In Taylor, read sections 6.2 to 6.4 for today, and 6.4 to 7.1 for Wednesday.

Y(x) = y(x) + αη(x)

The function η(x) is constrained to be 0 at the end points, η(x_{1}) = η(x_{2}) = 0, so that Y(x) satisfies the same boundary conditions as y(x), Y(x_{1}) = y_{1} = y(x_{1}), and Y(x_{2}) = y_{2} = y(x_{2}).
Except for possible smoothness criteria, η(x) is otherwise arbitrary.
The variable α is included to make the integral into a function of α, S = S(α), with the property that S is stationary when α = 0, that is (dS/dα)_{α=0} = 0.
Let's evaluate dS/dα.

S(α) = ∫_{1}^{2} f[y + αη, y' + αη', x] dx

dS/dα = ∫_{1}^{2} ∂f[y + αη, y' + αη', x]/∂α dx = ∫_{1}^{2} [(∂f/∂y)η + (∂f/∂y')η'] dx = 0.

Now, we're almost there. To put this in usable form, we must get both terms proportional to η (or η'). We can use integration by parts to convert the second term

∫_{1}^{2} (∂f/∂y')η' dx = [(∂f/∂y')η]_{1}^{2} - ∫_{1}^{2} η d(∂f/∂y')/dx dx = - ∫_{1}^{2} η d(∂f/∂y')/dx dx

where the term in square brackets evaluates to zero because η=0 at the endpoints. This gives us the result we want

dS/dα = [(∂f/∂y) + (d/dx)(∂f/∂y')]η dx = 0.

This result must hold for any η (for instance, η equal to the term in square brackets) and this can be guaranteed only if the term in square brackets is zero.

(∂f/∂y) + (d/dx)(∂f/∂y') = 0

This is known as the Euler-Lagrange equation.
Knowing the form of the function f(y,y',x) this yields a differential equation that, along with the boundary conditions y_{1} = y(x_{1}), and y_{2} = y(x_{2}), can be solved for y(x).

Let's return to the two earlier examples and apply the Euler-Lagrange equation to solve them.

On the x-y plane, find the shortest path between two points.

We determined that the length of a path from 1 to 2 is given by the integral

L = ∫_{1}^{2} ds = ∫_{1}^{2} sqrt(1 + y'(x)²) dx.

By inspection we see that f = sqrt(1 + y'(x)²). Applying the Euler-Lagrange equation to f we have

(∂f/∂y) + (d/dx)(∂f/∂y') = 0 + (d/dx)[y'/sqrt(1 + y'²)] = 0

Therefore y'/sqrt(1 + y'²) = C = constant or y'(x) = C /sqrt(1 - C²) = constant = m. Finally, integrating y'(x), we find y(x) = mx + b, the equation for a straight line. The endpoints determine m and b.

In principal y(x) = mx + b is the path that makes the length stationary. In this case it is obvious that this is the minimum of the integral, but how do we show that in the general situation?

The variables x and y are placeholders. The can stand for anything. In Lagrangian mechanics, x will become t (time), and y will become q (a "generalized coordinate"). It might be useful to write the Euler-Lagrange equation in terms of u and v, with u the independent variable, and v the dependent variable:

(∂f(v, v', u)/∂v) + (d/du)(∂f(v, v', u)/∂v') = 0

In the next example, we switch the roles of x and y.

© 2007 Robert Harr