# A question about the derivation of the Euler-Lagrange equation

• planck42
In summary: Euler-Lagrange equation is trying to find. In summary, the derivation of the Euler-Lagrange equation starts with defining two families of functions, which allows for any possible function that renders the functional stationary. The minimum value of I should occur when \frac{dI}{d\alpha} is zero at {\alpha}=0, which is what the Euler-Lagrange equation is looking for.
planck42
In the book Mathematical Methods for Engineers and Scientists 3, the derivation of the Euler-Lagrange equation starts roughly along the lines of this:

In order to minimize the functional $$I=\int_{x_1}^{x_2}{f(x,y,y')dx}$$, one should define two families of functions Y(x) and Y'(x), where Y(x) is $$y(x)+{\alpha}{\eta}(x)$$ and Y'(x) is the x-derivative of y(x), $$\alpha$$ is an arbitrary constant coefficient of $${\eta}(x)$$, which is an arbitrary function that must be zero at $$x_1$$ and $$x_2$$.

These steps aren't difficult to understand; it allows for any possible function that could render the functional stationary while maintaining the boundary conditions.

The minimum value of I should occur when $$\frac{dI}{d\alpha}$$ is zero at $${\alpha}=0$$

Whoa! Why are we dealing in $$\alpha$$'s all of a sudden? How does this minimize I and not the derivative with respect to x? Does working it out with x always produce a trivial solution? I need help with this one step!

Last edited:
I isn't a function of x (x is a dummy variable inside the integral, not a variable that you put into I).

On the other hand I DOES change as alpha changes (and as eta changes, but we're not going to worry about that)

Makes sense. However, I still have one question that I'll attempt to answer: why does $${\alpha}=0$$ produce an extreme, and not, for instance, $${\alpha}=1$$? My guess is that the value of $$\alpha$$ is irrelevant; any value of $$\alpha$$ can make the first derivative of I with respect to $$\alpha$$ vanish. Hence the simplest choice, $${\alpha}=0$$.

The point is that $y(x)$ is a function that minimizes (or extremizes) the integral.
So in $$y(x)+{\alpha}{\eta}(x)[/itex], the [tex]y(x)$$ is the sought after solution and $${\alpha}{\eta}(x)$$ is an arbitrary deviation from that solution.
Therefore, the minimum occurs at $${\alpha}=0$$

## 1. What is the Euler-Lagrange equation?

The Euler-Lagrange equation is a fundamental equation in calculus of variations that is used to find the function that minimizes or maximizes a given integral. It is named after the mathematicians Leonhard Euler and Joseph-Louis Lagrange who independently derived it.

## 2. How is the Euler-Lagrange equation derived?

The Euler-Lagrange equation is derived by setting the variation of the integral to zero and applying the calculus of variations. This involves taking the derivative of the integral with respect to the function and setting it equal to zero. The resulting differential equation is the Euler-Lagrange equation.

## 3. What is the significance of the Euler-Lagrange equation?

The Euler-Lagrange equation provides a necessary condition for a function to be a minimizer or maximizer of a given integral. This makes it a powerful tool in optimizing problems in fields such as physics, engineering, and economics.

## 4. In what situations is the Euler-Lagrange equation useful?

The Euler-Lagrange equation is useful in problems where the goal is to find the function that minimizes or maximizes a given integral. This can include finding optimal paths or trajectories, minimizing energy or cost, or maximizing profit or efficiency.

## 5. Are there any applications of the Euler-Lagrange equation in real life?

Yes, the Euler-Lagrange equation has many applications in real life. It has been used in physics to study the motion of particles and fields, in economics to optimize production and consumption, and in engineering to design efficient structures and systems. It is also used in computer graphics to simulate physical systems and in robotics to plan optimal movements.

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