CoV16

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Minimize $J(y) = \int_0^\infty \left(y^2 + y'^2 + (y''+y')^2\right)dx, y(0)=1, y'(0)=2, y(\infty)=0, y'(\infty)=0\,$

Let $F(x,y,y',y'')\,$ denote the integrand in the above. The Euler equation (see CoV4) for this problem is

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

$\implies\frac{d^2}{dx^2} 2(y''+y') - \frac{d}{dx}[2y'+2(y''+y')] + 2y = 0$

$\implies y^{(iv)} - 2y'' + y = 0$

The characteristic equation for this differential equation is $m 4 - 2 m 2 + 1 = 0$ which factors to $(m - 1) 2 (m + 1) 2 = 0$ ; so the roots are $m = 1,1, - 1, - 1$ . Thus, the general solution of the differential equation is $y = (c_1 + c_2 x)e^x + (c_3 + c_4 x)e^{-x}\,$

The conditions at infinity require that $c_1 = c_2 = 0\,$ . Thus, $y = (c_3 + c_4 x)e^{-x},\quad y' = \left((c_4 - c_3) - c_4 x\right)e^{-x}\,$

Applying the remaining conditions,

$c_3 = 1,\quad c_4 - c_3 = 2\,$ so $c_3 = 1, c_4 = 3\,$ and so the desired solution is

$y = (1 + 3x)e^{-x}\,$

Satisfying the Euler equation is a necessary condition for the given solution to minimize $J(y)\,$ ; proving that it actually does minimize $J(y)\,$ is a lot more difficult (and will not be dealt with here).

Calculus of Variations

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