Calculus of variations is an extremely useful and amazing tool in physics, math, computer science, and a variety of fields. Similar to how regular calculus is focused around functions and differentials, this field focuses on functionals and variations. A functional takes in a function and spits out a number. The following are examples of functionals.

(arc length of from to )

(normalization factor for i.e which is useful in quantum)

(value of at )

In some sense, this is just notational and should not be thought about too intensely.

A very important result of calculus is the ability to find minimums and maximums or more generally extrema. Now consider the differential of a function in regular calculus . The corresponding idea in calculus of variations is the variation of a functional denoted and defined . and here are functions. There exists an extremum for a function if and analogously a functional if . However, finding a function that minimizes a functional is a lot less straightforward than finding a number in a range. One cannot just systematically test all possibilities. In this case, it helps to talk about specific forms of functionals. The following is a VERY common functional.

The most common is arc length where

Let’s look at the variation of this functional.

Here, in order to preserve the boundary conditions for . We know that when is small, the following is true.

Here, we will use the following theorem (the proof is at the bottom of the post).

Theorem: Let be any continuous function such that . Assume then that, for all possible , the following is true.

This implies . (Highly recommend looking at the proof of this on the bottom before continuing).

Going back to our original integral, we see that and . This means . This is more formally written below and is called Euler’s equation.

However, sometimes not all variables are used so there exists 3 other forms of this.

If does not depend on , then so Euler’s equation reduces to the following.

If does not depend on , then so Euler’s equation becomes

If does not depend on , then we have to use some clever tricks to come to a simpler equation because still depends on . Let’s multiply the left side of the equation by which doesn’t change the value because it equals 0 anyway (I alternate between prime and differential notation and put terms in weird places but it will become clear why later).

From here, we make a simple change to the equation.

Because we know has no influence on , the first two terms form a chain rule.

Looking a little harder, it becomes clear that the last two terms are the result of a product rule.

Because it equals 0, we come to the final equation.

Altogether, this is a VERY powerful equation that is used in a huge range of physics and mathematics but it is understandable if there doesn’t seem to be any obvious use right now. This post was simply to establish a basis and derive the above equation. In the next post, I will give examples and uses of the equation which might make more clear what the purpose of the above steps were. In fact, I will be able to summarize 2 different past posts using this equation. If you want to get a head start, try proving that a line minimizes the distance between two points with this equation (Hint: use the arc length equation mentioned earlier and the -independent equation).

If you want to know more or see where I learned it from, read the book “Calculus of Variations” by Gelfand and Fomin (amazing book). You can find it in the Books section of this site.

Theorem: Let be any continuous function such that . Assume then that, for all possible , the following is true.

This implies

Proof:

Lemma: Let be any continuous function such that . Assume then that, for all possible , the following is true.

Then this implies

Proof of Lemma: Define as the constant such that the following is true.

Then let be defined as

A simple calculation shows that this function satisfies the boundary conditions. Now we see that the following integral reduces nicely.

We can also reduce it a different way.

We see that both of these statements can only hold true if because the square restricts the possibilities to all positive values. This means .

Now back to the original proof.

Define . Now let’s take the following integral by parts.

We can plug this back into our original integral.

From here, we can rearrange and apply our earlier lemma.

Differentiating both sides yields our desired result.

AnonymousDecember 10, 2017 / 1:01 pmsure here as say prof horia orasanu

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