This equation is useful when finding the critical point of the integral equation. Suppose that passes through the points and and it is a continuously differentiable function in . We want to find or .
Step 1. Define the extreme value .
Introducing a path function which consists of and , can be defined by
This function assumes that is the extreme value when it follows the path function .
Now, consider near the extreme value with the same start and end points. Then it represents another path function which consists of and . By Taylor Theorem,
It yields the following ,
Step 2. Apply the integration by parts to .
The second part of can be modified using the integration by parts.
This is because that since and have the same start and end points. Therefore is rewritten as
Step 3. Use the fact that .
Since is the extreme value, for the small enough . Therefore, it leads to
which is called Euler-Lagrange equation.
Example: Find the smallest distance between and points.
It is to find the smallest path among all the possible paths. From the step 1,
Now, step 2 and 3 lead to the following Euler-Lagrange equation.
where . It means that is of form for , , . Therefore,
This result it right because Euclidean distance is the smallest one between and points.