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How can I find a equivalent conditions for the following inequality $$ 0\leq x+\frac{1}{2}x(1-x)a\leq 1? $$

This is from a numerical analysis book, Finite-Volume Methods For Hyperbolic Problems. I have no idea why this can be equivalent to $$ 0\leq x\leq 1\qquad\text{and}~|a|\leq 2. $$

How can one come up with such conditions?

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I don't think they are equivalent. But the second one should imply the first one. The problem is that how can one come up with such condition? –  Jack Nov 9 '11 at 16:32
    
Let $f_a(x)$ be the function you wish to bound. To show $0 \leq f_a(x)$ and $f_a(x) -1 \leq 0$ by hand, just note that $f_a(x)$ (and hence $f_a(x) - 1$) are quadratic functions which open downwards. So finding such conditions really amounts to finding $x$-intercepts of quadratics. –  JavaMan Nov 9 '11 at 17:11
    
@JavaMan Not necessarily "opening downwards" -- if $a<0$. –  pharmine Nov 9 '11 at 17:26
    
They are not equivalent: take $x=\frac{4}{3}$, $a=3$. –  Arturo Magidin Nov 9 '11 at 19:32
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2 Answers 2

up vote 2 down vote accepted

The inequalities are not equivalent. The inequality above holds if $a=3$ and $x=\frac{4}{3}$: $$x + \frac{1}{2}x(1-x)a = \frac{4}{3} + \frac{1}{2}\left(\frac{4}{3}\right)\left(-\frac{1}{3}\right)(3) = \frac{4}{3} - \frac{2}{3} = \frac{2}{3},$$ but this clearly does not satisfy $0\leq x \leq 1$, $|a|\lt 2$.


Here's how I found the counterexample.

If $a=0$, then the inequality is equivalent to $0\leq x \leq 1$.

If $a\neq 0$, let's divide the inequality into two quadratic inequalities: $$0\leq x + \frac{1}{2}x(1-x)a \leq 1.$$

Consider first $$ 0 \leq x + \frac{1}{2}x(1-x)a = x\left(\left(1+\frac{a}{2}\right) - \frac{a}{2}x\right).$$

The product is nonnegative if and only if we have both $x\leq 0$ and $1+\frac{a}{2} \leq \frac{a}{2}x$; or $x\geq 0$ and $1 + \frac{a}{2}\geq \frac{a}{2}x$.

If $x\geq 0$, then we need $1+\frac{a}{2}\geq \frac{a}{2}x$.

  • If $a\gt 0$, then this is equivalent to $\frac{2}{a}+1\geq x$, so we need $0\leq x \leq 1+\frac{2}{a}$.
  • If $a\lt 0$, then this is equivalent to $\frac{2}{a}+1\leq x$; if $-2\leq a \lt 0$, then we just need $x\geq 0$. If $a\leq -2$, then we need $1+\frac{2}{a}\leq x$.

If $x\leq 0$, then we need $1 + \frac{a}{2}\leq \frac{a}{2}x$.

  • If $a\gt 0$, then this is equivalent to $\frac{2}{a}+1\leq x$, which is impossible if $x\leq 0$.
  • If $a\lt 0$, then we need $\frac{2}{a}+1\geq x$. If $a\leq -2$, then $1+\frac{2}{a}\geq 0\geq x$, so this holds. If $-2\leq a\lt 0$, then we need $x\leq 1+\frac{2}{a}$.

In summary:

If $a\neq 0$, then the left inequality holds if and only if one of the following happens:

  • $a\gt 0$ and $0\leq x \leq 1+\frac{2}{a}$; or
  • $-2\leq a \lt 0$ and $x\geq 0$; or
  • $-2\leq a\lt 0$ and $x\leq 1+\frac{2}{a}$; or
  • $a\leq -2$ and $1+\frac{2}{a}\leq x$; or
  • $a\leq -2$ and $x\leq 0$.

Now look at the right hand inequality $$\left(1+\frac{a}{2}\right)x - \frac{a}{2}x^2 \leq 1$$ which is equivalent to $$ax^2 - (2+a)x + 2\geq 0.$$ The two roots of this equation are $$\frac{2+a+|a-2|}{2a}\quad\text{and}\quad \frac{2+a-|a-2|}{2a};$$ Depending on the sign of $a-2$, one of the roots is $1$, the other root is $\frac{2}{a}$.

If $a\gt 0$, then the inequality is equivalent to not being between the two roots. If $a\lt 0$, then the inequality is equivalent to being between the two roots.

Thus:

Case 1. $a\leq -2$. We need either $x\leq 0$ and between $1$ and $\frac{2}{a}$; or $x\geq 1+\frac{2}{a}$ and between $1$ and $\frac{2}{a}$. So we must have either $\frac{2}{a}\leq x \leq 0$; or $1+\frac{2}{a}\leq x\leq 1$.

Case 2. $-2\leq a \lt 0$. Either $x\geq 0$ and between $\frac{2}{a}$ and $1$; or $x\leq 1 + \frac{2}{a}$ and between $1$ and $\frac{2}{a}$. That means, if $-2\leq a \lt 0$, we need either $0\leq x \leq 1$, or $\frac{2}{a}\leq x \leq 1+\frac{2}{a}$.

Case 3. $0\lt a\leq 2$. We need $0\leq x \leq 1+\frac{2}{a}$, and $x$ not between $1$ and $\frac{2}{a}$. This means we need $0\leq x \leq 1$ or $\frac{2}{a}\leq x \leq 1+\frac{2}{a}$.

Case 4. $2\leq a$. We need $0 \leq x \leq 1 + \frac{2}{a}$, and $x$ not between $\frac{2}{a}$ and $1$. This means we need $0\leq x \leq \frac{2}{a}$ or $1\leq x \leq 1 + \frac{2}{a}$. This is where the counterexample above came from.

In summary:

The inequality $$0 \leq x + \frac{1}{2}x(1-x)a\leq 1$$ holds if and only if:

  • $a=0$ and $0\leq x \leq 1$; or
  • $a\leq -2$ and $x\leq \frac{2}{a}$; or
  • $a\leq -2$ and $1+\frac{2}{a}\leq x \leq 1$; or
  • $-2\leq a \lt 0$ and $0\leq x \leq 1$; or
  • $-2\leq a \lt 0$ and $\frac{2}{a}\leq x \leq 1+\frac{2}{a}$; or
  • $0\lt a \leq 2$ and $0\leq x \leq 1$; or
  • $0\lt a\leq 2$, and $\frac{2}{a}\leq x \leq 1+\frac{2}{a}$; or
  • $2\leq a$ and $0\leq x\leq \frac{2}{a}$; or
  • $2\leq a$ and $1\leq x \leq 1+\frac{2}{a}$.

And I hope I didn't mess up the cases... but the example above, at least, is correct.

From this, we can deduce pharmine's result; for the inequality to hold for all $x$ with $0\leq x \leq 1$, we need either $a=0$, $-2\leq a\lt 0$, or $0\lt a\leq 2$. In all other cases, some values on $[0,1]$ will not satisfy the inequality.

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Hmmm... I'm somewhat worried by the fact that I cannot seem to derive pharmine's statement from this. I need to double check it. –  Arturo Magidin Nov 9 '11 at 21:02
    
Aha! Found the error. Fixing it. –  Arturo Magidin Nov 9 '11 at 21:45
    
For those who check the history: (i) I forgot to take into account the sign of $a$ when considering the conditions for the inequality involving the quadratic to hold; and (ii) I missed some values when $a\gt 0$ that do not lie between the roots but are positive anyway. –  Arturo Magidin Nov 9 '11 at 21:59
    
Thanks a lot for your elaboration. –  Jack Nov 15 '11 at 1:41
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We can show that the condition for the inequality $0\le x+\frac12x(1-x)a\le 1$ to hold for all $0\le x\le 1$ is $|a|\le 2$.

Proof. Let $f(x)=x+\frac12x(1-x)a$. The inequality holds for $a=0$; we assume $a\ne 0$ below. Note that $f(x)=-\frac12a\left(x-\frac{a+2}{2a}\right)^2+\frac{(a+2)^2}{8a}$ is a quadratic function of $x$, and $f(0)=0$, $f(1)=1$.

(i) If $0<\frac{a+2}{2a}<1$, that is, if $|a|>2$, the above condition is equivalent to $0\le f(\frac{a+2}{2a})\le 1$. This does not hold because $f(\frac{a+2}{2a})=\frac{(a+2)^2}{8a}$ is smaller than $0$ (if $a<-2$) or larger than $1$ (if $a>2$).

(ii) If $\frac{a+2}{2a}\le 0$ or $\frac{a+2}{2a}\ge 1$, that is, if $|a|\le 2$, the above condition is automatically satisfied. ($f(x)$ is monotone increasing in $[0,1]$.)

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