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I am reading this paper where it is given

enter image description here

where $x = g(s,\alpha)$ and $u$ is a function of $x$.

The above expression is for a given $\alpha$.

The first equation is straight forward from chain rule. However, I am not able to obtain the second equation.

When I apply chain rule, this is what I get

$$\frac{d^2u}{ds^2} = \frac{d}{ds}\Big(\frac{du}{ds}\Big) = \frac{d}{dx}\Big(\frac{du}{dx}\ g'\Big)\frac{dx}{ds}$$ $$\frac{d^2u}{ds^2} = \bigg[g' \frac{d}{dx}\Big(\frac{du}{dx}\Big) + \frac{d}{dx}\Big(\frac{dx}{ds}\Big) \frac{du}{dx}\bigg]g'$$

where $g' = \frac{dx}{ds}$

How do I go further to arrive at the above expression?

Rhinocerotidae
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1 Answers1

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First notice that by chain rule we have $$\dfrac d{dx} = \frac 1{g'(s,\alpha)}\dfrac d{ds}$$ and thus we have

\begin{align}\dfrac{d^2u}{dx^2} &= \dfrac d{dx}\left(\dfrac{du}{dx}\right)= \frac 1{g'(s,\alpha)}\dfrac d{ds}\left(\frac 1{g'(s,\alpha)}\dfrac {du}{ds}\right)\\ &= \frac 1{g'(s,\alpha)}\left(\dfrac d{ds}\left(\frac 1{g'(s,\alpha)}\right)\dfrac {du}{ds}+\frac 1{g'(s,\alpha)}\dfrac{d^2u}{ds^2}\right)\\ &= \frac 1{g'(s,\alpha)}\left(\dfrac {-g''(s,\alpha)}{g'(s,\alpha)^2}\dfrac {du}{ds}+\frac 1{g'(s,\alpha)}\dfrac{d^2u}{ds^2}\right).\end{align}

Ennar
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