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December 16

Proving the isoperimetric inequality with Lagrange multipliers applied to calculus of variations

I'm having a little bit of a conundrum here.

I am formulating the problem as maximizing $\underset{a}{\overset{b}{\int }}-x{y}^{\prime }dt$ subject to $\underset{a}{\overset{b}{\int }}\sqrt{x{\text{'}}^2+y{\text{'}}^2}dt=K$. If I take the functional derivative of the inside of each integral, multiply the right hand side's functional derivative by a Lagrange multiplier, equate components, and divide the equations, I end up with a tautology. More explicitly, I have $(y^{\prime },-x^{\prime })=-\frac{\lambda }{{\sqrt{x{\text{'}}^2+y{\text{'}}^2}}^3}(x^{″}(x{\text{'}}^2+y{\text{'}}^2)-x^{\prime }{x}^{\prime }{x}^{″}-x^{\prime }{y}^{″}{y}^{\prime },y^{″}(x{\text{'}}^2+y{\text{'}}^2)-y^{\prime }{x}^{\prime }{x}^{″}-y^{\prime }{y}^{″}{y}^{\prime })$ which gets simplified to $-y^{\prime }/x^{\prime }=\frac{x^{″}y{\text{'}}^2-x^{\prime }{y}^{\prime }{y}^{″}}{y^{″}x{\text{'}}^2-y^{\prime }{x}^{\prime }{x}^{″}}$, which is satisfied for all x and y such that the numerator and denominator aren't zero. This leads me to think that one must instead look at where they are zero, which occurs when $y^{″}{x}^{\prime }=y^{\prime }{x}^{″}$. This is satisfied only in the case $\ln (y^{\prime })=\ln (x^{\prime })+C$ which, however, does not exclude an ellipse.--Jasper Deng (talk) 03:49, 16 December 2019 (UTC)

I think the problem is you eliminated lambda. You get two equations but they are redundant, so just look at the first equation

$y^{\prime }=-\frac{\lambda }{{\sqrt{x{\text{'}}^2+y{\text{'}}^2}}^3}(x^{″}y{\text{'}}^2-x^{\prime }{y}^{″}{y}^{\prime }).$

Cancel y' and divide by -λ to get

$-\frac{1}{\lambda }=\frac{x^{″}{y}^{\prime }-x^{\prime }{y}^{″}}{{\sqrt{x{\text{'}}^2+y{\text{'}}^2}}^3},$

in other words curvature is constant, i.e. the curve is a circle. Btw, for those following along at home, the functional derivative in question is given in Euler–Lagrange equation#Several functions of single variable with single derivative but without the "=0". --RDBury (talk) 07:40, 16 December 2019 (UTC)

De Morgan's laws#Formal notation

For this,

$\mathrm{\neg }(P\wedge Q)⊢(\mathrm{\neg }P\vee \mathrm{\neg }Q)$

why its reverse direction,

$(\mathrm{\neg }P\vee \mathrm{\neg }Q)⊢\mathrm{\neg }(P\wedge Q)$

is not included? --Ans (talk) 14:02, 16 December 2019 (UTC)

Possibly it's because that is not a separate law, but rather a result obtained from the second law:

$(\mathrm{\neg }P\vee \mathrm{\neg }Q)=\mathrm{\neg }(\mathrm{\neg }((\mathrm{\neg }P)\vee (\mathrm{\neg }Q)))$

And by the law of negation of disjunction applied to the part in outermost brackets:

$⊢\mathrm{\neg }(\mathrm{\neg }(\mathrm{\neg }P)\wedge \mathrm{\neg }(\mathrm{\neg }Q)))=\mathrm{\neg }(P\wedge Q)$

CiaPan (talk) 14:13, 16 December 2019 (UTC)

In your third step, Are you are using $A⊢B$ to conclude $\mathrm{\neg }A⊢\mathrm{\neg }B$? RoxAsb (talk) 15:31, 16 December 2019 (UTC)

Template:Ping, your last step can imply $\mathrm{\neg }B$ from $\mathrm{\neg }A$ if $A=B$, but the law of negation of disjunction in sequent notation (in the article) is in the form, $A⊢B$, not the form, $A=B$. $A⊢B$ is not sufficient to imply $\mathrm{\neg }B$ from $\mathrm{\neg }A$ --Ans (talk) 05:25, 18 December 2019 (UTC)

If no any other opposed comments, I will add the reverse rules in the article, then. --Ans (talk) 05:49, 18 December 2019 (UTC)