Calculus of variations: Euler-Lagrange

In summary: You'll need to take the derivatives with respect to x, y, and the derivatives of x and y, and then set the resulting expression equal to zero. This will give you the two Euler-Lagrange equations for this specific problem. Then, for part (b), you need to use the two equations you found in part (a) and compare them to the given equation, (x-\alpha)\dot{x}+(y-\beta)\dot{y}=0, and see if you can manipulate them to match. This will show that the extremal curves satisfy this equation, as well as finding the values of \alpha and \beta.In summary, the problem is to find the Euler-Lagrange equations governing extrema of I
  • #1
jonz13
1
0
This is from a past paper (from a lecturer I don't particularly understand)

Homework Statement


a) {4 marks} Find the Euler-Lagrange equations governing extrema of [itex] I [/itex] subject to [itex] J=\text{constant} [/itex], where[tex]I=\int_{t_1}^{t_2}\text{d}t \frac{1}{2}(x\dot{y}-y\dot{x})=\int f(t,x,y,\dot{x},\dot{y})[/tex]
and[tex]J=\int_{t_1}^{t_2}\text{d}t (\dot{x}^2+\dot{y}^2)=\int g(t,x,y,\dot{x},\dot{y})[/tex]
b) {8 marks} show that for the problem in part a) the extremal curves satisfy [itex](x-\alpha)\dot{x}+(y-\beta)\dot{y}=0[/itex] where [itex]\alpha[/itex] and [itex]\beta[/itex] are constants.

Homework Equations


From an earlier part of the question I have two Euler-Lagrange equations (one differentiating w.r.t. [itex]y[/itex] aswell)[tex]\frac{\partial (f-\lambda g)}{\partial x}-\frac{\mathrm{d} }{\mathrm{d} t}\frac{\partial (f-\lambda g)}{\partial \dot{x}}=0[/tex]
and I think I can write, due to no dependence on [itex]t[/itex] (another one with [itex]y[/itex] again)[tex](f-\lambda g) - \dot{x}\frac{\partial (f-\lambda g)}{\partial \dot{x}}=\mathrm{constant}[/tex]

The Attempt at a Solution


For part a) I'm not particularly sure what I am being asked for, or if the equation above is the answer. for part b) I have tried subbing into the equations above and can get out linear equations for [itex]x(t) \text{ and } y(t)[/itex] and get a few dead ends, I'm not really sure what approach to use (a definite answer to part a) would probably help).
 
Physics news on Phys.org
  • #2
For part (a), you want to take the specific f and g you've been given and substitute them into the general Euler-Lagrange equation you cited as a relevant equation.
 

FAQ: Calculus of variations: Euler-Lagrange

What is the Euler-Lagrange equation?

The Euler-Lagrange equation is a fundamental equation in the calculus of variations, which is used to find the optimal path or curve for a given functional. It is derived from the principle of least action and is used to determine the stationary points of a functional.

How is the Euler-Lagrange equation derived?

The Euler-Lagrange equation is derived by applying the calculus of variations to a functional. This involves taking the functional and varying it with respect to the unknown function, setting the variation equal to zero, and solving for the unknown function. This process leads to the Euler-Lagrange equation.

What is the significance of the Euler-Lagrange equation?

The Euler-Lagrange equation is significant because it provides a necessary condition for a function to be a stationary point of a given functional. This means that the function satisfies the principle of least action and represents the optimal path or curve for the given problem. It is used in many areas of mathematics, physics, and engineering.

What are some real-world applications of the Euler-Lagrange equation?

The Euler-Lagrange equation is used in various real-world applications, such as in mechanics for optimizing the path of a moving object, in economics for optimizing production or cost functions, and in control theory for finding the optimal control path for a system. It is also used in quantum mechanics for calculating the wave function of a particle.

Are there any limitations to the use of the Euler-Lagrange equation?

While the Euler-Lagrange equation is a powerful tool for solving optimization problems, it does have some limitations. It can only be used for problems with well-defined boundary conditions and for which the functional is differentiable. It also does not guarantee a global minimum, and in some cases, multiple solutions may exist. Additional techniques may be needed to verify the optimality of the solution.

Back
Top