functional

riccardo24 · January 17th, 2013, 9:13 am

How in hell do you solve this one. f is a continuos function with some requir. and don't worry about that.the notation means H at time t starting from a. Inside the round brackets you have H at time t starting from u this means that you take the integral from u to t.how do you find the expression for H_t in function of t,f,\mu?thanks

Cuchulainn · January 17th, 2013, 12:43 pm

Nice!From a maths viewpoint this is a functional which is a higher-order function like in calculus of variations. Do you have a functional form for H?http://ergodicity.iamganesh.com/2006/08 ... unctions/I would do it using a combination of std::function and std::bind in C++ 11 (or Boost). Using delayed and binded functions and variables. In Mathematica I suspect it would be a piece of cake(?)This kind of stuff is on my TODO list.

Cuchulainn · January 17th, 2013, 1:18 pm

The equation itself is/looks like a Volterra integral equation of the 2nd kind. http://chesterrep.openrepository.com/cd ... er%204.pdf

Paul · January 17th, 2013, 1:32 pm

The notation is confusing. What is given, what is a parameter, what is independent, what is dependent,...?P

riccardo24 · January 17th, 2013, 2:13 pm

A Volterra Integral is what i thought at the beginning but it is not exactly in that form. anyway interesting chapteranyway there is a closed form. shouldn't be too difficult I think.H_t is what you want to find. f,a,\mu,t is given

Paul · January 17th, 2013, 2:16 pm

f(t) is a given function?t is an independent variable?mu is a constant parameter?What about a? A constant parameter? Or another variable?P

riccardo24 · January 17th, 2013, 2:19 pm

f(t) is a given functiona,\mu are constantst is independent and given

Paul · January 17th, 2013, 2:26 pm

Is it really just?Or something more subtle?P

Alan · January 17th, 2013, 2:49 pm

If I am reading it right, just (i) introduce some better/decent notation, (ii) differentiate w.r.t. 'a' and you get a very easy problem whose solution is something like H(t,a) = 1/[-mu (t-a) + 1/f(t)] (not quite: see later post)In other words, indicating the dependence on f in the OP is a canard. The unknown H only depends on f because of the factor in front of theexponential, but displaying that dependence only makes the problem look hard.

frenchX · January 17th, 2013, 3:09 pm

QuoteOriginally posted by: PaulIs it really just?Or something more subtle?Pin my case, I understood like you but I agree with Alan that the notations are really confusing. Transforming this into an ODE should be the best way. Or maybe some change of variable may help.

Cuchulainn · January 17th, 2013, 3:23 pm

QuoteOriginally posted by: riccardo24f(t) is a given functiona,\mu are constantst is independent and givenThat's the input. Do you want to find H(t)? Ans; Yes, OK. So, it's VIE, 2nd kind.QuoteTransforming this into an ODE should be the best waySolving an ODE is more error-prone than an integral equation. And you might need more assumptions (what if f(t) is only piecewise continuous etc.) . But why not. Quotemu is a constant parameter?When I saw it I thought it was a Lebesgue measure...

Paul · January 17th, 2013, 3:30 pm

f(t) is a red herring. I just don't understand the role of a.If it is the simple version I wrote down then it's trivial to turn it into something nicer.P

Alan · January 17th, 2013, 3:48 pm

I think it is just a time index; better to write a -> s; drop f dependence => H(t,s) => see my solution

Paul · January 17th, 2013, 3:55 pm

Have you plugged your solution back into the original? You need to check for consistency if you differentiated, especially if you have a two-dim problem.P