# derivation of Euler-Lagrange differential equation (elementary)

Let $[e,c]$ be a finite subinterval of $(a,b)$. Let the function $h:\mathbb{R}\to \mathbb{R}$ be chosen so that a) $h$ is twice differentiable^{}, b) $h(t)=0$ when $t\notin [e,c]$, c) $h(t)>0$ when $t\in (e,c)$, and d) ${\int}_{e}^{c}h(t)\mathit{d}t=1$.

Choose $f(\lambda ,t)=q(t)+\lambda h(t)$. It is easy to see that this function satisfies the requirements for $f$ laid out in the main entry. Then, we can write

$$g(\lambda ,x)={\int}_{a}^{b}L(t,q(t)+\lambda h(t),\dot{q}(t)+\lambda \dot{h}(t))\mathit{d}t$$ |

Let us split the integration into three parts — the integral from $a$ to $e$, the integral from $e$ to $c$, and the integral from $c$ to $b$. By the way $h$ was chosen, the integrand reduces to $L(t,q(t)(t),\dot{q}(t))$ when $t\in (a,e)$ or $t\in (c,b)$. Hence the pieces of the integral over the intervals $(a,e)$ and $(c,b)$ do not depend on $\lambda $ and we have

$$\frac{dg}{d\lambda}=\frac{d}{d\lambda}{\int}_{e}^{c}L(t,q(t)+\lambda h(t),\dot{q}(t)+\lambda \dot{h}(t))\mathit{d}t$$ |

Since $[e,c]$ is closed and bounded, it is compact. By our assumption, the derivative of the integrand is continuous^{}. Since continuous functions on compact sets are uniformly continuous, the derivative of the integrand is uniformly continuous. This imples that it is permissible to interchange differentiation^{} and integration:

$$\frac{dg}{d\lambda}={\int}_{e}^{c}\frac{d}{d\lambda}L(t,q(t)+\lambda h(t),\dot{q}(t)+\lambda \dot{h}(t))\mathit{d}t$$ |

Using the chain rule^{} (several variables) and setting $\lambda =0$, we have

$${\frac{dg}{d\lambda}|}_{\lambda =0}={\int}_{e}^{c}h(t)\frac{\partial L}{\partial q}(t,q(t),\dot{q}(t))+\dot{h}(t)\frac{\partial L}{\partial \dot{q}}(t,q(t),\dot{q}(t))\mathit{\hspace{1em}}dt$$ |

Integrating by parts and using the fact that $h$ was chosen so as to vanish at the endpoints $e$ qnd $c$, we find that

$${\frac{dg}{d\lambda}|}_{\lambda =0}={\int}_{e}^{c}h(t)\left[\frac{\partial L}{\partial q}(t,q(t),\dot{q}(t))-\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}(t,q(t),\dot{q}(t))\right)\right]\mathit{\hspace{1em}}dt={\int}_{e}^{c}h(t)EL(t)\mathit{d}t$$ |

(The last equals sign defines $EL$ as the quantity in the brackets in the first integral.)

I claim that requiring $dg/d\lambda =0$ for all finite intervals $[e,c]\subset (a,b)$ implies that the $EL(t)$ must equal zero for all $t\in [a,b]$. By our assumptions, $EL$ is a continuous function. Hence, for every ${t}_{0}\in (a,b)$ and every $\u03f5>0$, there must exist and $[e,c]\subset (a,b)$ such that ${t}_{0}\in [e,c]$ and ${t}_{1}\in [e,c]$ implies that $$. Therefore,

$$ |

Since this must be true for all $\u03f5>0$, it follows that $EL({t}_{0})=0$ for all ${t}_{0}\in (a,b)$. In other words, q satisfies the Euler-Lagrange equation.

Title | derivation of Euler-Lagrange differential equation^{} (elementary) |
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Canonical name | DerivationOfEulerLagrangeDifferentialEquationelementary |

Date of creation | 2013-03-22 14:45:35 |

Last modified on | 2013-03-22 14:45:35 |

Owner | rspuzio (6075) |

Last modified by | rspuzio (6075) |

Numerical id | 16 |

Author | rspuzio (6075) |

Entry type | Derivation |

Classification | msc 47A60 |