HA10-P1

[Victor Ivrii]

Problem 1 http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter10/S10.P.html#problem-10.P.1

[Jeremy Li 2]

We need to minimize the Lagrange functional. By subtracting the constraint multiplied by the Lagrange multiplier $\lambda$ from the original functional, we get:
\begin{equation}
\Phi \lbrack u \rbrack = \int_0^a(\rho gu\sqrt{1 + u'^2}-\lambda \sqrt{1+u'^2}) dx\\
= \int_0^a(\rho gu-\lambda)\sqrt{1 + u'^2} dx
\end{equation}

The variation of $\Phi$ is:
\begin{equation}
\delta\Phi=\int_0^a \left(\frac{\partial L}{\partial u} - \frac{\partial}{\partial x}\frac{\partial L}{\partial u'}\right)\delta u dx
\end{equation}
(The boundary term is zero since $\delta u|_0 = \delta u|_a = 0$)

We need to solve the Euler-Lagrange equation.
\begin{equation}
\frac{\partial L}{\partial u} - \frac{\partial}{\partial x}\frac{\partial L}{\partial u'} = 0
\end{equation}
for
\begin{equation}
L = (\rho gu-\lambda)\sqrt{1 + u'^2}
\end{equation}

Calculating the two terms...
\begin{equation}
\frac{\partial L}{\partial u} = \rho g\sqrt{1+u'^2}
\end{equation}
\begin{equation}
\frac{\partial L}{\partial u'} = (\rho gu-\lambda) \frac{1}{\sqrt{1+u'^2}} u'
\end{equation}
\begin{equation}
\frac{\partial}{\partial x}\frac{\partial L}{\partial u'} = \frac{\rho gu'^2}{\sqrt{1+u'^2}} + \frac{(\rho gu-\lambda)u''}{\sqrt{1+u'^2}} - \frac{(\rho gu-\lambda) u'^2 u''}{(1+u'^2)^{\frac{3}{2}}}
\end{equation}

Plugging these results into $(3)$:
\begin{equation}
\rho g \sqrt{1+u'^2} - \frac{\rho gu'^2}{\sqrt{1+u'^2}} - (\rho gu-\lambda)u''  \left( \frac{1}{\sqrt{1+u'^2}}-\frac{u'^2}{(1+u'^2)^{\frac{3}{2}}} \right) = 0
\end{equation}

Multiplying every term by $(1+u'^2)^{\frac{3}{2}}$:
\begin{equation}
\rho g(1+u'^2)^2 - \rho gu'^2 (1+u'^2) - (\rho gu - \lambda) u'' ((1+u'^2)-u'^2) = 0 \\
\rho g(1+u'^2)((1+u'^2)-u'^2) - (\rho gu - \lambda) u'' ((1+u'^2)-u'^2) = 0 \\
\rho g(1+u'^2)=(\rho gu - \lambda) u''
\end{equation}

Dividing by $\rho g$,
\begin{equation}
(1+u'^2)=\left(u - \frac{\lambda}{\rho g}\right) u''
\end{equation}

Doing some rearranging:
\begin{equation}
\frac{1}{1+u'^2} u''=\frac{1}{u - \frac{\lambda}{\rho g}}
\end{equation}

Multiplying both sides by u'
\begin{equation}
\frac{1}{1+u'^2} u' u''=\frac{1}{u - \frac{\lambda}{\rho g}} u'
\end{equation}

Integrating both sides with respect to $x$
\begin{equation}
\frac{1}{2} \ln{\left(1+u'^2\right)} = \ln{\left(u-\frac{\lambda}{\rho g}\right)} + C
\end{equation}

Exponentiating:
\begin{equation}
1+u'^2=A^2\left(u-\frac{\lambda}{\rho g}\right)^2
\end{equation}
By separation of variables
\begin{equation}
\frac{du}{dx}=\sqrt{A^2\left(u-\frac{\lambda}{\rho g}\right)^2-1}\\
dx=\frac{du}{\sqrt{A^2\left(u-\frac{\lambda}{\rho g}\right)^2-1}}
\end{equation}

We get
\begin{equation}
\cosh(Ax+B)=A\left(u-\frac{\lambda}{\rho g}\right)
\end{equation}

\begin{equation}
u=\frac{1}{A}\cosh(Ax+B)+\frac{\lambda}{\rho g}
\end{equation}

What remains is to apply the boundary conditions. I'll figure the rest out tomorrow.

[Victor Ivrii]

Right but a bit too complicated:

1) Since $\rho g$ is a constant factor, you may ignore it.

2) You don't need to write down the second order equation but only
$H :=y' L_{y'}-L = \mathrm{const}$ with $L= (y-\lambda) \sqrt{1+y'^2}$ we have $H=(y-\lambda)/\sqrt{1+y'^2}$. So we have the 1st order equation from the very beginning:
\begin{equation}
\sqrt{1+y'^2}= A(y-\lambda)\implies \frac{dy}{A^2(y-\lambda)^2-1}=dx
\end{equation}
and integration gives the same answer
\begin{equation}
y= \frac{1}{A}\cosh (A(x-B)) +\lambda.
\end{equation}
We have three parameters and three equations
\begin{equation}
y(x_1)=y_1,\quad y(x_2)=y_2,\quad \int _{x_1}^{x_2} \sqrt{1+y'^2}\,dx = L.
\end{equation}

Since on Quiz if this problem is selected all these equations will be given (f.e. $y(0)=0$, $y(2)=3$ and $\int _0^3 \sqrt{1+y'^2}\,dx = 25$) one should calculate the integral in the third equation explicitly. And may be even calculate $\int _{x_1}^{x_2} y\sqrt{1+y'^2}\,dx= E$