# CoV16

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 $m^{4}-2m^{2}+1=0$ 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).