1
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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
1
$\begingroup$

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 $x=b$. $\Box$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.