Stationary points of functional

  • Thread starter sardel
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  • #1
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Hello guys.

This is my first post at physics forums, so please be gentle :)

I am trying to understand functionals, so I am solving as many exercises from these lecture notes that I downloaded.

Homework Statement



Let [tex]f:[a,b]\rightarrow\mathbb{R}^k[/tex] be a continuous function and define

[tex]\Phi(\gamma)=\int_a^b f(t)\cdot\gamma(t)dt,[/tex]

where [tex]\gamma:[a,b]\rightarrow\mathbb{R}^k[/tex] is parametrized curve. Show that [tex]\Phi[/tex] is differentiable and equals its differential. Determine the stationary points of [tex]\Phi[/tex].

Homework Equations



Linearity: [tex] L(\alpha\gamma(t)+\beta\gamma'(t))=\alpha L(\gamma(t))+\beta L(\gamma'(t))[/tex]

The Attempt at a Solution



I can show that the functional is differentiable and equals its differential by showing linearity of [tex]\Phi[/tex].

However, I don't know how to find the stationary points of [tex]\Phi[/tex]. From what I understand, I have to find all [tex]\gamma\in C^1([a,b];\mathbb{R}^k)[/tex] such that

[tex]\Phi(h)=0[/tex]

for all [tex]h\in C^1_{0,0}([a,b];\mathbb{R}^k)[/tex].

I have no idea how to do this, as I can't see how to use something like the Euler-Lagrange equations. Any help is appreciated.

Thanks in advance,
Sardel
 

Answers and Replies

  • #2
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No love for functionals?

I should probably say that the definition of stationary points given in the lecture notes is

If [tex]\Phi[/tex] is a differential functional on [tex]D_{\Phi}=C^1([a,b];\mathbb{R}^k)[/tex] we call [tex]\gamma\in D_{\Phi}[/tex] a stationary point of [tex]\Phi[/tex] if the differential [tex]d\Phi_{\gamma}[/tex] vanishes on [tex]C^1_{0,0}([a,b];\mathbb{R}^k)[/tex], i.e. [tex]d\Phi_{\gamma}(h)=0[/tex] for all [tex]h\in C^1_{0,0}([a,b];\mathbb{R}^k)[/tex].

Further, [tex]C^1_{x,y}([a,b];\mathbb{R}^k)=\{\gamma\in C^1([a,b];\mathbb{R}^k)\mid \gamma(a)=x,\,\gamma(b)=y\}[/tex].

My problem is that the differential equals the functional itself, so [tex]d\Phi_{\gamma}(h)=\Phi(h)[/tex], which does not depend on [tex]\gamma[/tex]. Am I missing something?

Thanks,
Sardel
 
  • #3
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Can you give me the link to the lecture notes you download? I promess to have a look, but I don't understand many of the notations you're using...
 
  • #4
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The lecture notes are available here: http://www.math.ku.dk/~solovej/MATFYS/MatFys2.pdf" [Broken]. All the other chapters are available as MatFys1.pdf through MatFys6.pdf, and the directory listing can be viewed for the directory /MATFYS/

The exercise I am trying to solve is Exercise 2.6 and everything relevant should be in chapter 2.
 
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  • #6
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The Euler-Langrange equations are not directly applicable since [tex]\gamma[/tex] is maybe not C2. But I think you can apply the proof of the Euler-Lagrange equations.

If [tex]\Phi(h)=0[/tex] for every [tex]h\in C_{0,0}[/tex]. Then [tex]\Phi(h_1,0,0,...,0)=0[/tex]. Now apply the fundamental lemma of calculus...
 
  • #7
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Thank you.

By doing this, I end up with [tex]\gamma_1(t)=0[/tex] for all [tex]t\in[a,b][/tex] and analogously for [tex]\gamma_2,\ldots\gamma_k[/tex]. Making [tex](0,\ldots,0)[/tex] the only stationary point of [tex]\Phi[/tex]. Is this correct?
 
  • #8
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Oops... Actually, I think I get [tex]f_1(t)=0[/tex] for all [tex]t\in [a,b][/tex]. But I want to conclude somthing about [tex]\gamma[/tex], right? I think there is something fundamentally wrong with the way I am thinking about this.
 
  • #9
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No, there is nothing wrong. You indeed end up with f=0. Thus we can split up in cases. If f=0, then every point is stationary. If f is not 0, then no point is stationary...
 
  • #10
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Ah ok, thank you very much.
 

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