# Complex solution to Euler-Lagrange equation?

I'm currently working on Calculus of Variations and I came across an integral which I had to minimize. The integral I have to minimize is $$\int_0^1(1+y'^2)^2\,dx$$ After getting the Euler-Lagrange equation for this it has 3 solutions. One real and two imaginary solutions. Now my question is that are the complex solutions meaningful solutions since we're looking for a function that minimizes the integral, which is real.

Physically it would be meaningful to choose only the real solution but this is purely a mathematics course, so are complex solutions to EL-equation regarded as valid solutions?

Edit: The EL-equation I got is: $$4y''(1+3y'^2)=0$$ Which gives me $$y(x)=c_1x+c_2,\quad \text{or} \quad y(x)=\pm \frac{i}{\sqrt3}x+c_3$$

• Can you show more of your working and what are the three solutions? As you know, sometimes, we take complex solutions and turn them back into real ones. But it will depend on the particular form. – Simon S Jan 9 '15 at 17:03
• Here is a paper on complex calculus of variation. I am not sure if it will answer your question but it should be of some of worth to look it. Complex solutions tie into the Fenchel transform. – dustin Jan 9 '15 at 17:06
• I think that OP encountered a multiplier in final ODE that is strictly positive for any real-valued function $y(x)$. So it's the case when it can be thrown away and complex-valued solutions don't play here any role. – Evgeny Jan 9 '15 at 17:09

First of all, note how your second solution is merely a special case $c_1=\pm\frac i{\sqrt3}, c_2=c_3$ of the first solution. So you might as well pick $c_1=i$ to obtain a vanishing integral, which would be quite minimal. But you should stick with real values and thus plug your first solution into the integral and minimize it with respect to $c_1$ (spoiler: $c_1=0$).
However, once you do allow for complex values, $\int_0^1(1+y'^2)^2\,dx$ no longer makes sense for minimization, since that integral can become complex-valued and thus there is no order by which to determine what is "minimal". You need to involve absolute values, e.g. either $\int_0^1(1+|y'|^2)^2\,dx$ or $\int_0^1|1+y'^2|^2\,dx$. And since you're dealing with complex functions now, $y$'s complex conjugate $\bar y$ is an independent function (alternatively you can split $y$ into real and imaginary part to obtain to independent real valued functions), i.e. you also have to consider $-\frac d{dx}\frac{\partial}{\partial \bar y'}$ to obtain a second Euler-Lagrange equation, coupled to the other one.