Introduction to the Calculus of Variations


filed under : #math #math324

$$\newcommand{\lib}[2][]{\frac{\mathrm{d}#1}{\mathrm{d}#2}}$$

This is a supplement for my Math 324 class for the summer of 2018.

Suppose we have a functional $J(q) = \int_a^b L(t, q(t), \dot{q}(t))dt$, where $L$ is twice differentiable called the Lagrangian. Our goal is to find the extrema of $J$. That is we're trying to find continuous real curves on the domain $[a, b]$ that either maximize or minimize $J$. Suppose one of the extrema for $J$ is the curve $q(t)$. To better understand what properties the curve $q(t)$ satisfies we look at families of curves near $q(t)$ called variations. Given a continuous function $\eta(t)$ with $\eta(a) = \eta(b)$ a variation is the family of curves $g_{\epsilon} = q(t) + \epsilon \eta(t)$ parameterized by $\epsilon$ where $\epsilon$ takes on values near zero. Our first goal is to derive the Euler-Lagrange Equation for extrema curves $q(t)$.

The Euler-Lagrange Equation

Let $q(t)$ be an extrema for the functional $J$ defined above. Let $g_{\epsilon}(t) = q(t) + \epsilon\eta(t)$, where $\eta(a) = \eta(b) = 0$ be a variation for $q(t)$, and let $J_{\epsilon} = \int_a^b L(t,g_{\epsilon}(t), \dot{g_{\epsilon}}(t))\mathrm{d}t$. We see that

$$ \lib[J_{\epsilon}]{\epsilon} = \int_a^b \lib[]{\epsilon}L(t,g_{\epsilon}(t), \dot{g_{\epsilon}}(t)\mathrm{d}t. $$
We compute $\lib[L_{\epsilon}]{\epsilon}$ to be
$$ \begin{align*} \lib[L_{\epsilon}]{\epsilon} & = \lib[t]{\epsilon}\frac{\partial{L_{\epsilon}}}{\partial \epsilon} + \lib[g_{\epsilon}]{\epsilon}\frac{\partial{L_{\epsilon}}}{\partial g_\epsilon} + \lib[\dot{g}_{\epsilon}]{\epsilon} \frac{\partial{L_{\epsilon}}}{\partial \dot{g}_\epsilon}\\ & = \eta(t)\frac{\partial{L_{\epsilon}}}{\partial g_\epsilon} + \dot{\eta}(t) \frac{\partial{L_{\epsilon}}}{\partial \dot{g}_\epsilon}. \end{align*} $$
Since $\epsilon = 0$ is an extremim value for $J_{\epsilon}$, we have
$$ 0 = \lib[J_{\epsilon}]{\epsilon}{\bigg|}_{\epsilon = 0} = \int_a^b \eta(t)\frac{\partial{L}}{\partial{q}} + \dot{\eta}(t)\frac{\partial{L}}{\partial{\dot{q}}}\mathrm{d}t. $$
Recall that $\eta(a) = \eta(b) = 0$, so by integration by parts on the left side of the previous equation we get
$$ \lib[J_{\epsilon}]{\epsilon}{\bigg|}_{\epsilon = 0} = \int_a^b\left[\frac{\partial{L}}{\partial{q}} - \lib[]{t}\frac{\partial L}{\partial{\dot{q}}}\right]\eta(t)\mathrm{d}t. $$

Before we can move on, we need the following fundamental result.

Lemma. Suppose $f(t)$ is continuous on $[a,b]$ and
$$ \int_a^b \eta(t)f(t)\mathrm{d}t = 0 $$
for all continuous functions $\eta(t)$ which vanish at $a$ and $b$, then $f(t) = 0$ for all $t \in [a,b]$.
Proof. Without loss of generality, let's assume $f(c) > 0$ for some $c \in [a,b]$. Since $f$ is continuous on $[a,b]$, there exists $t_1$ and $t_2$ in $[a,b]$ such that $t_1 < c < t_2$ and $f(t)$ is strictly positive on $[t_1, t_2]$. Define $\eta(t)$ to be
$$ \eta(t) = \left\{\begin{array}{ll} (t-t_1)(t_2-t) & \mbox{for } t \in [t_1,t_2] \\ 0 & \mbox{otherwise} \end{array}\right.. $$
Now notice that $f(t)\eta(t) \ge 0$ for $t \in [a,b]$ and $f(t)\eta(t) > 0$ for $t \in [t_1, t_2]$. It directly follows that
$$ \int_a^b \eta(t)f(t)\mathrm{d}t = \int_{t_1}^{t_2}\eta(t)f(t)\mathrm{d}t > 0, $$
which is a contradiction. Therefore $f(t) = 0$ for $t \in [a,b]$

By using this lemma, we see that for $\lib[J_{\epsilon}]{\epsilon}{\bigg|}_{\epsilon = 0}$ to be zero, it must be the case that

$$ \frac{\partial{L}}{\partial{q}} - \lib[]{t}\frac{\partial L}{\partial{\dot{q}}} = 0. $$
This equation is called the Euler-Lagrange Equation. Let's see an example of how to use it.

Example 1. Suppose we want to minimize the length of the path joining two points. Of course, you probably can guess that it's going to be a straight line, but let 's see if we can derive this by using the Euler-Lagrange equation. Let's name the two points $(t_1,y_1)$ and $(t_2, y_2)$. We need to minimize the arclength function:
$$ s(q) = \int_{t_1}^{t_2}\sqrt{1+(\dot{q}(t))^2}\mathrm{d}t, $$
where we consider all continuous functions $q(t)$ with $q(t_1) = y_1$ and $q(t_2) = y_2$. Let's take $L(t,q,\dot{q})$ to be
$$ L(t,q,\dot{q}) = \sqrt{1+(\dot{q})^2}. $$
Since $\frac{\partial{L}}{\partial{q}} = 0$, we just have to solve for $q$ in
$$ \lib[]{t}\frac{\partial L}{\partial{\dot{q}}} = 0. $$
Plugging $L$ in, we get
$$ \lib[]{t}\left(\frac{\dot{q}}{\sqrt{1+\dot{q}^2}}\right) = 0. $$
Integrating both sides with respect to $t$, we get
$$ \frac{\dot{q}}{\sqrt{1+\dot{q}^2}} = C, $$
where $C$ is some constant. Solving for $\dot{q}$ we arrive at
$$ \dot{q} = \frac{C}{\sqrt{1-C^2}}. $$
Replacing the right hand side with $A$, and doing integration one more time we conclude that $q(t) = At + B$, where $A$ and $B$ are constants. Recall that $A$ and $B$ can be determined by the fact that $q(t_1) = y_1$ and $q(t_2) = y_2$ and this concludes the exercise.

The Beltrami Identity

Before we can do another example we need another equation called the Beltrami Identity. This equation is derived directly from the Euler-Lagrange equation by only assuming that $\frac{\partial{L}}{\partial{t}} = 0$, so let's jump right in!

Let's start off with the total derivative of $L$:

$$ \lib[L]{t} = \frac{\partial{L}}{\partial{q}}\dot{q} + \frac{\partial{L}}{\partial{\dot{q}}}\ddot{q}. $$
Remember that we assumed $\frac{\partial{L}}{\partial{t}} = 0$. Now using the Euler-Lagrange equation $\frac{\partial L}{\partial q} = \lib[]{t}\frac{\partial L}{\partial \dot{q}}$ observe
$$ \begin{align*}0 & = \lib[L]{t} - \frac{\partial{L}}{\partial{q}}\dot{q} - \frac{\partial{L}}{\partial{\dot{q}}}\ddot{q} \\ & = \lib[L]{t} - \lib[]{t}\frac{\partial{L}}{\partial{\dot{q}}}\dot{q} - \frac{\partial{L}}{\partial{\dot{q}}}\ddot{q} = \lib[]{t}\left(L - \dot{q}\frac{\partial{L}}{\partial{\dot{q}}}\right).\end{align*} $$
Therefore, we get by integration the Beltrami Identity:
$$ L - \dot{q}\frac{\partial{L}}{\partial{\dot{q}}} = C, $$
where $C$ is a constant.

Example 2. Suppose we want to minimize the surface area of a surface of revolution. For the set up let's have $q(t_1) = y_1$ and $q(t_2) = y_2$, and assume $q$ is rotated around the $t$-axis. We need to minimize the function
$$ A(q) = \int_{t_1}^{t_2} 2\pi q \sqrt{1+ (\dot{q})^2}\mathrm{d}t. $$
Therefore, we take $L(t,q, \dot{q})$ to be $2\pi q \sqrt{1+(\dot{q})^2}$. Using the Beltrami Identity we have
$$ 2\pi q\sqrt{1+(\dot{q})^2} - \dot{q}\frac{2\pi q \dot{q}}{\sqrt{1+(\dot{q})^2}} = C. $$
Solving for $\dot{q}$, we have
$$ \dot{q} = \sqrt{\frac{q^2 - C^2}{C^2}}. $$
We get
$$ \begin{align*}t & = \int\lib[t]{q}\mathrm{d}q = \int\frac{1}{\dot{q}}\mathrm{d}q \\ & = C\int \frac{1}{\sqrt{q^2-C^2}}\mathrm{d}q = C \cosh^{-1}\left(\frac{q}{C}\right) + B.\end{align*} $$
Therefore, $q = C\cosh\left(\frac{x-B}{C}\right)$, where $B$ and $C$ are constants that can be found by the restrictions on $q$ from the set up.