stationary action exercises done

materials/notes.ipynb

@@ -147,7 +147,7 @@
    "source": [
     "t, v0, t1, t2, g, m, e = symbols('t, v0, t1, t2, g, m, e')\n",
     "\n",
-    "    a_t = v0*t - 0.5 * g * t**2\n",
+    "a_t = v0*t - 0.5 * g * t**2\n",
     "\n",
     "def constants(expr):\n",
     "    return expr.subs(v0, 50.0).subs(g, 10.0).subs(m, 1.0)\n",

materials/notes.md

@@ -1,74 +0,0 @@
-# Principle of Least Action
-
-from Chapter 14 of Eriksson, Kenneth, Donald Estep, and Claes Johnson. 2004. *Applied Mathematics: Body and Soul*. Volume 2. Berlin, Heidelberg: Springer Berlin Heidelberg. <https://doi.org/10.1007/978-3-662-05798-8>.
-
-The principle of least action states that a system's trajectory $u(t)$ over an interval of time $[t_1, t_2]$ is such that the following integral, called the action integral, remains constant:
-$$
-I(u)=\int_{t_1}^{t_2} T(\dot{u}(t)) - V(u(t)) dt
-$$
-where $T(\dot{u}(t))$ is the kinetic energy and $V(u(t))$ is the potential energy.
-
-**Example 1.** Consider a ball that gets throw
-
-n into the air, under the influence of gravity. It's trajectory $u: [t_1, t_2] \to \mathcal{R}$ is a parabola between times $t_1$ and $t_2$, with $u(t_1) = u_1$ denoting the initial position, and $u(t_2) = u_2$ denoting the final position of the ball.
-The kinetic energy of the ball is  $T(\dot{u}(t)) = \frac{1}{2}m\dot{u}(t)^2$ and the potential energy is $V(u(t)) = mgu(t)$.
-
-We now explain what is meant with requiring that the integral $I(u)$ is stationary. Consider a possible trajectory  $u: [t_1, t_2] \to \mathcal{R}$ and consider its perturbed version $v: [t_1, t_2] \to \mathcal{R}$, given by $v(t) = u(t) + \epsilon w(t)$, or written equivalently, $v(t)=(u + \epsilon w)(t)$ where $w(t)$ is such that $w(t_1)=w(t_2)=0$, so that $v(t_1) = u(t_1)$ and $v(t_2) = u(t_2)$.
-For any such perturbation $w(t)$ the integral is stationary when $u(t)$ satisfies 
-$$
-\frac{d}{d\epsilon}I(u + \epsilon w) = 0 \text{ for } \epsilon = 0 \text{ (1)}
-$$
-**Remark:** Why can't we simplify $u + \epsilon w$ to $u$ in the above equation (since $\epsilon = 0$...)? The reason becomes clear when we apply the limit definition of the derivative, taking into account that $\epsilon = 0$:
-$$
-\frac{d}{d\epsilon}I(u + \epsilon w) = \lim_{\Delta \epsilon \to 0} \frac{I(u + \Delta \epsilon w) - I(u)}{\Delta \epsilon}
-$$
-The above expansion makes it clear that  $\frac{d}{d\epsilon}I(u) \neq \frac{d}{d\epsilon}I(u + \epsilon w)$ for $\epsilon = 0$. 
-
-Intuitively, constraint (1) resembles a constraint on a minimum (or maximum) point of a continuous function. Except that in this case, the minimum is a perturbed function (instead of a point).
-
-**Example 2.** Let's use the principle of least action to try to guess what the motion $u(t)$ of the ball introduced in Example 1, in the interval $[t_1=0, t_2=10]$. Assume that the initial position of the ball is $u_1 = 0$ and its initial speed is $\dot{u}_1 = 1$ (m/s) is,
-Let's take the following three possible trajectory functions:
-$$
-a(t) = \dot{u}_1 t - \frac{1}{2} g t^2 \\
-b(t) = 2a(t) \\
-c(t) = 0.5 a(t)
-$$
-And the corresponding derivatives:
-$$
-\dot{a}(t) = \dot{u}_1 - g t \\
-\dot{b}(t) = 2\dot{a}(t) \\
-\dot{c}(t) = 0.5 \dot{a}(t)
-$$
-
-
-They look like this:
-
-![1563522124353](assets/1563522124353.png)
-
-Now let's apply the test above to each one of them.
-
-
-
-Let $w(t)$ be a given function satisfying $w(t_1) = w( t_2 ) = 0$. Expanding the above constraint gives:
-$$
-\frac{d}{d\epsilon}I(u + \epsilon w) = \frac{d}{d\epsilon} \int_{t_1}^{t_2} T(\frac{d}{dt}(u + \epsilon w)(t)) - V((u + \epsilon w)(t)) dt \\
-\text{expand derivative} = \frac{d}{d\epsilon} \int_{t_1}^{t_2} T(\dot{u}(t) + \epsilon \dot{w}(t)) - V(u(t) + \epsilon w(t)) dt  \\
-\text{expand V and T} = \frac{d}{d\epsilon} \int_{t_1}^{t_2} \frac{1}{2}m(\dot{u}(t) + \epsilon \dot{w}(t))^2 - mg(u(t) + \epsilon w(t)) dt \\
-\text{omit (t)} = \frac{d}{d\epsilon} \int_{t_1}^{t_2} \frac{1}{2}m(\dot{u} + \epsilon \dot{w})^2 - mg(u + \epsilon w) dt \\
-\text{factor m} = m \frac{d}{d\epsilon} \int_{t_1}^{t_2} \frac{1}{2}(\dot{u} + \epsilon \dot{w})^2 - g(u + \epsilon w) dt \\
-\text{expand} = m \frac{d}{d\epsilon} \int_{t_1}^{t_2} \frac{1}{2} \dot{u}^2 + \dot{u} \epsilon \dot{w} + \frac{1}{2} \dot{w}^2 - gu - g\epsilon w dt \\
-\text{move der inside} = m \int_{t_1}^{t_2} \frac{d}{d\epsilon}\frac{1}{2} \dot{u}^2 + \frac{d}{d\epsilon}\dot{u} \epsilon \dot{w} + \frac{d}{d\epsilon} \frac{1}{2} \dot{w}^2 - \frac{d}{d\epsilon} gu - \frac{d}{d\epsilon}g\epsilon w dt \\
-\text{simplify derivative} = m \int_{t_1}^{t_2} \dot{u} \dot{w} - g w dt \\
-\text{simplify integral} = m \left[ \int_{t_1}^{t_2} \dot{u} \dot{w} dt - g \int_{t_1}^{t_2} w dt \right] \\
-\text{1st integral by parts} = m \left[  [\dot{u} w]_{t_1}^{t_2} - \int_{t_1}^{t_2} \ddot{u} w dt - g \int_{t_1}^{t_2} w dt \right] \\
-\text{apply w(t1)=w(t2)=0: } = m \left[ - \int_{t_1}^{t_2} \ddot{u} w dt - g \int_{t_1}^{t_2} w dt \right] \\ \\
-\text{join integrals and factor w: } =   \int_{t_1}^{t_2} (m\ddot{u} + mg) w dt \\
-$$
-So we get the simplified constraint
-$$
-\int_{t_1}^{t_2} (m\ddot{u} + mg) w dt = 0
-$$
-Since it applies to any given $w$, which can vary **arbitrarily** in the open interval $(t_1,t_2)$, the above constraint can only be satisfied if:
-$$
-m\ddot{u} + mg = 0 \Leftrightarrow m\ddot{u} = -mg
-$$