Euler-Lagrange Equation

Calculus of Variations


Introduction to Variational Calculus

Post under construction

Previous: The Euler-Lagrance Equations are well known from Analitic Mechanics and Minimum Action Principle, a very elegant alternative to Newton formulation of Mechanics. Here is exposed just as a mathematical tool within Calculus of Variations branch of Mathematical Analysis.

An important preliminar result

Given to real numbers a and b with \(a<b \in R\) and a continous function \(g(x): [a,b]\rightarrow \mathbb{R}\). If, \[\int_{a}^{b}\eta(x)g(x)dx=0 \qquad \forall \eta(x)\in \mathcal{C}^{2} \,, with\,, \eta(a)=\eta(b)=0\]

then, \[g(x)=0 \qquad \forall x\in [a,b]\]

This lemma is what is needed to demostrate that the Euler-Lagrange Equations are the solution to this other problem:

Theorem of Euler-Lagrange

Given a function \(f:[a,b]\rightarrow\mathbb{R}\) twice differentiable, so \(f(x)\in \mathcal{C}^{2}\), with prescribed boundary values \(f(a)=f_{a}\) and \(f(b)=f_{b}\) and a twice differentiable function \(g(x,f,f')\) on all its arguments. We define an integral as, \[I[x,f] = \int_{a}^{b}g(x,f,f')dx\]

Then, if \(I[x,f]\) have a minimum value for the function \(f\), then the function \(f\) satisfaces the following differential equation, \[\frac{\partial g}{\partial f} - \frac{d}{dx}\left( \frac{\partial g}{\partial f'} \right)=0\]

From a partical point of view the derivatives of the \(g\) function with respect to \(f\) and \(f'\) are done with the usual derivation rules.

This equation is well know by all the students who know Analytical Mechanics. In fact this result can be extended (both the Lemma and the Theorem) to a problem with several functions then,

Given a set of functions \(f_{i}:[a,b]\rightarrow\mathbb{R}\) twice differentiable, so \(f_{i}(x)\in \mathcal{C}^{2}\), with prescribed boundary values \(f_{i}(a)=f_{i,a}\) and \(f_{i}(b)=f_{i,b}\) and a twice differential function \(g(x,f_{1},...,f_{k},f'_{1},...,f'_{k})\) \[I[x,f_{1},...,f_{k},f'_{1},...,f'_{k}] = \int_{a}^{b}g(x,f_{1},...,f_{k},f'_{1},...,f'_{k})dx\]

Then, if \(I[x,f_{1},...,f_{k},f'_{1},...,f'_{k}]\) have a minimum value for the set function \(f_{i}\), then these function \(f_{i}\) satisfaces the following system of differential equations, \[\frac{\partial g}{\partial f_{1}} - \frac{d}{dx}\left( \frac{\partial g}{\partial f'_{1}} \right)=0\] \[\frac{\partial g}{\partial f_{2}} - \frac{d}{dx}\left( \frac{\partial g}{\partial f'_{2}} \right)=0\] \[...\] \[\frac{\partial g}{\partial f_{k}} - \frac{d}{dx}\left( \frac{\partial g}{\partial f'_{k}} \right)=0\]

The Isoperimetric problem

The previous result can be also extended to cases in which the function \(f(x)\) have to satisface an additional constrain like, \[C[x,f] = \int_{a}^{b}h(x,f,f')dx\]

where the function \(h\) and the value \(C\) are both known. This problem can be resolved given the Theorem of Euler-Lagrange and the method of Lagrange Multipliers.