Can Lagrange multipliers be used to find a function?

[jk22]

Problem statement : Let $f\in C^\infty ([-1;1])$ with $f(1)=f(-1)=0$ and $\int_{-1}^1f(x)dx=1$

Which curve has the lowest (maximal) absolute slope ?

Attempt :

Trying to minimize $f′(x)−\lambda f″(x)$ with Lagrange multipliers but to find f not x ?

I got $\frac{f^{(3)}(x)}{f″(x)}=(log⁡(f″(x)))′=1/\lambda$

$\Rightarrow f″(x)=Ae^{x/\lambda}\Rightarrow f(x)=\lambda^2Ae^{x/\lambda}+Bx+C$

Insterting the boundary and integral conditions gives A,B,C as function of $\lambda$.

Then remains to minimize wrt $\lambda$.

Is this approach valid ? I don't think since the result isn't symmetric.

 

[fresh_42]

Lowest maximal absolute slope, or lowest maximal slope?

 

[jk22]

Lowest maximal absolute slope, or lowest maximal slope?

Lowest maximal absolute slope (edited)

If it is lowest maximal slope then it could be asymmetric. It corresponds to the equation above. Thanks, now it is clear.

 

[fresh_42]

Looks like it is $1$, an infimum that isn't achieved by smooth approximations to the symmetric sawtooth wave. I get a lower bound of $0.25$, but that doesn't help. If I made no mistake, then $f(x)=\dfrac{\pi}{4}\cos\left(\dfrac{\pi}{2}x\right)$ results in a slope of $\approx 1.234$
We could try to prove that all solutions are symmetric to the $y-$axis and that they do not have a sign-change.

 

[jk22]

Yes it is obtained by $x\mapsto |x|$ in the function above. Getting $B=-\lambda A$ to obtain the smoothness at 0.

The second derivative never vanishes so the maximal slope is at the boundary ?

However I obtain a weird function for A $\frac{e^{(1/\lambda)}-1}{2(e^{1/\lambda}-1)(\lambda^3+\lambda^2)-3\lambda^2}$ giving a slope with an extremum of .355.

However this shall be wrong since the triangle solution has slope 1 and shall be the minimum but it is not analytic.

 

[fresh_42]

Yes it is obtained by $x\mapsto |x|$ in the function above. Getting $B=-\lambda A$ to obtain the smoothness at 0.

The second derivative never vanishes so the maximal slope is at the boundary ?

I think the maximal slope is everywhere, except at $0$. Chose $f(x)=1-|x|$ for all $x\in [-1,1]\backslash (-\varepsilon ,\varepsilon )$ and add something smooth on $[-\varepsilon ,\varepsilon ]$ with $f(0)=1,f'(0)=0$ and $f(-\varepsilon )=f(\varepsilon )=1-\delta$ and $f'(-\varepsilon )=1\, , \,f'(\varepsilon )=-1.$

 

[jk22]

Then the smooth part shall contain a slope bigger than 1 to balance the area.

 

[fresh_42]

Then the smooth part shall contain a slope bigger than 1 to balance the area.

Yes. Or the straight part can be a bit steeper than $1$. But with $\varepsilon, \delta \to 0$ we get the slope to $1$. Hence $1$ is an infimum, which cannot be taken.

 

[jk22]

The question is how to find the minimum inside analytic functions. But since that set is not closed the case you said happen.

However shouldn't the function be varied and not the coordinate x in the equation with Lagrange multipliers ?

 

[fresh_42]

I think we can find a family $\mathcal{F}\, : \,[0,1)\longrightarrow C^\infty ([-1,1])$ of functions with the required conditions such that $\mathcal{F}(0)=\dfrac{\pi}{4}\cos\left(\dfrac{\pi}{2}x\right)$ and $\lim_{t \to 1}\mathcal{F}(t)=1-|x|$. But why this is an optimal solution has to be proven. I don't think there is a unique solution. E.g. we can compensate the missing area in my model from post #6 with the left leg, the right, or both, and probably even with the cusp by setting the maximum $f(0)=1+\eta .$

 

[jk22]

Cos seems a good candidate maybe superposing with the line and joining at 0 could lead to a better slope ?

However there should be a method to find the extremal curve ?
I wanted to do it that way but I don't know if it is mathematically correct :
Let the operator (?!) with Lagrange multipliers
$J[f;\lambda_i]=f'(x)-\lambda_1f''(x)-\lambda_2(\int_{-1}^1f(s)ds-1)$

Then finding the extremum of J with variation for f and partial derivatives towards the lambdas ?

The problem is that the variation $\frac{d}{d\epsilon}J[f(x)+\epsilon\eta(x)]|_{\epsilon=0}$ makes f disappear and we get an equation for $\eta$ instead, those are affine functions. Does this mean that the functional shall be extremal towards the lambdas for all those eta's ?

 

[fresh_42]

Cos seems a good candidate maybe superposing with the line and joining at 0 could lead to a better slope ?

However there should be a method to find the extremal curve ?
I wanted to do it that way but I don't know if it is mathematically correct :
Let the operator (?!) with Lagrange multipliers
$J[f;\lambda_i]=f'(x)-\lambda_1f''(x)-\lambda_2(\int_{-1}^1f(s)ds-1)$

Then finding the extremum of J with variation for f and partial derivatives towards the lambdas ?

The problem is that the variation $\frac{d}{d\epsilon}J[f(x)+\epsilon\eta(x)]|_{\epsilon=0}$ makes f disappear and we get an equation for $\eta$ instead, those are affine functions. Does this mean that the functional shall be extremal towards the lambdas for all those eta's ?

The problem I see is, that we do not have a compact area of feasible functions. The solution I suggested lies on the boundary and isn't feasible. If we add the boundary, we will get $C^0$ and e.g. $1-|x|.$

If we want to optimize the problem on the open set of feasible functions, then where should we search? Which sequence of functions should we take? All we can hope for is to find the / a limit point $1+|x|$ but no algorithm will ever stop, so that we have no method to identify this limit point by algorithmic means.

It is probably easier to prove that $1-|x|$ is the / a solution and any sequence on $C^\infty$ which converges to it will do.

 

[jk22]

What about seeing it as an optimization problem with infinite series : $f(x)=\sum_{n=1}^\infty a_nx^n$.

Then it should be looked at x0 : $f''(x_0)=0$. I thought to see that as the infinite vector $(x_0^n)$ is orthogonal to the $(n(n-1)a_n)$. Thus we could take for the x nulls any vector orthogonalized to the second one with Gram-Schmidt process.

Then we shall minimize $f'(x_0)$ w.r.t to all the a n's.

Could this give a recurrence relation for the coefficients a n ?

 

[fresh_42]

What about seeing it as an optimization problem with infinite series : $f(x)=\sum_{n=1}^\infty a_nx^n$.

Then it should be looked at x0 : $f''(x_0)=0$. I thought to see that as the infinite vector $(x_0^n)$ is orthogonal to the $(n(n-1)a_n)$. Thus we could take for the x nulls any vector orthogonalized to the second one with Gram-Schmidt process.

Then we shall minimize $f'(x_0)$ w.r.t to all the a n's.

Could this give a recurrence relation for the coefficients a n ?

I see it as an optimization problem, but its solution is on the boundary outside the set of feasible functions. So there are many ways to define sequences that converge to the solution.
E.g. https://en.wikipedia.org/wiki/Triangle_wave

It is a bit like asking which point in $\{(x,y)\in\mathbb{R}^2\,|\,x^2+y^2<1\}$ is closest to $(1,0).$