# P r o j e c t s

N o t e s

## Euler-Lagrange Equation

This equation is useful when finding the critical point of the integral equation.
Suppose that y(x) passes through the points (x_1, y_1) and (x_2, y_2) and it is a continuously differentiable function in [x_1, x_2]. We want to find min int_{x_1}^{x_2}y(x)dx or max int_{x_1}^{x_2}y(x)dx.

Step 1. Define the extreme value F(y).

Introducing a path function f which consists of y(x) and y prime (x), F(y) can be defined by
\begin{align} F(y)= \int_{x_1}^{x_2} f(y(x), y^{\prime}(x), x) dx. \end{align}
This function F assumes that y is the extreme value when it follows the path function f.

Now, consider F(y+\delta y) near the extreme value F(y) with the same start and end points. Then it represents another path function f which consists of y(x)+\delta y(x) and y prime (x)+\delta y prime (x). By Taylor Theorem,
\begin{align} F(y+\delta y) &= \int_{x_1}^{x_2} f(y(x)+\delta y(x), y^{\prime}(x)+\delta y^{\prime}(x), x)dx \\ & \approx
\int_{x_1}^{x_2} f(y(x), y^{\prime}(x), x) + f_y \delta y + f_{y^{\prime}}\delta y^{\prime} + f_x 0\  dx \\ &=
F(y) + \int_{x_1}^{x_2} f_y \delta y + f_{y^{\prime}}\delta y^{\prime} dx
\end{align}
It yields the following delta F,
\begin{align} \delta F= F(y+\delta y) - F(y) \approx
\int_{x_1}^{x_2}f_y \delta y + f_{y^{\prime}}\delta y^{\prime} dx
\end{align}

# Step 2. Apply the integration by parts to delta F.

The second part of delta F can be modified using the integration by parts.
\begin{align} \int_{x_1}^{x_2} f_{y^{\prime}}\delta y^{\prime} dx = \left[ f_{y^{\prime}}\delta y \right]_{x_1}^{x_2} - \int_{x_1}^{x_2}\frac{d}{dx} \left( f_{y^{\prime}}\right) \delta y\  dx =  - \int_{x_1}^{x_2}\frac{d}{dx} \left( f_{y^{\prime}}\right) \delta y\  dx
\end{align}
This is because that delta y(x_1)=delta y(x_2)=0 since f(y(x), y prime (x), x) and f(y(x)+delta y(x), y prime (x) + delta y prime (x), x) have the same start and end points. Therefore, delta F is rewritten as
\begin{align} \delta F \approx
\int_{x_1}^{x_2} \left( f_y - \frac{d}{dx} f_{y^{\prime}} \right) \delta y\  dx
\end{align}

# Step 3. Use the fact that delta F(y)=0.

Since F(y) is the extreme value, delta F(y)=0 for the small enough delta y. Therefore, it leads to
\begin{gather} \delta F \approx
\int_{x_1}^{x_2} \left( f_y - \frac{d}{dx} f_{y^{\prime}} \right)\delta y\  dx = 0, \\
\color{red}{f_y - \frac{d}{dx} f_{y^{\prime}} = 0}
\end{gather}
which is called Euler-Lagrange equation.

Example.  Find the smallest distance between (x_1, y_1) and (x_2, y_2) points.

It is to find the smallest path among all the possible paths. From the step 1,
\begin{align} \min \int_{x_1}^{x_2} \sqrt{(dx)^2+(dy)^2} = \min \int_{x_1}^{x_2} \sqrt{1+y^{\prime 2}} dx = F(y). \end{align}
Now, step 2 and step 3 lead to the following Euler-Lagrange eqation.
\begin{align} & f_y - \frac{d}{dx} f_{y^{\prime}} = 0 - \frac{d}{dx}\left( \frac{y^{\prime}}{\sqrt{1+y^{\prime 2}}} \right) = 0 \\ & \Rightarrow  \frac{y^{\prime}}{\sqrt{1+y^{\prime 2}}} = C \\  & \Rightarrow y^{\prime} = \pm \sqrt{\frac{C^2}{1-C^2}}
\end{align}
where C in mathbb{R}. It means that y is of form ax+b for a, b in mathbb{R}, and y prime=a=frac{y_2-y_1}{x_2-x_1}. Therefore,
\begin{align} F(y) &= \min \int_{x_1}^{x_2} \sqrt{1+y^{\prime 2}} dx = \int_{x_1}^{x_2} \sqrt{1+a^2} dx =  \sqrt{1+a^2}(x_2-x_1) \\ \\ &=
\sqrt{1+ \frac{(y_2-y_1)^2}{(x_2-x_1)^2}}(x_2-x_1)=
\sqrt{ (x_2-x_1)^2 + (y_2-y_1)^2}
\end{align}
This result is right because Euclidean distance is the smallest one between (x_1, y_1) and (x_2, y_2) points.

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