CSR-DCF视频目标跟踪论文笔记（2）——关于滤波器Learning的推导（Augmented Lagrangian方法）

1. 论文基本信息

• 论文标题：Discriminative Correlation Filter with Channel and Spatial Reliability
• 作者：Alan Lukezic等
• 出处：CVPR，2017
• 文章链接：https://arxiv.org/abs/1611.08461
• 补充材料：https://www.semanticscholar.org/paper/Discriminative-Correlation-Filter-with-Channel-and-Luke%C5%BEi%C4%8D-Voj%C4%B1-%C5%99/7b485979c75b46d8c194868c0e70890f4a0f0ede
• 源码链接：https://github.com/alanlukezic/csr-dcf

这篇笔记主要针对滤波器求解的推导过程进行分析（拉格朗日乘子法），主要参考内容是原文的补充材料，关于论文其他部分创新点及其整体思路会在后续文章中进行分析。（笔记1的链接：http://blog.csdn.net/discoverer100/article/details/78182306）

2. 滤波器求解目标函数的构建

在多通道情况下，目标函数为

$$\begin{array}{c} \mathop {\arg \min }\limits_{\bf{h}} \sum\limits_{d = 1}^{{N_d}} {\left( {{{\left\| {{{\bf{f}}_d} \odot {{\bf{h}}_d} - {\bf{g}}} \right\|}^2} + \lambda {{\left\| {{{\bf{h}}_d}} \right\|}^2}} \right)} \\ {\rm{ = }}\mathop {\arg \min }\limits_{\bf{h}} \sum\limits_{d = 1}^{{N_d}} {\left( {{{\left\| {{\bf{\hat h}}_d^H{\rm{diag}}\left( {{{{\bf{\hat f}}}_d}} \right) - {{\hat g}_d}} \right\|}^2} + \lambda {{\left\| {{{{\bf{\hat h}}}_d}} \right\|}^2}} \right)} \end{array} \tag{1}$$
其中， $\bf{h}$表示滤波器， $d=1toN_d$表示 $N_d$个通道， $\bf{g}$表示期望的响应输出， $\lambda$表示正则项用于防止过拟合（关于正则项为什么可以防止过拟合可以参考： http://www.cnblogs.com/alexanderkun/p/6922428.html）

根据上述(1)式，为简化推导过程，将多通道情况改为单通道情况模式，则目标函数为

$$\begin{array}{c} \mathop {\arg \min }\limits_{\bf{h}} {\left\| {{\bf{f}} \odot {\bf{h}} - {\bf{g}}} \right\|^2} + \lambda {\left\| {\bf{h}} \right\|^2}\\ {\rm{ = }}\mathop {\arg \min }\limits_{\bf{h}} {\left\| {{{{\bf{\hat h}}}^H}{\rm{diag}}\left( {{\bf{\hat f}}} \right) - \hat g} \right\|^2} + \lambda {\left\| {{\bf{\hat h}}} \right\|^2} \end{array} \tag{2}$$
引入变量 $\bf{h}_c$并定义约束条件
$${{\bf{h}}_c} - {{\bf{h}}_m} = 0 \tag{3}$$
其中， ${{\bf{h}}_m} \equiv {\bf{m}} \odot {\bf{h}}$，而 $\bf{m}$表示论文中的空间置信图（spatial reliability map），也可以理解为一个mask，具体概念可以参考前面的一篇文章： http://blog.csdn.net/discoverer100/article/details/78182306，上述(3)式中引入的变量 $\bf{h}_c$可以先不理会其物理意义，它的主要作用是让算法能够收敛（论文原文表述：prohibits a closed-form solution），个人猜测：这里的下标命名为c，可能就是取constrained的第一个字母。

对(2)式引入上述约束条件，并进一步调整，得到最终的目标函数

$$\begin{array}{l} \mathop {\arg \min }\limits_{{{\bf{h}}_c},{{\bf{h}}_m}} {\left\| {{\bf{\hat h}}_c^H{\rm{diag}}\left( {{\bf{\hat f}}} \right) - \hat g} \right\|^2} + \frac{\lambda }{2}{\left\| {{{{\bf{\hat h}}}_m}} \right\|^2}\\ s.t.\;{{\bf{h}}_c} - {{\bf{h}}_m} = 0 \end{array} \tag{4}$$
上述的正则项前面多出了一个系数 $1/2$，其主要意图是求导数后系数可以变为 $1$，便于公式书写。

这样，公式(4)就是我们推导的起始表达式。

3. 构建Lagrange表达式

根据上述目标函数，以及Augmented Lagrangian方法（参考Distributed optimization and statistical learning via the alternating direction method of multipliers），构建Lagrang表达式，如下

$${\cal L}\left( {{{{\bf{\hat h}}}_c},{\bf{h}},{\bf{\hat I}}\left| {\bf{m}} \right.} \right) = {\left\| {{\bf{\hat h}}_c^H{\rm{diag}}\left( {{\bf{\hat f}}} \right) - \hat g} \right\|^2} + \frac{\lambda }{2}{\left\| {{{\bf{h}}_m}} \right\|^2} + \left[ {{{{\bf{\hat I}}}^H}\left( {{{{\bf{\hat h}}}_c} - {{{\bf{\hat h}}}_m}} \right) + \overline {{{{\bf{\hat I}}}^H}\left( {{{{\bf{\hat h}}}_c} - {{{\bf{\hat h}}}_m}} \right)} } \right] + \mu {\left\| {{{{\bf{\hat h}}}_c} - {{{\bf{\hat h}}}_m}} \right\|^2} \tag{5}$$
其中，字母 $\bf{I}$表示Lagrange乘数，字母上面的横杠表示 共轭矩阵，字母右上方的 $H$表示 共轭转置矩阵，因此有规律： ${{\bf{\bar A}}^T} = {{\bf{A}}^H}$（后面的推导中可能同时存在两种表示，需要留意）。将上述(5)式进行向量化表示，可得
$${\cal L}\left( {{{{\bf{\hat h}}}_c},{\bf{h}},{\bf{\hat I}}\left| {\bf{m}} \right.} \right) = {\left\| {{\bf{\hat h}}_c^H{\rm{diag}}\left( {{\bf{\hat f}}} \right) - \hat g} \right\|^2} + \frac{\lambda }{2}{\left\| {{{\bf{h}}_m}} \right\|^2} + \left[ {{{{\bf{\hat I}}}^H}\left( {{{{\bf{\hat h}}}_c} - \sqrt D {\bf{FMh}}} \right) + \overline {{{{\bf{\hat I}}}^H}\left( {{{{\bf{\hat h}}}_c} - \sqrt D {\bf{FMh}}} \right)} } \right] + \mu {\left\| {{{{\bf{\hat h}}}_c} - \sqrt D {\bf{FMh}}} \right\|^2} \tag{6}$$
不难看出，上述(5)式到(6)式，主要变化就是将变量 ${{{{\bf{\hat h}}}_m}}$的表达式替换为 ${\sqrt D {\bf{FMh}}}$，其中 $F$表示离散傅里叶变换矩阵，它相当于一个常量， $D$是 $F$的大小（ $F$是一个 $D \times D$的方阵）， $\bf{M}={{\rm{diag}}\left( {{\bf{m}}} \right)}$

将上述(6)式简单表述为四个项的和，为

$${\cal L}\left( {{{{\bf{\hat h}}}_c},{\bf{h}},{\bf{\hat I}}} \right) = {{\cal L}_1} + {{\cal L}_2} + {{\cal L}_3} + {{\cal L}_4} \tag{7}$$
其中，
$$\left\{ \begin{array}{l} {{\cal L}_1} = {\left\| {{\bf{\hat h}}_c^H{\rm{diag}}\left( {{\bf{\hat f}}} \right) - \hat g} \right\|^2} = \left( {{\bf{\hat h}}_c^H{\rm{diag}}\left( {{\bf{\hat f}}} \right) - {\bf{\hat g}}} \right){\overline {\left( {{\bf{\hat h}}_c^H{\rm{diag}}\left( {{\bf{\hat f}}} \right) - {\bf{\hat g}}} \right)} ^T}\\ {{\cal L}_2} = \frac{\lambda }{2}{\left\| {{{\bf{h}}_m}} \right\|^2}\\ {{\cal L}_3} = {{{\bf{\hat I}}}^H}\left( {{{{\bf{\hat h}}}_c} - \sqrt D {\bf{FMh}}} \right) + \overline {{{{\bf{\hat I}}}^H}\left( {{{{\bf{\hat h}}}_c} - \sqrt D {\bf{FMh}}} \right)} \\ {{\cal L}_4} = \mu {\left\| {{{{\bf{\hat h}}}_c} - \sqrt D {\bf{FMh}}} \right\|^2} = \mu \left( {{{{\bf{\hat h}}}_c} - \sqrt D {\bf{FMh}}} \right){\overline {\left( {{{{\bf{\hat h}}}_c} - \sqrt D {\bf{FMh}}} \right)} ^T} \end{array} \right. \tag{8}$$

4. 开始优化，首先对h_c求偏导数

对上述公式(4)的优化可以表述为下面的迭代过程

$$\begin{array}{l} {\bf{\hat h}}_c^{{\rm{opt}}} = \mathop {\arg \min }\limits_{{{\bf{h}}_c}} L\left( {{{{\bf{\hat h}}}_c},{\bf{h}},{\bf{\hat I}}} \right)\\ {{\bf{h}}^{{\rm{opt}}}} = \mathop {\arg \min }\limits_{\bf{h}} L\left( {{\bf{\hat h}}_c^{{\rm{opt}}},{\bf{h}},{\bf{\hat I}}} \right) \end{array} \tag{9}$$
现在看关于变量 ${{\bf{\hat h}}_c}$的优化，需要令满足 ${\nabla _{\overline {{{{\bf{\hat h}}}_c}} }}{\cal L} \equiv 0$，也就是
$${\nabla _{\overline {{{{\bf{\hat h}}}_c}} }}{{\cal L}_1} + {\nabla _{\overline {{{{\bf{\hat h}}}_c}} }}{{\cal L}_2} + {\nabla _{\overline {{{{\bf{\hat h}}}_c}} }}{{\cal L}_3} + {\nabla _{\overline {{{{\bf{\hat h}}}_c}} }}{{\cal L}_4} \equiv 0 \tag{10}$$
对各个分量求偏导数，有
$$\begin{aligned} {\nabla _{\overline {{{{\bf{\hat h}}}_c}} }}{{\cal L}_1} &= \frac{\partial }{{\partial \overline {{{{\bf{\hat h}}}_c}} }}\left[ {\left( {{\bf{\hat h}}_c^H{\rm{diag}}\left( {{\bf{\hat f}}} \right) - {\bf{\hat g}}} \right){{\overline {\left( {{\bf{\hat h}}_c^H{\rm{diag}}\left( {{\bf{\hat f}}} \right) - {\bf{\hat g}}} \right)} }^T}} \right]\\ &= \frac{\partial }{{\partial \overline {{{{\bf{\hat h}}}_c}} }}\left[ {{\bf{\hat h}}_c^H{\rm{diag}}\left( {{\bf{\hat f}}} \right){\rm{diag}}{{\left( {{\bf{\hat f}}} \right)}^H}{{{\bf{\hat h}}}_c} - {\bf{\hat h}}_c^H{\rm{diag}}\left( {{\bf{\hat f}}} \right){{{\bf{\hat g}}}^H} - {\bf{\hat g}}{\rm{diag}}{{\left( {{\bf{\hat f}}} \right)}^H}{{{\bf{\hat h}}}_c} + {\bf{\hat g}}{{{\bf{\hat g}}}^H}} \right]\\ &= {\rm{diag}}\left( {{\bf{\hat f}}} \right){\rm{diag}}{\left( {{\bf{\hat f}}} \right)^H}{{{\bf{\hat h}}}_c} - {\rm{diag}}\left( {{\bf{\hat f}}} \right){{{\bf{\hat g}}}^H}-0+0\\ &= {\rm{diag}}\left( {{\bf{\hat f}}} \right){\rm{diag}}{\left( {{\bf{\hat f}}} \right)^H}{{{\bf{\hat h}}}_c} - {\rm{diag}}\left( {{\bf{\hat f}}} \right){{{\bf{\hat g}}}^H} \end{aligned} \tag{11}$$

$$\begin{aligned} {\nabla _{\overline {{{{\bf{\hat h}}}_c}} }}{{\cal L}_{\rm{2}}} &= {\nabla _{\overline {{{{\bf{\hat h}}}_c}} }}\left[ {\frac{\lambda }{2}{{\left\| {{{\bf{h}}_m}} \right\|}^2}} \right]\\ &= {\rm{0}} \end{aligned} \tag{12}$$

$$\begin{aligned} {\nabla _{\overline {{{{\bf{\hat h}}}_c}} }}{{\cal L}_{\rm{3}}} &= \frac{\partial }{{\partial \overline {{{{\bf{\hat h}}}_c}} }}\left[ {{{{\bf{\hat I}}}^H}\left( {{{{\bf{\hat h}}}_c} - \sqrt D {\bf{FMh}}} \right) + \overline {{{{\bf{\hat I}}}^H}\left( {{{{\bf{\hat h}}}_c} - \sqrt D {\bf{FMh}}} \right)} } \right]\\ &= \frac{\partial }{{\partial \overline {{{{\bf{\hat h}}}_c}} }}\left[ {{{{\bf{\hat I}}}^H}{{{\bf{\hat h}}}_c} - {{{\bf{\hat I}}}^H}\sqrt D {\bf{FMh}} + {{{\bf{\hat I}}}^T}\overline {{{{\bf{\hat h}}}_c}} - {{{\bf{\hat I}}}^T}\overline {\sqrt D {\bf{FMh}}} } \right]\\ &= 0 - 0 + {{{\bf{\hat I}}}^T} - 0\\ &= {\bf{\hat I}} \end{aligned} \tag{13}$$

$$\begin{aligned} {\nabla _{\overline {{{{\bf{\hat h}}}_c}} }}{{\cal L}_{\rm{4}}} &= \frac{\partial }{{\partial \overline {{{{\bf{\hat h}}}_c}} }}\left[ {\mu \left( {{{{\bf{\hat h}}}_c} - \sqrt D {\bf{FMh}}} \right){{\overline {\left( {{{{\bf{\hat h}}}_c} - \sqrt D {\bf{FMh}}} \right)} }^T}} \right]\\ &= \frac{\partial }{{\partial \overline {{{{\bf{\hat h}}}_c}} }}\left[ {\mu \left( {{{{\bf{\hat h}}}_c} \cdot {\bf{\hat h}}_c^H - {{{\bf{\hat h}}}_c} \cdot \sqrt D {{\bf{h}}^H}{\bf{M}}{{\bf{F}}^H} - \sqrt D {\bf{FMh\hat h}}_c^H + \sqrt D {\bf{FMh}} \cdot \sqrt D {{\bf{h}}^H}{\bf{M}}{{\bf{F}}^H}} \right)} \right]\\ &= \frac{\partial }{{\partial \overline {{{{\bf{\hat h}}}_c}} }}\left[ {\mu \left( {{{{\bf{\hat h}}}_c} \cdot {\bf{\hat h}}_c^H - {{{\bf{\hat h}}}_c} \cdot \sqrt D {{\bf{h}}^H}{\bf{M}}{{\bf{F}}^H} - \sqrt D {\bf{FMh\hat h}}_c^H + D{\bf{FMh}}{{\bf{h}}^H}{\bf{M}}{{\bf{F}}^H}} \right)} \right]\\ &= \mu \left( {{{{\bf{\hat h}}}_c} - 0 - \sqrt D {\bf{FMh}} + 0} \right)\\ &= \mu {{{\bf{\hat h}}}_c} - \mu \sqrt D {\bf{FMh}} \end{aligned} \tag{14}$$

于是，上述公式(10)可以写成

$$\begin{array}{c} {\rm{diag}}\left( {{\bf{\hat f}}} \right){\rm{diag}}{\left( {{\bf{\hat f}}} \right)^H}{{{\bf{\hat h}}}_c} - {\rm{diag}}\left( {{\bf{\hat f}}} \right){{{\bf{\hat g}}}^H} + 0 + {\bf{\hat I}} + \mu \left( {{{{\bf{\hat h}}}_c} - 0 - \sqrt D {\bf{FMh}} + 0} \right) \equiv 0\\ {\rm{diag}}\left( {{\bf{\hat f}}} \right){\rm{diag}}{\left( {{\bf{\hat f}}} \right)^H}{{{\bf{\hat h}}}_c} - {\rm{diag}}\left( {{\bf{\hat f}}} \right){{{\bf{\hat g}}}^H} + {\bf{\hat I}} + \mu \left( {{{{\bf{\hat h}}}_c} - 0 - \sqrt D {\bf{FMh}} + 0} \right) \equiv 0 \end{array} \tag{15}$$
回顾公式(6)，我们曾将变量 ${{{{\bf{\hat h}}}_m}}$的表达式替换为 ${\sqrt D {\bf{FMh}}}$，现在我们将它替换回来，得
$${\rm{diag}}\left( {{\bf{\hat f}}} \right){\rm{diag}}{\left( {{\bf{\hat f}}} \right)^H}{{\bf{\hat h}}_c} - {\rm{diag}}\left( {{\bf{\hat f}}} \right){{\bf{\hat g}}^H} + {\bf{\hat I}} + \mu \left( {{{{\bf{\hat h}}}_c} - 0 - {{{\bf{\hat h}}}_m} + 0} \right) \equiv 0 \tag{16}$$
针对 ${{\bf{\hat h}}_c}$合并同类项，得
$${{\bf{\hat h}}_c} \cdot \left[ {{\rm{diag}}\left( {{\bf{\hat f}}} \right){\rm{diag}}{{\left( {{\bf{\hat f}}} \right)}^H} + \mu } \right] = \mu {{\bf{\hat h}}_m} + {\rm{diag}}\left( {{\bf{\hat f}}} \right){{\bf{\hat g}}^H} - {\bf{\hat I}} \tag{17}$$
于是，
$${{\bf{\hat h}}_c} = \frac{{{\rm{diag}}\left( {{\bf{\hat f}}} \right){{{\bf{\hat g}}}^H} + \mu {{{\bf{\hat h}}}_m} - {\bf{\hat I}}}}{{{\rm{diag}}\left( {{\bf{\hat f}}} \right){\rm{diag}}{{\left( {{\bf{\hat f}}} \right)}^H} + \mu }} \tag{18}$$
根据对角矩阵的乘法性质以及相关滤波基础概念，我们可以将上述(18)式的表达进行简化，得
$${{\bf{\hat h}}_c} = \frac{{{\bf{\hat f}} \odot {{{\bf{\hat g}}}^ * } + \mu {{{\bf{\hat h}}}_m} - {\bf{\hat I}}}}{{{\bf{\hat f}} \odot {{{\bf{\hat f}}}^ * } + \mu }} \tag{19}$$
由于右上角得星号表示共轭矩阵，为与作者原文表述一致，也可用顶部的横杠表示，有
$${{\bf{\hat h}}_c} = \frac{{{\bf{\hat f}} \odot \overline {{\bf{\hat g}}} + \mu {{{\bf{\hat h}}}_m} - {\bf{\hat I}}}}{{{\bf{\hat f}} \odot \overline {{\bf{\hat f}}} + \mu }} \tag{20}$$
这样就完成了对变量 ${{\bf{\hat h}}_c}$的最优化求解，它对应论文官方源码中的变量G（位于create_csr_filter.m中）

G = (Sxy + mu*H - L) ./ (Sxx + mu);

5. 对变量h求解偏导数

前面的公式(10)-(20)都是针对变量${{\bf{\hat h}}_c}$求解偏导数，由于论文提出的Lagrange表达式中含有两个变量，现在还需要针对变量$\overline {\bf{h}}$求解偏导数，也就是

$${\nabla _{\overline {\bf{h}} }}{{\cal L}_1} + {\nabla _{\overline {\bf{h}} }}{{\cal L}_2} + {\nabla _{\overline {\bf{h}} }}{{\cal L}_3} + {\nabla _{\overline {\bf{h}} }}{{\cal L}_4} \equiv 0 \tag{21}$$
首先回顾(8)式，也就是L1、L2、L3和L4这四个优化项
$$\left\{ \begin{array}{l} {{\cal L}_1} = {\left\| {{\bf{\hat h}}_c^H{\rm{diag}}\left( {{\bf{\hat f}}} \right) - \hat g} \right\|^2} = \left( {{\bf{\hat h}}_c^H{\rm{diag}}\left( {{\bf{\hat f}}} \right) - {\bf{\hat g}}} \right){\overline {\left( {{\bf{\hat h}}_c^H{\rm{diag}}\left( {{\bf{\hat f}}} \right) - {\bf{\hat g}}} \right)} ^T}\\ {{\cal L}_2} = \frac{\lambda }{2}{\left\| {{{\bf{h}}_m}} \right\|^2}\\ {{\cal L}_3} = {{{\bf{\hat I}}}^H}\left( {{{{\bf{\hat h}}}_c} - \sqrt D {\bf{FMh}}} \right) + \overline {{{{\bf{\hat I}}}^H}\left( {{{{\bf{\hat h}}}_c} - \sqrt D {\bf{FMh}}} \right)} \\ {{\cal L}_4} = \mu {\left\| {{{{\bf{\hat h}}}_c} - \sqrt D {\bf{FMh}}} \right\|^2} = \mu \left( {{{{\bf{\hat h}}}_c} - \sqrt D {\bf{FMh}}} \right){\overline {\left( {{{{\bf{\hat h}}}_c} - \sqrt D {\bf{FMh}}} \right)} ^T} \end{array} \right. \tag{8}$$
其关于变量 $\overline {\bf{h}}$的偏导数为
$$\begin{aligned} {\nabla _{\overline {\bf{h}} }}{{\cal L}_1} &= \frac{\partial }{{\partial \overline {\bf{h}} }}\left[ {\left( {{\bf{\hat h}}_c^H{\rm{diag}}\left( {{\bf{\hat f}}} \right) - {\bf{\hat g}}} \right){{\overline {\left( {{\bf{\hat h}}_c^H{\rm{diag}}\left( {{\bf{\hat f}}} \right) - {\bf{\hat g}}} \right)} }^T}} \right]\\ &= 0 \end{aligned} \tag{22}$$

$$\begin{aligned} {\nabla _{\overline {\bf{h}} }}{{\cal L}_{\rm{2}}} &= \frac{\partial }{{\partial \overline {\bf{h}} }}\left[ {\frac{\lambda }{2}{{\left\| {{{\bf{h}}_m}} \right\|}^2}} \right]\\ &= \frac{\partial }{{\partial \overline {\bf{h}} }}\left[ {\frac{\lambda }{2}{{\left\| {{\bf{m}} \odot {\bf{h}}} \right\|}^2}} \right]\\ &= \frac{\partial }{{\partial \overline {\bf{h}} }}\left[ {\frac{\lambda }{2}{{\overline {\left( {{\bf{Mh}}} \right)} }^T}\left( {{\bf{Mh}}} \right)} \right]\\ &= \frac{\partial }{{\partial \overline {\bf{h}} }}\left[ {\frac{\lambda }{2}{{\bf{h}}^H}{{\bf{M}}^H}{\bf{Mh}}} \right] \end{aligned} \tag{23}$$

观察(23)式，由于其中的$\bf {M}$表示论文定义的mask（也就是Spatial reliability map）的对角线元素，这个矩阵中的所有元素仅为0或者1，它们都是实数，因此(23)式中的${\bf {M}}^H \bf {M}$可以简化表示为$\bf M$，于是，

$$\begin{aligned} {\nabla _{\overline {\bf{h}} }}{{\cal L}_{\rm{2}}} &= \frac{\partial }{{\partial \overline {\bf{h}} }}\left[ {\frac{\lambda }{2}{{\bf{h}}^H}{\bf{Mh}}} \right]\\ &= \frac{\partial }{{\partial \overline {\bf{h}} }}\left[ {\frac{\lambda }{2}{{\left( {{{\left( {{{\bf{h}}^H}{\bf{Mh}}} \right)}^T}} \right)}^T}} \right]\\ &= \frac{\partial }{{\partial \overline {\bf{h}} }}\left[ {\frac{\lambda }{2}{{\left( {{{\bf{h}}^T}{{\bf{M}}^T}\overline {\bf{h}} } \right)}^T}} \right]\\ &= {\left( {\frac{\lambda }{2}\left( {{{\bf{h}}^T}{{\bf{M}}^T}} \right)} \right)^T}\\ &= \frac{\lambda }{2}{\bf{Mh}} \end{aligned} \tag{24}$$

$$\begin{aligned} {\nabla _{\overline {\bf{h}} }}{{\cal L}_{\rm{3}}} &= \frac{\partial }{{\partial \overline {\bf{h}} }}\left[ {{{{\bf{\hat I}}}^H}\left( {{{{\bf{\hat h}}}_c} - \sqrt D {\bf{FMh}}} \right) + \overline {{{{\bf{\hat I}}}^H}\left( {{{{\bf{\hat h}}}_c} - \sqrt D {\bf{FMh}}} \right)} } \right]\\ &= \frac{\partial }{{\partial \overline {\bf{h}} }}\left[ {{{{\bf{\hat I}}}^H}{{{\bf{\hat h}}}_c} - {{{\bf{\hat I}}}^H}\sqrt D {\bf{FMh}} + {{{\bf{\hat I}}}^T}\overline {{{{\bf{\hat h}}}_c}} - {{{\bf{\hat I}}}^T}\overline {\sqrt D {\bf{FMh}}} } \right]\\ &= {\rm{0}} - 0 + 0 - {{{\bf{\hat I}}}^T}\overline {\sqrt D {\bf{FM}}} \\ &= - \sqrt D \cdot {{{\bf{\hat I}}}^T} \cdot \overline {\bf{F}} \cdot \overline {\bf{M}} \\ &= - \sqrt D \cdot {\left( {{\bf{\hat I}}} \right)^T} \cdot {\left( {{{\bf{F}}^H}} \right)^T} \cdot {\left( {{{\bf{M}}^H}} \right)^T}\\ &= - \sqrt D {\left( {{{\bf{M}}^H}{{\bf{F}}^H}{\bf{\hat I}}} \right)^T}\\ &= - \sqrt D {\left( {{\bf{M}}{{\bf{F}}^H}{\bf{\hat I}}} \right)^T} \end{aligned} \tag{25}$$

注意：(25)式的求解结果与论文补充材料表述不一致，本人的求解结果多出了一个转置，原因还需要进一步确认。

上述(25)式中，利用了矩阵$\bf M$为实数矩阵的性质，所以有${\bf M}^H={\bf M}$，最后是L4项

$$\begin{aligned} {\nabla _{\overline {\bf{h}} }}{{\cal L}_{\rm{4}}} &= \frac{\partial }{{\partial \overline {\bf{h}} }}\left[ {\mu \left( {{{{\bf{\hat h}}}_c} - \sqrt D {\bf{FMh}}} \right){{\overline {\left( {{{{\bf{\hat h}}}_c} - \sqrt D {\bf{FMh}}} \right)} }^T}} \right]\\ &= \frac{\partial }{{\partial \overline {\bf{h}} }}\left[ {\mu \left( {{{{\bf{\hat h}}}_c} \cdot {\bf{\hat h}}_c^H - {{{\bf{\hat h}}}_c} \cdot \sqrt D {{\left( {\overline {\bf{F}} \cdot \overline {\bf{M}} \cdot \overline {\bf{h}} } \right)}^T} - \sqrt D {\bf{FMh\hat h}}_c^H + D{\bf{FMh}}{{\left( {\overline {\bf{F}} \cdot \overline {\bf{M}} \cdot \overline {\bf{h}} } \right)}^T}} \right)} \right]\\ &= \frac{\partial }{{\partial \overline {\bf{h}} }}\left[ {\mu \left( {{{{\bf{\hat h}}}_c} \cdot {\bf{\hat h}}_c^H - {{{\bf{\hat h}}}_c} \cdot \sqrt D {{\bf{h}}^H}{\bf{M}}{{\bf{F}}^H} - \sqrt D {\bf{FMh\hat h}}_c^H + D{\bf{FMh}}{{\bf{h}}^H}{{\bf{M}}^H}{{\bf{F}}^H}} \right)} \right]\\ &= 0 - \frac{\partial }{{\partial \overline {\bf{h}} }}\left[ {\mu \cdot {{{\bf{\hat h}}}_c} \cdot \sqrt D {{\bf{h}}^H}{\bf{M}}{{\bf{F}}^H}} \right] - 0 + \frac{\partial }{{\partial \overline {\bf{h}} }}\left[ {\mu \cdot D{\bf{FMh}}{{\bf{h}}^H}{{\bf{M}}^H}{{\bf{F}}^H}} \right]\\ &= - {\left[ {\frac{\partial }{{\partial {{\bf{h}}^H}}}\left[ {\mu \cdot {{{\bf{\hat h}}}_c} \cdot \sqrt D {{\bf{h}}^H}{\bf{M}}{{\bf{F}}^H}} \right]} \right]^T} + {\left[ {\frac{\partial }{{\partial {{\bf{h}}^H}}}\left[ {\mu \cdot D{\bf{FMh}}{{\bf{h}}^H}{{\bf{M}}^H}{{\bf{F}}^H}} \right]} \right]^T}\\ &= - {\left[ {\frac{\partial }{{\partial {{\bf{h}}^H}}}\left[ {\mu \cdot \sqrt D \cdot {{{\bf{\hat h}}}_c} \cdot {{\bf{h}}^H}{\bf{M}}{{\bf{F}}^H}} \right]} \right]^T} + {\left[ {\frac{\partial }{{\partial {{\bf{h}}^H}}}\left[ {\mu \cdot D{\bf{FMh}} \cdot {{\bf{h}}^H} \cdot {{\bf{M}}^H}{{\bf{F}}^H}} \right]} \right]^T}\\ &= - \mu \cdot \sqrt D \cdot {\left[ {{{\left( {{{{\bf{\hat h}}}_c}} \right)}^T} \cdot {{\left( {{\bf{M}}{{\bf{F}}^H}} \right)}^T}} \right]^T} + \mu \cdot D \cdot {\left[ {{{\left( {{\bf{FMh}}} \right)}^T} \cdot {{\left( {{{\bf{M}}^H}{{\bf{F}}^H}} \right)}^T}} \right]^T}\\ &= - \mu \cdot \sqrt D \cdot {\left[ {{{\left( {{\bf{M}}{{\bf{F}}^H}} \right)}^T}} \right]^T} \cdot {\left[ {{{\left( {{{{\bf{\hat h}}}_c}} \right)}^T}} \right]^T} + \mu \cdot D \cdot {\left[ {{{\left( {{{\bf{M}}^H}{{\bf{F}}^H}} \right)}^T}} \right]^T} \cdot {\left[ {{{\left( {{\bf{FMh}}} \right)}^T}} \right]^T}\\ &= - \mu \cdot \sqrt D \cdot {\bf{M}}{{\bf{F}}^H} \cdot {{{\bf{\hat h}}}_c} + \mu \cdot D \cdot {{\bf{M}}^H}{{\bf{F}}^H} \cdot {\bf{FMh}}\\ &= - \mu \cdot \sqrt D \cdot {\bf{M}}{{\bf{F}}^H} \cdot {{{\bf{\hat h}}}_c} + \mu \cdot D \cdot {{\bf{M}}^H}{\bf{IMh}}\\ &= - \mu \cdot \sqrt D \cdot {\bf{M}}{{\bf{F}}^H} \cdot {{{\bf{\hat h}}}_c} + \mu \cdot D \cdot \left( {{\bf{M}} \cdot {\bf{M}}} \right){\bf{h}}\\ &= - \mu \sqrt D {\bf{M}}{{\bf{F}}^H}{{{\bf{\hat h}}}_c} + \mu D{\bf{Mh}} \end{aligned} \tag{26}$$
上述(26)式末尾利用了 ${\bf M}{\bf M}={\bf M}$，这是因为 $\bf M$表示空间置信图（Spatial reliability map），根据论文中的定义，矩阵 $\bf M$中的元素要么为1，要么为0，因此 ${\bf M}{\bf M}={\bf M}$

将上述(22)-(26)式代入到(21)式中，有

$$\frac{\lambda }{{\rm{2}}}{\bf{Mh}} - \sqrt D {\bf{M}}{{\bf{F}}^H}{\bf{\hat I}} - \mu \sqrt D {\bf{M}}{{\bf{F}}^H}{{{\bf{\hat h}}}_c} + \mu D{\bf{Mh}} = 0 \tag{27}$$
对上式中的 ${\bf M}{\bf h}$进行合并同类项，有
$$\begin{aligned} \left( {\frac{\lambda }{{\rm{2}}}{\rm{ + }}\mu D} \right){\bf{Mh}} &= \sqrt D {\bf{M}}{{\bf{F}}^H}{\bf{\hat I}} + \mu \sqrt D {\bf{M}}{{\bf{F}}^H}{{{\bf{\hat h}}}_c}\\ {\bf{Mh}} &= \frac{{\sqrt D {\bf{M}}{{\bf{F}}^H}{\bf{\hat I}} + \mu \sqrt D {\bf{M}}{{\bf{F}}^H}{{{\bf{\hat h}}}_c}}}{{\frac{\lambda }{{\rm{2}}}{\rm{ + }}\mu D}}\\ {\bf{Mh}} &= {\bf{M}} \cdot \frac{{\sqrt D {{\bf{F}}^H}\left( {{\bf{\hat I}} + \mu {{{\bf{\hat h}}}_c}} \right)}}{{\frac{\lambda }{{\rm{2}}}{\rm{ + }}\mu D}}\\ {\bf{Mh}} &= {\bf{M}} \cdot \frac{{\frac{{\sqrt D }}{D}{{\bf{F}}^H}\left( {{\bf{\hat I}} + \mu {{{\bf{\hat h}}}_c}} \right)}}{{\frac{1}{D}\left( {\frac{\lambda }{{\rm{2}}}{\rm{ + }}\mu D} \right)}}\\ {\bf{Mh}} &= {\bf{M}} \cdot \frac{{\frac{1}{{\sqrt D }}{{\bf{F}}^H}\left( {{\bf{\hat I}} + \mu {{{\bf{\hat h}}}_c}} \right)}}{{\frac{\lambda }{{{\rm{2}}D}}{\rm{ + }}\mu }} \end{aligned} \tag{28}$$
现在简单回顾一下傅里叶逆变换的定义
$${{\cal F}^{ - 1}}\left( {{\bf{\hat x}}} \right) = \frac{1}{{\sqrt D }}{{\bf{F}}^H}{\bf{\hat x}} \tag{29}$$
据此，可以将上述(28)式表示为
$$\begin{aligned} {\bf{Mh}} = {\bf{M}} \cdot \frac{{{{\cal F}^{ - 1}}\left( {{\bf{\hat I}} + \mu {{{\bf{\hat h}}}_c}} \right)}}{{\frac{\lambda }{{{\rm{2}}D}}{\rm{ + }}\mu }}\\ {\rm{diag}}\left( {\bf{m}} \right){\bf{h}} = {\rm{diag}}\left( {\bf{m}} \right) \cdot \frac{{{{\cal F}^{ - 1}}\left( {{\bf{\hat I}} + \mu {{{\bf{\hat h}}}_c}} \right)}}{{\frac{\lambda }{{{\rm{2}}D}}{\rm{ + }}\mu }}\\ {\bf{m}} \odot {\bf{h}} = {\bf{m}} \odot \frac{{{{\cal F}^{ - 1}}\left( {{\bf{\hat I}} + \mu {{{\bf{\hat h}}}_c}} \right)}}{{\frac{\lambda }{{2D}} + \mu }} \end{aligned} \tag{30}$$
由于论文中最优化表达式的约束为 ${\bf{m}} \equiv {\bf{m}} \odot {\bf{h}}$（这里左边的 $\bf h$实际上也就是 ${\bf h}_m$），因此上述(30)式也可以表示为
$${\bf{h}} = {\bf{m}} \odot \frac{{{{\cal F}^{ - 1}}\left( {{\bf{\hat I}} + \mu {{{\bf{\hat h}}}_c}} \right)}}{{\frac{\lambda }{{2D}} + \mu }} \tag {31}$$
上述(31)式就对应论文原文中的公式(10)，相应的代码位于create_csr_filter.m中的变量H

H = fft2(real((1/(lambda + mu)) * bsxfun(@times, P, ifft2(mu*G + L))));

6. 小结

论文为求解最优的滤波器值，在原始目标函数的基础上，引入约束条件，利用Augmented Lagrangian方法构造最优化表达式，最后利用偏导数值为0的情况进行求解，这种利用拉格朗日最优化方法进行目标跟踪算法的建模思想值得我们学习，关于(25)式的转置问题，也欢迎大家共同讨论交流。