introduction to the calculus of variations (. Here $\Phi(f)$ is itself of the form $(\forall h\colon \Psi(f,h))\to \Theta(f)$, so $\neg\Phi(f)$ is $(\forall h\colon \Psi(f,h))\land\neg \Theta(f)$ it seems you supposed that $\neg\Phi(f)$ were $(\neg\forall h\colon \Psi(f,h))\to \Theta(f)$ (or equivalently $(\exists h\colon \neg\Psi(f,h))\to \Theta(f)$). Now we will apply the fundamental lemma of variational calculus to simplify the above necessary condition for. However, to qualify as a counterexample this would have to be shown for all allowed choices of $h$.Ībstractly, a counterexmple to a statement of the form $\forall f\colon \Phi(f)$ is a specific $f$ for which $\neg\Phi(f)$ (which thus proves $\exists f\colon \neg \Phi(f)$, the negation of $\forall f\colon \Phi(f)$). Theorem 1 (Fundamental Lemma of the Calculus of Variations). The lemma is then used to arrive at the Euler-Lagrange equatio. Instead, you merely present a single equation of the form $\int_a^bg(x)\,\mathrm dx=0$.Įven when using a crystal ball to interprete that, the closest to a counterexample is that you want $a=0$, $b=2\pi$, $f(x)=\sin x$ (which at least is non-zero), and have verified equation $(1)$ for a single specific case, namely $h(x)=\sin 2x$. This result is fundamental to the calculus of variations. Introduces the Fundamental Lemma of Variational Calculus and then proves it via contradiction. I am very pleased if I could get a some proof about this theorem. It is about the existence and uniqueness of the solutions 'in the large' of an equation of the form y F(x, y,y) y F ( x, y, y ). and show that $f$ is not identically zero. we get a function used in the proof above This concludes our proof of Fundamental Lemma of the Calculus of Variations. This theorem is on the page 16 in the book that I mentioned above.show that for all compactly supported smooth functions $h$ on $(a,b)$ equation $(1)$ holds, Fundamental lemma of calculus of variations with second derivative.specify a continuous function $f$ on $(a,b)$,.So in order to present a counter-example (if such existed), we'd have to $$\tag1\int _ x=0$$įor all compactly supported smooth functions $h$ on $( a, b )$, then $f$ is identically zero. If a continuous function $f$ on an open interval $( a, b )$ satisfies the equality Let us base our argument on the formulation of the lemma in Wikipedia:
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