Time derivatives in variational calculus

In summary, the conversation is discussing taking the variation of an energy of an object with energy density F in the OP varying in space. f(x,t) can be a displacement, x is a fixed coordinate not dependent on t. df(x,t)/dt can be a speed. The integrand has time dependent terms but the integral isn't over time, which isn't covered by variational calculus. The solution is to treat g(x,t) := \dot f(x,t) as simply another field in the integrand. The specific situation/problem is not clearly stated and more details are needed for a proper understanding.
  • #1
feynman1
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Taking the variation w.r.t f(x) of the integral over some x domain of F[f(x), f'(x), df(x)/dt], why doesn't df(x)/dt need to be taken a variational derivative and is treated as if it were constant?
 
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  • #2
Can you provide some more context about what problem you are working on or what class this came from?

These kinds of details help us decide on the level of response.
 
  • #3
jedishrfu said:
Can you provide some more context about what problem you are working on or what class this came from?

These kinds of details help us decide on the level of response.
Physics problem. It's one of taking the variation of an energy of an object with energy density F in the OP varying in space. f(x) can be a displacement, x can be a coordinate. df(x)/dt can be a speed. Clear?
 
  • #4
feynman1 said:
Taking the variation w.r.t f(x) of the integral over some x domain of F[f(x), f'(x), df(x)/dt], why doesn't df(x)/dt need to be taken a variational derivative and is treated as if it were constant?
From your notation, ##f## appears to be a function of only one variable; namely, x. In that case ##\frac{d(f(x))}{dt} = 0##.
For ##f## to have nonzero (partial) derivatives wrt x and t, it should use this notation: ##f(x, t)##. Then ##\frac{\partial f}{\partial x} = f_x((x, t)## and ##\frac{\partial f}{\partial t} = f_t((x, t)## would make sense.
 
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  • #5
Mark44 said:
From your notation, ##f## appears to be a function of only one variable; namely, x. In that case ##\frac{d(f(x))}{dt} = 0##.
For ##f## to have nonzero (partial) derivatives wrt x and t, it should use this notation: ##f(x, t)##. Then ##\frac{\partial f}{\partial x} = f_x((x, t)## and ##\frac{\partial f}{\partial t} = f_t((x, t)## would make sense.
right because I couldn't put a dot on top of f(x).
 
  • #6
Mark44 said:
##f## appears to be a function of only one variable; namely, x. In that case ##\frac{d(f(x))}{dt} = 0##.
Well, not if ##x## itself depends on ##t##.

To @feynman1: could you point us to a suitable reference? I'm not yet sure what you are asking.
 
  • #7
ergospherical said:
Well, not if ##x## itself depends on ##t##.

To @feynman1: could you point us to a suitable reference? I'm not yet sure what you are asking.
Sorry there's no direct reference as it's a work in progress. Physics problem. It's one of taking the variation of an energy of an object with energy density F in the OP varying in space. f(x,t) can be a displacement, x is a fixed coordinate not dependent on t. df(x,t)/dt can be a speed. Clear?
 
  • #8
Not clear at all, sorry. I can't help until there's a clear problem statement.
 
  • #10
I’m well aware of the calculus of variations! What I mean is that you have not adequately stated your problem.
 
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  • #11
ergospherical said:
Well, not if x itself depends on t.
Which would show up in a clearly posed question. If someone writes "f(x)" without anything further, the reasonable assumption is that f depends only on x, with no relationship to t.
 
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  • #12
Mark44 said:
Which would show up in a clearly posed question. If someone writes "f(x)" without anything further, the reasonable assumption is that f depends only on x, with no relationship to t.
sorry unable to edit the OP any more
 
  • #13
ergospherical said:
I’m well aware of the calculus of variations! What I mean is that you have not adequately stated your problem.
only difference from trivial cases: the integrand has time dependent terms but the integral isn't over time, which isn't covered by variational calculus. Clear?
 
  • #14
feynman1 said:
[...] because I couldn't put a dot on top of f(x). [...] only difference from trivial cases: the integrand has time dependent terms but the integral isn't over time, which isn't covered by variational calculus. Clear?
No, not clear. You said "Physics Problem" more than once, but with no further detail, as if we're supposed to know clairvoyantly which physics scenario you're talking about. If the integral isn't over time (but assuming ##t## and ##x## are independent variables), then you treat ##g(x,t) := \dot f(x,t)## as simply another field in the integrand.

[Aside: You need to spell out your specific situation/problem explicitly. If you won't put more effort into composing your questions, why should other people put more effort into helping you? ]
 
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  • #15
strangerep said:
[Aside: You need to spell out your specific situation/problem explicitly. If you won't put more effort into composing your questions, why should other people put more effort into helping you? ]
This...
 
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  • #16
v(x,t) is velocity, x is position, t is time. Kinetic energy of a body~ a volume integral of v(x,t)*v(x,t) over a domain (x). The time rate of the kinetic energy thus~ a volume integral of ##v(x,t)v_t(x, t)##. How to take the variational derivative of this kinetic energy rate w.r.t v(x,t)?
 

Related to Time derivatives in variational calculus

1. What is the concept of time derivatives in variational calculus?

Time derivatives in variational calculus refer to the rate of change of a function with respect to time. In variational calculus, this concept is used to find the optimal path or function that minimizes a certain functional over a given time interval.

2. How are time derivatives used in variational calculus?

Time derivatives are used in variational calculus to find the optimal path or function that minimizes a given functional. This is done by setting up an Euler-Lagrange equation, which involves taking the time derivative of the functional and setting it equal to zero.

3. What is the relationship between time derivatives and the Euler-Lagrange equation?

The Euler-Lagrange equation is a necessary condition for a function to be an extremum of a given functional. It is derived by setting the time derivative of the functional equal to zero, which represents the optimal path or function.

4. Can time derivatives be used in other areas of mathematics?

Yes, time derivatives have applications in other areas of mathematics such as differential equations, optimization, and control theory. They are also used in physics, particularly in the study of motion and dynamics.

5. Are there any limitations to using time derivatives in variational calculus?

One limitation of using time derivatives in variational calculus is that it assumes a continuous and differentiable function. This may not always be the case in real-world applications, and other methods may need to be used. Additionally, the use of time derivatives may lead to complex equations and solutions, making it challenging to solve for the optimal path or function.

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