I have the following problem in my Calculus of Variations course:
Find all smooth extremums if $a,b$ and $c$ are positive numbers
$$\min\int_0^1(ay^2+2byy'+cy'^2)\;dx,\;y(0)=0, y(1)=1$$
I have tried solving this for few days with the basic techniques by applying the Euler-Lagrange equation but still I get a very hairy calculations which make me start to doubt about the correctness of my solution. I would need a guiding light on this one. How to solve this?
Thank you!
With the Lagrangian
$$L(y,y') = ay^2 + 2byy' + cy'^2$$
the EL equation becomes
$$\frac{\partial L}{\partial y} = \frac{d}{dx}\frac{\partial L}{\partial y'}\implies 2ay + 2by' = \frac{d}{dx}[2by + 2cy']$$
which simplifies (for $c\not = 0$; if $c=0$ then there are no smooth solutions) to
$$y'' = \frac{a}{c}y \implies y(x) = A e^{\omega x} + B e^{-\omega x}$$
where $\omega^2 = a/c$ and $A,B$ are free constants (if $a=0$ the equation to solve becomes $y''=0$). The solution that satsify the boundary condtions $y(0)=0$ and $y(1) = 1$ is
$$y(x) = \frac{e^{\omega x} -e^{-\omega x}}{e^{\omega} - e^{-\omega}} = \frac{\sinh(\omega x)}{\sinh(\omega)}$$
If $a/c < 0$ then $\omega$ is imaginary. In this case you can use $\sinh(iz) = i\sin(z)$ to write down the solution (if you like to keep it real).
