The following problem optimization problem arose in a project I am working on with a student. I would like to minimize the quantity:

$$ M=\frac{1}{12} + \int_0^\frac{1}{2} \left( \tfrac{1}{2}-x \right)^2 Q(x)^2 \ dx - 4 \left[ \int_0^\frac{1}{2} \left( \tfrac{1}{2}-x \right) Q(x) \ dx\right]^2 $$ over all continuously differentiable functions $Q$ subject to the conditions that $Q(\tfrac{1}{2})=0$ and $$ \int_0^\frac{1}{2} Q(x)^2 \ dx = 1.$$

Is it possible to explicitly solve such a problem? (The presence of the constant $1/12$ should ensure that $M$ is positive for all $Q$.)

  • 2
    $\begingroup$ Calculus of variations. I have no idea what will happen when you write all the equations down and try to solve them. It also seems to me that without a $Q'$ term in your expression, the $Q(1/2)=0$ condition is worthless. You can just take a function that gives you the minimum, and put a little spike in it to satisfy the $Q(1/2)=0$ condition. $\endgroup$
    – Peter Shor
    Jan 6 '11 at 20:48

I'll change the variable $y=\frac 12-x$ to make typing easier. Since, as Peter already observed, the condition $Q(0)=0$ is worthless and since $\frac 1{12}$ is just an additive constant, we are just to minimize $\int_0^{1/2}y^2Q^2-4\left(\int_0^{1/2}yQ\right)^2$ under the condition $\int_0^{1/2}Q^2=1$. Since everything is about pure quadratic forms, we just need to find the least $a>0$ such that $\int_0^{1/2}y^2Q^2-4\left(\int_0^{1/2}yQ\right)^2+a^2\int_0^{1/2}Q^2\ge 0$. Then $-a^2$ ($+\frac 1{12}$, if you want to keep that term) is the answer. That is just a Cauchy-Schwarz type thingy and the answer is given by $\int_0^{1/2}\frac{4y^2}{y^2+a^2}dy=1$ with $Q$ proportional (with the coefficient to determine from the $\int_0^{1/2}Q^2=1$ condition) to $\frac {2y}{y^2+a^2}$ (so the condition that is worthless is automatic as well).

The integral determining $a$ is $2-4a\arctan\frac{1}{2a}$, so we need $a\arctan\frac 1{2a}=\frac 14$, which is something that one has to solve numerically. I've got $a\approx 0.2144889545$.


Here's an interesting/natural observation which might be useful. Let $U$ be a uniform$(0,1/2)$ random variable, so that $U$ has density function $f(x)=2$, $0 \lt x \lt 1/2$, and define a function $Y$ by $Y(x)=(1/2-x)Q(x)$, $0 \lt x \lt 1/2$. Then, $M=1/12 + \frac{1}{2}{\rm E}(Y^2 ) - {\rm E}^2 (Y)$. (Interestingly, a uniform$(0,1)$ random variable has variance equal to $1/12$.)


I tried calculus of variations. It's been a while since I've done calculus of variations, so I could be messing it up completely. Also, this isn't completely rigorous. There are ways of making calculus of variations rigorous, but I don't know them, and they're a lot harder than just doing calculations.

By my comment above, the $Q(\frac{1}{2})=0$ constraint is worthless, so we will ignore it.

Assume $Q$ is the minimum function (this is the first, and most important, non-rigorous step, since we are assuming a minimum exists). Now, let's plug in $Q+\epsilon$, where $Q$ and $\epsilon$ are both functions of $x$. We have

$$M=\frac{1}{12} + \int_0^\frac{1}{2} \left( \tfrac{1}{2}-x \right)^2 (Q+\epsilon)^2 \ dx - 4 \left[ \int_0^\frac{1}{2} \left( \tfrac{1}{2}-x \right) (Q+\epsilon) \ dx\right]^2 .$$

Extracting the first-order terms in $\epsilon(x)$, we get

$$\Delta M = 2\int_0^\frac{1}{2} \left( \tfrac{1}{2}-x \right)^2 Q \epsilon \ dx -8 \int_0^\frac{1}{2} \left( \tfrac{1}{2}-x \right) Q \ dx \ \int_0^\frac{1}{2} \left( \tfrac{1}{2}-x \right) \epsilon \ dx .$$

If we change $Q$ by adding an infinitessimal $\epsilon$, and keeping the normalization condition on $Q$, then $\Delta M$ has to be 0, or otherwise $Q$ wouldn't be a minimum. We still have to take care of the normalization condition $\int_0^\frac{1}{2} Q(x)^2 \ dx = 1$. We do this using Lagrange multipliers, and so we get the expression

$$2\int_0^\frac{1}{2} \left( \tfrac{1}{2}-x \right)^2 Q \epsilon \ dx -8 \int_0^\frac{1}{2} \left( \tfrac{1}{2}-x \right) Q \ dx \ \int_0^\frac{1}{2} \left( \tfrac{1}{2}-x \right) \epsilon \ dx + \lambda \int_0^\frac{1}{2} Q \epsilon\ dx .$$

This has to be zero for all functions $\epsilon(x)$. To simplify things further, let

$$C = \int_0^\frac{1}{2} \left( \tfrac{1}{2}-x \right) Q \ dx,$$

since it's a constant independent of $\epsilon$. Now, we have

$$2 \int_0^\frac{1}{2} \left( \tfrac{1}{2}-x \right)^2 Q \epsilon \ dx -8 C \ \int_0^\frac{1}{2} \left( \tfrac{1}{2}-x \right) \epsilon \ dx + \lambda \int_0^\frac{1}{2} Q \epsilon\ dx $$

or, putting everything in the same integral sign,

$$ \int_0^\frac{1}{2} 2 \left( \tfrac{1}{2}-x \right)^2 Q \epsilon -8 C \left( \tfrac{1}{2}-x \right) \epsilon + \lambda Q \epsilon\ dx $$

and for this to be $0$ for all functions $\epsilon$, we must have

$$Q = \frac{ (\frac{1}{2}-x)}{\alpha(\frac{1}{2}-x)^2+\beta}$$

for some $\alpha$, $\beta$.

Note that, if $\beta \neq 0$, we get $Q(\frac{1}{2}) = 0$, as desired.

A Maple or Mathematica program should at least let you calculate $\alpha$ and $\beta$ numerically.


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