# Calculus of variations question

• I
• Avatrin
In summary, the conversation discusses a function x^* that maximizes or minimizes an integral with specific endpoint conditions. It is stated that x^* must satisfy the Euler equation and the transversality condition, which ensures that the functional is maximized (or minimized). The condition \left(\frac{\partial F}{\partial \dot{x}}\right)_{t=t_1} \leq 0 is necessary to ensure that the value of the functional does not increase with allowed variations. This is equivalent to requiring that x^* maximizes the functional.
Avatrin
Hey, I have a theorem I cannot prove.

We have a function $x^*$ that maximizes or minimizes the integral:
$$\int^{t_1}_{t_0} F(t,x(t),\dot{x}(t))dt$$

Our end point conditions are:
$$x(t_0) = x_0, x(t_1) \geq x_1$$

I am told that $x^*$ has to satisfy the Euler equation. That I can fully understand since $x^*(t_1)$ can be equal to $x_1$. However, then it gives me the transversality condition:
$$\left(\frac{\partial F}{\partial \dot{x}}\right)_{t=t_1} \leq 0 \text{ ( = 0 if x^*(t_1) > x_1)}$$

I can understand the statement in the parentheses. However, I do not understand why $\left(\frac{\partial F}{\partial \dot{x}}\right)_{t=t_1}$ must be less than or equal to zero if $x^*(t_1) = x_1$. Why can it not be more than zero?

The original variation of the functional (call the functional ##\mathcal F##) is given by
$$\delta \mathcal F = \int_{t_0}^{t_1} \left( \frac{\partial F}{\partial x} \delta x + \frac{\partial F}{\partial \dot x} \delta \dot x\right) dt.$$
Integration by parts of the second term now leads to
$$\delta \mathcal F = \int_{t_0}^{t_1} \left( \frac{\partial F}{\partial x} - \frac{d}{dt} \frac{\partial F}{\partial \dot x} \right) \delta x \, dt + \left[\frac{\partial F}{\partial \dot x} \delta x\right]_{t=t_0}^{t_1}.$$
Since ##\delta x## is arbitrary, the Euler-Lagrange equation has to hold if ##x## is a stationary function of the functional. This, together with ##\delta x(t_0) = 0## leads to
$$\delta \mathcal F = \left. \frac{\partial F}{\partial \dot x}\right|_{t=t_1} \delta x(t_1).$$
It follows that to have a local stationary function, you must have
$$\left. \frac{\partial F}{\partial \dot x}\right|_{t=t_1} = 0.$$
However, there is also the option of having a function at the boundary ##x(t_1) = x_1##. If you are at the boundary, then only variations ##\delta x(t_1) \geq 0## are allowed and therefore ##\delta \mathcal F## can only be positive if
$$\left. \frac{\partial F}{\partial \dot x}\right|_{t=t_1} > 0$$
implying that the value of ##\mathcal F## would increase with the allowed variation. Your condition
$$\left. \frac{\partial F}{\partial \dot x}\right|_{t=t_1} \leq 0$$
is therefore equivalent to requiring that ##\delta \mathcal F \leq 0##, i.e., that ##x^*## maximises the functional (at least locally). For a minimisation, you would get the opposite inequality.

## 1. What is the purpose of calculus of variations?

The purpose of calculus of variations is to find the path, curve, surface, or function that minimizes or maximizes a given functional. This is useful in various fields such as physics, engineering, economics, and more.

## 2. How is calculus of variations different from traditional calculus?

Traditional calculus deals with finding the optimal value of a function at a specific point, while calculus of variations deals with finding the optimal curve or path for a given functional. It also involves using the Euler-Lagrange equation, rather than the usual derivative, to solve for the optimal curve.

## 3. What is the Euler-Lagrange equation?

The Euler-Lagrange equation is a necessary condition for finding the optimal curve or path in calculus of variations. It is derived from the principle of stationary action and involves taking the derivative of the functional with respect to the function and setting it equal to zero.

## 4. What are some real-world applications of calculus of variations?

Calculus of variations has many applications in fields such as physics, economics, and engineering. Some examples include finding the optimal path for a projectile, minimizing the surface area of a soap bubble, and maximizing profits for a company.

## 5. What are the limitations of calculus of variations?

Calculus of variations can only be used for continuous functions, and it may not always produce a unique solution. It also involves complex mathematical calculations, making it difficult to apply in some real-world situations. Additionally, it is limited to finding the optimal curve for a given functional and cannot be used for optimization problems with constraints.

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