# Calculus of variations for the functional $I[x]=\int_{t_0}^{t_1}\frac{1}{x} + \frac{(\dot{x})^2}{x} \ dt$

My question concerns the functional $$I[x]=\int_{-4}^{4}\frac{1}{x} + \frac{(\dot{x})^2}{x} \ dt$$ and what $x$ makes the functional stationary, where $(t, x)$ ranges from $(-4, 3)$ to $(4, 3)$. I have obtained the following from Euler-Lagrange: $2x\ddot{x}=(\dot{x})^2-1$. I am told that the correct answer has the form $x(t) = \frac{1}{c}-\frac{ct^2}{4}$ for some constant $c$.

Where am I going wrong? The DE I obtained seems way too complicated and I cannot think of a way of solving it. I am fairly new to Calc. of Variations, so a full solution would be much appreciated.

• You should show how you obtained the equation from Euler-Lagrange, because we have no super guessing powers. – Tom-Tom Feb 2 '18 at 23:01
• It's not too difficult, so I haven't included it. Assuming you know $\frac{d}{dt}\big(\frac{F}{d\dot{x}}\big) = \frac{dF}{dx}$, where $F$ is the integrand. – wrb98 Feb 2 '18 at 23:02
• OK, then. Where are you stuck ? – Tom-Tom Feb 2 '18 at 23:06
• Trying to solve the DE I provided above and get to the given form of the solution – wrb98 Feb 2 '18 at 23:28
• If this question was put on hold, it's because you didn't answer precisely to the question. You have to show what you have done explicitely : writing down the equations, name the theorems you use etc, even if that has failed. – Tom-Tom Feb 4 '18 at 22:29

## 1 Answer

The DGL, you have derive is correct. You question is how to solve it. For this it is import to know that given the fact that the DGL derives from a Lagrangian gives insights towards the solution.

In particular, in this case, the Lagrangian does not depend on $t$. With that you immediately can integrate the DGL once and reduce it to a first order DGL! In fact, what is called the `energy' $$E = \dot x \frac{\partial L}{\partial \dot x} - L = \frac{\dot x^2 -1}{x} \tag{1}$$ is conserved (that is it does not change over time).

This can be easily checked $$\frac{d}{dt} E = \frac{d}{dt}\frac{\dot x^2 -1}{x} =\frac{2 x \ddot x}{x} - \frac{\dot x (\dot x^2 -1)}{x^2} =\frac{\dot x (2 x \ddot x -\dot x^2 +1)}{x^2} = 0$$ due to the Euler-Lagrange equation.

So the remaining task is to integrate (1) once. The equation is separable and we obtain $$\int_3^x \frac{dy}{\sqrt{E y -1}}= \frac{2}{E} \Bigl(\sqrt{1+ Ex}-\sqrt{1+3E}\Bigr) = t+4\;. \tag{2}$$ The boundary condition $(4,3)$ immediately determines the constant $$E=-\frac{1}{4}\;.$$

Solving (2) for $x$ yields the result $$x= 4-\frac{1}{16} t^2\;.$$

• How did you make use of the boundary condition, because once 4 and 3 are substituted into (2), the LHS disappears. – wrb98 Feb 3 '18 at 13:36
• @Will: you can first solve for $x$ and then you will see that $E=-1/4$ is required to have the boundary condition $(3,4)$. – Fabian Feb 3 '18 at 13:39
• It becomes very messy if you square and rearrange and I end up with another solution for E. Why is it the case that when the boundary conditions are substituted into (2), we end up with 0 = 8? – wrb98 Feb 4 '18 at 12:47
• @Will: $\sqrt{\cdot}$ can (and should have) two values. They are $\pm\sqrt{\cdot}$. So the simpler (but conceptually more difficult) approach is to change the sign of $\sqrt{1+Ex}$ for the second boundary condition... – Fabian Feb 4 '18 at 15:13