An MCQ involving Rayleigh - Ritz method for the functional $$I(y) = \int_{0}^{1}(\frac{1}{2}(y^{'})^2 - y)dx$$
Let $y_\text{app}$ be polynomial approximation, involving only one coordinate function, for the functional $$I(y) = \int_0^1 \left(\frac{1}{2}(y^{'})^2 - y\right) \, dx ;\quad y(0) = 0,\ y(1) = 0$$ using the Rayleigh - Ritz method; here $y \in C^2[0, 1]$
If $y_e (x)$ is an exact extremizing function, then $y_e$ and $y_\text{app}$ are coincident at
$x = 0$ but not at remaining points in $[0, 1]$
$x = 1$ but not at remaining points in $[0, 1]$
$x = 0$ and $x = 1$, but not at other points in $[0, 1]$
all points in $x\in [0, 1]$
It can be shown that $y_e$ is quadratic by Euler Lagrange's method.
Somewhere I found a solution of this problem, there it was written that $y_{app}$ must be a quadratic polynomial, and it coincides with $y_e$ at two points viz., 0 and 1, so it will coincide $y_e$ at all points. I could not understand that why $y_{app}$ must be a quadratic polynomial? What is meant by an involvement of one coordinate function?
Please somebody tell. Thanks in advance.
I know how to find out $y_{app}$
take it of the form $ y= cx(1-x)$ i.e. a function vanishing on given boundary conditions. now calculate $y'$ and put it in the integral.
$I(y)=\frac{c^2-c}{6}$. Now by rayleigh Ritz find out $\frac{dI}{dc}$ and put it equal to $0$ to find out $c$. yow will get $y_{app}=\frac{x-x^2}{2}$. Further $y_e$ calculating by Euler Lagrange method comes out equal to $y_{app}$.
so d is answer.