--- title: "Calculus of variations" firstLetter: "C" publishDate: 2021-02-24 categories: - Mathematics - Physics date: 2021-02-24T18:50:06+01:00 draft: false markup: pandoc --- # Calculus of variations The **calculus of variations** lays the mathematical groundwork for [Lagrangian mechanics](/know/concept/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: $$\begin{aligned} 0 &= \delta J = \pdv{J}{\varepsilon} \Big|_{\varepsilon = 0} = \int \pdv{L}{\varepsilon} \dd{x} = \int \pdv{L}{f} \pdv{f}{\varepsilon} + \pdv{L}{f'} \pdv{f'}{\varepsilon} \dd{x} \\ &= \int \pdv{L}{f} \eta + \pdv{L}{f'} \eta' \dd{x} = \Big[ \pdv{L}{f'} \eta \Big]_{x_0}^{x_1} + \int \pdv{L}{f} \eta - \frac{d}{dx} \Big( \pdv{L}{f'} \Big) \eta \dd{x} \end{aligned}$$ 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 $f(x, 0)$. ## 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: $$\begin{aligned} 0 &= \pdv{J}{\varepsilon} \Big|_{\varepsilon = 0} = \int \pdv{L}{\varepsilon} \dd{x} = \int \sum_{n} \Big( \pdv{L}{f_n} \pdv{f_n}{\varepsilon} + \pdv{L}{f_n'} \pdv{f_n'}{\varepsilon} \Big) \dd{x} \\ &= \int \sum_{n} \Big( \pdv{L}{f_n} \eta_n + \pdv{L}{f_n'} \eta_n' \Big) \dd{x} \\ &= \Big[ \sum_{n} \pdv{L}{f_n'} \eta_n \Big]_{x_0}^{x_1} + \int \sum_{n} \eta_n \bigg( \pdv{L}{f_n} - \frac{d}{dx} \Big( \pdv{L}{f_n'} \Big) \bigg) \dd{x} \end{aligned}$$ 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: $$\begin{aligned} \boxed{ 0 = \pdv{L}{f_1} - \dv{x} \Big( \pdv{L}{f_1'} \Big) \quad \cdots \quad 0 = \pdv{L}{f_N} - \dv{x} \Big( \pdv{L}{f_N'} \Big) } \end{aligned}$$ ## 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: $$\begin{aligned} 0 &= \int \eta \bigg( \pdv{L}{f} + \sum_{n} (-1)^n \dv[n]{x} \Big( \pdv{L}{f^{(n)}} \Big) \bigg) \dd{x} \end{aligned}$$ Once again, because $\eta(x)$ is arbitrary, the Euler-Lagrange equation becomes: $$\begin{aligned} \boxed{ 0 = \pdv{L}{f} + \sum_{n} (-1)^n \dv[n]{x} \Big( \pdv{L}{f^{(n)}} \Big) } \end{aligned}$$ ## 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 $f_n(x)$: $$\begin{aligned} \phi_m(f_1, ..., f_N, x) = 0 \qquad \int_{x_0}^{x_1} \phi_m(f_1, ..., f_N, x) \dd{x} = C_m \end{aligned}$$ 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](/know/concept/lagrange-multiplier/) $\lambda_m$. 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. ## 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.