Functional Derivatives/Euler-Lagrange

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Homework Statement

Hi, I'm working on research and I hit a roadblock with something that should be very simple but I can't solve it because it gets so messy. If anyone can let me know how to do this, it would be greatly appreciated.

I have a functional T:
T = \int_{\lambda_{1}}^{\lambda_{2}} sqrt{\sum_{I=1}^{n}} \sum_{i=1}^{d} (\frac{d}{d \lambda}(\sum_{j=1}^{d} s(\lambda) R_{j}^{i}(\lambda)(q_{I}^{j}(\lambda) + a^{j}(\lambda))))^2} d \lambda

I need to take functional derivatives with respect to each function defining T and find when they are all concurrently zero. I believe, the Euler-Lagrange equation is able to do this?

I found what [tex]\frac{\partial{f}}{\partial{g}}[/tex] is, where g is just a place holder for [tex]s(\lambda), R_{j}^{i}(\lambda), a^{j}(\lambda), q_{I}^{j}(\lambda)[/tex]

Everytime I tried substituting say, df/dR, or df/ds I would get a huge 22-term, with 5 derivatives, impossible equation that I would have to take f with respected to, where [tex]T = \int_{\lambda_{1}}^{\lambda_{2}} f d \lambda.[/tex]

I'm looking at this euler-lagrange form:
[tex]\frac{\partial{f}}{\partial{x}} = \frac{d}{d \lambda} (\left \frac{\partial{f}}{\frac{d}{d \lambda}(\frac{\partial{f}}{\partial{x}})} \right) = 0 [/tex]
Where, each member [tex]s(\lambda), R_{j}^{i}(\lambda), a^{j}(\lambda), q_{I}^{j}(\lambda)[/tex] are in terms of "x".

Any tips, advice, ideas would be great.

Answers and Replies

  • #2
Homework Equations The equation defining T is given in the statement. The Euler-Lagrange equation is as follows: \frac{\partial{f}}{\partial{x}} = \frac{d}{d \lambda} (\left \frac{\partial{f}}{\frac{d}{d \lambda}(\frac{\partial{f}}{\partial{x}})} \right) = 0 The Attempt at a SolutionTo solve this problem, you will need to use the Euler-Lagrange equation to take functional derivatives of T with respect to each function defining it. This can be done by taking the partial derivative of T with respect to each of these functions and then setting them to equal zero. For example, for s(\lambda), the partial derivative would be: \frac{\partial{T}}{\partial{s(\lambda)}} = \int_{\lambda_{1}}^{\lambda_{2}} \left (\frac{d}{d \lambda}(\sum_{j=1}^{d} s(\lambda) R_{j}^{i}(\lambda)(q_{I}^{j}(\lambda) + a^{j}(\lambda))) \right )^2 d\lambdaSetting this equal to zero, we get: \frac{d}{d \lambda}(\sum_{j=1}^{d} s(\lambda) R_{j}^{i}(\lambda)(q_{I}^{j}(\lambda) + a^{j}(\lambda))) = 0 Similarly, you can take the partial derivative of T with respect to each of the other functions defining it, and then set each one equal to zero. This will give you a system of equations that can be solved to find the conditions for which all of the functional derivatives are zero.

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