To the best of my knowledge, the function that minimizes of the integral posed in calculus of variations must also satisfy the Euler-Lagrange equations. In other words, the Euler equations are a necessary condition for finding a minimizer.

One thing I do not understand, is, why do I see everyone treating it as if it were a sufficient condition to show optimality? I am solving some example problems in calculus of variations, and I am following the problems here: http://matematika.cuni.cz/dl/pyrih/variationProblems/variationProblems.pdf example

For example, look at problem 1.1 on page 2, which says:

Using the Euler equation find the extremals for the following functional: $\int_{a}^{b}12xy(x)+(\frac{\partial y(x)}{\partial x})^2dx$

then, as I am looking at the solution which is immediately below, it says "finally obtain the Euler equation for our functional"...and proceeds to solve the Euler equation as a means to solve the minimization problem to obtain the answer.

I do not understand why this is allowed, if the Euler equation is a necessary condition, but it is not sufficient.

Is the Euler equation necessary and sufficient? or just necessary?

An example of a resource that explicitly says it, says "Euler-Lagrange Equation and is a necessary, but not sufficient, condition for an extremal function", is here: http://www.maths.manchester.ac.uk/~wparnell/MT34032/34032_CalcVar

  • $\begingroup$ So, why do I see people solving these minimization problems as if it is sufficient (like the example I cited) ? Or is there a gap in my understanding about why they are allowed to solve it that way? $\endgroup$ – makansij Jan 3 '18 at 0:53
  • $\begingroup$ sure. here's an example taken from gilbert strang's textbook: ocw.mit.edu/courses/mathematics/…. look at the example on page 2 for the shortest path between two points. $\endgroup$ – makansij Jan 3 '18 at 1:00
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    $\begingroup$ Actually, this is a passage from Gilbert Strang's textbook Computational Science and Engineering. It's a real textbook, not just lecture notes or something. (Of course, Strang's writing style is fairly informal, so your point stands.) $\endgroup$ – littleO Jan 3 '18 at 1:09
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    $\begingroup$ We have an analogous situation in calculus when we minimize a differentiable function $f:\mathbb R^n \to \mathbb R$ by setting the derivative equal to $0$. The equation $\nabla f(x) = 0$ might have multiple solutions, and not all solutions correspond to extrema (some are saddle points). But, if it is known that a minimizer of $f$ exists, and if there is a unique point $x \in \mathbb R^n$ such that $\nabla f(x) = 0$, then this $x$ must be the minimizer of $f$. $\endgroup$ – littleO Jan 3 '18 at 1:13
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    $\begingroup$ Continuing littleO's comment: in the cited example, 1.1 on p.2, the functional is convex in $y$, which rules out the EL solution being a maximizer, and might, given further regularity properties (but I haven't checked) imply there does exist a minimizer. $\endgroup$ – kimchi lover Jan 3 '18 at 2:19

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