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- Given a functional
on with the boundary conditions and , we proceed by approximating the extremal curve by a polygonal line with segments and passing to the limit as the number of segments grows arbitrarily large.
Divide the interval into equal segments with endpoints and let . Rather than a smooth function we consider the polygonal line with vertices , where and . Accordingly, our functional becomes a real function of variables given by
Extremals of this new functional defined on the discrete points correspond to points where
Note that change of affects L not only at m but also at m-1 for the derivative of the 3rd argument.
Evaluating the partial derivative gives
Dividing the above equation by gives
and taking the limit as of the right-hand side of this expression yields
The left hand side of the previous equation is the functional derivative of the functional . A necessary condition for a differentiable functional to have an extremum on some function is that its functional derivative at that function vanishes, which is granted by the last equation. (en)
- The derivation of the one-dimensional Euler–Lagrange equation is one of the classic proofs in mathematics. It relies on the fundamental lemma of calculus of variations.
We wish to find a function which satisfies the boundary conditions , , and which extremizes the functional
We assume that is twice continuously differentiable. A weaker assumption can be used, but the proof becomes more difficult.
If extremizes the functional subject to the boundary conditions, then any slight perturbation of that preserves the boundary values must either increase or decrease .
Let be the result of such a perturbation of , where is small and is a differentiable function satisfying . Then define
We now wish to calculate the total derivative of with respect to ε.
The third line follows from the fact that does not depend on , i.e. .
When , has an extremum value, so that
The next step is to use integration by parts on the second term of the integrand, yielding
Using the boundary conditions ,
Applying the fundamental lemma of calculus of variations now yields the Euler–Lagrange equation (en)
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