let $J(\alpha)$ be a functional of the parameter $\alpha $ such that: \begin{equation}J(\alpha) = \int_{x_1}^{x_2}f\{y,y',z,z';x\}dx \end{equation} and let \begin{equation}f = f\{y,y',z,z';x\} \end{equation} where \begin{equation} y(\alpha,x) = y(0,x) + \alpha\eta_1(x) \end{equation} \begin{equation} z(\alpha,x) = z(0,x) + \alpha\eta_2(x) \end{equation} The constraint is: \begin{equation} g = g\{y_i;x\} = g\{y,z;x\} = 0 \end{equation} (for an example) $g = \sum\limits_{i} x^2_i -\rho^2 =0$ where $\rho = constant$ like a radius of a sphere.

From functions with several dependencies we get \begin{equation} \frac{\delta J}{\delta \alpha} = \int_{x_1}^{x_2}\left[\left(\frac{\delta f}{\delta y} - \frac{d}{dx}\frac{\delta f}{\delta y'}\right)\frac{\delta y}{\delta \alpha} + \left(\frac{\delta f}{\delta z} - \frac{d}{dx}\frac{\delta f}{\delta z'}\right)\frac{\delta z}{\delta \alpha}\right] dx \end{equation}

Now originally $\textbf{without}$ a constraint, we will have $\frac{\delta y}{\delta \alpha} = \eta_1(x)$ and $\frac{\delta z}{\delta \alpha} = \eta_2(x)$ each $\eta_i(x) $ is independent thus \begin{equation} \left(\frac{\delta f}{\delta y} - \frac{d}{dx}\frac{\delta f}{\delta y'} \right)= 0 \end{equation} and \begin{equation} \left(\frac{\delta f}{\delta z} - \frac{d}{dx}\frac{\delta f}{\delta z'} \right)= 0 \end{equation}

so that $\frac{\delta J}{\delta \alpha} = 0$ when $\alpha =0$, but now because of the constraint, "the variations $\frac{\delta y}{\delta \alpha}$ and $\frac{\delta z}{\delta \alpha}$ are no longer independent, so the expressions in parentheses do not separately vanish at $\alpha = 0$"


Why are $\frac{\delta y}{\delta \alpha}$ and $\frac{\delta z}{\delta \alpha}$ no longer independent?

I would really appreciate if you could help me understand it.

Here is the page from text book:(I am not really sure if I need to provide extra information)enter image description here

  • $\begingroup$ What does this constrain $g = g\{y_i;x\} = g\{y,z;x\}$ mean especially in light of the example $g=z+y$? Where/what is the "constrain"? $\endgroup$
    – hyportnex
    Jan 17, 2022 at 23:21
  • $\begingroup$ Which reference? Which page? $\endgroup$
    – Qmechanic
    Jan 18, 2022 at 3:40

1 Answer 1


The way I see it, if there exists an additional constraint $g{y_i;x}=0$, where $y_i$ are some functions of the parameter $x$ (imagine the $y_i$ to be coordinates that are parametrized by a variable $x$ for argument's sake), then at least one of the $y_i$ functions/coordinates will be expressed as a function of the remaining $y_j,\ j\ne i$ functions/coordinates (Consider for example the equation of the circle, in which $\rho^2=y_1^2+y_2^2$). Therefore, the derivatives with respect to this parametrization will be related (here $0=2y_1\frac{\partial y_1}{\partial a}+2y_2\frac{\partial y_2}{\partial a}$) and thus $\frac{\partial y_1}{\partial a}$ is a function of $\frac{\partial y_2}{\partial a}$. I hope that helps!!

  • $\begingroup$ Thank you for your answer Could we also write: $\frac{\delta y_i}{\delta \alpha} = \eta_i (x)$ ? Further: $h(x) = \frac{\delta y_1}{\delta \alpha} - \frac{2y_2}{2y_1} \frac{\delta y_2}{\delta \alpha}$? $\endgroup$
    – Reuben
    Jan 18, 2022 at 23:50
  • $\begingroup$ Hi @Reuben. I believe that your first question is correct and thus we can write $\frac{\delta y_i}{\delta \alpha}=\eta_i(x)$. However, about the second, I do not know how the $h(x)$ function is defined. Is it something you see in your book?? Have I missed something?? $\endgroup$
    – schris38
    Jan 19, 2022 at 13:40
  • 1
    $\begingroup$ I was just letting h(x) be a function of x such that $h(x) =\frac{\delta y_i}{\delta \alpha}$, but I guess that $\eta_i(x)$ is actually already a function of x $\endgroup$
    – Reuben
    Jan 19, 2022 at 15:32

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