Calculus of Variations and the Brachistochrone Problem

1. From Functions to Functionals

In standard calculus, you work with functions, which are maps from numbers to numbers. For example, $y=f(x)$ takes a number $x$ and returns a number $y$.

The calculus of variations deals with a different kind of object called a functional. A functional is a map from a function to a number.

Consider these examples:

• Arc Length: The total length $S$ of a curve described by the function $y(x)$ from $x_1$ to $x_2$ is given by the integral $S[y]={\int }_{x_1}^{x_2}\sqrt{1+(y^{\prime }(x){)}^2}dx$. The functional $S[y]$ takes the entire function $y(x)$ as input and outputs a single number, its length.
• Travel Time: The time $T$ it takes for a particle to travel along a path $y(x)$ with speed $v(x,y)$ is given by $T[y]=\int \frac{ds}{v}={\int }_{x_1}^{x_2}\frac{\sqrt{1+(y^{\prime }{)}^2}}{v(x,y(x))}dx$. Again, the functional $T[y]$ maps a path (a function) to a number (the travel time).

The fundamental problem of the calculus of variations is this: Given a functional that depends on a function and its derivatives, find the specific function that makes the functional's value an extremum (a minimum or a maximum).

2. The Euler-Lagrange Equation: A Condition for Extrema

How do we find the function $y(x)$ that extremizes a functional of the form $J[y]={\int }_{x_1}^{x_2}f(x,y,y^{\prime },\dots )dx$?

Let's draw an analogy with ordinary calculus. To find an extremum of a function $F(x)$, we find the point $x$ where its derivative is zero, $dF/dx=0$. For functionals, we do something similar. We assume we have the correct solution, $y(x)$, and we consider a family of "nearby" functions given by $Y(x)=y(x)+ϵ\eta (x)$. Here:

• $\eta (x)$ is an arbitrary, well-behaved function that vanishes at the endpoints ($\eta (x_1)=\eta (x_2)=0$) so that all varied paths start and end at the same points.
• $ϵ$ is a small parameter. When $ϵ=0$, we recover our true path $Y(x)=y(x)$.

By substituting $Y(x)$ into the functional $J$, it becomes a regular function of $ϵ$, i.e., $J(ϵ)$. The condition for an extremum is that the derivative with respect to $ϵ$ must be zero at $ϵ=0$:

Working through the mathematics of this condition, which involves differentiating under the integral sign and using integration by parts, leads to a remarkable result: the Euler-Lagrange equation.

The following video provides a clear, rigorous derivation. Given your preference for mathematical formalism, following these steps is crucial for a deep understanding.

The derivation shows that for the functional $J[y]={\int }_{x_1}^{x_2}f(x,y,y^{\prime })dx$ to be stationary, the function $y(x)$ must satisfy the Euler-Lagrange equation:

This is a second-order ordinary differential equation for $y(x)$. By solving it with the given boundary conditions, we can find the path that extremizes our functional.

3. First Application: The Shortest Path

Let's test this powerful equation on a simple problem: finding the shortest path between two points $(x_1,y_1)$ and $(x_2,y_2)$ in a plane. We intuitively know the answer is a straight line, but let's prove it.

The functional to minimize is the arc length:

Here, the integrand is $f(y,y^{\prime })=\sqrt{1+(y^{\prime }{)}^2}$. Notice that $f$ does not depend explicitly on $x$ or $y$.

Let's apply the Euler-Lagrange equation:

1. $\frac{\partial f}{\partial y}=0$ since $y$ does not appear in $f$.
2. $\frac{\partial f}{\partial y^{\prime }}=\frac{1}{2\sqrt{1+(y^{\prime }{)}^2}}\cdot (2y^{\prime })=\frac{y^{\prime }}{\sqrt{1+(y^{\prime }{)}^2}}$.

The Euler-Lagrange equation becomes:

This means the quantity inside the derivative must be a constant:

Solving for $y^{\prime }$ (by squaring both sides and rearranging) shows that $y^{\prime }$ must be a constant, let's call it $m$.

Integrating with respect to $x$ gives the equation of a straight line:

The constants $m$ and $b$ are determined by the boundary conditions that the line must pass through $(x_1,y_1)$ and $(x_2,y_2)$. This confirms our intuition and validates the method.

The Beltrami Identity: A Useful Shortcut

In the example above, and in many physics problems, the integrand $f$ does not depend explicitly on the variable of integration $x$. In such cases, there is a simplified "first integral" of the Euler-Lagrange equation, known as the Beltrami Identity (or Beltrami's formula):

Applying this to the shortest path problem ($f=\sqrt{1+(y^{\prime }{)}^2}$) gives:

This implies that $\sqrt{1+(y^{\prime }{)}^2}$ is constant, which in turn means $y^{\prime }$ must be constant. We arrive at the same conclusion much faster. Keep this identity in mind, as it will be very useful for our next problem.

4. The Brachistochrone Problem

Now we will tackle a famous historical problem posed by Johann Bernoulli in 1696: the brachistochrone problem. The name means "shortest time" in Greek. The challenge is to find the shape of a frictionless wire connecting two points, A and B, such that a bead sliding on it under gravity travels from A to B in the minimum possible time.

To solve this, we must first formulate the functional for the total travel time and then use the Euler-Lagrange equation to minimize it. The following video walks through the entire process, from physical setup to the final mathematical solution.

As the video explains, let's set up the coordinate system with the starting point at the origin (0,0) and the y-axis pointing downwards.

• Velocity: From conservation of energy, $\frac{1}{2}m{v}^2=mgy$, so the velocity at a depth $y$ is $v=\sqrt{2gy}$.
• Time element: The time to travel an infinitesimal arc length $ds$ is $dt=ds/v$.
• Functional: The total time to minimize is: $$T[y]=\int dt=\int \frac{ds}{v}={\int }_{x_1}^{x_2}\frac{\sqrt{1+(y^{\prime }{)}^2}}{\sqrt{2gy}}dx$$

The integrand, $f(y,y^{\prime })=\frac{\sqrt{1+(y^{\prime }{)}^2}}{\sqrt{2gy}}$, does not depend on $x$. This makes it a perfect candidate for the Beltrami Identity. Applying it leads to the differential equation:

where $C$ is a constant. This equation can be solved using a parameterization, and the solution is a remarkable curve: the cycloid.

The parametric equations for the cycloid are:

where $a$ is a constant related to the radius of the generating circle, and $\theta$ is the parameter. The constant $a$ is chosen to ensure the curve passes through the desired endpoint.

Conclusion

In this lesson, we have introduced the foundational concepts of the calculus of variations. You have learned how to shift your thinking from finding extremal points of functions to finding extremal paths (functions) of functionals.

Key Takeaways:

• A functional is a mapping from a function to a real number, often expressed as an integral involving the function and its derivatives.
• The condition for a functional $J[y]=\int f(x,y,y^{\prime })dx$ to be stationary is given by the Euler-Lagrange equation: $\frac{\partial f}{\partial y}-\frac{d}{dx}(\frac{\partial f}{\partial y^{\prime }})=0$.
• For cases where the integrand $f$ does not explicitly depend on $x$, the Beltrami identity provides a simpler, first-integral form of the solution: $f-y^{\prime }\frac{\partial f}{\partial y^{\prime }}=\text{constant}$.
• We applied this machinery to show that the shortest path between two points is a straight line and, more impressively, that the path of fastest descent under gravity (the brachistochrone) is a cycloid.

Preview of the Next Lesson:
The Euler-Lagrange equation you mastered today is not just a mathematical curiosity; it is the engine of analytical mechanics. In our next lesson, we will introduce the Principle of Least Action (also known as Hamilton's Principle). We will postulate that the true path taken by any classical mechanical system between two points in time is the one that minimizes a functional called the "action." By applying the Euler-Lagrange equation to the action functional, we will derive the fundamental equations of motion in a new, powerful form, marking our official entry into Lagrangian Mechanics.