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Below is extracted from page 8 of https://courses.maths.ox.ac.uk/node/view_material/44170:

 of

I have an issue with the line: "It is easy to see that this means both terms on the RHS must vanish...". I.e. I don't see why the first term of the RHS can't just be the negative of the other.

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  • $\begingroup$ This is supposed to be true for all functions $\eta$ so what happens to the first term when $\eta (a)=\eta (b) = 0$ but arbitrary otherwise? $\endgroup$ – AHusain Dec 11 '19 at 2:11
  • $\begingroup$ @AHusain The assumptions that $\eta (a)=\eta (b) = 0$ only comes after the above extract - this is what confuses me. Is this statement true: The Euler-Langrange equation is necessarily give a extremal function when fixed/natural boundary conditions are set. $\endgroup$ – helios321 Dec 11 '19 at 2:41
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Sketched proof:

  1. Assume that $y$ is stationary so that LHS. of eq. (15) vanishes.

  2. Start by considering $\eta$ that vanishes on the boundary. Then the first term on the RHS. of eq. (15) vanish. From the fundamental lemma of calculus of variations, it follows from that the Euler-Lagrange (EL) equations are satisfies on the open interval $]a,b[$, which by continuity can be extended to the closed interval $[a,b]$.

  3. Consider next $\eta$ that not necessarily vanishes at $x=a$ but vanishes at $x=b$. However, the second term on the RHS. of eq. (15) is still zero, cf. part 2. Therefore the first term must vanish as well. This can only happen in $2$ ways:

    1. fixed endpoint/essential/Dirichlet boundary condition at $x=a$, or
    2. natural boundary condition at $x=a$.
  4. There is a similar conclusion for the boundary condition at $x=b$. $\Box$

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