Jacobi's Theorem

From ProofWiki
Jump to navigation Jump to search


This article needs to be linked to other articles.
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding these links.
To discuss this page in more detail, feel free to use the talk page.
When this work has been completed, you may remove this instance of {{MissingLinks}} from the code.


Theorem

Let $\mathbf y = \sequence {y_i}_{1 \le i \le n}$, $\boldsymbol \alpha = \sequence {\alpha_i}_{1 \le i \le n}$, $\boldsymbol \beta = \sequence {\beta_i}_{1 \le i \le n}$ be vectors, where $\alpha_i$ and $ \beta_i$ are parameters.

Let $S = \map S {x, \mathbf y, \boldsymbol \alpha}$ be a a complete solution of the Hamilton-Jacobi equation.


Let:

$\begin {vmatrix} \dfrac {\partial^2 S} {\partial \alpha_i \partial y_k} \end{vmatrix} \ne 0$

where $\begin {vmatrix} \cdot \end{vmatrix}$ is a determinant.


Let:

$\dfrac {\partial S} {\partial \alpha_i} = \beta_i$


Then:

$p_i = \map {\dfrac {\partial S} {\partial y_i} } {x, \mathbf y, \boldsymbol \alpha}$
$y_i = \map {y_i} {x, \boldsymbol \alpha, \boldsymbol \beta}$

constitute a general solution of the canonical Euler's equations.


Proof 1

Consider the total derivative of $\dfrac {\partial S} {\partial \alpha_i}$ with respect to $x$:

\(\ds \frac \d {\d x} \frac {\partial S} {\partial \alpha_i}\) \(=\) \(\ds \frac {\partial^2 S} {\partial x \partial \alpha_i} + \sum_{j \mathop = 1}^n \frac {\partial^2 S} {\partial y_j \partial \alpha_i} \frac {\d y_j} {\d x} + \sum_{j \mathop = 1}^n \frac {\partial^2 S} {\partial \alpha_j \partial \alpha_i} \frac {\d \alpha_j} {\d x}\)
\(\ds \) \(=\) \(\ds \frac {\partial^2 S} {\partial x \partial \alpha_i} + \sum_{j \mathop = 1}^n \frac {\partial^2 S} {\partial y_j \partial \alpha_i} \frac {\d y_j} {\d x}\) $\alpha_j$ is a parameter, independent of $x$
\(\ds \) \(=\) \(\ds -\frac {\partial H} {\partial \alpha_i} + \sum_{j \mathop = 1}^n \frac {\partial^2 S} {\partial y_j \partial \alpha_i} \frac {\d y_j} {\d x}\) $S$ is a Solution to Hamilton-Jacobi Equation
\(\ds \) \(=\) \(\ds -\sum_{j \mathop = 1}^n \frac {\partial H} {\partial p_j} \frac {\partial p_j} {\partial \alpha_i} + \sum_{j \mathop = 1}^n \frac {\partial^2 S} {\partial y_j \partial \alpha_i} \frac {\d y_j} {\d x}\)
\(\ds \) \(=\) \(\ds -\sum_{j \mathop = 1}^n \frac {\partial H} {\partial p_j} \frac {\partial^2 S} {\partial \alpha_i \partial y_j} + \sum_{j \mathop = 1}^n \frac {\partial^2 S} {\partial y_j \partial \alpha_i} \frac {\d y_j} {\d x}\) Derivation of Hamilton-Jacobi Equation
\(\ds \) \(=\) \(\ds \sum_{j \mathop = 1}^n \map {\frac {\partial^2 S} {\partial x \partial \alpha_i} } {\frac {\d y_j} {\d x} - \frac {\partial H} {\partial p_j} }\)
\(\ds \) \(=\) \(\ds 0\) as $\dfrac {\d \beta_i} {\d x} = 0$
\(\ds \leadsto \ \ \) \(\ds \frac {\d y_j} {\d x}\) \(=\) \(\ds \frac {\partial H} {\partial p_j}\)


Next, consider the total derivative of $p_i$ with respect to $x$:

\(\ds \frac {\d p_i} {\d x}\) \(=\) \(\ds \frac \d {\d x} \frac {\partial S} {\partial y_i}\)
\(\ds \) \(=\) \(\ds \frac {\partial^2 S} {\partial x \partial y_i} + \sum_{j \mathop = 1}^n \frac {\partial^2 S} {\partial y_j \partial y_i} \frac {\d y_j} {\d x} + \sum_{j \mathop = 1}^n \frac {\partial^2 S} {\partial \alpha_j \partial y_i} \frac {\d \alpha_j} {\d x}\)
\(\ds \) \(=\) \(\ds \frac {\partial^2 S} {\partial x \partial y_i} + \sum_{j \mathop = 1}^n \frac {\partial^2 S} {\partial y_j \partial y_i} \frac {\d y_j} {\d x}\) $\alpha_j$ is a parameter, independent of $x$
\(\ds \) \(=\) \(\ds \frac {\partial^2 S} {\partial x \partial y_i} + \sum_{j \mathop = 1}^n \frac {\partial^2 S} {\partial y_j \partial y_i} \frac {\partial H} {\partial p_j}\) as $\dfrac {\d y_j} {\d x} = \dfrac {\partial H} {\partial p_j}$


On the other hand, the partial derivative of Hamilton-Jacobi equation yields:

\(\ds \frac {\partial^2 S} {\partial x \partial y_i}\) \(=\) \(\ds -\frac {\partial H} {\partial y_i} - \sum_{j \mathop = 1}^n \frac {\partial H} {\partial p_j} \frac {\partial p_j} {\partial y_i}\)
\(\ds \) \(=\) \(\ds -\frac {\partial H} {\partial y_i} - \sum_{j \mathop = 1}^n \frac {\partial H} {\partial p_j} \frac {\partial^2 S} {\partial y_i \partial y_j}\) Derivation of Hamilton-Jacobi Equation
\(\ds \leadsto \ \ \) \(\ds -\frac {\partial H} {\partial y_i}\) \(=\) \(\ds \frac {\partial^2 S} {\partial x \partial y_i} + \sum_{j \mathop = 1}^n \frac {\partial^2 S} {\partial y_i \partial y_j} \frac {\partial H} {\partial p_j}\)

By comparison of this and previous expressions:

$\dfrac {\d p_i} {\d x} = -\dfrac {\partial H} {\partial y_i}$

$\blacksquare$


Proof 2

Consider canonical Euler's equations:

$\dfrac {\d y_i} {\d x} = \dfrac {\partial H} {\partial p_i}$
$\dfrac {\d p_i} {\d x} = -\dfrac {\partial H} {\partial y_i}$

Apply a canonical transformation:

$\tuple {x, \mathbf y, \mathbf p, H} \to \tuple {x, \boldsymbol \alpha, \boldsymbol \beta, H^*}$

where $\Phi = S$.

By Conditions for Transformation to be Canonical:

$p_i = \dfrac {\partial S} {\partial y_i}$
$\beta_i = \dfrac {\partial S} {\partial \alpha_i}$
$H^* = H + \dfrac {\partial S} {\partial x}$

Because $S$ is a solution to the Hamilton-Jacobi equation:

$H^* = 0$

In these new coordinates canonical Euler's equations are:

$\dfrac {\d\alpha_i} {\d x} = \dfrac {\partial H^*} {\partial \beta_i}$
$\dfrac {\d \beta_i} {\d x} = -\dfrac {\partial H^*} {\partial \alpha_i}$

By $H^* = 0$:

$\dfrac {\d \alpha_i} {\d x} = 0$
$\dfrac {\d\beta_i} {\d x} = 0$

which imply that $ \alpha_i$ and $\beta_i$ are constant along each extremal.


This article, or a section of it, needs explaining.
In particular: the meaning of "extremal"
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it.
To discuss this page in more detail, feel free to use the talk page.
When this work has been completed, you may remove this instance of {{Explain}} from the code.


$\beta_i$ constancy provides with $n$ first integrals:

$\dfrac {\partial S} {\partial \alpha_i} = \beta_i$

Because $S = \map S {x, \mathbf y, \boldsymbol \alpha}$, the aforementioned set of first integrals is also a system of equations for functions $y_i$.

Thus, functions $y_i$ can be found.

Functions $p_i$ are found by the results of Conditions for Transformation to be Canonical:

$p_i = \dfrac {\partial} {\partial y_i} \map S {x, \mathbf y, \boldsymbol \alpha}$

Then:

$\map {y_i} {x, \boldsymbol \alpha, \boldsymbol \beta}$
$\map {p_i} {x, \boldsymbol \alpha, \boldsymbol \beta}$

are solutions to canonical Euler's equations.

$\blacksquare$


Source of Name

This entry was named for Carl Gustav Jacob Jacobi.


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Jacobi%27s_Theorem&oldid=446117"