# Euler's Equations with auxiliary conditions - "why is $\frac{\delta y}{\delta \alpha}$ and $\frac{\delta z}{\delta \alpha}$ no longer independent?"

let $$J(\alpha)$$ be a functional of the parameter $$\alpha$$ such that: $$$$J(\alpha) = \int_{x_1}^{x_2}f\{y,y',z,z';x\}dx$$$$ and let $$$$f = f\{y,y',z,z';x\}$$$$ where $$$$y(\alpha,x) = y(0,x) + \alpha\eta_1(x)$$$$ $$$$z(\alpha,x) = z(0,x) + \alpha\eta_2(x)$$$$ The constraint is: $$$$g = g\{y_i;x\} = g\{y,z;x\} = 0$$$$ (for an example) $$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 $$$$\frac{\delta J}{\delta \alpha} = \int_{x_1}^{x_2}\left[\left(\frac{\delta f}{\delta y} - \frac{d}{dx}\frac{\delta f}{\delta y'}\right)\frac{\delta y}{\delta \alpha} + \left(\frac{\delta f}{\delta z} - \frac{d}{dx}\frac{\delta f}{\delta z'}\right)\frac{\delta z}{\delta \alpha}\right] dx$$$$

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 $$$$\left(\frac{\delta f}{\delta y} - \frac{d}{dx}\frac{\delta f}{\delta y'} \right)= 0$$$$ and $$$$\left(\frac{\delta f}{\delta z} - \frac{d}{dx}\frac{\delta f}{\delta z'} \right)= 0$$$$

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 $$\alpha = 0$$"

$$\textbf{Question}$$

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)

• 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"? Jan 17, 2022 at 23:21
• Which reference? Which page? Jan 18, 2022 at 3:40

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!!
• Thank you for your answer Could we also write: $\frac{\delta y_i}{\delta \alpha} = \eta_i (x)$ ? Further: $h(x) = \frac{\delta y_1}{\delta \alpha} - \frac{2y_2}{2y_1} \frac{\delta y_2}{\delta \alpha}$? Jan 18, 2022 at 23:50
• 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?? Jan 19, 2022 at 13:40
• 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 Jan 19, 2022 at 15:32