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The Jacobi accessory equation has importance as a means of checking candidates for functional extrema. A book of mine ($\textit{Calculus of variations}$, by van Brunt) proves that we can find solutions to the Jacobi accessory equation by differentiating the general solution to the Euler-Lagrange equation; that is, if the latter has a general solution $y$ involving parameters $c_1, c_2$, then the functions $$u_1(x) = \frac{\partial y}{\partial c_1}, \quad u_2(x) = \frac{\partial y}{\partial c_2}$$ evaluated at some particular $(c_1, c_2)$ are solutions to the Jacobi accessory equation (given basic smoothness assumptions). However, van Brunt goes on to claim without proof that $u_1, u_2$ $\textit{form a basis for the solution space}$. Can anyone suggest how this might be proved?

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  • $\begingroup$ What is the Jacobi accessory equation? Is it the same as the Hamilton-Jacobi equation? $\endgroup$ – Raskolnikov Jun 1 '13 at 8:59
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Short answer: Because there are only two parameters involved.

Long answer: back to the derivation of the accessory equation. Consider the Euler-Lagrange equation for $$J[y] = \int^b_a F(x,y,y') \,dx,$$ where $y = y(x;c_1,c_2)$ is the minimizer, then $$\newcommand{\e}{\epsilon}g(\e) := J[y+\e v] = \int^b_a F(x,y+\e v,y'+\e v') \,dx.$$ Doing Taylor expansion at $\e= 0$: $$ g(\e) = g(0) + g'(0)\e + \frac{1}{2}g''(0)\e^2 + O(|\e|^3), $$ where $g(0)$ is just $J[y]$, $g'(0) = 0$ for the minimizer, now $$ g''(0) = \frac{1}{2}\int^b_a \left(v^2 \frac{\partial^2 F }{\partial y^2} + 2vv' \frac{\partial^2 F }{\partial y \partial y'}+ v'^2 \frac{\partial^2 F }{\partial y'^2}\right)\,dx, \tag{1} $$ and $$ J[y+\e v] - J[y] = \frac{1}{2}\e^2\big(g''(0)+ O(|\e|)\big). $$ Therefore for $y$ being the minimizer of the functional, $g''(0)$ must be greater than or equation to $0$. Now we fix this minimizer $y$, denote $$ f(x,v,v') := v^2 \frac{\partial^2 F }{\partial y^2} + 2vv' \frac{\partial^2 F }{\partial y \partial y'}+ v'^2 \frac{\partial^2 F }{\partial y'^2}, $$ and the Euler-Lagrange equation for (1) is $$ \frac{\partial f}{\partial v}-\frac{d}{dx}\frac{\partial f}{\partial v'}=0. $$ Simplifying leads to: $$ \frac{\partial^2 F }{\partial y'^2} v'' + \frac{d}{dx}\frac{\partial^2 F }{\partial y'^2} v' + \left(\frac{d}{dx}\frac{\partial^2 F }{\partial y \partial y'} -\frac{\partial^2 F }{\partial y^2} \right) v = 0. \tag{2} $$ Now this is a second order linear ODE in $v$ for a specific $y$. If the smoothness is assumed, by existence and uniqueness theorem for second order linear ode, the solution space is two dimensional, i.e., consists only two basis. If the author found two functions satisfying this equation, also these two functions are kind of easy to tell that they are not linearly dependent, then he could simply claimed they form a basis for the solution space. Say $u_1$ and $u_2$ are linearly dependent, for every $x$ we have the same $k$: $$ k\frac{\partial y}{\partial c_1} = \frac{\partial y}{\partial c_2}, $$ then $y = g(c_1 + kc_2)$, which says it is really a one parameter minimizer, not 2 parameters.

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