Lecture 21 and 22 (Apr 29)

Calculus of variation

Functional: I(y): function $\rightarrow \mathbb{R}$.

Example: Brachistochrone Problem:
$$
I(y)=\int_a^b \sqrt{\frac{1+(y')^2}{2gy}} dx \, .
$$

Euler-Lagrange equation

first order optimality condition. Necessary but not sufficient.

Euler's solution: let $N=\frac{\partial f}{\partial y}$ and $P=\frac{\partial f}{\partial y'}$,
$$
N- \frac{d}{dx} P =0 \, .
$$

Gradient Descent

$$
\min_x f(x)
$$
where $f:\mathbb{R} \rightarrow \mathbb{R}$.

Direct approach: Solve $f'(x)=0$ for $x$.

Iterative approach:
$$
x_{n+1}=x_n-\epsilon f'(x_n)
$$
for some initial guess $x_0$ and some small $\epsilon>0$.

Time-dependent approach: artificial time $t$,
$$
\frac{x_{n+1}-x_n}{\epsilon} = -f'(x_n) \, .
$$
Taking $\epsilon \rightarrow 0$, we have
$$
\frac{d}{dt} x(t) = -f'(x(t)) \, ,
$$
with the initial condition $x(t=0)=x_0$. The steady state solution will correspond to a local minimum.

$$
\frac{d}{dt} f(x(t)) = \frac{df}{dx} \cdot \frac{dx}{dt} = f' \cdot (-f') = -(f')^2 \le 0 \, .
$$

Gradient Descent for Functional

$$
\min_y I(y)
$$
for a functional $I$. We introduce an artificial time variable and solve
$$
\frac{\partial y}{\partial t} = - \nabla_y I
$$
with the initial condition $y(x,0)=y_0(x)$. The steady state solution solves $\nabla_y I =0$.

Reading Materials

Lecture Notes: p.110-116