CoV20

From Exampleproblems

Minimize $J(y)=\int 2\pi y \sqrt{1+y'^2} dx\,$

Our functional does not depend explicitly on x, so we can use the first integral $F - y' F_{y'} = c_1 \,$ .

$y \sqrt{1+y'^2} - \frac{y y'^2}{\sqrt{1+y'^2}} = c_1$

$\frac{y}{\sqrt{1+y'^2}} = c_1$

$y' = \frac{\sqrt{y^2-c_2^2}}{c_2^2}$

$dx = \int \frac{c_2 dy}{\sqrt{y^2-c_1^2}}$

Letting $y = c_1 \cosh t, dy = c_1 \sinh t dt \,$ , we get:

$dx = \int \frac{c_1^2 \sinh t dt}{c_1 \sinh t} = \int c_1 dt$

$x = c_1 t + c_2 \,$

$y = c_1 \operatorname{arccos} \frac{x-c_2}{c_1}$

This problem is equivalent to finding the equation of a hanging cable of uniform density with fixed endpoints. To minimize the total potential energy, we make our functional equal to the height of the cable multiplied by the arc length.

Calculus of Variations

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