calculus of variations problem

45 Views Asked by At

Find the general solution of the Euler equation corresponding to the functional $J[q]=\int_{a}^{b} f(t)\sqrt{1+q'^2} dt$ and investigate the special cases $ f(t)=\sqrt{t}$ and $f(t)=t$

What I got to work

$L(t,q,q')=t\sqrt{1+q'^2}$

$L_{q}(t,q,q')-\frac{d}{dt}\left ( L_{q'}(t,q,q') \right)=0$

$L_{q}(t,q,q')=0$

$L_{q'}(t,q,q')=t*\frac{1}{2\sqrt{1+q'^2}}*2q'=\frac{tq'}{\sqrt{1+q'^2}}$

$-\frac{d}{dt}\left ( L_{q'}(t,q,q') \right)=0$

$-\frac{d}{dt}\left (\frac{tq'}{\sqrt{1+q'^2}} \right)=0$

$\frac{q'\sqrt{1+q'^2}-tq'\frac{1}{2\sqrt{q'^2+1}}*2q'}{1+q'^2}=0$

HELP!!

We cannot solve this exercise when we start to get stuck, without answering thank you

1

There are 1 best solutions below

3
On

From

$$ -\frac{d}{dt}\left (\frac{tq'}{\sqrt{1+q'^2}} \right)=0 $$

we have

$$ \frac{tq'}{\sqrt{1+q'^2}}=C_0 $$

and following

$$ q'=\frac{\pm C_0}{t^2-C_0^2} $$

etc.

NOTE

If $f(t) = \sqrt{t}$ then

$$ q'=\frac{\pm C_0}{t-C_0^2} $$