1 Euler-Lagrange Equation
In Calculus of Variations and [[newtonian-mechanics]] the Euler-Lagrange equation are a system of second-order Ordinary Differential Equations whose solutions are stationary points of a given action [[functional|functional]]. A differentiable functional is stationary (i.e. the functional derivative is zero ) at its local extrema (minimum, maximum, inflection points) thus, the Euler-Lagrange equation is useful for solving optimization problems in which, given some functional, one seeks the function minimizing or maximizing it. This leads to solving the Euler-Lagrange equation.
In Lagrangian Mechanics, from Hamilton’s principle of stationary action, the evolution of a physical system is described by the solutions to the Euler equation for the action of the system. This is equivalent to Newton’s laws of motion with the added advantage that it takes the same form in any system of generalized coordinates.
1.1 Euler-Lagrange Equation
Suppose that we are given the (action) functional \[ \mathscr{A}[x(t)]=\int_{0}^{T}{\mathcal{L}(\dot{\boldsymbol{x}}, \boldsymbol{x},t)}~d{t} \] where we assume \(\boldsymbol{x}\in C^d\) is a \(d\)-times continuously differentiable function that maps time \(t\in[0,T]\) to \(\mathbb{R}^n\), and \(\mathcal{L}(\dot{\boldsymbol{x}},\boldsymbol{x},t):=\mathbb{R}^n\times \mathbb{R}^n\times [0,T]\mapsto \mathbb{R}\) is the Lagrangian. If we want to find the extremal (maximum, minimum or inflection point) of a classical real value function with finite dimensional domain, we typically calculate the derivatives and find where they equal zero. The following can be thought of intuitively as the infinite dimensional case. Note that in the following theorem that this is a system of \(n\)-equations.
Suppose \(\mathcal{L}\in C^2\) and \(\boldsymbol{x}:[0,T]\to \mathbb{R}^n\). Then if \(\boldsymbol{x}(t)\) is an extremal of \(\mathscr{A}\) we evaluate \[ - \frac{d}{dt}[\nabla_{\dot{x}}\mathcal{L}(\dot{\boldsymbol{x}}, \boldsymbol{x},t)]+\nabla_{x}L(\dot{\boldsymbol{x}}, \boldsymbol{x},t)=0, \] where \(\nabla\) denotes the Gradient.
PROOF: Suppose we have a smooth test function \(\phi(t)\in C^\infty\) where \(\phi(0)=\phi(T)=0\). Because \(\boldsymbol{x}\) is a local extrema of \(\mathscr{A}\) then if we perturb \(\boldsymbol{x}\) by function \(k\phi\) with small \(k\) the resulting functional \(\mathscr{A}(\boldsymbol{x}+k\phi)\) should be roughly unchanged, that is \[ \frac{d}{dk}[\mathscr{A}(\boldsymbol{x}+k\phi)]\Big|_{k=0}=0. \] Computing the derivative by Leibniz Integral Rule for each coordinate \(\boldsymbol{x}\) we have \[ \int_{0}^{T}{\frac{d\mathcal{L}(\dot{\boldsymbol{x}}_{i}+k \dot{\phi}_{i},\boldsymbol{x}_{i}+k\phi_{i},t)}{dk}}~d{t=0;\quad \forall i}. \] COMPLETE THE PROOF!
- Gateaux Derivative
1.1.1 Example: Diving
References: Functionals and Functional Derivatives Suppose we drop a weight suspended from a height. In cartesian coordinates we let the ground be at \(0\) and our weight at \(y\). Assume that the weight has mass \(m=2\) and assume that the gravitational constant \(g=2\). The potential and kinetic energy of the weight is given by \[ \begin{cases} K = mgh = 2 \cdot 2 \cdot y =4y \\ E = \frac{1}{2}mv^2=\dot{y}^2. \end{cases} \]