Differentiation in the geodesic problem

Question

Consider the geodesic problem on a flat plane. We have to find the extremal function $y$ such that: $$L[y] = \int_{a}^{b} \sqrt{1+(y')^2} \mathrm{d} x$$ is minimized, where $y(a) = y_1, y(b) = y_2$.

Applying the Euler-Lagrange equation to $L(x,y,y')$. $$\frac{\partial L}{\partial y} - \frac{\partial}{\partial x} \left(\frac{\partial L}{\partial y'} \right) = 0$$

When simplifying the above equation, we say that $$\frac{\partial L}{\partial y} = 0$$ as $L$ is not dependent on $y$.

My question is, how is that allowed? If $L$ is a function of $y'$, you will see a $\frac{d y'}{dy}$ term, which is equal to $$\frac{d y'}{dy}=\frac{\frac{dy'}{dx}}{\frac{dy}{dx}} = \frac{y''}{y'}$$

Am I incorrect in saying the above statement? What am I missing?

Answer 1

This is a notation problem.

Write $L(y) = \int_a^b Q(x,y(x), y'(x)) dt $, with $Q(f,g,h) = \sqrt{1+h^2}$.

Then Euler Lagrange says ${\partial Q(x,y(x),y'(x)) \over \partial g} = {d \over dx} H(x)$, where $H(x) = {\partial Q(x,y(x),y'(x)) \over \partial h} $.

Note that ${\partial Q(f,g,h) \over \partial g} = 0$.

Answer 2

This is a common question in calculus of variations, and the issue comes down to not understanding the notation used. Here, $L$ is a function of three variables, $L: \Bbb{R}^3 \to \Bbb{R}$, defined as \begin{align} L(\xi,\eta,\zeta) &= \sqrt{1+ \zeta^2} \end{align} What the Euler-Lagrange equations are saying is that a (sufficiently nice) function $y: [a,b] \to \Bbb{R}$ makes the "action" stationary if and only if for every $x \in [a,b]$, we have \begin{align} \dfrac{\partial L}{\partial \eta}\bigg|_{(x,y(x), y'(x))} - \dfrac{d}{dx}\left(\dfrac{\partial L}{\partial \zeta}\bigg|_{(x,y(x), y'(x))} \right) &= 0 \end{align} Or expressed in slightly different notation (which I think is the most precise way of expressing it) \begin{align} (\partial_2L)_{(x,y(x), y'(x))} - \dfrac{d}{ds}\bigg|_{s=x} \bigg((\partial_3L)_{(s,y(s), y'(s))} \bigg) &= 0 \end{align} Here is what the notation means. $L$ is a function of $3$ variables, so it has $3$ partial derivatives $\partial_1L, \partial_2L$ and $\partial_3L$. These are again functions $\Bbb{R}^3 \to \Bbb{R}$. So, the meaning of $(\partial_2L)_{(x,y(x),y'(x))}$ means that real number which is obtained when you apply the function $\partial_2L$ to the input $(x,y(x), y'(x)) \in \Bbb{R}^3$.

Similarly, $s\mapsto (\partial_3L)_{(s,y(s), y'(s))}$ is a function $\Bbb{R} \to \Bbb{R}$, and in the second term of the Euler-Lagrange equations, we want to consider the derivative of this single-variable function at the point $x$. Ultimately the key thing to remember with Calculus of variations and Euler-Lagrange equations is that you calculate partial derivatives FIRST, and then evaluate

For some reason people don't want to make this point explicit when teaching Euler-Lagrange equations and calculus of variations, which leads to questions like "how are the function and the derivative independent variables"? The answer is of course a function and its derivative are not independent... that would be extremely weird. The point is we are plugging in the curve only after performing the derivatives.

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In the above example, $L(\xi,\eta, \zeta) = \sqrt{1+\zeta^2}$. So, what is $\dfrac{\partial L}{\partial \zeta}$? Simple: \begin{align} \dfrac{\partial L}{\partial \zeta} \bigg|_{(\xi,\eta,\zeta)} &= \dfrac{\zeta}{\sqrt{1+ \zeta^2}}, \end{align} or in more precise notation, \begin{align} (\partial_3L)_{(\xi, \eta, \zeta)} &= \dfrac{\zeta}{\sqrt{1+ \zeta^2}}. \end{align} i.e the RHS is the value of the third partial derivative when evaluated at the point $(\xi,\eta,\zeta) \in \Bbb{R}^3$. For example, \begin{align} (\partial_3L)_{(1, 4, 8)} &= \dfrac{8}{\sqrt{1+ 8^2}} = \dfrac{8}{\sqrt{65}} \end{align} Now, suppose you have a function $y: [a,b] \to \Bbb{R}$. What is the value of the above partial derivative when evaluated at the point $(x,y(x), y'(x))$? Very simple: \begin{align} (\partial_3L)_{(x,y(x), y'(x))} &= \dfrac{y'(x)}{\sqrt{1+ (y'(x))^2}} \end{align} So, if for example $y(x) = x^2$ then $y'(x) = 2x$. So, \begin{align} (\partial_3L)_{(x,y(x), y'(x))} &= (\partial_3L)_{(x,x^2, 2x)} = \dfrac{2x}{\sqrt{1+ (2x)^2}} = \dfrac{2x}{\sqrt{1+4x^2}} \end{align}