Recently, i met an optimization problem
$$ \arg \min_{\mathbf{x}}\Vert \mathbf {Kx} - \mathbf{y} \Vert^2_2+\frac{\eta \Vert \mathbf{Dx} -\mathbf d \Vert_2^2 }{\Vert \mathbf{Dx} \Vert_2^2} $$
from that, $\mathbf{K, D, y} $ are considered known, and dimensions are $ \mathbf x\in \mathcal R^{n \times 1}$, $ \mathbf K, \mathbf D\in \mathcal R^{m \times n}$, $\mathbf y\in \mathcal R^{m \times 1}$, respectively.
My question is, how can i solve this equation and get $\mathbf x$ effectively?
Attemptation 1:
$$ \arg \min_{\mathbf{x}} \Vert \mathbf D \mathbf x \Vert^2_2 \cdot \Vert \mathbf{Kx} - \mathbf{y} \Vert^2_2+{ \eta \Vert \mathbf{Dx} -\mathbf d \Vert_2^2 } $$
let $f(\mathbf x)$ be the object, and do the gradient,
$$ \frac{\partial f(\mathbf x)}{\partial \mathbf x} = 2 \mathbf D^\top \mathbf D \mathbf x \cdot \mathbf x^\top \mathbf K^\top \mathbf K \mathbf x + 2 \mathbf x^\top \mathbf D^\top \mathbf D \mathbf x \cdot \mathbf K^\top \mathbf K \mathbf x - 2 \mathbf D^\top \mathbf D \mathbf x \cdot \mathbf y^\top \mathbf {Kx} - 2\mathbf x^\top \mathbf D^\top \mathbf D \mathbf K^\top \mathbf y + 2 \mathbf y^\top \mathbf y \cdot \mathbf D^\top \mathbf D \mathbf x $$
Unfortunately, I'm unable to solve this problem with my current knowledge. Is there anyone who could provide me with some advice? (The person I asked for help couldn't offer any helpful ideas either.)
I would greatly appreciate it if you could recommend any textbooks or literature on this particular topic.
$ \def\a{\alpha} \def\b{\beta} \def\g{\gamma} \def\t{\theta} \def\l{\lambda} \def\s{\sigma} \def\e{\varepsilon} \def\o{{\tt1}} \def\bb{\b^{-\o}} \def\bbb{\b^{-2}} \def\BR#1{\,\Big(#1\Big)} \def\LR#1{\left(#1\right)} \def\op#1{\operatorname{#1}} \def\trace#1{\op{Tr}\LR{#1}} \def\frob#1{\left\| #1 \right\|_F} \def\qiq{\quad\implies\quad} \def\p{\partial} \def\grad#1#2{\frac{\p #1}{\p #2}} \def\c#1{\color{red}{#1}} \def\CLR#1{\c{\LR{#1}}} \def\fracLR#1#2{\LR{\frac{#1}{#2}}} \def\gradLR#1#2{\LR{\grad{#1}{#2}}} $Rename the variables $(D,d)\to(A,b)\,$ so we don't confuse them with derivative operations, and define some auxiliary functions $$\eqalign{ \a &= \frob{Ax-b}^2 &\qiq d\a = \BR{2A^T\LR{Ax-b}}^Tdx \\ \b &= \frob{Ax}^2 &\qiq d\b = \BR{2A^T Ax}^Tdx \\ \g &= \frob{Kx-y}^2 &\qiq d\g = \BR{2K^T\LR{Kx-y}}^Tdx \\ }$$ Use these to rewrite the objective function and calculate its gradient $$\eqalign{ f &= \g + \eta\a\bb \\ df &= d\g + \eta\bb\,d\a - \eta\a\bbb\,d\b \\ &= 2\BR{K^T\LR{Kx-y} + \eta\bb A^T\LR{Ax-b} -\eta\a\bbb A^T{Ax}}^Tdx \\ &= 2\bb\BR{\b K^T\LR{Kx-y} +\eta A^T\LR{Ax-b} -\eta\a\bb A^T{Ax}}^Tdx \\ \grad fx &= 2\bb\BR{\b K^T\LR{Kx-y} +\eta A^T\LR{Ax-b} -\eta\a\bb A^T{Ax}} \\ }$$ Set this gradient to zero and solve for the optimal $x$ vector $$\eqalign{ &\BR{\b K^TK +\LR{\eta-\eta\a\bb}A^TA}\:x \;=\; \LR{\b K^Ty + \eta A^Tb} \\ &x \:=\, \BR{\b K^TK +\LR{\eta-\eta\a\bb}A^TA}^{-1}\LR{\b K^Ty + \eta A^Tb} \\ }$$ This looks like a solution, but it's not since there are variables on the RHS that are functions of $x$. However, it does suggest the following iterative scheme $$\eqalign{ &x_\o = \LR{K^TK}^{-\o}K^Ty \\ &{\rm repeat} \\ &\quad \a_k = \frob{Ax_k-b}^2 \\ &\quad \b_k = \frob{Ax_k}^2 \\ &\quad x_{k+\o} = \LR{\b_kK^TK +\LR{\eta-\eta\a_k\bb_k}A^TA}^{-\o}\LR{\b_kK^Ty + \eta A^Tb} \\ &\quad k=k+\o \\ &{\rm until\;\;converged}\\ }$$