Euler-Lagrange Equation

Calculus of variations attempts to solve the shortest path problem problem in a form of the following integral,

\[ \begin{equation} S = \int_{x_1}^{x_2}{f\left[y(x),y'(x),x\right] dx} \end{equation} \]

Generally,

\[\begin{split} \begin{align} S &= \int_{x_1}^{x_2}{f\left[u_1(v),u_1'(v),u_2(v),u_2'(v)...,v\right] dx}\\ S &= \int_{x_1}^{x_2}{f\left[\boldsymbol{u}(v),\boldsymbol{u}'(v),v\right] dx} \end{align} \end{split}\]

What we wish to find is \(y(x)\) which is a curve that minimizes \(S\).

Any function that satisfies the Euler-Lagrange equation (or shortly Lagrange equations) can construct the proper \(y(x)\). The Lagrange equation is written as,

\[\begin{align} \boxed{\frac{\partial f}{\partial y} - \frac{d}{dx}\frac{\partial f}{\partial y'} = 0} \end{align}\]

Proof

The method to find \(y(x)\) is weird because we are going to find the wrong answer/curve first. The wrong curve is in some form

\[\begin{align} Y'(x) = y(x) + \alpha\eta(x) \end{align}\]

• \(\eta(x)\) : some deviation from the correct curve \(y(x)\)

• \(\alpha\) : a coefficient of the deviation

You can see that the correct answer occurs when \(\alpha = 0\). We may also write \(S\) as a function with parameter \(\alpha\) such that \(S(\alpha)\) is minimum at \(S(0)\). The condition for a minimum, (more correct, an extrema) is \(dS/d\alpha=0\),

\[\begin{split} \begin{align} S(\alpha) &= \int_{x_1}^{x_2}{f(y+\alpha\eta, y'+\alpha\eta',x) \ dx} \\ \frac{dS}{d\alpha} &= \int_{x_1}^{x_2}{\frac{\partial f}{\partial \alpha} \ dx} \\ \frac{df}{d\alpha} &= \eta\frac{\partial f}{\partial y} + n'\frac{\partial f}{\partial y'}\nonumber\\ \frac{dS}{d\alpha} &= \int_{x_1}^{x_2}{\left(\eta\frac{\partial f}{\partial y} + n'\frac{\partial f}{\partial y'}\right) \ dx} = 0\\ \end{align} \end{split}\]

For the last line to be true, this must be true by integration by parts,

\[ \begin{align*} \int_{x_1}^{x_2}{\eta(x)\left(\frac{\partial f}{\partial y} - \frac{d}{dx}\frac{\partial f}{\partial y'}\right) \ dx} = 0 \end{align*} \]

Thus it is necessary for any arbitrary \(\eta(x) \ne 0\),

\[ \begin{equation} \boxed{\frac{\partial f}{\partial y} - \frac{d}{dx}\frac{\partial f}{\partial y'} = 0} \end{equation}\]

Properties

The two properties of the Lagrange equation are theorems that makes solving for the Lagrange equation much simpler,

**Theorem 1: ** : If \(f[x(t),x'(t),t]\) is independent on any of its zeroth order derivative parameters \(x\) such that,

$$ \frac{\partial f}{\partial x} = 0$$

: then its derivative along say \(x'\) is a constant \(C\),

$$ \boxed{\frac{\partial f}{\partial x'} = C} $$

: This theorem applies to all functions that satisfies the Lagrange equations.

**Theorem 2: ** : If \(f[x(t),x'(t),t]\) is independent on it’s free parameter \(t\) such that,

$$ \frac{df}{dt} = 0 $$

: then the rest of the Lagrange equation follow for some constant \(C\),

$$ \boxed{f - x'\frac{\partial f}{\partial x'} = C} $$

Example: Shortest Path

The shortest path between two points is,

\[ \begin{align}\begin{aligned} \begin{align} S &= \int_{x_1}^{x_2}{\sqrt{1+(y')^2}\ dx} \end{align} $$,\\Where $f \left(y, y', x\right) = \sqrt{1+(y')^2}$. Let's solve for the Lagrange equation,\end{aligned}\end{align} \]

\begin{gather} \frac{\partial f}{\partial y} = 0, \quad \frac{\partial f}{\partial y’} = \frac{y’}{(1+y’^2)^{1/2}} \nonumber\ \frac{y’}{(1+y’^2)^{1/2}} = C\ \end{gather} $\( For some constant \)C$.

\[\begin{split} \begin{align} y'^2 &= C^2(1+(y')^2) \nonumber\\ y' &= \text{const} \end{align} \end{split}\]

Let’s set the constant to be \(m\) such that \(y' = m\). We can integrate \(y'(x)\) to get \(y(x)\)

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\[ \begin{equation} \boxed{y(x) = mx + y_0} \end{equation} \]

Now you see that the shortest path is a straight line!

Multivariable

The Lagrange equation solves the integral,

\[\begin{split} \begin{gather} S = \int_{u_1}^{u_2}{f\left[x_1, x_2,...,x_1'(u), x_2'(u),... u\right] \ du} \\ \frac{\partial f}{\partial x_i} - \frac{d}{du}\frac{\partial f}{\partial x_i'} = 0 \\ \end{gather} \end{split}\]