Using Rayleigh-RHZ method find solution for the extremal of the functional,

$\int_{0}^{1} (2xy-y'^{2}-y^{2}) \ dx$

given $y(0)=0 \ \& \ y(1)=0$

Solution:

$I=\int_{0}^{1} F(x,y,y') \ dx$ where $F=2xy-y'^{2}-y^{2}$ --------------(1)

Assume trial solution,

$\bar{y}(x)=c_{0}+c_{1}x+c_{2}x^{2}$ --------------(2)

GIven, $y(0)=0,$ i.e., at $x=0$ in equation (2).

$\bar{y}(0)=c_{0}+c_{1}(0)+c_{2}(0)^{2}$

$\therefore \ c_{0}=0$

$y(1)=0,$ i.e., at $x=1$ in equation (2).

$\bar{y}(1)=c_{0}+c_{1}(1)+c_{2}(1)^{2}$

$0=c_{0}+c_{1}+c_{2}$

$0=0+c_{1}+c_{2}$

$0=c_{1}+c_{2}$ or $ \ c_{2}=-c_{1}$

$\therefore$ Equation (2) becomes,

$\bar{y}(x)=0+c_{1}x+c_{2}x^{2}$

$\bar{y}(x)=c_{1}x+c_{2}x^{2}$

$\bar{y}(x)=c_{1}x-c_{1}x^{2}$

$\bar{y}(x)=c_{1}x(1-x)$ --------------(3)

Differentiating (3) w.r.t. x,

$\bar{y}'(x)=c_{1}(1-2x)$ --------------(4)

Putting all the values in equation (1),

$I= \int_{0}^{1}(2x \bar{y} - \bar{y}^{2}- \bar{y}'^{2}) \ dx$

$I=\int_{0}^{1}[2xc_{1}x(1-x)-c_{1}^{2}x^{2}(1-x)^{2}-c_{1}^{2}(1-2x)^{2}] \ dx$

$I=\int_{0}^{1}[c_{1}(2x^{2}-2x^{3})-c_{1}^{2} x^{2} (1-2x+x^{2})-c_{1}^{2}(1-4x+4x^{2})] \ dx$

$I=c_{1} \int_{0}^{1}[(2x^{2}-2x^{3})-c_{1} (x^{2}-2x^{3}+x^{4})-c_{1}(1-4x+4x^{2})] \ dx$

$I=c_{1} \int_{0}^{1}[(2x^{2}-2x^{3})-c_{1} (x^{2}-2x^{3}+x^{4}+1-4x+4x^{2})] \ dx$

$I=c_{1} \int_{0}^{1}[(5x^{2}-2x^{3})-c_{1} (x^{2}-2x^{3}+x^{4}+1-4x)] \ dx$

$I=c_{1} \left \{ \left [ \cfrac{2x^{3}}{3}-\cfrac{2x^{4}}{4} \right ] - c_{1} \left[ \cfrac{5x^{3}}{3} - \cfrac{2x^{4}}{4} + \cfrac{x^{5}}{5} +x -\cfrac{4x^{2}}{2} \right ]\right \}_{0}^{1}$

$I=c_{1} \left \{ \left [ \cfrac{2}{3}-\cfrac{2}{4} \right ] - c_{1} \left[ \cfrac{5}{3} - \cfrac{2}{4} + \cfrac{1}{5} +1 -\cfrac{4}{2} \right ] - [0-0-c_{1}(0-0+0+0-0)] \right \}$

$I=c_{1} \left [ \left( \cfrac{1}{6} \right) - c_{1} \left( \cfrac{11}{30} \right) \right]$

$I=c_{1} \left [ \cfrac{1}{6} - c_{1} \cfrac{11}{30} \right]$

$I= \cfrac{c_{1}}{6} - \cfrac{11}{30} c_{1}^{2} $

Differentiating $I$ w.r.t. $c_{1}$, $\cfrac{dI}{dc_{1}}= \cfrac{1}{6}-\cfrac{11}{30} 2c_{1}$ $\cfrac{dI}{dc_{1}}= \cfrac{1}{6}-\cfrac{11}{15} c_{1}$ $\therefore \cfrac{dI}{dc_{1}}=0$ $0=\cfrac{1}{6}-\cfrac{11}{15}c_{1}$ $\cfrac{11}{15}c_{1}=\cfrac{1}{6}$ $c_{1}=\cfrac{15}{66}=\cfrac{5}{22}$ Therefore, equation (3) becomes $\bar{y}(x)=\cfrac{5}{22} x(1-x)$