Timeline for Calculus of variations for double sum with Lagrange multiplier
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 19, 2021 at 18:57 | vote | accept | J Bausch | ||
| Mar 19, 2021 at 16:58 | comment | added | André Schlichting | I did. Actually, I did not expect that the ansatz satisfies the Euler-Lagrange equation. I was quite surprised that one can do slightly better then just having all mass concentrated at $i=1$. It could by interesting to study the problem with the square-root replace by $f_i^\alpha$ for $\alpha \in (0,1)$, where the same strategy should give a solution. | |
| Mar 19, 2021 at 16:51 | answer | added | André Schlichting | timeline score: 1 | |
| Mar 19, 2021 at 15:05 | comment | added | J Bausch | Cool yeah. If you want to turn your comment into an answer I'll accept it :) | |
| Mar 19, 2021 at 15:01 | comment | added | Dabed | No, maybe I'm wrong, I just think there is no clash problem because $x$ doesn't matter in $S=\frac{\partial S}{\partial f(x)}$ with $\int_{x_i}^{x_f} L(x, f(x), f '(x))dx$ as shouldn't $i$ matter in $\frac{\partial S}{\partial f_i}$ with $S=\sum_{i=1}^\infty L(f_i) + \lambda$ so would say the notation $\delta f/\delta N_i$ is correct for the same reason but since the right way to solve it was as André Schlichting did I don't think there is a need to solve it in the original setting anymore. | |
| Mar 19, 2021 at 14:44 | comment | added | J Bausch | Great, indeed. I think that André's solution works! How did you come up with that ansatz for the $f_i$? | |
| Mar 19, 2021 at 14:34 | comment | added | J Bausch | @AndréSchlichting, this might be it. Let me check. | |
| Mar 19, 2021 at 14:32 | comment | added | J Bausch | @DanielD., I think your last comment is basically just that, $\delta f/\delta N_a$ to not clash with the sum's index; then rename $a$ to $i$ (so I think that your expression $\delta f/\delta N_i$ is actually incorrect). But maybe I'm wrong. I just treat the variation as a problem with an infinite number of independent variables. | |
| Mar 19, 2021 at 14:16 | comment | added | Dabed | For example for the functional of the Maxwell-Boltzmann statistics $f(N_i)=\sum_{i=1}^n[N_i\ln g_i-N_i\ln N_i + N_i-(\alpha+\beta\varepsilon_i) N_i]+ \alpha N +\beta E$ the variational derivative is $\frac{\delta f}{\delta N_i}=\sum_i\ln g_i-\ln N_i -(\alpha+\beta\varepsilon_i)=0\Rightarrow\frac{\partial f}{\partial N_i}=\ln g_i-\ln N_i -(\alpha+\beta\varepsilon_i) = 0$, first we do the derivate and then we select which index we want | |
| Mar 19, 2021 at 13:58 | comment | added | Dabed | As I see it for the continuum case $S=\int_{x_i}^{x_f} L(x, f(x), f '(x))dx+C$ we write $\frac{\partial S}{\partial f(x)}$ instead of $\frac{\partial S}{\partial f(y)}$ so for $S=\sum_{i=1}^\infty f_i \sum_{j=1}^i \sqrt{f_j} + \lambda(1-\sum_{i=1}^\infty f_i)=\sum_{i=1}^\infty [f_i \sum_{j=1}^i \sqrt{f_j}-\lambda f_i] + \lambda=\sum_{i=1}^\infty L(f_i) + \lambda$ one shouldn't take the functional derivative $\frac{\partial S}{\partial f(a)}$ but rather $\frac{\partial S}{\partial f(i)}$ at least I believe but we would be better if we could just put this on wolframalpha to know exactly | |
| Mar 19, 2021 at 12:13 | comment | added | André Schlichting | The continous analog $$\int_0^\infty f(x) \int_0^x \sqrt{f(y)} \,dy \, dx $$ has Dirac sequences at $0$ as minimizing sequences with value $0$. Regarding the discrete system as a regularization of the continuum one, one might think that $f_1=1$ is a candidate, giving value $1$ for the functional. One can do slightly better by setting $f_i = \gamma^{i-1} (1-\gamma)$ for $\gamma\in (0,1)$, which leads to the value $$\frac{\sqrt{1-\gamma}}{1-\gamma^{\frac{3}{2}}}.$$ The minimizer of this function gives $\gamma=\frac{2-\sqrt{3}}{2}$ with value approximately $0.978593$ as benchmark result. | |
| Mar 19, 2021 at 9:40 | comment | added | J Bausch | I think you might be using the same index for the sum and the derivative (i.e., try calculating $\delta S_\lambda / \delta f_a$ to have a distinct index from the sum's running index.) | |
| Mar 18, 2021 at 23:51 | comment | added | Dabed | $=\sum_{i=1}^\infty [\sum_{j=1}^i \sqrt{f_j}+f_i\frac{\delta}{\delta f_i}[ \sqrt{f_i}+\sum_{j=1}^{i-1} \sqrt{f_j}]-\lambda] =\sum_{i=1}^\infty [\sum_{j=1}^i \sqrt{f_j}+\frac{1}{2}\sqrt{f_i}-\lambda]\Rightarrow \sum_{j=1}^i \sqrt{f_j}+\frac{\sqrt{f_i}}{2}-\lambda=0 \Leftrightarrow f_i=[\frac{2}{3}(\lambda-\sum_{j=1}^{i-1}\sqrt{f_i})]^2 $ | |
| Mar 18, 2021 at 23:51 | comment | added | Dabed | I got something different $0 =\frac{\delta S_\lambda}{\delta f_i} =\frac{\delta}{\delta f_i}(\sum_{i=1}^\infty f_i \sum_{j=1}^i \sqrt{f_j} + \lambda\left(1-\sum_{i=1}^\infty f_i \right)) =\frac{\delta}{\delta f_i}(\sum_{i=1}^\infty [f_i \sum_{j=1}^i \sqrt{f_j}-\lambda f_i] + \lambda) =\sum_{i=1}^\infty \frac{\delta}{\delta f_i}[f_i \sum_{j=1}^i \sqrt{f_j}-\lambda f_i] =\sum_{i=1}^\infty [1\cdot\sum_{j=1}^i \sqrt{f_j}+f_i\frac{\delta}{\delta f_i}[ \sum_{j=1}^i \sqrt{f_j}]-\lambda]$ | |
| Mar 18, 2021 at 14:38 | history | edited | J Bausch | CC BY-SA 4.0 |
fixed typo in expression for lambda
|
| Mar 18, 2021 at 14:11 | history | edited | YCor | CC BY-SA 4.0 |
removed capitals from title
|
| Mar 18, 2021 at 13:19 | history | asked | J Bausch | CC BY-SA 4.0 |