In this Lagrangian problem, action is $$S = \int_{t_1}^{t_2} \sqrt{t}\sqrt{1+\dot{y}^2} \,\,dt$$ where $\dot{y} = dy/dt$ and $t_1$ and $t_2$ are some fixed points.

I tried to solve this problem using Euler-Lagrange:

$$L = \sqrt{t}\sqrt{1+\dot{y}^2},$$ so $\frac{\partial{L}}{\partial y} = 0$ and therefore $\frac{\partial L}{\partial \dot{y}} = K$ where $K$ is some constant.

Now after some substitutions, I get: $$2K\sqrt{t} + C = \sqrt{1+\dot{y}^2}$$ where $C$ is some constant.

We can square LHS and RHS and get: $$(2K\sqrt{t} + C)^2 -1 = \dot{y}^2.$$

And then we can substitute $u = \dot{y}$ and we get:

$$\sqrt{(2K\sqrt{t} + C)^2 -1}\,\,dx = dy.$$

But integrating this results in a weird answer, because I have a solution that says the answer is of form $2\sqrt{A(t-A)}+B$. What should be done to achieve this solution?

  • 1
    $\begingroup$ where are you getting dx from? $\endgroup$ – Jerry Schirmer Oct 31 '14 at 16:31

The Lagrangian provided is given by,


Clearly, $\partial L/\partial y = 0$. We compute the second term in the Euler-Lagrange equations:

$$\frac{d}{dt}\frac{\partial L}{\partial \dot{y}} = \frac{d}{dt} \left( \frac{t\dot{y}}{\sqrt{t+\dot{y}^2 t}} \right) = \frac{t(\dot{y} + \dot{y}^3 + 2t\ddot{y})}{2[t(1+\dot{y}^2)]^{3/2}} = 0$$

from which we can conclude,

$$\frac{t\dot{y}}{\sqrt{t+\dot{y}^2t}} = a, \quad a \in \mathbb{C}.$$

Re-arranging and squaring, we obtain $t^2 \dot{y}^2 = a^2 \left(t+t\dot{y}^2 \right)$. This is a second degree polynomial in $\dot{y}$, and solving for $\dot{y}$ we obtain,

$$\dot{y} = \pm \frac{a}{\sqrt{t-a^2}}$$

Now straightforwardly integrate by substitution:

$$y(t)=\pm \int dt\,\frac{a}{\sqrt{t-a^2}} = \pm 2a \sqrt{t-a^2} + C$$


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