Fig.1. A hanging chain.
Fig.2. Simple suspension bridge.
Fig.3. Keleti Railway Station (Budapest, Hungary).
Fig.4. Freely-hanging electric power cables.
Problem 1.
Determine the shape of a hanging chain.
Let the chain be described by the function $y(x)$, and let the
tension be described by the function $T(x)$. Consider a small piece of the chain, with
endpoints at $x$ and $x + dx$, as shown below.
Let the tension at $x$ pull downward at an angle $\theta_1$ with respect to the horizontal, and
let the tension at $x + dx$ pull upward at an angle $\theta_2$ with respect to the horizontal.
Balancing the horizontal and vertical forces on the small piece of chain gives
\begin{align}
T(x+dx)\cos(\theta_2)&=T(x)\cos(\theta_1),\nonumber \\
T(x+dx)\sin(\theta_2)&=T(x)\sin(\theta_1)+\frac{g\varrho dx}{\cos(\theta_1)},\label{eq:tension}
\end{align}
where $\varrho$ is the mass per unit length. The second term on the right is the weight of
the small piece, because $dx/ \cos(\theta_1)$ (or $dx/ \cos(\theta_2)$, which is essentially the same) is
its length.
Squaring and adding eqs. \eqref{eq:tension} gives \begin{equation*} [T(x+dx)]^2=[T(x)]^2+2T(x)g\varrho\tan(\theta_1)dx+O(dx^2). \end{equation*} From this
Since $\tan(\theta_1)=y'(x)$, we obtain \[ T'(x)=g\varrho y'(x). \] Integrating both sides we get \begin{equation} \label{eq:T_and_y} T(x)=g\varrho y(x)+c_1, \end{equation} where $c_1\in\mathbf{R}$ is an arbitrary constant.
Let’s see what we can extract from the first equation in eqs. \eqref{eq:tension}.
Using the trigonometrical identity \[ \cos^2(z)=\frac{1}{1+\tan^2(z)} \] it follows \[ \cos^2(\theta_1)=\frac{1}{1+(y'(x))^2}, \] and \[ \cos^2(\theta_2)=\frac{1}{1+(y'(x+dx))^2}. \] Subtituting these into the first equation of \eqref{eq:tension} we obtain \[ \frac{T^2(x+dx)}{1+(y'(x+dx))^2}=\frac{T^2(x)}{1+(y'(x))^2}, \] or equivalently, \begin{equation} \label{eq:Ty} \frac{T^2(x+dx)}{T^2(x)}=\frac{1+(y'(x+dx))^2}{1+(y'(x))^2}. \end{equation} Here \begin{align*} T^2(x+dx)&=[T(x)+T'(x)dx+O(dx^2)]^2\\ &=T^2(x)+2T(x)T'(x)dx+O(dx^2), \end{align*} and assuming the existence of $y''(x)$ \[ [y'(x+dx)]^2=[y'(x)]^2+2y'(x)y''(x)dx+O(dx^2). \] Substituting these values into \eqref{eq:Ty} we have \[ 2\frac{T'(x)}{T(x)}dx+O(dx^2)=2\frac{y'(x)y''(x)}{1+(y'(x))^2}dx+O(dx^2), \] that is, \[ \frac{T'(x)}{T(x)}-\frac{y'(x)y''(x)}{1+(y'(x))^2}=O(dx). \] Taking $dx\to 0$ we obtain \[ \frac{T'(x)}{T(x)}=\frac{y'(x)y''(x)}{1+(y'(x))^2}. \] Integrating both sides gives \[ \ln T(x)+c_2=\frac{1}{2}\ln(1+(y'(x))^2), \] where $c_2\in\mathbf{R}$ is an arbitrary constant. Exponentiating gives \begin{equation} \label{eq:Typrime} c_3^2T^2(x)=1+(y'(x))^2, \end{equation} where $c_3=e^{c_2}$.
We now consider eq. \eqref{eq:Typrime} and \eqref{eq:T_and_y}. Eliminating $T(x)$ yields $c_3^2(g\varrho y(x)+c_1)^2=1+(y'(x))^2$. We can rewrite this in the somewhat nicer form, \[ 1+(y'(x))^2=\alpha^2(y(x)+h)^2, \] where $\alpha=c_3g\varrho$, and $h=c_1/g\varrho$. We look for the non-constant solution. Separating the variables (it is enough to consider the $+$ sign) \[ \frac{1}{\sqrt{\alpha^2 (y+h)^2-1}}dy=1 dx. \] Integrating both sides (Hint: (a) $u:=\alpha (y+h)$ (b) $u:=\cosh(x)$ ) \[ \frac{1}{\alpha}\ln\left(\alpha(h+y)+\sqrt{\alpha^2(h+y)^2-1}\right)=x+a, \] where $a\in\mathbf{R}$ is an arbitrary constant.
From this \[ \alpha(h+y(x))+\sqrt{\alpha^2(h+y(x))^2-1}=e^{\alpha(x+a)}. \] Solving this equation for $h+y(x)$ we obtain \[ \bbox[lightblue,5px,border:2px solid red]{\color{#800000}{ \bf{y(x)+h=\frac{1}{\alpha}\cosh[\alpha(x+a)].}}} \]
Squaring and adding eqs. \eqref{eq:tension} gives \begin{equation*} [T(x+dx)]^2=[T(x)]^2+2T(x)g\varrho\tan(\theta_1)dx+O(dx^2). \end{equation*} From this
\begin{align*}
&\Step{1A}{[T(x+dx)-T(x)][T(x+dx)+T(x)]=2T(x)g\varrho\tan(\theta_1)dx+O(dx^2),}\\
&\Step{2A}{\frac{T(x+dx)-T(x)}{dx}[T(x+dx)+T(x)]=2T(x)g\varrho\tan(\theta_1)+O(dx),} \\
&\Step{3A}{\lim_{dx\to 0}\frac{T(x+dx)-T(x)}{dx}[T(x+dx)+T(x)]} \\
&\Step{4A}{\qquad\qquad\qquad\qquad\qquad\qquad\qquad=\lim_{dx\to 0}[2T(x)g\varrho\tan(\theta_1)+O(dx)],}\\
&\Step{5A}{\qquad T'(x)2T(x)=2T(x)g\varrho\tan(\theta_1),}\\
&\Step{6A}{\qquad T'(x)=g\varrho\tan(\theta_1).}\\
\end{align*}
Since $\tan(\theta_1)=y'(x)$, we obtain \[ T'(x)=g\varrho y'(x). \] Integrating both sides we get \begin{equation} \label{eq:T_and_y} T(x)=g\varrho y(x)+c_1, \end{equation} where $c_1\in\mathbf{R}$ is an arbitrary constant.
Let’s see what we can extract from the first equation in eqs. \eqref{eq:tension}.
Using the trigonometrical identity \[ \cos^2(z)=\frac{1}{1+\tan^2(z)} \] it follows \[ \cos^2(\theta_1)=\frac{1}{1+(y'(x))^2}, \] and \[ \cos^2(\theta_2)=\frac{1}{1+(y'(x+dx))^2}. \] Subtituting these into the first equation of \eqref{eq:tension} we obtain \[ \frac{T^2(x+dx)}{1+(y'(x+dx))^2}=\frac{T^2(x)}{1+(y'(x))^2}, \] or equivalently, \begin{equation} \label{eq:Ty} \frac{T^2(x+dx)}{T^2(x)}=\frac{1+(y'(x+dx))^2}{1+(y'(x))^2}. \end{equation} Here \begin{align*} T^2(x+dx)&=[T(x)+T'(x)dx+O(dx^2)]^2\\ &=T^2(x)+2T(x)T'(x)dx+O(dx^2), \end{align*} and assuming the existence of $y''(x)$ \[ [y'(x+dx)]^2=[y'(x)]^2+2y'(x)y''(x)dx+O(dx^2). \] Substituting these values into \eqref{eq:Ty} we have \[ 2\frac{T'(x)}{T(x)}dx+O(dx^2)=2\frac{y'(x)y''(x)}{1+(y'(x))^2}dx+O(dx^2), \] that is, \[ \frac{T'(x)}{T(x)}-\frac{y'(x)y''(x)}{1+(y'(x))^2}=O(dx). \] Taking $dx\to 0$ we obtain \[ \frac{T'(x)}{T(x)}=\frac{y'(x)y''(x)}{1+(y'(x))^2}. \] Integrating both sides gives \[ \ln T(x)+c_2=\frac{1}{2}\ln(1+(y'(x))^2), \] where $c_2\in\mathbf{R}$ is an arbitrary constant. Exponentiating gives \begin{equation} \label{eq:Typrime} c_3^2T^2(x)=1+(y'(x))^2, \end{equation} where $c_3=e^{c_2}$.
We now consider eq. \eqref{eq:Typrime} and \eqref{eq:T_and_y}. Eliminating $T(x)$ yields $c_3^2(g\varrho y(x)+c_1)^2=1+(y'(x))^2$. We can rewrite this in the somewhat nicer form, \[ 1+(y'(x))^2=\alpha^2(y(x)+h)^2, \] where $\alpha=c_3g\varrho$, and $h=c_1/g\varrho$. We look for the non-constant solution. Separating the variables (it is enough to consider the $+$ sign) \[ \frac{1}{\sqrt{\alpha^2 (y+h)^2-1}}dy=1 dx. \] Integrating both sides (Hint: (a) $u:=\alpha (y+h)$ (b) $u:=\cosh(x)$ ) \[ \frac{1}{\alpha}\ln\left(\alpha(h+y)+\sqrt{\alpha^2(h+y)^2-1}\right)=x+a, \] where $a\in\mathbf{R}$ is an arbitrary constant.
From this \[ \alpha(h+y(x))+\sqrt{\alpha^2(h+y(x))^2-1}=e^{\alpha(x+a)}. \] Solving this equation for $h+y(x)$ we obtain \[ \bbox[lightblue,5px,border:2px solid red]{\color{#800000}{ \bf{y(x)+h=\frac{1}{\alpha}\cosh[\alpha(x+a)].}}} \]
