Before delving into calculus of variations, let’s look at a more familiar topic. How would you find the local minimum in a function f(x)?

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First, you would derive         and then equalize       to 0. Then, finding the values of x satisfying this equation would be candidates for local min. As the solutions of         = 0  leads to the stationary points, further testing allows the determination of the nature of these points. In this problem, we are concerned with the single variable differential calculus. 

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Analogously, while finding the stationary functions of a functional      , which is a function of functions, we generally solve differential equations for the stationary function f(x). Techniques of calculus of variations are used to solve these equations.

 

To explain this concept further, we can look at an example. Let there be two points              and              on a xy plane. The problem is finding the shortest path linking A and B.

 

 

 

 

 

 

 

One could give the answer without doing any calculations as a straight line between A and B, yet we will approach the problem as a calculus of variations question. The length of a curve y = f(x) between two points A and B is given by the arc length formula:

 

 

 

 

Where 

By taking out dx, the following is reached:

 

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Inserting this into the integral, we end up with:

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A calculus of variations problem for this would be:

 

Find y = f(x) between points A and B such that the integral                                is minimized. The functional that we aim to minimize is I, and we try to minimize it with respect to y = f(x).

 

A general expression for calculus of variations is to find y = f(x) such that the following integral is stationary:

 

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Euler-Lagrange equations are a method used for finding the stationary functions of a functional. As these equations do not tell us about the nature of the stationary functions as maximizing or minimizing, they might be thought of as the analog of using         = 0 to find stationary points of the function f(x). 

 

While determining y = f(x), in addition to the functional being stationary, we ought to check the conditions:

 

                  and 

 

Derivation/Proof of Euler-Lagrange Equations:

 

Suppose y(x), which is called extremal, makes I stationary and satisfies the boundary conditions. 

 

A function         exists such that                                . It is a completely random function and has continuous second derivatives as all functions do in this proof.

 

We will also define         = y(x) +           . This represents a variation in the extremal y(x). As           is an arbitrary function,        can also represent any arbitrary function, and it also satisfies the boundary conditions. Because of the presence of the parameter     ,          is a family of curves.

 

To solve the problem, we will find the particular         that makes                                    stationary. I only depends on the parameter     because x gets integrated out from the definite integral. After inserting boundary values, the only variable remaining is    . Thus, making I stationary corresponds to making      = 0. The value of    that makes I stationary is   = 0 since we have supposed y(x) to already make I (  ) stationary. We can now use this fact to differentiate the integral expression for I, which will then lead us to the Euler-Lagrange equation. 

 

If                   , then                                               . Moving the derivative inside will give us the expression with partial derivative: 

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      and     are the only variables in F that depend on    . x is an independent variable by itself.

Using the chain rule for partial derivatives, the expression becomes:

 

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In order to find an expression for     , the definition of                                   will be used to derive

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The partial derivatives              and 

 

Plugging these expressions into the Equation 1 indicates:

 

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The second part of this integral can be integrated by parts as follows:

 

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Since                               ,                      . Thus, the Equation 2 becomes:

 

 

To end up with a factored expression, we will take     common. 

 

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When            , by definition,                    . This leads to the following simplified expression:

 

 

Because     is an arbitrary function, to guarantee this equation;

 

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This equation is called the Euler-Lagrange equation. This signifies if y(x) is an extremal of I, then y(x) must satisfy this Euler-Lagrange equation. This equation is a necessary condition for y(x) to make I stationary; it is not sufficient. Moreover, while the Euler-Lagrange equation indicates functions that make I stationary, it does not show whether the functional becomes maximum or minimum with the found function. One might then either infer or use other methods to reach the nature of the functions.

 

Calculus of variations is an intriguing topic to search about as it is crucial to solve problems which cannot be solved by only differential calculus. An example could be deriving the shape of the path between two paths of different heights that takes the minimum amount of time for a bead to slide under the force of gravity, which is known as the Brachistochrone problem.
 

References 

“Derivation of the Euler-Lagrange Equation | Calculus of Variations.” YouTube, Faculty of Khan, 16 July 2017, https://www.youtube.com/watch?v=sFqp2lCEvwM. Accessed 6 Feb. 2022. 

“Introduction to Calculus of Variations.” YouTube, Faculty of Khan, 9 July 2017, https://www.youtube.com/watch?v=6HeQc7CSkZs. Accessed 6 Feb. 2022. 

Sanchis, Gabriela R. “Historical Activities for Calculus - Module 3: Optimization – Galileo and the Brachistochrone Problem.” MAA, Mathematical Association of America, July 2014, https://www.maa.org/press/periodicals/convergence/historical-activities-for-calculus-module-3-optimization-galileo-and-the-brachistochrone-problem#:~:text=This%20is%20the%20Brachistochrone%20(%E2%80%9CShortest,)%2C%20among%20all%20possible%20curves.