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diff --git a/source/know/concept/calculus-of-variations/index.md b/source/know/concept/calculus-of-variations/index.md new file mode 100644 index 0000000..81d9a5f --- /dev/null +++ b/source/know/concept/calculus-of-variations/index.md @@ -0,0 +1,339 @@ +--- +title: "Calculus of variations" +date: 2021-02-24 +categories: +- Mathematics +- Physics +layout: "concept" +--- + +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: + +$$\begin{aligned} + 0 + &= \delta J + = \pdv{J}{\varepsilon} \Big|_{\varepsilon = 0} + = \int \pdv{L}{\varepsilon} \dd{x} + = \int \pdv{L}{f} \pdv{f}{\varepsilon} + \pdv{L}{f'} \pdv{f'}{\varepsilon} \dd{x} + \\ + &= \int \pdv{L}{f} \eta + \pdv{L}{f'} \eta' \dd{x} + = \Big[ \pdv{L}{f'} \eta \Big]_{x_0}^{x_1} + \int \pdv{L}{f} \eta - \dv{}{x}\Big( \pdv{L}{f'} \Big) \eta \dd{x} +\end{aligned}$$ + +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](/know/concept/lagrange-multiplier/) $\lambda_m$. +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. |