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Consider the following minimisation problem: $$\int_0^3\left(0.5\dot{x}^2-x\right)\,\mathrm{dt}$$ Subject to $x_0=0$ and $\dot{x}=0$. Using the Euler lagrange equation one can get: $$\frac{d^2x}{dt^2}=-1$$ The solution of the ODE is: $$x=-0.5t^2$$ and $$\dot{x}=-t$$ The integral is then: $$\int_0^3\left( 0.5t^2+0.5t^2 \right)\,\mathrm{dt}=\left.\frac{t^3}{3}\right|_0^3=9$$ So far so good. But what if $x=0$. Then the integral is simply zero. Or I can put $\dot{x}=0$ and $x=A$: $$\int_0^30.5A^2\,\mathrm{dt}=1.5A^2$$ Which can be both larger and smaller than 9. Why is the integral not maximised or minimised following the Euler-Lagrange equations. I can always choose a function that gives me a larger or smaller value. In other words, what does it mean to minimise a functional?

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Very broadly speaking, you typically cannot formulate a functional optimization problem without imposing some boundary conditions. Moreover the choice of the boundary conditions can have a marked effect on the character of the problem. The bluntest case I can think of is changing whether the solution is unique: typically the Dirichlet problem has a unique solution while the Neumann problem does not.

More narrowly speaking, the Euler-Lagrange equation comes from the variational problem with Dirichlet boundary conditions. With any other kind of boundary conditions, a different equation arises. The reason is that in deriving the Euler-Lagrange equation, one uses variations which satisfy a homogeneous Dirichlet boundary condition. From the point of view of functional optimization, we have to do this because we are comparing $F[u]$ to $F[u+\varepsilon v]$, and if $u+\varepsilon v$ doesn't satisfy the boundary condition then this doesn't make any sense. Because we use this type of variation, the integration by parts gives no boundary term, and the Euler-Lagrange equation comes entirely from the interior term.

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The Euler-Lagrange formula that was used in the OP applies to the case in which $x$ is fixed at both end points of the integral. This specific form does not apply to the case in which $x$ and $x'$ are fixed at one of the endpoints.

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