Euler Lagrange Equation


Euler-Lagrange Equation for $\int F(x, y, y’) dx$:

$$\frac{d}{dx} \left( \frac{\partial F}{\partial y’} \right) = \frac{\partial F}{\partial y}$$

– Dependent variable missing, E-L equation becomes:

$$\frac{\partial F}{\partial y’} = constant$$

– Independent variable missing, E-L equation becomes:

$$F – y’\frac{\partial F}{\partial y’} = constant$$

_____________

Several dependent variables:

E.g. $F = F (x, y_1 , y_2 , … , y_n , y’_1 , … , y’_n )$

$$\frac{\partial F}{\partial y_i} = \frac{d}{dx}\left( \frac{\partial F}{\partial y’_i} \right)$$

Several independent variables:

$$\frac{\partial F}{\partial y} = \sum_{i=1}^{n} \frac{\partial}{\partial x_i} \left( \frac{\partial F}{\partial y_{x_{i}}}\right), y_{x_{i}} = \frac{\partial y}{\partial x_i}$$

____________

Variable end points:

$$I = \int^b _a F(x,y,y’) dx$$

If lower end point a is fixed, we require:

$$\left. \frac{\partial F}{\partial y’} \right|_{x = b} = 0$$

If both ends can vary, we require:

$$\left. \frac{\partial F}{\partial y’} \right|_{x = a} = 0,$$

$$\left. \frac{\partial F}{\partial y’} \right|_{x = b} = 0$$

______________

Constraint

$$I = \int^b _a F(x,y,y’) dx$$

Constraint: $J = \int^b _a G(x,y,y’) dx$

\begin{array} {lcl} K & = & I + \lambda J \\ & = & \int^b _a F + \lambda G dx \end{array}

 

Back To Mathematics For An Undergraduate Physics Course


Sharing is caring:
Mini Physics

Administrator of Mini Physics. If you spot any errors or want to suggest improvements, please contact us.

Leave a Comment