I want to prove the following statement:

Suppose $\frac{1}{4}(\cos(\theta_1)+\cos(\theta_2))^2+\lambda^2(a\sin(\theta_1)+b\sin(\theta_2))^2\leq 1$ holds for all $\theta_1,\theta_2\in[-\pi,\pi]$, then we should have $\lambda^2(a^2+b^2)\leq \frac{1}{2}$.

This problem appears when I want to find the stability condition for a numerical scheme. I tried to use Lagrange multiplier, but it turns out to be very complicated. I have also tried to find some specific $\theta_1,\theta_2$, so that the first inequality can imply the second, but I failed to do so.

  • $\begingroup$ I'm just curious. What scheme is this for? $\endgroup$ – EpicMochi Oct 21 '14 at 13:28
  • $\begingroup$ @Petaro Lax-Friedrichs scheme for equation $u_t+au_x+bu_y=0$, with the same space step in $x$ and $y$. $\endgroup$ – John Oct 21 '14 at 13:31
  • $\begingroup$ It must be a basic algebraic trigonometric question, no matter how you found this problem. $\endgroup$ – Hardey Pandya Nov 15 '14 at 12:46
  • $\begingroup$ if $\theta_1 = \frac{\pi}{4}$ and $\theta_2 = -\frac{3\pi}{4}$.. and let $\lambda=1, a=1$ and $b=1$.. then these values contradicts the claim. $\endgroup$ – Hardey Pandya Nov 17 '14 at 14:10
  • 1
    $\begingroup$ @user91374 I don't think this contradicts the claim, since I want to show if the inequality holds for all $\theta_1,\theta_2$, we must have the condition. It's possible that even if the condition is not satisfied, we have some specific $\theta_1,\theta_2$ satisfied the inequality. $\endgroup$ – John Nov 19 '14 at 14:49

You want some bound for the expression $\lambda^2(a^2 + b^2)$.

We have

$$ \frac{1}{4}(\cos(\theta_1)+\cos(\theta_2))^2+\lambda^2(a\sin(\theta_1)+b\sin(\theta_2))^2\leq 1 $$

and with two other angles

$$ \frac{1}{4}(\cos(\theta_3)+\cos(\theta_4))^2+\lambda^2(a\sin(\theta_3)+b\sin(\theta_4))^2\leq 1 $$

Let us choose the angles such that $\sin(\theta_1)\sin(\theta_2) + \sin(\theta_3)\sin(\theta_4) = 0$. This is general, since it is necessary to remove terms $a \, b$.

Then we can add the two inequalities to give

$$ \frac{1}{4}((\cos(\theta_1)+\cos(\theta_2))^2 + (\cos(\theta_3)+\cos(\theta_4))^2) +\lambda^2(a^2(\sin^2(\theta_1) + \sin^2(\theta_3))+b^2 (\sin^2(\theta_2) + \sin^2(\theta_4)))\leq 2 $$

Let us now also impose $\sin^2(\theta_1) + \sin^2(\theta_3) - \sin^2(\theta_2) - \sin^2(\theta_4) = 0$. This is is general, since it is necessary to obtain the term $(a^2 +b^2)$.

Then we have

$$ \lambda^2(a^2 + b^2) < \frac{2 - \frac{1}{4}((\cos(\theta_1)+\cos(\theta_2))^2 + (\cos(\theta_3)+\cos(\theta_4))^2)}{\sin^2(\theta_1) + \sin^2(\theta_3) } $$

So the tightest bound will be obtained by minimizing the RHS subject to the two conditions $\sin^2(\theta_1) + \sin^2(\theta_3) - \sin^2(\theta_2) - \sin^2(\theta_4) = 0$ and $\sin(\theta_1)\sin(\theta_2) + \sin(\theta_3)\sin(\theta_4) = 0$.

The two conditions can be combined, replacing $\sin^2(\theta_1)$:

$\left[\sin^2(\theta_2) - \sin^2(\theta_3) \right]\left[ \sin^2(\theta_4) +\sin^2(\theta_2) \right]= 0$

This can only be observed for a) $\theta_2 = \theta_3$ or b) $\theta_2 = \pi \pm \theta_3$ or c) $\theta_2 = n \pi $ and $\theta_4 = m \pi$ $(n,m \in \cal Z)$. Let's look at these cases.

Case a) gives, for the second condition, $\sin(\theta_1) + \sin(\theta_4) = 0$, which is $\theta_1 = - \theta_4$, or $\theta_1 = \pi + \theta_4$. Inserting into the desired bound gives

$$ \lambda^2(a^2 + b^2) < \frac{2 - \frac{1}{2}(\cos(\theta_1)+\cos(\theta_2))^2 }{\sin^2(\theta_1) + \sin^2(\theta_2) } $$

The smallest value that the RHS can take is 1.

Case b) gives, for the second condition at $\theta_2 = \pi + \theta_3$ , $\sin(\theta_1) - \sin(\theta_4) = 0$, which is $\theta_1 = \theta_4$, or $\theta_1 = \pi - \theta_4$. Inserting into the desired bound gives

$$ \lambda^2(a^2 + b^2) < \frac{2 - \frac{1}{2}(\cos^2(\theta_1)+\cos^2(\theta_2)) }{\sin^2(\theta_1) + \sin^2(\theta_2) } $$

The smallest value that the RHS can take is 1.

Case c) gives, for the first condition, $\sin^2(\theta_1) + \sin^2(\theta_3) = 0$, which is $\theta_1 = n \pi $ and $\theta_3 = m \pi$ $(n,m \in \cal Z)$. Inserting into the desired bound gives a diverging RHS.

So in total, the tightest bound one can obtain for $\lambda^2(a^2+b^2)$ is

$$ \lambda^2(a^2+b^2)\leq 1 $$

So by adding specific inequalities, a tighter bound could not be found.


Here is an answer which shows the OP's claim.

Say A is the first condition and B is the second. The claim is that there is an implication $A \to B$. This is logically equivalent to the implication Not B $\to$ Not A. So one might show that one.

In full prose, the equivalent condition reads:

Suppose $\lambda^2(a^2+b^2) > \frac{1}{2}$. Then there exists some $\theta_1,\theta_2\in[-\pi,\pi]$ such that $\frac{1}{4}(\cos(\theta_1)+\cos(\theta_2))^2+\lambda^2(a\sin(\theta_1)+b\sin(\theta_2))^2 > 1$.

Let us instead write the assumption as $\lambda^2(a^2+b^2) \geq R > \frac{1}{2}$. The extra $R$ has been put in to show that indeed, for our argument, $R>1/2$ is necessary, and no smaller $R$ will be sufficient.


Since, by assumption, $\lambda^2 \geq R / (a^2+b^2)$, we can show $$\frac{1}{4}(\cos(\theta_1)+\cos(\theta_2))^2+R \frac{(a\sin(\theta_1)+b\sin(\theta_2))^2}{(a^2+b^2)} - 1> 0$$

Setting $b = q\, a$, this becomes

$$\frac{1}{4}(\cos(\theta_1)+\cos(\theta_2))^2+R \frac{(\sin(\theta_1)+q\sin(\theta_2))^2}{(1+q^2)} - 1> 0$$

Let us now set $\theta_2 = q \theta_1$. We will show that for this choice, the inequality can always be observed, given $R >1/2$. We obtain

$$\frac{1}{4}(\cos(\theta_1)+\cos(q \theta_1))^2+R \frac{(\sin(\theta_1)+q\sin(q \theta_1))^2 }{1+q^2} -1 > 0$$

Clearly, for $\theta_1 = 0$, the LHS $= 0$. Let us therefore choose a very small $\theta_1$. By a Taylor expansion of the LHS about $\theta_1 = 0$, we get

LHS = $(1+q^2)\, (R-1/2) \, \theta_1^2 \; + \cal O (\theta_1^4)$

So clearly, we have the above condition $R > 1/2$.

If this condition is observed we have that, for small enough $\theta_1$, the LHS $> 0$.

This proves the OP's claim. Moreover, it is shown that with the given line of argument, the $R$ in the bound cannot be made smaller.


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