2
$\begingroup$

The problem is stated as follows.

\begin{equation*} \underset{\|\mathbf{x}\|_2 \le \gamma}{\max}\quad\frac{1+|\mathbf{y}_1^H\mathbf{x}|^2}{1+|\mathbf{y}_2^H\mathbf{x}|^2}, \end{equation*}

where $\gamma > 0$ and $\mathbf{y}_1$, $\mathbf{y}_2 \in \mathbb C^n$ are given, and $\mathbf{x} \in \mathbb C^n$ is the optimization variable.

$\endgroup$
3
  • $\begingroup$ Are $\>{\bf y}_i^H\,{\bf x}\>$ meant to be scalar products? $\endgroup$ Mar 4, 2019 at 13:56
  • $\begingroup$ @ChristianBlatter It is the dot products between two vectors. $\endgroup$
    – Alex
    Mar 4, 2019 at 14:26
  • $\begingroup$ @RodrigodeAzevedo Yes, exactly. $\endgroup$
    – Alex
    Mar 5, 2019 at 9:06

2 Answers 2

0
$\begingroup$

Hint :

You can try writing $\bf y_2 = \bf a + \bf b$ with $\bf a=\frac{\langle \bf y_1, \bf y_2 \rangle}{\lVert \bf y_1\rVert^2} \bf y_1$ and then observe that $|\langle \bf y_2, \bf x\rangle|^2 = |\langle \bf a, \bf x\rangle|^2+|\langle \bf b, \bf x\rangle|^2$. You want the second term to be $0$ hence $\bf x$ and $\bf y_1$ to be collinear. Then you can write your problem as \begin{align*} \max_{x:\,|x|\cdot\lVert \bf y_1\rVert \leq \gamma} \frac{1+|x|^2 \cdot \lVert \bf y_1 \rVert^2}{1+|x|^2\cdot|\bf y_2^H\bf y_1|^2} \end{align*}

$\endgroup$
0
+50
$\begingroup$

$\color{brown}{\textbf{The task statement.}}$

Let $$\vec x=\binom{r\cos t}{r\sin t},\quad \vec y_1=\binom {p\cos\alpha}{p\sin\alpha},\quad \vec y_2=\binom {q\cos\beta}{q\sin\beta},\quad (p,q,r)\in [0,\infty),\quad (\alpha, \beta, t)\in (-\pi,\pi].\tag1$$ If given $\vec y_1=\dbinom ab,$ then $$p=\sqrt{a^2+b^2},$$ $$\alpha = -\pi\quad\text{if}\quad a=-p,\quad\text{and}\quad \alpha = 2 \arctan\frac b{p+a} \quad\text{otherwize}.$$ Similarly for $y_2$.

Then the task is to optimize the function $$f\left(\vec x\right)=\dfrac{1+p^2r^2\cos^2(t-\alpha)}{1+q^2r^2\cos^2(t-\beta)}\tag2$$ in the area $$r\in[0,\gamma].\tag3$$

$\color{brown}{\textbf{The special cases.}}$

Firstly, let us consider special cases, when $p=0$ or $q=0.$

If $\underline{(p=0)\ \wedge\ (q=0)}$ then the equation $(2)$ does not depend of $x,$ wherein $f\left(\vec x\right)=1.$

If $\underline{(p=0)}$ then the greatest value $f\left(\vec x\right)=1$ achieves if $r=0$ or $t\in\{\alpha\pm\dfrac12\pi,\alpha\pm\frac32\pi\}.$

If $\underline{(q=0)}$ then the greatest value $f\left(\vec x\right)=1+p^2\gamma^2$ achieves if $r=\gamma$ and $t\in\{\alpha, \alpha\pm\pi, \alpha\pm 2\pi\}.$

At the common case, the greatest value of $f\left(\vec x\right)$ achieves at the innner stationary points or in the bounds.

$\color{brown}{\textbf{The inner stationay points.}}$

The inner stationary points can be defined via the system $f'_r = f'_t=0,$ or \begin{cases} p^2r\cos^2(t-\alpha)(1+q^2r^2\cos^2(t-\beta))=(1+p^2r^2\cos^2(t-\alpha))q^2r\cos^2(t-\beta)\\[4pt] p^2r^2\sin2(t-\alpha)(1+q^2r^2\cos^2(t-\beta)) =(1+p^2r^2\cos^2(t-\alpha))q^2r^2\sin2(t-\beta) \tag4\end{cases}

Solution $r=0$ gives the stationary point $$f\left(\binom00\right)=1,\tag5$$ else there are

\begin{cases} p^2\cos^2(t-\alpha)=q^2\cos^2(t-\beta)\\[4pt] p^2\sin2(t-\alpha)=q^2\sin2(t-\beta), \end{cases} \begin{cases} p\cos(t-\alpha)=\pm q\cos(t-\beta)\\[4pt] p\sin(t-\alpha)=\pm q\sin(t-\beta), \end{cases} \begin{cases} p=q\\[4pt] \sin(t-\alpha)\cos(t-\beta)-\cos(t-\alpha)\sin(t-\beta)=0, \end{cases} $$p=q,\quad \cos(\alpha-\beta)=0,$$ therefore the case $$\binom{p}{\alpha}\in\binom{\{q\}}{\{\beta,\beta\pm\pi\}} $$ also gives the result $$f\left(\vec x\right)=1 \quad\forall \vec x.\tag6.$$

$\color{brown}{\textbf{The bounds.}}$

The bounds are defined by the conditons $(3),$ wherein zero case solution is given by $(5).$

If $\underline{r=\gamma}$ equation $(2)$ takes form of $$f\left(\binom{\gamma\cos t}{\gamma\sin t}\right)=\dfrac{1+p^2\gamma^2\cos^2(t-\alpha)}{1+q^2\gamma^2\cos^2(t-\beta)},\tag7$$ so the stationary points on the bound defines by the equation $$f\left(\binom{\gamma\cos t}{\gamma\sin t}\right)'_t=0,$$ $$p^2\sin2(t-\alpha)(1+q^2\gamma^2\cos^2(t-\beta)) =(1+p^2\gamma^2\cos^2(t-\alpha))q^2\sin2(t-\beta),$$ or $$p^2\sin(2t-2\alpha)(2+q^2\gamma^2(1+\cos(2t-2\beta)) =q^2\sin(2t-2\beta)(2+p^2\gamma^2(1+\cos(2t-2\alpha)),$$ $$p^2(2+q^2\gamma^2)\sin(2t-2\alpha)+p^2q^2\gamma^2\sin(2t-2\alpha)\cos(2t-2\beta)$$ $$=q^2(2+p^2\gamma^2)\sin(2t-2\beta)+p^2q^2\gamma^2\sin(2t-2\beta)\cos(2t-2\alpha)),$$

$$2p^2\sin(2t-2\alpha)-2q^2\sin(2t-2\beta) +p^2q^2\gamma^2(2\cos(2t-\alpha-\beta)\sin(\beta-\alpha)+\sin(\beta-\alpha))=0,$$ $$2p^2(\sin2t\cos2\alpha-\cos2t\sin2\alpha) -2q^2(\sin2t\cos2\beta-\cos2t\sin2\beta)$$ $$+p^2q^2\gamma^2\sin(\beta-\alpha)(2\cos2t\cos(\alpha+\beta)+2\sin2t\sin(\alpha+\beta)+1)=0,$$

$$2p^2(2\tau\cos2\alpha -(1-\tau^2)\sin2\alpha) -2q^2(2\tau\cos2\beta -(1-\tau^2)\sin2\beta)$$ $$+p^2q^2\gamma^2\sin(\beta-\alpha)(2(1-\tau^2)\cos(\alpha+\beta)+4\tau \sin(\alpha+\beta)+\tau^2+1)=0,$$ and unknown value $$\tau = \tan t\tag8$$ can be found from the square equation $$\begin{align} &(2p^2\sin2\alpha-2q^2\sin2\beta+p^2q^2\gamma^2\sin(\beta-\alpha)(1-2\cos(\alpha+\beta)))\tau^2\\ &+4(p^2\cos2\alpha-q^2\cos2\beta+p^2q^2\gamma^2\sin(\beta-\alpha)\sin(\alpha+\beta))\tau\\ &2q^2\cos2\beta-2p^2\cos2\alpha+p^2q^2\gamma^2\sin(\beta-\alpha)(2\cos(\alpha+\beta)+1)=0. \end{align}\tag9.$$

Looks more suitable to solve equations $(9),(8)$ and check the solutions on the maximum for given values of the five parameters.

$\endgroup$
2
  • $\begingroup$ Why did you choose $n=2$? $\endgroup$ Mar 17, 2019 at 7:25
  • $\begingroup$ @RodrigodeAzevedo Randomly. On the other hand, this case can be used as the start point for the general solution/. $\endgroup$ Mar 17, 2019 at 13:56

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .