empirical experiment with stationary motion

materials/notes.md

@@ -8,22 +8,67 @@ 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 thrown 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.
+**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 
 $$
-\frac{d}{d\epsilon}I(u + \epsilon w) = 0 \text{ for } \epsilon = 0
+\frac{d}{d\epsilon}I(u + \epsilon w) = 0 \text{ for } \epsilon = 0 \text{ (1)}
 $$
 **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  $\frac{d}{d\epsilon}I(u) \neq \frac{d}{d\epsilon}I(u + \epsilon w)$. 
+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:
 
-Intuitively, the above constraint resembles the 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).
+![1563522124353](assets/1563522124353.png)
 
-**Example 2.** Let's use the principle of least action to try to guess what the motion $u(t)$ of a ball that is thrown up at $\dot{u}_1 = 1$ (m/s) is, in the interval $[t_1=0, t_2=10]$. Let's take the following three possible trajectory functions:
-$a(t) = \dot{u}_1 + \frac{1}{2} g t^2  $
+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:
+$$
+m\ddot{u} + mg = 0 \Leftrightarrow m\ddot{u} = -mg
+$$