# Minimising Functional..... HELP (Functional Analysis)Watch

#1
Does anyone know how to start this problem (attached) i was given in a functional analysis module...
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10 years ago
#2
Any conditions on f?

Spoiler:
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If you define and then consider for a suitable choice of k, then I think you should get something usable.
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#3
nope, thats the whole question... only the boundary conditions given on u. im stuck on how to even start
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10 years ago
#4
Did you try the spoiler?
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#5
???
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10 years ago
#6
If you click on the

Spoiler:
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hidden text

button you will see some hidden text.

But in case it isn't working, here's what I said:

If you define and then consider for a suitable choice of k, then I think you should get something usable.
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#7
thanks alot for the help, but i still dont understand how that shows that u minimises the functional...
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#8
i think i need to start at the basics.. is the question asking, show that if you substitute f in for v, then F is minimal..
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10 years ago
#9
Well, forget the kx term for now. Take square the integrand, and integrate the 2v'(x)g(x) term by parts (integrating v', differentiating g). You should end up that looks a lot like your functional (+ a constant). But of course is minimized when v'(x) = -g(x).

I confess I haven't totally worked out how you add in the v(0)=v(1) condition. But if you know v'' = -f, then v'' is determined up to a Ax+B term, so you just need to do a little fiddling with that part (that was why I suggested the kx term, but I'm not sure it's the best approach).
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#10
how does the g(x) relate to the original question???
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10 years ago
#11
I've been inconsistent with case: I meant that g(x)=G(x)=.
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#12
I have squared the integrand and integrated by parts, which gives me,

INT(0,1)(V'(x)^2 + g(x)^2)dx + 2 v(1) INT(0,1)f(x)dx - 2 INT(0,1)f(x)v(x)dx

the last term looks like the last term of the functional which is good, im guessing the middle term vanishes when the boudary conditions are introduced but cant see exactly how that is done... but the first term doesnt look like (dv/dx)^2

and ideas....
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#13
sorry, i used the notation INT(0,1) to be the integral between o and 1 as i didnt know how to insert mathematical symbols
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#14
looking at it again, we have,

INT(0,1) (dv/dx)^2 - 2INT(0,1)f(x)v(x)dx +INT(0,1)(g(x)^2)dx +2v(1)g(1)

the first two terms are exactly what we need (the functional), but how can we get rid of the second 2 terms???
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10 years ago
#15
The term doesn't matter, it's a constant that doesn't depend on v. Minimising L(v) is identical to minimizing L(v)+const.

As far as showing the v(1)g(1) term cancels, what I'd like to do is show that v(1) = 0. But I haven't been able to see how to do this, and looking at the question again, I'm not sure that you can - I think you need to assume it.

E.g. suppose u = x(1-x). u' = 1-2x, u'' = -2, so f = 2. Then if v(x) = k (const), F(v)=-2k and so it's obvious u doesn't minimize the functional.

So I can only think that what you are meant to be doing is minimizing F(v) for v satisfying v(0)=v(1) = 0. In which case we're done.

Not 100% sure about this though - this isn't really the kind of thing I remember doing in funct analysis.
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#16
i suppose -u''=f along with the boundary conditions u(o)=u(1)=0 does minimise the functional then as for v=u, v(0)=v(1)=0...

But how is the g^2 term a constant as g is a function of f.
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10 years ago
#17
Yeah, but you're minimizing the functional w.r.t. v, not f. So f is a fixed function for the purposes of the question - it doesn't change.
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#18
your sooooo good at this, are you a lecturer of something...
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