The Euler-Lagrange Equation, or Euler's Equation

Definition 2   Let C^k[a, b] denote the set of continuous functions defined on the interval a≤x≤b which have their first k-derivatives also continuous on a≤x≤b.

The proof to follow requires the integrand F(x, y, y') to be twice differentiable with respect to each argument. What's more, the methods that we use in this module to solve problems in the calculus of variations will only find those solutions which are in C²[a, b]. More advanced techniques (i.e. beyond MATH0043) are designed to overcome this last restriction. This isn't just a technicality: discontinuous extremal functions are very important in optimal control problems, which arise in engineering applications.

Theorem 1   If I(Y) is an extremum of the functional

I(y) = $$\int_{a}^{b}$$ F(x, y, y') dx

defined on all functions y∈C²[a, b] such that y(a) = A, y(b) = B, then Y(x) satisfies the second order ordinary differential equation

$${\frac{{{\rm d}}}{{{\rm d} x}}} \left(\vphantom{ \frac{\partial F}{\partial y'} }\right. {\frac{{\partial F}}{{\partial y'}}} \left.\vphantom{ \frac{\partial F}{\partial y'} }\right) {\frac{{\partial F}}{{\partial y}}}$$ (1)

Definition 3   Equation () is the Euler-Lagrange equation, or sometimes just Euler's equation.

Warning 1   You might be wondering what ${\frac{{\partial F}}{{\partial y'}}}$ is suppose to mean: how can we differentiate with respect to a derivative? Think of it like this: F is given to you as a function of three variables, say F(u, v, w), and when we evaluate the functional I we plug in x, y(x), y'(x) for u, v, w and then integrate. The derivative ${\frac{{\partial F}}{{\partial y'}}}$ is just the partial derivative of F with respect to its second variable v. In other words, to find ${\frac{{\partial F}}{{\partial y'}}}$, just pretend y' is a variable.

Equally, there's an important difference between ${\frac{{{\rm d} F}}{{{\rm d} x}}}$ and ${\frac{{\partial F}}{{\partial x}}}$. The former is the derivative of F with respect to x, taking into account the fact that y = y(x) and y' = y'(x) are functions of x too. The latter is the partial derivative of F with respect to its first variable, so it's found by differentiating F with respect to x and pretending that y and y' are just variables and do not depend on x. Hopefully the next example makes this clear:

Example 1   Let F(x, y, y') = 2x + xyy' + y'² + y. Then

$${\frac{{\partial F}}{{\partial y'}}}$$ = xy + 2y'

$${\frac{{{\rm d}}}{{{\rm d} x}}} \left(\vphantom{ \frac{\partial F}{\partial y'} }\right. {\frac{{\partial F}}{{\partial y'}}} \left.\vphantom{ \frac{\partial F}{\partial y'} }\right)$$ = y + xy' + 2y'^′

$${\frac{{\partial F}}{{\partial y}}}$$ = xy' + 1

$${\frac{{\partial F}}{{\partial x}}}$$ = 2 + yy'

$${\frac{{{\rm d} F}}{{{\rm d} x}}}$$ = 2 + yy' + xy′² + xyy′′ +2y′′y′ + y′

and the Euler-Lagrange equation is

$$\qedhere$$

Warning 2   Y satisfying the Euler-Lagrange equation is a necessary, but not sufficient, condition for I(Y) to be an extremum. In other words, a function Y(x) may satisfy the Euler-Lagrange equation even when I(Y) is not an extremum.

Proof. Consider functions Y_ε(x) of the form

where η(x)∈C²[a, b] satisfies η(a) = η(b) = 0, so that Y_ε(a) = A and Y_ε(b) = B, i.e. Y_ε still satisfies the boundary conditions. Informally, Y_ε is a function which satisfies our boundary conditions and which is `near to' Y when ε is small.¹ I(Y_ε) depends on the value of ε, and we write I[ε] for the value of I(Y_ε):

I[ε] = $$\int^{b}_{a}$$ F(x, Y_ε, Y_ε') dx.

When ε = 0, the function I[ε] has an extremum and so

$${\frac{{{\rm d}I}}{{{\rm d}\epsilon}}}$$ = 0    when    ε = 0.

We can compute the derivative ${\frac{{{\rm d}I}}{{{\rm d}\epsilon}}}$ by differentiating under the integral sign:

$${\frac{{{\rm d}I}}{{{\rm d}\epsilon}}}$$ = $${\frac{{{\rm d}}}{{{\rm d}\epsilon}}}$$$$\int^{b}_{a}$$F(x, Y_ε, Y_ε') dx = $$\int^{b}_{a}$$$${\frac{{{\rm d}F}}{{{\rm d} \epsilon}}}$$(x, Y_ε, Y_ε') dx

We now use the multivariable chain rule to differentiate F with respect to ε. For a general three-variable function F(u(ε), v(ε), w(ε)) whose three arguments depend on ε, the chain rule tells us that

$${\frac{{{\rm d} F}}{{{\rm d} \epsilon}}} {\frac{{\partial F}}{{\partial u}}} {\frac{{{\rm d} u}}{{{\rm d} \epsilon}}} {\frac{{\partial F}}{{\partial v}}} {\frac{{{\rm d} v}}{{{\rm d} \epsilon}}} {\frac{{\partial F}}{{\partial w}}} {\frac{{{\rm d} w}}{{{\rm d} \epsilon}}}$$

In our case, the first argument x is independent of ε, so ${\frac{{{\rm d} x}}{{{\rm d} \epsilon}}}$ = 0, and since Y_ε = Y + εη we have ${\frac{{{\rm d} Y_\epsilon}}{{{\rm d} \epsilon}}}$ = η and ${\frac{{{\rm d} Y_\epsilon'}}{{{\rm d} \epsilon}}}$ = η'. Therefore

$${\frac{{{\rm d} F}}{{{\rm d} \epsilon}}}$$(x, Y_ε, Y_ε') = $${\frac{{\partial F}}{{\partial y}}}$$ η(x) + $${\frac{{\partial F}}{{\partial y'}}}$$ η'(x).

Recall that ${\frac{{{\rm d} I}}{{{\rm d} \epsilon}}}$ = 0 when ε = 0. Since Y₀ = Y and Y₀' = Y',

$$\int_{a}^{b} {\frac{{\partial F}}{{\partial y}}} {\frac{{\partial F}}{{\partial y'}}}$$ (2)

Integrating the second term in () by parts

$$\int_{a}^{b}$$$${\frac{{\partial F}}{{\partial y'}}}$$η'(x) dx = $$\left[\vphantom{ \frac{\partial F}{\partial y'}\eta(x) }\right.$$$${\frac{{\partial F}}{{\partial y'}}}$$η(x)$$\left.\vphantom{ \frac{\partial F}{\partial y'}\eta(x) }\right]^{b}_{a}$$ - $$\int_{a}^{b}$$$${\frac{{{\rm d}}}{{{\rm d} x}}}$$$$\left(\vphantom{ \frac{\partial F}{\partial y'} }\right.$$$${\frac{{\partial F}}{{\partial y'}}}$$$$\left.\vphantom{ \frac{\partial F}{\partial y'} }\right)$$η(x) dx.

The first term on the right hand side vanishes because η(a) = η(b) = 0. Substituting the second term into (),

$$\int^{b}_{a}$$ $$\left(\vphantom{ \frac{\partial F}{\partial y} - \frac{{\rm d}}{{\rm d} x} \frac{\partial F}{\partial y'}}\right.$$$${\frac{{\partial F}}{{\partial y}}}$$ - $${\frac{{{\rm d}}}{{{\rm d} x}}}$$$${\frac{{\partial F}}{{\partial y'}}}$$$$\left.\vphantom{ \frac{\partial F}{\partial y} - \frac{{\rm d}}{{\rm d} x} \frac{\partial F}{\partial y'}}\right)$$η(x) dx = 0.

The equation above holds for any η(x)∈C²[a, b] satisfying η(a) = η(b) = 0, so the fundamental lemma of calculus of variations (explained on the next page) tells us that Y(x) satisfies

$${\frac{{{\rm d}}}{{{\rm d} x}}}$$$$\left(\vphantom{ \frac{\partial F}{\partial y'} }\right.$$$${\frac{{\partial F}}{{\partial y'}}}$$$$\left.\vphantom{ \frac{\partial F}{\partial y'} }\right)$$ - $${\frac{{\partial F}}{{\partial y}}}$$ = 0.$$\qedhere$$

$\qedsymbol$

Definition 4   A solution of the Euler-Lagrange equation is called an extremal of the functional.²

Exercise 1   Find an extremal y(x) of the functional

I(y) = $$\int_{0}^{1}$$(y' - y)² dx,      y(0) = 0, y(1) = 2,    $$\left[\vphantom{ \mbox{Answer:} ~~ y(x)=2 \frac{\sinh{x}}{\sinh{1}} }\right.$$Answer:  y(x) = 2$${\frac{{\sinh{x}}}{{\sinh{1}}}}$$$$\left.\vphantom{ \mbox{Answer:} ~~ y(x)=2 \frac{\sinh{x}}{\sinh{1}} }\right]$$.$$\qedhere$$

Exercise 2   By considering y + g, where y is the solution from exercise 1 and g(x) is a variation in y(x) satisfying g(0) = g(1) = 0, and then considering I(y + g), show explicitly that y(x) minimizes I(y) in Exercise 1 above. (Hint: use integration by parts, and the Euler-Lagrange equation satisfied by y(x) to simplify the expression for I(y + g)).

Exercise 3   Prove that the straight line y = x is the curve giving the shortest distance between the points (0, 0) and (1, 1).

Exercise 4   Find an extremal function of

I[y] = $$\int_{1}^{2}$$x²(y')² + y dx,      y(1) = 1, y(2) = 1,    $$\left[\vphantom{ \mbox{Answer:}~~ y(x)=\frac{1}{2}\ln{x}+\frac{\ln{2}}{x}+1-\ln{2} }\right.$$Answer:  y(x) = $${\frac{{1}}{{2}}}$$lnx + $${\frac{{\ln{2}}}{{x}}}$$ +1 - ln2$$\left.\vphantom{ \mbox{Answer:}~~ y(x)=\frac{1}{2}\ln{x}+\frac{\ln{2}}{x}+1-\ln{2} }\right]$$.