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I want to find Euler-Lagrange equation for the following:

$$J(u) = \int \left( \frac{\psi(x) u + \dot{u}}{\psi(x)u - \dot{u}} \right)dx, \text{where} \ \psi(x) \ \text{is an explicit function of} \ x.$$

First, I have made the following substitution:

$$y = \frac{\dot{u}}{u} \implies \int \left( \frac{\psi(x) u + \dot{u}}{\psi(x)u - \dot{u}} \right)dx = \int \left( \frac{\psi(x) + y}{\psi(x) - y} \right)dx$$

This substitution should reduce the Euler-Lagrange equation to a first-order differential equation.

I know that the Euler-Lagrange equation, in general, looks like this:

$$\frac{d}{dt}(f_{\dot{x}}) - f_x = 0$$

Should the Euler-Lagrange for this particular functional look like this:

$$f(\psi(x), y, x) = \frac{\psi(x) + y}{\psi(x) - y} \implies \frac{d}{dx}(f_{\psi(x)}) - f_y = 0$$

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  • $\begingroup$ Is $\dot u = du/dx$? $\endgroup$ – David Nov 25 '14 at 1:10
  • $\begingroup$ Yes, $\dot u = du/dx$. $\endgroup$ – GaMbiTaaaa Nov 25 '14 at 1:25
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According to my calculations, this is ugly:

$$\dfrac{\partial L}{\partial u}=\dfrac{-2 \psi \dot u}{(\psi u-\dot u)^2},$$ $$\dfrac{\partial L}{\partial \dot u}=\dfrac{2 \psi u}{(\psi u-\dot u)^2},$$

with $L=\dfrac{\psi u+\dot u}{\psi u-\dot u}$. Euler-Lagrange equation says:

$$\dfrac{d}{dx}\dfrac{\partial L}{\partial \dot u}-\dfrac{\partial L}{\partial u}=0,$$

so:

$$\dfrac{d}{dx}\dfrac{\partial L}{\partial \dot u}=2\dfrac{-\psi u\dot u-\psi\dot u^2+2\psi u \ddot u-\psi u^2 \psi'-\psi^2 u \ddot u^2}{(\psi u-\dot u)^3}$$

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