Categories: Mathematics, Physics.

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.

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:

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

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 \(f(x, 0)\).

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:

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

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:

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

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:

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

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

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

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 \(x_1, x_2, ..., x_N\):

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

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 \(f_n(x)\):

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

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 all \(x\), while in the latter case \(\lambda_m\) is constant:

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

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.

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

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