How to find the curve extremizing a given functional?

Question

Given a functional $$I(y)=\int_1^2 {\frac {\sqrt{1+(y'(x))^2}}{x}}dx ,$$ with $y(1)=0$ and $y(2)=1$.

How to find the curve extremizing this functional?

Answer 1

For the functional $$ I[y] = \int\limits_a^b L(x, y, y') \, dx $$ you need to solve the Euler-Lagrange equation $$ 0 = \frac{\delta I}{\delta y} = \frac{\partial L}{\partial y} -\frac{d}{dx} \frac{\partial L}{\partial y'} = - \frac{d}{dx}\left(\frac{y'}{x\sqrt{1+(y')^2}}\right) $$ under the given conditions. That means \begin{align} c^2 x^2 (1+ (y')^2) &= (y')^2 \iff \\ y' &= \pm \frac{c x}{\sqrt{1-c^2 x^2}} \Rightarrow \\ y &= \mp \frac{\sqrt{1-c^2 x^2}}{c} + d \end{align} Those are scaled and $y$-translated half circles. Further the two conditions on $y$ should determine the integration constants $c$ and $d$. \begin{align} 0 &= \mp \frac{\sqrt{1-c^2}}{c} + d \\ 1 &= \mp \frac{\sqrt{1-4 c^2}}{c} + d \end{align} This means $$ d = \pm \frac{\sqrt{1-c^2}}{c} = 1 \pm \frac{\sqrt{1-4c^2}}{c} $$ which gives \begin{align} 1 &= \pm \left(\frac{\sqrt{1-c^2}}{c} - \frac{\sqrt{1-4c^2}}{c}\right) \\ &= \frac{1-c^2+1-4c^2-2\sqrt{(1-c^2)(1-4c^2)}}{c^2} \end{align} and $$ 1-3c^2 = 1-5c^2+4c^4 \iff 1 = 2c^2 \iff c = \pm 1 / \sqrt{2} $$ and this means I made an error somewhere, because $d$ would be imaginary.