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Given the second order ODE of a catenary curve $$y''=a\sqrt{1+(y')^2}$$

With initial conditions $y(0)=0$ and $y'(0)=0$

The equation was first rewritten as $2y''y'''=2a^{2}y'y''$ and then dividing both sides by $$2y''$$ and letting $z=y'$ we would get the equation $$z''=a^2z$$

intital conditions $z(0)=0$ and $z'(0)=a$

I came across this question while i was reading the following article.http://www.emis.de/journals/BAMV/conten/vol12/juangil.pdf I couldn't fully understand what the article meant such as how did they form the Differential equation of the Caterny curve and how did they managed to solve the differential equation. Could anyone Please explain. Thanks

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  • $\begingroup$ Which part or step in particular do you not understand? $\endgroup$
    – Simon S
    Mar 1, 2015 at 14:44
  • $\begingroup$ They solved the equation by squaring both sides of your first equation and taking the derivative of both sides with respect to $x$ (remember, $y = y(x)$ so use implicit differentiation). Then they used the change of variables, $z = y' \implies z' = y'' \implies z'' = y'''$, after dividing both sides by $2y''$ $\endgroup$
    – mattos
    Mar 1, 2015 at 14:44
  • $\begingroup$ The part where they tried to form the Differential Equation. $\endgroup$
    – ys wong
    Mar 1, 2015 at 14:45
  • $\begingroup$ But what about the differential equation confused you? Which line of the paper? $\endgroup$
    – mattos
    Mar 1, 2015 at 14:46
  • $\begingroup$ At the top of pg 252 where they mentioned'Since the sum of these forces must be zero, the slope of the tangential force is given...' $\endgroup$
    – ys wong
    Mar 1, 2015 at 14:49

1 Answer 1

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I'm starting from the "Since the sum of the forces..", since that is where the confusion starts.

Notice that the tangential force is the gradient of the Catenary (curve) and the slope can be found by using

$$\frac{\text{rise}}{\text{run}} = \frac{W}{H} = \frac{w}{h} \ \ \ (1)$$

Where $w$ and $h$ are the magnitudes of $W$ and $H$, respectively. But the paper also states that "the magnitude, $w$, is proportional to the length $s$ of the chain between the origin and the point $(x, y)$" i.e.

$$w \propto s \implies \frac{w}{h} = \frac{\mu s}{h} \ \ \ (2)$$

where $\mu$ is the weight density of the chain.

So we have the gradient of the Catenary, $y'(x)$, is given by

$$\begin{align} y'(x) &= \frac{\text{rise}}{\text{run}} \\ &= \frac{\mu s}{h} \ \ \ (3)\\ \end{align}$$

But remember, $s$ is the length of the chain from the point $(x, y)$ to the origin, also known as the arc length, which has an integral equation

$$s = \int_{0}^{x} \sqrt{1 + y'(t)^{2}} dt \ \ \ (4)$$

Substituting $(4)$ into $(3)$ gives

$$y'(x) = \frac{\mu}{h} \int_{0}^{x} \sqrt{1 + y'(t)^{2}} dt$$

Taking the derivative of both sides with respect to $x$ (the integral is just an application of the fundamental theorem of calculus) and setting $\alpha = \frac{\mu}{h}$, we get

$$y''(x) = \alpha \sqrt{1 + y'(x)^{2}}$$

They then solved the equation by squaring both sides of the equation and taking the derivative of both sides with respect to $x$, using implicit differentiation i.e.

$$\begin{align} y''^{2} &= \alpha^{2} (1 + y'^{2}) \\ \implies \frac{d}{dx} y''^{2} &= \frac{d}{dx} \alpha^{2} (1 + y'^{2}) \\ \implies 2y'' y''' &= 2 \alpha^{2} y' y'' \\ \end{align}$$

Dividing by $2y''$

$$\implies y''' = \alpha^{2} y' \ \ \ (5)$$

Then they used the change of variables,

$$z = y' \implies z' = y'' \implies z'' = y'''$$

Which makes $(5)$ become

$$z'' = \alpha^{2}z$$

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  • $\begingroup$ Yup Thanks for the answer. $\endgroup$
    – ys wong
    Mar 1, 2015 at 15:26
  • $\begingroup$ COuld u help me with this question as well. Thanks $\endgroup$
    – ys wong
    Mar 2, 2015 at 17:33
  • $\begingroup$ COuld u help me with this question as well. Thanks math.stackexchange.com/questions/1172103/… $\endgroup$
    – ys wong
    Mar 2, 2015 at 17:34

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