Welcome back. This week, we will take a short break from partial differential equations and have a brief foray into the calculus of variations, a field of mathematics that is concerned with optimizing functionals. A functional, essentially, is a real-valued function that takes functions in as inputs. For example, F[f] = ∫[0,1] f dx, is a functional that takes in a function, f, that is Riemann-integrable on [0,1] and outputs a number that represents the (signed) area under the graph of f.
The specific functional we are interested in today is the following, where L is some known continuously-differentiable function of three variables:
I[x(t)] = ∫[a,b] L(t, x, x’) dt.
Our inputs are a differentiable function x(t) and its derivative x’ = dx/dt. Let us suppose that x(t) is defined on the closed interval [a,b] with x(a) = xi and x(b) = xf. We then feed x(t) and x’(t) into our given function L(t, x, x’), which will yield an expression purely in terms of t. Lastly, we integrate that resultant expression over [a,b] to obtain a real number.
Now suppose that we have picked x(t) in such a way that the value of I[x(t)] is minimized. Our goal now is to determine an equation, which we will call the Euler-Lagrange equation, that this special x(t) must satisfy. Such an equation will be useful because it can allow us to determine specific formulas for x(t).
To derive the Euler-Lagrange equation, introduce some function η(t) that satisfies η(a) = η(b) = 0 and let ε be a real number variable that we can dial up and down. Then define the function X(t) as follows:
X(t) = x(t) + ε η(t).
Since η(a) = η(b) = 0, X(t) has the same values as x(t) on the boundary of our interval. Notice further that if ε = 0, then X(t) is equivalent to x(t), the function that we assumed to minimize our functional. Now plug X(t) into the functional to obtain:
I[X(t)] = ∫[a,b] L(t, X, X’) dt.
The expression inside the integral is a function of both t and ε, and because we are integrating with respect to t, the final value of I[X(t)] will depend on just the variable ε. If x(t) minimizes I[x(t)], that is the same as saying that dI/dε, the derivative of I with respect to ε, is equal to zero when ε = 0. So what we will do now is find a nice formula for “d/dε (I[X(t)])” and then see what happens to that expression when we plug in ε = 0 and set dI/dε = 0 at ε = 0.
Using the Leibniz rule for differentiating under the integral sign, we can bring the derivative operator in front of I[X(t)] inside the integral to obtain:
dI/dε = ∫[a,b] ∂/∂ε L(t, X, X’) dt.
Since the variable t does not depend on ε, we can rewrite the integrand (the expression inside the integral) using the multivariable chain rule to yield:
dI/dε = ∫[a,b] [(∂L/∂X) ∂X/∂ε + (∂L/∂X’) ∂X’/∂ε] dt.
Finally, we can substitute in derivatives and rewrite the second term in the integral on the right using integration by parts (this would be a nice exercise for the reader) to conclude that when ε = 0, then
∫[a,b] η(t) [∂L/∂x – d/dt (∂L/∂x’)] dt = 0.
The only way that the integral above can be equal to zero is if the expression inside the integral is equal to zero over [a,b]. Since η(t) is an arbitrary function that vanishes on the boundary of our domain, we conclude that the second factor inside the integral must be zero. This, indeed, is the Euler-Lagrange equation that x(t) must satisfy if x(t) minimizes I:
∂L/∂x – d/dt (∂L/∂x’) = 0.
The Euler-Lagrange equation has its most immediate application in classical mechanics when we are trying to determine the equation of motion of a particle of mass m. Whenever an object is in motion, it has kinetic energy, K, and potential energy, U. If we let L = K – U, we can call the following integral the action of the object:
Action = ∫ L dt.
We can note that the quantity L is often referred to as the Legrangian. According to the principle of least action, from time t = a to t = b, a particle will have the trajectory x(t) that minimizes its action. Since K = ½ m(x’)2 and U is often a function of x, finding the object’s equation of motion amounts to finding the function x(t) that minimizes the following functional.
I[x(t)] = ∫[a,b] L(x, x’) dt = ∫[a,b] ½ m(x’)2 – U(x) dt.
Using the Euler-Lagrange equation, we know that ∂L/∂x = – dU/dx is equivalent to d/dt(∂L/∂x’) = m x’’. In physics, the negative spatial derivative of potential energy is equivalent to the net force, F, acting on our object and the second derivative of the position function is defined to be our object’s acceleration, a. Therefore, we see that F = ma, Newton’s Second Law. Finding x(t) then amounts to manipulating Newton’s Second Law.
The Euler-Lagrange equation is thus highly useful because it provides us a path toward finding an object’s equation of motion. The content we covered today is also a perfect example of how math and physics come together. We began with the highly abstract notion of a function of a function, but ended up deriving Newton’s Second Law, which is the most important equation in classical physics. Next week, I am hoping that I will have the background research ready to present on Burgers’ equation. Until then, please take care.