In single-variable calculus, we find the minimum or maximum of a function $f(x)$ by finding where its derivative is zero. The input is a number, $x$, and the output is another number, $f(x)$. We are looking for a specific point that optimizes the function's value.

Now, let's level up. What if instead of finding a point, we needed to find an entire path or shape that minimizes some quantity? For instance, what is the shortest path between two points on a curved surface? The answer isn't a single point, but a whole curve.

This is where functionals come in. A functional is a mapping that takes a function as its input and returns a single real number. Think of it as a function of a function.

A common form for a functional, which we'll call $J$, takes a function $y(x)$ and is defined by an integral:

Our goal is to find the specific function $y(x)$ that makes the value of $J(y)$ a minimum or maximum. This is the core task of the calculus of variations.

How do we find the function $y(x)$ that makes a functional stationary (a minimum, maximum, or saddle point)? In standard calculus, we'd take the derivative and set it to zero. We need an equivalent tool for functionals.

Let's assume we have the optimal function, let's call it $y_{0}(x)$. Now, consider a slightly different function, which we can call an admissible variation. It's the original function plus a small deviation, $\eta(x)$, scaled by a tiny number, $\epsilon$. The function $\eta(x)$ must be zero at the endpoints, because we assume the start and end points of our path are fixed.

If we plug this new function $y(x, \epsilon)$ into our functional $J$, we get a value that depends on $\epsilon$. Since $y_0(x)$ is the optimal path, the value of $J$ should be at an extremum when $\epsilon = 0$. This means the derivative of $J$ with respect to $\epsilon$, evaluated at $\epsilon=0$, must be zero.

This derivative is called the first variation of the functional, denoted $\delta J$. Setting it to zero is the fundamental step in finding the optimal function. When we work through the calculus, this condition leads to a remarkable result.

After performing the differentiation and setting $\epsilon=0$, we arrive at an integral equation that must hold for any admissible variation function $\eta(x)$.

At first glance, this looks difficult. The integral of a product is zero. But how does that help us? The function $\eta(x)$ is our key. Since this equation must be true for any choice of $\eta(x)$, it forces the other part of the integrand to be zero everywhere.

This is the core idea of the Fundamental Lemma of Calculus of Variations (also known as the Du Bois-Reymond lemma). It states that if an integral of the form $\int_a^b f(x)g(x) \, dx = 0$ holds for all suitably smooth functions $g(x)$ that vanish at the endpoints, then the function $f(x)$ must be identically zero on the interval $(a, b)$.

Applying this powerful lemma to our integral gives us the celebrated Euler-Lagrange equation, which is a differential equation that the optimal function $y_0(x)$ must satisfy.

We've now seen how the problem of minimizing a functional is transformed. Instead of testing an infinite number of functions, we can solve a differential equation to find the one that does the job. This is the bridge from an integral problem to a differential one.