CoV6

From Exampleproblems

Derive the Euler-Lagrange equation from the attempt to minimize the functional

$T(y)=\int_a^b L(y,y',x)\,dx\,$

Let $h\,$ be a function such that $h(a)=0, h(b)=0\,$ .

Consider a small change in the function $y\,$ .

$T(y+\epsilon h) = \int_a^bL(y+\epsilon h, y'+\epsilon h', x)\,dx\,$

In general, the Taylor expansion for a function $g\,$ of three variables is

$g(x+\epsilon_1,y+\epsilon_2,z+\epsilon_3)=g(x,y,z)+\left(\epsilon_1\frac{\partial g}{\partial x} + \epsilon_2\frac{\partial g}{\partial y}+\epsilon_3\frac{\partial g}{\partial z}\right)\bigg|_{(x,y,z)} + ...$

$T(y+\epsilon h) =\int_a^b\left[L(y,y',x)+\frac{\partial L}{\partial y}\epsilon h(x) + \frac{\partial L}{\partial y'}\epsilon h'(x) + ...\right]\,dx\,$

$=T(y)+\epsilon\int_a^b\left(h(x)\frac{\partial L}{\partial y}+h'(x)\frac{\partial L}{\partial y'}\right)\,dx + ...\,$

$\int_a^bh(x)\frac{\partial L}{\partial y}\,dx + \int_a^b h'(x)\frac{\partial L}{\partial y'}\,dx=0\,$

$0=\int_a^bh(x)\frac{\partial L}{\partial y}\,dx + h(x)\frac{\partial L}{\partial y'}\bigg|_a^b - \int_a^bh(x)\frac{d}{dx}\frac{\partial L}{\partial y'}\,dx\,$

The evaluated term is identically zero because of the boundary conditions of $h\,$ .

$0=\int_a^bh(x)\left[\frac{\partial L}{\partial y}-\frac{d}{dx}\frac{\partial L}{\partial y'}\right]\,dx\,\forall h \implies \frac{\partial L}{\partial y}-\frac{d}{dx}\frac{\partial L}{\partial y'}=0\,$ .

Calculus of Variations

Main Page

Retrieved from "http://www.exampleproblems.com/wiki/index.php/CoV6"

Views

Personal tools

Create an account or log in

Search

Toolbox

Argan Oil
Natural Skin Care
Organic Skin Care