Find the minimun of $\int_0^{t_1} \frac{1 + \dot{x(t)}^2}{x(t)} dt, x(0) = 0, x(t_1) = x_1$

My attempt

We have: $$L(x,\dot{x}) = \frac{1 + \dot{x}^2}{x}$$


$$\frac{\partial L}{\partial x} = \frac{1 + \dot{x}^2}{-x^2}, \quad \frac{\partial L}{\partial \dot{x}} = 2\frac{\dot{x}}{x}, \quad \frac{d}{dt}\frac{\partial L}{\partial \dot{x}} = 2\frac{\ddot{x}}{x}$$

From Euler-Lagrange equation:

$$\frac{\partial L}{\partial x} - \frac{d}{dt}\frac{\partial L}{\partial \dot{x}} = 0 \iff 1 + \dot{x}^2 + 2x\ddot{x} = 0$$

I couldn't solve this ODE.

The book says that the answer for this question is $x(t) = Kt^2 - t, K = \frac{x_1 - t_1}{t_1^2}$

How could solve that ODE?

  • $\begingroup$ That solution does not solve the ODE you gave. $\endgroup$ – Paul Sep 29 '16 at 12:15

$$\frac{\partial L}{\partial x} = \frac{1 + \dot{x}^2}{-x^2}$$ Correct. $$\quad \frac{\partial L}{\partial \dot{x}} = 2\frac{\dot{x}}{x}$$ Correct $$\quad \frac{d}{dt}\frac{\partial L}{\partial \dot{x}} = 2\frac{\ddot{x}}{x}$$ NOT correct. You have to use the quotient rule, since $x$ depends on $t$ also, so that $$\quad \frac{d}{dt}\frac{\partial L}{\partial \dot{x}} = 2\frac{x\ddot{x}-\dot{x}^2}{x^2}$$

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