Euler Lagrange Equation Question

In summary: If you are holding x constant (as you should be in a partial derivative with respect to y) then why would you also be holding y constant in the full derivative?In summary, for the given function f(y,y',x) = 2yy' + 3x2y and y(x) = 3x4 - 2x +1, we can calculate ∂f/∂x = 6xy = 6x(3x4 - 2x +1) and df/dx = (3x2 + 2(y * d/dy(y') + y')) * (12x3 - 2). The full derivative is not simply the product of the partial derivative and
  • #1
HiggsBrozon
7
0

Homework Statement



Consider the function f(y,y',x) = 2yy' + 3x2y where y(x) = 3x4 - 2x +1. Compute ∂f/∂x and df/dx. Write both solutions of the variable x only.


Homework Equations



Euler Equation: ∂f/∂y - d/dx * ∂f/∂y' = 0



The Attempt at a Solution



Would I first just find ∂f/∂y and ∂f/∂y' as follows:

∂f/∂y = 2y' + 3x2
∂f/∂y' = 2y

and then insert y(x) and y'(x) into my two equations for ∂f/∂y and ∂f/∂y':

y(x) = 3x4 - 2x +1
y'(x) = 12x3 - 2

→ ∂f/∂y = 2(12x3 - 2) + 3x2
→ ∂f/∂y' = 2(3x4 - 2x +1)

This is where I begin to get lost. Would I plug these values into my Euler equation to find an expression for df/dx and ∂f/∂x?

I would attempt this but I want to see if I'm headed in the right direction or not first. Any help would be much appreciated.
 
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  • #2
HiggsBrozon said:
Would I first just find ∂f/∂y and ∂f/∂y' as follows:
You don't need to find those at all. You don't need to use the Euler-Lagrange equations either. Just calculate the things they ask for in the question. I think they are just testing your knowledge of how to do partial differentiation and the full derivative.
 
  • #3
Thanks for you response. I feel as if I made this problem more complicated than it should.

For ∂f/∂x:

∂f/∂x = 6xy = 6x(3x4 - 2x +1)

and df/dx:

df/dx = ∂f/∂y * dy/dx
= d/dy(3x2y + 2yy') * (12x3 - 2)
= (3x2 + 2(d/dy(yy'))) * (12x3 - 2)
= (3x2 + 2(y * d/dy(y') + y')) * (12x3 - 2)

However I'm a bit confused on how to take the derivative of d/dy(y')
Would I just treat y' as a fraction and cancel out dy's like so: y * d/dy(dy/dx) = d/dx * y = y'
 
  • #4
I think your answer for ∂f/∂x is correct. But I don't agree with your answer for df/dx. The line: df/dx = ∂f/∂y * dy/dx I think is not right. This is not the correct expression for the full derivative. p.s. you can realize this by thinking what is being held constant for ∂f/∂y
 
Last edited:
  • #5


Your approach is correct so far. To find ∂f/∂x and df/dx, you would plug in the values for ∂f/∂y and ∂f/∂y' that you have calculated into the Euler equation:

∂f/∂y - d/dx * ∂f/∂y' = 0

This will give you an expression for df/dx. Then, you can use the chain rule to express df/dx in terms of x only, using the given function y(x). This will give you the final solution for df/dx in terms of x only. Similarly, you can use the chain rule to express ∂f/∂x in terms of x only, using both y(x) and y'(x). This will give you the final solution for ∂f/∂x in terms of x only. Remember to simplify your expressions as much as possible to get a final solution.
 

1. What is the Euler-Lagrange equation?

The Euler-Lagrange equation is a mathematical equation that is used to find the stationary points of a functional. It is commonly used in the field of calculus of variations.

2. How is the Euler-Lagrange equation derived?

The Euler-Lagrange equation is derived by taking the derivative of the functional with respect to the function that minimizes the functional. This results in a second-order ordinary differential equation.

3. What is the significance of the Euler-Lagrange equation?

The Euler-Lagrange equation is significant in that it provides a way to find the function that minimizes a given functional. This is useful in many areas of science and engineering, including physics, economics, and optimization problems.

4. Can the Euler-Lagrange equation be used for both single-variable and multi-variable functions?

Yes, the Euler-Lagrange equation can be used for both single-variable and multi-variable functions. In the case of multi-variable functions, the equation is known as the Euler-Lagrange equation for multiple variables.

5. Are there any limitations to using the Euler-Lagrange equation?

While the Euler-Lagrange equation is a powerful tool for finding the stationary points of a functional, it does have some limitations. It may not always be possible to find an explicit solution to the equation, and it may also fail to find all possible solutions.

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