# A given functional is path independent if k equals to the one of the following:

The functional $$\int_0^1 (y^{\prime 2} + (y + 2y')y'' + kxyy' + y^2) ~dx,$$ $$y(0) = 0, ~y(1) = 1, ~y'(0) = 2, ~y'(1) = 3$$ is path independent if $k$ equals

(A) $1$

(B) $2$

(C) $3$

(D) $4$

I have used Euler's formula for extremizing the given functional and get $k=2$. But I am pretty sure that I have done mistake as I could not get to use the given conditions and also I will be grateful if someone explains what path independent really means and the appropriate formula to be used to tackle the problem.

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Slow down (asking lots of questions..., without accepting answers and/or asking for clarification and/or responding to requests for clarification). As suggested earlier, please type your problem statements, when, as in this case, the problem statement is rather brief. – amWhy Nov 19 '12 at 19:03
Also, you should really look for more descriptive titles; "problems related to ..." don't say much about the question at hand, and reduce the number of people who might help. – Pragabhava Nov 19 '12 at 19:10
Back to the problem, in what context is path independent used? – Pragabhava Nov 19 '12 at 19:11
@Pragabhava I have very little idea about the functional and its relation with path independence. I came across the problem with an entrance exam paper . – learner Nov 19 '12 at 19:18

Let $$F(x,y,y',y'') = \int_0^1 \big(y'^2 + (y + 2y')y'' + kxyy'+y^2\big)dx$$ Then, if $h$ is a $C^\infty(0,1)$ $$F\big(x,y+h,(y+h)',(y+h)''\big) - F(x,y,y',y'') = \int_0^1 \left\{(y'' + kxy' +2y)h + (2y' + 2y'' + kxy)h'+(y+2y')h''\right\}dx+O\big(\|h\|^2\big)$$ where $\|\cdot\|$ is the appropriate norm. Hence, the Frechet derivative is $$DF \cdot h = \int_0^1 \left\{y'' + kxy' + 2y - \frac{d}{dx}\left(2y' + 2y'' + kxy\right) + \frac{d^2}{dx^2}(y + 2 y')\right\}hdx + \mbox{ B.C.}$$ where $$\mbox{B.C.} = \big(2y' + 2y'' + kxy - \tfrac{d}{dx}(y + 2 y')\big)h\Big|_0^1 + (y + 2 y')h'\Big|_0^1 = 0$$ given that $$y(0) = y(1) = 0, \qquad y'(0) = y'(1) = 0.$$ The Euler-Lagrange equation is $$y'' + kxy' + 2y - \frac{d}{dx}\left(2y' + 2y'' + kxy\right) + \frac{d^2}{dx^2}(y + 2 y') = 0$$ and if $k = 2$, it can be written as \begin{multline} y'' + 2xy' + 2y - \frac{d}{dx}\left(2y' + 2y'' + 2xy\right) + \frac{d^2}{dx^2}(y + 2 y') = \\ y'' + \frac{d}{dx}\big(2xy\big) - 2y'' - 2y''' - \frac{d}{dx}\big(2xy\big) + y'' +2 y'' = 0 \end{multline} for all $y$. In terms of the functional \begin{multline} y'^2 + (y + 2y')y'' + 2xyy' + y^2 \\ = y'^2 + yy'' + 2y'y'' + \frac{d}{dx}\big(xy^2\big) = y'^2 + yy'' + \frac{d}{dx}\big(y'^2 + xy^2\big) = \frac{d}{dx}\big(yy' + y'^2 +xy^2\big) \end{multline} and then $$F(x,y,y',y'') = \int_0^1 \frac{d}{dx}\big(yy' + y'^2 +xy^2\big)dx = 0$$
Presumably, "path-dependent" means that the functional doesn't depend on the function $y(x)$.
Integration by parts transforms $yy''$ into $-y'^2$; the boundary term is fixed by the boundary conditions and hence independent of $y(x)$. Also, $2y'y''$ is the derivative of $y'^2$, so its integral is also fixed by the boundary conditions. That leaves only $kxyy'+y^2$, and this is the derivative of $xy^2$ for one of the given values of $k$.