CoV23

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Find the extremals of the functional $J(y)=\int_0^\pi\left[y'(x)\right]^2dx\,$ subject to the constraint $\int_0^\pi \left[ y(x)\right]^2dx=1, y(0)=y(\pi)=0\,$

Let $H(y,\lambda) = \int_0^\pi ([y'(x)]^2 + \lambda [y(x)]^2)\,dx$ where $\lambda\,$ is a Lagrange multiplier. The Euler equation is

$\frac{d}{dx}\frac{\partial (y'^2 + \lambda y^2)}{\partial y'} - \frac{\partial(y'^2 + \lambda y^2)}{\partial y} = 0$

$\implies y'' - \lambda y = 0$

(i) Try $\lambda > 0\,$ ; let $\lambda=\alpha^2\,$ . Then

$y'' - \alpha^2 y = 0\,$
$\implies y = a e^{\alpha x} + b e^{-\alpha x}$

$y(0) = 0 \implies a + b = 0;\quad y(\pi) = 0 \implies a e^{\alpha\pi} + b e^{-\alpha \pi} = 0$
But the solution for this is $a = b = 0\,$ , i.e., $y\equiv 0\,$ which does not satisfy the constraint $\int_0^\pi y^2\,dx = 1$

(ii) Try $\lambda = 0\,$ Then

$y'' = 0\,$
$\implies y = a x + b\,$

$y(0) = 0 \implies b = 0\,$ so $y = a x\,$ ; then $y(\pi) = 0 \implies a = 0 \implies y \equiv 0$ which, again, does not satisfy the constraint.

(iii) $\lambda < 0\,$ ; let $\lambda=-\alpha^2\,$ . Then

$y'' + \alpha^2 y = 0\,$
$\implies y = a \cos(\alpha x) + b \sin(\alpha x)$

$y(0) = 0 \implies a = 0$ ; $y(\pi) = 0 \implies b = 0$ or $\sin(\alpha \pi) = 0\,$

$b = 0 \implies y\equiv 0$ , so we reject this and take $\sin(\alpha \pi) = 0\,$ , which implies $\alpha = \pm 1, \pm 2, \dots$

Thus $y = b \sin(nx)\,$ where $n\,$ is any nonzero integer.

$b\,$ is found by imposing the constraint:

$\int_0^\pi b^2 \sin^2(nx)\,dx = 1 \implies \int_0^\pi b^2\, \frac{1-\cos(2nx)}{2}\,dx = 1 \implies\frac{b^2}{2} \left[x - \frac{\sin(2nx)}{2n}\right]_0^\pi = 1$
$\implies b = \pm\sqrt{2/\pi}$

Hence, the extremals of $J\,$ subject to the given constraint are
$y = \pm\sqrt{2/\pi}\sin(nx), n=\pm1, \pm2, \dots$

Calculus of Variations

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