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

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