$$ \begin{aligned} \mathbf{D}_{t} & \triangleq \underset{\mathbf{D} \in \mathcal{C}}{\arg \min } \frac{1}{t} \sum_{i=1}^{t} \frac{1}{2}\left\|\mathbf{x}_{i}-\mathbf{D} \boldsymbol{\alpha}_{i}\right\|_{2}^{2}+\lambda\left\|\boldsymbol{\alpha}_{i}\right\|_{1}, \\ &=\underset{\mathbf{D} \in \mathcal{C}}{\arg \min } \frac{1}{t}\left(\frac{1}{2} \operatorname{Tr}\left(\mathbf{D}^{T} \mathbf{D} \mathbf{A}_{t}\right)-\operatorname{Tr}\left(\mathbf{D}^{T} \mathbf{B}_{t}\right)\right) . \end{aligned} $$ where $x_i$ denotes the data vector, $\alpha_i$ denotes the corresponding sparse coefficient. $\mathcal{C}$ denotes the domain of solutions $D$. $$\begin{array}{l} \mathbf{A}_{t} \leftarrow \mathbf{A}_{t-1}+\boldsymbol{\alpha}_{t} \boldsymbol{\alpha}_{t}^{T} \\ \mathbf{B}_{t} \leftarrow \mathbf{B}_{t-1}+\mathbf{x}_{t} \boldsymbol{\alpha}_{t}^{T} \end{array}.$$
For solving the problem above, I tried to calculate its partial derivative with respect to D as $\frac{\partial L}{\partial D}=(DA+AD)^T-B=0$.
So what should I do to obtain the optimal $D$ by iteratively using this equation?