Calculus of Variation

Calculus of variation aims to determine a function that minimises/maximises a value: distance, time, energy etc.

Introduction

A Functional accepts a function to produce a scalar. For eg. Distance functional $I [y]$ calculates the distance traversed by the function $y$ from $x_{1}$ to $x_{2}$ . By setting the boundary condition of $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ we can get a finite value from the Functional.

\begin{aligned} I [y] & = \int_{x_{1}}^{x_{2}} \sqrt{(d x^{2} + d y^{2})} \\ = \int_{x_{1}}^{x_{2}} \sqrt{(1 + {(\frac{d y}{d x})}^{2})} d x \end{aligned}

It adds up small arc length $d s$ along $y (x)$ . Thus computing the distance effectively.

This particular integrand (Function being integrated) $\sqrt{(1 + {(\frac{d y}{d x})}^{2})}$ requires only the derivative of the function, but other types of Integrant might use $x$ , $y$ or $y^{'}$ . Thus we can make a general integrant given as $F$ below.

Suppose $y = f (x)$ is an extremal (minimises/maximises Functional $I$ ). The boundary condition is given as $y (x_{1}) = y_{1}$ and $y (x_{2}) = y_{2}$ .

I [y] = \int_{x_{1}}^{x_{2}} F (x, y, y^{'}) d x

Suppose we slightly perturb $y$ with an arbitrary function $η$ to get $\bar{y}$ , to ensure it satisfies a boundary condition we set $η (x_{1}) = 0$ and $η (x_{2}) = 0$ . Since $y$ is an extremal. The perturbation will cause a small difference to the Functional $I [\bar{y}]$ .

\bar{y} (x) = y (x) + ϵ \cdot η

If $y$ is truly extremal, then $I$ should be at the minimum/maximum when $ϵ = 0$ .

{\frac{d I [\bar{y}]}{d ϵ} |}_{ϵ = 0} = 0

Thus

\begin{aligned} {\frac{d}{d ϵ} \int_{x_{1}}^{x_{2}} F (x, \bar{y}, {\bar{y}}^{'}) d x |}_{ϵ = 0} & = 0 \\ {\int_{x_{1}}^{x_{2}} \frac{\partial}{\partial ϵ} F (x, \bar{y}, {\bar{y}}^{'}) |}_{ϵ = 0} d x & = 0 \\ \int_{x_{1}}^{x_{2}} {[\frac{\partial F}{\partial \bar{y}} \cdot \frac{\partial \bar{y}}{\partial ϵ} + \frac{\partial F}{\partial {\bar{y}}^{'}} \cdot \frac{\partial {\bar{y}}^{'}}{\partial ϵ}]}_{ϵ = 0} d x & = 0 \\ \int_{x_{1}}^{x_{2}} {[\frac{\partial F}{\partial \bar{y}} \cdot η + \frac{\partial F}{\partial {\bar{y}}^{'}} \cdot η^{'}]}_{ϵ = 0} d x & = 0 \end{aligned}

Next we evaluate the partial derivatives at $ϵ = 0$ thus replacing $\bar{y}$ with $y$ .

\int_{x_{1}}^{x_{2}} [\frac{\partial F}{\partial y} \cdot η + \frac{\partial F}{\partial y^{'}} \cdot η^{'}] d x = 0

Using integration by parts on the second segment of the equation. Let $u = \frac{\partial F}{\partial {\bar{y}}^{'}}$ and $v^{'} = η^{'}$ .

\int u v^{'} = u v - \int u^{'} v

Since $η (x_{2}) = η (x_{1}) = 0$ .

\begin{aligned} \int_{x_{1}}^{x_{2}} \frac{\partial F}{\partial y^{'}} \cdot η^{'} d x & = \frac{\partial F}{\partial \bar{y}} \cdot {[η]}_{x_{1}}^{x_{2}} - \int \frac{d}{d x} (\frac{\partial F}{\partial y}) \cdot η \\ = 0 - \int \frac{d}{d x} (\frac{\partial F}{\partial y}) \cdot η \end{aligned}

Thus plugging it back into the above equation.

\begin{aligned} \int_{x_{1}}^{x_{2}} [\frac{\partial F}{\partial y} \cdot η + \frac{\partial F}{\partial y^{'}} \cdot η^{'}] d x & = 0 \\ \int_{x_{1}}^{x_{2}} [\frac{\partial F}{\partial y} \cdot η - \frac{d}{d x} \frac{\partial F}{\partial y^{'}} \cdot η] d x & = 0 \\ \int_{x_{1}}^{x_{2}} [\frac{\partial F}{\partial y} - \frac{d}{d x} \frac{\partial F}{\partial y^{'}}] \cdot η d x & = 0 \end{aligned}

Since $η$ is an arbitrary function (with fixed boundary points). The only way for the integral to be 0. Is if

\frac{\partial F}{\partial y} - \frac{d}{d x} (\frac{\partial F}{\partial y^{'}}) = 0

This is the most famous Euler Lagrange Equation.