# Give the EL eqns of $\int_0^1 (\dot x_1^2 + \dot x_2^2 - k^2(x_1 + x_2)^2) dt$ and solve the coupled system of 2nd order ODEs

Find the EL eqns of $$\displaystyle \int_0^1 (\dot x_1^2 + \dot x_2^2 - k^2(x_1 + x_2)^2)\: dt$$ where $$x_1 = x_1(t), x_2=x_2(t)$$, $$k$$ constant and solve the system of equations

E-L eqns from: $$\displaystyle 0 = \frac{\partial f}{\partial x_i} - \frac{d}{dt}\left(\frac{\partial f}{\partial \dot{x_i}}\right)$$ and $$f(t,x_1,x_2, \dot x_1, \dot x_2) = \dot x_1^2 + \dot x_2^2 - k^2(x_1 + x_2)^2$$

I have \begin{align*} 0 &= \frac{\partial f}{\partial x_1} - \frac{d}{dt}\left(\frac{\partial f}{\partial\dot x_1 }\right)\\ &= -2k^2(x_1 + x_2) - \frac{\partial}{\partial t}2\dot x_1\\ \Longrightarrow 0 &= \ddot x_1 + k^2(x_1 + x_2)\\ \text{and } 0 &= \frac{\partial f}{\partial x_2} - \frac{d}{dt}\left(\frac{\partial f}{\partial \dot x_2}\right)\\ &= -2k^2(x_1 + x_2) - \frac{d}{dt}2\dot x_2\\ \Longrightarrow 0 &= \ddot x_2 + k^2(x_1 + x_2) \end{align*} Is

\begin{align*} 0 &= \ddot x_1 + k^2(x_1 + x_2)\\ 0 &= \ddot x_2 + k^2(x_1 + x_2) \end{align*}

the system of equations?

And how do I find the general solution to this system?

The system of equations you have looks correct to me. To solve the system, note that by setting: $$u=x_1+x_2,\quad v=x_1-x_2$$ We obtain the system (to obtain the first equation, add the two equations in your system, then subtract them for the second equation): $$\ddot{u}+2k^2 u=0$$ $$\ddot{v}=0$$ This is an uncoupled set of ODE's. After solving this, we can recover $$x_1(t)$$ and $$x_2(t)$$ using that: $$x_1=\frac{u+v}{2},\quad x_2=\frac{u-v}{2}$$
• very neat solution. The question asks for the general solution, however the integral is real so do you think I should take the real part of $x_1$ and $x_2$? Commented Sep 5, 2019 at 10:20
• If you express the general solution for the $u$ equation as a linear combinations of sines and cosines instead of complex exponentials, you don't have to worry about taking real parts, just take the arbitrary constants in the solution to be real. Commented Sep 5, 2019 at 10:23