Page 34 - 《应用声学》2024年第6期
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1210                                                                                2024 年 11 月

                                                                       ρ
                                      (t+1)
                 步骤4 更新辅助变量z                                         + (1 + µ g ||v|| 2 − z (t+1)  + τ  (t) 2
                                                                                                 )
                                                                       2               r       1
                                                                        (t)
                 z (t+1)                                             + λ (β (t+1)  + µ g ||v|| 2 − η + τ (t) )
                                                                        2                        2
                          R                                            ρ   (t+1)                (t) 2
                         ∑   (t)                     (t)             + (β      + µ g ||v|| 2 − η + τ 2  ) .  (52)
               = arg min    λ   (1 + µ g ||v (t) || 2 − z r + τ  )     2
                      z      1,r                     1
                                                                                        2
                         r=1                                       式(52)可转化为ℓ 2 + ℓ 结构的优化问题:
                    ρ          (t)         (t) 2                                        2
                  + (1 + µ g ||v  || 2 − z r + τ 1  )            (t+1)         (t+1)      ρ      (t+1) 2
                    2                                          v     =argmin ψ     ||v|| 2 + ||v−Γ v  || , (53)
                                                                                                      2
                                                                           v              2
                     R
                    ∑
                         (t)
                  +    κ (z r − G T  ˜ w (t+1) )                          (t+1)          (t)       (t)   (t)
                         r        0,r                          式 (53) 中,ψ      = µ g [2 + λ  + ρ(Rτ  + τ   +
                                                                                         2         1     2
                    r=1                                                       R
                    ρ        T   (t+1) 2                        (t+1)        ∑     (t)    (t+1) )]/[1+µ (1+R)],
                                                                                                     2
                  + (z r − G 0,r  ˜ w  ) .             (46)    β     +R−η)+      (λ 1,r  − ρz r      g
                    2                                                        r=1
                                                                                              2
             对式(46)做进一步化简:                                     Γ v (t+1)  = ( ˜ w (t+1)  − γ (t) /ρ)/[1 + µ (1 + R)]。根据
                                                                                              g
                                                               附录A,通过微分运算可求得其最优解:
                             R
                            ∑
              (t+1)                  ˆ (t) 2      ¯ (t+1) 2               
             z     = arg min    (z r − Γ z,r ) + (z r − Γ z,r  ) ,            (t+1)        (t+1)
                                                                                   ˆ
                          z                                                 Γ v  [Υ] + , ||Γ v  || 2 ̸= 0,
                            r=1                                   v (t+1)  =                             (54)
                                                       (47)                 0 U ,      ||Γ v (t+1) || 2 = 0,
                         (t)
                       ˆ
             式 (47) 中,Γ z,r = 1 + µ g ||v (t) || 2 + τ  (t)  + λ (t)  /ρ,  ˆ  (t+1)   (t+1)       (t+1) || 2 。
                                               1      1,r      其中,Υ = (||Γ v    || 2 − ψ   /ρ)/||Γ v
                                   (t)
             ¯ (t+1)   T   (t+1)  − κ r /ρ,r = 1, 2, · · · , R。将                        (t+1)
             Γ z,r  = G   ˜ w                                      步骤7 更新辅助变量τ
                       0,r                                                              1
             式(47) 拆分为 R 个独立子问题,容易求得每个子问                                          R
                                                                   (t+1)         ∑    (t)        (t+1)
             题的解:                                                 τ 1   = arg min    λ 1,r (1 + µ g ||v  || 2
                                                                             τ 1 >0
                                                                                  r=1
                                                                                          ρ
              z r (t+1)  = (Γ ˆ (t)  + Γ ¯ (t+1) )/2, r = 1, 2, · · · , R. (48)  − z (t+1)  + τ 1 ) + (1 + µ g ||v (t+1)
                        z,r
                              z,r
                                                                             r
                                                                                          2             || 2
                 步骤5 更新辅助变量Y           (t+1)                              − z (t+1)  + τ 1 ) .           (55)
                                                                                       2
                                                                             r
                                           T        (t+1)
                                                 ˜
                                         (t)
                (t+1)
              Y      = arg min δ||Y || 1 + σ  (Y − b ˜ w  )        易求得其最优解为
                           Y
                          ρ         (t+1)  2                        (t+1)  [
                                 ˜
                       + ||Y − b ˜ w    || .           (49)        τ     =  − 1 − µ g ||v (t+1)  || 2
                                         2
                          2                                         1
                 等价于求解如下问题:                                                   R                   ]
                                                                             ∑     (t+1)   (t)
                                                                           +     (z    − λ   /ρ)/R   .   (56)
                                      ρ        (t+1) 2                             r       1,r
                (t+1)
              Y     =arg min δ||Y || 1 + ||Y − Γ Y  || , (50)                r=1                    +
                                                    2
                          Y           2
                                                                                        (t+1)
                         (t+1)  ˜  (t+1)    (t)                    步骤8 更新辅助变量τ          2
             式 (50) 中,Γ       = b ˜ w   − σ   /ρ。式 (50) 为
                        Y
                                                                                 (t)
             典型的 ℓ 1 + ℓ 范数凸优化问题,可求得其软阈值                         τ (t+1)  = arg min λ (β (t+1)  + µ g ||v (t+1) || 2 − η
                         2
                         2                                        2              2
                                                                            τ 2 >0
             解  [17] :                                                   ρ
                                                                                                       2
                                                                 + τ 2 ) + (β (t+1)  + µ g ||v (t+1) || 2 − η + τ 2 ) . (57)
                                                                         2
                                (t+1)
                                        (t+1)
                  Y u (t+1)  = sign(Γ Y,u  )[|Γ Y,u  | − δ/ρ] + ,  类似步骤7,可求得最优解为
                          u = 1, 2, · · · , U,         (51)       (t+1)                             (t)
                                                                 τ     = [η − β (t+1)  − µ g ||v (t+1) || 2 − λ /ρ] + .
                                                                  2                                 2
             式(51)中,sign(·)表示符号函数。                                                                       (58)
                 步骤6 更新辅助变量v          (t+1)                                                       (t+1)  (t+1)
                                                                   步骤9 更新拉格朗日乘子变量{λ               1   , λ 2  ,
                  v (t+1)  =                                   ξ (t+1) , ς (t+1) , κ (t+1) , σ (t+1) , γ (t+1) }
                                       T       (t+1)
                                     (t)
                  arg min 2µ g ||v|| 2 + γ  (v − ˜ w  )             (t+1)   (t)            (t+1)     (t+1)
                       v                                           λ 1,r  = λ 1,r  + ρ(1 + µ g ||v  || 2 − z r
                     ρ       (t+1) 2                                           (t+1)
                   + ||v − ˜ w   || 2                                       + τ 1  ), r = 1, 2, · · · , R,  (59a)
                     2
                                                                    (t+1)   (t)     (t+1)      (t+1)
                      R                                            λ     = λ   + ρ(β     + µ g ||v
                     ∑   (t)              (t+1)   (t)               2       2                       || 2
                   +    λ 1,r (1 + µ g ||v|| 2 − z r  + τ 1  )                    (t+1)
                                                                            − η + τ    ),               (59b)
                     r=1                                                          2
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