diff options
author | Prefetch | 2021-02-24 21:27:41 +0100 |
---|---|---|
committer | Prefetch | 2021-02-24 21:27:41 +0100 |
commit | c157ad913aa9f975ea8c137e24175d134486f462 (patch) | |
tree | 2debabe1627e77085eed17f616d50076c2af54a9 /content/know/concept/calculus-of-variations/index.pdc | |
parent | b5608ab92a4f8a5140571acabf54e3c6bdebd0e4 (diff) |
Expand knowledge base
Diffstat (limited to 'content/know/concept/calculus-of-variations/index.pdc')
-rw-r--r-- | content/know/concept/calculus-of-variations/index.pdc | 236 |
1 files changed, 236 insertions, 0 deletions
diff --git a/content/know/concept/calculus-of-variations/index.pdc b/content/know/concept/calculus-of-variations/index.pdc new file mode 100644 index 0000000..fb043e0 --- /dev/null +++ b/content/know/concept/calculus-of-variations/index.pdc @@ -0,0 +1,236 @@ +--- +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. + +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 $f(x, \alpha\!=\!0)$ is the optimal route, then a slightly +different (and therefore worse) path between the same two points can be expressed +using the parameter $\alpha$: + +$$\begin{aligned} + f(x, \alpha) = f(x, 0) + \alpha \eta(x) + \qquad \mathrm{or} \qquad + \delta f = \alpha \eta(x) +\end{aligned}$$ + +Where $\eta(x)$ is an arbitrary differentiable deviation. +Since $f(x, \alpha)$ 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 $\alpha$, +a condition often written as $\delta J = 0$. +In the following, the integration limits have been omitted: + +$$\begin{aligned} + 0 + &= \delta J + = \pdv{J}{\alpha} \Big|_{\alpha = 0} + = \int \pdv{L}{\alpha} \dd{x} + = \int \pdv{L}{f} \pdv{f}{\alpha} + \pdv{L}{f'} \pdv{f'}{\alpha} \dd{x} + \\ + &= \int \pdv{L}{f} \eta + \pdv{L}{f'} \eta' \dd{x} + = \Big[ \pdv{L}{f'} \eta \Big]_{x_0}^{x_1} + \int \pdv{L}{f} \eta - \frac{d}{dx} \Big( \pdv{L}{f'} \Big) \eta \dd{x} +\end{aligned}$$ + +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 $\eta_n(x_0) = \eta_n(x_1) = 0$: + +$$\begin{aligned} + f_n(x, \alpha) = f_n(x, 0) + \alpha \eta_n(x) +\end{aligned}$$ + +The derivation procedure is identical to the case $N = 1$ from earlier: + +$$\begin{aligned} + 0 + &= \pdv{J}{\alpha} \Big|_{\alpha = 0} + = \int \pdv{L}{\alpha} \dd{x} + = \int \sum_{n} \Big( \pdv{L}{f_n} \pdv{f_n}{\alpha} + \pdv{L}{f_n'} \pdv{f_n'}{\alpha} \Big) \dd{x} + \\ + &= \int \sum_{n} \Big( \pdv{L}{f_n} \eta_n + \pdv{L}{f_n'} \eta_n' \Big) \dd{x} + \\ + &= \Big[ \sum_{n} \pdv{L}{f_n'} \eta_n \Big]_{x_0}^{x_1} + + \int \sum_{n} \eta_n \bigg( \pdv{L}{f_n} - \frac{d}{dx} \Big( \pdv{L}{f_n'} \Big) \bigg) \dd{x} +\end{aligned}$$ + +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: + +$$\begin{aligned} + 0 + &= \pdv{J}{\alpha} \Big|_{\alpha = 0} + = \int \pdv{L}{\alpha} \dd{x} + = \int \pdv{L}{f} \pdv{f}{\alpha} + \sum_{n} \pdv{L}{f^{(n)}} \pdv{f^{(n)}}{\alpha} \dd{x} + \\ + &= \int \pdv{L}{f} \eta + \sum_{n} \pdv{L}{f^{(n)}} \eta^{(n)} \dd{x} +\end{aligned}$$ + +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: + +$$\begin{aligned} + f(x, y; \alpha) = f(x, y; 0) + \alpha \eta(x, y) +\end{aligned}$$ + +The derivation procedure starts in the exact same way as before: + +$$\begin{aligned} + 0 + &= \pdv{J}{\alpha} \Big|_{\alpha = 0} + = \iint \pdv{L}{\alpha} \dd{x} \dd{y} + \\ + &= \iint \pdv{L}{f} \pdv{f}{\alpha} + \pdv{L}{f_x} \pdv{f_x}{\alpha} + \pdv{L}{f_y} \pdv{f_y}{\alpha} \dd{x} \dd{y} + \\ + &= \iint \pdv{L}{f} \eta + \pdv{L}{f_x} \eta_x + \pdv{L}{f_y} \eta_y \dd{x} \dd{y} +\end{aligned}$$ + +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 $x_1, x_2, ..., x_N$: + +$$\begin{aligned} + \boxed{ + 0 = \pdv{L}{f} - \sum_{n} \dv{x_n} \Big( \pdv{L}{f_{x_n}} \Big) + } +\end{aligned}$$ |