Timeline for prove that flat shape maximizes a functional
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Aug 10, 2012 at 12:27 | history | edited | Jon | CC BY-SA 3.0 |
Further clarification added.
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| Aug 9, 2012 at 12:17 | comment | added | Jon | Pierre, I am doing calculus of variations en.wikipedia.org/wiki/Calculus_of_variations and my conclusion is correct. In your case there are a couple of points to be fixed: 1) In your example you need a Lagrange multiplier that provides the right answer. 2) Your derivative is written incorrectly. It should eventually be $\frac{\delta Z_m[G]}{\delta G[\omega]}$. | |
| Aug 9, 2012 at 6:27 | comment | added | Pierre Robert | Consider the following case then: $\max_{y} \int y(x) w(x) dx $ subject to $\int y(x)^2dx=1$, which has the trivial solution $y(x)=w(x)$. However, following your approach yields $\delta y(x)=0$ and a flat shape appears. The problem is that the functional derivative is not what you write butinstead reads $$\delta Z_m[G]=\frac{A}{(G(\omega)+A)^2}e^{-imw}.$$ This complicates things greatly. Have I misunderstood the whole concept...? | |
| Aug 8, 2012 at 21:24 | comment | added | Jon | Yes,it is. For a given functional $\int_a^bL(q,q',t)dt$ the minimum is achieved through Euler-Lagrange equation en.wikipedia.org/wiki/Euler%E2%80%93Lagrange_equation. What makes your case easier is that there are no derivatives with respect to $\omega$. | |
| Aug 8, 2012 at 19:58 | comment | added | Pierre Robert | Jon, Is this really correct...? I thought that the functional derivative didn't inlcude the integral sign and the variation of G(w)...? In fact, with your reasoning, would a flat shap always be the maximizer to any functional as the functional derivative will always be zero? | |
| Aug 8, 2012 at 18:01 | history | edited | Jon | CC BY-SA 3.0 |
Expanded answer on OP request
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| Aug 8, 2012 at 15:55 | comment | added | Jon | Pierre, give me a few time to arrange an update to the answer. | |
| Aug 8, 2012 at 15:25 | comment | added | Pierre Robert | Jon, I dont get this...I took the functional derivative and obtained an equation system through a Lagrange multiplier. However, I get stuck since the functional derivative is horrible. Could you please share a few details how you obtained the condition you mentioned above? | |
| Jul 2, 2012 at 17:14 | history | edited | Jon | CC BY-SA 3.0 |
Expanded answer as required in the comment area.
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| Jul 1, 2012 at 12:23 | comment | added | user24799 | No, it should be $\bar{b}^*$ of course. Can ypu hint me how you obtained thr solution for $L=1$. Var calc is not my field and I only knpw the basic examples, like the braichostrone etc. My problem seems touger as I need to deal with multiplications of integrals. The pdf you sent doesn't seem to contains such examples. | |
| Jul 1, 2012 at 9:02 | comment | added | Jon | Is it write it correctly? I see a $\overline{b}_0^*$ multiplying Toeplitz matrix but should it be $\overline{b}^*$? Finally, it appears matrix $\overline{B}$ has all non-null elements. Is it so? As you may know, variation calculus applies to matrices as well. You can check math.uni-leipzig.de/~miersemann/variabook.pdf. | |
| Jul 1, 2012 at 6:06 | comment | added | user24799 | Jon, can you please hint me on thr direction I should take to resolve this problem. I do understand that the method should be var calc. But I do not know how it should be applied in this case. | |
| Jun 30, 2012 at 13:57 | comment | added | Jon | Pierre, you can extend this to any number of dimensions. Variation calculus has no limitations on this side. You have to apply it to each component and the conclusions are consistent. | |
| Jun 30, 2012 at 6:32 | comment | added | user24799 | Thanks, Can you please provide a few details as I dont know how to handle the case when there is not a single integral in var. Calc. Can I simply use the chain rule? Also, how about the following generalization: Let $b_k =\int G(\omega)\exp(-i\pi k)d\omega$ for $b=0,\ldots,L$. Define $\bar{B}=Toeplitz([b_0,\ldots,b_{L-1}])$ and $\bar{b}=[b_1,\ldots,b_L]$. Now I would like to minimize $I(G(\omega))=b_0-\bar{b}\bar{B}^{-1}\bar{b}_0^*$, which reduces to the original problem if $L=1$. Is the minimizer still flat for any $L$? | |
| Jun 28, 2012 at 7:00 | history | answered | Jon | CC BY-SA 3.0 |