A functional depending upon x and y'(x)

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In summary, the given problem involves finding the stationary path for a continuous differentiable function y(x) that satisfies a specific equation and minimizes the term O(ε) in the expansion of S[y + εh] - S[y]. After solving the equation and substituting the solution into the term, it is shown that the term O(ε) vanishes and the stationary path is given by y(x) = (n-2)(2n-1 -1)(1-x)/2n-1(n-1) + 1/(n-1)xn-1 + n-2/n-1. However, further guidance is needed on how to find the solution for y(x).
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mistereko
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Homework Statement



S[y] = [itex]\int[/itex]21dx ln(1 + xny'), y(1) = 1, y(2) = 21-n

where n > 1 is a constant integer, and y is a continuously differentiable function
for 1 ≤ x ≤ 2. Let h be a continuously differentiable function for 1 ≤ x ≤ 2 and ε
a constant. Let ∆ = S[y + εh] − S[y].

Show that if h(1) = h(2) = 0, the term O(ε) in this expansion vanishes if
y′(x) satisfies the equation

dy/dx = 1/c -1/xn


where c is a constant.
Solve this equation to show that the stationary path is

y(x) = (n-2)(2n-1 -1)(1-x)/2n-1(n-1) + 1/(n-1)xn-1 + n-2/n-1

Homework Equations





The Attempt at a Solution




Right, I've found S[y+ εh] -s[y] = [itex]\int[/itex]21dx (xnh(x)'/(1 + xny(x)')ε - ε/2 [itex]\int[/itex]21dx x2n(h(x)')2/(1+ xny(x)')2 + O(ε3)

I'm aslo pretty sure I've proved the O(ε) vanishes. I did this substituting y(x) = (n-2)(2n-1 -1)(1-x)/2n-1(n-1) + 1/(n-1)xn-1 + n-2/n-1 into ε term and got it to equal 0. I don't know how to find y(x) though. Through integration i got y(x) = 1/c - (1-n+1)/(-n+1)

I don't know how to progress from here or if I'm doing it correctly. Some guidance would be great! :)





 
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FAQ: A functional depending upon x and y'(x)

1. What does "A functional depending upon x and y'(x)" mean?

A functional depending upon x and y'(x) refers to a mathematical function that is expressed in terms of both a variable x and its derivative y'(x). This means that the output of the function is not only dependent on the value of x, but also on the rate of change of y with respect to x.

2. How is "A functional depending upon x and y'(x)" different from a regular function?

A functional depending upon x and y'(x) differs from a regular function in that it takes into account not only the value of x, but also the derivative of y with respect to x. This allows for a more precise and nuanced representation of the relationship between the variables.

3. Can you provide an example of "A functional depending upon x and y'(x)"?

One example of a functional depending upon x and y'(x) is the Euler-Lagrange equation in calculus of variations. This equation expresses a functional in terms of a function and its derivative, and is commonly used in optimization problems.

4. What are the applications of "A functional depending upon x and y'(x)"?

A functional depending upon x and y'(x) has various applications in mathematics, physics, and engineering. It is commonly used in differential equations, optimization problems, and variational calculus. It also has applications in fields such as mechanics, thermodynamics, and fluid dynamics.

5. How is "A functional depending upon x and y'(x)" used in scientific research?

In scientific research, "A functional depending upon x and y'(x)" can be used to model and analyze complex systems and phenomena. It allows for a more accurate and comprehensive understanding of the relationships between variables and can help researchers make predictions and draw conclusions. It is also used in the development of mathematical models and theories in various fields of study.

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