Calculus by variations Euler's eq

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In summary, the conversation is discussing a problem in classical mechanics involving the sale of stocks. The money made from selling the stocks can be represented as an integral with respect to time and the price per share depends on how the stocks are sold. The problem asks to find the optimal strategy for selling the stocks in a "bear market" scenario and to discuss the effects of a large B value. The attempted solution involves checking relevant equations and finding an auxiliary equation to account for the initial and final values of N.
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Homework Statement



This is for Classical mechanics 2

I'm not sure how to put partials in itex and the N with the dot beside it was supposed to be the derivative of N with respect to time

Suppose that you have N0 shares of stock and you want to make as money as you can by selling all of them in a single day. If N0 is large, and you sell your shares in small batches, the money you make can be written apporximately as an integral

[itex]\int dt N^{.}P(N,N^{.};t)[/itex] t1 to t2

where t1 and t2 are the opening and closing times for the stock exchange, N(t) is a smooth function that is approximately the number of shares you have sold at time t (which satisfies N(t1) = 0 and N(t2) = N0), N^{.} is the rate at which you sell the stock (in shares per hour, say) and P(N, N^{.};t) is the price per share as a function of time. The interesting thing is that the price depends on how you sell the shares. For example, if you selll them too fast, the price will drop. That is why P depends on N and N^{.}.

a) Suppose that P(N,N^{.};t) = P0 - Bt -CN^{.} for P0, B, C all positive (This is a "bear market" because of the -Bt term as the stock price is going down with time). Find N(t) and N^{.} that allow you to make the most money.

b) Discuss briefly what happens if B is too large.

Homework Equations



[itex]\frac{dP}{dN} - \frac{d}{dt}(\frac{dP}{dN^{.}} = 0[/itex]

The Attempt at a Solution



I've really got nothing for this because I'm unsure of how to start minus checking the relevant equation

[itex]\frac{dP}{dN} = 0 = \frac{d}{dt}\frac{dP}{dN^{.}}[/itex]

so

[itex]\frac{dP}{dN^{.}} = -C[/itex]

and i think there has to be an auxillary equation to do with the fact that N(t1) = 0 and N(t2) = N0

Homework Statement


Homework Equations


The Attempt at a Solution

 
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can anyone give me an idea
 

Related to Calculus by variations Euler's eq

1. What is Calculus by variations Euler's equation?

Calculus by variations Euler's equation is a mathematical concept that involves finding the extremum (maximum or minimum) of a functional by varying its inputs. It is named after the famous mathematician Leonhard Euler.

2. What is the significance of Calculus by variations Euler's equation?

Calculus by variations Euler's equation has many practical applications in physics and engineering, such as in the study of fluid dynamics and optimization problems. It also serves as the foundation for the calculus of variations, a branch of mathematics that deals with finding the optimal path or function for a given problem.

3. How is Euler's equation derived?

Euler's equation is derived using the calculus of variations, specifically the Euler-Lagrange equation. This involves setting up a functional (a mathematical expression involving a function) and finding the stationary points by varying the function. The resulting Euler-Lagrange equation is then solved to obtain Euler's equation.

4. What is the relationship between Euler's equation and the principle of least action?

Euler's equation is closely related to the principle of least action, which states that systems tend to move towards a state of minimal action. This principle is used in physics and is derived from Euler's equation, making it a fundamental concept in the study of motion and dynamics.

5. Are there any limitations to using Calculus by variations Euler's equation?

While Calculus by variations Euler's equation is a powerful tool for solving optimization problems, it does have some limitations. It is most effective for problems with well-defined and smooth functions, and may not work for more complex or discontinuous functions. Additionally, it can sometimes produce multiple solutions, so it is important to carefully analyze the results to determine the most appropriate solution.

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