simply redefining the constraint as $\phi_m^0 = \phi_m - C_m = 0$.

To solve this constrained optimization problem for $f_n(x)$, we introduce Lagrange multipliers $\lambda_m$. In the former case $\lambda_m(x)$ is a function of $x$, while in the latter case $\lambda_m$ is constant:

\begin{aligned} \int \lambda_m(x) \: \phi_m(\{f_n\}, x) \dd{x} = 0 \qquad \mathrm{or} \qquad \lambda_m \int \phi_m(\{f_n\}, x) \dd{x} = \lambda_m C_m \end{aligned}

The reason for this distinction in $\lambda_m$ is that we need to find the stationary points with respect to $\varepsilon$ of both constraint types. Written in the variational form, this is:

\begin{aligned} \delta \int \lambda_m \: \phi_m \dd{x} = 0 \end{aligned}

From this, we define a new Lagrangian $\Lambda$ for the functional $J$, with the contraints built in:

\begin{aligned} J[f_n] &= \int \Lambda(f_1, ..., f_N; f_1', ..., f_N'; \lambda_1, ..., \lambda_M; x) \dd{x} \\ &= \int L + \sum_{m} \lambda_m \phi_m \dd{x} \end{aligned}

Then we derive the Euler-Lagrange equation as usual for $\Lambda$ instead of $L$:

\begin{aligned} 0 &= \delta \int \Lambda \dd{x} = \int \pdv{\Lambda}{\varepsilon} \dd{x} = \int \sum_n \Big( \pdv{\Lambda}{f_n} \pdv{f_n}{\varepsilon} + \pdv{\Lambda}{f_n'} \pdv{f_n'}{\varepsilon} \Big) \dd{x} \\ &= \int \sum_n \Big( \pdv{\Lambda}{f_n} \eta_n + \pdv{\Lambda}{f_n'} \eta_n' \Big) \dd{x} \\ &= \Big[ \sum_n \pdv{\Lambda}{f_n'} \eta_n \Big]_{x_0}^{x_1} + \int \sum_n \eta_n \bigg( \pdv{\Lambda}{f_n} - \dv{}{x}\Big( \pdv{\Lambda}{f_n'} \Big) \bigg) \dd{x} \end{aligned}

Using the same logic as before, we end up with a set of Euler-Lagrange equations with $\Lambda$:

\begin{aligned} 0 = \pdv{\Lambda}{f_n} - \dv{}{x}\Big( \pdv{\Lambda}{f_n'} \Big) \end{aligned}

By inserting the definition of $\Lambda$, we then get the following. Recall that $\phi_m$ is holonomic, and thus independent of all derivatives $f_n'$:

\begin{aligned} \boxed{ 0 = \pdv{L}{f_n} - \dv{}{x}\Big( \pdv{L}{f_n'} \Big) + \sum_{m} \lambda_m \pdv{\phi_m}{f_n} } \end{aligned}

These are Lagrange’s equations of the first kind, with their second-kind counterparts being the earlier Euler-Lagrange equations. Note that there are $N$ separate equations, one for each $f_n$.

Due to the constraints $\phi_m$, the functions $f_n$ are not independent. This is solved by choosing $\lambda_m$ such that $M$ of the $N$ equations hold, i.e. solving a system of $M$ equations for $\lambda_m$:

\begin{aligned} \dv{}{x}\Big( \pdv{L}{f_n'} \Big) - \pdv{L}{f_n} = \sum_{m} \lambda_m \pdv{\phi_m}{f_n} \end{aligned}

And then the remaining $N - M$ equations can be solved in the normal unconstrained way.

## References

1. G.B. Arfken, H.J. Weber, Mathematical methods for physicists, 6th edition, 2005, Elsevier.
2. O. Bang, Applied mathematics for physicists: lecture notes, 2019, unpublished.