### Author Topic: HA10-P1  (Read 1123 times)

#### Victor Ivrii

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##### HA10-P1
« on: November 28, 2015, 11:36:40 AM »

#### Jeremy Li 2

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##### Re: HA10-P1
« Reply #1 on: November 29, 2015, 01:38:21 AM »
We need to minimize the Lagrange functional. By subtracting the constraint multiplied by the Lagrange multiplier $\lambda$ from the original functional, we get:

\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

The variation of $\Phi$ is:

\delta\Phi=\int_0^a \left(\frac{\partial L}{\partial u} - \frac{\partial}{\partial x}\frac{\partial L}{\partial u'}\right)\delta u dx

(The boundary term is zero since $\delta u|_0 = \delta u|_a = 0$)

We need to solve the Euler-Lagrange equation.

\frac{\partial L}{\partial u} - \frac{\partial}{\partial x}\frac{\partial L}{\partial u'} = 0

for

L = (\rho gu-\lambda)\sqrt{1 + u'^2}

Calculating the two terms...

\frac{\partial L}{\partial u} = \rho g\sqrt{1+u'^2}

\frac{\partial L}{\partial u'} = (\rho gu-\lambda) \frac{1}{\sqrt{1+u'^2}} u'

\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}}}

Plugging these results into $(3)$:

\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

Multiplying every term by $(1+u'^2)^{\frac{3}{2}}$:

\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''

Dividing by $\rho g$,

(1+u'^2)=\left(u - \frac{\lambda}{\rho g}\right) u''

Doing some rearranging:

\frac{1}{1+u'^2} u''=\frac{1}{u - \frac{\lambda}{\rho g}}

Multiplying both sides by u'

\frac{1}{1+u'^2} u' u''=\frac{1}{u - \frac{\lambda}{\rho g}} u'

Integrating both sides with respect to $x$

\frac{1}{2} \ln{\left(1+u'^2\right)} = \ln{\left(u-\frac{\lambda}{\rho g}\right)} + C

Exponentiating:

1+u'^2=A^2\left(u-\frac{\lambda}{\rho g}\right)^2

By separation of variables

\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}}

We get

\cosh(Ax+B)=A\left(u-\frac{\lambda}{\rho g}\right)

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

What remains is to apply the boundary conditions. I'll figure the rest out tomorrow.
« Last Edit: November 29, 2015, 04:04:22 AM by Jeremy Li 2 »

#### Victor Ivrii

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##### Re: HA10-P1
« Reply #2 on: November 29, 2015, 02:04:54 PM »
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:

\sqrt{1+y'^2}= A(y-\lambda)\implies \frac{dy}{A^2(y-\lambda)^2-1}=dx

and integration gives the same answer

y= \frac{1}{A}\cosh (A(x-B)) +\lambda.

We have three parameters and three equations

y(x_1)=y_1,\quad y(x_2)=y_2,\quad \int _{x_1}^{x_2} \sqrt{1+y'^2}\,dx = L.

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 $\int _{x_1}^{x_2} y\sqrt{1+y'^2}\,dx= E$
« Last Edit: December 02, 2015, 08:19:46 AM by Victor Ivrii »