# derivation of Euler-Lagrange differential equation (elementary)

Let $[e,c]$ be a finite subinterval of $(a,b)$. Let the function $h\colon\mathbb{R}\to\mathbb{R}$ 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) $\int_{e}^{c}h(t)\,dt=1$.

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

 $g(\lambda,x)=\int_{a}^{b}L(t,q(t)+\lambda h(t),\dot{q}(t)+\lambda\dot{h}(t))\,dt$

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

 ${dg\over d\lambda}={d\over d\lambda}\int_{e}^{c}L(t,q(t)+\lambda h(t),\dot{q}(% t)+\lambda\dot{h}(t))\,dt$

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:

 ${dg\over d\lambda}=\int_{e}^{c}{d\over d\lambda}L(t,q(t)+\lambda h(t),\dot{q}(% t)+\lambda\dot{h}(t))\,dt$

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

 ${dg\over d\lambda}\bigg{|}_{\lambda=0}=\int_{e}^{c}h(t){\partial L\over% \partial q}(t,q(t),\dot{q}(t))+\dot{h}(t)\frac{\partial L}{\partial\dot{q}}(t,% q(t),\dot{q}(t))\quad dt$

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

 ${dg\over d\lambda}\bigg{|}_{\lambda=0}=\int_{e}^{c}h(t)\left[{\partial L\over% \partial q}(t,q(t),\dot{q}(t))-{d\over dt}\left(\frac{\partial L}{\partial\dot% {q}}(t,q(t),\dot{q}(t))\right)\right]\quad dt=\int_{e}^{c}h(t)EL(t)\,dt$

(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 $\epsilon>0$, there must exist and $[e,c]\subset(a,b)$ such that $t_{0}\in[e,c]$ and $t_{1}\in[e,c]$ implies that $|EL(t_{0})-EL(t_{1})|<\epsilon$. Therefore,

 $\left|\quad{dg\over d\lambda}\bigg{|}_{\lambda=0}-EL(t_{0})\right|=\left|\int_% {e}^{c}h(t)\left(EL(t)-EL(t_{0})\right)\,dt\right|<\epsilon\left|\int_{e}^{c}h% (t)\,dt\right|=\epsilon$

Since this must be true for all $\epsilon>0$, 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) DerivationOfEulerLagrangeDifferentialEquationelementary 2013-03-22 14:45:35 2013-03-22 14:45:35 rspuzio (6075) rspuzio (6075) 16 rspuzio (6075) Derivation msc 47A60